lightbucket

Insolation, and a solar panel’s true power output

February 24, 2008 · 20 Comments

The standard figures quoted for a solar plant’s “peak power” and “installed capacity” assume that the solar panels are illuminated by 1 kilowatt per square metre of incoming sunlight. The average insolation in northern Europe is about ten times lower than this, so the electricity output from solar panels in northern Europe is around ten times lower than the figures in the data sheets.

Solar panel roof

Insolation (INcoming SOLar radiATION) is a measure of the solar energy striking a unit surface area in a unit time. It determines how much electrical power a photovoltaic solar panel will deliver – the more sunlight you have, the more power you get.

Mean solar power above the Earth’s atmosphere is 1366 Wm−2  [1], but sunlight is attenuated on its way through the atmosphere, so the insolation has fallen to 1000 Wm−2 by the time the light has reached the Earth’s surface. This is the incoming solar power on a clear day, at sea level, with the sun overhead. This illumination level has been adopted by the solar energy industry as a standard operating condition for quoting the peak output power of a solar panel.

The solar power industry has agreed a set of Standard Test Conditions for rating the output power of photovoltaic solar panels [2]. This standardised approach allows a like-for-like comparison to be made between products. The industry has chosen the figure of 1000 Wm−2 as the standard illumination level for measuring and quoting a solar panel’s power output. The electrical power produced by a panel when it is illuminated by 1 kWm−2 of solar radiation is called the “peak output power” of the panel. This is the figure that’s normally listed in product brochures and data sheets. Likewise, when the “installed capacity” of a solar power plant is quoted, this also refers to the peak power at an insolation level of 1 kWm−2.

Insolation Level In Europe

How does this theoretical “peak” insolation level of 1 kWm−2 compare with real insolation values? Figure 1 shows how insolation varies across Europe. It comes as no surprise to see that the south is a lot sunnier than the north.

Insolation Map of Europe
Figure 1.  Insolation map of Europe. The map shows the total solar energy falling on each square metre. Values depicted are worst case at optimum tilt.
Units are kilowatt-hours/m2/day.   (1 kWh/day = 41.7 W)

Looking at this in more detail, Table 1 shows monthly and annual average insolation levels for various cities in Europe. The data are from the NASA Atmospheric Science Data Center [3] at the NASA Langley Research Center. The numbers are day-night averages, they are affected by the number hours of daylight, cloud cover, and sun angle.

Table 1.  Insolation levels for European cities – monthly and annual averages
City Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year
W m−2
Limassol 105 136 189 250 285 325 325 299 255 185 126 96 215
Malaga 105 135 188 225 265 295 318 284 225 154 108 89 199
Athens 83 105 153 217 266 313 317 288 232 146 90 68 190
Lisbon 95 125 179 215 255 269 287 264 213 143 95 77 185
Madrid 80 115 170 201 244 272 296 263 205 128 82 66 177
Rome 74 105 155 203 249 285 295 264 201 128 83 65 176
Toulouse 58 89 133 168 201 215 244 211 170 103 66 52 143
Bern 46 74 114 150 196 211 237 206 153 91 53 38 131
Paris 37 68 109 165 204 201 223 192 139 83 47 30 125
Munich 44 75 118 165 202 194 214 186 133 83 43 33 124
Brussels 31 55 95 153 195 187 201 175 119 72 38 23 112
London 28 53 93 145 189 188 198 167 119 69 37 22 109
Hamburg 23 46 87 153 203 186 186 162 108 62 29 17 105
Dublin 23 45 82 138 183 179 179 142 112 60 32 18 99
Oslo 13 36 70 130 194 202 191 140 93 43 18 8 95
Edinburgh 18 39 78 133 180 181 172 142 101 50 25 13 94
Units are watts per square metre.
Ref: “NASA Surface Meteorology and Solar Energy Data Set for Renewable Energy Industry Use” [2]
Notes:
Data show monthly and annual averaged insolation incident on a horizontal surface.
Data show the insolation averaged over the 10-year period July 1983 – June 1993.
The highest (Limassol, July) and lowest (Oslo, December) monthly insolation values have been highlighted in yellow.
The global average insolation is approximately 250 watts per square metre.

