Page 127 - DJJ20063- Thermodynamics 1
P. 127
DJJ20063- Thermodynamics 1
The steady flow energy equation gives
C 2 −C 2
+
Q −W = m (h − h ) 2 1 ( + gZ − gZ )
2 1 2 2 1
Points to note :
i. The average velocity of flow of fluid through a turbine is normally high, and
the fluid passes quickly through the turbine. It may be assumed that,
because of this, heat energy does not have time to flow into or out of the
fluid during its passage through the turbine, and hence Q = 0 .
ii. Although velocities are high the difference between them is not large, and
the term representing the change in kinetic energy may be neglected.
iii. Potential energy is generally small enough to be neglected.
iv. W is the amount of external work energy produced per second.
The steady flow energy equation becomes
m
-W = (h − h 1 )
2
or W = (h − h 2 ) (3.27)
m
1
Example 3.6
A fluid flows through a turbine at the rate of 45 kg/min. Across the turbine the specific
enthalpy drop of the fluid is 580 kJ/kg and the turbine loss 2100 kJ/min in the form of
heat energy. Determine the power produced by the turbine, assuming that changes
in kinetic and potential energy may be neglected.
Solution to Example 3.6
The steady flow energy equation gives
C 2 −C 2
+
Q −W = m (h 2 − h 1 ) 2 1 ( + gZ 2 − gZ 1 )
2
118 | P a g e
