Following three questions refer to the information below. a coal-fired power plant produces and sells 10,000 mw electricity while operating at 40% thermal efficiency. the sale prioe of electric power is $0.04 per kwh. how much power does it reject to the atmosphere? how much power (in the form of heat rate) does it require from coal combustion?

Heat rejected=150000MW

Heat rate= 2.5

Step-by-step explanation:

Thermal efficiency is defined as work out divided by heat in.

Therefore

thermal efficiency = 40/100= 100000 MW/Heat in

Heat in = 250000MW

For a closed system undergoing a cycle the change in internal energy is zero. Therefore the work out of the system is equal to the net heat transfer of the system. Therefore

Work out= heat in - heat out

Therefore heat out = heat in - work out

Heat out = 250000 MW - 100000MW = 150000 MW which is the amount of heat rejected

The heat rate is defined as the inverse of the thermal efficiency expressed in kJ/kWh.Therefore

Heat rate = 1/thermal efficiency

Heat rate = 1/0.4 = 2.5 kJ/kWh

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step-by-step explanation: