Page 135 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

Feedback to Activity 3D

3.10 Neglecting the changes in potential energy, the steady flow energy equation is

  C 2 −C 2  
+
Q −W =  (h 2 − h 1 )  2 1  


  2   

Q is negative since heat is lost from the steam to the surroundings
 C 2 − C 2 
+
specific W = (h − h )  1 2  - Q
1 2  
 2 
2
( 16 − 37 2 )
= (2999-2530) + − 25
2x 10 3
= 434.443 kJ/kg

The steam flow rate = 324000/3600 = 90 kg/s

 W = 434.443 x 90

= 39099.97 kJ/s or kW

 39100 kW

 39.1 MW

3.11 The steady energy flow equation for nozzle gives

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

On simplification,

C = ( 2 h − h + 1 2
) C
2
1
2

3
2
3
= 2(1200x 10 - 900x10 ) + 200
= 800 m/s

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