Calculating BTU for Heating a Garage from 40 to 60°F in 4 Hours

In summary: So, the heat transfer through each wall in steady state is 1/R*A*dT. I'm not clear on the floor - is it insulated as well, under the radiant heating?The floor won't matter too much since you aren't going to make the garage much warmer than the ground under the floor.
  • #1
lurch85
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0
I have a 40x60 by 14 foot garage... 3 walls are R19... one wall maybe about a R10 (garage doors, somewhat well sealed) ceiling is R30...

(how would I calculate in BTU for raising the temperature from 40 degrees F to 60 degrees F in say four hours with an outdoor temperature of 20 degrees with a humidity of 60%?
It's been a long time since HS physics, and any abbreviations you use, please explain label (delta equals change, etc)(online calculators that I have found only figure in for it to be in a constant on/off cycle like a usual house, not a garage that I only want to heat once a week.)

Feel free to guesstimate any variables I may have missed

right now, with a underfloor heating working 15 minutes a day to keep the pipes from freezing, it keeps the garage at a pretty constant 40 degrees when it is 20 degrees out.

Feel free to just explain given a sealed box with no heat loss .. I am having a heck of a time trying to figure out any of this.. .. I am just trying to figure out how to go about this... any help would be appreciated.
 
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  • #2
A thermostat, some kind of simple heater, and a cheap timer?
 
  • #3
im trying to figure out how much more heat output i need to accomplish this feat.. I have underfloor heating in 1/3 of it, and it is a very slow process to heat the garage... (12 hours having it on went from 40 to 45 degrees)

trying to figure out how much more heat i need to accomplish this feat.
 
  • #4
I don't like using the salamander (oil fired unvented heater) in there..

Want to use hot water radiators with blowers
 
  • #5
It's pretty difficult to calculate the transient case because you need to know the thermal inertia of the space; how much thermal energy the structure and everything in it can hold. But 4 hours is a pretty long time to figure as a warm-up time, so once you calculate the steady state load and add a safety factor (say, 50%), you should be able to achieve what you are trying to get.

So, the heat transfer through each wall in steady state is 1/R*A*dT. I'm not clear on the floor - is it insulated as well, under the radiant heating? I suppose the floor won't matter too much since you aren't going to make the garage much warmer than the ground under the floor.
 
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  • #6
ummm... I took a simple physics class back in like 9th grade...

1/(radiancy?) times area times change in temperature?
(is trying to study radiancy, but honestly am not getting it)

I haven't even been able to figure out the heat required even if there is no heat loss at all...
 
  • #7
Russ had the right idea, assuming his 50% safety factor is in the ballpark for the effect of the thermal inertial of the walls, floor, and air. Since the current steady state operating point of your system is already established, it is simpler just to look at the incremental changes relative to the present steady state operation. You are trying to raise the inside temperature 20 F, with no change in the outside temperature. During the heat-up, the incremental temperature difference for heat loss is going to rise from 0 F to 20 F. So, as a worst case, you can just assume that the temperature difference is 20 F the entire time. This will give you an upper bound on the heat supply required. So, for each of the walls, you use a 20 F incremental temperature difference across the wall, and calculate the incremental heat loss over 4 hours using Russ's equation (20)A(3600 sec x 4)/R, where A is the area of the wall.

In lieu of Russ's 50% factor, now that you've established that the temperature difference across each wall is going to be 20 F higher, you can calculate the additional heat required by thermal inertia of the walls using the heat capacity of each wall, the mass of each wall, and an average temperature rise of 10 F. You can also do the same thing for the room air (although this probably won't be too significant).

Chet
 
  • #8
lurch85 said:
ummm... I took a simple physics class back in like 9th grade...

1/(radiancy?) times area times change in temperature?
(is trying to study radiancy, but honestly am not getting it)

I haven't even been able to figure out the heat required even if there is no heat loss at all...
That was 1/R-Value. You have all of the information in your first post, but I want you to take a swing at it. :wink:
 
  • #9
Chestermiller said:
Russ had the right idea, assuming his 50% safety factor is in the ballpark for the effect of the thermal inertial of the walls, floor, and air.
FYI, I actually picked it because of a lack of confidence in the inputs.
 
  • #10
lurch85 said:
...a garage that I only want to heat once a week...

