554 ENGINEERING THERMODYNAMICS

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\M-therm\Th12-1.pm5

∴ x 2 =

2104 521

809

−

= 0.728

∴ h 2 = hf 2 + x 2 hfg 2

= 151.5 + 0.728 × 2415.9 = 1910.27 kJ/kg

∴ Turbine work, Wturbine = h 1 – h 2 = 2802 – 1910.27 = 891.73 kJ/kg

Pump work, Wpump = hf 4 – hf 3 = vf (p 1 – p 2 )

=

0 001 28 0 06 10

1000

..bg−×^5

= 2.79 kJ/kg

[Q hf 4 = hf 3 + 2.79 = 151.5 + 2.79 = 154.29 kJ/kg]

∴ Net work, Wnet = Wturbine – Wpump

= 891.73 – 2.79 = 888.94 kJ/kg

Cycle efficiency =

W

Q

net

1

=

888 94

(^14)
.
hh− f

888.94
2802 154.29− = 0.3357 or 33.57%. (Ans.)
Work ratio = W
W
net
turbine
= 888 94
891.73

. = 0.997. (Ans.)

Specific steam consumption =^3600
Wnet

3600
888 94. =^ 4.049 kg/kWh. (Ans.)
+Example 12.6. In a Rankine cycle, the steam at inlet to turbine is saturated at a pres-
sure of 35 bar and the exhaust pressure is 0.2 bar. Determine :
(i)The pump work, (ii)The turbine work,
(iii)The Rankine efficiency, (iv)The condenser heat flow,
(v)The dryness at the end of expansion.
Assume flow rate of 9.5 kg/s.
Solution. Pressure and condition of steam, at inlet to the turbine,
p 1 = 35 bar, x = 1
Exhaust pressure, p 2 = 0.2 bar

Flow rate, (^) m& = 9.5 kg/s
(^51)
2
s
4
3
35 bar
0.2 bar
T
Fig. 12.10