The sunniest places in Europe, such as Limassol (average insolation 215 Wm−2), get about five times less solar power than the theoretical peak level assumed in solar panel specifications. The least sunny cities, such as Edinburgh (average insolation 94 Wm−2), receive nearly eleven times less solar radiation than the theoretical peak value.

As well as these geographical variations, there is also a large seasonal variation. Limassol gets three times less solar power in December than in July; Oslo gets twenty-five times less insolation in midwinter than in midsummer (onto a horizontal surface).

A Concrete Example

To see how this works in practice, let’s go through a concrete example, a domestic rooftop photovoltaic (PV) solar installation in northern Europe.

The example I’ve chosen is a residential property with thirty square metres of rooftop solar PV panels located in South East England. I’ll assume the panels operate at 19% efficiency, the highest efficiency commercially available. The manufacturer would quote the “peak power” or “installed capacity” of this 30m2 solar array as 5.7 kWp. This is the power the panels would generate if they were illuminated by 1000 Wm−2 of solar radiation. In fact, London and the South East get an average insolation of 109 Wm−2, about nine times less than the peak value. That means the 30m2 solar roof will generate not 5.7 kW, but 621 W of power averaged over the year.

There’s a very large seasonal variation as well. The panel generates more power during the summer, but much less during the winter, when energy needs are greater. The insolation for London in December is 22 Wm−2, so the power from this solar roof will be 125 W averaged over the month of December. The figures are summarised in Table 2.

Table 2.  Power output of a rooftop solar PV installation in London
  Power  
output 
Peak power, or “installed capacity”: 5700 W
Annual average power: 621 W
Monthly average power for July: 1129 W
Monthly average power for December: 125 W
These figures are based on the following example:
Mean insolation: 109 watts per square metre (London value);
Solar panel is horizontal;
Solar panel area: 30 square metres;
Solar panel efficiency: 19%

There’s a very large difference between the headline “peak power” figures quoted in the datasheets and the true average power output of a solar panel. It’s important not to confuse the “installed capacity” or “peak power” ratings provided by the solar panel manufacturers with the mean power that will actually be generated at a given location. In northern Europe, the mean power output is about ten times lower than the theoretical “peak power” rating.

In London in midwinter, a thirty square metre rooftop solar installation will generate an average of 125 W of electricity, enough for about ten low-energy light bulbs, or one incandescent bulb.

References

  1. The sun’s total and spectral irradiance for solar energy applications and solar radiation models, C.A. Gueymard, Solar Energy 76, 423–453 (2004)
  2. Handbook of Photovoltaic Science and Engineering, page 295, A. Luque and S. Hegedus, John Wiley and Sons (2000)
  3. Release 3 NASA Surface Meteorology and Solar Energy Data Set for Renewable Energy Industry Use, C.E. Whitlock et al., Rise & Shine 2000, the 26th Annual Conference of the Solar Energy Society of Canada Inc. and Solar, Oct. 21–24, 2000, Halifax, Nova Scotia, Canada. (requires login)

Categories: energy policy · solar photovoltaic
Tagged: , , ,

20 responses so far ↓

  • Fred Pittenger // March 16, 2008 at 11:11 pm

    Wow. PV Watts 4 kw array annual fixed
    Grand Junction Colorado 6108 kWh
    Brussels 2706 kWh.
    think I will keep installing in Colorado

  • Stuart Potter // March 19, 2008 at 11:06 am

    Sure those figures for London don’t look good, especially in winter but remember those figures are for a horizontal panel. If you had the panels inclined at say 45 degrees then the yearly output would rise by 20-30% and the winter figures would be at least 100% better. This would yeild annually about 7000 KWh which covers a typical house’s yearly usage if the excess can be sold back during the summer.

  • Steve Pluvia // April 8, 2008 at 6:47 pm

    Annual and monthly average power should be in KWh not watts;

    Once again your calculations are nonsensical.

  • lightbucket // April 8, 2008 at 6:55 pm

    Hi Steve,
    Welcome from the energy economics thread.

    > annual and monthly average power should be in KWh not watts

    I really don’t want to be rude, but your misunderstanding is rather more fundamental than I thought.

    kWh is a unit of energy, watt is a unit of power (i.e. energy per unit time). You don’t understand the difference between energy and power, which explains the rest of your problems.

