Vapor power cycles by Anupama.pptx

1. 1. In a simple Rankine cycle, dry saturated steam at 20 bar expands to a pressure
of 1 atmosphere. Determine (i) the pump work, (ii) turbine work, (iii) network
output, (iv) thermal efficiency, (v) quality of steam entering the condenser, and
(vi) specific steam consumption in kg/kWh and (vii) quality of steam entering
the condenser if the condenser pressure is reduced to 0.1 bar.

2. SOLUTION:
(a) Pump work:
𝑤𝑝 = 𝑣3(𝑝4 − 𝑝3)
From Table A-1 at 1 bar,
𝑣3 = 𝑣𝑓 = 0.0010434 𝑚3/𝑘𝑔.
𝑤𝑝 = 0.0010434 20 − 1 × 100
𝑤𝑝 = 1.982 𝑘𝐽/𝑘𝑔
Assuming mass flow rate of water=1kg/s; 𝑊
𝑝 = 1.982 𝑘𝑊
(b) Turbine work:
𝑤𝑇 = ℎ1 − ℎ2 −− −(1)
From Table A-1 at 20 bar,
ℎ1 = ℎ𝑔 = 2797.2 𝑘𝐽/𝑘𝑔.
ℎ2 = ℎ𝑓 + 𝑥ℎ𝑓𝑔
From Table A-1 at 1 bar,
ℎ𝑓 = 417.5
kJ
kg
; ℎ𝑓𝑔 = 2257.9
kJ
kg
∴ ℎ2= 417.5 + 𝑥 2257.9 −−−− −(2)

3. To find ‘x’:
Using isentropic process 1-2: 𝑠1 = 𝑠2
We know that, 𝑠1 = 𝑠𝑔 𝑎𝑡 20 𝑏𝑎𝑟.
From Table A-1 at 20 bar, 𝑠1 = 𝑠𝑔 = 6.3377
𝑘𝐽
𝑘𝑔𝐾
= 𝑠2
But, 𝑠2 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔
From Table A-1 at 1 bar, 𝑠𝑓 = 1.3027
𝑘𝐽
𝑘𝑔𝐾
; 𝑠𝑓𝑔 = 6.0571
𝑘𝐽
𝑘𝑔𝐾
𝑖. 𝑒. , 𝑠2 = 6.3377 = 1.3027 + 𝑥6.0571
∴ 𝑥 = 0.8312
Substituting ‘x’ in equation(2); ℎ2 = 417.5 + 0.8312 2257.9
∴ ℎ2= 2294.392 𝑘𝐽/𝑘𝑔

4. Substituting ‘ℎ1’ and ‘ℎ2’ in equation (1): 𝑤𝑇 = 2797.2 − 2294.392
∴ 𝑤𝑇= 502.8 𝑘𝐽/𝑘𝑔
(c) Net work output:
𝑊𝑁𝑒𝑡 = 𝑊𝑇 − 𝑊𝑃 = 502.8 − 1.982 = 500.818 𝑘𝑊
∴ 𝑊𝑁𝑒𝑡= 500.818 𝑘𝑊
(d) Thermal Efficiency:
𝜂𝑡ℎ𝑒𝑟𝑚𝑎𝑙=
𝑤𝑁𝑒𝑡
𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑
=
𝑤𝑁𝑒𝑡
ℎ1 − ℎ4
−−− −(3)
Assuming mass flow rate of steam=1kg/s; 𝑊𝑇= 502.8 𝑘𝑊

5. To find ℎ4: 𝑊𝑃 = (ℎ4 −ℎ3) = 1.982
From Table A-1 ℎ3 = ℎ𝑓 𝑎𝑡 1 𝑏𝑎𝑟= 417.5 kJ/kg
1.982 = (ℎ4 − 417.5)
ℎ4 = 419.482 𝑘𝐽/𝑘𝑔
Substituting in Equation (3): 𝜂𝑡ℎ𝑒𝑟𝑚𝑎𝑙=
𝑤𝑁𝑒𝑡
ℎ1−ℎ4
=
500.818𝑘𝐽/𝑠
2797.2−419.482 𝑘𝐽/𝑘𝑔×1𝑘𝑔/𝑠
𝜂𝑡ℎ𝑒𝑟𝑚𝑎𝑙= 0.2106 = 21.06%
(e) Quality of steam entering the condenser:
The steam enters the condenser at point ‘2’.
∴ 𝑥2 = 0.8312 = 83.12%

