# Homework Help: Heat Transfer

1. Feb 12, 2017

### encorelui2

1. The problem statement, all variables and given/known data

See attached figure. Ta>Tb. Show that the rate of energy conducted dQ/dt is 2*pi*L*k((Ta-Tb)/Ln(b/a))
2. Relevant equations

3. The attempt at a solution
I seem to be lost at deriving an equation for the medium area, A.
I understand the tansfer from low temp area to high. Pcond. is proportional to change in temp from H to L temp areas. I have attached my attempt at a soln.

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Last edited: Feb 12, 2017
2. Feb 12, 2017

### TSny

This forum requires you to show some work towards a solution before we can help you.

What have you learned about heat conduction that you think might be relevant? Do you know how to calculate the rate of heat flow through a thin slab of material? Do you know any formulas that might be helpful?

3. Feb 12, 2017

### TSny

Which parts of your notes shown below apply specifically to this problem?

4. Feb 13, 2017

### Staff: Mentor

If the radial temperature gradient is dT/dr, what is the rate of heat flow through the cylindrical surface at radius r (inside the conductor)?

5. Feb 13, 2017

### Staff: Mentor

@encorelui2 : An image of class notes does not comprise a solution attempt, particularly an image that is essentially illegible due to poor image quality and small handwriting (your personal handwriting may be clear as day to you, but it's close to a secret code to others when it's tiny and fuzzy and low contrast). Please provide an acceptable solution attempt or at least describe what approaches you've tried. If you use an image, be sure number every equation on the page so that helpers can refer to them in their responses.

If your images are not clear enough to read then you need to type in your attempt. Text-formatted math can be rendered using the icon tools in the edit panel header or using LaTeX syntax. It's much easier for helpers to read, quote, and comment on typed-in content and experience shows that you'll more, and more timely help responses that way.

Note that without a valid solution attempt your thread is in danger of being removed. See:
Hey! I posted here but now it's gone!