  • alan // July 16, 2008 at 9:40 pm

    “peak watts” are designated for the time when the devices are illuminated. Do the calculations quoted above figure in the dark hours in order to exaggerate the difference between peak power and mean power? I mean…I’m more than willing to agree that salespeople want to emphasize the maximum potential of the product. But the consumer would have to be pretty dim (pardon the pun) to think that solar panels will work in the dark.

  • lightbucket // July 17, 2008 at 9:20 am

    Hi Alan,
    “Peak watts” is just the the power when illuminated by 1 kW /sq. m. of sunlight, it doesn’t matter how long it’s illuminated for.

    The mean power is the average over day and night, summer and winter, sunny and cloudy days, etc, it tells you the average power you’ll get over a full year.

  • Howard // July 30, 2008 at 9:11 am

    I agree with Steve. The output of a solar panel should be presented in KWh as that is what consumers pay for. Yes KWh is energy and W is power, fine. So what. You buy a solar panel to save energy and money.
    The ‘headline’ rating is relevant as that power can be generated on a sunny day. The electronics and power output can occasionally reach that figure. Exceptionally clear skies can even exceed 1KW/m2
    I am a little confused by you map and table: London on the map shows between 1 and 1.5 KWh/m2/day (at optimal tilt). The table lists it as 109Wwhich is 109/41.7 KWH/m2/day = 2.6 (at horizontal). Optimal figures should be better than horizontal???
    Which is correct ?

  • lightbucket // July 30, 2008 at 9:47 am

    Hello Howard,

    The table shows insolation, which is a measure of solar power striking a unit area. Power has a specific physical meaning. It means energy per unit time.

    There are many units that power can be measured in, but the watt is the SI standard, and the one most commonly used.

    kWh is a unit of energy, not power. It simply cannot be used as a measure of power. It’s meaningless if used that way.

    It’s similar to the difference between “miles” and “miles per hour” when looking at the distance travelled and the speed of a car. You can’t measure speed in “miles”, and you can’t measure distance travelled in “miles per hour”. It’s the same issue with “power” and “energy”. This misunderstanding has come up several times now, I may do a blog post to explain it more fully.

    The map shows worst case at optimum tilt, not mean at optimum tilt, (i.e. the table and map are both correct) although I agree with your underlying point that it would make more sense for the map and the table to show exactly the same quantity. I’ll see if I can get a more suitable map, (showing average horizontal insolation) to go with the table.

  • Patrick // July 31, 2008 at 6:14 pm

    It seems to me that the only use for PV power is to power air conditioning apparatus :-) In July you can power a modest airco of a few KW when the sun is shining on it, and in winter, you can light a few fluorescent bulbs (only during daytime….)

  • lightbucket // July 31, 2008 at 7:25 pm

    Hello Patrick,
    Well, air conditioning isn’t the “only” use of PV power, but you have a point in that air conditioning is an interesting special case.

    Many energy drains are at their highest in winter, when solar energy is at its lowest, so you take a hit from each end – higher demand and lower supply.

    Air conditioning has the opposite behaviour: the demand is highest at about the same time that solar energy supply is highest – on a hot summer’s day. Air conditioning is exceptionally well matched to using solar power as the energy source.

  • Steven Grant // December 11, 2008 at 10:19 am

    Very Interesting article. I have been looking for a simple calculation like this for a few days now so glad i came accross this. I am currently trying to come up with a business model for a solar business model and trying to decide whether this is possible in Ireland. Would you have any inssue Lightbucket if i fired a few questions your way?

  • lightbucket // December 11, 2008 at 10:58 am

    Hello Steven,

    >>Would you have any inssue Lightbucket if i fired a few questions your way?

    Thanks for your interest, I’m always delighted to have a general discussion, but please note that I have no detailed knowledge of the financial incentives as they apply in Ireland.

    The main example of a viable solar energy industry in northern Europe is Germany’s, where they have specific policies in place to encourage renewables.

    As a first step, you might want to check how Germany’s incentives structure compares to Ireland’s, and see how that would affect the business model in the two cases.

