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Heat Transfer in Composting Pile

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data

    One dimensional system in vertical direction
    2 meter high composting pile @ 65 [tex]\circ[/tex]C
    Top of pile has wind @ 40 [tex]\circ[/tex]C and h = 50 W/m2*K
    Bottom of pile at ground temperature of 20 [tex]\circ[/tex]C
    Only conductive heat transfer WITHIN pile; no convection
    Volumetric biochemical heat generation, Q = 7 W/m3
    Compost has k = 0.1 W/m*K

    a.) Setup the governing equation and boundary conditions
    b.) Determine the temperature as a function of height from the ground
    c.) Calculate the maximum temperature in the pile
    d.) Calculate the top surface temperature of the pile

    2. Relevant equations

    General equation:​
    [tex]\rho c_{p}\frac{\partial T}{\partial t} + \rho c_{p}\frac{\partial}{\partial x} ( uT ) = k ( \frac{\partial^{2}T}{\partial x^{2}} ) + Q[/tex]​

    Fourier's Law​
    [tex]q_{x} = -k \frac{dT}{dx}[/tex]​

    Newton's Law of Cooling​
    [tex]q_{x} = h ( T_{s} - T_{\infty} )[/tex]​

    3. The attempt at a solution

    a.)
    There is no storage and no convection within the pile, so the general equation reduces to:

    [tex]\frac{d^{2}T}{dx^{2}} = -\frac{Q}{k} [/tex]​

    Boundary conditions:​
    At x = 0, T = 20 [tex]\circ[/tex]C​


    [tex] At \ x = L, -k \frac{dT}{dx} = h( T_{s} - T_{\infty} )[/tex]​
    Heat flux at the top surface due to conduction is equal to heat flux at the top surface due to wind convection​

    b.)
    Integrating once and using the second boundary condition gives:

    [tex]\frac{dT}{dx} = - \frac{Q}{k} ( x + C_{1} ) [/tex]

    [tex]C_{1}=\frac{h( T_{s} - T_{\infty})}{Q}-L [/tex]​

    [tex]C_{1}=\frac{(50 \ W/m^{2} \cdot K)(40^{\circ} C - 65^{\circ} C)}{(7 \ W/m^{3})}-(2 \ m)[/tex]​

    [tex]C_{1}=-177 \ m[/tex]​

    From what my professor said, this value for C(1) is incorrect (even the units). Haven't gotten to parts c.) and d.) yet. So I just need some using the boundary condition to find the first constant of integration.
     
    Last edited: Jan 24, 2010
  2. jcsd
  3. Jan 24, 2010 #2

    Mapes

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    Is it possible your professor has defined [itex]C_1[/itex] as

    [tex]\frac{dT}{dx}=-\frac{Q}{k}x+C_1[/tex]

    I would that's the more typical way to do the integration, though your way is fine too.
     
  4. Jan 24, 2010 #3
    Yeah, that is the way some of my peers have done it (edit: tried to do it). However, the answer he gave us was 80.06 W/K which I can't figure out. Our my temperatures correct?
     
    Last edited: Jan 24, 2010
  5. Jan 24, 2010 #4

    Mapes

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    Sorry, the answer for what? [itex]C_1[/itex]? [itex]C_1[/itex] can be defined multiple ways, so a number alone doesn't have any meaning without the definition.
     
  6. Jan 24, 2010 #5
    I realize that C1 can be defined multiple ways but the units are in W/k. I have a feeling he made a typo on the units though because the value for C1 works numerically for the following equation in order to determine the maximum temperature in part c.), however, the units don't make sense:

    [tex]T=-\frac{Q}{k}\frac{x^{2}}{2}+C_{1}x+C_{2}[/tex]

    where C2 = 20 degrees C found by using the first boundary condition.

    Max Temperature:​
    [tex]T_{max}, \ \frac{dT}{dx}=0[/tex]​

    [tex]0=-\frac{Q}{k}x+80.06[/tex]​

    [tex]x=1.14 \ m[/tex]​

    This answer, x, for Tmaxis another answer he gave us as being correct.

    It seems to me that C1 should be in Kelvin per meter not watts per Kelvin[/INDENT][/INDENT] Then again that could just be a typo although I still cannot figure out how to solve for C1 regardless of method to find the correct solution.
     
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