electric water heater

With that load you need a dedicated circuit from your consumer unit/ distribution board. The 7kW needs just under 30A, and 4mm cable unless it's a "long" run (say over 15m) or runs buried in thermal insulation - then you'll need 6mmsq minimum; the 9.6kw will pull 40A and would be better wired in 10mmsq (bit of a bummer to work with). No sensible easy way to reduce this load: water needs a lot of energy to heat it! If you don't need much temperature rise, try to find a 3kW inline heater instead.

HTH - Stefek

The only way to keep the current below the level of your trip is to work out the kW equivalent of your trip current and keep below that. For example, if your trip is 20 amps, watts = volts x amps = 20 x 240 = 4800 watts or 4.8 kW

There's no magic way of preventing a 9.6 kW heater from drawing 40 amps!

I suppose you could limit the power by reducing the voltage with a variac - but that would be expensive and would only have the same effect as using a lower power heater.

probably cheaper to buy two heaters & wire them in series

It will need a proper electrical installation on its own circuit. The trip and cable should be chosen for the load. If you don't have a suitable cable run now, you'll have to install one.

Christian.