# Solar map of France of sunshine and solar energy field

## Average sunshine duration in hours per year in France: from <1750h to> 2750h!

Example: if you live in the Bas-Rhin (North of Alsace) you will have less than 1750 hours of sunshine per year.

## Average energy potential in thermal kWh per year and per square meter: from 1220 kWh / m².year to more than 1760!

### Examples of photovoltaic solar production in the North and South of France

If you live in Bas-Rhin (Northern Alsace) you will have less than 1220 kWh of recoverable solar energy per year and per m². If you are in the Montpellier region, you will have 1620 to 1760 kWh/m² each year. This is around 40% more than in Alsace or the North of France.

These are figures for the raw solar energy received at ground level.

To get theequivalent photovoltaic electrical energy produced, these raw solar energy figures must be divided by approximately 6 (taking into account the yields of the panels and the injection inverter). Thus in the Montpellier region, each m² of solar panel receiving 1700 kWh per year will produce approximately 1700/6 = 285 kWh of electricity per year. An installation of 10 panels of 1.6 m², or 16 m², will therefore produce 285 * 16 = approximately 4500 kWh.

The same installation in Bas-Rhin will produce at best 1220/6 * 16 = 3250 kWh per year. We find the 40% more by doing 4500/3250 = 1.38, which is close to the 40% more.

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In order to take advantage of this solar potential, some companies offer to set up a photovoltaic shed

## Detailed example: thermal solar production in Alsace, in Bas-Rhin

### Solar power and energy data from maps

As read on the cards, we have a year:

• less than 1750 hours of sunshine, let's take 1500 hours arbitrarily.
• less than 1220 kWh/m2 of energy, let's take 1100kWh arbitrarily.

### Average powers per m2

We therefore have an average solar power when the sun shines of 1100/1500 = 733 W. Which is very correct (the world average is given for 1000W per m2).

For information, the average power over the year, including nights, would be 1100/8762 = 125 W.

At these values ​​to obtain the recoverable thermal energy, it is necessary to multiply by the efficiency of the solar panel (assuming that there are no other losses which is generally false) i.e. 70% for thermal and 15% for photovoltaics.

### Recoverable thermal energy per year

Each m2 would bring back 1100*0.7 = 770 thermal kWh per year.

We remind you that a liter of petroleum fuel has a calorific value of around 10 kWh. Assuming a boiler efficiency of 0.8, one m2 of panel will give the equivalent of 770 / (10 * 0.8) = 96,25L of fuel oil, or roughly, given the various estimates: 100L per m2 of panel.

Each m² of thermal solar panel therefore theoretically allows a energy saving of 100 L of fuel oil per year. Depending on your current fuel consumption, so you can estimate the number of m2 required to fully compensate your oil consumption.

We specify in theory because in practice, heating needs are highest when the sun shines least: in winter! Must therefore store this solar thermal energy to enjoy it in winter (concept of production/consumption phase shift). Which is not necessarily an easy thing to set up.

### Financial analysis of a solar thermal installation

Thus, an Alsatian house that consumed 2500 L of fuel oil will be needed, ideally, 2500 / 100 = 25m2 of panels (that's a lot, the m2 installed costs on average, with tank and installation, around € 1000 currently, price excluding aid and subsidies) and will save per year the equivalent of 2500 * 0.65 = 1625 € of fuel oil (this is not much compared to the price of the 25m2 installation which must be located, in 2010 and excluding subsidies, between 15 and 20 €) ...

We noted ideally because the reality is precisely not so ideal. Indeed; in winter, solar thermal energy (due to the fairly low heating temperatures reached) can only be used to heat domestic hot water or to supplement the conventional heating circuit (before the boiler). To switch to 100% solar, you need a low-temperature heated floor (or walls).

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This is why very few people use solar power for heating: the vast majority of installations aim to heat only domestic hot water (which constitutes on average approximately 10 to 15% of annual energy consumption). ).

### Conclusion: profitability of solar thermal still difficult in Northern Europe

This brief calculation shows that the profitability of solar thermal is difficult to achieve for people (the vast majority of us) who do not understand that the financial calculation. Subsidies and various aids, as explained in a page of this site, do not change much… (on the contrary!)

This will be the case as long as the non-renewability of fossil fuels and their induced pollution are not taken into account in their costs or as long as ecological aspects come second to financial aspects when making a purchase...The moral aspect should, ideally, also play in favor of the choice of non-fossil energies...Is this not the basis of an eco-friendly society and no longer just a petroleum-financial one?