This simplifies things greatly. How long do you plan on staying in the garage, for this one day of the week?
 
  • #11
A bit of experimenting could help. Would this be possible?
There really is no substitute for a situation like this as it's probably not similar enough to domestic heating situations for the tables and calculators to be reliable.
 
  • #12
Plan to stay in the garage at least half a day if i go through the trouble of heating it. I placed a vertical air modine style heater in... (a hot water (180 degree)radiator with a fan on it, that I have a hard time getting information on to figure out what kind of output it should be. (Dunham-Bush 0-300-GA) (300,000 btu?, but that designation is probably on steam, not 180 degree water, so maybe an output of 200,000?)
 
  • #13
also have 1000 sq feet of underfloor heating in it, but even with 24 hours of use, is difficult to get temp where I would like. I ran both the underfloor and the modine for 12 hours, with a 20 degree outside temperature, garage started at 40 degrees (the floor is on a timer for 15 minutes a day to keep it warm enough under the floor so the lines don't freeze)... and the best I had with both systems is 50 degrees. air inlet temperature of the modine was 40 degrees and outlet 55, (i don't know how many CFM, the motor is different, and like i said, i can't find information on it)

another day, used it for 12 hours, outdoor temp 40 degrees, starting temperature was 40 degrees and I got it up to 55 in 12 hours
 
  • #14
Doesn't that heater have a nameplate with a make and model number on it?

Did you do the calculation I suggested for the load?
 
  • #15
sorry.. been working the last few days.

using an outside temp of 20 and an intended temp of 60 degrees (40 degree diference)
garage is 40 x60 x14ft ... so the walls together ((40x2) + (60x2)) x14 x (1/19) x 40 degrees + (40x60 x (1/30) x40 degrees Ceiling
7073 + 3200
what do i do for the ceiling and the air itself?

the model is a dunham-bush 0-300-GA but i haven't been able to find any information at all!
 
  • #16
(i used r19 for the wall with the garage doors too for this calculation
 
  • #17
lurch85 said:
(i used r19 for the wall with the garage doors too for this calculation
How is the floor heated?
The floor heater numbers look kind of goofy:
900 seconds of operation yields an average temperature of 40°F over 24 hours
42300 seconds of operation yields an average temperature 45°F over 12 hours​
 
  • #18
I think I found the reason for all of my confusion about R... I am referring to the R value of the insulation itself, for instance, around here, northern Michigan, expected R values 4 walls are part 19 - 25, and ceiling. I'm thinking you are talking about. Only can access my phone at this point, so can't do much more research tonight
 
  • #19
The floor in the north third of the floor (13 foot by 60) has a grid of 1/2 inch pex aproximately 9 inches apart running 180 degree water through it. (not in it)..(it was a like 20 years ago, so i don't remember the spacing exatly)
 
  • #20
lurch85 said:
The floor in the north third of the floor (13 foot by 60) has a grid of 1/2 inch pex aproximately 9 inches apart running 180 degree water through it. (not in it)..(it was a like 20 years ago, so i don't remember the spacing exatly)
Do you know the fluid capacity of the tank, and its wattage rating?
 
  • #21
For a room that's 60 x40= 2400 square feet (length x width) in a cold climate they estimate 50 - 60 btu per foot. So a rough estimate from a hvac contractor would be 55 x 2400 = 132,000 btu. but that's not counting the ambient heat coming from the floor which keeps the room at a constant 40, this would be considered a mild climate. so the btus needed would be half. So roughly 66,000 btus. Other formulas calculate it by floor square feet 60 x 40 = 2400 and for a mild climate at 30 - 40 btu per foot which would be 84,000 btus. keep in mind these are figures for living spaces not storage spaces and these figures are also what's "recommended". So are most likely inflated for what you'd actually need. And these are for modern heating units.

addendum :
These are btu/ hours
 
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  • #22
Outdoor wood boiler Taylor the 750 have 165000 to 550000 btu max with 600 gallons of water in the stove itself
 
  • #23
With those units and the way they are sized... what would the usual heat time be? I know they figure for it to be on constantly... not intermittent quick heatings...
 