6. (f) Specific steam consumption in kg/kWh:
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑆𝑡𝑒𝑎𝑚 𝐶𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 =
𝑚 × 3600
𝑊𝑛𝑒𝑡
𝑘𝑔
𝑘𝑊ℎ
=
1 × 3600
500.818
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑆𝑡𝑒𝑎𝑚 𝐶𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 = 7.188 kg/kWh
(g) Quality of steam entering the condenser if the condenser pressure is reduced
to 0.1 bar:
Using isentropic process 1-2: 𝑠1 = 𝑠2; We know that, 𝑠1 = 𝑠𝑔 𝑎𝑡 20 𝑏𝑎𝑟.
From Table A-1 at 20 bar, 𝑠1 = 𝑠𝑔 = 6.3377
𝑘𝐽
𝑘𝑔𝐾
= 𝑠2
But, 𝑠2 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔
From Table A-1 at 0.1 bar, 𝑠𝑓 = 0.6493
𝑘𝐽
𝑘𝑔𝐾
; 𝑠𝑓𝑔 = 7.5081
𝑘𝐽
𝑘𝑔𝐾
𝑖. 𝑒. , 𝑠2 = 6.3377 = 0.6493 + 𝑥7.5081
∴ 𝑥 = 0.7576 = 75.76%

7. 2. In a simple Rankine cycle, steam conditions at the boiler exit are 10 bar and
300°C. The steam expands in the turbine to a pressure of 0.09 bar and is dry
saturated at turbine exit. The isentropic efficiency of the turbine is 0.86 and
that of the pump is 0.70. Determine (a) the condition of steam entering the
turbine (b) Turbine work per unit mass of steam (c) Actual pump work per
unit mass of water (d) Net work output and thermal efficiency of the cycle
and (e) quality of steam entering the condenser.

8. SOLUTION:
(a) Steam condition at turbine entry:
From Steam tables @10 bar,
𝑇𝑠𝑎𝑡 = 179.91℃.
Given steam is at 𝑇 = 300℃
𝑇𝑠𝑎𝑡(179.91℃) < 𝑇(300℃)
∴ 𝑻𝒉𝒆 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒕𝒆𝒂𝒎 𝒆𝒏𝒕𝒆𝒓𝒊𝒏𝒈 𝒕𝒉𝒆 𝒕𝒖𝒓𝒃𝒊𝒏𝒆 𝒊𝒔 𝒔𝒖𝒑𝒆𝒓𝒉𝒆𝒂𝒕𝒆𝒅.
(b) Turbine work per unit mass of steam:
Note: When the isentropic efficiency of the turbine is provided, then the
actual cycle should be analyzed. Hence, we need to find the actual turbine
work.
We have, 𝜂𝑇𝑢𝑟𝑏𝑖𝑛𝑒 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑊𝑜𝑟𝑘
𝐼𝑑𝑒𝑎𝑙 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑊𝑜𝑟𝑘
=
𝑤𝑇𝐴
𝑤𝑇𝐼

9. 0.86 =
𝑤𝑇𝐴
𝑤𝑇𝐼
𝑜𝑟 𝑤𝑇𝐴 = 0.86 × 𝑤𝑇𝐼
𝑖. 𝑒. , 𝑤𝑇𝐴 = 0.86 × ℎ1 − ℎ2 −−−−− −(1)
To find ℎ1: We know that the steam is superheated, hence, from Table A-3 (superheated
steam properties)
ℎ1 = ℎ 𝑎𝑡 10 𝑏𝑎𝑟 𝑎𝑛𝑑 300℃ = 3052.1 𝑘𝐽/𝑘𝑔
To find ℎ2:
First we have to find whether point ‘2’ is on saturated vapor line or in superheated region
or in wet region.
Wkt, 1-2 is isentropic process, i.e., 𝑠1 = 𝑠2
From Table A-3: 𝑠1 = 𝑠 𝑎𝑡 10 𝑏𝑎𝑟 𝑎𝑛𝑑 300℃ (𝑠𝑢𝑝𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑑) = 7.1251𝑘𝐽/𝑘𝑔𝐾
i.e., 𝑠1 = 𝑠2 = 7.1251 𝑘𝐽/𝑘𝑔𝐾

