TITLE.PM5

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)
−
− =

640. 
     
     
     
     467. 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.