570 ENGINEERING THERMODYNAMICS

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\M-therm\Th15-2.pm5

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)
−
−

640.1 417.5
2720 640.1
−
− = 0.107 kJ/kg of entering steam.
(Ans.)
At heater No. 2 :
m 2 h 2 + m 1 hf 1 + hf 4 = (m 1 + m 2 ) hf 2 + hf 2 ...(i)
At drain cooler :
(m 1 + m 2 ) hf 2 + hf 3 = hf 4 + (m 1 + m 2 ) hf 3
∴ hf 4 = (m 1 + m 2 ) (hf 2 – hf 3 ) + hf 3 ...(ii)
Inserting the value of hf 4 in eqn. (i), we get
m 2 h 2 + m 1 hf 1 + (m 1 + m 2 ) (hf 2 – hf 3 ) + hf 3 = (m 1 + m 2 ) hf 2 + hf 2
m 2 h 2 + m 1 hf 1 + (m 1 + m 2 )hf 2 – (m 1 + m 2 )hf 3 + hf 3 = (m 1 + m 2 )hf 2 + hf 2
m 2 h 2 + m 1 hf 1 – m 1 hf 3 – m 2 hf 3 + hf 3 = hf 2
m 2 (h 2 – hf 3 ) = (hf 2 – hf 3 ) – m 1 (hf 1 – hf 3 )
m 2 =
()()
()
hh mhh
hh
f f f f
f
23 13
3
1
2
−− −
−
= ()()
()
417.5 173.9 107 640.1 173.9
2450 173.9
−− −
−
0.
= 193 7
2276 1
.
.
= 0.085 kJ/kg. (Ans.)
Work done = 1 (h 0 – h 1 ) + (1 – m 1 ) (h 1 – h 2 ) + (1 – m 1 – m 2 ) (h 2 – h 3 )
= 1 (3115.3 – 2720) + (1 – 0.107) (2720 – 2450)

• 
  

  (1 – 0.107 – 0.085) (2450 – 2120)
  = 395.3 + 241.11 + 266.64 = 903.05 kJ/kg
  Heat supplied/kg = h 0 – hf 1
  = 3115.3 – 640.1 = 2475.2 kJ/kg
  ∴ Thermal efficiency of the cycle

  
  

  Work done
  Heat supplied

  
  

  903 05
  2475 2
  .

  
  

. = 0.3648 or 36.48%. (Ans.)
+Example 12.16. Steam at a pressure of 20 bar and 250°C enters a turbine and leaves it
finally at a pressure of 0.05 bar. Steam is bled off at pressures of 5.0, 1.5 and 0.3 bar.
Assuming (i) that the condensate is heated in each heater upto the saturation temperature of the
steam in that heater, (ii) that the drain water from each heater is cascaded through a trap into
the next heater on the low pressure side of it, (iii) that the combined drains from the heater
operating at 0.3 bar are cooled in a drain cooler to condenser temperature, calculate the
following :
(i)Mass of bled steam for each heater per kg of steam entering the turbine
(ii)Thermal efficiency of the cycle,