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Max temperature inside sphere generating heat

  1. Oct 25, 2014 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    Radioactive wastes are stored in a spherical stainless tank of inner diameter 1 m and 1 cm wall
    thickness. Heat is generated uniformly in the wastes at a rate of 30,000 W/m3. The outer
    surface of the tank is cooled by air at 300 K with a heat transfer coefficient of 100 W/m2 K.
    Determine the maximum temperature in the tank at steady state. Take the thermal
    conductivities of the wastes and steel tank as 2.1 W/m K and 15 W/m K, respectively.

    2. Relevant equations


    3. The attempt at a solution
    I don't really know how to do this, so I do what I know and start with a general energy balance. I assume one dimensional heat flow in the sphere.

    [tex] \frac {dE}{dt} = \dot Q_{r} - \dot Q_{r + \Delta r} + \dot e_{gen} A \Delta r [/tex]

    With the assumption of steady state (is this a valid assumption??), I get it reduced to
    [tex] \frac {1}{r^2} \frac {d}{dr} [r^2 \frac {dT}{dr}] + \frac {\dot e_{gen}}{k} = 0 [/tex]
    [tex] \int d[r^2 \frac {dT}{dr}] = - \frac {\dot e_{gen}}{k} \int_0^{r_{1} + t} r^2 dr [/tex]
    [tex] r^2 \frac {dT}{dr} = - \frac {\dot e_{gen}}{k} \frac {(r_{1} + t)^{3}}{3} [/tex]

    where ##r_{1}## is the inner radius of the sphere, and ##t## is the thickness of the sphere wall. However, there are two different thermal conductivities, so I don't know what to put as ##k## for this equation.

    However, I'm not really sure how to think of where the maximum temperature in the sphere is. Is this where ##\frac {dT}{dr} = 0##? I think since the generation of heat is greatest with the larger volume, then the further from the center, the larger the temperature, but I don't know how to go about showing this.
     
    Last edited: Oct 25, 2014
  2. jcsd
  3. Oct 25, 2014 #2

    mfb

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    Staff: Mentor

    That is a good approach.
    Sure, it is even given in the problem statement.
    It is useful to consider the steel wall and the waste separately.

    dT/dr will be zero at the point of maximal temperature, but you can simply look at the symmetry of the problem - if any part around the core would be hotter than the center, heat would flow inwards instead of outwards and the system would not be in equilibrium.
     
  4. Oct 27, 2014 #3

    Maylis

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    I'm not really making much progress on this problem. So I have two boundary conditions, ##T = T_{\infty}## at ##r = r_{1} + t##, and ##\frac {dT}{dr} = 0## at ##r = 0##

    So for ##0 \le r \le r_{1}##, I'll start back where I was
    [tex] \int d[r^2 \frac {dT}{dr}] = - \frac {\dot e_{gen}}{k_{w}} \int r^2 dr[/tex]
    [tex] r^2 \frac {dT}{dr} = - \frac {\dot e_{gen}}{k_{w}} \frac {r^3}{3} + C_{1} [/tex]
    [tex] \frac {dT}{dr} = - \frac {\dot e_{gen}}{k_{w}} \frac {1}{r^2} \frac {r^3}{3} + \frac {C_{1}}{r^2}[/tex]
    [tex] dT = [-\frac {\dot e_{gen}}{k_{w}} \frac {r}{3} + \frac {C_{1}}{r^2}]dr [/tex]
    [tex] T = -\frac {\dot e_{gen}}{k_{w}} \frac {r^2}{6} - \frac {C_{1}}{r} + C_{2} [/tex]

    Now I can do for ##r_{1} \le r \le r_{1} + t##, my final result is
    [tex]T = - \frac {\dot e_{gen}}{k_{s}} \frac {r^2}{6} - \frac {C_{3}}{r} + C_{4} [/tex]

    So my problem is with my expression ##\frac {dT}{dr} = - \frac {\dot e_{gen}}{k_{w}} \frac {1}{r^2} \frac {r^3}{3} + \frac {C_{1}}{r^2}##, if I try to evaluate this at ##r = 0##, then I will get ##\frac {dT}{dr} = \infty## due to the ##r## term in the denominator. I don't seem to have enough boundary conditions to solve for my 4 constants of integration from my two equations. I apply my boundary condition ##T = T_{\infty}## at ##r = r_{1} + t## and get
    [tex] T_{\infty} = - \frac {\dot e_{gen}}{k_{s}} \frac {r_{1} + t}{3} - \frac {C_{3}}{r_{1} + t} + C_{4} [/tex]
    With just this, I can't figure out both ##C_{3}## and ##C_{4}##. I then try to equate the equations for ##T## with each other, since they should both be valid at ##r = r_{1}##.

    [tex] -\frac {\dot e_{gen}}{k_{w}} \frac {r_{1}^2}{6} - \frac {C_{1}}{r+{1}} + C_{2} = - \frac {\dot e_{gen}}{k_{s}} \frac {r_{1}^2}{6} - \frac {C_{3}}{r_{1}} + C_{4}[/tex]

    I can subtract the constants ##C_{2}## and ##C_{4}## to make a new constant, ##C_{5}##

    [tex] -\frac {\dot e_{gen}}{k_{w}} \frac {r_{1}^2}{6} - \frac {C_{1}}{r_{1}} = - \frac {\dot e_{gen}}{k_{s}} \frac {r_{1}^2}{6} - \frac {C_{3}}{r_{1}} + C_{5}[/tex]
    But I still don't have a boundary condition at ##r = r_{1}## to help find any constant of integration, so I have exhausted all my possibilities that I can see from a general energy balance.

