help sizing an electric heater

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nizak

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Is anyone here familiar with the sizing of a fan forced electric heater for a garage?

Here are the particulars:
- 11ft ceilings
-650 sq ft
-R19 insulation/finished drywall
-maintain 45 degree temp.(just wanting to keep items from freezing,not trying to keep a comfortable temp for spending time out there)
-climate where temps routinely dip to single digits.


Homeowner is convinced that a 5000watt unit will suffice. Just from my limited experience with electric heat(baseboard type) it seems that 5000 is quite insufficient.

Thanks for any help.
 
Will there be closed cabinets or other storage on the outside walls? If so they will either need to be ventilated or the inside air temperature will need to be proportionally higher.
For occupant comfort infrared is in theory more efficient than forced air, but for content protection that may not hold.
But it can be aimed and it is not as susceptible to hot air rising to be wasted in the roof space.
 
I have put 5000 watt heaters in several garages, some about as big as your 650 sf. 45 degrees has never been hard to maintain in any of them, and they usually are much warmer then that even with single digits outdoors. Put heater up high and direct the blower downward, or even put in other fan(s) to circulate air. If you don't stir the air some you will be much warmer on the ceiling then on the floor, and can have "dead spots" much easier that maybe don't get warmed enough.
 
nizak said:
Here are the particulars:
- 11ft ceilings
-650 sq ft
-R19 insulation/finished drywall
-maintain 45 degree temp.(just wanting to keep items from freezing,not trying to keep a comfortable temp for spending time out there)
-climate where temps routinely dip to single digits.
Click to expand...
Here's a rough heat loss calculation, making a bunch of assumptions to simplify the calculation. Some assumptions are obviously wrong.

Assume walls and ceiling are 2x6 framing with 10% wood, 90% insulation. 10% R-6 + 90% R-20 = R-16 net.

Assume the floor heat loss matches the ceiling heat loss. Assume dimensions in plan are 25 ft x 25 ft, so wall area is 100 ft * 11 ft = 1100 ft^2. Floor and ceiling areas are 650 ft^2 each, for a total envelope area of 2400 ft^2.

Thus, a design temperature differential of 45 degrees F, an area of 2400 ft^2, and an R value of 16 gives a heat loss of 2400*45/16 = 6750 BTUs/hr. One BTU/hr is 0.293 watts, so the design heat loss is 2000 watts. Maybe double this for air infiltration and inaccuracy of the assumptions.

Cheers, Wayne
 
It would still be worth looking at alternate best sources.
If you assume 100% on time and power costs of $.20/kWh (not unreasonable in CA or many other states) running the heater will cost about $24 per day.
 
GoldDigger said:
It would still be worth looking at alternate best sources.
If you assume 100% on time and power costs of $.20/kWh (not unreasonable in CA or many other states) running the heater will cost about $24 per day.
Click to expand...
In NE the rate isn't all that high, and in winter months the rate actually drops for many customers. Maybe a summer rate of around $.10 but a winter rate of $.08, or in some cases first X number of kWhrs @ .08 and the next Y number of kWhrs @ .06...

The lines are loaded in summer months, in winter months they need incentive for seasonal loading just because line costs don't change even though energy use is down.

When the price of NG and propane are up electricity is usually cheaper to heat with here.
When they are in similar energy cost ranges, electric heat usually requires less maintenance cost then a gas fired system needs, not as much involved to break down in most cases.
 
The real equation takes up a sheet width of a 8.5" piece of paper.

This equation has never be presented here AFAIK.

It exists but I've never sought it out!

Yes, as mentioned there are sites that one could well use and plug in various information,
but over all you seem short on all that this true equation might require.

Sorry, I' m no help!
 
cadpoint said:
The real equation takes up a sheet width of a 8.5" piece of paper.

This equation has never be presented here AFAIK.

It exists but I've never sought it out!

Yes, as mentioned there are sites that one could well use and plug in various information,
but over all you seem short on all that this true equation might require.

Sorry, I' m no help!
Click to expand...
Instead of using a formula, you could always just solve the much shorter differential equation. :angel:
 
My equation is I have done this a few times before on similar sized garage and had no trouble keeping it even higher temperature then OP is asking to keep it.
 
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