MAHALAKSHMI
ENGINEERING COLLEGE
TIRUCHIRAPALLI-621213.
Department: Mechanical
Semester: III
Subject Code: ME2202
Subject Name: ENGG. THERMODYNAMICS
UNIT III
1) A rigid tank of 0.03 m 3 capacity contains wet vapour at 80 kPa. If the wet vapour mass is
12 kg, calculate the heat added and the quality of the mixture when the pressure inside
the tank reaches 7 MPa.
(AUC DEC’05)
Given:
Rigid tank
Volume (V1) = V2 = 0.03m 3 [rigid tank]
Pressure (P1) = 80 Kpa = 80 / 100 KN/ m 2 => 0.8 bar
Mass (m) = 12 Kg
Pressure (P2) = 7Mpa × 1000 = 7000/100 KN /m 2
= 70 bar
To find:
a) Heat added (Q)
b) Quality of mixture ( X1 & X2) (or) dryness fraction
Solution:
Step: 1
Specific volume (V) = V1 / m =0.03/ 12 =0.0025 m 3 /Kg
Step:2
From steam tables, corresponding to (P 1 =0.8 bar)
Read Vf, Vg, hf , hf g [pressure table pg: no: 9]
Vf = 0.001039 m 3/kg
Vg = 2.0869 m 3 / kg
hf = 391.7 KJ /kg
hf g = 2274.21 KJ /kg
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Step:3
Specific volume (V) = Vf + x1 Vhf g
0.0025 = 0.001039 + x1 [ 2.0869 – 0.001039]
0.0025 = 0.001039 + x1 (2.0858)
0.0025 - 0.001039 = 2.0858 x1
0.001461 = 2.0858 x1
0.001461/ 2.0858 = x1
x1 = 0.0007
Step :4
for rigid vessel (V1 =V2)
V2 = 0.0025 m 3 / kg
From steam tables , corresponding to (P2 = 70 bar) [pg:no:13]
Vf = 0.001351 m 3/kg
hf = 1267.4 KJ /kg
Vg = 0.027368 m 3 / kg
hf g = 1506.0 KJ /kg
Vg > V2
0.027368 > 0.0025
The steam is in wet condition
(V2) = Vf 2 + (x2 . Vg2)
0.0025 = 0.001351 + (x2) . [0.027368 – 0.001351]
0.0025 - 0.001351 = (x2) . (0.026017)
0.001149 = x2 . (0.026017)
0.001149 / 0.026017 = x2
x2 = 0.044
(h2) = hf 2 + (x2 . hf g2)
(h2) = 1267 .4 + [0.044 × 1506.0]
(h2) = 1333.9101 KJ/ kg
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Step:5
By first law of thermodynamics
Q = W + ( U)
Where
W =0 [for rigid vessel]
Q=
U
Total heat transfer (Q) = m [h2 – h1] – V [P2 – P1]
(Q) = 12 [1333.9101 – 393.2918] – 0.0025 [7000 – 80]
(Q) = 11287.4196 – 17.3
Q = 11270.1196 KJ
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o
2) Steam enters the turbine at 3 MPa and 400 C and is condensed at 10 kPa. Some
quantity of steam leaves the turbines at 0.6 MPa and enters open feed water heater.
Compute the fraction of the steam extracted per kg of steam and cycle thermal
efficiency. (AUC DEC’05)
Given:
Regenerative rankine cycle
Pressure (P1) = 3Mpa × 1000 KN / m 2 = 30 bar
100
Temp (T1) = 400oC
Pressure (P2) = 0.6 Mpa × 1000 KN / m 2 = 6 bar
100
Pressure (P3) = 10 Kpa = KN / m 2 = 0.1 bar
100
To find:
ή
Thermal efficiency ( )
Solution :
Step:1
From superheated steam table
At
P1 = 30 bar and T1 = 4000C
S1 = 6.925 KJ /Kg . K [pg.no :22]
H1 = 3232 . 5 KJ /Kg . K [pg.no :19]
S1 = S2 => 6.925 KJ /Kg . K
But at P2= bar => pressure table
Sg = 6.758 KJ /Kg . K [pg.no :11]
S1 > S2
6.925 > 6.758 so the steam is again in superheated state
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From mollier diagram, corresponding to S2= 6.925 KJ /Kg . K entropy at 6 bar the enthalpy is
found
H2 = 2630 KJ / Kg
Step:2
S1 = S3 => 6.925 KJ/kg.K
At P3 = 0.1 bar [pressure table pg.no:07]
Vf3 = 0.001010 m 3 /kg
hf3 = 191.8 KJ/kg
Vg3 = 14.675 m 3 /kg
hfg3 = 2392.9 KJ/kg
