I have a very simple electrical question. I have a portable heater which is rated at 12 V, 12.5 amps, which I have attached to a 12 V, 73 amp hour battery. I'm wanting to get a rough idea of how long this heater could run before completely draining the battery.
power consumption for portable heater
Apr 23, 2007
11 Replies
73/12.5 = ?
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Michael A. Terrell
Central Florida
Less than 6 hours - maybe a lot less.
And then assume an efficiency of about 80 percent, so (73/12.5)*.8
The amp-hour capacity of a battery is usually given assuming a 20 hour discharge time - your 73 AH battery could deliver 3.65 amps for 20 hours. Discharging much faster than the 20 hour rate will reduce the total amp-hours that you can get.
According to a table I have, you should get about 78% of the capacity, or 57 AH, which, at 12.5A, gives you 4.5 hours.
This also depends on your definition of "completely drains". The official definition is a discharge to 10.5 volts.
Peter Bennett, VE7CEI
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At that rate of discharge with a 73AH battery your effective heating efficiency will fall very rapidly.
Peter's calculations describe the math but in real life the battery will be getting hot, the internal resistance will go up rapidly and the power to the heater will fall quickly. You didn't say what kind of battery. Are you really concerned about how quickly it will drain the battery or how long you can get useful heat? My SWAG is 2 hours max. Maybe a little more if you cycle the on time to let the battery cool a bit.
Tom
"Sjouke Burry"
** So the missing 20% turns into what ?IR light ?
Cosmic radiation ?
Fairy dust ??
........ Phil
Heat in the battery and chemical changes and maybe a little fairy dust.
"Tom Biasi"
** You missed the point of the ambiguity - f****it.Plus all the wrong assumptions re the cell type in question.
...... Phil
As Peter suggests, look up Peukert's Law. You will get a lot fewer than 73 amp-hours from the battery when discharging at that rate.
In addition, keep in mind that the battery voltage will drop beginning almost immediately, which means that the heater will actually draw fewer than
12.5 amps given the fixed resistance of the load. This will work in the opposite direction of Peukert's factor, but the heater will give less heat as the voltage and current drop.Another consideration is that most lead-acid batteries cannot withstand more than a few total discharge cycles before they are permanently damaged. It would be good to avoid discharging to more than 50% capacity if you intend to keep the battery.
Chuck
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A heat pump would move more BTUs than a resistance heater... look for a DC RV AC
Thanks Phil, Your well thought out comments are always welcome.
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