December 17, 20196 yr 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?
December 17, 20196 yr 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.
December 17, 20196 yr 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.
December 17, 20196 yr Author 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
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