10. From Table A-1 (Saturated steam properties): If point ‘2’ would be on saturated
vapor line, then, 𝑠2 = 𝑠𝑔 𝑎𝑡 0.09 𝑏𝑎𝑟 = 8.1881
𝑘𝐽
𝑘𝑔𝐾
> 7.1251 𝑘𝐽/𝑘𝑔𝐾.
∴ 𝑝𝑜𝑖𝑛𝑡 2 𝑖𝑠 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑣𝑎𝑝𝑜𝑟 𝑑𝑜𝑚𝑒, 𝑖. 𝑒. , 𝑖𝑛 𝑡ℎ𝑒 𝑤𝑒𝑡 𝑟𝑒𝑔𝑖𝑜𝑛.
For wet steam at point ‘2’: ℎ2 = ℎ𝑓 + 𝑥ℎ𝑓𝑔 𝑎𝑡 0.09 𝑏𝑎𝑟
From Table A-1 at 0.09 bar, ℎ𝑓 = 183.3
𝑘𝐽
𝑘𝑔
𝑎𝑛𝑑 ℎ𝑓𝑔 = 2397.9
𝑘𝐽
𝑘𝑔
ℎ2 = 183.3 + 𝑥2397.9 --------(2)
To find ‘x’: 𝑠1 = 7.1251 = 𝑠2 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔𝑎𝑡 0.09 𝑏𝑎𝑟
From Table A-1 at 0.09 bar: 𝑠𝑓 = 0.6224
𝑘𝐽
𝑘𝑔𝐾
𝑎𝑛𝑑 𝑠𝑓𝑔 = 7.5657
𝑘𝐽
𝑘𝑔𝐾
∴ 7.1251 = 0.6224 + 𝑥7.5657
𝑥 = 0.859 ≅ 0.86
Substituting ‘x’ in equation (2): ℎ2 = 183.3 + 0.86 × 2397.9
ℎ2 = 2245.5𝑘𝐽/𝑘𝑔

11. Substituting ‘ℎ1’ and ‘ℎ2’ in equation (1): 𝑤𝑇𝐴= 0.86 × 3052.1 − 2245.5
𝑤𝑇𝐴= 0.86 × 3052.1 − 2245.5 = 693.676 𝑘𝐽/𝑘𝑔
(c) Actual pump work per unit mass of water:
We have, 𝜂𝑝𝑢𝑚𝑝 =
𝐼𝑑𝑒𝑎𝑙 𝑝𝑢𝑚𝑝 𝑊𝑜𝑟𝑘
𝐴𝑐𝑡𝑢𝑎𝑙 𝑝𝑢𝑚𝑝 𝑊𝑜𝑟𝑘
=
𝑤𝑃𝐼
𝑤𝑃𝐴
0.70 =
𝑤𝑃𝐼
𝑤𝑃𝐴
𝑖. 𝑒. , 𝑤𝑃𝐴 =
𝑤𝑃𝐼
0.7
−−−−− −(3)
𝑤𝑃𝐼 = ℎ4 − ℎ3 = 𝑣3(𝑝4 − 𝑝3)
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 𝐴 − 1, 𝑣3 = 𝑣𝑓 𝑎𝑡 0.09 𝑏𝑎𝑟 = 0.0010094 𝑚3
/𝑘𝑔
𝑤𝑃𝐼 = 0.0010094𝑚3/𝑘𝑔 × (10 − 0.09)100𝑘𝑁/𝑚2

