How to integrate Joule's First Law (Joule Heating)?by BucketOfFish Tags: current, heat, integration, joule heating 

#1
Jul2811, 12:34 PM

P: 50

1. The problem statement, all variables and given/known data
I know the current distribution through a rectangular cross section of metal. It varies only with respect to y. I need to find a heating function for the metal. (This is only the relevant part of a larger problem). 2. Relevant equations Q=I^{2}R 3. The attempt at a solution I attempted this problem twice and got two answers, depending on how I integrated. First I integrated current over the entire surface and plugged that in to Q=I^{2}R. On my second try, I found Q as a function of y and integrated that over the surface. The answers didn't match. Consider this: If a block with uniform current 4A is evaluated as a whole, the Joule heating equation gives Q=16R. However, if it were divided up into four pieces, each with 1A current, Joule heating would give Q=1R for each piece, for a total of Q=4R. So my question is: if I know the current through the material, how can I integrate in a way that makes sense? 



#2
Jul2811, 01:50 PM

P: 6

Hi,
In your last example, you're forgetting that if you slice the current into 4 individual currents, the cross section that each current is flowing through is 1/4 the total cross section which means the resistance for each current is 4times the total resistance. In the end you'll get the same Q. To answer your question, your second method (integrating Q) is correct. I'll go through this formally so that it's clear. Let's call the current distribution (current/area) = J. Let's also imagine slicing the cross section of the wire into little vertical slices of height, dy. The tiny amount of current flowing through each slice is therefore: J*W*dy where W is the width of the cross section. The resistance of each slice is given by: p*L/(W*dy) where p is the wire's resistivity and L is the wire's length. The tiny amount of heat generated, dQ, in our slice is given by using the Q=I^2*R equation and applying our expressions from the previous two steps. After a little algebra, you get. dQ = J^2*p*L*W*dy But p*L*W = R so in the end we find that the heat generated by our little slice is: dQ=J^2*R*dy The total heat is found by integrating... Q = Int[dQ]=Int[J^2*R*dy] I hope this helps. 



#3
Jul2811, 03:40 PM

P: 50

Wow, that makes a lot of sense! Thanks for your help!



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