556 ENGINEERING THERMODYNAMICS

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\M-therm\Th12-1.pm5

(i)Final condition of steam ; (ii)Rankine efficiency ;

(iii)Relative efficiency.

Solution. Power developed by the engine = 35 kW (I.P.)

Steam consumption = 284 kg/h

Condenser pressure = 0.14 bar

Steam inlet pressure = 15 bar, 250°C.

From steam tables :

At 15 bar, 250°C : h = 2923.3 kJ/kg, s = 6.709 kJ/kg K

At 0.14 bar : hf = 220 kJ/kg, hfg = 2376.6 kJ/kg,

sf = 0.737 kJ/kg K, sfg = 7.296 kJ/kg K

(i)Final condition of steam :

Since steam expands isentropically.

∴ s 1 = s 2 = sf 2 + x 2 sfg 2

6.709 = 0.737 + x 2 × 7.296

∴ x 2 = 6 709 0 737

7 296

..

.

− = 0.818~_0.82. (Ans.)

∴ h 2 = hf 2 + x 2 hfg 2 = 220 + 0.82 × 2376.6 = 2168.8 kJ/kg.

(ii)Rankine efficiency :

ηrankine =

hh

hhf

12

(^12)
−
− =
2923.3 2168.8
2923.3 220
−
− = 0.279 or 27.9%. (Ans.)
(iii)Relative efficiency :
ηthermal =
I.P.
mh h&ej 1 − f 2

35
284
3600
bg2923.3 220−
= 0.1641 or 16.41%
ηrelative = η
η
thermal
rankine
= 0.
0.
1641
279
= 0.588 or 58.8%. (Ans.)
Example 12.9. Calculate the fuel oil consumption required in a industrial steam plant to
generate 5000 kW at the turbine shaft. The calorific value of the fuel is 40000 kJ/kg and the
Rankine cycle efficiency is 50%. Assume appropriate values for isentropic turbine efficiency, boiler
heat transfer efficiency and combustion efficiency. (AMIE Summer, 2000)
Solution. Power to be generated at the turbine shaft, P = 5000 kW
The calorific value of the fuel, C = 40000 kJ/kg
Rankine cycle efficiency, ηrankine = 50%
Fuel oil combustion, mf :
Assume : ηturbine = 90% ; ηheat transfer = 85% ; ηcombustion = 98%
ηrankine = Shaftpower turbine
combustion
/
heat transfer
η
mCf××ηη×
or 0.5 =
5000 0 9
40000 0 85 0 98
/.
..
bg
mf×××