10-1
CHAPTER 10
The correspondence between the new problem set and the previous 4th edition
chapter 8 problem set.
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Old
new
new
1
2
3
5
6
7
9
15
16
10
11
49
47
13
14
38
52
8
New
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Old
new
new
18
new
19
20
21
new
new
23
24
new
new
25
26
27
32
28
33
34
New
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Old
35
41
42
46
54
56
39
55
53
new
43
51
17
29
30
31
36
40
44
57
The problems that are labeled advanced starts at number 53.
The English unit problems are:
New
61
62
63
64
65
66
67
68
69
70
Old
new
58
60
63
61
80
62
new
65
66
New
71
72
73
74
75
76
77
78
79
80
Old
67
new
68
new
69
70
72
new
79
84
New
81
82
83
84
85
86
87
Old
75
83
77
64
71
73
78
10-2
10.1
Calculate the reversible work and irreversibility for the process described in
Problem 5.18, assuming that the heat transfer is with the surroundings at 20°C.
C.V.: A + B. This is a control mass.
Continuity equation:
Energy:
m2 - (mA1 + mB1) = 0 ;
m2u2 - mA1uA1 - mB1uB1 = 1Q2 - 1W2
System: if VB ≥ 0 piston floats ⇒
PB = PB1 = const.
if VB = 0 then P2 < PB1 and v = VA/mtot see P-V diagram
P
State A1: Table B.1.1, x = 1
a
vA1 = 1.694 m3/kg, uA1 = 2506.1 kJ/kg
PB1
2
mA1 = VA/vA1 = 0.5903 kg
V2
State B1: Table B.1.2 sup. vapor
vB1 = 1.0315 m3/kg, uB1 = 2965.5 kJ/kg
mB1 = VB1/vB1 = 0.9695 kg =>
At (T2 , PB1)
m2 = mTOT = 1.56 kg
v2 = 0.7163 > v a = VA/mtot = 0.641 so VB2 > 0
so now state 2: P2 = PB1 = 300 kPa, T2 = 200 °C
=> u 2 = 2650.7 kJ/kg and V2 = m2 v2 = 1.56×0.7163 = 1.117 m3
(we could also have checked Ta at: 300 kPa, 0.641 m3/kg => T = 155 °C)
ac
1W2 =
1Q 2
⌠
⌡PBdVB = PB1(V2 - V1) = PB1(V2 - V1)tot = -264.82 kJ
B
= m2u2 - mA1uA1 - mB1uB1 + 1W2 = -484.7 kJ
From the results above we have :
sA1 = 7.3593, sB1 = 8.0329, s2 = 7.3115 kJ/kg K
rev
1W2 =
To(S2 - S1) - (U2 - U1) + 1Q2(1 - To/TH)
ac
= To(m2s2 - mA1sA1 - mB1sB1) + 1W2 - 1Q2To/TH
= 293.15 (1.5598 × 7.3115 - 0.5903 × 7.3593 - 0.9695 × 8.0329)
+ (-264.82) - (-484.7) × 293.15 / 293.15
= -213.3 - 264.82 + 484.7 = 6.6 kJ
1I 2
rev
ac
= 1W2 - 1W2 = 6.6 - (-264.82) = 271.4 kJ
10-3
10.2
Calculate the reversible work and irreversibility for the process described in
Problem 5.65, assuming that the heat transfer is with the surroundings at 20°C.
P
2
1
Linear spring gives
1W2
v
=⌠
⌡PdV = 2(P1 + P2)(V2 - V1)
1
1Q 2
= m(u2 - u1) + 1W2
Equation of state: PV = mRT
State 1: V1 = mRT1/P1 = 2 x 0.18892 x 673.15 /500 = 0.5087 m3
State 2: V2 = mRT2/P2 = 2 x 0.18892 x 313.15 /300 = 0.3944 m3
1W2
1
= 2(500 + 300)(0.3944 - 0.5087) = -45.72 kJ
From Figure 5.10: Cp(Tavg) = 45/44 = 1.023 ⇒ Cv = 0.83 = C p - R
For comparison the value from Table A.5 at 300 K is Cv = 0.653 kJ/kg K
1Q 2
rev
1W2 =
= mCv(T2 - T1) + 1W2 = 2 x 0.83(40 - 400) - 45.72 = -643.3 kJ
To(S2 - S1) - (U2 - U1) + 1Q2(1 - To/TH)
ac
= Tom(s2 - s1)+ 1W2 - 1Q2To/To
ac
= Tom[CP ln(T2 / T1) − R ln(P2 / P1)] + 1W2 - 1Q2
= 293.15 x 2 [ 1.023 ln(313/673) - 0.1889 ln(300/500)] - 45.72 + 643.3
= -402.6 - 45.72 + 643.3 = 195.0 kJ
1I 2
10.3
rev
ac
= 1W2 - 1W2 = 195.0 - (-45.72) = 240.7 kJ
The compressor in a refrigerator takes refrigerant R-134a in at 100 kPa, −20°C
and compresses it to 1 MPa, 40°C. With the room at 20°C find the minimum
compressor work.
Solution:
C.V. Compressor out to ambient. Minimum work in is the reversible work.
SSSF, 1 inlet and 2 exit energy Eq.:
wc = h1 - h2 + qrev
Entropy Eq.: s2 = s1 + ⌠
⌡dq/T + sgen = s1 + qrev/T0 + 0 => qrev = T0(s2- s1)
wc min = h1 - h2 + T0(s2- s1) = 387.22 - 420.25 + 293.15 × (1.7148 - 1.7665)
= -48.19 kJ/kg
10-4
10.4
Calculate the reversible work out of the two-stage turbine shown in Problem 6.41,
assuming the ambient is at 25°C. Compare this to the actual work which was
found to be 18.08 MW.
C.V. Turbine. SSSF, 1 inlet and 2 exits.
Use Eq. 10.12 for each flow stream with q = 0 for adiabatic turbine.
Table B.1: h1 = 3373.7, h2 = 2755.9, h3 = 251.4 + 0.9 x 2358.3 = 2373.9
s1 = 6.5966, s2 = 6.8382, s3 = 0.8319 + 0.9 x 7.0766 = 7.2009
.
.
.
.
.
.
.
Wrev = T0(m2s2 + m3s3 - m1s1) - (m2h2 + m3h3 - m1h1)
= 298.15(5 × 6.8382 + 15 × 7.2009 - 20 × 6.5966)
−(5 × 2755.9 + 15 × 2373.9 - 20 × 3373.7)
.
.
= 21.14 MW = Wac + Qrev = 3062.7 kW + 18084 kW
10.5
A household refrigerator has a freezer at TF and a cold space at TC from which
energy is removed and rejected to the ambient at T A as shown in Fig. P10.5.
.
Assume that the rate of heat transfer from the cold space, QC, is the same as from
.
the freezer, QF, find an expression for the minimum power into the heat pump.
.
Evaluate this power when T A = 20°C, TC = 5°C, TF = −10°C, and QF = 3 kW.
C.V. Refrigerator (heat pump), SSSF, no external flows except heat transfer.
.
.
.
.
Energy Eq.: QF + Qc + W = −QA (amount rejected to ambient)
Reversible gives minimum work in as from Eq. 10.1 or 10.9 on rate form.
TA . 
TA
.
. 

293.15

293.15
W = QF 1 −  + Qc 1 −  = 31 −
 + 31 −

263.15
278.15
TF 
TC




= -0.504 kW
10.6
(negative so work goes in)
An air compressor takes air in at the state of the surroundings 100 kPa, 300 K.
The air exits at 400 kPa, 200°C at the rate of 2 kg/s. Determine the minimum
compressor work input.
C.V. Compressor, SSSF, minimum work in is reversible work.
ψ1 = 0 at ambient conditions
s0 - s2 = sT°0 - sT°2 - R ln(P0/P2)
= 6.86926 - 7.3303 - 0.287 ln(100/400) = -0.06317
ψ2 = h2 - h0 + T0(s0 - s2) = 475.79 - 300.473 + 300(s0 - s2)
ψ2 = 156.365 kJ/kg
.
.
.
-WREV = m(ψ2-ψ1) = 312.73 kW = Wc
10-5
10.7
A supply of steam at 100 kPa, 150°C is needed in a hospital for cleaning purposes
at a rate of 15 kg/s. A supply of steam at 150 kPa, 250°C is available from a
boiler and tap water at 100 kPa, 15°C is also available. The two sources are then
mixed in an SSSF mixing chamber to generate the desired state as output.
Determine the rate of irreversibility of the mixing process.
1
3
2
B.1.1: h1 = 62.99 s1 = 0.2245
B.1.3: h2 = 2972.7 s2 = 7.8437
B.1.3: h3 = 2776.4 s3 = 7.6133
C.V. Mixing chamber, SSSF
.
.
.
Cont.:
m1 + m2 = m3
.
.
.
Energy: m1h1 + m2h2 = m3h3
.
.
.
.
Entropy: m1s1 + m2s2 + Sgen = m3s3
. .
2776.4 - 62.99
m2/m3 = (h3 - h1)/(h2 - h1) =
= 0.9325
2972.7 - 62.99
.
.
m2 = 13.988 kg/s, m1 = 1.012 kg/s
.
.
.
.
.
I = T0Sgen = T0(m3s3 - m1s1 - m2s2)
= 298.15 × (15 × 7.6133 - 1.012 × 0.2245 - 13.988 × 7.8437)
= 1269 kW
10.8
Two flows of air both at 200 kPa of equal flow rates mix in an insulated mixing
chamber. One flow is at 1500 K and the other is at 300 K. Find the irreversibility
in the process per kilogram of air flowing out.
1
3
2
C.V. Mixing chamber
.
.
.
.
Cont.:
m1 + m2 = m3 = 2m1
.
.
.
Energy: m1h1 + m1h2 = 2m1h3
Properties from Table A.7
h3 = (h1 + h2)/2 = (300.473 + 1635.8)/2 = 968.14 kJ/kg
° = 8.9867
⇒ sT3
.
.
.
.
Entropy Eq.:
m1s1 + m1s2 + Sgen = 2m1s3 ⇒
.
.
Sgen/2m1 = s3 - (s1 + s2)/2 = 8.9867 - (6.86926 - 8.61208)/2
= 1.246 kJ/kg K
10-6
10.9
A steam turbine receives steam at 6 MPa, 800°C. It has a heat loss of 49.7 kJ/kg
and an isentropic efficiency of 90%. For an exit pressure of 15 kPa and
surroundings at 20°C, find the actual work and the reversible work between the
inlet and the exit.
C.V. Reversible adiabatic turbine (isentropic)
wT = hi - he,s ;
se,s = si = 7.6566 kJ/kg K,
hi = 4132.7 kJ/kg
xe,s = (7.6566 - 0.7548)/7.2536 = 0.9515,
he,s = 225.91 + 0.9515×2373.14 = 2483.9
wT,s = 4132.7 - 2483.9 = 1648.79 kJ/kg
C.V. Actual turbine
wT,ac = ηwT,s = 1483.91 kJ/kg
= hi - he,ac - qloss ⇒
he,ac = hi - qloss - wT,ac = 4132.7 - 49.7 - 1483.91 = 2599.1
Actual exit state: P,h ⇒
sat. vap.,
se,ac = 8.0085
C.V. Reversible process, work from Eq.10.12
qR = T0(se,ac - si) = 293.15 × (8.0085 - 7.6566) = 103.15
kJ
kg
wR = hi - he,ac + qR = 4132.7 - 2599.1 + 103.16 = 1636.8 kJ/kg
10.10 A 2-kg piece of iron is heated from room temperature 25°C to 400°C by a heat
source at 600°C. What is the irreversibility in the process?
C.V. Iron out to 600°C source, which is a control mass.
Energy Eq.:
mFe(u2 - u1) = 1Q2 - 1W2
Process: Constant pressure
=>
1W2
= PmFe(v2 - v1)
⇒ 1Q2 = mFe(h2 - h1) = mFeC(T2 - T1)
1Q2
= 2 × 0.42 × (400 - 25) = 315 kJ
mFe(s2 - s1) = 1Q2/Tres + 1S2 gen
1S2 gen
= mFe(s2 - s1) - 1Q2/Tres = mFeC ln (T2/T1) - 1Q2/Tres
= 2 × 0.42 × ln
1I2
673.15 315
= 0.3233 kJ/K
298.15 873.15
= To (1S2 gen ) = 298.15 × 0.3233 = 96.4 kJ
10-7
10.11 A 2-kg/s flow of steam at 1 MPa, 700°C should be brought to 500°C by spraying
in liquid water at 1 MPa, 20°C in an SSSF setup. Find the rate of irreversibility,
assuming that surroundings are at 20°C.
C.V. Mixing chamber, SSSF. State 1 is superheated vapor in, state 2 is
compressed liquid in, and state 3 is flow out. No work or heat transfer.
•
•
2
•
m3 = m1 + m2
Cont.:
•
Energy:
1
•
3
•
m3h3 = m1h1 + m2h2
•
Entropy:
•
•
•
m3s3 = m1s1 + m2s2 + Sgen
Table B.1.3: h1 = 3923.1, s1 = 8.2731,
h3 = 3478.5, s3 = 7.7622,
For state 2 interpolate between, saturated liquid 20°C table B.1.1 and,
compressed liquid 5 MPa, 20°C from Table B.1.4: h2 = 84.9, s2 = 0.2964
•
•
x = m2/m1 = (h3 - h1)/(h2 - h3) = 0.13101
•
⇒ m2 = 2 × 0.131 = 0.262 kg/s ;
•
•
•
•
•
m3 = 2 + 0.262 = 2.262 kg/s
Sgen = m3s3 - m1s1 - m2s2 = 0.9342 kW/K
.
.
.
.
•
I = Wrev - Wac = Wrev = ToSgen = 293.15 × 0.9342 = 273.9 kW
10-8
10.12 Fresh water can be produced from saltwater by evaporation and subsequent
condensation. An example is shown in Fig. P10.12 where 150-kg/s saltwater, state
1, comes from the condenser in a large power plant. The water is throttled to the
saturated pressure in the flash evaporator and the vapor, state 2, is then condensed
by cooling with sea water. As the evaporation takes place below atmospheric
pressure, pumps must bring the liquid water flows back up to P0. Assume that the
saltwater has the same properties as pure water, the ambient is at 20°C and that
there are no external heat transfers. With the states as shown in the table below
find the irreversibility in the throttling valve and in the condenser.
State 1
2
3
T [°C] 30
25
25
.
.
.
C.V. Valve. m1 = mex = m2 + m3
Energy Eq.:
h1 = 125.77 ,
h1 = he ;
4
--
Entropy Eq.:
s1 = 0.4369 ,
5
23
6
--
7
17
s1 + sgen = se
P2 = Psat(T2 = T3) = 3.169 kPa
he = h1 ⇒ xe = (125.77 - 104.87)/2442.3 = 0.008558
⇒
se = 0.3673 + 0.008558 × 8.1905 = 0.4374
sgen = se - s1 = 0.4374 - 0.4369 = 0.000494
. .
I = mT0sgen = 150 × 293.15 × 0.000494 = 21.72 kW
C.V. Condenser.
h2 = 2547.2 , s2 = 8.558 ,
h5 = 96.5 ,
.
.
m2 = (1 - xe)m1 = 148.716 kg/s
s5 = 0.3392
h7 = 71.37 , s7 = 0.2535 , h8 = 83.96 , s8 = 0.2966
.
.
.
.
m2h2 + m7h7 = m2h5 + m7h8 ⇒
.
.
2547.2 - 96.5
kg
m7 = m2 × (h2 - h5)/(h8 - h7) = 148.716 ×
= 28948
83.96 - 71.37
s
.
.
.
.
.
m2s2 + m7s7 + Sgen = m2s5 + m7s8
.
.
.
.
I = T0Sgen = T0 [ m2(s5 - s2) + m7(s8 - s7)]
= 293.15[148.716(0.3392 - 8.558) + 28948(0.2966 - 0.2535)]
= 293.15 × 25.392 = 7444 kW
8
20
10-9
10.13 An air compressor receives atmospheric air at T0 = 17°C, 100 kPa, and
compresses it up to 1400 kPa. The compressor has an isentropic efficiency of 88%
and it loses energy by heat transfer to the atmosphere as 10% of the isentropic
work. Find the actual exit temperature and the reversible work.
C.V. Compressor
Isentropic: wc,in,s = he,s - hi ;
Table A.7:
se,s = si
Pr,e,s = Pr,i × (Pe/Pi) = 0.9917 × 14 = 13.884
⇒ he,s = 617.51
wc,in,s = 617.51 - 290.58 = 326.93
Actual: wc,in,ac = wc,in,s/ηc = 371.51 ;
qloss = 32.693
wc,in,ac + hi = he,ac + qloss
=> h e,ac = 290.58 + 371.51 - 32.693 = 629.4
=> Te,ac = 621 K
Reversible:
wrev = hi - he,ac + T0(se,ac - si)
= 290.58 - 629.4 + 290.15 × (7.6121 - 6.8357)
= -338.82 + 225.42 = -113.4 kJ/kg
Since qloss is also to the atmosphere it is not included as it will not be
reversible.
10.14 A piston/cylinder has forces on the piston so it keeps constant pressure. It contains
2 kg of ammonia at 1 MPa, 40°C and is now heated to 100°C by a reversible heat
engine that receives heat from a 200°C source. Find the work out of the heat engine.
C.V. Ammonia plus heat engine
Energy: mam(u2 - u1) = 1Q2,200 - WH.E. - 1W2,pist
Entropy: mam(s2 - s1) = 1Q2/Tres + 0
=>
1Q2
= mam(s2 - s1)Tres
Process: P = const. ⇒ 1W2 = P(v2 - v1)mam
WH.E. = mam(s2 - s1)Tres - mam(h2 - h1)
NH 3
⇑ QL
H.E.
⇒W
⇑ QH
200 C
Table B.2.2: h1 = 1508.5, s1 = 5.1778, h2 = 1664.3, s2 = 5.6342
WH.E. = 2 × [(5.6342 - 5.1778)473.15 - (1664.3 - 1508.5)] = 120.3 kJ
10-10
10.15 Air flows through a constant pressure heating device, shown in Fig. P10.15. It is
heated up in a reversible process with a work input of 200 kJ/kg air flowing. The
device exchanges heat with the ambient at 300 K. The air enters at 300 K, 400
kPa. Assuming constant specific heat develop an expression for the exit
temperature and solve for it.
C.V. Total out to T0
s1 + qrev
0 /T 0 = s2
rev
h1 + qrev
= h2
0 -w
⇒ qrev
0 = T0(s2 - s1)
h2 - h1 = T0(s2 - s1) - wrev
(same as Eq. 10.12)
Constant Cp gives: Cp(T2 - T1) = T0Cp ln (T2/T1) + 200
T2 - T0ln(T2/T1) = T1 + 200/Cp
T1 = 300 , Cp = 1.004 , T0 = 300
T2 - 300 ln (T2/300) = 300 + (200/1.004) = 499.3
At 600 K
LHS = 392 (too low)
At 800 K
Linear interpolation gives T2 = 790 K
LHS = 505.75
(LHS = 499.5 OK)
10.16 Air enters the turbocharger compressor (see Fig. P10.16), of an automotive engine
at 100 kPa, 30°C, and exits at 170 kPa. The air is cooled by 50°C in an intercooler
before entering the engine. The isentropic efficiency of the compressor is 75%.
Determine the temperature of the air entering the engine and the irreversibility of
the compression-cooling process.
a) Compressor. First ideal which is reversible adiabatic, constant s:
T2S = T1
P2
(P )
k-1
k
= 303.2
1
(170
)
100
0.286
= 352.9 K
wS = CP0(T1 - T2S) = 1.004(303.2 - 352.9) = -49.9
w = wS/ηS = -49.9/0.75 = -66.5 kJ/kg = CP(T1 - T2)
⇒ T2 = 369.5 K
T3(to engine) = T2 - ∆TINTERCOOLER = 369.5 - 50
= 319.5 K = 46.3°C
b) Irreversibility from Eq.10.13 with rev. work from Eq.10.12, (q = 0 at TH)
319.4
170
kJ
s3 - s1 = 1.004 ln
 - 0.287 ln  = -0.1001
kg K
303.2
100
i = T(s3 - s1) - (h3 - h1) - w = T(s3 - s1) - CP(T3 - T1) - CP(T1 - T2)
= 303.2(-0.1001) - 1.004(-50) = +19.8 kJ/kg
10-11
10.17 A car air-conditioning unit has a 0.5-kg aluminum storage cylinder that is sealed
with a valve and it contains 2 L of refrigerant R-134a at 500 kPa and both are at
room temperature 20°C. It is now installed in a car sitting outside where the whole
system cools down to ambient temperature at −10°C. What is the irreversibility of
this process?
C.V. Aluminum and R-134a
Energy Eq.:
mAl(u2 - u1)Al + mR(u2 - u1)R = 1Q2 - 1W2 (1W2 = 0)
Entropy Eq.:
mAL(s2 - s1)Al + mR(s2 - s1)R = 1Q2/T0 + 1S2 gen
(u2 - u1)Al = Cv,Al(T2 - T1) = 0.9(-10 - 20) = - 27 kJ/kg
(s2 - s1)Al = Cp,Al ln(T2/T1) = 0.9 ln(263.15/293.15) = -0.09716
Table B.5.2: v1 = 0.04226, u1 = 411.65 - 500×0.04226 = 390.5 kJ/kg,
s1 = 1.7342 kJ/kg K,
v2 = v1 = 0.04226 & T2
=>
mR134a = V/v1 = 0.0473 kg
x2 = (0.04226 - 0.000755)/0.09845 = 0.4216
u2 = 186.57 + 0.4216×185.7 = 264.9, s2 = 0.9507 + 0.4216×0.7812 = 1.2801
1Q2
1S2 gen
1I2
= 0.5 × (-27) + 0.0473(264.9 - 390.5) = - 19.44 kJ
= 0.5 (-0.09716) + 0.0473(1.2801 - 1.7342) +
19.44
= 0.003815 kJ/K
263.15
= T0 ( 1S2 gen ) = 263.15 × 0.003815 = 1.0 kJ
10.18 A steady combustion of natural gas yields 0.15 kg/s of products (having
approximately the same properties as air) at 1100°C, 100 kPa. The products are
passed through a heat exchanger and exit at 550°C. What is the maximum
theoretical power output from a cyclic heat engine operating on the heat rejected
from the combustion products, assuming that the ambient temperature is 20°C?
PROD.e
PROD.i
Heat
T
i
1100
o
o
o
Exchanger
( C)
1100 C
550 C
.
e
⇓ QH
.
550
Heat
⇒
W
Engine
NET
.
∆S
⇓ QL
o
T L= 20 C
TL
.
.
QH = mPRODCP0(TL - Te)PROD = 0.15 × 1.004(1100 - 550) = 82.83
.
.
.
∆SL = - ∆Si-e = mPRODCP0 ln(Ti/Te)PROD
= 0.15 × 1.004 ln(1373.15/823.15) = 0.077 kW/K
.
.
QL = TL ∆SL = 293.15 × 0.077 = 22.57 kW
.
.
.
WNET = QH - QL = 82.83 - 22.57 = 60.26 kW
20
s
10-12
10.19 A counterflowing heat exchanger cools air at 600 K, 400 kPa to 320 K using a
supply of water at 20°C, 200 kPa. The water flow rate is 0.1 kg/s and the air flow
rate is 1 kg/s. Assume this can be done in a reversible process by the use of heat
engines and neglect kinetic energy changes. Find the water exit temperature and
the power out of the heat engine(s).
2
1
⇓
H2O
⇒
AIR
·
W
•
•
•
•
m
as 1 + mH2Os3 = mas 2 + mH2Os4
⇓
3
C.V. Total
•
•
•
•
•
m
ah1 + mH2Oh3 = mah2 + mH2Oh4 + W
Table A.7: h1 = 607.316,
4
° = 6.93413,
Table A.7: h2 = 320.576, sT2
B.1.1:
(sgen = 0)
° = 7.57638
sT1
h3 = 83.96, s3 = 0.2966
• •
s4 = (m
a/mH2O)(s1 - s 2) + s3 = (1/0.1)(7.57638 - 6.93413) + 0.2966 = 6.7191
4: P4 = P3, s4 ⇒ Table B.1.2:
x4 = (6.7191-1.530)/5.597 = 0.9271,
h4 = 504.68 + 0.9271 × 2201.96 = 2546.1 kJ/kg,
•
•
T4 = 120.20°C
•
W = ma(h1 - h2) + mH2O(h3 - h4)
= 1(607.32 - 320.58) + 0.1(83.96 - 2546.1) = 40.53 kW
10.20 Water as saturated liquid at 200 kPa goes through a constant pressure heat
exchanger as shown in Fig. P10.20. The heat input is supplied from a reversible
heat pump extracting heat from the surroundings at 17°C. The water flow rate is 2
kg/min and the whole process is reversible, that is, there is no overall net entropy
change. If the heat pump receives 40 kW of work find the water exit state and the
increase in availability of the water.
C.V. Heat exchanger + heat pump.
.
.
.
.
.
.
.
.
.
m1 = m2 = 2 kg/min, m1h1 + Q0 + Win = m1h2, m1s1 + Q/T0 = m1s2
.
.
Substitute Q0 into energy equation and divide by m1
h1 - T0s1 + win = h2 - T0s2
LHS = 504.7 - 290.15 × 1.5301 + 40×60/2 = 1260.7 kJ/kg
State 2: P2 , h2 - T0s2 = 1260.7 kJ/kg
At sat. vap. hg - T0sg = 638.8 so state 2 is superheated vapor at 200 kPa.
At 600 C:
h2 - T0s2 = 3703.96 - 290.15 × 8.7769 = 1157.34 kJ/kg
At 700 C:
h2 - T0s2 = 3927.66 - 290.15 × 9.0194 = 1310.68 kJ/kg
Linear interpolation ⇒ T2 = 667°C
∆ψ = (h2 - T0s2) - (h1 - T0s1) = win = 1200 kJ/kg
= 1260.7 - 504.7 + 290.15 × 1.5301 ≈ 1200 kJ/kg
10-13
10.21 Calculate the irreversibility for the process described in Problem 6.63, assuming
that heat transfer is with the surroundings at 17°C.
I = To Sgen so apply 2nd law out to To = 17 oC
m2 s2 - m1s1 = misi + 1Q2 / To + 1S2gen
ToSgen = To ( m2 s2 - m1s1 - mi si ) - 1Q2
m1 = 0.90 kg
mi = 3.082 kg
m2 =3.982 kg
ToSgen = I = To [ m1 (s2 - s1) + mi (s2 - si)] - 1Q2
= 290.15[0.9(Cp ln
350
400
350
400
- R ln
) + 3.082(Cpln
- R ln
)]
290
300
600
500
- ( - 819.2 kJ)
= 290.15 (0.0956 - 1.4705) + 819.2
= 420.3 kJ
10.22 The high-temperature heat source for a cyclic heat engine is a SSSF heat
exchanger where R-134a enters at 80°C, saturated vapor, and exits at 80°C,
saturated liquid at a flow rate of 5 kg/s. Heat is rejected from the heat engine to a
SSSF heat exchanger where air enters at 150 kPa and ambient temperature 20°C,
and exits at 125 kPa, 70°C. The rate of irreversibility for the overall process is
175 kW. Calculate the mass flow rate of the air and the thermal efficiency of the
heat engine.
.
C.V. R-134a Heat Exchanger, mR134a = 5 kg/s
Inlet: T1 = 80oC, sat. vap. ‡ x1 = 1.0, Table B.5.1
h1 = hg = 429.189 kJ/kg, s1 = sg = 1.6862 kJ/kg-K
Exit: T2 = 80oC, sat. liq. ‡ x2 = 0.0
h2 = hf = 322.794 kJ/kg, s2 = sf = 1.3849 kJ/kg-K
C.V. Air Heat Exchanger, C p = 1.004 kJ/kg-K, R = 0.287 kJ/kg-K
Inlet: T3 = 20oC, P3 = 150 kPa
Exit: T4 = 70oC, P4 = 125 kPa
.
.
.
.
2nd Law: I = To Snet = mR134a (s2 - s1) + mair (s4 - s3)
s4 - s3 = Cpln(T4/T3) - Rln(P4/P3) = 0.2103 kJ/kg-K
. .
.
mair = [I - mR134a (s2 - s1)]/(s4 - s3) = 10.0 kg/s
.
.
.
.
.
1st Law for each line:
Q + mhin = mhex + W; W = 0
.
.
.
.
R-134a: 1Q2 = mR134a(h2 - h1) = -532 kW; QH = -1Q2
.
.
.
.
.
Air: 3Q4 = mair (h4 - h3) = mair (T4 - T3) = 501.8 kW; QL = 3Q4
.
.
.
.
.
Wnet = QH - QL = 30.2 kW; ηth = Wnet / QH = 0.057,
or 5.7%
10-14
10.23 A control mass gives out 10 kJ of energy in the form of
a.
Electrical work from a battery
b.
Mechanical work from a spring
c.
Heat transfer at 500°C
Find the change in availability of the control mass for each of the three cases.
a) ∆φ = -Wel = -10 kJ
b) ∆φ = -Wspring = -10 kJ
 298.15
c) ∆φ = -[1 - (T0/TH)] Qout = -1  10 = -6.14 kJ
 773.15
10.24 Calculate the availability of the water at the initial and final states of Problem
8.32, and the irreversibility of the process.
1: u1 = 83.94
s1= 0.2966
ac
1W2 =
2: u2 = 3124.3
s2= 7.7621
ac
1Q 2
0: uo = 104.86 ,
so= 0.3673
203 kJ
v1= 0.001
= 3243.4 v2= 0.354
vo= 0.001003 ,
TH = 873.15 K
φ = (u - Tos) - (uo - Toso) + Po( v - vo)
φ1 = (83.94 - 298.15×0.2966) - (104.86 - 298.15×0.3673)
+ 100 (0.001002 - 0.001003) = 0.159 kJ/kg
φ2 = (3124.3 - 298.15×7.7621) - (104.86 - 298.15×0.3673)
+ 100 (0.35411 - 0.001003) = 850 kJ/kg
ac
ac
1I2 = m(φ1 - φ2) + [1-(T0/TH)]1Q2 - 1W2 + Po( V2 - V1)
= -849.84 + (1 -
298.15
) 3243.4 - 203 + 100 (0.3541 - 0.001)
873.15
= -849.84 + 2135.9 - 203 + 35.31 = 1118. kJ
[(Sgen = 3.75 kJ/K
ToSgen = 1118 kJ
so OK]
10.25 A steady stream of R-22 at ambient temperature, 10°C, and at 750 kPa enters a
solar collector. The stream exits at 80°C, 700 kPa. Calculate the change in
availability of the R-22 between these two states.
R-22
i
e
Solar
Collector
Inlet (T,P) Table B.4.1 (liquid)
hi = 56.46 kJ/kg, si = 0.2173 kJ/kg K
Exit (T,P) Table B.4.2 (sup. vap.)
he = 305.91 kJ/kg, se = 1.0761 kJ/kg K
∆ψ ie = ψe - ψi = (he - hi) - T0(se - si) = (305.912 - 56.463)
- 283.2(1.0761 - 0.2173) = 6.237 kJ/kg
10-15
10.26 Nitrogen flows in a pipe with velocity 300 m/s at 500 kPa, 300°C. What is its
availability with respect to an ambient at 100 kPa, 20°C?
ψ = h1 - h0 + (1/2)V12 - T0(s1 - s0)
= Cp(T1 - T0) + (1/2)V12 - T0(Cp ln(T1/T0) - R ln(P1/P0))
= 1.0426(300 - 20) + (300 2/2000)

573.15
500
- 293.151.042 ln
 - 0.2968ln  = 271.96 kJ/kg

293.15
100
10.27 A 10-kg iron disk brake on a car is initially at 10°C. Suddenly the brake pad
hangs up, increasing the brake temperature by friction to 110°C while the car
maintains constant speed. Find the change in availability of the disk and the
energy depletion of the car’s gas tank due to this process alone. Assume that the
engine has a thermal efficiency of 35%.
All the friction work is turned into internal energy of the disk brake.
m(u2 - u1) = 1Q2 - 1W2 ⇒ 1Q2 = mFeCFe(T2 - T1)
1Q2
= 10 × 0.45 × (110 - 10) = 450 kJ
Neglect the work to the surroundings at P 0
∆φ = m(u2 - u1) - T0m(s2 - s1)
383.15
m(s2-s1) = mC ln(T2/T1) = 10 × 0.45 × ln 
 = 1.361 kJ/K
283.15
∆φ = 450 - 283.15 × 1.361 = 64.63 kJ
Wengine = ηthQgas = 1Q2 = Friction work
Qgas = 1Q2/ηth = 450/0.35 = 1285.7 kJ
10-16
10.28 A 1 kg block of copper at 350°C is quenched in a 10 kg oil bath initially at ambient
temperature of 20°C. Calculate the final uniform temperature (no heat transfer
to/from ambient) and the change of availability of the system (copper and oil).
C.V. Copper and oil.
Cco = 0.42
Coil = 1.8
m2u2 - m1u1 = 1Q2 - 1W2 = 0 = mcoCco(T2 - T1)co + (mC)oil(T2 - T1)oil
1 × 0.42 ( T2 - 350) + 10 × 1.8 (T2 - 20) = 0
18.42 T2 = 507
=>
T = 27.5 C = 300.65 K
For each mass copper and oil, we neglect work term (v = C) so Eq.10.22 is
(φ2 - φ1) = u2 - u1 - To(s2 - s1) = mC [(T2 - T1) - Toln (T2 / T1) ]
mcv(φ2 - φ1)cv + moil (φ2 - φ1)oil =
300.65
300.65
] + 10 × 1.8 [7.5 - 293.15 ln
]
623.15
293.15
= - 45.713 + 1.698 = - 44.0 kJ
= 0.42 × [(-322.5) - 293.15 ln
10.29 Calculate the availability of the system (aluminum plus gas) at the initial and final
states of Problem 8.74, and also the process irreversibility.
State 1:
T1 = 200 oC, v1 = V1/ m = 0.05 / 1.1186 = 0.0447
v2 = v1 × (2 / 1.5) × (298.15 / 473.15) = 0.03756
The metal does not change volume, so the combined is using Eq.10.22 as
State 2:
φ1 = mgasφgas + mAlφAl
= mgas[u1-uo-To(s1 - so)]cv + mgasPo(v1-vo) + mAl[u1-uo -To(s1-so)]Al
T
P
= mgasCv (T1 - To) - mgasTo [Cp ln 1 - R ln 1 ] + mgasPo (v1 - vo)
To
Po
+ mAl [C (T1 - To) - ToC ln (T1/To) ]Al
φ1 = 1.1186 [ 0.653(200-25) - 298.15 (0.842 ln
473.15
2000
- 0.18892 ln
)
298.15
100
+ 100 (0.0447 - 0.5633 ) ] + 4 × 0.90 [ 200 -25 - 298.15 ln
473.15
]
298.15
= 128.88 + 134.3 = 263.2 kJ
φ2 = 1.1186 [ 0.653(25 - 25) - 298.15 (0.842 ln
298.15
1500
- 0.18892 ln
)
298.15
100
+ 100 (0.03756 - 0.5633 ) ] + 4 × 0.9 [ 25 -25 - 298.15 ln
298.15
]
298.15
= 111.82 + 0 = 111.82 kJ
1I2 = φ1 - φ2 + [1-(T0/TH)] 1Q2 - 1W2AC + Pom( V2 - V1)
= 263.2 - 111.82 + 0 - (-14) + 100 × 1.1186 (0.03756 - 0.0447) = 164.58 kJ
[(Sgen = 0.552
ToSgen = 164.58
so OK ]
10-17
10.30 Consider the springtime melting of ice in the mountains, which gives cold water
running in a river at 2°C while the air temperature is 20°C. What is the
availability of the water (SSSF) relative to the temperature of the ambient?
ψ = h1 - h0 - T0(s1 - s0)
flow availability from Eq.10.19
Approximate both states as saturated liquid
ψ = 8.392 - 83.96 - 293.15(0.03044 - 0.2966) = 2.457 kJ/kg
Why is it positive? As the water is brought to 20°C it can be heated with qL
from a heat engine using qH from atmosphere TH = T0 thus giving out work.
10.31 Refrigerant R-12 at 30°C, 0.75 MPa enters a SSSF device and exits at 30°C, 100
kPa. Assume the process is isothermal and reversible. Find the change in
availability of the refrigerant.
hi = 64.539, si = 0.2397, he = 209.866, se = 0.8482
∆ψ = he - hi - T0(se - si) = 209.866 - 64.539
- 298.15(0.8482 - 0.2397) = -36.1 kJ/kg
10.32 A geothermal source provides 10 kg/s of hot water at 500 kPa, 150°C flowing into
a flash evaporator that separates vapor and liquid at 200 kPa. Find the three fluxes
of availability (inlet and two outlets) and the irreversibility rate.
C.V. Flash evaporator chamber. SSSF with no work or heat transfer.
.
.
.
Cont. Eq.:
m1 = m2 + m3 ;
Vap.
.
.
.
2
1
Energy Eq.:
m1h1 = m2h2 + m3h3
Liq.
.
.
.
.
3
Entropy Eq.: m1s1 + Sgen = m2s2 + m3s3
B.1.1:
B.1.2:
ho = 104.87, so = 0.3673, h1 = 632.18, s1= 1.8417
h2 = 2706.63, s2= 7.1271, h3 = 504.68, s3= 1.530
. .
h -h
h1 = xh2 + (1 - x) h3 => x = m2/m1 = 1 3 = 0.0579
h2 - h3
.
.
.
.
m2 = xm1 = 0.579 kg/s
m3 = (1-x)m1 = 9.421 kg/s
.
Sgen = 0.579 × 7.1271 + 9.421 × 1.53 - 10 × 1.8417 = 0.124 kW/K
Flow availability Eq.10.22: ψ = (h - Tos) - (ho - Toso) = h - ho - To(s - so)
ψ1 = 632.18 - 104.87 - 298.15 (1.8417 - 0.3673) = 87.72
ψ2 = 2706.63 - 104.87 - 298.15 (7.1271 - 0.3673) = 586.33
ψ3 = 504.68 - 104.87 - 298.15 (1.53 - 0.3673) = 53.15
.
.
.
m1 ψ1 = 877.2 kW
m2ψ2 = 339.5 kW
m3ψ3 = 500.7 kW
. .
.
.
I = m1 ψ1 - m2 ψ2 - m3ψ3 = 37 kW
10-18
10.33 Air flows at 1500 K, 100 kPa through a constant pressure heat exchanger giving
energy to a heat engine and comes out at 500 K. What is the constant temperature
the same heat transfer should be delivered at to provide the same availability?
C.V. Heat exchanger. SSSF, no work term.
qout = h1 - h2 = 1635.8 - 503.36 = 1132.4 kJ/kg
Heat exch
1
T
∆ψ = (1 - o ) qout = ψ1 - ψ2
TH
.
⇓ QH
H.E.
⇒
2
.
W
.
⇓ QL
ψ1 - ψ2 = h1 - h2 - To ( s1 - s2 )
= 1132.4 - 298.15 (8.612 - 7.3869) = 1132.4 - 356.3 = 767.1 kJ/kg
T
1 - o = (ψ1 - ψ2 ) / qout = 767.1 / 1132.4 = 0.6774
TH
To
= 0.3226 => TH = 924 K
TH
10.34 A wooden bucket (2 kg) with 10 kg hot liquid water, both at 85°C, is lowered 400
m down into a mineshaft. What is the availability of the bucket and water with
respect to the surface ambient at 20°C?
C.V. Bucket and water. Both thermal availability and potential energy terms.
v1 ≈ v0 for both wood and water so work to atm. is zero.
Use constant heat capacity table A.3 for wood and table B.1.1 (sat. liq.) for
water. From Eq.10.22
φ1 - φ0 = mwood[u1 - u0 - T0(s1- s0)] + mH2O[u1- u0- T0(s1- s0)] + mtotg(z1- z0)
273.15 + 85
] + 10[ 355.82 - 83.94
293.15
= 2[1.26(85 - 20) - 293.15× 1.26 ln
- 293(1.1342 - 0.2966)] + 12 × 9.807 × (-400) /1000
= 15.85 + 263.38 - 47.07 = 232.2 kJ
10-19
10.35 A rigid container with volume 200 L is divided into two equal volumes by a
partition. Both sides contains nitrogen, one side is at 2 MPa, 300°C, and the other
at 1 MPa, 50°C. The partition ruptures, and the nitrogen comes to a uniform state
at 100°C. Assuming the surroundings are at 25°C find the actual heat transfer and
the irreversibility in the process.
m2 = mA + mB =
PA1VA
RTA1
+
PB1VB
RTB1
=
2000×0.1
1000×0.1
+
0.2968×573.15 0.2968×323.15
= 1.176 + 1.043 = 2.219 kg
P2 = m2RT2/Vtot = 2.219 × 0.2968 × 373.15/0.2 = 1228.8 kPa
Energy Eq.:
1Q2
mA(u2 - u1)A + mB(u2 - u1)B = 1Q2 - 0/
= mACv(T2 - T1)A + mBCv(T2 - T1)B
= 1.176 × 0.745 × (100 - 300) + 1.043 × 0.745 × (100 - 50) = -136.4 kJ
mA(s2 - s1)A + mB(s2 - s1)B = 1Q2/Tsur + 1Ss gen
373.15
1228.8
(s2 - s1)A = 1.042 × ln
- 0.2968 × ln
= -0.09356
573.15
2000
373.15
1228.8
(s2 - s1)B = 1.042 × ln
- 0.2968 × ln
= 0.0887
323.15
1000
1S2,gen
= 1.176× (-0.09356) + 1.043× 0.0887 + 136.4/298.15 = 0.4396 kJ/K
I = (T0)(1S2,gen) = 131.08 kJ
10.36 An air compressor is used to charge an initially empty 200-L tank with air up to 5
MPa. The air inlet to the compressor is at 100 kPa, 17°C and the compressor
isentropic efficiency is 80%. Find the total compressor work and the change in
availability of the air.
C.V. Tank + compressor (constant inlet conditions) USUF, no heat transfer.
Continuity: m2 - m1 = min ( m1 = 0 )
Energy:
m2u2 = minhin - 1W2
Entropy:
m2s2 = minsin + 1S2 gen
Reversible compressor: 1S2 GEN = 0 ⇒ s2 = sin ⇒ Pr2 = Prin(P2/Pin) = 49.495
⇒ T2,s = 855, u2,s = 637.2
⇒ 1w2,s = hin - u2,s = -346.77
Actual compressor: 1w2,AC = 1w2,s/ηc = -433.46
u2,AC = hin - 1w2,AC = 723.89 ⇒ T2,AC = 958 K ⇒ v2 = RT2/P2 = 0.055
State 2 u, P
m2 = V2/v2 = 3.636 ⇒ 1W2 = m2(1w2,AC) = -1576 kJ
m2(φ2 - φ1) = m2[u2 - u1 + P0(v2 - v1) - T0(s2 - s1)]
= 3.636 [723.89 - 207.19 + 100(0.055 - 0.8323) - 290[8.0867 6.8352 - 0.287 ln(5000/100)] = 1460.3 kJ
10-20
10.37 Air enters a compressor at ambient conditions, 100 kPa, 300 K, and exits at 800
kPa. If the isentropic compressor efficiency is 85%, what is the second-law
efficiency of the compressor process?
800 kPa
T2s = 300(8)0.286 = 543.8 K
-Ws = 1.004(543.8 - 300) = 244.6 kJ/kg
2
100 kPa
2s
-Ws 244.6
-w =
=
= 287.8 K
0.85
ηs
300 K
1
-w
287.8
= 300 +
= 586.8 K
T 2 = T1 +
s
CP0
1.004
s2 - s1 = 1.004 ln(586.8/300) - 0.287 ln 8 = 0.07645
ψ2 - ψ1 = (h2 - h1) - T0(s2 - s1) = 287.8 - 300(0.07645) = 264.9 kJ/kg
2nd law efficiency:
η2nd Law = (ψ2 - ψ1)/(-w) = 264.9/287.8 = 0.92
10.38 Steam enters a turbine at 25 MPa, 550°C and exits at 5 MPa, 325°C at a flow rate
of 70 kg/s. Determine the total power output of the turbine, its isentropic
efficiency and the second law efficiency.
hi = 3335.6, si = 6.1765, he = 2996.5, se = 6.3289
Actual turbine: wT,ac = hi - he = 339.1 kJ/kg
Isentropic turbine: se,s = si ⇒ he,s = 2906.6
wT,s = hi - he,s = 429 kJ/kg
Rev. turbine: wrev = wT,ac + T0(se - si) = 339.1 + 45.44 = 384.54 kJ/kg
ηT = wT,ac/wT,s = 339.1/429 = 0.79
ηII = wT,ac/wrev = 339.1/384.54 = 0.88
10.39 A compressor is used to bring saturated water vapor at 1 MPa up to 17.5 MPa,
where the actual exit temperature is 650°C. Find the irreversibility and the
second-law efficiency.
Inlet state: Table B.1.2
hi = 2778.1,
si = 6.5864
Actual compressor Table B.1.3
h e,ac = 3693.9 , se,ac = 6.7356
Energy Eq. Actual compressor:
-wc,ac = he,ac - hi = 915.8 kJ/kg
From Eq.10.14: i = T0(se,ac - si) = 298.15 (6.7356 - 6.5864) = 44.48 kJ/kg
From Eq.10.13:
wrev = i + w c,ac = -915.8 + 44.48 = -871.32
ηII = -wrev/wc,ac = 871.32/915.8 = 0.951
10-21
10.40 A flow of steam at 10 MPa, 550°C goes through a two-stage turbine. The pressure
between the stages is 2 MPa and the second stage has an exit at 50 kPa. Assume
both stages have an isentropic efficiency of 85%. Find the second law efficiencies
for both stages of the turbine.
1
2
3
CV: T1, h1 = 3500.9 kJ/kg, s1 = 6.7561 kJ/kg K
s2s = s1 ⇒ h2s = 3017.9
wT1,s = h1 - h2s = 483 kJ/kg
Isentropic
T1
T2
Actual T1:
wT1,ac = ηT1 wT1,s = 410.55 = h 1 - h2ac
h2ac = h1 - wT1,ac = 3090.35,
s2ac = 6.8782
CV: T2, s3s = s2ac = 6.8782 ⇒ x3s = (6.8782-1.091)/6.5029 = 0.8899,
h3s = 340.47 + 0.8899 × 2305.4 = 2392.2 kJ/kg
wT2,s = h2ac - h3s = 698.15 ⇒
wT2,ac = ηT2 wT2,s = 593.4 kJ/kg
⇒ h3ac = 2496.9, x3ac = (2496.9 - 340.47)/2305.4 =0.9354,
s3ac = 1.091 + 0.9354 × 6.5029 = 7.1736 kJ/kg K
Actual T1:
iT1,ac = T0(s2ac-s1) = 298.15(6.8782 - 6.7561) = 36.4 kJ/kg
R
⇒ wT1 = wT1,ac + i = 447 kJ/kg,
R
ηII = wT1,ac/wT1 = 0.918
Actual T2: iT2,ac = T0(s3ac-s2ac) = 298.15(7.1736 - 6.8782) = 88.07 kJ/kg
R
⇒ wT2 = wT2,ac + iT2,ac = 681.5,
R
ηII = wT2,ac/wT2 = 0.871
10.41 Consider the two-stage turbine in the previous problem as a single turbine from
inlet to final actual exit and find its second-law efficiency.
From solution to Problem 10.40 we have
wT,ac = wT1,ac + wT2,ac = h1 - h3ac = 1004 kJ/kg
The actual compared to the reversible turbine has, Eq.10.14,
R
i = qR = T0(s3 - si) = qR
T1 + qT2 = 124.47 kJ/kg
wR = wT,ac + i = 1128.5 kJ/kg
ηII = wT,ac/wR = 0.89
10-22
10.42 The simple steam power plant shown in Problem 6.39 has a turbine with given
inlet and exit states. Find the availability at the turbine exit, state 6. Find the
second law efficiency for the turbine, neglecting kinetic energy at state 5.
h6 = 2393.2, s6 = 7.5501, h5 = 3404.3, s5 = 6.8953
ψ6 = h6 - h0 - T0(s6 - s0)
= 2393.2 - 104.89 - 298.15(6.8953 - 0.3674) = 146.79 kJ/kg
w rev = ψ5 - ψ6 = h5 - h6 - T0(s5 - s6) = 1206.3 kJ/kg.
ηII = wac/wrev = 0.838
wac = h5 - h6 = 1011.1 ;
10.43 A compressor takes in saturated vapor R-134a at −20°C and delivers it at 30°C,
0.4 MPa. Assuming that the compression is adiabatic, find the isentropic
efficiency and the second law efficiency.
Table B.5
hi = 386.08, si = 1.7395, he,ac = 423.22, se,ac = 1.7895
Ideal exit: Pe, se,s = si ⇒ he,s = 408.51
Isentropic compressor wc,s = he,s - hi = 22.43
Actual compressor wc,ac = he,ac - hi = 37.14
Reversible between inlet and actual exit
-wc,rev = hi - he,ac - T0(si - se,ac) = -37.14 - 298.15(1.7395 - 1.7895) = -22.23
ηTH,c = (wc,s/wc,ac) = (22.43/37.14) = 0.604
ηII = (wc,rev/wc,ac) = (22.23/37.14) = 0.599
10.44 Steam is supplied in a line at 3 MPa, 700°C. A turbine with an isentropic
efficiency of 85% is connected to the line by a valve and it exhausts to the
atmosphere at 100 kPa. If the steam is throttled down to 2 MPa before entering
the turbine find the actual turbine specific work. Find the change in availability
through the valve and the second law efficiency of the turbine.
Take C.V. as valve and a C.V. as the turbine.
Valve:
h2 = h1 = 3911.7 ,
s2 > s1 = 7.7571 ,
h2, P2 ⇒ s2 = 7.9425
ψ1 - ψ2 = h1−h2 −T0(s1-s2) = 0 -298.15(7.7571-7.9425) = 55.3 kJ/kg
So some potential work is lost in the throttling process.
Ideal turbine: s3 = s2 ⇒ h3s = 2929.13
wT,s = 982.57 kJ/kg
wT,ac = h2 - h3ac = ηwT,s = 835.2 kJ/kg
h3ac = 3911.7 - 835.2 = 3076.5
⇒ s3ac = 8.219 kJ/kg K
wrev = h2 - h3ac - T0(s2 - s3ac) = 835.2 - 298.15(7.9425 - 8.219)
= 917.63 kJ/kg
⇒
ηII = 835.2/917.63 = 0.91
10-23
10.45 The condenser in a refrigerator receives R-134a at 700 kPa, 50°C and it exits as
saturated liquid at 25°C. The flowrate is 0.1 kg/s and the condenser has air
flowing in at ambient 15°C and leaving at 35°C. Find the minimum flow rate of
air and the heat exchanger second-law efficiency.
C.V. Total heat exchanger.
4
3
AIR
2
1
R-134a
•
•
m
a = m1 ×
⇒
•
•
•
•
m
1h1 + mah3 = m1h2 + mah4
h1 - h2
h4 - h3
436.89 - 234.59
= 1.007 kg/s
1.004(35 - 15)
= 0.1 ×
ψ1 - ψ2 = h1 - h2 - T0(s1 - s2) = 436.89 - 234.59
- 288.15(1.7919 - 1.1201) = 8.7208
ψ4 - ψ3 = h4 - h3 - T0(s4 - s3)
= 1.004(35 - 15) - 288.15 × 1.004 × ln
•
•
ηII = ma(ψ4 - ψ3)/m1(ψ1 - ψ2) =
308.15
= +0.666
288.15
1.007(0.666)
= 0.77
0.1(8.7208)
10.46 A piston/cylinder arrangement has a load on the piston so it maintains constant
pressure. It contains 1 kg of steam at 500 kPa, 50% quality. Heat from a reservoir
at 700°C brings the steam to 600°C. Find the second-law efficiency for this
process. Note that no formula is given for this particular case so determine a
reasonable expression for it.
1: Table B.1.2
P 1 , x1
⇒
v1 = 0.001093 + 0.5×0.3738 = 0.188,
h1 = 640.21 + 0.5×2108.47 = 1694.5 , s1 = 1.8606 + 0.5×4.9606 = 4.341
2: P2 = P1,T2 ⇒ v2 = 0.8041, h2 = 3701.7 , s2 = 8.3521
m(u2 - u1) = 1Q2 - 1W2 = 1Q2 - P(V2 - V1)
1Q2
1W2
= m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1) = 2007.2 kJ
= Pm(v2 - v1) = 308.05 kJ
1W2 to atm =
P0m(v2 - v1) = 61.61 kJ
Useful work out = 1W2 - 1W2 to atm = 246.44
 298.15
∆φ reservoir = (1 - T0/Tres)1Q2 = 1  2007.2 = 1392.2
 973.15
ηII = Wnet/∆φ = 0.177
10-24
10.47 Air flows into a heat engine at ambient conditions 100 kPa, 300 K, as shown in
Fig. P10.47. Energy is supplied as 1200 kJ per kg air from a 1500 K source and in
some part of the process a heat transfer loss of 300 kJ/kg air happens at 750 K.
The air leaves the engine at 100 kPa, 800 K. Find the first and the second law
efficiencies.
C.V. Engine out to reservoirs
hi + q1500 = q750 + he + w
wac = 300.473 + 1200 - 300 - 822.202 = 378.27 kJ/kg
ηTH = w/q1500 = 0.3152
For second law efficiency also a q to/from ambient
si + (q1500/TH) + (q0/T0) = (q750/Tm) + se
q0 = T0(se - si) + (T0/Tm)q750 - (T0/TH)q1500

100 300
= 3007.88514 - 6.86925 - 0.287 ln  +
300
100 750

-(300/1500) 1200 = 184.764 kJ/kg
wrev = hi - he + q1500 - q750 + q0 = wac + q0 = 563.03 kJ/kg
ηII = wac/wrev = 378.27/563.03 = 0.672
10.48 Consider the high-pressure closed feedwater heater in the nuclear power plant
described in Problem 6.42. Determine its second-law efficiency.
For this case with no work the second law efficiency is from Eq. 10.25:
•
•
ηII = m
16(ψ18 - ψ16)/m17(ψ17 - ψ15)
Properties (taken from computer software):
h15 = 585
h16 = 565
h17 = 2593
h18 = 688
s15 = 1.728
s16 = 1.6603
s17 = 6.1918
s18 = 1.954
ψ18 - ψ16 = h18 - h16 - T0(s18 - s16) = 35.433
ψ17 - ψ15 = h17 - h15 - T0(s17 - s15) = 677.12
ηII = (75.6 × 35.433)/(4.662 × 677.12) = 0.85
10-25
10.49 Consider a gasoline engine for a car as an SSSF device where air and fuel enters
at the surrounding conditions 25°C, 100 kPa and leaves the engine exhaust
manifold at 1000 K, 100 kPa as products assumed to be air. The engine cooling
system removes 750 kJ/kg air through the engine to the ambient. For the analysis
take the fuel as air where the extra energy of 2200 kJ/kg of air released in the
combustion process, is added as heat transfer from a 1800 K reservoir. Find the
work out of the engine, the irreversibility per kilogram of air, and the first- and
second-law efficiencies.
2
1
AIR
C.V. Total out to reservoirs
•
•
•
•
•
m
ah1 + Q H = mah2 + W + Q out
•
•
•
•
•
m
as 1 + Q H/TH + S gen = mas s + Q out/T 0
° = 6.8631
h1 = 298.61
sT1
⇒ W·
·
·
⇓ Qout ⇑ QH
T0
TH
h2 = 1046.22
° = 8.1349
sT1
•
•
wac = W/m
a = h1 - h2 + qH - qout = 298.6 - 1046.22 + 2200 - 750 = 702.4 kJ/kg
ηTH = w/qH = 702.4/2200 = 0.319
sgen = s2 - s1 +
qout qH
750
2200
= 8.1349 - 6.8631 +
= 2.565 kJ/kg K
298.15 1800
T0 TH
itot = (T0)sgen = 764.8 kJ/kg
For reversible case have
sgen = 0 and qR0 from T0, no qout
qR
0,in = T0(s2 - s1) - (T0/TH)qH = 14.78 kJ/kg
wrev = h1 - h2 + qH + qR
0,in = wac + itot = 1467.2 kJ/kg
ηII = wac/wrev = 0.479
10.50 Air enters a steady-flow turbine at 1600 K and exhausts to the atmosphere at 1000
K. The second law efficiency is 85%. What is the turbine inlet pressure?
C.V.: Turbine, exits to atmosphere so assume Pe = 100 kPa
o
Inlet: Ti = 1600 K, Table A.7: hi = 1757.3 kJ/kg, si = 8.1349 kJ/kg K
o
Exit: Te = 1000 K, he = 1046.2 kJ/kg, se = 8.6905 kJ/kg K
1st Law: q + hi = he + w; q = 0
=> w = (hi - he) = 711.1 kJ/kg
2nd Law: ψi - ψe = w/η2ndLaw = 711.1/0.85 = 836.6 kJ/kg
ψi - ψe = (hi - he) - To(si - se) = 836.6 kJ/kg
hi - he = w = 711.1 kJ/kg, assume To = 25oC ‡ si - se = 0.4209 kJ/kg-K
o
o
si - se = se - si - R ln(Pi/Pe) = 0.4209
Pi = 3003 kPa
=> Pe/Pi = 30.03;
10-26
10.51 Consider the two turbines in Problem 9.72. What is the second-law efficiency for
the combined system?
From the solution to Problem 9.72 we have
•
•
s1 = 6.2828, s2 = 6.7240, s5 = 6.5233, m
T1/mtot = 0.6848
•
•
m
T1/mtot = 0.3152
=>
•
•
wac = W
net/mtot = 301.775 kJ/kg
C.V. Total system out to T0
•
• 
• 
rev
rev
rev
•
m
⇒ wqc + qrev
T1/mtoth1 + mT2/mtoth2 + q0 = h5 + w
0 =w
•
• 
• 
rev
•
m
T1/mtots 1 + mT2/mtots 2 + q0 = s5
⇒ qrev
0 = 298.15[6.5233 - 0.6848 × 6.2825 - 0.3152 × 6.724] = 30.242 kJ/kg
wrev = wac + qrev
0 = 301.775 + 30.242 = 332 kJ/kg
ηII = wac/wrev = 301.775/332 = 0.909
10.52 Air in a piston/cylinder arrangement is at 110 kPa, 25°C, with a volume of 50 L.
It goes through a reversible polytropic process to a final state of 700 kPa, 500 K,
and exchanges heat with the ambient at 25°C through a reversible device. Find the
total work (including the external device) and the heat transfer from the ambient.
C.V. Total out to ambient
ma(u2 - u1) = 1Q2 - 1W2,tot ,
ma(s2 - s1) = 1Q2/T0
ma = 110 × 0.05/0.287 × 298.15 = 0.0643 kg
1Q2
= T0ma(s2 - s1) = 298.15 × 0.0643[7.3869 - 6.8631
- 0.287 ln (700/110)] = -0.14 kJ
1W2,tot
= 1Q2 - ma(u2 - u1)
= -0.14 - 0.0643 × (359.844 - 213.037) = -9.58 kJ
10-27
Advanced Problems
10.53 Refrigerant-22 is flowing in a pipeline at 10°C, 600 kPa, with a velocity of 200
m/s, at a steady flowrate of 0.1 kg/s. It is desired to decelerate the fluid and
increase its pressure by installing a diffuser in the line (a diffuser is basically the
opposite of a nozzle in this respect). The R-22 exits the diffuser at 30°C, with a
velocity of 100 m/s. It may be assumed that the diffuser process is SSSF,
polytropic, and internally reversible. Determine the diffuser exit pressure and the
rate of irreversibility for the process.
C.V. Diffuser out to T0, Int. Rev. flow ⇒ sgen R-22 = 0/
•
•
2
2
m
1 = m2 , h1 + (1/2)V1 + q = h2 + (1/2)V2
s1 + ∫ dq/T + sgen = s2 = s1 + q/T0 + sgen
Rev. Polytropic Process:
w12 = 0/ = - ∫ v dP + (V12 - V22)/2
n
1
2
2
1
LHS = ∫ v dP = n-1(P2v2 - P1v1) = 2(V1 - V2) = 2(2002 - 1002)/1000 = 15 kJ/kg
P1v1n = P2v2n ⇒ n = ln(P2/P1)/ln(v1/v2)
Inlet state: v1 = 0.04018, s1 = 0.9295, h1 = 255.1
Exit state: T2,V2, ? ⇒ only P2 unknown.
⇒
v2 = v(P2), n = n(P2)
Trial and error on P2 to give LHS = 15:
P2 = 1 MPa ⇒ v2 = 0.0246,
1000
0.04018
n = ln 600  / ln 0.0246  = 1.0412




LHS = (1.0412/0.0412)(1000 × 0.0246 - 600 × 0.04018) = 12.43
LHS too small so P2 > 1.0 MPa
Assume sat. vapor @ 30°C, P2 = 1.1919 MPa
v2 = 0.01974,
LHS =
1.1919
0.04018
n = ln 0.6  /ln0.01974  = 0.9657

 

0.9657
(1191.9 × 0.01974 - 600 × 0.04018) = 16.33 > 15
-0.0343
P2 = 1.1 MPa ⇒ v2 = 0.021867, n = 0.9963, LHS = 14.596
Interpolate to get the final state:
P2 = 1.121 MPa, v2 = 0.021353, n = 0.9887, LHS = 15.01 OK
s2 = 0.8967 h2 = 260.56
Find q from energy equation (Assume T0 = T1 = 10°C)
1
q = h2 - h1 + 2(V22 - V12) = 260.56 - 255.1 - 15 = -9.54 kJ/kg
.
I• = m[T0(s2 - s1) - q ]= 0.1[283.15(0.8967 - 0.9295) + 9.54] = 0.0253 kW
10-28
10.54 A piston/cylinder contains ammonia at −20°C, quality 80%, and a volume of 10 L.
A force is now applied to the piston so it compresses the ammonia in an adiabatic
process to a volume of 5 L, where the piston is locked. Now heat transfer with the
ambient takes place so the ammonia reaches the temperature of the ambient at
20°C. Find the work and heat transfer. If it is done in a reversible process, how
much work and heat transfer would be involved?
P
Process 1 → 2 → 3
1: v1 = 0.49926 m3/kg, s1 = 4.5658 kJ/kg K
u1 = 1152.25 - 94.9 = 1057.4 kJ/kg
m = V1/v1 = 0.010/0.49926 = 0.02 kg
V
V2 V1
Assume reversible adiabatic compression
m(u2 - u1) = - 1W2 ;
m(s2 - s1) = 0, v2 = v1V2/V1 = 0.2496
2: s2 = s1, v1 ⇒ Trial and error on T2.
For 0°C, v2 ⇒ s = 4.6925,
For -2°C, v2 ⇒ s = 4.431
⇒ T2 = -1°C, x = 0.832, u2 = 1125 kJ/kg
1W2
= -m(u2 - u1) = - 1.345 kJ, (1Q2 = 0)
3: T3, v3 = v2 ⇒ P3 = 537 kPa, u3 = 1351.6, s3 = 5.382
2Q3
= m(u3 - u2) = 0.02(1351.6 - 1125) = 4.532 kJ
Qrev = mT0(s3 - s1) = 0.02×293.15(5.382 - 4.5658) = 4.785 kJ
Wrev = Qrev - m(u3 - u1) = 4.785 - 0.02(1351.6-1057.4) = -1.1 kJ
10-29
10.55 Consider the irreversible process in Problem 8.34. Assume that the process could
be done reversibly by adding heat engines/pumps between tanks A and B and the
cylinder. The total system is insulated, so there is no heat transfer to or from the
ambient. Find the final state, the work given out to the piston and the total work to
or from the heat engines/pumps.
C.V. Water mA + mB + heat engines. No Qexternal, only 1W2,cyl + WHE
m2 = mA1 + mB1 = 6 kg, m2u2 - mA1uA1 - mB1uB1 = -1W2,cyl - WHE
m2s2 - mA1sA1 - mB1sB1 = 0/ + 0/
vA1 = 0.06283 uA1 = 3448.5 sA1 = 7.3476 VA = 0.2513 m3
vB1 = 0.09053 uB1 = 2843.7 sB1 = 6.7428 VB = 0.1811 m3
m2s2 = 4×7.3476 + 2×6.7428 = 42.876 ⇒ s2 = 7.146 kJ/kg K
If P2 < Plift = 1.4 MPa then
V2’ = VA + VB = 0.4324 m3 , v2’ = 0.07207 m3/kg
(Plift , s2) ⇒ v2 = 0.20135 ⇒ V2 = 1.208 m3 > V2’
OK
⇒ P2 = Plift = 1.4 MPa u2 = 2874.2 kJ/kg
1W2,cyl =
Plift(V2 - VA - VB) = 1400×(1.208 - 0.4324) = 1085.84 kJ
WHE = mA1uA1 + mB1uB1 - m2u2 - 1W2,cyl
= 4 × 3447.8 + 2 × 2843.7 - 6 × 2874.2 - 1085.84 = 1147.6 kJ
10-30
10.56 Water in a piston/cylinder is at 100 kPa, 34°C, shown in Fig. P10.56. The cylinder
has stops mounted so Vmin = 0.01 m3 and Vmax = 0.5 m3. The piston is loaded
with a mass and outside P0, so a pressure inside of 5 MPa will float it. Heat of
15000 kJ from a 400°C source is added. Find the total change in availability of
the water and the total irreversibility.
P
1a
2
1
1b
v
CV water plus cyl. wall out to reservoir
m2 = m1 = m, m(u2- u1) = 1Q2 - 1W2
m(s2 - s1) = 1Q2/Tres + 1S2 gen
System eq: States must be on the 3 lines
Here Peq = 5 MPa, so since P1 < Peq ⇒ V1 = Vmin = 0.01 m3
1: v1 = 0.0010056 m3/kg, u1 = 142.49 kJ/kg, s1 = 0.4916 kJ/kg K
⇒ m = V1/v1 = 9.9443 kg
Heat added gives q = Q/m = 1508.4 kJ/kg so check if we go from 1 past 1a
(Peq,v1) and past 1b (Peq,vmax)
1a: T = 40°C, u1a = 166.95
⇒
1q1a
1b: T = 337.5°C, u1b= 2780.94
1q1b =
= u1a - u1 = 24.5 kJ/kg
⇒ as 1w1b = Peq(v1b - v1) = 246.4
h1b - u1 - Peqv1 = 2884.8
The state is seen to fall between 1a and 1b so P2 = Peq.
1w2
= Peq(v2 - v1), u2 - u1 = 1q2 - 1w2
⇒
u2 + Peqv2 = h2 = 1q2 + u1 + Peqv1 = 1655.9 < hg(P2) => T2 = 264°C
x2 = (1655.9 - 1154.21)/1640.12 = 0.3059,
v2 = 0.001286 + 0.3059 × 0.03815 = 0.01295,
s2 = 2.9201 + 0.3059 × 3.0532 = 3.8542
The non-flow availability change of the water (Eq.10.22)
m(φ2 - φ1) = m[u2 - u1 - T0(s2 - s1) + P0(v2 - v1)]
= 9.944[1591.15 - 142.49 - 298.15(3.8542 - 0.4916)
+ 100(0.01295 - 0.0010056)] = 9.944 × 447.3 = 4448 kJ
I = T(1S2 gen) = T [m(s2 - s1) - 1Q2/Tres ]=
298.15 [ 9.944(3.8542 - 0.4916) - 15000/673.15] = 3326 kJ
(Piston and atm. have an increase in availability)
10-31
10.57 A rock bed consists of 6000 kg granite and is at 70°C. A small house with lumped
mass of 12000 kg wood and 1000 kg iron is at 15°C. They are now brought to a
uniform final temperature with no external heat transfer.
a. For a reversible process, find the final temperature and the work done in the
process.
b.
If they are connected thermally by circulating water between the rock bed
and the house, find the final temperature and the irreversibility of the process,
assuming that surroundings are at 15°C.
a) For a reversible process a heat engine is installed between the rockbed and
the house. Take C.V. Total
mrock(u2 - u1) + mwood(u2 - u1) + mFe(u2 - u1) = -1W2
mrock(s2 - s1) + mwood(s2 - s1) + mFe(s2 - s1) = 0/
(mC)rockln
T2
T1
+ (mC)woodln
T2
T1
+ (mC)Feln
T2
T1
= 0/
6000 × 0.89 ln (T2/343.15) + 12000 × 1.26 ln (T2/288.15)
+ 1000 × 0.46 ln (T2/288.15) = /0
=>
T2 = 301.3 K
-1W2 = 6000 × 0.89(301.3 - 343.15)
+ (12000 × 1.26 + 460)(301.3 - 288.15)
⇒ 1W2 = 18602 kJ
b) No work, no Q irreversible process.
(mC)rock(T2 - 70) + (mCwood + mCFe)(T2 - 15) = 0/
T2 = 29.0°C = 302.2 K
Sgen = ∑mi(s2 - s1)i = 5340 ln
1I2
302.2
302.2
+ 15580 ln
= 63.13 kJ/K
343.15
288.15
= (T0)1S2,gen = 288.15 × 63.13 = 18191 kJ
10-32
10.58 Consider the heat engine in Problem 10.47. The exit temperature was given as
800 K, but what are the theoretical limits for this temperature? Find the lowest
and the highest, assuming that the heat transfers are as given. For an exit
temperature that is the average of the highest and lowest possible, find the firstand second-law efficiencies for the heat engine.
The lowest exhaust temperature will occur when the maxumum amount of
work is delivered which is a reversible process. Assume no other heat
transfers then
2nd law:
si + qH/TH + 0/ = se + qm/Tm
se - si = qH/TH - qm/Tm = s°Te - s°Ti - R ln(Pe/Pi)
s°Te = s°Ti + R ln(Pe/Pi) + qH/TH - qm/Tm
= 6.86926 + 0.287 ln(100/100) + 1200/1500 - 300/750
= 7.26926 ⇒ Te,min = 446 K
The maximum exhaust temperature occurs with no work out
hi + qH = qm + he ⇒ he = 300.473 + 1200 - 300 = 1200.5
Te,max = 1134 K
Te = (1/2)(Te,min + Te,max) = 790 K
see solution for 10.47 to follow calculations
he = 811.23,
s°Te = 7.8713
wac = 300.47 + 1200 - 300 - 811.23 = 389.24,
q0 = 300[7.8713 - 6.86926 - 0.287 ln
wrev = wac + q0 = 569.85,
ηTH = wac/qH = 0.32
100 300 1200
+
] = 180.61 kJ/kg
100 750 1500
ηII = wac/wrev = 389.24/569.85 = 0.683
10-33
10.59 Air in a piston/cylinder arrangement, shown in Fig. P10.59, is at 200 kPa, 300 K
with a volume of 0.5 m3. If the piston is at the stops, the volume is 1 m3 and a
pressure of 400 kPa is required to raise the piston. The air is then heated from the
initial state to 1500 K by a 1900 K reservoir. Find the total irreversibility in the
process assuming surroundings are at 20°C.
m2 = m1 = m = P1V1/RT1
P
2
Pstop
1
= 200 × 0.5/0.287 × 300 = 1.161 kg
m(u2 - u1) = 1Q2 - 1W2
1W 2
P1
Vstop
V1
⌠ dQ/T + 1S2 gen
m(s2 - s1) = ⌡
Process: P = P0 + α(V-V0)
if V ≤ Vstop
Information: Pstop = P0 + α(Vstop-V0)
Eq. of state ⇒ Tstop = T1PstopVstop/P1V1 = 1200 < T2
So the piston will hit the stops => V 2 = Vstop
1
1W2 = 2(P 1
1Q2 =
1
+ Pstop)(Vstop- V1) = 2(200 + 400)(1 - 0.5) = 150 kJ
M(u2 - u1) + 1W2 = 1.161(1205.25 - 214.36) + 150 = 1301 kJ
o
o
s2 - s1 = sT2 - sT1 -R ln(P2/P1) = 8.6121 - 6.8693 - 0.287 ln 2.5 = 1.48 kJ/kg K
Take control volume as total out to reservoir at T RES
1S2 gen tot =
1I2
m(s2 - s2) - 1Q2/TRES = 1.034 kJ/K
= T0(1S2 gen) = 293.15 × 1.034 = 303 kJ
10-34
10.60 Consider two rigid containers each of volume 1 m 3 containing air at 100 kPa, 400 K.
An internally reversible Carnot heat pump is then thermally connected between them
so it heats one up and cools the other down. In order to transfer heat at a reasonable
rate, the temperature difference between the working substance inside the heat pump
and the air in the containers is set to 20°C. The process stops when the air in the
coldest tank reaches 300 K. Find the final temperature of the air that is heated up, the
work input to the heat pump, and the overall second-law efficiency.
A
⇐ H.P. ⇐
QA
QB
⇑W
The high and the low temperatures in the heat
pump are TA+20 and TB-20, respectively.
Since TA and TB change during the process,
B
H.P.
the coefficient of performance changes, and so it must be integrated.
dUA = mAduA = mACvdTA = dQA dUB = mBduB = mBcvdTB = -dQB
Carnot heat pump:
⇒
dQA
dQB
=
TA + 20
TB - 20
=
-mACvdTA -dTA
=
mBCvdTB dTB
(dTA)/(TA + 20) = - (dTB)/(TB - 20)
⌠ dTA
TA2 + 20 ⌠ dTB
TB2 - 20

⇒
= ln
= -
= -ln
TA1 + 20 ⌡TB - 20
TB1 - 20
⌡TA + 20
TA1 = TB1 = 400 K, TB2 = 300 K ⇒
ln
TA2 + 20
420
= -ln
300 - 20
⇒ TA2 = 550 K
380
mA = mB = P1V1/RT1 = (100 × 1)/(0.287 × 400) = 0.871 kg
WH.P. = QA - QB = mCv(TA2 - TA1) + mCv(TB2 - TB1)
= 0.871 × 0.717 (550 - 400 + 300 - 400) = 31.2 kJ
Second law efficiency is total increase in air availability over the work input.
∆φ = mCv(T2 - T1) + P0(V2 - V1) - T0(s2 - s1)
∆φ A = QA - T0m(s2 - s1)A
∆φB = -QB - T0m(s2 - s1)B
∆φ A + ∆φ B = WH.P. - mT0[ (s2 - s1)A + (s2 - s1)B]
To find entropies we need the pressures (P2/P1) = T2/T1 for both A and B
s2 - s1 = Cpln(T2/T1) - R ln(T2/T1) = Cvln(T2/T1)
∆φ tot = WH.P. - mT0Cv[ln(T2/T1)A + ln(T2/T1)B]
 550
300
= 31.2 - 0.871 × 298.15 × 0.7175 × ln
+ ln
 = 25.47
400
 400
ηII = ∆φ tot/WH.P. = 25.47/31.2 = 0.816
10-35
ENGLISH UNIT PROBLEMS
10.61ECalculate the reversible work and irreversibility for the process described in
Problem 5.122, assuming that the heat transfer is with the surroundings at 68 F.
P
Linear spring gives
1
1
=⌠
⌡PdV = (P1 + P2)(V2 - V1)
2
1Q 2 = m(u2 - u1) + 1W2
Equation of state: PV = mRT
2
1W2
v
State 1: V1 = mRT1/P1 =
4 × 35.1 × (750 + 460)
= 16.85 ft3
70 × 144
State 2: V2 = mRT2/P2 =
4 × 35.1 × (75 + 460)
= 11.59 ft3
45 × 144
1W2
1
= 2(70 + 45)(11.59 – 16.85) x144/778 = -55.98 Btu
From Table C.7
Cp(Tavg) = [(6927-0)/(1200-537)]/M = 10.45/44.01 = 0.2347 Btu/lbm R
⇒ CV = Cp – R = 0.2375 – 35.10/778 = 0.1924
1Q 2
rev
1W2 =
= mCV(T2 - T1) + 1W2 = 4x 0.1924(75 - 750) - 55.98 = -575.46 Btu
ac
To(S2 - S1) - (U2 - U1) + 1Q2(1 - To/TH) = Tom(s2 - s1)+ 1W2 - 1Q2To/To
ac
= Tom[CP ln(T2 / T1) − R ln(P2 / P1)] + 1W2 - 1Q2
= 527.7 x 4 [ 0.2347 ln(535/1210) - 0.0451 ln(45/70)] - 55.98 + 575.46
= -362.24 - 55.98 + 575.46 = 157.2 Btu
1I 2
rev
ac
= 1W2 - 1W2 = 157.2 - (-55.98) = 213.2 Btu
10.62EThe compressor in a refrigerator takes refrigerant R-134a in at 15 lbf/in.2, 0 F and
compresses it to 125 lbf/in.2, 100 F. With the room at 70 F find the reversible heat
transfer and the minimum compressor work.
C.V. Compressor out to ambient. Minimum work in is the reversible work.
SSSF, 1 inlet and 2 exit energy Eq.:
wc = h1 - h2 + qrev
Entropy Eq.: s2 = s1 + ∫ dq/T + sgen = s1 + qrev/T0 + 0 => qrev = T0(s2- s1)
qrev = 529.67×(0.41262 - 0.42288) = -5.43 Btu/lbm
wc min = h1 - h2 + T0(s2- s1) = 167.193 - 181.059 - 5.43 = -19.3 Btu/lbm
10-36
10.63EA supply of steam at 14.7 lbf/in.2, 320 F is needed in a hospital for cleaning
purposes at a rate of 30 lbm/s. A supply of steam at 20 lbf/in. 2, 500 F is available
from a boiler and tap water at 14.7 lbf/in.2, 60 F is also available. The two sources
are then mixed in an SSSF mixing chamber to generate the desired state as output.
Determine the rate of irreversibility of the mixing process.
1
3
C.V. Mixing chamber
.
.
.
m1 + m2 = m3
.
.
.
m1h1 + m2h2 = m3h3
.
.
.
.
m1s1 + m2s2 + Sgen = m3s3
2
C.8.1 h1 = 28.08, s1 = 0.05555
C.8.2 h2 = 1286.8, s2 = 1.8919
C.8.2 h3 = 1202.1 , s3 = 1.828
. .
1202.1 - 28.08
m2/m3 = (h3 - h1)/(h2 - h1) =
= 0.9327
1286.8 - 28.08
.
.
m2 = 27.981 lbm/s , m1 = 2.019 lbm/s
.
.
.
.
.
I = T0Sgen = T0(m3s3 - m1s1 - m2s2)
= 536.67 × (30 × 1.828 - 2.019 × 0.05555 - 27.981 × 1.8919) = 961 Btu/s
10.64EA 4-lbm piece of iron is heated from room temperature 77 F to 750 F by a heat
source at 1100 F. What is the irreversibility in the process?
C.V. Iron out to 1100 F source, which is a control mass.
Energy Eq.:
mFe(u2 - u1) = 1Q2 - 1W2
Process: Constant pressure
=>
1W2
= PmFe(v2 - v1)
⇒ 1Q2 = mFe(h2 - h1) = mFeC(T2 - T1)
1Q2
= 4 × 0.107 × (750 - 77) = 288.04 Btu
mFe(s2 - s1) = 1Q2/Tres + 1S2 gen
1S2 gen
= mFe(s2 - s1) - 1Q2/Tres = mFeC ln (T2/T1) - 1Q2/Tres
= 4 × 0.107 × ln
1I2
1209.67 288.04
= 0.163 Btu/R
536.67 1559.67
= To (1S2 gen ) = 536.67 × 0.163 = 87.57 Btu
10-37
10.65EFresh water can be produced from saltwater by evaporation and subsequent
condensation. An example is shown in Fig. P10.12 where 300-lbm/s saltwater,
state 1, comes from the condenser in a large power plant. The water is throttled to
the saturated pressure in the flash evaporator and the vapor, state 2, is then
condensed by cooling with sea water. As the evaporation takes place below
atmospheric pressure, pumps must bring the liquid water flows back up to P0.
Assume that the saltwater has the same properties as pure water, the ambient is at
68 F, and that there are no external heat transfers. With the states as shown in the
table below find the irreversibility in the throttling valve and in the condenser.
State 1
T [F] 86
2
77
3
77
4
--
.
.
.
C.V. Valve: m1 = mex = m2 + m3,
Energy Eq.:
h1 = he ;
5
74
Entropy Eq.:
6
--
7
63
8
68
si + sgen = se
h1 = 54.08 si = 0.1043 P2 = Psat(T2 = T3) = 0.4641
he = h1 ⇒ xe = (54.08 - 45.08)/1050.0 = 0.008571
⇒ se = 0.08769 + 0.008571 × 1.9565 = 0.1045
sgen = se - si = 0.1045 - 0.1043 = 0.0002
. .
I = mT0sgen = 300 × 528 × 0.0002 = 31.68 Btu/s
C.V. Condenser.
h2 = 1095.1 , s2 = 2.044 ,
.
.
m2 = (1 - xe)mi = 297.44 lbm/s
h5 = 42.09 , s5 = 0.0821
h7 = 31.08 , s7 = 0.0613 , h8 = 36.09 , s8 = 0.0708
.
.
.
.
Energy Eq.: m2h2 + m7h7 = m2h5 + m7h8 ⇒
.
.
1095.1 - 42.09
lbm
m7 = m2 × (h2 - h5)/(h8 - h7) = 297.44
= 62516
36.09 - 31.08
s
.
.
.
.
.
Entropy Eq.: m2s2 + m7s7 + Sgen = m2s5 + m7s8
.
.
.
.
I = T0Sgen = T0[m2(s5 - s2) + m7(s8 - s7)]
= 528[297.44(0.0821 - 2.044) + 62516(0.0708 - 0.0613)]
= 528 × 10.354 = 5467 Btu/s
10-38
10.66EAir flows through a constant pressure heating device as shown in Fig. P10.15. It is
heated up in a reversible process with a work input of 85 Btu/lbm air flowing. The
device exchanges heat with the ambient at 540 R. The air enters at 540 R, 60
lbf/in.2. Assuming constant specific heat develop an expression for the exit
temperature and solve for it.
C.V. Total out to T0
Energy Eq.:
Entropy Eq.:
s1 + qrev
0 /T 0 = s2
rev
h1 + qrev
= h2
0 -w
⇒ qrev
0 = T0(s2 - s1)
h2 - h1 = T0(s2 - s1) - wrev
Constant Cp gives: Cp(T2 - T1) = T0Cp ln (T2/T1) + 85
T2 - T0ln(T2/T1) = T1 + 85/Cp
T1 = 540
Cp = 0.24
T0 = 540
T2 - 540 ln (T2/540) = 540 + (85/0.24) = 894.17
At 1400 R
LHS = 885.56 (too low)
Interpolate to get
T2 = 1414 R
At 1420 R
(LHS = 894.19
LHS = 897.9
OK)
10.67EAir enters the turbocharger compressor of an automotive engine at 14.7 lbf/in2, 90
F, and exits at 25 lbf/in2, as shown in Fig. P10.16. The air is cooled by 90 F in an
intercooler before entering the engine. The isentropic efficiency of the compressor
is 75%. Determine the temperature of the air entering the engine and the
irreversibility of the compression-cooling process.
2
3
Cooler
⇐.
Comp.
-Wc
1
⇓-Q. c
T3 = T2 - 90 F, P2 = 25 lbf/in2
ηs COMP = 0.75, P1 = 14.7 lbf/in2
T1 = 90 F = 550 R
T2s = T1
Air
P2
(P )
k-1
k
= 550
1
25
(14.7
)
0.286
= 640.2 R
-ws = CP0(T2s - T1) = 0.24(640.2 - 550) = 21.65
-w = - ws/ηs = 21.65/0.75 = 28.87 = CP0(T2 - T1) = 0.24(T2 - 550)
=>
T2 = 670.3 R
⇒ T3 = 670.3 - 90 = 580.3 R
b) Irreversibility from Eq.10.13 with rev. work from Eq.10.12, (q = 0 at TH)
s3 - s1 = 0.24 ln
580.3 53.34
25
ln
= -0.0235 Btu/lbm R
550
778
14.7
i = T(s3 - s1) - (h3 - h1) - w = T(s3 - s1) - CP(T3 - T1) - CP(T1 - T2)
= 550(-0.0235) - 0.24(-90) = +8.7 Btu/lbm
10-39
10.68ECalculate the irreversibility for the process described in Problem 6.101, assuming
that the heat transfer is with the surroundings at 61 F.
I = To Sgen
so apply 2nd law out to To = 61 F
m2 s2 - m1s1 = misi + 1Q2 / To + 1S2gen
ToSgen = To ( m2 s2 - m1s1 - mi si ) - 1Q2
m1 = 2.104 lbm
mi = 7.152 lbm
Use from table C.4: Cp = 0.24
m2 =9.256 lbm
R = 53.34 / 778 = 0.06856
ToSgen = I = To [ m1 (s2 - s1) + mi (s2 - si)] - 1Q2
= 520.7[2.104(Cp ln
630
60
630
60
- R ln ) +7.152(Cpln
- R ln )]
519.7
45
1100
75
- ( - 868.9)
= 520.7 (0.05569 - 0.8473) + 868.9
= 456.7 Btu
10.69EA control mass gives out 1000 Btu of energy in the form of
a.
Electrical work from a battery
b.
Mechanical work from a spring
c.
Heat transfer at 700 F
Find the change in availability of the control mass for each of the three cases.
a) ∆φ = -Wel = -1000 Btu
b) ∆φ = -Wspring = -1000 Btu
 T0

537 
c) ∆φ = -1 -  = Qout = -1  1000 = -537 Btu
 1160
 TH
10.70EA steady stream of R-22 at ambient temperature, 50 F, and at 110 lbf/in. 2 enters a
solar collector. The stream exits at 180 F, 100 lbf/in.2. Calculate the change in
availability of the R-22 between these two states.
R-22
i
e
Solar
50 F
110 lbf/in2
hi = 24.275
si = 0.0519
Collector
180 F
100 lbf/in2
he = 132.29
se = 0.2586
∆ψie = ψe - ψi
= (he - hi) - T0(se - si)
= (132.29 - 24.275)
-510(0.2586 - 0.0519)
= 2.6 Btu/lbm
10-40
10.71EA 20-lbm iron disk brake on a car is at 50 F. Suddenly the brake pad hangs up,
increasing the brake temperature by friction to 230 F while the car maintains
constant speed. Find the change in availability of the disk and the energy
depletion of the car’s gas tank due to this process alone. Assume that the engine
has a thermal efficiency of 35%.
All the friction work is turned into internal energy of the disk brake.
m(u2 - u1) = 1Q2 - 1W2 ⇒ 1Q2 = mFeCFe(T2 - T1)
1Q2
= 20 × 0.107 × (230 - 50) = 385.2 Btu
Neglect the work to the surroundings at P 0
690
m(s2 - s1) = mC ln(T2 / T1) = 20 × 0.107 × ln 510 = 0.6469
 
∆φ = m(u2 - u1) - T0m(s2 - s1) = 385.2 - 510 × 0.6469 = 55.28 Btu
Wengine = ηthQgas = 1Q2 = Friction work
Qgas = 1Q2/ηth = 385.2/0.35 = 1100 Btu
10.72ECalculate the availability of the system (aluminum plus gas) at the initial and final
states of Problem 8.113, and also the irreversibility.
State 1: T1 = 400 F v1 = 2/2.862 = 0.6988
P1 = 300 psi
Ideal gas v2 = v1(300 / 220)(537 / 860) = 0.595 ; vo = 8.904 = RTo / Po
The metal does not change volume so the terms as Eq.10.22 are added
φ1 = mgasφgas + mAlφAl
T
P
= mgasCv (T1 - To) - mgasTo [Cp ln 1 - R ln 1 ] + mgasPo (v1 - vo)
To
Po
+ mAl [C (T1 - To) - ToC ln (T1 / To) ]Al
860 35.1 300
φ1 = 2.862[0.156(400-77) - 537(0.201 ln
ln
)
537 778 14.7
144
860
+ 14.7 (0.6988 - 8.904 )(
) ] + 8 × 0.21 [ 400 -77 - 537 ln
]
778
537
= 143.96 + 117.78 = 261.74 Btu
220
537 35.1
ln
)
φ2 = 2.862[ 0.156(77 - 77) - 537 (0.201 ln
14.7
537 778
144
537
+ 14.7(0.595 - 8.904 )(
)] + 8 × 0.21 [ 77 -77 - 537 ln
]
778
537
= 122.91 + 0 = 122.91 Btu
1I2 = φ1 - φ2 + (1 - To/TH )1Q2 - 1W2AC + Pom( V2 - V1)
= 261.74 - 122.91 + 0 - (-14.29) + 14.7 × 2.862 ×
144
778
( 0.595 - 0.6988)
= 152.3 Btu
[(Sgen = 0.2837 Btu/R
ToSgen = 152.3
so OK ]
10-41
10.73EConsider the springtime melting of ice in the mountains, which gives cold water
running in a river at 34 F while the air temperature is 68 F. What is the
availability of the water (SSSF) relative to the temperature of the ambient?
ψ = h1 - h0 - T0(s1 - s0)
flow availability from Eq.10.19
Approximate both states as saturated liquid
ψ = 1.9973 - 36.088 - 527.67 × (0.00405 - 0.07081) = 1.136 Btu/lbm
Why is it positive? As the water is brought to 68 F it can be heated with qL
from a heat engine using qH from atmosphere TH = T0 thus giving out work.
10.74EA geothermal source provides 20 lbm/s of hot water at 80 lbf/in. 2, 300 F flowing
into a flash evaporator that separates vapor and liquid at 30 lbf/in.2. Find the three
fluxes of availability (inlet and two outlets) and the irreversibility rate.
C.V. Flash evaporator chamber. SSSF with no work or heat transfer.
.
.
.
Cont. Eq.:
m1 = m2 + m3 ;
Vap.
.
.
.
2
1
Energy Eq.:
m1h1 = m2h2 + m3h3
Liq.
.
.
.
.
3
Entropy Eq.: m1s1 + Sgen = m2s2 + m3s3
C.8.1:
ho = 45.08,
so = 0.08769, h1 = 269.73, s1= 0.4372
C.8.1:
h2 = 1164.3, s2= 1.6997, h3 = 218.9,
s3= 0.36815
. .
h -h
h1 = xh2 + (1 - x) h3 => x = m2/m1 = 1 3 = 0.05376
h2 - h3
.
.
.
.
m3 = (1-x)m1 = 18.925 lbm/s
m2 = xm1 = 1.075 lbm/s
.
Sgen = 0.579 × 7.1271 + 9.421 × 1.53 - 10 × 1.8417 = 0.124 kW/K
Flow availability Eq.10.22: ψ = (h - Tos) - (ho - Toso) = h - ho - To(s - so)
ψ1 = 269.73 - 45.08 - 537 × (0.4372 - 0.08769) = 36.963
ψ2 = 1164.3 - 45.08 - 537 × (1.6997 - 0.08769) = 253.57
ψ3 = 218.9 - -45.08 - 537 × (0.36815 - 0.08769) = 23.21
.
.
.
m1 ψ1 = 739.3 Btu/s m2ψ2 = 272.6 Btu/s m3ψ3 = 439.3 Btu/s
.
.
. .
I = m1 ψ1 - m2 ψ2 - m3ψ3 = 27.4 Btu/s
10-42
10.75EA wood bucket (4 lbm) with 20 lbm hot liquid water, both at 180 F, is lowered
1300 ft down into a mineshaft. What is the availability of the bucket and water
with respect to the surface ambient at 70 F?
v1 ≈ v0 for both wood and water
φ1 - φ0 = mwood[u1 - u0 -T0(s1 - s0)] + mH2O[u1 - u0 -T0(s1 - s0)] + mtotg(z1 -z0)
= 4[0.3(180 - 70) - 0.3 × 530 ln
640
] + 20[147.76 - 38.09
530
- 530(0.263 - 0.074)] + 24 × (g/gc) × (-1300 / 778)
= 12.05 + 199.3 - 40.1 = 171.25 Btu
10.76EAn air compressor is used to charge an initially empty 7-ft3 tank with air up to
750 lbf/in.2. The air inlet to the compressor is at 14.7 lbf/in.2, 60 F and the
compressor isentropic efficiency is 80%. Find the total compressor work and the
change in energy of the air.
C.V. Tank + compressor (constant inlet conditions)
Continuity:
Entropy:
m2 - 0 = min
Energy: m2u2 = minhin - 1W2
m2s2 = minsin + 1S2 GEN
To use isentropic efficiency we must calc. ideal device
Reversible compressor: 1S2 GEN = 0 ⇒ s2 = sin
⇒ Pr2 = Prin(P2/Pin) = 0.9767 × (750/14.7) = 49.832
⇒ T2,s = 1541 R u2,s = 274.49 Btu/lbm
⇒ 1w2,s = hin - u2,s = 124.38 - 274.49 = -150.11
Actual compressor: 1w2,AC = 1w2,s/ηc = -187.64 Btu/lbm
u2,AC = hin - 1w2,AC = 312 ⇒ T2,AC = 1729 R
Final state 2 u, P ⇒ v2 = RT2/P2 = 0.854 ft3/lbm
m2 = V2/v2 = 8.2 lbm
⇒ 1W2 = m2(1W2,AC) = -1539 Btu
m2(φ2 - φ1) = m2 [u2 - u1 + P0(v2 - v1) - T0(s2 - s1) ]
144
- 520 1.9311
778
[
= 8.2 312 - 88.733 + 14.7(0.854 - 13.103)
- 1.63074 -
53.34 750
ln
778 14.7
(
)] = 8.2 × 173.94 = 1426.3 Btu
10-43
10.77EA compressor is used to bring saturated water vapor at 150 lbf/in. 2 up to 2500
lbf/in.2, where the actual exit temperature is 1200 F. Find the irreversibility and
the second law efficiency.
Inlet state: Table C.8.1
hi = 1194.9 Btu/lbm, si = 1.5704 Btu/lbm R
Actual compressor C.8.2: he = 1587.7 Btu/lbm, se = 1.6101 Btu/lbm R
Energy Eq. actual compressor:
-wc,ac = he - hi = 392.8 Btu/lbm
Eq.10.14: i = T0(se - si) = 536.67 × (1.6101 - 1.5704) = 21.3 Btu/lbm
Eq.10.13:
wrev = i + w c,ac = 21.3 + 1194.9 - 1587.7 = -371.5 Btu/lbm
ηII = -wrev/(-wc,ac) = 371.5 / 392.8 = 0.946
10.78EThe simple steam power plant in Problem 6.91, shown in Fig P6.39 has a turbine
with given inlet and exit states. Find the availability at the turbine exit, state 6.
Find the second law efficiency for the turbine, neglecting kinetic energy at state 5.
Turbine from 6.91
C.8:
h6 = 1028.7
ho = 45.08
s6= 1.8053
h5 = 1455.6
s5= 1.6408
so= 0.08769
ψ6 = h6 - ho - To ( s6 - so ) = 61.26 Btu/lbm
wrev = ψ5 - ψ6 = h5 - h6 - To( s5 - s6 ) = 515.2 Btu/lbm
wAC = h5 - h6 = 426.9
ηII = wac / wrev = 426.9 / 515.2 = 0.829
10.79ESteam is supplied in a line at 450 lbf/in.2, 1200 F. A turbine with an isentropic
efficiency of 85% is connected to the line by a valve and it exhausts to the
atmosphere at 14.7 lbf/in.2. If the steam is throttled down to 300 lbf/in.2 before
entering the turbine find the actual turbine specific work. Find the change in
availability through the valve and the second law efficiency of the turbine.
C.V. Valve: h2 = h1 = 1630.8, s2 > s1, h2, P2 ⇒ s2 = 1.86348
Ideal turbine:
s3 = s2 ⇒ h3s = 1230.88
Actual turbine:
wT,ac = ηTwT,s = 339.93 Btu/lbm
wT,s = h2-h3s = 399.92
ψ2 - ψ1 = h2 - h1 - T0(s2 - s1)
= -536.67(1.86348 - 1.8192) = -23.76 Btu/lbm
h3ac = h2 - wTac = 1290.87 ⇒ s3ac = 1.9299
wrev = ψ2 - ψ3 = 1630.8 - 1290.87 - 536.67(1.86348 - 1.9299) = 375.58
ηII = wac/wrev = 339.93/375.58 = 0.905
10-44
10.80EA piston/cylinder arrangement has a load on the piston so it maintains constant
pressure. It contains 1 lbm of steam at 80 lbf/in.2, 50% quality. Heat from a
reservoir at 1300 F brings the steam to 1000 F. Find the second-law efficiency for
this process. Note that no formula is given for this particular case, so determine a
reasonable expression for it.
1: P1, x1 ⇒ v1 = 2.7458 h1 = 732.905 s1 = 1.0374
2: P2 = P1,T2 ⇒ v2 = 10.831 h2 = 1532.6 s2 = 1.9453
m(u2 - u1) = 1Q2 - 1W2 = 1Q2 - P(V2 - V1)
1Q2
1W2
= m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1) = 799.7 Btu
= Pm(v2 - v1) = 119.72 Btu
1W2 to atm =
P0m(v2 - v1) = 22 Btu
Useful work out = 1W2 - 1W2 to atm = 119.72 - 22 = 97.72 Btu

T0 

536.67 
∆φ reservoir = 1  1Q2 = 1  799.7 = 556 Btu
 1759.67
 Tres
nII = Wnet/∆φ = 0.176
10.81EAir flows into a heat engine at ambient conditions 14.7 lbf/in.2, 540 R, as shown
in Fig. P10.47. Energy is supplied as 540 Btu per lbm air from a 2700 R source
and in some part of the process a heat transfer loss of 135 Btu per lbm air happens
at 1350 R. The air leaves the engine at 14.7 lbf/in.2, 1440 R. Find the first- and
the second-law efficiencies.
C.V. Engine out to reservoirs
hi + qH = qL + he + w
wac = 129.18 + 540 - 135 - 353.483 = 180.7
ηTH = w/qH = 180.7/540 = 0.335
For second law efficiency also a q to/from ambient
si + (qH/TH) + (q0/T0) = (qloss/Tm) + se
q0 = T0[se - si + (qloss/Tm) - (qH/TH)]

135 540 
= 5401.88243 - 1.6398 +
 = 77.02
1350 2700

wrev = hi - he + qH - qloss + q0 = wac + q0 = 257.7
ηII = wac/wrev = 180.7/257.7 = 0.70
10-45
10.82EConsider a gasoline engine for a car as an SSSF device where air and fuel enters
at the surrounding conditions 77 F, 14.7 lbf/in.2 and leaves the engine exhaust
manifold at 1800 R, 14.7 lbf/in.2 as products assumed to be air. The engine
cooling system removes 320 Btu/lbm air through the engine to the ambient. For
the analysis take the fuel as air where the extra energy of 950 Btu/lbm of air
released in the combustion process, is added as heat transfer from a 3240 R
reservoir. Find the work out of the engine, the irreversibility per pound-mass of
air, and the first- and second-law efficiencies.
2
C.V. Total out to reservoirs
1
•
•
•
•
•
m
ah1 + Q H = mah2 + W + Q out
AIR
•
•
•
•
•
·
⇒ W
m
as 1 + Q H/TH + S gen = mas s + Q out/T 0
·
·
° = 1.63831
h1 = 128.381 sT1
Q
Q
⇓ out
T0
⇑ H
TH
h2 = 449.794
° = 1.94209
sT1
•
•
wac = W/m
a = h1 - h2 + qH - qout = 128.38 - 449.794 + 950 - 320 = 308.6 Btu/lbm
ηTH = w/qH = 308.6/950 = 0.325
itot = (T0)sgen = T0(s2 - s1) + qout- qHT0/TH
536.67
= 536.67(1.94209 - 1.63831) + 320 - 950  3240  = 325.67 Btu/lbm


