Sizing electric heater for garage

probably cant be done,,,,,,,,,,,,

how big is the amp capacity of the service?

you will probably be better off with a oil or propane heater

Anyone who has the numbers is going to need to know how cold it is outside-- is it likely to get to 20 below or is 32 the coldest it gets outside?

How much insulation and how many & what type of windows will also be needed if this is even doable. Hey- while you're at it- are those doors new, insulated doors, or 50 yr old aluminum shells?

All that said- I think you're going to want something in a kerosene, gas or oil heater. My WAG would be something in the 50K BTU range- that would be about 15000 watts- [62 amps of 240 if my memory and math serve me]

Electricity has its advantages, but IMO, heating a space up in a hurry isn't one of them. [though I will say I've never seen an electric furnace in action- if you're in the TN Valley that might be worth looking into]

Jim

The question is "How many Btus PER HOUR do I need?" :-)

Nick

I use a"monitor" k-1 heater 40000 btu,my space is 22x32 w/8ft ceilings,1 garage door this is a direct vent heater.they are also made to use lp/propane gas. my building is 2x4 construction as well insulated as possible cost for season runs average. 80 gallons heat is on at 46 degrees constant with time where i push to 64/66 degrees for hours or days at a time I'm very happy with the system

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This:

Thermal mass C = 0.5Btu/F/ft^2x2304ft^2 = 1152 Btu/F for 2304 ft^2 of 1/2" sheetrock.

At least (60F-32F)x1152Btu/F = 32,256, if we could raise the temperature instantly, with no loss of heat to the outdoors.

With (say) R30 insulation, we have something like this, in a fixed font:

--- R30/2304ft^2 = 0.013 F-h/Btu |--|-->|------------------www--------------- 32 F --- | I Btu/h | --- 1152 Btu/F --- | | -

This is equivalent to:

RC = 0.013x1152 = 15 hours and 60 = Tt + (32-Tt)e^(-1/RC) make I = 33,397 Btu/h, ie 9.8 kW.

Nick

Then again, if we insulate the INSIDE of the garage walls with (say) R9.2 1" foil-polyiso board, we have something like this:

T --- | R9.2/2304ft^2 = .004 0.013 |--|-->|-------------www---------------www--- 32 F --- | I Btu/h |Td --- 1152 Btu/F --- | | -

with the same equivalent circuit above. If T = 60 = Td + 0.004I and Td = Tt+(32-Tt)e^(-1/RC) = 32+0.000838I, I = 8856 Btu/h, ie 2.6 kW. After

OTOH, with no drywall, just an empty garage with R9.2 foamboard inside R30 insulation, we'd have something like 0.075lb/ft^3x24x24x12 = 518 pounds of air with C = 518x0.24Btu/F-lb = 124 Btu/F, like this:

RC = 0.017x124 = 2.1 hours and 60 = Tt + (32-Tt)e^(-1/RC) make I = (60-32)/(0.017(1-e^(1/2.1))) = 4372 Btu/h, ie 1.3 kW :-)

Nick

Easiest way is just to run the car for a while... sorry, couldn't resist. Find it hard to believe you want to do this electrically, unless, as another suggested, you have super cheap hydro power. Might want to put foamboard on the doors, improve seals if warranted.

Yikes!!!, Thanks Nick, I think. Well, you're way over my head. I'll go with your higher figure and recommend something in the 10 KW range. If the guy really wants to do this electrically, I want to be reasonably sure it'll be sufficient