12. 𝑤𝑃𝐼 = 1.0003154 𝑘𝐽/𝑘𝑔
substituting 𝑤𝑃𝐼 in equation (3): 𝑤𝑃𝐴=
1.0003154
0.7
= 1.429
𝑘𝐽
𝑘𝑔
(d) Net work output and thermal efficiency of the cycle :
𝑤𝑁𝑒𝑡 = 𝑤𝑇𝐴 − 𝑤𝑃𝐴
𝑤𝑁𝑒𝑡 = 693.676 − 1.429 = 692.246 𝑘𝐽/𝑘𝑔
Thermal efficiency, 𝜂𝑡ℎ𝑒𝑟𝑚𝑎𝑙=
𝑤𝑁𝑒𝑡
𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑
=
𝑤𝑁𝑒𝑡
ℎ1−ℎ4′
−−− −(4)
To find h4’ : 𝑤𝑃𝐴 = ℎ4′ − ℎ3 = 1.429
𝑘𝐽
𝑘𝑔
From Table A-1 ℎ3 = ℎ𝑓 𝑎𝑡 0.09 𝑏𝑎𝑟=183.3 kJ/kg
𝑖. 𝑒. , 𝑤𝑃𝐴 = 1.429 = ℎ4′ − 183.3
ℎ4′ = 184.729
𝑘𝐽
𝑘𝑔
C

13. Substituting in equation(4)
𝜂𝑡ℎ𝑒𝑟𝑚𝑎𝑙=
𝑤𝑁𝑒𝑡
𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑
=
692.246
3052.1 − 184.729
= 0.2414 = 24.14%
(e) Quality of steam entering the condenser:
𝜂𝑇𝑢𝑟𝑏𝑖𝑛𝑒 =
𝑤𝑇𝐴
𝑤𝑇𝐼
=
ℎ1 − ℎ2′
ℎ1 − ℎ2
0.86 =
3052.1 − ℎ2′
3052.1 − 2245.5
ℎ2′ = 2358.424 𝑘𝐽/𝑘𝑔

14. ℎ2′ = 2358.424
𝑘𝐽
𝑘𝑔
= ℎ𝑓 + 𝑥ℎ𝑓𝑔 𝑎𝑡 0.09 𝑏𝑎𝑟
From Table A-1 𝑎𝑡 0.09 𝑏𝑎𝑟 ℎ𝑓=183.3 kJ/kg and ℎ𝑓𝑔=2397.9 kJ/kg
ℎ2′ = 2358.424
𝑘𝐽
𝑘𝑔
= 183.3 + 𝑥 2397.9
The quality of steam entering the condenser , 𝑥 = 0.907 = 90.7%

15. 3. In a reheat steam cycle, the boiler exit conditions are 25 bar and 300° C. The
exit pressure of steam at the end of first stage is 5 bar. The steam is then
reheated to 300°C before expanding in the second turbine to 0.05 bar.
Assuming the high and low pressure turbines to have efficiencies of 87% and
85 % respectively, find (i) the thermal-energy input in the re-heater, (ii) the
cycle efficiency, and (iii) power output for a mass flow rate of 2 kg/s.

16. (a) Thermal-energy input in the re-heater:
Thermal-energy input in the re-heater= Heat supplied in the reheater
𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒ℎ𝑒𝑎𝑡𝑒𝑟 = ℎ5 − ℎ4′ −− −(1)
Given Efficiency of high pressure turbine=0.87;
𝜂𝐻𝑃𝑇 =
ℎ3 − ℎ4′
ℎ3 − ℎ4
−− − 2
At ‘3’ the steam is at 25 bar and 300°C.
From Table A-1 at 25 bar, 𝑇𝑠𝑎𝑡 = 223.94℃ < 300℃
∴ 𝐴𝑡 300℃, 25 𝑏𝑎𝑟, 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑚 𝑖𝑠 𝑠𝑢𝑝𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑑.
From Table A-3 at 25 bar and 300°C, ℎ3 = ℎ = 3010.4 kJ/kg
To locate point ‘4’ on T-s diagram: 3-4 is isentropic process, i.e., 𝑠3 = 𝑠4
From Table A-3: 𝑠3 = 𝑠 𝑎𝑡 25 𝑏𝑎𝑟 𝑎𝑛𝑑 300℃ (𝑠𝑢𝑝𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑑) = 6.6470𝑘𝐽/𝑘𝑔𝐾
i.e., 𝑠3 = 𝑠4 = 6.6470𝑘𝐽/𝑘𝑔𝐾