Sf3 = 0.649 KJ/kg.K
Sfg3 = 7.502 KJ/kg.K
S1 = S3 => Sf3 + (x3 × Sfg3)
6.925 = 0.649 + (x3 × 7.502)
6.925 - 0.649 = 7.502 x3
6.925 - 0.649 /7.502 = x3
x3 = 0.8365
h3 = hf3 + (x3 × hfg3)
h3 = 191.8 + (0.8365 × 2392.9)
h3 = 2193.6448 KJ/kg
h4 = hf3 => 191.8 KJ/Kg
h5 - h4 = Vf3 [P2 - P3]
h5 - h4 = 0.001010 [600 – 10]
h5 - h4 = 0.5959 KJ/Kg
h5= 0.5959 + h4
h5= 0.5959 + 191.8
h5= 192.3959 KJ/Kg
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Step:4
Amount of steam bleed
M = h6 – h5 / h2 – h5
h6 = hf = 670.4 KJ /Kg => [P2 =6 bar use pg.no:11]
M = 670.4 – 192.4 / 2630 – 192.4
M = 0.1960 KJ / Kg of steam
Step:5
At 6 bar
Vf = 0.001101 m 3 / kg
Wp 6-7 = h7 – h6 => Vf 2 [P1 – P2]
= 0.001101 [3000 – 600]
h7 – h6 = 2.6424 KJ /Kg
h7 = 2.6424 + h6
h7 = 2.6424 + 670.4
h7 = 673.0424 KJ /Kg
Step:6
Regenerative rankine cycle efficiency
ή
reg
= (h1- h7) – (l – m) (h3- hf 3)
(h1- h7)
= (3232.5 – 673.0424) – (1 – 0.1960) (2193.6448 – 191.8)
3232.5 – 673.0424
= 2559.4576 – 1609.4832
2559.4576
= 0.3711 × 100
ή
reg
= 37.11%
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3) 1 kg of steam initially dry saturated at 1.1 MPa expands in a cylinder following the law
PV1.13 = C. The pressure at the end of expansion is 0.1MPa. Determine
(i)
The final volume
(ii)
Final dryness fraction
(iii)
Work done
(iv)
The change in internal energy
(v)
The heat transferred.
(AUC DEC’06)
Given:
Poly tropic process
Mass (m) = 1Kg
Pressure (P1) =1.1 Mpa × 1000 => 1100 / 100 KN/m 2 = 11bar
“
(P2) =0.1 Mpa × 1000 => 100 / 100 KN/m 2 = 1 bar
PV1.13 = C
To find:
(i)
The final volume (V2)
(ii)
Final dryness fraction (x2)
(iii)
Work done (W)
(iv)
The change in internal energy (
(v)
The heat transferred (Q)
U)
Solution:
Step:1
From steam table at (P1 = 11 bar)
V1 = Vg1 = 0.17739 m 3 / kg
H1 = hg1 =2779.7 KJ / Kg
= [Pg.no:11]
Step:2
Polytropic process
V2 = [P1 /P2] 1/n × V1
V2 = [11/1]
1/1.13
x 0.17739
Final volume (V2) = 1.4808 m 3
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Step:3
From steam table at (P2 = 1 bar)
Hf 2 = 417.5 KJ/Kg
Vf 2= 0.001043 m 3 / kg
Hf g2 = 2257.9 KJ/Kg
Vg2 = 1.6938 m 3 / kg
Vfg2 = Vg2 - Vf 2 => 1.6938 – 0.001043
Vfg2 = 1.6927 m 3 / kg
V2= Vf2 + x2 . Vfg2
1.4808 = 0.001043 + X2 . (1.6927)
1.4808 - 0.001043 = 1.6927 X2
1.4808 - 0.001043 / 1.6927 = X2
Final dryness fraction (X2) = 0.8742
Step:4
Work done (W) = P1 V1 – P2 V2
n-1
= (1100 x 0.17739) –(100 x 1.4808)
1.13 – 1
(W) = 361.9153 KJ
Step:5
Change in internal energy (
(
U) = u1 – u2
U) = (h1 – h2) – (P1 V1 – P2 V2 )
Where
h2 = hf2 + x2 . hfg2
h2 = 417.5 + (0.8742 x 2257.9)
h2 = 2391.3561 KJ/Kg
(
U) = (h1 – h2) – (P1 V1 – P2 V2 )
(
U) = (2779.7 – 2391.3561) –(1100 x 0.17739 -100 x 1.4808)
(
U) = 388.3438 – 47.049
(
U) = 341.2948 KJ
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Step:6
By first law of thermodynamics
Heat transfer (Q) = W +
U
(Q) = 361.9153 + 341.2948
(Q) = 703.2101 KJ
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4) Steam at a pressure of 2.5 MPa and 500oC is expanded in a steam turbine to a
Condenser pressure of 0.05 MPa. Determine for Rankine cycle:
(i) The thermal efficiency of Rankine cycle
(ii) Specific steam consumption.