    I also thought to approach it with Fourier's law. I know that ##\dot Q## is constant, so using Newton's law of cooling, I got
    [tex] \dot Q = h_{\infty}A(T_{max} - T_{\infty}) [/tex]
    And from Fourier's law,
    [tex] \dot Q = -k_{w} 4 \pi r^2 \frac {dT}{dr} [/tex]
    I equate the two expressions,
    [tex]-k_{w} 4 \pi r^2 \frac {dT}{dr} = h_{\infty}A(T_{max} - T_{\infty}) [/tex]
    [tex] \int_{T_{max}}^{T_{1}} dT = -\frac {h_{\infty} \frac {d^2}{4}}{k_{w}}(T_{max} - T_{\infty}) \int_{0}^{r_{1}} \frac {1}{r^2} dr[/tex]

    Again, with this integration I get ##- \frac {1}{r}##, and at ##r = 0##, this goes to infinity. So I am stuck now
     
    Last edited: Oct 27, 2014
  5. Oct 27, 2014 #4
    It's easiest to work this problem backwards, starting at the outside.

    What is the total rate of heat generated within the sphere? All this heat has to flow through the air boundary layer outside the tank. Given the heat transfer coefficient, what is the temperature difference across the air boundary layer? Based on this temperature difference, what is the temperature at the outside of the steel shell?

    Chet
     
  6. Oct 27, 2014 #5

    mfb

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    Staff: Mentor

    Apart from a single, correct choice for the constant. But as Chestermiller said, start from the outside.

    The released energy in the steel is zero. You also know the total power flowing through the steel.

    Both things will allow to fix one constant. You can compute them one by one if you start from the outside.
     
  7. Oct 28, 2014 #6

    Maylis

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    How can I show from the general energy balance that the heat generated is equation to ##\dot Q##?

    [tex] \frac {dE}{dt} = \dot Q_{r} - \dot Q_{r + \Delta r} + \dot e_{gen}A \Delta r + \dot W_{s} [/tex]
    Assuming there is no shaft work,
    [tex] \rho \hat{c_{p}} A \Delta r \frac {dT}{dt} = \dot Q_{r} - \dot Q_{r + \Delta r} + \dot e_{gen}A \Delta r [/tex]

    With the steady state
    [tex]\lim_{\Delta r \rightarrow 0} \frac {1}{A \Delta r} (\dot Q_{r} - \dot Q_{r + \Delta r}) + \dot e_{gen} = 0[/tex]
    [tex] - \frac {d \dot Q}{A dr} + \dot e_{gen} = 0 [/tex]
    [tex] \frac {d \dot Q}{dV} = \dot e_{gen} [/tex]
    [tex] \int_{\dot Q} d \dot Q = \int_{V} \dot e_{gen} dV [/tex]
    I know this part is a little tricky with math, I know you can't do a ''regular'' integral over the heat flow, what's the deal with that?

    Anyways, I calculate the outside temperature of the shell to be 348 K, and I calculated the heat flow to be 15,708 W. Now, I need the inner temperature, so I suppose I will start with Fourier's law
    [tex] \dot Q_{steel} = -k_{s}A_{s} \frac {dT}{dr} [/tex]
    [tex] 15,708 = -15 \pi (1.02^{2} - 1^2) \frac {dT}{dr} [/tex]
    [tex] -8250.8 \int_{0.5}^{0.51} dr = \int_{T_{i}}^{348} dT [/tex]
    And I calculate the inner temperature of the steel wall, ##T_{i} = 430.5 K##.

    Then I do Fourier's Law for the waste
    [tex] \dot Q_{waste} = -k_{w}A_{w} \frac {dT}{dr} [/tex]
    [tex] 15,708 = -2.1 \pi (1^2) \frac {dT}{dr} [/tex]
    [tex] -2381 \int_{0}^{0.5} dr = \int_{T_{max}}^{430.5} dT[/tex]
    And I calculate ##T_{max} = 1621 K##
     
    Last edited: Oct 28, 2014
  8. Oct 28, 2014 #7
    Sorry, but the shell and the core are done incorrectly.

    For the shell, the equation should read:
    [tex]15708=-4\pi r^2k\frac{dT}{dr}[/tex]
    You need to integrate this equation with respect to r.

    For the core, not all the heat flow passes through each surface of constant r, only the heat flow coming from inside the surface of constant r. So,
    [tex]15708\left(\frac{r}{R}\right)^3=-4\pi r^2k_w\frac{dT}{dr}[/tex]
    Where R is the outer radius of the core (inner radius of the shell).
    You can get this same result from one of your earlier equations:
    [tex] r^2 \frac {dT}{dr} = - \frac {\dot e_{gen}}{k_{w}} \frac {r^3}{3} + C_{1} [/tex]
    This equation must hold at r = 0, so setting r = 0 gives C1=0
    Therefore,
    [tex] \frac {dT}{dr} = - \frac {\dot e_{gen}}{k_{w}} \frac {r}{3}[/tex]

    Chet
     
  9. Oct 28, 2014 #8

    Maylis

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    Okay, I did this after some more thought, and a little help from a TA to get started, but here is what I did. I just found out recently that I had to set the first constant of integration to 0.

    ImageUploadedByPhysics Forums1414495701.399266.jpg
     
    Last edited: Oct 28, 2014
  10. Oct 28, 2014 #9

    Maylis

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  11. Oct 28, 2014 #10
    Nice job.

    Chet
     
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