17. From Table A-1 (Saturated steam properties): If point ‘4’ would be on saturated
vapor line, then, 𝑠4 = 𝑠𝑔 𝑎𝑡 5 𝑏𝑎𝑟 = 6.8192𝑘𝐽/𝑘𝑔𝐾 > 6.6470𝑘𝐽/𝑘𝑔𝐾.
∴ 𝑝𝑜𝑖𝑛𝑡 4 𝑖𝑠 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑣𝑎𝑝𝑜𝑟 𝑑𝑜𝑚𝑒, 𝑖. 𝑒. , 𝑖𝑛 𝑡ℎ𝑒 𝑤𝑒𝑡 𝑟𝑒𝑔𝑖𝑜𝑛.
So, 𝑠4 = 𝑠3 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔 𝑎𝑡 5 𝑏𝑎𝑟
From Table A-1 at 5 bar: 𝑠𝑓 = 1.8604𝑘𝐽/𝑘𝑔𝐾; 𝑠𝑓𝑔 = 4.9588𝑘𝐽/𝑘𝑔𝐾
∴ 𝑠4= 𝑠3 = 6.6470 = 1.8604 + 𝑥4.9588
∴ 𝑥 = 0.9652
ℎ4 = ℎ𝑓 + 𝑥ℎ𝑓𝑔 𝑎𝑡 5 𝑏𝑎𝑟
From Table A-1 at 5 bar: ℎ𝑓 = 640.1𝑘𝐽/𝑘𝑔; ℎ𝑓𝑔 = 2107.4𝑘𝐽/𝑘𝑔
ℎ4 = 640.1 + (0.9652 × 2107.4)
∴ ℎ4= 2674.318 𝑘𝐽/𝑘𝑔

18. Substituting ‘ℎ3’ and ‘ℎ4’ in equation (2): 𝜂𝐻𝑃𝑇 =
ℎ3−ℎ4′
ℎ3−ℎ4
0.87 =
3010.4 − ℎ4′
3010.4 − 2674.318
∴ ℎ4′= 2718 𝑘𝐽/𝑘𝑔
From Table A-1 at 5 bar, 𝑇𝑠𝑎𝑡 = 151.85℃ < 300℃
∴ 𝐴𝑡 300℃, 5 𝑏𝑎𝑟, 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑚 𝑖𝑠 𝑠𝑢𝑝𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑑.
From Table A-3 at 5 bar and 300°C, ℎ5 = ℎ = 3064.8 kJ/kg
Substituting ‘ℎ4′’ and ‘ℎ5’ in equation (1):
𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒ℎ𝑒𝑎𝑡𝑒𝑟 = ℎ5 − ℎ4′
𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒ℎ𝑒𝑎𝑡𝑒𝑟 = 3064.8 − 2718 = 346.8𝑘𝐽/𝑘𝑔

19. (b) Cycle efficiency:
𝜂 =
𝑊𝑛𝑒𝑡
𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
=
𝑊𝐻𝑃𝑇 + 𝑊𝐿𝑃𝑇 − 𝑊𝑃
𝑄𝐵𝑜𝑖𝑙𝑒𝑟 + 𝑄𝑅𝑒ℎ𝑒𝑎𝑡𝑒𝑟
𝜂 =
ℎ3 − ℎ4′ + ℎ5 − ℎ6′ − 𝑣1(𝑝2 − 𝑝1)
(ℎ3 − ℎ2) + (ℎ5 − ℎ4′)
−−− − 3
To locate point ‘6’ on T-s diagram: 5-6 is isentropic process, i.e., 𝑠5 = 𝑠6
From Table A-3: 𝑠5 = 𝑠 𝑎𝑡 5 𝑏𝑎𝑟 𝑎𝑛𝑑 300℃ (𝑠𝑢𝑝𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑑) = 7.4614𝑘𝐽/𝑘𝑔𝐾
i.e., 𝑠5 = 𝑠6 = 7.4614𝑘𝐽/𝑘𝑔𝐾
From Table A-1 (Saturated steam properties): If point ‘6’ would be on saturated
vapor line, then, 𝑠6 = 𝑠𝑔 𝑎𝑡 0.05 𝑏𝑎𝑟 = 8.3960𝑘𝐽/𝑘𝑔𝐾 > 7.4614𝑘𝐽/𝑘𝑔𝐾.
∴ 𝑝𝑜𝑖𝑛𝑡 6 𝑖𝑠 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑣𝑎𝑝𝑜𝑟 𝑑𝑜𝑚𝑒, 𝑖. 𝑒. , 𝑖𝑛 𝑡ℎ𝑒 𝑤𝑒𝑡 𝑟𝑒𝑔𝑖𝑜𝑛.

