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How to make the connection between the energy released by the fission of uranium and the power of a nuclear plant ?

How to vary the power in a nuclear plant since the energy released by fission of uranium is a constant given by nature (physics) ?

(I know that the power is the derivated of the energy by the time.)

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The fission of one single atom releases an incomprehensibly tiny amount of energy. It's probably not always exactly the same amount, because there's more than one way a Uranium atom can split, but it probably all fits in to a narrow range.

Anyway, power is energy per unit time, so to control the power output, you need to control how many fission events happen in each second.

Control rods in a nuclear reactor have a somewhat delayed effect on the overall effective neutron multiplication factor of the fuel mass. That is, the average number of additional fission events that are caused by each fission event. If the factor (a.k.a., "$k$") is greater than 1, the power will increase (and keep increasing until some BAD THING happens), if $k<1$ then the power level will decrease, and if $k=1$ the power output should, in theory, stay constant.

It's tricky because the age of the fuel, and the temperature of the coolant, the presence (or not) of steam bubbles, and many other things also affect the $k$ value. So, keeping a contstant power level is kind of like balancing on a unicycle. You have to keep adjusting it all the time: can't stop an rest.

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The power of a power plant is how much electricity it can generate, not how much uranium it can split. All power plants are rated this way no matter how they generate the power. The Niagara Hydroelectric plant has a power output of 2675 MW. The Perry Street Plant using natural gas generates 20 MW of power here in Indianapolis. In any power plant other than pure solar, the heat generated in the power plant is used to make steam. The steam turns turbines and electricity is generated. This is the "power" usually stated when discussing power plants.

You are correct in the statement that the energy released by fission of uranium is a constant given by nature. However, I am sure you know the operators can control how much uranium is fissioning at any one time by the use of control rods.

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  • $\begingroup$ Thank you. Do I understand that if a nuclear plant wish to have the "optimal" power, it could decide to put control rods "just below" the limit of the nuclear explosion. Why don't they do that ? This is physics, so if they are "just below", with a automated (and working) system, there is no risk of explosion. Are nuclear plant power far below the physics limit before explosion (thus far below the maximum power that they could obtain) ? $\endgroup$ Commented Aug 30, 2020 at 15:42
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    $\begingroup$ @MathieuKrisztian, The control rods don't directly control the power output. What they control is closer to the rate of change of the power output. But really, it's even worse than that because there's a time lag: They really control what the rate of change of the power output will be in a few seconds from the moment when they are moved. Just part of the reason why designing, building, and operating the things is so ****ed complicated. $\endgroup$ Commented Aug 30, 2020 at 15:46
  • $\begingroup$ @Solomon Slow : thank you. $\endgroup$ Commented Aug 30, 2020 at 15:46
  • $\begingroup$ An easy way to think of the control rods is how much heat they regulate being generated. $\endgroup$
    – Rick
    Commented Aug 30, 2020 at 15:50
  • $\begingroup$ Modern light water reactors can't "explode", at least in the sense of a bomb. There is no "limit of explosion". I'm very curious where this idea came from. $\endgroup$ Commented Aug 31, 2020 at 4:09
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First you have to know the average amount of power released per fission. U235 releases about 193 MeV of recoverable energy per fission (i.e. not counting the neutrino energy). All of the other fissioning isotopes will release a slightly different amount of energy.

The next step is to find out the total number of fissions that are occurring in the reactor. This is the sum (or integral) of the neutron flux times the fission microscopic cross section times the fission isotope number densities times the volume.

The microscopic cross sections are pretty much a constant of nature, but they vary by temperature.

The number density of the fuel is going change with depletion. The U235 will deplete, but some of it will be replaced by plutonium. You will also get some fast fissions from U238.

The neutron flux is the parameter that will vary for different power levels. You operate the reactor to give the flux that will give you the desired power.

If you want any more detail than this, it is time to crack open a book on reactor physics.

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