If the steam pressure is reduced to 1 MPa and the temperature is kept same
500oC.Determine the thermal efficiency and the specific steam consumption.
Neglect feed pump work.
(AUC DEC’06)
Given:
Rankine cycle
P1 = 2.5Mpa x 1000 = 2500/100 KN / m 2
T1 = Tsup = 5000C
P2 = 0.05 Mpa x 1000 = 50 /100 KN/m 2
To find:
Case:1
(i) The thermal efficiency of Rankine cycle
(ii) Specific steam consumption.
Case:2
(i) The thermal efficiency of Rankine cycle
(ii) Specific steam consumption.
Solution:
case:1
Step:1
From super heated steam table at [P1 = 2.5 bar , T1= 5000C] (pg.no:22)
H1 = 3462.9 + 3460.6 / 2
H1 = 3461.75 KJ/Kg
S1 = 7.344 + 7.305 / 2
S1 = 7.3245 KJ/Kg.K
From steam table at (P2 = 0.5 bar)
Vf2 = 0.001030 m 3/ Kg
Vg2 = 3.2401 m 3/ Kg
hf2 = 340.6 KJ/Kg Sf2 = 1.091 KJ/Kg.k
hfg2 = 2305.4 KJ/Kg Sfg2 = 6.504 KJ/Kg.k
 S1 = S2 = Sf2 + X2 Sfg2
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7.3245 = 1.091 + (6.504 x X2)
7.3245 – 1.091 / 6.504 = X2
X2 = 0.9584
 H2 = hf2 + ( X2 hfg2)
H2 =340.6 + (0.9584 x 2305.4)
H2 = 2550.1188 KJ/Kg
 Heat supplied (Qs) = h1 – h2
(Qs) = 3461.75 – 340.6
(Qs) = 3121.15 KJ/Kg
 Work done (W) = h1 – h2
(W) =3461.75 – 2550.1188
(W) = 911.6312 KJ
 Thermal efficiency rankine = W/Qs = 911.6312 / 3121.15
ήth = 0.2920 x 100
ήth = 29.20 %
 Specific steam consumption
= 3600 / W =3600 / 911.6312
= 3.9489 Kg/Kw .hr
Case:2
Step:2
Condition:
 The steam pressure is reduced to
P1 = 1Mpa x 1000 =1000/100 KN /m 2 = 10 bar
T1 = 5000C
From super heated steam table at P1 = 10bar and T1 = 5000C
H1 = 3478.3 KJ/Kg [pg.no:19]
S1 = 7.763 KJ/Kg.K [pg.no:22]
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 From steam table P1 = 0.5bar
Vf2 = 0.001030 m 3 /kg
3
Vg2 = 3.2401 m /kg
hf2 = 340.6 KJ/kg Sf2 = 1.091 KJ/Kg.K
hfg2 = 2305.4 KJ/kg Sfg2 = 6.504 KJ/Kg.K
Sg2 = 7.595 KJ/Kg.K
S1 = S2 => 7.763 KJ / Kg.K
S2 > Sg2
7.763 > 7.595
=> steam is again super heated condition
Again from super heated table and [mollier diagram use]
Corresponding ( P2 = 0.5 bar and super heated steam 4000C meet)
H2 = 3279.0 KJ/Kg [pg.no: -18]
 Heat supplied (Qs) = h1 – h2
(Qs) = 3478.3 – 340.6
(Qs) = 3137.7 KJ/Kg
 Work done (W) = h1 – h2
(W) =3478.3 – 3279
(W) = 199.3 KJ
 Thermal efficiency rankine = W/Qs = 199.3 / 3137.7
ήth = 0.06351 x 100
ήth = 6.35 %
 Specific steam consumption
= 3600 / W =3600 / 199.3
= 18.0623 Kg/Kw .hr
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5) A cyclic steam power plant is to be designed for a steam temperature at turbine inlet
of633K and an exhaust pressure of 8 kPa. After isentropic expansion of steam in the
turbine, the moisture content at the turbine exhaust is not to exceed 157 o. Determine the
