Page 129 - DJJ20063- Thermodynamics 1
P. 129
DJJ20063- Thermodynamics 1
The steady flow energy equation gives
C 2 C 2
gZ + h + 1 + Q = gZ + h + 2 + W
1 1 2 2
2 2
and with the flow rate, m (kg/s) the equation may be written as
C 2 −C 2
+
Q −W = m (h − h ) 2 1 ( + gZ − gZ ) 3.28)
2 1 2 2 1
In applying this equation to the boiler, the following points should be noted :
i. Q is the amount of heat energy passing into the fluid per second
ii. W is zero since a boiler has no moving parts capable of affecting a work
transfer
iii. The kinetic energy is small as compared to the other terms and may
usually be neglected
iv. The potential energy is generally small enough to be neglected.
v. m (kg/s) is the rate of the flow of fluid.
Hence the equation is reduced to
Q = m (h − h 1 ) (3.29)
2
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