How to integrate Joule's First Law (Joule Heating)?

In summary, the two methods of solving the problem (integrating or finding Q as a function of y) give the same answer, depending on the shape of the current distribution.
  • #1
BucketOfFish
60
1

Homework Statement



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).

Homework Equations



Q=I2R

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 into Q=I2R. 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?
 
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  • #2
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 4-times 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
Wow, that makes a lot of sense! Thanks for your help!
 

1. What is Joule's First Law?

Joule's First Law, also known as Joule Heating, states that when an electric current passes through a conductor, heat is generated in the conductor at a rate proportional to the square of the current and the resistance of the conductor.

2. How is Joule's First Law related to energy conservation?

Joule's First Law is a manifestation of the Law of Conservation of Energy, which states that energy cannot be created or destroyed, only transformed from one form to another. In the case of Joule Heating, the electrical energy is transformed into heat energy.

3. What factors affect Joule Heating?

Joule Heating is affected by the amount of current flowing through the conductor, the resistance of the conductor, and the time duration of the current flow. In addition, the type of material and its temperature can also impact the amount of heat generated.

4. How is Joule Heating useful in practical applications?

Joule Heating has many practical applications, including electric heating devices such as toasters, hair dryers, and electric stoves. It is also used in industrial processes such as welding and melting metals. In addition, Joule Heating is used in scientific experiments to measure the electrical resistance of materials.

5. How can Joule Heating be minimized or controlled?

To minimize or control Joule Heating, the current flow through the conductor can be reduced, the resistance of the conductor can be decreased, or the time duration of the current flow can be shortened. Additionally, using materials with lower resistivity or increasing the surface area of the conductor can also help to reduce Joule Heating.

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