Thermodynamics

1. Thermodynamics
Tutorial 3
Dr. SUN Weimeng
swe3@np.edu.sg
Office no.:6460 8342

2. Question 1
Solution:
(a) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence

3. Question 1
2 2
1 2
1 2
1 2
2 2
in out
c c
mgZ mgZ m m
Q mh Q mh
 
   
1 2
2 1
159.1 / , 2803 /
( )
350 2.8(2803 159.1) 7752.92 /
in out
h kJ kg h kJ kg
Q Q m h h
kJ s
 
   
   
❶ compressed water at 38°C
❷dry saturated steam at 30 bar

4. Question 1
Solution:
(b) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 2
1 2
1 20; 0; ;
2 2
in in
c c
W Q mgZ mgZ m m   
1 2( )out outW m h h Q  

5. Question 1
❶ 450 °C and 30 bar superheated steam:
h1=3343 kJ/kg
❷ 0.42 bar and x=0.95 wet steam:
hf=323 kJ/kg; hg=2638 kJ/kg
2 (1 )
0.95 2638 0.05 323 2522.25 /
g fh xh x h
kJ kg
  
    
1 2( )
2.8(3343 2522.25) 130
2168.1 /
out outW m h h Q
kJ s
   
  


6. Question 2
Solution:
(a) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 2
1 2
1 20; 0; ;
2 2
in in
c c
W Q mgZ mgZ m m   
1 2( )out outW m h h Q  

7. Question 2
❶ 450 °C and 40 bar superheated steam:
h1=3330 kJ/kg
❷ 1.2 bar and x=0.95 wet steam:
hf=439 kJ/kg; hg=2683 kJ/kg
2 (1 )
0.95 2683 0.05 439 2570.8 /
g fh xh x h
kJ kg
  
    
1 2( )
2.8(3330 2570.8) 125
2000.76 /
out outW m h h Q
kJ s
   
  


8. Question 2
Solution:
(b) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 2
1 2
1 20; 0; 0; ;
2 2
in out in
c c
W W Q mgZ mgZ m m    
1 2( )outQ m h h 

9. Question 2
❶ Same as the 2nd stage of part (a):
h1=2570.8 kJ/kg
❷ 1.2 bar :
h2=hf=439 kJ/kg
1 2( )
2.8(2570.8 439)
5969.04 /
outQ m h h
kJ s
 
 


10. Question 3
Solution:
(a) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 2
1 2
1 20; 0; ;
2 2
in out
c c
W W mgZ mgZ m m   
2 1( )in outQ m h h Q  

11. Question 3
❶ 28 °C compressed water
h1=hf=117.3 kJ/kg
❷ 5 bar and x=0.68 wet steam:
hf=640 kJ/kg; hg=2749 kJ/kg
2 (1 )
0.68 2749 (1 0.68) 640 2074.12 /
g fh xh x h
kJ kg
  
     
2 1( )
1.2(2074.12 117.3) 30
2378.18 /
in outQ m h h Q
kJ s
  
  


12. Question 3
Solution:
(b) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
1 2 3
in outm m
m m m

 
 
22 2
31 2
1 1 1 2 2 2 3 3 3( ) ( ) ( )
2 2 2
in in out out
cc c
Q W m h gZ m h gZ Q W m h gZ           
22 2
31 2
1 1 2 2 3 3 1 2 3
0; 0; 0; 0;
;
2 2 2
in out in outW W Q Q
cc c
m gZ m gZ m gZ m m m
   
   
1 1 2 2 3 3m h m h m h 

13. Question 3
❶ Same as the 2nd stage of part (a):
h1=2074.12 kJ/kg
❷ 28 °C compressed water:
h2=hf=117.3 kJ/kg
❸ 75 °C compressed water:
h3=hf=313.9 kJ/kg
1 1 2 2 3 3
2 2
2
1.2 2074.12 117.3 (1.2 ) 313.9
10.74 /
m h m h m h
m m
m kg s
 
     

3 10.74 1.2 11.94 /m kg s   

14. Question 4
Solution:
(a) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
1 2 3
in outm m
m m m

 
 
22 2
31 2
1 1 1 2 2 2 3 3 3( ) ( ) ( )
2 2 2
in in out out
cc c
Q W m h gZ m h gZ Q W m h gZ           
22 2
31 2
1 1 2 2 3 3 1 2 3
0; 0; 0; 0;
;
2 2 2
in out in outW W Q Q
cc c
m gZ m gZ m gZ m m m
   
   
1 1 2 2 3 3m h m h m h 
1
2
3
m wet steam
m cold water
m hot water




15. Question 4
❶ 5 bar and x=0.62 wet steam:
hf=640 kJ/kg; hg=2749 kJ/kg
❷ 29 °C compressed water:
h2=hf=121.5 kJ/kg
❸ 70 °C compressed water:
h3=hf=293 kJ/kg
1 (1 )
0.62 2749 (1 0.62) 640
1947.58 /
g fh xh x h
kJ kg
  
    


16. Question 4
Take
Hence
Apply
Since
Hence
1 1 2 2 3 3m h m h m h 
1
2
m
x
m

1 2 3m m m 
1 2m xm
3 2(1 )m x m  
2 1 2 2 2 3(1 )xm h m h x m h   
1 2 3(1 )xh h x h  
3 2
1 3
293 121.5
0.1037
1947.58 293
h h
x
h h
 
  
 

17. Question 4
Solution:
(b) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
1 2 3
in outm m
m m m

 
 
22 2
31 2
1 1 1 2 2 2 3 3 3( ) ( ) ( )
2 2 2
in in out out
cc c
Q W m h gZ m h gZ Q W m h gZ           
22 2
31 2
1 1 2 2 3 3 1 2 3
0; 0; 0; 0;
;
2 2 2
in out in outW W Q Q
cc c
m gZ m gZ m gZ m m m
   
   
1 1 2 2 3 3m h m h m h 
1
2
3
m wet steam
m cold water
m hot water




18. Question 4
❶ 5 bar and x=0.62 wet steam:
hf=640 kJ/kg; hg=2749 kJ/kg
❷ 29 °C compressed water:
h2=hf=121.5 kJ/kg
1 (1 )
0.62 2749 (1 0.62) 640
1947.58 /
g fh xh x h
kJ kg
  
    


19. Question 4
Apply
Since
Hence
1 1 2 2 3 3m h m h m h 
1 2 3m m m 
3 0.6 5.2 5.8 /m kg s   
1 1 2 2
3
3
0.6 1947.58 5.2 121.5
5.8
310.4 /
m h m h
h
m
kJ kg
   
  

70 ; 293 /
?; 310.4 /
75 ; 313.9 /
f
f
f
t C h kJ kg
t h kJ kg
t C h kJ kg
 
 
 
70 310.4 293
75 70 313.9 293
74.16
t
t C
 

 
 

20. Thank you
Q & A