Heat Rejected by a Power Plant (from the Electric Output and the Thermal Efficiency)

A thermal plant, whether coal, gas, nuclear or biomass, turns only a fraction $\eta$ of the heat it makes into electricity. By the first law all the rest leaves as waste heat, into a river, a lake, the sea or the air through cooling towers. For an electric output $P_e$ at efficiency $\eta$, the heat input is $P_{th} = P_e/\eta$, and the heat rejected is that less $P_e$, which is $P_e(1 - \eta)/\eta$. It is the bookkeeping of one balance: heat in equals electricity out plus heat rejected.

Read the answer through the factor $(1 - \eta)/\eta$. At an efficiency of one third the plant rejects twice as much heat as it sends out as electricity: 1 000 MW of electricity goes with 2 000 MW of heat to the water. A plant with an efficiency above one half rejects less heat than the electricity it makes. The formula says how much heat, not where it goes; the cooling water's temperature rise follows from the flow it is given.

The page's example uses a published input. The EIA's average operating heat rate of nuclear plants in 2025 is 10 443 Btu per kilowatt-hour (Electric Power Annual, Table 8.1), which is an efficiency of $3\,412.142/10\,443 = 0.3267$. A plant sending out 980 MW of electricity then needs $980/0.3267 = 2\,999.7$ MW of heat and rejects 2 019.69 MW. The same table lists 2025 coal at 10 647 and natural gas at 7 781 Btu per kilowatt-hour, and each turns into an efficiency the same way.

Backwards works too. If 2 000 MW is rejected for 1 000 MW sent out, the heat in is 3 000 MW and $\eta = 1000/3000 = 0.33333$. With 3 000 MW in, 980 MW out as electricity and 2 020 MW rejected, $\eta = 980/3000 = 0.326667$. The efficiency has to lie between 0 and 1.