Insolation, and a solar panel’s true power output

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)

60 Responses to Insolation, and a solar panel’s true power output

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

  2. Stuart Potter

    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.

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

    Once again your calculations are nonsensical.

  4. lightbucket

    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.

  5. “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.

  6. lightbucket

    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.

  7. 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 ?

  8. lightbucket

    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.

  9. 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….)

  10. lightbucket

    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.

  11. 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?

  12. 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.

  13. 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

  14. 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.

  15. 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 ?

  16. 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.

  17. 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.

  18. 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.

  19. Fond of Beetles

    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.

  20. Alex, Tunbridge Wells

    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.

  21. Alex i’m really interested on what you have written. i have a case study for a model house in London with the panel facing south at 30% and i was wondering if you could either sent me the formula to calculate the insolation or if you have it just sent it to me.
    Thank you

  22. Alex, Tunbridge Wells

    Constantinos – can do, but not sure how. Can you cryptically give an e-mail here?

  23. I don’t think i can. However this is an email you can contact me: kcu07cc@reading.ac.uk. The 4th character is the number zero and not the letter O. thank you for your help. I really appreciate it!!!

  24. Hi Alex, I am also interested in the Excal sheet you have created. I think the easiest way to send it to everyone is to upload it onto a free file sharing site, for example http://www.megaupload.com

    Once it has been uploaded onto http://www.megaupload.com you will get a link which you can post here and we can all download it. Thanks and Merry Christmas to everyone! =)

  25. Hi,
    Alex i am also very interested in this calculator as it will aid me in my project and be able to convert the data to represent Edinburgh. If you have uploaded it to megaupload.com could you please tell me what it is labeld as, cheers!
    Also loving this blog! it is very useful and very helpful in understanding a tentative subject and i dont care how you represent your data because converting isn’t rocket science. Cheers Light bucket

  26. Hi Light bucket

    what is the best way to reach you (email)

    Sandeep

  27. Solar gain by tilting the panels only works for bright sunlight, on overcast days horizontal gives best yield, the only sure way to maximise output is mount the panel on a tracker, not really practical on most roofs..

    Re Stuart Potter’s comment, 30 sq-m of solar panel may supply a typical house’s annual power a typical house would not have 30 sq-m of south facing roof, my slightly larger than average south facing semi has less than 20 sq-m.

  28. The rule of thumb for the UK is 800kWh/year for each kWh installed capcity (30 deg slope, south facing, no shading). Typical domestic install is 1-2kW

  29. I followed the NASA link for your insolation table, which is not a deep link, but got to this page of figures for London – http://tinyurl.com/yg9aytp – which are quite different – much higher – from your figures. Can you clarify please? I know you are averaging over 24 hours, but in what sense does this help to calculate the total annual average kilowatt-hours generated? Paul gave a rule of thumb for the UK – 800kWh/year for each kWh installed capcity (30 deg slope, south facing, no shading). Both of you are arguing theoretically. What would be more convincing would be empirical data from actual installations. Jeremy Leggett for example claims an amount similar to Paul’s rule from the roof he used to have. This would seem to dispute your theoretical result.

  30. Hello David,

    >>but got to this page of figures for London – http://tinyurl.com/yg9aytp – which are quite different – much higher – from your figures. Can you clarify please?

    For the July peak, your numbers are about 4.8% higher than the ones in Table 1 above.
    They are averaged over a different time period, and are for a slightly different geographical area, which may account for the 5% discrepancy.

    >>I know you are averaging over 24 hours, but in what sense does this help to calculate the total annual average kilowatt-hours generated?

    The total annual energy is calculated by multiplying the average power by the total time.

    >>Paul gave a rule of thumb for the UK – 800kWh/year for each kWh installed capcity (30 deg slope, south facing, no shading).

    My Table 1 figure for London is 955kWh/year, about 19% higher than Paul’s rule of thumb.
    This is probably because Paul’s value is a UK average, whereas the London insolation is slightly higher than the UK average.

    >> Jeremy Leggett for example claims an amount similar to Paul’s rule from the roof he used to have. This would seem to dispute your theoretical result.

    I can’t see the discrepancy. My Table 1 value is 19% higher than Paul’s value, but I think that’s just because London has somewhat higher insolation than the UK average.

  31. >>What would be more convincing would be empirical data from actual installations.