  • Steven Grant // December 11, 2008 at 1:32 pm

    In the example above. The unit creates an annual average power of 621 watts. How can you work out the value of the power ie if you were to price it at say €0.15 per Kwh

  • lightbucket // December 11, 2008 at 2:25 pm

    621 watts of average power is 0.621 kWh of energy per hour,
    so the hourly price is 0.621×€0.15 = €0.09 worth of electrical energy per hour.

    Per year:

    Multiply the 0.621 kWh number by the number of hours in a year (which is 365×24=8760)

    to give you 0.621×8760 = 5440 kWh of energy per year.

    At a price of €0.15 per kWh, this 5440kWh has a cost of €816.

    i.e. the example unit will generate €816 worth of electrical energy per year, at your assumed price.

  • Kiko // January 3, 2009 at 3:18 am

    Is it true that the energy used to manufacture the photovoltaic cells and the solar panel assembly wil never be ‘recovered’ by the energy saved in the solar panel ’s lifetime ?

  • lightbucket // January 5, 2009 at 7:31 pm

    Hello Kiko,
    No, modern solar PV panels generate more energy over their lifetime than the energy needed to manufacture them. I’ve covered this issue in the post “Energy payback ratios for electricity generation”.

    In sunny regions (in this case, Denver, Colorado), the energy payback ratio is about 6. That is, the solar panel generates about 6 times more energy over its life than the energy used to manufacture it.

  • Chris Gueymard, PhD // March 13, 2009 at 7:03 pm

    The numbers in Table 1 are misleading because they represent an average power over a 24-hour day. This is the way climate-change physicists present their results (e.g, of “radiative forcing”), but should not be used in solar energy applications. For instance, using your data for Bern in June, this 211 W/m2 actually means 211 x 24 = 5064 Wh/m2. Since the daylength is then about 15.6 hr, the real average sunup power is 325 W/m2.
    Also, as noted in other comments, these values are for global radiation on a horizontal surface. For a latitude-tilt surface, the numbers in Table 1 would be similar in summer, but much higher in winter.

  • lightbucket // March 14, 2009 at 11:34 am

    Hello Chris,
    >>The numbers in Table 1 are misleading because they represent an average power over a 24-hour day.

    Yes, they’re a 24-hour average (and in fact averaged over the whole month). Why do you find that “misleading”? It tells you how much energy you’ll get over a full day.

    >>Since the daylength is then about 15.6 hr, the real average sunup power is 325 W/m2.

    Well, yes, the sunup power is higher than the average, and the nighttime power is zero.
    That’s why you take a 24-hour average to get the total energy you’ll collect over a full day. I still can’t see why you’d find that misleading.

    >> latitude-tilt surface, the numbers in Table 1 would be similar in summer, but much higher in winter.

    That’s right, a northern hemisphere solar panel should be tilted south for optimum operation, roughly to the midday sun angle for each month.

  • Fond of Beetles // March 17, 2009 at 3:33 am

    Why is it that every time I’m looking for a clarification of some issue when I’m reporting some story I find exactly what I’m looking for on your blog? It happens four or five times a year.

  • Alex, Tunbridge Wells // July 30, 2009 at 4:33 pm

    Thanks for the table. I really don’t care if you call it KWhrs / day, or KW Average. Converting is a just a function of a normal brain.

    The figures assume a flat surface. In London, in the summer, the solar latitude is 52+23=75 degrees. In other words, the sun is 15 degrees off the horizon at midday. Pointing the array directly at the sun (i.e at an angle of 75 degrees) will quadruple your power (Cos 0 / Cos 75). This angle will however reduce power in summer.

    I created a calculator which I think adjusts for this. It makes some mathematical assumptions (increase ratio at noon is the same as for the whole day) . It shows for London, with a panel at 45 degrees:

    Average power
    Jan 80
    Feb 113
    Mar 150
    Apr 190
    May 217
    Jun 207
    Jul 228
    Aug 219
    Sep 192
    Oct 147
    Nov 106
    Dec 74
    Average 160

    Increase the angle, and you increase your winter power and reduce your summer power. Best overall result is at 52 degrees. Best mid winter result is at 75 degrees. (Not many roofs are this steep – but if you want winter power, a south facing wall is better than a south facing roof with a 30 degree pitch). Of course, feed-in tariffs don’t take account of this.

    I can send you the Excel sheet if you want.

Leave a Comment