For reversible case have
sgen = 0 and qR0 from T0, no qout
qR
0,in = T0(s2 - s1) - (T0/TH)qH = itot - qout = 5.67 Btu/lbm
wrev = h1 - h2 + qH + qR
0,in = wac + itot = 634.3 Btu/lbm
ηII = wac/wrev = 0.486
10.83EConsider the two turbines in Problem 9.119, shown in Fig. P9.72. What is the
second-law efficiency of the combined system?
From the solution to 9.119 we have
•
•
s1 = 1.499 s2 = 1.582 s5 = 1.5651 m
T1/mtot = 0.525
•
•
m
T1/mtot = 0.475
•
•
wac = W
net/mtot = 146.317 Btu/lbm
C.V. Total system out to T0
rev
•
• 
• 
rev
rev
•
m
⇒ wqc + qrev
0 =w
T1/mtoth1 + mT2/mtoth2 + q0 = h5 + w
rev
•
• 
• 
•
m
T1/mtots 1 + mT2/mtots 2 + q0 = s5
⇒ qrev
0 = 536.67[1.5651 - 0.525 × 1.499 - 0.475 × 1.582] = 14.316 Btu/lbm
wrev = wac + qrev
0 = 146.317 + 14.316 = 160.63
ηII = wac/wrev = 146.317/160.63 = 0.911
10-46
10.84E(Adv.) Refrigerant-22 is flowing in a pipeline at 40 F, 80 lbf/in.2, with a velocity
of 650 ft/s, at a steady flowrate of 0.2 lbm/s. It is desired to decelerate the fluid
and increase its pressure by installing a diffuser in the line (a diffuser is basically
the opposite of a nozzle in this respect). The R-22 exits the diffuser at 80 F, with a
velocity of 160 ft/s. It may be assumed that the diffuser process is SSSF,
polytropic, and internally reversible. Determine the diffuser exit pressure and the
rate of irreversibility for the process.
Note: The exit velocity should be 320 ft/s at 80 F
C.V. Diffuser out to T0, Int. Rev. flow ⇒ sgen R-22 = 0/
2
2
•
•
m
1 = m2 , h1 + (1/2)V1 + q = h2 + (1/2)V2
s1 + ∫ dq/T + sgen = s2 = s1 + q/T0 + sgen
Rev. Process: w12 = 0/ = -∫ v dP + (V12 - V22)/2
Polytropic process:
n
LHS = ∫ v dP =
(P v - P1v1) = (V12 - V22)/2
n-1 2 2
= (1/2)(6502 - 3202)/32.174 = 4974.5 ft•lbf
P1v1n = P2v2n ⇒ n = ln(P2/P1)/ln(v1/v2)
Inlet state: v1 = 0.68782, s1 = 0.2211, h1 = 108.347
Exit state: T2,V2, ? ⇒ only P2 unknown.
⇒ v2 = v(P2), n = n(P2)
Trial and error on P 2 to give LHS = 4974.5 ft•lbf:
P2 = 125 psi ⇒ v2 = 0.46219
n = ln(125/80)/ln(0.68782/0.46219) = 1.12257
1.12257
LHS = 0.12257 (125 × 0.462187 - 80 × 0.68782) × 144 = 3624
P2 = 150 psi ⇒ v2 = 0.370548, n = 1.01627, LHS = 5006
Interpolate to get P2 = 149.4 psi. Take 150 psi as OK.
P2 = 150 psi ⇒ v2 = 0.370548, h2 = 111.556 s2 = 0.21484
Find q from energy equation (Assume T0 = T1 = 40 F)
q = h2 - h1 + (V22 - V12)/2 = 111.556 - 108.347 - (4974.5/778) = -3.185 Btu/lbm
•
i = •I/m
= T0(s2 - s1) - q = 499.67(0.21484 - 0.2211) + 3.185 = 0.057 Btu/lbm
10-47
10.85E(Adv.) Water in a piston/cylinder is at 14.7 lbf/in.2, 90 F, as shown in Fig.
P10.56. The cylinder has stops mounted so that Vmin = 0.36 ft3 and Vmax = 18 ft3.
The piston is loaded with a mass and outside P0, so a pressure inside of 700
lbf/in.2 will float it. Heat of 14000 Btu from a 750 F source is added. Find the
total change in availability of the water and the total irreversibility.
CV water plus cyl. wall out to reservoir
m2 = m1 = m, m(u2- u1) = 1Q2 - 1W2
m(s2 - s1) = 1Q2/Tres + 1S2 gen
System eq: States must be on the 3 lines
P
1a
2
1
1b
v
Here Peq = 700 lbf/in2, since P1 < Peq => V1 = Vmin = 0.36 ft3
1: v1 = 0.016099, u1 = 58.07, ⇒ m = V1/v1 = 22.362 lbm
Heat added gives q = Q/m = 626.062 Btu/lbm so check if we go from 1 past 1a
(Peq,v1) and past 1b (Peq,vmax)
1a: T ≈ 100 F, u1a ≅ 67.8
⇒
1b: T ≈ 600 F, u1b= 1180
=>
1q1a
= u1a - u1 ≅ 9.73
1q1b =
h1b - u1 - Peqv1 = 1221.8
The state is seen to fall between 1a and 1b so P2 = Peq.
1w2
= Peq(v2 - v1), u2 - u1 = 1q2 - 1w2
⇒
u2 + Peqv2 = h2 = 1q2 + u1 + Peqv1 = 686.22 < hg(P2)
x2 = 0.3132, v2 = 0.3043, s2 = 0.9044
T2 = 467 F,
1S2 gen
V2 = 6.805
= m(s2 - s1) - 1Q2/Tres = 22.362(0.9044 - 0.11165)
- 14000/1210 = 6.157 Btu/R
1I2
= T0(1S2,gen) = 3304 Btu
The non-flow availability change of the water, Eq.10.22
φ2 - φ1 = u2 - u1 + P0(v2 - v1) - T0(s2 - s1)
= 646.794 - 58.07 + 14.7(0.3043 - 0.016099) ×
- 536.67(0.9044 - 0.11165) = 164.06 Btu/lbm
m∆φ = 3669 Btu
144
778
10-48
10.86EA rock bed consists of 12000 lbm granite and is at 160 F. A small house with
lumped mass of 24000 lbm wood and 2000 lbm iron is at 60 F. They are now
brought to a uniform final temperature with no external heat transfer.
a.
For a reversible process, find the final temperature and the work done in
the process.
b.
If they are connected thermally by circulating water between the rock bed
and the house, find the final temperature and the irreversibility of the process
assuming that surroundings are at 60 F.
a) For a reversible process a heat engine is installed between the rockbed and
the house. Take C.V. Total
mrock(u2 - u1) + mwood(u2 - u1) + mFe(u2 - u1) = -1W2
mrock(s2 - s1) + mwood(s2 - s1) + mFe(s2 - s1) = 0/
(mC)rockln
T2
T1
+ (mC)woodln
T2
T1
+ (mC)Feln
T2
T1
= 0/
12000 × 0.243 ln (T2/619.67) + 24000 × 0.42 ln (T2/519.67)
+ 2000 × 0.107 ln (T2/519.67) = /0
⇒ T2 = 540.3 R
-1W2 = 12000 × 0.243(540.3 - 619.67)
+ (24000 × 0.42 + 314)(540.3 - 519.67)
⇒ 1W2 = 19078 Btu
b) No work, no Q irreversible process.
(mC)rock(T2 - 160) + (mCwood + mCFe)(T2 - 60) = 0/
T2 = 82 F = 541.7 R
Sgen = ∑mi(s2 - s1)i = 2916 ln
1I2
541.7
541.7
+ 10294 ln
= 35.26 Btu/R
619.67
519.67
= T0Sgen = 519.67 × 35.26 = 18324 Btu
10-49
10.87EAir in a piston/cylinder arrangement, shown in Fig. P10.59, is at 30 lbf/in.2, 540 R
with a volume of 20 ft3. If the piston is at the stops the volume is 40 ft3 and a
pressure of 60 lbf/in.2 is required. The air is then heated from the initial state to
2700 R by a 3400 R reservoir. Find the total irreversibility in the process
assuming surroundings are at 70 F.
m2 = m1 = m = P1V1/RT1
P
2
Pstop
= 30 × 20 × 144/(53.34 × 540) = 3.0 lbm
m(u2 - u1) = 1Q2 - 1W2
1
1W 2
P1
Vstop
V1
⌠ dQ/T + 1S2 gen
m(s2 - s1) = ⌡
Process: P = P0 + α(V-V0) if V ≤ Vstop
Information: Pstop = P0 + α(Vstop-V0)
Eq. of state ⇒ Tstop = T1PstopVstop/P1V1 = 2160 < T2
So the piston hits the stops => V 2 = Vstop => P2 = 2.5 P1
1
1W2 = 2(P 1
1Q2 =
1
+ Pstop)(Vstop- V1) = 2(30 + 60)(40 - 20) = 166.6 Btu
m(u2 - u1) + 1W2 = 3(518.165 - 92.16) + 166.6 = 1444.6 Btu
o
o
s2 - s1 = sT2 - sT1 -R ln(P2/P1)
= 2.0561 - 1.6398 - 53.34 × ln (2.5)/778 = 0.3535 Btu/lbm R
Take control volume as total out to reservoir at T RES
1S2 gen tot =
1I2
m(s2 - s2) - 1Q2/TRES = 0.6356 Btu/R
= T0(1S2 gen) = 530 × 0.6356 = 337 Btu
11-1
CHAPTER 11
The correspondence between the new problem set and the previous 4th edition
chapter 9 problem set.
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
94
95
96
97
98
99
Old
New
2
New
3
4a
4b
New
5
New
6
7
8
1
13
New
14
15 mod
16
17
New
18
New
New
19 mod
20 mod
22
New
23 mod
24 a,b
24 c,d mod
New
New
New
75
76
77
78
New
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
100
101
102
103
104
105
Old
25
New
26 new
27
New 28
New
30
31 mod
32
New
New
33
34
35
36
37
38
New
39
40
41
42
43
44 mod
44 mod
45
46
New
47
48
49
79
80
81
82
83
85
New
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
106
107
108
109
Old
50
51
52
53
New
54
55
56
57 mod
58 mod
New
New
59 mod
60
61 mod
New
New
New
62
63
New
64
65
66
67
68
New
69
New
70
71
87
88
89
91
11-2
The problems that are labeled advanced are:
New
Old
New
Old
New
Old
110
9
113
84
116
92
111
21
114
86
112
New
115
90
The English unit problems are:
New
Old
New
Old
New
Old
117
New
130
106
143
New
118
94
131
107
144
118
119
95 a
132
108
145
119
120
95 b
133
109
146
New
121
93
134
110
147
121
122
99
135
111
148
122
123
100
136
112
149
123
124
98
137
113
150
124
125
101 mod
138
114
151
125
126
102 mod
139
115 mod
152
127
127
103
140
116 mod
153
128
128
New
141
117 mod
154
130
129
105
142
New
155
96
11-3
11.1
A steam power plant as shown in Fig. 11.3 operating in a Rankine cycle has
saturated vapor at 3.5 MPa leaving the boiler. The turbine exhausts to the
condenser operating at 10 kPa. Find the specific work and heat transfer in each of
the ideal components and the cycle efficiency.
C.V. Pump Rev adiabatic
-wP = h2 - h1 ; s2 = s1
since incompressible it is easier to find work as
-w p = ∫ v dP = v1 (P2 - P1) = 0.00101 (3500 - 10) = 3.525 kJ/kg
=> h 2 = h1 - wp = 191.81 + 3.525 = 195.33
C.V. Boiler : qh = h3 - h2 = 2803.43 - 195.33 = 2608.1 kJ/kg
C.V. Turbine : wt = h3 - h4 ; s4 = s3
s4 = s3 = 6.1252 = 0.6492 + x4 (7.501) => x4 = 0.73
=> h 4 = 191.81 + 0.73 (2392.82) = 1938.57
wt = 2803.43 - 1938.57 = 864.9 kJ/kg
C.V. Condenser : qL = h4 - h1 = 1938.57 - 191.81 = 1746.8 kJ/kg
ηcycle = wnet / qH = (wt + wp) / qH = (864.9 - 3.5) / 2608.1 = 0.33
11.2
Consider a solar-energy-powered ideal Rankine cycle that uses water as the
working fluid. Saturated vapor leaves the solar collector at 175°C, and the
condenser pressure is 10 kPa. Determine the thermal efficiency of this cycle.
H2O ideal Rankine cycle
T
T3 = 175°C ⇒ P3 = PG 175°C = 892 kPa
CV: turbine s4 = s3 = 6.6256
= 0.6493 + x4 × 7.5009
x4 = 0.797
h4 = 191.83 + 0.797 × 2392.8 = 2098.3
wT = h3 - h4 = 2773.6 - 2098.3 = 675.3 kJ/kg
-wP = v1(P2 - P1) = 0.00101(892 - 10) = 0.89
wNET = wT + wP = 675.3 - 0.89 = 674.4 kJ/kg
h2 = h1 - wP = 191.83 + 0.89 = 192.72
qH = h3 - h2 = 2773.6 - 192.72 = 2580.9 kJ/kg
ηTH = wNET/qH = 674.4/2580.9 = 0.261
3
2
1
4
s
11-4
11.3
A utility runs a Rankine cycle with a water boiler at 3.5 MPa and the cycle has the
highest and lowest temperatures of 450°C and 45°C respectively. Find the plant
efficiency and the efficiency of a Carnot cycle with the same temperatures.
Solution:
1: 45oC , x = 0
=> h1 = 188.42 , v1 = 0.00101 , Psat = 9.6 kPa
3: 3.5 MPa , 450oC => h 3 = 3337.2 , s3 = 7.0051
C.V. Pump Rev adiabatic
-wP = h2 - h1 ; s2 = s1
since incompressible it is easier to find work as
-wp = ∫ v dP = v1 (P2 - P1) = 0.00101 (3500 - 9.6) = 3.525
=> h 2 = h1 - wp = 188.42 + 3.525 = 191.95
C.V. Boiler : qh = h3 - h2 = 3337.2 - 191.95 = 3145.3
C.V. Turbine : wt = h3 - h4 ; s4 = s3
s4 = s3 = 7.0051 = 0.6386 + x4 (7.5261)
=>
x4 = 0.8459
=> h 4 = 188.42 + 0.8459 (2394.77) = 2214.2
wt = 3337.2 - 2214.2 = 1123 kJ/kg
C.V. Condenser : qL = h4 - h1 = 2214.2 - 188.42 = 2025.78 kJ/kg
ηcycle = wnet / qH = (wt + wp) / qH = (1123 - 3.5) / 3145.3 = 0.356
273.15 + 45
= 0.56
ηcarnot = 1 - TL / TH = 1 273.15 + 450
11.4
A steam power plant operating in an ideal Rankine cycle has a high pressure of 5
MPa and a low pressure of 15 kPa. The turbine exhaust state should have a quality
of at least 95% and the turbine power generated should be 7.5 MW. Find the
necessary boiler exit temperature and the total mass flow rate.
C.V. Turbine
wT = h3 - h4; s4 = s3
4: 15 kPa, x4 = 0.95 => s4 = 7.6458 , h4 = 2480.4
3: s3 = s4, P3 ⇒ h3 = 4036.7, T3 = 758°C
wT = h3 - h4 = 4036.7 - 2480.4 = 1556.3
.
.
m = WT/wT = 7.5 × 1000/1556.3 = 4.82 kg/s
11-5
11.5
A supply of geothermal hot water is to be used as the energy source in an ideal
Rankine cycle, with R-134a as the cycle working fluid. Saturated vapor R-134a
leaves the boiler at a temperature of 85°C, and the condenser temperature is 40°C.
Calculate the thermal efficiency of this cycle.
CV: Pump (use R-134a Table B.5)
T
2
⌠ vdP ≈ v1(P2-P1)
-wP = h2 - h1 = ⌡
3
D
1
2
1
= 0.000873(2926.2 - 1017.0) = 1.67 kJ/kg
o
85 C
o
4
h2 = h1 - wP = 256.54 + 1.67 = 258.21 kJ/kg
40 C
CV: Boiler
s
qH = h3 - h2 = 428.10 - 258.21 = 169.89 kJ/kg
CV: Turbine
s4 = s3 = 1.6782 = 1.1909 + x4 × 0.5214 =>
x 4 = 0.9346
h4 = 256.54 + 0.9346 × 163.28 = 409.14
Energy Eq.:
wT = h3 - h4 = 428.1 - 409.14 = 18.96 kJ/kg
wNET = wT + wP = 18.96 - 1.67 = 17.29 kJ/kg
ηTH = wNET/qH = 17.29/169.89 = 0.102
11.6
Do Problem 11.5 with R-22 as the working fluid.
Same T-s diagram and analysis as in problem 11.5 with R-22
T
CV: Pump (use R-22 Table B.4)
2
⌠ vdP ≈ v1(P2-P1)
-wP = h2 - h1 = ⌡
D
1
= 0.000884(4037 - 1534) = 2.21 kJ/kg
2
1
3
o
85 C
o
4
h2 = h1 - wP = 94.27 + 2.21 = 96.48
40 C
s
CV: Boiler:
qH = h3 - h2 = 253.69 - 96.48 = 157.21 kJ/kg
CV: Turbine
s4 = s3 = 0.7918 = 0.3417 + x4 × 0.5329, =>
h4 = 94.27 + 0.8446 × 166.88 = 235.22
wT = h3 - h4 = 253.69 - 235.22 = 18.47 kJ/kg
ηTH = wNET/qH = (18.47 - 2.21)/157.21 = 0.1034
x4 = 0.8446
11-6
11.7
Do Problem 11.5 with ammonia as the working fluid.
Solution:
T
CV: Pump (use Ammonia Table B.2)
2
-w = h - h = ⌠
⌡ vdP = v (P -P )
P
2
1
1
1
2
1
= 0.001725(4608.6 - 1554.9) = 5.27 kJ/kg
h2 = h1 - wP = 371.43 + 5.27 = 376.7 kJ/kg
D
2
1
3
o
85 C
o
4
40 C
CV: Boiler
s
qH = h3 - h2 = 1447.8 - 376.7 = 1071.1 kJ/kg
CV: Turbine
s4 = s3 = 4.3901 = 1.3574 + x4 × 3.5083
h4 = 371.43 + 0.8643 × 1098.8 = 1321.13
Energy Eq.:
wT = h3 - h4 = 1447.8 - 1321.13 = 126.67 kJ/kg
wNET = wT + wP = 126.67 - 5.27 = 121.4 kJ/kg
ηTH = wNET/qH = 121.4/1071.1 = 0.113
=>
x4 = 0.8643
11-7
11.8
Consider the ammonia Rankine-cycle power plant shown in Fig. P11.8, a plant
that was designed to operate in a location where the ocean water temperature is
25°C near the surface and 5°C at some greater depth.
a. Determine the turbine power output and the pump power input for the cycle.
b. Determine the mass flow rate of water through each heat exchanger.
c. What is the thermal efficiency of this power plant?
a) Turbine
s2S = s1 = 5.0863 = 0.8779 + x2S × 4.3269
x2S = 0.9726
h2S = 227.08 + 0.9726 × 1225.09 = 1418.6
wST = h1 - h2S = 1460.29 - 1418.6 = 41.69
wT = ηS wST = 0.80 × 41.69 = 33.35 kJ/kg
T
1
3
2s
.
.
WT = mwT = 1000 × 33.35 = 33 350 kW
Pump: wSP ≈ -v3(P4 - P3) = -0.0016(857 - 615) = -0.387
wP = wSP/ηS = -0.387/0.80 = -0.484 kJ/kg
.
.
WP = mwP = 1000(-0.484) = -484 kW
b) h2 = h1 - wT = 1460.29 - 33.35 = 1426.94
.
Qto low T H2O = 1000(1426.94 - 227.08) = 1.1999×106 kW
1.1999×106
.
mlow T H2O =
= 142 840 kg/s
29.38 - 20.98
h4 = h3 - wP = 227.08 + 0.48 = 227.56
.
Qfrom high T H2O = 1000(1460.29 - 227.56) = 1.2327×106 kW
1.2327×106
.
mlow T H2O =
= 147 280 kg/s
104.87 - 96.50
c)
.
.
33 350 - 484
= 0.027
ηTH = WNET/QH =
1.2327×106
o
20 C
4
4s
2
o
10 C
s
11-8
11.9
Do Problem 11.8 with R-134a as the working fluid in the Rankine cycle.
Solution:
a) Turbine
s2S = s1 = 1.7183 = 1.0485 + x2S × 0.6733
T
x2S = 0.9948
h2S = 213.58 + 0.9948 × 190.65 = 403.24
wST = h1 - h2S = 409.84 - 403.24 = 6.6 kJ/kg
wT = ηS wST = 0.80 × 6.6 = 5.28 kJ/kg
1
o
20 C
4
4s
3
2s
2
.
.
WT = mwT = 5280 kW
Pump: wSP ≈ -v3(P4 - P3) = -0.000794(572.8 - 415.8) = -0.125 kJ/kg
.
.
WP = mwP = -156 kW
wP = wSP/ηS = -0.156 =>
b) h2 = h1 - wT = 409.84 - 5.28 = 404.56
.
Qto low T H2O = 1000(404.56 - 213.58) = 190980 kW
.
mlow T H2O =
190980
= 22736 kg/s
29.38 - 20.98
h4 = h3 - wP = 213.58 + 0.156 = 213.74
.
Qfrom high T H2O = 1000(409.84 - 213.74) = 196100 kW
.
mlow T H2O =
196100
= 23429 kg/s
104.87 - 96.50
.
.
5280 - 156
= 0.026
c) ηTH = WNET/QH =
196100
o
10 C
s
11-9
11.10 Consider the boiler in Problem 11.5 where the geothermal hot water brings the R134a to saturated vapor. Assume a counter flowing heat exchanger arrangement.
The geothermal water temperature should be equal to or greater than the R-134a
temperature at any location inside the heat exchanger. The point with the smallest
temperature difference between the source and the working fluid is called the
pinch point. If 2 kg/s of geothermal water is available at 95°C, what is the
maximum power output of this cycle for R-134a as the working fluid? (hint: split
the heat exchanger C.V. into two so the pinch point with ∆T = 0, T = 85°C
appears)
2 kg/s of water is available at 95 oC for the boiler. The restrictive factor is the
boiling temperature of 85° C. Therefore, break the process up from 2-3 into
two parts as shown in the diagram.
liquid
2
LIQUID
HEATER
R-134a
R-134a
BOILER
sat liq
.
-QBC at 85 oC
liquid
3 sat. vap
D
.
-QAB
B
C
H2O out
85o C
liq H2O
at 85 oC
A
liquid H2O
95 oC
Write the enrgy equation for the first section A-B and D-3:
.
.
-QAB = mH2O(hA - hB) = 2(397.94 - 355.88) = 84.12 kW
.
.
= mR134A(428.1 - 332.65) ⇒ mR134A = 0.8813 kg/s
To be sure that the boiling temp. is the restrictive factor, calculate TC from the
energy equation for the remaining section:
.
-QAC = 0.8813(332.65 - 258.21) = 65.60 kW = 2(355.88 - h C)
⇒ hC = 323.1, TC = 77.2°C > T2 OK
CV Pump:
-wP = v1(P2-P1) = 0.000873(2926.2 - 1017.0) = 1.67 kJ/kg
CV: Turbine
s4 = s3 = 1.6782 = 1.1909 + x4 × 0.5214 =>
x 4 = 0.9346
h4 = 256.54 + 0.9346 × 163.28 = 409.14
Energy Eq.:
wT = h3 - h4 = 428.1 - 409.14 = 18.96 kJ/kg
Cycle:
wNET = wT + wP = 18.96 - 1.67 = 17.29 kJ/kg
.
.
WNET = mR134AwNET = 0.8813 × 17.29 = 15.24 kW
11-10
11.11 Do the previous problem with R-22 as the working fluid.
A flow with 2 kg/s of water is available at 95 oC for the boiler. The restrictive
factor is the boiling temperature of 85oC. Therefore, break the process up
from 2-3 into two parts as shown in the diagram.
2
liquid
LIQUID
HEATER
R-22
3 sat. vap
D
sat liq
.
-QBC at 85 oC
H2O out
85o C
.
-QAB
B
C
liquid
R-22
BOILER
A
liquid H2O
95 oC
liq H2O
at 85 oC
.
.
-QAB = mH2O(hA - hB) = 2(397.94 - 355.88) = 84.12 kW
.
.
= mR-22(253.69 - 165.09) ⇒ mR-22 = 0.949 kg/s
To verify that TD = T3 is the restrictive factor, find TC.
.
-QAC = 0.949(165.09 - 96.48) = 65.11 = 2.0(355.88 - hC)
hC = 323.32 ⇒ TC = 77.2 oC OK
.
From Problem 11.6 then:
Wnet = 0.949 × (18.47 - 2.21) = 15.43 kW
11.12 The power plant in Problem 11.1 is modified to have a superheater section following
the boiler so the steam leaves the super heater at 3.5 MPa, 400°C. Find the specific
work and heat transfer in each of the ideal components and the cycle efficiency.
C.V. Tubine:
Energy:
wT,s = h3 - h4s;
Entropy:
s4s = s3 = 6.8404
⇒ x4s = (6.8404 - 0.6492)/7.501 = 0.8254 ;
h4s = 191.81 + 0.8254 × 2392.82 = 2166.8 kJ/kg
wT,s = 3222.24 - 2166.8 = 1055.4 kJ/kg
⌡v dP = v1(P2 - P1) = 0.00101(3500 - 10) = 3.525 kJ/kg
C.V. Pump: -wP = ⌠
⇒ h2 = h1 - wP = 191.81 + 3.525 = 195.34
C.V. Condenser: -qC = h4 - h1 = 2166.8 - 191.81 = 1975
C.V. Boiler:
qH = h3 - h2 = 3222.24 - 195.34 = 3027
ηCYCLE = wNET/qH = (1055.4 - 3.53)/3027 = 0.3475
11-11
11.13 A steam power plant has a steam generator exit at 4 MPa, 500°C and a condenser
exit temperature of 45°C. Assume all components are ideal and find the cycle
efficiency and the specific work and heat transfer in the components.
T
From the Rankine cycle we have the states:
1: 45°C x = 0/ , v1 = 0.00101 , h1 = 188.45
3
2: 4 MPa
3: 4 MPa, 500°C , h3 = 3445.3 , s3 = 7.0901
2
4
1
4: Psat(45°C) = 9.593
s
C.V. Turbine: s4 = s3 ⇒ x4 = (7.0901 - 0.6386)/7.5261 = 0.8572,
h4 = 188.42 + 0.8572 × 2394.77 = 2241.3
wT = h3 - h4 = 3445.3 - 2241.3 = 1204 kJ/kg
C.V. Pump: -wP = v1(P2 - P1) = 0.00101(4000 - 9.6) = 4.03 kJ/kg
-wP = h2 - h1 ⇒ h2 = 188.42 + 4.03 = 192.45 kJ/kg
C.V. Boiler:
qH = h3 - h2 = 3445.3 - 192.45 = 3252.8 kJ/kg
C.V. Condenser:
qL,out = h4 - h1 = 2241.3 - 188.42 = 2052.9 kJ/kg
ηTH = wnet/qH = (wT + wP)/qH = (1204 - 4.03)/3252.8 = 0.369
11.14 Consider an ideal Rankine cycle using water with a high-pressure side of the
cycle at a supercritical pressure. Such a cycle has a potential advantage of
minimizing local temperature differences between the fluids in the steam
generator, such as the instance in which the high-temperature energy source is the
hot exhaust gas from a gas-turbine engine. Calculate the thermal efficiency of the
cycle if the state entering the turbine is 25 MPa, 500°C, and the condenser
pressure is 5 kPa. What is the steam quality at the turbine exit?
s4 = s3 = 5.9592 = 0.4764 + x4 × 7.9187
3 25 MPa
o
500 C
T
x4 = 0.6924
Very low for a turbine exhaust
s1 = 0.4764 , h1 = 137.82
h4 = 1816 , h3 = 3162.4
s2 = s1 => h2 = 162.8
wNET = h3 - h4 - (h2 - h1) = 1321.4
qH = h3 - h2 = 2999.6, η = wNET/qH = 0.44
5 kPa
2
1
4
s
11-12
11.15 Steam enters the turbine of a power plant at 5 MPa and 400°C, and exhausts to the
condenser at 10 kPa. The turbine produces a power output of 20 000 kW with an
isentropic efficiency of 85%. What is the mass flow rate of steam around the
cycle and the rate of heat rejection in the condenser? Find the thermal efficiency
of the power plant and how does this compare with a Carnot cycle.
.
Solution:
WT = 20 000 kW and ηTs = 85 %
State 1 : T1 = 400oC , P1 = 5 MPa , superheated
h1 = 3195.6 kJ/kg , s1 = 6.6458 kJ/kgK
State 3 : P3 = P2 = 10 kPa , sat liq , x3 = 0
T3 = 45.8oC , h3 = hf = 191.8 kJ/kg , v3 = vf = 0.00101 m3/kg
C.V Turbine : 1st Law:
qT + h1 = h2 + wT ;
qT = 0
wT = h1 - h2 , Assume Turbine is isentropic
s2s = s1 = 6.6458 kJ/kgK , s2s = sf + x2s sfg , solve for x2s = 0.7994
h2s = hf + x2shfg = 1091.0 kJ/kg
wTs = h1 - h2s = 1091 kJ/kg , wT = ηTswTs = 927.3 kJ/kg
.
. WT
m=
= 21.568 kg/s , h2 = h1 - wT = 2268.3 kJ/kg
wT
qc + h2 = h3 + wc ; wc = 0
.
.
qc = h3 - h2 = -2076.5 kJ/kg , Qc = m qc = -44786 kW
C.V. Condenser: 1st Law :
C.V. Pump: Assume Isentropic
wps = - ∫ v dP = -v3(P4 - P3) = -5.04 kJ/kg
1st Law : qp + h3 = h4 + wp ; qp = 0
h4 = h3 - wp = 196.8 kJ/kg
C.V Boiler : 1st Law : qB + h4 = h1 + wB ; wB = 0
qB = h1 - h4 = 2998.8 kJ/kg (or qB = wT + wB - q c)
wnet = wT + wP = 922.3 kJ/kg
ηth = wnet / qB = 0.307
Carnot cycle : TH = T1 = 400oC , TL = T3 = 45.8oC
ηth =
T H - TL
TH
= 0.526
11-13
11.16 Consider an ideal steam reheat cycle where steam enters the high-pressure turbine
at 3.5 MPa, 400°C, and then expands to 0.8 MPa. It is then reheated to 400°C and
expands to 10 kPa in the low-pressure turbine. Calculate the cycle thermal
efficiency and the moisture content of the steam leaving the low-pressure turbine.
T
P3 = 3.5 MPa, T3 = T5 = 400 oC
3.5 MPa
o
3 5 400 C
P4 = P5 = 0.8 MPa, P6 = 10 kPa
4
2
From solution 11.12:
-wP = 3.525 kJ/kg,
1
h2 = 195.34
10 kPa
6
h3 = 3222.2, s3 = 6.8404
s
qH1 = h3-h2 = 3222.3-195.35 = 3026.9
s4 = s3 => h4 = 2851.6 ; h5 = 3267.1
qH = qH1 + h5 - h4 = 3026.9 +3267.1 - 2851.6 = 3442.4 kJ/kg
s6 = s5 = 7.5715 => x6 = (7.5715-0.6492)/7.501 = 0.9229 ; h6 = 2400
wT,tot = h3- h4 + h5 - h6 = 3222.2 - 2851.6 + 3267.1 - 2400 = 1237.8 kJ/kg
ηCYCLE = (1237.8 - 3.525)/3442.4 = 0.3586
11.17 The reheat pressure effect the operating variables and thus turbine performance.
Repeat Problem 11.16 twice, using 0.6 and 1.0 MPa for the reheat pressure.
3.5 MPa
o
3 5 400 C
T
P3 = 3.5 MPa, T3 = T5 = 400 oC
P4 = P5 = 0.6 MPa, P6 = 10 kPa
-wP = 3.525 kJ/kg,
h3 = 3222.2,
4
2
From solution 11.12:
1
h2 = 195.34
10 kPa
6
s3 = 6.8404
s
qH1 = h3-h2 = 3222.2-195.34 = 3026.9
wT,tot = h3- h4 + h5 - h6,
qH = qH1 + h5 - h4,
s4 = s3,
s6 = s5
For P4=P5 = 1 MPa: s5 = 7.465, state 4 is superheated vapor
For P4=P5 = 0.6 MPa: s5 = 7.7078, state 4 is superheated vapor
x6
0.9087
0.9410
P4=P5
h4
1
2900.7
0.6 2793.2
h5
3263.9
3270.3
h6
2366
2443.5
Notice the very small changes in efficiency.
wT
1219.5
1255.9
qH
3390.1
3504.0
ηCYCLE
0.3587
0.3574
11-14
11.18 The effect of a number of reheat stages on the ideal steam reheat cycle is to be
studied. Repeat Problem 11.16 using two reheat stages, one stage at 1.2 MPa and
the second at 0.2 MPa, instead of the single reheat stage at 0.8 MPa.
P4 = P5= 1.2 MPa, P6 = P7 = 0.2 MPa
3.5 MPa
o
3 5 7 400 C
T
From solution to 11.12,
-wP = 3.525 kJ/kg, h2 = 195.34
4
3: h3 = 3222.2, s3 = 6.8404
4: P4, s4 = s3 ⇒ sup. vap. h4 = 2940.9
2
1
5: h5 = 3260.7 , s5 = 7.3773
10 kPa
6
8
s
6: P6, s6 = s5 ⇒ sup. vap. h6 = 2811.2
7: h7 = 3276.5, s7 = 8.2217
8: P8, s8 = s7 ⇒ sup. vap. h8 = 2607.9
wT = (h3 - h4) + (h5 - h6) + (h7 - h8) = 281.4 + 449.5 + 668.6 = 1399.5 kJ/kg
qH = (h3 - h2) + (h5 - h4) + (h7 - h6) = 3027 + 319.8 + 465.3 = 3812.1 kJ/kg
ηTH = (wT + wP)/qH = (1399.5 - 3.53)/3812.1 = 0.366
11.19 A closed feedwater heater in a regenerative steam power cycle heats 20 kg/s of
water from 100°C, 20 MPa to 250°C, 20 MPa. The extraction steam from the
turbine enters the heater at 4 MPa, 275°C, and leaves as saturated liquid. What is
the required mass flow rate of the extraction steam?
3
2
From table B.1
1
4
B.1.4: 100°C, 20 MPa
h1 = 434.06
B.1.4: 250°C, 20 MPa
h2 = 1086.75
B.1.3: 4 MPa, 275°C
h3 = 2886.2
B.1.2: 4 MPa, sat. liq.
h4 = 1087.31
C.V. Feedwater Heater
.
.
.
.
Energy Eq.: m1h1 + m3h3 = m1h2 + m3h4
.
.
m3 = m1(h1-h2)/(h4-h3) = 7.257 kg/s
11-15
11.20 An open feedwater heater in a regenerative steam power cycle receives 20 kg/s of
water at 100°C, 2 MPa. The extraction steam from the turbine enters the heater at
2 MPa, 275°C, and all the feedwater leaves as saturated liquid. What is the
required mass flow rate of the extraction steam?
Solution:
2
C.V Feedwater heater
.
.
.
m1 + m2 = m3
.
.
.
.
.
m1h1 + m2h2 = m3h3 = (m1 + m2) h3
3
1
Table B.1.1 and Table B.1.4 at 100 C interpolate to 2 MPa: h1 = 420.4
Table B.1.3: h2 = 2963, Table B.1.2: h3 = 908.8
.
. h3 - h1
908.8 - 420.4
m2 = m1
= 20 ×
= 4.755 kg/s
2963 - 908.8
h2 - h3
Remark: h1 was interpolated between sat liq and compressed liquid at 5 MPa
both at 100oC. The saturated liquid value 419.02 could have been used with
very small error.
11.21 A power plant with one closed feedwater heater has a condenser temperature of
45°C, a maximum pressure of 5 MPa, and boiler exit temperature of 900°C.
Extraction steam at 1 MPa to the feedwater heater condenses and is pumped up to
the 5 MPa feedwater line where all the water goes to the boiler at 200°C. Find the
fraction of extraction steam flow and the two specific pump work inputs.
From turbine
s1 = 0.6387, h1 = 188.45
v1 = 0.00101
s4 = 2.1387, h4 = 762.81
T6 => h 6 = 853.9
6
s3 = sIN = 7.9593,
Pump 1
5
2
7
Pump 2
C.V. Turbine:
3
T3 = 573.8,
4
h3 = 3640.58
C.V. P1: wP1 = -v1(P2 - P1) = -5.04 ⇒ h2 = h1 - wP1 = 193.49
C.V. P2: wP2 = -v4(P7 - P4) = -4.508 ⇒ h7 = h4 - wP2 = 767.31
.
.
.
. .
. .
C.V. Tot: m6 = m1 + m3 ,
m6/m1 = 1 + x,
m3/m1 = x
(1 + x)h6 = xh3 - xwP2 - wP1 + h1 = xh3 - xwP2 + h2
. .
x = (h6 -h2)/(h3 + wP2 - h6) = 0.2364,
m3/m6 = 0.1912
1
From
condenser
11-16
11.22 A power plant with one open feedwater heater has a condenser temperature of
45°C, a maximum pressure of 5 MPa, and boiler exit temperature of 900°C.
Extraction steam at 1 MPa to the feedwater heater is mixed with the feedwater
line so the exit is saturated liquid into the second pump. Find the fraction of
extraction steam flow and the two specific pump work inputs.
Solution:
From turbine
To boiler
State out of boiler 5
6
Pump 1
1
h5 = 4378.82 , s5 = 7.9593
C.V. Turbine: s7 = s6 = s5
P6 , s6
=>
h6 = 3640.6,
4
3
Pump 2
2
T6 = 574oC
C.V Pump P1
-wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(1000 - 9.6) = 1.0 kJ/kg
=> h 2 = h1 - wP1 = 188.42 + 1.0 = 189.42 kJ/kg
.
.
C.V. Feedwater heater: Call m6 / mtot = x
Energy Eq.:
x=
(1 - x) h2 + x h6 = 1 h3
h3 - h2
h6 - h2
=
762.79 - 189.42
= 0.1661
3640.6 - 189.42
C.V Pump P2
-wP2 = h4 - h3 = v3(P4 - P3) = 0.001127(5000 - 1000) = 4.5 kJ/kg
From
condenser
11-17
11.23 A steam power plant operates with a boiler output of 20 kg/s steam at 2 MPa,
600°C. The condenser operates at 50°C dumping energy to a river that has an
average temperature of 20°C. There is one open feedwater heater with extraction
from the turbine at 600 kPa and its exit is saturated liquid. Find the mass flow rate
of the extraction flow. If the river water should not be heated more than 5°C how
much water should be pumped from the river to the heat exchanger (condenser)?
Solution: The setup is as shown in Fig. 11.10,
Condenser:
1: 50oC sat liq. v1 = 0.001012 , h1 = 209.31
2: 600 kPa
To
river
7 Ex turbine
From
s 2 = s1
3: 600 kPa
sat liq.
5: P , T
h3 = hf = 670.54
h5 = 3690.1
To pump 1
s5 = 7.7023
1
6: 600 kPa s6 = s5 => h 6 = 3270.0
CV P1
wP1 = -v1(P2 - P1) = -0.001012 (600 - 12.35) = -0.595
h2 = h1 - wP1 = 209.9
C.V FWH
x h6 + (1 -x) h2 = h3
x=
h3 - h2
h6 - h2
=
670.54 - 209.9
= 0.1505
3270.0 - 209.9
CV Turbine
s 7 = s6 = s5
=>
x 7 = 0.9493 ,
h7 = 2471.17 kJ/kg
qL = h7 - h1 = 2471.17 - 209.31 = 2261.86 kJ/kg
.
.
QL = (1 - x) mqL = 0.85 × 20 × 2261.86 = 38429 kW
.
.
= mH2O ∆hH2O = m (20.93) = 38429 kW
.
m = 1836 kg/s
CV Condenser + Heat Exchanger
.
QL .
.
0 = mH2O (s7 - s1) (1 - x) + Sg
TL
.
38429
Sg =
- 20 × 0.85 × (7.7023 - 0.7037) = 12.184 kW/K
293.15
river
11-18
11.24 Consider an ideal steam regenerative cycle in which steam enters the turbine at
3.5 MPa, 400°C, and exhausts to the condenser at 10 kPa. Steam is extracted from
the turbine at 0.8 MPa for an open feedwater heater. The feedwater leaves the
heater as saturated liquid. The appropriate pumps are used for the water leaving
the condenser and the feedwater heater. Calculate the thermal efficiency of the
cycle and the net work per kilogram of steam.
Solution:
C.V. Turbine
2nd Law
s7 = s6 = s5 = 6.8404 kJ/kg K
P6 , s6 => h6 = 2851 (sup vap)
C.V Pump P1
-wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(800 - 10) = 0.798 kJ/kg
=> h 2 = h1 - wP1 = 191.81 + 0.798 = 192.61
.
.
Call m6 / mtot = x
(1 - x) h2 + x h6 = 1 h3
x=
h3 - h2
h6 - h2
=
721.1 - 192.61
= 0.1988
2851 - 192.61
C.V Pump P2
-wP2 = h4 - h3 = v3(P4 - P3) = 0.001115(3500 - 800) = 3.01
h4 = h3 - wP2 = 721.1 + 3.01 = 724.1
CV Boiler
qH = h5 - w4 = 3222.2 + 724.1 = 2498.1
CV Turbine
.
.
WT / m5 = h5 - h6 + (1 - x) (h6 - h7)
s7 = s6 = s5 = 6.8404
=>
x7 =
6.8404 - 0.6492
7.501
x7 = 0.8254 => h7 = 191.81 + x7 2392.82 = 2166.8
wT = 3222.2 - 2851 + (1 - 0.1988) ( 2851 - 2166.8) = 919.38
wnet = wT + (1 - x) wP1 + wP2
= 919.38 + (1 - 0.1968)(-0.798) - 3.01 = 915.7 kJ/kg
ηcycle = wnet / qH = 915.7 / 2498.1 = 0.366
11-19
11.25 Repeat Problem 11.24, but assume a closed instead of an open feedwater heater.
A single pump is used to pump the water leaving the condenser up to the boiler
pressure of 3.5 MPa. Condensate from the feedwater heater is drained through a
trap to the condenser.
Solution:
C.V. Turbine, 2nd law:
BOILER
s4 = s5 = s6 = 6.8404 kJ/kg K
4
3
TURBINE.
h4 = 3222.24 , h5 = 2851
FW HTR
=> x 6 = (6.8404 - 0.6492)/7.501
5
6
= 0.8254
Trap 7
h6 = 191.81 + x6 2392.82 = 2166.8
2
P
1
COND.
Assume feedwater heater exit at the T of the condensing steam
C.V Pump
-wP = h2 - h1 = v1(P2 - P1) = 0.00101(3500 - 10) = 3.525
h2 = h1 - wP = 191.81 + 3.525 = 195.3
T3 = Tsat (P5) = 170.43 , h3 = hf = h7 = 721.1
C.V FWH
.
.
m5 / m3 = x ,
x=
h3 - h2
h5 - hf 800
Energy Eq.:
=
h2 + x h5 = h3 + h7 x
721.1 - 195.3
= 0.2469
2851 - 721.1
wT = h4 - h5 + (1 - x)(h5 - h6)
= 3222.2 - 2851 + 0.7531 ( 2851 - 2166.8) = 886.51
wnet = wT + wP = 886.51 - 3.53 = 883
qH = h4 - h3 = 3222.24 - 721.1 = 2501.1
ηcycle = wnet / qH = 883 / 2501.1 = 0.353
11-20
11.26 A steam power plant has high and low pressures of 25 MPa and 10 kPa, and one
open feedwater heater operating at 1 MPa with the exit as saturated liquid. The
maximum temperature is 800°C and the turbine has a total power output of 5
MW. Find the fraction of the flow for extraction to the feedwater and the total
condenser heat transfer rate.
The physical components and the T-s diagram is as shown in Fig. 11.10 in the
main text for one open feedwater heater. The same state numbering is used.
From the Steam Tables:
h5 = 4047.1, s5 = 6.9345, h1 = 191.83, v1 = 0.00101
h3 = 762.8 v3 = 0.001127
Pump P1: -wP1 = -v1(P2 - P1) = 0.00101 × 990 = 1 kJ/kg
h2 = h1 - wP1 = 192.83 kJ/kg
Turbine 5-6: s6 = s5 ⇒ h6 = 2948 kJ/kg
wT56 = h5 - h6 = 1099.1 kJ/kg
.
.
.
.
.
Feedwater Heater (mTOT = m5):
xm5h6 + (1 - x)m5h2 = m5h3
⇒ x=
h3 - h2
h6 - h2
=
762.8 - 192.83
= 0.2069
2948 - 192.83
To get state 7 into condenser consider turbine.
s7 = s6 = s5 ⇒ x7 = (6.9345 - 0.6493)/7.5009 = 0.83793
h7 = 191.81 + 0.83793 × 2392.82 = 2196.8 kJ/kg
Find specific turbine work to get total flow rate
.
.
.
.
.
WT = mTOTh5 - xmTOTh6 - (1 - x)mTOTh7 = mTOT × 1694.9
.
mTOT = 5000/1694.9 = 2.95 kg/s
.
.
QL = mTOT (1-x) (h7-h1) = 2.95 × 0.7931(2196.8-191.83) = 4691 kW
11-21
11.27 Do Problem 11.26 with a closed feedwater heater instead of an open and a drip
pump to add the extraction flow to the feed water line at 25 MPa. Assume the
temperature is 175°C after the drip pump flow is added to the line. One main
pump brings the water to 25 MPa from the condenser.
Solution:
v1 = 0.00101 , h1 = 191.83
From
condenser
From turbine 6
T4 = 175oC ; h4 = 741.16
3
4
h6 = 4047.08 , s6 = 6.9345
h6a = hf 1MPa = 762.79 ,
2
6b
v6a = 0.001127
Pump 2
Pump 1
6a
C.V Pump 1
-wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(25000 - 10) = 25.24 kJ/kg
=> h 2 = h1 - wP1 = 191.83 + 25.24 = 217.07
C.V Pump 2
-wP2 = h6b - h6a = v6a(P6b - P6a) = 0.001127(25000 - 1000) = 27.05 kJ/kg
Turbine section 1:
s 6 = s5
P6 = 1 MPa
=> h6 = 2948 kJ/kg
C.V FWH + P2
.
.
m6 / m4 = x
x h6 + (1 - x) h2 + x (-wP2)= h4
x=
Turbine:
h4 - h 2
h6 - h2 - wP2
=
741.16 - 217.07
= 0.19
2948 - 217.07 + 27.05
s7 = s6 = s5 & P7 = 10 kPa
=> x 7 =
1
6.9345 - 0.6493
= 0.83793
7.5009
.
.
WT = m5 [ h5 - h6 + (1 - x) (h6 - h7) ]
.
.
= m5 [ 4047.08 - 2948 + (2948 - 2196.8)] = m5 1707.6
.
.
.
WT = 5000 kW = m5 1707.6 => m5 = 2.928 kg/s
.
.
QL = m5(1 - x) (h7 - h1) = 2.928 × 0.81 (2196.8 - 191.83) = 4755 kW
11-22
11.28 Consider an ideal combined reheat and regenerative cycle in which steam enters
the high-pressure turbine at 3.5 MPa, 400°C, and is extracted to an open
feedwater heater at 0.8 MPa with exit as saturated liquid. The remainder of the
steam is reheated to 400°C at this pressure, 0.8 MPa, and is fed to the lowpressure turbine. The condenser pressure is 10 kPa. Calculate the thermal
efficiency of the cycle and the net work per kilogram of steam.
5: h5 = 3222.24,
s5 = 6.8404
7: h7 = 3267.07,
s7 = 7.5715
7
5
T1
3: h3 = hf = 721.1
C.V. T1
s 5 = s6
6
1-x
=> h6 = 2851
8
x
wT1 = h5 - h6 = 3222.2 - 2851 = 371.2
HTR
4
C.V. Pump 1
-wP1 = h2 - h1 = v1(P2 - P1)
T2
1-x
COND.
2
3
P
1
P
= 0.00101(800 - 10) = 0.798
=> h 2 = h1 - wP1 = 191.81 + 0.798 = 192.61
T
400 oC
5
7
C.V. FWH
x h6 + (1 - x) h2 = h3
h3 - h2
721.1 - 192.61
x=
=
= 0.1988
h6 - h2 2851 - 192.61
4
6
2 3
1
10 kPa
8
s
C.V. Pump 2
-wP2 = h4 - h3 = v3(P4 - P3) = 0.001115(3500 - 800) = 3.01
=> h 4 = h3 - wP2 = 721.1 + 3.01 = 724.11
qH = h5 - h4 + (1 - x)(h7 - h6 ) = 2498.13 + 333.35 = 2831.5
C.V. Turbine 2
s 7 = s8
=> x8 = (7.5715 - 0.6492)/7.501 = 0.92285
h8 = hf + x8 hfg = 191.81 + 0.92285 × 2392.82 = 2400.0
wT2 = h7 - h8 = 3267.07 - 2400.02 = 867.05
wnet = wT1 + (1 - x) wT2 + (1 - x) wP1 + wP2
= 371.2 + 694.7 - 0.64 - 3.01 = 1062.3 kJ/kg
ηcycle = wnet / qH = 1062.25 / 2831.5 = 0.375
11-23
11.29 An ideal steam power plant is designed to operate on the combined reheat and
regenerative cycle and to produce a net power output of 10 MW. Steam enters the
high-pressure turbine at 8 MPa, 550°C, and is expanded to 0.6 MPa, at which
pressure some of the steam is fed to an open feedwater heater, and the remainder
is reheated to 550°C. The reheated steam is then expanded in the low-pressure
turbine to 10 kPa. Determine the steam flow rate to the high-pressure turbine and
the power required to drive each of the pumps.
7
a)
T
5
5
550 oC
7
HI P LOW P
T1
T2
4
6
6 10 kPa
8
2 3
6a
1
8
COND.
HTR
s
4
2
3
P
P
1
b) -wP12 = 0.00101(600 - 10) = 0.6 kJ/kg
h2 = h1 - wP12 = 191.8 + 0.6 = 192.4
-wP34 = 0.00101(8000 - 600) = 8.1 kJ/kg
h4 = h3 - wP34 = 670.6 + 8.1 = 678.7 ;
s6 = s5 = 6.8778 ⇒ T6 = 182.32 oC
h5 = 3521.0,
h6 = 2810.0,
h7 = 3591.9, s8 = s7 = 8.1348 = 0.6493 + x8 ×7.5009 ⇒ x8 = 0.9979
h8 = 191.83 + 0.9979 × 2392.8 = 2579.7
CV: heater
Cont: m6a + m2 = m3 = 1 kg, 1st law:
m6a =
m6ah6 + m2h2 = m3h3
670.6 - 192.4
= 0.1827, m2 = m7 = 1 - m6a = 0.8173
2810.0 - 192.4
CV: turbine
wT = (h5 - h6) + (1 - m6a)(h7 - h8)
= 3521 - 2810 + 0.8173(3591.9 - 2579.7) = 1538.2
CV: pumps
wP = m2wP12 + m4wP34 = 0.8214×(-0.6) + 1×(-8.1) = -8.6 kJ/kg
wN = 1538.2 - 8.6 = 1529.6 kJ/kg (m5)
.
.
m5 = WN/wN = 10000/1529.6 = 6.53 kg/s
11-24
11.30 The low pressure turbine in a reheat and regenerative cycle receives 10 kg/s steam
at 600 kPa, 550°C. The turbine exhausts to a condenser operating at 10 kPa. The
condenser cooling water temperature is restricted to a maximum of 10°C increase
so what is the needed flow rate of the cooling water? The steam velocity in the
turbine-condenser connecting pipe is restricted to a maximum of 100 m/s, what is
the diameter of the connecting pipe?
Solution:
7
a)
T
5
550 oC
5
7
HI P LOW P
T1
T2
4
6
6 10 kPa
8
2 3
6a
∆T
COND.
1
8
HTR
s
4
2
1
V8 = 100 m/s, ∆TCOOL H2O = 10 oC
3
P
P
h7 = 3591.83, s7 = 8.13465
CV T2
=> s8 = s7 = 8.13465 = 0.6492 + x8 × 7.501 =>
x8 = 0.99793
v8 = vf + x8 vfg = 0.00101 + x8 × 14.67254 = 14.643
h8 = hf + x8 hfg = 191.81 + x8 × 2392.82 = 2579.7
CV CONDENSER:
.
.
QL = m(h8- h1) = 10(2579.7 - 191.81) = 23879 kW
∆ hH2O = h30 - h20 = 125.77 - 83.94 = 41.83 (=CP liq(∆T) = 4.184 × 10)
.
.
mH2O = QL/∆ hH2O = 23879 / 41.83 = 571 kg/s
Turbine - condenser connecting pipe:
.
.
m = ρAV = AV/v => A = mv/V = 10 × 14.643/100 = 1.464 m2
D = (4A/π)1/2 = 1.37 m
11-25
11.31 A steam power plant has a high pressure of 5 MPa and maintains 50°C in the
condenser. The boiler exit temperature is 600°C. All the components are ideal
except the turbine which has an actual exit state of saturated vapor at 50°C. Find
the cycle efficiency with the actual turbine and the turbine isentropic efficiency.
Simple Rankine cycle.
Boiler exit:
h3 = 3666.5 ,
s3 = 7.2588
4s: 50°C, s = s3 => x = (7.2588 - 0.7037)/7.3725 = 0.88913,
Ideal Turbine:
h4s = 209.31 + 0.88913 × 2382.75 = 2327.88
=>
wTs = h3 - h4s = 1338.62
Condenser exit:
h1 = 209.31 ,
Actual turbine:
wTac = h3 - h4ac = 1074.4
Actual turbine exit: h4ac = hg = 2592.1
ηT = wTac / wTs = 0.803: Isentropic Efficiency
Pump:
-wP = v1( P2 - P1) = 0.001012(5000-12.35) = 5.05 kJ/kg
h2 = h1 - wP = 209.31 + 5.05 = 214.36 kJ/kg
qH = h3 - h2 = 3666.5 - 214.36 = 3452.14 kJ/kg
ηcycle = (wTac + wP) / qH = 0.31: Cycle Efficiency
11.32 A steam power cycle has a high pressure of 3.5 MPa and a condenser exit
temperature of 45°C. The turbine efficiency is 85%, and other cycle components
are ideal. If the boiler superheats to 800°C, find the cycle thermal efficiency.
Basic Rankine cycle as shown in Figure 11.3 in the main text.
C.V. Turb.: wT = h3 - h4, s4 = s3 + sT,GEN
Ideal: s4 = s3 = 7.9134 =>
h4S = 2503.2,
wT,S = 1640.44
Actual: wT,AC = η × wT,S = 1394.38
C.V. Pump:
-wP = ∫ v dP ≈ v1(P2 - P1) = 3.52 kJ/kg
C.V. Boiler:
qH = h3 - h2 = h3 - h1 + wP = 3951.7
η = (wT,AC + wP)/qH = (1394.4 - 3.5)/3951.7 = 0.352
11-26
11.33 A steam power plant operates with with a high pressure of 5 MPa and has a boiler
exit temperature of of 600°C receiving heat from a 700°C source. The ambient at
20°C provides cooling for the condenser so it can maintain 45°C inside. All the
components are ideal except for the turbine which has an exit state with a quality
of 97%. Find the work and heat transfer in all components per kg water and the
turbine isentropic efficiency. Find the rate of entropy generation per kg water in
the boiler/heat source setup.
Take CV around each component (all SSSF) in standard Rankine Cycle.
1: v = 0.00101; h = 188.42, s = 0.6386 (saturated liquid at 45°C).
3: h = 3666.5, s = 7.2588 superheated vapor
4ac: h = 188.42 + 0.97 × 2394.8 = 2511.4 kJ/kg
CV Turbine: no heat transfer q = 0
wac = h3 - h4ac = 3666.5 - 2511.4 = 1155.1 kJ/kg
Ideal turbine: s4 = s3 = 7.2588 =>
x4s = 0.88,
h4s = 2295
ws = h3 - h4s = 3666.5 - 2295 = 1371.5,
Eff = wac / ws = 1155.1 / 1371.5 = 0.842
CV Condenser: no shaft work w = 0
qOut = h4ac - h1 = 2511.4 - 188.42 = 2323 kJ/kg
CV Pump: no heat transfer, q = 0 incompressible flow so v = constant
-w = v(P2- P1) = 0.00101(5000-9.59) = 5.04 kJ/kg
CV Boiler: no shaft work, w = 0
qH = h3 - h2 = h3 - h1 + wP = 3666.5 - 188.42 -5.04 = 3473 kJ/kg
s2 + (qH/ TH) + sGen = s3 and s2 = s1 (from pump analysis)
sGen = 7.2588 - 0.6386 - 3473/(700+273) = 3.05 kJ/kg K
11-27
11.34 Repeat Problem 11.26 assuming the turbine has an isentropic efficiency of 85%.
The physical components and the T-s diagram is as shown in Fig. 11.10 in the
main text for one open feedwater heater. The same state numbering is used.
From the Steam Tables:
h5 = 4047.1, s5 = 6.9345,
h1 = 191.83, v1 = 0.00101
h3 = 762.8 , v3 = 0.001127
Pump P1: -wP1 = -v1(P2 - P1) = 0.00101 × 990 = 1 kJ/kg
h2 = h1 - wP1 = 192.83 kJ/kg
Turbine 5-6: s6 = s5 ⇒ h6 = 2948 kJ/kg
wT56,s = h5 - h6 = 1099.1 kJ/kg ⇒ wT56,AC = 1099.1 × 0.85 = 934.24
wT56,AC = h5 - h6AC ⇒ h6AC = h5 - wT56,AC
= 4047.1 - 934.24 = 3112.86
.
.
.
.
.
Feedwater Heater (mTOT = m5):
xm5h6AC + (1 - x)m5h2 = m5h3
⇒
x=
h3 - h2
h6 - h2
=
762.8 - 192.83
= 0.1952
3112.86 - 192.83
To get the turbine work apply the efficiency to the whole turbine. (i.e. the first
section should be slightly different).
s7s = s6s = s5 ⇒ x7s = 0.83793 ,
h7s = 2196.8
wT57,s = h5 - h7s = 4047.1 - 2196.8 = 1850.3
wT57,AC = wT57,sηT = 1572.74 = h5 - h7AC => h 7AC = 2474.4
Find specific turbine work to get total flow rate
.
WT
.
5000
mTOT =
=
= 3.453 kg/s
xwT56 + (1-x)wT57 0.1952×934.24 + 0.8048×1572.74
.
.
QL = mTOT(1 - x)(h7 - h1) = 3.453 × 0.8048(2474.4 - 191.83) = 6343 kW
11-28
11.35 Steam leaves a power plant steam generator at 3.5 MPa, 400°C, and enters the
turbine at 3.4 MPa, 375°C. The isentropic turbine efficiency is 88%, and the
turbine exhaust pressure is 10 kPa. Condensate leaves the condenser and enters
the pump at 35°C, 10 kPa. The isentropic pump efficiency is 80%, and the
discharge pressure is 3.7 MPa. The feedwater enters the steam generator at 3.6
MPa, 30°C. Calculate the the thermal efficiency of the cycle and the entropy
generation for the process in the line between the steam generator exit and the
turbine inlet, assuming an ambient temperature of 25°C.
2
T
1
TURBINE.
ST.
GEN.
6
η = 0.88
5
4
5
5S
6 4
3
COND.
3.5 MPa
3.4 MPa
1
400 oC
375 oC
2
10 kPa
3S 3
s
P
1: h1 = 3222.3,
s1 = 6.8405 ,
2:
h2 = 3165.7,
s2 = 6.7675
3s: s3S = s2 ⇒ x3S = 0.8157, h3S = 2143.6
wT,S = h2 - h3S = 3165.7 - 2143.6 = 1022.1
wT,AC = ηwT,S = 899.4 kJ/kg,
3ac:
h3 = h2 - wT,AC = 2266.3
-wP,S = vf(P5 - P4) = 0.001006(3700 - 10) = 3.7 kJ/kg
-wP,AC = -wP,S/ηP = 4.6 kJ/kg
qH = h1 - h6 = 3222.3 - 129.0 = 3093.3 kJ/kg
η = wNET/qH = (899.4 - 4.6)/3093.3 = 0.289
C.V. Line from 1 to 2:
Energy Eq.:
w = 0/ ,
q = h2 - h1 = 3165.7 - 3222.3 = - 56.6 kJ/kg
Entropy Eq.: s1 + sgen + q/T0 = s2
=>
sgen = s2 - s1 -q/T0 = 6.7675 - 6.8405 - (-56.6/298.15) = 0.117 kJ/kg K
11-29
11.36 For the steam power plant described in Problem 11.1, assume the isentropic
efficiencies of the turbine and pump are 85% and 80%, respectively. Find the
component specific work and heat transfers and the cycle efficiency.
CV Pump, Rev & Adiabatic:
-wPs = h2s - h1 = v1(P2 - P1) = 0.00101(3500 - 10) = 3.525 ;
s2s = s1
-wPac = -wPs / ηP = 3.525/0.8 = 4.406 = h2a - h1
h2a = -wPac + h1 = 4.406 + 191.81 = 196.2 kJ/kg
CV Boiler:
qH = h3 - h2a = 2803.43 - 196.2 = 2607.2 kJ/kg
CV Turbine:
wTs = 2803.43 - 1938.57 = 864.9 kJ/kg
wTac = wTs × ηT = 735.2 = h3 - h4a
h4a = h3 - wTac = 2803.43 - 735.2 = 2068.2
CV Condensor:
qL = h4a - h1 = 2068.2 - 191.81 = 1876.4
ηcycle = (wTac + wPac) / qH = (735.2 - 4.41) / 2607.2 = 0.28
This compares to 0.33 for the ideal case.
11.37 A small steam power plant has a boiler exit of 3 MPa, 400°C while it maintains
50 kPa in the condenser. All the components are ideal except the turbine which
has an isentropic efficiency of 80% and it should deliver a shaft power of 9.0 MW
to an electric generator. Find the specific turbine work , the needed flow rate of
steam and the cycle efficiency.
CV Turbine (Ideal):
s4s = s3 = 6.9211, x4s = (6.9211 - 1.091)/6.5029 = 0.8965
h4s = 2407.35, h3 = 3230.8
=> wTs = h3 - h4s = 823.45
CV Turbine (Actual):
wTac = ηT × wTs = 658.76 = h3 - h4ac, => h 4ac = 2572
.
.
m = W / wTac = 9000/658.76 = 13.66 kg/s
C.V. Pump:
-wP = h2 - h1 = v1(P2 - P1) = 0.00103 (3000 - 50) = 3.04 kJ/kg
=> h 2 = h1 -wP = 340.47 + 3.04 = 343.51 kJ/kg
C.V. Boiler:
qH = h3 - h2 = 3230.8 - 343.51 = 2887.3 kJ/kg
ηcycle = (wTac + wP) / qH = (658.76 - 3.04) / 2887.3 = 0.227
11-30
11.38 In a particular reheat-cycle power plant, steam enters the high-pressure turbine at
5 MPa, 450°C and expands to 0.5 MPa, after which it is reheated to 450°C. The
steam is then expanded through the low-pressure turbine to 7.5 kPa. Liquid water
leaves the condenser at 30°C, is pumped to 5 MPa, and then returned to the steam
generator. Each turbine is adiabatic with an isentropic efficiency of 87% and the
pump efficiency is 82%. If the total power output of the turbines is 10 MW,
determine the mass flow rate of steam, the pump power input and the thermal
efficiency of the power plant.
5
5 MPa
T
2
2S
1
3
0.5 MPa
o
3
450 C
5
7.5 kPa
4S 4
LP
TURBINE.
HP
4
ηST1= ηST2= 0.87
6S 6
2
s
P
COND.
η SP= 0.82
a) s4S = s3 = 6.8185 = 1.8606 + x4S × 4.9606
1
=>
x 4S = 0.999
h4S = 640.21 + 0.999 × 2108.5 = 2746.6
wT1,S = h3 - h4S = 3316.1 - 2746.6 = 569.5 kJ/kg
wT1 = ηT1,S × wT1,S = 0.87 × 569.5 = 495.5 kJ/kg
h4ac = 3316.1 - 495.5 = 2820.6
s6S = s5 = 7.9406 = 0.5764 + x6S × 7.675
⇒
x6S = 0.9595
h6S = 168.79 + 0.9595 × 2406 = 2477.3 kJ/kg
wT2,S = h5 - h6S = 3377.9 - 2477.3 = 900.6
wT2 = 0.87 × 900.6 = 783.5
wT1+T2 = 783.5 + 495.5 = 1279 kJ/kg
.
.
m = WT/wT1+T2 = 10000/1279 = 7.82 kg/s
b)
-wP,S = (0.001004)(5000 - 7.5) = 5.01 kJ/kg
-wP = -wSP/ηSP = 5.01/0.82 = 6.11 kJ/kg
.
.
WP = wPm = -7.82 × 6.11 = -47.8 kW
c)
qH = (h3 - h2) + (h5 - h4)
= 3316.1 - 130.2 + (3377.9 - 2820.6) = 3743.2 kJ/kg
wN = 1279.0 - 6.11 = 1272.9 kJ/kg
ηTH = wN/qH = 1272.9/3743.2 = 0.34
6
11-31
11.39 A supercritical steam power plant has a high pressure of 30 MPa and an exit
condenser temperature of 50°C. The maximum temperature in the boiler is 1000°C
and the turbine exhaust is saturated vapor There is one open feedwater heater
receiving extraction from the turbine at 1MPa, and its exit is saturated liquid
flowing to pump 2. The isentropic efficiency for the first section and the overall
turbine are both 88.5%. Find the ratio of the extraction mass flow to total flow into
turbine. What is the boiler inlet temperature with and without the feedwater heater?
Basically a Rankine Cycle
1: 50 °C, 12.35 kPa,
h = 209.31, s = 0.7037
2: 30 MPa
3: 30 MPa, 1000 °C,
h = 4554.7, s = 7.2867
4AC: 50°C, x = 1, h = 2592.1
3
T
2b
2
1a
1b
1
30 MPa
1000 C
1 MPa
3b
3a
50 C
4s
4ac
s
a) C.V. Turbine Ideal: s4S = s3 ⇒ x4S = 0.8929,
h4S = 2336.8 =>
wT,S = h3 - h4S = 2217.86
Actual: wT,AC = h3-h4AC = 1962.6, η = wT,AC/wT,S = 0.885
b)
P2
1b
2b
m tot
3b
m1
1a
P1
1
1b: Sat liq. 179.91°C, h = 762.81
3a: 1 MPa, s = s3 -> h3a = 3149.09,
T3a = 345.96 -> wT1s = 1405.6
3b: 1 MPa, wT1ac = ηwT1s = 1243.96
wT1ac = h3-h3b => h 3b = 3310.74
1a: wP1 = -v1(P1a-P1) ≈ -1
h1a = h1 - wP1 = 210.31
.
.
.
.
C.V. Feedwater Heater: mTOTh1b = m1h3b + (mTOT - m1)h1a
. .
⇒ m1/mTOT = x = (h1b - h1a)/(h3b - h1a) = 0.178
.
.
.
.
c) C.V. Turbine: (mTOT)3 = (m1)3b + (mTOT - m1)4AC
.
.
.
.
.
_T=m
W
TOTh3 - m1h3b - (mTOT - m1)h4AC = 25 MW = mTOTwT
.
wT = h3-xh3b - (1-x)h4AC = 1834.7 => mTOT = 13.63 kg/s
d) C.V. No FWH, Pump Ideal:
-wP = h2S - h1, s2S = s1
Steam table ⇒ h2S = 240.1, T2S = 51.2°C
1 FWH, CV: P2. s2b = s1b = 2.1386 => T 2b = 183.9°C
11-32
11.40 In one type of nuclear power plant, heat is transferred in the nuclear reactor to
liquid sodium. The liquid sodium is then pumped through a heat exchanger where
heat is transferred to boiling water. Saturated vapor steam at 5 MPa exits this heat
exchanger and is then superheated to 600°C in an external gas-fired superheater.
The steam enters the turbine, which has one (open-type) feedwater extraction at
0.4 MPa. The isentropic turbine efficiency is 87%, and the condenser pressure is
7.5 kPa. Determine the heat transfer in the reactor and in the superheater to
produce a net power output of 1 MW.
5 MPa
T
6
6
600o C
SUP.
HT.
TURBINE.
Q
0.4 MPa
7
5
8
3
4
2 3
1
HTR.
REACT.
P
5
7s
4
COND.
2
7
7.5 kPa
8s 8
s
.
WNET = 1 MW , ηST = 0.87
1
P
-wP12 = 0.001008(400 - 7.5) = 0.4 kJ/kg
h2 = h1 - wP12 = 168.8 + 0.4 = 169.2
-wP34 = 0.001084(5000 - 400) = 5.0 kJ/kg
h4 = h3 - wP34 = 604.7 + 5.0 = 609.7
s7S = s6 = 7.2589, P7=0.4 MPa => T7S= 221.2 oC, h7S = 2904.5
h6 - h7 = ηST(h6 - h7S) ⇒ 3666.5 - h7 = 0.87(3666.5
- 2904.5) = 662.9 ⇒ h7 = 3003.6
s8S = s6 = 7.2589 = 0.5764 + x8S × 7.6750
x8S = 0.8707
h8S = 168.8 + 0.8707 × 2406.0 = 2263.7
h6 - h8 = ηST(h6 - h8S) ⇒ 3666.5 - h8 = 0.87(3666.5
- 2263.7) = 1220.4 ⇒ h8 = 2446.1
CV: heater
cont: m2 + m7 = m3 = 1.0 kg,
Energy Eq.:
m2h2 + m7h7 = m3h3
m7 = (604.7-169.2)/(3003.6-169.2) = 0.1536
CV: turbine
wT = (h6 - h7) + (1 - m7)(h7 - h8)
= 3666.5-3003.6 + 0.8464(3003.6-2446.1) = 1134.8 kJ/kg
11-33
CV: pumps
wP = m1wP12 + m3wP34 = 0.8464(-0.4) + 1(-5.0) = -5.3 kJ/kg
.
wNET = 1134.8 - 5.3 = 1129.5 => m = 1000/1129.5 = 0.885 kg/s
CV: reactor
.
.
QREACT = m(h5 - h4) = 0.885(2794.3 - 609.7) = 1933 kW
CV: superheater
.
QSUP = 0.885(h6 - h5) = 0.885(3666.5 - 2794.3) = 746 kW
11.41 A cogenerating steam power plant, as in Fig. 11.17, operates with a boiler output
of 25 kg/s steam at 7 MPa, 500°C. The condenser operates at 7.5 kPa and the
process heat is extracted as 5 kg/s from the turbine at 500 kPa, state 6 and after
use is returned as saturated liquid at 100 kPa, state 8. Assume all components are
ideal and find the temperature after pump 1, the total turbine output and the total
process heat transfer.
Pump 1:
Inlet state is saturated liquid:
h1 = 168.79, v1 = 0.001008
wP1 = -∫ v dP = -v1 ( P2 - P1) = -0.001008(7000-7.5) = -7.05 kJ/kg
- wP1 = h2 - h1 => h2 = h1 - wP1 = 175.84, T5 = 42°C
Turbine:
h5 = 3410.3, s5 = 6.7975
P6, s6 = s5 => x6 = 0.9952, h6 = 2738.6
P7, s7 = s5 => x7 = 0.8106, h7 = 2119.0
.
.
.
WT = m5 ( h5 - h6) + 0.80m5 ( h6 - h7) = 25 (3410.3 - 2738.6)
+ 20 (2738.6 - 2119) = 16792.5 + 12392 = 29.185 MW
.
.
Qproc = m6(h6 - h8) = 5(2738.6 - 417.46) = 11.606 MW
11-34
11.42 A 10 kg/s steady supply of saturated-vapor steam at 500 kPa is required for drying
a wood pulp slurry in a paper mill. It is decided to supply this steam by
cogeneration, that is, the steam supply will be the exhaust from a steam turbine.
Water at 20°C, 100 kPa, is pumped to a pressure of 5 MPa and then fed to a steam
generator. It may be assumed that the isentropic efficiency of the pump is 75%,
and that of the turbine is 85%. What is the steam temperature exiting the steam
generator? What is the additional heat transfer rate to the steam generator beyond
what would have been required to produce only the desired steam supply? What is
the difference in net power?
State 3: P3 = 5 MPa
State 4: P4 = 500 kPa, sat. vap. -> x4 = 1.0, T4 = 151.9°C
h4 = hg = 2748.7 kJ/kg, s4 = sg = 6.8212 kJ/kg-K
State 1: T1 = 20°C, P1 = 100 kPa
h1 = hf = 83.94 kJ/kg, v1 = vf = 0.001002 m3/kg
State 2: P2 = 5 MPa
(a) C.V.: Turbine, Assume reversible & adiabatic, s3 = s4s
s3 = s4s = sf + x4s sfg = 1.8606 + x4s4.9606
h4s = hf + x4s hfg = 640.2 + x4s2108.5
ηts =
wt
=
h 3 - h4
= 0.85
wts h3 - h4s
Trial and Error to find T3 @ P3 = 5 MPa
Assume T3 = 400°C -> s3 = 6.6458 kJ/kg-K, h3 = 3195.6 kJ/kg
s4s = s3 -> x4s = 0.9646,
h4s = hf + x4s hfg = 2674.1 kJ/kg
h3 - h 4
= 0.857 ≈ ηts; T3 = 400°C
h3 - h4s
(b) With Cogeneration; C.V. Pump
wPs = -∫ vdP = -v1( P2- P1) = -4.91 kJ/kg
wPw = wPs / ηPs = -6.55 kJ/kg
st
1 Law: q + h1 = h2 + w; q = 0; h2 = h1 - wPw = 90.5 kJ/kg
C.V. Steam Generator: qw = h3 - h2 = 3105.1 kJ/kg
Without Cogeneration; C.V. Pump: wPs = -v1( P4- P1) = -0.4 kJ/kg
wPw/o = wPs / ηPs = -0.53 kJ/kg; h2 = h1 - wPw/o = 84.5 kJ/kg
C.V. Steam Generator: qw/o = h4- h2 = 2664.2 kJ/kg
.
Additional Heat Transfer: qw - qw/o = 440.9 kJ/kg; Qextra = 4409 kW
Difference in Net Power: w diff = (wt + wPw) - wPw/o,
wt = h3 - h4 = 446.9 kJ/kg
.
wdiff = 440.9 kJ/kg, Wdiff = 4409 kW
11-35
11.43 An industrial application has the following steam requirement: one 10-kg/s stream
at a pressure of 0.5 MPa and one 5-kg/s stream at 1.4 MPa (both saturated or
slightly superheated vapor). It is obtained by cogeneration, whereby a highpressure boiler supplies steam at 10 MPa, 500°C to a turbine. The required
amount is withdrawn at 1.4 MPa, and the remainder is expanded in the lowpressure end of the turbine to 0.5 MPa providing the second required steam flow.
Assuming both turbine sections have an isentropic efficiency of 85%, determine
the following.
a. The power output of the turbine and the heat transfer rate in the boiler.
b. Compute the rates needed were the steam generated in a low-pressure boiler
without cogeneration. Assume that for each, 20°C liquid water is pumped to the
required pressure and fed to a boiler.
o
BOILER.
2
.
-W P
3
.
QH
P
1
10 MPa, 500 C
ηs = 0.85
o
20 C
.
W HPT
HP TURB.
4
MPa
{1.4
5 kg/s
H 2O IN
ηs = 0.85
LP TURB.
5
.
W LPT
MPa
{0.5
10 kg/s
a) high-pressure turbine
s4S = s3 = 6.5966 ⇒ T4S = 219.9 oC, h4S = 2852.6
wS HPT = h3 - h4S = 3373.7 - 2852.6 = 521.1
wHPT = ηSwS HPT = 0.85 × 521.1 = 442.9
h4 = h3 - w = 3373.7-442.9 = 2930.8
⇒ T4 = 251.6°C, s4 = 6.7533
low-pressure turbine
s5S = s4 = 6.7533 = 1.8607 + x5S × 4.9606, x5S = 0.9863
h5S = 640.23 + 0.9863 × 2108.5 = 2719.8
wS LPT = h4 - h5S = 2930.8 - 2719.8 = 211.0
wLPT = ηSwS LPT = 0.85 × 211.0 = 179.4 kJ/kg
STEAM
STEAM
11-36
h5 = h4 - w = 2930.8 - 179.4 = 2751.4 > hG OK
.
WTURB = 15 × 442.9 + 10 × 179.4 = 8438 kW
.
b) -wP = 15[0.001002(10000 - 2.3)] = 150.3 kW
h2 = h1 - wP = 83.96 + 10.02 = 94.0
.
.
QH = m1(h3 - h2) = 15(3373.7 - 94.0) = 49196 kW
This is to be compared to the amount of heat required to supply 5 kg/s of 1.4
MPa sat. vap. plus 10 kg/s of 0.5 MPa sat. vap. from 20 oC water.
1
5 kg/s
o
20 C
2
3
P
5 kg/s
SAT. VAP.
AT 1.4 MPa
.
Q 56
10 kg/s
o
20 C
4
5
P
.
-W P2
6
10 kg/s
SAT. VAP.
AT 0.5 MPa
h2 = h1 - wP = 83.96 - 0.001002(1400 - 2.3) = 85.4
.
.
Q
= m1(h3 - h2) = 5(2790.0 - 85.4) = 13523 kW
2 3
.
-WP1 = 5 × 14.0 = 7 kW
h5 = h4 - wP2 = 83.96 + 0.001002(500 - 2.3) = 84.5
.
.
Q = m4(h6 - h5) = 10(2748.7 - 84.5) = 26642 kW
5 6
.
-WP2 = 10 × 0.5 = 5 kW
.
Total QH = 13523 + 26642 = 40165 kW
11-37
11.44 In a cogenerating steam power plant the turbine receives steam from a highpressure steam drum and a low-pressure steam drum as shown in Fig. P11.44. The
condenser is made as two closed heat exchangers used to heat water running in a
separate loop for district heating. The high-temperature heater adds 30 MW and
the low-temperature heaters adds 31 MW to the district heating water flow. Find
the power cogenerated by the turbine and the temperature in the return line to the
deaerator.
Assume a reversible turbine
.
.
.
.
.
m1h1 + m2h2 = m3h3 + m4h4 + WT
.
.
.
m1s1 + m2s2 = mTOTsmix
h1 = 3445.9
2
.
WT
1
Turbine
s1 = 6.9108
h2 = 2855.4 s2 = 7.0592
.
mTOT = 27 kg/s ⇒ sMIX = 6.9383
3
s3 = sMIX ⇒ h3 = 2632.4 x3 = 0.966
s4 = sMIX ⇒ h4 = 2413.5 x4 = 0.899
.
WT = 22 × 3445.9 + 5 × 2855.4 - 13 × 2632.4 - 14 × 2413.5
= 22077 kW = 22 MW
.
.
District heating line QTOT = m(h95 - h60) = 60935 kW
OK, this matches close enough
.
.
.
.
C.V. Both heaters:
m3h3 + m4h4 - QTOT = mTOThEX
13 × 2632.4 - 14 × 2413.5 - 60935 = 7075.2 = 27 × hEX
hEX = 262 ≈ hf ⇒ TEX = 62.5°C
4
11-38
11.45 A boiler delivers steam at 10 MPa, 550°C to a two-stage turbine as shown in Fig.
11.17. After the first stage, 25% of the steam is extracted at 1.4 MPa for a process
application and returned at 1 MPa, 90°C to the feedwater line. The remainder of
the steam continues through the low-pressure turbine stage, which exhausts to the
condenser at 10 kPa. One pump brings the feedwater to 1 MPa and a second pump
brings it to 10 MPa. Assume the first and second stages in the steam turbine have
isentropic efficiencies of 85% and 80% and that both pumps are ideal. If the
process application requires 5 MW of power, how much power can then be
cogenerated by the turbine?
h5 = 3500.9, s5 = 6.7567
5
4s: s6S = s5 ⇒ h6S = 2932.1
T1
wT1,S = h5 - h6S = 568.8
6
⇒ wT1,AC = 483.5
7
h6AC = h5 - wT1,AC = 3017.4
4
6ac: P6, h6AC ⇒ s6AC = 6.9129
7s: s7S = s6AC ⇒ h7S = 2189.9
wT2,S = h6AC - h7S = 827.5
wT2,AC = 622 = h 6AC - h7AC
T2
Boiler
P2
8
Process
heat
5 MW
3
2 P1
⇒
h7AC = 2355.4
1
8: h8 = 377.6 qPROC = h6AC - h8 = 2639.8
.
.
.
m6 = Q/qPROC = 5000/2639.8 = 1.894 kg/s = 0.25 mTOT
.
.
.
.
.
⇒ mTOT = m5 = 7.576 kg/s, m7 = m5 - m6 = 5.682 kg/s
.
.
.
.
WT = m5h5 - m6h6AC - m7h7AC = 7424 kW
C
11-39
11.46 Consider an ideal air-standard Brayton cycle in which the air into the compressor
is at 100 kPa, 20°C, and the pressure ratio across the compressor is 12:1. The
maximum temperature in the cycle is 1100°C, and the air flow rate is 10 kg/s.
Assume constant specific heat for the air, value from Table A.5. Determine the
compressor work, the turbine work, and the thermal efficiency of the cycle.
P
T
P
2
P1 = 100 kPa
s
T1 = 20 oC
3
s
s
P
2
1
3
P
s
4
P2
4
P1
P
1
s
v
= 12
T3 = 1100 oC
.
m = 10 kg/s
k-1
P2 k
a) T2 = T1  = 293.2(12)0.286 = 596.8 K
P1
-wC = -1w2 = CP0(T2 - T1) = 1.004(596.8 - 293.2) = 304.8 kJ/kg
k-1
P4 k
 1 0.286
T4 = T3  = 1373.2 
= 674.7 K
12
P3
wT = CP0(T3 - T4) = 1.004(1373.2 - 674.7) = 701.3 kJ/kg
.
.
.
.
WC = mwC = -3048 kW,
WT = mwT = 7013 kW
qH = CP0(T3 - T2) = 1.004(1373.2 - 596.8) = 779.5 kJ/kg
ηTH = wNET/qH = (701.3 - 304.8)/779.5 = 0.509
b)
v4 = RT4/P4 =
0.287 × 674.7
= 1.9364 m3/kg
100
v2 = RT2/P2 =
0.287 × 596.8
= 0.1427
1200
mep =
wNET
v4 - v2
=
396.5
= 221 kPa
1.9364 - 0.1427
Too low for a reciprocating machine (compared with corresponding values for
Otto and Diesel cycles.)
11-40
11.47 Repeat Problem 11.46, but assume variable specific heat for the air, table A.7.
a) From A.7:
s 2 = s1
h1 = 293.6,
Pr1 = 1.0286
⇒ Pr2 = Pr1 × (P2/P1) = 1.0286 × 12 = 12.343
⇒ T2 = 590 K, h2 = 597.3 kJ/kg
-wC = -1w2 = h2 - h1 = 597.3 - 293.6 = 303.7 kJ/kg
From A.7:
h3 = 1483.3,
Pr3 = 333.59
s4 = s3 ⇒ Pr4 = Pr3 × (P4/P3) = 333.59 × (1/12) = 27.8
⇒ T4 = 735 K, h4 = 751.2 kJ/kg
wT = h3 - h4 = 1483.2 - 751.2 = 732 kJ/kg
.
.
.
.
⇒ WC = mwC = -3037 kW, WT = mwT = 7320 kW
qH = h3 - h2 = 1483.2 - 597.3 = 885.9 kJ/kg
wNET = 732 - 303.7 = 428.3 kJ/kg
ηTH = wNET/qH = 428.3/885.9 = 0.483
v4 = (0.287 × 735)/100 = 2.1095 m3/kg
b)
v2 = (0.287 × 590)/1200 = 0.1411
mep = 428.3/(2.1095 - 0.1411) = 217.6 kPa
11.48 An ideal regenerator is incorporated into the ideal air-standard Brayton cycle of
Problem 11.46. Find the thermal efficiency of the cycle with this modification.
T
3
Problem 11.46 + ideal regen. From 11.46:
wT = 701.3, wC = -304.8 kJ/kg
s
x
2
s
wNET = 396.5 kJ/kg
4
y
Ideal regen.:
TX = T4 = 674.7 K
qH = h3 - hX = 1.004(1373.2 - 674.7)
1
s
= 701.3 kJ/kg = wT
ηTH = wNET/qH = 396.5/701.3 = 0.565
11-41
11.49 A Brayton cycle inlet is at 300 K, 100 kPa and the combustion adds 670 kJ/kg.
The maximum temperature is 1200 K due to material considerations. What is the
maximum allowed compression ratio? For this calculate the net work and cycle
efficiency assuming variable specific heat for the air, table A.7.
Combustion: h3 = h2 + qH;
2w3 = 0 and Tmax = T3 = 1200 K
h2 = h3 - qH = 1277.8 - 670 = 607.8
T2 ≈ 600 K; Pr2 = 13.0923 ; T1 = 300 K; Pr1 = 1.1146
Ideal Compression: P2 / P1 = Pr2 / Pr1 = 11.75
Ideal Expansion:
Pr4 = Pr3 / (P3 / P4) = 191.174 / 11.75 = 16.27
T4 ≈ 636 K, h4 = 645.7
linear interpolation
wT = h3 - h4 = 1277.8 - 645.7 = 632.1
-wC = h2 - h1 = 607.8 - 300.47 = 307.3
wnet = wT + wC = 632.1 - 307.3 = 324.8
η = wnet / qH = 324.8 / 670 = 0.485
11.50 A large stationary Brayton cycle gas-turbine power plant delivers a power output
of 100 MW to an electric generator. The minimum temperature in the cycle is 300
K, and the maximum temperature is 1600 K. The minimum pressure in the cycle
is 100 kPa, and the compressor pressure ratio is 14 to 1. Calculate the power
output of the turbine. What fraction of the turbine output is required to drive the
compressor? What is the thermal efficiency of the cycle?
T
3
T1 = 300 K, P2/P1 = 14, T3 = 1600 K
a) Assume const CP0:
⇒ T2 = T1(P2/P1
k-1
)k
s
s2 = s1
2
= 300(14)
0.286
= 638.1 K
-wC = -w12 = h2 - h1 = CP0(T2 - T1)
s
1
= 1.004 (638.1 - 300) = 339.5 kJ/kg
Also, s4 = s3 → T4 = T3(P4/P3
k-1
)k
4
s
= 1600 (1/14)0.286 = 752.2 K
wT = w34 = h3 − h4 = CP0(T3 − T4) = 1.004 (1600 − 752.2) = 851.2 kJ/kg
wNET = 851.2 - 339.5 = 511.7 kJ/kg
.
.
m = WNET/wNET = 100000/511.7 = 195.4 kg/s
.
.
WT = mwT = 195.4 × 851.2 = 166.32 MW
-wC/wT = 339.5/851.2 = 0.399
b) qH = CP0(T3 - T2) = 1.004 (1600 - 638.1) = 965.7 kJ/kg
ηTH = wNET/qH = 511.7/965.7 = 0.530
11-42
11.51 Repeat Problem 11.50, but assume that the compressor has an isentropic
efficiency of 85% and the turbine an isentropic efficiency of 88%.
Same as problem 11.50 , except ηSC = 0.85 & ηST = 0.88
P1 = 100 kPa, T1 = 300 K, P2 = 1400 kPa, T3 = 1600 K
a) From solution 11.50: T2S = 638.1 K,
wSC = -339.5 kJ/kg
⇒ -wC = -wSC/ηSC = 339.5/0.85 = 399.4 kJ/kg = CP0(T2-T1)
⇒ T2 = T1 - wc/CP0 = 300 +
399.4
= 697.8 K
1.004
From solution 11.50: T4S = 752.2 K, wST = 851.2 kJ/kg
⇒ wT = ηST wST = 0.88 × 851.2 = 749.1 kJ/kg = C P0(T3-T4)
⇒ T4 = T3 - wT/CP0 = 1600 -
749.1
= 853.9 K
1.004
wNET = 749.1 - 399.4 = 349.7 kJ/kg
.
.
m = WNET/wNET = 100000/349.7 = 286.0 kg/s
.
.
wT = mwT = 286.0×749.1 = 214.2 MW
-wC/wT = 399.4/749.1 = 0.533
b)
qH = CP0(T3 - T2) = 1.004(1600 - 697.8) = 905.8 kJ/kg
ηTH = wN/qH = 349.7/905.8 = 0.386
11.52 Repeat Problem 11.51, but include a regenerator with 75% efficiency in the cycle.
T
3
x'
2s 2
4
x
⇒ TX = 814.9 K
4s
1
s
b)
Same as 11.51, but with a regenerator
hX - h2 TX - T2
TX - 697.8
=
=
ηREG = 0.75 =
T4 - T2 853.9 - 697.8
'
hX - h2
a) Turbine and compressor work not affected by
regenerator.
qH = CP0(T3 - TX) = 1.004(1600 - 814.9) = 788.2 kJ/kg
ηTH = wNET/qH = 349.7/788.2 = 0.444
11-43
11.53 A gas turbine with air as the working fluid has two ideal turbine sections, as
shown in Fig. P11.53, the first of which drives the ideal compressor, with the
second producing the power output. The compressor input is at 290 K, 100 kPa,
and the exit is at 450 kPa. A fraction of flow, x, bypasses the burner and the rest
(1 − x) goes through the burner where 1200 kJ/kg is added by combustion. The
two flows then mix before entering the first turbine and continue through the
second turbine, with exhaust at 100 kPa. If the mixing should result in a
temperature of 1000 K into the first turbine find the fraction x. Find the required
pressure and temperature into the second turbine and its specific power output.
C.V.Comp.: -wC = h2 - h1; s2 = s1
Pr2 = Pr1(P2/P1) = 0.9899(450/100) = 4.4545, Tr2 = 445
h2 = 446.74,
-wC = 446.74 - 290.43 = 156.3
C.V.Burner: h3 = h2 + qH = 446.74 + 1200 = 1646.74 kJ/kg
⇒ T3 = 1509 K
C.V.Mixing chamber: (1 - x)h3 + xh2 = hMIX = 1046.22 kJ/kg
h3 - hMIX
1646.74 - 1046.22
= 0.50
1646.74 - 446.74
h3 - h2
.
.
.
WT1 = WC,in ⇒ wT1 = -wC = 156.3 = h 3 - h4
x=
=
h4 = 1046.22 - 156.3 = 889.9 ⇒ T4 = 861 K
P4 = (Pr4/PrMIX)PMIX = (51/91.65) × 450 = 250.4 kPa
s4 = s5 ⇒ Pr5 = Pr4(P5/P4) = 51(100/250.4) = 20.367
h5 = 688.2 T5 = 676 K
wT2 = h4 - h5 = 889.9 - 688.2 = 201.7 kJ/kg
11-44
11.54 The gas-turbine cycle shown in Fig. P11.54 is used as an automotive engine. In
the first turbine, the gas expands to pressure P , just low enough for this turbine
5
to drive the compressor. The gas is then expanded through the second turbine
connected to the drive wheels. The data for the engine are shown in the figure and
assume that all processes are ideal. Determine the intermediate pressure P , the
5
net specific work output of the engine, and the mass flow rate through the engine.
Find also the air temperature entering the burner T , and the thermal efficiency of
3
the engine.
k-1
P2 k
a) s2 = s1 ⇒ T2 = T1  = 300(6)0.286 = 500.8 K
P1
-wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg
wT1 = -wC = 201.6 = C P0(T4 - T5) = 1.004(1600 - T5)
⇒ T5 = 1399.2 K
k-1
T5 k
1399.23.5
s5 = s4 ⇒ P5 = P4  = 600
 = 375 kPa
 1600 
T4
k-1
P6 k
1000.286
b) s6 = s5 ⇒ T6 =T5  = 1399.2 
= 958.8 K
375
P5
wT2 = CP0(T5 - T6) = 1.004(1399.2 - 958.8) = 442.2 kJ/kg
.
.
m = WNET/wT2 = 150/442.2 = 0.339 kg/s
c) Ideal regenerator ⇒ T3 = T6 = 958.8 K
qH = CP0(T4 - T3) = 1.004(1600 - 958.8) = 643.8 kJ/kg
ηTH = wNET/qH = 442.2/643.8 = 0.687
11-45
11.55 Repeat Problem 11.54, but assume that the compressor has an efficiency of 82%,
that both turbines have efficiencies of 87%, and that the regenerator efficiency is
70%.
k-1
P2 k
a) From solution 11.54: T2 = T1  = 300(6)0.286 = 500.8 K
P1
-wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg
-wC = -wSC/ηSC = 201.6/0.82 = 245.8 kJ/kg = wT1
= CP0(T4 - T5) = 1.004(1600 - T5) ⇒ T5 = 1355.2 K
wST1 = wT1/ηST1 = 245.8/0.87 = 282.5
= CP0(T4 - T5S) = 1.004(1600 - T5S)
s5S = s4 ⇒ P5 = P4(T5S/T4
k
k-1
)
= 600(
⇒
T5S = 1318.6 K
1318.6 3.5
) = 304.9 kPa
1600
b) P6 = 100 kPa, s6S = s5
k-1
P6 k
 100 0.286
T6S = T5  = 1355.2

= 985.2K
304.9
P5
wST2 = CP0(T5-T6S) = 1.004(1355.2- 985.2) = 371.5 kJ/kg
wT2 = ηST2 × wST2 = 0.87 × 371.5 = 323.2 kJ/kg
323.2 = CP0(T5-T6) = 1.004(1355.2 -T6) ⇒ T6 = 1033.3K
.
.
m = WNET/wNET = 150/323.2 = 0.464 kg/s
c)
wC = 245.8 = C P0(T2 - T1) = 1.004(T2 – 300) ⇒ T2 = 544.8 K
ηREG =
h3 - h2
h6 - h2
=
T 3 - T2
T3 - 544.8
=
= 0.7
T6 - T2 1033.3 - 544.8
⇒ T3 = 886.8 K
qH = CP0(T4 - T3) = 1.004(1600 – 886.8) = 716 kJ/kg
ηTH = wNET/qH = 323.2/716 = 0.451
11-46
11.56 Repeat the questions in Problem 11.54 when we assume that friction causes
pressure drops in the burner and on both sides of the regenerator. In each case, the
pressure drop is estimated to be 2% of the inlet pressure to that component of the
system, so P3 = 588 kPa, P4 = 0.98 P3 and P6 = 102 kPa.
k-1
P2 k
a) From solution 11.54: T2 = T1  = 300(6)0.286 = 500.8 K
P1
-wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg
P3 = 0.98 × 600 = 588 kPa, P4 = 0.98 × 588 = 576.2 kPa
s5 = s4 ⇒ P5 = P4(T5S/T4
k
k-1
)
= 576.2(
1399.2 3.5
) = 360.4 kPa
1600
b) P6 = 100/0.98 = 102 kPa, s6S = s5
k-1
P6 k
 102 0.286
T6 = T5  = 1399.2

= 975.2K
292.8
P5
wST2 = CP0(T5-T6) = 1.004(1399.2 - 975.2) = 425.7 kJ/kg
.
.
m = WNET/wNET = 150/425.7 = 0.352 kg/s
c) T3 = T6 = 975.2 K
qH = CP0(T4 - T3) = 1.004 (1600 - 975.2) = 627.3 kJ/kg
ηTH = wNET/qH = 425.7/627.3 = 0.678
11-47
11.57 Consider an ideal gas-turbine cycle with two stages of compression and two
stages of expansion. The pressure ratio across each compressor stage and each
turbine stage is 8 to 1. The pressure at the entrance to the first compressor is 100
kPa, the temperature entering each compressor is 20°C, and the temperature
entering each turbine is 1100°C. An ideal regenerator is also incorporated into the
cycle. Determine the compressor work, the turbine work, and the thermal
efficiency of the cycle.
10
REG
I.C.
CC
1
5
2
COMP
9
6
4
COMP
TURB
TURB
7
8
CC
P2/P1 = P4/P3 = P6/P7 = P8/P9 = 8.0
T
8
6
P1 = 100 kPa
T1 = T3 = 20 oC, T6 = T8 = 1100 oC
5
7
Assume const. specific heat
s2 = s1 and s4 = s3
4
k-1
P2 k
⇒ T4 = T2 = T1  = 293.2(8)0.286 = 531.4 K
P1
3
2
9
10
1
s
Total -wC = 2 × (-w12) = 2CP0(T2 - T1) = 2 × 1.004(531.4 - 293.2) = 478.1 kJ/kg
k-1
P7
10.286
Also s6 = s7 and s8 = s9: ⇒ T7 = T9 = T6  k = 1373.2 
= 757.6 K
8
P6
Total wT = 2 × w67 = 2CP0(T6 - T7) = 2 × 1.004(1373.2 - 756.7) = 1235.5 kJ/kg
wNET = 1235.5 - 478.1 = 757.4 kJ/kg
Ideal regenerator: T5 = T9, T10 = T4
⇒ qH = (h6 - h5) + (h8 - h7) = 2CP0(T6 - T5)
= 2 × 1.004(1373.2 - 757.6) = 1235.5 kJ/kg
ηTH = wNET/qH = 757.4/1235.5 = 0.613
11-48
11.58 Repeat Problem 11.57, but assume that each compressor stage and each turbine
stage has an isentropic efficiency of 85%. Also assume that the regenerator has an
efficiency of 70%.
T
T1 = T3 = 20 oC, T6 = T8 = 1100 oC
8
6
P2/P1 = P4/P3 = P6/P7 = P8P9 = 8.0
9
5
From solution 11.57:
T4S = T2S = 531.4 K, wSC = 478.1 kJ/kg
4
4S
T7S = T9S = 757.6 K, wST = 1235.5 kJ/kg
⇒ -wC = -wSC/ηSC = 478.1/0.85 = 562.5 kJ/kg
7S
2S
3
1
wT = ηST wST = 0.85 × 1235.5 = 1050.2 kJ/kg
w67 = w89 = 1050.2/2 = 525.1 kJ/kg
281.3
= 573.5 K
1.004
T7 = T9 = T6 - (+w67/CP0) = 1373.2 -
525.1
= 849.9 K
1.004
ηREG =
h5 - h4
h9 - h4
=
T 5 - T4
T9 - T 4
=
T5 - 573.5
849.9 - 573.5
= 0.7 ⇒ T5 = 767 K
qH = CP0(T6 - T5) + CP0(T8 - T7)
= 1.004(1373.2 - 767.0) + 1.004(1373.2 - 849.9) = 1133.5 kJ/kg
wNET = wT + wC = 1050.2 - 562.5 = 487.7
ηTH = wNET/qH = 487.7/1133.5 = 0.430
9S
2
-w12 = -w34 = 562.5/2 = 281.3 kJ/kg
T2 = T4 = T1 + (-w12/CP0) = 293.2 +
7
s
11-49
11.59 A gas turbine cycle has two stages of compression, with an intercooler between
the stages. Air enters the first stage at 100 kPa, 300 K. The pressure ratio across
each compressor stage is 5 to 1, and each stage has an isentropic efficiency of
82%. Air exits the intercooler at 330 K. The maximum cycle temperature is 1500
K, and the cycle has a single turbine stage with an isentropic efficiency of 86%.
The cycle also includes a regenerator with an efficiency of 80%. Calculate the
temperature at the exit of each compressor stage, the second-law efficiency of the
turbine and the cycle thermal efficiency.
State 1: P1 = 100 kPa, T1 = 300 K
State 7: P7 = Po = 100 kPa
State 3: T3 = 330 K; State 6: T6 = 1500 K, P6 = P4
P2 = 5 P1 = 500 kPa;
P4 = 5 P3 = 2500 kPa
Ideal compression T2s = T1 (P2/P1)(k-1)/k = 475.4 K
st
1 Law: q + hi = he + w; q = 0
=>
wc1 = h1 - h2 = CP (T1 - T2)
wc1 s = CP (T1 - T2s) = -176.0 kJ/kg,
wc1 = wc1 s/ η = -214.6
T2 = T1 - wc1/CP = 513.9 K
T4s = T3 (P4/P3)(k-1)/k = 475.4 K
wc2 s = CP (T3 - T4s) = -193.6 kJ/kg;
wc2 = -236.1 kJ/kg
T4 = T3 - wc2 / CP = 565.2 K
Ideal Turbine (reversible and adiabatic)
T7s = T6(P7/P6)(k-1)/k = 597.4 K => w Ts = CP (T6 - T7s) = 905.8 kJ/kg
st
1 Law Turbine:
q + h6 = h7 + w;
q=0
wT = h6 - h7 = CP (T6 - T7) = ηTs wTs = 0.86 × 905.8 = 779.0 kJ/kg
T7 = T6 - wT/ CP = 1500 - 779/1.004 = 723.7 K
s6 - s7 = CP ln
T6
T7
- R ln
P6
P7
= -0.1925 kJ/kg K
ψ6 - ψ7 = (h6 - h7) - To(s6 - s7) = 779.0 - 298.15(-0.1925) = 836.8 kJ/kg
η2nd Law =
d)
wT
ψ6-ψ7
ηth = qH / wnet ;
= 779.0 / 836.8 = 0.931
wnet = wT + wc1 + wc2 = 328.3 kJ/kg
st
1 Law Combustor: q + hi = he + w;
w=0
qc = h6 - h5 = CP (T6 - T5)
Regenerator: ηreg =
T 5 - T4
T7 - T4
= 0.8
qH = qc = 810.7 kJ/kg;
->
T5 = 692.1 K
ηth = 0.405
11-50
11.60 A two-stage air compressor has an intercooler between the two stages as shown in
Fig. P11.60. The inlet state is 100 kPa, 290 K, and the final exit pressure is 1.6
MPa. Assume that the constant pressure intercooler cools the air to the inlet
temperature, T3 = T1. It can be shown, see Problem 9.130, that the optimal
pressure, P2 = (P1P4)1/2, for minimum total compressor work. Find the specific
compressor works and the intercooler heat transfer for the optimal P2.
Optimal intercooler pressure P2 =
100 × 1600 = 400 kPa
1: h1 = 290.43, Pr1 = 0.9899
C.V.: C1 -wC1 = h2 - h1, s2 = s1
⇒ Pr2 = Pr1(P2/P1) = 3.9596, T2 = 430, h2 = 431.95
-wC1 = 431.95 - 290.43 = 141.5 kJ/kg
C.V.Cooler: T3 = T1 ⇒ h3 = h1
qOUT = h2 - h3 = h2 - h1 = -wC1 = 141.5 kJ/kg
C.V.: C2 T3 = T1, s4 = s3
⇒ Pr4 = Pr3(P4/P3) = Pr1(P2/P1) = 3.9596, T4 = T2
Thus we get -wC2 = -wC1 = 141.5 kJ/kg
11.61 Repeat Problem 11.60 when the intercooler brings the air to T = 320 K. The
3
corrected formula for the optimal pressure is P =[ P P (T /T )n/(n-1)]1/2 see
2
1 4
3
1
Problem 9.131, where n is the exponent in the assumed polytropic process.
The polytropic process has n = k (isentropic) so n/(n - 1) = 1.4/0.4 = 3.5
P2 = 400
(320/290)3.5 = 475.2 kPa
C.V. C1: s2 = s1 ⇒ Pr2 = Pr1(P2/P1) = 0.9899(475.2/100)
= 4.704 ⇒ T2 = 452 K, h2 = 453.75
-wC1 = h2 - h1 = 453.75 - 290.43 = 163.3 kJ/kg
C.V. Cooler:
qOUT = h2 - h3 = 453.75 - 320.576 = 133.2 kJ/kg
C.V. C2: s4 = s3 ⇒ Pr4 = Pr3(P4/P3) = 1.3972(1600/475.2) = 4.704
⇒
T4 = T2 = 452 K,
h4 = 453.75
-wC2 = h4 - h3 = 453.75 - 320.576 = 133.2 kJ/kg
11-51
11.62 Consider an ideal air-standard Ericsson cycle that has an ideal regenerator as
shown in Fig. P11.62. The high pressure is 1 MPa and the cycle efficiency is
70%. Heat is rejected in the cycle at a temperature of 300 K, and the cycle
pressure at the beginning of the isothermal compression process is 100 kPa.
Determine the high temperature, the compressor work, and the turbine work per
kilogram of air.
T
P
P2 = P3 = 1 MPa
P
2
3
3
T 4
P1 = 100 kPa
T
T
P
1
P
q = -4q1 (ideal reg.)
2 3
P
4
⇒ qH = 3q4 & wT = qH
2
T
T1 = T2 = 300 K
1
s
v
rp = P2/P1 = 10
ηTH = ηCARNOT TH. = 1 - TL/TH = 0.7 ⇒ T3 = T4 = TH = 1000 K
P2
1000
⌡v dP = RT1ln  = 0.287 × 300 × ln
 = 198.25
qL = -wC = ⌠
 100 
P1
⌠v dP = -RT ln(P /P ) = 660.8 kJ/kg
wT = qH = -⌡
3
4 3
11.63 An air-standard Ericsson cycle has an ideal regenerator. Heat is supplied at
1000°C and heat is rejected at 20°C. Pressure at the beginning of the isothermal
compression process is 70 kPa. The heat added is 600 kJ/kg. Find the compressor
work, the turbine work, and the cycle efficiency.
See the cycle diagrams in solution to 11.62:
T3 = T4 = 1273.15 K, P1 = 70 kPa,
T1 = T2 = 293.15 K,
Ideal regenerator: 2q3 = -4q1, wT = qH = 600 kJ/kg
ηTH = ηCARNOT = 1 - 293.15/1273.15 = 0.7697
wNET = ηTHqH = 0.7697 × 600 = 461.82
qL = -wC = 600 - 461.82 = 138.2
qH = 600 kJ/kg
11-52
11.64 Consider an ideal air-standard cycle for a gas-turbine, jet propulsion unit, such as
that shown in Fig. 11.27. The pressure and temperature entering the compressor
are 90 kPa, 290 K. The pressure ratio across the compressor is 14 to 1, and the
turbine inlet temperature is 1500 K. When the air leaves the turbine, it enters the
nozzle and expands to 90 kPa. Determine the pressure at the nozzle inlet and the
velocity of the air leaving the nozzle.
2
T
3
3
BURN
P
s4
s
5
2
COMP
TURB
P = 90 kPa
5
1
1
4
s
NOZ
C.V. Comp.: s2 = s1 ⇒ Pr2 = Pr1 × (P2/P1) = 0.9899 × 14 = 13.8586
h2 = 617.2, T2 = 609 K
-wC = h2 - h1 = 617.2 - 290.43 = 326.8 kJ/kg
C.V. Turb.: wT = h3 - h4 = -wC ⇒
h4 = h3 + wC = 1635.8 - 326.8 = 1309
⇒ Pr4 = 209.1, T4 = 1227
P4 = Pr4 × (P3/Pr3) = 209.1(1260/483.155) = 545.3 kPa
C.V. Nozzle: s2 = s1 ⇒ Pr5 = Pr4 × (P5/P4) = 209.1(90/545.3) = 34.51
=> T5 = 778 K,
h5 = 798.1 kJ/kg
(1/2)V25 = h4 - h5 = 510.9 ⇒
V5 =
2 × 1000 × 510.9 = 1010.8 m/s
11-53
11.65 The turbine in a jet engine receives air at 1250 K, 1.5 MPa. It exhausts to a nozzle
at 250 kPa, which in turn exhausts to the atmosphere at 100 kPa. The isentropic
efficiency of the turbine is 85% and the nozzle efficiency is 95%. Find the nozzle
inlet temperature and the nozzle exit velocity. Assume negligible kinetic energy
out of the turbine.
C.V. Turb.: hi = 1336.7, Pri = 226, se = si
⇒ Pre = Pri × (Pe/Pi) = 226(250/1500) = 37.667
Te = 796, he = 817.9, wT,s = 1336.7 - 817.9 = 518.8
wT,AC = wT,s × ηT = 441 = he,AC - hi ⇒ he,AC = 895.7
⇒ Te,AC = 866 K, Pre,AC = 52.21
C.V. Nozzle: (1/2)V2e = hi - he; se = si
⇒ Pre = Pri × (Pe/Pi) = 52.21(100/250) = 20.884
⇒ Te,s = 681 K, he,s = 693.1 kJ/kg
2
= hi - he,s = 895.7 - 693.1 = 202.6 kJ/kg
(1/2)Ve,s
2
2
= (1/2)Ve,s
× ηNOZ = 192.47 kJ/kg
(1/2)Ve,AC
Ve,AC =
2 × 1000 × 192.47 = 620 m/s
11-54
11.66 Repeat Problem 11.64, but assume that the isentropic compressor efficiency is
87%, the isentropic turbine efficiency is 89%, and the isentropic nozzle efficiency
is 96%.
Same as 11.64 except
ηSC = 0.87, ηST = 0.89, ηSN = 0.96
T
Assume const. specific heat
P1 = P5 = 90 kPa, T1 = 290 K
3
4
5
2S
2
5S
P2/P1 = 14, T3 = 1500 K
From solution 11.64:
T2S = 609 K, wSC = 326.8 kJ/kg
1
s
⇒ wC = wSC/ηSC = 326.8/0.87 = 375.6 kJ/kg
wT = wC = 375.6 = h 3 - h4 ⇒ h4 = 1635.8 - 375.6 = 1260.2, T4 = 1185 K
wST = wT/ηST = 375.6/0.89 = 422.0 kJ/kg = 1635.8 - h4s
⇒ h4s = 1213.8, T4s = 1145, Pr4s = 158.06
s4S= s3 ⇒ P4 = P3 Pr4s/Pr3 = 1260×158.06/483.155 = 412 kPa
s5S = s4 ⇒ Pr5 = Pr4 P5/P4 = 181.6×90/412 = 39.67
⇒ T5S = 807, h5S = 829.8,
2
/2000 = h4 − h5S = 1260.2 - 829.8 = 430.4
V5S
2
/2000 = 0.96 × 430.4 = 413.2 kJ/kg
=> V25/2000 = ηSN V5s
⇒ V5 = 2000*413.2 = 909 m/s
11-55
11.67 Consider an air standard jet engine cycle operating in a 280K, 100 kPa
environment. The compressor requires a shaft power input of 4000 kW. Air enters
the turbine state 3 at 1600 K, 2 MPa, at the rate of 9 kg/s, and the isentropic
efficiency of the turbine is 85%. Determine the pressure and temperature entering
the nozzle at state 4. If the nozzle efficiency is 95%, determine the temperature and
velocity exiting the nozzle at state 5.
.
.
.
C.V. Shaft: WT = m(h3 - h4) = WC
.
.
h3 - h4 = WC / m = 4000/9 = 444.4
h4 = 1757.3 – 444.4 = 1135.8
wTa = wC = 444.4
⇒ wTs = wTa / η = 522.82 = h 3 - h4s
h4s = 1234.5 ⇒ Pr = 168.28,
T4s ≈ 1163 K
P4 = (Pr4 / Pr3) P3 = (168.28/ 634.967)2000 = 530 kPa
4a: 530 kPa, h = 1312.85, T ≈ 1230 K, Pr4 = 211.92
5: 100 kPa ⇒ Pr5s = Pr4(100/530) = 40
h5s = 830.95
2
⇒ 0.5V5s = h4a - h5s = 481.9 kJ/kg
2
2
0.5V5a = η(0.5V5s) = 457.808
2
⇒ V5a= 957 m/s
h5a = h4 - 0.5V5a = 1312.85 – 457.808 = 855 ⇒
T5a ≈ 830 K
11-56
11.68 A jet aircraft is flying at an altitude of 4900 m, where the ambient pressure is
approximately 55 kPa and the ambient temperature is −18°C. The velocity of the
aircraft is 280 m/s, the pressure ratio across the compressor is 14:1 and the cycle
maximum temperature is 1450 K. Assume the inlet flow goes through a diffuser
to zero relative velocity at state 1. Find the temperature and pressure at state 1 and
the velocity (relative to the aircraft) of the air leaving the engine at 55 kPa.
3
T
4
Ambient
TX = -18oC = 255.2 K, PX = 55 kPa = P5
5
also VX = 280 m/s
2
1
Assume that the air at this state is reversibly
decelerated to zero velocity and then enters the
comp. at 1.
P amb
X
P2/P1 = 14 & T3 = 1450 K
s
2
T 1 = TX +
VX
2 × 1000
= 255.2 +
(280)2
= 294.3 K
2 × 1000 × 1.0035
k
 T1k-1
294.33.5
 = 90.5 kPa
P1 = PX  = 55
255.2
TX
Then, T2 = T1 (P2/P1
k-1
)k
= 294.3(14)0.286 = 626.0 K
-wC = -w12 = CP0(T2-T1) = 1.004(1450 - T4)
⇒ T4 = 1118.3 K
P3 = P2 = 14 × 90.5 = 1267 kPa
P4 =
k
k-1
P3 (T4/T3)
T5 = T4 (P5/P4
V25
2 × 1000
k-1
)k
= 1267(1118.3/1450)3.5 = 510 kPa
= 1118.3(55/510)
0.286
= 591.5 K
= CP0(T4 - T5) = 1.004(1118.3 - 591.5) = 528.7 kJ/kg
⇒ V5 = 1028 m/s
11-57
11.69 Air flows into a gasoline engine at 95 kPa, 300 K. The air is then compressed with
a volumetric compression ratio of 8;1. In the combustion process 1300 kJ/kg of
energy is released as the fuel burns. Find the temperature and pressure after
combustion.
Compression 1 to 2: s2 = s1 ⇒ vr2 = vr1/8 = 179.49/8 = 22.436,
T2 = 673 K, u2 = 491.5, Pr2 = 20
P2 = Pr2(P1/Pr1) = 20(95/1.1146) = 1705 kPa
Compression 2 to 3: u3 = u2 + qH = 491.5 + 1300 = 1791.5
T3 = 2118 K
P3 = P2 × (T3/T2) = 1705(2118/673) = 5366 kPa
11.70 A gasoline engine has a volumetric compression ratio of 9. The state before
compression is 290 K, 90 kPa, and the peak cycle temperature is 1800 K. Find the
pressure after expansion, the cycle net work and the cycle efficiency using
properties from Table A.7.
Use table A.7 and interpolation.
Compression 1 to 2: s2 = s1 ⇒ vr2 = vr1(v2/v1)
vr2 = 196.37/9 = 21.819 ⇒ T2 ≅ 680 K,
Pr2 ≅ 20.784,
P2 = P1(Pr2/Pr1) = 90(20.784/0.995) = 1880 kPa
w = u1 - u2 = 207.19 - 496.94 = -289.75 kJ/kg
1 2
Combustion 2 to 3:
qH = u3 - u2 = 1486.33 - 496.94 = 989.39 kJ/kg
P3 = P2(T3/T2) = 1880(1800/680) = 4976 kPa
Expansion 3 to 4:
s4 = s3 ⇒ vr4 = vr3 × 9 = 1.143 × 9 = 10.278
⇒ T4 = 889 K, Pr4 = 57.773, u4 = 665.8
P4 = P3(Pr4/Pr3) = 4976(57.773/1051) = 273.5 kPa
w = u3 - u4 = 1486.33 - 665.8 = 820.5 kJ/kg
3 4
wNET = 3w4 + 1w2 = 820.5 - 289.75 = 530.8 kJ/kg
η = wNET/qH = 530.8/989.39 = 0.536
u2 = 496.94
11-58
11.71 To approximate an actual spark-ignition engine consider an air-standard Otto
cycle that has a heat addition of 1800 kJ/kg of air, a compression ratio of 7, and a
pressure and temperature at the beginning of the compression process of 90 kPa,
10°C. Assuming constant specific heat, with the value from Table A.5, determine
the maximum pressure and temperature of the cycle, the thermal efficiency of the
cycle and the mean effective pressure.
P
T
3
s
v
v
4
v
1
2
s
2
s
3
qH = 1800 kJ
s
v1/v2 = 7
4
P1 = 90 kPa
T1 = 10 oC
v
1
v
s
k
a) P2 = P1(v1/v2) = 90(7)1.4 = 1372 kPa
T2 = T1(P2/P1)(v2/v1)= 283.15 × 15.245 ×
1
= 616.6 K
7
T3 = T2 + qH/CV0 = 616.6 + 1800/0.717 = 3127 K
P3 = P2T3/T2= 1372 × 3127 / 616.6 = 6958 kPa
b)
ηTH = 1 - T1/T2 = 1 - 283.15/616.5 = 0.541
c)
wNET = ηTH × qH = 0.544 × 1800 = 979.2 kJ/kg
v1 = RT1/P1 = (0.287 × 283.2)/90 = 0.9029
v2 = (1/7) v1 = 0.1290 m3/kg
mep =
wNET
v1-v2
=
979.2
= 1265 kPa
0.9029 - 0.129
11-59
11.72 Repeat Problem 11.71, but assume variable specific heat. The ideal gas air tables,
Table A.7, are recommended for this calculation (and the specific heat from Fig.
5.10 at high temperature).
Table A.7 is used with interpolation.
a)
T1 = 283.15 K,
u1 = 202.3, vr1 = 207.94
s2 = s1 ⇒ vr2 = vr1(v2/v1) = 207.94(1/7) = 29.705
⇒ T2 = 606.7 K, u2 = 440.2 =>
u3 = 440.2 + 1800 = 2240.2 =>
-w12 = u2 - u1 = 237.9,
T3 = 2575.5 K ,
vr3 = 0.3402
P3 = 90 × 7 × 2575.5 / 283.15 = 5730 kPa
b)
vr4 = vr3 × 7 = 2.3814
3w 4
Ë
=>
T 4 = 1437 K;
u4 = 1147
= u3 - u4 = 2240.2 - 1147 = 1093.2
wnet = 1093.2 - 237.9 = 855.3 kJ/kg
ηTH = wnet / qH = 855.3 / 1800 = 0.475
(c)
mep = 855.3 / (0.9029 - 0.129) = 1105 kPa
11.73 A gasoline engine takes air in at 290 K, 90 kPa and then compresses it. The
combustion adds 1000 kJ/kg to the air after which the temperature is 2050 K. Use
the cold air properties (i.e. constant heat capacities at 300 K) and find the
compression ratio, the compression specific work and the highest pressure in the
cycle.
Standard Otto Cycle
T3 = 2050 K u2 = u3 - qH
T2 = T3 - qH / Cvo = 2050 - 1000 / 0.717 = 655.3 K
P2 = P1(T2 / T1)k/(k-1) = 90(655.3/290) 3.5 = 1561 kPa
CR = v1 / v2 = (T2 / T1)1/(k-1) = (655.3 / 290) 2.5 = 7.67
-1w2 = u2 - u1 = Cvo( T2 - T1) = 0.717(655.3 - 290) = 262 kJ / kg
P3 = P2T3 / T2 = 1561 × 2050 / 655.3 = 4883 kPa
11-60
11.74 Answer the same three questions for the previous problem, but use variable heat
capacities (use table A.7).
T3 = 2050 K u3 = 1725.714
u2 = u3 - qH = 1725.714 - 1000 = 725.714
⇒ T2 = 960.56
vr2 = 8.216
vr1 = 195.361 =>
v1 / v2 = vr1 / vr2 = 195.361 / 8.216 = 23.78
-1w2 = u2 - u1 = 725.714 - 207.19 = 518.5 kJ/kg
P3=P2T3 / T2 = P1( T3 / T1)( v1 / v3) = 90×(2050 / 290)×23.78 = 15129 kPa
11.75 When methanol produced from coal is considered as an alternative fuel to
gasoline for automotive engines, it is recognized that the engine can be designed
with a higher compression ratio, say 10 instead of 7, but that the energy release
with combustion for a stoichiometric mixture with air is slightly smaller, about
1700 kJ/kg. Repeat Problem 11.71 using these values.
same as 11.71 except v1/v2 = 10 & qH = 1700 kJ
k
a) P2 = P1(v1/v2) = 90(10)1.4 = 2260.7 kPa
k-1
T2 = T1(v1/v2)
= 283.15(10)0.4 = 711.2 K
T3 = T2 + qH / Cvo = 711.2 + 1700 / 0.7165 = 3084 K
P3 = P2(T3 / T2) = 2260.7×3084 / 711.2 = 9803 kPa
ηTH = 1 - T1/T2 = 1 - 283.15/711.2 = 0.602
wnet = ηTH × qH = 0.6 × 1700 = 1023.4 kJ/kg
v1 = RT1/P1 = 0.287×283.15/90= 0.9029, v2 = v1/10= 0.0903
mep = wNET/(v1 - v2)= 1023.4 / (0.9029 - 0.0903) = 1255 kPa
11-61
11.76 It is found experimentally that the power stroke expansion in an internal
combustion engine can be approximated with a polytropic process with a value of
the polytropic exponent n somewhat larger than the specific heat ratio k. Repeat
Problem 11.71 but assume that the expansion process is reversible and polytropic
(instead of the isentropic expansion in the Otto cycle) with n equal to 1.50.
See solution to 11.71 except for process 3 to 4.
T3 = 3127 K, P3 = 6.958 MPa
v3 = RT3/P3 = v2 = 0.129 m3/kg, v4 = v1 = 0.9029
Process:
Pv1.5 = constant.
P4 = P3(v3/v4)1.5 = 6958 (1/7)1.5 = 375.7 kPa
T4 = T3(v3/v4)0.5 = 3127(1/7)0.5 = 1181.9 k
w =⌠
⌡Pdv = 1-1.4(T2 - T1) = -0.4 (606.6 -283.15)= -239.3
R
0.287
1 2
w =⌠
⌡Pdv = R(T4 - T3)/(1 - 1.5)
3 4
= -0.287(1181.9-3127)/0.5 = 1116.5
wNET = 1116.5 - 239.3 = 877.2
ηCYCLE = wNET/qH = 877.2/1800 = 0.487
mep = wNET/(v1 - v2) = 877.2/(0.9029 - 0.129) = 1133 kPa
Note a smaller wNET, ηCYCLE, mep compared to an ideal cycle.
11-62
11.77 In the Otto cycle all the heat transfer qH occurs at constant volume. It is more
realistic to assume that part of qH occurs after the piston has started its downward
motion in the expansion stroke. Therefore, consider a cycle identical to the Otto
cycle, except that the first two-thirds of the total qH occurs at constant volume and
the last one-third occurs at constant pressure. Assume that the total qH is 2100
kJ/kg, that the state at the beginning of the compression process is 90 kPa, 20°C,
and that the compression ratio is 9. Calculate the maximum pressure and
temperature and the thermal efficiency of this cycle. Compare the results with
those of a conventional Otto cycle having the same given variables.
P
3
T
4
3
s
P1 = 90 kPa, T1 = 20oC
s
rV = v1/v2 = 7
v
5
2
4
2
s
s
1
5
= 1400 kJ/kg;
1
s
v
b)
a) q23 = (2/3) × 2100
v
q34 = 2100/3 = 700
P2 = P1(v1/v2)k = 90(9)1.4 = 1951 kPa
T2 = T1(v1/v2)k-1 = 293.15(9)0.4 = 706 K
T3 = T2 + q23/CV0 = 706 + 1400/0.717 = 2660 K
P3 = P2T3/T2 = 1951(2660/706) = 7350.8 kPa = P4
T4 = T3 + q34/CP0 = 2660 + 700/1.004 = 3357 K
P4 T1 7350.8 293.15
= = × =
×
= 7.131
90
3357
v4 v4 P1 T4
v5
v1
T5 = T4(v4/v5)k-1 = 3357(1/7.131)0.4 = 1530 K
qL = CV0(T5-T1) = 0.717(1530 - 293.15) = 886.2 kJ/kg
ηTH = 1 - qL/qH = 1 - 886.2/2100 = 0.578
Std. Otto Cycle:
ηTH = 1 - (9)-0.4 = 0.585, small difference
11-63
11.78 A diesel engine has a compression ratio of 20:1 with an inlet of 95 kPa, 290 K,
state 1, with volume 0.5 L. The maximum cycle temperature is 1800 K. Find the
maximum pressure, the net specific work and the thermal efficiency.
P2 = 95×(20) 1.4 = 6297.5 kPa
T2 = T1(v1 / v2)k-1 = 290×200.4 = 961 K
P3 = P2(T3 / T2) = 6297.5 × (1800 / 961) = 11796 kPa
-1w2 = u2 - u1 ≈ Cvo( T2 - T1) = 0.717(961 - 290) = 481.1 kJ / kg
T4 = T3( v3 / v4)0.4 = 1800(1 / 20) 0.4 = 543 K
3w 4
= u3 - u4 ≈ Cvo(T3 - T2) = 0.717(1800 - 543) = 901.3 kJ / kg
wnet = 3w4 + 1w2 = 901.3 - 481.1 = 420.2 kJ / kg
η = wnet / qH = 1- (1 / 20) 0.4 = 0.698
( = 420.2 / 0.717(1800 - 961))
11.79 A diesel engine has a bore of 0.1 m, a stroke of 0.11 m and a compression ratio of
19:1 running at 2000 RPM (revolutions per minute). Each cycle takes two
revolutions and has a mean effective pressure of 1400 kPa. With a total of 6
cylinders find the engine power in kW and horsepower, hp.
Power from mean effective pressure.
mep = wnet / (vmax - vmin) -> wnet = mep(vmax - vmin)
∆V = πBore2 × 0.25 × S = π × 0.12 × 0.25 × 0.11 = 0.000864 m3
W = mep(Vmax - Vmin) = 1400× π × 0.25 × 0.12 × 0.11 = 1.2096
Work per Cylinders per Power Stroke
.
W = W × n × RPM × 0.5 (cycles / min)×(min / 60 s)×(kJ / cycle)
= 1.2096 × 6 × 2000 × 0.5 × (1/60) = 121 kW = 162 hp
11-64
11.80 At the beginning of compression in a diesel cycle T = 300 K, P = 200 kPa and
after combustion (heat addition) is complete T = 1500 K and P = 7.0 MPa. Find
the compression ratio, the thermal efficiency and the mean effective pressure.
P2 = P3 = 7000 kPa => v1 / v2 = (P2/P11/
) k = (7000 / 200)0.7143 = 12.67
T2 = T1(P2 / P1)(k-1) / k = 300(7000 / 200) 0.2857 = 828.4
v3 / v2 = T3 / T2 = 1500 / 828.4 = 1.81
v4 / v3 = v1 / v3 = (v1 / v2)( v2 / v3) = 12.67 / 1.81 = 7
T4 = T3(v3 / v4)k-1 = (1500 / 7) 0.4 = 688.7
qL = Cvo(T4 - T1) = 0.717(688.7 - 300) = 278.5
qH = h3 - h2 ≈ Cpo(T3 - T2) = 1.004(1500 - 828.4) = 674
η = 1 - qL / qH = 1- 278.5 / 674 = 0.587
wnet = qnet = 674 - 278.5 = 395.5
vmax = v1 = R T1 / P1 = 0.287×300 / 200 = 0.4305 m3 / kg
vmin = vmax / (v1 / v2) = 0.4305 / 12.67 = 0.034 m3 / kg
mep = wnet / (vmax - vmin) = 395.5 / (0.4305 - 0.034) = 997 kPa
11.81 Consider an ideal air-standard diesel cycle in which the state before the compression
process is 95 kPa, 290 K, and the compression ratio is 20. Find the maximum
temperature (by iteration) in the cycle to have a thermal efficiency of 60%?
P1 = 95 kPa,
Diesel cycle:
k-1
T2 = T1(v1/v2)
T1 = 290 K,
v1/v2 = 20, ηTH = 0.6
= 290(20)0.4 = 961.2 K
v1 = 0.287 × 290/95 = 0.876 = v4,
v2 = 0.876 / 20 = 0.0438
v3 = v2(T3/T2) = 0.043883(T3/961.2) = 0.0000456 T3
k-1
T3 = T4 (v4/v3)
ηTH = 0.60 = 1 -
0.876
=(
)0.4 ⇒ T4 = 0.019345 T1.4
3
0.0000456 T3
T 4 - T1
k(T3 - T2)
=1-
0.019345 × T1.4
- 290
3
1.4(T3 - 961.2)
⇒ 0.019345 × T1.4
- 0.56 × T3 + 248.272 = 0
3
3050: LHS = +1.06
T3 = 3040 K
3040: LHS = -0.036,
11-65
11.82 Consider an ideal Stirling-cycle engine in which the state at the beginning of the
isothermal compression process is 100 kPa, 25°C, the compression ratio is 6, and
the maximum temperature in the cycle is 1100°C. Calculate the maximum cycle
pressure and the thermal efficiency of the cycle with and without regenerators.
T
P
3
T 4
3
T
v
4
2
v
v
T
P1 = 100 kPa
v
v1/v2 = 6
2
1
Ideal Stirling cycle
T1 = T2 = 25 oC
T
1
s
v
T3 = T4 = 1100 oC
a) T1 = T2 ⇒ P2 = P1(v1/v2) = 100 × 6 = 600 kPa
V2 = V3 ⇒ P3 = P2T3/T2 = 600×1373.2/298.2 = 2763 kPa
b) w34 = q34 = RT3 ln(v4/v3) = 0.287×1373.2×ln6 = 706.1 kJ/kg
q23 = CV0(T3 - T2) = 0.7165(1100 - 25) = 770.2 kJ/kg
w12 = q12 = -RT1 ln(v1/v2) = -0.287× 298.2 ln(6) = -153.3 kJ/kg
wNET = 706.1 - 153.3 = 552.8 kJ/kg
ηNO REGEN =
552.8
= 0.374,
706.1+770.2
ηWITH REGEN =
552.8
= 0.783
706.1
11.83 An air-standard Stirling cycle uses helium as the working fluid. The isothermal
compression brings helium from 100 kPa, 37°C to 600 kPa. The expansion takes
place at 1200 K and there is no regenerator. Find the work and heat transfer in all
of the 4 processes per kg helium and the thermal cycle efficiency.
Helium table A.5:
R = 2.077 ,
Compression/expansion:
Cvo = 3.1156
v4 / v3 = v1 / v2 = P2 / P1 = 600 / 100 = 6
1 -> 2 -1w2 = -q12 = ∫ P dv = R T1ln(v1 / v2) = RT1ln (P2 /P1)
= 2.077 × 310 × ln6 = 1153.7 kJ/kg
2 -> 3 : 2w3 = 0;
3 -> 4:
q23 = Cvo(T3 - T2) = 3.1156(1200 - 310) = 2773 kJ/kg
v4
w
=
q
=
R
T
ln
= 2.077×1200 ln6 = 4465.8 kJ/kg
3 4
34
3 v
3
4 -> 1 4w1 = 0;
q41 = Cvo(T4 - T1) = -2773 kJ/kg
ηcycle = (1w2+3w4)/(q23+q34) = (-1153.7+4465.8) /(2773 +4465.8) = 0.458
11-66
11.84 Consider an ideal air-standard Stirling cycle with an ideal regenerator. The
minimum pressure and temperature in the cycle are 100 kPa, 25°C, the
compression ratio is 10, and the maximum temperature in the cycle is 1000°C.
Analyze each of the four processes in this cycle for work and heat transfer, and
determine the overall performance of the engine.
Ideal Stirling cycle diagram as in solution 11.82, with
P1 = 100 kPa,
T1 = T2 = 25oC,
From 1-2 at const T:
v1/v2 = 10,
T3 = T4 = 1000oC
w = 1q2 = T1(s2 - s1)
1 2
= -RT1ln(v1/v2) = -0.287 × 298.2 × ln(10) = -197.1 kJ/kg
From 2-3 at const V:
w = 0/
2 3
q23 = CV0(T3 - T2) = 0.7165(1000 - 25) = 698.6
From 3-4 at const T;
= +RT3 × ln
w = 3q4 = T3(s4 - s3)
3 4
v4
v3
From 4-1 at const V;
= 0.287 × 1237.2 × ln(10) = 841.4 kJ/kg
w = 0/
4 1
q41 = CV0(T1 - T4) = 0.7165(25 - 1000) = -698.6 kJ/kg
wNET = -197.1 + 0 + 841.4 + 0 = 644.3 kJ/kg
Since q23 is supplied by -q41 (regenerator)
qh = q34 = 841.4 kJ/kg,
NOTE:
ηTH =
qH = q34 = RT3 × ln(10),
wNET
qH
=
644.3
= 0.766
841.4
qL = -q12 = RT1 × ln(10)
ηTH = (qH - qL)/qH = (T3 - T1)/T3 = (975/1273.2) = 0.766 = Carnot efficiency
11-67
11.85 The air-standard Carnot cycle was not shown in the text; show the T–s diagram
for this cycle. In an air-standard Carnot cycle the low temperature is 280 K and
the efficiency is 60%. If the pressure before compression and after heat rejection
is 100 kPa, find the high temperature and the pressure just before heat addition.
η = 0.6 = 1 - TH/TL
⇒ TH = TL/0.4 = 700 K
T
H
T
2
3
P1 = 100 kPa
T
L
1
P2 = P1(TH/TL)k-1 = 2.47 MPa
1
4
s
[or P2 = P1(Pr2/Pr1) = 2.645 MPa]
11.86 Air in a piston/cylinder goes through a Carnot cycle in which T = 26.8°C and the
L
total cycle efficiency is η = 2/3. Find T , the specific work and volume ratio in
H
the adiabatic expansion for constant C , C . Repeat the calculation for variable
p v
heat capacities.
Carnot cycle: Same T-s diagram as in previous problem.
η = 1 - TL/TH = 2/3 ⇒ TH = 3 × TL = 3 × 300 = 900 K
Adiabatic expansion 3 to 4:
w = (P4v4 - P3v3/(1 - k) =
3 4
Pvk = constant
R
(T - T ) = u3 - u4
1-k 4 3
= Cv(T3 - T4) = 0.7165(900 - 300) = 429.9 kJ/kg
v4/v3 = (T3/T4)1/(k - 1) = 32.5 = 15.6
For variable Cp, Cv we get, TH = 3 × TL = 900 K
w = u3 - u4 = 674.824 - 214.364 = 460.5 kJ/kg
3 4
v4/v3 = vr4/vr3 = 179.49/9.9169 = 18.1
11-68
11.87 Consider an ideal refrigeration cycle that has a condenser temperature of 45°C
and an evaporator temperature of −15°C. Determine the coefficient of
performance of this refrigerator for the working fluids R-12 and R-22.
T
Ideal Ref. Cycle
Tcond = 45 oC = T3
Tevap = -15 oC
2
3
4
1
s
h1, kJ/kg
R-12
180.97
R-22
244.13
s 2 = s1
0.7051
0.9505
P2, MPa
1.0843
1.729
T2, oC
54.7
74.4
212.63
289.26
h3=h4, kJ/kg
79.71
100.98
wC = h2-h1
31.66
45.13
qL = h1-h4
101.26
143.15
β =qL/wC
3.198
3.172
h2, kJ/kg
11.88 The environmentally safe refrigerant R-134a is one of the replacements for R-12
in refrigeration systems. Repeat Problem 11.87 using R-134a and compare the
result with that for R-12.
T
h1 = 389.2, s2 = s1 = 1.7354
h3 = 264.11, P3 = P2 = 1.16 MPa
At 1 MPa, T2 = 45.9, h2 = 426.8
At 1.2 MPa,T2 = 53.3, h2 = 430.7
⇒ T2 = 51.8, h2 = 429.9
2
o
45 C
o
-15 C
-wC = h2 - h1 = 429.9 - 389.2
= 40.7 kJ/kg
qL = h1 - h4 = h1 - h3 = 389.2 - 264.11 = 125.1 kJ/kg
β = qL/(-wC) = 125.1/40.7 = 3.07
3
4
1
s
11-69
11.89 A refrigerator using R-22 is powered by a small natural gas fired heat engine with
a thermal efficiency of 25%. The R-22 condenses at 40°C and it evaporates at 20°C and the cycle is standard. Find the two specific heat transfers in the
refrigeration cycle. What is the overall coefficient of performance as QL/Q1?
Evaporator: Inlet State is sat. liq-vap with
h4 = h3 =94.272 kJ / kg
The exit state is sat. vap. with h = 242.055 kJ / kg
qL = h1 - h4 = h1 - h3 = 147.78 kJ / kg
Compressor: Inlet State 1 and Exit State 2 about 1.6 MPa
wC = h2 - h1 ;
2:
T2 ≈ 70°C
s2 = s1 = 0.9593 kJ / kg K
h2 = 287.17 kJ / kg
wC = h2 - h1 = 45.11 kJ / kg
Condenser: Brings it to sat liq at state 3
qH = h2 - h3 = 287.17 - 94.272 = 192.9 kJ / kg
Overall Refrigerator:
β = qL / wC = 147.78 / 45.11 = 3.276
Heat Engine:
.
.
.
.
WHE = ηHEQ1 = WC = QL / β
.
.
QL / Q1 = ηβ = 0.25×3.276 = 0.819
11.90 A refrigerator with R-12 as the working fluid has a minimum temperature of
−10°C and a maximum pressure of 1 MPa. Assume an ideal refrigeration cycle as
in Fig. 11.32. Find the specific heat transfer from the cold space and that to the
hot space, and the coefficient of performance.
h1 = 183.058, s1 = 0.7014, h3 = 76.155
s2 = s1 & P2 ⇒ h2 ≈ 210.16
qL = h1 - h4 = h1 - h3 = 183.058 - 76.155 = 106.9 kJ/kg
qH = h2 - h3 = 210.16 - 76.155 = 134 kJ/kg
b = qL/(-wc) = qL/(qH - qL) = 3.945
11-70
11.91 A refrigerator in a meat warehouse must keep a low temperature of -15°C and the
outside temperature is 20°C. It uses R-12 as the refrigerant which must remove 5
kW from the cold space. Find the flow rate of the R-12 needed assuming a
standard vapor compression refrigeration cycle with a condenser at 20°C.
Basic refrigeration cycle:
T1 = T4 = -15°C,
T3 = 20°C
Table B.3:
h4 = h3 = 54.27 ;
h1 = hg = 180.974
.
.
.
QL = mR-12 × qL = mR-12(h1 - h4)
qL = 180.974 - 54.874 = 126.1 kJ/kg
.
mR-12 = 5.0 / 126.1 = 0.03965 kg / s
11.92 A refrigerator with R-12 as the working fluid has a minimum temperature of
−10°C and a maximum pressure of 1 MPa. The actual adiabatic compressor exit
temperature is 60°C. Assume no pressure loss in the heat exchangers. Find the
specific heat transfer from the cold space and that to the hot space, the coefficient
of performance and the isentropic efficiency of the compressor.
State 1: Inlet to compressor, sat. vap. -10°C,
h1 = 183.058, s1 = 0.7014
State 2: Actual compressor exit, h2AC = 217.81
State 3: Exit condenser, sat. liq. 1MPa, h3 = 76.155
State 4: Exit valve, h4 = h3
C.V. Evaporator: qL = h1 - h4 = h1 - h3 = 106.9 kJ/kg
C.V. Ideal Compressor: wC,S = h2,S - h1, s2,S = s1
State 2s: T2,S = 49.66, h2,S = 209.9, wC,S = 26.842
C.V. Actual Compressor: wC = h2,AC - h1 = 34.752 kJ/kg
β = qL/wC = 3.076, ηC = wC,S/wC = 0.7724
C.V. Condenser: qH = h2,AC - h3 = 141.66 kJ/kg
11-71
11.93 Consider an ideal heat pump that has a condenser temperature of 50°C and an
evaporator temperature of 0°C. Determine the coefficient of performance of this
heat pump for the working fluids R-12, R-22, and ammonia.
T
2
Ideal Heat Pump:
Tcond = 50 oC = T3
3
Tevap = 0 oC
4
h1, kJ/kg
s 2 = s1
P2, MPa
T2, oC
h2, kJ/kg
h3=h4, kJ/kg
wC = h2-h1
qH = h2-h3
1
s
R-12
R-22
NH3
187.53
0.6965
1.2193
56.7
249.95
0.9269
1.9423
72.2
1442.32
5.3313
2.0333
115.6
211.95
84.94
24.42
127.01
284.25
107.85
34.3
176.4
1672.84
421.58
230.52
1251.26
5.201
5.143
5.428
β′ =qH/wC
11.94 The air conditioner in a car uses R-134a and the compressor power input is 1.5 kW
bringing the R-134a from 201.7 kPa to 1200 kPa by compression. The cold space is
a heat exchanger that cools atmospheric air from the outside down to 10°C and
blows it into the car. What is the mass flow rate of the R-134a and what is the low
temperature heat transfer rate. How much is the mass flow rate of air at 10°C?
Standard Refrigeration Cycle
Table B.5
h1 = 392.285;
s1 = 1.7319 ;
h4 = h3 = 266
C.V. Compressor (assume ideal)
.
.
m1 = m2
wC = h2 - h1; s2 = s1 + sgen
P2, s = s1 -> h2 = 429.5 -> wC = 37.2
.
.
.
m wC = WC -> m = 1.5 / 37.2 = 0.0403 kg /s
C.V. Evaporator
.
.
QL = m(h1 - h4) = 0.0405(392.28 - 266) = 5.21 kW
C.V. Air Cooler
.
.
.
mair∆hair = QL ≈ mairCp∆T
.
.
mair = QL / (Cp∆T) = 5.21 / (1.004×20) = 0.26 kg / s
11-72
11.95 A refrigerator using R-134a is located in a 20°C room. Consider the cycle to be
ideal, except that the compressor is neither adiabatic nor reversible. Saturated
vapor at -20°C enters the compressor, and the R-134a exits the compressor at
50°C. The condenser temperature is 40°C. The mass flow rate of refrigerant
around the cycle is 0.2 kg/s, and the coefficient of performance is measured and
found to be 2.3. Find the power input to the compressor and the rate of entropy
generation in the compressor process.
Table B.5: P2 = P3 = Psat 40C = 1017 kPa,
s2 ≈ 1.7472 , h2 ≈ 430.87 ;
h4 = h3 = 256.54 kJ/kg
s1 = 1.7395 , h1 = 386.08
β = qL / wC -> wC = qL / β = (h1- h4) / β = (386.08 - 256.54) / 2.3 = 56.32
.
.
WC = m wC = 11.26 kW
C.V. Compressor h1 + wC + q = h 2 ->
qin = h2 - h1 - wC = 430.87 - 386.08 - 56.32 = -11.53
i.e. a heat loss
s1 + ∫ dQ/T + sgen = s2
sgen = s2 - s1 - q / To = 1.7472 - 1.7395 + (11.53 / 293.15) = 0.047
.
.
Sgen = m sgen = 0.2 × 0.047 = 0.0094 kW / K
11.96 A small heat pump unit is used to heat water for a hot-water supply. Assume that
the unit uses R-22 and operates on the ideal refrigeration cycle. The evaporator
temperature is 15°C and the condenser temperature is 60°C. If the amount of hot
water needed is 0.1 kg/s, determine the amount of energy saved by using the heat
pump instead of directly heating the water from 15 to 60°C.
T
Ideal R-22 heat pump
o
o
T1 = 15 C, T3 = 60 C
2
h1 = 255.02, s2 = s1 = 0.9062
3
P2 = P3 = 2.427 MPa, h3 = 122.18
T2 = 78.4 oC, h2 = 282.86
4
wC = h2 - h1 = 27.84
qH = h2 - h3 = 160.68
To heat 0.1 kg/s of water from 15 oC to 60 oC,
.
.
QH2O = m(∆h) = 0.1(251.11 - 62.98) = 18.81 kW
Using the heat pump
.
.
WIN = QH2O(wC/qH) = 18.81(27.84/160.68) = 3.26 kW
a saving of 15.55 kW
1
s
11-73
11.97 The refrigerant R-22 is used as the working fluid in a conventional heat pump
cycle. Saturated vapor enters the compressor of this unit at 10°C; its exit
temperature from the compressor is measured and found to be 85°C. If the
isentropic efficiency of the compressor is estimated to be 70%, what is the
coefficient of performance of the heat pump?
T
R-22 heat pump:
2
o
TEVAP = 10 C, ηS COMP = 0.70
2S
o
3
T2 = 85 C
Isentropic compressor:
s2S = s1 = 0.9129
4
but P2 unknown. Trial & error.
1
s
Assume P2 = 2.11 MPa
At P2,s2S: T2S = 72.1oC, h2S = 281.59, At P2,T2 : h2 = 293.78
calculate ηS COMP = (h2S - h1)/(h2 - h1) =
281.59 - 253.42
= 0.698 ≈ 0.70
293.28 - 253.42
OK ⇒ P2 = 2.11 MPa = P3
⇒ T3 = 53.7 oC, h3 = 112.99, wC = h2 - h1 = 40.36
qH = h2 - h3 = 180.79,
β′ = qH/wC = 4.48
11-74
11.98 In an actual refrigeration cycle using R-12 as the working fluid, the refrigerant flow
rate is 0.05 kg/s. Vapor enters the compressor at 150 kPa, −10°C, and leaves at 1.2
MPa, 75°C. The power input to the compressor is measured and found be 2.4 kW.
The refrigerant enters the expansion valve at 1.15 MPa, 40°C, and leaves the
evaporator at 175 kPa, −15°C. Determine the entropy generation in the compression
process, the refrigeration capacity and the coefficient of performance for this cycle.
Actual ref. cycle
2
T
P1 = 0.15 MPa, P2 = 1.2 MPa
1: comp. inlet
T1 = -10 oC, T2 = 75 oC
3
P3 = 1.15 MPa, P5 = 0.175 MPa
4
5
T3 = 40 oC, T5 = -15 oC 5: evap. exit
.
.
WCOMP = 2.4 kW in,
m = 0.05 kg/s
Table B.3
1
s
h1 = 184.619, s1 = 0.7318, h2 = 226.543, s2 = 0.7404
CV Compressor: h1 + qCOMP = h2 + wCOMP ;
s1 + ∫ dq/T + sgen = s2
wCOMP = -2.4/0.05 = -48.0 kJ/kg
qCOMP = h2 + wCOMP - h1 = 226.5 - 48.0 - 184.6 = -6.1 kJ/kg
sgen = s2 - s1 - q / To = 0.7404 - 0.7318 + 6.1/298.15 = 0.029 kJ / kg K
b)
qL = h5 - h4 = 181.024 - 74.527 = 106.5 kJ/kg
.
.
⇒ QL = mqL = 0.05 × 106.5 = 5.325 kW
c)
β = qL/-wCOMP = 106.5/48.0 = 2.219
11.99 Consider a small ammonia absorption refrigeration cycle that is powered by solar
energy and is to be used as an air conditioner. Saturated vapor ammonia leaves
the generator at 50°C, and saturated vapor leaves the evaporator at 10°C. If 7000
kJ of heat is required in the generator (solar collector) per kilogram of ammonia
vapor generated, determine the overall performance of this system.
T
NH3 absorption cycle:
o
sat. vapor at 50 oC exits the generator
50 C
o
sat. vapor at 10 C exits the evaporator
qH = qGEN = 7000 kJ/kg NH3 out of gen. 10 oC
GEN.
EXIT
1
2
EVAP
EXIT
qL = h2 - h1 = hG 10oC - hF 50oC
= 1452.2 - 421.6 = 1030.6 kJ/kg
⇒ qL/qH = 1030.6/7000 = 0.147
s
11-75
11.100 The performance of an ammonia absorption cycle refrigerator is to be compared
with that of a similar vapor-compression system. Consider an absorption system
having an evaporator temperature of −10°C and a condenser temperature of 50°C.
The generator temperature in this system is 150°C. In this cycle 0.42 kJ is
transferred to the ammonia in the evaporator for each kilojoule transferred from
the high-temperature source to the ammonia solution in the generator. To make
the comparison, assume that a reservoir is available at 150°C, and that heat is
transferred from this reservoir to a reversible engine that rejects heat to the
surroundings at 25°C. This work is then used to drive an ideal vapor-compression
system with ammonia as the refrigerant. Compare the amount of refrigeration that
can be achieved per kilojoule from the high-temperature source with the 0.42 kJ
that can be achieved in the absorption system.
o
T
T H = 50 C
o
T'
H = 150 C
QH
Q'
H = 1 kJ
WC
REV.
H.E.
2
3
3
COND.
2
1
4
COMP.
s
Q L'
o
T1 = -10 C
EVAP.
1
o
T L'= 25 C
QL
4
o
T L = -10 C
h1 = 1430.8 , s1 = 5.4673
h4 = h3 = 421.48
298.2
′ ′
ηTH = 1 - TL/TH = 1 = 0.295
423.2
For the rev. heat engine:
′
⇒ WC = ηTH Q = 0.295 kJ
H
For the NH3 refrig. cycle:
s2 = s1 = 5.4673
P2 = P3 = 2033 kPa ,
=>
T2 ≈ 135°C
Use 2000 kPa Table
h2 ≈ 1724
wC = h2 - h1 = 1724 - 1430.8 = 293.2
qL = h1 - h4 = 1430.8 - 421.48 = 1009.3
β = qL/wC = 1009.3 / 293.2 = 3.44
⇒ QL = βwC = 3.44 × 0.295 = 1.015 kJ
based on assumption of ideal heat engine & refrig. cycle.
11-76
11.101 A heat exchanger is incorporated into an ideal air-standard refrigeration cycle, as
shown in Fig. P11.101. It may be assumed that both the compression and the
expansion are reversible adiabatic processes in this ideal case. Determine the
coefficient of performance for the cycle.
q
6
T
L
2
3
1
2
qH
5
3
1
4
6
4
COMP
5
EXP
s
T1 = T3 = 15 oC = 288.2 K, P1 = 100 kPa, P2 = 1.4 MPa
T4 = T6 = -50 oC = 223.2 K, s2 = s1
⇒ T2 = T1(P2/P1
k-1
)k
= 288.2(1400/100)0.286 = 613 K
wC = -w12 = CP0(T2 - T1) = 1.0035(613 - 288.2) = 326 kJ/kg
k-1
s5 = s4 ⇒ T5= T4(P5/P4) k = 223.2(100/1400)
0.286
= 104.9 K
wE = CP0(T4 - T5) = 1.0035(223.2 - 104.9) = 118.7 kJ/kg
wNET = 118.7 - 326.0 = -207.3 kJ/kg
qL = CP0(T6 - T5) = 1.0035(223.2 - 104.9) = 118.7 kJ/kg
β = qL/wNET = 118.7/207.3 = 0.573
11.102 Repeat Problem 11.101, but assume an isentropic efficiency of 75% for both the
compressor and the expander.
2
From solution 11.101 :
T2S = 613 K, wSC = 326 kJ/kg
2S
T
T5S = 104.9 K, wSE = 118.7
⇒ wC = wSC / ηSC = 326/0.75 = 434.6 kJ/kg
wE = ηSE × wSE = 0.75 × 118.7 = 89.0 kJ/kg
= CP0(T4-T5) = 1.004(223.2 -T5)
⇒ T5 = 134.5 K
wNET = 89.0 - 434.6 = -345.6 kJ/kg
qL = CP0(T6 - T5) = 1.004(223.2 - 134.5) = 89.0 kJ/kg
β = qL/(-wNET) = 89.0/345.6 = 0.258
3
1
4
6
5S
5
s
11-77
11.103 Repeat Problems 11.101 and 11.102, but assume that helium is the cycle working
fluid instead of air. Discuss the significance of the results.
a) Problem 11.101 , except for Helium
14000.40
 100 0.40
T2 = 288.2
 = 828.2 K, T5 = 223.2
 = 77.7 K
 100 
1400
wC = -w12 = 5.1926(828.2 - 288.2) = 2804.1 kJ/kg
wE = 5.1926(223.2 - 77.7) = 755.5 kJ/kg
wNET = 755.5 - 2804.1 = -2048.6 kJ/kg
qL = 5.1926(223.2 - 77.7) = 755.5 kJ/kg
β = 755.5/2048.6 = 0.369
b) Problem 11.102 , except for Helium
From part a): T2S = 828.2 K, wSC = 2804.1
T5S = 77.7 K, wSE = 755.5
⇒ wC = wSC/ηSC = 2804.1/0.75 = 3738.8 kJ/kg
wE = ηSE × wSE = 0.75 × 755.5 = 566.6 kJ/kg
wNET = 566.6 - 3738.8 = -3172.2 kJ/kg
wE = 566.6 = 5.1926(223.2 - T5) ⇒ T5 = 114.1 K
qL = 5.1926(223.2 - 114.1) = 566.6 kJ/kg
β = 566.6/3172.2 = 0.179
11-78
11.104 A binary system power plant uses mercury for the high-temperature cycle and
water for the low-temperature cycle, as shown in Fig. 11.39. The temperatures
and pressures are shown in the corresponding T–s diagram. The maximum
temperature in the steam cycle is where the steam leaves the superheater at point 4
where it is 500°C. Determine the ratio of the mass flow rate of mercury to the
mass flow rate of water in the heat exchanger that condenses mercury and boils
the water and the thermal efficiency of this ideal cycle.
The following saturation properties for mercury are known
Tg, °C
P, MPa
0.04
1.60
hf, kJ/kg hg, kJ/kg sf kJ/kgK
309
562
42.21
75.37
335.64
364.04
0.1034
0.1498
sg, kJ/kgK
0.6073
0.4954
a) For the mercury cycle:
sd = sc = 0.4954 = 0.1034 + xd × 0.5039, xd = 0.7779
hb = ha - wP HG ≈ ha ( since vF is very small)
qH = hc - ha = 364.04 - 42.21 = 321.83 kJ/kg
qL = hd - ha = 270.48 - 42.21 = 228.27 kJ/kg
For the steam cycle:
s5 = s4 = 7.0097 = 0.6493 + x5 × 7.5009, x5 = 0.8480
h5 = 191.83 + 0.848 × 2392.8 = 2220.8
-wP ≈ v1(P2 - P1) = 0.00101(4688 - 10) = 4.7 kJ/kg
h2 = h1 - wP = 191.8 + 4.7 = 196.5
qH (from Hg) = h3 - h2 = 2769.9 - 196.5 = 2600.4
qH (ext. source) = h4 - h3 = 3437.4 - 2796.9 = 640.5
CV: Hg condenser - H2O boiler:
mHg/mH2O =
b)
1st law: mHg(hd - ha) = mH2O(h3 - h2)
2796.9 - 196.5
= 11.392
270.48 - 42.21
qH TOTAL = (mHg/mH2O)(hc - hb) + (h4 - h3) (for 1 kg H2O)
= 11.392 × 321.83 + 640.5 = 4306.8 kJ
All qL is from the H2O condenser:
qL = h5 - h1 = 2220.8 - 191.8 = 2029.0 kJ
wNET = qH - qL = 4306.8 - 2029.0 = 2277.8 kJ
ηTH = wNET/qH = 2277.8/4306.8 = 0.529
11-79
11.105 A Rankine steam power plant should operate with a high pressure of 3 MPa, a low
pressure of 10 kPa, and the boiler exit temperature should be 500°C. The
available high-temperature source is the exhaust of 175 kg/s air at 600°C from a
gas turbine. If the boiler operates as a counterflowing heat exchanger where the
temperature difference at the pinch point is 20°C, find the maximum water mass
flow rate possible and the air exit temperature.
T
-wP = h2 - h1 = v1(P2 - P1)
= 0.00101(3000 - 10) = 3.02
h2 = h1 - wP = 191.83 + 3.02 = 194.85
3
2a
h3 = 3456.5, hair,in = 903.16
2
State 2a: T2a = TSAT = 233.9 °C
1
h2a = 1008.42
s
hair(T2a + 20) = 531.28
Air temperature should be
253.9°C at the point where
the water is at state 2a.
C.V.
Section 2a-3,
. i-a
.
mH O(h3 - h2a) = mair(hi - ha)
e
a
i
HEAT EXCH
2
2a
3
2
.
903.16 - 531.28
mH O = 175
= 26.584 kg/s
3456.5
- 1008.42
2
.
.
Take C.V. Total: mH O(h3 - h2) = mair(hi - he)
2
.
.
⇒ he = hi - mH O(h3 - h2)/mair
2
= 903.6 - 26.584(3456.5 - 194.85)/175 = 408.13
⇒ Te = 406.7 K = 133.6 °C, Te > T2 = 46.5 °C OK.
11.106 A simple Rankine cycle with R-22 as the working fluid is to be used as a
bottoming cycle for an electrical generating facility driven by the exhaust gas
from a Diesel engine as the high temperature energy source in the R-22 boiler.
Diesel inlet conditions are 100 kPa, 20°C, the compression ratio is 20, and the
maximum temperature in the cycle is 2800°C. Saturated vapor R-22 leaves the
bottoming cycle boiler at 110°C, and the condenser temperature is 30°C. The
power output of the Diesel engine is 1 MW. Assuming ideal cycles throughout,
determine
a. The flow rate required in the diesel engine.
b. The power output of the bottoming cycle, assuming that the diesel exhaust is
cooled to 200°C in the R-22 boiler.
11-80
T
T
3
s
v
7
4
2
s
v
1
DIESEL
AIR-STD
CYCLE
IDEAL R-12
RANKINE
BOTTOMING
CYCLE
6
5
8
s
s
P1 = 100 kPa, T1 = 20 oC, T7 = 110 oC, x7 = 1.0
.
v1/v2 = 20, T3 = 2800oC, T5 = T8 = 30oC, WDIESEL = 1.0 MW
a)
T2 = T1(v1/v2)k-1 = 293.2(20)0.4 = 971.8 K
P2 = P1(v1/v2)k = 100(20)1.4 = 6629 kPa
qH = CP0(T3 - T2) = 1.004(3073.2 - 971.8) = 2109.8 kJ/kg
v1 =
0.287 × 293.2
0.8415
= 0.8415, v2 =
= 0.04208
20
100
v3 = v2(T3/T2) = 0.04208(3073.2/971.8) = 0.13307
v3k-1
0.133 070.4
 = 1469.6 K
T4 = T3  = 3073.2
 0.8415 
v4
qL = 0.717(293.2 - 1469.6) = -843.5
wNET = 2109.8 - 843.5 = 1266.3 kJ/kg
.
.
mAIR = WNET/wNET = 1000/1266.3 = 0.79 kg/s
b)
s8 = s7 = 0.60758 = 0.2399 + x8 × 0.4454, x8 = 0.8255
h8 = 64.59 + 0.8255 × 135.03 = 176.1
wT = h7 - h8 = 198.0 - 176.1 = 21.9 kJ/kg
-wP = v5(P6 - P5) = 0.000774(3978.5 - 744.9) = 2.50
h6 = h5 - wP = 64.6 + 2.5 = 67.1
qH = h7 - h6 = 198.0 - 67.1 = 130.9 kJ/kg
.
QH available from Diesel exhaust cooled to 200 oC:
.
QH = 0.79 × 0.717(1469.6 - 473.2) = 564 kW
.
.
⇒ mR-12 = QH/qH = 564/130.9 = 4.309 kg/s
.
WR-12 = 4.309(21.9 - 2.5) = 83.6 kW
11-81
11.107 For a cryogenic experiment heat should be removed from a space at 75 K to a
reservoir at 180 K. A heat pump is designed to use nitrogen and methane in a
cascade arrangement (see Fig. 11.41), where the high temperature of the nitrogen
condensation is at 10 K higher than the low-temperature evaporation of the
methane. The two other phase changes take place at the listed reservoir
temperatures. Find the saturation temperatures in the heat exchanger between the
two cycles that gives the best coefficient of performance for the overall system.
The nitrogen cycle is the bottom cycle and the methane cycle is the top cycle.
Both std. refrigeration cycles.
THm = 180 K = T 3m ,
TLN = 75 K = T 4N = T1N
TLm = T4m = T1m = T3N - 10, Trial and error on T3N or TLm.
For each cycle we have,
-wC = h2 - h1, s2 = s1,
-qH = h2 - h3,
qL = h1 - h4 = h1 - h3
Nitrogen: T4 = T1 = 75 K ⇒ h1 = 74.867 s1 = 5.4609
T3
120
115
110
N2
a)
b)
c)
h3
-17.605
-34.308
-48.446
P2
2.5125
1.9388
1.4672
h2
202.96
188.35
173.88
-wc
128.1
113.5
99.0
-qH
220.57
222.66
222.33
qL
92.47
109.18
123.31
-qH
540.8
581.6
630.2
qL
221.5
212.7
203.4
Methane: T3 = 180 K ⇒ h3 = -0.5 P2 = 3.28655 MPa
CH4
a)
b)
c)
T4
110
105
100
h1
221
212.2
202.9
s1
9.548
9.691
9.851
h2
540.3
581.1
629.7
-wc
319.3
368.9
426.8
The heat exchanger that connects the cycles transfers a Q
.
.
.
.
. .
QHn = qHn mn = QLm = qLm mm => mm/mn = qHn/qLm
The overall unit then has
.
.
.
.
.
QL 75 K = mn qLn ; Wtot in = - (mnwcn + mmwcm)
.
.
. .
β = QL 75 K/Wtot in = qLn/[-wcn -(mm/mn)wcm]
Case
a)
b)
c)
. .
mm/mn
0.996
1.047
1.093
. .
wcn+(mm/mn)wcm
446.06
499.65
565.49
β
0.207
0.219
0.218
A maximum coeff. of performance is between case b) and c).
11-82
11.108 A cascade system is composed of two ideal refrigeration cycles, as shown in Fig.
11.41. The high-temperature cycle uses R-22. Saturated liquid leaves the
condenser at 40°C, and saturated vapor leaves the heat exchanger at −20°C. The
low-temperature cycle uses a different refrigerant, R-23 (Fig. G.3 or the
software). Saturated vapor leaves the evaporator at −80°C, and saturated liquid
leaves the heat exchanger at −10°C. Calculate the ratio of the mass flow rates
through the two cycles and the coefficient of performance of the system.
COND
2'
3'
T3' = 40o C
x '= 0.0
3
C
R-22
o
T1' = -20 C 1'
x1'= 1.0
4'
2
3
T3 = -10o C
x = 0.0
3
T1 = -80o C
1
x1 = 1.0
R-23
C
4
EVAP
T
T
2'
2
3'
3
4'
1
4
1'
s
1′
2′
3′
4′
T,oC
-20
71
40
-20
s
P
h
s
0.245
1.534
1.534
242.1
289.0
94.3
94.3
0.9593
0.9593
1
2
3
4
T,oC
-80
50
-10
-80
P
h
s
0.12
1.90
1.90
0.12
330
405
185
185
1.76
1.76
h′1 - h′4 242.1 - 94.3
. .
a) m/m′ =
=
= 0.672
405 - 185
h2 - h3
. .
b) QL/m = h1 - h4 = 330 - 185 = 145
.
.
. .
- WTOT/m = (h2 - h1) + (m′/m)(h′2 - h′1)
= (405 - 330) + (1/0.672)(289 - 242.1) = 144.8
.
β = QL/-WTOT = 145/144.8 = 1.0
11-83
11.109 Consider an ideal dual-loop heat-powered refrigeration cycle using R-12 as the
working fluid, as shown in Fig. P11.109. Saturated vapor at 105°C leaves the
boiler and expands in the turbine to the condenser pressure. Saturated vapor at −
15°C leaves the evaporator and is compressed to the condenser pressure. The ratio
of the flows through the two loops is such that the turbine produces just enough
power to drive the compressor. The two exiting streams mix together and enter
the condenser. Saturated liquid leaving the condenser at 45°C is then separated
into two streams in the necessary proportions. Determine the ratio of mass flow
rate through the power loop to that through the refrigeration loop. Find also the
performance of the cycle, in terms of the ratio Q /Q
L H.
T
1
TURB.
COMP.
7
2
6
BOIL.
COND.
5
3
P
E
V
A
P
.
.
QL
6
5
2
3
7
4
4
1
s
o
T3 = 45 C sat. liq.
T1 = -15 oC sat. vap.
P5 = P6 = 3.6509 MPa; T6 = 105 oC sat. vap.; P2 = P3 = P7 = 1.0843 MPa
h1 = 180.846; h3 = h4 = 76.647; h6 = 206.099
s7 = s6 = 0.6319 = 0.2875 + x7 × 0.3931; x7 = 0.8761
h7 = 79.647 + 0.8761 × 125.074 = 189.226
s2 = s1 = 0.7046, P2 => T2 = 54.7 oC, h2 = 212.401
a) CV: turbine + compressor
.
.
.
.
.
.
.
.
Cont: m1 = m2, m6 = m7 ; 1st law: m1h1 + m6h6 = m2h2 + m7h7
. .
m6/m1 = (212.401 - 180.846)/(206.099 - 189.226) = 1.870
b) CV: pump
-wP = v3(P5 - P3) = 0.000811(3651 - 1084) = 2.082 kJ/kg
h5 = h3 - wP = 81.729
.
.
CV: evaporator ⇒ QL = m1(h1 - h4)
.
.
CV: boiler ⇒ QH = m6(h6 - h5)
.
.
QL m1(h1 - h4)
180.846 - 79.647
=
= 0.435
β= . = .
QH m6(h6 - h5) 1.87(206.099 - 81.729)
11-84
11.110 Find the availability of the water at all four states in the Rankine cycle described
in Problem 11.12. Assume that the high-temperature source is 500°C and the lowtemperature reservoir is at 25°C. Determine the flow of availability in or out of
the reservoirs per kilogram of steam flowing in the cycle. What is the overall
cycle second law efficiency?
Reference State: 100 kPa, 25°C, s0 = 0.3674, h0 = 104.89
ψ1 = h1 - h0 - T0(s1 - s0)
= 191.83 - 104.89 - 298.15(0.6493 - 0.3674) = 2.89
ψ2 = 195.35 - 104.89 - 298.15(0.6493 - 0.3674) = ψ1 + 3.525 = 6.42
ψ3 = 3222.3 - 104.89 - 298.15(6.8405 - 0.3674) = 1187.5
ψ4 = ψ3 - wT,s = 131.96
∆ψ H = (1 - T0/TH)qH = 0.6144 × 3027 = 1859.7
∆ψ L = (1 - T0/T0)qC = 0/
ηII = wNET/∆ψH = (1055.5 - 3.53)/1859.7 = 0.5657
Notice— TH > T3, TL < T4 = T1 so cycle is externally irreversible. Both qH
and qC over finite ∆T.
11-85
11.111 The effect of a number of open feedwater heaters on the thermal efficiency of an
ideal cycle is to be studied. Steam leaves the steam generator at 20 MPa, 600°C,
and the cycle has a condenser pressure of 10 kPa. Determine the thermal
efficiency for each of the following cases. A: No feedwater heater. B: One
feedwater heater operating at 1 MPa. C: Two feedwater heaters, one operating at
3 MPa and the other at 0.2 MPa.
a) no feed water heater
3
2
wP = ⌠
⌡ vdP
1
≈ 0.00101(20000 - 10)
= 20.2 kJ/kg
h2 = h1 + wP = 191.8 + 20.2 = 212.0
ST.
GEN.
TURBINE.
4
COND.
s4 = s3 = 6.5048
2
= 0.6493 + x 4 × 7.5009
1
P
x4 = 0.78064
h4 = 191.83 + 0.780 64 × 2392.8
20 MPa
o
3 600 C
T
= 2059.7
wT = h3 - h4 = 3537.6 - 2059.7
= 1477.9 kJ/kg
wN = wT - wP = 1477.9 - 20.2 = 1457.7
10 kPa
2
qH = h3 - h2 = 3537.6 - 212.0 = 3325.6
4
1
s
ηTH =
wN
qH
=
1457.7
= 0.438
3325.6
b) one feedwater heater
wP12 = 0.00101(1000 - 10)
= 1.0 kJ/kg
h2 = h1 + wP12 = 191.8 + 1.0 = 192.8
5
ST.
GEN.
TURBINE.
6
wP34 = 0.001127 (20000- 1000)
HTR.
= 21.4 kJ/kg
h4 = h3 + wP34 = 762.8 + 21.4 = 784.2
s6 = s5 = 6.5048
= 2.1387 + x6 × 4.4478
7
3
COND.
1
P
4
2
P
11-86
x6 = 0.9816
20 MPa
T
o
h6 = 762.8 + 0.9816 × 2015.3 = 2741.1
1 MPa
CV: heater
const: m3 = m6 + m2 = 1.0 kg
4
1st law: m6h6 + m2h2 = m3h3
m6 =
600 C
5
6
2 3
1
762.8 - 192.8
= 0.2237
2741.1 - 192.8
10 kPa
7
s
m2 = 0.7763, h7 = 2059.7 ( = h4 of part a) )
CV: turbine
wT = (h5 - h6) + m2(h6 - h7)
= (3537.6 - 2741.1) + 0.7763(2741.1 - 2059.7) = 1325.5 kJ/kg
CV: pumps
wP = m1wP12 + m3wP34 = 0.7763(1.0) + 1(21.4) = 22.2 kJ/kg
wN = 1325.5 - 22.2 = 1303.3 kJ/kg
CV: steam generator
qH = h5 - h4 = 3537.6 - 784.2 = 2753.4 kJ/kg
ηTH = wN/qH = 1303.3/2753.4 = 0.473
c) two feedwater heaters
wP12 = 0.00101 ×
(200 - 10)
= 0.2 kJ/kg
h2 = wP12 + h1
7
= 191.8 + 0.2
= 192.0
wP34 = 0.001061 ×
8
9
10
HP
HTR
(3000 - 200)
= 3.0 kJ/kg
h4 = h3 + wP34
= 504.7 + 3.0
= 507.7
TURBINE.
ST.
GEN.
LP
HTR
1
5
3
P
6
COND.
P
4
P
2
11-87
T
wP56 = 0.001217(20000 - 3000)
80 MPa
o
600 C
7
= 20.7 kJ/kg
h6 = h5 + wP56 = 1008.4 + 20.7 = 1029.1
3 MPa
6
s8 = s7 = 6.5048 T8 = 293.2 oC

