| Re: [GMCnet] Induction cooktop run off an inverter [message #325054 is a reply to message #325026] |
Tue, 17 October 2017 16:02   |
 |
RF_Burns
Messages: 2277
Registered: June 2008
Location: S. Ontario, Canada
Karma:
|
Senior Member |
|
|
Consider the power you are consuming from your batteries at these power levels. For easy calculating lets use 1800 watts load and 100Amp/hr battery, we can scale it from there.
1800 watts @ 85% efficiency = ~2120watts input to the inverter. your fully charged battery at 12.6v will with internal battery resistance, connector and wire resistance, and fuse resistance, easily drop to 11.5v at the inverter terminals. 2120/11.5v = ~185amps. Inverters typically will shut-down below 10.5V input, at that point the input current is over 200amps.
Now you don't want to discharge your lead acid battery below 50%. Your 100Amp/hr rating is for a 20C rate, or 5amps over 20 hrs. At 185amps, its capacity will drop like a stone, likely down to 50% or less, so now you have only 50Amp/hrs and at 50% discharge is now only 25amp/hrs. so 25Amp/hrs @ 185Amps = about 8 minutes till you are discharged to the battery damage point.
So you have 6 x 6v batteries and you have lots of power... so that's about 3 x 150 = 450amp/hrs, so you have about 40 minutes to cook your stew.
Then it will take 10-15 hours to recharge the batteries.
High current, deep discharge will severely shorten the life your your batteries.
IMHO Better to run your generator for a few minutes and keep your batteries for lighter loads
Bruce Hislop
ON Canada
77PB, 455 Dick P. rebuilt, DynamicEFI EBL EFI & ESC.
1 ton front end
http://www.gmcmhphotos.com/photos/showphoto.php?photo=29001
My Staff says I never listen to them, or something like that
|
|
|
|
GMCforum
Search
Help
Register
Login
Home
