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?

Question

Mathematics

Caylee Pace

14 July, 14:06

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?

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Answers (1)

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Russell · Answer 1

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