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Heat Transfer Problem

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations

    Governing Equation : [itex] \frac{1}{r^2} \frac{d}{dr} (kr^2\frac{dT}{dr}) + q''' = 0[/itex]

    Temperature: [itex] T(r) = -\frac{q'''}{6k}r^{2} - \frac{C_1}{r} + C_2[/itex]

    3. The attempt at a solution

    For part a) all I did was multiply the equation provided in the question by the formula for the volume of a sphere to get [itex]q[/itex].

    The result is: [itex]q = \frac{4}{3} {\pi}{R_i} {q_0}''' [{R_i}^2 - {r}^2][/itex]

    Is this correct, or should I be getting an answer only in terms of [itex]R_i[/itex]?

    For part b) I know that I have to use the temperature equation I stated above. My result is:

    [tex] T(R_i) = -\frac{q'''}{6k}{R_i}^{2} - \frac{C_1}{R_i} + C_2[/tex]

    I'm unsure how to proceed. Do I apply boundary conditions?
     
  2. jcsd
  3. Oct 4, 2011 #2
    If this is in the wrong section, can a mod please move this to the engineering section?
     
  4. Oct 4, 2011 #3
    This ought to be in another section. But anyway, I'll comment.

    The heat generation is given as a function of r that tends to zero at the outer radius of the fuel. This implies that there must be an integration performed to get the result. If it were a constant, you could merely multiply by the volume to get the answer.

    Anent to the second question. The equation solved is the Poisson equation. It is steady state with a non-homgenous term for the internal heat generation. There are two constants of integration that must be evaluated from the boundary conditions.
    From what is stated I am not sure whether your problem covers two different materials, one for the cladding and the other for the fuel. If that is the case, you will have two domains. At the interface of the two domains, you will have to have continuity of heat flux and temperature. Could you more clearly state the boundary conditions.

    Another bit of confusion is the boundary condition on the exterior. Is it the forced convection condition or is it a constant temperature condition?
     
  5. Oct 4, 2011 #4
    I reattempted part a), in which I integrated the equation for heat generation from [itex]0[/itex] to [itex]R_i[/itex]. Then, I multiplied the result by the cross-sectional area of the sphere:

    [tex]\int_{0}^{R_i} q^{'''}(r) dr = \int_{0}^{R_i} q^{'''}_0 [1 - \frac{r^2}{{R_i}^2}]dr [/tex]

    [tex]q^{''}(r)= q^{'''}_0 [r - \frac{r^3}{3{R_i}^2}|_{0}^{R_i} [/tex]

    [tex]q^{''}(r) = \frac{{2q^{'''}_0}R_i}{3} [/tex]

    [tex]q = q^{''} A_s[/tex]

    [tex]q = \frac{2 \pi {q^{'''}_0}{R_i}^3}{3}[/tex]

    Is this correct?
     
  6. Oct 5, 2011 #5
    What are the units of q'''? If they are energy per unit time and volume, you need to put the volume element in the integral rather than multiplying it by a surface area after the integration.

    It seems to me that the initial integral should be:

    E = integral (q'''(r) * dV) from 0 to R, where E is energy and R is outer radius of pellet. You need the dV which is a function of r.

    Are you taking a course in nuclear engineering or mechanical engineering?
     
  7. Oct 5, 2011 #6
    If you consider the sphere to be a solid of revolution generated by rotating a circle about a y axis you can start with the equation x**2 + y**2 = R**2.

    Then dV = pi * (R**2 - x**2)dx

    Let dx be dr and x be r. If you integrate the dV from 0 to R and double it, you get the volume of a sphere. You double it because the integration limits cover 1/2 of the volume. Only reason I'm mentioning this is to demonstrate it is correct for dV.

    I think this (dV) should be multiplied by the heat generation term and the result put in the integrand and integrated.
     
  8. Oct 5, 2011 #7
    Thanks for the help. I'm going to try it out and post what I get later on.

    And I'm taking a heat transfer course in Mech. Eng.
     
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