Topic: Power Plant Engineering
A simple steam power cycle receives 1,00,000 kJ/min as heat transfer from hot combustion gases and rejects 66,000 kJ/min as heat transfer to the environment. If the pump power required is 1400 kJ/min the thermal efficiency of the cycle and turbine power output is
- A. 51.5% and 590 kW
- B. 34% and 590 kW
- C. 51.5% and 566.6 kW
- D. 34% and 566.6 kW
Answer: B. 34% and 590 kW
Explanation: The heat supplied is $Q_S = 1,00,000$ kJ/min and heat rejected is $Q_R = 66,000$ kJ/min. Thermal efficiency is $\eta_{th} = 1 - (Q_R / Q_S) = 1 - (66000 / 100000) = 0.34 = 34\%$. The net work output is $W_{net} = Q_S - Q_R = 34,000$ kJ/min. Since $W_{net} = W_{turbine} - W_{pump}$, we have $W_{turbine} = W_{net} + W_{pump} = 34,000 + 1,400 = 35,400$ kJ/min. Converting this to kW: $35,400 / 60 = 590$ kW.