# Sizing electric heater for garage

Don't know if there is a formula for this and I'm not looking for a typical heat loss calculation. The scenario is this: Garage is 24' X 24' with 12' ceiling. There is sheetrock and insulation in the ceiling and three walls. The garage has two standard overhead doors. Temperature is aprox 32 degrees. I want to raise the temperature to 60 degrees in about one hour. How many BTU's do I need?
Thanks for any help, Roy
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
I'm on the same wavelength as you, it seems impractical at best. The building is fairly new, and well insulated, no windows, with foam filled aluminum doors. The outdoor temp is 32 degrees.
wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

The question is "How many Btus PER HOUR do I need?" :-)
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
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
edward pearl painting in maine
wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
RBM wrote:

This:
http://www.northerntool.com/webapp/wcs/stores/servlet/product_6970_200316365_200316365
NG and LP versions, and it's a proper vented unit so it doesn't release combustion gasses and moisture into the heated space.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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:
0.013 -----------www---------- 32->60 F | | | Tt = 32+0.013I | --- --- 1152 Btu/F - --- | | | | - -
RC = 0.013x1152 = 15 hours and 60 = Tt + (32-Tt)e^(-1/RC) make I = 33,397 Btu/h, ie 9.8 kW.
Nick
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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 1 hour, the air is 60 F and the drywall behind the insulation is 39 F.
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:
0.017 ------------------------www---- 32 F | | | Tt = 32+0.017I | --- --- 124 Btu/F - --- | | | | - -
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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
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
wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

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.