Dear Friends and Colleagues,

Most of you, I’m sure, are aware that newspapers and popular journals regularly engage in propagating comforting myths about photovoltaics (PV). They do this by reporting only the rated capacity (also called peak capacity) of an installation. For instance, they will report that 100 MW of power has been installed somewhere. Those who know that a fairly large power station has a capacity of 1000 MW may be impressed.

However, with PV it is not the rated capacity that tells the truth of the matter, but rather the output which can be expected from the installation. The output from a 100 MW of rated capacity, at a sunny site in Spain for example, would be about 14 MW.

This seems to escape nearly all journalists, yet determining the output is simple. It depends on only two things the rated output and the amount of insolation (the ‘sunshine’) at the site. There is a need to inform journalists whenever they have written a report which only gives the rated capacity of a PV installation.

Two simple calculations are included in the spreadsheet.. Although Ken’s calculation method is longer, it does show the area of module required (provided one knows the efficiency of the modules). Whichever calculation is used, the spreadsheet makes the task easy (and I tentatively suggest that it could be urged upon recalcitrant journalists!).

On further rumination, your calculation of PV output based on rated capacity and insolation, proved rather useful. I put the calculation on a small spreadsheet. There I first show your calculation, for a 100 MW installation, but based on a 200 W/m2 insolation. I start by using the same 80% ‘loss’ factor that you used. Then I repeat the calculation assuming a 70% ‘loss’ factor.

This confirmed the following points:

a) That your calculation gives exactly the same result as the Rule of thumb, provide that the ‘loss’ factor used is 70% (i.e. the same as the Rule of thumb).

b) That it does not matter what the efficiency of the modules is. The output remains the same.

c) That your calculation involves eight steps whereas the rule of thumb requires four (your calculation has the possible advantage of also showing the area of the module, provided the correct module efficiency is put in).

d) That by changing the insolation to 2000 kWh/m2/yr = 2000 / 8760 x 1000 = 228.3 W/m2, i.e. the insolation you used in the example that you gave in your email, your result of 18.3 MW is obtained.

Of course the question arises whether the 70% ‘loss’ factor — for electricity delivered to the grid — is appropriate to account for the reduction in output which occurs because PV does not operate as well as the recorded levels of insolation might suggest at such times as when the sun is just over the horizon, or just about to sink below it, and also because of such factors as dust interference, and conversion from DC to AC.

However the arguments for 70% were presented at length in the paper A Rule of Thumb for PV Annual Capacity Factors, in the October 2004 issue of the OPT Journal — which incidentally relied heavily on the data given to me by Tom Hansen.

Kindest regards, Andrew.