greatest allowable steam pressure at the turbine inlet, and calculate the Rankine cycle
efficiency for these steam conditions. Estimate also the mean temperature of heat
addition.(AUC DEC’07)
Solution:
Rankine cycle
T1 = 633 K
After isentropic expansion ( X2) = 85/100 =>0.85
Exhaust pressure (P2) = 8Kpa / 100 =0.08 bar
To find:
a) Thermal efficiency (
ήth )
b) Mean temp (Tmean)
Solution:
Step:1
From steam table ( P2 = 0.8 bar) [pg.no:07]
Sf2 = 0.593 KJ/Kg.K Vf2 = 0.001008 m 3 /kg hf2 = 173.9 KJ/Kg
Sfg2 = 7.637 KJ/Kg.K Vg2 =18.105 m 3 /kg hfg2 = 2403.2 KJ/Kg
Sg2 = 8.230 KJ/Kg.K
hg2 = 2577.1 KJ/Kg
 S2 = Sf2 + (x2 Sfg2)
S2 = 0.593 + (0.25 x 7.637)
S2 = 7.0844 KJ/Kg.K
 h2 = hf2 + (x2 hfg2)
h2 =173.9 + (0.85 x 2403.2)
h2 = 2216.62 KJ/Kg
S1 = S2 = 7.0844 KJ/Kg.K
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S1 = 7.0844 => compare steam table [Pg.no: -22]
6.960 – 7.130 KJ/Kg.K
Select 20 bar
H1 = 3138.6 + 3248.7 / 2 => at 20 bar [pg.no:20]
H1 = 3193.65 KJ/Kg
H3 = hf2 => 173.9 KJ/Kg
 pump work (Wp) = Vf2 ( P1 – P2)
(P1 = 20 bar x 100 = 2000 KN /m 2)
(Wp) = 0.001008 (2000 – 8)
(Wp) = 2.0079 KJ/Kg
 heat supplied (Qs) = h1 (hf2 + Wp)
(Qs) = 3193.65 – (173.9 + 2.0079)
(Qs) = 3017.7421 KJ/Kg
 work done (W) = (h1 – h2) – Wp
(W) = (3193.65 – 2216.62)
(W) = 975 .0221 KJ/Kg
 Thermal efficiency = W/Qs = 975 .0221 / 3017.7421
ήth = 0.3230 x 100
ήth = 32.30 %
 Mean temp of heat addition (Tmean)
= h1 – h3 / s 1 – s 3 = 3193.65 – 173.9 / 7.0844 – 0.593
(Tmean) = 465 K
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6) 3 kg of steam at 18 bar occupy a volume of 0.2550 m 3. During a constant volume
process, the heat rejected is 1320 kJ. Determine final internal energy. Find dryness
fraction and pressure, change in entropy and work done.
(AUC MAY’08)
Given:
Constant volume process
V1 = 0.2550 m 3
P1 = 18 bar x 100 = 1800 KN / m 2
M1 = 3 kg
Heat rejected (Q) =1320 KJ
To find:
a) Internal energy (u2)
b) Initial dryness fraction (X)
c) Work done (W)
Solution:
Step:1
Specific volume (V1 ) = V / m
= 0.2550 / 3
(V1 ) = 0.085 m 3 / kg
Step:2
From steam table , corresponding to 18 bar
Vf1 = 0.00168 m 3 / kg
hf1 = 884.5 KJ/kg
Vg1 = 0.11033 m 3 / kg
hfg1 = 1910.3 KJ/Kg
Sf1 = 2.398 KJ/Kg.K
Sfg1 = 3.977 KJ/Kg.k
V1 = X1 x Vg1
0.085 = X1 x 0.11033
0.085 / 0.11033 = X1
X1 = 0.7704
H1 = hf1 + (X1 hfg1)
H1 = 884.5 + (0.7704 x 1910.3)
H1 = 2356.2257 KJ/kg.K
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S1 = Sf1 + (X1 Sfg1)
S1 = 2.398 + (0.7704 x 3.977)
S1 = 5.4618 KJ/Kg.k
Step:3
For closed vessel (V1 = V2)
For constant volume process (W) = 0
By first law of thermodynamics
Q=W+
u
Where
Q = u2 – u1
Q = m (u2 – (n1 – P1 V1))
1320 = 3 (u2 – (2356.2257 – (1800 x 0.085))
