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Electrical heat gain

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Posted by Curbie on April 12, 2009, 4:10 pm
Is there a standard formula used to calculate heat gain from
electrical devices. When considering devices like an oven, OR an
incandescent light, OR a drill press, it wouldn't seem so, but I
really don't know?



Posted by RamRod Sword of Baal on April 12, 2009, 8:06 pm


Lights 3.4 BTUs per watt per hour or 3412 BTUs per Kilowatt per hour
Electrical appliances that heat 3.4 BTUs per watt per hour or 3412 BTUs per
Kilowatt per hour x time on

Note fluorescent light wattage should be multiplied by 1.25 then multiplied
by 3.4 BTUs per watt to include the heat from the ballast

IE a toaster is maybe 2 KW but only runs for say 4 minutes, hence 2 Kw for 4
minutes would around 455 BTUs

Electric motors in room

1/4 HP efficiency 64% = 640 BTUs
1/2 HP  efficiency 70% = 1,280 BTUs
1 HP efficiency 79% =  2540 BTUs

These numbers are for 1 hour operation. Again you need to know how long the
motor is going to run for, 1/2 an hour would indicate 1/2 the BTUs

Ovens are Kw multiplied 3412 that you need to know how long the actual
element is on for in that hour, as the thermostat switches the element on
and off.

So if you have a 2 KW element and it is on for 1/2 the hour, then it would
be around 3412 BTUs. Time on for the element will change as to how high you
have the thermostat set and how often you open the oven door.

There would be some variation if the kitchen is very cold, compared to a hot
summers day. It is better to over estimate the heat gain than under estimate
it if you are looking at what size cooling equipment that is required.

It is NOT a good idea to wildly overestimate the cooling load, as it can
effect the humidity in the conditioned space.

IE do not install a 5 HP system if the load indicates 3 HP.

In a home I find it better to let the kitchen heat up a bit rather than use
the full load of the cooking appliances in selection of the cooling
equipment, as most of the time these appliances are not running, so it is
easy to oversize the cooling equipment if sizing up the cooling unit for all
day operation

Usually one does not have all the cook top elements on at once plus the
oven, yes it may happen at Christmas, but is it worth a few thousand dollars
extra to cool the kitchen for one day a year. Again you must consider not
over sizing the system for the times the cooking equipment is not running.

A good reputable air conditioning company should be able to advise you.

Mind you the cheapest (and smallest) is not usually the best selection when
getting quotes.

I am assuming here you are looking at a ducted system that does more than
one room, but the heat loads are still accurate for single rooms.

Posted by Curbie on April 12, 2009, 9:27 pm

Just what I was looking for, I think I have human, electrical, and the
last is solar heat gain. I won't know exactly what I need there until
I finish some TMY data conversions which should take a few more days,
Know anything about solar heat gain?

Thanks a million.


Posted by Neon John on April 12, 2009, 11:01 pm
 I stopped reading that BS when he got to the part that made
fluorescent lights generate more heat than electrical input.

The answer to your question is simple.  With the one exception of a
heat pump moving heat from the outside, the heat gain of any
electrical device is exactly 1.  One watt of heat out for one watt of
electricity in.  To convert to BTU, multiply watts by 3.415179 BTU/hr.



Posted by Fred F. on April 13, 2009, 1:33 am
 Neon John:

Here is a question (it might be a dumb question), From your answer you
a basically saying that all electricity is converted to heat, is there
anything else that electricity is converted to or is 100% of wattage
input end up as heat?

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