1. In a steam power plant 1.5 kg of water is supplied per second to the boiler. The enthalpy and velocity ofwater entering the boiler are 800 kJ/kg and 10 m/s. Heat at the rate of 2200 kJ per kg of water is suppliedto the water. The steam after passing through the turbine comes out with a velocity of 50 m/s andenthalpy of 2520 kJ/kg. The boiler inlet is 5 m above the turbine exit. The heat loss from the boiler is1800 kJ/min and from the turbine 600 kJ/min. Determine the power capacity of the turbine, consideringboiler and turbine as single unit
Answers
Explanation:
Enthalpy of water entering the boiler,
h₁
Velocity of water entering the boiler,
800kJ/kg
C_{1} = 5
m/s
Enthalpy of steam at the outlet of the turbine,
h₂ = 2520 kJ/kg
Velocity of steam at the ourtlet of the turbine,
C₂ = 50 m/s
Elevation difference,
(Z_{1} - Z_{2}) = 4 m
Net heat added to the water in the boiler,
h₁ 2200-202180 kJ/kg
Power developed by the turbine:
Using the flow equation,
h_{1}+\frac{C_{1}^{2}}{2}+Z_{1}g+Q=h_{2}+\frac{C_{2}^{2}}{2}+Z_{2}g+W
W=(h_{1}-h_{2})-(\frac{C_{1}^{2}}{2}-\frac{C_{2}^{2}}{2}) +
(Z₁ Z₂) g+Q
- (800 - 2520) +
1000
-
2
+
4x9.81
1000
+
2180
= -1720 +1000 (12.5 - 1250) +
39.24
1000
+2180
=-1720-1.2375 +0.03924 + 2180
= 458.8kJ/kg = 458.8kJ/s = 458.8 kW
h_{1}+\frac{C_{1}^{2}}{2}+Z_{1}g+Q=h_{2}+\frac{C_{2}^{2}}{2}+Z_{2}g+W
W=(h_{1}-h_{2})-(\frac{C_{1}^{2}}{2}-\frac{C_{2}^{2}}{2}) +
(Z₁ Z₂) g+Q
- (800 - 2520) +
1000
-
2
+
4x9.81
1000
+
2180
= -1720 +1000 (12.5 - 1250) +
39.24
1000
+2180
=-1720-1.2375 +0.03924 + 2180
= 458.8kJ/kg = 458.8kJ/s = 458.8 kW