The load factors: nuclear and wind

What are the load factors of a wind power plant and a nuclear power plant?

How many wind turbines does it take to produce energy for a nuclear reactor?

Definition: the load factor is the effective annual average load in relation to the nominal load of the installation. This quantity is very important in calculating the profitability of an energy installation., whether renewable, nuclear or fossil.

Here are the French average figures for wind and nuclear energy.

In the case of nuclear power: the load factor is between 78 and 80%.

In the case of wind power: the load factor is within 20%.

In other words, a wind turbine is running at its rated power than 1 / 5 time.

To produce the energy equivalent of a 1,300 GW nuclear reactor (or at worst 0,78 * 1,300 = 1,014GW average effective), it is necessary to install not 1,053 GW of wind turbine but 1,014 / 20% = 5,070 GW.

The average power of future wind turbines built in France being 2 to 3MW, a nuclear reactor is replaceable at best: 5070 / 3 1690 = wind.

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1 1690 nuclear reactor = 3MW large wind turbines.

In 2005, there were 59 reactors for 19 nuclear power plants in France. To obtain energy autonomy (or rather to do without nuclear), it is nearly 100 wind turbines of 000MW that would have to be built ... and this assuming that we know how to store energy for peak hours … Which is, currently, far from being the case.

These figures are all the more important, as 3MW is a very large power for “onshore” wind power, most current wind turbines making between 0,750 and 1,5MW.

More:
- Is nuclear cogeneration possible?
- France Map of nuclear power plants
- Map of nuclear power plants worldwide
- Forum nuclear energy
- Followed by the nuclear accident in Japan following the earthquake of March 11, 2011
- All your questions about nuclear energy to a nuclear specialist
- The power of a nuclear reactor
- The efficiency of a nuclear power plant
- Nuclear and wind load factor
- Equivalence wind, solar and nuclear
- Key figures for wind power in France and Germany
- Complete file on wind energy
- The tidal: marine wind

Source of the figures on the load factor: Jacques Percebois, in "C dans l'Air" of 24/11/06, Center for Research in Economics and Energy Law.

8 comments on “Load factors: nuclear and wind”

  1. Hello everybody
    I don't quite understand the usefulness of the load factor!
    If I compare a nuclear power plant and a 1Kw wind turbine, the plant will produce 80% or 0,8Kw and the wind turbine will be 20% or 0,2Kw !!!
    Is it this ?
    Thank you for your answers
    Philippe

    1. That's right, but in kWh.

      The 2 will produce 0.8 kWh and 0.2 kWh per hour and per installed kW during the X hours of operation.

      Over a year 8740 hours, that makes 0.8 * 8740 = 7000 kWh / kW for nuclear power and 0.2 * 8740 = 1750 kWh / kW.

      1 kW nuclear therefore produces 4 times more than 1 kW wind power.

      For solar it's worse given the day / night cycle and the weather.

      Thus a 1 kW photovoltaic installation in the North of France will produce around 1000 kWh per year… We therefore have a real effective charge bill of 1000/8740 = 11%.
      But it would have to be corrected by day-night cycles since we can't do much about that, so we would have a corrected factor of 22%.

      Intermittent production is THE big problem with renewable energies.

      1. It seems to me that the load factor only makes sense for a wind turbine if you connect the wind turbine to the grid. If the wind turbine is isolated, and is coupled for example to an electrolyser. which produces storable hydrogen, the problem of intermittency is resolved. However, there remains the cost per MW installed.
        But this is a false problem because I do not think that the current produced by the nuclear power plant has the same value at night as during the day. It seemed to me that man was also intermittent, since naturally he sleeps at night and by isolating himself well he then needs very little energy (in our latitudes).

  2. Hello Christopher,
    Thank you for your reply.

    Can I deduce that the load factor is profitability (your example on photovoltaics)? or I'm wrong ?
    Philippe

    1. It is not exactly the profitability but it is linked: a PV installation in the south of France will be about twice as profitable (quickly) as in the North at the same installation price in € / Wp…

  3. It's clearer to me!
    So we can say that the load factor of a nuclear power plant is relatively constant regardless of its location, while wind and solar can have different factors depending on their location?

    So nuclear seems the ideal means of production in relation to its surface area and its regularity?
    Basically a Swiss clock! Hey French!

    So why separate from nuclear?
    Have a good day,
    Philip,

    1. That's right !

      Nuclear power has advantages (stability of production, high power, little CO2, etc.) but it poses other problems: waste treatment (which we do not yet know how to fully treat ... despite 50 years of R&D), fuel supply (Mali), aging plants (originally they were supposed to last only 20 years ...), popular fear linked to a dangerous image ... Also there is research on nuclear fusion which is progressing more or less rapidly (nuclear plant at 0 waste) .... which became obsolete all the current fission plants ...

      See you soon

  4. the load factor of nuclear power plants takes into account the long periods of compulsory maintenance or the reactors are shut down and also the periods of shutdown that should have been made to meet the new security requirements imposed by the nuclear safety authority following the fukushima accident

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