at P8 = 3 MPa  h8 = 2974.8
s9 = s8 = 6.5048 = 1.5301 + x9 × 5.5970
4 5
2 3
1
8 0.2 MPa
10 kPa
9
10
s
x9 = 0.8888 => h9 = 504.7 + 0.888 × 2201.9 = 2461.8
CV: high pressure heater
cont:
m5 = m4 + m8 = 1.0 kg ;
m8 =
1st law: m5h5 = m4h4 + m8h8
1008.4 - 507.7
= 0.2030
2974.8 - 507.7
m4 = 0.7970
CV: low pressure heater
cont:
m9 + m2 = m3 = m4 ;
m9 =
1st law: m9h9 + m2h2 = m3h3
0.7970(504.7 - 192.0)
= 0.1098
2461.8 - 192.0
m2 = 0.7970 - 0.1098 = 0.6872
CV: turbine
wT = (h7 - h8) + (1 - m8)(h8 - h9) + (1 - m8 - m9)(h9 - h10)
= (3537.6 - 2974.8) + 0.797(2974.8 - 2461.8)
+ 0.6872(2461.8 - 2059.7) = 1248.0 kJ/kg
CV: pumps
wP = m1wP12 + m3wP34 + m5wP56
= 0.6872(0.2) + 0.797(3.0) + 1(20.7) = 23.2 kJ/kg
wN = 1248.0 - 23.2 = 1224.8 kJ/kg
CV: steam generator
qH = h7 - h6 = 3537.6 - 1029.1 = 2508.5 kJ/kg
ηTH = wN/qH = 1224.8/2508.5 = 0.488
11-88
11.112 Find the availability of the water at all the states in the steam power plant
described in Problem 11.36. Assume the heat source in the boiler is at 600°C and
the low-temperature reservoir is at 25°C. Give the second law efficiency of all the
components.
From solution to 11.1 and 11.36 :
States
1 - 2a - 3 - 4a - 1
1:
hf = 191.81,
3:
h3 = 2803.43,
s3 = 6.1252
4a:
h4a = 2068.2,
s4a = 6.5313,
ψ = h - ho - To(s - so)
sf = 0.6492
2a: h2a = 196.2,
s2a ≈ 0.6628
x4a = 0.7842
ho = 104.89, so = 0.3674
ψ1= 191.81 - 104.89 - 298.15(0.6492 - 0.3674) = 2.90
ψ2a = 196.2 - 104.89 - 298.15(0.6628 - 0.3674) = 3.24
ψ3 = 2803.43 - 104.89 - 298.15(6.1252 - 0.3674) = 981.85
ψ4a = 2068.2 - 104.89 - 298.15(6.5313 - 0.3674) = 125.54
ηII Pump = (ψ2a - ψ1) / wp ac = (3.24 - 2.9) / 4.406 = 0.077
ηII Boiler = (ψ3 - ψ2a) / [(1- To/TH) qH]
= (981.85 - 3.24) / [0.658×2607.2] = 0.57
ηII Turbine = wT ac / (ψ3 - ψ4a) = 735.2 / (981.85 - 125.54) = 0.859
ηII Cond = ∆ψamb / (ψ4a - ψ1) = 0
11-89
11.113 The power plant shown in Fig. 11.40 combines a gas-turbine cycle and a steamturbine cycle. The following data are known for the gas-turbine cycle. Air enters
the compressor at 100 kPa, 25°C, the compressor pressure ratio is 14, and the
isentropic compressor efficiency is 87%; the heater input rate is 60 MW; the
turbine inlet temperature is 1250°C, the exhaust pressure is 100 kPa, and the
isentropic turbine efficiency is 87%; the cycle exhaust temperature from the heat
exchanger is 200°C. The following data are known for the steam-turbine cycle.
The pump inlet state is saturated liquid at 10 kPa, the pump exit pressure is 12.5
MPa, and the isentropic pump efficiency is 85%; turbine inlet temperature is
500°C and the isentropic turbine efficiency is 87%. Determine
a. The mass flow rate of air in the gas-turbine cycle.
b. The mass flow rate of water in the steam cycle.
c. The overall thermal efficiency of the combined cycle.
.
QH = 60 MW
P2 /P1 = 14
HEAT
2
3
T3 = 1250o C
η = 0.87
SC
.
GAS
TURB
COMP
W NET CT
P4 = 100 kPa
AIR
5
1
4
ST
P 1= 100 kPa
T1 = 25o C
T5 = 200 oC HEAT EXCH
o
7
6
SP
T7 = 500 C
P6 = P7 = 12.5 MPa
.
9
P
WST
STEAM
TURB
H 2O
η = 0.85
η = 0.87
ηST= 0.87
8
COND
a) From Air Tables, A.7: Pr1 = 1.0913,
P8 = P9 = 10 kPa
h1 = 298.66,
h5 = 475.84
s2 = s1 ⇒ Pr2S = Pr1(P2/P1) = 1.0913 × 14 = 15.2782
T2S = 629 K, h2S = 634.48
wSC = h1 - h2S = 298.66 - 634.48 = -335.82 kJ/kg
wC = wSC/ηSC = -335.82/0.87 = -386 = h1 - h2
At T3 = 1523.2 K: Pr3 = 515.493, h3 = 1663.91
⇒ h2 = 684.66
11-90
.
.
mAIR = QH/(h3 - h2) =
b)
60 000
= 61.27 kg/s
1663.91 - 684.66
Pr4S = Pr3(P4/P3) = 515.493(1/14) = 36.8209
=> T4S = 791 K, h4S = 812.68
wST = h3 - h4S = 1663.91 - 812.68 = 851.23
wT = ηST × wST = 0.87 × 851.23 = 740.57 = h 3 - h4 => h4 = 923.34 kJ/kg
Steam cycle: -wSP ≈ 0.00101(12500 - 10) = 12.615
-wP = - wSP/ηSP = 12.615/0.85 = 14.84
h6 = h9 - wP = 191.83 + 14.84 = 206.67
At 12.5 MPa, 500 oC: h7 = 3341.7, s7 = 6.4617
h4 - h5
923.34 - 475.84
.
=
m
=
61.27
= 8.746 kg/s
O
AIR h - h
3341.7 - 206.67
7
6
2
.
mH
c) s8S = s7 = 6.4617 = 0.6492 + x8S × 7.501, x8S = 0.7749
h8S = 191.81 + 0.7749 × 2392.8 = 2046.0
wST = h7 - h8S = 3341.7 - 2046.0 = 1295.7
wT = ηST × wST = 0.87 × 1295.7 = 1127.3 kJ/kg
.
.
.
WNET = m(wT+wC)AIR +m(wT+wP)H O
2
= 61.27(740.57 - 386.0) + 8.746(1127.3 - 14.84)
= 21725 + 9730 = 31455 kW = 31.455 MW
.
.
ηTH = WNET/QH = 31.455/60 = 0.524
11-91
11.114 For Problem 11.105, determine the change of availability of the water flow and
that of the air flow. Use these to determine a second law efficiency for the boiler
heat exchanger.
From solution to 11.105 :
.
mH2O = 26.584 kg/s, h2 = 194.85, s2 = 0.6587
h3 = 3456.5, s3 = 7.2338, s°
= 7.9820, s°Te = 7.1762
Ti
hi = 903.16, he = 408.13
ψ3 - ψ2 = h3 - h2 - T0(s3 - s2) = 1301.28 kJ/kg
ψi - ψe = hi - he - T0(s°Ti - s°Te) = 254.78 kJ/kg
.
(ψ3 - ψ2)mH2O 1301.28 × 26.584
.
ηII =
=
= 0.776
(ψi - ψe)mair
254.78 × 175
11-92
11.115 One means of improving the performance of a refrigeration system that operates
over a wide temperature range is to use a two-stage compressor. Consider an ideal
refrigeration system of this type that uses R-12 as the working fluid, as shown in
Fig. P11.115. Saturated liquid leaves the condenser at 40°C and is throttled to
−20°C. The liquid and vapor at this temperature are separated, and the liquid is
throttled to the evaporator temperature, −70°C. Vapor leaving the evaporator is
compressed to the saturation pressure corresponding to −20°C, after which it is
mixed with the vapor leaving the flash chamber. It may be assumed that both the
flash chamber and the mixing chamber are well insulated to prevent heat transfer
from the ambient. Vapor leaving the mixing chamber is compressed in the second
stage of the compressor to the saturation pressure corresponding to the condenser
temperature, 40°C. Determine
a. The coefficient of performance of the system.
b. The coefficient of performance of a simple ideal refrigeration cycle operating
over the same condenser and evaporator ranges as those of the two-stage
compressor unit studied in this problem.
ROOM
.
+QH
COND
5
T
5
1
SAT.LIQ.
o
40 C
40 C
o
-20 C
COMP.
ST.2
4
o
1
6
2
o
SAT.VAP.
o
-20 C
FLASH
MIX.CHAM
-70 C
2
6 SAT.LIQ.
o
9
-20 C
COMP.
ST.1
EVAP
.
R-12 ref. with
2-stage
compression
7
-QL
SAT.VAP.
o
-70 C
9
8
s
CHAMBER
3
8
7
4
3
COLD SPACE
CV: expansion valve, upper loop
h2 = h1 = 74.527 = 17.8 + x2 × 160.81; x2 = 0.353
m3 = x2m2 = x2m1 = 0.353 kg ( for m1=1 kg)
m6 = m1 - m3 = 0.647 kg
CV: expansion valve, lower loop
h7 = h6 = 17.8 = -26.1 + x7 × 181.64, x7 = 0.242
11-93
QL = m3(h8 - h7) = 0.647(155.536 - 17.8)
qL = 89.1 kJ/kg-m1
CV: 1st stage compressor
s8 = s9 = 0.7744, P9 = PSAT -20 oC= 0.1509 MPa
⇒ T9 = 9 oC, h9 = 196.3 kJ/kg
CV: mixing chamber (assume const. press. mixing)
1st law: m6h9 + m3h3 = m1h4
or h4 = 0.647 × 196.3 + 0.353 × 178.61 = 190.06 kJ/kg
h4, P4 = 0.1509 MPa ⇒ T4 = -1.0 oC, s4 = 0.7515
CV: 2nd stage compressor P4 = 0.1509 MPa = P9 = P3
P5 = Psat 40oC = 0.9607 MPa, s5 = s4 ⇒ T5 = 70oC, h5 = 225.8
CV: condenser
1st law: -qH = h1 - h5 = 74.527 - 225.8 = -151.27
β 2 stage = qL/(qH - qL) = 89.1/(151.27 - 89.1) = 1.433
b) 1 stage compression
h3 = h4 = 74.527
T
h1 = 155.536
qL = h1 - h4 = 81.0 kJ/kg
s1 = s2 = 0.7744 

