I am interested to know how large a solar panel one would need to produce pure oxygen and pure hydrogen from sea water.

I already found the required energy here:

If you assume perfect efficiency.....in terms of power, this is 3.67 kW for one hour.

  • 1
    I don't think you mean hydrolise. Electrolysis, perhaps? (by the way, we do have a site dedicated to Chemistry)
    – 410 gone
    Dec 14 '15 at 14:10
  • Thanks - the word is 'electrolise'. (it is also possible I might be better served on another site - let's wait and see)
    – Hugh Jones
    Dec 14 '15 at 14:35

The flippant answer is "it depends how long you're prepared to wait". But, let's assume that you want to electrolyse a litre of water an in hour.

You've identified above that the energy needed to electrolyse one litre of water is 3.67kWh, or 3.67kW for one hour (assuming no losses in the process). So what we need is a solar system capable of delivering 3.67kW.

Let's assume that we're in a temperate latitude, that the sky is clear, and that it's the middle of the day. That should give us around a 1kW/m^2 of energy, in the form of light, reaching the ground. Decent (but not bleeding-edge) photovoltaic panels are around 15% efficient, so we'll get around 150W out of them for each square metre of area. That means that in this scenario, we'd need 24m^2 of solar panels.

That's a lot. It gets worse if you want to do this continuously, rather than as a one-off, because the average power of sunlight reaching the ground in temperate climes, including cloudy days and including nighttime, is around 250kW/m^2, one quarter of the peak value. So then you need 100m^2 of solar panels.

Let's look at it another way, and say that we have 20m^2 of rooftop available. That would mean that you produced an average of 20 x 250 x 0.15 = 750W, which works out to splitting around 5 litres of water per day.


Looking at it from systems I can buy rather than theory, if you want 1 litre per hour you need 3.67kW of hydrolysis output from your system, but the hydrolosis is not 100% efficient. There are two main options, and they're both a bit over 50% efficient:

Wikipedia says

Reported working efficiencies are in the range 60-75% for alkaline and 65–90% for PEM.

You will need 3.67kW/0.60 = 6.1kW of input power. But you will have other losses, so should aim for 7kW or more.

In the US that looks to cost about $US10,000 as a DIY kit, or $OZ11,000 here. In Australia it's common to put slightly more panels on than your inverter is rated for so that you get peak power for longer. I would try to put 7.5kW or 8kW of panels on that 7kW inverter (almost all of them will do that, it's a question of maximum input voltage) so that you get 7kW out of the system more often.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy