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Thermal Diffusion Problem - Cylindrical Rod

  1. May 4, 2014 #1
    1. The problem statement, all variables and given/known data

    29lfsiu.png

    Part (a): Under what circumstances do these equations work?
    Part (b): Find Temperature as a function of x, in steady state.
    Part (c): Show heat loss is proportional to r^1.5
    Part (d):Assume A = 0, show solution works, and find γ.


    2. Relevant equations



    3. The attempt at a solution

    Part(a)
    dQ = T dS works only for reversible processes.
    dW = -p dV works only for reversible processes.

    Part (b)
    In steady state, ##\frac{\partial T}{\partial t} = 0##.
    DE becomes:
    [tex]k\frac{\partial^2T}{\partial x^2} = \frac{A}{r}(T-T_0)[/tex]

    Solving, one obtains:

    [tex]T = (T_1-T_0)e^{-\sqrt{\frac{A}{rk}}x} + T_0[/tex]

    Part (c)

    [tex]P \propto \pi r^2\int R dx \propto r^{\frac{3}{2}}[/tex]

    Part (d)

    The DE now becomes:

    [tex]\rho C_v \frac{\partial T}{\partial t} = k\frac{\partial ^2 T}{\partial x^2} [/tex]

    I have subtituted ##T = Bt^{-\frac{1}{2}}e^{-\frac{x^2}{\gamma t}}## and it is a solution, but I can't seem to find an expression for ##\gamma##, as it cancels out on both sides.
     
  2. jcsd
  3. May 4, 2014 #2
    It would have been easier to do part c by evaluating the heat flux at x = 0, and then multiplying by the cross sectional area. The heat flux is just -κdT/dx.
    It doesn't seem possible that it would cancel from both sides. Show us what you did.

    Chet
     
  4. May 4, 2014 #3
    Starting:

    [tex]\rho C_v \left[-\frac{1}{2} t^{-\frac{3}{2}} + t^{-\frac{1}{2}} (\frac{x^2}{\gamma t^2}) \right] e^{-\frac{x^2}{\gamma t}} = k \frac{\partial}{\partial x} \left[ t^{-\frac{1}{2}}(\frac{-2x}{\gamma t})e^{-\frac{x^2}{\gamma t}}\right][/tex]

    [tex]\rho C_v \left[ t^{-\frac{1}{2}} \left( \frac{x^2}{\gamma t^2}\right) - \frac{1}{2} t^{-\frac{3}{2}} \right] e^{-\frac{x^2}{\gamma t}} = kt^{-\frac{1}{2}}\left(-\frac{2}{\gamma t}\right)\left[e^{-\frac{x^2}{\gamma t}} + x\left(\frac{-2x}{\gamma t}\right)e^{-\frac{x^2}{\gamma t}}\right][/tex]

    Multiply throughout by ##t^{\frac{3}{2}}##:

    [tex]\rho C_v \left[ \frac{x^2}{\gamma t} - \frac{1}{2}\right] = -2k\left[1 - \frac{2x^2}{\gamma t}\right][/tex]

    This implies that ##\rho C_v = 4k##
     
    Last edited: May 4, 2014
  5. May 4, 2014 #4
    You were missing a gamma in the denominator. The equation should read:

    [tex]\rho C_v \left[ \frac{x^2}{\gamma t} - \frac{1}{2}\right] = -\frac{2k}{γ}\left[1 - \frac{2x^2}{\gamma t}\right][/tex]

    Chet
     
  6. May 4, 2014 #5
    Thanks alot! For part (a), dQ = T dS only holds for reversible processes, right? Since the definition is ##dS = \frac{dQ_{rev}}{T}##.

    But for dW = -p dV, must the process strictly be reversible?
     
  7. May 4, 2014 #6
    It depends on what you are calling p. In an non-quasistatic process, the pressure within the system is not uniform, so the is some question as to what pressure to use. If you use the value of the pressure at the interface between the system and the surroundings, then this equation gives you the correct result for dW. Also, based on the the sign of this equation, this is the work that the surroundings do on the system.

    Chet (see my PF Blog on the 1st and 2nd Laws of Thermodynamics)
     
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