thermodynamics t3
TRANSCRIPT
Question 1Solution:(a) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
Question 12 21 2
1 2
1 2
2 2
in out
c cmgZ 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 hkJ s
❶ compressed water at 38°C❷dry saturated steam at 30 bar
Question 1Solution:(b) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 21 2
1 20; 0; ;2 2in inc cW Q mgZ mgZ m m
1 2( )out outW m h h Q
Question 1❶ 450 °C and 30 bar superheated steam:
h1=3343 kJ/kg❷ 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) 1302168.1 /
out outW m h h Q
kJ s
Question 2Solution:(a) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 21 2
1 20; 0; ;2 2in inc cW Q mgZ mgZ m m
1 2( )out outW m h h Q
Question 2❶ 450 °C and 40 bar superheated steam:
h1=3330 kJ/kg❷ 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) 1252000.76 /
out outW m h h Q
kJ s
Question 2Solution:(b) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 21 2
1 20; 0; 0; ;2 2in out inc cW W Q mgZ mgZ m m
1 2( )outQ m h h
Question 2❶ Same as the 2nd stage of part (a):
h1=2570.8 kJ/kg❷ 1.2 bar :
h2=hf=439 kJ/kg
1 2( )2.8(2570.8 439)5969.04 /
outQ m h h
kJ s
Question 3Solution:(a) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
2 21 2
1 20; 0; ;2 2in outc cW W mgZ mgZ m m
2 1( )in outQ m h h Q
Question 3❶ 28 °C compressed water
h1=hf=117.3 kJ/kg❷ 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) 302378.18 /
in outQ m h h Q
kJ s
Question 3Solution:(b) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
1 2 3
in outm m
m m m
22 231 2
1 1 1 2 2 2 3 3 3( ) ( ) ( )2 2 2in in out out
cc cQ W m h gZ m h gZ Q W m h gZ
22 231 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 cm gZ m gZ m gZ m m m
1 1 2 2 3 3m h m h m h
Question 3❶ Same as the 2nd stage of part (a):
h1=2074.12 kJ/kg❷ 28 °C compressed water:
h2=hf=117.3 kJ/kg❸ 75 °C compressed water:
h3=hf=313.9 kJ/kg1 1 2 2 3 3
2 2
2
1.2 2074.12 117.3 (1.2 ) 313.910.74 /
m h m h m hm m
m kg s
3 10.74 1.2 11.94 /m kg s
Question 4Solution:(a) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
1 2 3
in outm m
m m m
22 231 2
1 1 1 2 2 2 3 3 3( ) ( ) ( )2 2 2in in out out
cc cQ W m h gZ m h gZ Q W m h gZ
22 231 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 cm gZ m gZ m gZ m m m
1 1 2 2 3 3m h m h m h
1
2
3
m wet steamm cold waterm hot water
Question 4❶ 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) 6401947.58 /
g fh xh x h
kJ kg
Question 4Take
Hence Apply
Since
Hence
1 1 2 2 3 3m h m h m h
1
2
mxm
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 0.10371947.58 293
h hxh h
Question 4Solution:(b) Apply the continuity flow equation
Apply the steady flow energy
Since
Hence
1 2 3
in outm m
m m m
22 231 2
1 1 1 2 2 2 3 3 3( ) ( ) ( )2 2 2in in out out
cc cQ W m h gZ m h gZ Q W m h gZ
22 231 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 cm gZ m gZ m gZ m m m
1 1 2 2 3 3m h m h m h
1
2
3
m wet steamm cold waterm hot water
Question 4❶ 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) 6401947.58 /
g fh xh x h
kJ kg
Question 4Apply
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 23
3
0.6 1947.58 5.2 121.55.8
310.4 /
m h m hhm
kJ kg
70 ; 293 /?; 310.4 /
75 ; 313.9 /
f
f
f
t C h kJ kgt h kJ kg
t C h kJ kg
70 310.4 29375 70 313.9 293
74.16
t
t C
Thank youQ & A