568 ENGINEERING THERMODYNAMICS
dharm
\M-therm\Th15-2.pm5(^0) T u r b i n e
1 kg
30 bar,
400 Cº 3 bar
m 1
m 1
1 kg
130 Cº
Heater Drain cooler
Hot well
(1 – m ) 1
m 1
1 kg
(1 – m ) 1
(^12) 0.04 bar
(a)
0
2
h
h 0
h 1
h 2
Saturation line
0.04 bar
3 bar
30 bar
s
1
(b)
Fig. 12.18
(i) Mass of steam used, m 1 :
Heat lost by the steam = Heat gained by water.
Taking the feed-heater and drain-cooler combined, we have :
m 1 (h 1 – hf 2 ) = 1 × 4.186 (130 – 27)
or m 1 (2700 – 121.5) = 4.186 (130 – 27)
∴ m 1 =
4.186 (130 27)
(2700 121.5)
−
− = 0.1672 kg. (Ans.)
(ii) Thermal efficiency of the cycle :
Work done per kg of steam
= 1(h 0 – h 1 ) + (1 – m 1 ) (h 1 – h 2 )