PV Geyser Element System Sizing and Savings
Energy cannot be created or destroyed but only converted from one form to another. So there is nothing special or magical about a PV geyser element. If we assume it is 100% efficient then it has a “COP” of 1. In other words, if it is a 1kW element it will consume 1kW of electricity to produce 1kW of thermal output. This power must come from a PV panel.
The best efficiency of PV panels are around 17% (this is of course under lab conditions with the cell at 25°C which will seldom happen in real life). If we have 3 x 300W perfectly facing PV panels on the roof we should in theory be able to get a maximum of about 5 hours of 900W output (annual average – winter you will have less because of less sunlight hours and in summer you will have more. Again this is mismatched to when the most hot water is needed). We therefore have a theoretical maximum of 4.5kWh but with panel temperature derating (0.4%/°C with 20°C NOCT) you will typically get only 4.1kWh. Next you need a MPPT to ensure optimal power transfer from the panels to the element. A good MPPT is typically be around 95% efficient and therefore the power available to the element will be around 3.9kWh.
The thermal standing loss on a 200L Kwikot B-rated geyser is about 1.5kWh (in real life with piping connected it could easily be double). You therefore have a maximum of 3.9-1.5=2.4kWh left to heat the water with which for 200L will give you a temperature increase of 11°C (so if the incoming water in the tank was 15°C you now will have only 26°C). Of course if you only use 100L/day from the system you only need to heat 100L plus standing losses then your temperature increase will be about double and you end up with 37°C. The element controller will use Eskom to do the rest of the work and get the tank to 60°C (SANS 151 requires water to be stored at 60°C for Legionella disease prevention).
This 900W system has a physical footprint of 6 m2 (3 x 300W panels).