    I’ve got them right here:

    Rooftop solar power in the UK – real world data

  32. Ah, I’ve used those figures myself on my own blog. I loved the one for the insolvency centre – if we all did that we would all go bankrupt! I’ve just asked CAT if they have monitored figures for their installations and they do not. I know the ECI did a survey but that may be out of date. Would be nice to see more figures in line with what we can expect for the FITs. I assume you think that PV has an almost zero capacity credit then?

  33. >>I assume you think that PV has an almost zero capacity credit then?

    I haven’t looked at the capacity credit of PV in this post.
    In regions where the peak electrical load comes from daytime air conditioning, the capacity credit can be nearly equal to the peak daytime power output of the PV.

  34. Solar photovoltaics on the UK power system have no capacity credit. As this source suggests it’s “because the triad peak periods are always on cold winter evenings after dark when there is no sunshine. Many other places (like New York, Paris and Vancouver) will be the same when their peak demand is in mid-winter but in countries with tropical climates, where the peak demand is in mid-summer, due to air-conditioning loads (like Florida, California and Greece), this will obviously not apply and photovoltaics may then have nearly 100% capacity factors.”

    More generally, in oral evidence to the Commons Environment Select Committee, Malcolm Wicks, then energy minister at DBERR, said they will expect to allow a capacity credit for renewables of around 10%-20%. To which Nigel Lawson bitingly replied: “That is an unusually optimistic estimate.” Well, I dispute that as an average figure: wind is generally assumed to be around 40%. Solar will be therefore close to zero.

    This does not mean it is worthless – clearly, when generating, it is displacing something. Any proper model of how we move to over 20% renewable energy penetration has to account for what credit will be available at all times of the year, and wind and solar are a good match in general – in a highly varied scenario.

  35. As I wrote over here http://www.energynumbers.info/german-pv in the section “So what should Britain do?”, PV can make a positive contribution to UK energy security. Capacity Credit was once a useful back-of-envelope proxy for energy security 20 years ago, when there were few computers around, and almost all our electricity generation was by thermal plant. These days, it’s really past its sell-by date.

    It was a proxy for the Loss of Load Expectation – a probabilistic estimate of energy security. Clearly a system X + some PV is less likely to lose load than system X without PV, but you need a good energy model to tell you by how much.

    Looking at a putative 2030 grid, with say 60% mean energy coming from wind power, and 5% from PV, that PV gives you a lot of energy security, because insolation is negatively correlated with wind, over the year. And getting, on average, 5% of your electricity from PV, means your installed PV capacity is about 50% of mean demand. So if we had another long hot summer like 1976, we’d have plenty of PV to see us through the windless times.

    Regards,
    Andrew

  36. David, thanks for that link.
    It argues that storage is pointless for grid-connected wind and PV, because they have zero marginal cost and would be the last to be stored. The energy that got stored would have come from the last generating units to be brought on line, i.e. those with highest marginal cost.
    It’s a subtle and non-obvious point.

  37. Andrew,
    I haven’t looked at capacity credit of PV on this blog, but I’ve looked at wind in the post “The capacity credit of wind power”.

    In the comments there, jack cadogan makes a similar point, that capacity credit is a rather simplistic view by modern modelling standards.

  38. I agree with Jack Cadogan on that.

    The back-of-envelope capacity credit for wind, from a large number of peer-reviewed studies, is equal to the lower number of: square root of the installed capacity in gigawatts; and mean power. So, with 1GWe mean power from 4GWp installed capacity, capacity credit is 1GW. With 36GWp installed capacity, capacity credit is 6GW, and mean power will be about 13GWe, assuming most of the installed capacity is offshore.

    But, of course, the geographic distribution of the installed wind capacity matters a lot. Having 9-12GWp at Dogger Bank alone will skew things a bit…

  39. I have just found Nottinghamshire’s Hockerton Housing Project’s published output for wind and pv. I put the PV figures in a spreadsheet. They have an installation rated at 7.6kW. It has produced on average for four complete years 698W-h per year per installed kW – a total of 21.232kWh for four years and eight months.

    They say the installation cost £40,650. This translates to a payback period of around 20 years assuming 38p per kWh, ie the current FIT level.

  40. Or 81.6 years at 15p per unit:

    total kWh: 21.232
    over months: 56
    average kWh per month: 0.37914285714
    installed rated kWp: 7.6
    average monthly kWh per rated kWp: 0.04988721805
    multiply by number of hours in year 8760: 437.01203008
    @ 15p per kWh: £65.55
    Value of full array: £498.19
    Cost of array: £40,650
    Years to pay back investment @ 15p per kWh: 81.60

  41. Your energy calculations look to be out by a factor of over 1000. Hockerton has produced about 725,000Wh/y/kW. Not 698Wh/y/kW. And in 4 years, 8 months, it’s generated about 26MWh (26 198 kWh to be precise). At 15p/kWh, payback time is about 49 years; yes, I agree with your figure of 20 years at 38p/kWh.

    That’s a capacity factor of 8.4% which is probably a bit lower than I’d have expected (my guess would have been 9-11%). That’ll partly be because they face SSW at 25 degrees.

    Costs of £5k/kWp installed is a fair bit higher than I’d have expected, too, which might partly be because they’ve got it wired up as 6 arrays with 6 inverters. Also, the panels are mono-crystalline, which puts them at the top of the market: they should last 25-30 years, and at the end of that time, it will be possible to take out the individual cells (the expensive bit) and re-encapsulate them for at least another 30 years lifetime. A modern thin-film should be about 50% the price of mono, per Wp, and give better performance at lower light levels. Albeit they probably won’t be generating in 50 years time, whereas most of the mono cells will be.

    In conclusion: Hockerton – interesting, but not particularly representative of much apart from Hockerton.

  42. Anish Wastrad

    Hi Lightbucket,

    Can you inform me on where I can find information on solar Insolation, its measurement and how it affects the output of PV panels.

  43. From what we’re seeing here in Ireland solar panels are more than sufficient to provide hot water for one week for a family in a normal residential property

    And its going to get better. Looking good ;-)

    Gary

  44. Gary, sorry, this discussion is about solar electricity. I think you’re referring to solar water heating panels.

  45. We should all be looking at the environment and what we can do to help the situation. It’s great that you are promoting solar energy and its be benefits.
    Keep up the good work

  46. Very interesting material you have about solar power. Just wondering about the different panels and the difference between the flat solar panels v the tubular panels for domestic hot water. One works from direct sunlight and the other one works off daylight. Can u tell me the pros and cons of both. What is the output of the panels. Do pv panels create more energy /m2. I know every panel is different. Also does the circulator pump for the HW panels run the whole time i.e. Loosing heat during the night.
    Finally from reading your material are solar panels worth considering installing if they only generate small amounts of energy. I live on the west coast of Ireland and we don’t get much sun

  47. If anyone manages to put more than 20% wind power on the (UK) Grid then it is likely to go unstable and you get lots of “black lights”. So 36% is pie-in-the-sky and technically infeasible. A further problem with wind power is it requires a Kw of “spinning reserve” for each Kw on-line with the (serious) fluctuations placing a demand on plant (alternators) that can supply fast “ramping power”, which stresses the grid infrastructure since it must flow from the ramping source around the system.
    As far as I am aware grid-tied PV based systems do not show this sort of toxic behavior, in part because you have a box of electronics (inverter/controller) acting as an electronic flywheel between sun and grid.
    I am interest if anyone has practical experience of operating AC modular type arrays. I see references to a “plug-and-play” unit from the 1990s now out of production (SunSine TM 300) that appears to provide 115 or 240 Volt AC single phase output without any batteries. It came with a micro-inverter per PV panel and they self synced. Sounds like the sort of concept who’s time has come. is anyone doing anything like this now?

  48. The issue of spinning reserve is totally different from the issue of grid voltage and current stability. I’m not quite sure which you want an answer to. Inverters can do the same job for wind as for PV, both of which fluctuate and both of which are DC. ….Some modules can come with their own inverters. Arrays of the same module type should be connected to the same inverter for maximum efficiency. The outgoing metering and charge controller dynamically adjusts output to suit grid frequency, which itself fluctuates constantly. All of these technical problems have been solved. A good supplier should advise you. If you’re looking for a resource, check the Good Energy weblog “Which solar supplier?” at http://good-energy.typepad.com/greenenergyrepublic/solar-panels/.