But in 2023, new installations of thermal solar panels have become anecdotal. The fall in prices (and the end of subsidies) on photovoltaic solar panels make photovoltaic solar panels largely competitive. It is therefore today more interesting to carry out a photovoltaic installation than a much more complex installation of thermal solar panels.

Global warming will also change solar radiation maps, and perhaps much faster than we think…to be continued!

#### 12 comments on “Solar map of France of sunshine and solar energy deposit”

1. Jean Luc says:

Bravo for this honest demonstration of solar. You have reasons aids are not a viable solution. (Financially and culturally, it distorts the perception of energies and their cost).
also thank you for the info. Greeting.

2. Fanfoué Bacouni says:

Well done for this study. Concrete and pragmatic, this is the basis of any successful business ...
However, I think it would be interesting to take into account two additional factors to change the way we look at solar energy.
1st point the altitude. 1 hour of sunshine in Nice provides less solar energy than an hour of sunshine in Gap. The 800 meters less atmosphere makes a big difference. This largely compensates for the difference in latitude. By integrating this parameter, the Pyrenees would become much more suitable for the installation of panels.
It may seem futile, but the connection to the network in a mountain village is much more expensive than in the plain. auto production then becomes more interesting.
2nd point: the temperature. with the development of solar systems with Stirling engines, it is the temperature differential that is sought. Once again the altitude becomes an advantage.
We have a solar energy deposit in France which is much larger than what we are willing to admit!

3. Clement says:

A housing estate in Canada (Drake Landing) is heated in the winter with the heat accumulated in the summer (and stored in tubes driven into the ground): yet the sunshine there (and the winter temperatures) are certainly less favorable. just in France.
So why limit yourself to thinking of hot water used immediately (or very quickly, within 1 or 2 days)? Lack of imagination, cost problem,…?

4. Bouyer says:

Hello, I do not understand your calculations concerning the power of sunshine in the lower rhine: 1100/1500 = 733?

1. Hello, this is simply the estimation calculation of the average solar power when the Sun shines per m2: we divide the energy in kWh by the hours of sunshine. We therefore obtain Watts.

Happy New Year 2022 (and good luck)

Hello,
Your calculation is wrong… If you divide the energy potential (kwh/m²/year) by the number of hours of sunshine to find out the energy produced by your solar panels, then the more hours of sunshine there are, the less your solar panel produces energy...
The error comes from a misunderstanding of the unit kWh. This is the energy produced for 1 hour. To know the energy produced by your solar panels during a year, you have to multiply it (and not divide it) by the number of hours of sunshine.

2. The calculation is correct, the kWh has nothing to do with the hour, it is an arbitrary unit… We can produce 1 kWh in 10 hours with 100W of average power…

5. Lykcnarf says:

No, I confirm what Adrien says, your calculation concerning the average power does not really make sense. Exactly, as you say, the kWh hour has nothing to do with the hour, so why divide it with the number of hours of sunshine??? If I follow your reasoning, in the PACA region it would give arbitrarily 2800h of sunshine and 1800kWh, which gives 642W/m² (a lower power per m² than the Bas Rhin even though it is the sunniest region in France ??). You only need to look at an irradiation map to realize that the further you go down towards the equator, the more the irradiation increases. Thanks to the solar field map you have the answer, no need to divide anything, it indirectly takes into account the number of hours of sunshine. Good day.

1. There is not always an error since we are talking about average annual power per hour of sunlight (i.e. neither when the weather is bad nor when it is dark).

So in PACA there are many more hours of sunshine than in Alsace, but that does not mean that there is the same gain in energy radiated at ground level. In winter, the sun is low, even in the south.

So in winter, the radiation is low in the 2 regions even if there are more days of good weather in the south. Mathematically this lowers the average energy radiated per hour in the south.
The reasoning is good, the calculation is good (it is trivial)

You will find a more precise radiation map here: https://www.econologie.com/carte-solaire-irradiation-dni-france/ or here https://www.econologie.com/forums/solaire-thermique/carte-precise-du-rayonnement-solaire-en-france-dni-france-t7232.html

2. ps: if you want to compare the 2 regions, then just take the average hourly power per year…1200 kWh/8760h = 137 W and 1800 kWh/8760h = 205 W

There is therefore, on an hourly average, 205/137 = 1.5 = 50% more solar energy radiated in PACA than in Alsace.

But in terms of power per hour of sunlight, the previous reasoning remains correct.