  • #24
lurch85 said:
using an outside temp of 20 and an intended temp of 60 degrees (40 degree diference)
garage is 40 x60 x14ft ... so the walls together ((40x2) + (60x2)) x14 x (1/19) x 40 degrees + (40x60 x (1/30) x40 degrees Ceiling
7073 + 3200
I'm getting 5895+3200=9095 BTU/hr (2.7 kW)

That seems low. The insulation values may be overly optimistic.

What about windows? Doors? You said the garage doors are insulated at R19 - that seems unlikely.
what do i do for the ceiling and the air itself?
This is a steady-state calculation. Those temperatures aren't changing so they have no impact on the result.

However, depending on how "tight" the building is, you may want to figure on air infiltration from outside -- up to half an air change per hour.
the model is a dunham-bush 0-300-GA but i haven't been able to find any information at all!
Yeah, I can't find any information on it either.
 
  • #25
R 7-10 is probably more likely... 1 1/2 inch of low grade foam
 
  • #26
lurch85 said:
Outdoor wood boiler Taylor the 750 have 165000 to 550000 btu max with 600 gallons of water in the stove itself
So you have an outdoor, wood fed stove, and the garage heat is on a timer? This problem gets more interesting every minute!
Taylor Outside Wood Stoves
...
T 750 H

165,000 BTU/hr. Max. 550,000 BTU/hr.
12--24 hr. burn time
600 gal. of water
heats 4,000 sq. ft.
...

Code:
550,000 btu/hr * 0.293 watts         161,000 watts
                 -----------   = 
                 1 btu/hr

161 kw source. :bugeye:
wow
You put my 5 kw home heat sources to shame.
 
  • #27
If you have (165000 btu input) 80% efficiency you are out putting 132000, according to what I have read you need 165000 / 0.8 = ( 206250 btu input) for 80% efficiency. So if you have a 60x40= 2400 sqft space needs 132,000 btu/h output, you need 13200 / 0.8 = 165000 btu/h . As far as time needed that will depend on a lot of different factors every thing that's cold in the space will "suck" up some of the heat do to thermodynamic equilibrium.

http://www.wikihow.com/Calculate-BTU-Per-Square-Foot

https://en.wikipedia.org/wiki/Thermodynamic_equilibrium
 
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  • #28
The only reason I have it on a timer, is to make sure the pipes don't freeze. I have a switch hooked up so I can either run it on a thermostat, or on the timer to keep it from freezing
 
  • #29
lurch85 said:
The only reason I have it on a timer, is to make sure the pipes don't freeze. I have a switch hooked up so I can either run it on a thermostat, or on the timer to keep it from freezing

I'm now more curious as to how often you have to throw wood into your stove.

ps. On a trivial side-note, I determined that a 100% efficient thermal exchange unit, with your wood stove running at 100%, would heat the air in your garage to 60°F in 79 seconds.
I know you asked for "4 hours", but there are still numerous unknown variables.
I will try and determine the thickness of your concrete floor at a later time, as both I, and my cats, are out of food, and I need to make a run to the store.
 
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  • #30
It also heats the house
 
  • #31
As to the wattage, heat only gets to the garage whatever can be transferred through a one-inch text line at 22gpm
 
  • #32
By the way, the wood box is 32 inches x 32 inches x 36 inches. You can get as big of a piece of wood in there as anybody could ever lift
 
  • #33
Question for lurch85, just to get a better picture, is there a radiator/heat exchanger in the garage? Or are you totally relying on heat transfer from the pipes running under the floor? 165000 btu input would be more then adequate to heat that space if the source was direct (ie above the floor). Please paint a detailed picture of what you are working with.
 
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  • #34
The north 3rd of the concrete has half inch X running under the floor at what I think is 9 inches apart. I also have a forced air round radiator with a blower in the in the center. I have not been able to find out the heat output. I am probably going to take another heat exchanger/ plenum and put a blower behind it and make my own. What I'm trying to figure out, is if I should have one big one, or two mediums, or some sort of an idea on how many BTUs I need before I go buying anything. A heat exchanger that should be able to put off 200000 BTU and put a blower behind it, but if I would need more than that, I may be better off putting into smaller ones into 2 corners.
 
  • #35
The short answer is 200000 btu should be plenty. 60x40=2400x55= (132000 btu/h output) 132000 / 0.8= 165000 btu/h input.
 

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