1320 / 3 = u2 – 2203.2257
440 = u2 – 2203.2257
440 + 2203.2257 = u2
u2 = 2643.2257 KJ/Kg
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7) A steam power plant runs on a single regenerative heating process. The s team enters
the turbine at 30 bar and 400oC and steam fraction is withdrawn at 5 bar. The remaining
steam exhausts at 0.10 bar to the condenser. Calculate the efficiency, steam fraction
and steam rate of the power plant. Neglect pumps work. (AUC MAY’08)
Given:
Single regenerative heating process
P1 = 30 bar
T1 = 4000C
With draw pressure (P2 ) = 5 bar
Exhaust pressure to the condenser (P3 ) = 0.1 bar
To find:
a) Efficiency
b) Steam fraction
c) Steam rate of the power plant
Solution:
Step:1
At 30 bar and T1 =4000C => from super heated steam table
H1 =3232.5 KJ/Kg
S1 = 6.925 KJ/Kg.K
P2 = 5 bar => from pressure table [pg.no:10]
Tsat = 151.80C
hf2 =640.1 KJ/kg
3
Vf2 = 0.001093 m / kg
hfg2 = 2107.4 KJ/kg
Vg2 = 0.37466 m 3 / kg
hg2 = 2747.5 KJ/kg
Sf2 =1.860 KJ/kg.K
Sfg2 = 4.959 KJ/kg.K
Sg2 = 6.819 KJ/kg.K
P3 = 0.1 bar => from pressure table [pg.no:07]
Tsat = 45.830C
Sf3 =0.649 KJ/kg.K
Hf3 =191.8 KJ/kg
Sfg3 = 7.052 KJ/kg.K
Hfg3 = 2392.9 KJ/kg
Sg3 = 8.151 KJ/kg.K
Hg3 =2584.7 KJ/kg
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Step:2
1 – 2 => isentropic expansion
S1 = S2 = 6.925 KJ/kg.K
S2 > Sg2 => at 5 bar
6.925 > 6.819
So, steam is at super heat condition
At 5 bar and S2 = 6.925 KJ /Kg.K
Use mollier diagram => interaction point
P2 = 5 bar => T= 1700 C
:. H2 = 2794 KJ/Kg
Step :3
3 – 2 Isentropic expansion
S2 = S3 = 6.925 KJ/Kg.K
S3 < Sg3
6.925 < 8.151 KJ/kg => at P3 = 0.1 bar
:. Steam is wet condition
:. S3 = Sf3 + (X3 . Sfg3)
6.925 = 0.649 + (X3 . 7.502)
6.925 – 0.649 / 7.502 = X3
X3 = 0.8365
 H3 = hf3 + ( X3 . hfg3)
H3 = 191.8 + (0.8365 x 2392.9)
H3 = 2193.6448
 H4 =hf3
H4 = 191.8 KJ/kg
Step:5
 Mass of steam bled (m)
M= hf2 – hf4 / h2 – h4 => 640.1 – 191.8/ 2794 – 191.8
M= 0.1722 KJ/kg of steam
 Work done by the turbine with generation (W reg)
W reg = (h1 – h2) + ( l – m) (h2 – h3 )
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W reg = (3232.5 – 2794) + (1 – 0.1722) (2794 – 2193.64)
W reg = 935.4740 KJ/Kg
 Efficiency of cycle with regeneration (
(
ή
reg
(
ή
reg
(
ή
reg
ή
)
reg
) = W / h1 – h2 = (935.4740 / 3232.5 – 640.1) x 100
) = 0.3608 x 100
) = 36.08%
 Steam rate with regeneration
SSCreg = 3600 / W REG = 3600 / 935.4740
= 3.8483 Kg/Kw - hr
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8) Ten kg of water at 45oC is heated at a constant pressure of 10 bar until it becomes
Superheated vapour at 300oC. Find the changes in volume, enthalpy, internal energy
(AUC DEC’09)
and entropy.
