Originally Posted by wesslvm
Checked it out this morning, working just fine.
Thanks for all the operating experience.
Now I'll have to learn what all the "parasitic" DC loads are and learn what I can do to minimize them when off of shore power.
Can't really justify solar panels when I have a generator at the ready, but any thoughts on amount of draw (in amps) the parasitic DC loads are? Maybe I should save that for another thread?
Very glad to hear that your fridge is still good.
You could rig up something to allow you to power just your fridge and not any of the other parasitic devices if that would buy you much. Rather than estimating numbers, I think you need to use an ammeter to determine how much current is actually being drawn. I would start by pulling the fuse for the fridge and see how much the "parasites" are using. Make sure you turn off everything you can (like the thermostat for instance).
Then put the fridge fuse back in and look at current draw for that. It's going to be in three parts:
- Power for the electronic controls (~ constant)
- Power for the gas solenoid
- Power for any fans.
Since the second two will not be drawn except for say 30% of the time (but will be bigger numbers) you would have to figure out how to measure on vs. off currents for those.
Once you know what the current draw will be, you can calculate how much battery you need.
As an example lets just use 1 amp continuous times 168 hours per week =168 amp-hours.
Let's use 4 more amps (total 5) when those things are on: 4 * 168 * 30% = 201 amp-hours.
So that's a total of 369 amp-hours. Since you don't want to discharge your batteries more than 50% of capacity, you need 369/.5 = 738 amp-hours of battery capacity. (Currents are low so no derating due to higher currents per Peukert's Law.)
So 8 batteries of 100 amp-hour capacity rating should do you just fine! (No, I don't think that's practical.) Hopefully someone like Lou (Herk7769) will check my math. He's the real expert on this stuff.