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Thermal Conductivity
Last Post 31 Mar 2013 06:15 PM by charnathan. 2 Replies.
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dbros
New Member
Posts:1
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| 13 Feb 2013 04:25 AM |
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I'm using Q = [2*Pi*k*L*(T1-T2)]/[Ln(r2/r1)] to find heat loss from a pipe. The units for k are W/m*K, the m, meter refers to the thickness of the pipe, the direction of heat transfer. But the L in the equation is the length of the pipe, so I have two units of length, in different dimensions, that don't cancel or match up.
What do I do to get them out of there and have W at the end?
I suspect I'm over thinking this, I know it's not too difficult. Any help is greatly appreciated.
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Niel
Basic Member
Posts:193
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| 13 Feb 2013 03:29 PM |
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Where do you take the ID and the OD of the pipe into account?
Niel Leon
engineering.com |
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charnathan
New Member
Posts:4
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| 31 Mar 2013 06:15 PM |
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m is not the thickness of the pipe, but the unit of length. since you are listing the thermal conductivity in SI units, it is the meter. The thickness of the pipe is accounted for the in derivation that gives you (2*pi*L/(ln(r2/r1))).
so your answer for Q, which is actually a Q dot, will result in W, or power.
Actual numbers are irrelevant to the question. |
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