From: J. Forster (jfor@quik.com)
Date: Mon Jun 21 2004 - 12:28:41 PDT
Rather than answer your questions directly, here's how to figure out the answers:
Power drawn by the A/C:
Either look at the plate for the wattage or estimate by: (BTU capacity) / (Energy
Efficiency Rating [EER}) = Watts
Typical EERs range from 7 for very old units to 12 for the newest, most efficient
ones. You will need extra wattage from your inverter (2 to 3 times) to start the
compressor.
For example, a 7000 BTU with an EER of 10 draws 700 watts.
**********
Inverter:
The power required by the load is the output power of the inverter.
To get the input power: (Output Power) / (Inverter Efficiency) = Input power
For example, 700 Watts / 90% = 778 Watts. That is what comes from the battery.
The inverter specs will likely list several efficiencies ( often: no load, half
load, full load. Use half load in this estimate)
*******
Batteries:
Assuming a worst case condition, that the A/C will run 100% of the time, you will
need a useful capacity of:
(Inverter Input Wattage) x (Number of Hours) = Watt-Hours of battery capacity
Continuing the example: (778) x (8) = 6224 Watt-Hours.
BUT, you should not discharge your batteries all the way. The spec sheets will list
"Depth of Discharge" as some %. I'm going to use 50%, but you should look it up for
your batteries. Some are 'Deep Discharge', some are not. The depth of discharge
affects battery life.
Battery Capacity required = (Watt-Hours needed) / (Depth of Discharge)
Continuing the example: (6224) / (50%) = 12,448 Watt-Hours capacity
This capacity is in Watt-Hours. To determine the battery capacity:
Amp-Hours = (Watt-Hours) / (Battery Voltage)
More example, for a 12 Volt battery: 12,448 / 12 = 1037 A-H; for a 24 Volt battery
12,448 / 24 = 518 A-H
NOTE: In general, a 24 volt system would be somewhat more efficient than a 12 Volt
one. 36 V and 48 V are used in some applications. Note also, the battery A-H for a
24 V system is 1/2 that for a 12 V, but 2 x as many calls are required.
If you used 250 A-H 6 Volt golf cart batteries, you'd need roughly 8 of them. Two
banks of 4 in series.
Charging:
Charging Amperage = (Watts Output) / (Voltage)
Charging Time = 1.1 x (Battery Capacity) / (Charging Amps)
The 1.1 is to allow for an extra 10% due to battery losses.
So, if your charger develops 3000 watts = 24 Volts @ 125 amps
Charging time = 1,1 x (518) / (125) = 4.6 Hours.
Check the battery specs for charging recommendations though.
BTW: As an aside, a fully charged car battery (12 V @ 60 A-H) has about the same
electrical stored energy as 1 pint of gasoline used to run a small generator.
-John
Recovry4x4@aol.com wrote:
> This is a bit off topic but definitely related. I have an S-318 shelter with
> a 7000BTU window shaker A/C unit installed. In the past I've used my little
> MEP-105 1500 watt genset to run theA/C unit. A friend of mine (who is now MIA
> here in the states) had a similar A/C unit installed in a sleeper on an over
> the road truck. He had a 3000 watt inverter that he used to run the A/C. He also
> had 5 groups of 2, golf cart batteries used for storage and to run the A/C at
> night. My question is how many of these 6 volt golf cart batteries would it
> take to run this A/C through a similar type of inverter for 6 to 8 hours. Next,
> what kind of amps and time would it take to recharge them? I've just been
> given a 3KW, 28Volt unit unit that says it develops 107 amps. I'm pretty sure
> that the generator is underrated. I like the idea of running the A/C at night
> without the generator but want some educated opinions on how this might work.
> Thanks,
> Kenny
>
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