P2 = 0.9607 MPa
2
40o C
3
o
-70 C
4
⇒ T2 = 80.9 oC, h2 = 234.0
qH = h2 - h3 = 234.0 - 74.527 = 159.47 kJ/kg
β 1 stage = qL/(qH - qL) = 81.0/(159.47 - 81.0) = 1.032
1
s
11-94
11.116 A jet ejector, a device with no moving parts, functions as the equivalent of a
coupled turbine-compressor unit (see Problems 9.82 and 9.90). Thus, the turbinecompressor in the dual-loop cycle of Fig. P11.109 could be replaced by a jet
ejector. The primary stream of the jet ejector enters from the boiler, the secondary
stream enters from the evaporator, and the discharge flows to the condenser.
Alternatively, a jet ejector may be used with water as the working fluid. The
purpose of the device is to chill water, usually for an air-conditioning system. In
this application the physical setup is as shown in Fig. P11.116. Using the data
given on the diagram, evaluate the performance of this cycle in terms of the ratio
Q /Q .
L H
a. Assume an ideal cycle.
b. Assume an ejector efficiency of 20% (see Problem 9.90).
T
2
VAP o
150 C
2
JET
EJECT.
BOIL.
.
QH
3
11
30 C
HP
P.
10
1
3
4
9 5,10
8 7 6
VAP
o
10 C
COND.
o
11
1'
2'
1
s
4
o
5
FLASH
CH.
6
9
o
LIQ
o
10 C
7
20 C
CHILL
.
QL
8
LP
P.
T1 = T7 = 10 C
T2 = 150 oC
T4 = 30 oC
T9 = 20 oC
Assume T5 = T10
(from mixing streams 4 & 9).
P3 = P4 = P5 = P8 = P9 = P10= PG 30 oC = 4.246 kPa
P11 = P2 = PG 150oC = 475.8 kPa,
P1 = P6 = P7 = PG 10oC = 1.2276 kPa
.
.
.
.
.
.
.
.
Cont: m1 + m9 = m5 + m10, m5 = m6 = m7 + m1
.
.
.
.
.
.
.
.
m7 = m8 = m9, m10 = m11 = m2, m3 = m4
.
.
.
a) m1 + m2 = m3; ideal jet ejector
s′1 = s1 & s′2 = s2 (1' & 2' at P3 = P4)
.
.
then, m1(h′1 - h1) = m2(h2 - h′2)
From s′2 = s2 = 0.4369 + x′2 × 8.0164; x′2 = 0.7985
h′2 = 125.79 + 0.7985 × 2430.5 = 2066.5
11-95
From s′1 = s1 = 8.9008 ⇒ T′1 = 112 °C, h′1 = 2710.4
2746.5 - 2066.5
. .
⇒ m1/m2 =
= 3.5677
2710.4 - 2519.8
Also h4 = 125.79, h7 = 42.01, h9 = 83.96
Mixing of streams 4 & 9 ⇒ 5 & 10:
.
.
.
.
.
.
(m1 + m2)h4 + m7h9 = (m7 + m1 + m2)h5 = 10
. .
.
.
Flash chamber (since h6 = h5) :
(m7+m1)h5 = 10 = m1h1 + m7h1
.
⇒ using the primary stream m2 = 1 kg/s:
.
.
4.5677 × 125.79 + m7 × 83.96 = (m7 + 4.5677)h5
.
.
& (m7 + 3.5677)h5 = 3.5677 × 2519.8 + m7 × 42.01
.
Solving, m7 = 202.627 & h5 = 84.88
LP pump: -wLP P = 0.0010(4.246 - 1.2276) = 0.003
h8 = h7 - wLP P = 42.01 + 0.003 = 42.01
.
.
.
Chiller: QL = m7(h9-h8) = 202.627(83.96 - 42.01) = 8500 kW (for m2 = 1)
HP pump: -wHP P = 0.001002(475.8 - 4.246) = 0.47
h11 = h10 - wHP P = 84.88 + 0.47 = 85.35
.
.
Boiler:
Q11 = m11(h2 - h11) = 1(2746.5 - 85.35) = 2661.1 kW
. .
⇒ QL/QH = 8500/2661.1 = 3.194
. .
. .
b) Jet eject. eff. = (m1/m2)ACT/(m1/m2)IDEAL = 0.20
⇒
(m. 1/m. 2)ACT = 0.2 × 3.5677 = 0.7135
.
.
.
using m2 = 1 kg/s: 1.7135 × 125.79 + m7 × 83.96 = (m7 + 1.7135)h5
.
.
& (m7 + 0.7135)h5 = 0.7135 × 2519.8 + m7 × 42.01
.
Solving, m7 = 39.762 & h5 = h10 = 85.69
.
Then, QL = 39.762(83.96 - 42.01) = 1668 kW
h11 = 85.69 + 0.47 = 86.16
.
QH = 1(2746.5 - 86.16) = 2660.3 kW
. .
& QL/QH = 1668/2660.3 = 0.627
11-96
English Unit Problems
11.117E A steam power plant, as shown in Fig. 11.3, operating in a Rankine cycle has
saturated vapor at 600 lbf/in.2 leaving the boiler. The turbine exhausts to the
condenser operating at 2 lbf/in.2. Find the specific work and heat transfer in each
of the ideal components and the cycle efficiency.
1: h1 = 93.73, v1 = 0.01623, 3: h3 = hg = 1204.06, s3 = sg = 1.4464
144
= 1.8 Btu/lbm
C.V. Pump: -wP = ∫ v dP = v1(P2 - P1) = 0.01623(600 - 2)
778
h2 = h1 - wP = 93.73 + 1.8 = 95.81 Btu/lbm
C.V. Boiler: qH = h3 - h2 = 1204.06 - 95.53 = 1108.53 Btu/lbm
C.V. Tubine: wT = h3 - h4, s4 = s3
s4 = s3 = 1.4464 = 0.1744 + x4 × 1.7461
=> x 4 = 0.7285,
h4 = 93.73 + 0.7285 × 1022.2 = 838.42
wT = 1204.06 - 838.42 = 365.63 Btu/lbm
ηCYCLE = (wT + wP)/qH = (365.63 - 1.8)/1108.53 = 0.33
C.V. Condenser:
qL = h4 - h1 = 838.42 - 93.73 = 744.69 Btu/lbm
11.118E Consider a solar-energy-powered ideal Rankine cycle that uses water as the
working fluid. Saturated vapor leaves the solar collector at 350 F, and the
condenser pressure is 1 lbf/in. 2. Determine the thermal efficiency of this cycle.
T
H2O ideal Rankine cycle
CV: turbine
s4 = s3 = 1.5793 = 0.13266 + x4 × 1.8453
x4 = 0.784
3
2
1
4
h4 = 69.74 + 0.784 × 1036.0 = 881.9
wT = h3 - h4 = 1193.1 - 881.9 = 311.2 Btu/lbm
-wP = ∫ vdP ≈ v1(P2 - P1) = 0.016136(67 - 1) 144 / 778 = 0.2
⇒ wNET = wT + wP = 311.2 - 0.2 = 311
h2 = h1 - wP = 69.7 + 0.2 = 69.9 Btu/lbm
qH = h3 - h2 = 1193.1 - 69.9 = 1123.2 Btu/lbm
ηTH = wNET/qH = 311/1123.2 = 0.277
s
11-97
11.119E A supply of geothermal hot water is to be used as the energy source in an ideal
Rankine cycle, with R-134a as the cycle working fluid. Saturated vapor R-134a
leaves the boiler at a temperature of 180 F, and the condenser temperature is 100
F. Calculate the thermal efficiency of this cycle.
T
a) From the R-134a tables,
h1 = 108.86 v1 = 0.01387
D
P1 = 138.93 P2 = P3 = 400.4
h3 = 184.36 s3 = 0.402
2
1
3
4
CV. Pump:
-wP = v1(P2 - P1)
= 0.01387(400.4-138.93)
180 F
100 F
s
144
= 0.671 Btu/lbm = h2 - h1
778
h2 = h1 - wP = 108.86 + 0.671 = 109.53
CV: Turbine
s4 = s3 ⇒ x4 = (0.402 - 0.2819)/0.1272 = 0.9442
h4 = 176.08, wT = h3 - h4 = 8.276 btu/lbm
CV: Boiler
qH = h3 - h2 = 184.36 - 109.53 = 74.83 Btu/lbm
ηTH = (wT + w12)/qH = (8.276 - 0.671)/74.83 = 0.102
11.120E Do Problem 11.119 with R-22 as the working fluid.
Same diagram as in problem 11.119, now from the R-22 tables,
h1 = 39.267, v1 = 0.01404, P1 = 210.6,
P2 = P3 = 554.8, h3 = 110.07, s3 = 0.1913
CV: Pump
144
= 0.894 Btu/lbm
-wP = v1(P2-P1) = -0.01404 (554.8-210.6)
778
h2 = h1 - wP = 39.267 + 0.894 = 40.16
CV: Turbine s4 = s3
⇒ x4 = (0.1913 - 0.07942)/0.13014 = 0.9442
h4 = 101.885, wT = h3 - h4 = 8.185 Btu/lbm
CV: Boiler
qH = h3 - h2 = 110.07 - 40.16 = 69.91 Btu/lbm
ηTH = (wT + w12)/qH = (8.185 - 0.894)/157.21 = 0.104
11-98
11.121E The power plant in Problem 11.117 is modified to have a superheater section
following the boiler so the steam leaves the super heater at 600 lbf/in.2, 700 F.
Find the specific work and heat transfer in each of the ideal components and the
cycle efficiency.
h3 = 1350.6, s3 = 1.5871, h1 = 94.01, v1 = 0.01623
⌠v dP = v (P - P )
C.V. Pump: -wP = ⌡
2 2
1
= 0.01623(600 - 2)(144 / 778) = 1.8 Btu/lbm
h2 = h1 - wP = 95.81 Btu/lbm
C.V. Boiler: qH = h3 - h2 = 1350.6 - 95.81 = 1254.79 Btu/lbm
C.V. Tubine: wT = h3 - h4, s4 = s3
⇒ x4 = 0.8093, h4 = 921.23
wT = 1350.6 - 921.23 = 429.37 Btu/lbm
ηCYCLE = (wT + wP)/qH = (429.37 - 1.8)/1254.79 = 0.341
C.V. Condenser:
qL = h4 - h1 = 921.23 - 94.01 = 827.22 Btu/lbm
11.122E
Consider a simple ideal Rankine cycle using water at a supercritical
pressure. Such a cycle has a potential advantage of minimizing local temperature
differences between the fluids in the steam generator, such as the instance in
which the high-temperature energy source is the hot exhaust gas from a gasturbine engine. Calculate the thermal efficiency of the cycle if the state entering
the turbine is 3500 lbf/in.2, 1100 F, and the condenser pressure is 1 lbf/in.2. What
is the steam quality at the turbine exit?
s4 = s3 = 1.5193 = 0.1326 + x4×1.8453
x4 = 0.7515
3500 lbf/in2
3
1100 F
T
very low for a turbine exhaust
h4 = 848.3, h1 = 69.73, h3 = 1496
s2 = s1 = 0.1326 => h2 = 80.13
qH = h3 - h2 = 1415.9 Btu/lbm
1 lbf/in2
2
1
4
s
wnet = h3 - h4 - (h2 - h1) = 637.3 Btu/lbm
η = wnet/qH = 637.3/1415.9 = 0.45
11-99
11.123E
Consider an ideal steam reheat cycle in which the steam enters the highpressure turbine at 600 lbf/in.2, 700 F, and then expands to 120 lbf/in.2. It is then
reheated to 700 F and expands to 2 lbf/in.2 in the low-pressure turbine. Calculate
the thermal efficiency of the cycle and the moisture content of the steam leaving
the low-pressure turbine.
Basic cycle as in 11.121 plus reheat.
From solution 11.121:
-wP = 1.8 Btu/lbm, h2 = 95.81
T
3 5
h3 = 1350.6, s3 = 1.5871
2
qH1 = h3-h2 = 1350.6-95.81 = 1254.79
4
1
6
CV T1: wT1 = h3 - h4; s4 = s3 =>
x4 = 0.9986, h4 = 1190 Btu/lbm
s
CV T2: wT2 = h5 - h6; s6 = s5 = 1.7825, h5 = 1378.2
=> x 6 = 0.9213, h6 = 1035.7, wT2 = 1378.2 - 1035.7 = 342.5
wT,tot = wT1 + wT2 = 160.6 + 342.5 = 503.1 Btu/lbm
qH = qH1 + h5 - h4 = 1254.79 + 1378.2 - 1190 = 1443 Btu/lbm
ηCYCLE = wT,tot/qH = 503.1/1443 = 0.349
11.124E
A closed feedwater heater in a regenerative steam power cycle heats 40
lbm/s of water from 200 F, 2000 lbf/in.2 to 450 F, 2000 lbf/in.2. The extraction
steam from the turbine enters the heater at 500 lbf/in.2, 550 F and leaves as
saturated liquid. What is the required mass flow rate of the extraction steam?
3
1
2
From the steam tables:
h1 = 172.6 h2 = 431.14 all
h3 = 1266.6 h4 = 449.5 Btu/lbm
4
C.V.
Feedwater
Heater
.
.
m3 = m1(h1 - h2)/(h4 - h3)
= 40
lbm
172.6 - 431.14
= 12.656
s
449.5 - 1266.6
11-100
11.125E
Consider an ideal steam regenerative cycle in which steam enters the
turbine at 600 lbf/in.2, 700 F, and exhausts to the condenser at 2 lbf/in.2. Steam is
extracted from the turbine at 120 lbf/in. 2 for an open feedwater heater. The
feedwater leaves the heater as saturated liquid. The appropriate pumps are used
for the water leaving the condenser and the feedwater heater. Calculate the
thermal efficiency of the cycle and the net work per pound-mass of steam.
5
From solution 11.121:
h5 = 1350.6, s5 = 1.5871
h7 = 921.23, h1 = 94.01
2nd law:
s6 = s5 =>
TURBINE.
ST.
GEN.
6
x6 = 0.999
7
4
FW
HTR
h6 = 1190 Btu/lbm
P
Also: h3 = 312.66
COND.
P
2
1
3
T
wP12 = 0.01623(120 - 2)144/778 = 0.35
600 psi
5
wP34 = 0.0179(600 - 120)144/778 = 1.59
⇒ h2 = 94.36,
120 psi
4
3
2
h4 = 314.25
6
7
1
CV: feedwater heater
cont : m3 = m6 + m2 = 1 lbm,
2 psi
s
1st law: m6h6 + m2h2 = m3h3
m6(1190) + (1 - m6)(94.36) = 312.66
⇒ m6 = 0.1992, m2 = 0.8008
CV: turbine
wT = (h5 - h6) + (1 - m6)(h6 - h7)
= (1350.6 - 1190) + 0.8008(1190 - 921.2) = 375.9 Btu/lbm
CV: pumps
wP = m2wP12 + m3wP34 = 0.8008 × 0.35 + 1 × 1.59 = 1.87 Btu/lbm
wNET = wT - wP = 375.9 - 1.87 = 374 Btu/lbm
CV: steam generator
qH = h5 - h4 = 1350.6 - 314.25 = 1036.35 Btu/lbm
ηTH = wNET/qH = 374/1036.35 = 0.361
11-101
11.126E
Consider an ideal combined reheat and regenerative cycle in which steam
enters the high-pressure turbine at 500 lbf/in.2, 700 F, and is extracted to an open
feedwater heater at 120 lbf/in.2 with exit as saturated liquid. The remainder of the
steam is reheated to 700 F at this pressure, 120 lbf/in.2, and is fed to the lowpressure turbine. The condenser pressure is 2 lbf/in.2. Calculate the thermal
efficiency of the cycle and the net work per pound-mass of steam.
5: h5 = 1356.66,
s5 = 1.6112
7: h7 = 1378.17,
s7 = 1.7825
7
5
T1
3: h3 = hf = 312.59, v3 = 0.01788
s 5 = s6
6
1-x
C.V. T1
= 146.9 Btu/lbm
8
x
=> h6 = 1209.76
wT1 = h5 - h6 = 1356.66 - 1209.76
T2
4
HTR
P
1-x
COND.
2
3
1
P
C.V. Pump 1
-wP1 = h2 - h1 = v1(P2 - P1)
= 0.01623(120 - 2) = 0.354
=> h 2 = h1 - wP1 = 93.73 + 0.354 = 94.08 Btu/lbm
T
700 F
C.V. FWH
x h6 + (1 - x) h2 = h3
5
7
4
h3 - h2
312.59 - 94.08
x=
=
= 0.1958
1209.76
- 94.08
h6 - h2
C.V. Pump 2
6
2 psi
2 3
1
8
s
-wP2 = h4 - h3 = v3(P4 - P3) = 0.01788(500 - 120)(144/778) = 1.26 Btu/lbm
=> h 4 = h3 - wP2 = 312.59 + 1.26 = 313.85 Btu/lbm
qH = h5 - h4 + (1 - x)(h7 - h6 ) = 1042.81 + 135.43 = 1178.2 Btu/lbm
C.V. Turbine 2
s 7 = s8
=> x8 = (1.7825 - 0.1744)/1.746 = 0.921
h8 = hf + x8 hfg = 93.73 + 0.921 × 1022.2 = 1035.2
wT2 = h7 - h8 = 1378.17 - 1035.2 = 342.97
wnet = wT1 + (1 - x) wT2 + (1 - x) wP1 + wP2
= 146.9 + 275.8 - 0.285 - 1.26 = 421.15 kJ/kg
ηcycle = wnet / qH = 421.15 / 1178.2 = 0.357
11-102
11.127E
A steam power cycle has a high pressure of 500 lbf/in. 2 and a condenser
exit temperature of 110 F. The turbine efficiency is 85%, and other cycle
components are ideal. If the boiler superheats to 1400 F, find the cycle thermal
efficiency.
C.V. Turb.: wT = h3 - h4, s4 = s3 + sT,GEN
Ideal: s4S = s3 = 1.8704 => x4S = 0.9519, h4S = 1059.7 Btu/lbm
=> wT,S = 1741 - 1059.7 = 681.3 Btu/lbm
Actual:
wT,AC = η × wT,S = 579.1 Btu/lbm
C.V. Pump: -wP = ⌠
⌡vdP ≈ v1(P2 - P1)
= 0.016166(500-1.28)144/778 = 1.49 Btu/lbm
C.V. Boiler:
qH = h3 - h2 = h3 - h1 + wP = 1661.5 Btu/lbm
C.V. Condenser:
qL = h4ac - h1 = 1161.9 - 78.01 = 1083.9
η = (wT,AC - wP)/qH = (579.1 - 1.49)/1661.5 = 0.348
11.128E
The steam power cycle in Problem 11.117 has an isentropic efficiency of
the turbine of 85% and that for the pump it is 80%. Find the cycle efficiency and
the specific work and heat transfer in the components.
States numbered as in fig 9.3 of text.
CV Pump: -wP,S = v1(P2-P1) = 0.01623(600 - 2)144/778 = 1.8 Btu/lbm
⇒ -wP,AC = 1.8/0.8 = 2.245 Btu/lbm
h2 = h1 - wP,AC = 93.73 + 2.245 = 95.975
CV Turbine: wT,S = h3 - h4s ,
s4 = s3 = 1.4464 Btu/lbm R
⇒ wT,S = 1204.06 - 838.42 = 365.64 Btu/lbm
wT,AC = h3 - h4AC = 365.64 × 0.85 = 310.78 ⇒ h4AC = 893.3
CV Boiler:
qH = h3 - h2 = 1204.06 - 95.975 = 1108.1 Btu/lbm
qL = h4AC - h1 = 893.3 - 93.73 = 799.5 Btu/lbm
ηCYCLE = (wT + wP)/qH = (310.78 - 2.245)/1108.1 = 0.278
Compared to (365.64-1.8)/1108.5 = 0.33 in the ideal case.
11-103
11.129E
Steam leaves a power plant steam generator at 500 lbf/in.2, 650 F, and
enters the turbine at 490 lbf/in.2, 625 F. The isentropic turbine efficiency is 88%,
and the turbine exhaust pressure is 1.7 lbf/in.2. Condensate leaves the condenser
and enters the pump at 110 F, 1.7 lbf/in.2. The isentropic pump efficiency is 80%,
and the discharge pressure is 520 lbf/in.2. The feedwater enters the steam
generator at 510 lbf/in.2, 100 F. Calculate the thermal efficiency of the cycle and
the entropy generation of the flow in the line between the steam generator exit
and the turbine inlet, assuming an ambient temperature of 77 F.
2
T
1
500 lbf/in
2
1
650 F
625 F
TURBINE.
ST.
GEN.
ηsT =
2
0.88
5
5s
6 4
3
6
COND.
5
3s 3
s
ηST = 0.88, ηSP = 0.80
4
P
1.7 lbf/in
h1 = 1328.0, h2 = 1314.0
s3S = s2 = 1.5752 = 0.16483 + x3S×1.7686
=>
x 3S = 0.79745
h3S = 88.1 + 0.797 45×1025.4 = 905.8
wST = h2 - h3S = 1314.0 - 905.8 = 408.2
wT = ηSTwST = 0.88×408.2 = 359.2 Btu/lbm
h3 = h2 - wT = 1314.0 - 359.2 = 954.8
-wSP = 0.016166(520-1.7)
144
= 1.55 Btu/lbm
778
-wp = -wSP/ηSP = 1.55/0.80 = 1.94 Btu/lbm
qH = h1 - h6 = 1328.0 - 68.1 = 1259.9 Btu/lbm
ηTH = wNET/qH = (359.2 - 1.94)/1259.9 = 0.284
C.V. Line from 1 to 2:
Energy Eq.:
w = 0/ ,
q = h2 - h1 = 1314 - 1328 = - 14 Btu/lbm
Entropy Eq.: s1 + sgen + q/T0 = s2
=>
sgen = s2 - s1 -q/T0 = 1.5752 - 1.586 - (-14/536.7) = 0.0153 Btu/lbm R
2
11-104
11.130E
In one type of nuclear power plant, heat is transferred in the nuclear
reactor to liquid sodium. The liquid sodium is then pumped through a heat
exchanger where heat is transferred to boiling water. Saturated vapor steam at 700
lbf/in.2 exits this heat exchanger and is then superheated to 1100 F in an external
gas-fired superheater. The steam enters the turbine, which has one (open-type)
feedwater extraction at 60 lbf/in.2. The isentropic turbine efficiency is 87%, and
the condenser pressure is 1 lbf/in.2. Determine the heat transfer in the reactor and
in the superheater to produce a net power output of 1000 Btu/s.
700 lbf/in 2
6
SUP.
HT.
60 lbf/in 2
7
5
8
HTR.
3
4
1100 F
TURBINE.
Q
REACT.
6
T
P
COND.
2
1
P
4
5
7s
2 3
1
7
1 lbf/in 2
8s 8
s
.
WNET = 1000 Btu/s, ηST = 0.87
-wP12 = 0.016136(60 - 1)144/778 = 0.18 Btu/lbm
h2 = h1 - wP12 = 69.73 + 0.18 = 69.91 Btu/lbm
-wP34 = 0.017378(700 - 60)144/778 = 2.06 Btu/lbm
h4 = h3 - wP34 = 262.24 + 2.06 = 264.3 Btu/lbm
s7S=s6 = 1.7682, P7
=> T7S = 500.8 F,
h7S = 1283.4
h7 = h6 - ηST(h6 - h7S) = 1625.8 - 0.87(1625.8 - 1283.4) = 1327.9
s8S = s6 = 1.7682 = 0.13264 + x8S × 1.8453 =>
x 8S = 0.8863
h8S = 69.73 + 0.8863 × 1036 = 987.9 Btu/lbm
h8 = h6 - ηST(h6 - h8S) = 1625.8 - 0.87(1625.8 - 987.9) = 1070.8
CV: heater: cont: m2 + m7 = m3 = 1.0 lbm, 1st law: m2h2 + m7h7 = m3h3
m7 = (262.24-69.91) / (1327.9-69.91) = 0.1529
CV: turbine: wT = (h6 - h7) + (1 - m7)(h7 - h8)
= 1625.8-1327.9 + 0.8471(1327.9-1070.8) = 515.7 Btu/lbm
CV pumps: wP = m1wP12 + m3wP34 = -(0.8471×0.18 + 1×2.06) = -2.2 Btu/lbm
.
wNET = 515.7 - 2.2 = 513.5 Btu/lbm => m = 1000/513.5 = 1.947 lbm/s
.
.
CV: reactor
QREACT = m(h5-h4) = 1.947(1202 - 264.3) = 1825.7 Btu/s
.
.
CV: superheater QSUP = m(h6 - h5) = 1.947(1625.8 - 1202) = 825 Btu/s
11-105
11.131E
A boiler delivers steam at 1500 lbf/in.2, 1000 F to a two-stage turbine as
shown in Fig. 11.17. After the first stage, 25% of the steam is extracted at 200
lbf/in.2 for a process application and returned at 150 lbf/in. 2, 190 F to the
feedwater line. The remainder of the steam continues through the low-pressure
turbine stage, which exhausts to the condenser at 2 lbf/in. 2. One pump brings the
feedwater to 150 lbf/in.2 and a second pump brings it to 1500 lbf/in.2. Assume the
first and second stages in the steam turbine have isentropic efficiencies of 85%
and 80% and that both pumps are ideal. If the process application requires 5000
Btu/s of power, how much power can then be cogenerated by the turbine?
3: h3 = 1490.3, s3 = 1.6001
4s: s4S = s3 ⇒ h4S = 1246.6
3
wT1,S = h3 - h4S = 243.7
T2
T1
⇒ wT1,AC = 207.15 Btu/lbm
B
4
h4AC = h3 - wT1,AC = 1283.16
4ac: P4, h4AC ⇒ s4AC = 1.6384
5s: s5S = s4AC ⇒ h5S = 951.3
wT2,S = h4AC - h5S = 331.9
5
2
P2
7
1
Proc.
5000
Btu/s
6
P1
wT2,AC = 265.5 = h 4AC - h5AC
⇒ h5AC = 1017.7 Btu/lbm
7: h7 = 158.02 ; qPROC = h4AC - h7 = 1125.1 Btu/lbm
.
.
.
m4 = Q/qPROC = 5000/1125.1 = 4.444 lbm/s = 0.25 mTOT
.
.
.
.
.
⇒ mTOT = m3 = 17.776 lbm/s,
m5 = m3 - m4 = 13.332 lbm/s
.
.
.
.
WT = m3h3 - m4h4AC - m5h5AC = 7221 Btu/s
C
11-106
11.132E
A large stationary Brayton cycle gas-turbine power plant delivers a power
output of 100000 hp to an electric generator. The minimum temperature in the
cycle is 540 R, and the maximum temperature is 2900 R. The minimum pressure
in the cycle is 1 atm, and the compressor pressure ratio is 14 to 1. Calculate the
power output of the turbine, the fraction of the turbine output required to drive the
compressor and the thermal efficiency of the cycle?
Brayton:
.
wNET = 100 000 hp
3
T
s
P1 = 1 atm, T1 = 540 R
P2/P1 = 14, T3 = 2900 R
2
s
a) Assume const CP0:
1
4
s 2 = s1
s
P2
(P )
→ T2 = T 1
k-1
k
= 540(14)0.286 = 1148.6 R
1
-wC = -w12 = h2 - h1 = CP0(T2-T1)
= 0.24(1148.6-540) = 146.1 Btu/lbm
P4
(P )
As s4 = s3 → T4 = T3
k-1
k
= 2900(
3
1 0.286
)
= 1363.3 R
14
wT = w34 = h3 - h4 = CP0(T3-T4)
= 0.24(2900-1363.3) = 368.8 Btu/lbm
wNET = wT + wC = 368.8 - 146.1 = 222.7 Btu/lbm
.
.
m = WNET/wNET = 100 000×2544/222.7 = 1 142 344 lbm/h
.
.
WT = mwT = 165 600 hp,
-wC/wT = 0.396
b)
qH = CP0(T3-T2) = 0.24(2900-1148.6) = 420.3 Btu/lbm
ηTH = wNET/qH = 222.7/420.3 = 0.530
11-107
11.133E
An ideal regenerator is incorporated into the ideal air-standard Brayton
cycle of Problem 11.132. Calculate the cycle thermal efficiency with this
modification.
T
Problem 9.108 + ideal regen., where :
wT = 368.8, wC = 146.1 Btu/lbm
3
s
wNET = 222.7 Btu/lbm
x
2
Ideal regen.: TX = T4 = 1363.3 R
4
y
s
qH = h3 - hX = 0.24(2900 - 1363.3)
1
= 368.8 Btu/lbm = w T
ηTH = wNET/qH = 222.7/368.8 = 0.604
s
11.134E
Consider an ideal gas-turbine cycle with two stages of compression and
two stages of expansion. The pressure ratio across each compressor stage and
each turbine stage is 8 to 1. The pressure at the entrance to the first compressor is
14 lbf/in.2, the temperature entering each compressor is 70 F, and the temperature
entering each turbine is 2000 F. An ideal regenerator is also incorporated into the
cycle. Determine the compressor work, the turbine work, and the thermal
efficiency of the cycle.
10
REG
I.C.
CC
1
9
5
2
6
4
COMP
COMP
TURB
TURB
7
8
CC
P2/P1 = P4/P3 = P6/P7 = P8/P9 = 8.0
T
8
6
P1 = 14 lbf/in2
T1 = T3 = 70 F, T6 = T8 = 2000 F
Assume const. specific heat
s2 = s1 and s4 = s3
T4 = T2 = T1(P2/P1
k-1
)k
= 529.67(8)0.2857 = 959.4 R
5
7
4
3
2
9
10
1
s
Total compressorwork
-wC = 2 ×(-w12) = 2CP0(T2 - T1) = 2 × 0.24(959.4 - 529.67) = 206.3 Btu/lbm
11-108
Also s6 = s7 and s8 = s9
k-1
P7 k
10.2857
⇒ T7 = T9 = T6  = 2459.67 
= 1357.9 R
8
P6
Total turbinework
wT = 2× w67 = 2CP0(T6 - T7) = 2 × 0.24(2459.67 - 1357.9) = 528.85 Btu/lbm
wNET = 528.85 - 206.3 = 322.55 Btu/lbm
Ideal regenerator: T5 = T9, T10 = T4
⇒ qH = (h6 - h5) + (h8 - h7) = 2CP0(T6 - T5)
= 2 × 0.24(2459.67 - 1357.9) = wT = 528.85 Btu/lbm
ηTH = wNET/qH = 322.55/528.85 = 0.61
11.135E
Repeat Problem 11.134, but assume that each compressor stage and each
turbine stage has an isentropic efficiency of 85%. Also assume that the
regenerator has an efficiency of 70%.
T4S = T2S = 959.4 R, -wCS = 206.3
T
8
6
T7S = T9S = 1357.9 R, wTS = 528.85
⇒
9
5
-wC = -wSC/ηSC = 242.7 Btu/lbm
4
-w12 = -w34 = 242.7/2 = 121.35
4S
T2 = T4 = T1 + (-w12/CP0)
3
7S
7
9S
2S 2
1
= 529.67 + 121.35/0.24 = 1035.3 R
s
wT = ηT wTS = 449.5
T7 = T9 = T6 - (+w67/CP0) = 2459.67 - 449.5/2×0.24 = 1523 R
ηREG =
h5 - h4
h9 - h4
=
T 5 - T4
T5 - 1035.3
=
= 0.7
T9 - T4 1523 - 1035.3
⇒ T5 = 1376.7 R
qH = CP0(T6 - T5) + CP0(T8 - T7)
= 0.24(2459.67 - 1376.7) + 0.24(2459.67 - 1523) = 484.7 Btu/lbm
wNET = wT + wC = 449.5 - 242.7 = 206.8 Btu/lbm
ηTH = wNET/qH = 206.8/484.7 = 0.427
11-109
11.136E
An air-standard Ericsson cycle has an ideal regenerator as shown in Fig.
P11.62. Heat is supplied at 1800 F and heat is rejected at 68 F. Pressure at the
beginning of the isothermal compression process is 10 lbf/in.2. The heat added is
275 Btu/lbm. Find the compressor work, the turbine work, and the cycle
efficiency.
T
P
T4 = T3 = 1800 F
P
2
3
3
T 4
P1 = 10 lbf/in2
T
T
P
1
P
q = -4q1 (ideal reg.)
P
4
2 3
⇒ qH = 3q4 &
2
T
T1 = T2 = 68 F = 527.7R
1
v
s
wT = qH = 275 Btu/lbm
ηTH = ηCARNOT TH. = 1 - TL/TH = 1 - 527.7/2349.7 = 0.775
wnet = ηTH qH = 0.775 × 275 = 213.13 Btu/lbm
qL = -wC = 275 - 213.13 = 61.88 Btu/lbm
11.137E
The turbine in a jet engine receives air at 2200 R, 220 lbf/in.2. It exhausts
to a nozzle at 35 lbf/in.2, which in turn exhausts to the atmosphere at 14.7 lbf/in.2.
The isentropic efficiency of the turbine is 85% and the nozzle efficiency is 95%.
Find the nozzle inlet temperature and the nozzle exit velocity. Assume negligible
kinetic energy out of the turbine.
C.V. Turb.: hi = 560.588, Pri = 206.092, se = si
⇒ Pre = Pri × (Pe/Pi) = 206.092(35/220) = 32.787
Te = 1381, he = 338.13, wT,s = 560.588 - 338.13 = 222.46
wT,AC = wT,s × ηT = 189.09 = he,AC - hi ⇒ he,AC = 371.5
⇒ Te,AC = 1508.4 R, Pre,AC = 45.917
C.V. Nozzle: (1/2)V2e = hi - he; se = si
⇒ Pre = Pri × (Pe/Pi) = 45.917(14.7/35) = 19.285
⇒ Te,s = 1199.6 R, he,s = 291.3 Btu/lbm
2
= hi - he,s = 371.5 - 291.3 = 80.2 Btu/lbm
(1/2)Ve,s
2
2
= (1/2)Ve,s
× ηNOZ = 76.19 Btu/lbm
(1/2)Ve,AC
Ve,AC =
2 × 25037 × 76.19 = 1953 ft/s
11-110
11.138E
Air flows into a gasoline engine at 14 lbf/in.2, 540 R. The air is then
compressed with a volumetric compression ratio of 8;1. In the combustion
process 560 Btu/lbm of energy is released as the fuel burns. Find the temperature
and pressure after combustion.
⇒ vr2 = vr1/8 = 179.49/8 = 22.436
Compression 1 to 2: s2 = s1
T2 = 1212 R, u2 = 211.31 Btu/lbm, Pr2 = 20
P2 = Pr2(P1/Pr1) = 20(14/1.1146) = 251.2 lbf/in2
Compression 2 to 3:
u3 = u2 + qH = 211.31 + 560 = 771.3, T3 = 3817 R
P3 = P2 × (T3/T2) = 251.2(3817/1212) = 791 lbf/in2
11.139E
To approximate an actual spark-ignition engine consider an air-standard
Otto cycle that has a heat addition of 800 Btu/lbm of air, a compression ratio of 7,
and a pressure and temperature at the beginning of the compression process of 13
lbf/in.2, 50 F. Assuming constant specific heat, with the value from Table C.4,
determine the maximum pressure and temperature of the cycle, the thermal
efficiency of the cycle and the mean effective pressure.
P
3
qH = 800 Btu
s
v1/v2 = 7
4
P1 = 13 lbf/in2, T1 = 50 F
T
3
s
v
v
4
v
1
2
s
2
s
1
v
a)
v1 = RT1/P1
v
s
= 53.34×510/13×144
= 14.532 ft3/lbm
v2 = v1/7 = 2.076 ft3/lbm
k
P2 = P1(v1/v2) = 13(7)1.4 = 198.2 lbf/in2
k-1
T2 = T1(v1/v2)
= 510(7)0.4 = 1110.7 R
T3 = T2 + qH/CV0 = 1110.7 + 800/0.171 = 5789 R
P3 = P2T3/T2= 198.2 × 5789/1110.7 = 1033 lbf/in2
b)
ηTH = 1 - (T1/T2) = 1 - 510/1110.7 = 0.541
c)
wNET = ηTH × qH = 0.541 × 800 = 432.8 Btu
mep =
wNET
v1-v2
=
432.8×778
= 188 lbf/in2
(14.532-2.076)×144
11-111
11.140E
In the Otto cycle all the heat transfer qH occurs at constant volume. It is
more realistic to assume that part of qH occurs after the piston has started its
downwards motion in the expansion stroke. Therefore consider a cycle identical
to the Otto cycle, except that the first two-thirds of the total qH occurs at constant
volume and the last one-third occurs at constant pressure. Assume the total qH is
700 Btu/lbm, that the state at the beginning of the compression process is 13
lbf/in.2, 68 F, and that the compression ratio is 9. Calculate the maximum
pressure and temperature and the thermal efficiency of this cycle. Compare the
results with those of a conventional Otto cycle having the same given variables.
P
3
T
4
3
s
P1 = 13, T1 = 527.67 R
s
rV = v1/v2 = 7
5
2
Btu
q23 = ×700 = 466.7
3
lbm
1
Btu
q34 = ×700 = 233.3
3
lbm
v
5
2
4
2
s
s
1
v
1
s
v
k
P2 = P1(v1/v2) = 13(9)1.4 = 281.8 lbf/in2
k-1
T2 = T1(v1/v2)
= 527.67(9)0.4 = 1270.7 R
T3 = T2 + q23/CV0 = 1270.7 + 466.7/0.171 = 4000 R
P3 = P2(T3/T2) = 281.8 × 4000/1270.7 = 887.1 lbf/in2 = P4
T4 = T3 + q34/CP0 = 4000 + 233.3/0.24 = 4972 R
v5
v4
=
v1
v4
= (P4/P1) × (T1/T4) =
k-1
T5 = T4(v4/v5)
88.1 527.67
×
= 7.242
13
4972
= 4972(1/7.242)
0.4
= 2252 R
qL = CV0(T5-T1) = 0.171(2252 - 527.67) = 294.9 Btu/lbm
ηTH = 1 - qL/qH = 1 - 294.9/700 = 0.579
Std Otto cycle: ηTH = 1 - (9)-0.4 = 0.585
11-112
11.141E
It is found experimentally that the power stroke expansion in an internal
combustion engine can be approximated with a polytropic process with a value of
the polytropic exponent n somewhat larger than the specific heat ratio k. Repeat
Problem 11.139 but assume the expansion process is reversible and polytropic
(instead of the isentropic expansion in the Otto cycle) with n equal to 1.50.
See solution to 11.139 :
T3 = 5789 R,
P3 = 1033 lbf/in2
v3 = RT3/P3 = v2 = 2.076 ft3/lbm, v4 = v1 = 14.532
except for process 3 to 4:
Pv1.5 = constant.
P4 = P3(v3/v4)1.5 = 1033(1/7)1.5 = 55.78 lbf/in2
T4 = T3(v3/v4)0.5 = 5789(1/7)0.5 = 2188 R
1w2
= ∫ P dv = R(T2 - T1)/(1 - 1.4)
= -53.34(1110.7 - 510)/(0.4×778) = -102.96 Btu/lbm
3w4
= ∫ P dv = R(T4 - T3)/(1 - 1.5)
= -53.34(2188 - 5789)/(0.5×778) = 493.8 Btu/lbm
wNET = 493.8 - 102.96 = 390.84 Btu/lbm
ηCYCLE = wNET/qH = 390.84/700 = 0.488
mep = wNET/(v1-v2) = 390.84×778/(14.532 - 2.076) = 169.5 lbf/in2
Notice a smaller wNET, ηCYCLE, mep compared to ideal cycle.
11.142E
A diesel engine has a bore of 4 in., a stroke of 4.3 in. and a compression
ratio of 19:1 running at 2000 RPM (revolutions per minute). Each cycle takes two
revolutions and has a mean effective pressure of 200 lbf/in. 2. With a total of 6
cylinders find the engine power in Btu/s and horsepower, hp.
Power from mean effective pressure.
mep = wnet / (vmax - vmin)
->
wnet = mep(vmax - vmin)
∆V = πBore2 × 0.25 × S = π × 42 × 0.25 × 4.3 = 54.035 in3
W = mep(Vmax - Vmin) = 200×54.035/(12×778) = 1.1575 Btu
Work per Cylinders per Power Stroke
.
W = W×n×RPM×0.5 (cycles / min)×(min / 60 s)×(Btu / cycle)
= 1.1575×6×2000×0.5×(1/60) = 115.75 Btu/s
= 115.75 × 3600/2544.43 hp = 164 hp
11-113
11.143E
At the beginning of compression in a diesel cycle T = 540 R, P = 30
2
lbf/in. and the state after combustion (heat addition) is 2600 R and 1000
lbf/in.2. Find the compression ratio, the thermal efficiency and the mean effective
pressure.
P2 = P3 = 1000 lbf/in2
T2 = T1(P2/P1)(k-1)/k = 540(1000/30) 0.2857 = 1470.6
v1/v2 = (P2/P1)1/k = (1000/30)0.7143 = 12.24
v3/v2 = T3/T2 = 2600/1470.6 = 1.768
v4/v3 = v1/v3 = (v1/v2)(v2/v3) = 12.24/1.768 = 6.923
T4 = T3(v3/v4)k-1 = 2600*6.923-0.4 = 1199
qL = CV(T4 - T1) = 0.171(1199-540) = 112.7
qH = h3 - h2 = CP (T3 - T2) = 0.24(2600 - 1470.6) = 271.1
η = 1 - qL/qH = 1 - 112.7 / 271.1 = 0.5843
wnet = qnet = 271.1 - 112.7 = 158.4
vmax = v1 = RT1/P1 = 53.34 × 540/(30 × 144) = 6.6675
vmin = vmax(v1/v2) = 6.6675 / 12.24 = 0.545
mep = [158.4/(6.6675 - 0.545)] × (778/144) = 139.8 lbf/in2
11-114
11.144E
Consider an ideal air-standard diesel cycle where the state before the
compression process is 14 lbf/in.2, 63 F and the compression ratio is 20. Find the
maximum temperature(by iteration) in the cycle to have a thermal efficiency of
60%.
Diesel cycle: P1= 14, T1 = 522.67 R, v1/v2 = 20, ηTH = 0.60
k-1
T2 = T1(v1/v2)
v1 =
= 522.67(20)0.4 = 1732.4 R
53.34×522.67
13.829
= 13.829, v2 =
= 0.6915
20
14×144
v3 = v2×T3/T2 = 0.6915×T3/1732.4 = 0.000399 T3
T3
T4
=
v4
13.829
(v ) =(0.000399
)
T
k-1
3
3
ηTH = 0.60 = 1 ⇒
0.4
T4-T1
k(T3-T2)
1.4
→ T4 = 0.0153 T3
1.4
=1-
0.0153 T3 -522.67
1.4(T3-1732.4)
1.4
0.0153 T3 - 0.56 T3 + 447.5 = 0
5100R: LHS = -35.54,
5500R: LHS = 5.04,
Linear interpolation, T3 = 5455 R
5450R: LHS = -0.5
11-115
11.145E
Consider an ideal Stirling-cycle engine in which the pressure and
temperature at the beginning of the isothermal compression process are 14.7
lbf/in.2, 80 F, the compression ratio is 6, and the maximum temperature in the
cycle is 2000 F. Calculate the maximum pressure in the cycle and the thermal
efficiency of the cycle with and without regenerators.
T
P
3
T 4
3
T
v
2
P1 = 14.7 lbf/in2
4
v
v
T
Ideal Stirling cycle
T1 = T2 = 80 F
v1
v
v2
2
1
T
1
s
v
=6
T3 = T4 = 2000 F
T1 = T2 → P2 = P1× v1/v2 = 14.7×6 = 88.2
V2 = V3 → P3 = P2× T3/T2 = 88.2×
2460
= 401.8 lbf/in2
540
w34 = q34 = RT3 ln (v4/v3)
= (53.34/778) × 2460 ln 6 = 302.2 Btu/lbm
q23 = CV0(T3-T2) = 0.171(2460-540) = 328.3 Btu/lbm
w12 = q12 = -RT1 ln
v1
v2
=-
53.34
×540 ln 6 = -66.3 Btu/lbm
778
wNET = 302.2 - 66.3 = 235.9 Btu/lbm
ηNO REGEN =
235.9
= 0.374,
302.2+328.3
ηWITH REGEN =
235.9
= 0.781
302.2
11-116
11.146E
An ideal air-standard Stirling cycle uses helium as working fluid. The
2
2
isothermal compression brings the helium from 15 lbf/in. , 70 F to 90 lbf/in. .
The exspansion takes place at 2100 R and there is no regenerator. Find the work
and heat transfer in all four processes per lbm helium and the cycle efficiency.
Substance helium C.4:
R = 386 ft-lbf/lbmR
CV = 0.753
v4/v3 = v1/v2 = P2/P1 = 90/15 = 6
1 -> 2:
-1w2 = -1q2 = ∫ P dV = 386 × 530 × ln(6)/778 = 471.15 Btu/lbm
2 -> 3:
2w 3
= 0; 2q3 = CP (T3 - T2) = 0.753(2100 - 530) = 1182.2
3 -> 4:
3w 4
= 3q4 = RT3ln(v4/v3) = 386 × 2100 × ln(6)/778 = 1866.8
4 -> 1:
4w 1
= 0;
4q 1
= CP (T4 - T1) = -1182.2
ηCycle = wnet/ qH = (-471.15 + 1866.0) / (1182.2 + 1866.8) = 0.458
11.147E
The air-standard Carnot cycle was not shown in the text; show the T–s
diagram for this cycle. In an air-standard Carnot cycle the low temperature is 500
R and the efficiency is 60%. If the pressure before compression and after heat
rejection is 14.7 lbf/in.2, find the high temperature and the pressure just before
heat addition.
η = 0.6 = 1 - TH/TL
⇒ TH = TL/0.4 = 500/0.4 = 1250 R
P2 = P1(TH/TL
1
k-1
)
= 363.2 lbf/in
= 14.7(
1250 3.5
)
500
T
H
T
L
T
2
3
1
2
[or P2 = P1(Pr2/Pr1) = 14.7 × 22.48 / 0.8515 = 388 lbf/in2]
4
s
11-117
11.148E
Air in a piston/cylinder goes through a Carnot cycle in which TL = 80.3 F
and the total cycle efficiency is η = 2/3. Find TH, the specific work and volume
ratio in the adiabatic expansion for constant Cp, Cv. Repeat the calculation for
variable heat capacities.
Carnot cycle: Same process diagram as in previous problem.
η = 1 - TL/TH = 2/3 ⇒ TH = 3 × TL = 3 × 540 = 1620 R
Pvk = constant
Adiabatic expansion 3 to 4:
w = (P4v4 - P3v3)/(1 - k) = [R/(1-k)](T4 - T3) = u3 - u4
3 4
= Cv(T3 - T4) = 0.171(1620 - 540) = 184.68 Btu/lbm
v4/v3 = (T3/T4)1/(k - 1) = 32.5 = 15.6
For variable Cp, Cv we get, TH = 3 × TL = 1620 R
w = u3 - u4 = 290.13 - 92.16 = 197.97 Btu/lbm
3 4
v4/v3 = vr4/vr3 = 179.49/9.9289 = 18.1
11.149E
Consider an ideal refrigeration cycle that has a condenser temperature of
110 F and an evaporator temperature of 5 F. Determine the coefficient of
performance of this refrigerator for the working fluids R-12 and R-22.
T
Ideal Ref. Cycle
Tcond = 110 F = T3
2
3
Tevap = 5 F
4
Use Table C.10 for R-22
Use computer table for R-12
1
s
h1, Btu/lbm
R-12
77.803
R-22
104.954
s 2 = s1
0.16843
0.22705
P2, lbf/in2
151.11
241.04
T2 , F
127.29
161.87
h2, Btu/lbm
91.107
123.904
h3=h4, Btu/lbm
33.531
42.446
13.3
18.95
44.27
62.51
3.33
3.30
-wC = h2-h1
qL = h1-h4
β =qL/(-wC)
11-118
11.150E
The environmentally safe refrigerant R-134a is one of the replacements for
R-12 in refrigeration systems. Repeat Problem 11.149 using R-134a and compare
the result with that for R-12.
T
Ideal Ref. Cycle
Tcond = 110 F
2
3
Tevap = 5 F
h1 = 167.325 Btu/lbm
1
4
s2 = s1 = 0.4145 Btu/lbm R
s
P2 = 161.124 lbf/in2, T2 = 122.2 F,
h2 = 184.44 Btu/lbm
h3=h4 = 112.455 Btu/lbm
-wC = h2-h1 = 17.115 Btu/lbm ; qL = h1-h4 = 54.87 Btu/lbm
β =qL/(-wC) = 3.206
11.151E
Consider an ideal heat pump that has a condenser temperature of 120 F
and an evaporator temperature of 30 F. Determine the coefficient of performance
of this heat pump for the working fluids R-12, R-22, and ammonia.
T
2
Ideal Heat Pump
Tcond = 120 F
3
Tevap = 30 F
Use Table C.9 for NH3
Use Table C.10 for R-22
Use computer table for R-12
R-12
h1, Btu/lbm
80.42
1
4
s
R-22
107.28
NH3
619.58
s 2 = s1
0.1665
0.2218
1.2769
P2, lbf/in2
172.3
274.6
286.5
T2 , F
132.2
160.4
239.4
h2, Btu/lbm
91.0
122.17
719.5
h3=h4, Btu/lbm
36.011
45.71
178.83
-wC = h2-h1
10.58
14.89
99.92
qH = h2-h3
54.995
76.46
540.67
5.198
5.135
5.411
β′ =qH/(-wC)
11-119
11.152E
The refrigerant R-22 is used as the working fluid in a conventional heat
pump cycle. Saturated vapor enters the compressor of this unit at 50 F; its exit
temperature from the compressor is measured and found to be 185 F. If the
isentropic efficiency of the compressor is estimated to be 70%, what is the
coefficient of performance of the heat pump?
T
2
R-22 heat pump: T2 = 185 F
2S
TEVAP = 50 F, ηS COMP = 0.70
3
Isentropic compressor:
s2S = s1 = 0.2180
1
4
but P2 unknown. Trial & error.
s
Assume P2 = 307 lbf/in2
At P2,s2S: T2S = 162 F, h2S = 121.07, At P2,T2: h2 = 126.24
calculate ηS COMP=
h2S - h1
h2 - h1
=
121.07 - 108.95
= 0.701 ≈ 0.70
126.24 - 108.95
OK ⇒ P2 = 307 lbf/in2 = P3 ⇒ T3 = 128.8 F, h3 = 48.66
-wC = h2 - h1 = 17.29,
qH = h2 - h3 = 77.58,
β′ = qH/(-wC) = 4.49
11.153E
Consider a small ammonia absorption refrigeration cycle that is powered
by solar energy and is to be used as an air conditioner. Saturated vapor ammonia
leaves the generator at 120 F, and saturated vapor leaves the evaporator at 50 F. If
3000 Btu of heat is required in the generator (solar collector) per pound-mass of
ammonia vapor generated, determine the overall performance of this system.
NH3 absorption cycle:
sat. vapor at 120 F exits the generator.
Sat. vapor at 50 F exits the evaporator
qH = qGEN = 3000 Btu/lbm NH3
T
GEN.
EXIT
120F
50 F
out of generator.
1
2
EVAP
EXIT
s
qL = h2 - h1 = hG 50 F - hF 120 F = 624.28 - 178.79
= 445.49 Btu/lbm ⇒ qL/qH = 445.49/3000 = 0.1485
11-120
11.154E
Consider an ideal dual-loop heat-powered refrigeration cycle using R-12
as the working fluid, as shown in Fig. P11.109. Saturated vapor at 220 F leaves
the boiler and expands in the turbine to the condenser pressure. Saturated vapor at
0 F leaves the evaporator and is compressed to the condenser pressure. The ratio
of the flows through the two loops is such that the turbine produces just enough
power to drive the compressor. The two exiting streams mix together and enter
the condenser. Saturated liquid leaving the condenser at 110 F is then separated
into two streams in the necessary proportions. Determine the ratio of mass flow
rate through the power loop to that through the refrigeration loop. Find also the
performance of the cycle, in terms of the ratio QL /QH.
T
1
TURB.
COMP.
7
2
6
BOIL.
COND.
5
E
V
A
P
.
.
QL
6
5
2
3
7
4
1
4
s
3
P
T
F
P
2
h
Btu/lbm
s
Btu/lbm R
Computer tables for
properties.
77.271
168.88
P2=P3=PSAT at 110 F
168.88
P5=P6=PSAT at 220 F
0.067 45
s2=s1=0.168 88
1
0
lbf/in
23.849
2
-
151.11
3
110
151.11
33.531
4
0
23.849
33.531
5
6
220
524.43
524.43
7
110
151.11
89.036
-wP = 0.0129(524.4 - 151.1)
h2=91.277
0.067 45
0.151 49
0.151 49
144
= 0.894
778
h5 = 33.531 + 0.894 = 34.425 Btu/lbm
(1-x7) =
0.162 79 - 0.151 49 0.011 30
=
= 0.1187
0.095 34
0.095 34
h7 = 87.844 - 0.1187(54.313) = 81.397 Btu/lbm
Pump work:
-wP = h5-h3
≈ v5(P5-P3)
11-121
CV: turbine + compressor
.
.
.
.
Continuity Eq.: m1 = m2,
m6 = m7
.
.
.
.
Energy Eq.:
m1h1 + m6h6 = m2h2 + m7h7
89.036-81.397 7.639
. .
m1/m6 =
=
= 0.545,
91.277-77.271 14.006
. .
m6/m1 = 1.833
-wP = v3(P5-P3), h5 = h3 - wP
.
.
.
.
CV evaporator: QL = m1(h1-h4),
CV boiler: QH = m6(h6-h5)
.
m1(h1-h4)
. .
77.271-33.531
=
= 0.436
⇒ β = QL/QH = .
m6(h6-h5) 1.833(89.036-34.425)
CV: pump:
11.155E
(Adv.) Find the availability of the water at all four states in the Rankine
cycle described in Problem 11.121. Assume the high-temperature source is 900 F
and the low-temperature reservoir is at 65 F. Determine the flow of availability in
or out of the reservoirs per pound-mass of steam flowing in the cycle. What is the
overall cycle second law efficiency?
Ref. state 14.7 lbf/in2, 77°F, h0 = 45.08 Btu/lbm,
s0 = 0.08774 Btu/lbm R
From solution to 11.121:
s1 = 0.17497,
h1 = 94.01,
s2 = 0.175 = s1,
h2 = 95.81,
s4 = s3 = 1.5871 Btu/lbm R
h3 = 1350.6,
h4 = 921.23
ψ = h - h0 - T0(s - s0)
ψ1 = 94.01 - 45.08 - 536.67(0.17497 - 0.08774) = 2.116
ψ2 = 3.92, ψ3 = 500.86, ψ4 = 71.49 Btu/lbm
∆ψ H = (1 - T0/TH)qH = 0.6054 × 1254.79 = 759.65 Btu/lbm
∆ψ L = (1 - T0/T0)qC = 0/
ηII = wNET/∆ψH = (429.37 - 1.8)/759.65 = 0.563
Notice TH > T3, TL < T4 = T1, so cycle is externally irreversible. Both qH and
qC over finite ∆T.
12-1
CHAPTER 12
The correspondence between the new problem set and the previous 4th edition
chapter 11 problem set.
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Old
New
New
New
1
New
3
new
new
new
new
new
new
5
new
4
new
new
new
6
7
new
new
12
8a mod
8b mod
10
new
11
13
new
New
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Old
15
new
new
16
17
new
new
20
new
new
24
new
26
new
23
32
40
37
46
28
new
new
39
new
new
new
35
new
27
21
New
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
The problems that are labeled advanced start at number 74.
Old
new
22
25
38
50
new
43
44
47
48
49
52
41
9
14
33 mod
34 mod
36
42
51
12-2
The English unit problems are:
New
Old
81
82
83
84
85
86
87
88
89
new
new
new
new
new
77
78
79
82
New
90
91
92
93
94
95
96
97
98
99
Old
80
84
86
85 mod
87
91
93
94
88
97
New
100
101
102
103
Old
96
99
83
98
12-3
12.1
A gas mixture at 20°C, 125 kPa is 50% N2, 30% H2O and 20% O2 on a mole
basis. Find the mass fractions, the mixture gas constant and the volume for 5 kg of
mixture.
From Eq. 12.3: ci = yi Mi/ ∑ yjMj
MMIX = ∑ yjMj = 0.5×28.013 + 0.3×18.015 + 0.2×31.999
= 14.0065 + 5.4045 + 6.3998 = 25.811
cN2 = 14.0065 / 25.811 = 0.5427, cH2O = 5.4045 / 25.811 = 0.2094
cO2 = 6.3998 / 25.811 = 0.2479, sums to 1 OK
−
RMIX = R/MMIX = 8.3145 / 25.811 = 0.3221 kJ/kg K
V = mRMIX T/P = 5×0.3221×393.15/125 = 5.065 m3
12.2
A 100 m3 storage tank with fuel gases is at 20°C, 100 kPa containing a mixture of
acetylene C2H2, propane C3H8 and butane C4H10. A test shows the partial
pressure of the C2H2 is 15 kPa and that of C3H8 is 65 kPa. How much mass is
there of each component?
Assume ideal gases, then the ratio of partial to total pressure is the mole
fraction, y = P/Ptot
yC2H2 = 15/100 = 0.15, yC3H8 = 65/100 = 0.65, yC4H10 = 20/100 = 0.20
−
ntot = PV/RT = 100 × 100 / (8.31451 × 293.15) = 4.1027 kmoles
mC2H2 = (nM)C2H2 = yC2H2ntotMC2H2 = 0.15×4.1027×26.038 = 16.024 kg
mC3H8 = (nM)C3H8 = yC3H8ntotMC3H8 = 0.65×4.1027×44.097 = 117.597 kg
mC4H10 = (nM)C4H10 = yC4H10ntotMC4H10 = 0.20×4.1027×58.124 = 47.693 kg
12.3
A mixture of 60% N2, 30% Ar and 10% O2 on a mole basis is in a cylinder at
250kPa, 310 K and volume 0.5 m3. Find the mass fractions and the mass of argon.
From Eq. 12.3:
ci = yi Mi/ ∑ yjMj
MMIX = ∑ yjMj = 0.6×28.013 + 0.3×39.948 + 0.1×31.999 = 31.992
cN2 = (0.6×28.013) / 31.992 = 0.5254, cAr = (0.3×39.948) / 31.992 = 0.3746
cO2 = (0.1×31.999) / 31.992 = 0.1, sums to 1 OK
−
RMIX = R/MMIX = 8.3145 / 31.992 = 0.260 kJ/kg K
mMIX = PV/(RMIX T) = 250×0.5 / 0.26×310 = 1.551 kg
mAr = cAr × mMIX = 0.3746×1.551 = 0.581 kg
12-4
12.4
A carbureted internal combustion engine is converted to run on methane gas
(natural gas). The air-fuel ratio in the cylinder is to be 20 to 1 on a mass basis.
How many moles of oxygen per mole of methane are there in the cylinder?
The mass ratio mAIR/mCH = 20, so relate mass and mole
4
nAIR
nCH4
→
12.5
=
(mm )× M
AIR
CH4
nO2
nCH4
CH4/MAIR
n = m/M
= 20× 16.04/28.97 = 11.0735
= 0.21×11.0735 = 2.325 mole O2/mole CH4
Weighing of masses gives a mixture at 60°C, 225 kPa with 0.5 kg O2, 1.5 kg N2
and 0.5 kg CH4. Find the partial pressures of each component, the mixture
specific volume (mass basis), mixture molecular weight and the total volume.
From Eq. 12.4: yi = (mi /Mi) / ∑ mj/Mj
ntot = ∑ mj/Mj = (0.5/31.999) + (1.5/28.013) + (0.5/16.04)
= 0.015625 + 0.053546 + 0.031172 = 0.100343
yO2 = 0.015625/0.100343 = 0.1557,
yN2 = 0.053546/0.100343 = 0.5336,
yCH4 = 0.031172/0.100343 = 0.3107
PO2 = yO2 Ptot = 0.1557×225 = 35 kPa, PN2 = yN2 Ptot = 0.5336×225 = 120 kPa
PCH4 = yCH4 Ptot = 0.3107×225 = 70 kPa
−
Vtot = ntot RT/P = 0.100343 × 8.31451 × 333.15 / 225 = 1.235 m3
v = Vtot/mtot = 1.235 / (0.5 + 1.5 + 0.5) = 0.494 m3/kg
MMIX = ∑ yjMj = mtot/ntot = 2.5/0.100343 = 24.914
12-5
12.6
At a certain point in a coal gasification process, a sample of the gas is taken and
stored in a 1-L cylinder. An analysis of the mixture yields the following results:
Component
H2
CO
CO2 N2
Percent by volume
25
40
15
20
Determine the mass fractions and total mass in the cylinder at 100 kPa, 20°C.
How much heat transfer must be transferred to heat the sample at constant volume
from the initial state to 100°C?
Volume fractions same as mole fractions so From Eq. 12.3: ci = yi Mi/ ∑ yjMj
yi
× Mi
= kgi/kg
= ci
kmoli/kmol
× kgi/kmoli
= kgi/kmol
= kgi/kg
H2
0.25
× 2.016
= 0.504/23.9121
= 0.0211
CO
CO2
0.40
0.15
× 28.01
× 44.01
= 11.204/23.9121
= 6.6015/23.9121
= 0.4685
= 0.2761
N2
0.20
× 28.013
= 5.6026/23.9121
= 0.2343
MMIX = 23.9121
−
RMIX = R/MMIX = 8.3145/23.9121 = 0.34771 kJ/kg/K
m = PV/RT = 100×10-3/0.34771× 293.15 = 9.81×10-4 kg
CV0 MIX = ∑ ci×CV0 i = 0.0211×10.0849 + 0.4685×0.7445
+ 0.2761×0.6529 + 0.2343×0.7448 = 0.9164 kJ/kg K
Q 12 = ∆U12 = mCV0(T2-T1) = 9.81×10-4× 0.9164×(100-20) = 0.0719 kJ
12.7
A pipe, cross sectional area 0.1 m2, carries a flow of 75% O2 and 25% N2 by
mole with a velocity of 25 m/s at 200 kPa, 290 K. To install and operate a mass
flow meter it is necessary to know the mixture density and the gas constant. What
are they? What mass flow rate should the meter then show?
From Eq. 12.3: ci = yi Mi/ ∑ yjMj
MMIX = ∑ yjMj = 0.75×31.999 + 0.25×28.013 = 31.0025
−
RMIX = R/MMIX = 8.3145 / 31.0025 = 0.2682 kJ/kg K
v = RMIX T/P = 0.2682× 290 / 200 = 0.38889 m3/kg
ρ = 1/v = 2.5714 kg/m3
.
m = ρAV = 2.5714×0.1×25 = 6.429 kg/s
12-6
12.8
A pipe flows 0.05 kmole a second mixture with mole fractions of 40% CO2 and
60% N2 at 400 kPa, 300 K. Heating tape is wrapped around a section of pipe with
insulation added and 2 kW electrical power is heating the pipe flow. Find the
mixture exit temperature.
C.V. Pipe heating section. Assume no heat loss to the outside, ideal gases.
.
. - .−
.
Energy Eq.: Q = m(he − hi) = n(he − hi) = nCP mix(Te − Ti)
−
−
CP mix = ∑ yi Ci = 0.4 × 37.056 + 0.6 × 29.189 = 32.336 kJ/kmole
. .−
Te = Ti + Q / nCP mix = 300 + 2/(0.05 × 32.336) = 301.2 K
12.9
A rigid insulated vessel contains 0.4 kmol of oxygen at 200 kPa, 280 K separated
by a membrane from 0.6 kmol carbon dioxide at 400 kPa, 360 K. The membrane
is removed and the mixture comes to a uniform state. Find the final temperature
and pressure of the mixture.
C.V. Total vessel. Control mass with two different initial states.
Mass:
n = nO2 + nCO2 = 0.4 + 0.6 = 1.0 kmole
Process: V = constant (rigid)
=> W = 0, insulated => Q = 0
−
−
Energy: U2 - U1 = 0 - 0 = n O2 CV O2(T2 - T1 O2) + nCO2CV CO2(T2 - T1 CO2)
Initial state from ideal gas Table A.5
−
−
CV O2 = 0.6618×31.999 = 21.177, CV CO2 = 0.6529×44.01 = 28.73 kJ/kmole
−
O2 : VO2 = nRT1/P = 0.4 × 8.3145 × 280/200 = 4.656 m3,
−
CO2 : VCO2 = nRT1/P = 0.6 × 8.3145 × 360/400 = 4.49 m3
−
−
−
−
nO2 CV O2 T2 + nCO2CV CO2 T2 = nO2 CV O2 T1 O2 + nCO2CV CO2 T1 CO2
(8.4708 + 17.238 ) T2 = 2371.8 + 6205.68 = 8577.48 kJ => T2 = 333.6 K
−
P2 = nRT2/V = 1 × 8.3145 × 333.6 / 9.146 = 303.3 kPa
12.10 An insulated gas turbine receives a mixture of 10% CO 2, 10% H2O and 80% N2
on a mole basis at 1000 K, 500 kPa. The volume flow rate is 2 m 3/s and its
exhaust is at 700 K, 100 kPa. Find the power output in kW using constant specific
heat from A.5 at 300 K.
C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0.
.
. - .−
.
Energy Eq.:
WT = m(hi − he) = n(hi − he) = nCP mix(Ti − Te)
.
. −
−
PV = nRT => n = PV / RT = 500×2/(8.3145×1000) = 0.1203 kmole/s
−
−
CP mix = ∑ yi Ci = 0.1 × 44.01 × 0.842 + 0.1 × 18.015 × 1.872
+ 0.8 × 28.013 × 1.042 = 30.43 kJ/kmol K
.
WT = 0.1203 × 30.43 (1000 − 700) = 1098 kW
12-7
12.11 Solve Problem 12.10 using the values of enthalpy from Table A.8.
C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0.
.
. - .
.
- Energy Eq.:
WT = m(hi − he) = n(hi − he) = n [ ∑yj (hi − he)j ]
.
. −
−
PV = nRT => n = PV / RT = 500×2/(8.3145×1000) = 0.1203 kmole/s
.
WT = 0.1203[0.1(33397 - 17754) + 0.1(26000 - 14190) + 0.8(21463 - 11937)]
= 1247 kW
12.12 Consider Problem 12.10 and find the value for the mixture heat capacity, mole
basis and the mixture ratio of specific heats, kmix, both estimated at 850 K from
values (differences) of h in Table A.8. With these values make an estimate for the
reversible adiabatic exit temperature of the turbine at 100 kPa.
−
We will find the individual heat capacities by : CP i = (h900 -h800)/(900 - 800)
−
−
CP CO2 = (28030 - 22806)/100 = 52.24; CP H2O = (21937 - 18002)/100 = 39.35
−
CP N2 = (18223 - 15046)/100 = 31.77
−
−
CP mix = ∑ yi CP i = 0.1 × 52.24 + 0.1 × 39.35 + 0.8 × 31.77 = 34.575 kJ/kmol
−
−
−
−
−
CV mix = CP mix - R = 34.575 - 8.3145 = 26.26,
k = CP mix/CV mix = 1.3166
Rev. ad. turbine
Te =
=> s = constant. Assume const. avg. heat capacities:
k-1
Ti(Pe/Pi) k =
0.3166
1000 (100/500)1.3166 = 1000 (0.2)0.2405 = 679 K
12.13 The gas mixture from Problem 12.6 is compressed in a reversible adiabatic
process from the initial state in the sample cylinder to a volume of 0.2 L.
Determine the final temperature of the mixture and the work done during the
process.
CP0 MIX = CV0 MIX + RMIX = 0.9164 + 0.34771 = 1.2641
→ k = CP0/CV0 = 1.2641/0.9164 = 1.379
(VV )
→ T2 = T 1
1 k-1
2
= 293.15
(0.21 )
0.3805
= 539.5 K
W12 = - ∆U12 = -mCV0(T2-T1)
= - 9.81×10-4 × 0.9164×(539.5-293.15) = - 0.22 kJ
12-8
12.14 Three SSSF flows are mixed in an adiabatic chamber at 150 kPa. Flow one is 2
kg/s of O2 at 340 K, flow two is 4 kg/s of N2 at 280 K and flow three is 3 kg/s of
CO2 at 310 K. All flows are at 150 kPa the same as the total exit pressure. Find
the exit temperature and the rate of entropy generation in the process.
C.V. Mixing chamber, no heat transfer, no work.
.
.
.
.
Continuity: m1 + m2 + m3 = m4
.
.
.
.
Energy: m1h1 + m2h2 + m3h3 = m4h4
.
.
.
.
.
Entropy: m1s1 + m2s2 + m3s3 + Sgen = m4s4
1
O2
N2
CO2
2
3
4
mix
Assume ideal gases and since T is close to 300 K use heat capacity from A.5
.
.
.
m1CP O2(T1 - T4) + m2CP N2(T2 - T4) + m3CP CO2(T3 - T4)= 0
2 × 0.922 × 340 + 4 × 1.042 × 280 + 3 × 0.842 × 310
= (2 × 0.922 + 4 × 1.042 + 3 × 0.842 ) T4
=>
2577.06 = 8.538 T 4
=>
T4 = 301.83 K
State 4 is a mixture so the component exit pressure is the partial pressure. For
each component se − si = CP ln(Te / Ti) − R ln(Pe / Pi) and the pressure
ratio is
Pe / Pi = y P4 / Pi = y for each.
n=∑
m 2
4
3
=
+
+
= 0.0625 + 0.1428 + 0.06817 = 0.2735
M 32 28.013 44.01
0.0625
0.1428
0.06817
= 0.2285, yN2 =
= 0.5222, yCO2 =
= 0.2493
yO2 =
0.2735
0.2735
0.2735
.
.
.
.
Sgen = m1(s4 - s1) + m2(s4 - s2) + m3(s4 - s3)
=
2 [ 0.922 ln(301.83/340) - 0.2598 ln(0.2285)]
+ 4 [ 1.042 ln(301.83/280) - 0.2968 ln(0.5222)]
+ 3 [ 0.842 ln(301.83/310) - 0.1889 ln(0.2493)]
= 0.5475 + 1.084 + 0.2399 = 1.871 kW/K
12-9
12.15 Carbon dioxide gas at 320 K is mixed with nitrogen at 280 K in a SSSF insulated
mixing chamber. Both flows are at 100 kPa and the mole ratio of carbon dioxide
to nitrogen is 2;1. Find the exit temperature and the total entropy generation per
mole of the exit mixture.
.
.
CV mixing chamber, SSSF. The inlet ratio is nCO = 2 nN and assume no
2
2
external heat transfer, no work involved.
.
.
.
.
Continuity: nCO + 2nN = nex = 3nN ;
2
2
2
. . Energy Eq.: nN (hN + 2hCO ) = 3nN hmix ex
2
2
2
2
- Take 300K as reference and write h = h300 + CPmix(T-300).
CP N (Ti N -300) + 2CP CO (Ti CO -300) = 3CP mix(Tmix ex-300)
2
2
2
2
CP mix = ∑ yiCP i = (29.178 + 2 37.05)/3 = 34.42
3CP mixTmix ex = CP N2Ti N2 + 2CP CO2Ti CO2 = 31882
Tmix ex = 308.7 K; Pex N = Ptot/3; Pex CO = 2Ptot/3
2
2
.
. ... - . - Sgen = nexsex-(ns)iCO -(ns)iN = nN (se-si)N + 2nN (se-si)CO
2
2
2
2
2
2
Tex −
Tex
.
.
−
Sgen/3nN2 = [CPN2ln
- Rln yN2 + 2CPCO2ln
- 2 Rln yCO2]/3
TiN2
TiCO2
= [2.8485 + 9.1343 - 2.6607+6.742]/3 = 5.35 kJ/kmol mix K
12.16 A piston/cylinder contains 0.5 kg argon and 0.5 kg hydrogen at 300 K, 100 kPa.
The mixture is compressed in an adiabatic process to 400 kPa by an external force
on the piston. Find the final temperature, the work and the heat transfer in the
process.
C.V. Mixture in cylinder. Control mass with adiabatic process: 1Q2 = 0
Cont.Eq.:
m 2 = m1 = m ;
Entropy Eq.:
Energy Eq.:
m(u2 − u1) = − 1W2
m(s2 − s1) = ∫ dQ/T + 1S2 gen = 0 + 0
assume reversible.
Rmix = ∑ ciRi = 0.5 × 0.2081 + 0.5 × 4.1243 = 2.1662
CP mix = ∑ ciCPi = 0.5 × 0.52 + 0.5 × 14.209 = 7.3645
CV = CP - R = 7.3645-2.1662 = 5.1983,
k = CP / CV = 1.417
s2 = s1 => T2 = T1(P2/P1)(k-1)/k = 300 (400/100)0.2943 = 451 K
1W2
= m(u1 − u2) = mCV (T1 − T2) = 1×5.1983(300 - 451) = -784.9 kJ
12-10
12.17 Natural gas as a mixture of 75% methane and 25% ethane by volume is flowing to
a compressor at 17°C, 100 kPa. The reversible adiabatic compressor brings the
flow to 250 kPa. Find the exit temperature and the needed work per kg flow.
C.V. Compressor. SSSF, adiabatic q = 0, reversible sgen = 0
Energy Eq.:
-w = hex - hin ;
Entropy Eq.:
Process: reversible => sgen = 0 =>
si + sgen = si = se
se = si
Assume ideal gas mixture and constant heat capacity, so we need k and CP
CP = ∑ yi CPi = 0.75 × 2.254 × 16.043 + 0.25 × 1.766 × 30.07 = 40.3966
Mmix = 0.75 × 16.043 + 0.25 × 30.07 = 19.55
−
Cp = CP/ Mmix = 2.066 , R = R/ Mmix = 8.31451 / 19.55 = 0.4253
Cv = Cp - R = 1.6407 ,
k = Cp/ Cv = 1.259
Te = Ti (Pe/ Pi)(k-1)/k = 290 (250/100) 0.2057 = 350.2 K
wc in = CP (Te- Ti) = 2.066 (350.2 – 290) = 124.3 kJ/kg
12.18 Take Problem 12.15 with inlet temperature of 1400 K for the carbon dioxide and
300 K for the nitrogen. First estimate the exit temperature with the specific heats
from Table A.5 and use this to start iterations using A.8 to find the exit
temperature.
.
.
CV mixing chamber, SSSF. The inlet ratio is nCO = 2 nN and assume no
2
2
external heat transfer, no work involved.
CP CO2 = 44.01 × 0.842 = 37.06 CP N2 = 28.013 × 1.042 = 29.189
.
. . .
Cont. Eq.: 0 = Σnin - Σnex; Energy Eq.: 0 = Σnin hin - Σnex hex
. . 0 = 2nN2 CP CO2(Tin- Tex) CO2 + nN2 CP N2 (Tin- Tex) N2
0 = 2 × 37.06 × (1400-Tex) + 29.189 × (300-Tex)
0 = 103768 + 8756.7 – 103.309 Tex
_ Tex = 1089 K
.
.
. From Table A.8: Σnin hin = nN2 [2 × 55895 + 1 × 54] = nN2 × 111844
.
.
. @ 1000K
: Σnex hex = nN2 [2 × 33397 + 21463] = nN2 × 88257
.
.
. @ 1100K
: Σnex hex = nN2 [2 × 38885 + 24760] = nN2 × 102530
.
.
. @ 1200K
: Σnex hex = nN2 [2 × 44473 + 28109] = nN2 × 117055
Now linear interpolation between 1100 K and 1200 K
Tex = 1100 + 100 ×
111844-102530
= 1164 K
117055-102530
12-11
12.19 A mixture of 60% helium and 40% nitrogen by volume enters a turbine at 1 MPa,
800 K at a rate of 2 kg/s. The adiabatic turbine has an exit pressure of 100 kPa
and an isentropic efficiency of 85%. Find the turbine work.
Assume ideal gas mixture and take CV as turbine.
wT s = hi - hes, ses = si,
Tes = Ti(Pe/Pi)(k-1)/k
CP mix = 0.6× 5.1926× 4.003 + 0.4× 1.0416× 28.013 = 24.143
− (k-1)/k = R/CP mix = 8.31451/24.143 = 0.3444
Mmix= 0.6×4.003 + 0.4×28.013 = 13.607, CP=CP/Mmix= 1.774
Tes = 800(100/1000)0.3444 = 362 K,
wTs = CP(Ti-Tes) = 777
.
wT ac = ηwTs = 660.5 kJ/kg; W = mwTs = 1321 kW
12.20 A mixture of 50% carbon dioxide and 50% water by mass is brought from 1500
K, 1 MPa to 500 K, 200 kPa in a polytropic process through a SSSF device. Find
the necessary heat transfer and work involved using values from Table A.5.
Process Pvn = constant
leading to
n ln(v2/v1) = ln(P1/P2); v = RT/P
n = ln(1000/200)/ln(500 × 1000/200 × 1500) =3.1507
Rmix = ciRi = 0.5 × 0.18892 + 0.5 × 0.46152 = 0.32522
∑
CP mix = ciCPi = 0.5 × 0.8418 + 0.5 × 1.8723 = 1.357
∑
⌠vdP = - n (P v -P v )= - nR (T -T ) = 476.4 kJ/kg
w = -⌡
n-1 e e i i
n-1 e i
q = he-hi + w = CP(Te-Ti) + w = -880.6 kJ/kg
12.21 Solve Problem 12.20 using specific heats CP = ∆h/∆T, from Table A.8 at 1000 K.
38885-28030
= 1.223
(1100-900)×44.01
30190-21937
CP H2O =
= 2.2906
(1100-900)×18.015
CP CO2 =
CP mix = Σ ciCPi = 0.5 × 1.223 + 0.5 × 2.2906 = 1.7618
Rmix = Σ ciRi = 0.5 × 0.18892 + 0.5 × 0.46152 = 0.32522
Process Pvn = C
=>
n = ln(P1/P2) / ln(v2/v1) = 3.1507
⌠vdP = - n (P v -P v )= - nR (T -T ) = 476.4 kJ/kg
w = -⌡
n-1 e e i i
n-1 e i
q = he-hi + w = 1.7618(500-1500) + 476.4 = -1285.4 kJ/kg
12-12
12.22 A 50/50 (by mole) gas mixture of methane CH 4 and ethylene C2H4 is contained
in a cylinder/piston at the initial state 480 kPa, 330 K, 1.05 m3. The piston is now
moved, compressing the mixture in a reversible, polytropic process to the final
state 260 K, 0.03 m3. Calculate the final pressure, the polytropic exponent, the
work and heat transfer and net entropy change for the process.
Ideal gas mixture: CH4, C2H4, 50% each by mol => yCH4 = yC2H4 = 0.5
−
n = P1V1/RT1 = 480 × 1.05 / (8.31451 × 330) = 0.18369 kmol
State 1:
Cv mix = ∑ yi Mi Cvi = 31.477 kJ/kmol-K
V1 T2
= 13236 kPa
State 2: T2 = 260 K, V2 = 0.03 m3, Ideal gas => P2 = P1
V T
2
1
V1n-1
T2
V1
=   , ln = (n-1) ln
=> n = 0.933
T1 V2
T1
V2
1
=>
1W2 = ∫ P dv = 1-n(P 2V2- P1V1) = -1595.7 kJ
- Energy Eq.:
1Q2 = n(u2- u1) + 1W2 = n Cv mix (T2- T1) + 1W2
Process: PVn = constant,
T2
= 0.18369 × 31.477(260-330) -1595.7 = -2000.4 kJ
- s2- s1 = Cv mix ln (T2 / T1) + R ln (V2 / V1) = -37.065 kJ/kmol-K
2nd Law:
∆ Snet = n(s-2- -s1) - 1Q2 /To
with
T0 = 260 K
∆Snet = 0.18369 × (-37.065) - (-2000.4)/260 = 0.8854 kJ/K
12.23 A mixture of 2 kg oxygen and 2 kg of argon is in an insulated piston cylinder
arrangement at 100 kPa, 300 K. The piston now compresses the mixture to half its
initial volume. Find the final pressure, temperature and the piston work.
C.V. Mixture. Control mass, boundary work and no Q, assume reversible.
Energy Eq.: u2 - u1 = 1q2 - 1w2 = - 1w2 ; Entropy Eq.: s2-s1 = 0 + 0 = 0
Process: constant s
=>
Pvk = constant, v2 = v1/2,
Assume ideal gases (T1 >> TC ) and use kmix and Cv mix for properties.
Rmix = Σ ciRi = 0.5 × 0.25983 + 0.5 × 0.20813 = 0.234 kJ/kg mix
CPmix = Σ ciCPi = 0.5 × 0.9216 + 0.5 × 0.5203 = 0.721 kJ/kg mix
Cvmix = CPmix -Rmix = 0.487, kmix = CPmix/Cvmix = 1.4805
P2 = P1(v1/v2)k = P1(2)k = 100(2)1.4805 = 279 kPa
T2 = T1(v1/v2)k-1 = T1(2)k-1 = 300(2)0.4805 = 418.6 K
1W2
= mtot (u1-u2) = mtot Cv(T1-T2) = 4×0.487 (300 - 418.6) = -231 kJ
12-13
12.24 Two insulated tanks A and B are connected by a valve. Tank A has a volume of 1
m3 and initially contains argon at 300 kPa, 10°C. Tank B has a volume of 2 m3
and initially contains ethane at 200 kPa, 50°C. The valve is opened and remains
open until the resulting gas mixture comes to a uniform state. Determine the final
pressure and temperature.
C.V. Tanks A + B. Control mass no W, no Q.
Energy Eq.: U2-U1 = 0 = nArCV0(T2-TA1) + nC H CVO(T2-TB1)
2 6
−
nAr = PA1VA/RTA1 = (300 × 1) / (8.31451 × 283.15) = 0.1274 kmol
−
nC2H6 = PB1VB/RTB1 = (200 × 2) / (8.31451 × 323.15) = 0.1489 kmol
Continuity Eq.:
Energy Eq.:
n2 = nAr + nC
2H6
= 0.2763 kmol
0.1274 × 39.948 × 0.3122 (T2 - 283.2)
+ 0.1489 × 30.07 × 1.4897 (T2 - 323.2) = 0
Solving, T2 = 315.5 K
−
P2 = n2RT2/(VA+VB) = 0.2763×8.3145×315.5/3 = 242 kPa
12-14
12.25 Reconsider the Problem 12.24, but let the tanks have a small amount of heat
transfer so the final mixture is at 400 K. Find the final pressure, the heat transfer
and the entropy change for the process.
C.V. Both tanks. Control mass with mixing and heating of two ideal gases.
−
nAr = PA1VA/RTA1 =
300×1
= 0.1274 kmol
8.3145×283.2
−
nC2H6 = PB1VB/RTB1 =
Continuity Eq.:
Energy Eq.:
200×2
= 0.1489 kmol
8.3145×323.2
n2 = nAr + nC
2H6
= 0.2763 kmol
U2-U1 = nArCV0(T2-TA1) + nC H CVO(T2-TB1) = 1Q2
2 6
−
P2 = n2RT2/(VA+VB) = 0.2763×8.3145×400 / 3 = 306.3 kPa
1Q 2
= 0.1274×39.948×0.3122(400 - 283.15)
+ 0.1489×30.07×1.4897(400 - 323.15) = 698.3 kJ
∆SSYS = nAr∆SAr + nC2H6∆SC2H6
∆SSURR = -1Q2/TSURR;
yAr = 0.1274/0.2763 = 0.4611
T2 − yArP2
∆SAr = CP Ar ln
- R ln
TA1
PA1
= 39.948×0.5203 ln
0.4611×306.3
400
- 8.3145 ln
283.15
300
= 13.445 kJ/kmol K
∆SC
T2 − yC2H6P2
= CC2H6 ln
- R ln
2H6
TB1
PB1
= 30.07×1.7662 ln
0.5389×306.3
400
- 8.3145 ln
323.15
200
= 12.9270 kJ/kmol K
Assume the surroundings are at 400 K (it heats the gas)
∆SNET = nAr∆SAr + nC2H6∆SC2H6 + ∆SSURR
= 0.1274×13.445 + 0.1489×12.9270 - 698.3/400
= 1.892 kJ/K
12-15
12.26 A piston/cylinder contains helium at 110 kPa at ambient temperature 20°C, and
initial volume of 20 L as shown in Fig. P12.26. The stops are mounted to give a
maximum volume of 25 L and the nitrogen line conditions are 300 kPa, 30°C.
The valve is now opened which allows nitrogen to flow in and mix with the
helium. The valve is closed when the pressure inside reaches 200 kPa, at which
point the temperature inside is 40°C. Is this process consistent with the second
law of thermodynamics?
P1 = 110 kPa, T1 = 20 oC, V1 = 20 L, Vmax = 25 L = V2
P2 = 200 kPa, T2 = 40 oC, Pi = 300 kPa, Ti = 30 oC
Const. P to stops, then const. V = Vmax
Qcv = U2 - U1 + Wcv - nihi, Wcv = P1(V2 - V1)
= n2h2 - n1h1 - nihi - (P2 - P1)V2
= nA(hA2 - hAi) + nB(hB2 - hB1) - (P2 - P1)V2
nB = n1 = P1V1/RT1 = 110×0.02/8.3145×293.2 = 0.0009 kmol
n2 = nA + nB = P2V2/RT2 = 200×0.025/8.3145×313.2 = 0.00192 kmol,
nA = n2 - nB = 0.00102 kmol
Qcv = 0.00102×28.013×1.0416(40-30) + 0.0009×4.003×5.1926(40-20)
- (200 - 110) 0.025
= 0.298 + 0.374 - 2.25 = - 1.578 kJ
Sgen = n2s2 - n1s1 - nisi - Qcv/T0
= nA(sA2 - sAi) + nB(sB2 - sB1) - Qcv/T0
yA2 = 0.00102/0.00192 = 0.5313, yB2 = 0.4687
313.2 - 0.5313*200
sA2 - sAi = 29.178 ln
- R ln
= 9.5763
303.2
300
313.2 - 0.4687*200
- R ln
= 2.7015
sB2 - sB1 = 20.786 ln
293.2
110
Sgen = 0.00102×9.5763 + 0.0009×2.7015 + 1.578/293.2
= 0.0176 kJ/K > 0
Satisfies 2nd law.
12-16
12.27 Repeat Problem 12.17 for an isentropic compressor efficiency of 82%.
C.V. Compressor. SSSF, adiabatic q = 0, reversible sgen = 0
Energy Eq.:
-w = hex - hin ;
Entropy Eq.:
si + sgen = si = se
Process: For ideal compressor reversible => sgen = 0 =>
se = si
Assume ideal gas mixture and constant heat capacity, so we need k and CP
CP = ∑ yi CPi = 0.75 × 2.254 × 16.043 + 0.25 × 1.766 × 30.07 = 40.3966
Mmix = 0.75 × 16.043 + 0.25 × 30.07 = 19.55
−
Cp = CP/ Mmix = 2.066 , R = R/ Mmix = 8.31451 / 19.55 = 0.4253
Cv = Cp - R = 1.6407 ,
k = Cp/ Cv = 1.259
Te = Ti (Pe/ Pi)(k-1)/k = 290 (250/100) 0.2057 = 350.2 K
wc in = CP (Te- Ti) = 2.066 (350.2 – 290) = 124.3 kJ/kg
wc actual = wc in/η = 124.3/0.82 = 151.6 kJ/kg = Cp (Te actual - Ti)
=> Te actual = T + w c actual/CP = 290 + 151.6 / 2.066 = 363.4 K
12.28 A spherical balloon has an initial diameter of 1 m and contains argon gas at 200
kPa, 40°C. The balloon is connected by a valve to a 500-L rigid tank containing
carbon dioxide at 100 kPa, 100°C. The valve is opened, and eventually the balloon
and tank reach a uniform state in which the pressure is 185 kPa. The balloon
pressure is directly proportional to its diameter. Take the balloon and tank as a
control volume, and calculate the final temperature and the heat transfer for the
process.
PA1VA1
π
200×0.5236
A
=
= 1.606 kg
VA1 = 13 = 0.5236, mA1 =
RTA1
6
0.208 13×313.2
mB1 = PB1VB1/RTB1 = 100×0.50/0.18892×373.2 = 0.709 kg
-1/3
(PP ) = 0.5236(185
)
200
-1/3
P2VA2 = PA1VA1 → VA2 = VA1
B
CO 2
2: Uniform ideal gas mix. :
2
3
3
= 0.4144 m3
A1
P2(VA2+VB) = (mARA+mBRB)T2
T2 = 185(0.4144+0.50) / (1.606×0.20813 + 0.709×0.18892) = 361.3 K
W12 =
P2VA2-PA1VA1
1-(-1/3)
=
185×0.4144-200×0.5236
= -21.0 kJ
(4/3)
Q = mACV0A(T2-TA1) + mBCV0B(T2-TB1) + W12
= 1.606×0.3122(361.3-313.2) + 0.709×0.6529(361.3-373.2) - 21.0
= 18.6 - 21.0 = -2.4 kJ
12-17
12.29 A large SSSF air separation plant takes in ambient air (79% N2, 21% O2 by
volume) at 100 kPa, 20°C, at a rate of 1 kmol/s. It discharges a stream of pure O2
gas at 200 kPa, 100°C, and a stream of pure N 2 gas at 100 kPa, 20°C. The plant
operates on an electrical power input of 2000 kW. Calculate the net rate of
entropy change for the process.
Air 79 % N2
21 % O2
P1 = 100 kPa
1
P2 = 200 kPa
2
pure O 2
T2 = 100 oC
pure N 2
3
T1 = 20 oC
.
n1 = 1 kmol/s
P3 = 100 kPa
.
-WIN = 2000 kW
T3 = 20 oC
.
.
.
.
.
.
.
Sgen = Σ ni∆s-i - QCV/T0 = (n2-s2 + n3-s3 - n1-s1) - QCV/T0
.
.
.
. . . QCV = Σ n∆hi + WCV = nO2CP0 O2(T2-T1) + nN2CP0 N2(T3-T1) + WCV
= 0.21 ×32 × 0.9216 ×(100-20) + 0 - 2000 = -1505 kW
.
[
373.2
200
Σ ni∆s-i = 0.21 29.49 ln 293.2 - 8.3145 ln 21
]
+ 0.79 [0 - 8.3145 ln (100/79)]= -3.9894 kW/K
.
Sgen = 1505/293.2 - 3.9894 = 1.1436 kW/K
12-18
12.30 An insulated vertical cylinder is fitted with a frictionless constant loaded piston of
cross sectional area 0.1 m 2 and the initial cylinder height of 1.0 m. The cylinder
contains methane gas at 300 K, 150 kPa, and also inside is a 5-L capsule
containing neon gas at 300 K, 500 kPa. The capsule now breaks, and the two
gases mix together in a constant pressure process. What is the final temperature,
final cylinder height and the net entropy change for the process.
Ap = 0.1 m2, h = 1.0 m =>
Vtot = Va1 + Vb1 = 0.1 m3
Methane: M = 16.04 kg/kmol, Cp = 2.2537 kJ/kg-K, R = 0.51835 kJ/kg-K
Neon: M = 20.183 kg/kmol, Cp = 1.0299 kJ/kg-K, R = 0.41195 kJ/kg-K
State 1: Methane, Tal = 300 K, Pa1 = 150 kPa, Va1 = Vtot- Vb1 = 0.095 m3
Neon, Tbl = 300 K, Pb1 = 500 kPa, Vb1 = 5 L
ma =
mb =
PalVal
= 0.0916 kg, na =
RaTal
PblVbl
RbTbl
= 0.0202 kg, nb =
ma
Ma
mb
Mb
= 0.00571 kmol
= 0.001 kmol
State 2: Mix, P2mix = Pa1 = 150 kPa
Energy Eq: 1Q2 = ma(ua2-ua1) + mb( ub2 - ub1) + 1W2;
1Q2
=0
= ∫ P dV = P2(V2-V1) tot; P2V2 = mtotRmixT2 = (maRa + mbRb)T2
1W2
Assume Constant Specific Heat
0 = maCpa(T2-Ta1) + mbCpb(T2-Tb1 + (maRa + mbRb)T2 - P2V1 tot
Solving for T2 = 293.3 K
V2 =
(maRa + mbRb)T2
P2
Entropy Eq.:
ya =
na
na+nb
= 0.1087 m3,
h2 =
V2
Ap
∆Snet = ma(sa2- sa1) + mb(sb2-sb1) = 0.851,
= 1.087 m
1Q 2
T0
yb = 1 - ya = 0.149
T2
yaP2
= 0.02503 kJ/kg-K
sa2 - sa1 = Cpa ln - Raln
Ta1
Pa1
T2
ybP2
- Rb ln
= 1.2535 kJ/kg-K
sb2 - sb1 = Cpb ln
Tb1
Pb1
∆Snet = 0.0276 kJ/K
;
1Q2
=0
12-19
12.31 The only known sources of helium are the atmosphere (mole fraction approximately
5 × 10−6) and natural gas. A large unit is being constructed to separate 100 m3/s of
natural gas, assumed to be 0.001 He mole fraction and 0.999 CH 4. The gas enters the
unit at 150 kPa, 10°C. Pure helium exits at 100 kPa, 20°C, and pure methane exits at
150 kPa, 30°C. Any heat transfer is with the surroundings at 20°C. Is an electrical
power input of 3000 kW sufficient to drive this unit?
P2 = 100 kPa
1
2
He
0.999 CH4
T = 20 oC
2
0.001 He
at 150 kPa, 10 oC
.
V1 = 100 m3/s
CH 4
3
P3 = 140 kPa
.
-WCV = 3000 kW
T3 = 30 oC
.
.
n1 = P1V1/RT1 = 150×100/(8.3145×283.2) = 6.37 kmol/s
.
.
.
=> n2 = 0.001; n1 = 0.006 37; n3 = 6.3636
CP CH4= 16.04×2.2537 = 36.149
CP He= 4.003×5.1926 = 20.786,
.
.
.
. . - . . . QCV = n2h2 + n3h3 - n1h1 + WCV = n2CP0 He(T2-T1) + n3CP0 CH (T3-T1) + WCV
4
= 0.00637×20.786(20-10)+6.3636×36.149(30-10)+(-3000) = +1600 kW
.
.
.
.
.
Sgen = n2-s2 + n3-s3 - n1-s1 - QCV/T0
[
= 0.00637 20.786 ln
[
293.2
100
- 8.3145 ln
283.2
0.001×150
+ 6.3636 36.149 ln
]
303.2
140
- 8.3145 ln
283.2
0.999×150
] - 1600/293.2
= +13.498 kW/K > 0
12.32 A steady flow of 0.01 kmol/s of 50% carbon dioxide and 50% water at 1200K and
200 kPa is used in a heat exchanger where 300 kW is extracted from the flow.
Find the flow exit temperature and the rate of change of entropy using Table A.8.
C.V. Heat exchanger, SSSF, 1 inlet, 1 exit, no work.
Continuity Eq.:
yCO2 = yH2O = 0.5
. - . .
.
.
Energy Eq.: Q = m(he − hi) = n(he − hi) => he = hi + Q/n
Inlet state: Table A.8 hi = 0.5 × 44473 + 0.5 × 34506 = 39489.5 kJ/kmol
. .
Exit state: he = hi + Q/n = 39489.5 - 300/0.01 = 9489.5 kJ/kmol
Trial and error for T with h values from Table A.8
@500 K he = 0.5(8305 + 6922) = 7613.5
@600 K he = 0.5(12906 + 10499) = 11702.5
Interpolate to have the right h:
T = 545.9 K
12-20
12.33 An insulated rigid 2 m3 tank A contains CO2 gas at 200°C, 1MPa. An uninsulated
rigid 1 m3 tank B contains ethane, C2H6, gas at 200 kPa, room temperature 20°C.
The two are connected by a one-way check valve that will allow gas from A to B,
but not from B to A. The valve is opened and gas flows from A to B until the
pressure in B reaches 500 kPa and the valve is closed. The mixture in B is kept at
room temperature due to heat transfer. Find the total number of moles and the
ethane mole fraction at the final state in B. Find the final temperature and pressure
in tank A and the heat transfer to/from tank B.
Tank A: VA = 2 m3, state A1 : CO2, TA1 = 200°C = 473.2 K, PA1 = 1 MPa
C v0 CO2 = 0.653 × 44.01 = 28.74, C P0 CO2 = 0.842 × 44.01 = 37.06
Tank B: VB = 1 m3, state B1: C2H6, TB1 = 20°C = 293.2 K, PB1 = 200 kPa
Slow Flow A to B to PB2 = 500 kPa and assume TB2 = TB1 = T0
PB1VB = nB1RTB1 ,
yC2H6 B2 =
nB1 =
PB1VB
RTB1
nB2 mix =
nB1
nB2
=
=
PB1
PB2
PB2VB = nB2 mix RTB2
=
200
= 0.400
500
200 × 1
= 0.08204 kmol
R × 293.2
PB2VB
RTB2
=
500 × 1
= 0.2051 kmol
R × 293.2
nCO2 B2 = 0.2051 – 0.08201 = 0.12306 kmol
PA1VA
1000×2
= 0.50833 kmol ; nA2 = nA1 - nCO2 B2 = 0.38527 kmol
RTA1 R×473.2
= 0 = n u- n u+ n hC.V. A: All CO Energy Eq.: Q
nA1 =
=
2
CV A
A2
A2
A1
A1
ave
ave
0 = nA2 C v0 CO2TA2 - nA1 C v0 CO2TA1 + nave C P0 CO2 ( TA1 + TA2)/2
0 = 28.74( 0.38527× TA2 - 0.50833×473.2) + 0.12306 × 37.06(473.2 + TA2)/2
=>
PA2 =
TA2 = 436.9 K
nA2 RTA2
VA
=
0.38527 × R × 436.9
= 700 kPa
2
C.V. B:
QCV B + nBi h Bi ave = (nu-) B2 - (nu-) B1 = (nu-)CO2 B2 + (nu-)C2H6 B2 - (nu-)C2H6 B1
QCV B = 0.12306 × 28.74 × 293.2 + 0 – 0.12306 × 37.06 (473.2 + 436.9)/2
= -1037.8 kJ
12-21
12.34 A tank has two sides initially separated by a diaphragm. Side A contains 1 kg of
water and side B contains 1.2 kg of air, both at 20°C, 100 kPa. The diaphragm is
now broken and the whole tank is heated to 600°C by a 700°C reservoir. Find the
final total pressure, heat transfer and total entropy generation.
C.V. Total tank out to reservoir.
Energy Eq.:
U2-U1 = ma(u2-u1)a + mv(u2-u1)v = 1Q2
Entropy Eq.: S2-S1 = ma(s2-s1)a + mv(s2-s1)v = 1Q2/Tres + Sgen
Volume:
V2 = VA + VB = mvvv1 + mava1 = 0.001 + 1.009 = 1.01 m3
vv2 = V2/mv = 1.01, T2 => P2v = 400 kPa
va2 = V2/ma = 0.8417, T2 => P2a = mRT2/V2 = 297.7 kPa
P2tot = P2v + P2a = 697.7 kPa
Water table B.1:
u1 = 83.95, u2 = 3300, s1 = 0.2966, s2 = 8.4558
Air table A.7: u1 = 293, u2 = 652.3, sT1 = 2.492, sT2 = 3.628
From energy equation we have
1Q2
= 1(3300 - 83.95) + 1.2(652.3 - 293) = 3647.2 kJ
From the entropy equation we have
Sgen = 1(8.4558 - 0.2966) + 1.2[3.628 - 2.492 - 0.287×ln(301.6/100)]
- 3647.2 / 973.2 = 5.4 kJ/K
12-22
12.35 A 0.2 m3 insulated, rigid vessel is divided into two equal parts A and B by an
insulated partition, as shown in Fig. P12.35. The partition will support a pressure
difference of 400 kPa before breaking. Side A contains methane and side B
contains carbon dioxide. Both sides are initially at 1 MPa, 30°C. A valve on side
B is opened, and carbon dioxide flows out. The carbon dioxide that remains in B
is assumed to undergo a reversible adiabatic expansion while there is flow out.
Eventually the partition breaks, and the valve is closed. Calculate the net entropy
change for the process that begins when the valve is closed.
A
CH 4
∆PMAX = 400 kPa,
B
CO 2
PA1 = PB1 = 1 MPa
VA1 = VB1 = 0.1 m3
TA1 = TB1 = 30 oC = 303.2 K
CO2 inside B: sB2 = sB1 to PB2 = 600 kPa (PA2 = 1000 kPa)
0.289
600 1.289
= 270.4 K
For CO2, k = 1.289 => TB2 = 303.2
1000
nB2 = PB2VB2/RTB2 = 600×0.1/8.3145×270.4 = 0.026 688
(
)
nA2 = nA1 = 1000×0.1/8.3145×303.2 = 0.039 668 kmol
Q23 = 0 = n 3u-3 - ni2u-i2 + 0 = n A2CV0 A(T3-TA2) + nB2CV0 B(T3-TB2) = 0
∑
i
0.039 668×16.04×1.736(T3-303.2) + 0.026 688×44.01×0.653(T3-270.4) = 0
Solve
T3 = 289.8 K
P3 = 0.066356×8.3145×289.8/0.2 = 799.4 kPa
PA3 = 0.5978×799.4 = 477.9 kPa ,
PB3 = P3 - PA3 = 321.5 kPa
-s - -s = 16.04×2.254 ln(289.8) - 8.3145 ln 477.9 = 4.505 kJ/kmol K
A3
A2
303.2
1000
-s - -s = 44.01×0.842 ln(289.8) - 8.3145 ln 321.5 = 7.7546 kJ/kmol K
B3
B2
270.4
600
∆SNET = 0.039668×4.505 + 0.026688×7.7546 = +0.3857 kJ/K
12-23
12.36 Consider 100 m3 of atmospheric air which is an air–water vapor mixture at 100
kPa, 15°C, and 40% relative humidity. Find the mass of water and the humidity
ratio. What is the dew point of the mixture?
Air-vap P = 100 kPa, T= 15 oC, φ = 40%
Pg = Psat15 = 1.705 kPa
mv =
P vV
RvT
=
=>
Pv = φ Pg = 0.4×1.705 = 0.682 kPa
0.682×100
= 0.513 kg
0.461×288.15
Pa = Ptot- Pv1 = 100 – 0.682 = 99.32
ma =
w1 =
P aV
RaT
mv
ma
=
=
99.32×100
= 120.1 kg
0.287×288.15
0.513
= 0.0043
120.1
Tdew is T when Pv = Pg = 0.682 kPa;
T = 1.4 oC
12.37 The products of combustion are flowing through a SSSF heat exchanger with 12%
CO2, 13% H2O and 75% N2 on a volume basis at the rate 0.1 kg/s and 100 kPa.
What is the dew-point temperature? If the mixture is cooled 10°C below the dewpoint temperature, how long will it take to collect 10 kg of liquid water?
yH
2O
= 0.13; PH2O = 0.13×100 = 13 kPa, TDEW = 50.95 oC
Cool to 40.95 oC → PG = 7.805 kPa
yH
2O
nH
= 7.805/100 = nH2O(v)/(nH2O(v) + 0.87)
2O(v)
= 0.07365 per kmol mix in
→ nLIQ = 0.13 - 0.07365 = 0.05635
MMIX IN = 0.12×44.01 + 0.13×18.015 + 0.75×28.013 = 28.63 kg/kmol
.
.
nMIX IN = mTOTAL/MMIX IN = 0.1/28.63 = 0.003493 kmol/s
.
nLIQ COND = 0.003 493×0.05635 = 0.000 197 kmol/s
.
or mLIQ COND = 0.000 197 × 18.015 = 0.003 55 kg/s
For 10 kg, takes ~ 47 minutes
12-24
12.38 A new high-efficiency home heating system includes an air-to-air heat exchanger
which uses energy from outgoing stale air to heat the fresh incoming air. If the
outside ambient temperature is −10°C and the relative humidity is 30%, how
much water will have to be added to the incoming air, if it flows in at the rate of 1
m3/s and must eventually be conditioned to 20°C and 40% relative humidity?
Outside ambient air:
PV1 = φ1PG1 = 0.30×0.2602 = 0.078 kPa
Assuming P1 = P2 = 100 kPa, => PA1 = 100 - 0.078 = 99.922 kPa
.
PA1V1
99.922×1
0.078
.
mA =
=
= 1.3228 kg/s w1 = 0.622×
= 0.00049
99.922
RAT1 0.287×263.2
Conditioned to :
T2 = 20 oC, φ2 = 0.40
PV2 = φ2PG2 = 0.40×2.339 = 0.9356 =>
w2 = 0.622×
0.9356
= 0.00587
99.064
.
.
mLIQ IN = mA(w2-w1) = 1.3228(0.00587 - 0.00049) = 0.00712 kg/s = 25.6 kg/h
12.39 Consider 100 m3 of atmospheric air at 100 kPa, 25°C, and 80% relative humidity.
Assume this is brought into a basement room where it cools to 15°C, 100 kPa.
How much liquid water will condense out?
Pg = Psat25 = 3.169 kPa => Pv = φ Pg = 0.8 × 3.169 = 2.535 kPa
P vV
2.535×100
mv1 =
=
= 1.844 kg
RvT 0.461×298.15
Pg15°C = 1.705 < Pv1 => State 2 is saturated φ2 = 100% , Pv2 = Pg2 = 1.705
P vV
1.705×100
mv2 =
=
= 1.2835 kg
RvT 0.461×288.15
mliq = mv1 - mv2 = 1.844 – 1.2835 = 0.56 kg
12.40 A flow of 2 kg/s completely dry air at T1, 100 kPa is cooled down to 10°C by
spraying liquid water at 10°C, 100 kPa into it so it becomes saturated moist air at
10°C. The process is SSSF with no external heat transfer or work. Find the exit
moist air humidity ratio and the flow rate of liquid water. Find also the dry air
inlet temperature T1.
2: saturated
Pv = Pg = 1.2276 kPa
and
hfg (10°C)= 2477.7 kJ/kg
w2 = 0.622 × 1.2276/ (100 - 1.2276) = 0.00773
C.V. Box
.
.
.
.
.
ma + mliq = m a(1 + w2)
=> mliq = w2 ma = 0.0155kg/s
.
.
.
ma ha1 - mliq hf = ma (ha2 + w2 hg2)
ha1 - ha2 = Cpa (T1- T2) = w2 hg2- w2 hf = w2 hfg = 19.15
=>
T1 = 29.1°C
12-25
12.41 A piston/cylinder has 100 kg of saturated moist air at 100 kPa, 5°C. If it is heated
to 45°C in an isobaric process, find 1Q2 and the final relative humidity. If it is
compressed from the initial state to 200 kPa in an isothermal process, find the
mass of water condensing.
Energy Eq.:
m(u2-u1) = Q12 - W12,
Initial state 1: φ1 = 100%, h̃1 = 39 kJ/kg dry air, w1 = 0.054
ma = mtot/(1+ w1) = 99.463 kg, mv1 = w1ma = 0.537 kg
Case a: P = constant => W12 = mP(v2-v1) =>
Q12 = m(u2-u1) + W12 = m(h2-h1) = ma(h̃2 - h̃1)
State 2: w2 = w1, T2 => h̃2 = 79 , φ2 = 9.75%
Q12 = 99.463 (79 - 39) = 3979 kJ.
Case b: T = constant & φ2 = 100% => Pv = Pg = 0.8721 kPa
w2 = 0.622 Pv2/Pa2 = 0.622 Pg2/(Ptot2-Pg2) = 0.002697
mv2 = w2 ma = 0.268 kg, mliq = mv1 - mv2 = 0.269 kg
12.42 A flow moist air at 100 kPa, 40°C, 40% relative humidity is cooled to 15°C in a
constant pressure SSSF device. Find the humidity ratio of the inlet and the exit
flow, and the heat transfer in the device per kg dry air.
.
.
.
C.V. Cooler.
mv1 = mliq + mv2
Psychrometric chart: State 2: T < Tdew = 23°C => φ2 = 100%
. .
. .
~
mv2/ma = ω2 = 0.0107 , h̃2 = 62
mv1/ma = ω1 = 0.018, h1 = 106;
.
.
mliq/ma = ω1- ω2 = 0.0073 , h̃f = 62.98
.
.
.
.
m q- = m h̃ - m h - m h̃ =>
a
out
a 1
liq f
a
2
q-out = h̃1 - (ω1- ω2) hf - h̃2 = 106 – 0.0073 × 62.98 – 62
= 43.54 kJ/kg-dry air
Tables:
Pg1 = 7.384 kPa
ω1 = 0.0189
Pv1 = 2.954
Pv2 = 1.705 kPa = Pg2 => ω2 = 0.0108
hv1 = 2574.3 ,
hv2 = 2528.9 ,
hf = 62.98
q-out = CP (T1 - T2) + ω1hv1 - ω2 hv2 - (ω1- ω2) hf
= 0.717(40 - 15) + 0.0189 × 2574.3 - 0.0108 × 2528.9 - 0.0073 × 62.98
= 38.8 kJ/kg dry air
12-26
12.43 Ambient moist air enters a steady-flow air-conditioning unit at 102 kPa, 30°C,
with a 60% relative humidity. The volume flow rate entering the unit is 100 L/s.
The moist air leaves the unit at 95 kPa, 15°C, with a relative humidity of 100%.
Liquid condensate also leaves the unit at 15°C. Determine the rate of heat transfer
for this process.
State 1:
PV1 = φ1PG1 = 0.60×4.246 = 2.5476
w1 = 0.622 × 2.5476/(102 - 2.5476) = 0.01593
.
PA1V1
99.45×0.1
.
=
= 0.1143 kg/s
mA =
RAT1 0.287×303.2
PV2 = PG2 = 1.705, w2 = 0.622 × 1.705/(95 - 1.705) = 0.01137
.
.
.
.
.
.
Energy Eq.:
QCV + mAhA1 + mV1hV1 = mAhA2 + mV2hA2 + m3hL3
.
.
QCV/mA = CP0A(T2-T1) + w2hV2 - w1hV1 + (w1-w2)hL3
= 1.004(15-30) + 0.01137×2528.9 - 0.01593×2556.2
+ 0.00456×63.0 = -26.732 kJ/kg air
.
Q CV = 0.1143(-26.73) = -3.055 kW
12.44 A steady supply of 1.0 m3/s air at 25°C, 100 kPa, 50% relative humidity is needed
to heat a building in the winter. The outdoor ambient is at 10°C, 100 kPa, 50%
relative humidity. What are the required liquid water input and heat transfer rates
for this purpose?
Air: R = 0.287 kJ/kg K, Cp = 1.004 kJ/kg-K
State 1: T1 = 10°C,
φ1 = 50%, P1 = 100 kPa
Pg1= 1.2276 kPa,
Pv1= φ1Pg1 = 0.6138 kPa,
Pa1 = P1- Pv1 = 99.39 kPa => ω1 = 0.622 Pv1/Pa1 = 0.003841
.
State 2: T2 = 25°C, P2 = 100 kPa, φ2 = 50%, V a2 = 1 m3/s
Pg2 = 3.169 kPa,
Pv2= φ2Pg2 = 1.5845 kPa,
P = P2 - Pv2 = 98.415 kPa, ω2 = 0.622 Pv2/Pa2 = 0.010014
. a2
.
m a2 = Pa2 V a2/RTa2 = 98.415 × 1/(0.287 × 298.15) = 1.15 kg/s
Steam tables B.1.1:
hv1 = 2519.7 kJ/kg,
hv2 = 2547.2 kJ/kg
State 3: Assume: Liq. Water at T3 = 25°C, hf3 = 104.9 kJ/kg
.
.
.
.
.
Conservation of Mass: ma1 = ma2, mf3 = mv2 - mv1
.
.
mf3= ma2(ω2 - ω1) = 1.15 × 0.006173 = 0.0071 kg/s
.
.
.
.
.
.
1stLaw:
Q + ma1ha1 + mv1hv1 + mf3hf3 = ma2ha2 +mv2hv2
.
.
mf3
.
Q
. = Cp(T2- T1) + ω2hv2 - ω1hv1 - . hf3 =>
Q = 34.76 kW
ma
ma
12-27
12.45 Consider a 500-L rigid tank containing an air–water vapor mixture at 100 kPa,
35°C, with a 70% relative humidity. The system is cooled until the water just
begins to condense. Determine the final temperature in the tank and the heat
transfer for the process.
Pv1 = φPG1 = 0.7×5.628 = 3.9396 kPa
Since mv = const & V = const & also Pv = PG2:
PG2 = Pv1× T2/T1 = 3.9396× T2/308.2 = 0.01278 T2
Assume T2 = 30oC:
0.01278×303.2 = 3.875 =/ 4.246 = PG 30C
Assume T2 = 25oC:
0.01278×298.2 = 3.811 =/ 3.169 = PG 25C
interpolating → T2 = 28.2 oC
w2 = w1 = 0.622
3.9396
= 0.025 51
(100-3.9369)
ma = Pa1V/RaT1 = (100-3.94)×0.5/(0.287×308.2) = 0.543 kg
1st law: Q12 = U2-U1 = ma(ua2-ua1) + mv(uv2-uv1)
= 0.717(28.2 - 35) + 0.02551 (2414.2 - 2423.4) = -5.11 kJ/kg
→ Q12 = 0.543(-5.11) = -2.77 kJ
12.46 Air in a piston/cylinder is at 35°C, 100 kPa and a relative humidity of 80%. It is
now compressed to a pressure of 500 kPa in a constant temperature process. Find
the final relative and specific humidity and the volume ratio V2/V1.
Check to see if the second state is saturated or not. First assume no water is
condensed
1: w1 = 0.029
2: w2 = 0.622 Pv2/(P2-Pv2)
w2 = w1 => Pv2 = 22.568 > Pg = 5.628 kPa
Conclusion is state 2 is saturated
φ2 = 100%, w2 = 0.622 Pg/(P2-Pg) = 0.00699
To get the volume ratio, write the ideal gas law for the vapor phases
V2 = Va2 + Vv2 + Vf2 = (maRa + mv2Rv)T/P2 + mliqvf
V1 = Va1 + Vv1 = (maRa + mv1Rv)T/P1
Take the ratio and devide through with maRaT/P2 to get
V2/V1 = (P1/P2)[1 + 0.622w2 + (w1-w2)P2vf/RaT] / [1+0.622 w1]
= 0.1973
The liquid contribution is nearly zero (=0.000126) in the numerator.
12-28
12.47 A 300-L rigid vessel initially contains moist air at 150 kPa, 40°C, with a relative
humidity of 10%. A supply line connected to this vessel by a valve carries steam
at 600 kPa, 200°C. The valve is opened, and steam flows into the vessel until the
relative humidity of the resultant moist air mixture is 90%. Then the valve is
closed. Sufficient heat is transferred from the vessel so the temperature remains at
40°C during the process. Determine the heat transfer for the process, the mass of
steam entering the vessel, and the final pressure inside the vessel.
i
Pv1 = φ1PG1 = 0.1×7.384 = 0.7384 kPa
H 2O
Pv2 = 0.9×7.384 = 6.6456 kPa
Pa2 = Pa1 = 150 - 0.738 = 149.262 kPa
AIR
+
H 2O
w1 = 0.622 ×
0.7384
= 0.003 08
149.262
w2 = 0.622 ×
6.6456
= 0.0277
149.262
ma = 149.262×0.3/(0.287×313.2) = 0.5 kg
P2 = 149.262 + 6.6456 = 155.9 kPa
mvi = 0.5(0.0277 - 0.00308) = 0.0123 kg
uv1 = uv2 ≈ uG at 40 oC and
ua2 = ua1
QCV = ma(ua2 - ua1) + mv2uv2 - mv1uv1 - mvihi
= mvi(uG at T - hi) = 0.0123(2430.1 - 2850.1) = -5.15 kJ
12-29
12.48 A combination air cooler and dehumidification unit receives outside ambient air
at 35°C, 100 kPa, 90% relative humidity. The moist air is first cooled to a low
temperature T2 to condense the proper amount of water, assume all the liquid
leaves at T2. The moist air is then heated and leaves the unit at 20°C, 100 kPa,
relative humidity 30% with volume flow rate of 0.01 m3/s. Find the temperature
T2, the mass of liquid per kilogram of dry air and the overall heat transfer rate.
MIX
IN
1
2
COOL
MIX
OUT
HEAT
2'
.
-Q C LIQ
OUT
CV
.
QH
a) Pv1 = φ1PG1 = 0.9 × 5.628 = 5.0652 kPa
w1 = 0.622 ×
5.0652
= 0.033 19
100-5.0652
Pv3 = φ3PG3 = 0.3 × 2.339 = 0.7017 kPa
0.7017
w2 = w3 = 0.622 ×
= 0.0044
100-0.7017
.
.
mLIQ 2′/ma = w1 - w2 = 0.033 19 - 0.0044 = 0.028 79 kg/kg air
PG2 = Pv3 = 0.7017 kPa
→
T2 = 1.7 oC
b) For a C.V. around the entire unit
.
.
.
QCV = QH + QC
Net heat transfer, 1st law:
.
.
.
.
QCV/ma = (ha3-ha1) + w3hv3 - w1hv1 + mL2′ hL2′/ma
= 1.004(20-35) + 0.0044×2538.1 - 0.033 19×2565.3 + 0.028 79×7.28
= -88.82 kJ/kg air
.
Pa3V3 (100-0.7017)×0.01
.
ma =
=
= 0.0118 kg/s
RaT3
0.287×293.2
.
QCV = 0.0118(-88.82) = -1.05 kW
12-30
12.49 A rigid container, 10 m3 in volume, contains moist air at 45°C, 100 kPa, φ = 40%.
The container is now cooled to 5°C. Neglect the volume of any liquid that might be
present and find the final mass of water vapor, final total pressure and the heat transfer.
CV container.
m 2 = m1 ;
m2u2 - m1u1 = 1Q2
1: 45°C, φ = 40% => w1 = 0.0236 ,
Tdew = 27.7°C
Final state T2 < Tdew so condensation, φ2 = 100%
Pv1 = 0.4 Pg = 0.4*9.593 = 3.837 kPa,
Pa1 = Ptot - Pv1 = 97.51 kPa
ma = Pa1V/RT1 = 10.679 kg, mv1 = w1 ma = 0.252 kg
Pv2 = Pg2 = 0.8721 kPa,
Pa2 = Pa1T2/T1 = 85.25 kPa
P2 = Pa2 + Pv2 = 86.12 kPa
mv2 = Pv2V/RvT2 = 0.06794 kg
(=V/vg= 0.06797 steam table)
mf2 = mv1 - mv2 = 0.184 kg
The heat transfer from the energy equation becomes
1Q 2
= ma(u2-u1)a + mv2ug2 + mf2uf2 - mv1ug1
= ma Cv(T2-T1) + mv2 2382.3 + mf2 20.97 - mv1 2436.8
= -306.06 + 161.853 + 3.858 - 614.07 = - 754.4 kJ
12.50 A saturated air-water vapor mixture at 20 oC, 100 kPa, is contained in a 5-m3
closed tank in equilibrium with 1 kg of liquid water. The tank is heated to 80oC. Is
there any liquid water in the final state? Find the heat transfer for the process.
a) Since VLIQ = mLIQvF ≈ 0.001 m3, VGAS ≈ V
AIR
+
VAP
LIQ
ma =
Pa1V
RaT1
At state 2:
φ1 = 1.00 → Pv1 = PG1 = 2.339 kPa
Q12
w1 = 0.622× 2.339 /(100 - 2.339) = 0.0149
=
97.661×4.999
= 5.802 kg
0.287×293.2
Pa2 = 97.661×
=>
mv1 = w1ma = 0.086 kg
353.2 4.999
×
= 117.623 kPa
293.2
5
wMAX 2 = 0.622 × 47.39 / 117.623 = 0.2506
But
w2 ACTUAL =
0.086 + 1.0
= 0.1872 < wMAX 2
5.802
→ No liquid at 2
Q12 = ma (ua2-ua1) + mv2 uv2 - mv1uv1 - mliq 1uliq 1
= 5.802 × 0.717(80 - 20) + 1.086 × 2482.2 - 0.086 × 2402.9 - 1 × 84.0
= 249.6 + 2695.7 - 206.65 - 84 = 2655 kJ
12-31
12.51 An air-water vapor mixture enters a steady flow heater humidifier unit at state 1:
10°C, 10% relative humidity, at the rate of 1 m3/s. A second air-vapor stream
enters the unit at state 2: 20°C, 20% relative humidity, at the rate of 2 m3/s.
Liquid water enters at state 3: 10°C, at the rate of 400 kg per hour. A single airvapor flow exits the unit at state 4: 40°C. Calculate the relative humidity of the
exit flow and the rate of heat transfer to the unit.
Assume: P = 100 kPa
.
State 1: T1 = 10°C, φ1 = 10%, Va1 = 1 m3/s
Pv1= φ1Pg1 = 0.1228 kPa,
Pa1 = P - Pv1 = 99.877 kPa
.
Pv1
Pa1Va1
.
ω1 = 0.622
= 0.000765, ma1 =
= 1.2288 kg/s
Pa1
RTa1
.
.
mv1 = ω1m a1 = 0.00094 kg/s, hv1 = hg1 = 2519.7 kJ/kg
.
State 2: T2 = 20°C, φ2 = 20%, Va2 = 2 m3/s
Pg1= 1.2276 kPa,
Pv2= φ Pg2 = 0.4677 kPa, Pa2 = P - Pv2 = 99.532 kPa
.
Pv2
Pa2 V a2
.
ω2 = 0.622
= 0.002923, m a2 =
= 2.3656 kg/s
Pa2
RTa2
.
.
mv2 = ω2ma2 = 0.00691 kg/s, hv2 = hg2 = 2538.1 kJ/kg
.
State 3: Liquid. T3 = 10°C, mf3 = 400 kg/hr = 0.1111 kg/s, hf3 = 42 kJ/kg
Pg2 = 2.3385 kPa,
State 4: T4 = 40°C
.
.
.
m a4 = m a2 + m a1 = 3.5944,
.
.
.
.
mv4 = mv1 + mv2 + mf3 = 0.11896 kg/s
Continuity Eq. air:
Continuity Eq. water:
.
m v4
Pv4
ω4 = . = 0.0331 = 0.622
m a4
P-Pv4
Pg4 = 7.384 kPa,
φ4 =
Pv4
Pg4
→ Pv4 = 5.052 kPa
= 0.684, hv4 = hg4 = 2574.3 kJ/kg
.
.
.
.
.
.
.
.
1stLaw: Q + ma1ha1 + mv1hv1 + ma2ha2 +mv2hv2 + mf3hf3 = ma4ha4 +mv4hv4
.
Q = 1.004(3.5944 × 40 - 1.2288 × 10 - 2.3656 × 20) + 0.11896 × 2574.3
- 0.00094 × 2519.7 - 0.00691 × 2538.1 - 0.1111 × 42.0
= 366 kW
12-32
12.52 In a hot and dry climate, air enters an air-conditioner unit at 100 kPa, 40°C, and
5% relative humidity, at the steady rate of 1.0 m3/s. Liquid water at 20°C is
sprayed into the air in the AC unit at the rate 20 kg/hour, and heat is rejected from
the unit at at the rate 20 kW. The exit pressure is 100 kPa. What are the exit
temperature and relative humidity?
.
State 1: T1 = 40°C, P1 = 100 kPa, φ1 = 5%, Va1 = 1 m3/s
Pg1= 7.3837 kPa, Pv1= φ1Pg1 = 0.369 kPa, Pa1 = P- Pv1 = 99.63 kPa
.
Pv1
Pa1Va1
.
ω1 = 0.622
= 0.0023, ma1 =
= 1.108 kg/s, hv1 = 2574.3 kJ/kg
Pa1
RTa1
.
State 2 : Liq. Water. 20°C, mf2 = 20 kg/hr = 0.00556 kg/s, hf2 = 83.9 kJ/kg
.
.
.
.
.
Conservation of Mass: ma1 = ma3, mv1 + m12 = mv3
.
.
ω3 = (mf2 / ma1) + ω1 = ( 0.00556/1.108 ) + 0.0023 = 0.0073
State 3 : P3 = 100 kPa and Pv3 = P3ω3/(0.622 + ω3) = 1.16 kPa
.
.
.
.
.
.
.
1stLaw: Q + ma1ha1 + mv1hv1 + mf2hf2 = ma3ha3 +mv3hv3; Q = - 20 kW
.
. .
(ha3-ha1) + ω3hv3 = ω1hv1 + (mf2hf2 - Q )/ma1 = 24.39; Unknowns: ha3, hv3
Trial and Error for T3; T3 = 301.6 K, Pg3 = 3.901 kPa , φ3 =
Pv3
Pg3
= 0.297
12.53 A water-filled reactor of 1 m3 is at 20 MPa, 360°C and located inside an insulated
containment room of 100 m3 that contains air at 100 kPa and 25°C. Due to a
failure the reactor ruptures and the water fills the containment room. Find the final
pressure.
CV Total container.
mv(u2-u1) + ma(u2-u1) = 1Q2 - 1W2 = 0
Initial water:
Initial air:
v1 = 0.0018226, u1 = 1702.8, mv = V/v = 548.67 kg
ma = PV/RT = 100×99/0.287×298.2 = 115.7 kg
Substitute into energy equation
548.67 (u2 - 1702.8) + 115.7×0.717 (T2 - 25) = 0
u2 + 0.1511 T2 = 1706.6
Trial and error 2-phase
T = 150 LHS = 1546
&
v2 = V2/mv = 0.18226 m3/kg
(T guess,
v2 => x2 => u 2 => LHS)
T = 160 LHS = 1820.2
T = 155 LHS = 1678.1 => T = 156 C LHS = 1705.7 OK
x2 = 0.5372, Psat = 557.5 kPa
Pa2 = Pa1V1T2/V2T1 = 100× 99 × 429.15 / (100×298.15) = 142.5 kPa
=>
P2 = Pa2 + Psat = 700 kPa.
12-33
12.54 Use the psychrometric chart to find the missing property of: φ, ω, Twet, Tdry
a. Tdry = 25°C, φ =80%
b. Tdry =15°C, φ =100%
c.Tdry = 20°C, and ω = 0.008
d. Tdry = 25°C, Twet = 23°C
a. 25°C, φ = 80%
=> ω = 0.016; Twet = 22.3°C
b. 15°C, φ = 100%
=> ω = 0.0106; Twet = 15°C
c. 20°C, ω = 0.008
=> φ = 57%; Twet = 14.4°C
d. 25°C, Twet = 23°C
=>
ω = 0.017; φ = 86%
12.55 Use the psychrometric chart to find the missing property of: φ, ω, Twet, Tdry
a. φ = 50%, ω = 0.012
b. Twet =15°C, φ = 60%.
c. ω = 0.008 and Twet =17°C
d. Tdry = 10°C, ω = 0.006
a. φ = 50%, ω = 0.012
=> Tdry = 23.5°C, Twet = 20.6 °C
b. Twet = 15°C, φ =60%
=> Tdry = 20.2°C, ω = 0.0086
c. ω = 0.008, Twet =17°C => Tdry = 27.2°C, φ = 37%
d. Tdry= 10°C, ω = 0.006 => φ = 80%, Twet = 8.2°C
12.56 For each of the states in Problem 12.55 find the dew point temperature.
The dew point is the state with the same humidity ratio (abs humidity ω) and
completely saturated φ = 100%. From psychrometric chart:
a.
Tdew = 16.8°C
c. Tdew = 10.9°C
b.
Tdew = 12°C
d. Tdew = 6.5°C
Finding the solution from the tables is done for cases a,c and d as
Eq.12.21 solve:
Pv = Pg = ωPtot /[ω + 0.622] = Psat(Tdew) in B.1.1
For case b use energy Eq. 12.23 to find ω1 first from Tad sat = Twet.
12-34
12.57 One means of air-conditioning hot summer air is by evaporative cooling, which is
a process similar to the SSSF adiabatic saturation process. Consider outdoor
ambient air at 35°C, 100 kPa, 30% relative humidity. What is the maximum
amount of cooling that can be achieved by such a technique? What disadvantage
is there to this approach? Solve the problem using a first law analysis and repeat it
using the psychrometric chart, Fig. F.5.
Cooled
Ambient
Air
1
2
Air
P1 = P2 = 100 kPa
T1 = 35 oC, φ1 = 30%
Pv1 = φ1Pg1 = 0.30×5.628 = 1.6884
3
Liquid
ω1 = 0.622×1.6884/98.31 = 0.01068
For adiabatic saturation (Max. cooling is for φ2 = 1), 1st law, Eq.12.23
ω1 (hv1 - hf2) = Cp(T2 - T1) + ω2 hfg2
φ2 = 1
&
ω2 = 0.622 × PG2/(P2 - PG2)
Only one unknown: T2 . Trial and error on energy equation:
CpT2 + ω2 hfg2 + ω1 hf2 = CpT1 + ω1hv1
= 1.004 × 35 + 0.01068 × 2565.3 = 62.537
T2 = 20 oC: PG2 = 2.339, hf2 = 83.94 , hfg2 = 2454.12
=> ω2 = 0.622 × 2.339/ 97.661 = 0.0149
LHS = 1.004 × 20 + 0.0149 × 2454.1 + 0.01068 × 83.94 = 57.543
T2 = 25 oC: PG2 = 3.169, hf2 = 104.87 , hfg2 = 2442.3
=> ω2 = 0.622 × 3.169/ 96.831 = 0.02036
LHS = 1.004 × 25 + 0.02036 × 2442.3 + 0.01068 × 104.87 = 75.945
linear interpolation:
b)
T2 = 21.4 oC
chart F.5 : Ad. sat. T ≈ WetBulbTemperature ≈ 21.5 oC
12-35
12.58 Use the formulas and the steam tables to find the missing property of: φ, ω, and
Tdry, total pressure is 100 kPa; repeat the answers using the psychrometric chart
a. φ = 50%, ω = 0.010
b. Twet =15°C, φ = 50%
c. Tdry = 25°C, Twet = 21°C
a. From Eq.12.21 Pv = P ω /(0.622 + ω) = 100 × 0.01/0.632 = 1.582 kPa
=> T = 25°C
From Eq.12.18 Pg = Pv/φ = 1.582/0.5 = 3.165 kPa
b. Assume Twet is adiabatic saturation T and use energy Eq.12.23
At 15°C: Pg = 1.705 => ω = 0.622 × 1.705/(100 - 1.705) = 0.01079
LHS = ω1 (hv1 - hf2) + CpT1 = RHS = CpT2 + ω2 hfg2
RHS = 1.004×15 + 0.01079 × 2465.93 = 41.667 kJ/kg
ω1 = 0.622 φPg/(100 - φPg)
where Pg is at T1.
LHS25C = 49.98, LHS20C = 38.3
=>
Trial and error.
T = 21.4°C, ω1 = 0.008
c. At 21°C:: Pg = 2.505 => ω2 = 0.622 × 2.505/(100 - 2.505) = 0.016
hf2 = 88.126
and
hfg2 = 2451.76 kJ/kg,
hv1 = 2547.17
From Eq.12.23: ω1 = [Cp(T2-T1) + ω2 hfg2 ]/(hv1 - hf2) = 0.0143
Pv = P ω /(0.622 + ω) = 2.247,
φ = 2.247/3.169 = 0.71
Using the psychrometric chart F.5:
a: Tdry = 25.3 °C
b. Tdry = 21.6°C, ω = 0.008
c: ω = 0.0141, φ = 71-72%
12.59 Compare the weather two places where it is cloudy and breezy. At beach A it is
20°C, 103.5 kPa, relative humidity 90% and beach B has 25°C, 99 kPa, relative
humidity 20%. Suppose you just took a swim and came out of the water. Where
would you feel more comfortable and why?
Your skin being wet and air is flowing over it will feel Twet. With the small
difference in pressure from 100 kPa use the psychrometric chart.
A: 20°C, φ = 90%
=> Twet = 18.7°C
B: 25°C, φ = 20%
=> Twet = 12.3°C
At beach A it is comfortable, at B it feels chilly.
12-36
12.60 Ambient air at 100 kPa, 30°C, 40% relative humidity goes through a constant
pressure heat exchanger in a SSSF process. In one case it is heated to 45°C and in
another case it is cooled until it reaches saturation. For both cases find the exit
relative humidity and the amount of heat transfer per kilogram dry air.
.
.
.
.
CV heat exchanger:
mAi = mAe, mvi = mve,
we = wi
(ha + whv)i + q = (h a + whv)e = h̃e,
Using the psychrometric chart:
i:
q = h̃e - h̃i
wi = 0.0104, h̃i = 76
Case I) e: Te = 45 oC, we = wi => h̃e = 92,
φe = 17%, q = 92-76 = 16 kJ/kg dry air
Case II) e: w e = wi, φe = 100% => h̃e = 61, Te = 14.5oC
q = 61-76 = -15 kJ/kg dry air
12.61 A flow, 0.2 kg/s dry air, of moist air at 40°C, 50% relative humidity flows from
the outside state 1 down into a basement where it cools to 16°C, state 2. Then it
flows up to the living room where it is heated to 25°C, state 3. Find the dew point
for state 1, any amount of liquid that may appear, the heat transfer that takes place
in the basement and the relative humidity in the living room at state 3.
Solve using psychrometric chart:
a) Tdew = 27.2 (w = w1, φ = 100%)
w1 = 0.0232, h̃1 = 118.2
b) T2 < Tdew so we have φ2 = 100% liquid water appear in the basement.
=> w2 = 0.0114
h̃2 = 64.4 and from steam tbl. hf = 67.17
.
.
mliq = mair (w1-w2) = 0.2(0.0232-0.0114) = 0.00236 kg/s
.
.
.
c) Energy equation: mair h̃1 = mliq hf + mair h̃2 + Qout
Qout = 0.2[118.2 - 64.4 - 0.0118×67.17] = 10.6 kW
d)
w3 = w2 = 0.0114 & 25°C
=>
φ3 = 58%.
If you solve by the formula's and the tables the numbers are:
Pg40 = 7.384 ; Pv1 = 0.5×7.384 = 3.692 kPa
w1 = 0.622 × 3.692 / (100 - 3.692) = 0.02384
Pv1 = Pg (Tdew) => Tdew 1 = 27.5 °C
2: φ = 100%, Pv2 = Pg2 = 1.832 kPa, w2 = 0.622×1.832/98.168 = 0.0116
.
.
mliq = mair (w1-w2) = 0.2×0.01223 = 0.00245 kg/s
3:
w3 = w2 => Pv3 = Pv2 = 1.832 & Pg3 = 3.169
φ3 = Pv/Pg = 1.832/3.169 = 57.8%
12-37
12.62 A flow of air at 5°C, φ = 90%, is brought into a house, where it is conditioned to
25°C, 60% relative humidity. This is done in a SSSF process with a combined
heater-evaporator where any liquid water is at 10°C. Find any flow of liquid, and
the necessary heat transfer, both per kilogram dry air flowing. Find the dew point
for the final mixture.
CV heater and evaporator. Use psychrometric chart.
Inlet: w1 = 0.0048, h̃1 = 37.5, hf = 42.01
Exit: w2 = 0.0118, h̃2 = 75, Tdew = 16.5°C
From these numbers we see that water and heat must be added. Continuity eq.
and energy equation give
.
.
mLIQ IN/mA = w2 - w1 = 0.007 kg/kg dry air
q = h̃2 - h̃1 - (w2-w1)hf = 37.3 kJ/kg dry air
12.63 Atmospheric air at 35°C, relative humidity of 10%, is too warm and also too dry.
An air conditioner should deliver air at 21°C and 50% relative humidity in the
amount of 3600 m3 per hour. Sketch a setup to accomplish this, find any amount
of liquid (at 20°C) that is needed or discarded and any heat transfer.
CV air conditioner. Check from psychrometric chart.
Inlet:
w1 = 0.00305, h̃mix,1 = 62.8, hf,20 = 83.96
Exit:
w2 = 0.0076, h̃mix,2 = 60.2
Water must be added. Continuity and energy equations
.
.
.
.
.
.
.
mA(1+w1) + mliq = mA(1+w2); mAh̃1mix + mliqhf + QCV = mAh̃2mix
.
.
mA = PaV/RT = (101.325 - 1.25)1/0.287×294.15 = 1.1845 kg/s
.
.
mliq = mA(w2 - w1) = 0.00539 kg/s = 19.4 kg/h
.
QCV = 1.1845(60.2 - 62.8 - 0.00455×83.96) = - 3.53 kW
Liquid water
Cooler
Exit
Inlet
12-38
12.64 In a car’s defrost/defog system atmospheric air, 21°C, relative humidity 80%, is
taken in and cooled such that liquid water drips out. The now dryer air is heated to
41°C and then blown onto the windshield, where it should have a maximum of
10% relative humidity to remove water from the windshield. Find the dew point
of the atmospheric air, specific humidity of air onto the windshield, the lowest
temperature and the specific heat transfer in the cooler.
Solve using the psychrometric chart
Air inlet: T = 21°C, φ = 80%
=> w1 = 0.0124, Tdew = 17.3°C
Air exit: T = 41°C, φ = 10%
=> w3 = 0.0044, Tdew = 1.9°C
To remove enough water we must cool to the exit T dew, followed by heating
to Tex. The enthalpies are
h̃1 = 72,
h̃2 = 32.5, hf(1.9°C) = 8
CV cooler:
.
.
mliq /mair = w1-w3 = 0.0124 - 0.0044 = 0.008 kg liq/kg air
.
q = Q/mair = h̃2 + (w1 - w3) hf - h̃1
= 32.5 + 0.008×8 - 72 = -39.4 kJ/kg dry air
If the steam and air tables are used the numbers are
1: Pg1 = 2.505, Pv1 = 2.004 => w1 = 0.01259
hg1 = 2539.9, ha1 = 294.3 => h̃1 = 326.3
3: Pg3 = 7.826, Pv3 = 0.783 => w3 = 0.00486
2: wg3 = w3 => T2 = T3dew = 3.3°C, hf2 = 13.77
hg2 = 2507.4, ha2 = 276.56 => h̃2 = 288.75
.
.
mliq /mair = 0.00773, q = 288.75 + 0.00773× 13.77 - 326.3 = -37.45 kJ/kg air
12.65 Two moist air streams with 85% relative humidity, both flowing at a rate of 0.1
kg/s of dry air are mixed in a SSSF setup. One inlet flowstream is at 32.5°C and
the other at 16°C. Find the exit relative humidity.
1: w1 = 0.0266, h̃1 = 120
2: w2 = 0.0094, h̃2 = 60
CV mixing chamber.
Continuity Eq. water:
Energy Eq.:
.
.
.
mair w1 + mair w2 = 2mair wex;
.
.
.
mair h̃1 + mair h̃2 = 2mair h̃ex
wex = (w1 + w2)/2 = 0.018 ;
exit: wex, h̃ex
=>
h̃ex = (h̃1 + h̃2)/2 = 90
Tex = 24.5°C, φ = 94%
12-39
12.66 A flow of moist air at 21°C, 60% relative humidity should be produced from
mixing of two different moist air flows. Flow 1 is at 10°C, relative humidity 80%
and flow 2 is at 32°C and has Twet = 27°C. The mixing chamber can be followed
by a heater or a cooler. No liquid water is added and P = 100 kPa. Find the two
controls one is the ratio of the two mass flow rates ma1/ma2 and the other is the
heat transfer in the heater/cooler per kg dry air.
1:
w1 = 0.006 , h̃ 1 = 45
2:
w2 = 0.0208 , h̃ 2 = 105
4:
w4 = 0.0091 , h̃ 4 = 64 , Tdew 4 = 12.5°C
C.V : Total Setup
.
.
.
.
ma1 w1 + ma2 w2 = (ma1 + ma2) w4
.
.
.
.
.
ma1 h̃ 1 + ma2 h̃ 2 + Qa1 = (ma1 + ma2) h̃ 4
w4- w2
.
.
x = ma1/ma2 => x w1 + w2 = (1+x) w4 => x =
= 3.773
w -w
1
4
.
.
.
q̃ = Qa1/(ma1 + ma2) = h̃ 4 - [x/(1+x)] h̃ 1 - [1/(1+x)] h̃2
= 64 – 0.7905 × 45 − 0.2095 × 105 = 6.43 kJ/kg-dry air
12.67 Consider two states of atmospheric air. (1) 35°C, Twet = 18°C and (2) 26.5°C, φ =
60%. Suggest a system of devices that will allow air in a SSSF process to change
from (1) to (2) and from (2) to (1). Heaters, coolers (de)humidifiers, liquid traps
etc. are available and any liquid/solid flowing is assumed to be at the lowest
temperature seen in the process. Find the specific and relative humidity for state
1, dew point for state 2 and the heat transfer per kilogram dry air in each
component in the systems.
Use the psychrometric chart
1: w1= 0.006, h̃1 = 70.5, φ1 = 18%, Tdew = 6.5°C, h̃dew = 42
2: w2= 0.013, h̃2 = 79.4, φ2 = 60%, Tdew = 18°C, h̃dew = 71
Since w2 > w1 water must be added in process I to II and removed in the
process II to I. Water can only be removed by cooling below dew point
temperature so
I to II: Adiab. sat I to Dew,II, then heater from Dew,II to II
II to I: Cool to Dew,I then heat Dew,I to I
The first one can be done because Tdew II = Tad sat I
I to II: q = h̃II - h̃dewII = 79.4 - 71 = 8.4 kJ/kg air
II to I: qcool = h̃II - h̃dewI - (w2-w1)hf(at TdewI)
= 79.4 - 0.007 × 27.29 = 37.2 kJ/kg air
qheat = h̃I - h̃dewI = 70.5 - 42 = 28.5 kJ/kg air
12-40
12.68 An insulated tank has an air inlet, ω1 = 0.0084, and an outlet, T2 = 22°C, φ2 =
90% both at 100 kPa. A third line sprays 0.25 kg/s of water at 80°C, 100 kPa. For
a SSSF operation find the outlet specific humidity, the mass flow rate of air
needed and the required air inlet temperature, T 1.
Take CV tank in SSSF. Continuity and energy equations are:
.
.
.
.
.
.
Continuity: m3 + ma w1 = ma w2;
Energy Eq.: m3hf + ma h̃1 = ma h̃2
All state properties are known except T1. From the psychrometric chart we get
State 2: w2 = 0.015, h̃2 = 79.5 State 3: hf = 334.91 (steam tbl)
.
.
ma = m3/(w2-w1) = 0.25/(0.015-0.0084) = 37.88 kg/s
h̃1 = h̃2 - (w2-w1)hf = 79.5 - 0.0066 × 334.91 = 77.3
Chart (w1, h̃1) =>
T1 = 36.5°C
12.69 You have just washed your hair and now blow dry it in a room with 23°C, φ =
60%, (1). The dryer, 500 W, heats the air to 49°C, (2), blows it through your hair
where the air becomes saturated (3), and then flows on to hit a window where it
cools to 15°C (4). Find the relative humidity at state 2, the heat transfer per
kilogram of dry air in the dryer, the air flow rate, and the amount of water
condensed on the window, if any.
The blowdryer heats the air at constant specific humidity to 2 and it then goes
through an adiabatic saturation process to state 3, finally cooling to 4.
1: 23°C, 60% rel hum => w1 = 0.0104, h̃1 = 69
2: w2 = w1, T2 => φ2 = 15%, h̃2 = 95
CV. 1 to 2:
w 2 = w1;
CV. 2 to 3:
q = h̃2 - h̃1 = 95 - 69 = 26 kJ/kg dry air
.
ma = Q/q = 0.5/26 = 0.01923 kg/s
.
.
.
.
.
w3 - w2 = mliq /ma ; ma h̃2 + mliq hf = ma h̃3
3: φ = 100% => T3 = Twet,2 = 24.8°C, w3 = 0.0198
4: φ = 100%, T4 => w4 = 0.01065
.
.
mliq = (w3-w4)ma = (0.0198-0.01065)×0.01923 = 0.176 g/s
If the steam tables and formula's are used then we get
hg1 = 2543.5, hg2 = 2590.3, Pg1 = 2.837, Pv1 = 1.7022,
Pg2 = 11.8, w1 = 0.01077, w2 = w1 , Pv2 = Pv1
φ2 = Pv2/Pg2 = 14.4%, hf3 = 114,
Trial and error for adiabatic saturation temperature.
T3 = 25°C, w3 = 0.02, Pv4 =Pg4 = 1.705 kPa,
w4 = 0.622×1.705/(100-1.705) = 0.0108
12-41
12.70 A water-cooling tower for a power plant cools 45°C liquid water by evaporation.
The tower receives air at 19.5°C, φ = 30%, 100 kPa that is blown through/over the
water such that it leaves the tower at 25°C, φ = 70%. The remaining liquid water
flows back to the condenser at 30°C having given off 1 MW. Find the mass flow
rate of air, and the amount of water that evaporates.
CV Total cooling tower, SSSF.
.
.
Water in air: win + mevap/ma = wex
.
.
.
.
.
Energy: ma h̃in + m1 h45 = ma h̃ex + (m1 - mevap) h30
Inlet: 19.5°C, 30% rel hum => win = 0.0041, h̃in = 50
Exit : 25°C, 70% rel hum => wex = 0.0138, h̃ex = 80
Take the two water flow difference to mean the 1 MW
.
.
.
.
Q = m1 h45 - (m1 - mevap) h30 = 1 MW
.
.
.
ma(h̃ex-h̃in) = ma(80-50) = 1000 kW => ma = 33.33 kg/s
.
.
mevap = (wex - win) ma = 0.0097 × 33.33 = 0.323 kg/s
The needed make-up water flow could be added to give a slightly different
meaning to the 1 MW.
12.71 An indoor pool evaporates 1.512 kg/h of water, which is removed by a
dehumidifier to maintain 21°C, φ = 70% in the room. The dehumidifier, shown in
Fig. P12.71, is a refrigeration cycle in which air flowing over the evaporator cools
such that liquid water drops out, and the air continues flowing over the condenser.
For an air flow rate of 0.1 kg/s the unit requires 1.4 kW input to a motor driving a
fan and the compressor and it has a coefficient of performance, β = QL/ Wc = 2.0.
Find the state of the air as it returns to the room and the compressor work input.
The unit must remove 1.512 kg/h liquid to keep steady state in the room. As
water condenses out state 2 is saturated.
1: 21 °C, 70% => w1 = 0.0108, h̃1 = 68.5
.
.
.
.
CV 1 to 2: mliq = ma(w1 - w2) => w2 = w1 - mliq /ma
qL = h̃1 - h̃2 - (w1 - w2) hf2
w2 = 0.0108 - 1.512/3600×0.1 = 0.0066
2: w2, 100% => T2 = 8°C, h̃2 =45, hf2 = 33.6
qL = 68.5 - 45 - 0.0042×33.6 = 23.36 kJ/kg dry air
. .
CV Total system : h̃3 = h̃1 + Wel/ma - (w1-w2) hf
= 68.5 + 14 - 0.14 = 82.36 kJ/kg dry air
3: w3 = w2, h̃3 => T3 = 46°C,
.
.
Wc = ma qL/ β = 1.165 kW
φ3 = 11-12%
12-42
12.72 To refresh air in a room, a counterflow heat exchanger, see Fig. P12.72, is mounted
in the wall, drawing in outside air at 0.5°C, 80% relative humidity and pushing out
room air, 40°C, 50% relative humidity. Assume an exchange of 3 kg/min dry air in
a SSSF device, and also that the room air exits the heat exchanger to the atmosphere
at 23°C. Find the net amount of water removed from the room, any liquid flow in
the heat exchanger and (T, φ) for the fresh air entering the room.
State 1: w1 = 0.0232, h̃1 = 119.2, Tdew,1 = 27°C
The room air is cooled to 23°C < Tdew1 so liquid will form in the exit flow
channel and state 2 is saturated.
2: 23°C, φ = 100% => w2 = 0.0178, h̃2 = 88, hf2 = 96.52
3: 0.5°C, φ = 80% => w3 = 0.0032, h̃3 = 29.2
CV 1 to 2:
.
.
mliq,2 = ma (w1 - w2) = 3 (0.0232 - 0.0178) = 0.0162 kg/min
.
.
.
CV room: mv,out = ma (w1 - w4) = ma (w1 - w3)
= 3(0.0232-0.0032) = 0.06 kg/min
CV Heat exchanger:
ma(h̃4 - h̃3) = ma(h̃1 - h̃2) - mliqhf2
h̃4 = h̃3 + h̃1 - h̃2 - (w1-w2) hf2 = 29.2 + 119.2 - 88 - 0.0054×96.52
= 59.9 kJ/kg dry air
4: w4 = w3, h̃4 => T4 = 32.5°C, φ = 12%
12-43
12.73 Steam power plants often utilize large cooling towers to cool the condenser
cooling water so it can be recirculated; see Fig. P12.73. The process is essentially
evaporative adiabatic cooling, in which part of the water is lost and must therefore
be replenished. Consider the setup shown in Fig. P12.73, in which 1000 kg/s of
warm water at 32°C from the condenser enters the top of the cooling tower and
the cooled water leaves the bottom at 20°C. The moist ambient air enters the
bottom at 100 kPa, dry bulb temperature of 18°C and a wet bulb temperature of
10°C. The moist air leaves the tower at 95 kPa, 30°C, and relative humidity of
85%. Determine the required mass flow rate of dry air, and the fraction of the
incoming water that evaporates and is lost.
Air + vap.
P4 = 95 kPa
4
1
T1 = 32 oC
T4 = 30 oC
.
φ4 = 0.85
m1 = 1000 kg/s
LIQ H 2O
P2 = 100 kPa
T2 = 18 oC
T3 = 20 oC
3
WBT2 = 10 oC
2
Air + vap.
2'
2
o
18 C
w2′ = 0.622×
LIQ IN
w2 =
(ha2′-ha2) + w2′hFG2′
hv2 - hF 2′
=
φ2′ = 1.0
T2′ = 10 oC,
1.2276
= 0.00773
100-1.2276
1.0035(10-18) + 0.00773×2477.7
= 0.00446
2534.4 - 42.0
3.609
Pv4 = 0.85 × 4.246 = 3.609, w4 = 0.622 ×
= 0.02456
95-3.609
.
.
.
.
.
.
.
m1 + mv2 = m3 + mv4
Cons. mass: ma2 = ma4 = ma,
.
.
.
. .
or
m3 = m1 + ma(w2-w4) and set
r = m1/ma
.
.
.
.
.
.
1st law: m1h1 + maha2 + mv2hv2 = m3h3 + maha4 + mv4hv4
or
r h1 + (ha2-ha4) + w2hv2 = (r + w2 - w4)h3 + w4hv4
r(h1-h3) = ha4 - ha2 + w4hv4 - w2hv2 - (w4-w2)h3
r(134.15-83.96) = 1.004(30-18) + 0.024 56×2556.3
- 0.004 46×2534.4 - 0.0201×83.96
. .
.
r = m1/ma = 1.232 →
ma = 811.7 kg/s
.
.
.
m3 = m1 + ma(w2-w4) = 1000 - 811.7×0.0201 = 983.7
. .
∆m/m1 = 0.0163
12-44
12.74 A semipermeable membrane is used for the partial removal of oxygen from air
that is blown through a grain elevator storage facility. Ambient air (79% nitrogen,
21% oxygen on a mole basis) is compressed to an appropriate pressure, cooled to
ambient temperature 25°C, and then fed through a bundle of hollow polymer
fibers that selectively absorb oxygen, so the mixture leaving at 120 kPa, 25°C,
contains only 5% oxygen. The absorbed oxygen is bled off through the fiber walls
at 40 kPa, 25°C, to a vacuum pump. Assume the process to be reversible and
adiabatic and determine the minimum inlet air pressure to the fiber bundle.
A = N2 ;
0.79 A
+
0.21 B
P1 = ?
1
B = O2
2
3
0.79 A
+
0.0416 B
(0.79/0.95)
MIX
5 %
O2
P2 = 120 kPa
P3 = 40 kPa
All T = 25 oC
0.1684 B
Let sA1 = sB1 = 0 at T = 25 oC & P1
-s
MIX 1 = 0 + 0 - y A1R ln yA1 - yB1R ln yB1
-s
MIX 2 = 0 + 0 - R ln (P 2/P1) - yA2R ln yA2 - yB2R ln yB2
Pure B: -s3 = 0 - R ln (P3/P1)
.
.
.
For n -s = n -s + n -s
1 1
2 2
3 3
[
R -0.8316 ln (P2/P1) - 0.79 ln 0.95 - 0.0416 ln 0.05
]
- 0.1684 ln (P3/P1) + 0.79 ln 0.79 + 0.21 ln 0.21 = 0
0.8316 ln (P2/P1) + 0.1684 ln (P3/P1) = -0.3488 + 4.6025 - ln P1 = -0.3488
P1 min = 141 kPa
For P1 > P1 min →
.
∆S > 0
12-45
12.75 A 100-L insulated tank contains N2 gas at 200 kPa and ambient temperature 25°
C. The tank is connected by a valve to a supply line flowing CO2 at 1.2 MPa, 90°
C. A mixture of 50% N2, 50% CO2 by mole should be obtained by opening the
valve and allowing CO2 flow in to an appropriate pressure is reached and close
the valve. What is the pressure? The tank eventually cools to ambient
temperature. Find the net entropy change for the overall process.
CO 2 i
V = 100 L, P1 = 200 kPa, T1 = T0 = 25 oC
N
Pi = 1.2 MPa, Ti = 90 oC
At state 2: yN2 = yCO2 = 0.50
n2 CO2 = n2 N2 = n1 N2 = P1V/RT1
2
= 200*0.1/8.3145*298.2 = 0.00807 kmol
n2 = 0.01614 kmol
1st law: nihi = n2u-2 - n1u-1 , for const specific heats
niCPoiTi = (niCVoi+n1CVo1)T2 - n1CVo1T1
But ni = n1
→ CPoiTi = CVoiT2 + CVo1(T2-T1)
44.01×0.842×363.2 = 44.01×0.653 T2 + 28.013×0.745(T2-298.2)
T2 = 396.7 K
P2 = n2RT2/V = 0.01614×8.3145×396.7/0.1 = 532 kPa
Cool to T3 = T0 = 298.2 K
P3 = P2× T3/T2 = 532× 298.2/396.7 = 400 kPa
Q23 = n2CVo2(T3-T2) = 0.016 14(0.5×28.013×0.745
+ 0.5×44.01×0.653)(298.2-396.7) = -39.4 kJ
∆SNET = n3-s3 - n1-s1 - ni-si - Q23/T0
= n [(s- ) - -s ] + n [(s- ) - s- ] - Q /T
i
CO2 3
i
1
N2 3
= 0.00807(44.01×0.8418 ln
1
23
0
0.5×400
298.2
- 8.3145 ln
)
363.2
1200
+ 0.00807×( - 8.3145 ln
0.5×400
-39.4
)298.2
200
= +0.0613 + 0 + 0.1321 = +0.1934 kJ/K
12-46
12.76 A cylinder/piston loaded with a linear spring contains saturated moist air at 120
kPa, 0.1 m3 volume and also 0.01 kg of liquid water, all at ambient temperature
20°C. The piston area is 0.2 m2, and the spring constant is 20 kN/m. This cylinder
is attached by a valve to a line flowing dry air at 800 kPa, 80°C. The valve is
opened, and air flows into the cylinder until the pressure reaches 200 kPa, at
which point the temperature is 40°C. Determine the relative humidity at the final
state, the mass of air entering the cylinder and the work done during the process.
P1 = 120 kPa, T1 = 20 oC = T0, V1 = 0.1 m3,
mLIQ 1 = 0.01 kg, AP = 0.2 m2, ks = 20 kN/m
Pi = 800 kPa P2 = 200 kPa
A+V
Ti = 80 oC
i
T2 = 40 oC
2
P2 = P1 + (ks/Ap)(V2-V1)
200 = 120 + (20/0.22)(V2-0.1) → V2 = 0.26 m3
DRY AIR
φ1 = 1.0
mv1 =
(or
Pv1V1
RvT1
=
w1 = 0.622×2.339/117.66 = 0.012 36)
2.339×0.1
= 0.001 73 ( = w1mA1 )
0.461 52×293.2
Assume no liquid at state 2
mv2 = mv1 + mL1 = 0.01173 kg
→ Pv2 =
mvRvT2 0.0011 73×0.461 52×313.2
=
= 6.521 kPa
V2
0.26
a) φ2 =
6.521
= 0.883
7.384
b) mA1 =
mA2 =
PA1V1
RAT1
=
117.66×0.1
= 0.1398
0.287×293.2
193.479×0.26
= 0.5596
0.287×313.2
mAi = mA2 - mA1 = 0.4198 kg
1
1
c) WCV = ⌠
⌡ PdV = (P1+P2)(V2-V1) = (120+200)(0.26-0.1) = 25.6 kJ
2
2
12-47
12.77 Consider the previous problem and additionally determine the heat transfer. Show
that the process does not violate the second law.
a) QCV = mA2hA2 - mA1hA1 - mAihAi + mv2hv2 -mv1hv1 - mL1hL1 - P2V2
+ P1V1 + WCV
= 0.5596×1.004×313.2 - 0.1398×1.004×293.2
- 0.4198×1.004×353.2 + 0.001173×2574.3
- 0.00173×2538.4 – 0.01 ×83.9 - 200×0.26 + 120×0.1 + 25.6
= -3.48 kJ
b) ∆SCV = S2 - S1 = mA2sA2 - mA1sA1 + mvsv2 - mv1sv1 - mL1sL1
∆SSURR = - QCV/T0 - mAisAi
∆SNET = mA2sA2 - mA1sA1 - mAisAi + mvsv2- mv1sv1 - mL1sL1 - QCV/T0
= mA2(sA2-sAi) + mA1(sAi-sA1) + mvsv2- mv1sv1 - mL1sL1 - QCV/T0
sA2 - sAi = 1.004 ln
313.2
193.479
- 0.287 ln
= +0.2866
353.2
800
sAi - sA1 = 1.004 ln
313.2
800
- 0.287 ln
= -0.3633
293.2
117.66
sv2 = sG2 - Rv ln φ2 = 8.2569 - 0.46152 ln 0.883 = 8.3143
sv1 = sG1 - Rv ln φ1 = 8.6671 – 0 = 8.6671
sL1 = sF1 = 0.2966
→ ∆SNET = 0.5596 (+0.2868) + 0.1398 (-0.3633) + 0.01173×8.3143 –
0.00173×8.6671 – 0.01×0.2966 + 3.48/293.2
= +0.201 kJ/K
12-48
12.78 The air-conditioning by evaporative cooling in Problem 12.57 is modified by
adding a dehumidification process before the water spray cooling process. This
dehumidification is achieved as shown in Fig. P12.78 by using a desiccant
material, which absorbs water on one side of a rotating drum heat exchanger. The
desiccant is regenerated by heating on the other side of the drum to drive the
water out. The pressure is 100 kPa everywhere and other properties are on the
diagram. Calculate the relative humidity of the cool air supplied to the room at
state 4, and the heat transfer per unit mass of air that needs to be supplied to the
heater unit.
States as noted on Fig. P12.78, text page 466.
At state 1, 35 oC: PV1 = φ1PG1 = 0.30×5.628 = 1.6884
w1 = 0.622×1.6884/98.31 = 0.010 68
At T3 = 25 oC:
w3 = w2 = w1/2 = 0.00534
Evaporative cooling process to state 4, where T4 = 20oC
As in Eq. 12.23:
w3(hV3-hL4) = CP0A(T4-T3) + w4hFG4
0.005 34(2547.2-83.9) = 1.004(20-25) + w4×2454.2
w4 = 0.0074 = 0.622×PV4/(100-PV4)
PV4 = 1.176 kPa, φ4 = 1.176/2.339 = 0.503
At T5 = 25 oC, w5 = w4 = 0.0074
Evaporative cooling process to state 6, where T6 = 20oC
w5(hV5-hL6) = CP0A(T6-T5) + w6hFG6
0.0074(2547.2-83.9) = 1.004(20-25) + w6×2454.2
For adiabatic heat exchanger,
.
.
.
.
.
mA2 = mA3 = mA6 = mA7 = mA,
=>
w6 = 0.009 47
Also w2 = w3, w6 = w7
hA2 + w2hV2 + hA6 + w6hV6 = hA3 + w3hV3 + hA7 + w7hV7
or
CP0AT7 + w6(hV7-hV6) = CP0A(T2+T6-T3) + w2(hV2-hV3)
1.004 T7 + 0.009 47(hV7 - 2538.1) = 1.004(60 + 20 - 25)
+ 0.005 34(2609.6-2547.2) = 55.526
By trial and error, T7 = 54.7 oC, hV7 = 2600.3
For the heater 7-8, w8 = w7,
. .
Q/mA = CP0A(T8-T7) + w7(hV8-hV7)
= 1.004(80-54.7) + 0.009 47(2643.7-2600.3) = 25.8 kJ/kg dry air
12-49
12.79 A vertical cylinder is fitted with a piston held in place by a pin, as shown in Fig.
P12.79. The initial volume is 200 L and the cylinder contains moist air at 100 kPa,
25°C, with wet-bulb temperature of 15°C. The pin is removed, and at the same
time a valve on the bottom of the cylinder is opened, allowing the mixture to flow
out. A cylinder pressure of 150 kPa is required to balance the piston. The valve is
closed when the cylinder volume reaches 100 L, at which point the temperature is
that of the surroundings, 15°C.
a.
Is there any liquid water in the cylinder at the final state?
b.
Calculate the heat transfer to the cylinder during the process.
c.
Take a control volume around the cylinder, calculate the entropy change
of the control volume and that of the surroundings.
From Fig. F.5, at T1 = 25 oC & WBT1 = 15 oC
a) w1 = 0.006 67
for saturation at state 2, where P2 = PEXT = 150 kPa,
T2 = 15 oC; Max PV2 = PG2 = 1.705 kPa
wMAX 2 = 0.622×1.705/(150-1.705) = 0.00715 > w 1
→ not saturated at 2, no liquid
b) w1 = 0.006 67 = 0.622×
PV1
100-PV1
→ PV1 = 1.061 kPa
mA1 =
(100-1.061)×0.2
= 0.2312 kg, mV1 = w1mA1 = 0.00154 kg
0.287×298.2
mA2 =
(150-1.592)×0.1
= 0.1794 kg, mV2 = w2mA2 = 0.0012 kg
0.287×288.2
WCV = ⌠
⌡ PEXTdV = PEXT(V2-V1) = 150(0.1-0.2) = -15 kJ
QCV = mA2uA2 - mA1uA1 + mV2uV2 - mV1uV1
+ (mAEhAE AVE + mVEhVE AVE) + WCV
= 0.1794×0.717×288.2 - 0.2312×0.717×298.2
+ 0.0012×2396.0 - 0.00154×2409.8
+ 0.0518×1.004(298.2 +288.2)/2
+ 0.00034(2547.2+2528.9)/2 - 15.0 = -12.1 kJ
c) ∆SNET = mA2(sA2- sAE AVE) + mA1(sAE AVE - sA1)
+ mV2(sV2-sVE AVE) + mV1(sVE AVE - sV1) - QCV/T0
12-50
Assume that the mixture exiting is throttled across the value to 100 kPa and
then discharged. Therefore, since the composition is constant.
PAE = PA1 = 98.94 kPa,
PVE = PV1 = 1.061 kPa
Also, PA2 = 148.41 kPa & PV2 = 1.592 kPa
Using TAE ≈ TVE ≈ 293.2 K & T0 = 15 oC = 288.2 K
∆SNET = 0.1794( 1.004 ln
288.2
148.425
- 0.287 ln
)
293.2
98.94
+ 0.2312( 1.004 ln
293.2
-0)
298.2
+ 0.0012( 1.8723 ln
288.2
1.575
- 0.461 52 ln
)
293.2
1.05
+ 0.001 54( 1.8723 ln
= +0.0138 kJ/K
293.2
- 0 ) − (−12.1)/288.2
298.2
12-51
12.80 Ambient air is at a condition of 100 kPa, 35°C, 50% relative humidity. A steady
stream of air at 100 kPa, 23°C, 70% relative humidity, is to be produced by first
cooling one stream to an appropriate temperature to condense out the proper
amount of water and then mix this stream adiabatically with the second one at
ambient conditions. What is the ratio of the two flow rates? To what temperature
must the first stream be cooled?
P1 = P2 = 100 kPa
1
5
2
T1 = T2 = 35 oC
MIX
COOL
.
-Q COOL
3
4
.
-Q MIX= 0
LIQ H 2O
φ1 = φ2 = 0.50, φ4 = 1.0
P5 = 100, T5 = 23 oC
φ5 = 0.70
Pv1 = Pv2 = 0.5×5.628 = 2.814 kPa => w1 = w2 = 0.622×
Pv5 = 0.7×2.837 = 1.9859 kPa => w5 = 0.622×
C.V.: Mixing chamber:
Call the mass flow ratio
2.814
= 0.0180
100-2.814
1.9859
= 0.0126
100-1.9859
r = ma2/ma1
cons. mass:
w1 + r w4 =(1+ r)w5
Energy Eq.:
ha1 + w1hv1 + rha4 + rw4hv4 = (1+r)h a5 + (1+r)w5hv5
→ 0.018 + rw4 = (1+r) 0.0126
or
ma2
0.018-0.0126
=
,
r=
0.0126-w4
ma1
with
w4 = 0.622 ×
PG4
100-PG4
1.004×308.2 + 0.018×2565.3 + r×1.004×T4 + r w4hv4
= (1+r)×1.004 × 296.2 + (1+r)×0.0126×2543.6
or
[
]
r 1.004×T4 + w4hG4 - 329.3 + 26.2 = 0
Assume T4 = 5 oC
→ PG4 = 0.8721, hG4 = 2510.5
w4 = 0.622×0.8721/(100-0.8721) = 0.0055
r = ma2/ma1 =
0.018-0.0126
= 0.7606
0.0126-0.0055
0.7606[1.004×278.2 + 0.0055×2510.5 - 329.6] + 26.2 = -1.42 ≈ 0
OK
12-52
ENGLISH UNIT PROBLEMS
12.81EA gas mixture at 70 F, 18 lbf/in.2 is 50% N2, 30% H2O and 20% O2 on a mole
basis. Find the mass fractions, the mixture gas constant and the volume for 10 lbm
of mixture.
From Eq. 12.3: ci = yi Mi/ ∑ yjMj
MMIX = ∑ yjMj = 0.5 × 28.013 + 0.3 × 18.015 + 0.2 × 31.999
= 14.0065 + 5.4045 + 6.3998 = 25.811
cN2 = 14.0065 / 25.811 = 0.5427, cH2O = 5.4045 / 25.811 = 0.2094
cO2 = 6.3998 / 25.811 = 0.2479, sums to 1 OK
−
RMIX = R/MMIX = 1545.36 / 25.811 = 59.87 lbf ft/lbm R
V = mRMIX T/P = 10 × 59.87 × 710 / 18 × 144 = 164 ft3
12.82EWeighing of masses gives a mixture at 80 F, 35 lbf/in.2 with 1 lbm O2, 3 lbm N2
and 1 lbm CH4. Find the partial pressures of each component, the mixture specific
volume (mass basis), mixture molecular weight and the total volume.
From Eq. 12.4: yi = (mi /Mi) / ∑ mj/Mj
ntot = ∑ mj/Mj = (1/31.999) + (3/28.013) + (1/16.04)
= 0.031251 + 0.107093 + 0.062344 = 0.200688
yO2 = 0.031251/0.200688 = 0.1557,
yN2 = 0.107093/0.200688 = 0.5336,
yCH4 = 0.062344/0.200688 = 0.3107
PO2 = yO2 Ptot = 0.1557 × 35 = 5.45 lbf/in.2,
PN2 = yN2 Ptot = 0.5336 × 35 = 18.676 lbf/in.2,
PCH4 = yCH4 Ptot = 0.3107 × 35 = 10.875 lbf/in.2
−
Vtot = ntot RT/P = 0.200688 × 1545 × 539.7 / (35 × 144) = 33.2 ft3
v = Vtot/mtot = 33.2 / (1 + 3 + 1) = 6.64 ft3/lbm
MMIX = ∑ yjMj = mtot/ntot = 5/0.200688 = 24.914
12-53
12.83EA pipe flows 0.15 lbmol a second mixture with molefractions of 40% CO2 and
60% N2 at 60 lbf/in.2, 540 R. Heating tape is wrapped around a section of pipe
with insulation added and 2 Btu/s electrical power is heating the pipe flow. Find
the mixture exit temperature.
C.V. Pipe heating section. Assume no heat loss to the outside, ideal gases.
.
. - .−
.
Energy Eq.: Q = m(he − hi) = n(he − hi) = nCP mix(Te − Ti)
−
−
CP mix = ∑ yi Ci = 0.4 × 8.846 + 0.6 × 6.975 = 7.723 Btu/lbmole
. .−
Te = Ti + Q / nCP mix = 540 + 2/(0.15 × 7.723) = 541.7 R
12.84EAn insulated gas turbine receives a mixture of 10% CO 2, 10% H2O and 80% N2
on a mole basis at 1800 R, 75 lbf/in.2. The volume flow rate is 70 ft3/s and its
exhaust is at 1300 R, 15 lbf/in.2. Find the power output in Btu/s using constant
specific heat from C.4 at 540 R.
C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0.
.
. - .−
.
Energy Eq.:
WT = m(hi − he) = n(hi − he) = nCP mix(Ti − Te)
.
. −
−
PV = nRT => n = PV / RT = 75 × 144 × 70 / (1545.4 × 1800) = 0.272 lbmole/s
−
−
CP mix = ∑yi Ci = 0.1 × 44.01 × 0.201 + 0.1 × 18.015 × 0.447
+ 0.8 × 28.013 × 0.249 = 7.27 Btu/lbmol R
.
WT = 0.272 × 72.7 × (1800 − 1300) = 988.7 Btu/s
12.85ESolve Problem 12.84 using the values of enthalpy from Table C.7
C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0.
.
. - .
Energy Eq.:
WT = m(hi − he) = n(hi − he)
.
. −
−
PV = nRT => n = PV / RT = 75 × 144 × 70 / (1545.4 × 1800) = 0.272 lbmole/s
.
WT = 0.272 × [0.1(14358 − 8121) + 0.1(11178 − 6468.5) + 0.8(9227 − 5431)]
= 1123.7 Btu/s
12-54
12.86ECarbon dioxide gas at 580 R is mixed with nitrogen at 500 R in a SSSF insulated
mixing chamber. Both flows are at 14.7 lbf/in.2 and the mole ratio of carbon
dioxide to nitrogen is 2;1. Find the exit temperature and the total entropy
generation per mole of the exit mixture.
.
.
CV mixing chamber, SSSF. The inlet ratio is nCO2 = 2 nN2 and assume no
external heat transfer, no work involved.
.
.
.
.
. . nCO + 2nN = nex = 3nN ; nN (hN + 2hCO ) = 3nN hmix ex
2
2
2
2
2
2
2
- Take 540R as reference and write h = h540 + CPmix(T-540).
CP N (Ti N -540) + 2CP CO (Ti CO -540) = 3CP mix(Tmix ex -540)
2
2
2
2
CP mix = ∑yiCP i = (29.178 + 2×37.05)/3 = 8.2718
3CP mixTmix ex = CP N Ti N + 2CP CO Ti CO = 13837
2
2
2
2
Tmix ex = 557.6 R; Pex N = Ptot/3; Pex CO = 2Ptot/3
2
2
.
. ... - . - Sgen = nexsex-(ns)iCO - (ns)iN = nN (se - si)N + 2nN (se - si)CO
2
2
2
2
2
2
Tex
Tex
.
.
Sgen/nN = CPN ln
- Rln yN + 2CPCO ln
- 2 Rln yCO
2
2 TiN
2
2 TiCO
2
2
2
= 0.7575 + 2.1817 - 0.7038 + 1.6104 = 3.846 Btu/lbmol N2 R
12.87EA mixture of 60% helium and 40% nitrogen by volume enters a turbine at 150
lbf/in.2, 1500 R at a rate of 4 lbm/s. The adiabatic turbine has an exit pressure of
15 lbf/in.2 and an isentropic efficiency of 85%. Find the turbine work.
Assume ideal gas mixture and take CV as turbine.
wT s = hi-hes,
ses = si,
Tes = Ti(Pe/Pi)(k-1)/k
CP mix = 0.6× 1.25× 4.003 + 0.4× 0.248× 28.013 = 5.7811
- (k-1)/k = R/CP mix = 1545/(5.7811×778) = 0.3435
Mmix= 0.6×4.003 + 0.4×28.013 = 13.607, CP=CP/Mmix= 0.4249
Tes= 1500(15/150)0.3435 = 680 R, wTs = CP(Ti-Tes) = 348.4
.
.
wT ac = wTs = 296.1 Btu/lbm; W = mwTs = 1184 Btu/s
12-55
12.88EA mixture of 50% carbon dioxide and 50% water by mass is brought from 2800
R, 150 lbf/in.2 to 900 R, 30 lbf/in.2 in a polytropic process through a SSSF device.
Find the necessary heat transfer and work involved using values from C.4.
Process Pvn = constant
leading to
n ln(v2/v1) = ln(P1/P2); v = RT/P
n = ln(150/30)/ln(900 × 150/30 × 2800) =3.3922
Rmix = ΣciRi = (0.5 × 35.1 + 0.5 × 85.76)/778 = 0.07767
CP mix = ΣciCPi = 0.5 × 0.203 + 0.5 × 0.445 = 0.324
⌠vdP = - n (P v -P v ) = - nR (T -T ) = 209.3 Btu
w = -⌡
n-1 e e i i
n-1 e i
lbm
q = he-hi + w = CP(Te-Ti) + w = -406.3 Btu/lbm
12.89EA mixture of 4 lbm oxygen and 4 lbm of argon is in an insulated piston cylinder
arrangement at 14.7 lbf/in.2, 540 R. The piston now compresses the mixture to
half its initial volume. Find the final pressure, temperature and the piston work.
Since T1 >> TC assume ideal gases.
u2-u1 = 1q2 - 1w2 = - 1w2 ;
s2-s1 = 0
Pvk = constant, v2 = v1/2
P2 = P1(v1/v2)k = P1(2)k; T2 = T1(v1/v2)k-1 = T1(2)k-1
Find kmix to get P2,T2 and Cv mix for u2-u1
Rmix = ΣciRi = (0.5 × 48.28 + 0.5 × 38.68)/778 = 0.055887
CPmix = ΣciCPi = 0.5 × 0.219 + 0.5 × 0.1253 = 0.17215
Cvmix = CPmix -Rmix = 0.11626, kmix = CPmix/Cvmix = 1.4807
P2 = 14.7(2)1.4805= 41.03 lbf/in2, T2 = 540*20.4805= 753.5 R
w = u1-u2= Cv(T1-T2)= 0.11626(540-753.5) = -24.82 Btu/lbm
1 2
1
W2 = mtot 1w2 = 8 (-24.82) = -198.6 Btu
12-56
12.90ETwo insulated tanks A and B are connected by a valve. Tank A has a volume of
30 ft3 and initially contains argon at 50 lbf/in.2, 50 F. Tank B has a volume of 60
ft3 and initially contains ethane at 30 lbf/in.2, 120 F. The valve is opened and
remains open until the resulting gas mixture comes to a uniform state. Find the
final pressure and temperature and the entropy change for the process.
a) 1st law: U2-U1 = 0 = nArCV0(T2-TA1) + nC H CVO(T2-TB1)
2 6
50×144×30
nAr = PA1VA/RTA1 =
= 0.2743 lbmol
1545×509.7
30×144×60
=
P
V
/R
T
=
= 0.2894 lbmol
H
B1 B
B1 1545×579.7
2 6
nC
n2 = nAr + nC
H
2 6
= 0.5637 lbmol
0.2743×39.948×0.0756(T2-509.7)
+ 0.2894×30.07×0.361(T2-509.7) = 0
Solving, T2 = 565.1 R
0.5637×1545×565.1
= 38 lbf/in2
P2 = n2RT2/(VA+VB) =
90×144
b) ∆SSURR = 0
→ ∆SNET = ∆SSYS = nAr∆SAr + nC H ∆SC
2 6
H
2 6
yAr = 0.2743/0.5637 = 0.4866
T2 - yArP2
∆SAr = CP Ar ln
- R ln
TA1
PA1
= 39.948×0.1253 ln
565.1 1545 0.4866×38
ln
= 2.4919 Btu/lbmol R
509.7 778
50
T2 - yC2H6P2
- R ln
∆SC2H6 = CC2H6 ln
TB1
PB1
= 30.07×0.427 ln
565.1 1545 0.5134×38
ln
579.7 778
30
= 0.5270 Btu/lbmol R
∆SNET = 0.2743×2.4919 + 0.2894×0.5270 = 0.836 Btu/R
12-57
12.91EA large SSSF air separation plant takes in ambient air (79% N2, 21% O2 by
volume) at 14.7 lbf/in.2, 70 F, at a rate of 2 lb mol/s. It discharges a stream of pure
O2 gas at 30 lbf/in.2, 200 F, and a stream of pure N2 gas at 14.7 lbf/in.2, 70 F. The
plant operates on an electrical power input of 2000 kW. Calculate the net rate of
entropy change for the process.
Air 79 % N2
1
2
pure O 2
21 % O2
pure N 2
P1 = 14.7
3
P2 = 30
T2 = 200 F
P3 = 14.7
.
T1 = 70 F
T3 = 70 F
-WIN = 2000 kW
.
n1 = 2 lbmol/s
.
.
QCV .
QCV .
dSNET
.
.
=+ ni∆s-i = + (n2-s2 + n3-s3 - n1-s1)
dt
T0 i
T0
.
.
.
. . . QCV = Σn∆hi + WCV = nO CP0 O (T2-T1) + nN CP0 N (T3-T1) + WCV
∑
2
2
2
2
= 0.21×2× [32×0.213×(200-70)] + 0 - 2000×3412/3600
= +382.6 - 1895.6 = -1513 Btu/s
660 1545
30
.
Σni∆s-i = 0.21×2 32×0.219 ln
ln
530 778 0.21×14.7
[
[
+ 0.79×2 0 -
1545
14.7
ln
778 0.79×14.7
]
= -1.9906 Btu/R s
dSNET
dt
=+
1513
- 1.9906 = 0.864 Btu/R s
530
]
12-58
12.92EA tank has two sides initially separated by a diaphragm. Side A contains 2 lbm of
water and side B contains 2.4 lbm of air, both at 68 F, 14.7 lbf/in.2. The
diaphragm is now broken and the whole tank is heated to 1100 F by a 1300 F
reservoir. Find the final total pressure, heat transfer, and total entropy generation.
U2-U1 = ma(u2-u1)a + mv(u2-u1)v = 1Q2
S2-S1 = ma(s2-s1)a + mv(s2-s1)v = ⌠
⌡ 1Q2/T + Sgen
V2 = VA + VB = mvvv1 + mava1 = 0.0321 + 31.911 = 31.944 ft3
vv2 = V2/mv = 15.9718, T2 => P2v = 58.7 lbf/in2
va2 = V2/ma = 13.3098, T2 => P2a = mRT2/V2 = 43.415 lbf/in2
P2tot = P2v + P2a = 102 lbf/in2
Water: u1 = 36.08, u2 = 1414.3, s1 = 0.0708, s2 = 2.011
Air: u1 = 90.05, u2 = 278.23, sT1 = 1.6342, sT2 = 1.9036
Q = 2(1414.3 - 36.08) + 2.4(278.23 - 90.05) = 3208 Btu
1 2
Sgen = 2(2.011-0.0708) + 2.4[1.9036 - 1.6342
- (53.34/778)×ln(43.415/14.7)] - 3208/1760
= 3.8804 + 0.4684 - 1.823 = 2.526 Btu/R
12.93EConsider a volume of 2000 ft3 that contains an air-water vapor mixture at 14.7
lbf/in.2, 60 F, and 40% relative humidity. Find the mass of water and the humidity
ratio. What is the dew point of the mixture?
Air-vap P = 14.7 lbf/in.2, T= 60 F, φ = 40%
Pg = Psat60 = 0.256 lbf/in.2
Pv = φ Pg = 0.4 × 0.256 = 0.1024 lbf/in.2
mv1 =
P vV
RvT
=
0.1024 × 144 × 2000
= 0.661 lbm
85.76 × 520
Pa = Ptot- Pv1 = 14.7 – 0.1024 = 14.598
ma =
w1 =
P aV
RaT
mv
ma
=
=
14.598 × 144 × 2000
= 151.576 lbm
53.34 ×⊇520
0.661
= 0.00436
151.576
Tdew is T when Pg(Tdew) = 0.1024;
T = 35.5 F
12-59
12.94EConsider a 10-ft3 rigid tank containing an air-water vapor mixture at 14.7 lbf/in.2,
90 F, with a 70% relative humidity. The system is cooled until the water just
begins to condense. Determine the final temperature in the tank and the heat
transfer for the process.
Pv1 = φPG1 = 0.7 × 0.6988 = 0.489 lbf/in2
Since mv = const & V = const & also Pv = PG2:
PG2 = Pv1×T2/T1 = 0.489×T2/549.7
For T2 = 80 F:
0.489×539.7/549.7 = 0.4801 =/ 0.5073 ( = PG at 80 F )
For T2 = 70 F:
0.489×529.7/549.7 = 0.4712 =/ 0.3632 ( = PG at 70 F )
interpolating → T2 = 78.0 F
w2 = w1 = 0.622
ma =
Pa1V
RaT1
=
0.489
= 0.0214
(14.7-0.489)
14.211×144×10
= 0.698 lbm
53.34×549.7
1st law:
Q12 = U2-U1 = ma(ua2-ua1) + mv(uv2-uv1)
= 0.698[0.171(78 - 90) + 0.0214(1036.3 - 1040.2)]
= 0.698(-2.135 Btu/lbm air) = -1.49 Btu
12.95EAir in a piston/cylinder is at 95 F, 15 lbf/in.2 and a relative humidity of 80%. It is
now compressed to a pressure of 75 lbf/in.2 in a constant temperature process.
Find the final relative and specific humidity and the volume ratio V2/V1.
Check if the second state is saturated or not. First assume no water is condensed
1:
Pv1= φ1PG1 = 0.66,
w1 = 0.622×0.66/14.34 = 0.0286
2: w2 = 0.622 Pv2/(P2-Pv2) = w1 => Pv2 = 3.297 > Pg = 0.825 lbf/in2
Conclusion is state 2 is saturated
φ2 = 100%, w2 = 0.622 Pg/(P2-Pg) = 0.00692
To get the volume ratio, write the ideal gas law for the vapor phases
V2 = Va2 + Vv2 + Vf2 = (maRa + mv2Rv)T/P2 + mliqvf
V1 = Va1 + Vv1 = (maRa + mv1Rv)T/P1
Take the ratio and devide through with maRaT/P2 to get
V2/V1 = (P1/P2)[1 + 0.622w2 + (w1-w2)P2vf/RaT]/(1+0.622 w1) = 0.1974
The liquid contribution is nearly zero (=0.000127) in the numerator.
12-60
12.96EA 10-ft3 rigid vessel initially contains moist air at 20 lbf/in.2, 100 F, with a
relative humidity of 10%. A supply line connected to this vessel by a valve carries
steam at 100 lbf/in.2, 400 F. The valve is opened, and steam flows into the vessel
until the relative humidity of the resultant moist air mixture is 90%. Then the
valve is closed. Sufficient heat is transferred from the vessel so the temperature
remains at 100 F during the process. Determine the heat transfer for the process,
the mass of steam entering the vessel, and the final pressure inside the vessel.
i
Air-vap mix: P1 = 20 lbf/in2, T1 = 560 R
H 2O
φ1 = 0.10, T2 = 560 R, φ2 = 0.90
Pv1 = φ1PG1 = 0.1×0.9503 = 0.095 lbf/in2
AIR
+
H 2O
Pv2 = 0.9×0.9503 = 0.8553 lbf/in2
Pa2 = Pa1 = P1 - Pv1 = 20 - 0.095 = 19.905
w1 = 0.622×0.095/19.905 = 0.002 96
w2 = 0.622×0.8553/19.905 = 0.026 64
w=
mv
ma
→ mvi = ma(w2-w1), ma =
19.905×144×10
= 0.96 lbm
53.34×560
P2 = 19.905 + 0.855 = 20.76 lbf/in2
mvi = 0.96(0.02664 - 0.00296) = 0.0227 lbm
CV: vessel
QCV = ma(ua2-ua1) + mv2uv2 - mv1uv1 - mvihi
uv ≈ uG at T → uv1 = uv2 = uG at 100 F, ua2 = ua1
→ QCV = mvi(uG at T - hi) = 0.0227(1043.5-1227.5) = -4.18 Btu
12-61
12.97EA water-filled reactor of 50 ft3 is at 2000 lbf/in.2, 550 F and located inside an
insulated containment room of 5000 ft3 that has air at 1 atm. and 77 F. Due to a
failure the reactor ruptures and the water fills the containment room. Find the final
pressure.
CV Total container.
Energy:
mv(u2-u1) + ma(u2-u1) = 1Q2 - 1W2 = 0
Initial water: v1 = 0.021407, u1 = 539.24,
Initial air:
mv = V/v = 2335.7 lbm
ma = PV/RT = 14.7×4950×144/53.34×536.67 = 366.04 lbm
Substitute into energy equation
2335.7 (u2-539.24) + 366.04×0.171 (T2-77) = 0
u2 + 0.0268 T2 = 541.3
&
v2 = V2/mv = 2.1407 ft3/lbm
Trial and error 2-phase (T guess, v2 => x2 => u 2 => LHS)
T = 300 LHS = 550.789
T = 290 LHS = 506.05
T2 = 298 F, x2 = 0.3198, Psat = 65 lbf/in2, LHS=541.5 OK
Pa2 = Pa1V1T2/V2T1 = 14.7×4950×757.7/5000×536.67 = 20.55 lbf/in2
=> P2 = Pa2 + Psat = 85.55 lbf/in2
12.98EAtmospheric air at 95 F, relative humidity of 10%, is too warm and also too dry.
An air conditioner should deliver air at 70 F and 50% relative humidity in the
amount of 3600 ft3 per hour. Sketch a setup to accomplish this, find any amount
of liquid (at 68 F) that is needed or discarded and any heat transfer.
CV air conditioner. Check from psychrometric chart, inlet 1, exit 2.
In: w1 = 0.0034, h̃mix,1 = 27, hf,68 = 36.08, Ex: w2 = 0.0078, h̃mix,2 = 25.4
Water must be added. Continuity and energy equations
.
.
.
.
.
.
.
mA(1 + w1) + mliq = mA(1 + w2) & mAh1mix + mliqhf + QCV = mAh2mix
.
.
mtot = PVtot /RT = 14.7×3600×144/53.34×529.67 = 270 lbm/h
.
.
mA = mtot/(1+w2) = 267.91 lbm/h
.
.
mliq = mA(w2 - w1) = 267.91(0.0078 - 0.0034) = 1.179 lbm/h
.
QCV = 267.91(25.4-27-0.0044×36.08) = - 471.2 Btu/h
If the properties are calculated from the tables then
Pg1 = 0.8246, hg1 = 1102.9,
Pv1 = φ1 Pg1 = 0.08246,
Pg2 = 0.36324,
hg2 = 1092
Pv2 = φ2 Pg2 = 0.1816
w1 = 0.622Pv1/(Ptot -Pv1) = 0.0035, w2 = 0.00778
h̃2-h̃1 = 0.24(70-95) + 0.00778×1092 - 0.0035×1102.9 = -1.364
.
mliq = 267.91(0.00778 - 0.0035) = 1.147 lbm/h
.
QCV = 267.91(-1.364 - 0.00428×36.08) = - 406.8 Btu/h
12-62
12.99ETwo moist air streams with 85% relative humidity, both flowing at a rate of 0.2
lbm/s of dry air are mixed in a SSSF setup. One inlet flowstream is at 90 F and
the other at 61 F. Find the exit relative humidity.
CV mixing chamber.
.
.
.
ma w1 + ma w2 = 2ma wex;
.
.
.
ma h1 + ma h2 = 2ma hex
1: w1 = 0.0262, h1 = 50.4 2: w2 = 0.0097, h2 = 25.2
wex = (w1 + w2)/2 = 0.018 ; hex = (h1 + h2)/2 = 37.8
Exit: wex, hex => Tex = 75 F,
φ = 95%
12.100E An indoor pool evaporates 3 lbm/h of water, which is removed by a
dehumidifier to maintain 70 F, Φ = 70% in the room. The dehumidifier is a
refrigeration cycle in which air flowing over the evaporator cools such that liquid
water drops out, and the air continues flowing over the condenser, as shown in
Fig. P12.71. For an air flow rate of 0.2 lbm/s the unit requires 1.2 Btu/s input to a
motor driving a fan and the compressor and it has a coefficient of performance, β
= Q_L /W_c = 2.0. Find the state of the air after the evaporator, T2, ω2, Φ2 and the
heat rejected. Find the state of the air as it returns to the room and the compressor
work input.
The unit must remove 3 lbm/h liquid to keep steady state in the room. As
water condenses out state 2 is saturated.
1: 70 F, 70% => w1 = 0.011, h1 = 28.8
.
.
.
.
CV 1 to 2: mliq = ma(w1 - w2) => w2 = w1 - mliq /ma
qL = h1 - h2 - (w1 - w2) hf2
w2 = 0.011 - 3/(3600 × 0.2) = 0.0068
2: w2, 100% => T2 = 47 F, h̃2 = 18.8, hf2 = 15.04
qL = 28.8 - 18.8 - 0.0042×15.04 = 9.94 Btu/lbm dry air
. .
CV Total system : h̃3 = h̃1 + Wel/ma - (w1-w2) hf
= 28.8 + 1.2/0.2 - 0.0632 = 34.74 Btu/lbm dry air
3: w3 = w2, h3 => T3 = 112 F, φ3 = 12%
.
.
Wc = ma qL/ β = 1 Btu/s
12-63
12.101E To refresh air in a room, a counterflow heat exchanger is mounted in the wall, as
shown in Fig. P12.72. It draws in outside air at 33 F, 80% relative humidity and
draws room air, 104 F, 50% relative humidity, out. Assume an exchange of 6
lbm/min dry air in a SSSF device, and also that the room air exits the heat
exchanger to the atmosphere at 72 F. Find the net amount of water removed from
room, any liquid flow in the heat exchanger and (T, φ) for the fresh air entering
the room.
State 1: w1 = 0.0236, h1 = 51, Tdew,1 = 81.5 F
The room air is cooled to 72 F < Tdew1 so liquid will form in the exit flow
channel and state 2 is saturated.
2: 72 F, φ = 100% => w2 = 0.017, h2 = 36, hf2 = 40.09
3: 33 F, φ = 80% => w3 = 0.0031, h3 = 11.3
.
.
CV 1 to 2: mliq,2 = ma (w1 - w2) = 6 (0.0236 - 0.017) = 0.04 lbm/min
.
.
.
CV room: mv,out = ma (w1 - w4) = ma (w1 - w3)
= 6(0.0236-0.0031) = 0.123 lbm/min
.
.
.
CV Heat exchanger: ma(h̃4 - h̃3) = ma(h̃1 - h̃2) - mliq hf2
h4 = h3 + h1 - h2 - (w1-w2) hf2
= 11.3 + 51 - 36 - 0.0066 × 40.09 = 26 Btu/lbm dry air
4: w4 = w3, h4 => T4 = 94 F, φ = 9%
12-64
12.102E A 4-ft3 insulated tank contains nitrogen gas at 30 lbf/in.2 and ambient
temperature 77 F. The tank is connected by a valve to a supply line flowing
carbon dioxide at 180 lbf/in.2, 190 F. A mixture of 50 mole percent nitrogen and
50 mole percent carbon dioxide is to be obtained by opening the valve and
allowing flow into the tank until an appropriate pressure is reached and the valve
is closed. What is the pressure? The tank eventually cools to ambient temperature.
Calculate the net entropy change for the overall process.
V = 4 ft3, P1 = 30 lbf/in2, T1 = T0 = 77 F
CO 2 i
Pi = 180 lbf/in.2, Ti = 190 F
N
n2 CO
2
2
At state 2: yN = yCO = 0.50
2
2
= n2 N = n1 N = P1V/RT1
2
2
= 30×4×144/(1545×536.67) = 0.02084 lbmol
n2 = 0.04168 lbmol
1st law: nihi = n2u-2 - n1u-1 , for const specific heats
niCPoiTi = (niCVoi+n1CVo1)T2 - n1CVo1T1
But ni = n1
→ CPoiTi = CVoiT2 + CVo1(T2-T1)
44.01×0.201×649.67 = 44.01×0.156 T2 + 28.013×0.178(T2-536.67)
T2 = 710.9 R
P2 = n2RT2/V = 0.04168×1545×710.9/4×144 = 79.48 lbf/in2
Cool to T3 = T0 = 77 F = 536.67 R
P3 = P2× T3/T2 = 79.48× 536.67/710.9 = 60 lbf/in2
Q23 = n2CVo2(T3-T2) = 0.04168(0.5×28.013×0.178
+ 0.5×44.01×0.156)(536.67-710.9) = - 43.0 Btu
∆SNET = n3-s3 - n1-s1 - ni-si - Q23/T0
= ni[(s-CO )3 - -si] + n1[(s-N )3 - s-1] - Q23/T0
2
2
= 0.02084(44.01×0.201 ln
+ 0.02084×( - 1.98589 ln
0.5×60
536.67
- 1.98589 ln
)
649.67
180
0.5×60
−43.0
)−
30
536.67
= +0.03893 + 0 + 0.0801 = +0.119 Btu/R
12-65
12.103E Ambient air is at a condition of 14.7 lbf/in.2, 95 F, 50% relative humidity. A
steady stream of air at 14.7 lbf/in.2, 73 F, 70% relative humidity, is to be
produced by first cooling one stream to an appropriate temperature to condense
out the proper amount of water and then mix this stream adiabatically with the
second one at ambient conditions. What is the ratio of the two flow rates? To what
temperature must the first stream be cooled?
1
P1 = P2 = P5 = 14.7 lbf/in2
5
2
T1 = T2 = 95 F
MIX
COOL
.
-Q COOL
3
φ1 = φ2 = 0.50, φ4 = 1.0
.
-Q MIX= 0
4
T5 = 73 F, φ5 = 0.70
LIQ H 2O
Pv1 = Pv2 = 0.5×0.8246 = 0.4123
w1 = w2 = 0.622×
0.4123
= 0.0179
14.7-0.4123
Pv5 = 0.7×0.4064 = 0.2845
=>
w5 = 0.622×
0.2845
= 0.0123
14.7-0.2845
MIX: Call the mass flow ratio r = ma2/ma1
Conservation of water mass:
w1 + r w4 =(1+ r)w5
Energy Eq.:
ha1 + w1hv1 + rha4 + rw4hv4 = (1+r)h a5 + (1+r)w5hv5
→
or
r=
0.0179 + rw4 = (1+r) 0.0123
0.0179-0.0123
, with
0.0123-w4
w4 = 0.622×
PG4
14.7-PG4
0.24×555 + 0.0179×1107.2 + r×0.24×T4 + rw4hv4
= (1+r)×0.24×533 + (1+r)×0.0123×1093.3
[
Assume T4 = 40 F
w4 = 0.622×
ma2
ma1
]
r 0.24×T4 + w4hG4 - 141.4 + 11.66 = 0
or
=
→ PG4 = 0.121 66, hG4 = 1078.9
0.121 66
= 0.0052
14.7-0.121 66
0.0179-0.0123
= 0.7887
0.0123-0.0052
0.7887[0.24×500 + 0.0052×1078.9 - 141.4] + 11.66 = -0.29 ≈ 0
=> T4 = 40 F
OK
13-1
CHAPTER 13
The correspondence between the new problem set and the previous 4th edition
chapter 10 problem set:
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Old
1
10
2
3
4
5
6
7
9
12
13
14
15
16
17
18
19
21
22
23
24
New
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
Old
25
new
26 mod
27
28
new
30
31 mod
new
35
36
37
39
new
40
new
41
38
45
44
47
New
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
Old
48
new
49
new
new
51
53
55
56
61
63
new
65
66
new
67
68
new
new
Old
52
70 mod
57
58
62
New
72
73
74
Old
new
new
new
The problems that are labeled advanced are:
New
62
63
64
65
66
Old
8
29
20
46
50
New
67
68
69
70
71
13-2
The English-unit problems are:
New
75
76
77
78
79
80
81
82
Old
74
75
76
78
79
80
82
83
New
83
84
85
86
87
88
89
90
Old
84
81
86 mod
87
88
89
90
91
New
91
92
93
94
95
96
97
Old
92
93
96
97
98
99
101
mod indicates a modification from the previous problem that changes the solution
but otherwise is the same type problem.
13-3
The following table gives the values for the compressibility, enthalpy departure and the
entropy departure along the saturated liquid-vapor boundary. These are used for all the
problems using generalized charts as the figures are very difficult to read accurately
(consistently) along the saturated liquid line. It is suggested that the instructor hands out
copies of this page or let the students use the computer for homework solutions.
Tr
Pr
Zf
Zg
d(h/RT)f
d(h/RT)g
d(s/R)f
d(s/R)g
0.96
0.78
0.14
0.54
3.65
1.39
3.45
1.10
0.94
0.69
0.12
0.59
3.81
1.19
3.74
0.94
0.92
0.61
0.10
0.64
3.95
1.03
4.00
0.82
0.90
0.53
0.09
0.67
4.07
0.90
4.25
0.72
0.88
0.46
0.08
0.70
4.17
0.78
4.49
0.64
0.86
0.40
0.07
0.73
4.26
0.69
4.73
0.57
0.84
0.35
0.06
0.76
4.35
0.60
4.97
0.50
0.82
0.30
0.05
0.79
4.43
0.52
5.22
0.45
0.80
0.25
0.04
0.81
4.51
0.46
5.46
0.39
0.78
0.21
0.035
0.83
4.58
0.40
5.72
0.35
0.76
0.18
0.03
0.85
4.65
0.34
5.98
0.31
0.74
0.15
0.025
0.87
4.72
0.29
6.26
0.27
0.72
0.12
0.02
0.88
4.79
0.25
6.54
0.23
0.70
0.10
0.017
0.90
4.85
0.21
6.83
0.20
0.68
0.08
0.014
0.91
4.92
0.18
7.14
0.17
0.66
0.06
0.01
0.92
4.98
0.15
7.47
0.15
0.64
0.05
0.009
0.94
5.04
0.12
7.81
0.12
0.60
0.03
0.005
0.95
5.16
0.08
8.56
0.08
0.58
0.02
0.004
0.96
5.22
0.06
8.97
0.07
0.54
0.01
0.002
0.98
5.34
0.03
9.87
0.04
0.52
0.0007
0.0014
0.98
5.41
0.02
10.38
0.03
13-4
13.1
A special application requires R-12 at −140°C. It is known that the triple-point
temperature is −157°C. Find the pressure and specific volume of the saturated
vapor at the required condition.
The lowest temperature in Table B.3 for R-12 is -90oC, so it must be extended
to -140oC using the Clapeyron Eq. 13.7 integrated as in example 13.1
Table B.3:
at T1 = -90oC = 183.2 K,
P1 = 2.8 kPa.
8.3145
= 0.068 76
120.914
P hfg (T - T1) 189.748 (133.2 - 183.2)
ln =
=
= -5.6543
P1 R T × T
0.068 76 133.2 × 183.2
R=
1
P = 2.8 exp(-5.6543) = 0.0098 kPa
13-5
13.2
In a Carnot heat engine, the heat addition changes the working fluid from
saturated liquid to saturated vapor at T, P. The heat rejection process occurs at
lower temperature and pressure (T − ∆T), (P − ∆P). The cycle takes place in a
piston cylinder arrangement where the work is boundary work. Apply both the
first and second law with simple approximations for the integral equal to work.
Then show that the relation between ∆P and ∆T results in the Clapeyron equation
in the limit ∆T → dT.
T
1
T
T −∆T
P
P
2
P-∆P
qH = TsFG;
4
s
sfg at T
qL = (T-∆T)sFG ;
2
T
P-∆P
3
4
1
P
3
vfg at T
T-∆T
v
wNET = qH - qL = ∆TsFG
Problem similar to development in section 13.1 for shaft work, here boundary
movement work, w = ⌠
⌡ Pdv
3
4
wNET = P(v2-v1) + ⌠
⌡ Pdv + (P-∆P)(v4-v3) + ⌠
⌡ Pdv
2
1
Approximating,
3
∆P
⌠
⌡ Pdv ≈ (P - ) (v3-v2);
2
2
4
⌠
⌡ Pdv ≈
1
& collecting terms,
v2+v3
v1+v4
wNET ≈ ∆P[(
)
-(
)]
2
2
(the smaller the ∆P, the better the approximation)
⇒
sFG
∆P
≈
v +v
v +v
∆T
( 22 3) - ( 12 4)
In the limit as ∆T → 0:
v3 → v2 = vG ,
lim ∆P dPSAT sFG
& ∆T→0 =
=
dT
vFG
∆T
v4 → v1 = vF
∆P
(P - 2 ) (v1-v4)
13-6
13.3
Ice (solid water) at −3°C, 100 kPa is compressed isothermally until it becomes
liquid. Find the required pressure.
Water, triple point T = 0.01oC , P = 0.6113 kPa
Table B.1.1: vF = 0.001, hF = 0.01 kJ/kg,
Tabel B.1.5:
vI = 0.001 0908, hI = -333.4 kJ/kg
dPIF
Clapeyron
dT
∆P ≈
dPIF
dT
=
hF-hI
(vF-vI)T
=
333.4
= -13 442
-0.0000908×273.16
∆T = -13 442(-3 - 0.01) = 40 460 kPa
P = Ptp + ∆P = 40 461 kPa
13.4
Calculate the values hFG and sFG for nitrogen at 70 K and at 110 K from the
Clapeyron equation, using the necessary pressure and specific volume values
from Table B.6.1.
dPG hFG sFG
=
=
Clapeyron equation Eq.13.7:
dT TvFG vFG
For N2 at 70 K, using values for PG from Table B.6 at 75 K and 65 K, and
also vFG at 70 K,
∆PG
hFG ≈ T(vG-vF)
∆Τ
= 70(0.525 015)
(76.1-17.41
) = 215.7 kJ/kg (207.8)
75-65
sFG = hFG/T = 3.081 kJ/kg K (2.97)
Comparison not very close because PG not linear function of T. Using 71 K &
69 K,
hFG = 70(0.525 015)
(44.56-33.24
) = 208.0
71-69
At 110 K, hFG ≈ 110(0.014 342)
sFG =
(1938.8-1084.2
) = 134.82 kJ/kg (134.17)
115-105
134.82
= 1.226 kJ/kg K (1.22)
110
13-7
13.5
Using thermodynamic data for water from Tables B.1.1 and B.1.5, estimate the
freezing temperature of liquid water at a pressure of 30 Mpa.
P
30 MPa
H 2O
dPIF
dT
=
hIF
TvIF
≈ const
At the triple point,
vIF = vF - vI = 0.001 000 - 0.001 090 8
T.P.
= -0.000 090 8 m3/kg
T
hIF = hF - hI = 0.01 - (-333.40) = 333.41 kJ/kg
dPIF
dT
=
333.41
= -13 442 kPa/K
273.16(-0.000 090 8)
⇒ at P = 30 MPa,
T ≈ 0.01 +
13.6
(30 000-0.6)
= -2.2 oC
(-13 442)
Helium boils at 4.22 K at atmospheric pressure, 101.3 kPa, with hFG = 83.3
kJ/kmol. By pumping a vacuum over liquid helium, the pressure can be lowered
and it may then boil at a lower temperature. Estimate the necessary pressure to
produce a boiling temperature of 1 K and one of 0.5 K.
Helium at 4.22 K: P1 = 0.1013 MPa, hFG = 83.3 kJ/kmol
dPSAT
dT
=
hFG
TvFG
≈
hFGPSAT
RT2
⇒
ln
P2
P1
=
hFG 1
1
[
− ]
R T1 T2
For T2 = 1.0 K:
ln
P2
101.3
For T2 = 0.5 K:
ln
P2
101.3
=
83.3
1
1
[
−
]
8.3145 4.22 1.0
=
83.3
1
1
[
−
]
8.3145 4.22 0.5
=> P2 = 0.048 kPa = 48 Pa
P2 = 2.1601×10-6 kPa = 2.1601×10-3 Pa
13-8
13.7
A certain refrigerant vapor enters an SSSF constant pressure condenser at 150
kPa, 70°C, at a rate of 1.5 kg/s, and it exits as saturated liquid. Calculate the rate
of heat transfer from the condenser. It may be assumed that the vapor is an ideal
gas, and also that at saturation, vf ! vg. The following quantities are known for
this refrigerant:
ln Pg = 8.15 - 1000/T CP = 0.7 kJ/kg K
with pressure in kPa and temperature in K. The molecular weight is 100.
Refrigerant: State 1 T1 = 70oC P1 = 150 kPa
State 2 P2 = 150 kPa
x2 = 1.0
State 3
P3 = 150 kPa x3 = 0.0
ln (150) = 8.15 - 1000/T2 => T2 = 318.5 K = 45.3oC = T3
q13 = h3 -h1 = (h3 -h2) + (h2 -h1) = - hFG T3 + CP0(T2 -T1)
dPG
dT
=
hFG
TvFG
d ln PG
dT
,
dPG
RT
vFG ≈ vG =
,
PG
dT
= PG
d ln PG
dT
=
hFG
RT2
PG
= +1000/T2 = hFG/RT2
hFG = 1000 × R = 1000 × 8.3145/100 = 83.15
q13 = -83.15 + 0.7(45.3 - 70) = -100.44 kJ/kg
.
QCOND = 1.5(-100.44) = -150.6 kW
13.8
A container has a double wall where the wall cavity is filled with carbon dioxide
at room temperature and pressure. When the container is filled with a cryogenic
liquid at 100 K the carbon dioxide will freeze so the wall cavity has a mixture of
solid and vapor carbon dioxide at the sublimation pressure. Assume that we do
not have data for CO2 at 100 K, but it is known that at −90°C: Psat = 38.1 kPa,
hIG = 574.5 kJ/kg. Estimate the pressure in the wall cavity at 100 K.
For CO2 space: at T1 = -90 oC = 183.2 K , P1 = 38.1 kPa, hIG = 574.5 kJ/kg
For T2 = TcO2 = 100 K: Clapeyron
ln
P2
P1
=
hIG
R
dPSUB
dT
=
hIG
TvIG
≈
hIGPSUB
RT2
1
1
574.5
1
1
[183.2
−
]
=
[
−
] = -13.81
100
0.188 92 183.2 100
or P2 = P1 × 1.005×10-6
⇒ P2 = 3.83×10-5 kPa = 3.83×10-2 Pa
13-9
13.9
Small solid particles formed in combustion should be investigated. We would like
to know the sublimation pressure as a function of temperature. The only
information available is T, hFG for boiling at 101.3 kPa and T, hIF for melting at
101.3 kPa. Develop a procedure that will allow a determination of the sublimation
pressure, Psat(T).
P
TNBP = normal boiling pt T.
Solid
TNMP = normal melting pt T.
101.3 kPa
PTP
TTP = triple point T.
Liquid
Vap.
T TP T
1) TTP ≈ TNMP
T
T NBP
NMP
TTP
P TP
⌠h
 FG
⌠
⌡ (1/PSAT) dPSAT ≈ ⌡ RT2 dT
2)
0.1013 MPa
TNMP
Since hFG ≈ const ≈ hFG NBP we ca perform the integral over temperature
ln
PTP
0.1013
≈
hFG NBP
R
[T 1
-
NBP
1
]
TTP
→
get PTP
3) hIG at TP = hG - hI = (hG - hF) + (hF - hI) ≈ hFG NBP + hIF NMP
Assume hIG ≈ const. again we can evaluate the integral
ln
PSUB
PTP
P SUB
T
⌠h
h
 IG dT ≈ IG[ 1 − 1 ]
= ⌠
(1/P
)
dP
≈
⌡
SUB
SUB
R TTP T
⌡ RT2
P TP
or PSUB = fn(T)
TTP
13-10
13.10 Derive expressions for ( ∂T/∂v)u and for (∂h/∂s)v that do not contain the properties
h, u, or s.
∂P
P - T( )v
∂T
∂u
∂u
∂T
(∂v )u = - (∂v)T (∂T)v = C
v
/
(see Eqs. 13.33 and 13.34)
∂h
∂P
∂T
As dh = Tds + vdP => ( )v = T + v( )v = T - v( )s
∂s
∂s
∂v
∂P
T( )v
∂T
∂s
∂s
∂T
(∂v )s = - (∂v)T (∂T)v = - C
v
/
But
∂h
(Eq.13.20)
(Eq.13.22)
∂P
(∂s )v = T + vT
( )
C ∂T v
⇒
v
13.11 Derive expressions for ( ∂h/∂v)T and for (∂h/∂T )v that do not contain the
properties h, u, or s.
Find
(∂h∂v)T and (∂h∂T)v
dh = Tds + vdP
⇒
and use Eq.13.22
(∂h∂v)T = T(∂s∂v)T + v(∂P
)
∂v T
=T
∂P
+
v
(∂P
)
(
)
∂T v
∂v T
Also for the second first derivative use Eq.13.28
∂s
∂P
=
C
+
v
(∂h∂T)v = T(∂T
)v + v(∂P
)
(
)
v
∂T v
∂T v
13.12 Develop an expression for the variation in temperature with pressure in a constant
entropy process, (∂T/∂P)s, that only includes the properties P–v–T and the specific
heat, Cp.
∂s
∂T
(∂P)s = ∂s
(∂P)T
∂s
(∂T)P
∂v
∂v
)
∂T P T ∂v
==
( )
(CP/T) CP ∂T P
-(
{(∂P)T = -(∂T)P, Maxwell relation Eq. 13.23 and the other is Eq.13.27}
13-11
13.13 Determine the volume expansivity, α P, and the isothermal compressibility, β T, for
water at 20°C, 5 MPa and at 300°C, and 15 MPa using the steam tables.
Water at 20oC, 5 MPa (compressed liquid)
1 ∂v
1 ∆v
≈ ( )P
(
)
P
v ∂T
v ∆T
αP =
Estimate by finite difference.
Using values at 0oC, 20oC and 40oC,
αP ≈
1
0.001 0056 - 0.000 9977
= 0.000 1976/oC
0.000 9995
40 - 0
βT = -
1 ∂v
1 ∆v
≈
(
)
( )
v ∂P T
v ∆P T
Using values at saturation, 5 MPa and 10 MPa,
βT ≈ -
1
0.000 9972 - 0.001 0022
= 0.000 50 MPa-1
0.000 9995
10 - 0.0023
Water at 300oC, 15 MPa (compressed liquid)
αP ≈
1
0.001 4724 - 0.001 3084
= 0.002 977/oC
0.001 377
320 - 280
βT ≈ -
1
0.001 3596 - 0.001 3972
= 0.002 731 MPa-1
0.001 377
20 - 10
13.14 Sound waves propagate through a media as pressure waves that cause the media
to go through isentropic compression and expansion processes. The speed of
sound c is defined by c 2 = (∂P/∂ ρ )s and it can be related to the adiabatic
compressibility, which for liquid ethanol at 20°C is 940 µm2/N. Find the speed of
sound at this temperature.
c2 =
∂P
=
−v
(∂P
)
(
) =
∂ρ s
∂v s
1
2
1 ∂v
ρ
v ∂P s
-
From Table A.4 for ethanol,
⇒
c=
(940×101
-12
)
×783
( )
=
1
β sρ
ρ = 783 kg/m3
1/2
= 1166 m/s
13-12
13.15 Consider the speed of sound as defined in Problem 13.14. Calculate the speed of
sound for liquid water at 20°C, 2.5 MPa and for water vapor at 200°C, 300 kPa
using the steam tables.
c2 =
From problem 13.14 :
∂P
=
-v
(∂P
)
(
)
∂ρ s
∂v s
2
Liquid water at 20oC, 2.5 MPa, assume
∆P
≈
(∂P
)
(
)
∂v s
∆v
T
Using saturated liquid at 20oC and compressed liquid at 20oC, 5 MPa,
9995
5-0.0023
(0.001 002+0.000
)
(
) = 2.002×10
2
0.000 9995-0.001 002
2
c2 = -
=>
6
c = 1415 m/s
Superheated vapor water at 200oC, 300 kPa
v = 0.7163, s = 7.3115
At P = 200 kPa & s = 7.3115: T = 157oC,
v = 0.9766
At P = 400 kPa & s = 7.3115: T = 233.8oC, v = 0.5754
c2 = -(0.7163)2
=>
0.400-0.200
(0.5754-0.9766
) = 0.2558×10
6
c = 506 m/s
13.16 Find the speed of sound for air at 20°C, 100 kPa using the definition in Problem
13.14 and relations for polytropic processes in ideal gases.
From problem 13.14 :
c2 =
∂P
=
-v
(∂P
)
(
)
∂ρ s
∂v s
2
For ideal gas and isentropic process, Pvk = const
P = Cv-k ⇒
∂P
= -kCv-k-1 = -kPv-1
∂v
c2 = -v2(-kPv-1) = kPv = kRT
c = kRT =
1.4×0.287×293.15×1000 = 343.2 m/s
13-13
13.17 A cylinder fitted with a piston contains liquid methanol at 20°C, 100 kPa and
volume 10 L. The piston is moved, compressing the methanol to 20 MPa at
constant temperature. Calculate the work required for this process. The isothermal
compressibility of liquid methanol at 20°C is 1220 (µm)2/N.
2
2
⌠ ∂v
⌠
w12 = ⌠
⌡ Pdv = 
⌡ P ∂P T dPT = -⌡ vβ T PdPT
( )
1
1
For v ≈ constant & βT ≈ constant the integral can be evaluated
w12 = -
vβ T
2
(P
2
2
2
- P1
)
For liquid methanol, from Table A.4: ρ = 787 m3/kg
V1 = 10 L,
W12 =
m = 0.01×787 = 7.87 kg
0.01×1220
(20)2 - (0.1)2
2
[
] = 2440 J = 2.44 kJ
13.18 A piston/cylinder contains 5 kg of butane gas at 500 K, 5 MPa. The butane
expands in a reversible polytropic process with polytropic exponent, n = 1.05,
until the final pressure is 3 MPa. Determine the final temperature and the work
done during the process.
C4H10 m = 5 kg T1 = 500 K P1 = 5 MPa
Rev. polytropic process (n=1.05), P2 = 3 MPa
Tr1 =
500
5
= 1.176, Pr1 =
= 1.316
425.2
3.8
V1 =
n
P1V1
From Fig. D.1:
Z1 = 0.68
5×0.68×0.14304×500
= 0.0486 m3
5000
=
Z2Tr2 =
n
P2V2
=>
P2V2
mRTC
=
V2 = 0.0486
1
1.05
(5/3)
= 0.0791 m3
3000×0.0791
= 0.7803
5×0.143 04×425.2
at Pr2 = 3/3.8 = 0.789
Trial & error: Tr2 = 1.062, Z2 = 0.735
2
W12 = ⌠
⌡ PdV =
1
P2V2 - P1V1
1-n
=
=>
T2 = 451.6 K
3000×0.0791 - 5000×0.0486
= 114 kJ
1 - 1.05
13-14
13.19 Show that the two expressions for the Joule–Thomson coefficient µJ given by Eq.
13.54 are valid.
µj = (
Also
µj =
∂T
∂h
=
)
(
)
∂P h
∂P T
/(∂h∂T)P = [T(∂v∂T)P - v]/C
v=
ZRT
,
P
P
∂v
(∂T)P =
RT ∂Z
[ZRT
+
( ) - v]/C
P
P ∂T P
2
=
P
ZR RT ∂Z
+
( )
P
P ∂T P
RT2 ∂Z
( )
CPP ∂T P
13.20 A 200-L rigid tank contains propane at 9 MPa, 280°C. The propane is then
allowed to cool to 50°C as heat is transferred with the surroundings. Determine
the quality at the final state and the mass of liquid in the tank, using the
generalized compressibility chart, Fig. D.1.
Propane C3H8: V = 0.2 m3, P1 = 9 MPa, T1 = 280oC = 553.2 K
cool to T2 = 50 oC = 323.2 K
From Table A.2:
TC = 369.8 K,
PC = 4.25 MPa
9
553.2
= 2.118, Tr1 =
= 1.496 From Fig. D.1: Z1 = 0.825
4.25
369.8
Z1RT1 0.825×0.188 55×553.2
v2 = v1 =
=
= 0.00956
P1
9 000
Pr1 =
From Fig. D.1 at Tr2 = 0.874,
PG2 = 0.45 × 4250 = 1912 kPa
vG2 = 0.71 × 0.188 55 × 323.2/1912 = 0.02263
vF2 = 0.075 ×0.188 55× 323.2/1912 = 0.00239
0.00956 = 0.002 39 + x 2(0.02263 - 0.00239)
mLIQ 2 = (1-0.354)×0.2/0.00956 = 13.51 kg
=>
x 2 = 0.354
13-15
13.21 A rigid tank contains 5 kg of ethylene at 3 MPa, 30°C. It is cooled until the
ethylene reaches the saturated vapor curve. What is the final temperature?
V = const m = 5 kg
P1 = 3 MPa T1 = 30 oC = 303.2 K
C2 H4
T
cool to x2 = 1.0
1
Pr1 =
2
Fig. D.1:
v
Pr2 = Pr1
Z2Tr2
Z1Tr1
Trial & error:
Tr2
ZG2
0.866
0.72
3
= 0.595,
5.04
= 0.595
Pr2
Tr1 =
303.2
= 1.074
282.4
Z1 = 0.82
ZG2Tr2
0.82×1.074
= 0.6756 ZG2Tr2
Pr2 CALC
0.42
0.421
~ OK
=> T2 = 244.6 K
13.22 Two uninsulated tanks of equal volume are connected by a valve. One tank
contains a gas at a moderate pressure P 1, and the other tank is evacuated. The
valve is opened and remains open for a long time. Is the final pressure P2 greater
than, equal to, or less than P1/2?
VA = VB ⇒ V2 = 2V1, T2 = T1 = T
V1 Z2 mRT
1 Z2
=
=
2 Z1
P1 V2 Z1 mRT
P2
If T < TB, Z2 > Z1
A
GAS
P2
1
⇒
>
P1 2
If T > TB, Z2 < Z1 ⇒
P2
P1
Z
1
1.0
<
1
2
B
EVAC.
T > TB
2
2
P2
T < TB
1
P1 P
13-16
13.23 Show that van der Waals equation can be written as a cubic equation in the
compressibility factor involving the reduced pressure and reduced temperature as
27 Pr2
27 Pr 
Z–
Z3 – (
+ 1 ) Z2 + 
=0
8Tr
64 T2
512 Tr 3
r

