Borgnakke and Sonntag
5.61
A rigid tank is divided into two rooms by a membrane, both containing water, shown
in Fig. P5.61. Room A is at 200 kPa, v = 0.5 m3/kg, VA = 1 m3, and room B contains
3.5 kg at 0.5 MPa, 400°C. The membrane now ruptures and heat transfer takes place
so the water comes to a uniform state at 100°C. Find the heat transfer during the
process.
Solution:
A
C.V.: Both rooms A and B in tank.
Continuity Eq.:
m2 = mA1 + mB1 ;
Energy Eq.:
m2u2 - mA1uA1 - mB1uB1 = 1Q2 - 1W2
State 1A: (P, v) Table B.1.2,
B
mA1 = VA/vA1 = 1/0.5 = 2 kg
v – vf
0.5 - 0.001061
=
= 0.564
xA1 = v
0.88467
fg
uA1 = uf + x ufg = 504.47 + 0.564 × 2025.02 = 1646.6 kJ/kg
State 1B: Table B.1.3, vB1 = 0.6173, uB1 = 2963.2, VB = mB1vB1 = 2.16 m3
Process constant total volume:
Vtot = VA + VB = 3.16 m3 and 1W2 = 0/
m2 = mA1 + mB1 = 5.5 kg =>
v2 = Vtot/m2 = 0.5746 m3/kg
State 2: T2 , v2 ⇒ Table B.1.1 two-phase as v2 < vg
v2 – vf
0.5746 – 0.001044
x2 = v
=
= 0.343 ,
1.67185
fg
u2 = uf + x ufg = 418.91 + 0.343 × 2087.58= 1134.95 kJ/kg
Heat transfer is from the energy equation
1Q2 = m2u2 - mA1uA1 - mB1uB1 = -7421 kJ
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Borgnakke and Sonntag
5.134
An air pistol contains compressed air in a small cylinder, shown in Fig. P5.134.
Assume that the volume is 1 cm3, pressure is 1 MPa, and the temperature is 27°C
when armed. A bullet, m = 15 g, acts as a piston initially held by a pin (trigger); when
released, the air expands in an isothermal process (T = constant). If the air pressure is
0.1 MPa in the cylinder as the bullet leaves the gun, find
a.
The final volume and the mass of air.
b.
The work done by the air and work done on the atmosphere.
The work to the bullet and the bullet exit velocity.
c.
Solution:
C.V. Air.
Air ideal gas: mair = P1V1/RT1 = 1000 × 10-6/(0.287 × 300) = 1.17×10-5 kg
Process: PV = const = P1V1 = P2V2 ⇒ V2 = V1P1/P2 = 10 cm3
⌠P1V1
dV = P1V1 ln (V2/V1) = 2.303 J
1W2 = ⌠
⌡PdV = 
⌡ V
-6
1W2,ATM = P0(V2 - V1) = 101 × (10 − 1) × 10 kJ = 0.909 J
1
Wbullet = 1W2 - 1W2,ATM = 1.394 J = 2 mbullet(Vexit)2
Vexit = (2Wbullet/mB)1/2 = (2 × 1.394/0.015)1/2 = 13.63 m/s
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Borgnakke and Sonntag
6.138
A steam engine based on a turbine is shown in Fig. P6.138. The boiler tank has a
volume of 100 L and initially contains saturated liquid with a very small amount
of vapor at 100 kPa. Heat is now added by the burner, and the pressure regulator
does not open before the boiler pressure reaches 700 kPa, which it keeps constant.
The saturated vapor enters the turbine at 700 kPa and is discharged to the
atmosphere as saturated vapor at 100 kPa. The burner is turned off when no more
liquid is present in the boiler. Find the total turbine work and the total heat
transfer to the boiler for this process.
Solution:
C.V. Boiler tank. Heat transfer, no work and flow out.