20. For wet steam at point ‘6’: ℎ6 = ℎ𝑓 + 𝑥ℎ𝑓𝑔 𝑎𝑡 0.05 𝑏𝑎𝑟
From Table A-1 at 0.05 bar, ℎ𝑓 = 137.8𝑘𝐽/𝑘𝑔𝑎𝑛𝑑 ℎ𝑓𝑔 = 2423.8𝑘𝐽/𝑘𝑔
ℎ6 = 137.8 + 𝑥2423.8 --------(4)
To find ‘x’: 𝑠5 = 7.4614 = 𝑠6 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔𝑎𝑡 0.05 𝑏𝑎𝑟
From Table A-1 at 0.05 bar: 𝑠𝑓 = 0.4763𝑘𝐽/𝑘𝑔𝐾𝑎𝑛𝑑 𝑠𝑓𝑔 = 7.9197𝑘𝐽/𝑘𝑔𝐾
∴ 7.4614 = 0.4763 + 𝑥7.9197
𝑥 = 0.8819 ≅ 0.882
Substituting ‘x’ in equation (4): ℎ6 = 137.8 + (0.882 × 2423.8)
ℎ6 = 2275.59𝑘𝐽/𝑘𝑔

21. To find ‘h6’’: Given Efficiency of high pressure turbine=0.85;
𝜂𝐿𝑃𝑇 = 0.85 =
ℎ5 − ℎ6′
ℎ5 − ℎ6
0.85 =
3064.8 − ℎ6′
3064.8 − 2275.59
ℎ6′ = 2393.97 𝑘𝐽/𝑘𝑔
To find ‘h2’: we have, Pump work= 𝑣1 𝑝2 − 𝑝1 =(ℎ2 − ℎ1)
From Table A-1 at 0.05 bar ℎ𝑓 = ℎ1 = 137.8
𝑘𝐽
𝑘𝑔
; 𝑣𝑓 = 𝑣1 = 0.0010052 𝑘𝐽/𝑘𝑔𝐾
∴ 𝑃𝑢𝑚𝑝 𝑊𝑜𝑟𝑘 = 0.0010052 25 − 0.05 × 100 = (ℎ2 − 137.8)
∴ ℎ2 = 140.308 𝑘𝐽/𝑘𝑔

22. Substituting in equation (3): 𝜂 =
ℎ3−ℎ4′ + ℎ5−ℎ6′ − 𝑣1(𝑝2−𝑝1)
(ℎ3−ℎ2)+(ℎ5−ℎ4′)
=
3010.4 − 2718 + 3064.8 − 2393.97 − 0.0010052(25 − 0.05) × 100
(3010.4 − 140.308) + (3064.8 − 2718)
=
960.7221
3216.892
𝜂 = 0.2986 = 29.86%
(c) power output for a mass flow rate of 2 kg/s:
𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑤𝑛𝑒𝑡 × 𝑚
𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑊𝐻𝑃𝑇 + 𝑊𝐿𝑃𝑇 − 𝑊𝑃 × 𝑚
𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 = 292.4 + 670.83 − 2.5079
𝑘𝐽
𝑘𝑔
× 2
𝑘𝑔
𝑠
=1921.4 kW

23. 4. Steam leaves the boiler and enters the turbine at 45 bar absolute and 450°C.
After expansion to 4 bar absolute some of the steam is extracted from the
turbine for the purpose of heating the feed water in an open heater. The
condenser pressure is 0.25 bar. Determine the mass of steam extracted per kg
of steam leaving the boiler and thermal efficiency of the cycle.

