Heat Loss question


Old 03-01-19, 06:30 AM
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Join Date: Mar 2019
Location: United States
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Heat Loss question

I'm new to this. I am wondering and don't know the answer. Could someone explain to me:

If the R-Value, the temperature differential, the time, and the area are all the same for both a vertical wall and a horizontal flat roof is the heat loss thru each of them the same?

I've always been told that heat is usually lost faster thru the surfaces above us versus the horizontal surface next to us, and even less for the surface (floor) under us.
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Old 03-01-19, 09:22 AM
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A vertical wall and a ceiling will almost never truly see the same temperatures. Heat rises so unless you have a decent, continuous airflow the ceiling will be warmer that the bottom of a wall. In my living room with 12ft ceiling height it is noticeably warmer when I go up on a ladder to change a light bulb.
Old 03-01-19, 09:22 AM
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I have two answers to help explain the differences.

Yes, The heat loss is the same through both horizontal and vertical surfaces when you only measure the 'radiated' heat loss. This is because heat radiates in all directions equally and is not affected by the air it passes through.

Yes, The heat loss is greater thru the ceiling, when compared to the walls and floors as you measure the 'convected (convection) heat loss in a room. This is because warm air rises toward the ceiling due to convection properties of air.

The essential difference between the two, is which method you are using to measure heat loss: radiation or convection. There are no practical applications for radiation effects in home construction.

Interestingly, infra red space heaters can heat a nearby wall to the point of catching fire, while convection space heaters will more quickly heat a entire room. That is as good an example as I can give.
Old 03-04-19, 07:36 AM
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Join Date: Mar 2019
Location: United States
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guys - thanks for the replies. much appreciated.

followup question: is the equation for convective heat loss Rate of Heat Loss (BTU/hr) = [Area of surface*temp difference between the two surfaces (inside and outside)*time] / R-Value

If it is, can you tell me the equation for the radiation heat loss?

Old 03-04-19, 09:39 AM
Join Date: Oct 2008
Location: New England
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I think your equation for conductive heat loss is good but trying to calculate radiant heat loss would be a challenge. Understanding that it is part of the process is probably the best you can do, but actual btus/hr would have too many variables.

The one detail about radiant heat loss I keep in mind is the rate is a function of the temperature difference raised to the 4th power. That 4th power is enormous and emphasizes how much radiation can contribute.


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