# Homework Help: Thermal Physics problem: Heat transfer coefficient?

1. Apr 13, 2005

### rune

I'm having a bit of trouble with a Thermal Physics excercise. 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?

2. Apr 13, 2005

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,

3. Apr 13, 2005

### rune

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: Apr 13, 2005