# Need help calculating (Estimating?) the heat transfer coefficient or r-value based on differences in temperature

Hello,
I've bought an infrared thermometer to help find areas in my house that are letting too much heat through. I was wondering if it would be
possible to estimate the heat transfer coefficient for any material based only on measures of the four following variables: Exterior temperature (outside ambient) Exterior surface temperature (surface temperature of window, door, wall, etc from the outside) Interior temperature (outside ambient) Interior surface temperature (surface temperature of window, door, wall, etc from the inside) I've searched the internet in vain. Does anyone know if this is possible?
Thanks!
Garrett Cook
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Hello,
I've bought an infrared thermometer to help find areas in my house that are letting too much heat through. I was wondering if it would be possible to estimate the heat transfer coefficient for any material based only on measures of the four following variables: Exterior temperature (outside ambient) Exterior surface temperature (surface temperature of window, door, wall, etc from the outside) Interior temperature (outside ambient) Interior surface temperature (surface temperature of window, door, wall, etc from the inside) I've searched the internet in vain. Does anyone know if this is possible?
Thanks!
Garrett Cook
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Hello,
I've bought an infrared thermometer to help find areas in my house that are letting too much heat through. I was wondering if it would be possible to estimate the heat transfer coefficient for any material based only on measures of the four following variables: Exterior temperature (outside ambient) Exterior surface temperature (surface temperature of window, door, wall, etc from the outside) Interior temperature (outside ambient) Interior surface temperature (surface temperature of window, door, wall, etc from the inside) I've searched the internet in vain. Does anyone know if this is possible?
Thanks!
Garrett Cook
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Dear g.cook.a:

Take your heating bill. Estimate the amount of energy lost on an average day. U = Q' / /\T Q' = rate of heat loss to ambient /\T = temperature difference from inside air to outside air. U = overall heat transfer coefficient
The other numbers you measure will partition the heat transfer coefficient from the room to the wall (forced and natural convection), through the wall (conduction and convection if not solid or is not filled with insulation), and from the wall to the outside air (forced and natural convection again).
1/U = 1/u_inner + 1/u_wall + 1/u_outer And you can write a "U = Q' / /\T" for each segment of the heat transfer path as well.
What kind of heating do you have? What kind of landscaping do you have? How much insulation is in your attic?
David A. Smith
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| Hello, | | I've bought an infrared thermometer to help find areas in my house that | are letting too much heat through.
Very sensible.
| I was wondering if it would be | possible to estimate the heat transfer coefficient for any material | based only on measures of the four following variables: | Exterior temperature (outside ambient) | Exterior surface temperature (surface temperature of window, door, | wall, etc from the outside) | Interior temperature (outside ambient) | Interior surface temperature (surface temperature of window, door, | wall, etc from the inside) | I've searched the internet in vain. Does anyone know if this is | possible? | | Thanks! | | Garrett Cook
Well, let's face it. Even if you found a window with outside surface temperature 32 degrees F and a nice comfortable inside surface temperature of 68 degrees F, are you going to replace the window? I wouldn't. Maybe you were looking for this: http://hyperphysics.phy-astr.gsu.edu/hbase/tables/thrcn.html

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On 10 Dec 2006 17:12:19 -0800, snipped-for-privacy@gmail.com wrote:

The thermal resistance values associated with building materials are calculated from the following factors: 1) temperature on one surface 2) temperature on opposite surface 3) area of surface 4) thickness of material between two faces under consideration.
and here's the kicker.... 5) the amount of power transmitted between the two surfaces
In shorthand form, 5) = 3) X [ 2) - 1) ] / 4) times a constant for a particular material.
You can easily see you are missing information about how much power is transmitted. To measure insulation, it helps to put a heater of known power on one surface, then see what happens on the other surface.
Brian Whatcott Altus OK
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Brian Whatcott wrote:

Thanks for your all's responses. The problem I've got is that I don't really know what the materials are and I don't have any type of heating equipment that I can really use to test it. I didn't know if, based on the difference in surface temperatures and ambient temperatures (and maybe the individual differences between those), one could estimate thermal conductivity or something like that for the material. Something like this seems like it might work: dTs/dTa someheatconstant, where dTs is the difference in surface temperatures and dTa is the difference in ambient temperatures. Is such an equation valid?
Garrett
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On 11 Dec 2006 18:04:29 -0800, snipped-for-privacy@gmx.net wrote:

Your efforts towards an analytical method which suits the data you mentioned are (in my view) doomed by the inadequate data set.
But here is a practical, practical way out of your difficulty.
Take a panel of KNOWN thermal conductivity. Place the panel at a representative spot in the house. It might be in a door or window aperture (You would blank off the surrounding aperture, naturally.)
Ensure that the internal temperature is pretty uniform as between your benchmark panel and the surrounding structures. With a fan?
Then, given the internal surface temperature is constant between the known panel and the unknown structure's interior surfaces, the external temperatures of the various elements WILL give you an analytical basis for estimating structural thermal conductivities.
Good luck
Brian Whatcott Altus OK