24. 450°C

25. (a) mass of steam extracted per kg of steam leaving the boiler:
Referring to the figure and Writing mass and energy balance equation for
FWH:
𝑚ℎ6 + 1 − 𝑚 ℎ2 = 1ℎ3
𝑚ℎ6 + ℎ2 − 𝑚ℎ2 = ℎ3
𝑚 ℎ6 − ℎ2 = ℎ3 − ℎ2
𝑚 =
ℎ3 − ℎ2
ℎ6 − ℎ2
−−− −(1)
To find ‘h6’: First we have to locate point ‘6’ on the T-s plot
At point ‘5’, steam is at 45 bar and 450°C.
From Table A-1 at 45 bar and 450°C, 𝑇𝑠𝑎𝑡 = 257.41℃ < 450℃
Hence, at point ‘5’, steam is superheated.
From Table A-3 at 45 bar 450°C (superheated),
𝑠5 = 𝑠 = 6.8702𝑘𝐽/𝑘𝑔𝐾(𝐵𝑦 𝑖𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛) = 𝑠6

26. From Table A-1 (Saturated steam properties): If point ‘6’ would be on saturated
vapor line, then, 𝑠6 = 𝑠𝑔 𝑎𝑡 4 𝑏𝑎𝑟 = 6.8943𝑘𝐽/𝑘𝑔𝐾 ≅ 6.8702𝑘𝐽/𝑘𝑔𝐾.
∴ 𝑝𝑜𝑖𝑛𝑡 6 𝑖𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑 𝑣𝑎𝑝𝑜𝑟 𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑚 𝑖𝑠 𝑑𝑟𝑦 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑.
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 𝐴 − 1 𝑎𝑡 4 𝑏𝑎𝑟, ℎ6 = ℎ𝑔 = 2737.6 𝑘𝐽/𝑘𝑔
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 𝐴 − 1 𝑎𝑡 4 𝑏𝑎𝑟, ℎ3 = ℎ𝑓 = 604.7 𝑘𝐽/𝑘𝑔
To find ‘h2’: we have 𝑝𝑢𝑚𝑝 𝑤𝑜𝑟𝑘 = 𝑣1 𝑝2 − 𝑝1 = ℎ2 − ℎ1
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 𝐴 − 1 𝑎𝑡 0.25 𝑏𝑎𝑟, ℎ1 = ℎ𝑓 = 272.0𝑘𝐽/𝑘𝑔; 𝑣1 = 𝑣𝑓 = 0.0010199 𝑘𝐽/𝑘𝑔𝐾
∴ 𝑝𝑢𝑚𝑝 𝑤𝑜𝑟𝑘 = 0.0010199 4 − 0.25 100 = ℎ2 − 272.0
ℎ2 = 272.382 𝑘𝐽/𝑘𝑔
Substituting in equation (1): 𝑚 =
ℎ3−ℎ2
ℎ6−ℎ2
=
604.7−272.382
2737.6−272.382
𝑚 = 0.135kg/kg of steam

27. (b) Thermal efficiency of the cycle:
𝜂 =
𝑊𝑛𝑒𝑡
𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
=
𝑊𝑇 − (𝑊𝑃1+𝑊𝑃2)
𝑄𝐵𝑜𝑖𝑙𝑒𝑟
=
1 ℎ5 − ℎ6 + 1 − 𝑚 ℎ6 − ℎ7 + [ 1 − 𝑚 ℎ2 − ℎ1 + 1 ℎ4 − ℎ3 ]
ℎ5 − ℎ4
− − 2
Turbine work, 𝑊𝑇 = 1 ℎ5 − ℎ6 + 1 − 𝑚 ℎ6 − ℎ7
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 𝐴 − 3 𝑎𝑡 45 𝑏𝑎𝑟 𝑎𝑛𝑑 450℃, ℎ5 = ℎ𝑔 = 3323.2 𝑘𝐽/𝑘𝑔
To locate point ‘7’ on T-s diagram:
From isentropic process 6-7, 𝑠6 = 𝑠7 𝑎𝑛𝑑
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 𝐴 − 1 𝑎𝑡 4 𝑏𝑎𝑟, 𝑠6 = 𝑠𝑔 = 6.8943 𝑘𝐽/𝑘𝑔𝐾

