Page 132 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

Often C1 is negligible compared with C2. In this case the equation becomes

  C 2  
+
0 = m   (h − h )  2  


 2 1  2   

C 2
or 2 = (h − h 2 )
1
2

or C = ( 2 h − h 2 ) (3.30)
1
2

Example 3.8

Fluid with a specific enthalpy of 2800 kJ/kg enters a horizontal nozzle with negligible

velocity at the rate of 14 kg/s. At the outlet from the nozzle the specific enthalpy and

3
specific volume of the fluid are 2250 kJ/kg and 1.25 m /kg respectively. Assuming an

adiabatic flow, determine the required outlet area of the nozzle.

Solution to Example 3.8
The steady flow energy equation gives

  C 2 −C 2  
+
Q −W = m   (h 2 − h 1 )  2 1   ( + gZ 2 − gZ 1 )


  2  



When applied to the nozzle, this becomes

  C 2 −C 2  
+
h
0 = m   ( − h 1 )   2 1  

2
  2   


Since the inlet C1 is negligible, this may be written as
2
C = ( 2 h − h )
2 1 2
C2 = √ [2(2800 – 22500]

= √ 1100
= 33.167 m/s

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