TITLE.PM5

VAPOUR POWER CYCLES 553

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\M-therm\Th12-1.pm5

Actual pump work =

8072.

ηpump

=

8 072

08

.

.

= 10.09 kJ/kg

Specific work (Wnet) = 1275 – 10.09 = 1264.91 kJ / kg. (Ans.)

Thermal efficiency =

W

Q

net

1

where, Q 1 = h 1 – hf 4

But hf 4 = hf 3 + pump work = 191.9 + 10.09 = 202 kJ/kg

∴ Thermal efficiency, ηth=

1264 91

3642 202

.
− = 0.368 or^ 36.8 %. (Ans.)
Example 12.5. A simple Rankine cycle works between pressures 28 bar and 0.06 bar, the
initial condition of steam being dry saturated. Calculate the cycle efficiency, work ratio and
specific steam consumption.
Solution.

Fig. 12.9

From steam tables,

At 28 bar : h 1 = 2802 kJ/kg, s 1 = 6.2104 kJ/kg K

At 0.06 bar : hf 2 = hf 3 = 151.5 kJ/kg, hfg 2 = 2415.9 kJ/kg,

sf 2 = 0.521 kJ/kg K, sfg 2 = 7.809 kJ/kg K

vf = 0.001 m^3 /kg

Considering turbine process 1-2, we have :

s 1 = s 2

6.2104 = sf 2 + x 2 sfg 2 = 0.521 + x 2 × 7.809