The yield of a nuclear plant

QUESTION: How efficient is a nuclear power plant?

ANSWER : the performance of a nuclear power plant is around 30%.

EXPLANATION: This means that 70% of the “atomic” energy resulting from the fission of uranium 235 is “wasted” as heat in the cooling towers.

For a power plant with 2 1,3 GW electric reactors, this corresponds to a heat loss of the order of 6 GW and an atomic power of 8,6 GW.

These 6 GW are "evacuated" in the cooling towers of French power plants, there is one tower per reactor (so you can easily know the number of reactors in a power plant by counting the number of towers).

The heating needs of a modern house are roughly (smoothed over the year) of 60 W per m2. Either for a house of 100 m2, 6 kW.

The “lost” thermal energy of a single power plant with 2 reactors therefore corresponds to the heating of a million houses!

Read also: Stirling cogeneration to wood pellets by Sunmachine

Assuming (which is not the case but it is for the image) that this energy was recoverable in the form of cogeneration, 14 16 to nuclear reactors would be enough to heat the whole of France without any consumption of heating or electrical or fuel oil or gas!

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3 comments on "The efficiency of a nuclear power plant"

  1. the number of "cooling towers" does not correspond to the number of reactors, eg the "Bugey" generating station
    has 4 reactors "REP of 900 MW"; two reactors (bugey 2 and 3) do not have cooling towers and two rep reactors of 900 mw (bugey 4 and 5) have two cooling towers per reactor; Another example: Tricastin power plant, two cooling towers for
    4 reactors, therefore, no ratio between the "number of cooling towers / number of reactor".

  2. We should install a dozen small nuclear reactors in the middle of Paris. They would only be started in winter and the heat "lost" (70%) would be sent to the district heating network. As I live in the province, I do not care much about the risk of a nuclear accident in Paris.

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