Question 3.11: A steam engine is governed by varying the initial steam pres...
A steam engine is governed by varying the initial steam pressure, used 6 kg/kWh when developing 60 kW and 4.65 kg/kWh when developing 150 kW. If steam supplied is 96% dry at 15 bar and a back pressure of 0.36 bar, determine the steam consumption per hour and indicated thermal efficiency when developing 120 kW.
Refer Fig. 37.
Steam consumption/hour developing 60 kW
= 60 × 6 = 360 kg/h
Steam consumption/hour developing 150 kW
= 150 × 4.65 = 697.5 kg/h
In throttle governing, Willan’s law holds good,
i.e., m = a + b × I.P.
360 = a + b × 60 …(i)
697.5 = a + b × 150 . ..(ii)
[m = Steam consumption (kg/h)
a = Steam consumption at no-load (kg/h)
b = Slope of Willan’s line. ]
Subtracting (i) from (ii), we get
337.5 = 90 b
∴ b = \frac{337.5}{90} = 3.75
Inserting the value of b in (i), we get
360 = a + 3.75 × 60
∴ a = 135.
Steam consumption for developing 120 kW
= 135 + 3.75 × 120 = 585 kg/h.
Heat supplied/kg = h_{1} – h_{f_2}
From steam tables :
At 15 bar : h_{f_1} = 844.7 kJ/kg, h_{fg_1} = 1945.2 kJ/kg.
At 0.36 bar : h_{f_2} = 307.1 kJ/kg.
∴ Heat supplied/kg = (844.7 + 0.96 × 1945.2) – 307.1 = 2405 kJ/kg.
Indicated thermal efficiency (developing 120 kW)
η_{th(I)} = \frac{I.P.}{m_{s} (h_{1} – h_{f_2})} = \frac{120}{\frac{585}{3600} × 2405} = 0.3070 or 30.7%.
