Issue 079 - Energy generation - Energy efficiency and waste heat
How much waste heat would a 702 MW fuel-cell plant produce?
American Electric Power plans a 702-megawatt solid-oxide fuel-cell facility near Cheyenne, Wyoming, with Google buying its output under a 20-year agreement. How much heat would flow through the plant alongside the electricity?
Energy and infrastructureAbout 1 minute
Sources checked October 16, 2026. Figures and circumstances may have changed.
The problem
While producing 702 MW of electricity near full output, approximately how much waste heat would the fuel-cell plant produce?
Estimate the electrical efficiency, infer the chemical-energy input, and treat the remaining energy as heat for this scale comparison.
The guess unit is megawatts of thermal power. As extensions, turn that heat into one day's energy, home-heating equivalents, and the amount of water it could evaporate.
This is an order-of-magnitude estimate. The Fermi reference collects useful facts and equations.
Grounding facts
AEP announced a roughly $4 billion, 702 MW fuel-cell project near Cheyenne under a 20-year power-purchase agreement with Google. Construction is planned to start in 2026, with full operation expected by the end of 2028. The project is planned around Bloom Energy technology and includes carbon-capture integration.
Bloom's 2026 Energy Server datasheet lists cumulative net AC electrical efficiency of 65% to 53% on a lower-heating-value basis. A round 60% is a useful central estimate. The same datasheet reports no water use during normal operation, so the evaporation extension below is only a physical comparison, not a description of the planned cooling system.
The U.S. Geological Survey gives water's heat of vaporization as about 2,260 kJ/kg, or 2.26 MJ per liter at roughly one kilogram per liter.
EIA's 2020 household data put average natural-gas space-heating use at 44.9 million Btu per year nationally and 63.0 million Btu in Wyoming. Wyoming's January average was 11.3 million Btu. Those correspond to about 1.5 kW nationally averaged across a year, 2.1 kW for Wyoming across a year, and 4.5 kW during a Wyoming January.
After checking sources
Checked answer and calculation
electrical output = 702 MW
electrical efficiency ~= 60% = 0.60
chemical-energy input
= 702 MW / 0.60
~= 1,170 MW
waste heat
= 1,170 MW - 702 MW
~= 468 MW
~= 470 MWThe central answer is about 470 MW of waste heat: firmly in the hundreds of megawatts. Using Bloom's full 53% to 65% efficiency range gives roughly 620 MW to 380 MW. This simple energy balance treats all non-electric input as heat; how much is recoverable at a useful temperature is a separate engineering question.
daily waste heat
~= 468 MW x 24 hours
~= 11,200 MWh
~= 11.2 GWh of heat
~= 4.0 x 10^13 J
water evaporated
~= 4.0 x 10^13 J / 2.26 x 10^6 J/kg
~= 1.8 x 10^7 kg/day
~= 18 million liters/day
~= 4.7 million gallons/dayThe hypothetical evaporative heat sink would therefore consume about 18 million liters per day. Heating water to its evaporation temperature and real cooling-system losses would increase the requirement. Again, Bloom's datasheet says the Energy Server itself uses no water during normal operation.
Wyoming annual-average home-heating power
~= 63 million Btu/year
~= 18,500 kWh/year
~= 2.1 kW average
annual-average heating equivalents
~= 468,000 kW / 2.1 kW/home
~= 220,000 Wyoming homes
cold-month heating equivalents
~= 468,000 kW / 4.5 kW/home
~= 100,000 Wyoming homesThe home comparison depends heavily on weather and building efficiency. The waste heat is equivalent to the annual-average space-heating energy of roughly 220,000 Wyoming natural-gas-heated homes, but closer to 100,000 homes during a cold month. Physical proximity, distribution losses, temperature, and the need for heat at the same time would sharply limit actual district-heating use.
annual electricity at 95% availability
~= 702 MW x 8,760 hours/year x 0.95
~= 5.8 TWh/year
household-electricity equivalents
~= 5.8 billion kWh / 10,791 kWh/home-year
~= 540,000 homesAt 95% annual availability, the plant would generate about 5.8 TWh per year, equivalent to the annual electricity purchases of roughly 540,000 average U.S. residential customers. That is a scale comparison, not an estimate of the number of homes the Google agreement will serve.
Before checking sources
Matt's original estimate
This is Matt's original, unverified estimate, preserved before the source check.
I assumed that the chemical energy process that produces energy by the fuel cells is about 25% efficient, and that it takes about 500W to heat the average home, and that the heat of evaporation of a liter of water is about 3 * 10^8 J.
Starting from the top, about 700 MW at 25% efficiency (and assuming most of the rest of the energy is waste heat energy) means waste heat is about 2100 MW of heat energy (or 2.1 GW). daily, that's 500 GWH of waste heat energy.
if it takes about 500 W to heat the average household, that's about 12 KWH daily. Dividing that into 500 GWH, I estimate that's enough waste heat to warm about 42 million homes.
if all of that waste heat were to be used to evaporate water, 500 GWH * 3660 seconds/hour gives about 1,830,000 GJ or 1.8 * 10^15 J of waste heat to remove daily. If the energy required to evaporate a liter of water is about 3 * 10^8 (complete guess on my part) then it would take about 6 million kg or liters of water daily to deal with the waste heat.
Reasoning score
Matt's reasoning score: 50 / 100
Useful facts: 0/30. The 25% efficiency was well below Bloom's published range, 500 W understated household heating demand, and the heat-of-vaporization peg was about 130 times too high.
Reasoning approach: 30/30. The energy-balance structure was sound: infer fuel input from efficiency, subtract electrical output, then divide heat by household demand or vaporization energy.
Math: 0/10. Multiplying 2.1 GW by 24 hours gives about 50 GWh, not 500 GWh, creating a tenfold error in the daily heat and home comparison.
Final estimate: 20/30. The 2.1 GW primary estimate was about 4.5 times the checked 470 MW, within an order of magnitude. The water estimate also landed within a factor of three because the input errors partly canceled, while the home count was much too high.
Post-check reflection
Matt's reflection
My assumption about battery efficiency was too low by less than half, energy to heat the home was low by half, and about heat of evaporation was way too high by about 100*, and my math was sloppy. I think it would be helpful to remember the heat of vaporization and the household heating daily requirement as memory pegs.
Recommended memory peg
A modern solid-oxide fuel cell converts roughly 55% to 65% of fuel energy into electricity, so electric output and waste heat are often on the same order. Water takes about 2.3 MJ/kg to evaporate. A gas-heated U.S. home averages roughly 1.5 kW of space heat across a year, while a cold Wyoming month is closer to 4 to 5 kW.