Solar Panels – the math

May 6, 2009 at 1:00 pm 3 comments

On another social media network, the question was posed…”If every single rooftop in the country was covered in PVs, I’ve heard that we would generate enough electrical energy not to need any other source of electrical power! But, has anyone done the maths?”

The questioner was from the UK. Many people immediately jumped on the issue as a dumb idea because of many other logistical issues, but no one “did the math”. Anyone that knows me at all knows that I tend to “do the math” first then look at the resulting implications.

Basis and Assumptions:

On average the sun provides about 1000 watts per square meter (at sea level, higher as you go up in elevation, but a convenient number for my purpose…)

Current solar panels are currently less than 30% efficient. We’ll use 30% because it makes the math easier. We’ll add “windage” later.

Let’s be generous – given Britain’s famous weather – and say you can generate electricity from all panels at this peak efficiency for 10 hours per day, 365 days a year.

Current consumption in UK is nearly 400TwH per year.

The Math

Solar panels (at 30% efficiency) generate 300 watts per square meter. So for each hour of sunlight, they generate 300 watt-hours or 0.3 KwH.

Over a 10 hour period, each square meter of solar panel can generate 3 KwH of electricity. Over the course of a year, each square meter could produce just over 1 MwH of electricity.

To generate 400 TwH of electricity would require almost 400 square kilometers of solarpanels.

If, on the average, one could put 2 square meters of PV’s in the most optimal south facing position on the roof of a building, then you would need 200,000,000 buildings.

Given my rather positive assertions related to both efficiency, and available sunlight, I would double that for a realistic scenario. SO, you would need to put PV’s on 400,000,000 buildings


Solar panels are not a panacea that will solve all of our problems. My scenario above ignores the complex grid and energy storage structures that would be required to move electricity from such a dispersed generation to concentrated population centers and industrial applications and storing summer generation for use in winter. I’m sure any utility engineer could add dozens more considerations that I’ve not mentioned.

I believe that all of the low carbon emission options must be explored and applied to the maximum extent feasible to lower both dependence on non-domestic sources of fuel and GHG impact on our planet. But, we must maintain a balanced application of all of these technologies in order to maintain a society we all want to live in.


Entry filed under: Basics. Tags: , , .

A Parable of Power Definitions and other matters


  • 1. Michael  |  June 24, 2009 at 2:18 pm

    It’s nice to see someone “do the math” at least in this case.

    I suppose another way to think about it is if solar energy from roof panels on one house would totally cover the needs of that house, then at least we could provide energy for all homes. Guessing that wouldn’t pan out. 🙂 I’ve always kind of figured that a 30% efficiency is nowhere near adequate.

    I’ve been a bit curious lately about other forms of energy capture / storage / conversion a person could put together for their home. Anyway, enjoying your blog so far, keep up the good work! –Michael

    • 2. margaretharding  |  June 24, 2009 at 8:39 pm

      Thanks! most current PV’s are only 15% efficient so I was being very generous. Household applications are a good idea, but unless there is a government subsidy program (tax rebate, etc) they usually won’t pay for themselves before exceeding design lifetime.

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