  49. If the total load ever approaches the total spinning capacity of the Grid at any time then the voltage and frequency will become unstable as the on-line plant attempts to service the over-load and falls out of sync.
    If this is not true then the power generation industry is wasting lots of money and effort as it performs simulations to determine the static and dynamic stability margins on an ongoing basis.
    Since the transmission network has finite capacity, anything that causes surge flow in the network; such as the ramping flows associated with windmill generated power is detrimental to the stability of the system.
    It is vell established that for every Mw of wind-electricity on-line, another Mw of fast slewing no-wind spinning reserve is required. My view is that it would be more sensible to just provide that Mw of power from a non-wind resource primarily Nuclear.
    Even in the US with it’s continental wide scale, the wind-power fluctuations show no-zero correlation . That is the wind power dips are seen to occur on each end of continent wide feeders. In the UK with it’s small base line ,by comparison, dip and surge are likely to occur instantaneously throughout the grid.

    Any sensible size wind machine would need to output AC if only from the ability to alleviate transmission energy drop, by transmitting at high voltage. The other real concern with any horizontal axis machine is the size of the safety zone required around it, I certainly would not have one within a 100 yards of my house.

    As for Photovoltaics: at my location (54 North) it just is not viable. Even where I optimize for maximum output in winter. The cash value of the output I get [1.8 Kwh/sq m/day] (or its cash value as a reduction in my electricity bill) will not even repay the interest on the value of my investment, even without allowing for the 40 sunless days between November 1 and March 1 [based on Met Office Data]. Something that cuts across the hype of the many.

  50. Anubhav Kothari

    I am quite new to the world of solar panels and their setup, so my understanding is limited.

    I am planning to setup a Solar PV system with 10 kWp DC output. The average yearly insolation in my region for fixed-TILT panel is 5.25 kWh/m-sq/day. I want to find out the total kWh that my solar PV system will generate in the year. Is it a simple multiplication of 10*5.25*365 = 19,162 kWh of DC output or am I doing this wrong.

  51. Anubhav,

    Indeed your calculation is too simple. The rated output of any panel is for a reference incident irradiance of 1000 W/m2. With your daily solar resource figure of 5.25 kWh/m2, your average 24-hour irradiance is only about 220 W/m2. Therefore, your annual output will be about 5 times less than what you assumed. For more accurate estimates, you would need to use free or commercial software (e.g., PVwatts or PVsys), which take into consideration many more variables and derating factors.

  52. See http://project.mesor.net/ © EC

    The EC-funded project MESoR (Management and Exploitation of Solar Resource Knowledge) is a comprehensive portal giving user-friendly access to several free sources of solar energy data using a map-based GUI. The sources include NASA, PVGIS, SoDa, HelioClim, NCEP, SWERA, MetoNorm and MeteoTest plus others. Users input their locations and the output format can be chosen. Map layers can be shown in Google Earth. Work is on-going to combine and harmonize the various data sources and provide quality control to improve the reliability of the final figures.

  53. Anubhav

    Chris is right about your calculation being too simple but you are not as far out as he suggests, your calculation is right as far as it goes but needs some additional correction factors.

    I’m assuming your insolation figure is for a horizontal surface, (this is the normal figure given for insolation levels) in which case your value equates to 1916KWh/m2/year. For comparison in the UK the average (where I am) is taken as nominal 1000KWh/m-sq/year, which is used as the basis for PV generation calculations.

    An important point to bear in mind is that the KWp figure for the panel is under specific conditions, vis; Ta = 25deg.C and irradiance = 1KW/m2 incident normal to the panel (~= direct sun on a clear day) and relates to operation at maximum efficiency. In practice these conditions will never be met and the array efficiency will be reduced by; lower irradiance levels (non-normal incidence or clouds), raised ambient temperatures and increased reflections at low angles of incidence (morning and evening).

    In the UK, with lower average irradiance levels than yours, these parameters introduce a 0.8 correction factor but with your, almost double, insolation levels your correction factor will be higher, at a guess between 0.85 and 0.9. However at optimum tilt angle (~35deg. in UK) there is an effective increase in annual insolation of about 8% to 1080KW/m2/y, so in UK out put from panels (@ 35deg. tilt, with no shading and facing due south) would be:
    PV-Array KWp * 1080 * 0.8 = 8,640KWh/y
    Therefore, for your system using my above assumptions output woul be
    Approx = 10(KWp) * 1916(KWh/m-sq/year) * 1.08 (tilt gain) * 0.875(loss factor)
    ~= 18,100KWh/y

    Hope this helps Good luck

  54. Basic outline calculations show PV is unlikely to be cost effective in Ireland.
    Very detailed calculations are not requires as shown below, plus the figures look even worse than I originally believed.

