How can you calculate how much energy a PV panel (known size and wattage) produces in a day given the exact insolation in kWh/m2/day?


4 Answers 4


Irradiance is defined as the power of electromagnetic radiation per unit area incident on a surface. In the example you've given, that should be kW/m², not kWh/m²/day.

Energy is defined as power expended over time, with the kilowatt-hour (kWh) being a unit of energy equivalent to one kilowatt of power expended for one hour. Consequently, one kilowatt expended for 30 minutes would be 0.5kWh, ( as would 500 watts expended for one hour) and 1kW expended for one day would be 24kWh.

Then, being irradiated by 1kW/m² falling on, and normal to the surface of a perfect 1 square meter PV array, it could supply 1000 watts of powwr to a load for as long as the irradiation persisted.

  • \$\begingroup\$ His unit of kWh/m^2/day is correct and appropriate in this context. You have rederived this in your answer. The units of " watts of powwr to a load for as long as the irradiation persisted." IS wh/day. Go from there. \$\endgroup\$
    – Russell McMahon
    Jun 8, 2014 at 11:10
  • \$\begingroup\$ Since irradiance is defined as incident power per unit area and his definition differs substantially from that, I must disagree with your evaluation of the correctness of his unit. \$\endgroup\$
    – EM Fields
    Jun 8, 2014 at 12:06
  • \$\begingroup\$ It deep-ends on whether one wishes to dwell on the English language aspects or the electronics. You are correct re the units of irradiance. But his subject line and text both mention energy and the units he used give energy per day per metre^2. Rather than correcting his units to those of irradiance it would, arguably, be better to correct his terminology to read "insolation". As I have now done. Changing the units to kWh/m^2 foundationally alters the technical basis of the question. \$\endgroup\$
    – Russell McMahon
    Jun 8, 2014 at 13:34

Your question assumes several ideal parameters - but is capable of an exact answer IF the idealising assumptions are correct. (They won't be in real life :-) ).

PV panels are usually rated at Watts out into an ideal load at 1000 W/m^2 at 25C at AM 1.5.
For meaning of AM 1.5 see Wikipedia = Air Mass - solar energy. Briefly - AM1.5 is defined as the standardised solar insolation conditions when a certain mass of air lies in a defined atmospheric path between the panel and the outer edge of the atmosphere.

Assume AM1.5 conditions (everyone tends to).
Assume 25C operation for now (usually panel temperature will be 50C + at full output and Watts out will be perhaps 10% down on 25C value.
Assume an ideally matched load for now.


... energy a PV panel (known size and wattage) produces in a day given the exact irradiance in kWh/m2/day?

Say panel power = Wmp
Panel size irrelevant.
Assume the kWh/m^2/day = SSH (sunshine hours) that you state is correct.

  • Energy out = Wmp x SSH

Non idealities:

SSH = kWh/m^2/day available to the panel will be lower than that available at the site due to non ideal panel conditions. A clean tracking panel will be close.

Panel max power point load impedance varies with insolation and temperature and panel cleanliness. A MPPT (maximum power point tracking) controller will attempt to optimise watts actually delivered. eg a 12V system panel typically has a Voc of > 20V and a Vmp of 18V. A lead acid "12V" may need between 12V and about 14V at the battery so without MPPT max efficiency of transfer is <= 14V/18V =~ 78%. Minimum efficiency is <= 12V/18V ~= 67%.
Those efficiencies are solely due to panel to load mismatch. A battery will have an energy out / energy in efficiency which varies with chemistry and circumstance but may be in the 50-75% range. (eg a NiCd single cell has a max terminal voltage on charge of 1.45V but delivers energy at 1.0 - 1.2 V over most of its capacity range. In solar PV systems it is usual to provide ~= 2V/cell Vmp to charge NimH batteries. So battery energy efficiency max JUST because of voltage mismatch = 1.0/2v = 50% when battery is discharged and 1.2/2 =~ 60% average. Even when providing the necessary maximum 1.45V per cell when charging the efficiency is <= 1.45V/2V ~= 73% due to voltage mismatch alone.

To get kWh/m^2/day fir a typical day in a given month GarBling: Gaisma city_name and look at table that shows kWh/m^2 by month.
eg searching for: gaisma vladivostok provides - http://www.gaisma.com/en/location/vladivostok.html And the 4th table down shows kWh/m^2/day to be a maximum of 5.25 kWh on an average May day and a minimum of 1.76 kWh on an average December day.
So an eg 20Wmp panel would produce 20 x 1.76 = 35.2 Wh/day on an avergae May day at 25C AM 1.5 and optimum orientation all day.


As stated, the question cannot be answered - there is not enough information.

If, in addition, you know the peak irradiance, and can assume that the known PV wattage is produced at that irradiance, you can calculate the unit's efficiency.

To a first approximation, you simply multiply the total irradiance by the efficiency to get total energy.

Of course, it's never that simple. As the angle of the sun changes, the intensity at different wavelengths changes, due to both scattering and absorption. This will change the apparent efficiency. Also, the efficiency of a PV cell changes with irradiance level and temperature. Monocrystalline silicon, for instance, has higher efficiency at high irradiance levels, and works best in bright sunlight. Amorphous silicon, on the other hand works better at lower levels.


There is plenty of information given: You're just missing a time of exposure for completeness or the deviation from how your insolation is measured vs. how a panel is measured.

You simply take the STC (Standard Test Conditions) panel rating which specifies the temperature, irradiance and angle.

The specific data of a solar module are measured under standard test conditions. Standard test conditions are defined as the solar irradiation of one kilowatt (kW) per square metre, a module temperature of 25 degrees Celsius and a solar irradiation angle of 45 degrees.

Once you have your time of exposure you can now simply determine the insolation under STC and then scale appropriately. OR you can determine the scaled time value and assume that the STC conditions are met.

If you assume that the Insolation measurement is taken under the same conditions at the STC ratings of the panel. Simply multiplying the Insolation value by the panel area and it's Wpeak will yield your W*Hr rating (Energy) for the day.

Insolation divided by STC irradiance = $$ \frac{(kWhr)(m)^{-2}{d}^{-1}}{(kW)(m)^{-2}} = \frac{(hr)}{(d)}$$

since you are talking about one day, if you divide your Insolation by STC Irradiance you get a scaled time result.

But this assume sthat the angles and temepratures are the same as STC.

  • \$\begingroup\$ He specified kW h / m^2/day so time is effectively included. \$\endgroup\$
    – Russell McMahon
    Jun 9, 2014 at 13:35
  • \$\begingroup\$ yes he did @RussellMcMahon but STC does NOT include time, see kW/m^2 - irradiation NOT Insolation. \$\endgroup\$ Jun 9, 2014 at 13:45
  • \$\begingroup\$ you are correct about what you are talking about, but it was not what he asked. He used units of energy which is insolation but termed it irradiance. You can choose to correct his English or choose to change the actual variable he asked about. His phrase " the exact name_of_unit_here in kWh/m2/day" is independent of whether AM 1.5 applies and independent of season and site and atmospheric conditions and. As asked the question is much closer to answerable than if you change it to a quite different one. | Useful Gaisma \$\endgroup\$
    – Russell McMahon
    Jun 10, 2014 at 8:10
  • \$\begingroup\$ @RussellMcMahon If he has measured Insolation (an assumption I am making) then the only thing missing is the deviance from STC. All other factors are implied like efficiency etc. and there is no need to measure/estimate or guess AM. That is built into the panel ratings. I've modified my answer to expound on this. Hopefully this is clearer. \$\endgroup\$ Jun 10, 2014 at 15:48

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