Given:
Constant pressure process
Mass (m) =10 kg
(T1) = 450C
(P1) = (P2) = 10 bar x 100 => 1000 Kpa
(T2) = 300oC
To find:
a)
V => change in volume
b)
H => change in enthalpy
c)
U => change in internal energy
d)
S => change in entropy
Solution:
Step:1
From steam table at T1 = 450C [use temp table] (pg.no:02)
Vf = V1 = 0.001010 m 3 / kg
Sf = S1 = 0.638 KJ/Kg.K
Hf = h1 = 188.4 KJ /Kg
Step:2
From steam table P2 = 10 bar and T2 = 3000C
Use super heated steam table
V2 = 0.2580 m 3 / kg [pg.no:16]
H2 = 3052.1 KJ/Kg [pg.no:19]
S2 = 7.125 KJ/Kg.K [pg.no:22]
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Step:3
 Change in volume (
V) = m (V2 – V1)
= 10 (0.258 – 0.001008)
(
V) = 2.57 m 3
 Change in enthalpy ( H) = m(h2 – h1)
= 10 (3052.1 – 188.4)
( H) = 28637 KJ
 Change in internal energy ( u) = m [(h2 – h1) + P2 (V2 –V1)]
( u) = 10 [(3052.1 – 188.4) + 1000 (0.258 – 0.001008)]
( u) = 31206.92 KJ
 Change in entropy ( s) = m(S2 – S1)
= 10 (7.125 – 0.638)
( s) = 64.87 KJ/Kg.K
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9) A steam power plant running on rankine cycle has steam entering H P turbine at 20
Mpa, 5000C and leaving L P turbine at 90% dryness. Considering condenser pressure of
0.005Mpa and reheating occurring upto the temp of 5000C
Determine
a) The pressure at which steam leaves H P turbine
b) The thermal efficiency
c) Work done
(AUC MAY 2011)
Given:
Rankine cycle
H.P turbine
P1 = 20 Mpa
=> 200 bar
T1 = 5000C
L.P turbine
Dryness (X4) = 0.9
Condenser pressure (P3) = 0.005 Mpa
(P3) = 0.05 bar
To find:
a) The pressure at which steam leaves H P turbine (P2)
b) The thermal efficiency
c) Work done (W)
Solution:
Step:1
P1 = 200 bar & T1 = 5000C => from super heated steam table
H1 = 3241.1 KJ/Kg
[pg.no:20]
S1 = 6.146 KJ / Kg.K [pg.no:23]
P4 = 0.05 bar => from pressure table
Vf4 = 0.001005 m 3 /kg
3
Vg4 = 28.194 m /kg
hf4 = 137.8 KJ/Kg Sf4 = 0.476 KJ/Kg.K
hfg4 = 2423.8 KJ/Kg Sfg4 =7.920KJ/Kg.K
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 H4 = hf4 + (X4 x hfg4)
H4 = 137.8 + (0.9 x 2423.8)
H4 = 2319.22 KJ/Kg
 S4 = Sf4 + (X4 x Sfg4)
S4 = 0.476 + (0.9 x 7.920)
S4 = 7.604 KJ/Kg.K
 S1 =S2 = 6.146 KJ/Kg.K
S4 =S3 = 7.604 KJ/Kg.K
S3 =7.604 KJ/Kg.K & T = 5000C => use super heated steam table
P2 = 14 bar [pg.no:22]
H3 = 3473.9 KJ/Kg
H5 = hf4
H5 = 137.8 KJ/Kg
 P2 = 14 bar => use pressure steam table
Hf2 = 830.1 KJ/Kg
Sf2 = 2.284 KJ/Kg.K
Hfg2 = 1957.7 KJ/Kg
Sfg2 = 4.181 KJ/Kg.K
Hg2 = 2787.8 KJ/Kg
Sg2 = 6.465 KJ/Kg.K
 S2 = Sf2 + (X2 . Sfg2)
6.146 = 2.284 + ( X2 . 4.181)
6.146 – 2.284/4.181 = X2
X2 = 0.92
 H2 = hf2 + (X2 x hfg2)
H2 = 830.1 + (0.92 x 1957)
H2 = 2631.184 KJ/Kg
 H6 = hf4 + Vf4 (P1 – P2)
H6 = 137.8 + 0.001005 (20x1000 – 0.005 x1000)
H6 = 157.8949 KJ/Kg
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(
ή
rankine
) = (h1 – h2) + (h3 – h4) – (h6 – h5) / (h1 – h6) + (h3 – h2)
= (3241.1 – 2631.184) + (3473.9 – 2319.22) – (107.8949 -137.8)
(3241.1 – 2631.184) + (3473.9 – 2613.184)
= 609.916 + 1154.68 – 20.0949
3083.2051 + 842.716
= 1744.5011
3925.9211
= 0.4443 x 100
(
ή
rankine
) = 44.43%
Work done by the pump (W p) = Vf4 (P1 –P2)
= 0.001005 (20000 – 1400)
(W p) = 18.693 KJ/Kg
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10) A rankine cycle works between 40 bar and 0.2 bar with saturated steam at turbine inlet
determine cycle efficiency and the ratio of pump work and turbine work
[AUC DEC 11]
Given:
Rankine cycle
P1 = 40 bar x 100 = 4000 KN/m 2
P2 = 0.2 bar x 100 = 20 KN/m 2
To find:
a) Cycle efficiency
b) The ratio of pump work and turbine work
Solution:
Step:1
From steam tables, at 40 bar
H1 = hg1 = 2800.3 KJ/Kg
Sg1 = S1 = 6.069 KJ/Kg.K
Step:2
From steam tables at 0.2 bar
Vf2 = 0.0010107 m 3 / kg
Vg2 = 7.6498 m 3 / kg
hf 2 = 251.5 KJ/kg
sf2 = 0.832 KJ/Kg.K
hfg 2 = 2358.4 KJ/kg
sfg2 = 7.077 KJ/Kg.K
hg 2 = 2609.9 KJ/kg
sg2 = 7.909 KJ/Kg.K
Step:3
Pump work (Wp) = Vf2 (P1 –P2)
(Wp) = 0.001017 (4000 – 20)
(Wp) = 4.0476 KJ/Kg
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Step:4
Turbine work (W T ) = h1 – h2
Where
H2 = ?
