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Length of slab for infinite temperature

  1. Oct 28, 2014 #1

    Maylis

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    Gold Member

    1. The problem statement, all variables and given/known data
    An explosive is to be stored in large slabs of thickness 2L clad on both sides with a protective sheath. The rate at which heat is generated within the explosive is temperature-dependent and can be approximated by the linear relation ##\dot Q_{gen} = a + b(T - T_{\infty})##, where ##T_{\infty}## is the prevailing ambient air temperature. If the overall heat transfer coefficient between the slab surface and the ambient air is U, show that the condition for an explosion (or the temperature inside the slab to become infinite) is ##L = (\frac {k}{b})^{1/2} tan^{-1} \Big [ \frac {U}{(kb)^{1/2}} \Big ]##. Determine the slab thickness if k = 0.9 W/m K , U = 0.2 W/m2 K, a = 60 W/m3 K, and b=6.0 W/m3 K.

    2. Relevant equations


    3. The attempt at a solution
    This is an extremely strangely worded problem in my opinion. I can't tell if the overall heat transfer coefficient is only counting for convective heat transfer, since it is saying that it is between the slab surface the the ambient air, so this so called ''overall'' heat transfer coefficient is ignoring conduction in the slab. I am also assuming this explosive is just continuously exploding, hence a steady state process, which also seems very stupid. I don't think it should be steady state, because the generation of heat depends on the temperature, which depends on the position, hence there is a non-uniform generation of heat.

    [tex] \frac {dE}{dt} = \dot Q_{in} - \dot Q_{out} + \dot Q_{gen} [/tex]
    I am saying that no heat is flowing it, and steady state, so ##\dot Q_{out} = \dot Q_{gen}##. However, there is conduction in the slab whose value of ##\dot Q## is continuously changing.

    I know ##\dot Q = UA(T_{s} - T_{\infty})##, where ##T_{s}## is the temperature of the surface of the slab. I would like to assume that the temperature is uniform in the slab, such that the ##T## term in the equation for the generation of heat is equal to ##T_{s}##, and also I don't know what the Area term would be since no information is given on the area.

    Then equating the two, I get
    [tex] a + b(T - T_{\infty}) = UA(T_{s} - T_{\infty})[/tex]

    Anyways, I can't see anyway to equate these so that I get some crazy inverse tangent function. I think it somehow comes out of the integral

    [tex] \int \frac {dx}{a^2 + x^2} = \frac {1}{a} tan^{-1} \Big (\frac {x}{a} \Big ) [/tex]
    Of course I can know what a and x should be to fit them to the equation, just that I don't know how I will get them in that form to be able to integrate it. I would need my integral to look like this

    [tex] \int \frac {d(\frac {U}{k})^2}{\frac {b}{k} + (\frac {U}{k})^2} [/tex]

    How the heck will I ever get something like this??
    I suppose as ##T \rightarrow \infty##, I will get
    [tex] a + b(T - T_{\infty}) = UA(T - T_{\infty}) [/tex]
    since a is negligible, then it is cancelled and the two temperature terms are cancelled, leaving
    [tex] A = \frac {b}{U} [/tex]
    So I put this into Fourier's Law
    [tex] \dot Q = -k \Big ( \frac {b}{U} \Big ) \frac {dT}{dL} [/tex]
    But I still don't see how to get it into that form I need it to be in.
     
    Last edited: Oct 28, 2014
  2. jcsd
  3. Oct 28, 2014 #2

    Maylis

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    Gold Member

    I solved it, what a tedious math exercise...
     
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