Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Thermal Physics problem: Heat transfer coefficient?

  1. Apr 13, 2005 #1
    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 [tex]T(r)[/tex] in a hollow cyllinder with length [tex]L[/tex] (where [tex]r[/tex] is the distance from the axis of the cyllinder, so that the cyllinder is symmetric about this axis), inner radius [tex]r_1[/tex], outer radius [tex]r_2[/tex], inner wall is kept at temperature [tex]T_1[/tex] and outer wall temperature is kept at [tex]T_2[/tex] (temperature of the surroundings), after equilibrium has been reached. The material of the cyllinder has heat transfer coefficient [tex]H[/tex]. Also, [tex]T_2<T_1[/tex]. (It also asks explicitly if the result depends on [tex]H[/tex].)

    Any tips or hints to get me started?
     
  2. jcsd
  3. Apr 13, 2005 #2
    Thermal Conductivity coefficient is what you're looking for. The equation to use is:
    [tex] q = -(kA)\frac{dt}{dx} [/tex]

    where

    [tex] q [/tex] is heat flow rate is watts.

    [tex] k [/tex] is thermal conductivity in [tex] \frac{W}{m \Dot K} [/tex]

    [tex] A [/tex] is cross sectional area normal to flow in [tex] m^2 [/tex]

    [tex] \frac{dt}{dx} [/tex] is the temperature gradient in [tex] \frac{K}{m} [/tex]

    Regards,

    Nenad
     
  4. Apr 13, 2005 #3
    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 [tex]A=2 \pi rL[/tex], and then using your equation: [tex]\frac{dQ}{dt}=-(HA)\frac{dT}{dr}=C[/tex] , 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 [tex]T_1[/tex] on the inner wall, right?).
    For simplicity, I then set [tex]C'=-\frac{C}{H2 \pi L}[/tex], and the resulting differential equation is:
    [tex]dT=\frac{C'}{r}dr[/tex]
    which gives:
    [tex]T(r)=C'\ln(r)+T_0[/tex]
    solving for [tex]C'[/tex] and [tex]T_0[/tex] using given initial conditions ([tex]T(r_1)=T_1[/tex] and [tex]T(r_2)=T_2[/tex]) gives:
    [tex]T(r)=\frac{T_2-T_1}{\ln(\frac{r_2}{r_1})}\ln(\frac{r}{r_1})+T_1[/tex]
    Thus [tex]T(r)[/tex] doesn't depend on [tex]H[/tex].
     
    Last edited: Apr 13, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook