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Heat transfer in a cylindrical tube |
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| Jul15-12, 11:03 AM | #1 |
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Heat transfer in a cylindrical tube
1. The problem statement, all variables and given/known data
I don't really understand something in my textbook. It says the heat Resistor (how to say that in English? I hope it's alright like this...) trough a cylindrical tube (from the inside to the outside or revers) would be: R(heatrisistor)=1/2∏Lλ * ln (R1/R2) And then my textbook gives the following picture (well oke a but nicer, but that's what I make of it) But if I were to follow this I get a negative R every time (since R1 is smaller than R2 and so the ln gets negative right?), which I find very strange... So I searched trough other textbooks that I have and they had no information about it and searched trough the internet (I'm not very good at that ^^') and found nothing I can use. So my question is: Can someone explain to me, nice and simple, what I should do with the R1 and the R2? (I probably am misunderstanding my textbook... I'm sure of it ^^') Well thanks for reading all of this anyway, hope someone can help me with this. :) [ 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution |
| Jul15-12, 11:33 AM | #2 |
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I can't see your diagram but I think there is a mistake. R1 and R2 should be the other way around. See...
http://wiki.answers.com/Q/Formula_fo...loss_in_a_pipe Q= 2 pi k L(T1-T2)/ln(r2/r1) where k=conductivity of material of which pipe is made T1= temperature of inside layer pipe T2=temperature outside layer of pipe L= length of pipe r1= inner radius r2=outer radius ln=natural logrithem |
| Jul15-12, 12:13 PM | #3 |
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thank you so much! I can defiantly do something with that =D =D =D. I can be so stupid sometimes ^^'
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