# power consumption for portable heater

Discussion in 'Electronic Basics' started by [email protected], Apr 23, 2007.

1. ### Guest

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.

2. ### Michael A. TerrellGuest

73/12.5 = ?

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Michael A. Terrell
Central Florida

3. ### Homer J SimpsonGuest

Less than 6 hours - maybe a lot less.

4. ### Sjouke BurryGuest

And then assume an efficiency of about 80 percent,
so (73/12.5)*.8

5. ### Peter BennettGuest

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.

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Peter Bennett, VE7CEI
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6. ### Tom BiasiGuest

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

7. ### Phil AllisonGuest

"Sjouke Burry"

** So the missing 20% turns into what ?

IR light ?

Fairy dust ??

......... Phil

8. ### Tom BiasiGuest

Heat in the battery and chemical changes and maybe a little fairy dust.

9. ### Phil AllisonGuest

"Tom Biasi"

** You missed the point of the ambiguity - fuckwit.

Plus all the wrong assumptions re the cell type in question.

....... Phil

10. ### ChuckGuest

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

11. ### BobGGuest

A heat pump would move more BTUs than a resistance heater... look for
a DC RV AC

12. ### Tom BiasiGuest

Thanks Phil,
Your well thought out comments are always welcome.