Thermal Physics problem: Heat transfer coefficient?

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SUMMARY

The discussion centers on the calculation of temperature distribution T(r) in a hollow cylinder using the heat transfer coefficient H. The relevant equation for heat flow rate is q = -(kA)(dt/dx), where k is thermal conductivity, A is the cross-sectional area, and dt/dx is the temperature gradient. The user successfully derived the temperature distribution as T(r) = (T2 - T1) / ln(r2/r1) * ln(r/r1) + T1, demonstrating that T(r) does not depend on H. This conclusion is critical for understanding heat transfer in cylindrical geometries.

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  • Familiarity with heat transfer concepts, specifically the heat transfer coefficient
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rune
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I'm having a bit of trouble with a Thermal Physics exercise. I was wondering if anyone knew what the definition of the heat transfer coefficient was? I can't find anything about it in my textbook (Thermal Physics, 2nd ed., C.B.P. Finn).

More specifically, the assignment at hand says to find the temperature T(r) in a hollow cyllinder with length L (where r is the distance from the axis of the cyllinder, so that the cyllinder is symmetric about this axis), inner radius r_1, outer radius r_2, inner wall is kept at temperature T_1 and outer wall temperature is kept at T_2 (temperature of the surroundings), after equilibrium has been reached. The material of the cyllinder has heat transfer coefficient H. Also, T_2<T_1. (It also asks explicitly if the result depends on H.)

Any tips or hints to get me started?
 
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Thermal Conductivity coefficient is what you're looking for. The equation to use is:
q = -(kA)\frac{dt}{dx}

where

q is heat flow rate is watts.

k is thermal conductivity in \frac{W}{m \Dot K}

A is cross sectional area normal to flow in m^2

\frac{dt}{dx} is the temperature gradient in \frac{K}{m}

Regards,

Nenad
 
First of all, thanks for the help Nenad.
I guess I should've mentioned the text was in Norwegian, and heat transfer coefficient seemed like the most direct translation of the wording in the text. Another thing I think i mis-translated was that it said stationary conditions, not equillibrium.

I think I got it right thanks to your help though:
First setting A=2 \pi rL, and then using your equation: \frac{dQ}{dt}=-(HA)\frac{dT}{dr}=C , where I set the heat flow rate to a constant since there are stationary conditions (but not zero since that would be absurd as long as we are using energy to maintain T_1 on the inner wall, right?).
For simplicity, I then set C'=-\frac{C}{H2 \pi L}, and the resulting differential equation is:
dT=\frac{C'}{r}dr
which gives:
T(r)=C'\ln(r)+T_0
solving for C' and T_0 using given initial conditions (T(r_1)=T_1 and T(r_2)=T_2) gives:
T(r)=\frac{T_2-T_1}{\ln(\frac{r_2}{r_1})}\ln(\frac{r}{r_1})+T_1
Thus T(r) doesn't depend on H.
 
Last edited:

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