Solving a Short Heat Question with q = Q/A - Area Calculation Help

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The discussion focuses on solving a heat transfer problem using the equation q = Q/A, where the main challenge is calculating the area of a cylinder. Participants clarify that the surface area of a cylinder, excluding the ends, is given by the formula 2πrh. The conversation emphasizes the need to calculate both the surface area and the total power output based on the cylinder's dimensions, specifically its radius and length. A participant expresses initial confusion but ultimately finds clarity in the calculations. The thread concludes with a resolution to the problem at hand.
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Homework Statement



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Homework Equations



q = Q/A where q = heat flux, Q = rate of heat transfer, A = area

The Attempt at a Solution



Well I guess my main problem is figuring out the area. I know the answer to the question, but I can't seem to calculate it right.

Thanks everyone
 
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What is the total power output (in W), letting r be the radius and L be the length of the cylinder? What is the surface area of a cylinder, ignoring the ends (which are negligible for a long cylinder)?
 
Mapes said:
What is the total power output (in W), letting r be the radius and L be the length of the cylinder? What is the surface area of a cylinder, ignoring the ends (which are negligible for a long cylinder)?

I don't know, those are the only variables given in the question.
 
Studious_stud said:
I don't know, those are the only variables given in the question.

Yes, I'm asking you to calculate both quantities as part of solving the problem.
 
Mapes said:
Yes, I'm asking you to calculate both quantities as part of solving the problem.

Well, I know the surface area is 2pi rh. However, I don't think I've ever seen a formula relating power to radius or length..
 
You're given the power output per volume, yes? And the volume of the cylinder is easily calculated.
 
Mapes said:
You're given the power output per volume, yes? And the volume of the cylinder is easily calculated.

Ahh, I got it now. Thanks!
 
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