TITLE.PM5

(Ann) #1

572 ENGINEERING THERMODYNAMICS


dharm
\M-therm\Th15-2.pm5

At 0.3 bar : hf 3 = 289.3 kJ/kg
At 0.05 bar : hf 4 = 137.8 kJ/kg.
(i)Mass of bled steam for each heater per kg of steam :
Using heat balance equation :
At heater No. 1 :
m 1 h 1 + hf 2 = m 1 hf 1 + hf 1

∴ m 1 =

hh
hh

f f
f

12

(^11)

− =





    1. 2600 640.




11
1



= 0.088 kJ/kg of entering steam. (Ans.)
At heater No. 2 :
m 2 h 2 + hf 3 + m 1 hf 1 = hf 2 + (m 1 + m 2 )hf 2

m 2 =

()()
()

hh mhh
hh

f f f f
f

23 12
2

1
2

+− −

=
()()
()

467.1 289.3 088 640.1 467.1
2430 467.1

−− −

0.
= 162 57
1962 9

.
.
= 0.0828 kJ/kg of entering steam. (Ans.)
At heater No. 3 :
m 3 h 3 + hf 5 + (m 1 + m 2 )hf 2 = hf 3 + (m 1 + m 2 + m 3 )hf 3 ...(i)
At drain cooler :
(m 1 + m 2 + m 3 )hf 3 + hf 4 = hf 5 + (m 1 + m 2 + m 3 )hf 4
∴ hf 5 = (m 1 + m 2 + m 3 ) (hf 3 – hf 4 ) + hf 4 ...(ii)
Inserting the value of hf 5 in eqn. (i), we get
m 3 h 3 + (m 1 + m 2 + m 3 ) (hf 3 – hf 4 ) + hf 4 + (m 1 + m 2 ) hf 2 = hf 3 + (m 1 + m 2 + m 3 )hf 3

∴ m 3 =

()()()hh mmhh
hh

f f f f
f

34 24
4

12
3

−−+ −

=

()( )()
()

289.3 137.8 088 0828 467.1 137.8
2210 137.8

−− + −









=
151.5 56.24
2072.2


= 0.046 kJ/kg of entering steam. (Ans.)

Work done/kg (neglecting pump work)
= (h 0 – h 1 ) + (1 – m 1 ) (h 1 – h 2 ) + (1 – m 1 – m 2 ) (h 2 – h 3 ) + (1 – m 1 – m 2 – m 3 ) (h 3 – h 4 )
= (2905 – 2600) + (1 – 0.088) (2600 – 2430) + (1 – 0.088 – 0.0828) (2430 – 2210)
+ (1 – 0.088 – 0.0828 – 0.046) (2210 – 2000)
= 305 + 155.04 + 182.42 + 164.47 = 806.93 kJ/kg
Heat supplied/kg = h 0 – hf 1 = 2905 – 640.1 = 2264.9 kJ/kg.
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