 S1 = S2 => 6.069 KJ/Kg.K
S1 = S2 + (X2 . Sfg2)
6.069 = 0.832 + (X2 . 7.077)
6.069 – 0.832 / 7.077 = X2
0.7399 = X2
 H2 = hf2 + (X2 . hfg2)
note: P2 = bar use mollier diagram
h2 = 251.5 + (0.7399 x 2358.4)
h2 = 2010 KJ/Kg
h2 = 1996.48016 KJ/Kg
 Turbine work (W T) = h1 – h2
(W T) = (2800.3 – 2010)
(W T) = 790.3 Kw
 The ratio of pump work and turbine work = Wp / W T = 4.0476 / 790.3= 0.005
 Cycle efficiency (
ήrank) = W
net
/ Q1
W net = W T – Wp
(W T) = 790.3 – 4.0476
(W T) = 786.2524
Q1 = h1 – h4
Where
Hf2 = h3 => h3 = 251.5 KJ/kg
Wp = h4 – h3
:. H4 = 4.0476 + 251.5
H4 = 255.476 KJ/kg
Q1 = h1 – h4 => 2800.3 – 255.476
Q1 = 2544.7524 KJ
ή
cy cle
= W net / Q1 = 786.2524 / 2544.7524
ή
cy cle
= 0.3089 x 100=30.89%
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11) One kg of steam at 10 bar exists at the following condition (i) wet and 0.8 dry (ii) dry and
saturated (iii) at a temp of 199.90 C . determine the enthalpy, specific volume , density ,
internal energy and entropy in each case take Cps = 2.25 KJ/kg (AUC JUNE 2012)
Given:
P1 = 10 bar x 100 = 1000 KN/m 2
X = 0.8
T= 199.90 C
Cps = 2.25 KJ/kg
To find:
Enthalpy (h)
specific volume (V)
density (P)
internal energy
entropy
Solution:
Step:1
From steam table (P1) = 10 bar
Ts = 179.90C
hf = 762.6 KJ/Kg
sf=2.138 KJ/Kg.K
Vf = 0.001127 m 3/kg hfg = 2013.6 KJ/Kg
Vg = 0.19430 m 3/kg hg = 2776.2 KJ/Kg
sfg=4.445 KJ/Kg.K
sfg=6.583 KJ/Kg.K
Condition:
When steam is 0.8 bar
 H1 = hf1 + (x . hfg)
H1 = 762.6 + (0.8 x 2013.6)
H1 = 2373.48 KJ/Kg
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 V1 = X . Vg1
V1 = 0.8 x 0.19430
V1 = 0.15544 m 3 / kg
 ρ = 1 / V1 = 1/ 0.15544
ρ = 6.4333 Kg/m 3
 U = h1 – P x V1
U = 2373.48 – (1000 x 0.15544)
U = 2218.08 KJ/Kg
 S=Sf1 + (X1 x Sfg1)
S= 2.138 + (0.8 x 4.445)
S=5.694 KJ/Kg.K
Condition:2
When steam is dry and saturated
Hg1 = h
 H = 2776.2 KJ/kg
Vg1 = V1
 Vg1 = 0.19430 m 3 / kg
 ρ = 1 / V1 = 1/0.19430
ρ = 5.1466 Kg/m 3
 U = h – P1V1
U = 2776.2 – (1000 x 0.19430)
U = 2581.9 KJ/Kg
 S1 = Sg1
S1 = 6.583 KJ/Kg.K
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Condition:3
When steam is superheated
 H= hg1 + Cp (T – Ts)
= 2776.2 + 2.25 (199.9 – 179.9)
H= 2821.2 KJ/Kg
 V1 = Vg1 (T+ 273 / Ts + 273)
=0.19430 (199.9 + 273 / 179.9 + 273)
V1 = 0.2028 m 3 / Kg
 U = h –P1V1
U= 2821.2 – (1000 x 0.2028)
U= 2618.3197 KJ/Kg
 S = Sg1 + CPs ln (T+ 273 / Ts + 273)
S = 6.583 + 2.25 ln (427.9 / 452.9)
S = 6.6802 KJ/Kg.K
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12) In a steam generator compressed liquid water at 10 Mpa , 300C enters a 30mm dia tube
at the rate of 3 lit/sec, steam at 9 Mpa, 4000C exists the tube. Find the rate of heat
transfer to the water
[AUC JUNE 2012]
Given:
P1 = 10 Mpa x 10 = 100 bar
Temp of water (Tw) = 300C
