1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Thermal Physics problem: Heat transfer coefficient?
Loading...