heat loss via electric "test"?

chris22

So,.. in an apt house there are so many variables, Is it a top unit? how much insulation in the ceiling? Is it a middle floor, no heat loss from floor or ceiling and maybe only one or two walls to the exterior, If the windows are lowE or other window variable, if the unit has been air sealed via blower door etc, in an older building have the areas between the studs had cellulose blown into them?

So ( I know you can't do this most of the year for jobs, I'm talking for maybe your own properties)

  I understand that any electric heater puts out something like 5,300 btu's and hr,.Ok,..if you wait for a very cold day and say, if you can keep the unit at 62 degrees with two units running on constant,..and if you can get to 72 with one set at that temp which now it cycles on and off, I'm thinking that .you then can tell that the unit needs between 12-15000 btu's an hour?

"IF" you have the luxury of doing this, wouldn't it be more accurate than the calculations that (I think) may have been set up when there weren't so many variables as there are now?

Brad White

Yes, Sort of

What you describe is a technique I use when determining AC capacity in a room, most often a laboratory. To determine if a room can accept a new piece of equipment, incubator, analyzer, etc., we put in an electric heater but with a Watt check meter and rheostat to monitor actual Watts input to get BTUs per hour.  (If the room AC responds and keeps temperature, we have enough airflow, is the idea.)



So yes, you are on to it, in part. What you will have to do is measure the concurrent indoor and outdoor temperatures and, the hard part, discount any solar gains. If you can do this across a night, better yet.  The idea is that the temperature be maintained at as precise a level as you can for as long as you can, steady-state.



You will have to extrapolate for infiltration (varies with wind, stack effects, temperature), residual heat from building mass and internal gains (things you cannot count upon when you need them).  And even though you keep say, 70F in the space, you actually may not have to add heat until it is below 60F, the "balance point", which also varies.



You would continue to extrapolate for your design temperature, suggesting that if your presumed heat loss at 40F outdoors and 70F indoors is "Q" and your design is 10F outdoors,  then your design heat loss would be Q x 2.



So in short, yes you can but with caveats.

Ironman

The Formula...

For converting electric heat is: Watts x 3.413 = Btus. Use actual wattage, not rated, as this can vary significantly.