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Calculating Load for Cooling

Calculating Load for Cooling

Calculating Load for Cooling

(OP)
If I have a 120v circuit with 10A and one with 16A which totals 26A.  But when I calculate for required total BTU/hr for cooling CFM, it doesn't come out to what my conditions currently are.

But since both of the circuits have a common Nuetral, would the total amps for these 2 circuits be calculated as the total (26) divided by 2 X 1.73  or 22.5 amps?

Representing the total actual heat load in the room? Assuming no lights, people or other loads?

RE: Calculating Load for Cooling

You need to calculate the "real" power for heat load.  It may be less than the "apparent" power of 26 x 120.

RE: Calculating Load for Cooling

(OP)
You are correct,  our typical pwer factor on thgis circuit is 0.96. Basically  the load is still 25 amps

RE: Calculating Load for Cooling

It's 25A x 120V then.  Ignore Mr. Neutral.

What kind of room is it?
How warm are adjacent rooms?
How well is this room insulated from those rooms?

Why do you think the calculated wattage is seeming low?

If you are not actually turning the air over and the room is well insulated a 100W bulb will get it pretty dang warm, let alone, your 3kW.

More numbers?

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

RE: Calculating Load for Cooling

(OP)
Electrically  120 x 25  is correct as you state. I asked a different question/related to this on the HVAC/R forum and even the 1 deg F cooler adjoining spaces could make a difference.  The room is a telecom closet and very small.  The walls are only sound bat insulated 5/8 sheetrock. We are only talking about 4amps or 480w difference in the calcs, so the result is not an issue.  I always like to understand why the calcs would provide different results, that is why I asked the question.

You also pointed out correctly
How warm are adjacent rooms?
How well is this room insulated from those rooms?
It is enough to make this small difference.

I did set up a calculator for solar gain for exterior walls and roof.  I guess as a point of excersice, an adjoing room heat loss/gain calculation could be set up, since all adjoining interior walls/floors/overheads are built exactly the same.

Thanks for the nail on the head reply

RE: Calculating Load for Cooling

You bet.

My office is highly insulated.  I built it that way because the neighboring business ran a nasty table saw often.  I have walls that are 9" thick packed with insulation.  My door is gasketed.  My office is a space about 30 x 13 or 400sqft.   If I leave my PC on, ~100W, overnight during the summer the next day it will be uncomfortably hot and I have to run my AC.  If I leave it OFF then then next day my office is comfortably cool.

In the winter I leave my PC on, sometimes two, so the next day my office is habitable.  Hence my personal heat load experience.

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

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