# [rescue] Cooling (Long Message, sorry)

Joshua D Boyd jdboyd at cs.millersville.edu
Mon Apr 15 21:41:14 CDT 2002

```On Mon, Apr 15, 2002 at 10:21:49PM -0400, Dave McGuire wrote:

>   To be more exact about it, you can get a clamp-on ammeter and
> measure the current consumption of each device.  Multiply this by your
> (measured) line voltage to get power in watts.  Count on all of that
> being turned into heat.  Apply the watts->BTUs conversion formula
> (I don't remember what that is) and there's how much heat you're
> producing in the room that needs to be removed.

Hmm.  1 BTU is between (I say between because there is more than one BTU, and
I don't remeber which is used for AC) 1054 and 1060 Joules, with between 1054
and 1055 being most common.

Now, when talking about heating cooling devices, we really mean BTUs per
hour.  Devide that by 3600 to get seconds.

So, 1 BTU hour = 1055 Joules per hour = 1055 / 3600 Joules/second ~= 0.29 J/s.

Now, a watt is a volt * amp = Joules per coulomb times coloubs per second.
The coulombs cancel, and we are left with joules per second (J/s).

So, x BTUs per hour = 0.29*x J/s, or more simply, x Watts = 3.45 BTUs/h.

Hope that helps.  I didn't have written formula handy, but it was easy enough
to derive.

--
Joshua D. Boyd

```

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