Dia (D) = 30mm / 1000 => 0.03
(V1) =3 / 1000 = 0.003 m 3/sec
P2 = 9Mpa x 10 => 90 bar
Ttube (T2) = 4000C
To find:
Rate of heat transfer to the water (Q)
Solution:
Step:1
From steam table corresponding to 300C
V1 = Vf1 = 0.001004 m 3 / kg
Hf1 = h1 = 125.7 KJ/Kg
Step:2
 Mass flow rate of steam (m)
M = V / Vf1 = 0.003 / 0.001004
M= 2.9880 Kg/s
Step:3
 Area of tube (A)
(A) = /4 x d2 => /4x (0.03)2
(A) = 7.0685 x 10-4 m 2
Step:4
 Inlet velocity (C1)
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Page 30
C1 = m x V1 / A
= 2.9880 x 0.001004 / 7.0685 x 10-4 m 2
C1 = 4.2440 m/s
Step:5
 From steam tables corresponding to P2 = 9 bar and T2 = 4000C
V2 = 0.02993 m 3/Kg
H2 = 3121.2 KJ/Kg
Step:6
 Final velocity
C2 = m x V2 /A = 2.9880 x 0.02993 / 7.0685 x 10-4
C2 = 126.5202 m/s
Step:7
M
h1 + C12 / 2000 + Z1 g + Q = m h2 + C22 / 2000 + Z2 g
+W
M h1 + C12 / 2000 +Q = m h2 + C22 / 2000
2.9880 125.7 + (4.2440)2 / 2000 +Q = 2.9880 3121.2 + 126.5202 / 2000
375.6185 + Q = 9350.0606
Q = 9350.0606 – 375.6185
Q = 8974.4418 KJ/sec
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13) A steam turbine receive steam at pressure 20 bar superheated at 300 0C the exhaust
pressure 0.07 bar and expansion taken place isentropic calculated (i) heat rejected
(ii) heat supplied (iii) work done (iv) thermal efficiency [AUC JUNE 2013]
Given:
Isentropic process – rankine cycle
(P1) = 20 bar x 100 = 2000 KN/m 2
(T1) = 3000C + 273 = 573K
(P2) = 0.07 bar x 100 = 7 KN/m 2
To find:
a) heat supplied (Q)
b) heat rejected (Qr)
c) work done (W)
d) thermal efficiency (
ήther)
Solution:
Step:1
From super heated steam table at P1 = 20 bar and T1 = 3000C
H1 = 3025.0 KJ/Kg => [pg.no:19]
S1 = 6.770 KJ/Kg.K => [pg.no:22]
Step:2
From steam table P2 = 0.07 bar => [pg.no:07]
Vf2 = 0.001007 m 3 /kg
3
Vg2 = 20.531 m /kg
hf2 = 163.4 KJ/Kg Sf2 = 0.559 KJ/Kg.K
hfg2 = 2409.2 KJ/Kg Sfg2 =7.718 KJ/Kg.K
hg2 = 2572.6 KJ/Kg Sg2 =8.277 KJ/Kg.K
Step:3
 S1 = S2
S2 = Sf2 + ( X2 x Sfg2)
6.770 = 0.559 + (X2 x 7.718)
6.770 – 0.559 / 7.718 = X2
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Page 32
X2 = 0.8047
H2 = hf2 + (X2 . hfg2)
H2 = 163.4 + (0.8047 x 2409.2)
H2 = 2102.1848 KJ/Kg
Step;4
Work done (Wp) = Vf2 (P1 – P2)
Wp = 0.001007 (2000- 7)
Wp = 2.0069 KJ/Kg
Step:5
Heat supplied (Qs) = h1 - (hf2 + Wp)
(Qs) = 3025.0 – (163.4 + 2.0069)
(Qs) = 2859.5931 Kg/KJ
Step:6
Heat rejected (Q R ) = h2 –hf2 (or) h2 – h3
(QR ) =2102.1848 – 163.4
(QR ) = 1938.7848 Kg/KJ
Step:7
Work done (W) = Qs - QR
(W) = 2859.5931 – 1938.7840
(W) = 920.8084 KJ/Kg
Step:8
Thermal efficiency (
ήthermal) = W/Qs
= 920.8084 / 2859.5931
= 0.3220 x 100
(
ήthermal) = 32.20%
Step:9
Specific steam consumption
= 3600 / W
=> 3600 / 920.8084
= 3.9096 Kg/Kw - h
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