RBM rbm2(remove wrote:

... Garage is 24' X 24' with 12' ceiling. There is sheetrock and insulation
in the ceiling and three walls...

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.

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