Pr
van der Waals equation, Eq.13.55:
2 2
27 R TC
a=
64 PC
P=
RT a
v-b v2
b=
RTC
8PC
v2(v-b)
multiply equation by
P
Get:
v3 - (b +
Multiply by
Get:
RT 2 a
ab
)v +( )v- =0
P
P
P
Pv
P3
3 3 and substitute Z = RT
R T
bP
aP
abP2
Z3 – ( + 1) Z2 + ( 2 2) Z – ( 3 3) = 0
RT
RT
R T
Substitute for a and b, get:
27 Pr 
27 Pr2
Z–
Z3 – (
+ 1 ) Z2 + 
=0
8Tr
64 T2
512 Tr 3
r

Pr
Where
Pr =
P
,
Pc
Tr =
T
Tc
13-17
13.24 Develop expressions for isothermal changes in enthalpy and in entropy for both
van der Waals equation and Redlich-Kwong equation of state.
RT a
van der Waals equation of state:
P=
−
v-b v2
R
=
(∂P
)
∂T v v-b
RT RT a
-P=
(∂u∂v)T = T(∂P
)
−
+
v
v-b v-b v
∂T
2
2
2
⌠
∂P
⌠a
1 1

P]dv
=
(u2-u1)T = 
[T
dv
=
a(
− )
2
⌡
v1 v2
∂T v
⌡v
( )
1
1
1 1
(h2-h1)T = (u2-u1)T + P2v2 - P1v1 = P2v2 − P1v1 + a( − )
v1 v2
2
2
v2-b
⌠ ∂P
⌠ R

dv
=
dv
=
R
ln
(s2-s1)T = 
⌡ ∂T v
⌡ v-b
v1-b
( )
( )
1
1
Redlich-Kwong equation of state:
R
a
=
(∂P
)
+
v
v-b 2v(v+b)T
∂T
P=
RT
a
v-b v(v+b)T1/2
3/2
2
v2+b v2
⌠
3a
-3a
(u2-u1)T = 
dv =
ln[(
)(v +b)]
v2
⌡ 2v(v+b)T1/2
2bT1/2
1
1
(h2-h1)T = P2v2 - P1v1 -
3a
ln
2bT1/2
[(vv+b)(vv+b)]
2
2
2
1
2
⌠ R
a/2
(s2-s1)T = 
+
⌡ v-b v(v+b)T3/2 dv
[
]
1
= R ln
(vv -b-b)- 2bTa ln[(vv+b)(vv+b)]
2
1
2
3/2
1
2
1
13-18
13.25 Determine the reduced Boyle temperature as predicted by an equation of state (the
experimentally observed value for most substances is about 2.5), using the van der
Waals equation and the Redlich–Kwong equation. Note: It is helpful to use Eqs.
13.47 and 13.48 in addition to Eq. 13.46
lim ∂Z
( ) =0
P→0 ∂P T
The Boyle temp. is that T at which
RT
lim Z-1 1 lim
lim ∂Z
But P→0( )T = P→0
=
(
v- )
P→0
P-0 RT
P
∂P
van der Waals:
P=
multiply by
v-b =
RT a
−
v-b v2
v-b
, get
P
RT a(v-b)
or
P
Pv2
v-
RT
a(1-b/v)
=b−
P
Pv
&
a(1-0)
lim ∂Z
RT × P→0( )T = b −
=0
RT
∂P
or
TBoyle =
Redlich-Kwong:
only at TBoyle
a 27
= T = 3.375 TC
Rb 8 C
P=
RT
a
−
v-b v(v+b)T1/2
as in the first part, get
RT
a(1-b/v)
v=b−
P
Pv(1+b/v)T1/2
a(1-0)
lim ∂Z
& RT × P→0( )T = b −
=0
∂P
Pv(1+0)T1/2
5/2
or
3/2
TBoyle =
TBoyle = (
only at TBoyle
2
PC
a 0.427 48 R TC
=
×
Rb
RPC
0.08 664 R TC
0.427 48 2/3
) TC = 2.9 TC
0.086 64
13-19
13.26 Consider a straight line connecting the point P = 0, Z = 1 to the critical point
P = PC, Z = ZC on a Z versus P compressibility diagram. This straight line will be
tangent to one particular isotherm at low pressure. The experimentally determined
value is about 0.8 TC. Determine what value of reduced temperature is predicted
by an equation of state, using the van der Waals equation and the Redlich–Kwong
equation. See also note for Problem 13.25.
slope =
ZC - 1
1.0
Z
PC - 0
lim ∂Z
( ) for T = T′
P→0 ∂P T
From solution 13.25
C.P.
ZC
But also equals
0
PC P
1 lim
RT
lim Z-1
lim ∂Z
=
(
)
=
(
v
−
)
P→0 ∂P T
P→0 P-0
RT P→0
P
VDW: using solution 13.25:
ZC - 1
1
a
lim ∂Z
=
(
)
=
[
b
−
]
P→0 ∂P T
PC
RT′
RT′
1-Z
( P C)(RT′)2 + bRT′ − a = 0
or
C
2
3
Substituting ZC = ,
8
2
27 R TC
a=
,
64 PC
b=
2
40 T′r + 8 T′r − 27 = 0
RTC
8PC
solving,
Redlich-Kwong: using solution 13.25,
ZC-1
1
a
lim ∂Z
=
(
)
=
[
b
] or
P→0 ∂P T
PC RT′
RT′ 3/2
′
Tr = 0.727
1-Z
( P C)R2T′ 5/2 + bRT′ 3/2 - a = 0
C
5/2
Substitute
get
1
ZC = ,
3
a = 0.42748
R2TC
PC
,
b = 0.08664
2 5/2
3/2
T′r + 0.086 64 T′r − 0.427 48 = 0
3
solving,
T′r = 0.787
RTC
PC
13-20
13.27 Determine the 2nd virial coefficient B(T) using the van der Waals equation and
the Redlich–Kwong equation of state. Find also its value at the critical point (the
−
experimentally observed value is about -0.34 RTC/PC).
−
RT −
lim
where Eq.13.47: α =
−v
From Eq.13.51: B(T) = - P→0 α
P
−
−
RT a
v -b
van der Waals:
P = − - −2
which we can multiply by
, get
P
v -b v
−
−
− -b)
R
T
a(v
RT
a(1-b/v− )
−
−
v -b=
− −2
or v −
=b−
P
P
Pv−
Pv
Taking the limit for P -> 0 then we get :
−
RTC 1 27 TC
−
B(T) = b − a/RT =
( −
)
PC 8 64 T
where a,b are from Eq.13.59. At T = TC then we have
−
−
RTC 19
RTC
B(TC) =
( - ) = −0.297
64
PC
PC
For Redlich Kwong the result becomes
−
a(1- b/v− )
RT
−
=b−
v−
− ) T1/2
P
Pv− (1 + b/v
=>
Now substitute Eqs.13.61 and 13.62 for a and b,
−
RTC
TC3/2
B(T) =
0.08664 - 0.42748  
T
PC
[
a
B(T) = b − − 3/2
RT
]
and evaluated at TC it becomes
−
−
RTC
RTC
B(TC) =
0.08664 - 0.42748 = −0.341
PC
PC
[
]
13-21
13.28 One early attempt to improve on the van der Waals equation of state was an
expression of the form
RT a
P=
v-b v2T
Solve for the constants a, b, and v using the same procedure as for the van der
c
Waals equation.
From the equation of state take the first two derivatives of P with v:
∂P
RT
2a
(∂v )T = - (v-b)
and
2+ 3
vT
∂ 2P
2RT 6a
(∂v2)T = - (v-b)
3- 4
vT
Since both these derivatives are zero at the critical point:
RT
2a
2RT 6a
and =0
2+ 3 =0
(v-b) v T
(v-b)3 v4T
RTC
a
− 2
Also,
PC =
vC-b v T
C C
solving these three equations:
3
vC = 3b,
2
27 R TC
a=
,
64 PC
b=
RTC
8PC
13.29 Use the equation of state from the previous problem and determine the Boyle
temperature.
RT
a
P=
− 2
v-b v T
v-b
RT a(1-b/v)
Multiplying by
gives: v − b =
−
P
P
PvT
Using solution from 13.25 for TBoyle:
RT
a(1-0)
a
(
v
−
)
=
b
−
=
b
−
= 0 at TBoyle
P→0
P
RT×T
RT2
lim
3
or
TBoyle =
a
=
Rb
2
27 R TC 1 8PC
=
64 PC R RTC
27
T
8 C
13-22
13.30 Calculate the difference in internal energy of the ideal-gas value and the real-gas
value for carbon dioxide at the state 20°C, 1 MPa, as determined using the virial
equation of state, including second virial coefficient terms. For carbon dioxide we
have: B = -0.128 m3/kmol, T(dB/dT) = 0.266 m3/kmol, both at 20°C.
virial eq.:
P=
⌠v
u-u = -
⌡∞
[
*
∂P
RT BRT
+ 2 ;
v
v
R
RT dB
(∂T) v = v + BR
2 + 2 (dT )
v
v
⌠v RT2 dB
∂P
dB
(∂T) v - P dv = - ⌡∞ v2 (dT ) dv = - RT
T( )
v
dT
]
[
]
[
]
Solution of virial equation (quadratic formula):
−
−
RT 8.3145×293.15
− 1 RT
−
v=
1 + 1 + 4BP/RT where:
=
= 2.43737
2 P
1000
P
[
]
− 1
v = × 2.43737
2
[1 +
1 + 4(-0.128)/2.43737
] = 2.3018 m /kmol
3
Using the minus-sign root of the quadratic formula results in a compressibility
factor < 0.5, which is not consistent with such a truncated equation of state.
u-u* =
-8.3145 × 293.15
0.266 = - 281.7 kJ/kmol
2.3018
[
]
13.31 Refrigerant-123, dichlorotrifluoroethane, which is currently under development as
a potential replacement for environmentally hazardous refrigerants, undergoes an
isothermal SSSF process in which the R-123 enters a heat exchanger as saturated
liquid at 40°C and exits at 100 kPa. Calculate the heat transfer per kilogram of R123, using the generalized charts, Fig. D.2
R-123: M = 152.93, TC = 456.9 K, PC = 3.67 MPa
1
T1 = T2 = 40 oC, x1 = 0
P2 = 100 kPa
2
Heat Exchanger
SSSF
From D.1: P1 = 0.084×3670 = 308 kPa
Tr1 = Tr2 = 313.2/456.9 = 0.685,
From Fig. D.2:
Pr1 = 0.084,
Pr2 = 0.1/3.67 = 0.027
(h* − h)1/RTC = 4.9
P2 < P1 with no work done, so process is irreversibel.
Energy Eq.:
q + h1 = h2,
From Fig. D.2:
Entropy Eq.:
s1 + ∫ dq/T + sgen = s2, sgen > 0
(h*-h)2/RTC = 0.056
h2 - h1 = 8.3145 × 456.9 [-0.056 + 0 + 4.901]/152.93= 120.4 kJ/kg
13-23
13.32 Calculate the heat transfer during the process described in Problem 13.18.
From solution 13.18,
V1 = 0.0486 m3, V2 = 0.0791 m3,
Tr1 = 1.176, Pr1 = 1.316 ;
From Fig. D.2:
*
W12 = 114 kJ
Tr2 = 1.062, Pr2 = 0.789, T2 = 451.6 K
(h*- h)1 = 1.30 RTC ,
(h*- h)2 = 0.90 RTC
*
h2 - h1 = 1.7164(451.6 - 500) = -83.1 kJ/kg
8.3145×425.2
(-0.90 + 1.30) = -58.8 kJ/kg
58.124
U2 - U1 = m(h2 - h1) - P2V2 + P1V1
h2 - h1 = -83.1 +
= 5(-58.8) - 3000×0.0791 + 5000×0.0486 = -288.3 kJ
Q12 = U2 - U1 + W12 = -174.3 kJ
13.33 Saturated vapor R-22 at 30°C is throttled to 200 kPa in an SSSF process.
Calculate the exit temperature assuming no changes in the kinetic energy, using
the generalized charts, Fig. D.2 and the R-22 tables, Table B.4
R-22 throttling process 1st law:
*
*
*
*
h2-h1 = (h2-h2) + (h2-h1) + (h1-h1) = 0
a) Generalized Chart, Fig. D.2, R = 8.31451/86.469 = 0.096156
303.2
*
= 0.821 => (h1-h1) = 0.096156 × 369.3 (0.53) = 18.82
Tr1 =
369.3
For CP0, use h values from Table B.4 at low pressure.
CP0 ≈ 278.115 - 271.594) / (30 - 20) = 0.6521 kJ/kg K
*
Substituting: (h2-h2) + 0.6521(T2-30) + 18.82 = 0
at Pr2 = 200/4970 = 0.040
Assume T2 = 5.0 oC => Tr2 =278.2/369.3 = 0.753
*
(h2-h2) = RT × 0.07 = 0.096156 × 369.3 (0.07) = 2.49
Substituting :
-2.49 + 0.6521(5.0-30) + 18.82 = -0.03 ≈ 0
⇒ T2 = 5.0 oC
b)
R-22 tables, B.4: at T1 = 30 oC, x1 = 1.0
=> h1 = 259.12
h2 = h1 = 259.12 , P2 = 0.2 MPa => T2 = 4.7 oC
13-24
13.34 250-L tank contains propane at 30°C, 90% quality. The tank is heated to 300°C.
Calculate the heat transfer during the process.
V = 250 L = 0.25 m3
T1 = 30 oC = 303.2 K, x1 = 0.90
C3 H8
T
2
Heat to T2 = 300 oC = 573.2 K
1
v
M = 44.094, TC = 369.8 K, PC = 4.25 MPa
R = 0.188 55, CP0 = 1.6794
Tr1 = 0.82 → Fig. D.1:
Z1 = (1- x1) Zf1 + x1 Zg1 = 0.1 × 0.05 + 0.9 × 0.785 = 0.711
*
Fig D.2:
SAT
Pr
m=
h1-h1
RTc
SAT
= 0.30 P1
= 0.1 × 4.43 + 0.9 × 0.52 = 0.911
= 1.275 MPa
1275×0.25
= 7.842 kg
0.711×0.188 55×303.2
Pr2 =
7.842×Z2×0.188 55×573.2
at Tr2 = 1.55
0.25×4250
=
Z2
1.254
Trial and error on Pr2
Pr2 = 0.743 => P2 = 3.158 MPa, Z2 = 0.94 , (h*- h)2 = 0.35 RTC
*
*
(h2-h1) = 1.6794(300-30)
= 453.4 kJ/kg
*
(h1-h1) = 0.911×0.188 55×369.8 = 63.5 kJ/kg
*
(h2-h2) = 0.35×0.188 55×369.8 = 24.4 kJ/kg
Q12 = m(h2-h1) - (P2-P1)V = 7.842(-24.4+453.4+63.5) - (3158-1275)×0.25
= +3862 - 471 = 3391 kJ
13-25
13.35 The new refrigerant fluid R-123 (see Table A.2) is used in a refrigeration system
that operates in the ideal refrigeration cycle, except the compressor is neither
reversible nor adiabatic. Saturated vapor at -26.5°C enters the compressor and
superheated vapor exits at 65°C. Heat is rejected from the compressor as 1 kW
and the R-123 flow rate is 0.1 kg/s. Saturated liquid exits the condenser at 37.5°C.
Specific heat for R-123 is Cp =0.6 kJ/kg. Find the coefficient of performance.
R-123: Tc = 456.9 K, Pc = 3.67 MPa, M = 152.93 kg/kmol, R = 0.05438 kJ/kg K
State 1: T1 = -26.5oC = 246.7 K, sat vap., x1 = 1.0
Tr1 = 0.54, Fig D.1, Pr1 = 0.01, P1 = Pr1Pc = 37 kPa
*
Fig. D.2,
h1-h1 = 0.03 RTC = 0.8 kJ/kg
State 2: T2 = 65oC = 338.2 K
State 3: T = 37.5oC = 310.7 K, sat. liq., x = 0
3
3
Tr3 = 0.68, Fig. D.1: Pr3 = 0.08, P3 = Pr3Pc = 294 kPa
P2 = P3 = 294 kPa, Pr2 = 0.080, Tr2 = 0.74,
*
Fig. D.2:
h2-h2 = 0.25 RTC = 6.2 kJ/kg
*
h3-h3 = 4.92 RTC = 122.2 kJ/kg
State 4: T4 = T1 = 246.7 K,
h4 = h3
1st Law Evaporator: qL + h4 = h1 + w; w = 0,
*
qL = h1 - h3 = (h1 − h1)
*
*
*
h4 = h3
*
+ (h1 − h3) + (h3 − h3)
*
h1 − h3 = Cp(T1 - T3) = -38.4 kJ/kg, qL = -0.8 – 38.4 + 122.2 = 83.0 kJ/kg
.
.
1st Law Compressor: q + h1 = h2 + wc; Q = -1.0 kW, m = 0.1 kg/s
*
wc = h1 - h2 + q; h1 - h2 = (h1 − h1)
*
*
*
h1 − h2 = CP (T1 - T2) = -54.9 kJ/kg,
wc = -0.8 –54.9 + 6.2 – 10.0 = -59.5 kJ/kg
β = qL/wc = 83.0/59.5 = 1.395
*
*
+ (h1 − h2) + (h2 − h2)
13-26
13.36 A cylinder contains ethylene, C2H4, at 1.536 MPa, −13°C. It is now compressed
in a reversible isobaric (constant P) process to saturated liquid. Find the specific
work and heat transfer.
Ethylene C2H4 P1 = 1.536 MPa = P2 , T1 = -13oC = 260.2 K
State 2: saturated liquid, x2 = 0.0
Tr1 =
260.2
1.536
= 0.921 Pr1 = Pr2 =
= 0.305
282.4
5.04
*
From Figs. D.1, D.2: Z1 = 0.85 , (h1-h1)/RTc = 0.40
Z1RT1 0.85×0.29637×260.2
v1 =
=
= 0.042675
P1
1536
*
(h1-h1) = 0.296 37×282.4×0.40 = 33.5
From Figs. D.1, D.2: T2 = 0.824×282.4 = 232.7 K
*
Z2 = 0.05 , (h2-h2)/RTc = 4.42
v2 =
Z2RT2 0.05×0.29637×232.7
=
= 0.002245
P2
1536
*
(h2-h2) = 0.296 37×282.4×4.42 = 369.9
*
*
(h2-h1) = CP0(T2-T1) = 1.5482(232.7-260.2) = -42.6
w12 = ⌠
⌡ Pdv = P(v2-v1) = 1536(0.002 245-0.042 675) = -62.1 kJ/kg
q12 = (u2-u1) + w12 = (h2-h1) = -369.9 - 42.6 + 33.5 = -379 kJ/kg
13-27
13.37 A piston/cylinder initially contains propane at T = -7°C, quality 50%, and volume
10L. A valve connecting the cylinder to a line flowing nitrogen gas at T = 20°C, P
= 1 MPa is opened and nitrogen flows in. When the valve is closed the cylinder
contains a gas mixture of 50% nitrogen, 50% propane on a mole basis at T =
20°C, P = 500 kPa. What is the cylinder volume at the final state and how much
heat transfer took place?
State 1: Propane, T1 = -7oC, x1 = 0.5, V1 = 10 L
Tcp = 369.8 K, Pcp = 4.25 kPa, Cpp = 1.679 kJ/kg-K, M = 44.097 kg/kmol
Fig. D.1: Tr1 = 0.72, Pr1 = 0.12, P1 = Pr1Pc = 510 kPa
Fig. D.1: Zf1 = 0.020, Zg1 = 0.88, Z1 = (1 - x1)Zf + x1Zg = 0.45
−
n1 = P1V1/(Z1RT1) = 510 × 0.01/(0.45 × 8.3145 × 266.2) = 0.00512 kmol
−* −
− −*
−*
−
h 1 = h 1o + CPP(T1 - To) + (h 1 - h 1 ) ; h 1o = 0,
−* −
−
−* −
−
(h 1-h 1)f /RTc = 4.79, (h 1-h 1)g /RTc = 0.25
−* −
−* −
−* −
h 1 - h 1 = (1 - x1) (h 1 - h 1)f + x1 (h 1 - h 1)g = 7748 kJ/kmol
−
h 1 = 0 + 1.679 × 44.094(-7 - 20) - 7748 = -9747 kJ/kmol
Inlet: Nitrogen, Ti = 20oC, Pi = 1.0 MPa,
Tcn = 126.2 K, Pcn = 3.39 MPa, Cpn = 1.042 kJ/kg-K, M = 28.013 kg/kmol
−* −
Tri = 2.323, Pri = 0.295, h i -h i = 0.06 × 8.3145 × 126.2 = 62.96 kJ/kmol
− −* −
− −*
−*
h i = h io + CPn (Ti - To) + (h i - h i ) ; h io = 0, Ti - To = 0
State 2: 50% Propane, 50% Nitrogen by mol, T2 = 20oC, P2 = 500 kPa
Tcmix = ∑yiTci = 248 K, Pcmix = ∑yiPci = 3.82 MPa
−* − −
Tr2 = 1.182, Pr2 = 0.131, Z2 = 0.97, (h 2-h 2)/RTc = 0.06
−
−* −
− −*
−*
h 2 = h 2o + CPmix(T2 - To) + (h 2 - h 2 ) ;; h 2o = 0, T2 - To = 0
−
a) ni = n1 => n 2 = n1 + ni = 0.1024, V2 = n2Z2RT2/P2 = 0.0484 m3
b) 1st Law: Qcv + nihi = n2u-2 - n21u-21 + Wcv;
u- = h - PvWcv = (P1 + P2)(V2 - V1)/2 = 19.88 kJ
Qcv = n2h2 - n1h1 - nihi - P2V2 + P1V1 + Wcv
−
−
h i = -62.96 kJ/kmol, h 2 = -123.7 kJ/kmol, Qcv = 50.03 kJ
13-28
13.38 An ordinary lighter is nearly full of liquid propane with a small amount of vapor,
the volume is 5 cm3 and temperature is 23°C. The propane is now discharged
slowly such that heat transfer keeps the propane and valve flow at 23°C. Find the
initial pressure and mass of propane and the total heat transfer to empty the
lighter.
Propane C3H8
T1 = 23oC = 296.2 K = constant,
V1 = 5 cm3 = 5×10-6 m3,
x1 = 0.0
Tr1 = 296.2/369.8 = 0.804
From Figs. D.1 and D.2,
P1 = PG T1 = 0.25×4.25 = 1.063 MPa,
Z1 = 0.04
*
(h1-h1) = 0.188 55×369.8×4.51 = 314.5
m1 =
P1V1
1063×5×10-6
=
= 0.00238 kg
Z1RT1 0.04×0.188 55×296.2
State 2: Assume vapor at 100 kPa, 23oC
Therefore, m2 much smaller than m1 ( ∼ 9.0 × 10-6 kg)
QCV = m2u2 - m1u1 + mehe
= m2h2 - m1h1 - (P2-P1)V + (m1-m2)he
= m2(h2-he) + m1(he-h1) - (P2-P1)V
(he - h1) = 0 + 0 + 314.5
QCV = ≈ 0 + 0.00238(314.5) - (100-1063)×5×10-6 = 0.753 kJ
13-29
13.39 An uninsulated piston/cylinder contains propene, C3H6, at ambient temperature,
19°C, with a quality of 50% and a volume of 10 L. The propene now expands
very slowly until the pressure in the cylinder drops to 460 kPa. Calculate the mass
of propene, the work and heat transfer for this process.
Propene C3H6:
T1 = 19oC = 292.2 K,
x1 = 0.50,
From Fig. D.1:
Tr1 = 292.2/364.9 = 0.80,
Pr1 = Pr sat = 0.25,
P1 = 0.25 × 4.6 = 1.15 MPa
Z1 = 0.5 × 0.04 + 0.5 × 0.805 = 0.4225
From D.1:
m=
P1V1
Z1RT1
V1 = 10 L
=
1150×0.010
= 0.471 kg
0.4225×0.197 58×292.2
Assume reversible and isothermal process (slow, no friction, not insulated)
Q12 = m(u2-u1) + W12
2
2
W12 = ⌠
⌡ PdV (cannot integrate);
Q12 = ⌠
⌡ TdS = Tm(s2-s1)
1
1
From Figs. D.2 and D.3:
*
h1-h1 = 0.19758 × 364.9(0.5 × 4.51 + 0.5 × 0.46) = 179.2
*
(s1-s1) = 0.197 58 (0.5 × 5.46 + 0.5 × 0.39) = 0.5779
*
*
(h2-h1) = 0 and
* *
(s2-s1) = 0 - 0.197 58 ln
460
= + 0.1829
1161
At Tr2 = 0.80, Pr2 = 0.10, from D.1, D.2 and D.3, Z2 = 0.93
*
(h2-h2) = 0.197 58 × 364.9 × 0.16 = 11.5
*
(s2-s2) = 0.197 58 × 0.13 = 0.0257
(s2-s1) = -0.0257 + 0.1829 + 0.5779 = 0.7351
q12 = 292.2 × 0.7351 = 214.8 kJ/kg
=>
Q12 = m q12 = 101.2 kJ
(h2-h1) = -11.5 + 0 + 179.2 = 167.7
u2-u1 = (h2-h1) + RT(Z1-Z2) = 167.7 + 0.197 58×292.2(0.4225-0.93) = 138.4
w12 = q12 - (u2-u1) = 214.8 - 138.4 = 76.4 kJ/kg
W12 = m w12 = 36.0 kJ
13-30
13.40 A 200-L rigid tank contains propane at 400 K, 3.5 MPa. A valve is opened, and
propane flows out until half the initial mass has escaped, at which point the valve
is closed. During this process the mass remaining inside the tank expands
according to the relation Pv1.4 = constant. Calculate the heat transfer to the tank
during the process.
C3H8: V = 200 L, T1 = 400 K, P1 = 3.5 MPa
Flow out to m2 = m1/2 ; Pv1.4 = const inside
Tr1 =
400
3.5
= 1.082, Pr1 =
= 0.824 Fig D.1: Z1 = 0.74
369.8
4.25
0.74×0.188 55×400
= 0.01594, v2 = 2v1 = 0.03188
3500
0.2
1
m1 =
= 12.55 kg, m2 = m1 = 6.275 kg,
0.015 94
2
v1 =
P 2 = P1
(vv )
1 1.4
=
2
3500
= 1326 kPa
21.4
Trial & error: saturated with
1.326
Pr2 =
= 0.312  T2 = 0.826×369.8 = 305.5 K &
4.25

P2v2 = Z2RT2
 Z2 = 1326×0.03188 = 0.734
0.188 55×305.5
Z2 = ZF2 + x2(ZG2 - ZF2) = 0.734 = 0.05 + x2(0.78-0.05)
=>
x2 = 0.937
*
(h1-h1) = 0.188 55×369.8(0.9) = 62.8
*
*
(h2-h1) = 1.6794(305.5-400) = -158.7
*
[
*
(h2-h2) = (h2-hF2) - x2hFG2 = 0.188 55×369.8 4.41 - 0.937(4.41-0.55)
= 55.3
1st law: QCV = m2h2 - m1h1 + (P1-P2)V + mehe AVE
*
Let h1 = 0 then
*
*
*
*
h1 = 0 + (h1-h1) = -62.8
*
h2 = h1 + (h2-h1) + (h2-h2) = 0 - 158.7 – 55.3 = -214.0
he AVE = (h1+h2)/2 = -138.4
QCV = 6.275(-214.0) - 12.55(-62.8)
+ (3500-1326)×0.2 + 6.275(-138.4) = -981.4 kJ
]
13-31
13.41 A newly developed compound is being considered for use as the working fluid in
a small Rankine-cycle power plant driven by a supply of waste heat. Assume the
cycle is ideal, with saturated vapor at 200°C entering the turbine and saturated
liquid at 20°C exiting the condenser. The only properties known for this
compound are molecular weight of 80 kg/kmol, ideal gas heat capacity CPO= 0.80
kJ/kg K and TC= 500 K, PC= 5 MPa. Calculate the work input, per kilogram, to
the pump and the cycle thermal efficiency.
T1 = 200oC = 473.2 K, x1 = 1.0
1
.
WT
Turbine
.
QH
Ht.
Exch
T3 = 20oC = 293.2 K, x3 = 0.0
Properties known:
M = 80, CPO = 0.8 kJ/kg K
2
TC = 500 K, PC = 5.0 MPa
Cond
Tr1 =
P
4
.
-W P
473.2
293.2
= 0.946 , Tr3 =
= 0.586
500
500
3
From Fig. D.1,
Pr1 = 0.72, P1 = 0.72×5 = 3.6 MPa = P4
Pr3 = 0.023, P3 = 0.115 MPa = P2 , ZF3 = 0.004
vF3 =
ZF3RT3
P3
=
0.004×8.3145×293.2
= 0.00106
115×80
4
⌡ vdP ≈ vF3(P4 -P3) = -0.00106(3600-115) = -3.7 kJ/kg
wP = - ⌠
3
qH + h4 = h1 , but h3 = h4 + wP =>
q H = (h1-h3) + wP
From Fig. D.2:
*
(h1-h1) = 0.103 93×500×1.25 = 64.9
*
(h3-h3) = 0.103 93×500×5.2 = 270.2
*
*
(h1-h3) = CP0(T1-T3) = 0.80(200-20) = 144.0
(h1-h3) = -64.9 + 144.0 + 270.2 = 349.3
13-32
qH = 349.3 + (-3.7) = 345.6 kJ/kg
*
*
*
*
Turbine, (s2 - s1) = 0 = -(s2 - s2)+(s2 - s1) + (s1 - s1)
From Fig. D.3,
*
(s1-s1) = 0.10393×0.99 = 0.1029
* *
(s2-s1) = 0.80 ln
293.2
115
- 0.103 93 ln
= -0.0250
473.2
3600
Substituting,
*
*
s2-s2 = +0.1029 - 0.0250 = 0.0779 = (s2-sF2) - x2sFG2
0.0779 = 0.103 93×8.85 - x2×0.103 93(8.85-0.06)
*
*
(h2-h2) = (h2-hF2) - x2hFG2
From Fig. D.2,
hFG2 = 0.10393×500(5.2-0.07) = 266.6
*
(h2-h2) = 270.2 -0.922 × 266.6 = 25.0
wT = (h1-h2) = -64.9 + 144.0 + 25.0 = 104.1 kJ/kg
ηTH =
wNET
qH
=
104.1-3.7
= 0.29
345.6
=> x 2 = 0.922
13-33
13.42 A geothermal power plant on the Raft river uses isobutane as the working fluid.
The fluid enters the reversible adiabatic turbine, as shown in Fig. P13.42, at
160°C, 5.475 MPa and the condenser exit condition is saturated liquid at 33°C.
Isobutane has the properties Tc= 408.14 K, Pc= 3.65 MPa, CP0= 1.664 kJ/kg K
and ratio of specific heats k = 1.094 with a molecular weight as 58.124. Find the
specific turbine work and the specific pump work.
Turbine inlet: T1 = 160oC , P1 = 5.475 MPa
Condenser exit: T3 = 33oC , x3 = 0.0, Tr3 = 306.2 / 408.1 = 0.75
From Fig. D.1:
=> P2 = P3 = 0.16 × 3.65 = 0.584 MPa
Pr3 = 0.16, Z3 = 0.03
Tr1 = 433.2 / 408.1 = 1.061,
Pr1 = 5.475 / 3.65 = 1.50
From Fig. D.2 & D.3:
*
(h1-h1) = 0.143 05×408.1×2.84 = 165.8
*
(s1-s1) = 0.143 05×2.15 = 0.3076
* *
(s2-s1) = 1.664 ln
*
306.2
0.584
- 0.143 05 ln
= -0.2572
433.2
5.475
*
(s2-s2) = (s2-sF2) - x2sFG2
= 0.143 05×6.12 - x2×0.143 05(6.12-0.29) = 0.8755 - x2×0.8340
(s2-s1) = 0 = -0.8755 + x2×0.8340 - 0.2572 + 0.3076
*
=>
x2 = 0.99
*
(h2-h1) = CP0(T2-T1) = 1.664(306.2 - 433.2) = -211.3
From Fig. D.2:,
*
*
(h2-h2) = (h2-hF2) - x2hFG2 = 0.143 05×408.1[4.69-0.99(4.69-0.32)]
= 273.8 − 0.99 × 255.1 = 21.3
Turbine:
wT = (h1-h2) = -165.8 + 211.3 + 21.3 = 66.8 kJ/kg
Pump:
vF3 =
ZF3RT3
P3
=
0.03×0.143 05×306.2
= 0.00225
584
wP = - ∫ v dP ≈ vF3(P4 -P3) = -0.00225 (5475-584) = -11.0 kJ/kg
13-34
13.43 Carbon dioxide collected from a fermentation process at 5°C, 100 kPa should be
brought to 243 K, 4 MPa in an SSSF process. Find the minimum amount of work
required and the heat transfer. What devices are needed to accomplish this change
of state?
278.2
100
Tri =
= 0.915 Pri =
= 0.0136
304.1
7380
From D.2 and D.3 : (h*-h) /RTC = 0.02, (s*-s)ri/R = 0.01
ri
Tre =
243
4
= 0.80, Pre =
= 0.542
304.1
7.38
From D.2 and D.3:
*
(h*-h) /RTC = 4.5
re
*
*
(s*-s)re/R = 4.74
*
(hi-he) = - (hi -hi) + (hi -he) + (he-he)
= - 0.188 92×304.1×0.01 + 0.8418(278.2-243)
+ 0.188 92×304.1×4.5
*
* *
= 287.6 kJ/kg
*
(si-se) = - (si -si) + (si -se) + (se-se)
= - 0.188 92×0.01 + 0.8418 ln(278.2/243)
- 0.188 92 ln(0.1/4) + 0.188 92×4.74 = 1.7044 kJ/kg K
wrev = (hi-he) -T0(si-se) = 287.6 - 278.2(1.7044) = -186.6 kJ/kg
qrev = (he-hi) + wrev = -287.6 -186.6 = -474.2 kJ/kg
13-35
13.44 An insulated cylinder fitted with a frictionless piston contains saturated-vapor
carbon dioxide at 0oC, at which point the cylinder volume is 20 L. The external
force on the piston is now slowly decreased, allowing the carbon dioxide to
expand until the temperature reaches - 30 oC. Calculate the work done by the
CO2during this process.
CO2: Tc = 304.1 K, Pc = 7.38 MPa, Cp = 0.842 kJ/kg-K, R = 0.18892 kJ/kg K
State 1: T1 = 0oC, sat. vap., x1 = 1.0, V1 = 20 L
Tr1 = 0.9, P1 = Pr1Pc = 0.53 × 7380 = 3911 kPa, Z1 = Zg = 0.67
*
(h1 − h1)g
P 1V 1
*
= 0.9 RTC, (s1 − s1)g/R = 0.72, m =
= 2.262 kg
Z1RT1
State 2: T2 = -30oC
Tr2 = 0.8, P2 = Pr2Pc = 0.25 × 7380 = 1845 kPa
2nd Law:
∆Snet =m(s2 − s1) − 1Q2/T ;
*
s2 - s1 = (s2 − s2)
*
*
s2 − s1 = CP ln
T2
T1
*
*
1Q 2
− R ln
*
*
*
P2
*
P1
= 0.044 kJ/kg-K, s1 − s1 = 0.136 kJ/kg-K
*
*
(s2 − s2)f = 5.46 R, (s2 − s2)g = 0.39 R
*
*
(s2 − s2) = (1-x2)(s2 − s2)f + x2 (s2 − s2)g
1Q 2
∆Snet = 0
+ (s2 − s1) + (s1 − s1) = 0
s2 - s2 = 0.180 kJ/kg K,
1st Law:
= 0,
‡ x2 = 0.889
= m(u2 − u1) + 1W2 ; 1Q2 = 0,
u = h - Pv
Z2 = (1 - x2)Zf + x2Zg = 0.111 × 0.04 + 0.889 × 0.81 = 0.725;
*
(h2 - h1) = (h2 − h2)
*
*
*
*
+ (h2 − h1) + (h1 − h1)
*
*
h2 − h1 = Cp(T2 - T1) = -25.3 kJ/kg, (h1 − h1) = 51.7 kJ/kg
*
(h2 − h2)f
= 4.51 RTC ,
*
*
*
(h2 − h2)g = 0.46 RTC
*
(h2 − h2) = (1 - x2)(h2 − h2)f + x2 (h2 − h2)g =
52.2 kJ/kg
h2 - h1 = -52.2 – 25.3 + 51.7 = -25.8 kJ/kg
u2 - u1 = (h2 - h1) - Z2RT2 + Z1RT1 = -25.8 – 0.725 × 0.18892 × 243.2
+ 0.67 × 0.18892 × 273.2 = -24.5 kJ/kg
1W2 = 55.4 kJ
13-36
13.45 An evacuated 100-L rigid tank is connected to a line flowing R-142b gas,
chlorodifluoroethane, at 2 MPa, 100°C. The valve is opened, allowing the gas to
flow into the tank for a period of time and then it is closed. Eventually, the tank
cools to ambient temperature, 20°C, at which point it contains 50% liquid, 50%
vapor, by volume. Calculate the quality at the final state and the heat transfer for
the process. The ideal-gas specific heat of R-142b is C = 0.787 kJ/kg K.
Rigid tank V = 100 L, m1 = 0
Line: R-142b CH3CClF2
M = 100.495, TC = 410.3 K, PC = 4.25 MPa, CP0 = 0.787 kJ/kg K
−
R = R/M = 8.31451 / 100.495 = 0.082 73 kJ/kg K
Line Pi = 2 MPa, Ti = 100 oC, Flow in to T2 = T0 = 20oC
VLIQ 2 = VVAP 2 = 50 L
Continuity: mi = m2 ; Energy:
QCV + mihi = m2u2 = m2h2 - P2V
From D.2 at i: Pri = 2 / 4.25 = 0.471,
Tri = 373.15 / 410.3 = 0.91
*
(hi -hi) = 0.082 73×410.3×0.72 = 24.4
*
*
(h2-hi ) = CP0(T2-Ti) = 0.787(20-100) = -63.0
From D.2: Tr2 =
293.2
= 0.715 => P2 = 0.115×4250 = 489 kPa
410.3
sat. liq.: ZF = 0.02, (h*-hF) = RTC×4.85 = 164.6
sat. vap.: ZG = 0.88, (h*-hG) = RTC×0.25 = 8.5
mLIQ 2 =
mVAP 2 =
P2VLIQ 2
ZFRT2
=
P2VVAP 2
ZGRT2
489×0.050
= 50.4 kg
0.02×0.082 73×293.2
= 1.15 kg,
m2 = 51.55 kg
x2 = mVAP 2/m2 = 0.0223
*
*
*
(h2-h2) = (1-x2)(h2-hF2) + x2(h2-hG2) = 0.9777 × 164.6 + 0.0223 × 8.5 = 161.1
QCV = m2(h2-hi) - P2V = 51.55(-161.1-63.0+24.4) - 489×0.10
= -10 343 kJ
13-37
13.46 A cylinder fitted with a movable piston contains propane, initially at 67oC and 50
% quality, at which point the volume is 2 L. The piston has a cross-sectional area
of 0.2 m2. The external force on the piston is now gradually reduced to a final
value of 85 kN, during which process the propane expands to ambient
temperature, 4oC. Any heat transfer to the propane during this process comes
from a constant-temperature reservoir at 67oC, while any heat transfer from the
propane goes to the ambient. It is claimed that the propane does 30 kJ of work
during the process. Does this violate the second law?
+Q from Tres = 67oC
C3 H8
-Q to Environment To = 4oC
Fext
Fext 2 = 85 kN
Propane: Tc= 369.8 K, Pc = 4.25 MPa, R = 0.18855 kJ/kg K, Cp = 1.679 kJ/kg K
State 1: T1 = 67oC = 340.2 K, x1 = 0.5, V1 = 2.0 L
Tr1 = 0.92, Fig D.1, Pr1 = 0.61, P1 = Pr1Pc = 2.592 MPa
Zf1 = 0.10, Zg1 = 0.64, Z1 = (1 - x1)Zf1 + x1Zg1 = 0.37
P 1V 1
*
*
m=
= 0.218 kg, (h1 − h1)f = 3.95 RTc , (h1 − h1)g = 1.03 RTc
Z1RT1
*
*
(s1 − s1)f = 4.0 R , (s1 − s1)g = 0.82 R
State 2: T2 = 4oC = 277.2 K, Fext 2 = 85 kN
sat
sat
Tr2 = 0.75, P2 = Pr2 Pc = 0.165 × 4250 = 701 kPa
P2 = Fext 2/Ap = 425 kPa,
Pr2 = 0.10, Z2 = 0.92,
sat
P2 < P2 ‡ State 2 is a vapor
V2 = mZ2RT2/P2 = 0.0247 m3
*
h2 − h2 = 0.18 RTc =12.6 kJ/kg,
1st Law: 1Q2 = m(u2 - u1) + 1W2;
1Q 2
*
s2 − s2 = 0.16 R = 0.0302 kJ/kg K
1W2
= 30 kJ, u = h - Pv
= m(h2 - h1) - P2V2 + P1V1 + 1W2
*
(h2 - h1) = (h2 − h2)
*
*
*
*
+ (h2 − h1) + (h1 − h1)
*
*
(h1 − h1) = (1 - x1)(h1 − h1)f + x1 (h1 − h1)g =
*
173.6 kJ/kg
*
h2 − h1 = Cp(T2 - T1) = -105.8 kJ/kg
1Q 2
= 0.218 (-12.6 - 105.8 + 173.6) - 425×0.0247 + 2592×0.002 + 30
= 36.7 kJ
13-38
1Q 2
∆Snet = m(s2 − s1) −
; Tres = 67oC = 340.2 K
T
2nd Law:
*
s2 - s1 = (s2 − s2)
*
*
*
*
+ (s2 − s1) + (s1 − s1)
*
*
s1 − s1 = (1 - x1)(s1 − s1)f + x1 (s1 − s1)g = 0.4544 kJ/kg-K
*
*
s2 − s1 = Cpln
T2
T1
- R ln
P2
P1
= -0.0030 kJ/kg K
∆Snet = 0.218 (-0.0302-0.0030+0.4544) – 36.7/340.2 = -0.0161 kJ/K;
∆Snet < 0
Process is Impossible
13.47 Consider the following equation of state, expressed in terms of reduced pressure
and temperature:
Pr
6
(
1- 2)
Z=1+
14 Tr
Tr
What does this equation predict for enthalpy departure from the ideal gas value at
the state Pr = 0.4, Tr = 0.9 ? What does this equation predict for the reduced
Boyle temperature?
Pr
Pv
6
a)
Z=
=1+
(
1 - 2)
RT
14 Tr
Tr
3
6Tc2
RT RTc
v=
+
(1 - T2 ) ;
P 14Pc
∂v 
R 12RTc
  = +
∂T p P 14PcT3
3
RTc 18RTc
∂v 
v-T  =
∂T p 14Pc 14PcT3
P
RTc
∂v 
18
h - h* = ⌠
⌡ [v - T   ] dP =
(
1 − 2) Pr = 0.606 RTc
14
∂T p
Tr
0
∂Z
1
6
b)   =
(
1 - 2)
14P
∂P T
Tr
c Tr
(1 - T62) = 0
r
‡ Tr =
=>
6 = 2.45
Lim ∂Z
P→0 ∂P T = 0 at Tboyle
13-39
13.48 Saturated liquid ethane at 2.44 MPa enters (SSSF) a heat exchanger and is
brought to 611 K at constant pressure, after which it enters a reversible adiabatic
turbine where it expands to 100 kPa. Find the heat transfer in the heat exchanger,
the turbine exit temperature and turbine work.
From D.2,
Pr1 = 2.44/4.88 = 0.50 , Tr1 = 0.89, T1 = 0.89×305.4 = 271.8 K
*
(h1-h1) = 0.2765×305.4×4.12 = 347.9
*
*
(h2-h1) = 1.766 (611 - 271.8) = 599.0
Pr2 = 0.50 , Tr2 = 611/305.4 = 2.00
From D.2:
*
(h2-h2) = RTc × 0.14 = 0.2765×305.4×0.14 = 11.8
q = (h2-h1) = -11.8 + 599.0 + 347.9 = 935.1 kJ/kg
From D.3,
*
(s2-s2) = 0.2765×0.05 = 0.0138
* *
(s3-s2) = 1.766 ln
T3
- 0.2765 ln
611
Assume T3 = 368 K , Tr3 = 1.205
100
2440
at Pr3 = 0.020
* *
(s3-s2) = -0.8954 + 0.8833 = -0.0121
From D.3,
*
(s3-s3) = 0.2765×0.01 = 0.0028
(s3-s2) = -0.0028 - 0.0121 + 0.0138 ≈ 0
ΟΚ
Therefore, T3 = 368 K
From D.2,
*
(h3-h3) = 0.2765×305.4×0.01 = 0.8
w = (h2-h3) = -11.8 + 1.766 (611 - 368) + 0.8 = 418.1 kJ/kg
13-40
13.49 A flow of oxygen at 230 K, 5 MPa is throttled to 100 kPa in an SSSF process.
Find the exit temperature and the entropy generation.
Tri =
230
5
= 1.488, Pri =
= 0.992,
154.6
5.04
Pre =
0.1
= 0.02
5.04
*
From D.2:
(hi -hi) = 0.2598×154.6×0.50 = 20.1
*
First law:
*
*
*
(he-hi) = 0 = - (h e-he) + (he-hi ) + (hi -hi)
*
Assume Te = 208 K , Tre = 1.345:
*
(he-hi ) = 0.9216(208 - 230) = -20.3
*
From D.2:
(he-he) = 0.2598×154.6×0.01 = 0.4
(he-hi) = -0.4 -20.3 + 20.1 ≈ 0 OK
Check first law
=>
Te = 208 K
From D.3,
*
(si -si) = 0.2598×0.25 = 0.0649
* *
(se-si ) = 0.9216 ln
and
*
(se-se) = 0.2598×0.01 = 0.0026
208
0.1
- 0.2598 ln
= 0.9238
230
5
(se-si) = -0.0026 + 0.9238 + 0.0649 = 0.9861 kJ/kg K
13.50 A cylinder contains ethylene, C2H4, at 1.536 MPa, −13°C. It is now compressed
isothermally in a reversible process to 5.12 MPa. Find the specific work and heat
transfer.
Ethylene C2H4 P1 = 1.536 MPa , T2 = T1 = -13oC = 260.2 K
Tr2 = Tr1 = 260.2 / 282.4 = 0.921 , Pr1 = 1.536 / 5.04 = 0.305
From D.1, D.2 and D.3:
Z1 = 0.85
*
(h1-h1) = 0.2964×282.4×0.40 = 33.5 and
From D.1, D.2 and D.3:
Z2 = 0.17 ,
Pr2 = 5.12/5.04 = 1.016 (comp. liquid)
*
(h2-h2) = 0.2964×282.4×4.0 = 334.8 and
Ideal gas:
*
*
(h2-h1) = 0
and
*
(s1-s1) = 0.2964×0.30 = 0.0889
*
(s2-s2) = 0.2964×3.6 = 1.067
* *
(s2-s1) = 0 - 0.2964 ln
5.12
= -0.3568
1.536
q12 = T(s2-s1) = 260.2(-1.067 - 0.3568 + 0.0889) = -347.3 kJ/kg
(h2-h1) = -334.8 + 0 + 33.5 = -301.3
(u2-u1) = (h2-h1) - RT(Z2-Z1) = -301.3 - 0.2964×260.2(0.17-0.85) = -248.9
w12 = q12 - (u2-u1) = -347.3 + 248.9 = -98.4 kJ/kg
13-41
13.51 A control mass of 10 kg butane gas initially at 80°C, 500 kPa, is compressed in a
reversible isothermal process to one-fifth of its initial volume. What is the heat
transfer in the process?
T1 = 80 oC, P1 = 500 kPa
Butane C4H10: m = 10 kg,
Compressed, reversible T = const, to V2 = V1/5
353.2
500
= 0.831, Pr1 =
= 0.132
425.2
3800
Tr1 =
From D.1 and D.3:
Z1 = 0.92,
*
(s1- s1) = 0.143×0.16 = 0.0230
Z1RT1 0.92×0.143×353.2
v1 =
=
= 0.09296 m3/kg
P1
500
v2 = v1/5 = 0.01859 m3/kg
At Tr2 = Tr1 = 0.831
From D.1: PG = 0.325×3800 = 1235 kPa
sat. liq.: ZF = 0.05,
(s*-sF) = R×5.08 = 0.7266
(s*-sG) = R×0.475 = 0.0680
sat. vap.: ZG = 0.775,
Therefore
vF =
0.05×0.143×353.2
= 0.00205
1235
vG =
0.775×0.143×353.2
= 0.0317
1235
Since vF < v2 < vG → x2 = (v2-vF)/(vG-vF) = 0.5578
*
*
*
(s2-s2) = (1-x2)(s2-sF2) + x2(s2-sG2)
= 0.4422×0.7266 + 0.5578×0.0680 = 0.3592
* *
(s2-s1) = CP0 ln (T2/T1) - R ln (P2/P1) = 0 - 0.143 ln (1235/500) = -0.1293
(s2-s1) = -0.3592 - 0.1293 + 0.0230 = -0.4655
Q12 = Tm(s2-s1) = 353.2×10(-0.4655) = -1644 kJ
13-42
13.52 An uninsulated compressor delivers ethylene, C2H4, to a pipe, D = 10 cm, at
10.24 MPa, 94°C and velocity 30 m/s. The ethylene enters the compressor at 6.4
MPa, 20.5°C and the work input required is 300 kJ/kg. Find the mass flow rate,
the total heat transfer and entropy generation, assuming the surroundings are at
25°C.
293.7
6.4
= 1.040 , Pri =
= 1.270
282.4
5.04
From D.2 and D.3,
Tri =
*
(hi -hi) = 0.296 37×282.4×2.65 = 221.8
*
(si -si) = 0.296 37×2.08 = 0.6164
367.2
10.24
= 1.30 , Pre =
= 2.032 => From D.1:
282.4
5.04
ZeRTe 0.69×0.296 37×367.2
ve =
=
= 0.0073 m3/kg
Pe
10 240
Tre =
π 2
Ae = De = 0.007 85 m2
4
=>
Ze = 0.69
. AeVe 0.007 85×30
m=
=
= 32.26 kg/s
ve
0.0073
From D.2 and D.3,
*
(he-he) = 0.296 37×282.4×1.6 = 133.9
*
(se-se) = 0.296 37×0.90 = 0.2667
*
*
(he-hi ) = 1.5482(367.2-293.7) = 113.8
* *
(se-si ) = 1.5482 ln
367.2
10.24
- 0.296 37 ln
= 0.2065
293.7
6.4
(he-hi) = -133.9 + 113.8 + 221.8 = 201.7 kJ/kg
(se-si) = -0.2667 + 0.2065 + 0.6164 = 0.5562 kJ/kg K
First law:
302
- 300 = -97.9 kJ/kg
q = (he-hi) + KEe + w = 201.7 +
2×1000
.
.
Qcv = mq = 32.26(-97.9) = -3158 kW
.
Q
.
.
cv
3158
Sgen = −
+ m(se - si) = +
+ 32.26(0.5562) = 28.53 kW/K
298.2
T0
13-43
13.53 A distributor of bottled propane, C3H8, needs to bring propane from 350 K, 100 kPa
to saturated liquid at 290 K in an SSSF process. If this should be accomplished in a
reversible
. .setup given the surroundings at 300 K, find the ratio of the volume flow
rates Vin/Vout, the heat transfer and the work involved in the process.
From Table A.2: Tri =
350
= 0.946 ,
369.8
Pri =
0.1
= 0.024
4.25
From D.1, D.2 and D.3,
Zi = 0.99
*
(hi -hi) = 0.1886×369.8×0.03 = 2.1
*
(si -si) = 0.1886×0.02 = 0.0038
290
= 0.784,
369.8
From D.1, D.2 and D.3,
Tre =
Pre = 0.22 , Pe = 0.22×4.25 = 0.935 MPa
and
Ze = 0.036
*
(he-he) = 0.1886×369.8×4.57 = 318.6
*
(se-se) = 0.1886×5.66 = 1.0672
*
*
(he-hi ) = 1.679(290 - 350) = -100.8
* *
(se-si )
= 1.679 ln
290
0.935
- 0.1886 ln
= -0.7373
350
0.1
(he-hi) = -318.6 - 100.8 + 2.1 = -417.3
(se-si) = -1.0672 - 0.7373 + 0.0038 = -1.8007
.
ZiTi/Pi
Vin
0.99 350 0.935
. =
=
×
×
= 310.3
0.1
Vout ZeTe/Pe 0.036 290
wrev = (hi-he) -T0(si-se) = 417.3 - 300(1.8007) = -122.9 kJ/kg
qrev = (he-hi) + wrev = -417.3 –122.9 = -540.2 kJ/kg
13-44
13.54 Saturated-liquid ethane at T1 = 14°C is throttled into a SSSF mixing chamber at
the rate of 0.25 kmol/s. Argon gas at T2 = 25°C, P2 = 800 kPa, enters the
chamber at the rate of 0.75 kmol/s. Heat is transferred to the chamber from a heat
source at a constant temperature of 150 oC at a rate such that a gas mixture exits
the chamber at T3 = 120oC, P3 = 800 kPa. Find the rate of heat transfer and the
rate of entropy generation.
.
Argon, Ta2 = 25oC, P2 = 800 kPa, n2 = 0.75 kmol/s
Tca = 150 K, Pca = 4.87 MPa, Ma = 39.948 kg/kmol, Cpa = 0.52 kJ/kg K
ha3 - ha2 = MaCpa(T3 - Ta2) = 1973.4 kJ/kmol
.
Inlet: Ethane, Tb1 = 14oC, sat. liq., xb1 = 0, n1 = 0.25 kmol/s
Tcb = 305.4 K, Pcb = 4.88 MPa, Mb = 30.07 kg/kmol, Cpb = 1.766 kJ/kg-K
Tr1 = 0.94, Pb1 = Pr1Pcb = 0.69 × 4880 = 3367 kPa
∗
−
−
−∗ −
h b1 − h b1 = 3.81 RTcb = 9674.5 kJ/kmol, -sb1 − -sb1 = 3.74 R = 31.1
−∗ −∗
h b3 - h b1 = MbCpb(T3 - Tb1) = 5629.6 kJ/kmol
Exit: Mix, T3 = 120oC, P3 = 800 kPa consider this an ideal gas mixture.
. . . . . . SSSF Energy Eq.: n1hb1 + n2ha2 +Q = n3h3 = n1hb3 + n2ha3
. . . Q = n1(hb3 - hb1) + n2(ha3 - ha2) = 0.25 (5629.6 + 9674.5) + 0.75(1973.4)
= 5306 kW
.
.
.
.
Entropy Eq.: Sgen = n1(s-b3 − -sb1) + n2(s-a3 − -sa2) - Q/TH ;
. .
. .
ya = n2/ntot = 0.75; yb = n1/ntot = 0.25
T
yaP3
-s − -s = M C ln 3 - −
R
ln
= 8.14 kJ/kmol-K
a3
a2
a pa T
P
a2
TH = 150oC
a2
T
ybP3
∗
-s − -s = M C ln 3 - −
R
ln
+ -sb1 − -sb1 =
b3
b1
b pb T
Pb1
b1
= 40.172 + 31.1 = 71.27 kJ/kmol K
.
Sgen = 0.25 × 71.27 + 0.75 × 8.14 - 5306 / 423 = 11.38 kW/K
13-45
13.55 One kilogram per second water enters a solar collector at 40°C and exits at 190°C,
as shown in Fig. P13.55. The hot water is sprayed into a direct-contact heat
exchanger (no mixing of the two fluids) used to boil the liquid butane. Pure
saturated-vapor butane exits at the top at 80°C and is fed to the turbine. If the
butane condenser temperature is 30°C and the turbine and pump isentropic
efficiencies are each 80%, determine the net power output of the cycle.
H2O cycle: solar energy input raises 1 kg/s of liquid H2O from 40oC to 190oC.
Therefore, corresponding heat input to the butane in the heat exchanger is
.
.
QH = m(hF 190 C-hF 40 C)H2O = 1(807.62-167.57) = 640.05 kW
C4H10 cycle
1
.
WT
Turbine
.
QH
ηST = ηSP = 0.80
Ht.
Exch
2
Cond
P
4
Tr3 =
T1 = 80 oC, x1 = 1.0 ; T3 = 30 oC, x3 = 0.0
353.2
= 0.831
425.2
From D.1, D.2 and D.3:
P1 = 0.325×3800 = 1235 kPa
Tr1 =
*
.
-W P
3
(h1-h1) = 0.143 04×425.2×0.56 = 34.1
*
(s1-s1) = 0.143 04×0.475 = 0.0680
303.2
= 0.713
425.2
From D.1, D.2 and D.3: P3 = 0.113×3800 = 429 kPa
sat. liq.: (h*-hF) = RTC×4.81 = 292.5 ;
(s*-sF) = R×6.64 = 0.950
sat. vap.: (h*-hG) = RTC×0.235 = 14.3 ;
(s*-sG) = R×0.22 = 0.031
Because of the combination of properties of C 4H10 (particularly the large CP0
/R), s1 is larger than sG at T3. To demonstrate,
* *
(s1-sG3) = 1.7164 ln
353.2
1235
- 0.143 04 ln
= 0.1107
303.2
429
(s1-sG3) = -0.0680 + 0.1107 + 0.031 = +0.0737 kJ/kg K
13-46
T
so that T2S will be > T3, as shown in the T-s
1
diagram. A number of other heavy hydrocarbons
also exhibit this behavior.
Assume T2S = 315 K, Tr2S = 0.741
2s 2
3
s
From D.2 and D.3:
*
(h2S-h2S) = RTC×0.21 = 12.8
* *
(s1-s2S) = 1.7164 ln
and
*
(s2S-s2S) = R×0.19 = 0.027
353.2
1235
- 0.143 04 ln
= +0.0453
315
429
(s1-s2S) = -0.0680 + 0.0453 + 0.027 ≈ 0
⇒ T2S = 315 K
*
*
(h1-h2S) = 1.7164(353.2-315) = 65.6
wST = h1-h2S = -34.1 + 65.6 + 12.8= 44.3 kJ/kg
wT = ηS×wST = 0.80×44.3 = 35.4 kJ/kg
At state 3,
v3 =
0.019×0.143 04×303.2
= 0.001 92 m3/kg
429
-wSP ≈ v3(P4-P3) = 0.001 92(1235-429) = 1.55 kJ/kg
-wP =
-wSP
ηSP
=
1.55
= 1.94 kJ/kg
0.8
wNET = wT + wP = 35.4 - 1.94= 33.46 kJ/kg
For the heat exchanger,
.
.
QH = 640.05 = mC4H10(h1-h4)
But
h1-h4 = h1-h3+wP
*
*
*
*
h1-h3 = (h1-h1) + (h1-h3) + (h3-h3)
= -34.1 + 1.716(80 - 30) + 292.5 = 344.2 kJ/kg
Therefore,
640.05
.
mC4H10 =
= 1.87 kg/s
344.2-1.94
.
.
WNET = mC4H10wNET = 1.87 × 33.46 = 62.57 kW
13-47
13.56 A line with a steady supply of octane, C8H18, is at 400°C, 3 MPa. What is your
best estimate for the availability in an SSSF setup where changes in potential and
kinetic energies may be neglected?
Availability of Octane at
3
= 1.205,
2.49
From D.2 and D.3,
Pri =
Tri =
Ti = 400 oC, Pi = 3 MPa
673.2
= 1.184
568.8
*
(h1-h1) = 0.072 79×568.8×1.13 = 46.8 ;
*
(s1-s1) = 0.072 79×0.69 = 0.05
Exit state in equilibrium with the surroundings, assume T0 = 298.2 K, P0 =
100 kPa
298.2
0.1
= 0.524 , Pr0 =
= 0.040
568.8
2.49
From D.2 and D.3,
Tr0 =
*
(h0-h0) = RTC×5.4 = 223.6
*
and
*
(s0-s0) = R×10.37 = 0.755
*
(hi -h0) = 1.7113(673.2-298.2) = 641.7
* *
(si -s0) = 1.7113 ln
673.2
3
- 0.072 79 ln
= 1.1459
298.2
0.1
(hi-h0) = -46.8 + 641.7 + 223.6 = 818.5
(si-s0) = -0.05 + 1.1459 + 0.755 = 1.8509
ϕi = wrev = (hi-h0) - T0(si-s0) = 818.5 - 298.2(1.8509) = 266.6 kJ/kg
13-48
13.57 A piston/cylinder contains ethane gas, initially at 500 kPa, 100 L, and at ambient
temperature, 0°C. The piston is now moved, compressing the ethane until it is at
20°C, with a quality of 50%. The work required is 25% more than would have been
required for a reversible polytropic process between the same initial and final states.
Calculate the heat transfer and the net entropy change for the process.
Ethane: Tc = 305.4 K, Pc = 4.88 MPa,
R = 0.2765 kJ/kg-K, Cp = 1.766 kJ/kg K
State 1: Tr1 = 0.895, Pr1 = 0.102 ‡ Z1 = 0.95
v1 = Z1RT1/P1 = 0.1435 m3/kg, m1 = V1/v1 = 0.697 kg
*
*
(h1 − h1) = 0.13RTc = 11.0 kJ/kg, (s1 − s1) = 0.09 R = 0.025 kJ/kg K
State 2: T2 = 20oC, x2 = 0.5, 1W2 = 1.25Wrev
Tr2 = 0.96, Pr2 = 0.78, P2 = Pr2Pc = 3806 kPa
Zf2 = 0.14, Zg2 = 0.54, Z2 = (1 - x2)Zf + x2Zg = 0.34
*
(h2 − h2) = (1 - x2) 3.65 RTc + x2 (1.39 RTc) = 212.8 kJ/kg
*
(s2 − s2) = (1 - x2) 3.45 R + x2 × 1.10 R = 0.629 kJ/kg K
v2 = Z2RT2/P2 = 0.0072 m3/kg,
n
n
P1V1 = P2V2 ,
ln
Wrev = ∫ P dV =
P2
P1
= n ln
V2 = mv2 = 0.005 m3
V1
V2
‡
n = 0.6783
P 2V 2 - P 1V 1
= -96.3 kJ, 1W2 = 1.25Wrev = -120.4 kJ
1-n
a) 1st Law: 1Q2 = m(u2 - u1) + 1W2; u = h - Pv
u2 - u1 = (h2 - h1) - (P2v2 - P1v1)
*
h2 - h1 = = (h2 − h2)
*
*
*
*
+ (h2 − h1) + (h1 − h1)
*
(h2 − h1) = Cp(T2 - T1) = 35.3 kJ/kg,
h2 - h1 = -212.8 +35.3 + 11.0 = -166.5 kJ/kg, u2 - u1 = -122.2 kJ/kg
1Q2 = 0.697(-122.2) - 120.4 = -205.6 kJ
b) 2nd Law: ∆Snet = m(s2 - s1) - 1Q2 /To; To = 0oC
*
s2 - s1 = (s2 − s2)
*
*
*
*
+ (s2 − s1) + (s1 − s1)
*
(s2 − s1) = Cp ln(T2 / T1) − R ln(P2 / P1) = -0.436 kJ/kg K,
205.6
∆Snet = 0.697(-0.629 - 0.436 + 0.025) +
= 0.028 kJ/K
273.2
13-49
13.58 The environmentally safe refrigerant R-152a (see Problem 13.65) is to be
evaluated as the working fluid for a heat pump system that will heat winter
households in two different climates. In the colder climate the cycle evaporator
temperature is −20°C, and the more moderate climate the evaporator temperature
is 0°C. In both climates the cycle condenser temperature is 30°C. For this study
assume all processes are ideal. Determine the cycle coefficient of performance for
the two climates.
R152a difluoroethane.
From 13.65: CP0 = 0.996
Ideal Heat Pump TH = 30 oC
M = 66.05, R = 0.125 88, T C = 386.4 K, PC = 4.52 MPa
From A.2:
CASE I: T1 = -20 oC
T
303.2
= 0.785
386.4
Pr3 = Pr2 = 0.22 =>
2
Tr3 =
3
1
4
Sat.liq.:
v
CASE I)
CASE II: T1 = 0 oC
P3 = P2 = 994 kPa
*
h3 - h3 = 4.56×RTC = 221.8
T1 = -20 oC = 253.2 K, Tr1 = 0.655, Pr1 = 0.058 → P1 = 262 kPa
*
h1 - h1 = 0.14×RTC = 6.8
and
*
s1 - s1 = 0.14×R = 0.0176
Assume T2 = 307 K, Tr2 = 0.795 given Pr2 = 0.22
*
From D.2, D.3: s2 - s2 = 0.34×R = 0.0428 ;
*
*
s2 - s1 = 0.996 ln
*
h2 - h2 = 0.40×RTc = 19.5
307
994
- 0.125 88 ln
= 0.0241
253.2
262
s2 - s1 = -0.0428 + 0.0241 + 0.0176 = -0.001 ≈ 0 OK
⇒ h2 - h1 = -19.5 + 0.996(307-253.2) + 6.8 = 40.9
h2 - h3 = -19.5 + 0.996(307-303.2) + 221.8 = 206.1
β=
qH
wIN
=
h2 - h3
h2 - h1
=
206.1
= 5.04
40.9
13-50
case II) T1 = 0 oC = 273.2 K, Tr1 = 0.707 => Pr1 = 0.106, P1 = 479 kPa
*
h1 - h1 = 0.22×RTC = 10.7
*
and
s1 - s1 = 0.21×R = 0.0264
Assume T2 = 305 K, Tr2 = 0.789
*
s2 - s2 = 0.35×R = 0.0441
*
*
s2 - s1 = 0.996 ln
*
and
h2 - h2 = 0.38×RTC = 18.5
305.0
994
- 0.125 88 ln
= 0.0178
273.2
479
s2 - s1 = -0.0441 + 0.0178 + 0.0264 = 0.0001 ≈ 0 OK
h2 - h1 = -18.5 + 0.996(305.0-273.2) + 10.7 = 23.9
h2 - h3 = -18.5 + 0.996(305.0-303.2) + 221.8 = 205.1
β=
h2 - h3
h2 - h1
=
205.1
= 8.58
23.9
13.59 Repeat the calculation for the coefficient of performance of the heat pump in the
two climates as described in Problem 13.58 using R-12 as the working fluid.
Compare the two results.
Problem the same as 13.58 , exept working fluid is R-12
For R-12: At T3 = 30 oC:
CASE I)
T1 = -20 oC,
P3 = P2 = 0.745 MPa, h3 = 64.6
h1 = 178.7,
s1 = 0.7087
 T2 = 39.8 oC
At s2 = s1 & P2 → 
 h2 = 206.8
⇒β=
CASE II:
h2 - h3
h2 - h1
=
T1 = 0oC,
At s2 = s1 & P2
⇒
206.8 - 64.6 142.2
=
= 5.06
206.8 - 178.7 28.1
β=
h1 = 187.5,
s1 = 0.6965
→ T = 34.6oC , h = 203.0
203.0-64.6 138.4
=
= 8.93
203.0-187.5 15.5
13-51
13.60 One kmol/s of saturated liquid methane, CH4, at 1 MPa and 2 kmol/s of ethane,
C2H6, at 250°C, 1 MPa are fed to a mixing chamber with the resultant mixture
exiting at 50°C, 1 MPa. Assume that Kay’s rule applies to the mixture and
determine the heat transfer in the process.
Control volume the mixing chamber, inlet CH4 is 1, inlet C2H6 is 2 and the
exit state is 3. Energy equation is
.
. . . QCV = n3 h3 - n1 h1 - n2 h2
Select the ideal gas reference temperature to be T3 and use the
generalized charts for all three states.
Pr1 = Prsat = 1/4.60 = 0.2174 => Trsat = 0.783,
T1 = 0.783 × 190.4 = 149.1 K,
∆h1 = 4.57
Pr2 = 1/4.88 = 0.205, Tr2 = 523/305.4 = 1.713
h1 = C1(T1 - T3) - ∆h1 RTc = 36.15(149.1 - 323.2) - 4.57 × 8.3145 × 190.4
= -13528 kJ/kmol
h2 = C2(T2 - T3) - ∆h2 RTc = 53.11(250 - 50) - 0.08 × 8.3145 × 305.4
= 10419 kJ/kmol
Tcmix = (1 × 190.4 + 2 × 305.4)/3 = 267.1 K
Pcmix = (1 × 4.60 + 2 × 4.88)/3 = 4.79 MPa
Tr3 = 323.2/267.1 = 1.21 , Pr3 = 1/4.79 = 0.21
h3 = 0 - 0.15 × 267.1 × 8.3145 = - 333 kJ/kmol
.
QCV = 3(-333) - 1(-13528) - 2(10419) = - 8309 kW
13.61 Consider the following reference state conditions: the entropy of real saturated
liquid methane at −100°C is to be taken as 100 kJ/kmol K, and the entropy of
hypothetical ideal gas ethane at −100°C is to be taken as 200 kJ/kmol K.
Calculate the entropy per kmol of a real gas mixture of 50% methane, 50% ethane
(mole basis) at 20°C, 4 MPa, in terms of the specified reference state values, and
assuming Kay’s rule for the real mixture behavior.
CH : T = -100 oC, -s
= 100 kJ/kmol K
4
0
LIQ 0
*
C2H6: T0 = -100 oC, P0 = 1 MPa, -s0 = 200 kJ/kmol K
Also for CH4: TC = 190.4 K, PC = 4.60 MPa
For a 50% mixture:
Tcmix = 0.5 × 190.4 + 0.5 × 305.4 = 247.9 K
Pcmix = 0.5 × 4.60 + 0.5 × 4.88 = 4.74 MPa
13-52
IG MIX at T0(=-100 oC), P0(=1 MPa):
CH4: Tr0 = 0.91 ,
PG = 0.57 × 4.60 = 2.622 MPa
-* -s*
0 CH4 = s LIQ 0 P + (s -s LIQ)at P - R ln (P 0/PG)
G
G
= 100 + 4.01×8.3145 - 8.3145 ln (1/2.622) = 141.36
-s*
= 0.5×141.36 + 0.5×200 - 8.3145(0.5 ln 0.5 + 0.5 ln 0.5) = 176.44
0 MIX
CP0 MIX = 0.5×16.04×2.254 + 0.5×30.07×1.766 = 44.629
293.2
4
-s*
=
176.44
+
44.629
ln
8.3145
ln
= 188.41 kJ/kmol K
TP MIX
173.2
1
For the mixture at T,P: Tr = 1.183, Pr = 0.844
-s*
- -sTP MIX = 0.4363×8.3145 = 3.63 kJ/kmol K
TP MIX
Therefore,
-s
= 188.41 - 3.63 = 184.78 kJ/kmol K
TP MIX
An alternative is to form the ideal gas mixture at T,P instead of at T0,P0 :
T P
-s
- *-s- ) + C
-s*
=
+
(s
ln
R
ln
LIQ 0
LIQ
P0 CH4
TP CH4
T0
PG
P ,T
G 0
at P , T
G 0
= 100 + 33.34 + 16.04×2.254 ln
293.2
4
- 8.3145 ln
173.2
2.6
= 100 + 33.34 + 19.03 - 3.53 = 148.84
293.2
4
-s*
= 200 + 30.07×1.766 ln
- 8.3145 ln
TP C2H6
173.2
1
= 200 + 27.96 - 11.53 = 216.43
-s*
TP MIX = 0.5×148.84 + 0.5×216.43
- 8.3145(0.5 ln 0.5 + 0.5 ln 0.5) = 188.41
-s
= 188.41 - 3.63 = 184.78 kJ/kmol K
TP MIX
13-53
Advanced Problems
13.62 An experiment is conducted at −100°C inside a rigid sealed tank containing liquid
R-22 with a small amount of vapor at the top. When the experiment is done the
container and the R-22 warms up to room temperature of 20°C. What is the
pressure inside the tank during the experiment? If the pressure at room
temperature should not exceed 1 MPa, what is the maximum percent of liquid by
volume that can be used during the experiment?
T
R-22 tables Go to -70 oC
a) For hFG ≈ const &
2
vFG ≈ vG ≈ RT/PG
hFG 1 1
ln
≈
PG0 R T0 T1
PG1
[
]
o
20 C
o
-100 C
1
extrapolating from -70 oC
v
(Table B.4.1) to TAVE = -85 oC, hFG ≈ 256.5
8.3145
= 0.096 15
86.469
For T0 = 203.2 K & T1 = 173.2 K
Also R =
PG1
256.5
(20.5)= 0.096
[ 1 - 1 ]
15 203.2 173.2
ln
PG1 = 2.107 kPa
b) Extrapolating vF from -70 oC to T1 = -100 oC
vF1 ≈ 0.000 634
Also vG1 ≈ RT1/PG1 =
0.096 15×173.2
= 7.9037
2.107
Since v1 = v2 ≈ vF2 = 0.000 824
0.000 824 = 0.000 634 + x 1×7.9031 => x 1 = 2.404×10-5
VLIQ 1
m
= (1-x1)vF1 = 0.000 634,
% LIQ, by vol. =
VVAP 1
m
= x1vG1 = 0.000 190
0.000 634
×100 = 76.9 %
0.000 824
13-54
13.63 Determine the low-pressure Joule–Thomson inversion temperature from the
condition in Eq. 13.54 as predicted by an equation of state, using the van der
Waals equation and the Redlich–Kwong equation.
∂v
T( ) P - v
∂T
∂T
µ j = ( )n =
, But
CP
∂P
a) van der Waal VDW:
∂P
∂v
(∂T)P = -
∂v
(∂T)P = -
1
CP
[
RT
2a
[- (v-b)2 + v3]
v2
and then rearranging,
v2
1
b
-RT(
)(
)+ 2a
1-b/v 1-b/v
1 2 2a
) -v
RT(
1-b/v
]
lim
Then
or
∂P
(∂v )T
R
(v-b
)
Substitute into µj eq'n & multiply by
µj =
(∂T)v
µ =
P→0 j
TINV =
1
2a
[-b + RT
]= 0
CP
only at TINV
2a 27
= T = 6.75 TC
Rb 4 C
1
∂P
R
2a
=
(∂T)v v-b + v(v+b)T3/2
b) R-K:
∂P
-RT
a(2v+b)
(∂v )T = (v-b)2 + v2(v+b)
2 1/2
T
Substituting as in part a)
1
µj =
1
CP
[
2a
bRT
(1-b/v)2 (1+b/v)2 T1/2
RT
a(2+b/v)
(1-b/v)2 v(1+b/v)2 T1/2
]
then
lim
µ =
P→0 j
or
1
5 a
[
b]= 0 only at TINV
CP
2 RT3/2
TINV = (
5a 2/3
) = 5.3 TC
2Rb
13-55
13.64 Suppose the following information is available for a given pure substance:
vx
Known:
T
PSAT = fn(T)
5 4
6
3
1
2
vapor P = fn(T,v)
values of vF, PC, & TC
vapor CV at vx
s
Outline the procedure that should be followed to develop a table of
thermodynamic properties comparable to Tables 1,2 and 3 of the steam tables.
1) from vapor pressure curve, get P SAT at each T.
2) vF values are known.
3) reference state: T0 & sat. liquid. Set
h0 = 0, s0 = 0, u0 = 0 - P0v0
4) from eq'n of state, find v1 = vG at T0
then vFG(T0) = vG(T0) - vF(T0)
5) from clapeyron eq'n, using slope of vapor pressure
curve & vFG, find hFG(T0) then h1 = hG(T0) = 0 + h FG(T0)
6) u1 = uG(T0) = h1 - P1v1; uFG(T0) = u1 - u0
hFG(T0)
7) sFG(T0) =
; s1 = sG(T0) = 0 + sFG(T0)
T0
8) find P2 from eq'n of state ( T2 = T0, v2 = vx )
2
⌠
∂P
9) u2 - u1 = [T( )v - P]dvT
⌡
∂T
1
evaluate from eq'n of state then h2 = u2 + P2v2
2
⌠ ∂P
10) s2 - s1 =  ( )v dvT, find s2.
⌡ ∂T
1
3
11) u3 - u2 = ⌠
⌡ Cvx dT, find u3.
2
12) find P3 from eq'n of state ( T3, v3 = vx )
13) h3 = u3 + P3v3
13-56
3
14) s3 - s2 = ⌠
⌡ (Cvx/T) dT, find s3
2
15) at any point 4 (T4=T3) given by desired P4, find v4 from eq'n of state.
16) find P5 = PSAT at T5 = T3 from vap. pressure.
17) find v5 = vG(T5) from eq'n of state.
18) find u5, h5, s5 as in steps 9 & 10.
19) find hFG(T5) from Clapeyron eq'n as in step 5.
20) h6 = hF(T5) = h5 - hFG(T5)
21) sFG(T5) =
hFG(T5)
, s6 = sF(T5) = sF - sFG(T5)
T5
22) u6 = h6 - P6v6, uFG(T5) = u5 - u6
23) pick a different T (instead of T3) and repeat steps
11 to 22. For T ⇒ T5 steps 16 to 22 are eliminated.
13-57
13.65 The refrigerant R-152a, difluoroethane, is tested by the following procedure. A
10-L evacuated tank is connected to a line flowing saturated-vapor R-152a at
40°C. The valve is then opened, and the fluid flows in rapidly, so that the process
is essentially adiabatic. The valve is to be closed when the pressure reaches a
certain value P2, and the tank will then be disconnected from the line. After a
period of time, the temperature inside the tank will return to ambient temperature,
25°C, through heat transfer with the surroundings. At this time, the pressure
inside the tank must be 500 kPa. What is the pressure P2 at which the valve
should be closed during the filling process? The ideal gas specific heat of R-152a
is CP0 = 0.996 kJ/kg K.
R-152a CHF2CH3 : A.2:
M = 66.05, TC = 386.4 K, PC = 4.52 MPa,
T3 = T0 = 25oC, P3 = 500 kPa, R = R/M = 8.3145/66.05 = 0.12588
Tr3 = 298.2/386.4 = 0.772,
Pr3 = 500/4520 = 0.111
From D.1 and D.2 at 3: Z3 = 0.92,
⇒
m3 = m2 = mi =
Filling process:
or
P 3V
Z3RT3
Energy Eq.:
=
(h*-h)3 = 0.19 RTC
500×0.010
= 0.145 kg
0.92×0.125 88×298.2
hi = u2 = h2 - Z2RT2
*
*
(h2-h2) + CP0(T2-Ti) + (hi -hi) - P2V/m2 = 0
From D.2 with Tri = 313.2/386.4 = 0.811,
*
(hi -hi) = 0.125 88×386.4×0.49 = 23.8 ;
Pi = 0.276×4520 = 1248 kPa
Assume P2 = 575 kPa, Pr2 = 0.127
Now assume T2 = 339 K, Tr2 = 0.877 =>
⇒
From D.1: Z2 = 0.93
Z2T2 0.93×339
Z3T3 0.92×298.2
=
= 0.5483 ≈
=
= 0.5487
P2
P3
575
500
⇒ T2 = 339 K is the correct T2 for the assumed P2 of 575 kPa. Now
check the 1st law to see if 575 kPa is the correct P 2.
From D.2,
*
h2-h2 = 0.125 88×386.4×0.17 = 8.3
Substituting into 1st law,
-8.3 + 0.996(339-313.2) + 23.8 -
575×0.010
= +1.5 ≈ 0
0.1456
⇒ P2 = 575 kPa
(Note: for P2 = 580 kPa, T2 = 342 K,
1st law sum = +4.2)
13-58
13.66 An insulated cylinder has a piston loaded with a linear spring (spring constant of
600 kN/m) and held by a pin, as shown in Fig. P13.66. The cylinder crosssectional area is 0.2 m2, the initial volume is 0.1 m3, and it contains carbon
dioxide at 2.5 MPa, 0°C. The piston mass and outside atmosphere add a force per
unit area of 250 kPa, and the spring force would be zero at a cylinder volume of
0.05 m3. Now the pin is pulled out; what is the final pressure inside the cylinder?
AP = 0.2 m3, VFs = 0 = 0.05 m3 , ks = 600 kN/m
P0 = 250 kPa, P1 = 2.5 MPa, V1 = 0.1 m3, T1 = 0oC
R = 0.1889, C = 0.842
ks
Any point: PEXT = P0 + 2 (V-V0)
Ap
600
Pin pulled: PEXT = 250 + 2 (0.1-0.05) = 1000 kPa
0.2
But P1 = 2500 kPa
P0
CO2
Q = 0
⇒ Irr. process, expansion to 1000 < P2 < 2500
At 2: P2 = 250 +
600
(V -0.05) = 15 000 V2 - 500
0.22 2
2
1
W12 = ⌠
⌡ PEXTdV = 2(PEXT 1 + PEXT 2)(V2-V1)
1
Q12 = 0 = m(h2-h1) - P2V2 + P1V1 + W12
*
*
h2-h1 = (h2-h2) + CP0(T2-T1) + (h1-h1)
Tr1 =
273.2
2.5
= 0.898, Pr1 =
= 0.339 ,
304.1
7.38
Also P1V1 = mZ1RT1
From D.2,
m=
2500×0.1
= 5.91 kg
0.82×0.188 92×273.2
*
(h1-h1) = 0.49×0.1889×304.1 = 28.2
Assume P2 = 1655 kPa
Z2Tr2 =
=>
From Fig. D.1, Z1 = 0.82
P2V2
mRTC
=
=>
V2 =
1655×0.1437
= 0.7004
5.91×0.188 92×304.1
At Tr2 = 0.824 and Pr2 = 0.224 =>
T2 = 250.6 K
1655+500
= 0.1437 m3
15 000
Z2 = 0.850 , ZTr = 0.7004 ~ OK
13-59
*
(h2-h2) = 0.37 × 0.1889 × 304.1 = 21.3
*
*
(h2-h1) = 0.842(250.6 - 273.2) = -19.0
W12 = 0.5 (1000 + 1655) (0.1437-0.10) = 58.0 kJ
Q12 = 5.91(-21.3-19.0+28.2) - 1655×0.1437 + 250 + 58.0
= -1.3 ≈ 0 OK (Note: for P2 = 1660 kPa, Q12 = +2.7)
P2 = 1655 kPa
13.67 A 10- m3 storage tank contains methane at low temperature. The pressure inside is
700 kPa, and the tank contains 25% liquid and 75% vapor, on a volume basis. The
tank warms very slowly because heat is transferred from the ambient.
a. What is the temperature of the methane when the pressure reaches 10 MPa?
b. Calculate the heat transferred in the process, using the generalized charts.
c. Repeat parts (a) and (b), using the methane tables, Table B.7. Discuss the
differences in the results.
CH4: V = 10 m3, P1 = 700 kPa
VLIQ 1 = 2.5 m3, VVAP 1 = 7.5 m3
0.70
10
= 0.152, Pr2 =
= 2.174
4.60
4.60
From D.1: ZF1 = 0.025, ZG1 = 0.87 &
a) Pr1 =
T1 = 0.74 × 190.4 = 140.9 K
vF1 =
0.025×0.518 35×140.9
= 0.00261
700
0.87×0.518 35×140.9
= 0.0908
700
2.5
7.5
mLIQ 1 =
= 957.9 kg, mVAP 1 =
= 82.6 kg
0.00261
0.0908
vG1 =
Total m = 1040.3 kg
v2 = v1 =
Z2×0.518 35×190.4×Tr2
V
10
=
= 0.00961 =
m 1040.5
10 000
or Z2Tr2 = 0.9737 at Pr2 = 2.174
By trial and error
13-60
Tr2 = 1.334 & Z2 = 0.73, T2 = 1.334×190.4 = 254.0 K
b) 1st law:
Q12 = m(u2-u1) = m(h2-h1) - V(P2-P1)
Using D.2 & x1 =
*
82.6
= 0.0794
1040.5
*
(h1-h1) = (h1-hF1) - x1hFG1
[
= 0.518 35×190.4 4.72-0.0794(4.72-0.29)
*
] = 431.1
*
(h2-h1) = 2.2537(254.0-140.9) = 254.9
*
(h2-h2) = 0.518 35×190.4(1.47) = 145.1
(h2-h1) = -145.1 + 254.9 + 431.1 = 540.9 kJ/kg
Q12 = 1040.5(540.9) - 10(10 000-700) = 469 806 kJ
c) Using Table B.7 for CH4
T1 = TSAT 1 = 141.7 K,
vF1 = 0.002 675,
uF1 = -178.47
vG1 = 0.090 45 , uG1 = 199.84
mLIQ 1 =
2.5
7.5
= 934.6, mVAP 1 =
= 82.9
0.002 675
0.090 45
Total mass
m = 1017.5 kg
and
v2 =
10
= 0.009 828 m3/kg
1017.5
 T2 = 259.1 K
At v2 & P2 = 10 MPa → 
 u2 = 296.11
Q12 = m(u2-u1) = 1017.5×296.11 - 934.6(-178.47) - 82.9(199.84)
= 451 523 kJ
13-61
13.68 Calculate the difference in entropy of the ideal-gas value and the real-gas value
for carbon dioxide at the state 20°C, 1 MPa, as determined using the virial
equation of state. Use numerical values given in Problem 13.30.
CO2 at T = 20oC, P = 1 MPa
RT/P*
RT/P*
⌠ ∂P
*
sP* - sP = 
⌡ ∂T v dv ; ID Gas,
⌠R
*
P
sP* - sP =  dv = R ln *
⌡v
P
( )
v(P)
v(P)
RT/P*
*
Therefore, at P:
sP - sP = -R ln
⌠ ∂P
P

