Jump to content

A few calculations for an inverter


isetech

Recommended Posts

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.

Selection:

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. 

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...