FIRST LAW OF THERMODYNAMICS 169dharm
/M-therm/Th4-4.pm520 kJ/s, calculate the power developed by the turbine. Consider the boiler and turbine as single
system.
Solution. Enthalpy of water entering the boiler, h 1 = 800 kJ/kg
Velocity of water entering the boiler, C 1 = 5 m/s
Enthalpy of steam at the outlet of the turbine, h 2 = 2520 kJ/kg
Velocity of steam at the outlet of the turbine, C 2 = 50 m/s
Elevation difference, (Z 1 – Z 2 ) = 4 m
Net heat added to the water in the boiler, Q = 2200 – 20 = 2180 kJ/kg
Power developed by the turbine :
Using the flow equation,
h 1 + C^12
2 + Z^1 g + Q = h^2 +C 22
2 + Z^2 g + W∴ W = (h 1 – h 2 ) +
CC 12 22
22 −F
HGI
KJ
+ (Z 1 – Z 2 ) g + Q= (800 – 2520) +
1
10005
250
24981
100022
L −
NMO
QP+ ×. + 2180= – 1720 +
1
1000
12.5 1250^39
1000
()−+.24 + 2180
= – 1720 – 1.2375 + 0.03924 + 2180
= 458.8 kJ/kg = 458.8 kJ/s = 458.8 kW
Hence, power developed by the turbine = 458.8 kW. (Ans.)
Example 4.39. A turbine, operating under steady-flow conditions, receives 4500 kg of
steam per hour. The steam enters the turbine at a velocity of 2800 m/min, an elevation of 5.5 m
and a specific enthalpy of 2800 kJ/kg. It leaves the turbine at a velocity of 5600 m/min, an elevation
of 1.5 m and a specific enthalpy of 2300 kJ/kg. Heat losses from the turbine to the surroundings
amout to 16000 kJ/h.
Determine the power output of the turbine.
Solution. Quantity of steam supplied to the turbine, m = 4500 kg/h
Steam velocity at the entrance to the turbine, C 1 = 2800 m/min
Elevation at the entrance, Z 1 = 5.5 m
Specific enthalpy at the entrance, h 1 = 2800 kJ/g
Steam velocity at the exit, C 2 = 5600 m/min
Elevation at the exit, Z 2 = 1.5 m
Specific enthalpy at the exit, h 2 = 2300 kJ/kg
Heat losses from the turbine to the surroundings, Q = – 16000 kJ/h
Applying the steady flow energy equation at entry (1) and exit (2)m h 1 C^1 Zg2
++ 2 1
F
HGI
KJ- Q = m (^) h 2 C^2 Zg
2
++ 2 2
F
HG
I
KJ - W
∴ Q – W = mh h
()CC ( )ZZg
21 2
2
2
2
−+ 2 21
F −
HG
I
KJ
+−
L
N
M
M
O
Q
P
P