+
dv
P* ⌡ ∂T v
( )
v(P)
virial:
P=
R BR RT dB
= +
(∂P
)
+
( )
v
v v
∂T
v dT
RT BRT
+ 2
and
v
v
2
2
Integrating,
P
RT
dB
- sP = -R ln * + R ln * + R B + T
dT
P
P v
RT
dB 1
= R ln
+ B+T
Pv
dT v
Using values for CO2 from solution 13.30,
[
*
sP
[
(
( )](
( )) ]
1 P*
v RT
)
1
-s* - -s = 8.3145 ln 2.437 37 + -0.128 + 0.266
P
P
2.3018
2.3018
= 0.9743 kJ/kmol K
[
(
)
]
13.69 Carbon dioxide gas enters a turbine at 5 MPa, 100°C, and exits at 1 MPa. If the
isentropic efficiency of the turbine is 75%, determine the exit temperature and the
second-law efficiency.
CO2 turbine: ηS = w/wS = 0.75
inlet: T1 = 100oC, P1 = 5 MPa, exhaust: P2 = 1 MPa
5
373.2
1
= 0.678, Tr1 =
= 1.227, Pr2 =
= 0.136
7.38
304.1
7.38
From D.2 and D.3,
a) Pr1 =
*
(h1-h1) = 0.188 92×304.1×0.52 = 29.9
*
(s1-s1) = 0.188 92×0.30 = 0.0567
13-62
Assume T2S = 253 K, Tr2S = 0.832
From D.2 and D.3:
*
(h2S-h2S) = RTC×0.20 = 11.5
*
(s2S-s2S) = R×0.17 = 0.0321
*
*
(s2S-s1) = 0.8418 ln
253
1
- 0.188 92 ln = -0.0232
373.2
5
(s2S-s1) = -0.0321 - 0.0232 + 0.0567 ≈ 0
⇒ T2S = 253 K
*
*
(h2S-h1) = 0.8418(253-373.2) = -101.2
wS = (h1-h2S) = -29.9 + 101.2 + 11.5 = 82.8 kJ/kg
*
*
*
*
w = ηS×wS = 0.75×82.8 = 62.1 kJ/kg = (h 1-h1) + (h1-h2) + (h2-h2)
Assume T2 = 275 K, Tr2 = 0.904
*
*
(h1-h2) = 0.8418(373.2-275) = 82.7
From D.2 and D.3,
*
(h2-h2) = RTC×0.17 = 9.8 ;
*
(s2-s2) = R×0.13 = 0.0245
Substituting,
w = -29.9 + 82.7 + 9.8 = 62.7 ≈ 62.1
⇒ T2 = 275 K
* *
b) (s2-s1) = 0.8418 ln
275
1
- 0.188 92 ln = +0.0470
373.2
5
(s2-s1) = -0.0245 + 0.0470 + 0.0567 = +0.0792
Assuming T0 = 25 oC,
(ϕ1-ϕ2) = (h1 - h2) - T0(s1 - s2) = 62.1 + 298.2(0.0792) = 85.7 kJ/kg
η2nd Law =
w
62.1
=
= 0.725
ϕ1-ϕ2 85.7
13-63
13.70 A 4- m3 uninsulated storage tank, initially evacuated, is connected to a line
flowing ethane gas at 10 MPa, 100°C. The valve is opened, and ethane flows into
the tank for a period of time, after which the valve is closed. Eventually, the
whole system cools to ambient temperature, 0°C, at which time the it contains
one-fourth liquid and three-fourths vapor, by volume. For the overall process,
calculate the heat transfer from the tank and the net change of entropy.
Rigid tank V = 4 m3, m1 = 0
Line: C2H6 at Pi = 10 MPa, Ti = 100 oC
Flow in, then cool to
M = 30.07,
T2 = T0 = 0 oC, VLIQ 2 = 1 m3 & VVAP 2 = 3 m3
R = 0.2765, CP0 = 1.766
10
= 2.049,
4.88
From D.2 and D.3,
Pri =
Tri =
373.2
= 1.225
305.4
*
(hi -hi) = 0.2765×305.4×2.0 = 168.9 and
Tr2 =
*
(si -si) = 0.2765×1.22 = 0.3373
273.2
= 0.895
305.4
From D.1, D.2 and D.3,
P2 = PG = 0.51×4880 = 2489 kPa
sat. liq.: ZF = 0.087 ; (h*-hF) = RTC×4.09 = 345.4 ; (s*-sF) = R×4.3 = 1.189
sat. vap. : ZG = 0.68 ; (h*-hG) = RTC×0.87 = 73.5 ; (s*-sG) = R×0.70 = 0.193
mLIQ 2 =
2489×1
= 378.7 kg
0.087×0.2765×273.2
mVAP 2 =
2489×3
= 145.4 kg
0.68×0.2765×273.2
m2 = 524.1 kg
=>
x2 =
145.4
= 0.277
524.1
1st law:
*
*
*
*
QCV = m2u2 - mihi = m2(h2-hi) - P2V = m2[(h2-h2) + (h2-hi ) + (hi -hi)]- P2V
*
*
(h2-hi ) = 1.7662(0-100) = -176.6
*
*
*
(h2-h2) = (1-x2)(h2-hF2) + x2(h2-hG2)
= 0.723 × 345.4 + 0.277 × 73.5 = 270.1
QCV = 524.1[-270.1 - 176.6 + 168.9]- 2489 × 4 = -155 551 kJ
13-64
∆SNET = m2(s2-si) - QCV/T0
*
* *
*
(s2-si) = (s2-s2) + (s2-si ) + (si -si)
* *
(s2-si ) = 1.7662 ln
*
273.2
2.489
- 0.2765 ln
= -0.1664
373.2
10
*
*
(s2-s2) = (1-x2)(s2-sF2) + x2(s2-sG2)
= 0.723 ×1.189 + 0.277 × 0.193 = 0.9131
(s2-si) = -0.9131 - 0.1664 + 0.3373 = -0.7422
∆SNET = 524.1(-0.7422) -
-155 551
= 180.4 kJ/K
273.2
13.71 The environmentally safe refrigerant R-142b (see Problem 13.45) is to be
evaluated as the working fluid in a portable, closed-cycle power plant, as shown
in Fig. P13.71. The air-cooled condenser temperature is fixed at 50°C, and the
maximum cycle temperature is fixed at 180°C, because of concerns about thermal
stability. The isentropic efficiency of the expansion engine is estimated to be
80%, and the minimum allowable quality of the fluid exiting the expansion engine
is 90%. Calculate the heat transfer from the condenser, assuming saturated liquid
at the exit. Determine the maximum cycle pressure, based on the specifications
listed
T
1
R-142b CH3CClF2
M = 100.495, TC = 410.3 K, PC = 4.25 MPa
4
CP0 = 0.787 kJ/kg K
3
R = 0.082 73 kJ/kg K
2S2
s
T1 = 180 oC, T2 = T3 = 50 oC, x2 = 0.90, ηS EXP ENG = 0.80
a) Tr2 =
323.2
= 0.788
410.3
From D.1, D.2 and D.3: P2 = P3 = 0.23×4250 = 978 kPa
sat. liq.: (h*-hF) = RTC×4.55 = 154.4 ;
(s*-sF) = R×5.62 = 0.4649
sat. vap.: (h*-hG) = RTC×0.42 = 14.3 ;
(s*-sG) = R×0.37 = 0.0306
Condenser, 1st law
qCOND = h3 - h2 = hF - (hF + x2hFG) = -x2hFG
= -0.90(154.4 - 14.3) = -126.1 kJ/kg
13-65
ηS EXP ENGINE =
b) expansion engine, efficiency
w12
w12S
1st law & 2nd law, ideal engine:
w12S = h1 - h2S,
s2S - s1 = 0
1st law & 2nd law, real engine:
w12 = h1 - h2,
s2 - s1 > 0
*
*
*
*
(h1-h2) = (h1-h2S) = 0.787(180-50) = +102.3
* *
(s1-s2)
=
* *
(s1-s2S)
P1
453.2
= 0.787 ln
- 0.08273 ln
323.2
978
= 0.266 05 - 0.08273 ln
*
P1
978
*
*
(h2-h2) = (1-x2)(h2-hF2) + x2S(h2-hG2) = 0.1 × 154.4 + 0.9 × 14.3 = 28.3
*
*
*
(h2S-h2S) = (1-x2S)(h2S-hF2) + x2S(h2S-hG2) = 154.4 - x2S×140.1
*
*
*
(s2S-s2S) = (1-x2S)(s2S-sF2) + x2S(s2S-sG2) = 0.4649 - x2S×0.4343
Substituting,
*
(s1-s1)
+ 0.266 05 - 0.082 73 ln
P1
+ 0.4649 - x2S × 0.4343 = 0 (Eq.1)
978
Also
*
(h1-h1) + 102.3 + 28.3
*
(h1-h1)
+ 102.3 + 154.4 - x2S×140.1
= 0.80 (Eq.2)
P1(kPa)
453.2
, Tr1 =
= 1.105
Both Pr1 =
410.3
4250
Trial & Error solution:
Assume P1 = 9300 kPa, Pr1 = 2.188
From D.2 and D.3:
*
(h1-h1) = RTC×3.05 = 103.5 and
*
(s1-s1) = R×2.10 = 0.1737
Substituting into eq. 1,
-0.1737 + 0.730 75 - 0.1863 - x2S×0.4343 = 0
=>
Substituting into eq. 2,
27.1
-103.4 + 130.5
=
= 0.806 ≈ 0.8
-103.4 + 256.7 - 0.8537 ×⊇140.1 33.6
⇒ P1 = 9300 kPa
x2S = 0.8537
13-66
13.72 The refrigerant fluid R-21 (see Table A.2) is to be used as the working fluid in a
solar energy powered Rankine-cycle type power plant. Saturated liquid R-21
enters the pump, state 1, at 25oC, and saturated vapor enters the turbine, state 3, at
88oC. For R-21 : CP0 = 0.582 kJ/kg K and find the boiler heat transfer q23 and
the thermal efficiency of the cycle.
R-21: Tc = 451.6 K, Pc = 5.18 MPa,
M = 102.925 kg/kmol, R = 0.08079 kJ/kg-K
State 1: T1 = 25oC, x1 = 0
From D.1 and D.2: Tr1 = 0.66, Pr1 = 0.06, P1 = Pr1Pc = 311 kPa
Z1 = 0.01 ,
*
(h1 − h1)f = 4.98 RTc ,
v1 = Z1RT1/P1 = 0.000775 m3/kg
State 3: T3 = 88oC, x3 = 1.0
Tr3 = 0.8, Pr3 = 0.25, P3 = Pr3Pc = 1295 kPa
*
*
(h3 − h3)g = 0.46 RTc , (s3 − s3)g = 0.39 R
=> qH = h3 - h2
a) 1st Law Boiler: qH + h2 = h3 + w; w = 0
1st Law Pump: q + h1 = h2 + wp; q = 0
wp = -∫ v dP = -v1(P2 - P1) = -0.000775(1295 - 311) = -0.76 kJ/kg
*
h3 - h1 = (h3 − h3)
*
*
*
*
+ (h3 − h1) + (h1 − h1)
*
h3 − h1 = CP (T3 - T1) = 0.582(88 - 25) = 36.7 kJ/kg,
h3 - h1 = 36.7 + RTc( -0.46 + 4.98) = 201.6 kJ/kg
h2 = h1 - wp;
qH = h3 - h1 + wp = 201.6 - 0.76 = 200.8 kJ/kg
b) ηth = wnet/qH; wnet = wt + wp
1st Law Turbine: q + h3 = h4 + wt; q = 0
*
wt = h3 - h4 = (h3 − h3)
*
*
*
*
h3 − h4 = Cp(T3 - T4) = 36.7 kJ/kg;
*
*
+ (h3 − h4) + (h4 − h4)
*
T4 = T1
*
h4 − h4 = (1 - x4)(h4 − h4)f + x4 (h4 − h4)g; x4 = ?
2nd Law Turbine:
∆Snet = m(s3 - s4) - Q/T;
*
s3 - s4 = (s3 − s3)
*
*
*
*
Assume Isentropic
*
+ (s3 − s4) + (s4 − s4) = 0
T3
(s3 − s4) = CP ln T
4
− R ln
P3
P4
= -0.0033 kJ/kg K
13-67
*
s4 − s4 = 0.0348 kJ/kg-K,
*
*
*
s4 − s4 = (1 - x4)(s4 − s4)f + x4 (s4 − s4)g = 7.47 R - x4(7.47 - 0.15) R
Tr4 = Tr1 = 0.66,
*
*
(s4 − s4)f = 7.47 R, (s4 − s4)g = 0.15 R
0.0348 = 7.47 R - x4(7.47 - 0.15) R
*
(h4 − h4)f = 4.98 RTc ,
*
h4 − h4 = 11.5,
=>
x 4 = 0.962
*
(h4 − h4)g = 0.15 RTc
wt = h3 - h4 = 31.5 kJ/kg;
wnet = 30.7 kJ/kg
ηth = wnet/qH = 0.153
13.73 A cylinder/piston contains a gas mixture, 50% CO2 and 50% C2H6 (mole basis)
at 700 kPa, 35°C, at which point the cylinder volume is 5 L. The mixture is now
compressed to 5.5 MPa in a reversible isothermal process. Calculate the heat
transfer and work for the process, using the following model for the gas mixture:
a. Ideal gas mixture.
b. Kay’s rule and the generalized charts.
c. The van der Waals equation of state.
a) Ideal gas mixture
U2 - U1 = mCv mix(T2 - T1) = 0
Q12 = W12 = ⌠
⌡ P dV = P1V1 ln(V2/V1) = - P1V1 ln(P2/P1)
= - 700 × 0.005 ln(5500/700) = -7.71 kJ
b) Kay's rule
Tcmix = 0.5 × 304.1 + 0.5 × 305.4 = 304.75 K
Pcmix = 0.5 × 7.38 + 0.5 × 4.88 = 6.13 MPa
Tr1 = 308.15/304.75 = 1.011, P r1 = 0.7/6.13 = 0.1142
Z1 = 0.96, ∆h1 = 0.12, ∆s1 = 0.08
n = P1V1/Z1R T1 =
700*0.005
= 0.00142 kmol
0.962*8.3145*308.15
Tr2 = Tr1 , Pr2 = 5.5/6.13 = 0.897,
Z2 = 0.58, ∆h2 = 1.35, ∆s2 = 1.0
- - h2 - h1 = (h2 - h1) - R Tc(∆h2 - ∆h1)
= 0 - 8.3145 × 304.75(1.35 - 0.12) = - 3117
- u-2 - u-1 = h2 - h1 + RT(Z1 - Z2) = - 3117
13-68
+ 8.3145 × 308.15(0.96 - 0.58) = -2143 kJ/kmol
Q12 = nT(s-2 - -s1)T = 0.00142 × 308.15 × 8.3145[ 0 - ln(5.5/0.7) -1.0
W12 = Q12
+ 0.08 ] = - 10.85 kJ
- n(u- - u- ) = -10.85 - 0.00142(-2143) = - 7.81 kJ
2
1
c) van der waal's equation
For CO2 :
b = R Tc/8Pc = 8.3145 × 304.1/8 × 7380 = 0.04282
a = 27 Pc b2 = 27 × 7380 × 0.042822 = 365.45
For C2H6 :
b = R Tc/8Pc = 8.3145 × 305.4/8 × 4880 = 0.06504
a = 27 Pc b2 = 27 × 4880 × 0.065042 = 557.41
amix = (0.5 365.45 + 0.5 557.41)2 = 456.384
bmix = 0.5 × 0.04282 + 0.5 × 0.06504 = 0.05393
8.3145*308.2 456.384
- - 2 - 700 = 0
v-1 - 0.05393
v1
By trial and error: v- = 3.5329 m3/kmol
1
8.3145*308.2 456.384
- - 2 - 5500 = 0
v-2 - 0.05393
v2
By trial and error: v-2 = 0.2815 m3/kmol
n = V1/v-1 = 0.005/3.5329 = 0.00142
v-2 - b
Q12 = nT(s2 - s1)T = n R T ln v -b
1
= 0.00142 × 8.3145 × 308.2 ln
0.2815 - 0.05392
= - 9.93 kJ
3.5329 - 0.05392
U2-U1 = 0.00142 × 456.39(3.5329-1 - 0.2815-1) = -2.12 kJ
Q12 = U2-U1 + W12 => W12 = -9.93 -(-2.12) = -7.81 kJ
13-69
13.74 A gas mixture of a known composition is frequently required for different
purposes, e.g., in the calibration of gas analyzers. It is desired to prepare a gas
mixture of 80% ethylene and 20% carbon dioxide (mole basis) at 10 MPa, 25°C
in an uninsulated, rigid 50-L tank. The tank is initially to contain CO2 at 25°C
and some pressure P1. The valve to a line flowing C2H4 at 25°C, 10 MPa, is now
opened slightly, and remains open until the tank reaches 10 MPa, at which point
the temperature can be assumed to be 25°C. Assume that the gas mixture so
prepared can be represented by Kay’s rule and the generalized charts. Given the
desired final state, what is the initial pressure of the carbon dioxide, P1?
Determine the heat transfer and the net entropy change for the process of charging
ethylene into the tank.
Pi =10 MPa A
A = C2H4, B = CO2
T1 = 25 oC
P2 = 10 MPa, T2 = 25 oC
yA2 = 0.8, yB2 = 0.2
o
Ti = 25 C
V=0.05 m 3
a) Mixture at 2 :
PC2 = 0.8 × 5.04 + 0.2 × 7.38 = 5.508 MPa
TC2 = 0.8 × 282.4 + 0.2 × 304.1 = 286.7 K
Tr2 = 298.15/286.7 = 1.040; Pr2 = 10/5.508 = 1.816
D.1 :
n2 =
Z2 = 0.32
P 2V
10 000×0.05
= 0.6302 kmol
- =
Z2RT2 0.32×8.3145×298.2
nA2 = ni = 0.8 n2 = 0.5042 kmol C2H4
nB2 = n1 = 0.2 n2 = 0.1260 kmol CO2
298.2
= 0.981
304.1
n1ZB1RT1 0.126 ZB1× 8.3145×298.2
Pr1 =
=
= 0.8466 ZB1
PCBV
7380×0.05
Tr1 =
By trial & error: Pr1 = 0.618 & ZB1 = 0.73
`
⇒ P1 = 0.618 × 7.38 = 4.56 MPa
b) 1st law: QCV + nihi = n2u-2 - n1u-1 = n2h2 - n1h1 - (P2-P1)V
- -*
- -*
- -*
or QCV = n2(h2-h2) - n1(h1-h1) - ni(hi-hi ) - (P2-P1)V
B
13-70
-* -* -*
(since Ti = T1 = T2, hi = h1 = h2)
-* (h1-h1) = 0.83×8.3145×304.1 = 2099 kJ/kmol
-* (h2-h2) = 3.40×8.3145×286.7 = 8105 kJ/kmol
Tri =
298.2
10
= 1.056, Pri =
= 1.984
282.4
5.04
-* (hi -hi) = 3.35×8.3145×282.4 = 7866 kJ/kmol
QCV = 0.6302(-8105) - 0.126(-2099) - 0.5042(-7866) - (10 000-4560)×0.05
= -1149 kJ
∆SCV = n2-s2 - n1-s1 , ∆SSURR = - QCV/T0 - ni-si
∆SNET = n2-s2 - n1-s1 - QCV/T0 - ni-si
*
*
Let -sA0 = -sB0 = 0 at T0 = 25 oC, P0 = 0.1 MPa
*
Then -sMIX 0 = -8.3145 (0.8 ln 0.8 + 0.2 ln 0.2) = 4.161 kJ/kmol K
-s = -s* + (s-* -s-* ) + (s- -s-* )
B0
P1 T1 P0 T0 B
1
1 P1 T1 B
= 0 + (0-8.3145 ln
4.56
) - 0.60 × 8.3145 = -36.75 kJ/kmol K
0.1
-s = -s* + (s-* -s-* ) + (s- -s-* )
A0
Pi Ti P0 T0 A
i
i Pi Ti A
= 0 + (0-8.3145 ln
10
) - 2.44×8.3145 = -58.58 kJ/kmol K
0.1
-s = -s*
-*
-*
- -*
MIX 0 + (s P2 T2-s P0 T0)MIX + (s 2-s P2 T2)MIX
2
= 4.161 + (0-8.3145 ln
10
) - 2.551×8.3145 = -55.34 kJ/kmol K
0.1
∆SNET = 0.6302(-55.33) - 0.126(-36.75) - 0.5042(-58.58) + 1149/298.2
= +3.15 kJ/K
13-71
English Unit Problems
13.75EA special application requires R-22 at −150 F. It is known that the triple-point
temperature is less than −150 F. Find the pressure and specific volume of the
saturated vapor at the required condition.
The lowest temperature in Table C.10 for R-22 is -100 F, so it must be
extended to -150 F using the Clapeyron eqn. At T1 = -100 F = 359.7 R ,
P1 = 2.398 lbf/in.2
and
R=
1.9859
= 0.022 97 Btu/lbm R
86.469
P hfg (T-T1)
107.9 (309.7-359.7)
ln =
=
= -2.1084
P1 R T × T
0.022 97 309.7 × 359.7
1
P = 0.2912 lbf/in.2
vg =
RT 0.022 97 × 778 × 309.7
=
= 132 ft3/lbm
Pg
0.2912 × 144
13.76EIce (solid water) at 27 F, 1 atm is compressed isothermally until it becomes liquid.
Find the required pressure.
Water, triple point T = 32.02 F = 491.69 R, P = 0.088 67 lbf/in.2
vF = 0.016 022
vI = 0.017 473
hF = 0.00
dPIF
dT
∆P ≈
=
hI = -143.34
hF-hI
(vF-vI)T
dPIF
dT
=
143.34×778.2
= -1085.8
-0.001 451×491.69×144
∆T = -1085.8(27-32.02) = 5450.7 lbf/in.2
P = Ptp + ∆P = 5451 lbf/in.2
13-72
13.77E Using thermodynamic data for water from Tables C.8.1 and C.8.3, estimate the
freezing temperature of liquid water at a pressure of 5000 lbf/in.2.
P
H 2O
T.P.
dPIF
dT
=
hIF
TvIF
≈ constant
At the triple point,
vIF = vF - vI = 0.016 022 - 0.017 473
T
= -0.001 451 ft3/lbm
hIF = hF - hI = 0.0 - (-143.34) = 143.34 Btu/lbm
dPIF
dT
=
143.34
778.2
×
= -1085.8 lbf/in.2 R
491.69(-0.001 451) 144
⇒ at P = 5000 lbf/in2,
T ≈ 32.02 +
(5000-0.09)
= 27.4 F
(-1085.8)
13.78E Determine the volume expansivity, α P, and the isothermal compressibility, β T ,
for water at 50 F, 500 lbf/in.2 and at 500 F, 1500 lbf/in.2 using the steam tables.
Water at 50 F, 500 lbf/in.2 (compressed liquid)
1 ∂v
1 ∆v
≈
(
)
( )
v ∂T P v ∆T P
αP =
Using values at 32 F, 50 F and 100 F
αP ≈
1
0.016 106 - 0.015 994
= 0.000 103 F-1
0.015 998
100 - 32
βT = -
1 ∂v
1 ∆v
≈
(
)
( )
v ∂P T
v ∆P T
Using values at saturation, 500 and 1000 lbf/in.2
1
0.015 971 - 0.016 024
βT ≈ = 0.000 0033 in.2/lbf
0.015 998
1000 - 0.178
Water at 500 F, 1500 lbf/in.2 (compressed liquid)
1
0.021 579 - 0.019 264
αP ≈
= 0.001 143 F-1
0.020 245
550 - 450
βT ≈ -
1
0.020 139 - 0.020 357
= 0.000 0108 in.2/lbf
0.020 245
2000 - 1000
13-73
13.79E Sound waves propagate through a media as pressure waves that causes the media
to go through isentropic compression and expansion processes. The speed of
sound c is defined by c2 = (∂P/∂ρ)s and it can be related to the adiabatic
compressibility, which for liquid ethanol at 70 F is 6.4 in.2/lbf. Find the speed of
sound at this temperature.
∂P
=
-v
(∂P
)
(
)=
∂ρ s
∂v s
c2 =
1
2
1 ∂v
ρ
v ∂P s
-
( )
32.174×144
(6.4×10
)
×48.9
1/2
-6
1
β sρ
ρ = 48.9 lbm/ft3
From Table C.3 for ethanol,
⇒c=
=
= 3848 ft/s
13.80E Consider the speed of sound as defined in Problem 13.79. Calculate the speed of
sound for liquid water at 50 F, 250 lbf/in.2 and for water vapor at 400 F, 80
lbf/in.2 using the steam tables.
From problem 13.79 :
∂P
=
-v
(∂P
)
(
)
∂ρ s
∂v s
c2 =
2
Liquid water at 50 F, 250 lbf/in.2
Assume
∆P
≈( )
(∂P
)
s
∂v
∆v
T
Using saturated liquid at 50 F and compressed liquid at 50 F, 500 lbf/in.2,
(500-0.18)×144×32.174
(0.016024+0.015998
)
(
) = 22.832×10
2
0.015998-0.016024
c2 = -
2
c = 4778 ft/s
Superheated vapor water at 400 F, 80 lbf/in.2
v = 6.217, s = 1.6790
At P = 60 lbf/in.2 & s = 1.6790: T = 343.8 F, v = 7.7471
At P = 100 lbf/in.2 & s = 1.6790: T = 446.2 F, v = 5.2394
c2 = -(6.217)2
c = 1690 ft/s
((100-60)×144×32.174
) = 2.856×10
5.2394-7.7471
6
6
13-74
13.81E A cylinder fitted with a piston contains liquid methanol at 70 F, 15 lbf/in.2 and
volume 1 ft3. The piston is moved, compressing the methanol to 3000 lbf/in.2 at
constant temperature. Calculate the work required for this process. The isothermal
compressibility of liquid methanol at 70 F is 8.3 × 10−3 in.2/lbf.
2
2
⌠ ∂v
⌠
w12 = ⌠
⌡ Pdv = 
⌡ P ∂P T dPT = -⌡ vβ T PdPT
( )
1
1
For v ≈ const & β T ≈ const.
=>
w 12 = -
vβ T
(P
2
2
2
2
- P1
)
For liquid methanol, from Table C.3 : ρ = 49.1 lbm/ft3
m = 1.0 × 49.1 = 49.1 lbm
V1 = 1.0 ft3,
W12 =
1.0×0.0083
(3000)2 - (15)2 ×144 = 37 349 ft lbf = 48.0 Btu
2
[
]
13.82E A piston/cylinder contains 10 lbm of butane gas at 900 R, 750 lbf/in.2. The butane
expands in a reversible polytropic process with polytropic exponent, n = 1.05,
until the final pressure is 450 lbf/in.2. Determine the final temperature and the
work done during the process.
C4H10 ,
P1 = 750 lbf/in.2
m = 10 lbm , T1 = 900 R ,
Rev. polytropic process (n=1.05), P2 = 450 lbf/in.2
Tr1 =
900
750
= 1.176, Pr1 =
= 1.361
765.4
551
V1 =
n
P1V1
10 × 0.67 × 26.58 × 900
750 × 144
=
Z2Tr2 =
n
P2V2
750
→ V2 = 1.484
450
P2V2
mRTC
Z1 = 0.67
= 1.484 ft3
( )
=
=> From Fig. D.1 :
1
1.05
450 × 144 × 2.414
10 × 26.58 × 765.4
= 2.414 ft3
= 0.7688
at Pr2 = 450/551 = 0.817
Trial & error:
2
W12 = ⌠
⌡ PdV =
1
Tr2 = 1.068,
P2V2 - P1V1
1-n
=
Z2 = 0.72
=>
T2 = 817.4 R
750 × 1.484
(450 × 2.4141 --1.05
) × 144
= 98.8 Btu
778
13-75
13.83E A 7-ft3 rigid tank contains propane at 1300 lbf/in.2, 540 F. The propane is then
allowed to cool to 120 F as heat is transferred with the surroundings. Determine
the quality at the final state and the mass of liquid in the tank, using the
generalized compressibility charts.
Propane C3H8:
V = 7.0 ft3, P1 = 1300 lbf/in.2, T1 = 540 F = 1000 R
cool to T2 = 120 F = 580 R
From Table C.1 : TC = 665.6 R, PC = 616 lbf/in.2
1300
1000
= 2.110, Tr1 =
= 1.502
616
665.6
From D.1: Z1 = 0.83
Pr1 =
Z1RT1 0.83 × 35.04 × 1000
v2 = v1 =
=
P1
1300 × 144
= 0.1554 ft3/lbm
From D.1 at Tr2 = 0.871, saturated => PG2 = 0.43 × 616 = 265 lbf/in.2
vG2 =
0.715 × 35.04 × 580
265 × 144
= 0.3808
vF2 =
0.075 × 35.04 × 580
265 × 144
= 0.0399
0.1554 = 0.0399 + x 2(0.3781-0.0399)
mLIQ 2 = (1 - 0.3388)
7.0
= 29.8 lbm
0.1554
=>
x2 = 0.3388
13-76
13.84E A rigid tank contains 5 lbm of ethylene at 450 lbf/in.2, 90 F. It is cooled until the
ethylene reaches the saturated vapor curve. What is the final temperature?
C2H4 m = 5 lbm
C 2H 4
P1 = 450 lbf/in2, T1 = 90 F = 249.7 R
T
1
Pr1 =
450
= 0.616,
731
Fig. D.1
2
Tr1 =
549.7
= 1.082
508.3
⇒ Z1 = 0.82
v
Pr2 = Pr1
Z2Tr2
Z1Tr1
= 0.616
ZG2Tr2
0.82 ×⊇1.082
Trial & error:
Tr2
ZG2
Pr2
Pr2 CALC
0.871 0.715
0.43
0.432
= 0.6943 ZG2Tr2
~ OK
=>
T2 = 442.7 R
13.85E Calculate the difference in internal energy of the ideal-gas value and the real-gas
value for carbon dioxide at the state 70 F, 150 lbf/in. 2, as determined using the
virial equation of state. For carbon dioxide at 70 F,
B = -2.036 ft3/lb mol , T(dB/dT) = 4.236 ft3/lb mol
CO2 at 70 F, 150 lbf/in2
Solution:
virial:
P=
R BR RT dB
=
(∂P
)
+
+
( )
∂T v v v
v dT
RT BRT
+ 2 ;
v
v
2
2
⌠∞
⌠∞ RT2 dB
∂P
RT2 dB


- P dv = u-u = - ⌡ T
⌡v v2 dT dv = - v dT
∂T v
v
*
[( ) ]
( )
B = -2.036 ft3/lbmol
T
(dB
)= 4.236 ft /lbmol
dT
3
-v = 1 RT 1 + 1+4BP/R
T
2 P
RT 1545×529.7
But
=
= 37.8883
P
150×144
[
]
[
v- = 0.5×37.8883 1 +
1+4(-2.036)/37.8883
] = 35.7294 ft /lbmol
-1.9859×529.7
4.236 = -123.9 Btu/lbmol
u--u-* =
35.7294
3
13-77
13.86E Calculate the heat transfer during the process described in Problem 13.81.
From solution 13.82,
V1 = 1.473 ft3,
Tr1 = 1.176,
V2 = 2.396 ft3,
Pr1 = 1.361,
*
From D.1:
(hRT-h)
C
*
W12 = 98.8 Btu
Tr2 = 1.068,
Pr2 = 0.817, T2 = 817.4 R
*
= 1.36,
1
(hRT-h)
C
2
= 0.95
*
h2 - h1 = 0.415 (817.4 - 900) = -34.3 Btu/lbm
26.58×765.4
(-0.95 + 1.36) = -23.6 Btu/lbm
778
U2-U1 = m(h2-h1) - P2V2 + P1V1
h2 - h1 = -34.3 +
450×144×2.414 750×144×1.484
+
= -231.1 Btu
778
778
Q12 = U2-U1 + W12 = -132.3 Btu
= 10(-23.6) -
13-78
13.87E Saturated vapor R-22 at 90 F is throttled to 30 lbf/in.2 in a SSSF process.
Calculate the exit temperature assuming no changes in the kinetic energy, using
the generalized charts, Fig. D.2 and repeat using the R-22 tables, Table C.10.
R-22 throttling process
T1 = 90 F, x1 = 1.00, P2 = 30 lbf/in.2
*
Energy Eq.:
*
*
*
h2-h1 = (h2-h2) + (h2-h1) + (h1-h1) = 0
Generalized charts,
Tr1 =
549.7
= 0.827
664.7
1.9859×664.7
(0.55) = 8.40
86.469
To get CP0, use h values from Table C.10 at low pressure.
*
From D.2:
(h1-h1) =
CP0 ≈
121.867-118.724
= 0.1572
100-80
Substituting into energy Eq.:
at Pr2 =
*
(h2-h2) + 0.1572 (T2 - 30) + 8.40 = 0
30
= 0.042
721
Assume T2 = 43.4 F = 503.1 R => Tr2 =
503.4
= 0.757
664.7
1.9859×664.7
(0.07) = 1.07
86.469
Substituting,
*
(h2-h2) =
-1.07 + 0.1572(43.4 - 90) + 8.40 = 0.005 ≈ 0
⇒ T2 = 43.4 F
b) R-22 tables, C.10:
T1 = 90 F, x1 = 1.0
=> h 1 = 111.62
h2 = h1 = 111.62 , P2 = 30 lbf/in.2 => T2 = 42.1 F
13-79
13.88E A 10-ft3 tank contains propane at 90 F, 90% quality. The tank is heated to 600 F.
Calculate the heat transfer during the process.
V = 10 ft3
T1 = 90 F = 549.7 R, x1 = 0.90
C 3H 8
T
Heat to T2 = 600 F = 1059.7 R
2
M = 44.094, TC = 665.6 R
1
PC = 616 lbf/in.2
v
R = 35.04, CP0 = 0.407
Tr1 = 0.826 Figs. D.1 and D.2 → Z1 = 0.1 × 0.053 + 0.9 × 0.78 = 0.707,
*
h1-h1
RTc
SAT
Pr
m=
= 0.1 × 4.4 + 0.9 × 0.55 = 0.935
= 0.31 ;
SAT
P1
= 0.31 × 616 = 191 lbf/in.2
191 × 144 × 10
0.707 × 35.04 × 549.7
Pr2 =
= 20.2 lbm
20.2 × Z 2 × 35.04 × 1059.7
616 × 144 × 10
=
Z2
1.183
at Tr2 = 1.592
 Pr2 = 0.79
Trial & error: 
 Z2 = 0.94
*
P2 = 490 lbf/in.2
*
h2-h2
RTc
= 0.36
*
(h2-h1) = 0.407(600-90) = 207.6 Btu/lbm
*
(h1-h1) = 0.935 × 35.04 × 665.9/ 778 = 28.0
*
(h2-h2) = 0.36 × 35.04 × 665.9/ 778 = 10.8
Q12 = m(h2-h1) - (P2-P1)V
= 20.2 (-10.8 + 207.6 + 28.0) - (490 - 191) ×
= +4541 - 553 = 3988 Btu
144 × 60
778
13-80
13.89E A newly developed compound is being considered for use as the working fluid in
a small Rankine-cycle power plant driven by a supply of waste heat. Assume the
cycle is ideal, with saturated vapor at 400 F entering the turbine and saturated
liquid at 70 F exiting the condenser. The only properties known for this
compound are molecular weight of 80 lbm/lbmol, ideal gas heat capacity
Cpo = 0.20 Btu/lbm R and Tc = 900 R, Pc = 750 lbf/in.2. Calculate the work
input, per lbm, to the pump and the cycle thermal efficiency.
T1 = 400 F = 860 R
1
.
WT
Turbine
.
QH
x1 = 1.0
T3 = 70 F = 530 R
Ht.
Exch
2
Cond
Properties known:
M = 80, CPO = 0.2 Btu/lbm R
3
TC = 900 R, PC = 750 lbf/in.2
P
4
.
-W P
x3 = 0.0
Tr1 =
860
530
= 0.956, Tr3 =
= 0.589
900
900
From D.1: Pr1 = 0.76, P1 = 0.76 × 750 = 570 lbf/in.2 = P4
Pr3 = 0.025, P3 = 19 lbf/in.2 = P2, ZF3 = 0.0045
vF3 = ZF3RT3/P3 =
0.0045 × 1545 × 530
= 0.0168
19 × 144 × 80
4
⌠ vdP ≈ v (P -P ) = -0.0168 (570 - 19) ×
wP = - ⌡
F3 4
3
3
144
= -1.71 Btu/lbm
778
qH + h4 = h1 , but h3 = h4 + wP => q H = (h1-h3) + wP
From D.2,
*
(h1-h1) = (1.9859/80) × 900 × 1.34 = 30.0
*
(h3-h3) = (1.9859/80) × 900 × 5.2 = 116.1
*
*
(h1-h3) = CP0(T1-T3) = 0.2(400-70) = 66.0
(h1-h3) = -30.0 + 66.0 + 116.1 = 152.1
qH = 152.1 + (-1.71) = 150.4 Btu/lbm
13-81
*
* *
*
Turbine, (s2-s1) = 0 = -(s2-s2)+(s2-s1)+(s1-s1)
From D.3,
*
(s1-s1) = (1.9859/80) × 1.06 = 0.0263
* *
(s2-s1) = 0.20 ln
530
19
- 0.024 82 ln
= -0.0124
860
570
Substituting,
*
*
(s2-s2) = - 0.0124 + 0.0263 = 0.0139 = (s2-sF2) - x2sFG2
0.0139 = 0.024 82 × 8.77 - x2 × 0.024 82 (8.77 - 0.075)
=> x 2 = 0.9444
*
*
(h2-h2) = (h2-hF2) - x2hFG2
From D.2,
hfg2 = 0.024 82 × 900 (5.2 − 0.07) = 114.6
*
(h2-h2) = 116.1 - 0.9444 × 114.6 = 7.9
wT = (h1-h2) = -30.0 + 66.0 + 7.9 = 43.9 Btu/lbm
ηTH =
wNET
qH
=
43.9 - 1.7
= 0.281
150.4
13-82
13.90E A 7-ft3 rigid tank contains propane at 730 R, 500 lbf/in.2. A valve is opened, and
propane flows out until half the initial mass has escaped, at which point the valve
is closed. During this process the mass remaining inside the tank expands
according to the relation Pv1.4 = constant. Calculate the heat transfer to the tank
during the process.
C3H8: V = 7.0 ft3, T1 = 400 K, P1 = 500 lbf/in.2
Pv1.4 = const inside
Flow out to m2 = m1/2;
Tr1 =
730
500
= 1.097, Pr1 =
= 0.812
665.6
616
0.76 × 35.04 × 730
= 0.270,
500 × 144
m1 = 7.0/0.270 = 25.92 lbm,
v1 =
P 2 = P1
v1
(v )
1.4
=
2
=> From D.1:
Z1 = 0.76
v2 = 2v1 = 0.54
m2 = m1/2 = 12.96 lbm,
500
= 189.5 lbf/in.2
21.4
Trial & error: saturated with
189.5
Pr2 =
= 0.308  T2 = 0.825 × 665.6 = 549.4 R &
616

P2v2 = Z2RT2
 Z2 = 189.5 × 144 × 0.54 = 0.764
35.04 × 549.4
Z2 = ZF2 + x2(ZG2 - ZF2)
0.764 = 0.052 + x 2(0.78 - 0.052) =>
x2 = 0.978
*
(h1-h1) = 35.04 × 665.6 × 0.85 / 778 = 25.5
*
*
(h2-h1) = 0.407 (549.4 - 730) = -73.4
*
*
(h2-h2) = (h2-hF2) - x2hFG2 =
18.7
1st law:
[4.41 - 0.978(4.41 - 0.54)] =
QCV = m2h2 - m1h1 + (P1-P2)V + mehe AVE
*
Let h1 = 0 then:
*
35.04 × 665.6
778
*
*
*
h1 = 0 + (h1-h1) = -25.5
*
h2 = h1 + (h2-h1) + (h2-h2) = 0 - 73.4 - 18.7 = -92.1
he AVE = (h1 + h2)/2 = -58.8
QCV = 12.96 (-92.1) - 25.92(-25.5) + (500-189.5)×7.0×
= -892 Btu
144
+ 12.96(-58.8)
778
13-83
13.91E A geothermal power plant on the Raft river uses isobutane as the working fluid as
shown in Fig. P13.42. The fluid enters the reversible adiabatic turbine at
320 F, 805 lbf/in.2 and the condenser exit condition is saturated liquid at 91 F.
Isobutane has the properties Tc = 734.65 R, Pc = 537 lbf/in.2, Cpo = 0.3974
Btu/lbm R and ratio of specific heats k = 1.094 with a molecular weight as
58.124. Find the specific turbine work and the specific pump work.
R = 26.58 ft lbf/lbm R = 0.034 166 Btu/lbm R
Turbine inlet: T1 = 320 F , P1 = 805 lbf/in.2
Condenser exit: T3 = 91 F , x3 = 0.0 ;
Tr3 = 550.7 / 734.7 = 0.75
Pr3 = 0.165, Z3 = 0.0275
From D.1 :
P2 = P3 = 0.165 × 537 = 88.6 lbf/in.2
Tr1 = 779.7 / 734.7 = 1.061,
Pr1 = 805 / 537 = 1.499
From D.2 and D.3,
*
(h1-h1) = 0.034 166 × 734.7 × 2.85 = 71.5
*
(s1-s1) = 0.034 166 × 2.15 = 0.0735
* *
(s2-s1) = 0.3974 ln
*
550.7
88.6
- 0.034 166 ln
= -0.0628
779.7
805
*
(s2-s2) = (s2-sF2) - x2sFG2 = 0.034 166 × 6.12 - x2× 0.034 166(6.12 - 0.29)
= 0.2090 - x2× 0.1992
(s2-s1) = 0 = -0.2090 + x2 × 0.1992-0.0628 + 0.0735
*
=>
x2 = 0.9955
*
(h2-h1) = CP0(T2-T1) = 0.3974(550.7-779.7) = -91.0
From D.2,
*
*
(h2-h2) = (h2-hF2) - x2hFG2 = 0.034 166×734.7[4.69 - 0.9955(4.69 - 0.32)]
= 117.7 − 0.9955 × 109.7 = 8.5
Turbine:
Pump
wT = (h1-h2) = -71.5 + 91.0 + 8.5 = 28.0 Btu/lbm
vF3 =
4
ZF3RT3
P3
=
0.0275 ×⊇26.58 × 550.7
= 0.031 55
88.6 ×⊇144
144
wP = - ⌠
⌡ vdP ≈ vF3(P4 -P3) = -0.031 55(805-88.6) ×
= -4.2 Btu/lbm
778
3
13-84
13.92E Carbon dioxide collected from a fermentation process at 40 F, 15 lbf/in.2 should
be brought to 438 R, 590 lbf/in.2 in a SSSF process. Find the minimum amount of
work required and the heat transfer. What devices are needed to accomplish this
change of state?
35.1
R=
= 0.045 12 Btu/lbm R
778
Tri =
500
= 0.913 ,
547.4
438
= 0.80,
547.4
From D.2 and D.3:
*
15
= 0.014
1070
h*-h
(RT ) = 0.02,
C
From D.2 and D.3:
Tre =
Pri =
Pre =
590
= 0.551
1070
*
he-he = 4.50 RTc,
*
*
s*-s
( R ) = 0.01 R
*
se-se = 4.70 R
*
(hi-he) = - (hi -hi) + (hi -he) + (he-he)
= - 0.045 12 × 547.4 × 0.02 + 0.203(500 - 438) + 0.045 12 × 547.4 × 4.50
= 123.2 Btu/lbm
*
* *
*
(si-se) = - (si -si) + (si -se) + (se-se)
= - 0.045 12×0.01 + 0.203 ln
15
500
- 0.045 12 ln
+ 0.045 12×4.70
590
438
= 0.4042 Btu/lbm R
wrev = (hi-he) -T0(si-se) = 123.2 - 500(0.4042) = -78.4 Btu/lbm
qrev = (he-hi) + wrev = -123.2 - 78.9 = -202.1 Btu/lbm
13-85
13.93E A control mass of 10 lbm butane gas initially at 180 F, 75 lbf/in.2, is compressed
in a reversible isothermal process to one-fifth of its initial volume. What is the
heat transfer in the process?
T1 = 180 F, P1 = 75 lbf/in.2
Butane C4H10: m = 10 lbm,
Compressed, reversible T = const, to V2 = V1/5
Tr1 =
640
75
= 0.836, Pr1 =
= 0.136 => From D.1 and D.3: Z 1 = 0.92
765.4
551
*
(s1-s1) = 0.16 ×
26.58
= 0.0055
778
Z1RT1 0.92 ×⊇26.58 ×⊇640
v1 =
=
= 1.449 ft3/lbm
P1
75 ×⊇144
v2 = v1/5 = 0.2898 ft3/lbm
At Tr2 = Tr1 = 0.836
From D.1 and D.3:
sat. liq.:
sat. vap.:
PG = 0.34 × 551 =187 lbf/in.2
(s*-sF) = R × 5.02 = 0.1715
ZF = 0.058 ;
(s*-sG) = R × 0.49 = 0.0167
ZG = 0.765 ;
Therefore
vF =
0.058 ×⊇26.58 ×⊇640
= 0.0366
187 ×⊇144
vG =
0.77 ×⊇26.58 ×⊇640
= 0.4864
187 ×⊇144
Since vF < v2 < vG → two-phase
*
*
x2 =
v2-vF
vG-vF
= 0.563
*
(s2-s2) = (1-x2)(s2-sF2) + x2(s2-sG2)
= 0.437 × 0.1715 + 0.563 × 0.0167 = 0.0843
* *
(s2-s1)
= CP0 ln
T2
T1
- R ln
P2
P1
=0-
26.58
187
× ln
= -0.0312
778
75
(s2-s1) = -0.0843 - 0.0312 + 0.0055 = -0.110 Btu/lbm R
Q12 = Tm(s2-s1) = 640×10(-0.110) = -704 Btu
13-86
13.94E A cylinder contains ethylene, C2H4, at 222.6 lbf/in.2, 8 F. It is now compressed
isothermally in a reversible process to 742 lbf/in.2. Find the specific work and
heat transfer.
Ethylene C2H4 P1 = 222.6 lbf/in.2, T2 = T1 = 8 F = 467.7 R
State 2: P2 = 742 lbf/in.2
R = 55.07 ft lbf/lbm R = 0.070 78 Btu/lbm R
467.7
222.6
Tr2 = Tr1 =
= 0.920 Pr1 =
= 0.305
508.3
731
From D.1, D.2 and D.3:
Z1 = 0.85 ,
h*-h
RTC
( )
1
= 0.40
*
(h1-h1) = 0.070 78 × 508.3 × 0.40 = 14.4
*
(s1-s1) = 0.070 78 × 0.30 = 0.0212
742
= 1.015 (comp. liquid)
Pr2 =
731
From D.1, D.2 and D.3:
Z2 = 0.17
*
(h2-h2) = 0.070 78 × 508.3 × 4.0 = 143.9
*
(s2-s2) = 0.070 78 × 3.6 = 0.2548
*
*
(h2-h1) = 0
* *
(s2-s1) = 0 - 0.070 78 ln
742
= -0.0852
222.6
q12 = T(s2-s1) = 467.7(-0.2548 - 0.0852 + 0.0212) = -149.1 Btu/lbm
(h2-h1) = -143.9 + 0 + 14.4 = -129.5
(u2-u1) = (h2-h1) - RT(Z2-Z1)
= -129.5 - 0.070 78 × 467.7 (0.17 - 0.85) = -107.0
w12 = q12 - (u2-u1) = -149.1 + 107.0 = -42.1 Btu/lbm
13-87
13.95E A cylinder contains ethylene, C 2H4, at 222.6 lbf/in.2, 8 F. It is now compressed in
a reversible isobaric (constant P) process to saturated liquid. Find the specific
work and heat transfer.
Ethylene C2H4 P1 = 222.6 lbf/in.2 = P2, T1 = 8 F = 467.7 R
State 2: saturated liquid, x2 = 0.0
R = 55.07 ft lbf/lbm R = 0.070 78 Btu/lbm R
467.7
222.6
= 0.920 Pr1 = Pr2 =
= 0.305
Tr1 =
508.3
731
From D.1 and D.2:
v1 =
Z1RT1
P1
=
Z1 = 0.85 ,
h*-h
RTC
( )
1
= 0.40
0.85 × 55.07 × 467.7
= 0.683
222.6 × 144
*
(h1-h1) = 0.070 78 × 508.3 × 0.40 = 14.4
From D.1 and D.2:
Z2 = 0.05 ,
v2 =
Z2RT2
P2
h*-h
RTC
T2 = 0.822 × 508.3 = 417.8 R
( )
=
2
= 4.42
0.05 × 55.07 × 417.8
= 0.035 89
222.6 × 144
*
(h2-h2) = 0.070 78 × 508.3 × 4.42 = 159.0
*
*
(h2-h1) = CP0(T2-T1) = 0.411(417.8 - 467.7) = -20.5
2
144
w12 = ⌠
⌡ Pdv = P(v2-v1) = 222.6(0.035 89 - 0.683) ×
= -26.7 Btu/lbm
778
1
q12 = (u2 - u1) + w12 = (h2-h1) = -159.0 - 20.5 + 14.4 = -165.1 Btu/lbm
13-88
13.96E A distributor of bottled propane, C3H8, needs to bring propane from 630 R, 14.7
lbf/in.2 to saturated liquid at 520 R in a SSSF process. If this should be
accomplished in a reversible setup given the surroundings at 540 R, find the ratio
of the volume flow rates V.in / V.out, the heat transfer and the work involved in the
process.
R = 35.04/778 = 0.045 04 Btu/lbm R
630
14.7
Tri =
= 0.947 Pri =
= 0.024
665.6
616
From D.1, D.2 and D.3 : Zi = 0.99
*
(hi -hi) = 0.045 03 × 665.6 × 0.03 = 0.9
*
(si -si) = 0.045 04 × 0.02 = 0.0009
Tre = 520/665.6 = 0.781,
From D.1, D.2 and D.3 :
Pre = = 0.21 , Pe = 0.21 × 616 = 129 lbf/in.2
Ze = 0.035
*
(he-he) = 0.045 04 × 665.6 × 4.58 = 137.3
*
(se-se) = 0.045 04 × 5.72 = 0.2576
*
*
(he-hi ) = 0.407 (520 - 630) = -44.8
* *
(se-si ) = 0.407 ln
520
132
- 0.045 04 ln
= -0.1770
630
14.7
(he-hi) = -137.3 - 44.8 + 0.9 = -181.2
(se-si) = -0.2576 - 0.1759 + 0.0009 = -0.4326
.
ZiTi/Pi
Vin
0.99 630 129
. =
=
×
×
= 300.7
Vout ZeTe/Pe 0.035 520 14.7
wrev = (hi-he) -T0(si-se) = 181.2 - 540(0.4326) = -52.4 Btu/lbm
qrev = (he-hi) + wrev = -181.2 - 52.4 = -233.6 Btu/lbm
13-89
13.97E A line with a steady supply of octane, C8H18, is at 750 F, 440 lbf/in.2. What is
your best estimate for the availability in an SSSF setup where changes in potential
and kinetic energies may be neglected?
Availability of Octane at
Ti = 750 F, Pi = 440 lbf/in.2
R = 13.53 ft lbf/lbm R = 0.017 39 Btu/lbm R
440
1209.7
Pri =
= 1.219, Tri =
= 1.182
361
1023.8
From D.2 and D.3:
*
(h1-h1) = 0.017 39 × 1023.8 × 1.15 = 20.5
*
(s1-s1) = 0.017 39 × 0.71 = 0.0123
Exit state in equilibrium with the surroundings
Assume T0 = 77 F, P0 = 14.7 lbf/in.2
536.7
14.7
= 0.524 , Pr0 =
= 0.041
1023.8
361
From D.2 and D.3:
Tr0 =
*
(h0-h0) = RTC × 5.41 = 96.3
*
and
*
(s0-s0) = R × 10.38 = 0.1805
*
(hi -h0) = 0.409(1209.7 - 536.7) = 275.3
* *
(si -s0) = 0.409 ln
1209.7
440
- 0.017 39 ln
= 0.2733
536.7
14.7
(hi-h0) = -20.5 + 275.3 + 96.3 = 351.1
(si-s0) = -0.0123 + 0.2733 + 0.1805 = 0.4415
ψi = wrev = (hi-h0) - T0(si-s0) = 351.1 - 536.7 (0.4415) = 114.1 Btu/lbm
14-1
CHAPTER 14
The correspondence between the new problem set and the previous 4th edition
chapter 12 problem set.
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Old
new
3
new
new
1
2
6
4
5
10
7
8
9
11
12
13
new
15
16
18
19
new
20
21
22
23
24
25
26
27
New
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Old
28
new
29
30
31
32
33
34
36
37
39
40 mod
42
17
43
44
46 mod
49
50
47
48
52
new
56
57
new
58
59
67
65
New
61
62
63
64
65
66
67
68
69
70
71
72
73
74
The problems that are labeled advanced start at number 66.
Old
new
66
69
70
71
14
38
41
54
60
61
62
68
new
14-2
The English unit problems are:
New
Old
75
76
77
78
79
72
73
74
76
77
New
80
81
82
83
84
85
86
87
88
89
Old
78
79
80
81
82
83
84
85
86
87
New
90
91
92
93
94
95
96
97
98
Old
88
89
90
92 mod
96
97
98
100
101
14-3
14.1
Calculate the theoretical air-fuel ratio on a mass and mole basis for the
combustion of ethanol, C2H5OH.
Reaction Eq.: C2H5OH + νO2(O2 + 3.76N2) => aCO 2+ bH2O + cN2
Balance C: 2 = a
Balance H: 6 = 2b
=> b = 3
=> νO2 = 3
Balance O: 1 + 2νO2 = 2a + b = 4 + 3 = 7
(air/fuel)mol = νO2(1 + 3.76)/1 = 3 × 4.76 = 14.28
(air/fuel)mass = (νO2MO2 + νN2 MN2)/MFuel
= (3×31.999 + 11.28×28.013)/46.069 = 8.943
14.2
In a combustion process with decane, C10H22, and air, the dry product mole
fractions are 86.9% N2, 1.163% O2, 10.975% CO2 and 0.954% CO. Find the
equivalence ratio and the percent theoretical air of the reactants.
x C10H22 + φ νO (O2 + 3.76 N2) → a H2O + b CO2 + c CO + d N2 + e O2
2
Stoichiometric combustion: φ = 1, c = 0, e = 0,
C balance: b = 10x
H balance: a = 22x/2 = 11x, O balance: 2 νO = a + 2b = 11x + 20x = 31x
2
νO = 15.5x, νN = 58.28x
2
=> (A/F)s = (νO
2
Actual combustion:
C balance:
d = 86.9
2
→ φ νO × 3.76 = 86.9
2
10x = 10.975 + 0.954 = 11.929
+ν
N2
)/x = 73.78
→ φ νO = 23.11
2
=> x = 1.1929
(A/F)ac = φ ν O × 4.76/1.1929 = 92.22
2
φ = (A/F)ac / (A/F)s = 92.22 / 73.78 = 1.25
14.3
or 125%
A certain fuel oil has the composition C10H22. If this fuel is burned with 150%
theoretical air, what is the composition of the products of combustion?
C10H22 + φ νO (O2 + 3.76 N2) → a H2O + b CO2 + c N2 + d O2
2
Stoichiometric combustion: φ = 1, d = 0,
H balance: a = 22/2 = 11,
=> νO2 = 15.5
O balance: 2 νO = a + 2b = 11 + 20 = 31
Actual case:
H balance: a = 11, C balance:
O2 balance:
C balance: b = 10
2
φ = 1.5 => νO2 = 1.5×15.5 = 23.25
b = 10,
N balance: c = 23.25×3.76 = 87.42
d = 23.25 - 10 - 11/2 = 7.75 (excess oxygen)
14-4
14.4
Natural gas B from Table 14.2 is burned with 20% excess air. Determine the
composition of the products.
The reaction equation (stoich. complete comb.) with the fuel composition is:
60.1 CH4 + 14.8 C2H6 + 13.4 C3H8 + 4.2 C4H10 + 7.5 N2
+ νO2 (O2 + 3.76 N2) →
a H2O + b CO2 + c N2
C balance:
60.1 + 2×14.8 + 3×13.4 + 4×4.2 = b = 146.7
H balance:
4×60.1 + 6×14.8 + 8×13.4 + 10×4.2 = 2a = 478.4 => a = 239.2
O balance:
2 νO2 = a + 2b = 239.2 + 2×146.7
N2 balance:
7.5 + 3.76 νO2 = c = 1008.8
20% excess air: νO2 = 1.2×266.3 = 319.56
so now more O2 and N2
Extra oxygen: d = 319.56 - 266.3 = 53.26,
c = 7.5 + 3.76×319.56 = 1209
239.2 H2O + 146.7 CO2 + 1209 N2 + 53.26 O2
Products:
14.5
=> νO2 = 266.3
A Pennsylvania coal contains 74.2% C, 5.1% H, 6.7% O, (dry basis, mass
percent) plus ash and small percentages of N and S. This coal is fed into a gasifier
along with oxygen and steam, as shown in Fig. P14.5. The exiting product gas
composition is measured on a mole basis to: 39.9% CO, 30.8% H2, 11.4% CO2,
16.4% H2O plus small percentages of CH4, N2, and H2S. How many kilograms of
coal are required to produce 100 kmol of product gas? How much oxygen and
steam are required?
Number of kmol per 100 kg coal:
C : n = 74.2/12.01 = 6.178
H2: n = 5.1/2.016 = 2.530
O2: n = 6.7/31.999 = 0.209
(6.178 C + 2.53 H2 + 0.209 O2)x + y H2O + z O2 in and
39.9 CO + 30.8 H2 + 11.4 CO2 + 16.4 H2O out
in 100 kmol of mix out
→
C balance:
6.178 x = 39.9 + 11.4
H2 balance:
2.53×8.304 + y = 30.8 + 16.4
O2 balance: 0.209 × 8.304 +
→ y = 26.191
26.191
39.9
16.4
+z=
+ 11.4 +
→ z = 24.719
2
2
2
Therefore, for 100 kmol of mixture out
require:
x = 8.304
830.4 kg of coal
26.191 kmol of steam
24.719 kmol of oxygen
14-5
14.6
Repeat Problem 14.5 for a certain Utah coal that contains, according to the coal
analysis, 68.2% C, 4.8% H, 15.7% O on a mass basis. The exiting product gas
contains 30.9% CO, 26.7% H2, 15.9% CO2 and 25.7% H2O on a mole basis.
Number of kmol per 100 kg coal:
C : 68.2/12.01 = 5.679
H2: 4.8/2.016 = 2.381
O2: 15.7/32.00 = 0.491
(5.679 C + 2.381 H2 + 0.491 O2)x + y H2O + z O2 in
30.9 CO + 26.7 H2 + 15.9 CO2 + 25.7 H2O out
in 100 kmol of mix out
→
C:
5.679x = 30.9 + 15.9
x = 8.241
H 2:
2.381 × 8.241 + y = 26.7 + 25.7
O 2:
0.491 × 8.241 +
→ y = 32.778
32.778
30.9
25.7
+z=
+ 15.9 +
2
2
2
→
z = 23.765
Therefore, for 100 kmol of mixture out,
require: 824.1 kg of coal
32.778 kmol of steam
23.765 kmol of oxygen
14.7
A sample of pine bark has the following ultimate analysis on a dry basis, percent
by mass: 5.6% H, 53.4% C, 0.1% S, 0.1% N, 37.9% O and 2.9% ash. This bark
will be used as a fuel by burning it with 100% theoretical air in a furnace.
Determine the air–fuel ratio on a mass basis.
Converting the Bark Analysis from a mass basis:
Substance
S
H2
C
O2
N2
c/M =
0.1/32
5.6/2
53.4/12
37.9/32
0.1/28
kmol / 100 kg coal
0.003
2.80
4.45
1.184
0.004
Product
SO2
H 2O
CO2
oxygen required
0.003
1.40
4.45
--
--
Combustion requires: 0.003 + 1.40 + 4.45 = 5.853 kmol O2 there is in the
bark 1.184 kmol O2 so the net from air is 4.669 kmol O2
AF = (4.669 + 4.669 × 3.76) ×
28.97
kg air
= 6.44
100
kg bark
14-6
14.8
Liquid propane is burned with dry air. A volumetric analysis of the products of
combustion yields the following volume percent composition on a dry basis: 8.6%
CO2, 0.6% CO, 7.2% O2 and 83.6% N2. Determine the percent of theoretical air
used in this combustion process.
a C3H8 + b O2 + c N2 → 8.6 CO2 + 0.6 CO + d H2O + 7.2 O2 + 83.6 N2
C balance:
3a = 8.6 + 0.6 = 9.2 ⇒ a = 3.067
H2 balance:
4a = d ⇒ d = 12.267
N2 balance:
c = 83.6
O2 balance:
b = 8.6 +
Air-Fuel ratio =
0.6 12.267
+
+ 7.2 = 22.234
2
2
22.234 + 83.6
= 34.51
3.067
Theoretical:
C3H8 + 5 O2 + 18.8 N2 → 3 CO2 + 4 H2O + 18.8 N2
⇒ theo. A-F ratio =
% theoretical air =
14.9
5 + 18.8
= 23.8
1
34.51
× 100 % = 145 %
23.8
A fuel, CxHy, is burned with dry air and the product composition is measured on a
dry basis to be: 9.6% CO2, 7.3% O2 and 83.1% N2. Find the fuel composition
(x/y) and the percent theoretical air used.
νFuCxHy + νO O2 + 3.76νO N2 → 9.6CO2 + 7.3O2 + 83.1N2 + νH OH2O
2
2
2
N2 balance: 3.76νO = 83.1 ⇒ νO = 22.101
2
2
1
O2 balance: νO = 9.6 + 7.3 + 2 νH
2O
2
H balance: νFu y = 2 νH
2O
⇒ νH
2O
= 10.402
= 20.804
C balance: νFu x = 9.6
Fuel composition ratio = x/y = 9.6/20.804 = 0.461
Theoretical air =
νO AC
2
=
22.101
νO stoich 9.6 + 1 × 29.804
2
4
= 1.493
14-7
14.10 Many coals from the western United States have a high moisture content.
Consider the following sample of Wyoming coal, for which the ultimate analysis
on an as-received basis is, by mass:
Component Moisture
H
C S N O
Ash
% mass
28.9
3.5 48.6 0.5 0.7 12.0
5.8
This coal is burned in the steam generator of a large power plant with
150% theoretical air. Determine the air–fuel ratio on a mass basis.
Converting from mass analysis:
Substance
S
H2
C
O2
N2
c/M =
0.5/32
3.5/2
4.86/12
12/32
0.7/28
kmol / 100 kg coal
0.0156
1.75
4.05
0.375
0.025
Product
SO2
H 2O
CO2
oxygen required
0.0156
0.875
4.05
--
--
Combustion requires then oxygen as: 0.0156 + 0.875 + 4.05 = 4.9406
The coal does include 0.375 O2 so only 4.5656 O2 from air/100 kg coal
AF = 1.5 × (4.5656 + 4.5656 × 3.76) × 28.97/100 = 9.444 kg air/kg coal
14.11 Pentane is burned with 120% theoretical air in a constant pressure process at 100
kPa. The products are cooled to ambient temperature, 20°C. How much mass of
water is condensed per kilogram of fuel? Repeat the answer, assuming that the air
used in the combustion has a relative humidity of 90%.
C5H12 + 1.2 × 8 (O2 + 3.76 N2) → 5 CO2 + 6 H2O + 0.96 O2 + 36.1 N2
Products cooled to 20oC, 100 kPa, so for H2O at 20°C: Pg = 2.339 kPa
yH2O MAX = Pg/P =
Therefore,
nH2O MAX
2.339
=
=> n H2O MAX = 1.007 < n H2O
100
nH2O MAX + 42.06
nH2O VAP = 1.007,
mH2O LIQ =
nH2O LIQ = 6 - 1.007 = 4.993
4.993 × 18.015
= 1.247 kg/kg fuel
72.151
Pv1 = 0.9 × 2.339 = 2.105 kPa
nH2O IN = 0.013375 ×
=>
w1 = 0.622 ×
2.105
= 0.013375
97.895
28.97
× (9.6 + 36.1) = 0.983 kmol
18.015
nH2O OUT = 0.983 + 6 = 6.983 => nH2O LIQ = 6.983 - 1.007 = 5.976 kmol
nH2O LIQ =
5.976 × 18.015
= 1.492 kg/kg fuel
72.151
14-8
14.12 The coal gasifier in an integrated gasification combined cycle (IGCC) power plant
produces a gas mixture with the following volumetric percent composition:
Product
CH4
% vol.
0.3
H
2
29.6
CO
CO2
41.0
10.0
N
2
0.8
H 2O
H 2S
NH3
17.0
1.1
0.2
This gas is cooled to 40°C, 3 MPa, and the H2S and NH are removed in
3
water scrubbers. Assuming that the resulting mixture, which is sent to the
combustors, is saturated with water, determine the mixture composition and the
theoretical air–fuel ratio in the combustors.
CH4
H2
CO
CO2
N2
n
0.3
yH2O =
29.6
nV
,
nV+81.7
where
41.0
10.0
0.8
81.7
nV = number of moles of water vapor
Cool to 40°C PG = 7.384, P = 3000 kPa
nV
7.384
=
yH2O MAX =
3000 nV+81.7
a) Mixture composition:
CH4
H2
0.3 kmol
29.6
→
nV = 0.2016
CO
CO2
N2
H2O(v)
41.0
10.0
0.8
0.2016
81.9016 kmol (from 100 kmol of the original gas mixture)
0.3 CH4 + 0.6 O2 → 0.3 CO2 + 0.6 H2O
29.6 H2 + 14.8 O2 → 29.6 H2O
41 CO + 20.5 O2 → 41 CO2
⇒ Number of moles of O2 = 0.6 + 14.8 + 20.5 = 35.9
Number of moles of air = 35.9 + 3.76 × 35.9 (N2)
Air Fuel ratio =
28.97(35.9 + 3.76(35.9))
0.3(16) + 29.6(2) + 41(28) + 10(44) + 0.8(28) + 0.2016(18)
= 2.95 kg air/kg fuel
14-9
14.13 The hot exhaust gas from an internal combustion engine is analyzed and found to
have the following percent composition on a volumetric basis at the engine
exhaust manifold. 10% CO2, 2% CO, 13% H2O, 3% O2 and 72% N2. This gas is
fed to an exhaust gas reactor and mixed with a certain amount of air to eliminate
the carbon monoxide, as shown in Fig. P14.13. It has been determined that a mole
fraction of 10% oxygen in the mixture at state 3 will ensure that no CO remains.
What must the ratio of flows be entering the reactor?
1 Exh. gas
Exhaust gas at state 1: CO2 10 %, H2O 13%,
Reactor
CO 2%, O2 3%, N2 72%
CO = 0 %, O2 = 10 %
Exhaust gas at state 3:
3 gas
2 Air
⇒ 0.02 CO + x O2 + 3.76x N2 → 0.02 CO2 + (x-0.01) O2 + 3.76x N2
At 3: CO2 = 0.10 + 0.02 = 0.12,
H2O = 0.13
O2 = (x-0.01) + 0.03 + x + 0.02
N2 = 0.72 + 3.76x
or nTOT = 0.12 + 0.13 + x + 0.02 + 0.72 + 3.76x = 0.99 + 4.76x
yO2 = 0.16 =
or
x+0.02
0.99+4.76x
air 2
Exh. Gas 1
=
→
x = 0.151
4.76x
kmol air
= 0.718
1
kmol Exh. gas
14.14 Methanol, CH3OH, is burned with 200% theoretical air in an engine and the
products are brought to 100 kPa, 30°C. How much water is condensed per
kilogram of fuel?
CH3OH + νO {O2 + 3.76 N2}→ CO2 + 2 H2O + 3.76 νO N2
2
Stoichiometric νO
2
2S
= 1.5 ⇒ νO
2 AC
=3
Actual products: CO2 + 2 H2O + 1.5 O2 +11.28 N2
Psat(30°C) = 4.246 kPa
⇒ yH2O = 0.04246 =
νH2O
1 + νH2O + 1.5 + 11.28
⇒ νH2O = 0.611 ⇒ ∆ν H2O cond = 2 - 0.611 = 1.389
MFu = 32.042
∆MH2O 1.389 × 18
kg H2O
=
= 0.781
kg fuel
MFu
32.042
out
14-10
14.15 The output gas mixture of a certain air–blown coal gasifier has the composition of
producer gas as listed in Table 14.2. Consider the combustion of this gas with
120% theoretical air at 100 kPa pressure. Determine the dew point of the products
and find how many kilograms of water will be condensed per kilogram of fuel if
the products are cooled 10°C below the dew-point temperature.
{3 CH4 + 14 H2 + 50.9 N2 + 0.6 O2 + 27 CO + 4.5 CO2}
+ 31.1 O2 + 116.9 N2 → 34.5 CO2 + 20 H2O + 5.2 O2 + 167.8 N2
Products:
yH2O = yH2O MAX = PG/100 =
⇒ PG = 8.79 kPa →
20
34.5 + 20 + 5.2 + 167.8
TDEW PT = 43.2°C
At T = 33.2°C, P G = 5.13 kPa
yH2O =
mH2O LIQ =
nH2O
5.13
=
100 nH2O+34.5+5.2+167.8
→
nH2O = 11.22
8.78(18)
3(16) +14(2) +50.9(28) +0.6(32) +27(28) +4.5(44)
= 0.0639 kg/kg fuel
14.16 Pentene, C5H10 is burned with pure oxygen in an SSSF process. The products at
one point are brought to 700 K and used in a heat exchanger, where they are
cooled to 25°C. Find the specific heat transfer in the heat exchanger.
C5H10 + νO O2 → 5 CO2 + 5 H2O ⇒ νO = 7.5
2
2
. . 5 nF hCO + 5 nF hH
2
.
. -°
+
Q
=
5
n
F hf CO
O
2
2
. + (5 - x) nF h°liq H
2O
Find x: yH
2O max
=
Pg(25°)
Ptot
= 0.0313 =
. + (x) nF h°vap H
2O
x
⇒ x = 0.1614
5+x
Out of the 5 H2O only 0.1614 still vapor.
.
Q
-°
-°
-°
-°
⋅n = -5 ∆hCO2,700 + (5-x)(hf liq - hf vap - ∆h700) + x(hf vap - hf vap - ∆h700)
F
= -5(17761) + 4.84(-44011-14184) - 0.16(14184) = -372738 kJ/kmol Fu
14-11
14.17 Butane gas and 200% theoretical air, both at 25C, enter a SSSF combustor. The
products of combustion exits at 1000 K. Calculate the heat transfer from the
combustor per kmol of butane burned.
C4H10 + (1/φ) νO2(O2 + 3.76 N2) → a CO2 + b H2O + c N2 + d O2
First we need to find the stoichiometric air ( φ = 1, d = 0 )
C balance: 4 = a,
H balance: 10 = 2b =>
b=5
O balance: 2νO2 = 2a + b = 8 + 5 = 13 => νO2 = 6.5
Now we can do the actual air: (1/φ) = 2 => νO2 = 2 × 6.5 = 13
N balance: c = 3.76 νO2 = 48.88, O balance: d = 13 - 6.5 = 6.5
o
o
o
q = HR - HP = HR - HP - ∆HP = M(-HRP) - ∆HP
o
Table 14.3: HRP = -45714,
The rest of the values are from Table A.8
−
−
−
−
∆h CO2 = 33397, ∆h N2 = 21463, ∆h O2 = 22703, ∆h H2O = 26000 kJ/kmol
q = 58.124(45714) - (4 × 33397 + 5 × 26000 + 48.88 × 21463 + 6.5 × 22703)
= 2 657 080 - 1 460 269 = 1 196 811 kJ/kmol butane
14.18 Liquid pentane is burned with dry air and the products are measured on a dry
basis as: 10.1% CO2, 0.2% CO, 5.9% O2 remainder N2. Find the enthalpy of
formation for the fuel and the actual equivalence ratio.
νFuC5H12 + νO O2 + 3.76 νO N2 →
2
2
x H2O + 10.1 CO2 + 0.2 CO + 5.9 O2 + 83.8 N2
Balance of C: 5 νFu = 10.1 + 0.2 ⇒ νFu = 2.06
Balance of H: 12 νFu = 2 x ⇒ x = 6 νFu = 12.36
Balance of O: 2 νO = x + 20.2 + 0.2 + 2 × 5.9 ⇒ νO = 22.28
2
2
Balance of N: 2 × 3.76 νO = 83.8 × 2 ⇒ νO = 22.287 ⇒ OK
2
2
νO for 1 kmol fuel = 10.816
2
φ = 1, C5H12 + 8 O2 + 8 × 3.76 N2 → 6 H2O + 5 CO2 + 30.08 N2
°
°
°
-°
-°
-°
HRP = HP - HR = 6 hf H2O + 5 hf CO2 - hf fuel
°
14.3: HRP = 44983 × 72.151
⇒
φ = νO2 stoich/νO2 AC = 8/10.816 = 0.74
-°
hf fuel = -172 998 kJ/kmol
14-12
14.19 A rigid vessel initially contains 2 kmol of carbon and 2 kmol of oxygen at 25°C,
200 kPa. Combustion occurs, and the resulting products consist of 1 kmol of
carbon dioxide, 1 kmol of carbon monoxide, and excess oxygen at a temperature
of 1000 K. Determine the final pressure in the vessel and the heat transfer from
the vessel during the process.
1
2 C + 2 O2 → 1 CO2 + 1 CO + 2 O2
Process
P 2 = P1 ×
V = constant,
n2T2
n1T1
= 200 ×
C: solid, n1(GAS) = 2, n2(GAS) = 2.5
2.5 × 1000
= 838.4 kPa
2 × 298.2
H1 = 0
H2 = 1(-393522 + 33397) + 1(-110527 + 21686)
+ (1/2)(0 + 22703) = -437615 kJ
Q = (U2 - U1) = (H2 - H1) - n2RT2 + n1RT1
1 2
= (-437615 - 0) - 8.3145(2.5 × 1000 - 2 × 298.2) = -453442 kJ
14.20 In a test of rocket propellant performance, liquid hydrazine (N2H4) at 100 kPa,
25°C, and oxygen gas at 100 kPa, 25°C, are fed to a combustion chamber in the
ratio of 0.5 kg O2/kg N2H4. The heat transfer from the chamber to the
surroundings is estimated to be 100 kJ/kg N2H4. Determine the temperature of the
products exiting the chamber. Assume that only H2O, H2, and N2 are present. The
enthalpy of formation of liquid hydrazine is +50417 kJ/kmol.
1
o
Liq. N2H4: 100 kPa, 25 C
Gas O2: 100 kPa, 25oC
Products
2
.
.
.
.
mO2/mN2H4 = 0.5 = 32nO2/32nN2H4
Energy Eq.:
3
Comb.
Chamber
and
. .
Q/mN2H4 = -100 kJ/kg
QCV = HP - HR = -100 × 32.045 = -3205 kJ/kmol fuel
1
1 N2H4 + 2 O2 → H2O + H2 + N2
1
HR = 1(50417) + 2(0) = 50417 kJ
HP = -241826 + ∆hH2O + ∆hH2 + ∆hN2
∆HP = ∆hH2O + ∆hH2 + ∆hN2 = 241826 + 50417 -3205 = 289038
Table A.8 : 2800 K ∆HP = 282141
Interpolate to get
3000 K ∆HP = 307988
TP = 2854 K
14-13
14.21 Repeat the previous problem, but assume that saturated-liquid oxygen at 90 K is
used instead of 25°C oxygen gas in the combustion process. Use the generalized
charts to determine the properties of liquid oxygen.
Problem same as 14.20, except oxygen enters at 2 as saturated liquid at 90 K.
. .
.
.
.
.
mO2/mN2H4 = 0.5 = 32nO2/32nN2H4 and
Q/mN2H4 = -100 kJ/kg
Energy Eq.:
QCV = HP - HR = -100 × 32.045 = -3205 kJ/kmol fuel
1
1 N2H4 + 2 O2 → H2O + H2 + N2
Reaction equation:
At 90 K, Tr2 = 90/154.6 = 0.582 ⇒
~
∆hf = 5.2
-* Figure D.2, (h - h) = 8.3145 × 154.6 × 5.2 = 6684 kJ/kmol
∆hAT 2 = -6684 + 0.922 × 32(90 - 298.15) = -12825 kJ/kmol
1
HR = 50417 + 2(0 - 12825) = 44005 kJ,
1st law: ∆hP = ∆hH
2O
HP° = -241826
+ ∆hH + ∆hN = Qcv + HR - H°P = 282626
2
2
From Table A.8, ∆HP 2800K = 282141, ∆HP 3000K = 307988
Therefore, T P = 2804 K
14.22 The combustion of heptane C7H16 takes place in a SSSF burner where fuel and
air are added as gases at P , T . The mixture has 125% theoretical air and the
o o
products are going through a heat exchanger where they are cooled to 600 K.Find
the heat transfer from the heat exchanger per kmol of heptane burned.
The reaction equation for stiochiometric ratio is:
C7H16 + vO2 (O2 + 3.76 N2)
=> 7CO2 + 8 H2O + vO2 × 3.76 N2
So the balance (C and H was done in equation) of oxygen gives vO2 = 7 + 4 =
11, and actual one is 11×1.25 = 13.75. Now the actual reaction equation is:
C7H16 + 13.75 O2 + 51.7 N2
=> 7CO2 + 8 H2O + 51.7 N2+ 2.75 O2
To find the heat transfer take control volume as combustion chamber and heat
exchanger
Q = HPo + ∆HP - HRo = HRPo + ∆HP
Q = 100.205(-44922) + 7(12906) + 8(10499) + 51.7(8894) + 2.75(9245)
HR + Q = HP =>
= -4 501 409 + 90342 + 83922 + 459 819.8 + 25423.75
= - 3 841 831 kJ/kmol fuel
14-14
14.23 Ethene, C2H4, and propane, C3H8, in a 1;1 mole ratio as gases are burned with
120% theoretical air in a gas turbine. Fuel is added at 25°C, 1 MPa and the air
comes from the atmosphere, 25°C, 100 kPa through a compressor to 1 MPa and
mixed with the fuel. The turbine work is such that the exit temperature is 800 K
with an exit pressure of 100 kPa. Find the mixture temperature before
combustion, and also the work, assuming an adiabatic turbine.
φ=1
C2H4 + C3H8 + 8 O2 + 30.08 N2 → 5 CO2 + 6 H2O + 30.08 N2
φ = 1.2
C2H4 + C3H8 + 9.6 O2 + 36.096 N2
→ 5 CO2 + 6 H2O + 1.6 O2 + 36.096 N2
45.696 kmol air per 2 kmol fuel
C.V. Compressor (air flow)
Energy Eq.:
wc = h2 - h1
Entropy Eq.:
s2 = s1
⇒
Pr = Pr × P2/P1 = 13.573 ⇒ T2 air = 570.8 K
2
1
wc = 576.44 - 298.34 = 278.1 kJ/kg = 8056.6 kJ/kmol air
C.V. Mixing Chanber (no change in composition)
. . . nairhair in + nFu1h1 in + nFu2h2 in = (SAME)exit
CP F1 + CP F2Texit - T0 = 45.696 CP airT2 air - Texit
C2H4: CP F1 = 43.43, C3H8: CP F2 = 74.06, CP air = 29.07
45.696CP airT2 + CP F1 + CP F2T0
Texit =
= 548.7 K
CP F1 + CP F2 + 45.696 CP air
6
Dew Point Products: yH
2O
PH
2O
= 5 + 6 + 1.6 + 36.096 = 0.1232
= yH OPtot = 123.2 kPa ⇒ Tdew = 105.5°C
2
C.V. Turb. + combustor + mixer + compressor (no Q)
wnet
= Hin - Hout = HR - HP 800
(800°K out so no liquid H2O)
-°
-°
= hf C2H4 + hf C3H8 - 5 hCO - 6 hH O - 1.6 hO - 36.096 hN
2
2
kJ
= 2 576 541 2 kmol Fu
wT= wnet + wcomp = 2 944 695
kJ
2 kmol Fu
2
2
14-15
14.24 One alternative to using petroleum or natural gas as fuels is ethanol (C 2H5OH),
which is commonly produced from grain by fermentation. Consider a combustion
process in which liquid ethanol is burned with 120% theoretical air in an SSSF
process. The reactants enter the combustion chamber at 25°C, and the products
exit at 60°C, 100 kPa. Calculate the heat transfer per kilomole of ethanol, using
the enthalpy of formation of ethanol gas plus the generalized charts.
C2H5OH + 1.2 × 3 (O2 + 3.76 N2) → 2CO2 + 3H2O + 0.6O2 + 13.54N2
-0
Fuel: hf = -235000 kJ/kmol for IG
yH2O MIX =
Products at 60°C, 100 kPa
nV MAX
19.94
=
=> n V MAX = 4.0 > 3 ⇒ No liq.
100
nV MAX+2+0.6+13.54
Tr = 298.2 / 513.9 = 0.58
⇒
∆hf = 5.23
from Figure D.2
-* (h - h)LIQ = 8.3145 × 513.9 × 5.23 = 22347 kJ/kmol
HR = 1(-235000 - 22347) + 0 + 0 = -257 347 kJ
HP = 2(-393522 + 1327) + 3(-241826 + 1178)
+ 0.6(0+1032) + 13.54(0+1020) = -1 491 904 kJ
QCV = HP - HR = -1 234 557 kJ
14.25 Another alternative to using petroleum or natural gas as fuels is methanol,
CH3OH, which can be produced from coal. Both methanol and ethanol have been
used in automotive engines. Repeat the previous problem using liquid methanol as
the fuel instead of ethanol.
CH3OH + 1.2 × 1.5 (O2 + 3.76 N2) → 1 CO2 + 2 H2O + 0.3 O2 + 6.77 N2
React at 25 oC, Prod at 60 oC = 333.2 K, 100 kPa
yH2O MAX =
nV MAX
19.94
=
=> n V MAX = 2.0 > 2 ⇒ No liq.
100
nV MAX+1+0.3+6.77
-o
CH3OH: hf = -201 300 kJ, TC = 512.6 K
~
Tr = 298.15 / 512.6 = 0.582 ⇒ ∆hf = 5.22 Figure D.2
- (h*-h)LIQ = 8.3145 × 512.6 × 5.22 = 22 248
HR = 1 hLIQ = -201 300 - 22 248 = -223 548 kJ
HP = 1(-393 522 + 1327) + 2(-241 826 + 1178)
+ 0.3(1032) + 6.77(1020) = -866 276 kJ
Q = HP - HR = -642 728 kJ
14-16
14.26 Another alternative fuel to be seriously considered is hydrogen. It can be
produced from water by various techniques that are under extensive study. Its
biggest problem at the present time is cost, storage, and safety. Repeat Problem
14.24 using hydrogen gas as the fuel instead of ethanol.
H2 + 1.2 × 0.5 O2 + 1.2 × 3.76 × 0.5 N2 → 1 H2O + 0.1 O2 + 2.256 N2
Products at 60°C, 100 kPa
yH2O MAX =
nV MAX
19.94
=
100
nV MAX+0.1+2.256
Solving, nV MAX = 0.587 < 1 => nV = 0.587, nLIQ = 0.413
HR = 0 + 0 + 0 = 0
HP = 0.413(-285830 + 18.015(251.1 - 104)) + 0.587(-241826 + 1178)
+ 0.1(0 + 1032) + 2.256(0 + 1020) = -255 816 kJ
Q CV = HP - HR = -255 816 kJ
14.27 Hydrogen peroxide, H2O2, enters a gas generator at 25°C, 500 kPa at the rate of
0.1 kg/s and is decomposed to steam and oxygen exiting at 800 K, 500 kPa. The
resulting mixture is expanded through a turbine to atmospheric pressure, 100 kPa,
as shown in Fig. P14.27. Determine the power output of the turbine, and the heat
transfer rate in the gas generator. The enthalpy of formation of liquid H2O2 is −
187 583 kJ/kmol.
.
mH2O2
1
0.1
.
H2O2 → H2O + O2
nH2O2 =
=
= 0.00294 kmol/s
2
34.015
M
.
.
nMIX = nH2O2× 1.5 = 0.00441 kmol/s
2
1
CP0 MIX = × 1.872 × 18.015 + × 0.922 × 31.999 = 32.317
3
3
CV0 MIX = 32.317 - 8.3145 = 24.0 => kMIX = 32.317/24.0 = 1.3464
CV: turbine. Assume reversible → s3 = s2
P3
k-1
k
100 0.2573
T3 = T2( ) = 800(
)
= 528.8 K
500
P2
w = CP0(T2 - T3) = 32.317(800 - 528.8) = 8765 kJ/kmol
.
WCV = 0.00441 × 8765 = 38.66 kW
CV: Gas Generator
.
H1 = 0.00294(-187 583 + 0) = -551.49
.
H2 = 0.00294(-241 826 + 18002) + 0.00147(0 + 15836) = -634.76
.
.
.
QCV = H2 - H1 = -634.76 + 551.49 = -83.27 kW
14-17
14.28 In a new high-efficiency furnace, natural gas, assumed to be 90% methane and
10% ethane (by volume) and 110% theoretical air each enter at 25°C, 100 kPa,
and the products (assumed to be 100% gaseous) exit the furnace at 40°C, 100 kPa.
What is the heat transfer for this process? Compare this to an older furnace where
the products exit at 250°C, 100 kPa.
0.90CH4 + 0.10C2 H6
Prod.
40 oC
100 kPa
Furnace
o
25 C 110% Air
0.9 CH4 + 0.1 C2H6 + 1.1 × 2.15 O2 + 3.76 × 2.365 N2
→ 1.1 CO2 + 2.1 H2O + 0.215 O2 + 8.892 N2
HR = 0.9(-74873) + 0.1(-84740) = -75860 kJ
HP = 1.1(-393 522 + 562) + 2.1(-241 826 + 504) + 0.215(441) + 8.892(437)
= -935052 kJ assuming all gas
QCV = HP - HR = -859 192 kJ/kmol fuel
b) TP = 250 oC
HP = 1.1(-393 522 + 9346) + 2.1(-241 826 + 7740)
+ 0.215(6808) + 8.892(6597) = -854 050 kJ
QCV = HP - HR = -778 190 kJ/kmol fuel
14.29 Repeat the previous problem, but take into account the actual phase behavior of
the products exiting the furnace.
Same as 14.28, except check products for saturation at 40 oC, 100 kPa
yV MAX =
nV MAX
7.384
=
=>
100
nV MAX+10.207
nV = 0.814,
Solving, nV MAX = 0.814
nLIQ = 2.1 - 0.814 = 1.286
HLIQ = 1.286[-285 830 + 18.015(167.6 - 104.9)] = -366125 kJ
HGAS = 1.1(-393 522 + 562) + 0.814(-241826 + 504 )
+ 0.215(441) + 8.892(437) = -624711 kJ
QCV = HP - HR = (-366 125 - 624 711) + 75860 = -914 976 kJ/kmol fuel
b) no liquid, answer the same as 14.28.
14-18
14.30 Methane, CH4, is burned in an SSSF process with two different oxidizers: Case
Pure oxygen, O2 and case B: A mixture of O2 + x Ar. The reactants are
A:
supplied at T0, P0 and the products should for both cases be at 1800 K. Find the
required equivalence ratio in case (A) and the amount of Argon, x, for a
stoichiometric ratio in case (B).
a) Stoichiometric has ν = 2, actual has:
CH4 + νO2 → CO2 + 2H2O + (ν - 2)O2
°
°
°
HR = HP + ∆HP 1800 => ∆HP 1800= - HRP = 50010 × 16.04 = 802 160
∆h
= 79432, ∆ hH O= 62693, ∆ hO = 51674
CO
2
2
2
∆HP 1800= 101470 + ν 51674 = 802160
⇒
ν = 13.56,
φ = 6.78
CH4 + 2 O2 + 2x Ar → CO2 + 2H2O + 2x Ar
b)
∆HP 1800= 79432 + 2 × 62693
+ 2x × 0.5203 × 39.948(1800 - 298) = 204818 + x62438
⇒
802160 = 204818 + x 62438
x = 9.567
14.31 Butane gas at 25°C is mixed with 150% theoretical air at 600 K and is burned in
an adiabatic SSSF combustor. What is the temperature of the products exiting the
combustor?
o
25 C GAS
150% Air
600 K
C4 H10
Adiab.
Comb.
Prod.
at Tp
QCV = 0
C4H10 + 1.5×6.5 (O2 + 3.76 N2 )→ 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2
Reactants:
hAIR = 9.75(9245) + 36.66(8894) = 416193 kJ ;
-o
hC4H10 = hf IG = -126 200 kJ
=>
H R = +289 993 kJ
-*
-*
-*
-*
HP = 4(-393522 + ∆hCO2) + 5(-241826 + ∆hH2O) + 3.25 ∆hO2 + 36.66 ∆hN2
Energy Eq.:
HP - HR = 0
-*
-*
-*
-*
4 ∆hCO2 + 5 ∆hH2O + 3.25 ∆hO2 + 36.66 ∆hN2 = 3 073 211
Trial and Error:
LHS2000 K = 2 980 000,
Linear interpolation to match RHS
LHS2200 K = 3 369 866
=> TP = 2048 K
14-19
14.32 In a rocket, hydrogen is burned with air, both reactants supplied as gases at Po, To.
The combustion is adiabatic and the mixture is stoichiometeric (100% theoretical
air). Find the products dew point and the adiabatic flame temperature (~2500 K).
The reaction equation is:
H2 + vO2 (O2 + 3.76 N2) => H2O + 3.76 vO2 N2
The balance of hydrogen is done, now for oxygen we need v O2 = 0.5 and
thus we have 1.88 for nitrogen.
yv = 1/(1+1.88) = 0.3472 => Pv = 101.325 × 0.3472 = 35.18 kPa = Pg
Table B.1.2: Tdew = 72.6 C.
HR = HP => 0 = -241826 + ∆hwater + 1.88 ∆hnitrogen
Find now from table A.13 the two enthalpy terms
At 2400 K : ∆HP = 93741 + 1.88 × 70640 = 226544
At 2600 K : ∆HP = 104520 + 1.88 × 77963 = 251090.4
Then interpolate to hit 241826 to give
T = 2524.5 K
14.33 Liquid butane at 25°C is mixed with 150% theoretical air at 600 K and is burned
in an adiabatic SSSF combustor. Use the generalized charts for the liquid fuel and
find the temperature of the products exiting the combustor.
o
TC = 425.2 K
25 C LIQ. C4 H10
Prod.
Adiab.
TR = 0.701
150% Air
Comb.
at Tp
600 K
QCV = 0
-o
hC4H10 = hf IG + (hLIQ - h*)
see Fig. D.2
= -126200 + (-4.85 × 8.3145 × 425.2) = -143 346 kJ
C4H10 + 1.5 × 6.5 O2 + 3.76 × 9.75 N2
→ 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2
hAIR = 9.75(9245) + 36.66(8894) = 416 193 kJ
⇒
HR = 416 193- 143 346 = +272 847 kJ
-*
-*
-*
-*
HP = 4(-393522 + ∆hCO2) + 5(-241826 + ∆hH2O) + 3.25 ∆hO2 + 36.66 ∆hN2
Energy Eq.:
HP - HR = 0
-*
-*
-*
-*
4 ∆hCO2 + 5 ∆hH2O + 3.25 ∆hO2 + 36.66 ∆hN2 = 3 056 065
Trial and Error:
LHS2000 K = 2 980 000,
Linear interpolation to match RHS
LHS2200 K = 3 369 866
=> TP = 2039 K
14-20
14.34 A stoichiometric mixture of benzene, C6H6, and air is mixed from the reactants
flowing at 25°C, 100 kPa. Find the adiabatic flame temperature. What is the error
if constant specific heat at T0 for the products from Table A.5 are used?
C6H6 + νO O2 + 3.76 νO N2 → 6CO2 + 3H2O + 3.76 νO N2
2
2
2
νO = 6 + 3/2 = 7.5 ⇒ νN = 28.2
2
2
°
°
°
HP=HP + ∆HP = HR = HR ⇒ ∆HP = -HRP = 40576 × 78.114 = 3169554
∆HP 2600K= 3280600,
∆HP 2400K= 2968000
⇒
TAD= 2529 K
∑νiC-Pi = 6 × 37.048 + 3 × 33.73 + 28.2 × 29.178 = 1146.3
∆T = ∆HP/∑νiCPi = 2765
⇒
TAD= 3063 K, 21% high
14.35 Liquid n-butane at T0, is sprayed into a gas turbine with primary air flowing at 1.0
MPa, 400 K in a stoichiometric ratio. After complete combustion, the products are
at the adiabatic flame temperature, which is too high, so secondary air at 1.0 MPa,
400 K is added, with the resulting mixture being at 1400 K. Show that Tad > 1400
K and find the ratio of secondary to primary air flow.
C.V. Combustion Chamber.
C4H10 + 6.5 O2 + 6.5 × 3.76 N2 → 5 H2O + 4 CO2 + 24.44 N2
Primary
Air
Fuel
Combustion
Chamber
Air
Secondary
To turbine
Mixing
TAD
1400 K
Energy Eq.: Hair + Hfuel = HR = HP
°
°
HP + ∆HP = HR + ∆HR
⇒
°
°
∆HP = HR + ∆HR - HP
∆HP = 45344 × 58.124 + 6.5(3.76 × 2971 + 3027) = 2727861
∆HP 1400 = 5.43491 + 4.55895 + 24.44 × 34936 = 1294871 < ∆HP
Try TAD > 1400: ∆HP = 2658263 @2400 K,
C.V. Mixing Ch.
∆HP = 2940312 @2600 K
Air Second: νO2 sO2 + 3.76 N2
∆ΗP + νO2 second ∆Hair = ∆HP 1400 + νO2 second ∆Hair 1400
⇒ νO2 second=
∆HP - ∆HP 1400
∆Hair 1400 - ∆Hair 400
=
ratio = νO2 sec/νO2 prim = 9.3/6.5 = 1.43
1432990
= 9.3
168317 - 14198
14-21
14.36 Consider the gas mixture fed to the combustors in the integrated gasification
combined cycle power plant, as described in Problem 14.12. If the adiabatic flame
temperature should be limited to 1500 K, what percent theoretical air should be
used in the combustors?
Product
CH4
% vol.
0.3
H
2
29.6
CO
CO2
41.0
10.0
N
2
0.8
H 2O
H 2S
NH3
17.0
1.1
0.2
Mixture may be saturated with water so the gases are ( H 2S and NH3 out)
CH4
H2
CO
CO2
N2
n
0.3
29.6
41.0
10.0
0.8
81.7
yV MAX = 7.384/3000 = nV/(nV + 81.7)
Solving, nV = 0.2 kmol, rest condensed
{0.3 CH4 + 29.6 H2 + 41.0 CO + 10.0 CO2 + 0.8 N2
+ 0.2 H2O + 35.9x O2 + 3.76 × 35.9x N2}→
51.3 CO2 + 30.4 H2O + 35.9(x - 1) O2 + (135.0x + 0.8) N2
For the fuel gas mixture,
nCP0 MIX = 0.3 × 16.04 × 2.2537 + 29.6 × 2.016 × 14.2091
+ 41.0 × 28.01 × 1.0413 + 10.0 × 44.01 × 0.8418
+ 0.8 × 28.013 × 1.0416 + 0.2 × 18.015 × 1.8723 = 2455.157
-0
nhf MIX = 0.3(-74873) + 29.6(0) + 41.0(-110527)
+ 10.0(-393522) + 0.8(0) + 0.2(-241826) = -8537654 kJ
At 40°C, for the fuel mixture:
HMIX = -8537654 + 2455.157(40 - 25) = -8500827 kJ
Assume air enters at 25°C: hAIR = 0
Products at 1500 K:
HP = 51.3(-393522 + 61705) + 30.4(-241826 + 48149)
+ 35.9(x - 1)(0 + 40600) + (135x + 0.8)(0 + 38405)
= -24336806 + 6642215x
1st law: HP = HR = HMIX
x=
+24336809 - 8500827
= 2.384
6642215
or
238 % theo. air
14-22
14.37 Acetylene gas at 25°C, 100 kPa is fed to the head of a cutting torch. Calculate the
adiabatic flame temperature if the acetylene is burned with
a.
100% theoretical air at 25°C.
b.
100% theoretical oxygen at 25°C.
C2H4 + 2.5 O2 + 2.5 × 3.76 N2 → 2 CO2 + 1 H2O + 9.4 N2
a)
-o
HR = hf C2H2 = +226 731 kJ
-*
-*
-*
HP = 2(-393 522 + ∆hCO2) + 1(-241 826 + ∆hH2O) + 9.4 ∆hN2
-*
-*
-*
2 ∆hCO2 + 1 ∆hH2O + 9.4 ∆hN2 = 1 255 601 kJ
QCV = HP - HR = 0 ⇒
Trial and Error A.8: LHS2800 = 1 198 369,
Linear interpolation: TPROD = 2909 K
LHS3000 = 1 303 775
C2H2 + 2.5 O2 → 2 CO2 + H2O
b)
HR = +226 731 kJ ;
⇒
-*
-*
HP = 2(-393 522 + ∆hCO2) + 1(-241 826 + ∆hH2O)
-*
-*
2 ∆hCO2 + 1 ∆hH2O = 1 255 601
2 × 343 782 + 302 295 = 989 859
At 6000 K (limit of A.8)
2 × 337 288 + 296 243 = 970 819
At 5900 K
or 19040/100 K change
Difference, extrapolating
TPROD ≈ 6000 + 265 742/190.40 ≈ 7400 K
14.38 Ethene, C2H4, burns with 150% theoretical air in an SSSF constant-pressure
process with reactants entering at P0, T0. Find the adiabatic flame temperature.
C2H4 + 3(O2 + 3.76N2) → 2CO2 + 2H2O + 11.28N2
C2H4 + 4.5(O2 + 3.76N2) → 2CO2 + 2H2O + 1.5 O2 + 16.92N2
yH
2O
= 2/(2 + 2 + 1.5 + 16.92) = 0.0892
=> Pv = 9.041 ⇒ TDEW = 43.8°C
°
HP = HP + 2∆hCO + 2∆hH
2
°
-°
HR = hf Fu
2O
+ 1.5∆hO + 16.92∆hN
2
°
2
°
∆HP + HP = HR
°
kJ
⇒ ∆HP = -HRP = 28.054 × 47158 = 1322970.5 kmol Fu
Initial guess based on N2 from A.8
∆HP(2000) = 1366982,
T1 = 2100 K
∆HP(1900) = 1278398
=> TAD ≅ 1950 K
14-23
14.39 Solid carbon is burned with stoichiometric air in an SSSF process. The reactants
at T0, P0 are heated in a preheater to T2 = 500 K as shown in Fig. P14.39, with the
energy given by the product gases before flowing to a second heat exchanger,
which they leave at T0. Find the temperature of the products T4, and the heat
transfer per kmol of fuel (4 to 5) in the second heat exchanger.
Control volume: Total minus last heat exchanger.
C + O2 + 3.76N2 → CO2 + 3.76N2
Energy Eq.:
°
°
HR = HR = HP = HP + ∆HP = hf CO + ∆hCO + 3.76∆hN
3
hf CO = -393 522, ∆HP
3
2
=381 436, ∆HP
3 2400
2
2
⇒ T3 = Tad.flame = 2461 K
Control volume: Total. Then energy equation:
°
°
−
HR + Q = HP
kJ
− -°
-°
Q = HRP = hf CO2 - 0 = -393 522 kmol fuel
= 401 339
3 2500
2
14-24
14.40 A study is to be made using liquid ammonia as the fuel in a gas-turbine engine.
Consider the compression and combustion processes of this engine.
a. Air enters the compressor at 100 kPa, 25°C, and is compressed to 1600 kPa,
where the isentropic compressor efficiency is 87%. Determine the exit
temperature and the work input per kilomole.
b. Two kilomoles of liquid ammonia at 25°C and x times theoretical air from the
compressor enter the combustion chamber. What is x if the adiabatic flame
temperature is to be fixed at 1600 K?
Air
P1 = 100 kPa
1
2
P2 = 1600 kPa
COMP.
T1 = 25 oC
ηS COMP = 0.87
-W
a) ideal compressor process (adiabatic reversible):
P2
s2S = s1 ⇒ T2S = T1( )
P1
k-1
k
= 298.2(
1600 0.286
) = 659 K
100
-wS = CP0(T2S - T1) = 1.004(659 - 298.2) = 362.2
Real process:
-w = -wS/ηS = 362.2/0.87 = 416.3 kJ/kg
T2 = T1 - w/CP0 = 298.2 + 416.3/1.004 = 713 K
Also -w = 416.3 × 28.97 = 12060 kJ/kmol
b)
o
2 liq NH3, 25 C
COMP
Air 100 kPa, 25 oC
-W
COMB.
CHAMBER
Prod.
PP = 1600 kPa
TP = 1600 K
-Q = 0
2 NH3 + 1.5x O2 + 5.64x N2 → 3 H2O + 1.5(x - 1) O2 + (5.64x + 1) N2
Using Tables A.16 and A.2,
hNH3 = -45720 + 17.031(298.36 - 1530.04) = -66697 kJ/kmol
HR = 2(-66697) + 0 = -133394 kJ
-W = 12060 × 7.14x = 86108 x kJ
HP = 3(-241826 + 52907) + 1.5(x - 1)(0 + 44267) + (5.64x + 1)(0 + 41904)
= 302739x - 591254
1st law: HR = HP + W = -133394 = 302739 x - 591254 - 86108 x
=>
x = 2.11
14-25
14.41 A closed, insulated container is charged with a stoichiometric ratio of oxygen and
hydrogen at 25°C and 150 kPa. After combustion, liquid water at 25°C is sprayed
in such that the final temperature is 1200 K. What is the final pressure?
1
H2 + 2O2 → H2O
 
P: 1 H2O + xiH2O
°
3U2 - U1 = xihi = xihf liq = 1 + xiHP - HR - 1 + xiRTP + 2RTR
°
HR = φ, HP = -241826 + 34506 = -207320, hf liq= -285830
Substitute
xi( -285830 + 207320 + 8.3145 × 1200) =
3
-207320 - 8.3145 1200 - 2×298.15= -213579
xi = 3.116
P1V1 = nRRT1, P2V1 = npRTp
P2 =
P11 + xiTP
3
2
T1
=
150 × 4.116 × 1200
3
2×
= 1657 kPa
298.15
14.42 Wet biomass waste from a food-processing plant is fed to a catalytic reactor,
where in an SSSF process it is converted into a low-energy fuel gas suitable for
firing the processing plant boilers. The fuel gas has a composition of 50%
methane, 45% carbon dioxide, and 5% hydrogen on a volumetric basis. Determine
the lower heating value of this fuel gas mixture per unit volume.
For 1 kmol fuel gas,
0.5 CH4 + 0.45 CO2 + 0.05 H2 + 1.025 O2
→ (0.5 + 0.45) CO2 + 1.05 H2O
The lower heating value is with water vapor in the products. Since the
0.45 CO2 cancels,
hRP = 0.5(-393522) + 1.05(-241826) - 0.5(-74873) - 0.05(0)
= -413242 kJ/kmol fuel gas
With
n
100
= P/RT =
= 0.04033 kmol/m3
V
8.3145 × 298.2
LHV = +413242 × 0.04033 = 16 666 kJ/m3
14-26
14.43 Determine the lower heating value of the gas generated from coal as described in
Problem 14.12. Do not include the components removed by the water scrubbers.
The gas from problem 14.12 is saturated with water vapor. Lower heating
value LHV has water as vapor.
°
°
°
LHV = -HRP = HP - HR
Only CH4, H2 and CO contributes. From 14.12 the gas mixture after the
scrubbers has ∑νi = 81.9 of composition:
0.3CH4 + 29.6H2 + 41CO + 10CO2 + 0.8N2 + 0.2016H2O
-°
-°
-°
LHV = -[0.3HRPCH4 + 29.6HRPH2 + 41HRPCO]/81.9
= -[0.3(-50010 × 16.043) + 29.6(-241826)
+ 41(-393522 + 110527)]/81.9
kJ
= 232009 kmol gas
14.44 Propylbenzene, C9H12, is listed in Table 14.3, but not in table A.9. No molecular
weight is listed in the book. Find the molecular weight, the enthalpy of formation
for the liquid fuel and the enthalpy of evaporation.
^ = 9 × 12.011 + 12 × 1.0079 = 120.194 Hfg = 384 kJ/kg
M
C9H12 + 12 O2 → 9 CO2 + 6 H2O
-°
-° -°
-°
-° -°
HRP = νihfi - hfFu ⇒ hfFu = νihfi - hRP
∑
∑
P
P
-°
-°
-°
^ (-41219)
hfFu = 9hfCO2 + 6hfH2O g + M
liq
Fu H2O vap
= -38383.5 kJ/kmol
14.45 Determine the higher heating value of the sample Wyoming coal as specified in
Problem 14.10.
-0
Note: For SO2 , hf = -296842 kJ/kmol
From solution 14.10 , for 100 kg of coal there are 4.05 kmol CO2, 1.75 kmol
H2O, 0.0156 kmol SO2 in the products. Therefore,
hRP0 = 4.05(-393522) + 1.75(-285830) + 0.0156(-296842)
= -2098597 kJ/100 kg coal
So that HHV = +20986 kJ/kg coal
14-27
14.46 Consider natural gas A and natural gas D, both of which are listed in Table 14.2.
Calculate the enthalpy of combustion of each gas at 25°C, assuming that the
products include vapor water. Repeat the answer for liquid water in the products.
Natural Gas A
0.939 CH4 + 0.036 C2H6 + 0.012 C3H8 + 0.013 C4H10
+ 2.1485 O2 + 3.76 × 2.1485 N2 → 1.099CO2 + 2.099 H2O + 8.0784 N2
HR = 0.939(-74878) + 0.036(-84740) + 0.012(-103900)
+ 0.013(-126200) = -76244 kJ
a) vapor H2O
HP = 1.099(-393522) + 2.099(-241826) = -940074
hRP = HP - HR = -863830 kJ/kmol
b) Liq. H2O
HP = 1.099(-393522) + 2.099(-285830) = -1032438
hRP = -956194 kJ/kmol
Natural Gas D:
0.543 CH4 + 0.163 C2H6 + 0.162 C3H8 + 0.074 C4H10
+ 0.058 N2 + O2 + N2
→
1.651 CO2 + 2.593 H2O + N2
HR = 0.543(-74873) + 0.163(-84740) + 0.162(-130900)
+ 0.074(-126200) = -80639 kJ
a) vapor H2O
HP = 1.651(-393522) + 2.593(-241826) = -1276760 kJ
hRP = -1196121 kJ/kmol
b) Liq. H2O
HP = 1.651(-393522) + 2.593(-285830) = -1390862 kJ
hRP = -1310223 kJ/kmol
14-28
14.47 Blast furnace gas in a steel mill is available at 250°C to be burned for the
generation of steam. The composition of this gas is, on a volumetric basis,
Component
CH4 H2 CO CO2 N2 H2O
Percent by volume
0.1 2.4 23.3 14.4 56.4 3.4
Find the lower heating value (kJ/m3) of this gas at 250°C and ambient pressure.
Of the six components in the gas mixture, only the first 3 contribute to the
heating value. These are, per kmol of mixture:
0.024 H2,
0.001 CH4,
0.233 CO
For these components,
0.024 H2 + 0.001 CH4 + 0.233 CO + 0.1305 O2 → 0.026 H2O + 0.234 CO2
The remainder need not be included in the calculation, as the contributions to
reactants and products cancel. For the lower HV(water as vapor) at 250°C
hRP = 0.026(-241826+7742) + 0.234(-393522+9348) - 0.024(0+6558)
- 0.001(-74873+2.254×16.04(250-25)) - 0.233(-110527+6625) –
kJ
0.1305(0+6810) = -72 573 kmol fuel
−
v-0 = R To/Po = 8.3145 × 523.2/100 = 43.5015 m3/kmol
LHV = +72573 / 43.5015 = 1668 kJ/m3
14.48 The enthalpy of formation of magnesium oxide, MgO(s), is −601827 kJ/kmol at
25°C. The melting point of magnesium oxide is approximately 3000 K, and the
increase in enthalpy between 298 and 3000 K is 128449 kJ/kmol. The enthalpy of
sublimation at 3000 K is estimated at 418000 kJ/kmol, and the specific heat of
magnesium oxide vapor above 3000 K is estimated at 37.24 kJ/kmol K.
a. Determine the enthalpy of combustion per kilogram of magnesium.
b. Estimate the adiabatic flame temperature when magnesium is burned with
theoretical oxygen.
a)
1
Mg + O2 → MgO(s)
2
∆hCOMB = ∆hCOMB/M = h°f /M = -601827/24.32 = -24746 kJ/kg
b) assume TR = 25°C and also that T P > 3000 K,
1st law: QCV = HP - HR = 0,
(MgO = vapor phase)
but HR = 0
⇒ HP = h°f + (h3000 - h298)SOL + ∆hSUB + CP VAP(TP - 3000)
= -601827 + 128449 + 418000 + 37.24(TP - 3000) = 0
Solving, TP = 4487 K
14-29
14.49 A rigid container is charged with butene, C4H8, and air in a stoichiometric ratio at
P0, T0. The charge burns in a short time with no heat transfer to state 2. The
products then cool with time to 1200 K, state 3. Find the final pressure, P3, the
total heat transfer, 1Q3, and the temperature immediately after combustion, T2.
The reaction equation is, having used C and H atom balances:
C4H8 + νO O2 + 3.76 N2 → 4 CO2 + 4 H2O + 3.76 νO N2

