Heat conduction in hollow cilinder.

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Homework Help Overview

The discussion revolves around the heat conduction in a hollow cylinder characterized by its inner and outer radii, thermal conductivity, length, and temperature conditions at both ends. Participants are tasked with determining the temperature gradient and heat flow in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between heat flow and changing area in the cylinder, questioning how to mathematically express the temperature gradient given the varying cross-sectional area. There is discussion about using an intermediate temperature and equating powers across different radii. Others suggest integrating the heat flux equation to find the temperature profile, while some express confusion about the integration results and the implications of the temperature gradient.

Discussion Status

The discussion is active, with participants sharing different approaches to the problem. Some have provided guidance on integrating the heat flux equation, while others are questioning the correctness of their mathematical manipulations. There is no explicit consensus yet, but several productive lines of inquiry are being explored.

Contextual Notes

Participants are navigating the complexities of the problem, including the assumption of a linear temperature profile and the implications of the integration results. There is acknowledgment of potential oversights in the mathematical approach, particularly concerning the units involved in logarithmic expressions.

aranud
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Homework Statement


Given is a hollow cilinder with inner radius R1 and outer radius R2. The heat conductivity of the material is k. The cilinder has length l and an inner temperature of T1 and outer temperature T2. Determine the temperature gradient in the cilinder and the heat flow that leaks away radially.

Homework Equations


Power = k*Area(Temperatureinside-Temperatureoutside)/(thickness)
Area = 2*pi*R*l

The Attempt at a Solution


The reason I can't solve this problem is the changing area when going from inside to outside the cilinder. Obviously the heat flow going from the inside to the outside should be minus the heat flow from outside to inside, since the areas are different I think the only way this is possible if the temperature gradient falls off quicker going inside causing the heat flow to be the equal despite the smaller area.
So much for intuition though since I've tried fruitlessly to write this down mathematically.
I also think the trick is to use an intermediate temperature T and equate the powers going through an area between R2 and R1 since we've used that one before.
 
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aranud said:
Power = k*Area(Temperatureinside-Temperatureoutside)/(thickness)

This equation assumes that the temperature profile is linear, which isn't the case here. A more general equation is \mathrm{Heat~flux}=kA(x)(dT/dx), or since you're working with radius r, \mathrm{Heat~flux}=kA(r)(dT/dr). You should be able to find T(r) through integration.
 
When I integrate the formula you've given me I get T = (heat flux)/(k*2pi*l)*ln(R)
To find the heat flux should i simply fill in T1 and R1? I feel like I've missed something since the problem seems a bit simple this way.
 
aranud said:
When I integrate the formula you've given me I get T = (heat flux)/(k*2pi*l)*ln(R)
To find the heat flux should i simply fill in T1 and R1? I feel like I've missed something since the problem seems a bit simple this way.

Check your integration. A lone ln(R) can't be correct; you can't take the logarithm of a parameter with units.
 
You're right, is this better? T = T1+ heatflux/(k*2*pi*l)*ln(r/R1)
It seems to describe the right heat gradient but how do I get the heatflux from this?
 
You know the temperature at the outer radius too.
 
Ah yes of course:blushing: My brain is all but melted right now because of all the studying I guess. Thanks for the help.
 

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