Power Plant Engineering

Solution. From Mollier diagram
H
1
= Total heat of steam at point 1 = 3386.24 kJ/kg
H
2
= Total heat of steam at point 2 = 2006.2 kJ/kg
H
w2
= Total Heat of water at point 2 = 121.42 kJ/kg
(a) Thermal efficiency = 
1
2
1
2
(H
H )
(H
H
)
−
−
w
= 
(3386.24
2006.2)
(3386.24 121.42)
−
−
= 42.27%
(b) Specific steam consumption is the amount of steam in kg per kW-hr.
Now 1 kW-hr = 3600 kJ
Specific steam consumption = 
1
2
3600
(H
H )
−
= 
3600
(3386.24
2006.2)
−
= 2.61 kg/kW-hr
(c) Hourly steam consumption = 2.61 × Kilowatts
= 2.61 × 50,000 = 1.305 Tonnes/hr Ans.
Example 3. A steam power plant, operating with one regenerative feed water heating is run at
the initial steam conditions of 35.0 bar and 440°C with exhaust pressure of 0.040 bar. Steam is bled
from the turbine for feed water heating at a pressure of 1.226 bar. Determine
(1) Specific heat consumption
(2) Thermal efficiency of the cycle
(3) Economy percentage compared with the cycle of a simple condensing power plant.
Solution. From Mollier diagrams and steam table,
H
1
= 3314 kJ/kg
H
2
= 2560 kJ/kg
H
3
= 2100 kJ/kg
H
w2
= 439.43 kJ/kg
H
w3
= 121.42 kJ/kg
From the heat balance for the feed water heater
m(H
2
– H
w2
) = (1 – m)(H
w2
– H
w3
)
m(2560 – 439.43) = (1 – m)(439.43 – 121.42)

---

FUNDAMENTAL OF POWER PLANT
31
On solving, we get m = 0.1304 kg
Total work done = 1 × (H
1
– H
2
) + (1 – m)(H
2
– H
3
)
= (3314 – 2460) + (1 – 0.1304)(2560 – 2100)
= 1154 kJ/kg
(1) Specific steam consumption = 
3600
1154
= 3.12Kg/kW-hr. Ans.
(2) Thermal efficiency = 
1154
(3314
439.43)
−
= 40.15%. Ans.
(3) With out regeneration feed water heating the work done will be
H
1
– H
2
= 3314 – 2100 = 1214 kJ/kg
Steam consumption = 
3600
1214
= 2.94 kg/kW-hr
Without regeneration heating, the thermal efficiency
η
= 
1
3
1
3
(H
H )
(H
H
)
−
−
w
Now from the steam tables
H
w3
= 121.42 kJ/kg
η
= 
(3314
2100)
(3314 121.42)
−
−
= 0.38
Increase in thermal efficiency due to regeneration feed water heating is
= 
(0.4015
0.38)
0.4015
−
= 5.5 % Ans.