2
2
Counting now the oxygen atoms we get νO = 6.
2
C.V. analysis gives:
U2 - U1 = Q - W = Q = H2 - H1 - P2V2 + P1V1
= H2 - H1 - Rn2T2 - n1T1
°
°
°
°
^ HRP + ∆H
H2 - H1 = HP 1200 - HR = HP - HR + ∆HP = M
P
= -2542590 + 950055 = -1592535
^ = 56.108
Where M
n1= 1 + 6 × 4.76 = 29.56,
and
n2 = 4 + 4 + 6 × 3.76 = 30.56,
Τable A.8 at 1200 K:
∆hCO = 44473,
2
∆hH O=34506,
2
∆hN =28109.
2
Now solving for the heat transfer:
kJ
Q = -1592535 - 8.3145(30.56 × 1200 - 29.56 × 298.15) = -1824164 kmol fuel
To get the pressure, assume ideal gases:
n2RT2
n2T2
P2 =
= P1
= 421.6 kPa
V2
n1T1
Before heat transfer takes place we have constant U so:
U1 - U1 = 0 = H1 - H1 - n2RT1 + n1RT1
a
a
a
°
Now split the enthalpy H1 = HP + ∆HPT1 and arrange things with the
a
 a
unknowns on LHS and knowns on RHS:
°
∆HP- n2RT = HR - HP - n1RT1= 2 542 590 - 73278 = 2 469 312
Trial and error leads to:
LHS (3000 K) = 3 209 254 - 30.56 × 8.31451 × 3000 = 2 446 980
LHS (3000 K) = 3 471 331 - 30.56 × 8.31451 × 3200 = 2 658 238
linear interpolation
T = 3021 K
14-30
14.50 In an engine a mixture of liquid octane and ethanol, mole ratio 9;1, and
stoichiometric air are taken in at T0, P0. In the engine the enthalpy of combustion
is used so that 30% goes out as work, 30% goes out as heat loss and the rest goes
out the exhaust. Find the work and heat transfer per kilogram of fuel mixture and
also the exhaust temperature.
0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2
→ 8.4 H2O + 7.4 CO2 + 43.428 N2
For 0.9 octane + 0.1 ethanol
°
°
kJ
-°
HRP mix= 0.9HRP C8H18 + 0.1HRP C2H5OH = -4690690.3 kmol
^
^ + 0.1 M
^ = 107.414
M
= 0.9 M
mix
oct
alc
Energy:
-°
-°
hin + qin = hex + ωex = hex + ∆hex + ωex
-° -°
-°
hex - hin = HRP mix ⇒
-°
ωex + ∆hex - qin = -HRP mix
kJ
kJ
-°
ωex= -qin = 0.3-HRP= 1407207 kmol = 13101 kg Fu
kJ
-°
∆hprod = ∆hex = 0.4-HRP= 1 876 276 kmol Fu
∆hprod = 8.4 ∆hH
2O
+ 7.4∆hCO + 43.428 ∆hN
2
2
⇒ satisfied for T = 1216 K
14.51 Consider the same situation as in the previous problem. Find the dew point
temperature of the products. If the products in the exhaust are cooled to 10°C, find
the mass of water condensed per kilogram of fuel mixture.
Reaction equation with 0.9 octane and 0.1 ethanol is
0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2
→ 8.4 H2O + 7.4 CO2 + 43.428 N2
yH
=
O
2
PH
2O
8.4
= 0.1418
8.4 + 7.4 + 43.428
⇒ Tdew= 52.9 °C
= yH OPtot = 14.3 kPa
2
10 °C ⇒ PH
2O
= 1.2276 ⇒
yH
2O
= 0.012276 =
kmol
x
x + 7.4 + 43.428
⇒ x = 0.6317 ⇒ ∆ν H O= -7.77 kmol Fu mix
2
-∆ν H O × 18.015
mH
2
=
O cond
2
107.414
kmol
= 1.303 kmol Fu mix
14-31
14.52 Calculate the irreversibility for the process described in Problem 14.19.
From solution 14.19,
Reactants: 2 C(s) + 2 O2 at 25°C, 200 kPa
1
Product mixture: 1 CO2 + 1 CO + O2 at 1000 K, 838.4 kPa
2
& 1Q2 = -453442 kJ
Reactants:
SR = 2(5.740) + 2(205.148 - 8.31451 ln
200
) = 410.250 kJ/K
100
Products:
ni
yi
-s°
i
S-i
yiP
-Rln
P0
CO2
1.0
0.40
269.299
-10.061
259.238
CO
O2
1.0
0.5
0.40
0.20
234.538
243.579
-10.061
-4.298
224.477
239.281
SP = 1.0(259.238) + 1.0(224.477) + 0.5(239.281) = 603.355 kJ/K
I = T0(SP - SR) - 1Q2
= 298.15(603.355 - 410.250) - (-453442) = +511 016 kJ
14-32
14.53 Pentane gas at 25°C, 150 kPa enters an insulated SSSF combustion chamber.
Sufficient excess air to hold the combustion products temperature to 1800 K
enters separately at 500 K, 150 kPa. Calculate the percent theoretical air required
and the irreversibility of the process per kmol of pentane burned.
C5H12 + 8XO2 + 30.08XN2 ‡ 5CO2 + 6H2O + 8(X-1)O2 + 30.08XN2
1st Law: Qcv + HR = HP + Wcv; Wcv = 0 , Qcv = 0 => HR = HP
−o
−
Reactants: C5H12 : h f from A.9 and ∆h 500 for O2 and N2 from A.8
−o
−
−
HR =(h f )C5H12 + 8X ∆h O2 + 30.08X ∆h N2
= -146 500 + 8X 6086 + 30.08 X 5911 = 226 491 X - 146 500
−o
−
−o
−
−
−
HP = 5(h f + ∆h )CO2 + 6(h f + ∆h )H2O + 8(X-1) ∆h O2 + 30.08 X ∆h N2
= 5(-393522 + 79432) + 6(-241826 + 62693) + 8(X-1) 51674
+ 30.08 X 48979 = 1 886 680 X - 3 058 640
Energy Eq. solve for X;
=>
X = 1.754
o
b) Reactants: Pi = 150 kPa, Po = 100 kPa, −s f
ni
yi
−s o / −s o
f
500
− yiPi
- R ln
Po
Si
C5H12
1
1
348.945
-3.371
352.316
O2
8X
0.21
220.693
9.604
211.089
N2
30.08 X
0.79
206.74
1.411
208.151
kJ
kmol K
SR = ∑ niSi = 14410.249 kJ/K
Products: Pe = 150 kPa, Po = 100 kPa
ni
yi
−s o
1800
− yiPe
- R ln
Po
Si
CO2
5
0.0716
302.969
18.550
321.519
H2O
6
0.086
259.452
17.027
276.479
O2
8(X-1)
0.0864
264.797
16.988
281.785
N2
30.08X
0.756
248.304
-1.045
247.259
SP = ∑ niSi = 17732.073 kJ/K;
I = To(SP - SR) = 298.15(17 732.07 - 14 410.25) = 990 MJ
kJ
kmol K
14-33
14.54 Consider the combustion of methanol, CH3OH, with 25% excess air. The
combustion products are passed through a heat exchanger and exit at 200 kPa,
40°C. Calculate the absolute entropy of the products exiting the heat exchanger
per kilomole of methanol burned, using appropriate reference states as needed.
CH3OH + 1.25 × 1.5 (O2 + 3.76 N2) → CO2 + 2 H2O + 0.38 O2 + 7.05 N2
Products exit at 40 oC, 200 kPa, check for saturation:
PG
nV MAX
7.384
yV MAX =
=
=
200
P
nV MAX + 1 + 0.38 + 7.05
nV = nV MAX = 0.323
Gas mixture:
ni
nLIQ = 1.677
yi
-s°
i
- yiP
-Rln
P0
Si
CO2
1.0
0.1143
215.633
+12.270
227.903
H 2O
0.323
0.0369
190.485
+21.671
212.156
O2
0.38
0.0434
206.592
+20.322
226.914
N2
7.05
0.8054
193.039
-3.964
189.075
SGAS MIX =
∑ n S- = 1715.64 kJ/K
i i
-s
= 69.950 + 18.015(0.5725 - 0.3674) = 73.645
LIQ
SLIQ = 1.677 × 73.645 = 123.50 kJ/K
SPROD = 1715.64 + 123.50 = 1839.14 kJ/K
14-34
14.55 The turbine in Problem 14.23 is adiabatic. Is it reversible, irreversible, or
impossible?
Inlet to the turbine is the exit from the mixing of air and fuel at 1 MPa.
From solution to 14.23, we have:
CP C H = 43.43, CP C H = 74.06, Tturbine,in = 548.7 K
2 2
3 8
C2H4 + C3H8 + 9.6 O2 + 36.096 N2
→ 5 CO2 +6 H2O + 1.6 O2 + 36.096 N2
⌠dQ
Sex - Sin =  + Sgen = Sgen ≥ φ
⌡T
Inlet: 1 MPa, 548.7 K
-° SFu = Si + CP Fu lnT/T0
-s°
i
- yiP
-Rln
P0
Si
ni
yi
C2H4
1
0.02097
245.82
12.989
258.809
C3H8
1
0.02097
315.09
12.989
328.079
O2
9.6
0.2013
223.497
-5.816
217.681
N2
36.096
0.7568
209.388
-16.828
192.56
Sin = 258.809 + 328.079 + 9.6 × 217.681 + 36.096 × 192.56 = 9627.3
ni
yi
-s°
i
- yiP
-Rln
P0
S-i
CO2
5
0.1027
257.496
18.925
276.421
H 2O
6
0.1232
223.826
17.409
241.235
O2
1.6
0.0329
235.92
28.399
264.319
N2
36.096
0.7413
221.016
2.489
223.505
Sex
= 5 × 276.421 + 6 × 241.235 1.6 × 264.319
kJ
+ 36.096 × 223.505 = 11320 2kmol Fu K
Sgen
kJ
= Sex - Sin = 1693 2kmol Fu K > 0
Possible, but one should check the state after combustion to account for
generation by combustion alone and then the turbine expansion separately.
14-35
14.56 Saturated liquid butane enters an insulated constant pressure combustion chamber
at 25°C, and x times theoretical oxygen gas enters at the same P and T. The
combustion products exit at 3400 K. With complete combustion find x. What is
the pressure at the chamber exit? and what is the irreversibility of the process?
Butane: T1 = To = 25oC, sat liq., x1 = 0, Tc = 425.2 K, Pc = 3.8 MPa
Fig. D.1: Tr1 = 0.7, Pr1 = 0.1, P1 = Pr1Pc = 380 kPa
− * − 
−
−
−* −
h 1 − h 1 f = 4.85 RTc , (s 1 - s 1)f = 6.8 R
Oxygen: T2 = To = 25oC, X - Theoretical O2
Products: T3 = 3400 K, Assumes complete combustion
Figs. D.2 and D.3:
C4H10 + 6.5XO2 ‡ 4CO2 + 5H2O + 6.5(X-1)O2
1st Law: Q
cv + HR = HP + Wcv; Qcv = 0, Wcv = 0
−o
−
HR = n(h f + ∆h )C4H10 = 1(-126200 + -17146) = -143346 kJ
Products: CO2
−o
−
n(h f + ∆h )CO2 = 4(-393522 + 177836) = -862744 kJ
H2O
−o
−
n(h f + ∆h )H2O = 5(-241826 + 149073) = -463765 kJ
O2
−o
−
n(h f + ∆h )O2 = 6.5(X-1)(0 + 114101) = (X-1)741656.5 kJ
HP =
−o
−
∑ni (h f + ∆h )i = 741656.5X - 2068165.5
HP = HR solve for X;
X = 2.594
Assume that the exit pressure equals the inlet pressure:
−s
C
Pe = Pi = 380 kPa
P1
P1
−s o - −
− * - −s ) ] ; −s = [−s o - −
=
[
R
ln
(s
R
ln
]
1 f
O2
f
1
4H10
Po
Po
SR = SC4H10 + SO2 = [306.647 - 11.10 - 56.539]
+ [205.48 - 11.10] × 6.5 × 2.594 = 3516.45 kJ/K
Products:
ni
yi
−s o
i
− yiPe
− R ln
Po
Si
CO2
4
0.2065
341.988
2.016
344.004
H2O
5
0.2582
293.550
0.158
293.708
O2
10.368
0.5353
289.499
-5.904
283.595
SP = ∑ niSi = 5784.87 kJ/K;
I = To(SP - SR) = 298.15 (5784.87 - 3516.45) = 676329 kJ
kJ
kmol K
14-36
14.57 An inventor claims to have built a device that will take 0.001 kg/s of water from
the faucet at 10°C, 100 kPa, and produce separate streams of hydrogen and
oxygen gas, each at 400 K, 175 kPa. It is stated that this device operates in a 25°C
room on 10-kW electrical power input. How do you evaluate this claim?
Liq H2O
H2 gas
10oC, 100 kPa
0.001 kg/s
O2 gas
.
-WCV = 10 kW
 each at
 400 K
 175 kPa
T0 = 25 oC
1
H 2O → H 2 + 2 O 2
1
Hi - He = [-285830 + 18.015(42.01 - 104.89)] - 2961 - 2 (3027) = -291437 kJ
(Si - Se) = [69.950 + 18.015(0.151 - 0.3674)] - (139.219 - 8.3145 ln 1.75)
1
- 2 (213.873 - 8.3145 ln 1.75) = -173.124
WREV = (Hi - He) - T0(Si - Se) = -291437 - 298.15(-173.124) = -239820 kJ
.
WREV = (0.001/18.015)(-239820) = -13.31 kW
.
.
.
I = WREV - WCV = -13.31 - (-10) < 0
Impossible
14.58 Two kilomoles of ammonia are burned in an SSSF process with x kmol of oxygen.
The products, consisting of H2O, N2, and the excess O2, exit at 200°C, 7 MPa.
a. Calculate x if half the water in the products is condensed.
b. Calculate the absolute entropy of the products at the exit conditions.
2NH3 + xO2 → 3H2O + N2 + (x - 1.5)O2
Products at 200 oC, 7 MPa with
a) yH2O VAP = PG/P =
nH2O LIQ = nH2O VAP = 1.5
1.5538
1.5
=
7
1.5 + 1 + x - 1.5
b) SPROD = SGAS MIX + SH2O LIQ
=>
x = 5.757
Gas mixture:
ni
yi
-s°
i
-Rln(yiP/P0)
S-i
H 2O
1.5
0.222
204.595
-22.810
181.785
O2
4.257
0.630
218.985
-31.482
187.503
N2
1.0
0.148
205.110
-19.439
185.671
SGAS MIX = 1.5(181.785) + 4.257(187.503) + 1.0(185.67) = 1256.55 kJ/K
SH2O LIQ = 1.5[69.950 + 18.015(2.3223 - 0.3674)] = 157.75 kJ/K
SPROD = 1256.55 + 157.75 = 1414.3 kJ/K
14-37
14.59 Consider the SSSF combustion of propane at 25°C with air at 400 K. The
products exit the combustion chamber at 1200 K. It may be assumed that the
combustion efficiency is 90%, and that 95% of the carbon in the propane burns to
form carbon dioxide; the remaining 5% forms carbon monoxide. Determine the
ideal fuel–air ratio and the heat transfer from the combustion chamber.
Ideal combustion process, assumed adiabatic, excess air to keep 1200 K out.
C3H8 + 5x O2 + 18.8x N2 → 3 CO2 + 4 H2O + 5(x - 1) O2 + 18.8x N2
HR = -103900 +5x(0 + 3027) + 18.8x(0 + 2971) = -103900 + 70990x
HP = 3(-393522 + 44473) + 4(-241826 + 34506)
+ 5(x - 1)(0 + 29761) + 18.8x(0 + 28109) = -2025232 + 677254x
1st law: HP - HR = 0
Solving, x = 3.169
FAIDEAL = 1/(23.8 × 3.169) = 0.01326
b)
FAACTUAL = 0.01326/0.90 = 0.01473
C3H8 + 14.26 O2 + 53.62 N2
→ 2.85 CO2 + 0.15 CO + 4 H2O + 9.335 O2 + 53.62 N2
HR = -103900 + 14.26(0 + 3027) + 53.62(0 + 2971) = +98570 kJ
HP = 2.85(-393522 + 44473) + 0.15(-110527 + 28427) + 4(-241826 + 34506)
+ 9.335(0 + 29761) + 53.62(0 + 28109) = -51361 kJ
QCV = HP - HR = -149931 kJ
14-38
14.60 Graphite, C, at P0, T0 is burned with air coming in at P0, 500 K in a ratio so the
products exit at P0, 1200 K. Find the equivalence ratio, the percent theoretical air,
and the total irreversibility.
C + φb(O2 + 3.76 N2) → CO2 + ( φ - 1) O2 + 3.76 φ N2
°
°
HP = HR ⇒ ∆HP 1200 - ∆HR = HR - HP
44473 + ( φ - 1) 29761 + 3.76φ28109 - φ( 6086 + 3.76×5911)
= 0 - (-393522) ⇒ φ = 3.536
Sgen = SP - SR =
∑ν-s
°
- R ln( y) 
P-R
R: yO = 0.21, yN = 0.79
2
2
P: yO = 0.1507, yN = 0.79, yCO = 0.0593
2
2
2
°
SP = 279.39 + 2.536 × 250.011 + 13.295 × 234.227 = 4027.5
°
SR = 5.74 + 3.536(220.693 + 3.76 × 206.74) = 3534.8
For the pressure correction the term with the nitrogen drops out (same y).
R
∑-ν ln(y) = 8.3145(2.8235 + 1.8927 - 1.5606) = 26.236
P-R
Sgen = 4027.5 - 3534.8 + 26.236 = 518.94
kJ
I = T0 Sgen = 154721 kmol C
14-39
14.61 A gasoline engine is converted to run on propane. Assume the propane enters the
engine at 25°C, at the rate 40 kg/h. Only 90% theoretical air enters at 25°C such
that 90% of the C burns to form CO2, and 10% of the C burns to form CO. The
combustion products also include H2O, H2 and N2, exit the exhaust at 1000 K.
Heat loss from the engine (primarily to the cooling water) is 120 kW. What is the
power output of the engine? What is the thermal efficiency?
.
Propane: T = 25oC, m = 40 kg/hr, M = 44.094 kg/kmol
1
Air: T2 = 25oC, 90% theoretical Air produces 90% CO 2, 10% CO
Products: T3 = 1000 K, CO2, CO, H2O, H2, N2
C3H8 + 4.5O2 + 16.92N2 ‡ 2.7 CO2 + 0.3CO + 3.3H2O + 0.7H2 + 16.92N2
.
.
nC3H8 = m/(M×3600) = 0.000252 kmol/s
.
.
.
1st Law: Q + H = H + W ; Q = -120 kW
R
P
−o
HR = nC3H8 h f = -103 900 kJ
Products:
CO2
-
−o
−
nCO2(h f + ∆h ) = 2.7(-393522 + 33397) = -972337.5 kJ
CO
-
−o
−
nCO(h f + ∆h ) = 0.3(-110527 + 21686) = -26652 kJ
H 2O
-
−o
−
nH2O(h f + ∆h ) = 3.3(-241826 + 26000) = -712226 kJ
H2
-
−o
−
nH2(h f + ∆h ) = 0.7(0 + 20663) = 14464.1 kJ
N2
-
−o
−
nN2(h f + ∆h ) = 16.92(0 + 21463) = 363154 kJ
−o
−
HP = ∑ni (h f + ∆h )i = -1 333 598 kJ
.
. .
W = Q + n(HR - HP) = 189.9 kW
C3H8: Table 14.3
HRPo = -50343 kJ/kg
.
.
HHV = nC3H8 M(-HRPo) = 559.4 kW
.
.
ηth = W/HHV = 0.339
14-40
14.62 A small air-cooled gasoline engine is tested, and the output is found to be 1.0 kW.
The temperature of the products is measured to 600 K. The products are analyzed
on a dry volumetric basis, with the result: 11.4% CO2, 2.9% CO, 1.6% O2 and
84.1% N2. The fuel may be considered to be liquid octane. The fuel and air enter
the engine at 25°C, and the flow rate of fuel to the engine is 1.5 × 10-4 kg/s.
Determine the rate of heat transfer from the engine and its thermal efficiency.
a C8H18 + b O2 + 3.76b N2
→ 11.4 CO2 + 2.9 CO + c H2O + 1.6 O2 + 84.1 N2
84.1
1
= 22.37,
a = (11.4 + 2.9) = 1.788
3.76
8
c = 9a = 16.088
C8H18 + 12.5 O2 + 47.1 N2
b=
→ 6.38 CO2 + 1.62 CO + 9 H2O + 0.89 O2 + 47.1 N2
-0
HR = hf C8H18 = -250105 kJ
HP = 6.38(-393522 + 15788) + 1.62(-110527 + 10781)
+ 9(-241826 + 12700) + 0.89(0 + 11187)
+ 47.1(0 + 10712) = -4119174 kJ
H P - HR = -4119174 - (-250105) = -3869069 kJ
.
.
HP - HR = (0.00015/114.23)(-3869069) = -5.081 kW
.
QCV = -5.081 + 1.0 = -4.081 kW
.
QH = 0.00015(47893) = 7.184 kW
.
.
ηTH = WNET/QH = 1.0/7.184 = 0.139
14.63 A gasoline engine uses liquid octane and air, both supplied at P , T , in a
0 0
stoichiometric ratio. The products (complete combustion) flow out of the exhaust
valve at 1100 K. Assume that the heat loss carried away by the cooling water, at
100°C, is equal to the work output. Find the efficiency of the engine expressed as
(work/lower heating value) and the second law efficiency.
C8H18 + νO O2 + 3.76 N2 → 8 CO2 + 9 H2O + 47 N2

2
2 νO = 16 + 9 ⇒ νO = 12.5
2
2
kJ
kJ
LHV = 44425 kg fuel ⇒ LHV = 5.07476×106 kmol fuel
∆HP 1100 = 8 × 38885 + 9 × 30190 + 47 × 24760 = 1746510
C.V. Total engine
14-41
Hin = Hex + W + Qloss = Hex + 2 W
°
⇒ 2 W = Hin - Hex= HR - Hν= -HRP + ∆HR- ∆HP 1100
= 5.07476×106 + 0 - 1746510 = 3328250
kJ
W = 1.664×106 kmol fuel
1.664×106
W
ηth =
=
= 0.328
LHV 5.07476×106
Find entropies in and out:
-°
inlet: SFu
= 360.575
1
SO° = 205.148 - 8.3145 ln
= 218.12
4.76
2
3.76
= 193.57
SN° = 191.609 - 8.3145 ln
4.76
2
-°
= 360.575 + 12.5 × 218.12 + 47 × 193.57 = 12185
Sin
8
- °
exit: SCO
= 275.528 - 8.3145 ln = 292.82
64
2
9
SH° O = 236.732 - 8.3145 ln = 253.04
64
2
47
SN° = 231.314 - 8.1345 ln = 233.88
64
2
-°
= 8 × 292.82 + 9 × 253.04 + 47 × 233.88 = 15612
Sex
Assume the same Q loss out to 100°C reservoir in the reversible case and
rev
compute Q0 :
rev
Sin + Q0 /T0 = Sex + Qloss/Tres
rev
Q0
= T0( Sex - Sin) + Qloss T0/Tres
= 298.15(15612 - 12185) + 1.664×106 × 298.15/373.15
kJ
= 2.351×106 kmol fuel
rev
Hin + Q0 = Hex + Wrev + Qloss
rev
rev
kJ
⇒ Wrev = Hin - Hex - Qloss + Q0 = Wac + Q0 = 4.015×106 kmol fuel
ηII = Wac/ Wrev =1.664×106/4.015×106 = 0.414
14-42
14.64 In Example 14.16, a basic hydrogen–oxygen fuel cell reaction was analyzed at
25°C, 100 kPa. Repeat this calculation, assuming that the fuel cell operates on air
at 25°C, 100 kPa, instead of on pure oxygen at this state.
Anode: 2 H2
→ 4 e- + 4 H+
Cathode: 4 H+ + 4 e- → 1 O2 + 2 H2O
Overall: 2 H2 + 1 O2 → 2 H2O
Example 14.16: ∆G25°C = -474283 kJ (for pure O2)
For PO2 = 0.21 × 0.1:
SO = 205.148 – 8.3145 ln0.21 = 218.124
2
∆S = 2(69.950) – 2(130.678) – 1(218.124) = -339.58 kJ/K
∆G25°C = -571660 – 298.15(-339.58) = -470414 kJ
E° = +470414/(96487 × 4) = 1.219 V
14-43
14.65 Consider a methane-oxygen fuel cell in which the reaction at the anode is
CH4 + 2H2O CO2 + 8e- + 8H+
The electrons produced by the reaction flow through the external load, and the
positive ions migrate through the electrolyte to the cathode, where the reaction is
8 e- + 8 H+ + 2 O2 4 H2O
a. Calculate the reversible work and the reversible EMS for the fuel cell operating
at 25°C, 100 kPa.
b. Repeat part (a), but assume that the fuel cell operates at 600 K instead of at room
temperature.
CH4 + 2H2O → CO2 + 8e- + 8H+
and 8e- + 8H+ + 2CO2 → 4H2O
Overall
CH4 + 2O2 → CO2 + 2H2O
a) 25 oC assume all liquid H2O and all comp. at 100 kPa
0
∆H25 C = -393522 + 2(-285830) – (-74873) – 0 = -890 309 kJ
0
∆S25 C = 213.795 + 2(69.950) – 186.251 – 2(205.148) = - 242.852 kJ/K
0
∆G25 C = -890309 – 298.15(-242.852) = - 817 903 kJ
Wrev = -∆G0 = +817903 kJ
-∆G0
+817903
0
=
= 1.06 V
E =
96485 × 8 96485 × 8
b) 600 K assume all gas H2O and all comp. at 100 kPa
0
∆H600 K = 1(-393522 + 12906) + 2(-241826 + 10499) - 2(0 + 9245)
- 1[-74873 + 16.04 × 2.2537(600 - 298.2)] = -797797 kJ
0
∆S600 K = 1(243.284) + 2(213.051)
- 1(186.251 + 16.04 × 2.2537 ln
600
) - 2(226.450)
298.2
= 4.9606 kJ/K
0
0
0
∆G600 K = ∆H600 K - T∆S600 K = -797797 - 600(4.9606) = -800773 kJ
Wrev = +800773 kJ
E0=
+800773
= 1.037 V
96485 × 8
14-44
Advanced Problems
14.66 A gas mixture of 50% ethane and 50% propane by volume enters a combustion
chamber at 350 K, 10 MPa. Determine the enthalpy per kilomole of this mixture
relative to the thermochemical base of enthalpy using Kay’s rule.
-*
hMIX O = 0.5(-84740) + 0.5(-103900) = -94320 kJ/kmol
CP0 MIX = 0.5 × 30.07 × 1.7662 + 0.5 × 44.097 × 1.67 = 63.583
-*
-*
h350 - h298 = 63.583(350 - 298.2) = 3294 kJ/kmol
Kay’s rule:
TC MIX = 0.5 × 305.4 + 0.5 × 369.8 = 337.6 K
PC MIX = 0.5 × 4.88 + 0.5 × 4.25 = 4.565 MPa
Tr = 350/337.6 = 1.037,
P r = 10/4.565 = 2.19
-* From Fig. D.2: h - h = 8.3145 × 337.6 × 3.53 = 9909 kJ/kmol
hMIX 350K,10MPa = -94320 + 3294 - 9909 = -100935 kJ/kmol
14-45
14.67 A mixture of 80% ethane and 20% methane on a mole basis is throttled from 10
MPa, 65°C, to 100 kPa and is fed to a combustion chamber where it undergoes
complete combustion with air, which enters at 100 kPa, 600 K. The amount of air
is such that the products of combustion exit at 100 kPa, 1200 K. Assume that the
combustion process is adiabatic and that all components behave as ideal gases
except the fuel mixture, which behaves according to the generalized charts, with
Kay’s rule for the pseudocritical constants. Determine the percentage of
theoretical air used in the process and the dew-point temperature of the products.
Reaction equation:
-0
Fuel mix:
hf FUEL = 0.2(-74873) + 0.8(-84740) = -82767 kJ/kmol
CP0 FUEL = 0.2 × 2.2537 × 16.04 + 0.8 × 1.7662 × 30.07 = 49.718
-*
∆hFUEL = 49.718(65 - 25) = 1989 kJ/kmol
TCA = 305.4 K, TCB = 190.4 K ⇒ Tc mix=282.4 K
PCA = 4.88, PCB=4.60 ⇒ Pc mix= 4.824 MPa
Tr = 338.2/282.4 = 1.198, P r = 10/4.824 = 2.073
- (h* - h)FUEL IN = 8.31451 × 282.4 × 2.18 = 5119
kJ
⇒ hFUEL IN = -82767 + 1989 - 5119 = -85897 kmol
1st law:
1.8(-393522 + 44473) + 2.8(-241826 + 34506)
+ 3.2(x - 1)(29761) + (12.03x)(28109)
+ 85897 - (3.2x)(9245) - (12.03x)(8894) = 0
a)
x = 4.104 or 410.4 %
b) nP = 1.8 + 2.8 + 3.2(4.104 - 1) + 12.03 × 4.104 = 63.904
yH2O = 2.8/63.904 = PV/100 ;
PV = 4.38 kPa, T = 30.5°C
14-46
14.68 Gaseous propane mixes with air, both supplied at 500 K, 0.1 MPa. The mixture
goes into a combustion chamber and products of combustion exit at 1300 K, 0.1
MPa. The products analyzed on a dry basis are 11.42% CO2, 0.79% CO, 2.68%
O2, and 85.11% N2 on a volume basis. Find the equivalence ratio and the heat
transfer per kmol of fuel.
C3H8 + α O2 + 3.76 α N2 → β CO2 + γ H2O + 3.76 α N2
β = 3, γ = 4, α = β +
γ
=5
2
( A/F) S = 4.76α = 23.8
φ = ( A/F) /( A/F) S = 1.1123
% Theoretical Air = 111.23 %
°
hP = hP + ∑νj∆h(1300 K)
°
q = hP - hR = hP - hR + ∑νj∆h(1300 K) = -∆H + ∑νj∆h(1300 K)
kJ
∑νj∆h(1300 K) = 983230.7 kmol Fu
kJ
q = -1182480 kmol Fu
14-47
14.69 A closed rigid container is charged with propene, C H , and 150% theoretical air
3 6
at 100 kPa, 298 K. The mixture is ignited and burns with complete combustion.
Heat is transferred to a reservoir at 500 K so the final temperature of the products
is 700 K. Find the final pressure, the heat transfer per kmole fuel and the total
entropy generated per kmol fuel in the process.
C3H6 + νO O2 + 3.76 N2 → 3 CO2 + 3 H2O + x N2

2
2 νO = 6 + 3 = 9
Oxygen, O2 , balance:
Actual Combustion:
2
φ = 1.5 ⇒ νO
2 ac
⇒
νO = 4.5
2
= 1.5 × 4.5 = 6.75
C3H6 + 6.75 O2 + 25.38 N2 → 3 CO2 + 3 H2O + 25.38 N2 + 2.25 O2
P 2 = P1
npT2
nRT1
= 100 ×
33.63 × 700
= 238.3 kPa
33.13 × 298.15
kJ
∆HP 700 = 3×17754 + 3×14190 + 25.38×11937 + 2.25×12499 = 426916 kmol fuel
U2 - U1 = 1Q2 - 0 = H2 - H1 - n2R
 T2 + n1R
 T1
°
Q = HRP + ∆HP 700 - nPR
 T2 + n1R
 T1
1 2
= -45780 × 42.081 + 426916 - 33.63 × 8.3145 × 700
kJ
+ 33.13 × 8.3145 × 298.15 = -1.613×106 kmol fuel
-s°
Reactants:
ni
yi
-Rln(yiP/P0)
i
S-i
C3H8
1.0
0.0302
267.066
29.104
296.17
O2
6.75
0.2037
205.143
13.228
218.376
N2
25.38 0.7661
191.609
2.216
189.393
kJ
S1= 296.17 + 6.75 × 218.376 + 25.38 × 189.393 = 6577 kmol fuel K
-s°
Products:
ni
yi
-Rln(yiP/P0)
S-i
i
CO2
3
0.0892
250.752
+12.875
263.627
H 2O
3
0.0892
218.739
+12.875
231.614
O2
2.25
0.0669
231.465
+15.266
246.731
N2
25.38
0.7547
216.865
- 4.88
211.985
S2 = 3(263.627 + 231.614) + 2.25 × 246.731 + 25.38 × 211.985
= 7421 kJ/kmol fuel K
1.613×106
kJ
S
=
S
S
Q
/T
=
7421
6577
+
=
4070
kmol
fuel K
1 2 gen
2
1 1 2 res
500
14-48
14.70 Consider one cylinder of a spark-ignition, internal-combustion engine. Before the
compression stroke, the cylinder is filled with a mixture of air and methane.
Assume that 110% theoretical air has been used, that the state before compression
is 100 kPa, 25°C. The compression ratio of the engine is 9 to 1.
a. Determine the pressure and temperature after compression, assuming a
reversible adiabatic process.
b. Assume that complete combustion takes place while the piston is at top dead
center (at minimum volume) in an adiabatic process. Determine the temperature
and pressure after combustion, and the increase in entropy during the
combustion process.
c. What is the irreversibility for this process?
1 CH4 + 1.1 × 2 O2 + 3.76 × 2.2 N2 → 1 CO2 + 2 H2O + 0.2 O2 + 2 H2O
P1 = 100 kPa, T1 = 298.2 K,
V2/V1 = 1/8, Rev. Ad. s2 = s1
Assume T2 ~ 650 K → TAVE ~ 475 K
Table A.6: CP0 CH4 = 44.887, CP0 O2 = 30.890, CP0 N2 = 29.415
CP0 MIX = (1 × 44.887 + 2.2 × 30.890 + 8.27 × 29.415)/11.47 = 31.047
- CV0 MIX = CP0 - R = 22.732,
k = CP0/CV0 = 1.366
a) T2 = T1(V1/V2)k-1 = 298.2 (9)0.366 = 666.4 K (avg OK)
P2 = P1(V1/V2)k = 100 (9)1.366 = 2011 kPa
b) comb. 2-3 const. vol., Q = 0
Q = 0 = (H3 - H2) - R(n3T3 - n2T2)
2 3
-0
H2 = 1 hf CH4 + n2 CP0 MIX (T2 - T1)
H2 = -74873 + 11.47 × 31.047(666.4 -298.2) = +56246 kJ
-*
-*
-*
-*
H3 = 1(-393522 + ∆hCO2) + 2(-241826 + ∆hH2O) + 0.2 ∆hO2 + 8.27 ∆hN2
Substituting,
-*
-*
-*
-*
1 ∆hCO2 + 2 ∆hH2O + 0.2 ∆hO2 + 8.27 ∆hN2 - 95.366 T3 - 869868 = 0
Trial & error: T3 = 2907 K
1 × 147072 + 2 × 121377 + 0.2 × 94315 + 8.27 × 89274
- 95.366 × 2907 - 869868 ≈ 0 OK
P 3 = P2
n3T3
n2T2
= P2
T3
T2
= 2011 ×
2907
= 8772 kPa
666.4
14-49
c) state 1
REAC
ni
yi
-s°
i
- R ln(yiP/P0)
S-i
CH4
1
0.0872
186.251
+20.283
206.534
O2
2.2
0.1918
205.148
+13.730
218.878
N2
8.27
0.7210
191.609
+2.720
194.329
11.47
S2 = S1 = ∑ niSi = 2295.17 kJ/K
state 3
PROD
ni
yi
-s°
i
- R ln(yiP/P0)
S-i
CO2
1
0.0872
332.213
-16.916
315.297
H 2O
2
0.1744
284.753
-22.680
262.073
O2
0.2
0.0174
283.213
-3.516
279.697
N2
8.27
0.7210
265.726
-34.480
231.246
11.47
S3 = ∑ niSi = 2807.79 kJ/K
I = T0(S3 - S2) = 298.2(2807.79 - 2295.17) = 152860 kJ
14-50
14.71 Consider the combustion process described in Problem 14.67.
a. Calculate the absolute entropy of the fuel mixture before it is throttled into the
combustion chamber.
b. Calculate the irreversibility for the overall process.
From solution 14.67 , fuel mixture 0.8 C2H6 + 0.2 CH4 at 65°C, 10 MPa
CP0 FUEL = 49.718 kJ/kmol K. Using Kay’s rule: Tr1 = 1.198, Pr1 = 2.073
and x = 410.4 % theoretical air
or 13.13 O2 + 49.36 N2
in at 600 K, 100 kPa
and 1.8CO2 + 2.8H2O + 9.93O2 + 49.36N2
out at 100 kPa, 1200 K
*
a) -s0 FUEL = 0.2(186.251) + 0.8(229.597)
- 8.3145(0.2 ln 0.2 + 0.8 ln 0.8) = 225.088
*
∆sTP = 49.718 ln
338.2
10
- 8.3145 ln
= -32.031
298.2
0.1
*
(s- -s-)FUEL = 1.37 × 8.3145 = 11.391
From Fig. D.3:
-s
= 225.088 - 32.031 - 11.391 = 181.66 kJ/kmol K
FUEL
b) Air at 600 K, 100 kPa
ni
yi
-s°
i
-Rln(yiP/P0)
S-i
O2
13.13
0.21
226.45
+12.976
239.426
N2
49.36
0.79
212.177
+1.96
214.137
SAIR = ∑ niSi = 13713.47 kJ/K
SR = 181.66 + 13713.47 = 13895.1 kJ/K
Products at 1200 K, 100 kPa
PROD
ni
yi
-so
i
-Rln(yiP/P0)
S-i
CO2
1.8
0.0282
279.390
+29.669
309.059
H 2O
2.8
0.0438
240.485
+26.008
266.493
O2
9.93
0.1554
250.011
+15.479
265.490
N2
49.36
0.7726
234.227
+2.145
236.372
SP = ∑ niSi = 15606.1 kJ/K
I = T0(SP - SR) - QCV = 298.15(15606.1 - 13895.1) + 0 = 510132 kJ
14-51
14.72 Liquid acetylene, C2H2, is stored in a high-pressure storage tank at ambient
temperature, 25°C. The liquid is fed to an insulated combustor/steam boiler at the
steady rate of 1 kg/s, along with 140% theoretical oxygen, O2, which enters at 500
K, as shown in Fig. P14.72. The combustion products exit the unit at 500 kPa, 350
K. Liquid water enters the boiler at 10°C, at the rate of 15 kg/s, and superheated
steam exits at 200 kPa.
a.Calculate the absolute entropy, per kmol, of liquid acetylene at the storage tank state.
b. Determine the phase(s) of the combustion products exiting the combustor boiler
unit, and the amount of each, if more than one.
c. Determine the temperature of the steam at the boiler exit.
-°
a) C2H2: SIG 25°C = 200.958
TR = 298.2/308.3 = 0.967 => From Fig. D.1:
PR = 0.82
1
P1 = 0.82 × 6.14 = 5.03 MPa,
Sliq T
1 P1
= S°T
0 P0
1
- ( S* - S) 1 = 3.33R = 27.687
- + ∆T - R lnP1/P° + ( S - S*) P
kJ
1 T1
= 140.695 kmol K
b) 1 C2H2 + 1.4 × 2.5 O2 → 2 CO2 + 1 H2O + 1 O2
H1 = 226731 + (-3.56 × R × 308.3) = 217605 kJ
H2 = 3.5(0 + 6086) = 21301 kJ
Products T3 = 350 K = 76.8°C ⇒ PG = 41.8 kPa
yV max =
PG
P
=
nV max
41.8
= 0.0836 =
⇒ nV max = 0.2737 = n V gas mix
500
nV max + 2 + 1
⇒ nliq = 1 - 0.2737 = 0.7263
Gas Mix = 2 CO2 + 0.2737 H2O + 1 O2
c) Hliq = 0.7263(-285830 + 18.015(321.5 - 104.9)) = -204764 kJ
3
Hgas mix = 2(-393522 + 2036) + 0.2737(-241826 + 1756) + 1541
3
= -847138 kJ
H3 = Hliq + Hgas mix = -204 764 -847 138 = -1 051 902 kJ
3
3
H3 - H1 - H2 = -1 290 808 kJ
.
.
.
.
or H3 - H1 - H2 = -1290808/26.038 = -49574 kW = mH Oh4 - h5
2
h5 = 42.01 +
49574
= 3346.9 ⇒ T5 = 433.4 °C
15
14-52
14.73 Natural gas (approximate it as methane) at a ratio of 0.3 kg/s is burned with 250%
theoretical air in a combustor at 1 MPa where the reactants are supplied at T0.
Steam at 1 MPa, 450°C at a rate of 2.5 kg/s is added to the products before they
enter an adiabatic turbine with an exhaust pressure of 150 kPa. Determine the
turbine inlet temperature and the turbine work assuming the turbine is reversible.
CH4 + νO O2 + 3.76 N2 → CO2 + 2 H2O + 7.52 N2

2
2 νO = 2 + 2 ⇒ ν O = 2
2
Actual νO = 2 × 2.5 = 5
=>
2
2
CH4 + 5 O2 + 18.8 N2 → CO2 + 2 H2O + 3 O2 + 18.8 N2
C.V. combustor and mixing chamber
HR + nH OhH O in = HP ex
2
2
.
.
nH2O mH2OMFu 2.5 × 16.043
kmol steam
nH O = .
= .
=
= 7.421 kmol fuel
2
nFu mFuMH2O 0.3 × 18.015
Energy equation becomes
nH Ohex - hinH O + ∆hCO + 2∆hH O + 3∆hO + 18.8∆hN ex

2
2
2
2
2
2
°
= -HRP = 50010 × 16.043 = 802 310
hex - hinH O = ∆hH O ex - 15072.5, so then:
2
2
kJ
∆hCO2 + 9.421∆hH2O + 3∆hO2 + 18.8∆hN2ex = 914163 kmol fuel
Trial and error on Tex
Tex = 1000 K ⇒ LHS = 749956 ;
Tex = 1200 K ⇒ LHS = 987286
Tex = 1100 K ⇒ LHS = 867429
Tex ≅ 1139 K = Tin turbine
If air then Tex turbine ≈ 700 K and Tavg≈ 920 K. Find CP mix between 900 and
1000K. From Table A.8:
=>
53.67 + 9.421(40.63) + 3(34.62) + 18.8(32.4)
CP mix = ∑ niCPi/∑ ni =
32.221
= 35.673 kJ/kmol K
CV mix = CP mix - R = 27.3587, kmix= 1.304
Tex turbine = 1139 (150 / 1000)0.2331= 732 K
∆H732 = 19370.6 + 9.421(15410) + 3(13567) + 18.8(12932) = 448371
kJ
wT = Hin - Hex = ∆Hin - ∆Hex = 914163 - 448371 = 465792 kmol fuel
.
.
•
^
WT = nFuwT = m
Fuw T/MFu = (0.3 × 465792)16.043 = 8710 kW
14-53
14.74 Liquid hexane enters a combustion chamber at 31°C, 200 kPa, at the rate 1 kmol/s
200% theoretical air enters separately at 500 K, 200 kPa, and the combustion
products exit at 1000 K, 200 kPa. The specific heat of ideal gas hexane is C =
p
143 kJ/kmol K.
Hexane:
Tc = 507.5 K, Pc = 3010 kPa
Tr1 = 0.6, Fig. D.1: Prg = 0.028, Pg1 = Pr1Pc = 84.47 kPa
−* −
−
−
*
Figs D.2 and D.3:
(h 1 - h 1) f = 5.16 RTc , (s− 1 - −s 1) f = 8.56 R
Air: T2 = 500 K, P2 = 200 kPa, 200% theoretical air
Products: T3 = 1000 K, P3 = 200 kPa
−
−o −* −
−* −*
−* −
a) h C6H14 = h f - (h 1 - h 1)f + (h 1 - h 0) + (h 0 - h 0)
−* −
−* −* −
−o
h 0 - h 0 = 0 , h 1 - h 0 = CP (T1 - To) = 858 kJ/kmol, hf = -167300 kJ/kmol
−* −
−
h 1 - h 1 = 5.16 × 8.3145 × 507.5 = 21773 kJ/kmol, h C6H14 = -188215 kJ/kmol
T1 − P1
o
*
−
−s
−s
=
+
C
ln
- R ln + (−s 1 - −s 1)
C6H14
P
To
To
Po
T1 − P1
o
−
−s
+
C
ln
- R ln = 387.979 + 2.85 – 5.763 = 385.066 kJ/kmol-K
P
To
To
Po
−s * - −s = 8.56×8.3145 = 71.172 kJ/kmol-K, −s
1
C6H14 = 313.894 kJ/kmol-K
1
b) C6H14 + 19O2 + 71.44N2 ‡ 6CO2 + 7H2O + 9.5O2 + 71.44N2
T3
Tc prod = ∑yiTci = 179.3 K, Tr3 =
= 5.58 ‡ Ideal Gas
Tc prod
c) 1st Law: Q + HR = HP + W;
W=0
=>
Q = HP - H R
−
−
−
HR = (h )C6H14 + 19∆h O2 + 71.44 ∆h N2
= -188 215 + 19 6086 + 71.44 5911 = 349701 kJ/kmol fuel
−o
−
−o
−
−o
−
−o
−
HP = 6(h f + ∆h )CO2 = 7(h f + ∆h )H2O + 9.5 (h f + ∆h )O2 + 71.44(h f + ∆h )N2
CO2
-
(−h of + ∆h−) = (-393522 + 33397) = -360125 kJ/kmol
H2O
-
(−h of + ∆h−) = (-241826 + 26000) = -215826 kJ/kmol
O2
-
(−h of + ∆h−) = (0 + 22703) = 22703 kJ/kmol
14-54
N2
(−h of + ∆h−) = (0 + 21463) = 21463 kJ/kmol
-
.
HP = -1922537 kJ; Q = -2272238 kW
.
.
.
d) I = To n (SP - SR) - Q; To = 25oC
o
o
− yO2P2
− yN2P2
SR = (−s )C6H14 + 19 (−s 500 - Rln
)O2 + 71.44(s− 500 - Rln
)
Po
Po N2
(−s )C6H14 = 313.894 kJ/kmol K, (−s 500)O2 = 220.693 kJ/kmol K
o
(−s 500)N2 = 206.740 kJ/kmol K,
o
.
SR = 19141.9 kW/K
yO2 = 0.21, yN2 = 0.79
Products:
ni
yi
−s o
i
Si
− yiPe
− R ln
Po
(kJ/kmol-K)
CO2
6
0.0639
269.299
17.105
286.404
H2O
7
0.0745
232.739
15.829
248.568
O2
9.5
0.1011
243.579
13.291
256.87
N2
71.44 0.7605 228.171 -3.487
SP = ∑ ni−s i = 21950.1 kJ/K;
.
.
.
I = To n (SP - SR) - Q = 3109628 kW
224.684
14-55
English unit problems
14.75E
Pentane is burned with 120% theoretical air in a constant pressure process at
14.7 lbf/in2. The products are cooled to ambient temperature, 70 F. How much
mass of water is condensed per pound-mass of fuel? Repeat the answer,
assuming that the air used in the combustion has a relative humidity of 90%.
C5H12 + 1.2 × 8 (O2 + 3.76 N2) →
5 CO2 + 6 H2O + 0.96 O2 + 36.1 N2
Products cooled to 70 F, 14.7 lbf/in2
a) for H2O at 70 F: PG = 0.3632 lbf/in2
PG
nH2O MAX
0.3632
yH2O MAX =
=
=
14.7
P
nH2O MAX + 42.06
Solving, nH2O MAX = 1.066 < n H2O
Therefore, nH2O VAP = 1.066, nH2O LIQ = 6 - 1.066 = 4.934
mH2O LIQ =
4.934 × 18.015
= 1.232 lbm/lbm fuel
72.151
b) Pv1 = 0.9 × 0.3632 = 0.3269 lbf/in2
w1 = 0.622 ×
0.3269
= 0.014 147
14.373
nH2O IN = 0.014147 ×
28.97
× (9.6 + 36.1) = 1.040 lbmol
18.015
nH2O OUT = 1.04 + 6 = 7.04
nH2O LIQ = 7.04 - 1.066 = 5.974 lb mol
nH2O LIQ =
5.974 × 18.015
= 1.492 lbm/lbm fuel
72.151
14-56
14.76E The output gas mixture of a certain air-blown coal gasifier has the composition of
producer gas as listed in Table 14.2.
2 Consider the combustion of this gas with
120% theoretical air at 14.7 lbf/in. pressure. Find the dew point of the products
and the mass of water condensed per pound-mass of fuel if the products are
cooled 20 F below the dew point temperature?
a) {3 CH4 + 14 H2 + 50.9 N2 + 0.6 O2 + 27 CO + 4.5 CO2}
+ 31.1 O2 + 116.9 N2 → 34.5 CO2 + 20 H2O + 5.2 O2 + 167.8 N2
Products:
yH2O = yH2O MAX =
PG
14.7
=
20
34.5 + 20 + 5.2 + 167.8
⇒ PG = 1.2923 lbf/in2 → TDEW PT = 110.4 F
b) At T = 90.4 F, PG = 0.7089 lbf/in2
nH2O
0.7089
yH2O =
=
14.7
nH2O + 34.5 + 5.2 + 167.8
=>
n H2O = 10.51
⇒ nH2O LIQ = 20 - 10.51 = 9.49 lb mol
mH2O LIQ =
9.49(18)
3(16)+14(2)+50.9(28)+0.6(32)+27(28)+4.5(44)
= 0.069 lbm/lbm fuel
14.77E Pentene, C5H10 is burned with pure oxygen in an SSSF process. The products at
one point are brought to 1300 R and used in a heat exchanger, where they are
cooled to 77 F. Find the specific heat transfer in the heat exchanger.
C5H10 + νO O2 → 5 CO2 + 5 H2O, stoichiometric: νO = 7.5
2
2
Heat exchanger in at 1300 R, out at 77 F, so some water will condense.
5 H2O → (5 - x)H2Oliq + x H2Ovap
yH
2Omax
=
Pg 77
Ptot
=
0.464
x
= 0.03158 =
⇒ x = 0.163
14.696
5+x
Q⋅
q= .
= 5hex - hinCO + 5hex - hinH O - (5 -x)hfg H O
nfuel
2
2 vap
2
Btu
= -164340 lb mol fuel
14-57
14.78E A rigid vessel initially contains 2 pound mole of carbon and 2 pound mole of
oxygen at 77 F, 30 lbf/in. 2. Combustion occurs, and the resulting products consist
of 1 pound mole of carbon dioxide, 1 pound mole of carbon monoxide, and
excess oxygen at a temperature of 1800 R. Determine the final pressure in the
vessel and the heat transfer from the vessel during the process.
1
2 C + 2 O2 → 1 CO2 + 1 CO + O2
2
V = constant, C: solid, n1(GAS) = 2, n2(GAS) = 2.5
P 2 = P1 ×
n2T2
n1T1
= 30 ×
2.5 × 1800
lbf
= 125.8 2
in
2 × 536.7
H1 = 0
1
H2 = 1(-169184 + 14358) + 1(-47518 + 9323) + 2(0 + 9761) = -188141 Btu
Q = (U2-U1) = (H2-H1) - n2RT2 + n1RT1
1 2
= (-188141 - 0) - 1.98589(2.5 × 1800 - 2 × 536.67) = -194945 Btu
14.79E In a test of rocket propellant performance, liquid hydrazine (N2H4) at 14.7
lbf/in.2, 77 F, and oxygen gas at 14.7 lbf/in.2, 77 F, are fed to a combustion
chamber in the ratio of 0.5 lbm O2/lbm N2H4. The heat transfer from the chamber
to the surroundings is estimated to be 45 Btu/lbm N2H4. Determine the
temperature of the products exiting the chamber. Assume that only H2O, H2, and
N2 are present. The enthalpy of formation of liquid hydrazine is +21647 Btu/lb
mole.
1
N2H4
3
Comb.
Chamber
O2
Products
2
1
N2H4 + 2 O2 → H2O + H2 + N2
⋅ ≅ (m
⋅ × 32)/(m
⋅ × 32) = 0.5 ;
n⋅ O /n
Fu
O
Fu
2
2
.
⋅ = 45
QCV/m
Fu
Btu
QCV = -45 × 32.045 = -1442 lb mol fu
.
C.V. combustion chamber:
n⋅ Fuh1 + n⋅ O h2 + QCV = n⋅ toth3
⇒
2
- or - H1 + H2 + QCV = HP
3
°
=>
°
°
HR + QCV = HP + ∆HP
3
°
Btu
∆HP = HR - HP + QCV = 21647 + 103966 - 1442 = 124171 lb mol fuel
3
Trial and error on T3:
T3 = 5000 ⇒ ∆HP = 120071,
T3 = 5200 ⇒ ∆HP = 126224
⇒ T3 = 5133 R
14-58
14.80E Repeat the previous problem, but assume that saturated-liquid oxygen at 170 R is
used instead of 77 F oxygen gas in the combustion process. Use the generalized
charts to determine the properties of liquid oxygen.
Problem the same as 14.79, except oxygen enters at 2 as
saturated liquid at 170 R.
At 170 R, Tr2 =
170
−
= 0.61 ⇒ ∆h f = 5.1
278.6
From Fig. D.2:
-* (h - h) = 1.98589 × 278.6 × 5.1 = 2822 Btu/lbmol
°
°
∆HP = HR + ∆HR - HP + QCV = 21647 - 0.5(2822)
3
+ 0.5(0.219)(170 - 536.67)(32) + 103966 - 1442 = 121475
With ∆HP
3 5000 R
= 120071, ∆HP
⇒ T3 = 5045 R
3 5200 R
= 126224
14-59
14.81E Ethene, C2H4, and propane, C3H 8 , in a 1;1 mole ratio as gases are burned with
120% theoretical air in a gas turbine. Fuel is added at 77 F, 150 lbf/in.2 and the air
comes from the atmosphere, 77 F, 15 lbf/in.2 through a compressor to 150 lbf/in.2
and mixed with the fuel. The turbine work is such that the exit temperature is
1500 R with an exit pressure of 14.7 lbf/in. 2. Find the mixture temperature before
combustion, and also the work, assuming an adiabatic turbine.
C2H4 + C3H8 + νO2(O2+3.76N2) → 5CO2 + 6H2O + νN2N2
φ = 1 ⇒ νO2 = 8
φ = 1.2 ⇒ νO2 = 9.6
so we have 45.696 lbmol air per 2 lbmol fuel
C2H4 + C3H8 + 9.6(O2 + 3.76N2) → 5CO2 + 6H2O + 1.6 O2 + 36.096N2
C.V. Compressor (air flow)
wc,in = h2 - h1
s2 = s1 ⇒ Pr1 = 1.0907
Pr2 = Pr1P2/P1 = 10.907 ⇒ T2 air = 1027.3 R
wc,in = 247.81 - 128.38 = 119.53 Btu/lbm = 3462.4 Btu/lbmol air
C.V. Mixing chamber
• •
•
nairhair in + nfu1hfu1 + nfu2hfu2 = (same)exit
(CPF1 + CPF2)(Tex - T0) = 45.696 CP air(T2 air - Tex)
CPF1 = 11.53, CPF2 = 17.95, CP air = 6.953
45.696CP airT2 + (CPF1 + CPF2)T0
= 985.6 R = Tin combust
Tex =
CPF1 + CPF2 + 45.696CP air
Turbine work: take C.V. total and subtract compressor work.
Wtotal = Hin - Hout = HR - HP,1500
= h°f F1 + h°f F2 - 5hCO2 - 6hH2O - 36.096hN2 - 1.6hO2
= 22557 + (-44669) - 5(10557 - 169184)
- 6(8306 - 103966) - 36.096 × 6925 - 1.6 × 7297.5
= 1083342 Btu/2 lbmol Fuel
wT = wtot + wc,in = 1083342 + 3462.4 × 45.696
= 1 241 560 Btu/2 lbmol fuel
14-60
14.82E One alternative to using petroleum or natural gas as fuels is ethanol (C2H5OH),
which is commonly produced from grain by fermentation. Consider a combustion
process in which liquid ethanol is burned with 120% theoretical air in an SSSF
process. The reactants enter the combustion chamber at 77 F, and the products
exit at 140 F, 14.7 lbf/in.2. Calculate the heat transfer per pound mole of ethanol,
using the enthalpy of formation of ethanol gas plus the generalized tables or
charts.
C2H5OH + 1.2 × 3(O2 + 3.76N2) → 2CO2 + 3H2O + 0.6O2 + 13.54N2
Products at 140 F, 14.7 lbf/in2,
yH2O = 2.892/14.7 = nv/(2 + 0.6 + nv + 13.54)
nv = 3.953 > 3 ⇒ no condensation
h°f = -101032 Btu/lbmol as gas
Tr = 536.67/925 = 0.58 ⇒ D.2:
∆h/RTc = 5.23
HR = -101032 - 5.23 × 1.98589 × 925 + 0 + 0 = -110639 Btu/ lbmol fuel
HP = 2(-169184 + 570) + 3(-47518 + 506.5)
+ 0.6(443.7) + 13.54(438.4) = -472060 Btu/ lbmol fuel
QCV = HP - HR = -361421 Btu/lbmol fuel
14-61
14.83E Hydrogen peroxide, H2O2, enters a gas generator at 77 F, 75 lbf/in. 2 at the rate of
0.2 lbm/s and is decomposed to steam and oxygen exiting at 1500 R, 75 lbf/in.2.
The resulting mixture is expanded through a turbine to atmospheric pressure, 14.7
lbf/in.2, as shown in Fig. P14.27. Determine the power output of the turbine, and
the heat transfer rate in the gas generator. The enthalpy of formation of liquid H2
O2 is −80541 Btu/lb mol.
H2O2
•
→H
1
2
O + 2 O2
•
nFu = mFu/MFu = 0.2/34.015 = 0.00588 lbmol/s
•
•
nex,mix = 1.5 × nFu = 0.00882 lbmol/s
2
1
Cp mix = 3 × 0.445 × 18.015 + 3 × 0.219 × 31.999 = 7.6804
Cv mix = Cp mix - 1.98588 = 5.6945 ;
kmix = Cp mix/Cv mix = 1.3487
Reversible turbine
T3 = T2 × (P3/P2)(k-1)/k = 1500 × (14.7/75)0.2585 = 984.3 R
- w = Cp(T2 - T3) = 7.6804(1500 - 984.3) = 3960.8 Btu/lbmol
•
•
WCV = nmix × w = 0.00882 × 3960.8 = 34.9 Btu/s
C.V. Gas generator
•
•
•
QCV = H2 - H1 = 0.00588 × (-103966 + 8306) + 0.00294(7297.5)
- 0.00588(-80541) = -67.45 Btu/s
14-62
14.84E In a new high-efficiency furnace, natural gas, assumed to be 90% methane and
10% ethane
(by volume) and 110% theoretical air each enter at 77 F, 14.7
2
lbf/in. , and2 the products (assumed to be 100% gaseous) exit the furnace at 100 F,
14.7 lbf/in. . What is the heat transfer for this process?
2 Compare this to an older
furnace where the products exit at 450 F, 14.7 lbf/in. .
0.90CH 4+ 0.10C H
2 6
Prod.
Furnace
77 F 110% Air
100 F
2
14.7 lbf/in
HR = 0.9(-32190) + 0.1(-36432) = -32614 Btu
0.9 CH4 + 0.1 C2H6 + 1.1 × 2.15 O2 + 3.76 × 2.365 N2
→ 1.1 CO2 + 2.1 H2O + 0.215 O2 + 8.892 N2
a) TP = 100 F
HP = 1.1(-169184 + 206) + 2.1(-103966 + 185)
+ 0.215(162) + 8.892(160) = -402360 Btu, assuming all gas
QCV = HP - HR = -369746 Btu/lb mol fuel
b) TP = 450 F
HP = 1.1(-169184 + 3674) + 2.1(-103966 + 3057)
+ 0.215(2688) + 8.892(2610) = -370184 Btu
QCV = HP - HR = -337570 Btu/lb mol fuel
14.85E Repeat the previous problem, but take into account the actual phase behavior of
the products exiting the furnace.
Same as 14.84 , except possible condensation.
a) 100 F, 14.7 lbf/in2
yv max = 0.9503 / 14.7 = nv max/[nv max + 10.207]
nv max = 0.705 ⇒ nv = 0.705; nliq = 2.1 - 0.705 = 1.395
Hliq = 1.395[-122885 + 18.015(68.05 - 45.09)] = -170847
Hgas = 1.1(-169184 + 206) + 0.705(-103966 + 185)
+ 0.215(162) + 8.892(160) = -257584
HP = Hliq + Hgas = -428431
QCV = HP - HR = -395817 Btu/lbmol fuel
b) Older furnace has no condensation, so same as in 14.84.
14-63
14.86E Methane, CH , is burned in an SSSF process with two different oxidizers: A.
Pure oxygen, 4O2 and B a mixture of O2 + x Ar. The reactants are supplied at T ,
0
P and the products in are at 3200 R both cases. Find the required equivalence
0
ratio in case A and the amount of Argon, x, for a stoichiometric ratio in case B.
a) CH4 + νO2 → CO2 + 2H2O + (ν - 2)O2
νs = 2 for stoichiometric mixture.
HP 3200 = HR° = HP° + ∆HP 3200
∆hCO2 = 33579, ∆hH2O = 26479, ∆hO2 = 21860
° = (50010/2.326) × 16.04 = 344867
∆HP = HR° - H°P = -HRP
= 33579 + 2 × 26479 + (ν - 2)21860 = 42817 + ν × 21860
ν = 13.8175 ⇒ φ = ν/νs = 6.91
b) CH4 + 2O2 + 2xAr → CO2 + 2H2O + 2xAr
∆HP = 33579 + 2 × 26479 + 2 × 0.1253 × 39.948 × (3200 - 536.67)
= 86537 + x × 26662.5 = 344867 ⇒ x = 9.689
14.87E Butane gas at 77 F is mixed with 150% theoretical air at 1000 R and is burned in
an adiabatic SSSF combustor. What is the temperature of the products exiting the
combustor?
C4H10 + 1.5 × 6.5(O2 + 3.76N2) → 4CO2 + 5H2O + 3.25 O2 + 36.66N2
HR = HR° + ∆Hair,in
HP = HP° + 4∆hCO + 5∆hH
2
2O
+ 3.25∆hO + 36.66∆hN
2
2
HP = HR ⇒ ∆HP = HR° - H°P + ∆Hair,in
45714
° + ∆H
∆HP = -HRP
= 2.326 × 58.124 + 9.75 × 3366 + 36.66 × 3251
air,in
= 1294339 Btu/lbmol fuel
= 4∆hCO2 + 5∆hH2O + 3.25∆hO2 + 36.66∆hN2 at Tad
∆HP,3600R = 1281185
∆HP,3800R = 1374068
Tad = 3628 R
14-64
14.88E Liquid n-butane
at T , is sprayed into a gas turbine with primary air flowing at
2
150 lbf/in. , 700 R 0in a stoichiometric ratio. After complete combustion, the
products are at the
2 adiabatic flame temperature, which is too high so secondary
air at 150 lbf/in. , 700 R is added, with the resulting mixture being at 2500 R.
Show that T > 2500 R and find the ratio of secondary to primary air flow.
ad
C4H10 + 6.5(O2 + 3.76N2) → 4CO2 + 5H2O + 24.44N2
2nd.air
Fuel
Primary
air
Tad
MIXING
COMBUSTOR
2500 R
C.V. Combustor
HR = Hair + HFu = HP = HP° + ∆HP = HR° + ∆HR
° + ∆H
∆HP = HR° − H°P + ∆HR = -HRP
R
= 45344 × 58.124/2.326 + 6.5(1158 + 3.76 × 1138)
= 1168433 Btu/lbmol fuel
∆HP,2500R = 4 × 23755 + 5 × 18478 + 24.44 × 14855 = 550466
∆HP > ∆HP,2500R ⇒ Tad > 2500 R (If iteration Tad≅ 4400 R)
C.V. Mixing chamber
∆HP + νO2 2nd∆Hair,700 = ∆HP,2500R + νO2 2nd∆Hair,2500R
νO2 2nd =
∆HP - ∆HP,2500
∆Hair,2500 - ∆Hair,700
=
1168433 - 550466
= 9.344
71571 - 5437
Ratio = νO2 2nd/νO2 Prim. = 9.344/6.5 = 1.44
14-65
2
14.89E Acetylene gas at 77 F, 14.7 lbf/in. is fed to the head of a cutting torch. Calculate
the adiabatic flame temperature if the acetylene is burned with 100% theoretical
air at 77 F. Repeat the answer for 100% theoretical oxygen at 77 F.
a) C2H4 + 2.5 O2 + 2.5 × 3.76 N2 → 2 CO2 + 1 H2O + 9.4 N2
-o
HR = hf C2H2 = 97477 Btu
-*
-*
-*
HP = 2(-169184 + ∆hCO2) + 1(-103966 + ∆hH2O) + 9.4 ∆hN2
QCV = HP - HR = 0
-*
-*
-*
⇒ 2 ∆hCO2 + 1 ∆hH2O + 9.4 ∆hN2 = 539811 Btu
Trial and Error: TPROD = 5236 R
2 × 147196 + 121488 + 9.4 × 89348 = 1255751 OK
b) C2H2 + 2.5 O2 → 2 CO2 + H2O
HR = 97477 Btu
-*
-*
HP = 2(-169184 + ∆hCO2) + 1(-103966 + ∆hH2O)
-*
-*
⇒ 2 ∆hCO2 + 1 ∆hH2O = 539811
At 10000 R (limit of C.7): 2 × 135426 + 118440 = 389292
At 9500 R: 2 × 127734 + 111289 = 366757
or 4507/100 R change,
TPROD ≈ 10000 +
Difference, extrapolating
150519
≈ 13340 R
45.07
14.90E Ethene, C2H 4 , burns with 150% theoretical air in an SSSF constant-pressure
process with reactants entering at P , T . Find the adiabatic flame temperature.
0 0
C2H4 + 1.5 × 3(O2 + 3.76N2) → 2CO2 + 2H2O + 1.5O2 + 16.92N2
HP = HR = HP° + ∆HP = HR° ⇒
° = 28.054 × 47158/2.326 = 568775
∆HP = HR° - H°P = -HRP
∆HP = 2∆hCO2 + 2∆hH2O + 1.5∆hO2 + 16.92∆hN2
Trial and error on Tad...
∆HP,3400R = 545437
Tad = 3510 R
∆HP,3600R = 587736
14-66
14.91E Solid carbon is burned with stoichiometric air in an SSSF process, as shown in
Fig. P14.39. The reactants at T , P are heated in a preheater to T = 900 R with
0 0before flowing to a second heat
2 exchanger,
the energy given by the products
which they leave at T . Find the temperature of the products T , and the heat
0 (4 to 5) in the second heat exchanger. 4
transfer per lb mol of fuel
a) Following the flow we have: Inlet T1, after preheater T2, after mixing and
combustion chamber T3, after preheater T4, after last heat exchanger T5 = T1.
b) Products out of preheater T 4. Control volume: Total minus last heat
exchanger.
C + O2 + 3.76N2 → CO2 + 3.76N2
Energy Eq.:
°
°
HR = HR = HP = HP + ∆HP = hf CO + ∆hCO + 3.76∆hN
3
h°f CO = -169184, ∆HP
3
3 4400
2
2
= 167764, ∆HP
3 4600
2
2
= 177277
⇒ T3 = Tad.flame = 4430 R
c) Control volume total. Then energy equation:
°
°
HR + q- = HP
Btu
° = h-°f CO - 0 = -169184 lbmol fuel
q- = HRP
2
14.92E A closed, insulated container is charged
with a stoichiometric ratio of oxygen and
2
hydrogen at 77 F and 20 lbf/in. . After combustion, liquid water at 77 F is
sprayed in such that the final temperature is 2100 R. What is the final pressure?
1
H2 + 2O2 → H2O ;
P: 1 H2O + xiH2O
3-°
U2 - U1 = xihi = xihf liq = 1 + xiHP - HR - 1 + xiRTP + 2RTR
HR = φ, HP = -103966 + 14218.5 = -89747.5, h°f liq= -122885
Substitute
xi(-122885 + 89747.5 + 1.98588 × 2100)
3
= -89747.5 - 1.98588(2100 - 2 × 536.67)= -92319.2
xi = 3.187
P1V1 = nRRT1, P2V2 = npRTp
⇒ P2 =
P1(1 + xi)TP
1.5 T1
=
20(4.187)(2100)
= 218.5 lbf/in2
1.5(536.67)
14-67
14.93E Blast furnace gas in a steel mill is available at 500 F to be burned for the
generation of steam. The composition of this gas is, on a volumetric basis,
Component
CH4 H2 CO CO2 N2 H2O
Percent by volume
0.1 2.4 23.3 14.4 56.4 3.4
3
Find the lower heating value (Btu/ft ) of this gas at 500 F and P .
0
Of the six components in the gas mixture, only the first 3 contribute to the
heating value. These are, per lb mol of mixture:
0.024 H2, 0.001 CH4, 0.233 CO
For these components,
0.024 H2 + 0.001 CH4 + 0.233 CO + 0.1305 O2 → 0.026 H2 + 0.234 CO2
The remainder need not be included in the calculation, as the contributions to
reactants and products cancel. For the lower HV(water vapor) at 500 F
hRP = 0.026(-103966 + 3488) + 0.234(-169184 + 4229) - 0.024(0 + 2101)
- 0.001(-32190 + 0.538 × 16.04(500-77)) - 0.233(-47518 + 2981)
- 0.1305(0 + 3069) = -31257 Btu/lb mol fuel
RT0 1545 × 536.7
-v =
=
= 391.47 ft3/lb mol
0
P0
14.7 × 144
LHV = 31 680 / 391.47 = 79.85 Btu/ft3
14-68
14.94E Two pound moles of ammonia are burned in an SSSF process with x lb mol of
oxygen. The2 products, consisting of H2O, N2, and the excess O2, exit at 400 F,
1000 lbf/in. .
a. Calculate x if half the water in the products is condensed.
b. Calculate the absolute entropy of the products at the exit conditions.
2NH3 + xO2 → 3H2O + N2 + (x - 1.5)O2
Products at 400 F, 1000 lbf/in2 with nH2O LIQ= nH2O VAP= 1.5
a) yH2O VAP =
PG
P
=
247.1
1.5
=
1000 1.5 + 1 + x - 1.5
x = 5.070
b) SPROD = SGAS MIX + SH2O LIQ
Gas mixture:
ni
yi
-s°
i
yiP
-Rln
Si
P0
H 2O
1.5
0.2471
48.939
-5.604
43.335
O2
3.57
0.5881
52.366
-7.326
45.040
N2
1.0
0.1648
49.049
-4.800
44.249
SGAS MIX = 1.5(43.335) + 3.57(45.040) + 1.0(44.249) = 270.04 Btu/R
SH2O LIQ = 1.5[16.707 + 18.015(0.5647 - 0.0877)] = 37.95 Btu/R
SPROD = 270.04 + 37.95 = 307.99 Btu/R
14-69
14.95E Graphite, C, at P , T is burned with air coming in at P , 900 R in a ratio so the
0 R. Find the equivalence ratio, the
0 percent theoretical air
products exit at P0 , 2200
0
and the total irreversibility.
C + φ(O2 + 3.76N2) → CO2 + (φ - 1)O2 + 3.76φN2
°
HP = HR ⇒ ∆HP,2200 - ∆HR = HR° - H°P = -HRP
19659 + (φ - 1)13136 + 3.76 × φ × 12407 - φ(2616 + 3.76 × 2541)
= 0 - (-169184)
⇒ 6523 + φ × 47616.2 = 169184 ⇒ φ = 3.416
or 342 % theoretical air
Equivalence ratio = 1/φ = 0.293
S gen = sP - sR =
∑ν (s° - R-ln y ) - ∑ν (s° - R-ln y )
i
i
i
i
P
R: yO = 0.21
2
i
i
R
yN = 0.79
2
P: yO = 0.1507
2
yN = 0.79
2
yCO = 0.0593
Sgen = 66.952 + 5.610 + 2.416(59.844 + 3.758)
+ 3.76 × 3.416(56.066 + 0.468) - 1.371
- 3.416(52.686 + 3.099 + 3.76(49.353 + 0.468))
= 120.5 Btu/lbmol C × R
I = T0Sgen = 64677 Btu/lbmol C
2
14-70
14.96E A small air-cooled gasoline engine is tested, and the output is found to be 2.0 hp.
The temperature of the products is measured and found to be 730 F. The products
are analyzed on a dry volumetric basis, with the following result 11.4% CO 2,
2.9% CO, 1.6% O2 and 84.1% N2 . The fuel may be considered to be liquid
octane. The fuel and air enter the engine at 77 F, and the flow rate of fuel to the
engine is 1.8 lbm/h. Determine the rate of heat transfer from the engine and its
thermal efficiency.
a C8H18 + b O2 + 3.76b N2 → 11.4 CO2 + 2.9 CO + c H2O + 1.6 O2 + 84.1 N2
b = 84.1/3.76 = 22.37,
a = (1/8)(11.4 + 2.9) = 1.788,
C8H18 + 12.5 O2 + 47.1 N2
→ 6.38 CO2 + 1.62 CO + 9 H2O + 0.89 O2 + 47.1 N2
-°
a) HR = hf C8H18 = -107526 Btu
HP = 6.38(-169184 + 6807) + 1.62(-47518 + 4647)
+ 9(-103966 + 5475) + 0.89(0+4822)
+ 47.1(0 + 4617) = -1770092 Btu
HP-HR = -1770092 - (-107526) = -1662566 Btu
. .
1.8
HP-HR =
(-1662566) = -26198 Btu/h
114.23
.
QCV = -26198 + 2.0(2544) = -21110 Btu/h
.
b) QH = 1.8 × 20590 = 37062
.
WNET = 2.0 × 2544 = 5088 ;
ηTH =
5088
= 0.137
37062
c = 9a = 16.088
14-71
14.97E A gasoline engine uses liquid octane and air, both supplied at P , T , in a
0
stoichiometric ratio. The products (complete combustion) flow out of 0the exhaust
valve at 2000 R. Assume that the heat loss carried away by the cooling water, at
200 F, is equal to the work output. Find the efficiency of the engine expressed as
(work/lower heating value) and the second law efficiency.
C8H18 + 12.5(O2 + 3.76N2) → 8CO2 + 9H2O + 47N2
LHV = 44425 × 114.232/2.326 = 2181753 Btu/lbmol fuel
∆HP,2000 = 8 × 16982 + 9 × 13183 + 47 × 10804 = 643470
C.V. Total engine: Hin = Hex + W + Qloss = Hex + 2W
W = (Hin - Hex)/2 = (HR - HP)/2 = (-H°RP - ∆HP,2000)/2
= (2181753 - 643470)/2 = 769142 Btu/lbmol fuel
ηTH = W/LHV = 769142/2181753 = 0.353
For 2nd law efficiency we must find reversible work
Sin = -sfuel + 12.5(s-O + 3.76s-N )
2
2
= 86.122 + 12.5 [48.973 - 1.98589 ln (1/4.76)]
+ 47[45.739 - 1.98589 ln (3.76/4.76)]
= 2908.8 Btu/lbmol fuel × R
Sex = 8sCO + 9sH
2
2O
+ 47s-N = 8[65.677 - 1.98589 ln (8/64)]
2
+ 9[56.619 - 1.98589 ln (9/64)]
+ 47[55.302 - 1.98589 ln (47/64)]
= 3731.1 Btu/lbmol fuel × R
Assume the same Qloss out to 200 F = 659.67 R reservoir and compute Qrev
:
0
/T0 = Sex + Qloss/Tres
Sin + Qrev
0
Qrev
= T0(Sex - Sin) + QlossT0/Tres
0
= 536.67(3731.1 - 2908.8) + 769142*536.67/659.67
= 1067034 Btu/lbmol fuel
Wrev = Hin - Hex - Qloss + Qrev
= Wac + Qrev
0
0
= 769142 + 1067034 = 1836176
ηII = Wac/Wrev = 769142/1836176 = 0.419
14-72
14.98EIn Example 14.16, a basic hydrogen–oxygen fuel cell reaction was analyzed at
25°C, 100 kPa. Repeat this calculation, assuming that the fuel cell operates on air
at 77 F, 14.7 lbf/in.2, instead of on pure oxygen at this state.
Anode: 2 H2
→ 4 e- + 4 H+
Cathode: 4 H+ + 4 e- → 1 O2 + 2 H2O
Overall: 2 H2 + 1 O2 → 2 H2O
Example 14.16 : ∆G25°C = -474283 kJ/kmol
Or ∆G77 F = -203904 Btu/lbmol
PO = yO2 × P = 0.21 × 14.7 = 3.087 lbf/in2
2
-s = 48.973 – 1.98589 ln 0.21 = 52.072
O
2
∆S = 2(16.707) – 2(31.186) – 1(52.072) = - 81.03 Btu/R
∆H = 2(-122885) – 2(0) – 1(0) = -245770 Btu
∆G77 F = -245770 – 536.67(-81.03) = 202 284 Btu
E° = -∆G/N0e ne = 202284 × 2.326/(96485 × 4) = 1.219 V
15-1
CHAPTER 15
The correspondence between the new problem set and the previous 4th edition
chapter 13 problem set.
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Old
1 mod
2 mod
new
11
new
12
new
13
14
new
16
new
17
new
15
new
18
20
new
21
New
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Old
22
19
23
24
25
27
28
new
29
30
new
new
31
32
33
35
34
new
36
37 mod
New
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
Old
38
39
40
41
42
43
45
47
49 mod
50
51
52
new
new
new
44
48
The problems that are labeled advanced start at number 53.
The English unit problems are:
New
58
59
60
61
62
63
Old
53
new
58
59
60
61
New
64
65
66
67
68
69
Old
62
63
64
new
new
65
New
70
71
72
73
74
75
Old
67
68
69
70
71
72
15-2
15.1
Carbon dioxide at 15 MPa is injected into the top of a 5-km deep well in
connection with an enhanced oil-recovery process. The fluid column standing in
the well is at a uniform temperature of 40°C. What is the pressure at the bottom of
the well assuming ideal gas behavior?
Z1
(Z1-Z2) = 5000 m, P1 = 15 MPa
1
T = 40 oC = const
CO 2
Z2
Equilibrium at constant T
2
-wREV = 0 = ∆g + ∆PE = RT ln (P2/P1) + g(Z2-Z1) = 0
ln (P2/P1) =
9.807×5000
= 0.8287
1000×0.188 92×313.2
P2 = 15 exp(0.8287) = 34.36 MPa
15.2
Consider a 2-km-deep gas well containing a gas mixture of methane and ethane at
a uniform temperature of 30°C. The pressure at the top of the well is 14 MPa, and
the composition on a mole basis is 90% methane, 10% ethane. Determine the
pressure and composition at the bottom of the well, assuming an ideal gas
mixture.
Z1
(Z1-Z2) = 2000 m, Let A = CH4, B = C2H6
1
GAS
MIX
A+B
2
P1 = 14 MPa, yA1 = 0.90, yB1 = 0.10
T = 30 oC = const
Z2
From section 15.1, for A to be at equilibrium between
1 and 2:
WREV = 0 = n A(GA1-GA2) + nAMAg(Z1-Z2)
Similarly, for B: WREV = 0 = n B(GB1-GB2) + nBMBg(Z1-Z2)
Using eq. 15.10 for A: RT ln (PA2/PA1) = MAg(Z1-Z2)
with a similar expression for B. Now, ideal gas mixture,
Substituting:
ln
yA2P2
yA1P1
=
MAg(Z1-Z2)
RT
ln (yA2P2) = ln(0.9×14) +
and
ln
PA1 = yA1P, etc.
yB2P2
yB1P1
=
16.04×9.807(2000)
= 2.6585
1000×8.3145×303.2
=> y A2P2 = 14.2748
ln (yB2P2) = ln(0.1×14) +
=>
30.07×9.807(2000)
= 0.570 43
1000×8.3145×303.2
y B2P2 = (1-yA2)P2 = 1.76903
Solving: P2 = 16.044 MPa &
yA2 = 0.8897
MBg(Z1-Z2)
RT
15-3
15.3
Using the same assumptions as those in developing Eq. d in Example 15.1:
Develop an expression for pressure at the bottom of a deep column of liquid in
terms of the isothermal compressibility, β T. For liquid water at 20°C, β T=0.0005
[1/MPa]. Use the result of the first question to estimate the pressure in the Pacific
ocean at the depth of 3 km.
d gT = v° (1-β TP) dPT
d gT + g dz = 0
v° (1-β TP) dPT + g dz = 0 and integrate
⌠P (1-β P) dP = + g ⌠+Hdz
T
T
⌡P 0
v° ⌡0
P (1 -
⌠
⌡dz
⌡v°(1-β TP) dPT = - g ⌠
1
g
P - P0 - β T [P2 - P02] = H
2
v°
=>
1
1
g
β P) = P0 - β T P02 + H
2 T
2
v°
v° = vf 20°C = 0.001002; H = 3000 m , g = 9.80665 m/s2; βT = 0.0005 1/MPa
1
1
P (1 - 2 × 0.0005P) = 0.101 - 2 × 0.0005 × 0.1012
+ [9.80665 × 3000/0.001002] × 10-6
= 29.462 MPa,
which is close to P
P = 29.682 MPa
15.4
Calculate the equilibrium constant for the reaction O2 <=> 2O at temperatures of
298 K and 6000 K.
Reaction
O2
<=>
2O
At 25 oC (298.15 K):
-0
-0
∆H0 = 2hf O - 1hf O2 = 2(249 170) - 1(0) = 498 340 kJ/kmol
0
0
∆S0 = 2s-O - 1s-O2 = 2(161.059) - 1(205.148) = 116.97 kJ/kmol K
∆G0 = ∆H0 - T∆S0 = 498 340 - 298.15×116.97 = 463 465 kJ/kmol
∆G0
463 465
= -186.961
ln K = - - = RT
8.3145×298.15
At 6000 K:
∆H0 = 2(249 170+121 264) - (0+224 210) = 516 658 kJ/kmol
∆S0 = 2(224.597) -1(313.457) = 135.737 kJ/kmol K
∆G0 = 516 658 - 6000×135.737 = -297 764 kJ/kmol
ln K =
+297 764
= +5.969
8.3145×6000
15-4
15.5
Calculate the equilibrium constant for the reaction H2 <=> 2H at a temperature of
2000 K, using properties from Table A.8. Compare the result with the value listed
in Table A.10.
∆hH = 52942
sH = 188.419
−o
hf = 0
∆hH = 35375
sH = 154.279
−o
h f = 217999
2
2
∆G0 = ∆H - T∆S = HRHS - HLHS – T (S0RHS - S0LHS)
= 2 × (35375+217999) – 52943 – 2000(2×154.279-182.419)
= 213528
ln K = -∆G0/RT = -213528 / (8.3145×2000) = -12.8407
Table A.10 ln K = -12.841 OK
15.6
Plot to scale the values of lnK versus 1/T for the reaction 2CO2 <=> 2CO + O 2.
Write an equation for lnK as a function of temperature.
2 CO2 ⇔ 2 CO + 1 O2
T(K)
104 ×
2000
2400
2800
3200
3600
5.000
4.167
3.571
3.125
2.778
1
T
For the range
below ~ 5000 K,
ln K ≈ A + B/T
Using values at
2000 K & 5000 K
A = 19.5056
B = -65 543 K
ln K
T(K)
-13.266
-7.715
-3.781
-0.853
1.408
4000
4500
5000
5500
6000
1
T
2.500
2.222
2.000
1.818
1.667
ln K
104 ×
8
3.204
4.985
6.397
7.542
8.488
almost
4
linear
0
-4
-8
-12
0
1
2
3
4
5
4 1
10 x _
T
15-5
15.7
Calculate the equilibrium constant for the reaction:2CO <=> 2CO + O at 3000
2
2
K using values from Table A.8 and compare the result to Table A.10.
∆hCO = 93504
-0
hf CO = -110527
sCO = 273.607
∆hCO = 152853
-0
hf CO = -393522
sCO = 334.17
∆hO = 98013
sO = 284.466
2
2
2
2
2
∆G0 = ∆H - T∆S = 2 HCO + HO – 2 HCO - T (2sCO + sO - 2sCO )
2
2
2
2
= 2 (93504 – 110527) + 98013 + 0 – 2(152853-393522)
-3000(2×273.607 + 284.466 - 2×334.17) = 55285
ln K = -∆G0/RT = -55285/ (8.31451×3000) = -2.2164
Table A.10 ln K = -2.217 OK
15.8
Pure oxygen is heated from 25°C to 3200 K in an SSSF process at a constant
pressure of 200 kPa. Find the exit composition and the heat transfer.
The only reaction will be the dissociation of the oxygen
O2 ⇔ 2O ; K(3200) = 0.046467
Look at initially 1 mol Oxygen and shift reaction with x
nO2 = 1 - x; nO = 2x; ntot = 1 + x; yi = ni/ntot
y2O P
4x2 1 + x
8x2
2-1
K= ( ) =
2=
y0 Po
(1 + x)2 1 - x
1 - x2
2
K/8
⇒ x = 0.07599; y0 = 0.859; y0 = 0.141
1 + K/8
2
q = n0 exh0 ex + n0exhOex - h0 in = (1 + x)(y 0 h0 + y0hO) - 0
x2 =
2
2
h0 = 106022;
2
2
hO = 249170 + 60767
2
2
⇒
q = q/32 = 4532 kJ/kg (=3316.5 if no reaction)
q = 145015 kJ/kmol O2
15-6
15.9
Pure oxygen is heated from 25°C, 100 kPa to 3200 K in a constant volume
container. Find the final pressure, composition, and the heat transfer.
As oxygen is heated it dissociates
O2 ⇔ 2O
ln Keq = -3.069
C. V. Heater: U2 - U1 = 1Q2 = H2 - H1 - P2v + P1v
- Per mole O2: 1q2 = h2 - h1 + R(T1 - (n2/n1)T2)
Shift x in reaction 1 to have final composition: (1 - x)O 2 + 2xO
n1 = 1
yO
22
n2 = 1 - x + 2x = 1 + x
= (1 - x)/(1 + x) ; yO2 = 2x/(1 + x)
Ideal gas and V2 = V1 ⇒ P2 = P1n2T2/n1T1
Keq = e
⇒
-3.069
⇒ P2/Po = (1 + x)T2/T1
y2O P2
2x 2 1 + x 1 + x T2
= ( )=(
) (
)(
)( )
1+x 1-x
1
y02 Po
T1
4x2 T1 -3.069
= e
= 0.00433 ⇒ x=0.0324
1 - x T2
(nO )2 = 0.9676, (nO)2 = 0.0648, n2 = 1.0324
2
q = 0.9676(106022) + 0.0648(249170 + 60767) - Ø
1 2
+ 8.3145(298.15 - 1.0324(3200)) = 97681 kJ/kmolO2
yO 2=0.9676/1.0324 = 0.937; yO2=0.0648/1.0324 = 0.0628
2
15.10 Nitrogen gas, N2 , is heated to 4000 K, 10 kPa. What fraction of the N2 is
dissociated to N at this state?
N2 <=> 2 N
Initial
1
0
Change
-x
2x
Equil.
1-x
2x
yN2 =
@ T = 4000 K, lnK = -12.671
K = 3.14x10-6
ntot = 1 - x + 2x = 1 + x
2x
1-x
, yN =
1+x
1+x
2
yN  P 2-1
K=
  ; =>
yN2 Po
nN2 = 0.99911,
3.14x10 -6 =
nN = 0.00177,
4x2  10 
 
1 - x2 100
yN2 = 0.9982,
=> x = 0.000886
yN = 0.0018
15-7
15.11 Hydrogen gas is heated from room temperature to 4000 K, 500 kPa, at which state
the diatomic species has partially dissociated to the monatomic form. Determine
the equilibrium composition at this state.
H2 ⇔ 2H
Equil.
-x +2x
nH2 = 1 - x
nH = 0 + 2x
n =1+ x
(2x)2
P 2-1
K=
(
)
(1-x)(1+x) P0
at 4000 K:
2.545
x2
= 0.127 25 =
4×(500/100)
1-x2
ln K = 0.934 => K = 2.545
Solving,
x = 0.3360
nH2 = 0.664, nH = 0.672, ntot = 1.336
yH2 = 0.497, yH = 0.503
15.12 Consider the chemical equilibrium involving H2O, H2, CO, and CO2, and no other
substances. Show that the equilibrium constant at any temperature can be found
using values from Table A.10 only.
The reaction between the species is
(1)
1 CO2 + 1 H2 ⇔ 1 CO + 1 H2O
From A.10 we find:
(2) 2 H2O ⇔ 2 H2 + 1 O2,
(3) 2 CO2 ⇔ 2 CO + 1 O2
Form (1) from (2) and (3) as:
(1) =
From A.10
1
1
(3) - (2)
2
2
ln K
=>
ln K = (ln K3 - ln K2)/2
at any T is listed for reactions (2) and (3)
15-8
15.13 One kilomole Ar and one kilomole O2 is heated up at a constant pressure of 100
kPa to 3200 K, where it comes to equilibrium. Find the final mole fractions for
Ar, O2, and O.
The only equilibrium reaction listed in the book is dissociation of O2.
So assuming that we find in Table A.10:
ln(K) = -3.072
Ar + O2 ⇒ Ar + (1 - x)O2 + 2x O
The atom balance already shown in above equation can also be done as
Species
Ar
O2
Start
1
0
Change
0
-x
2x
Total
1
1-x
2x
1
The total number of moles is
O
ntot = 1 + 1-x + 2x = 2 + x
so
yAr = 1/(2 + x); yO = 1 - x/(2 + x); yO = 2x/(2 + x)
2
and the definition of the equilibrium constant (Ptot = Po) becomes
K=e
-3.072
y2O
4x2
= 0.04633 =
=
y02 (2 + x)(1 - x)
The equation to solve becomes from the last expression
(K + 4)x2 + Kx - 2K = 0
If that is solved we get
x = -0.0057 ± 0.1514 = 0.1457;
yO = 0.1358;
y02 = 0.3981;
x must be positive
yAr = 0.4661
15-9
15.14 A piston/cylinder contains 0.1 kmol hydrogen and 0.1 kmol Ar gas at 25°C, 200
kPa. It is heated up in a constant pressure process so the mole fraction of atomic
hydrogen is 10%. Find the final temperature and the heat transfer needed.
When gas is heated up H2 splits partly into H as
H2 ⇔ 2H
Component
and the gas is diluted with Ar
H2
Ar
H
Initial
0.1
0.1
0
Shift
-x
0
2x
Final
0.1-x 0.1
2x
yH = 0.1 = 2x/(0.2+x) ⇒ 2x = 0.02+0.1x
Total = 0.2 + x
⇒
x = 0.010526
⇒ ntot = 0.21053
yH2 = 0.425 = [(0.1-x)/(0.2+x)];
yAr = 1 – rest = 0.475
Do the equilibrium constant:
2
K(T) =
yH
yH2
0.01
(PP0)2-1 = (0.425
) × (200
) = 0.047059
100
ln (K) = -3.056 so from Table A.10 interpolate to get T = 3110 K
To do the energy eq., we look up the enthalpies in Table A.8 at 3110K
hH2 = 92829.1; hH = 217999 + 58447.4 = 276445.4 (= hf + ∆h)
hAr = CP(3110–298.15) = 20.7863 × (3110-298.13) = 58447.9
(same as ∆h for H)
Now get the total number of moles to get
nH = 0.021053;
nH2 = ntot ×
1-x
= 0.08947;
2+x
nAr = 0.1
Since pressure is constant W = P∆V and Q becomes differences in h
Q = n∆h = 0.08947 × 92829.1 – 0 + 0.021053 × 276446.4
– 0 + 0.1 × 58447.9
= 19970 kJ
15-10
15.15 Air (assumed to be 79% nitrogen and 21% oxygen) is heated in an SSSF process
at a constant pressure of 100 kPa, and some NO is formed. At what temperature
will the mole fraction of N.O be 0.001?
0.79 N2 + 0.21 O2 heated at 100 kPa, forms NO
N2 + O2 ⇔ 2 NO
nN2 = 0.79 - x
-x -x
nO2 = 0.21 - x
+2x
nNO = 0
+ 2x
ntot = 1.0
At exit, yNO = 0.001 =
2x
1.0
⇒
x = 0.0005
⇒ nN2 = 0.7895, nO2 = 0.2095
2
K=
yNO
yN2yO2
-6
10
(PP0)0 = 0.7895×0.2095
= 6.046×10-6
From Table A.10,
T = 1444 K
or ln K = -12.016
15-11
15.16 Saturated liquid butane enters an insulated constant pressure combustion chamber
at 25°C, and x times theoretical oxygen gas enters at the same pressure and
temperature. The combustion products exit at 3400 K. Assuming that the products
are a chemical equilibrium gas mixture that includes CO, what is x?
Butane: T1 = 25oC, sat. liq., x1 = 0, Tc = 425.2 K, Pc = 3.8 Mpa
Tr1 = 0.7, Figs. D.1and D.2, Pr1 = 0.10, P1 = Pr1Pc = 380 kPa
− * − 
h 1 − h 1f = 4.85 RTc
Oxygen: T = 25oC, X * theoretical air
Products: T3 = 3400 K
C4H10 + 6.5XO2 => ?CO 2 + 5H2O + ?O2
2
2CO2 <=> 2CO + O2
Initial
4
0
6.5(X-1)
Change
-2a
2a
a
Equil.
4-2a
2a
6.5(X-1) + a
nCO = 4 - 2a, nCO = 2a, nO = 6.5(X-1) + a, nH2O = 5, ntot = 2.5 + a + 6.5X
2
2
2a
4 - 2a
6.5(x - 1) + a
yCO =
, yCO2 =
, yO2 =
2.5 + a +6.5X
2.5 + a +6.5X
2.5 + a +6.5X
2
yCOyO P12+1-2  a 2 6.5X - 6.5 + a P1
K= 2 2 
=  
 
2-a 6.5X - 2.5 + a Po
Po


y CO
2
@ T3 = 3400 K,
ln(K) = 0.346,
K = 1.4134
 a 2 6.5X - 6.5 + a
1.4134 =   
 (3.76)
2-a 6.5X - 2.5 + a
Equation 1.
Need a second equation:
1st Law:
Qcv + HR = HP + Wcv; Qcv = 0, Wcv = 0
−o
−
HR = (h f + ∆h )C H = (-126200 - 17146) = -143346 kJ
4 10
Products @ 3400 K:
−o
−
−o
−
−o
−
−o
−
HP = n(h f + ∆h )CO2 + n(h f + ∆h )CO + n(h f + ∆h )O2 + n(h f + ∆h )H2O
= (4 - 2a)(-393522 + 177836) + 2a(-110527 + 108440)
+ [6.5(X - 1) + a](0 + 114101) + 5(-241826 + 149073)
= -463765 kJ/kmol
H P = HR
=>
1924820 = 541299a + 741656.5 X
Two equations and two unknowns, solve for X and a.
a ≅ 0.87, X ≅ 1.96
Equation 2.
15-12
15.17 The combustion products from burning pentane, C5H12, with pure oxygen in a
stoichiometric ratio exists at 2400 K, 100 kPa. Consider the dissociation of only
CO2 and find the equilibrium mole fraction of CO.
C5H12 + 8 O2 → 5 CO2 + 6 H2O
2 CO2 ⇔ 2 CO + 1 O2
At 2400K,
ln K = -7.715
K = 4.461x10-4
Initial
Change
Equil.
5
-2z
5-2z
0
+2z
2z
0
+z
z
Assuming P = Po = 0.1 MPa, and ntot = 5 + z + 6 = 11 + z
y2COyO
K=
2
y2CO
2
P
 2z 2  z 
( 0) = 
 
(1) = 4.461x10-4 ;
P
5 - 2z 11 + z
Trial & Error (compute LHS for various values of z):
nCO = 4.418;
2
nCO = 0.582;
nO = 0.291
2
z = 0.291
=>
yCO = 0.0515
15.18 Find the equilibrium constant for the reaction 2NO + O ⇔ 2NO from the
2
2
elementary reactions in Table A.10 to answer which of the nitrogen oxides, NO or
NO , is the more stable at ambient conditions? What about at 2000 K?
2
2 NO + O2 ⇔ 2 NO2 (1)
But N2 + O2 ⇔ 2 NO (2)
N2 + 2 O2 ⇔ 2 NO2 (3)
Reaction 1 = Reaction 3 - Reaction 2
⇒
0
0
0
∆G1 = ∆G3 - ∆G2
At 25 oC, from Table A.10:
or
=>
ln K1 = ln K3 - ln K2
ln K1 = -41.355 - (-69.868) = +28.513
K1 = 2.416×1012
an extremely large number, which means reaction 1 tends to go very strongly
from left to right.
At 2000 K:
ln K1 = -19.136 - (-7.825) = -11.311
or K1 = 1.224×10-5
meaning that reaction 1 tends to go quite strongly from right to left.
15-13
15.19 Methane in equilibrium with carbon and hydrogen as: CH4 ⇔ C + 2H2. has lnK =
-0.3362 at 800 K. For a mixture at 100 kPa, 800 K find the equilibrium mole
fractions of all components (CH 4, C, and H2 neglect hydrogen dissociation).Redo
the molefractions for a mixture state of 200 kPa, 800 K.
CH4 ⇔ C + 2H2
ln K = - 0.3362
nCH = 1-x; nC = x; nH = 2x; ntot = 1+2x
4
2
x   4x 
1+2x
×

 (1+2x)  100

=
× 
  1-x   100
 1+2x 
2
y2 yC
H2
yCH
4
 P 1+2-1
= K = e -0.3362
P 
 o
2
2
x3
0.71448
=
=
= 0.17862
2
4
(1+2x) (1-x)
Solve by trial and error or successive substitutions
x = 0.5
LHS = 0.0625
x = 0.6
LHS = 0.11157
x = 0.7
LHS = 0.198495
x ≈ 0.67
LHS = 0.16645
Interpolate
LHS = 0.17787 ⇒ x = 0.682
x = 0.6814
 1-x 
yCH4 = 
 = 0.1345
1+2x
 x 
yC = 
 = 0.2885
1+2x
 2x 
yH2 = 
 = 0.577
1+2x
 P  200
P  = 100 = 2
 o  
⇒
x3
0.71448
=
= 0.044655
2
(1+2x) (1-x) 4 × 22
x = 0.4
LHS = 0.03292
x = 0.45
x = 0.44
LHS = 0.04304
x = 0.446
 1-x 
 = 0.293;
yCH4 = 
1+2x
LHS = 0.045895
 x 
 2x 
yC = 
 = 0.236;yH = 
 = 0.471
2 1+2x
1+2x
15-14
15.20 A mixture of 1 kmol carbon dioxide, 2 kmol carbon monoxide, and 2 kmol
oxygen, at 25°C, 150 kPa, is heated in a constant pressure SSSF process to 3000
K. Assuming that only these same substances are present in the exiting chemical
equilibrium mixture, determine the composition of that mixture.
initial mix:
Equil. mix: CO2, CO, O2 at
Const.
1 CO2, 2 CO,
T = 3000 K,
Pressure
2 O2
P = 150 kPa
Q
Reaction
2 CO2
initial
change
equil.
1
-2x
(1-2x)
⇔ 2 CO
+
2
+2x
(2+2x)
O2
2
+x
(2+x)
From A.10 at 3000 K: K = 0.1090
1
For each n > 0 ⇒ -1 < x < +2
2
yCOyO2 P 1
1+x 2 2+x 150
K= 2
(P0) = 4(1-2x
) (5+x)(100)
y
CO2
or
1+x
(1-2x
) (2+x
)= 0.018 167,
5+x
2
nCO2 = 2.042
nCO = 0.958
Trial & error: x = -0.521
y
= 0.4559
nO2 = 1.479   CO2
= 0.2139
 y
nTOT = 4.479   CO
 yO2 = 0.3302
15.21 Repeat the previous problem for an initial mixture that also includes 2 kmol of
nitrogen, which does not dissociate during the process.
Same as Prob. 15.20 except also 2 kmol of inert N2
Equil.: nCO2 = (1-2x), nCO = (2+2x), n O2 = (2+x), n N2 = 2
2
yCOyO2 P 1
1+x 2 2+x 150
(P0) = 4(1-2x
) (7+x)(100)
K= 2
y
CO2
or (
1+x 2 2+x
) ( )= 0.018167 Trial & error: x = -0.464
1-2x 7+x
nCO2 = 1.928
nCO = 1.072
  yCO2 = 0.295
  y = 0.164
nTOT = 6.536   CO
nO2 = 1.536
nN2 = 2.0
yO2 = 0.235
yN2 = 0.306
15-15
15.22 Complete combustion of hydrogen and pure oxygen in a stoichiometric ratio at
P , T to form water would result in a computed adiabatic flame temperature of
0 0
4990 K for an SSSF setup.
a. How should the adiabatic flame temperature be found if the equilibrium reaction
2H + O ⇔ H O is considered? Disregard all other possible reactions
2
2
2
(dissociations) and show the final equation(s) to be solved.
b. Find the equilibrium composition at 3800 K, again disregarding all other reactions.
c. Which other reactions should be considered and which components will be
present in the final mixture?
2H2 + O2 ⇔ 2H2O
a)
Species
HP = HR = HoP + ∆HP = HoR = Ø
y2H
Keq =
2O
P
( )
y2H yO P0
2 2
-1
,
H2
O2
H 2O
Initial
2
1
Ø
Shift
-2x
-x
2x
Final
2-2x
1-x
2x
ntot = 2-2x + 1-x + 2x = 3-x
Hp = (2-2x)∆hH + (1-x)∆hO + 2x(hofH2O + ∆hH2O) = Ø
2
2
4x2 (3-x)2 3-x x2(3-x)
=
= Keq(T)
Keq =
(3-x)2 (2-2x)2 1-x (1-x)3
hofH2O = -241826; ∆hH (T), ∆hO (T), ∆hH O(T)
2
2
(1)
(2)
2
Trial and Error (solve for x,T) using Eqs. (1) and (2).
yO = 0.15;
2
b) At 3800 K
yH = 0.29; yH
2
2O
= 0.56]
Keq = e1.906 (Reaction is times -1 of table)
x2(3-x)(1-x)-3 = e1.906 = 6.726 ⇒ x ≅ 0.5306
yH
2
=
O
2x
1-x
2-2x
= 0.43; yO =
= 0.19; yH =
= 0.38
3-x
3-x
2 3-x
2
c) Other possible reactions from table A.10
H2 ↔ 2 H
O2 ↔ 2 O
2 H2O ↔ H2 + 2 OH
15-16
15.23 Gasification of char (primarily carbon) with steam following coal pyrolysis yields
a gas mixture of 1 kmol CO and 1 kmol H2. We wish to upgrade the hydrogen
content of this syngas fuel mixture, so it is fed to an appropriate catalytic reactor
along with 1 kmol of H 2O. Exiting the reactor is a chemical equilibrium gas
mixture of CO, H 2, H2O, and CO2 at 600 K, 500 kPa. Determine the equilibrium
composition. Note: see Problem 15.12.
1 CO + 1 H 2
Chem. Equil. Mix
CO, H2, H2O, CO2
1 H 2O
600 K
500 kPa
(1) 1 CO + 1 H2O ⇔ 1 CO2 + 1 H2
-x
-x
+x
+x
(2) 2 H2O ⇔ 2 H2 + 1 O2
(1) =
(3) 2 CO2 ⇔ 2 CO + 1 O2
1
1
(2) - (3)
2
2
ln K1 =
1
[-85.79-(-92.49)]= +3.35,
2
K1 = 28.503
Equilibrium:
nCO = 1-x,
∑ n = 3,
28.503 =
nH2O = 1-x,
nCO2 = 0 + x,
n H2 = 1 + x
yCO2yH2 P
yCO2yH2
0
K=
( ) =y y
yCOyH2O P0
CO H2O
x(1+x)
→ x = 0.7794
(1-x)2
n
y
CO
0.2206
0.0735
H2O 0.2206
0.0735
%
7.35
7.35
CO2
0.7794
0.2598
26.0
H2
1.7794
0.5932
59.3
15-17
15.24 A gas mixture of 1 kmol carbon monoxide, 1 kmol nitrogen, and 1 kmol oxygen
at 25°C, 150 kPa, is heated in a constant pressure SSSF process. The exit mixture
can be assumed to be in chemical equilibrium with CO 2, CO, O2, and N2 present. The
mole fraction of CO2 at this point is 0.176. Calculate the heat transfer for the
process.
initial mix:
Equil. mix:
Const.
1 CO, 1 O2,
CO2, CO, O2, N2
Pressure
1 N2
yCO2 = 0.176
Q
reaction
2 CO2
initial
change
equil.
0
+2x
2x
yCO2 = 0.176 =
2x
3-x
⇔
2 CO
P = 150 kPa
+
1
-2x
(1-2x)
⇒
O2
also,
1
-x
(1-x)
N2
1
0
1
x = 0.242 65
nCO2 = 0.4853 nO2 = 0.7574  yCO2 = 0.176
yO2 = 0.2747

nCO = 0.5147 nN2 = 1
  yCO = 0.1867
2
yCOyO2 P 1 0.18672×0.2747 150
K= 2
(P0) = 0.1762 (100) = 0.4635
y
CO2
From A.10, TPROD = 3213 K
From A.9, HR = -110 527 kJ
HP = 0.4853(-393 522 + 166 134) + 0.5147(-110 527 + 101 447)
+ 0.7574(0 + 106 545) + 1(0 + 100 617)
= +66 284 kJ
QCV = HP - HR = 66 284 - (-110 527) = +176 811 kJ
15-18
15.25 A rigid container initially contains 2 kmol of carbon monoxide and 2 kmol of
oxygen at 25°C, 100 kPa. The content is then heated to 3000 K at which point an
equilibrium mixture of CO2, CO, and O2 exists. Disregard other possible species
and determine the final pressure, the equilibrium composition and the heat
transfer for the process.
2 CO + 2 O2 ⇔ 2 CO2 + O2
Initial 2
2
Species:
CO
-2x+x
Final 2-2x
2-x
2x : ntot = 2-2x + 2-x + 2x = 4-x
yO =
2-x
;
4-x
2-2x
;
4-x
CO2
0
Shift -2x
yCO =
O2
2
2x
yCO =
2
2x
4-x
U2 - U1 = 1Q2 = H2 - H1 - P2v + P1v
= (2 - 2x)hCO 2 + (2 - x)hO
22
+ 2xhCO - 2hofCO - 2hofO
2
2
2
- R(4 - x)T2 + 4RT1
Keq = e2.217 =
y2CO
P
2
2 -1
2 (P )
y0 yCO o
2
=
4x2 4-x 4T1
4(1-x)2 2-x (4-x)T2
x 2 1 1 T2 2.217
⇒ ( )
=
e
= 23.092
1-x 2-x 4 T1
x = 0.8382;
yCO = 0.102;
yO = 0.368;
2
yCO = 0.53
2
P2 = P1(4-x)T2/4T1 = 100(3.1618)(3000/4)(298.15) = 795.4 kPa
Q = 0.3236(-110527 + 93504) + 1.1618(98013) + 1.6764(-393522
1 2
+ 152853) - 2(-110527) - 2(Ø) + 8.3145(4(298.15) - 3000(3.1618))
= -142991 kJ
15-19
15.26 One approach to using hydrocarbon fuels in a fuel cell is to “reform” the
hydrocarbon to obtain hydrogen, which is then fed to the fuel cell. As a part of the
analysis of such a procedure, consider the reforming section
a. Determine the equilibrium constant for this reaction at a temperature of 800 K.
b. One kilomole each of methane and water are fed to a catalytic reformer. A
mixture of CH4, H 2O, H2, and CO exits in chemical equilibrium at 800 K, 100
kPa; determine the equilibrium composition of this mixture.
For CH4, use CP0 at ave. temp., 550 K. Table A.6, CP0 = 49.316 kJ/kmol K
-0 -0
a) h800 K = hf + CP0∆T = -74 873 + 49.316(800-298.2) = -50 126 kJ/kmol
800
-s0
800 K = 186.251 + 49.316 ln 298.2 = 234.918 kJ/kmol K
CH4 + H2O ⇔ 3H2 + CO
For
0
∆H800 K = 3(0+14 681) + 1(-110 527+15 174) - 1(-50 126)
- 1(-241 826+18 002) = +222 640 kJ
0
∆S800 K = 3(159.554) + 1(227.277) - 1(234.918) - 1(223.826) = +247.195 kJ/K
∆G0 = ∆H0 - T∆S0 = 222 640 - 800(247.195) = +24 884 kJ
∆G0
-24 884
ln K = - - =
= -3.7411
=>
K = 0.0237
RT 8.3144×800
b)
CH4
+H2O
+
CO
⇔ 3 H2
initial
change
equil.
1
-x
(1-x)
1
-x
(1-x)
0
+3x
3x
0
+x
x
nTOTAL = 2 + 2x
3
K=
yH2yCO
P 2
(3x)3x
100 2
(
)
=
(
)
yCH4yH2O P0
(1-x)(1-x)(2+2x)2 100
or
x
x
(1-x
) (1+x
)
or
x2
=
1-x2
2
2
=
4×0.0237
= 0.003 51
27×1
0.003 51 = 0.059 25
 y
y
= 0.7095
y
= 0.2365
 y

