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A few calculations for an inverter


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Did my home and I think it is right ... thought I would share to see if I got it right.

I want to install and inverter only ... no solar (online type ... maybe a mecer king would be a good selection)

The inverter will be used as a backup system for an office. 

The installation: 

AC mains 

230 VAC 

18 amps (a constant 18 amps load ...measured) 

4140 watts 

backup time required ... about 3 hours.

DC section:

4140 / 0.7 = 5914.28/ 1000 5.9 KVA  (an 8 kva would be required) 

0.7 used for inverter losses.

48 VDC 

86.25 amps 

for a 3 hr backup rounded to 90 amps x 3 = 270 Amp/hr 

The customer will be supplying the batteries and he want s lead acid ... so we need to divide the amp/hr 0.5 

270/.5 = 540 Amp/hr 

If we were looking at lithium then it would just be the amp/hr x VDC 

270x48= 12960 w/hr divided by 1000 = 12.96 kw/hr battery.


1 x 8 kva mecer king in inverter 

3 x 5,5kw/hr batteries 

I also considered a 3 phase inverter ... but decided to stick with the single phase to keep it simple. 

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