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VAPOUR POWER CYCLES 551

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Example 12.3. In a steam turbine steam at 20 bar, 360°C is expanded to 0.08 bar. It then
enters a condenser, where it is condensed to saturated liquid water. The pump feeds back the
water into the boiler. Assume ideal processes, find per kg of steam the net work and the cycle
efficiency.

T

s

p = 0.08 bar 2

p = 20 bar 1

5

4

(^32)
1(360 C)º
Fig. 12.7
Solution. Boiler pressure, p 1 = 20 bar (360°C)
Condenser pressure, p 2 = 0.08 bar
From steam tables :
At 20 bar (p 1 ), 360°C : h 1 = 3159.3 kJ/kg
s 1 = 6.9917 kJ/kg-K
At 0.08 bar (p 2 ) : h 3 = hfbgp 2 = 173.88 kJ/kg,
s 3 = sfbgp 2 = 0.5926 kJ/kg-K
hfgbgp 2 = 2403.1 kJ/kg, sgbgp 2 = 8.2287 kJ/kg-K
vfbgp 2 = 0.001008 m^3 /kg ∴ sfgbgp 2 = 7.6361 kJ/kg-K
Now s 1 = s 2
6.9917 =sfbgp 2 + x 2 sfgbgp 2 = 0.5926 + x 2 × 7.6361
∴ x 2 = 0.69917 0.5926
7.6361
− = 0.838
∴ h 2 = hfbgp 2 + x 2 hfgbgp 2
= 173.88 + 0.838 × 2403.1 = 2187.68 kJ/kg.
Net work, Wnet :
Wnet = Wturbine – Wpump
Wpump = hf 4 – hfbgp 2 (= hf 3 ) = vfbgp 2 (p 1 – p 2 )
= 0.00108 (m^3 /kg) × (20 – 0.08) × 100 kN/m^2
= 2.008 kJ/kg