= 2.473 
nCH4 = 0.7635
Solving,
nH2O = 0.7635
CH4
= 0.3087
nH2
H2O
= 0.3087
nCO
nTOT
H2
= 0.2870
CO
= 0.0956
x = 0.2365
15-20
15.27 In a test of a gas-turbine combustor, saturated-liquid methane at 115 K is to be
burned with excess air to hold the adiabatic flame temperature to 1600 K. It is
assumed that the products consist of a mixture of CO , H O, N , O , and NO in
2 2
2 2
chemical equilibrium. Determine the percent excess air used in the combustion,
and the percentage of NO in the products.
CH4 + 2x O2 + 7.52x N2
→ 1 CO2 + 2 H2O + (2x-2) O2 + 7.52x N2
Then
N2 + O2 = 2 NO
init
ch.
equil.
7.52x 2x-2
0
-a
-a +2a
(7.52x-a) (2x-2-a) 2a
Also CO2 H2O
1
0
1
2
0
2
nTOT = 1 + 9.52x
1600 K: ln K = -10.55, K = 2.628×10-5
2
2
yNO
yNO
P 0
4a2
-5
2.628×10 K =
( ) = y y = (7.52x-a)(2x-2-a)
yN2yO2 P0
N2 O2
From A.9 and B.7,
HR = 1[-74 873 + 16.043(-274.7-624.1)]+ 0 + 0 = -89 292 kJ
(Air assumed 25 oC)
HP = 1(-393 522 + 67 569) + 2(-241 826 + 52 907)
+ (7.52x-a)(41 904) + (2x-2-a)(44 267) + 2a(90 291 + 43 319)
= -792 325 + 403 652 x + 181 049 a
Assume a ~ 0, then from
HP - HR = 0 → x = 1.7417
Subst.
2.628×10-5
a2
=
,
(13.098-a)(1.483-a)
4
get a ≈ 0.0113
Use this a in 1st law
x=
703 042 - 181 049×0.0113
= 1.7366
403 652
-5
2.628×10
a2
=
, a = 0.0112
⇒
(13.059-a)(1.4732-a)
4
% excess air = 73.7 %
% NO =
2×0.0112×100
= 0.128 %
1+9.52×1.7366
⇒
x = 1.7366
15-21
15.28 The van't Hoff equation
o
∆H
d ln K = − 2 dTPo
RT
relates the chemical equilibrium constant K to the enthalpy of reaction ∆Ho. From
the value of K in Table A.10 for the dissociation of hydrogen at 2000 K and the
value of ∆Ho calculated from Table A.8 at 2000 K use van’t Hoff equation to
predict the constant at 2400 K.
H2 ⇔ 2H
∆H° = 2 × (35375+217999) – 52942 = 453806
lnK2000 = -12.841;
Assume ∆H° is constant and integrate the Van’t Hoff equation
2000
∆H° 1
1
⌠
- 2
lnK2400 - lnK2000 = ⌡(∆H°/RT )dT = - - (
)
R T2400 T2000
2400
1
lnK2400 = lnK2000 + ∆H° (
-
1
)/R
T2400 T2000
= -12.841 + 453806 (
6-5
) / 8.31451 = -12.841 + 4.548
12000
= -8.293
Table A.10 lists –8.280 (∆H° not exactly constant)
15-22
15.29 Catalytic gas generators are frequently used to decompose a liquid, providing a
desired gas mixture (spacecraft control systems, fuel cell gas supply, and so
forth). Consider feeding pure liquid hydrazine, N H , to a gas generator, from
2 4
which exits a gas mixture of N , H , and NH in chemical equilibrium at 100°C,
2 2
3
350 kPa. Calculate the mole fractions of the species in the equilibrium mixture.
Initially, 2 N2H4 → 1 N2 + 1 H2 + 2 NH3
then,
N2
initial
change
equil.
+
1
-x
(1-x)
K=
3
yN2yH2
(PP0)
-2
⇔
1
-3x
(1-3x)
2
yNH3
3 H2
=
2 NH3
2
+2x
(2+2x)
nTOTAL = (4-2x)
(2+2x)2(4-2x)2 350 -2
( )
(1-x)(1-3x)3 100
At 100 oC = 373.2 K, for NH3 use A.5
CP0 = 17.03×2.130 = 36.276
-0
hNH3 = -45 720 + 36.276(373.2-298.2) = -42 999 kJ/kmol
-s0 = 192.572 + 36.276 ln 373.2 = 200.71 kJ/kmol K
NH3
298.2
Using A.8,
0
∆H100 C = 2(-42 999) - 1(0+2188) - 3(0+2179) = -94 723 kJ
0
∆S100 C = 2(200.711) - 1(198.155) - 3(137.196) = -208.321 kJ/K
0
∆G100 C = ∆H0 - T∆S0 = -94 723 - 373.2(-208.321) = -16 978 kJ
∆G0
+16 978
ln K = - - =
= 5.4716
RT 8.3145×373.2
=>
Therefore,
[
(1+x)(2-x)
(1-3x)
]
2
237.84×3.52
1
=
= 182.096
(1-x)(1-3x)
16
By trial and error, x = 0.226
y = 0.2181
nN2 = 0.774 nNH3 = 2.452   N2
 y = 0.0908
nH2 = 0.322 nTOT = 3.518   H2
 yNH3 = 0.6911
K = 237.84
15-23
15.30 Acetylene gas at 25°C is burned with 140% theoretical air, which enters the
burner at 25°C, 100 kPa, 80% relative humidity. The combustion products form a
mixture of CO2, H 2O, N2, O2, and NO in chemical equilibrium at 2200 K, 100
kPa. This mixture is then cooled to 1000 K very rapidly, so that the composition
does not change. Determine the mole fraction of NO in the products and the heat
transfer for the overall process.
C2H2 + 3.5 O2 + 13.16 N2 + water →
2 CO2 + 1 H2O + 1 O2 + 13.16 N2 + water
water: PV = 0.8×3.169 = 2.535 kPa
nV = nA PV/PA = (3.5+13.16) 2.535/97.465 = 0.433
So, total H2O in products is 1.433.
a) reaction: N2 + O2 <-> 2 NO
change : -x
-x
+2x
at 2200 K, from A.10:
Equil. products:
K = 0.001 074
nCO2 = 2,
nH2O = 1.433, nO2 = 1-x,
nN2 = 13.16-x, nNO = 0+2x,
n TOT = 17.593
(2x)2
= 0.001 074 => x = 0.0576
K=
(1-x)(13.16-x)
yNO =
2×0.0576
= 0.006 55
17.593
b) Final products (same composition) at 1000 K
HR = 1(226 731) + 0.433(-241 826) = 122 020 kJ
HP = 2(-393 522 + 33 397) + 1.433(-241 826+26 000)
+ 0.9424(0+22 703) + 13.1024(0+21 463) + 0.1152(90 291+22 229)
= -713 954 kJ
QCV = HP - HR = -835 974 kJ
15-24
15.31 The equilibrium reaction as: CH4 ⇔ C + 2H2. has ln K = -0.3362 at 800 K and
lnK = -4.607 at 600 K. By noting the relation of K to temperature show how you
would interpolate ln K in (1/T) to find K at 700 K and compare that to a linear
interpolation.
A.10:
ln K = - 0.3362 at 800K
ln K = -4.607 at 600K
1
1
800
-1
700 800
700
lnK700 = lnK800 +
× (-4.607+0.3362) = -0.3362 +
×(-4.2708)
1
1
800
-1
600 800
600
= -2.1665
Linear interpolation:
lnK700 = lnK600 +
= -4.607 +
700 - 600
(lnK800 - lnK600)
800 -600
1
(-0.3362 + 4.607) = -2.4716
2
o
15.32 Use the information in Problem 15.31 to estimate the enthalpy of reaction, ∆H ,
at 700 K using Van’t Hoff equation with finite differences for the derivatives.
- 2
dlnK = [∆H°/RT ]dT
or
solve for ∆H°
- 2 dlnK - 2 ∆lnK
∆H° = RT
= RT
dT
∆T
2
= 8.31451 × 700 ×
-0.3362 + 4.607
= 86998 kJ/kmol
800 - 600
[Remark: compare this to A.8 values + A.5, A.9,
∆H° = HC + 2HH - HCH = 0.61 × 12 × (700-298) + 2 × 11730
2
4
– 2.254 × 16.04 × (700-298) - (-74873) = 86739 ]
15-25
15.33 A step in the production of a synthetic liquid fuel from organic waste matter is the
following conversion process: 1 kmol of ethylene gas (converted from the waste)
at 25°C, 5 MPa, and 2 kmol of steam at 300°C, 5 MPa, enter a catalytic reactor.
An ideal gas mixture of ethanol, ethylene, and water in chemical equilibrium
leaves the reactor at 700 K, 5 MPa. Determine the composition of the mixture and
the heat transfer for the reactor.
IG chem. equil.
o
1
C
H
mixture
25 C, 5 MPa
2 4
C2H5OH, C2H4, H2O
2 H 2O
700 K, 5 MPa
300 oC, 5 MPa
A.6 at ~ 500 K:
1 C2H4 + 1 H2O ⇔ 1 C2H5OH
init
1
2
0
CP0 C2H4 = 62.3
ch.
equil.
-x
(1-x)
-x
(2-x)
+x
x
0
a) ∆H700 K = 1(-235 000 + 115(700-298.2)) - 1(+52 467 + 62.3(700-298.2))
- 1(-241 826 + 14 190) = -38 656 kJ
0
∆S700 K = 1(282.444 + 115 ln
700
700
) - 1(219.330 + 62.3 ln
) - 1(218.739)
298.2
298.2
= -110.655 kJ/K
0
∆G700 K = ∆H0 - T∆S0 = +38 803 kJ
-∆G0
ln K = - = -6.667
RT
⇒
=>
K = 0.001 272 =
yC2H5OH
(PP )
yC2H4yH2O
-1
0
5.0
(1-xx )(3-x
)
= 0.001272×
= 0.0636
2-x
0.1
By trial and error: x = 0.0404 => C 2H5OH: n = 0.0404, y = 0.01371
C2H4: n = 0.9596, y = 0.3242,
b) Reactants:
H2O: n = 1.9596 , y = 0.6621
C2H4:
Tr = 298.2/282.4 = 1.056, P r = 5/5.04 = 0.992
- A.15: (h*-h) = 1.30×8.3145×282.4 = 3062 kJ
⇒ HC2H4 = 1(+52 467 - 3062) = +49 405 kJ
H2O, LIQ Ref. + St. Table:
HH2O = 2(-285830 + 18.015(2924.5-104.9)) = -470 070 kJ
HPROD = 0.0404(-235 000 + 115(700-298.2)) + 0.9596(+52 467
+ 62.3(700-298.2)) + 1.9596(-241 826 + 14 190) = -379 335 kJ
QCV = HP - HR = +41 330 kJ
15-26
15.34 Methane at 25°C, 100 kPa, is burned with 200% theoretical oxygen at 400 K, 100
kPa, in an adiabatic SSSF process, and the products of combustion exit at 100
kPa. Assume that the only significant dissociation reaction in the products is that
of carbon dioxide going to carbon monoxide and oxygen. Determine the
equilibrium composition of the products and also their temperature at the
combustor exit.
CH4 + 4O2 → CO2 + 2H2O + 2O2
Combustion:
Dissociation:
2 CO2 ⇔
initial
change
equil.
1
-2x
1-2x
2 CO + O2
,H2O
0
+2x
2x
2
0
2
2
+x
2+x
inert
nTOT = 5+x
2
yCOyO2 P
x 2 2+x P
(P0)= (0.5-x
) (5+x)(P0)
Equil. Eq'n: K = 2
y
CO2
or
x
K
(0.5-x
) (2+x
)
=
5+x (P/P0)
(1)
2
1st law: HP - HR = 0
(1-2x)(-393 522 + ∆hCO2) + 2x(-110 527 + ∆hCO)
+ 2(-241 826 + ∆hH2O) + (2+x)∆hO2 - 1(-74 873) - 4(3027) = 0
or (1-2x)∆hCO2 + 2x∆hCO + 2∆hH2O + (2+x)∆hO2 + 565 990x - 814 409 = 0
Assume TP = 3256 K. From A.10:
K = 0.6053
Solving (1) by trial & error, x = 0.2712
Substituting x and the ∆h values from A.8 (at 3256 K) into (2)
0.4576×168 821 + 0.5424×103 054 + 2×140 914
+ 2.2712×108 278 + 565 990×0.2712 - 814 409 ≈ 0 OK
TP = 3256 K & x = 0.2712
nCO2 = 0.4576, nCO = 0.5424, nH2O = 2.0, nO2 = 2.2712
15-27
15.35 Calculate the irreversibility for the adiabatic combustion process described in the
previous problem.
From solution of Prob. 15.34 , it is found that the product mixture consists of
0.4576 CO2, 0.5424 CO, 2.0 H2O & 2.2712 O2 at 3256 K, 100 kPa. The
reactants include
1 CH4 at 25 oC, 100 kPa and 4 O2 at 400 K, 100 kPa.
Reactants:
SR = 1(186.251) + 4(213.873) = 1041.74 kJ/K
Products:
ni
yi
-s0
i
- yiP
-R ln
P0
-*
Si
CO2
0.4576
0.0868
339.278
+20.322
359.600
CO
H 2O
0.5424
2.0
0.1029
0.3794
276.660
291.099
+18.907
+8.058
295.567
299.157
O2
2.2712
0.4309
287.749
+7.000
294.749
SP = 0.4576(359.600) + 0.5424(295.567) + 2.0(299.157)
+ 2.2712(294.749) = 1592.62 kJ/K
I = T0(SP-SR) - QCV = 298.15(1592.62 - 1041.74) - 0 = 164 245 kJ
15.36 In rich (too much fuel) combustion the excess fuel may be broken down to give
H2 and CO may form. In the products at 1200 K, 200 kPa the reaction called the
water gas reaction may take place: CO2 + H2 ⇔ H2O + CO
Find the equilibrium constant for this reaction from the elementary reactions.
(1)
1 CO2 + 1 H2 ⇔ 1 CO + 1 H2O
(2)
2 H 2O ⇔ 2 H 2 + 1 O 2,
(1) =
1
1
(3) - (2)
2
2
=>
(3)
2 CO2 ⇔ 2 CO + 1 O2
ln K = (ln K3 - ln K2)/2
From A.10 at 1200 K for reactions (2) and (3)
ln K2 = - 36.363, ln K3 = - 35.736
=> ln K = 0.3135
=> K = 1.3682
15-28
15.37 An important step in the manufacture of chemical fertilizer is the production of
ammonia, according to the reaction: N 2 + 3H2 ⇔ 2NH3
a. Calculate the equilibrium constant for this reaction at 150°C.
b. For an initial composition of 25% nitrogen, 75% hydrogen, on a mole basis,
calculate the equilibrium composition at 150°C, 5 MPa.
1N2 + 3H2 <=> 2NH3 at 150 oC
-o
hNH3 150 C = -45 720 + 2.13×17.031(150-25) = -41 186
423.2
-so
NH3 150 C = 192.572 + 2.13×17.031 ln 298.2 = 205.272
o
∆H150 C = 2(-41 186) - 1(0+3649) - 3(0+3636) = -96 929 kJ
0
∆S150 C = 2(205.272) - 1(201.829) - 3(140.860) = -213.865 kJ/K
0
∆G150 C = -96 929 - 423.2(-213.865) = -6421 kJ/kmol
ln K =
+6421
= 1.8248, K = 6.202
8.3144×423.2
b) nNH3 = 2x, nN2 = 1-x, nH2 = 3-3x
2
K=
yNH3
3
yN2yH2
P
P0
( )
2
-2
(2x)222(2-x)2 P
=
33(1-x)4
P0
( )
2
27
5 2
× 6.202 × ( ) = 26165
16
0.1
or
(1-xx ) (2-x
)
1-x
or
(1-xx )(2-x
) = 161.755
1-x
→ Trial & Error:
x = 0.9215
=
-2
NH3
n
1.843
y
0.8544
N2
0.0785
0.0364
H2
0.2355
0.1092
15-29
15.38 A space heating unit in Alaska uses propane combustion is the heat supply. Liquid
propane comes from an outside tank at -44°C and the air supply is also taken in
from the outside at -44°C. The airflow regulator is misadjusted, such that only
90% of the theoretical air enters the combustion chamber resulting in incomplete
combustion. The products exit at 1000 K as a chemical equilibrium gas mixture
including only CO2, CO, H2O, H2, and N2. Find the composition of the products.
Propane: Liquid, T1 = -44oC = 229.2 K
Air: T2 = -44oC = 229.2 K, 90% Theoretical Air
Products: T3 = 1000 K, CO2, CO, H2O, H2, N2
Theoretical Air:
C3H8 + 5O2 + 18.8N2 => 3CO2 + 4H2O + 18.8N2
90% Theoretical Air:
C3H8 + 4.5O2 + 16.92N2 => aCO2 + bCO + cH2O + dH2 + 16.92N2
Carbon:
a+b=3
Oxygen:
2a + b + c = 9 Where: 2 ≤ a ≤ 3
Hydrogen: c + d = 4
CO + H2O
Reaction:
↔ CO2
+ H2
Initial:
b
c
a
d
Change:
-x
-x
x
x
Equil:
b-x
c-x
a+x
d+x
Chose an Initial guess such as: a = 2, b = 1, c = 4, d = 0
Note: A different initial choice of constants will produce a different value for
x, but will result in the same number of moles for each product.
nCO2 = 2 + x, n CO = 1 - x, nH2O = 4 - x, nH2 = x, nN2 = 16.92
The reaction can be broken down into two known reactions to find K
(1)
2CO2 ↔ 2CO + O2 @ 1000 K ln(K1) = -47.052
(2)
2H2O ↔ 2H2 + O2 @ 1000 K ln(K2) = -46.321
For the overall reaction: lnK = (ln(K2) - ln(K1))/2 = 0.3655; K = 1.4412
K=
yCO2yH2  P 1+1-1-1 yCO2yH2
(2 + x)x
=
= 1.4412 =
P 
yCOyH2O  o
yCOyH2O
(1 − 4)(4 − x)
=>
x = 0.6462
nCO2 = 2.6462
nCO = 0.3538
nH2O = 3.3538
nH2 = 0.6462
nN2 = 16.92
15-30
15.39 One kilomole of carbon dioxide, CO2 , and 1 kmol of hydrogen, H2 at room
temperature, 200 kPa is heated to 1200 K at 200 kPa. Use the water gas reaction,
see problem 15.36, to determine the mole fraction of CO. Neglect dissociations of
H2 and O2.
1 CO2 + 1 H2 ⇔ 1 CO + 1 H2O
Initial
1
1
0
0
Shift
-x
-x
+x
+x
Total
1-x
1-x
x
x;
ntot = 2
yH2O = yCO = x/2, yH2 = yCO2 = (1-x)/2
From solution to problem 15.36, K = 1.3682
(x/2)(x/2)
x2
=K=
1-x 1-x
(1-x)2
( )( )
2
2
=>
x
= 1.1697
1-x
x = 1.1697 / 2.1697 = 0.5391
yH2O = yCO = x/2 = 0.27,
yH2 = yCO2 = (1-x)/2 = 0.23
15-31
15.40 Consider the production of a synthetic fuel (methanol) from coal. A gas mixture
of 50% CO and 50% H leaves a coal gasifier at 500 K, 1 MPa, and enters a
2
catalytic converter. A gas mixture of methanol, CO and H2 in chemical
equilibrium with the reaction: CO + 2H 2 <=> CH3OH leaves the converter at the
same temperature and pressure, where it is known that ln K = -5.119.
a. Calculate the equilibrium composition of the mixture leaving the converter.
b. Would it be more desirable to operate the converter at ambient pressure?
1 CO
1 H2
a)
Equil.
Mix
CONVERTER
CH3OH, CO, H2
500 K, 1 MPa
2 H2 ⇔
CH3OH
1
1
0
change
-x
-2x
+x
equil.
(1-x)
(1-2x)
x
Reaction:
CO
initial
K=
+
yCH3OH P -2
x 2-2x 2 P -2
2 ( 0) = (1-x)(1-2x) ( 0)
P
y y P
=>
CO H2
ln K = -5.119,
K = 0.005 98
x(1-x) 0.005 98 1 2
=
(0.1) = 0.1495
4
(1-2x)2
nCH3OH = x = 0.1045,
yCH3OH
x(1-x) K P 2
= ( )
(1-2x)2 4 P0
=>
x = 0.1045
n CO = 1-x = 0.8955,
= 0.0583,
yCO = 0.5000,
b) For P = 0.1 MPa
x(1-x) 0.005 98 0.1 2
=
(0.1) = 0.001 495
4
(1-2x)2
x is much smaller (~ 0.0015) not good
n H2 = 1 - 2x = 0.791
yH2 = 0.4417
15-32
15.41 Consider the following coal gasifier proposed for supplying a syngas fuel to a gas
turbine power plant. Fifty kilograms per second of dry coal (represented as 48 kg
C plus 2 kg H) enter the gasifier, along with 4.76 kmol/s of air and 2 kmol/s of
steam. The output stream from this unit is a gas mixture containing H2, CO, N2,
CH4, and CO2 in chemical equilibrium at 900 K, 1 MPa.
a. Set up the reaction and equilibrium equation(s) for this system, and calculate
the appropriate equilibrium constant(s).
b. Determine the composition of the gas mixture leaving the gasifier.
a) Entering the gasifier:
4 C + 1 H2 + 1 O2 + 3.76 N2 + 2 H2O
Since the chem. equil. outlet mixture contains no C, O2 or H2O, we must first
consider “preliminary” reaction (or reactions) to eliminate those substances in
terms of substances that are assumed to be present at equilibrium. One
possibility is
4 C + 1 O2 + 2 H2O → 4 CO + 2 H2
such that the "initial" composition for the equilibrium reaction is
4 CO + 3 H2 + 3.76 N2
(or convert equal amounts of CO and H2 to half of CH4 and CO2 - also
present at equilibrium. The final answer will be the same.)
reaction
2 CO
initial
change
equil.
4
-2x
(4-2x)
+
2 H2
3
-2x
(3-2x)
CH4
0
+x
x
+
CO2
0
+x
x
also N2
3.76
0
3.76
nTOT = 10.76 - 2x
For CH4 at 600 K (formula in Table A.6), CP0 = 52.22
At 900 K
-0
hCH4 = -74 873 + 52.22(900 - 298.2) = -43 446 kJ/kmol
-s0 = 186.251 + 52.22 ln (900 / 298.2) = 243.936 kJ/kmol K
CH4
(The integrated-equation values are -43 656 and 240.259)
0
∆H900 K = 1(-43 447) + 1(-393 522 + 28 030)
- 2(-110 527 + 18 397) - 2(0 + 17 657) = -259 993 kJ
0
∆S900 K = 1(243.935) + 1(263.646)
- 2(231.074) - 2(163.060) = -280.687 kJ/K
0
∆G900 K = -259 993 - 900(-280.687) = -7375 kJ
15-33
ln K =
b)
+7375
= 0.9856,
8.3145×900
K=
K = 2.679
yCH4yCO2 P -2 x×x×(10.76-2x)2 P -2
(P0) = (4-2x)2(3-2x)2 (P0)
2
2
y y
CO H2
or
x(10.76-2x) P
1
= 0 K=
2.679 = 16.368
(4-2x)(3-2x) P
0.1
By trial & error,
x = 1.2781
nCO = 1.444,
nH2 = 0.444, nCH4 = nCO2 = 1.278,
nN2 = 3.76
yCO = 0.176,
yH2 = 0.054, yCH4 = yCO2 = 0.156,
yN2 = 0.458
15-34
15.42 Ethane is burned with 150% theoretical air in a gas turbine combustor. The
products exiting consist of a mixture of CO , H O, O , N , and NO in chemical
2 2
2 2
equilibrium at 1800 K, 1 MPa. Determine the mole fraction of NO in the
products. Is it reasonable to ignore CO in the products?
Combustion:
C2H6 + 5.25 O2 + 19.74 N2 → 2 CO2 + 3 H2O + 1.75 O2 + 19.74 N2
Products at 1800 K, 1 MPa
Equilibrium mixture: CO2, H2O, O2, N2, NO
+
N2
initial
change
equil.
19.74
-x
19.74-x
O2
2 NO
1.75
-x
1.75-x
0
+2x
2x
Equil. comp. nCO2 = 2, nO2 = 1.75-x, nNO = 2x , nH2O = 3, nN2 = 19.74-x
2
yNO
P 0
4x2
( ) = (19.74-x)(1.75-x)
K = 1.192×10 =
yN2yO2 P0
-4
Solving, x = 0.031 75
yNO =
2×0.031 75
= 0.0024
26.49
2 CO2 ⇔
b)
initial
2
2 CO +
O2
0
0
change -2a
+2a
+2x
equil. 2-2a
2a
2x
2
yCOyO2 P 1
2a 2 1.75-x+a 1
-8
K = 4.194×10 = 2
(P0) =(2-2a
) ( 26.49+a )×0.1
y
CO2
This equation should be solved simultaneously with the equation solved in
part a) (modified to include the unknown a). Since x was found to be small
and also a will be very small, the two are practically independent. Therefore,
use the value x = 0.031 75 in the equation above, and solve for a.
a
75+a
0.1
(1-a
) (1.75-0.031
)
= ( )×4.194×10-8
26.49+a
1.0
2
Solving, a = 0.000 254 or yCO = 1.92×10-5 negligible for most applications.
15-35
15.43 One kilomole of liquid oxygen, O2, at 93 K, and x kmol of gaseous hydrogen, H ,
2
at 25°C, are fed to an SSSF combustion chamber. x is greater than 2, such that
there is excess hydrogen for the combustion process. There is a heat loss from the
chamber of 1000 kJ per kmol of reactants. Products exit the chamber at chemical
equilibrium at 3800 K, 400 kPa, and are assumed to include only H2O, H2, and O.
a. Determine the equilibrium composition of the products and also x, the amount of
H2 entering the combustion chamber.
b. Should another substance(s) have been included in part (a) as being present in
the products? Justify your answer.
x H2 + 1 O2 → 2 H2O + (x-2) H2
(1) 1 H2O ⇔ 1 H2 + 1 O
shift
-a
+a
Equil 2-a
x-2+a
+a
and a > 0
a
a < 2 and
(2) 2 H2O ⇔ 2 H2 + 1 O2
(3)
ln K2 = -1.906
1 O2 ⇔ 2 O
ln K3 = -0.017
ln K1 = 0.5( ln K2 + ln K3 ) = -0.9615
Equil.:
ntot = x + a
=>
K1 = 0.3823
K1
(x-2+a)a 0.3823
=
=
= 0.95575
1
(2-a)(x+a)
4
(P/Po)
1st law: Q + HR = HP , Q = (1+x)(-1000) kJ
-*
Table A.8: ∆hIG = -5980 kJ/kmol
[or = 0.922 × 32(93 - 298.2) = - 6054 kJ/kmol ]
Fig. D.2: Tr = 93/154.6 = 0.601, ∆hf = -5.16×R×154.6 = -6633
-*
HR = x(0 + 0) + 1(0 + ∆hIG + ∆hf) = 1(-5980 - 6633) = - 12613 kJ
HP = (2-a)(-241 826 + 171 981) + (x-2+a)(0 + 119077)
+ a(249170 + 73424) = 119077 x + 511516 a - 377844
= Q + HR = -1000 - 1000 x - 12613
Rearrange eq. to:
x + 4.2599 a = 3.03331
Substitute it into the equilibrium eq.:
Solve a = 0.198, LHS = 0.09547,
yH2O =
2-a
= 0.755,
x+a
yH2 =
Other substances and reactions:
H2 <=> 2 H,
(1.03331 + 5.2599 a) a
= 0.095575
(2-a)(3.03331-3.2599 a)
x = 2.1898
x-2+a
= 0.162,
x+a
yO =
a
= 0.083
x+a
2 H2O <=> H2 + 2 OH,
: ln K = 0.201,
O2 <=> 2 O, :
All are significant as K's are of order 1.
ln K = -0.984,
ln K = -0.017
15-36
15.44 Butane is burned with 200% theoretical air, and the products of combustion, an
equilibrium mixture containing only CO , H O, O , N , NO, and NO , exit from
2 2
2 2
2
the combustion chamber at 1400 K, 2 MPa. Determine the equilibrium
composition at this state.
Combustion:
C4H10 + 13 O2 + 48.9 N2 → 4 CO2 + 5 H2O + 6.5 O2 + 48.9 N2
Dissociation:
N2 + O2 ⇔ 2 NO
1)
change -a
-a
N2 + 2O2 ⇔ 2 NO2
2)
+2a
change
-b
-2b
At equilibrium:
nN2 = 48.9-a-b
nH2O = 5
nNO = 2a
nCO2 = 4
nNO2 = 2b
nO2 = 6.5-a-2b
+2b
nTOT = 64.4-b
At 1400 K, from A.10:
K1 =
K1 = 3.761×10-6, K2 = 9.026×10-10
(2a)2
;
(48.9-a-b)(6.5-a-2b)
K2 =
(2b)2(64.4-b)
P -1
(
)
2
(6.5-a-2b) (48.9-a-b) P0
As K1 and K2 are both very small, with K2 << K1, the unknowns a & b will
both be very small, with b << a. From the equilibrium eq.s, for a first trial
a~
1
2
1
K1×48.9×6.5 ~ 0.0173 ;b ~ ×6.5
2
2 48.9
K 2× ×
~ 0.000 38
0.1 64.4
Then by trial & error,
3.761×10-6
a2
=
= 0.940 25×10-6
(48.9-a-b)(6.5-a-2b)
4
2
b (64.4-b)
=
(6.5-a-2b)2(48.9-a-b)
9.026×10-10×(
4
2
)
0.1
= 45.13×10-10
Solving, a = 0.017 27, b = 0.000 379
nCO2 = 4 ,
nH2O = 5 ,
yCO2 = 0.062 11 , yH2O = 0.077 64 ,
nNO = 0.034 54 ,
nNO2 = 0.000 76
yNO = 0.000 55 ,
yNO2 = 0.000 01
nN2 = 48.882 ,
nO2 = 6.482 ,
yN2 = 0.759 04 , yO2 = 0.100 65
15-37
15.45 A mixture of 1 kmol water and 1 kmol oxygen at 400 K is heated to 3000 K, 200
kPa, in an SSSF process. Determine the equilibrium composition at the outlet of
the heat exchanger, assuming that the mixture consists of H2O, H2, O2, and OH.
Reactions and equilibrium eq'ns the same as in example 15.8 (but different
initial composition).
At equil.: nH2O = 1-2a-2b, nH2 = 2a+b, nO2 = 1+a
nOH = 2b,
nTOT = 2+a+b
Since T = 3000 K is the same, the two equilibrium constants are the same:
K1 = 0.002 062,
K2 = 0.002 893
The two equilibrium equations are
K1 =
2a+b
1+a P
( );
1-2a-2b 2+a+b P0
K2 =
2 P
2a+b
2b
(
)
( )
2+a+b 1-2a-2b P0
which must be solved simultaneously for a & b. If solving manually, it
simplifies the solution to divide the first by the second, which leaves a
quadratic equation in a & b - can solve for one in terms of the other using the
quadratic formula (with the root that gives all positive moles). This reduces
the problem to solving one equation in one unknown, by trial & error.
Solving =>
b = 0.116,
a = -0.038
=>
nH2O = 0.844, nH2 = 0.0398, nO2 = 0.962, nOH = 0.232, nTOT = 2.0778
yH2O = 0.4062, yH2 = 0.0191, yO2 = 0.4630, yOH = 0.1117
15-38
15.46 One kilomole of air (assumed to be 78% nitrogen, 21% oxygen, and 1% argon) at
room temperature is heated to 4000 K, 200 kPa. Find the equilibrium composition
at this state, assuming that only N2, O2, NO, O, and Ar are present.
1 kmol air (0.78 N2, 0.21 O2, 0.01 Ar) heated to
4000 K, 200 kPa.
Equil.:
1)
N2 + O2 ⇔ 2 NO
change -a -a
nN2 = 0.78-a
+2a
nO2 = 0.21-a-b
nAr = 0.01
2)
O2 ⇔ 2 O
nO = 2b
change -b +2b
nNO = 2a
ntot = 1+b
4a2
200 0
(
)
K1 = 0.0895 =
(0.78-a)(0.21-a-b) 100
4b2
200
K2 = 2.221 =
(
)
(1+b)(0.21-a-b) 100
Divide 1st eq'n by 2nd and solve for a as function(b), using
K1 P
X = ( 0)= 0.0806
K2 P
Get
Xb2
a=
[-1+
2(1+b)
1+
4×0.78(1+b)
Xb2
(1)
]
Also
K2
b2
=
= 0.277 63
(1+b)(0.21-a-b) 4(P/P0)
Assume
b = 0.1280
From (1), get a = 0.0296
Then, check a & b in (2) ⇒ OK
Therefore,
nN2 = 0.7504 nO = 0.2560 yN2 = 0.6652 yO = 0.2269
nO2 = 0.0524 nNO = 0.0592 yO2 = 0.0465 yNO = 0.0525
nAr = 0.01
yAr = 0.0089
(2)
15-39
15.47 Acetylene gas and x times theoretical air (x > 1) at room temperature and 500 kPa
are burned at constant pressure in an adiabatic SSSF process. The flame
temperature is 2600 K, and the combustion products are assumed to consist of N 2,
O2, CO2, H2O, CO, and NO. Determine the value of x.
Combustion:
C2H2 + 2.5x O2 + 9.4x N2 → 2 CO2 + H2O + 2.5(x-1)O2 + 9.4xN2
Eq. products 2600 K, 500 kPa: N2, O2, CO2, H2O, CO & NO
2 Reactions:
1)
2 CO2 ⇔ 2 CO + O2
change -2a
Equil. Comp.:
+2a
+a
nN2 = 9.4x-b ,
nO2 = 2.5x - 2.5 + a - b ,
N2 + O2 ⇔ 2 NO
2)
change -b
nH2O = 1 ,
-b
+2b
nCO = 2a ,
nNO = 2b
nCO2 = 2 - 2a , nTOT = 11.9x + 0.5 + a
At 2600 K, from A.10: K1 = 3.721×10-3,
K2 = 4.913×10-3
K1
3.721×10-3
a 2 2.5x-2.5+a-b
=
=
(
) (11.9x+0.5+a )
1-a
5
(P/P0)
(2b)2
K2 = 4.913×10 =
(9.4-b)(2.5x-2.5+a-b)
-3
Also, from the 1st law: HP - HR = 0 where
HR = 1(+226 731) + 0 + 0 = +226 731 kJ
HP = (9.4x-b)(0+77 963) + (2.5x-2.5+a-b)(0+82 225)
+ (2-2a)(-393 522+128 074) + 1(-241 826+104 520)
+ 2a(-110 527+78 679) + 2b(90 291+80 034)
Substituting,
EQ1:
988 415x + 549 425a + 180 462b - 1 100 496 = 0
which results in a set of 3 equations in the 3 unknowns x, a, b. Assume x =
1.07 Then
EQ2:
3.721×10-3
a 2 0.175+a-b
=( ) (
)
1-a 13.233+a
5
4.913×10-3
b2
EQ3:
=
4
(10.058-b)(0.175+a+b)
Solving, a = 0.1595,
b = 0.0585
Then checking in EQ1,
988 415×1.07 + 549 425×0.1595 + 180 462×0.0585 - 1 100 496 ≈ 0
Therefore, x = 1.07
15-40
15.48 One kilomole of water vapor at 100 kPa, 400 K, is heated to 3000 K in a constant
pressure SSSF process. Determine the final composition, assuming that H 2O, H2,
H, O2, and OH are present at equilibrium.
Reactions:
1)
2 H 2O ⇔ 2 H 2 + O2
change -2a
3)
+2a
2)
2 H2O ⇔ H2 + 2 OH
+a change
-2b
+b
+2b
H2 ⇔ 2 H
change -c
+2c
At equilibrium (3000 K, 100 kPa)
nH2O =
1-2a-2b
nO2 =
a
nH =
2c
nH2 =
nTOT =
1+a+b+c
2a+b-c
nOH =
2b
K1
2.062×10-3
2a+b-c 2
a
=
=
(
)
(
)
0
1-2a-2b
1+a+b+c
1
(P/P )
K2
(P/P0)
=
2
2.893×10-3
2a+b-c
2b
=(
)(
)
1+a+b+c 1-2a-2b
1
K3
2.496×10-2
(2a)2
=
=
1
(2a+b-c)(1+a+b+c)
(P/P0)
These three equations must be solved simultaneously for a, b & c:
a = 0.0622,
and
b = 0.0570,
c = 0.0327
nH2O = 0.7616 yH2O = 0.6611
nH2 = 0.1487 yH2 = 0.1291
nO2 = 0.0622 yO2 = 0.0540
nOH = 0.1140 yOH = 0.0990
nH = 0.0654 yH = 0.0568
15-41
15.49 Operation of an MHD converter requires an electrically conducting gas. It is
proposed to use helium gas “seeded” with 1.0 mole percent cesium, as shown in
Fig. P15.49. The cesium is partly ionized (Cs ⇔ Cs` + e−) by heating the mixture
to 1800 K, 1 MPa, in a nuclear reactor to provide free electrons. No helium is
ionized in this process, so that the mixture entering the converter consists of He,
Cs, Cs`, and e−. Determine the mole fraction of electrons in the mixture at 1800 K,
where ln K = 1.402 for the cesium ionization reaction described.
+
Reaction: CS ⇔ CS + e-,
init
0.01
change
-x
Also He
0
0
0.99
+x
+x
0
x
x
Equil (0.01-x)
ln K = 1.402
=>
0.99 ;
total: 1 + x
K=
ye-yCs+ P
x
=
0
0.01-x
yCs P
or
x
x
(0.01-x
)(1+x
)= 4.0633 / (1/0.1) = 0.40633
( )(
Quadratic equation:
K = 4.0633
x
)(1+x
)(PP )
0
x = 0.009767
⇒
ye- = x/(1+x) = 0.00967
15.50 One kilomole of argon gas at room temperature is heated to 20000 K, 100 kPa.
Assume that the plasma in this condition consists of an equilibrium mixture of Ar,
Ar`, Ar``, and e− according to the simultaneous reactions
1) Ar ⇔ Ar+ + e-
2) Ar+ ⇔ Ar++ + e-
The ionization equilibrium constants for these reactions at 20000 K have been
calculated from spectroscopic data as ln K1 = 3.11 and ln K2 = -4.92. Determine
the equilibrium composition of the plasma.
1) Ar ⇔ Ar+ + e-
2) Ar+ ⇔ Ar++ + e-
ch. -a
ch. -b
+a
+a
+b
+b
Equil. Comp.: nAr = 1-a, nAr+ = a-b, nAr++ = b, ne- = a+b, nTOT = 1+a+b
K1 =
yAr+ ye- P
(a-b)(a+b)
=
(1) = 22.421
0
(1-a)(1+a+b)
yAr P
K2 =
yAr++ye- P
b(a+b)
=
(1) = 0.0073
0
(a-b)(1+a+b)
yAr+ P
( )
( )
By trial & error:
a = 0.978 57,
b = 0.014 13
nAr = 0.02143, nAr+ = 0.96444, nAr++ = 0.01413, ne- = a0.9927
yAr = 0.0107,
yAr+ = 0.484,
yAr++ = 0.0071,
ye- = 0.4982
15-42
15.51 Plot to scale the equilibrium composition of nitrogen at 10 kPa over the
temperature range 5000 K to 15000 K, assuming that N 2 , N, N` , and e − are
present. For the ionization reaction N ⇔ N` + e−, the ionization equilibrium
constant K has been calculated from spectroscopic data as
T [K]
10000 12 000 14 000 16 000
100K 6.26 10-2 1.51
15.1
92
1) N2 ⇔ 2N
2) N ⇔ N+ + e-
change -a
+2a
change -b
+b +b
Equil. Comp.: nN2 = 1-a, nN = 2a-b, nN+ = b,
ne- = b
2
yN
(2a-b)
(PP )= (1-a)(1+a+b)
(PP )
EQ1:
K1 =
EQ2:
yN+ ye- P
b2
P
K2 =
=
0
yN P
(2a-b)(1+a+b) P0
yN2
0
0
( )
For T < 10 000 K:
( )
b ~ 0 so neglect EQ2:
To extrapolate K1 above 6000 K:
⇒ K1 =
ln K1 ≈ 16.845 -
4a2 10
( )
(1-a2) 100
118 260
T
(from values at 5000 K & 6000 K)
T(K)
5000
6000
7000
8000
10000
K1
0.0011
0.0570
0.9519
7.866
151.26
For T > 10 000 K:
T(K)
a
a ≈ 1.0
K2
0.0524
0.3532
0.8391
0.9755
0.9987
=>
b
yN
0.0996
0.5220
0.9125
0.9876
0.9993
yN2
0.9004
0.4780
0.0875
0.0124
0.0007
b2
10
b2
K2 =
( ) = (4-b2) 0.1
(2-b)(2+b) 100
yN
yN2
10 000
6.26×10-4
0.1577
0.8538
0.0731
12 000
1.51×10-2
0.151
0.92
0.7244
0.4862
0.2659
1.5512
1.8994
0.1264
0.0258
0.4368
0.4871
14 000
16 000
Note that b ≈ 0 is not a very good approximation in the vicinity of 10 000 K.
In this region, it would be better to solve the original set simultaneously for a
& b. The answer would be approximately the same.
15-43
15.52 Hydrides are rare earth metals, M, that have the ability to react with hydrogen to
form a different substance MH with a release of energy. The hydrogen can then
x
be released, the reaction reversed, by heat addition to the MH . In this reaction
x
only the hydrogen is a gas so the formula developed for the chemical equilibrium
is inappropriate. Show that the proper expression to be used instead of Eq. 15.34
is
ln (PH2/Po) = ∆G0/RT
when the reaction is scaled to 1 kmol of H2.
1
M + x H2 <=> MH x
2
At equilibrium GP = GR , assume g of the solid is a function of T only.
0
0
-0
g-MHx = hMHx - Ts-MHx = g-MHx ,
0
0
-0
g-M = hM - Ts-M = g-M
0
0
-0
g-H2 = hH2 - Ts-H2 + RT ln(PH2/Po) = g-H2 + RT ln(PH2/Po)
GP = GR:
1
0
1
0
g-MHx = g-M + x g-H2 = g-M + x[g-H2 + RT ln(PH2/Po)]
2
2
0
0
0
0
0
∆G0 = g-MHx - g-M - x g-H2/2 = g-MHx - g-M
Scale to 1 mole of hydrogen
~ 0 = (g-0 - g-0 )/(x/2) = R
∆G
T ln(PH2/Po)
MHx
M
which is the desired result.
15-44
Advanced Problems
15.53 Repeat Problem 15.1 using the generalized charts, instead of ideal gas behavior.
(Z1-Z2) = 5000 m, P1 = 15 MPa
Z1
1
T = 40 oC = const
313.2
15
Tr =
= 1.03, Pr1 =
= 2.033
304.1
7.38
Equilibrium: -wREV = 0 = ∆g + ∆PE
CO 2
Z2
2
g2 - g1 = h2 - h1 -T(s2 - s1) = g(Z1 - Z2) =
9.807×5000
= 49.04 kJ/kg
1000
From Figures D.2 and D.3,
*
*
h1 - h1 = RTc× 3.54 = 203.4 kJ/kg ; s1 - s1 = R × 2.61 = 0.4931 kJ/kg K
*
*
h2 - h 1 = 0 ;
*
*
s2 - s1 = 0 - R ln(P2 /P1) = 0.18892 ln( P2 /15)
Trial and error. Assume P2 = 55 MPa (Pr2 = 55/7.38 = 7.45)
*
*
h2 - h2 = RTc× 3.60 = 206.8 kJ/kg ; s2 - s2 = R × 2.14 = 0.4043 kJ/kg K
∆g = -206.8 + 0 + 203.4 - 313.2[-0.4043 - 0.18892 ln(55/15) + 0.4931] = 45.7
Too low so assume P2 = 60 MPa (Pr2 = 60/7.38 = 8.13)
*
*
h2 - h2 = RTc× 3.57 = 205.1 kJ/kg ; s2 - s2 = R × 2.11 = 0.3986 kJ/kg K
∆g = -205.1 + 0 + 203.4 - 313.2[-0.3986 - 0.18892 ln(60/15) + 0.4931] = 50.7
Make linear interpolation
⇒
P2 = 58 MPa
15.54 Derive the van’t Hoff equation given in problem 15.28, using Eqs.15.12 and
15.15. Note: the d(g-/T) at constant P for each component can be expressed using
the relations in Eqs. 13.18 and 13.19.
0
0
0
0
Eq. 15.12: ∆G0 = vC g-C + vD g-D - vA g-A - vB g-B
Eq. 15.15: lnK = ∆G0/RT
Eq. 13.19:
∆G0 = ∆Η° - T ∆S0
1
dlnK
d ∆G0
1 dG0 ∆G0
dG0
0
=(
)=+
=
[∆G – T
]
dT
dT R-T
RT dT R-T2 R-T2
dT
1
dg0
0
= - 2 [∆G + T ∆S ]
used Eq.13.19
= - -s
dT
RT
1
= - 2 ∆Η0
RT
15-45
15.55 A coal gasifier produces a mixture of 1 CO and 2H2 that is then fed to a catalytic
converter to produce methane. A chemical-equilibrium gas mixture containing
CH4, CO, H2, and H2O exits the reactor at 600 K, 600 kPa. Determine the mole
fraction of methane in the mixture.
CO
+ 3H2 ↔ CH4 + H2O
Initial
1
2
0
0
Change
-x
-3x
x
x
Equil.
1-x
2-3x
x
x
n = (1 - x) + (2 - 3x) + x + x = 3 - 2x
yCH4yH2O  P (1+1-1-3) = -2
x2
 P -2
K=
=


 
3
P
(1-x)(2-3x)3 Po
yCOyH2  o
−
lnK = - ∆Go/R; ∆Go = ∆Ho - T∆So
−o −
−o
−
HP = nCH4 [h f + CP(T - To)] + nH2O (h f + ∆h )
= [-74873 + 2.254×16.04(600 - 298.15)] + (-241826 + 10499) = -295290
−o
−
HR = nCO (h f + ∆h )
+ nH2 (−h f + ∆h−) = 1(-110527 + 8942) + 3(0 + 8799)
o
= -75188 kJ
o
∆Η600 = HP - HR = -295290 - (-75188) = -220102 kJ
o
o
−
(s− T)CH4 = −s To + CPln(T/To) = 186.251 + 2.254×16.04 ln(600/298.2) = 211.549
o
(s− T )H2O = 213.051 kJ/kmol-K;
SP = 424.6 kJ/K
o
o
(s− T)CO = 218.321 kJ/kmol-K, (s− T)H2 = 151.078 kJ/kmol-K
o
o
o
o
o
∆S600 = SP - SR = (n−s T)CH4 + (n−s T )H2O - (ns− T)CO - (ns− T)H2
= (211.549 + 213.051) - (218.321 + 3 × 151.078) = -246.955 kJ/K
∆Go = ∆Ho - T∆So = -220 102 - 600(-246.955) = -71929 kJ,
K = 1.827×106
lnK = -(-71915)/(8.31451×600) = 14.418
=>
Solve for x,
yCH4 = 0.4
x = 0.6667,
ntot = 1.6667,
15-46
15.56 Dry air is heated from 25°C to 4000 K in a 100-kPa constant-pressure process.
List the possible reactions that may take place and determine the equilibrium
composition. Find the required heat transfer.
Air assumed to be 21% oxygen and 79% nitrogen by volume.
From the elementary reactions we have at 4000 K (A.10)
2
(1) O2 <=> 2 O
K1 = 2.221 = y O/yO2
(2) N2 <=> 2 N
K2 = 3.141 × 10-6 = yN/yN2
2
2
(3) N2 + O2 <=> 2 NO
K3 = 0.08955 = yNO/yN2 yO2
Call the shifts a,b,c respectively so we get
nO2 = 0.21-a-c, nO = 2a, nN2 = 0.79-b-c, nN = 2b,
nNO = 2c, ntot = 1+a+b
From which the molefractions are formed and substituted into the three
equilibrium equations. The result is
2
K1 = 2.221 = y O/yO2 = 4a2/[(1+a+b)(0.21-a-c)]
2
K2 = 3.141 × 10-6 = yN/yN2 = 4b2/[(1+a+b)(0.79-b-c)]
2
K3 = 0.08955 = y NO/yN2 yO2 = 4c2/[(0.79-b-c)(0.21-a-c)]
which gives 3 eqs. for the unknowns (a,b,c). Trial and error assume b = c = 0
solve for a from K 1 then for c from K3 and finally given the (a,c) solve for b
from K2. The order chosen according to expected magnitude K1>K3>K2
a = 0.15, b = 0.000832, c = 0.0244 =>
nO2 = 0.0356, nO = 0.3, nN2 = 0.765, nN = 0.00167, nNO = 0.049
Q = Hex - Hin = nO2∆hO2 + nN2∆hN2 + nO(hfO + ∆hO)
+ nN(hfN + ∆hN) + nNO(hfNO + ∆hNO) - 0
= 0.0356 × 138705 + 0.765 × 130027 + 0.3(249170 + 77675)
+ 0.00167(472680 + 77532) + 0.049(90291 + 132671)
= 214 306 kJ/kmol air
[If no reac. Q = nO2∆hO2 + nN2∆hN2 = 131 849 kJ/kmol air]
15-47
15.57 Methane is burned with theoretical oxygen in an SSSF process, and the products
exit the combustion chamber at 3200 K, 700 kPa. Calculate the equilibrium
composition at this state, assuming that only CO , CO, H O, H , O , and OH are
2
2
2 2
present.
CH4 + 2 O2 → CO2 + 2 H2O
Combustion:
Dissociation reactions:
2 H 2O ⇔ 2 H 2 + O2
1)
change -2a
+2a +a
2)
2 H2O ⇔ H2 + 2 OH
change -2b
+b +2b
2 CO2 ⇔ 2 CO + O2
3)
change -2c
+2c +c
At equilibrium:
NH2O =
2-2a-2b
nO2 =
a+c
nCO2 =
1-2c
NH2 =
nOH =
2b
nCO =
2c
nTOT =
3+a+b+c
2a+b
Products at 3200 K, 700 kPa
K1 = 0.007 328 = (
2a+b 2
a+c
700
)
(
)(
)
2-2a-2b 3+a+b+c 100
K2 = 0.012 265 = (
2
2b
2a+b
700
)
(
)(
)
2-2a-2b 3+a+b+c 100
K3 = 0.426 135 = (
2c 2
) ( a+c )(700)
1-2c 3+a+b+c 100
These 3 equations must be solved simultaneously for a, b, & c. If solving by
hand divide the first equation by the second, and solve for c = fn(a,b). This
reduces the solution to 2 equations in 2 unknowns. Solving,
a = 0.024, b = 0.1455, c = 0.236
Substance:
H 2O
H2
O2
OH
CO2
CO
n
1.661
0.1935
0.260
0.291
0.528
0.472
y
0.4877
0.0568
0.0764
0.0855
0.1550
0.1386
15-48
ENGLISH UNIT PROBLEMS
15.58ECarbon dioxide at 2200 lbf/in.2 is injected into the top of a 3-mi deep well in
connection with an enhanced oil recovery process. The fluid column standing in
the well is at a uniform temperature of 100 F. What is the pressure at the bottom
of the well assuming ideal gas behavior?
Z1
(Z1-Z2) = 3 miles = 15 840 ft
1
P1 = 2200 lbf/in2, T = 100 F = const
CO 2
Z2
Equilibrium and ideal gas beahvior
2
-wREV = 0 = ∆g + ∆PE = RT ln (P2/P1) + g(Z2-Z1) = 0
ln (P2/P1) =
32.2×15 840
= 0.8063
32.2×35.1×559.7
P2 = 2200 exp(0.8063) = 4927 lbf/in2
15.59E Calculate the equilibrium constant for the reaction O2 <=> 2O at temperatures of
537 R and 10 000 R.
Find the change in Gibbs function at the two T’s from Table C.7:
∆H0 = 2(249 170+121 264) - (0+224 210) = 516 658 kJ/kmol
∆S0 = 2(224.597) -1(313.457) = 135.737 kJ/kmol K
∆G0 = 516 658 - 6000×135.737 = -297 764 kJ/kmol
ln K =
537 R:
+297 764
= +5.969
8.3145×6000
-0
-0
∆H0 = 2hf O - 1hf O2 = 2 × 107 124 = 214 248 ;
0
0
∆S0 = 2s-O - 1s-O2 = 2 × 38.442 - 48.973 = 27.911
∆G0 = ∆H0 - T∆S0 = 214 248 - 537 × 27.911 = 199 260
∆G0
+199 260
- = - 185.85
ln K = - - =
RT 1.98589×537
-0
-0
10 000 R: ∆H0 = 2hf O - 1hf O2 = 2 (107 124 + 47897) - 87997 = 222 045;
0
0
∆S0 = 2s-O - 1s-O2 = 2 × 53.210 - 74.034 = 32.386
∆G0 = ∆H0 - T∆S0 = 222 045 - 10 000 × 32.386 = -101 815
ln K = - ∆G0/RT = 101 815 / (1.98589 × 10 000) = +5.127
15-49
15.60E Pure oxygen is heated from 77 F to 5300 F in an SSSF process at a constant
pressure of 30 lbf/in.2. Find the exit composition and the heat transfer.
The only reaction will be the dissociation of the oxygen
O2 ⇔ 2O ;
K(5300 F) = K(3200 K) = 0.046467
Look at initially 1 mol Oxygen and shift the above reaction with x
nO2 = 1 - x; nO = 2x; ntot = 1 + x; yi = ni/ntot
y2O P
4x2 1 + x
8x2
K=
( )2-1 =
2
=
y02 Po
(1 + x)2 1 - x
1 - x2
K/8
⇒ x = 0.07599; y02 = 0.859; y0 = 0.141
1 + K/8
q = n02exh02ex + n0exhOex - h02in = (1 + x)(y 02h02 + y0hO) - 0
h02 = 45 581; hO = 107124 + 26125 = 133 249
q = 1.076(0.859 × 45581 + 0.141 × 133249) = 62345 Btu/lbmol O2
q = q/32 = 1948 Btu/lbm (=1424 if no dissociation)
x2 =
15.61E Pure oxygen is heated from 77 F, 14.7 lbf/in.2 to 5300 F in a constant volume
container. Find the final pressure, composition, and the heat transfer
As oxygen is heated it dissociates
O2 ⇔ 2O
C. V. Heater:
Per mole O2:
ln Keq = -3.069
U2 - U1 = 1Q2 = H2 - H1 - P2v + P1v
- 1q2 = h2 - h1 + R(T1 - (n2/n1)T2)
Shift x in reaction final composition:
n1 = 1
(1 - x)O2 + 2xO
n2 = 1 - x + 2x = 1 + x
yO2 = (1 - x)/(1 + x) ;
yO = 2x/(1 + x)
Ideal gas and V2 = V1 ⇒ P2 = P1n2T2/n1T1 ⇒ P2/Po = (1 + x)T2/T1
Keq = e
⇒
-3.069
y2O P2
2x 2 1 + x 1 + x T2
=
( )=(
) (
)(
)( )
1+x 1-x
1
y02 Po
T1
4x2 T1 -3.069
= e
= 0.00433 ⇒ x=0.0324
1 - x T2
(nO2)2 = 0.9676,
(nO)2 = 0.0648,
n2 = 1.0324
1q2 = 0.9676(45581) + 0.0648(107124 + 26125) - Ø
+ 1.98589(536.67 - 1.0324 × 5760) = 41996 Btu/lbmol O2
yO2=0.9676 / 1.0324 = 0.937;
yO =0.0648/1.0324 = 0.0628
15-50
15.62E Air (assumed to be 79% nitrogen and 21% oxygen) is heated in an SSSF process
at a constant pressure of 14.7 lbf/in.2, and some NO is formed. At what
temperature will the mole fraction of NO be 0.001?
heated at 14.7 lbf/in2, forms NO
0.79N2 + 0.21O2
At exit, yNO = 0.001
N2 + O2 ⇔ 2 NO
nN2 = 0.79 - x
-x
nO2 = 0.21 - x
-x
+2x
nNO = 0
+ 2x
n = 1.0
2x
⇒
1.0
yNO = 0.001 =
x = 0.0005
2
K=
yNO
⇒
nN2 = 0.7895,
nO2 = 0.2095
-6
yN2yO2
10
(PP0)0 = 0.7895×0.2095
= 6.046×10-6
From Table A.10,
or
ln K = -12.016
T = 1444 K = 2600 R
15.63E The combustion products from burning pentane, C5H12, with pure oxygen in a
stoichiometric ratio exists at 4400 R. Consider the dissociation of only CO2 and
find the equilibrium mole fraction of CO.
C5H12 + 8 O2 → 5 CO2 + 6 H2O
2 CO2 ⇔ 2 CO + 1 O2
At 4400 R,
ln K = -7.226
K = 7.272x10-4
Initial 5
Change
Equil. 5-2z
0
-2z
2z
Assuming P = Po = 0.1 MPa,
y2COyO
K=
P
2z 2 z
( 0) = (
)(
)(1); T & E : z = 0.2673
5 - 2z 5 + z
P
2
y2CO
2
nCO = 4.4654;
2
nCO = 0.5346;
yCO = 0.1015
nO = 0.2673
2
0
+2z
z
+z
15-51
15.64E A gas mixture of 1 pound mol carbon monoxide, 1 pound mol nitrogen, and 1
pound mol oxygen at 77 F, 20 lbf/in.2, is heated in a constant pressure SSSF
process. The exit mixture can be assumed to be in chemical equilibrium with
CO2, CO, O2, and N2 present. The mole fraction of CO2 at this point is 0.176.
Calculate the heat transfer for the process.
initial mix:
1 CO, 1 O2,
Equil. mix:
CO2, CO, O2, N2
Const.
Pressure
1 N2
yCO2 = 0.176
Q
reaction
2 CO2⇔
initial
change
equil.
0
+2x
2x
yCO2 = 0.176 =
2x
3-x
2 CO
P = 20 lbf/in2
+
1
-2x
(1-2x)
⇒
O2
also,
1
-x
(1-x)
N2
1
0
1
x = 0.242 65
y
= 0.176
nCO2 = 0.4853 nO2 = 0.757 35  CO2
  yCO = 0.186 67
nCO = 0.5147 nN2 = 1
 y = 0.274 67
 O2
2
yCOyO2 P 1 0.186 672×0.274 67 20
(P0) =
(14.504)= 0.426 07
K= 2
0.1762
y
CO2
Since Table A.10 corresponds to a pressure P0 of 100 kPa, which is 14.504
lbf/in2. Then, from A.10, TPROD = 3200 K = 5760 R
HR = -47 518 Btu
HP = 0.4853(-169 184+71 075) + 0.5147(-47 518+43 406)
+ 0.757 35(0+45 581) + 1(0+43 050) = +27 842 Btu
QCV = HP - HR = 27 842 - (-47 518) = +75 360 Btu
15-52
15.65EIn a test of a gas-turbine combustor, saturated-liquid methane at 210 R is to be
burned with excess air to hold the adiabatic flame temperature to 2880 R. It is
assumed that the products consist of a mixture of CO2, H2O, N2, O2, and NO in
chemical equilibrium. Determine the percent excess air used in the combustion,
and the percentage of NO in the products.
CH4 + 2x O2 + 7.52x N2
→ 1 CO2 + 2 H2O + (2x-2) O2 + 7.52x N2
Then
N2
init
ch.
equil.
+
O2 ⇔ 2 NO
7.52x
2x-2
-a
-a
(7.52x-a) (2x-2-a)
Also CO2 H2O
0
+2a
2a
1
0
1
2
0
2
nTOT = 1 + 9.52x
2880 R:
ln K = -10.55,
K = 2.628×10-5
2
2
yNO
yNO
P 0
4a2
2.628×10-5 K =
(
)
=
=
yN2yO2 P0
yN2yO2 (7.52x-a)(2x-2-a)
HR = 1[-32 190 + (-1854-4300)]+ 0 + 0 (assume 77 F) = -38 344 Btu
HP = 1(-169 184 + 29 049) + 2(-103 966 + 22 746)
+ (7.52x-a)(18 015) + (2x-2-a)(19 031) + 2a(38 818 + 18 624)
= -340 639 + 173 535 x + 77 838 a
Assume
a ~ 0, then from
→ x = 1.742
HP - HR = 0
Subst.
2.628×10-5
a2
=
,
(13.1-a)(1.484-a)
4
a ≈ 0.01125
get
Use this a in 1st law
x=
302 295 - 77 838×0.01125
= 1.737
173 535
-5
⇒
⇒
2.628×10
a2
=
, a = 0.0112
(13.062-a)(1.474-a)
4
x = 1.737
% NO =
% excess air = 73.7 %
2×0.0112×100
= 0.128 %
1+9.52×1.737
15-53
15.66E Acetylene gas at 77 F is burned with 140% theoretical air, which enters the burner
at 77 F, 14.7 lbf/in.2, 80% relative humidity. The combustion products form a
mixture of CO2, H2O, N2, O2, and NO in chemical equilibrium at 3500 F, 14.7
lbf/in.2. This mixture is then cooled to 1340 F very rapidly, so that the
composition does not change. Determine the mole fraction of NO in the products
and the heat transfer for the overall process.
C2H2 + 3.5 O2 + 13.16 N2 + water →
2 CO2 + 1 H2O + 1 O2 + 13.16 N2 + water
Water:
PV = 0.8×0.46 = 0.368 lbf/in2
PV
0.368
nV = nA× = (3.5+13.16)×
= 0.428
14.332
PA
So, total H2O in products is : 1 + nV = 1.428.
a)
reaction:
N2 + O2 ⇔ 2 NO
change :
-x
-x
+2x
at 3500F = 3960 R (=2200 K), from A.10:
K = 0.001 074
Equilibrium products:
nCO2 = 2,
nH2O = 1.428, nO2 = 1-x,
nN2 = 13.16-x,
K=
nNO = 0+2x,
n TOT = 17.588
(2x)2
= 0.001 074
(1-x)(13.16-x)
By trial and error,
yNO =
x = 0.0576
2×0.0576
= 0.006 55
17.588
b) Final products (same composition) at 1340 F = 1800 R
HR = 1(97 476) + 0.428(-103 966) = 52 979 Btu
HP = 2(-169 184+14 358) + 1.428(-103 966+11 178)
+ 0.9424(0+9761) + 13.1024(0+9227) + 0.1152(38 818+9557)
= -306 486 Btu
QCV = HP - HR = -359 465 Btu
15-54
15.67E The equilibrium reaction with methane as CH4 ⇔ C + 2H2 has ln K = -0.3362 at
1440 R and ln K = -4.607 at 1080 R. By noting the relation of K to temperature,
show how you would interpolate lnK in (1/T) to find K at 1260 R and compare
that to a linear interpolation.
ln K = -0.3362 at 1440 R
ln K = -4.607 at 1080 R
1
1
1260 1440
lnK1260 = lnK1440 +
× (-4.607+0.3362)
1
1
1080 1440
1440
-1
1260
= -0.3362 +
× (-4.2708) = -2.1665
1440
-1
1080
Linear interpolation:
lnK1260 = lnK1080 +
= -4.607 +
1260 - 1080
(lnK1440 - lnK1080)
1440 - 1080
1
(-0.3362 + 4.607) = -2.4716
2
o
15.68E Use the information in problem 15.67 to estimate the enthalpy of reaction, ∆H ,
at 1260 R using the van’t Hoff equation (see problem 15.28) with finite
differences for the derivatives.
- 2
dlnK = [∆H°/RT ]dT or
solve for ∆H°
- 2 dlnK - 2 ∆lnK
∆H° = RT
= RT
dT
∆T
2
= 1.98589 × 1260 ×
-0.3362 + 4.607
= 37403 Btu/lb mol
1440 - 1080
[Remark: compare this to C.7 values + C.4, C.12,
∆H° = HC + 2HH - HCH = 0.146 × 12 × (1260-537) + 2 × 5044
2
4
– 0.538 × 16.043 × (1260-537) -(-32190) = 37304 ]
15-55
15.69E An important step in the manufacture of chemical fertilizer is the production of
ammonia, according to the reaction: N 2 + 3H2 ⇔ 2NH3
a. Calculate the equilibrium constant for this reaction at 300 F.
b. For an initial composition of 25% nitrogen, 75% hydrogen, on a mole basis,
calculate the equilibrium composition at 300 F, 750 lbf/in.2.
1N2 + 3H2 <=> 2NH3 at 300 F
-o
a) hNH3 300 F = -19 656 + 0.509×17.031(300-77) = -17723
760
-so
NH3 300 F = 45.969 + 0.509×17.031 ln 537 = 48.980
o
∆H300 F = 2(-17723) - 1(0 + 1557) - 3(0 + 1552) = -41 659 Btu
0
∆S300 F = 2(48.98) - 1(48.164) - 3(33.60) = -51.0 Btu/R
0
∆G300 F = -41 659 - 760(-51.0) = -2899 Btu
ln K =
b)
+2899
= 1.9208,
1.98589×760
nNH3 = 2x,
nN2 = 1-x,
2
K=
yNH3
3
yN2yH2
(PP )
-2
0
2
=
2
(1-xx ) (2-x
)
1-x
or
(1-xx )(2-x
)= 173.16
1-x
→ Trial & Error:
x = 0.9242
nH2 = 3-3x
(2x)222(2-x)2 P
33(1-x)4
P0
or
=
K = 6.826
( )
-2
27
750 2
×6.826× (
) = 29985
16
14.7
NH3
n
1.848
y
0.8591
N2
0.0758
0.0352
H2
0.2273
0.1057
15-56
15.70E Ethane is burned with 150% theoretical air in a gas turbine combustor. The
products exiting consist of a mixture of CO2, H2O, O2, N2, and NO in chemical
equilibrium at 2800 F, 150 lbf/in.2. Determine the mole fraction of NO in the
products. Is it reasonable to ignore CO in the products?
Combustion:
C2H6 + 5.25 O2 + 19.74 N2 → 2 CO2 + 3 H2O + 1.75 O2 + 19.74 N2
Products at 2800 F, 150 lbf/in2
a) Equilibrium mixture: CO2, H2O, O2, N2, NO
+
N2
initial
change
equil.
O2
19.74
-x
19.74-x
⇔
1.75
-x
1.75-x
2 NO
0
+2x
2x
Equil. comp. nCO2 = 2, nH2O = 3, nO2 = 1.75-x, nN2 = 19.74-x, nNO = 2x
2
yNO
P 0
4x2
( ) = (19.74-x)(1.75-x)
K = 1.283×10 =
yN2yO2 P0
-4
Solving,
x = 0.032 95
yNO =
2×0.032 95
= 0.002 49
26.49
b)
2 CO2
initial
change
equil.
⇔
2 CO
2
-2a
2-2a
0
+2a
2a
+
O2
0
+2x
2x
2
yCOyO2 P 1
2a 2 1.75-x+a
150
-8
K = 5.259×10 = 2
(P0) = (2-2a
) ( 26.49+a )(14.504
)
y
CO2
Since Table A.10 corresponds to a pressure P0 of 100 kPa, which is 14.504
lbf/in2. This equation should be solved simultaneously with the equation
solved in part a) (modified to include the unknown a). Since x was found to
be small and also a will be very small, the two are practically independent.
Therefore, use the value x = 0.032 95 in the equation above, and solve for a.
a
95+a
14.504
(1-a
) (1.75-0.032
)
=(
)×5.259×10-8
26.49+a
150
2
Solving, a = 0.000 28
or yCO = 2.1×10-5 negligible for most applications.
15-57
15.71E One pound mole of air (assumed to be 78% nitrogen, 21% oxygen, and 1% argon)
at room temperature is heated to 7200 R, 30 lbf/in.2. Find the equilibrium
composition at this state, assuming that only N2, O2, NO, O, and Ar are present.
1 lbmol air (0.78 N2, 0.21 O2, 0.01 Ar) heated to 7200 R, 30 lbf/in2.
N2 + O2 ⇔ 2 NO
1)
change -a
-a
+2a
2)
change
Equil.:
nAr =
nN2 =
0.78-a
nO2 =
0.21-a-b
O2 ⇔ 2 O
nO =
-b
+2b
0.01
nNO =
2a
2b
n =
1+b
4a2
30 0
(
)
K1 = 0.0895 =
(0.78-a)(0.21-a-b) 14.504
4b2
30
K2 = 2.221 =
(
)
14.504
(1+b)(0.21-a-b)
Divide 1st eq'n by 2nd and solve for a as function(b), using
K1 P
X=
( ) = 0.083 35
K2 P0
Get
Xb2
a=
[-1 +
2(1+b)
Also
1+
4×0.78(1+b)
Xb2
(1)
]
K2
b2
=
= 0.268 44
(1+b)(0.21-a-b) 4(P/P0)
Assume b = 0.1269
(2)
From (1), get a = 0.0299
Then, check a & b in (2) ⇒ OK
Therefore,
Subst.
N2
O2
Ar
O
NO
n
0.7501
0.0532
0.01
0.2538
0.0598
y
0.6656
0.0472
0.0089
0.2252
0.0531
15-58
15.72E Dry air is heated from 77 F to 7200 R in a 14.7 lbf/in.2 constant-pressure process.
List the possible reactions that may take place and determine the equilibrium
composition. Find the required heat transfer.
Air assumed to be 21% oxygen and 79% nitrogen by volume.
From the elementary reactions at 4000 K = 7200 R (A.10):
2
(1) O2 <=> 2 O
K1 = 2.221 = y O/yO2
(2) N2 <=> 2 N
K2 = 3.141 × 10-6 = yN/yN2
2
(3) N2 + O2 <=> 2 NO
2
K3 = 0.08955 = yNO/yN2 yO2
Call the shifts a,b,c respectively so we get
nO2 = 0.21-a-c,
nNO = 2c,
nO = 2a,
nN2 = 0.79-b-c,
nN = 2b,
ntot = 1+a+b
From which the molefractions are formed and substituted into the three
equilibrium equations. The result is corrected for 1 atm = 14.7 lbf/in2 =
101.325 kPa versus the tables 100 kPa
2
K1 = 2.1511 = y O/yO2 = 4a2/[(1+a+b)(0.21-a-c)]
2
K2 = 3.042×10-6 = yN/yN2 = 4b2/[(1+a+b)(0.79-b-c)]
2
K3 = 0.08955 = y NO/yN2 yO2 = 4c2/[(0.79-b-c)(0.21-a-c)]
which give 3 eqs. for the unknowns (a,b,c). Trial and error assume b = c = 0
solve for a from K 1 then for c from K3 and finally given the (a,c) solve for b
from K2. The order chosen according to expected magnitude K1>K3>K2
a = 0.15, b = 0.000832, c = 0.0244 =>
nO2 = 0.0356, nO = 0.3, nN2 = 0.765, nN = 0.00167, nNO = 0.049
Q = Hex - Hin = nO2∆hO2 + nN2∆hN2 + nO(hfO + ∆hO)
+ nN(hfN + ∆hN) + nNO(hfNO + ∆hNO) - 0
= 0.0356×59632 + 0.765×55902 + 0.3(107124 + 33394)
+ 0.00167(203216 + 33333) + 0.049(38818 + 57038)
= 92 135 Btu/lbmol air
[If no reaction: Q = nO2∆hO2 + nN2∆hN2 = 56 685 Btu/lbmol air]
15-59
15.73E Acetylene gas and x times theoretical air (x > 1) at room temperature and 75
lbf/in.2 are burned at constant pressure in an adiabatic SSSF process. The flame
temperature is 4600 R, and the combustion products are assumed to consist of N2,
O2, CO2, H2O, CO, and NO. Determine the value of x.
Combustion:
C2H2 + 2.5x O2 + 9.4x N2 → 2 CO2 + H2O + 2.5(x-1)O2 + 9.4x N2
Eq. products 4600 R, 75 lbf/in2: N2, O2, CO2, H2O, CO, NO
2 Reactions:
1)
2 CO2 ⇔ 2 CO + O2
change -2a
+2a
+a
2)
change
N2 + O2 ⇔ 2 NO
-b
-b
Equil. Comp.:
9.4x-b
nN2 =
nCO2 =
2-2a
nO2 =
2.5x-2.5+a-b
nH2O =
1
nCO =
2a
nNO =
2b
nTOT =
11.9x+0.5+a
+2b
At 4600 R, from A.17: K1 = 2.359×10-3, K2 = 4.249×10-3
K1
2.359×10-3
a 2 2.5x-2.5+a-b
= 4.622×10-4 =( ) (
)
0 =
1-a 11.9x+0.5+a
5.103
(P/P )
(2b)2
K2 = 4.249×10 =
(9.4-b)(2.5x-2.5+a-b)
-3
Also, from the 1st law: HP - HR = 0 where
HR = 1(+97 477) + 0 + 0 = +97 497 Btu
HP = (9.4x-b)(0+32 817) + (2.5x-2.5+a-b)(0+34 605)
+ (2-2a)(-169 184+53 885) + 1(-103 966+43 899)
+ 2a(-47 518+33 122) + 2b(38 818+31 161)
Substituting,
394 992 × x + 236 411 × a + 72 536 × b - 377 178 = 97477
which results in a set of 3 equations in the 3 unknowns x,a,b. Trial and error
solution from the last eq. and the ones for K1 and K2. The result is
x = 1.12 , a = 0.1182, b = 0.05963
15-60
15.74E One pound mole of water vapor at 14.7 lbf/in. 2, 720 R, is heated to 5400 R in a
constant pressure SSSF process. Determine the final composition, assuming that
H2O, H2, H, O2, and OH are present at equilibrium.
Reactions:
1)
2 H 2O ⇔ 2 H 2 + O2
change -2a
3)
+2a
+a
2)
2 H2O ⇔ H2+ 2 OH
change
-2b
+b
H2 ⇔ 2 H
change -c +2c
At equilibrium (5400 R, 14.7 lbf/in2)
nH2O = 1-2a-2b
nOH =
2b
nH2 =
2a+b-c
nH =
2c
nO2 =
a
nTOT =
1+a+b+c
K1
2.062×10-3
2a+b-c 2
a
=
=
(
)
(
)
1-2a-2b 1+a+b+c
1.03
(P/P0)
K2
2
2.893×10-3
2a+b-c
2b
=(
)(
)
0 =
1+a+b+c 1-2a-2b
1.03
(P/P )
K3
2.496×10-2
(2a)2
=
=
1.03
(2a+b-c)(1+a+b+c)
(P/P0)
These three equations must be solved simultaneously for
a, b & c:
and
a = 0.0622, b = 0.0570, c = 0.0327
nH2O = 0.7616 yH2O = 0.6611
nH2 = 0.1487 yH2 = 0.1291
nO2 = 0.0622 yO2 = 0.0540
+2b
15-61
15.75E Methane is burned with theoretical oxygen in an SSSF process, and the products
exit the combustion chamber at 5300 F, 100 lbf/in.2. Calculate the equilibrium
composition at this state, assuming that only CO2, CO, H2O, H2, O2, and OH are
present.
Combustion:
1)
Dissociation reactions:
2 H 2O ⇔ 2 H 2 + O2
change -2a
2)
At equilibrium:
nH2O =
+2a +a
2 H2O ⇔ H2 + 2 OH
change -2b
3)
CH4 + 2 O2 → CO2 + 2 H2O
+b +2b
2 CO2 ⇔ 2 CO + O2
change -2c
+2c +c
2-2a-2b
nH2 =
2a+b
nO2 =
a+c
nOH =
2b
nCO2 =
1-2c
nCO =
2c
nTOT =
3+a+b+c
Products at 5300F, 100 lbf/in2
K1 = 0.007 328 = (
2a+b 2
) ( a+c )( 100 )
2-2a-2b 3+a+b+c 14.504
K2 = 0.012 265 = (
2
2b
2a+b
100
)
(
)(
)
2-2a-2b 3+a+b+c 14.504
K3 = 0.426 135 = (
2c 2
a+c
100
)
(
)(
)
1-2c 3+a+b+c 14.504
These 3 equations must be solved simultaneously for a, b, & c. If solving by
hand divide the first equation by the second, and solve for c = fn(a,b). This
reduces the solution to 2 equations in 2 unknowns. Solving,
a = 0.0245, b = 0.1460, c = 0.2365
Substance:
H2O
H2
O2
OH
CO2
CO
n
1.659
0.195
0.260
0.292
0.527
0.473
y
0.4871
0.0573
0.0763
0.0857
0.1547
0.1389
16-1
CHAPTER 16
Notice that most of the solutions are done using the computer tables, which includes the
steam tables, air table, compressible flow table and the normal shock table. This
significantly reduces the amount of time it will take to solve a problem, so this should be
considered in problem assignments and exams.
Changes of problems from the 4th edition Chapter 14 to the new Chapter 16 are:
New
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
New
Old
1
new
2
new
new
3
new
new
5
6
7
new
new
new
new
8
9 mod skip a)
10
11
12
Old
New
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
New
Old
13
14
new
15
16
17
18
19
20
21
22
23
24
25
26
new
27
28
29
new
Old
New
41
42
43
44E
45E
46E
47E
48E
49E
50E
51E
52E
53E
54E
55E
56E
57E
Old
30
31
4
39
38
new
new
40
new
new
new
43
44
45
46
47
new
New
Old
16-2
16.1
Steam leaves a nozzle with a pressure of 500 kPa, a temperature of 350°C, and a
velocity of 250 m/s. What is the isentropic stagnation pressure and temperature?
h0 = h1 + V21/2 = 3167.7 + 2502/2000 = 3198.4 kJ/kg
s0 = s1 = 7.6329 kJ/kg K
Computer software: (ho, so) ⇒ To = 365°° C, Po = 556 kPa
16.2
An object from space enters the earth’s upper atmosphere at 5 kPa, 100 K with a
relative velocity of 2000 m/s or more . Estimate the object`s surface temperature.
ho1 - h1 = V21/2 = 20002/2000 = 2000 kJ/kg
ho1 = h1 + 2000 = 100 + 2000 = 2100 kJ/kg
=>
T = 1875 K
The value for h from ideal gas table A.7 was estimated since the lowest T in the
table is 200 K.
16.3
The products of combustion of a jet engine leave the engine with a velocity relative
to the plane of 400 m/s, a temperature of 480°C, and a pressure of 75 kPa.
Assuming that k = 1.32, Cp = 1.15 kJ/kg K for the products, determine the
stagnation pressure and temperature of the products relative to the airplane.
ho1 - h1 = V21/2 = 4002/2000 = 80 kJ/kg
To1 - T1 = (ho1 - h1)/Cp = 80/1.15 = 69.6 K
To1 = 480 + 273.15 + 69.6 = 823 K
Po1 = P1(To1/T1)k/(k-1) = 75(823/753.15)4.125 = 108 kPa
16.4
A meteorite melts and burn up at temperatures of 3000 K. If it hits air at 5 kPa, 50
K how high a velocity should it have to experience such a temperature?
Assume we have a stagnation T = 3000 K
h1 + V21/2 = hstagn.
Use table A.7, hstagn. = 3525.36, h1 = 50
V21/2 = 3525.36 – 50 = 3475.4 kJ/kg
V1 =
2 × 3475.4 × 1000 = 2636 m/s
(remember convert to J/kg = m2/s2)
16-3
16.5
I drive down the highway at 110 km/h on a day with 25°C, 101.3 kPa. I put my
hand , cross sectional area 0.01 m2, flat out the window. What is the force on my
hand and what temperature do I feel?
The air stagnates on the hand surface :
h1 + V21/2 = hstagn.
Use constant heat capacity
Tstagn. = T1 +
V21/2
0.5 × 1102 × (1000/3600)2
= 25 +
= 25.465°°C
Cp
1004
Assume a reversible adiabatic compression
Pstagn. = P1 (Tstagn./T1)k/(k-1) = 101.3 (298.615/298.15)3.5
= 101.85 kPa
16.6
Air leaves a compressor in a pipe with a stagnation temperature and pressure of
150°C, 300 kPa, and a velocity of 125 m/s. The pipe has a cross-sectional area of
0.02 m2. Determine the static temperature and pressure and the mass flow rate.
ho1 - h1 = V21/2 = 1252/2000 = 7.8125 kJ/kg
To1 - T1 = (ho1 - h1)/Cp = 7.8125/1.004 = 7.8 K
T1 = To1 - ∆T = 150 - 7.8 = 142.2 °C = 415.4 K
P1 = Po1(T1/To1)k/(k-1) = 300(415.4/423.15)3.5 = 281 kPa
AV P1AV1 281.2(0.02)(125)
.
m = ρAV =
=
=
= 5.9 kg/s
v
0.287(415.4)
RT1
16.7
A stagnation pressure of 108 kPa is measured for an air flow where the pressure is
100 kPa and 20°C in the approach flow. What is the incomming velocity?
Assume a reversible adiabatic compression
108 0.2857
)
= 299.667 K
100
To1 = T1 × (Po1/P1)(k-1)/k = 293.15 × (
V21/2 = ho1 - h1 = Cp (To1 - T1) = 6.543
V1 =
2 × 6.543 × 1000 = 114.4 m/s
16-4
16.8
A jet engine receives a flow of 150 m/s air at 75 kPa, 5°C across an area of 0.6 m 2
with an exit flow at 450 m/s, 75 kPa, 600 K. Find the mass flow rate and thrust.
.
m = ρAV; ideal gas ρ = P/RT
75
) × 0.6 × 150 = 0.9395 × 0.6 × 150
0.287 × 278.15
.
m = (P/RT)AV = (
= 84.555 kg/s
.
Fnet = m (Vex-Vin) = 84.555 × (450-150) = 25367 N
16.9
A water cannon sprays 1 kg/s liquid water at a velocity of 100 m/s horizontally
out from a nozzle. It is driven by a pump that receives the water from a tank at
15°C, 100 kPa. Neglect elevation differences and the kinetic energy of the water
flow in the pump and hose to the nozzle. Find the nozzle exit area, the required
pressure out of the pump and the horizontal force needed to hold the cannon.
0.001001
.
.
m = ρAV = AV/v ⇒ A= mv/V = 1
= 1.0 × 10-5 m2
100
.
.
. 2
_
W
= mwp = mv(Pex - Pin) = mVex
/2
p
2
Pex = Pin + Vex
/2v = 100 + 1002/2 × 1000 × 0.001 = 150 kPa
.
F = mVex = 1 × 100 = 100 N
16.10 An irrigation pump takes water from a lake and discharges it through a nozzle as
shown in Fig. P16.10. At the pump exit the pressure is 700 kPa, and the
temperature is 20°C. The nozzle is located 10 m above the pump and the
atmospheric pressure is 100 kPa. Assuming reversible flow through the system
determine the velocity of the water leaving the nozzle.
Assume we can neglect kinetic energy in the pipe in and out of the pump.
Incompressible flow so Bernoulli's equation applies (V1 ≅ V2 ≅ V3 ≅ 0)
v(P3 - P2)+ (V23 - V22)/2 + g(Z3 - Z2) = 0
P 3 = P2 -
g(Z3 - Z2)
v
= 700 -
9.807(10)
= 602 kPa
1000(0.001002)
V24/2 = v(P3 - P4)
⇒ V4 =
2v(P3 - P4) =
2 × 0.001002 × 502.1 × 1000
= 31.72 m/s
16-5
16.11 A water turbine using nozzles is located at the bottom of Hoover Dam 175 m
below the surface of Lake Mead. The water enters the nozzles at a stagnation
pressure corresponding to the column of water above it minus 20% due to friction.
The temperature is 15°C and the water leaves at standard atmospheric pressure. If
the flow through the nozzle is reversible and adiabatic, determine the velocity and
kinetic energy per kilogram of water leaving the nozzle.
∆P = ρg∆Z =
g∆Z
9.807 × 175
=
= 1714.5 kPa
v
0.001001 × 1000
∆Pac = 0.8 ∆P = 1371.6 kPa
2
/2 ⇒ Vex =
v∆P = Vex
Vex =
2v∆P
2 × 0.001001 × 1000 × 1371.6 = 62.4 m/s
2
Vex
/2 = v∆P = 1.373 kJ/kg
16.12 A water tower on a farm holds 1 m3 liquid water at 20°C, 100 kPa in a tank on top
of a 5 m tall tower. A pipe leads to the ground level with a tap that can open a 1.5
cm diameter hole. Neglect friction and pipe losses and estimate the time it will
take to empty the tank for water.
Bernoulli Equation:
Pe = Pi; Vi = 0; Ze = 0; Zi = H
V2e /2 = gZi ⇒
Ve =
2gZ =
2 × 9.807 × 5 = 9.9 m/s
.
m = ρAVe = AVe/v = ∆m/∆t
∆m = V/v; A = πD2/4 = π × 0.0152 / 4 = 1.77 × 10-4 m2
⇒
∆t =
∆t = ∆mv/AVe = V/AVe
1
= 571.6 sec = 9.53 min
1.77 × 10 -4 × 9.9
16-6
16.13 Find the speed of sound for air at 100 kPa at the two temperatures 0°C and 30°C.
Repeat the answer for carbon dioxide and argon gases.
From eq. 16.28 we have
c0 = kRT =
c30 =
1.4 × 0.287 × 273.15 × 1000 = 331 m/s
1.4 × 0.287 × 303.15 × 1000 = 349 m/s
For Carbon Dioxide: R = 0.1889, k = 1.289
c0 =
1.289 × 0.1889 × 273.15 × 1000 = 257.9 m/s
c30 =
1.289 × 0.1889 × 303.15 × 1000 = 271.7 m/s
For Argon: R = 0.2081, k = 1.667
c0 =
1.667 × 0.2081 × 273.15 × 1000 = 307.8 m/s
c30 =
1.667 × 0.2081 × 303.15 × 1000 = 324.3 m/s
16.14 If the sound of thunder is heard 5 seconds after seing the lightning and the
weather is 20°C how far away is the lightning taking place?
The sound travels with the speed of sound in air (ideal gas).
L = c × t = kRT × t =
1.4 × 0.287 × 293.15 × 1000 × 5 = 1716 m
16.15 Estimate the speed of sound for steam directly from Eq. 16.28 and the steam
tables for a state of 6 MPa, 400°C. Use table values at 5 and 7 MPa at the same
entropy as the wanted state. Eq. 16.28 is then done by finite difference. Find also
the answer for the speed of sound assuming steam is an ideal gas.
c2 = (
δP
∆P
)
=
(
)
δρ s
∆ρ s
State 6 MPa, 400°C ⇒ s = 6.5407
7 MPa, s ⇒ v = 0.04205; ρ = 1/v = 23.777
5 MPa, s ⇒ v = 0.05467; ρ = 1/v = 18.2909
c2 =
7000 - 5000
= 364.56 × 1000 ⇒ c = 603.8 m/s
23.777 - 18.2909
Cp =
14190 - 10499
= 36.91;
100
k = Cp/Cv = 1.2908;
c = kRT =
Cv = Cp - R = 36.91 - 8.31451 = 28.595
R = 0.4615
1.2908 × 0.4615 × 673.15 × 1000 = 633.2 m/s
16-7
16.16 A convergent-divergent nozzle has a throat diameter of 0.05 m and an exit
diameter of 0.1 m. The inlet stagnation state is 500 kPa, 500 K. Find the back
pressure that will lead to the maximum possible flow rate and the mass flow rate
for three different gases as: air; hydrogen or carbon dioxide.
There is a maximum possible flow when M=1 at the throat,
T* =
2
2 k
2 1
To; P* = Po ( )k-1; ρ* =ρ o (
)k-1
k+1
k+1
k+1
.
m = ρ *A*V = ρ*A*c = P*A* k/RT*
A* = πD2/4 = 0.001963 m2
ρ*
.
m
k
T*
P*
a)
1.400
416.7
264.1
448.2
2.209
1.944
b)
1.409
415.1
263.4
1704.5
0.154
0.515
c)
1.289
436.9
273.9
348.9
3.318
2.273
c
AE/A* = (DE/D*)2 = 4. There are 2 possible solutions corresponding to points
c and d in Fig. 16.13 and Fig. 16.17. For these we have
ME
PE/Po
ME
PE/Po
a)
0.1466
0.985
2.940
0.0298
b)
0.1464
0.985
2.956
0.0293
c)
0.1483
0.986
2.757
0.0367
PB = PE ≅ 0.985 × 500 = 492.5 kPa all cases point c
PB = PE = a) 14.9 b) 14.65 c) 18.35 kPa, point d
16-8
16.17 Air is expanded in a nozzle from 2 MPa, 600 K, to 200 kPa. The mass flow rate
through the nozzle is 5 kg/s. Assume the flow is reversible and adiabatic and
determine the throat and exit areas for the nozzle.
 2 k
P * = Po 
k-1
k+1
Mach #
= 2 × 0.5283 = 1.056 MPa
Velocity
Area
Density
T* = To × 2/(k+1) = 600 × 0.8333 = 500 K
v* = RT*/P* = 0.287 × 500/1056
2.0 MPa
P
0.2 MPa
= 0.1359 m3/kg
kRT* = 1.4 × 1000 × 0.287 × 500 = 448.2 m/s
.
A* = mv*/c* = 5 × 0.1359/448.2 = 0.00152 m2
c* =
P2/Po = 200/2000 = 0.1 ⇒ M*2 = 1.701 = V2/c*
V2 = 1.701 × 448.2 = 762.4 m/s
T2 = 500 × 0.5176 = 258.8 K
v2 = RT2/P2 = 0.287 × 258.8/200 = 0.3714 m3/kg
.
A2 = mv2/V2 = 5 × 0.3714/762.4 = 0.002435 m2
16.18 Consider the nozzle of Problem 16.17 and determine what back pressure will
cause a normal shock to stand in the exit plane of the nozzle. This is case g in Fig.
16.17. What is the mass flow rate under these conditions?
PE/Po = 200/2000 = 0.1 ; ME = 2.1591 = Mx
My = 0.5529 ; Py/Px = 5.275
Py = 5.275 × 200 = 1055 kPa
.
m = 5 kg/s same as in Problem 16.9
16-9
16.19 At what Mach number will the normal shock occur in the nozzle of Problem
16.17 if the back pressure is 1.4 MPa? (trial and error on Mx)
Relate the inlet and exit conditions to the shock conditions with reversible
flow before and after the shock. It becomes trial and error.
Assume Mx = 1.8 ⇒
My = 0.6165 ;
Poy/Pox = 0.8127
AE/A*x = A2/A* = 0.002435/0.001516 = 1.6062
Ax/A*x = 1.439 ; Ax/A*y = 1.1694
AE/A*y = (AE/A*x)(Ax/A*y)/(Ax/A*x) =
1.6062(1.1694)
= 1.3053
1.439
⇒ ME = 0.5189 ; PE/Poy = 0.8323
PE = (PE/Poy)(Poy/Pox)Pox = 0.8323 × 0.8127 × 2000 = 1353 kPa
So select the mach number a little less
Mx = 1.7 ⇒ My = 0.64055 ; Poy/Pox = 0.85573
Ax/A*x = 1.3376 ; Ax/A*y = 1.1446
AE/A*y = (AE/A*x)(Ax/A*y)/(Ax/A*x) =
1.6062(1.1446)
= 1.3744
1.3376
⇒ ME = 0.482 ; PE/Poy = 0.853
PE = (PE/Poy)(Poy/Pox)Pox = 0.853 × 0.85573 × 2000 = 1459.9 kPa
Now interpolate between the two
Mx = 1.756 ⇒ My = 0.6266 ; Poy/Pox = 0.832
Ax/A*x = 1.3926 ; Ax/A*y = 1.1586
AE/A*y = 1.6062 × 1.1586/1.3926 = 1.3363
⇒ ME = 0.5 ; PE/Poy = 0.843
PE = 0.843 × 0.832 × 2000 = 1402.7 kPa OK
16-10
16.20 Consider the nozzle of Problem 16.17. What back pressure will be required to
cause subsonic flow throughout the entire nozzle with M = 1 at the throat?
AE/A* = 0.002435/0.001516 = 1.6062
ME = 0.395 ;
PE/Po = 0.898
PE = 0.898 × 2000 = 1796 kPa
16.21 Determine the mass flow rate through the nozzle of Problem 16.17 for a back
pressure of 1.9 MPa.
PE/Pox = 1.9/2.0 = 0.95 ⇒ ME = 0.268
TE = (T/To)To = 0.9854 × 600 = 591.2 K
cE =
kRTE =
1.4 × 1000 × 0.287 × 591.2 = 487.4 m/s
VE = MEcE = 0.268 × 487.4 = 130.6 m/s vE
= RT/P = 0.287 × 591.2/1900 = 0.0893 m3/kg
.
m = AEVE/vE = 0.002435 × 130.6/0.0893 = 3.561 kg/s
16.22 At what Mach number will the normal shock occur in the nozzle of Problem
16.16 flowing with air if the back pressure is halfway between the pressures at c
and d in Fig. 16.17?
First find the two pressure that will give exit at c and d. See solution to 16.8 a)
2
AE/A* = (DE/D*) = 4 ⇒
PE = 492.5 (c) 16.9 (d)
PE = (492.5 + 14.9)/2 = 253.7 kPa
Assume ME = 2.4 ⇒ My = 0.5231 ; Poy/Pox = 0.54015
Ax/A*x = 2.4031 ; Ax/A*y = 1.298
AE/A*y = (AE/A*x)(Ax/A*y) / (Ax/A*x) = 4 × 1.298/2.4031 = 2.1605
⇒ ME = 0.2807 ; PE/Poy = 0.94675
PE = (PE/Poy)(Poy/Pox)Pox = 0.94675 × 0.54015 × 500 = 255.7 kPa
Repeat if Mx = 2.5 ⇒ PE = 233.8 kPa
Interpolate to match the desired pressure =>
Mx = 2.41
16-11
16.23 A convergent nozzle has minimum area of 0.1 m2 and receives air at 175 kPa,
1000 K flowing with 100 m/s. What is the back pressure that will produce the
maximum flow rate and find that flow rate?
P*
2 k
=(
)k-1 = 0.528
Po k+1
Critical Pressure Ratio
Find Po:
h0 = h1 + V21/2 = 1046.22 + 1002/2000 = 1051.22
T0 = Ti + 4.4 = 1004.4 from table A.7
P0 = Pi (T0/Ti)k/(k-1) = 175 × (1004.4/1000)3.5 = 177.71
P* = 0.528 Po = 0.528 × 177.71 = 93.83 kPa
T* = 0.8333 To = 836.97 K
ρ* =
P*
93.83
=
= 0.3906
RT* 0.287 × 836.97
V = c = kRT* = 1.4 × 1000 × 0.287 × 836.97 = 579.9 m/s
.
m = ρAV = 0.3906 × 0.1 × 579.9 = 22.65 kg/s
16.24 A convergent-divergent nozzle has a throat area of 100 mm 2 and an exit area of
175 mm2. The inlet flow is helium at a total pressure of 1 MPa, stagnation
temperature of 375 K. What is the back pressure that will give sonic condition at
the throat, but subsonic everywhere else?
AE/A* = 175/100 = 1.75 ; kHe = 1.667
ME = 0.348 ; PE/PO = 0.906
PE = 0.906 × 1000 = 906 kPa
16-12
16.25 A nozzle is designed assuming reversible adiabatic flow with an exit Mach
number of 2.6 while flowing air with a stagnation pressure and temperature of 2
MPa and 150°C, respectively. The mass flow rate is 5 kg/s, and k may be assumed
to be 1.40 and constant.
a. Determine the exit pressure, temperature, and area, and the throat area.
b. Suppose that the back pressure at the nozzle exit is raised to 1.4 MPa, and that
the flow remains isentropic except for a normal shock wave. Determine the exit
Mach number and temperature, and the mass flow rate through the nozzle.
(a)
From Table A.11:
ME = 2.6
PE = 2.0 × 0.05012 = 0.1002 MPa
T* = 423.15 × 0.8333 = 352.7 K
P* = 2.0 × 0.5283 = 1.057 MPa
c* =
1.4 × 1000 × 0.287 × 352.7 = 376.5 m/s
v* = 0.287 × 352.7/1057 = 0.0958 m3/kg
A* = 5 × 0.0958/376.5 = 1.272 × 10-3m2
AE = 1.272 × 10-3 × 2.896 = 3.68 × 10-3m2
TE = 423.15 × 0.42517 = 179.9 K
Assume Mx = 2 then
My = 0.57735, Poy/Pox = 0.72088, AE/A*x = 2.896
Ax/A*x = 1.6875, Ax/A*y = 1.2225,
AE/A*y = 2.896 × 1.2225/1.6875 = 2.098
⇒ ME = 0.293, PE/Poy = 0.94171
PE = 0.94171 × 0.72088 × 2.0 = 1.357 MPa, OK
TE = 0.98298 × 423.15 = 416 K,
.
m = 5 kg/s
16-13
16.26 A jet plane travels through the air with a speed of 1000 km/h at an altitude of 6
km, where the pressure is 40 kPa and the temperature is −12°C. Consider the inlet
diffuser of the engine where air leaves with a velocity of 100 m/s. Determine the
pressure and temperature leaving the diffuser, and the ratio of inlet to exit area of
the diffuser, assuming the flow to be reversible and adiabatic.
v 1 = 0.287 × 261.15/40 = 1.874
V = 1000 km/h = 277.8 m/s,
h1 = 261.48, Pr1 = 0.6862
ho1 = 261.48 + 277.8 2/2000 = 300.07
⇒ To1 = 299.7 K, Pro1 = 1.1107
Po1 = 40 × 1.1107/0.6862 = 64.74 kPa
h2 = 300.07 - 1002/2000 = 295.07
⇒
T2 = 294.7 K, Pr2 = 1.0462
P2 = 64.74 × 1.0462/1.1107 = 61 kPa
v2 = 0.287 × 294.7/61 = 1.386
A1/A2 = (1.874/1.386)(100/277.8) = 0.487
16.27 A 1-m3 insulated tank contains air at 1 MPa, 560 K. The tank is now discharged
through a small convergent nozzle to the atmosphere at 100 kPa. The nozzle has
an exit area of 2 × 10−5 m2.
a. Find the initial mass flow rate out of the tank.
b. Find the mass flow rate when half the mass has been discharged.
c. Find the mass of air in the tank and the mass flow rate out of the tank when
the nozzle flow changes to become subsonic.
a. The back pressure ratio:
PB/Po1 = 100/1000 = 0.1 < (P*/Po)crit = 0.5283
so the initial flow is choked with the maximum possible flow rate.
ME = 1 ; PE = 0.5283 × 1000 = 528.3 kPa
TE = T* = 0.8333 × 560 = 466.7 K
VE = c =
kRT* =
1.4 × 1000 × 0.287 × 466.7 = 433 m/s
vE = RT*/PE = 0.287 × 466.7/528.3 = 0.2535 m3/kg
16-14
.
m1 = AVE/vE = 2 × 10-5 × 433/0.2535 = 0.0342 kg/s
b. The initial mass is
m1 = P1V/RT1 = 1000 × 1/0.287 × 560 = 6.222 kg
with a mass at state 2 as m2 = m1/2 = 3.111 kg.
Assume an adiabatic reversible expansion of the mass that remains in the tank.
P2 = P1(v1/v2)k = 100 × 0.51.4 = 378.9 kPa
T2 = T1(v1/v2)k-1 = 560 × 0.50.4 = 424 K
The pressure ratio is still less than critical and the flow thus choked.
PB/Po2 = 100/378.9 = 0.264 < (P*/Po)crit
ME = 1 ; PE = 0.5283 × 378.9 = 200.2 kPa
TE = T* = 0.8333 × 424 = 353.7 K
VE = c =
kRT* =
1.4 × 1000 × 0.287 × 353.7 = 377 m/s
-5
2×10 (377)(200.2)
.
= 0.0149 kg/s
m2 = AVEPE/RTE =
0.287(353.7)
c. The flow changes to subsonic when the pressure ratio reaches critical.
PB/Po3 = 0.5283 ⇒ Po3 = 189.3 kPa
v1/v3 = (Po3/P1)1/k = (189.3/1000)0.7143 = 0.3046
m3 = m1v1/v3 = 1.895 kg
T3 = T1(v1/v3)k-1 = 560 × 0.30460.4 = 348 K
PE = PB = 100 kPa ; ME = 1
TE = 0.8333 × 348 = 290 K ; VE =
-5
kRTE = 341.4 m/s
2×10 (341.4)(100)
.
= 0.0082 kg/s
m3 = AVEPE/RTE =
0.287(290)
16-15
16.28 A 1-m3 uninsulated tank contains air at 1 MPa, 560 K. The tank is now
discharged through a small convergent nozzle to the atmosphere at 100 kPa while
heat transfer from some source keeps the air temperature in the tank at 560 K. The
nozzle has an exit area of 2 × 10−5 m2.
a. Find the initial mass flow rate out of the tank.
b. Find the mass flow rate when half the mass has been discharged.
c. Find the mass of air in the tank and the mass flow rate out of the tank when the
nozzle flow changes to become subsonic.
a.
Same solution as in 16.27 a)
b.
From solution 16.27 b) we have m2 = m1/2 = 3.111 kg
P2 = P1/2 = 500 kPa ; T2 = T1 ;
PB/P2 = 100/500 = 0.2 < (P*/Po)crit
The flow is choked and the velocity is:
VE = c =
kRT* = 433 m/s from 16.27 a)
PE = 0.5283 × 500 = 264.2 kPa ; ME = 1
-5
2×10 (433)(264.2)
.
= 0.01708 kg/s
m2 = AVEPE/RTE =
0.287(466.7)
c. Flow changes to subsonic when the pressure ratio reaches critical.
PB/Po = 0.5283 ; P3 = 189.3 kPa
m3 = m1P3/P1 = 1.178 kg ; T3 = T1 ⇒ VE = 433 m/s
-5
2×10 (433)(189.3)
.
= 0.01224 kg/s
m3 = AVEPE/RTE =
0.287(466.7)
16.29 The products of combustion enter a convergent nozzle of a jet engine at a total
pressure of 125 kPa, and a total temperature of 650°C. The atmospheric pressure
is 45 kPa and the flow is adiabatic with a rate of 25 kg/s. Determine the exit area
of the nozzle.
The critical pressure: Pcrit = P2 = 125 × 0.5283 = 66 kPa > Pamb
The flow is then choked.
V 2 = c2 =
T2 = 923.15 × 0.8333 = 769.3 K
1.4 × 1000 × 0.287 × 769.3 = 556 m/s
v2 = 0.287 × 769.3/66 = 3.3453
A2 = 25 × 3.3453/556 = 0.1504 m2
16-16
16.30 Air is expanded in a nozzle from 700 kPa, 200°C, to 150 kPa in a nozzle having
an efficiency of 90%. The mass flow rate is 4 kg/s. Determine the exit area of the
nozzle, the exit velocity, and the increase of entropy per kilogram of air. Compare
these results with those of a reversible adiabatic nozzle.
T2s = T1(P2/P1)(k-1)/k = 473.2 (150/700)0.286 = 304.6 K
V2s2 = 2 × 1000 × 1.0035(473.2 - 304.6) = 338400
V22 = 0.9 × 338400 ⇒ V2 = 552 m/s
h2 + V22/2 = h1 ⇒ T2 = T1 - V22/2Cp
T2 = 473.2 - 5522/(2 × 1000 × 1.004) = 321.4 K ;
v2 = 0.287 × 321.4/150 = 0.6149 m3/kg
A2 = 4 × 0.6149/552 = 0.00446 m2 = 4460 mm2
321.4
150
s2 - s1 = 1.0035 ln
 - 0.287 ln  = 0.0539 kJ/kg K
473.2


700
16.31 Repeat Problem 16.26 assuming a diffuser efficiency of 80%.
Same as problem 16.26, except
ηD = 0.80. We thus have from 16.26
h3 - h1
ho1 - h1
=
h3 - 261.48
300.07 - 261.48
h
01
02
= 0.8
3
2
⇒ h3 = 292.35, Pr3 = 1.0129
1
Po2 = P3 = 40 × 1.0129/0.6862 = 59.04 kPa
s
Pro2 = Pro1 = 1.1107
h2 = 300.07 - 1002/2000 = 295.07
⇒ T2 = 294.7 K,
P2 = 59.04 × 1.0462/1.1107 = 55.6 kPa
v2 = 0.287 × 294.7/55.6 = 1.521
A1/A2 = (1.874/1.521)(100/277.8) = 0.444
Pr2 = 1.0462
16-17
16.32 Consider the diffuser of a supersonic aircraft flying at M = 1.4 at such an altitude
that the temperature is −20°C, and the atmospheric pressure is 50 kPa. Consider
two possible ways in which the diffuser might operate, and for each case calculate
the throat area required for a flow of 50 kg/s.
a. The diffuser operates as reversible adiabatic with subsonic exit velocity.
b. A normal shock stands at the entrance to the diffuser. Except for the normal
shock the flow is reversible and adiabatic, and the exit velocity is subsonic. This
is shown in Fig. P16.32.
a. Assume a convergent-divergent diffuser with M = 1 at the throat.
Relate the inlet state to the sonic state
P1/Po = 0.31424 ; P*/Po1 = 0.5283
P* =
c* =
0.5283
0.8333
50 = 84 kPa ; T* =
; 253.2 = 293.7 K
0.31424
0.71839
kRT* =
1.4 × 1000 × 0.287 × 293.7 = 343.5 m/s
v* = RT*/P* = 0.287 × 293.7/84 = 1.0035
.
A* = mv*/c* = 50 × 1.0035/343.5 = 0.1461 m2
b.
Across the shock we have
My = 0.7397 ; Py = 50 × 2.12 = 106 kPa ;
Ty = 253.2 × 1.2547 = 317.7 K
P* =
0.5283
× 106 = 80.6 kPa
0.6952
T* =
0.8333
× 317.7 = 293.7 K, c* = 343.5 m/s
0.9011
v* = 0.287 × 293.7/80.6 = 1.0458
A* = 50 × 1.0458/343.5 = 0.1522 m2
16-18
16.33 Air enters a diffuser with a velocity of 200 m/s, a static pressure of 70 kPa, and a
temperature of −6°C. The velocity leaving the diffuser is 60 m/s and the static
pressure at the diffuser exit is 80 kPa. Determine the static temperature at the
diffuser exit and the diffuser efficiency. Compare the stagnation pressures at the
inlet and the exit.
To1 = T1 + V21/2Cp = 267.15 + 2002/(2000 × 1.004) = 287.1 K
To2 = To1 ⇒ T2 = To2 - V22/2Cp = 287.1 - 602/(2000 × 1.004) = 285.3 K
To1 - T1
k-1 Po1 - P1
=
k
P1
⇒
Po1 - P1 = 18.25 ⇒ Po1 = 88.3 kPa
k - 1 Po2 - P2
=
k
P2
⇒
Po2 - P2 = 1.77 ⇒ Po2 = 81.8 kPa
T1
To2 - T2
T2
ex
Ts = T1 (Po2/P1)k-1/k = 267.15 × 1.0454 = 279.3 K
ex
ηD =
Ts - T1
To1 - T1
=
279.3 - 267.15
= 0.608
287.1 - 267.15
16.34 Steam at a pressure of 1 MPa and temperature of 400°C expands in a nozzle to a
pressure of 200 kPa. The nozzle efficiency is 90% and the mass flow rate is 10
kg/s. Determine the nozzle exit area and the exit velocity.
First do the ideal reversible adiabatic nozzle
s2s= s1= 7.4651 ⇒ T2s = 190.4°C ; h2s = 2851, h1= 3263.9
Now the actual nozzle can be calculated
h1 - h2ac = ηD(h1 - h2s) = 0.9(3263.9 - 2851) = 371.6
h2ac = 2892.3 , T2 = 210.9°C, v2 = 1.1062
V2 = 2000(3263.9 - 2892.3) = 862 m/s
.
A2 = mv2/V2 = 10 × 1.1062/862 = 0.01283 m2
16-19
16.35 Steam at 800 kPa, 350°C flows through a convergent-divergent nozzle that has a
throat area of 350 mm2. The pressure at the exit plane is 150 kPa and the exit
velocity is 800 m/s. The flow from the nozzle entrance to the throat is reversible
and adiabatic. Determine the exit area of the nozzle, the overall nozzle efficiency,
and the entropy generation in the process.
ho1 = 3161.7, so1 = 7.4089
P*/Po1 = (2/(k+1))k/(k-1) = 0.54099 ⇒ P* = 432.7 kPa
At *: (P*,s* = so1) ⇒ h* = 2999.3, v* = 0.5687
∆ h = V2/2 ⇒ V* = 2000(3161.7-2999.3) = 569.9 m/s
.
m = AV*/v* = 350 × 10-6 × 569.9/0.5687 = 0.3507 kg/s
he = ho1 - V2e /2 = 3161.7 - 8002/2 × 1000 = 2841.7
Exit: Pe, he: ve = 1.395, se =7.576
.
Ae = mve/Ve = 0.3507 × 1.395/800 = 6.115 × 10-4 m2
sgen = se - so1 = 7.576 - 7.4089 = 0.167 kJ/kg K
16.36 Air at 150 kPa, 290 K expands to the atmosphere at 100 kPa through a convergent
nozzle with exit area of 0.01 m2. Assume an ideal nozzle. What is the percent
error in mass flow rate if the flow is assumed incompressible?
Pe k-1
Te = Ti ( ) k = 258.28 K
Pi
V2e /2 = hi - he = Cp (Ti - Te) = 1.004 (290 - 258.28) = 31.83
Ve = 252.3 m/s;
ve =
RTe
Pe
=
0.287 × 258.28
= 0.7412
100
0.01 × 252.3
.
m = AVe / ve =
= 3.4 kg/s
0.7413
Incompressible Flow: vi = RT/P = 0.287 × 290/150 = 0.55487 m3/kg
V2e /2 = v ∆P = vi (Pi - Pe) = 0.55487 (150 - 100) = 27.74 kJ/kg
.
=> Ve = 235 m/s => m = AVe / vi = 0.01 × 235 / 0.55487 = 4.23 kg/s
.
mincompressible 4.23
.
=
= 1.25
about 25% overestimation.
3.4
mcompressible
16-20
16.37 A sharp-edged orifice is used to measure the flow of air in a pipe. The pipe
diameter is 100 mm and the diameter of the orifice is 25 mm. Upstream of the
orifice, the absolute pressure is 150 kPa and the temperature is 35°C. The pressure
drop across the orifice is 15 kPa, and the coefficient of discharge is 0.62.
Determine the mass flow rate in the pipeline.
k-1∆P
0.4 15
∆T = Ti  = 308.15 ×
×
= 8.8
1.4 150
 k  Pi
vi = RTi/Pi = 0.5896
Pe = 135 kPa, Te = 299.35, ve = 0.6364
.
.
mi = me ⇒ Vi = Ve(De/Di)2 vi/ve = 0.0579
hi - he = V2e (1 - 0.05792)/2 = Cp(Ti - Te)
e
Vs =
2 × 1000 × 1.0035 × 8.8/(1 - 0.0579)2 = 133.1 m/s
.
m = 0.62 (π/4) (0.025)(2133.1/0.6364) = 0.06365 kg/s
16.38 A critical nozzle is used for the accurate measurement of the flow rate of air.
Exhaust from a car engine is diluted with air so its temperature is 50°C at a total
pressure of 100 kPa. It flows through the nozzle with throat area of 700 mm2 by
suction from a blower. Find the needed suction pressure that will lead to critical
flow in the nozzle, the mass flow rate and the blower work, assuming the blower
exit is at atmospheric pressure 100 kPa.
P* = 0.5283 Po = 52.83 kPa, T* = 0.8333 To = 269.3 K
v* = RT*/P* = 0.287 × 269.3/52.83 = 1.463
c* = kRT* = 1.4 × 1000 × 0.287 × 269.3 = 328.9 m/s
.
m = Ac*/v* = 700 × 10-6 × 328.9/1.463 = 0.157 kg/s
16-21
16.39 A convergent nozzle is used to measure the flow of air to an engine. The
atmosphere is at 100 kPa, 25°C. The nozzle used has a minimum area of 2000
mm2 and the coefficient of discharge is 0.95. A pressure difference across the
nozzle is measured to 2.5 kPa. Find the mass flow rate assuming incompressible
flow. Also find the mass flow rate assuming compressible adiabatic flow.
Assume Vi ≅ 0,
vi = RTi/Pi = 0.287 × 298.15/100 = 0.8557
Ve,s2/2 = hi - he,s = vi(Pi - Pe) = 2.1393 kJ/kg
Ve,s =
2 × 1000 × 2.1393 = 65.41 m/s
.
ms = AVe,s/vi = 2000 × 10-6 × 65.41/0.8557 = 0.153 kg/s
.
.
ma = CDms = 0.1454 kg/s
Te,s = Ti (Pe/Pi)(k-1)/k = 298.15(97.5/100)0.2857 = 296 K
∆h = Cp∆T = 1.0035 × 2.15 = 2.1575 = Ve,s2/2
Ve,s =
2 × 1000 × 2.1575 = 65.69 m/s
ve,s = 0.287 × 296/97.5 = 0.8713
.
ms = AVe,s/ve,s = 2000 × 10-6 × 65.69/0.8713 = 0.1508 kg/s
.
.
ma = CDms = 0.1433 kg/s
16-22
16.40 A convergent nozzle with exit diameter of 2 cm has an air inlet flow of 20°C, 101
kPa (stagnation conditions). The nozzle has an isentropic efficiency of 95% and the
pressure drop is measured to 50 cm water column. Find the mass flow rate
assuming compressible adiabatic flow. Repeat calculation for incompressible flow.
Convert ∆P to kPa:
∆P = 50 cm H2O = 0.5 × 9.8064 = 4.903 kPa
T0 = 20°C = 293.15 K
P0 = 101 kPa
Assume inlet Vi = 0
Pe = P0 - ∆P = 101 - 4.903 = 96.097 kPa
Pe k-1
96.097 0.2857
Te = T0 ( ) k = 293.15 ×(
)
= 289.01
101
P0
V2e /2 = hi - he = Cp (Ti - Te) = 1.004 × (293.15 - 289.01)
= 4.1545 kJ/kg = 4254.5 J/kg
Ve 2ac/2 = η Ve2s/2 = 0.95 × 4154.5 = 3946.78
Te ac = Ti ρe ac =
Pe
RTp
Ve 2ac/2
Cp
=
= 293.15 -
=> Ve = 91.15 m/s
⇒ Ve ac = 88.85 m/s
3.9468
= 289.2
1.0035
96.097
= 1.158 kg/m3
0.287 × 289.2
π
.
m = ρAV = 1.158 × × 0.022 × 88.85 = 0.0323 kg/s
4
16.41 Steam at 600 kPa, 300°C is fed to a set of convergent nozzles in a steam turbine.
The total nozzle exit area is 0.005 m 2 and they have a discharge coefficient of
0.94. The mass flow rate should be estimated from the measurement of the
pressure drop across the nozzles, which is measured to be 200 kPa. Determine the
mass flow rate.
se,s = si = 7.3724 kJ/kg K ;
(Pe, se,s) ⇒ he,s = 2961 kJ/kg
ve,s = 0.5932, hi = 3061.6
Ve,s =
2 × 1000(3061.6 - 2961) = 448.55 m/s
.
m = 0.005 × 448.55/0.5932 = 3.781 kg/s
16-23
16.42 The coefficient of discharge of a sharp-edged orifice is determined at one set of
conditions by use of an accurately calibrated gasometer. The orifice has a
diameter of 20 mm and the pipe diameter is 50 mm. The absolute upstream
pressure is 200 kPa and the pressure drop across the orifice is 82 mm of mercury.
The temperature of the air entering the orifice is 25°C and the mass flow rate
measured with the gasometer is 2.4 kg/min. What is the coefficient of discharge
of the orifice at these conditions?
∆P = 82 × 101.325/760 = 10.93 kPa
k-1
0.4
∆T = Ti   ∆P/Pi = 298.15 ×
× 10.93/200 = 4.66
1.4
k 
vi = RTi/Pi = 0.4278, ve = RTe/Pe = 0.4455
Vi = VeAevi/Aive = 0.1536 Ve
(V2e - V2i )/2 = V2e (1 - 0.15362)/2 = hi - he = Cp∆T
Ve =
2 × 1000 × 1.004 × 4.66/(1 - 0.15362) = 97.9 m/s
π
.
m = AeVe/ve = × 0.022 × 97.9/0.4455 = 0.069 kg/s
4
CD = 2.4/60 × 0.069 = 0.58
16-24
16.43 (Adv.) Atmospheric air is at 20°C, 100 kPa with zero velocity. An adiabatic
reversible compressor takes atmospheric air in through a pipe with cross-sectional
area of 0.1 m2 at a rate of 1 kg/s. It is compressed up to a measured stagnation
pressure of 500 kPa and leaves through a pipe with cross-sectional area of 0.01
m2. What is the required compressor work and the air velocity, static pressure and
temperature in the exit pipeline?
C.V. compressor out to standing air and exit to stagnation point.
.
.
.
_
m ho1 + W
= m(h + V2/2)ex = mho,ex
c
.
. o
mso1 = msex ⇒ Pr,o,ex = Pr,o1
(Pst,ex/Po1) = 1.028(500/100) = 5.14
⇒ To,ex = 463,
ho,ex = 465.38, ho1 = 209.45
.
_
=
m
(ho,ex - ho1) = 1(465.38 - 209.45) = 255.9 kW
W
c
Pex = Po,ex(Tex/To,ex)k/(k-1)
2
Tex = To,ex - Vex
/2Cp
.
m = 1 kg/s = (ρAV)ex = PexAVex/RTex
.
Now select 1 unknown amongst Pex, Tex, Vex and write the continuity eq. m
and solve the nonlinear equation. Say, use Tex then
Vex =
2Cp(To,ex - Tex)
o
.
m = 1 kg/s = Pex(Tex/To,ex)k/k-1A
2Cp(To,ex - Tex)/RTex
solve for Tex/To,ex (close to 1)
Tex = 462.6 K ⇒ Vex= 28.3 m/s, Pex = 498.6 kPa
16-25
English Unit Problems
16.44E Steam leaves a nozzle with a velocity of 800 ft/s. The stagnation pressure is 100
lbf/in2, and the stagnation temperature is 500 F. What is the static pressure and
temperature?
h1 = ho1 -
V21/2gc=
8002
Btu
1279.1 × 778 = 1266.3
lbm
2 × 32.174
s1 = s0 = 1.7085 Btu/lbm R
(h, s) Computer table ⇒ P1 = 88 lbf/in.2, T = 466 F
16.45E Air leaves the compressor of a jet engine at a temperature of 300 F, a pressure of
45 lbf/in2, and a velocity of 400 ft/s. Determine the isentropic stagnation
temperature and pressure.
ho1 - h1 = V21/2gc = 4002/2 × 32.174 × 778 = 3.2 Btu/lbm
To1 - T - 1 = (ho1 - h1)/Cp = 3.2/0.24 = 13.3
To1 = T + ∆T = 300 + 13.3 = 313.3 F = 773 R
k
Po1 = P1 To1/T1k-1 = 45(773/759.67)3.5 = 47.82 lbf/in2


16.46E A meteorite melts and burn up at temperatures of 5500 R. If it hits air at 0.75
lbf/in.2, 90 R how high a velocity should it have to reach such temperature?
Assume we have a stagnation T = 5500 R
h1 + V21/2 = hstagn.
Extrapolating from table C.6,
hstagn. = 1546.5, h1 = 21.4
V21/2 = 1546.5 – 21.4 = 1525.1 Btu/lbm
V1 =
2 × 32.174 × 778 × 1525.1 = 8738 ft/s
16-26
16.47E A jet engine receives a flow of 500 ft/s air at 10 lbf/in.2, 40 F inlet area of 7 ft2
with an exit at 1500 ft/s, 10 lbf/in.2, 1100 R. Find the mass flow rate and thrust.
.
m = ρAV; ideal gas ρ = P/RT
.
m = (P/RT)AV =
10 × 144
× 7 × 500 = 189.1 lbm/s
53.34 × 499.7
.
Fnet = m (Vex - Vin) = 189.1 × (1500 - 500) / 32.174 = 5877 lbf
16.48E A water turbine using nozzles is located at the bottom of Hoover Dam 575 ft
below the surface of Lake Mead. The water enters the nozzles at a stagnation
pressure corresponding to the column of water above it minus 20% due to
friction. The temperature is 60 F and the water leaves at standard atmospheric
pressure. If the flow through the nozzle is reversible and adiabatic, determine the
velocity and kinetic energy per kilogram of water leaving the nozzle.
∆P =
ρg∆Z
= g(∆Z/v)/gc = 575/(0.016035 × 144) = 249 lbf/in.2
gc
∆Pac = 0.8∆P = 199.2 lbf/in.2
Vex =
2v∆ P =
2g∆ Z =
and Bernoulli
v∆P = Vex2/2
2× 32.174× 575 = 192.4 ft/s
2
Vex
/2 = v∆P = g∆Z/gc = 575/778 = 0.739 Btu/lbm
16.49E Find the speed of sound for air at 15 lbf/in.2, at the two temperatures of 32 F and
90 F. Repeat the answer for carbon dioxide and argon gases.
From eq. 16.28 we have
c32 =
c90 =
kRT =
1.4 × 32.174 × 53.34 × 491.7 = 1087 ft/s
1.4 × 32.174 × 53.34 × 549.7 = 1149 ft/s
For Carbon Dioxide: R = 35.1, k = 1.289
c32 =
1.289 × 32.174 × 35.1 × 491.7 = 846 ft/s
c90 =
1.289 × 32.174 × 35.1 × 549.7 = 894.5 ft/s
For Argon: R = 38.68, k = 1.667
c32 =
1.667 × 32.174 × 38.68 × 491.7 = 1010 ft/s
c90 =
1.667 × 32.174 × 38.68 × 549.7 = 1068 ft/s
16-27
16.50E Air is expanded in a nozzle from 300 lbf/in.2, 1100 R to 30 lbf/in. 2. The mass
flow rate through the nozzle is 10 lbm/s. Assume the flow is reversible and
adiabatic and determine the throat and exit areas for the nozzle.
 2 k
P * = Po 
k-1
k+1
Mach #
Velocity
Area
Density
= 300 × 0.5283 = 158.5 lbf/in.2.
T* = To× 2/(k+1) = 1100 × 0.8333 = 916.6 R
v* = RT*/P* = 53.34 × 916.6/(158.5 × 144)
300 psia
P
30 psia
= 2.1421 ft3/lbm
kRT* = 1.4 × 32.174 × 53.34 × 916.6 = 1484 ft/s
.
A* = mv*/c* = 10 × 2.1421/1484 = 0.0144 ft2
c* =
P2/Po = 30/300 = 0.1
Table A.11 ⇒
M*2 = 1.701 = V2/c*
V2 = 1.701 × 1484 = 2524 ft/s
T2 = 916.6 × 0.5176 = 474.4 R
v2 = RT2/P2 = 53.34 × 474.4/(30 × 144) = 5.8579 ft3/lbm
.
A2 = mv2/V2 = 10 × 5.8579 / 2524 = 0.0232 ft2
16-28
16.51E A convergent nozzle has a minimum area of 1 ft2 and receives air at 25 lbf/in.2,
1800 R flowing with 330 ft/s. What is the back pressure that will produce the
maximum flow rate and find that flow rate?
P*
2 k
=(
)k-1 = 0.528
Po k+1
Find Po:
Critical Pressure Ratio
Cp = (463.445 - 449.794)/50 = 0.273
h0 = h1 + V21/2
T0 = 1800 +
from table C.6
⇒ T0 = Ti + V2/2Cp
3302/2
= 1807.97 => T * = 0.8333 To = 1506.6 R
32.174 × 778 × 0.273
P0 = Pi (T0/Ti)k/(k-1) = 25 × (1807.97/1800)3.5 = 25.39
P* = 0.528 Po = 0.528 × 25.39 = 13.406 lbf/in2
13.406 × 144
P*
ρ =
=
= 0.024
RT* 53.34 × 1506.6
*
V = c = kRT* = 1.4 × 53.34 × 1506.6 × 32.174 = 1902.6 ft/s
.
m = ρAV = 0.024 × 1 × 1902.6 = 45.66 lbm/s
16.52E A jet plane travels through the air with a speed of 600 mi/h at an altitude of
20000 ft, where the pressure is 5.75 lbf/in. 2 and the temperature is 25 F.
Consider the diffuser of the engine where air leaves at with a velocity of 300
ft/s. Determine the pressure and temperature leaving the diffuser, and the ratio of
inlet to exit area of the diffuser, assuming the flow to be reversible and
adiabatic.
V = 600 mi/h = 880 ft/s
v1 = 53.34 × 484.67/(5.75 × 144) = 31.223 , h1 = 115.91, Pr1 = 0.7637
ho1 = 115.91 + 880 2/(2 × 32.174 × 778) = 131.38 Btu/lbm
Pro1 = 1.1822 ⇒ Po1 =
5.75 × 1.1822
= 8.9 lbf/in.2
0.7637
h2 = 131.38 - 3002/(2 × 32.174 × 778) = 129.58
T2 = 542 R, Pr2 = 1.1267 => P2 = 8.9 × 1.1267/1.1822 = 8.48 lbf/in.2
v2 = 53.34 × 542/(8.482 × 144) = 23.67 ft3/lbm
A1/A2 = (31.223/23.67)(300/880) = 0.45
16-29
16.53E The products of combustion enter a nozzle of a jet engine at a total pressure of
18 lbf/in.2, and a total temperature of 1200 F. The atmospheric pressure is 6.75
lbf/in.2. The nozzle is convergent, and the mass flow rate is 50 lbm/s. Assume
the flow is adiabatic. Determine the exit area of the nozzle.
Pcrit = P2 = 18 × 0.5283 = 9.5 lbf/in.2 > Pamb
The flow is then choked.
T2 = 1660 × 0.8333 = 1382 R
V 2 = c2 =
1.4 × 32.174 × 53.34 × 1382 = 1822 ft/s
v2 = 53.34 × 1382/9.5 × 144 = 53.9
A2 = 50 × 53.9/1822 = 1.479 ft2
16.54E Repeat Problem 16.52 assuming a diffuser efficiency of 80%.
From solution to 16.52
h1 = 115.91, ho1 = 131.38, Pr1 = 0.7637, v1 = 31.223
h2 = 129.58, Pr2 = 1.1267, Pro1 = 1.1822
ηD = (h3 - h1)/(ho1 - h1) = 0.8
⇒
h3 = 128.29, Pr3 = 1.088
Po2=P3 = 5.75 × 1.088/0.7637 = 8.192 lbf/in.2
P2 = Po2Pr2/Pro1 = 8.192 × 1.1267/1.1822 = 7.807 lbf/in.2
T2 = 542 R ⇒ v2 =
53.34 × 542
× 144 = 25.716 ft3/lbm
7.807
A1/A2 = v1V2/v2V1 = 31.223 × 300/(25.716 × 880) = 0.414
16-30
16.55E A 50-ft3 uninsulated tank contains air at 150 lbf/in.2, 1000 R. The tank is now
discharged through a small convergent nozzle to the atmosphere at 14.7 lbf/in.2
while heat transfer from some source keeps the air temperature in the tank at
1000 R. The nozzle has an exit area of 2 × 10−4 ft2.
a. Find the initial mass flow rate out of the tank.
b. Find the mass flow rate when half the mass has been discharged.
c. Find the mass of air in the tank and the mass flow rate out of the tank
when the nozzle flow changes to become subsonic.
PB/Po = 14.7/150 = 0.098 < (P*/Po)crit = 0.5283
a. The flow is choked, max possible flow rate
ME =1 ;
PE = 0.5283 × 150 = 79.245 lbf/in.2
TE = T* = 0.8333 × 1000 = 833.3 R
VE = c =
kRT* =
1.4 × 53.34 × 833.3 × 32.174 = 1415 ft/s
vE = RT*/PE = 53.34 × 833.3/(79.245 × 144) = 3.895 ft3/lbm
.
Mass flow rate is : m1 = AVE/vE = 2 × 10-4 × 1415/3.895 = 0.0727 lbm/s
b. m1 = P1V/RT1 = 150 × 50 × 144/53.34 × 1000 = 20.247 lbm
m2 = m1/2 = 10.124 lbm,
P2=P1/2 = 75 lbf/in.2 ;
T2 = T1
PB/P2= 14.7/75 = 0.196 < (P*/Po)crit
The flow is choked and the velocity is the same as in 1)
PE = 0.5283 × 75 = 39.623 lbf/in.2 ; ME =1
2 × 10-4 × 1415 × 39.623 × 144
.
= 0.0303 lbm/s
m2 = AVEPE/RTE =
53.34 × 1000
c. Flow changes to subsonic when the pressure ratio reaches critical.
PB/Po = 0.5283 P3 = 27.825 lbf/in.2
m3 = m1P3/P1 = 3.756 lbm ; T3 = T1 ⇒ VE = 1415 ft/s
2 × 10-4 × 1415 × 27.825 × 144
.
= 0.02125 lbm/s
m3 = AVEPE/RTE =
53.34 × 1000
16-31
16.56E Air enters a diffuser with a velocity of 600 ft/s, a static pressure of 10 lbf/in.2,
and a temperature of 20 F. The velocity leaving the diffuser is 200 ft/s and the
static pressure at the diffuser exit is 11.7 lbf/in. 2 . Determine the static
temperature at the diffuser exit and the diffuser efficiency. Compare the
stagnation pressures at the inlet and the exit.
To1 = T1 +
To2 = To1
V21
2gcCp
= 480 + 6002/(2 × 32.174 × 778 × 0.24) = 510 R
⇒
T2 = To2 - V22/2Cp = 510 - 2002/(2 × 32.174 × 0.24 × 778) = 506.7 R
To2 - T2
T2
=
k-1 Po2 - P2
⇒ Po2 - P2 = 0.267 ⇒ Po2 = 11.97 lbf/in.2
k
P2
Tex,s = T1 (Po2/P1)(k-1)/k = 480 × 1.0528 = 505.3 R
ηD =
Tex,s - T1
To1 - T1
=
505.3 - 480
= 0.844
51 - 480
16-32
16.57E A convergent nozzle with exit diameter of 1 in. has an air inlet flow of 68 F,
14.7 lbf/in.2 (stagnation conditions). The nozzle has an isentropic efficiency of
95% and the pressure drop is measured to 20 in. water column. Find the mass
flow rate assuming compressible adiabatic flow. Repeat calculation for
incompressible flow.
Convert ∆P to lbf/in2
∆P = 20 in H2O = 20 × 0.03613 = 0.7226 lbf/in2
T0 = 68 F = 527.7 R
P0 = 14.7 lbf/in2
Assume inlet Vi = 0
Pe = P0 - ∆P = 14.7 - 0.7226 = 13.977 lbf/in2
Pe k-1
13.977 0.2857
Te = T0 ( ) k = 527.7 ×(
)
= 520.15
14.7
P0
V2e /2 = hi - he = Cp (Ti - Te) = 0.24 × (527.7 - 520.15) = 1.812 Btu/lbm
Ve 2ac/2 = η V2e /2 = 0.95 × 1.812 = 1.7214 Btu/lbm
⇒ Ve ac =
2 × 32.174 × 1.7214 × 778 = 293.6 ft/s
Te ac = Ti -
Ve 2ac/2
ρe ac =
Pe
RTe ac
Cp
=
= 527.7 -
1.7214
= 520.53
0.24
13.977 × 144
= 0.07249 lbm/ft3
53.34 × 520.53
π
1
.
m = ρAV = 0.07249 × × ( )2 × 293.6 = 0.116 lbm/s
4 12
P0
14.7 × 144
= 0.0752 lbm/ft3
53.34 × 527.7
Incompressible:
ρi =
V2e /2 = vi (Pi - Pe) =
∆P 0.7226 × 144
=
= 1.7785 Btu/lbm
ρi 0.0752 × 778
RT0
=
Ve 2ac/2 = η V2e /2 = 0.95 × 1.7785 = 1.6896 Btu/lbm
⇒ Ve ac =
2 × 32.174 × 1.6896 × 778 = 290.84 ft/s
π
.
1
m = ρAV = 0.0752 × × ( )2 × 290.84 = 0.119 lbm/s
4 12