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Homework Help: Solve the heat equation

  1. Dec 8, 2017 #1
    1. The problem statement, all variables and given/known data
    show ## \rho c_m \frac{\partial T}{\partial t} = \kappa \frac{\partial^2T}{\partial x^2} -\frac{2}{a}R(T)##
    where ##R(T)=A(T-T_0) ##

    a) Obtain an expression for T as a function of x for the case of an infinitely long rod whose hot end has temperature ##T_m##
    b) Show that the heat that can be transported away by a long rod of radius a is proportional to ##a^{\frac{3}{2}}##, provided A is independent of a.

    2. Relevant equations

    3. The attempt at a solution
    so for part a) I got
    ##T=T_0 +(T_m-T_0)e^{-\sqrt{\frac{2A}{a}}x}##

    Then for b) I thought that the rate of heat transfer will be ##2 \pi a R(T)## but this gives something that is proportional to a not ##a^{\frac{3}{2}}##

    Many thanks in advance
  2. jcsd
  3. Dec 8, 2017 #2


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    Science Advisor

    a) is not right. Is your exponential term dimensionally correct?
    b) 2πaR(T) is the rate per unit length. What do you get when you integrate it over x?
  4. Dec 8, 2017 #3
    What is the temperature gradient at x = 0? If you multiply this by the cross sectional area and thermal conductivity, what do you get for the rate of heat flow into the rod?
  5. Dec 8, 2017 #4
    Sorry, i just typed this wrong my answer is $$T=T_0 +(T_m-T_0)e^{-\sqrt{\frac{2A}{a\kappa}x}}$$

    integrating over x i get the rate of heat loss is $$2\pi \sqrt{\frac{\kappa}{2A}}a^{\frac{3}{2}} (T_m - T_0)(e^{-\sqrt{\frac{2A}{a\kappa}l}}-1)$$ Does this seem right

    The question then asks for what l is this approximation valid. Any ideas?

    Many thanks
    Last edited by a moderator: Dec 8, 2017
  6. Dec 8, 2017 #5
    I see that you decided not to follow the approach I suggested in post #3. So, I guess I'll answer my own questions.

    The corrected equation for the temperature is: $$T=T_0 +(T_m-T_0)e^{-\sqrt{\frac{2A}{\kappa a}}x}$$
    The temperature gradient at location x is $$\frac{dT}{dx}=-(T_m-T_0)\sqrt{\frac{2A}{\kappa a}}e^{-\sqrt{\frac{2A}{\kappa a}}x}$$
    So the temperature gradient at x = 0 is: $$\frac{dT}{dx}=-(T_m-T_0)\sqrt{\frac{2A}{\kappa a}}$$
    So the heat flux at x = 0 is: $$q(0)=-\kappa\left(\frac{dT}{dx}\right)_{x=0}=(T_m-T_0)\sqrt{\frac{2A\kappa}{ a}}$$
    So the rate of heat loss is: $$Q=\pi a^2q(0)=\pi a^2(T_m-T_0)\sqrt{\frac{2A\kappa}{ a}}$$This is proportional to ##a^{3/2}##

    Your answer in post #4 is not consistent with this.
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