    I set up PVwatts v1 and ran it for the two nearest locations to me Birr and Balmullet, picked minimum size system of 4Kw(p) at fixed tilt (optimum) and with the recommended derateing factor of 0.77, and a a slightly inflated utility price of 0.181 Euro/Kwh based on my total charge divided by total units used.

    Running PVwatts v.1 for both locations; produced almost the same numbers: 2866 and 2466 Kwh/year.
    I reduce the numbers to give the specific power yield in Kwh/Kw(p)installed per year: 716.5 and 616.5.

    Inserting the higher number into a costing spread sheet and presuming 2% interest over 10 years ; shows that at 716.5 Kwh/Kw/year if the capital cost Euro per Watt(p) installed, exceeds 1.64 ; the installation looses money. The picture is worse for the lower power yield figure and this has not included maintenance or land charges.

    As currently I am unable to find PV panels below around $US3/Watt(p) even if I ship them direct from China I conclude it would not be cost effective.

  55. Specific power yields vary widely from one part of the UK to another; in the case of Oban the figure is similar to here, while in Gatwick it would be some250Kwh/Kw(p) per year. All figures from PVwatts v.1 In the end the only way is to put up a trial panel and log the data.

  56. Comments posted above suggest (so far as I can see) that the total annual average benefit from 35 degree angled pv collectors in the UK, compared to horizontal collectors, will be slight – presumably because while it increases efficiency in the winter, there is less energy available to be harvested in the winter, so the overall difference in energy collected won’t be all that much. Have I misunderstood?

    If one takes account of the fact that there is much more cloud in the winter, then presumably the gains from optimal angles become even less.

    Finally, will horizontal collectors will be able to collect sunshine from further east and further west than angled ones? In other words, for longer during the day. I am an historian not a mathematician and I can’t do the 3-D geometry, but it seems to me intuitively that if I look at surfaces angled to the south, they get less strong sunshine on them in the morning and the evening than flat surfaces (are they in shade for longer). Also, I would have thought that the critical angle of reflection on the surface of the glass on the front of the collectors would mean that there might be more reflection of sunlight from angled collectors than from flat ones. But this may all be completely mistaken – O level physics was my peak, and that was a long time ago.

    I am interested in this because I have a huge area of flat roof on a building that I help to run, but no ability to mount angled collectors on it, because of weight on the structure from the mounting brackets, and wind (which might well pick up and carry away any angled collectors that were not very firmly fixed indeed).

    Finally, can anyone advise me on alternative forms of pv collectors – thin film, tiles, and so on – that would be better at mounting on a flat surface than the standard commercial offerings.

  57. Hello Harold,
    Roughly speaking, the optimum panel angle is south-facing, tilted at the angle of your latitude if averaged over the year.
    For the summer (when most of the energy comes in) in London the optimum angle is about 30° from the horizontal, so a horizontal panel doesn’t lose much (about 15%) compared to an optimally tilted panel during the summer.

    There’s a nice graph here showing optimum panel angle for every latitude and every day of the year:
    Solar Panel Angles for Various Latitudes

    Many solar PV panels have anti-reflection coatings.

  58. It seems to me that the only use of photovoltaic power conditioning device :-) In July, you can feed a few small airco KW, when the sun shines, and in winter you can turn on a pair of fluorescent lamps (only during the days.

  59. hi , the main demand for electric is by and large determined by our waking hours unless using storage radiators and their like, also a mean average over 24 hours is not really the only factor that could or should be used, a daylight isolation ie 8 hours of fullsun light and the dusk & dawn hours better indicate power performance. Given that many potential users have peak demand in daytime when power is at its greatest and also given that batteries may store excess power would help many who are considering this form of free &green energy to arrive at more practical effieciency calculations

  60. I think many of the potential users of PV are by nature a determined lot and many have a mind to DIY
    The commercial cost vs DIY would be a useful thread to develop ..I myself am going ahead later this year on a self build Chalet and a self build house both in dept 82 Midi Pyrenees so any comments and sagely advice is really most welcome.
    I am perhaps misguided but it seems with some careful planning the PV or PV/wind combo can produce hassle free electric at a saving let alone at an acceptable cost .In SW France we have power cuts courtesy of EDF on a oh so regular basis.
    I have my hand forced to some extent as EDF do not connect new self build houses to the grid if they are 100m +away as my 2 plots are
    SUN IS SHINING APLENTY!! So lets see…… I will post my failures and successes as they unfold. Seems like a great blog site with many knowledgeable guys and gals out there.

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