Continuity Eq.6.15:
m2 - m1 = − me
Energy Eq.6.16:
m2u2 - m1u1 = QCV - mehe
State 1: Table B.1.1, 100 kPa => v1 = 0.001 043, u1 = 417.36 kJ/kg
=> m1 = V/v1 = 0.1/0.001 043 = 95.877 kg
State 2: Table B.1.1, 700 kPa => v2 = vg = 0.2729, u2 = 2572.5 kJ/kg
=>
m2 = V/vg = 0.1/0.2729 = 0.366 kg,
Exit state: Table B.1.1, 700 kPa => he = 2763.5 kJ/kg
From continuity eq.: me = m1 - m2 = 95.511 kg
QCV = m2u2 - m1u1 + mehe
= 0.366 × 2572.5 - 95.877 × 417.36 + 95.511 × 2763.5
= 224 871 kJ = 224.9 MJ
C.V. Turbine, steady state, inlet state is boiler tank exit state.
Turbine exit state: Table B.1.1, 100 kPa => he = 2675.5 kJ/kg
Wturb = me (hin- hex) = 95.511 × (2763.5 - 2675.5) = 8405 kJ
W
cb
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Borgnakke and Sonntag
8.171
Water in a piston/cylinder is at 1 MPa, 500°C. There are two stops, a lower one at
which Vmin = 1 m3 and an upper one at Vmax = 3 m3. The piston is loaded with a
mass and outside atmosphere such that it floats when the pressure is 500 kPa. This
setup is now cooled to 100°C by rejecting heat to the surroundings at 20°C. Find
the total entropy generated in the process.
C.V. Water.
v1 = 0.35411 m3/kg, u1 = 3124.3, s1 = 7.7621
m =V/v1 = 3/0.35411 = 8.472 kg
Initial state: Table B.1.3:
T
P
1000
500
v=C
1
1
2
2
s
v
Final state: 100°C and on line in P-V diagram.
Notice the following: vg(500 kPa) = 0.3749 > v1,
v1 = vg(154°C)
Tsat(500 kPa) = 152°C > T2 , so now piston hits bottom stops.
State 2: v2 = vbot = Vbot/m = 0.118 m3/kg,
x2 = (0.118 − 0.001044)/1.67185 = 0.0699,
u2 = 418.91 + 0.0699×2087.58 = 564.98 kJ/kg,
s2 = 1.3068 + 0.0699×6.048 = 1.73 kJ/kg K
Now we can do the work and then the heat transfer from the energy equation
PdV = 500(V2 - V1) = -1000 kJ (1w2 = -118 kJ/kg)
1W2 = ⌠
⌡
1Q2 = m(u2 - u1) + 1W2 = -22683.4 kJ (1q2 = -2677.5 kJ/kg)
Take C.V. total out to where we have 20°C:
m(s2 - s1) = 1Q2/T0 + Sgen ⇒
Sgen = m(s2 - s1) − 1Q2/T0 = 8.472 (1.73 - 7.7621) + 22683 / 293.15
= 26.27 kJ/K
( = ∆Swater + ∆Ssur )
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Borgnakke and Sonntag
8.178
A cylinder fitted with a piston contains 0.5 kg of R-134a at 60°C, with a quality
of 50 percent. The R-134a now expands in an internally reversible polytropic
process to ambient temperature, 20°C at which point the quality is 100 percent.
Any heat transfer is with a constant-temperature source, which is at 60°C. Find
the polytropic exponent n and show that this process satisfies the second law of
thermodynamics.
Solution:
C.V.: R-134a, Internally Reversible, Polytropic Expansion: PVn = Const.