28. From Table A-1 (Saturated steam properties): If point ‘7’ would be on saturated
vapor line, then, 𝑠7 = 𝑠𝑔 𝑎𝑡 0.25 𝑏𝑎𝑟 = 7.8423 𝑘𝐽/𝑘𝑔𝐾 > 6.8943 𝑘𝐽/𝑘𝑔𝐾.
∴ 𝑝𝑜𝑖𝑛𝑡 7 𝑖𝑠 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑣𝑎𝑝𝑜𝑟 𝑑𝑜𝑚𝑒, 𝑖. 𝑒. , 𝑖𝑛 𝑡ℎ𝑒 𝑤𝑒𝑡 𝑟𝑒𝑔𝑖𝑜𝑛.
For wet steam at point ‘7’: ℎ7 = ℎ𝑓 + 𝑥ℎ𝑓𝑔 𝑎𝑡 0.25 𝑏𝑎𝑟
From Table A-1 at 0.25 bar, ℎ𝑓 = 272.0
𝑘𝐽
𝑘𝑔
𝑎𝑛𝑑 ℎ𝑓𝑔 = 2346.4
𝑘𝐽
𝑘𝑔
ℎ7 = 272 + 𝑥2346.4 --------(3)
To find ‘x’: 𝑠6 = 6.8943 = 𝑠7 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔𝑎𝑡 0.25 𝑏𝑎𝑟
From Table A-1 at 0.25 bar: 𝑠𝑓 = 0.8933
𝑘𝐽
𝑘𝑔𝐾
𝑎𝑛𝑑 𝑠𝑓𝑔 = 6.9390
𝑘𝐽
𝑘𝑔𝐾
∴ 6.8943 = 0.8933 + 𝑥6.9390
𝑥 = 0.864
Substituting ‘x’ in equation (3): ℎ7 = 272 + 0.864 × 2346.4
ℎ7 = 2301.2𝑘𝐽/𝑘𝑔

29. Turbine work, 𝑊𝑇 = 1 ℎ5 − ℎ6 + 1 − 𝑚 ℎ6 − ℎ7
𝑊𝑇 = 1 3323.2 − 2737.6 + 1 − 0.135 2737.6 − 2301.2
𝑊𝑇 = 585.6 + 377.486
𝑊𝑇 = 963.086 𝑘𝐽
Pump Work, (𝑊𝑃1+𝑊𝑃2) = [ 1 − 𝑚 ℎ2 − ℎ1 + 1 ℎ4 − ℎ3 ]
ℎ4 − ℎ3 = 𝑣3 𝑝4 − 𝑝3 100
ℎ4 − 604.7 = 0.0010839 45 − 4 100
ℎ4 = 609.144 𝑘𝐽/𝑘𝑔
𝑊
𝑝𝑢𝑚𝑝 = [ 1 − 0.135 272.382 − 272 + 1 609.144 − 604.7 ]
𝑊
𝑝𝑢𝑚𝑝 = 0.33043 + 4.444 = 4.774 𝑘𝐽
Heat Supplied, 𝑄𝐵𝑜𝑖𝑙𝑒𝑟 = ℎ5 − ℎ4
𝑄𝐵𝑜𝑖𝑙𝑒𝑟 = 3323.2 − 609.144 = 2714.056𝑘𝐽/𝑘𝑔 × 1𝑘𝑔
𝑄𝐵𝑜𝑖𝑙𝑒𝑟 = 2714.056 kJ

30. 𝜂 =
𝑊𝑇 − (𝑊𝑃1+𝑊𝑃2)
𝑄𝐵𝑜𝑖𝑙𝑒𝑟
=
963.086 − 4.774
2714.056
= 0.353 = 35.3%