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All the examples I've been looking at assume you know the temperature on the outside of the pipe and are being used to calculate intermediate temperatures, i.e., at a pipe/insulation interface.

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- Thread starter bill nye scienceguy!
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- #1

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All the examples I've been looking at assume you know the temperature on the outside of the pipe and are being used to calculate intermediate temperatures, i.e., at a pipe/insulation interface.

- #2

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Ive been thinking of the general enthusiasm concerning ground heat. So I wanted to understand how much energy transfer one can expect in the ideal case.

Looking at the case with a axial symmetry I was only able to find one formula, developed by Ingersoll:

q = L (T_1 - T_2) * k

Where L is the length of line which is kept at T_2, situated in the ground, with heat T_1.. k is thermal conductivity.

This would mean that if a ground heat line is 100m, kept at 1*C, and T_1 is 11*C of a marble "infinite" reservoir with k=2w/mK, the q would be 2000Watts..

But this doesnt make any sense. So I also would like to see any further explanation.

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