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?

Answer: 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