Cont.Eq.: m2 = m1 = m ;
Entropy Eq.:
Energy Eq.:
m(u2 − u1) = 1Q2 − 1W2
m(s2 − s1) = ∫ dQ/T + 1S2 gen
State 1: T1 = 60oC, x1 = 0.5, Table B.5.1: P1 = Pg = 1681.8 kPa,
v1 = vf + x1vfg = 0.000951 + 0.5×0.010511 = 0.006207 m3/kg
s1 = sf + x1sfg = 1.2857 + 0.5×0.4182 = 1.4948 kJ/kg K,
u1 = uf + x1ufg = 286.19 + 0.5×121.66 = 347.1 kJ/kg
State 2: T2 = 20oC, x2 = 1.0, P2 = Pg = 572.8 kPa, Table B.5.1
v2 = vg = 0.03606 m3/kg, s2 = sg = 1.7183 kJ/kg-K
u2 = ug = 389.19 kJ/kg
P1
v2
P1 v2n
Process: PVn = Const. => P = v  => n = ln P / ln v = 0.6122
2  1
2
1
1W2 = ∫ PdV =
P2V2 - P1V1
1-n
= 0.5(572.8 × 0.03606 - 1681.8 × 0.006207)/(1 - 0.6122) = 13.2 kJ
2nd Law for C.V.: R-134a plus wall out to source:
QH
∆Snet = m(s2 − s1) − T ;
Check ∆Snet > 0
H
QH = 1Q2 = m(u2 − u1) + 1W2 = 34.2 kJ
∆Snet = 0.5(1.7183 - 1.4948) - 34.2/333.15 = 0.0092 kJ/K,
∆Snet > 0 Process Satisfies 2nd Law
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Borgnakke and Sonntag
9.159
Air at 100 kPa, 17°C is compressed to 400 kPa after which it is expanded through
a nozzle back to the atmosphere. The compressor and the nozzle are both
reversible and adiabatic and kinetic energy in and out of the compressor can be
neglected. Find the compressor work and its exit temperature and find the nozzle
exit velocity.
Solution:
1
air
.
W
i
2
-W
Turbine
Adiabatic : q = 0.
Reversible: sgen = 0
e
1=3
Energy Eq.6.13:
Entropy Eq.9.8:
Separate control volumes around
compressor and nozzle. For ideal
compressor we have
inlet : 1 and exit : 2
h1 + 0 = wC + h2;
s1 + 0/T + 0 = s2
- wC = h2 - h1 , s2 = s1
Properties Table A.5 air: CPo = 1.004 kJ/kg K, R = 0.287 kJ/kg K, k = 1.4
Process gives constant s (isentropic) which with constant CPo gives Eq.8.32
=>
⇒
k-1
T2 = T1( P2/P1) k
= 290 (400/100) 0.2857 = 430.9 K
−wC = CPo(T2 – T1) = 1.004 (430.9 – 290) = 141.46 kJ/kg
The ideal nozzle then expands back down to P1 (constant s) so state 3 equals
state 1. The energy equation has no work but kinetic energy and gives:
1 2
2V
⇒
V3 =
= h2 - h1 = -wC = 141 460 J/kg (remember conversion to J)
2×141460 = 531.9 m/s
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Borgnakke and Sonntag
9.160
Assume both the compressor and the nozzle in Problem 9.37 have an isentropic
efficiency of 90% the rest being unchanged. Find the actual compressor work and
its exit temperature and find the actual nozzle exit velocity.
1
T
2
3
P2 = P3
C.V. Ideal compressor, inlet: 1 exit: 2
3
-W
1 4 5
5
Energy Eq.6.13:
Entropy Eq.9.8:
P1
Adiabatic : q = 0.
Reversible: sgen = 0
s
h1 + 0 = wC + h2;
s1 + 0/T + 0 = s2
- wCs = h2 - h1 , s2 = s1
Properties use air Table A.5:
kJ
kJ
CPo = 1.004 kg K, R = 0.287 kg K, k = 1.4,
Process gives constant s (isentropic) which with constant CPo gives Eq.8.32
=>
⇒
k-1
T2 = T1( P2/P1) k = 290 (400/100) 0.2857 = 430.9 K
−wCs = CPo(T2 – T1) = 1.004 (430.9 – 290) = 141.46 kJ/kg
The ideal nozzle then expands back down to state 1 (constant s). The actual
compressor discharges at state 3 however, so we have:
wC = wCs/ηC = -157.18 ⇒ T3 = T1 - wC/Cp = 446.6 K
Nozzle receives air at 3 and exhausts at 5. We must do the ideal (exit at 4)
first.
s4 = s3 ⇒ Eq.8.32:
1 2
2 Vs
T4 = T3 (P4/P3
k-1
)k
= 300.5 K
1
= Cp(T3 - T4) = 146.68 ⇒ 2 V2ac = 132 kJ/kg ⇒ Vac = 513.8 m/s
If we need it, the actual nozzle exit (5) can be found:
T5 = T3 - V2ac/2Cp = 315 K
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