Good Day.

A little bit of theory...

If I boiled my 2000w kettle for 1 hour, theoretically the Kwh usage would be 2Kwh, right?

In theory does that mean that my 2.4Kwh pylontech battery would sustain this kettle for 1 hour, with (theoretically) 0.4Kwh of power left in it? 

what would be the real world actuality of this theory?   

There are other losses involved.

So your inverter should have around 93% efficiency turning DC-AC then you also have other wire losses and the like. 

But just taking the inverters 93% efficiency (assuming no peukert factor since you are using LFP) 

It would take around 2150Wh * 0.93 = 2000Wh 

So 2150Wh from your batteries would be required to keep a 2KW load on for 1 hour, of course there are other losses as well.

1 hour ago, jasonvanwyk said:

In theory does that mean that my 2.4Kwh pylontech battery would sustain this kettle for 1 hour

The recommended discharge current for a US 2000 is 25amps, so that is a constant load of 1200W. You would need two of the US2000s to run that kettle. 

Thanks Jaco - as always, there is so much more technical stuff to consider, than what meets the untrained eye:) This definitely helps to understand usage better