2nd order differential equation in Reaction Engineering

  1. May 8, 2011 #1
    There's a catalyst pellet in a reactor and I'm supposed to prove an equation for the maximum temperature which is:

    Tmax=Ts +[(-H)*(D*Cas)]/k

    First thing I did was develop an Energy Balance across the spherical catalyst and I got the following equation:

    (1/r^2)*d/dr(r^2*k*dT/dr) + (-H)*(-Rxn) = 0

    expanding i get:

    d^2T/dr^2 + (2/r)*dT/dr + (-H)*(-Rxn)/k = 0

    Boundary Conditions: at r=0, dT/dr =0 AND at r=R, T=Ts (at the catalyst surface i.e. r=R the temperature T = Ts (catalyst surface temperature))

    Using the boundary conditions and integrating factor I got
    T=Ts + [(-H)(-Rxn)*(R^2-r^2)]/(6*k) ......(eqn 1)

    Now for the mole balance across the catalyst I got (where Ca is the concentration):

    d^2Ca/dr^2 + (2/r)*dCa/dr - (k/D)*Ca =0 ...... (eqn 2)

    Boundary Condition: at r=R, Ca=Cas

    Can someone confirm the equation I got for temperature (eqn 1) and also for the concentration (eqn 2) how do I go about to solve it??

    I tried another way to solve it, as suggested by my instructor, which is by using the un-expanded forms of the energy and mole balance equations which are:

    (1/r^2)*d/dr(r^2*k*dT/dr) + (-H)*(-Rxn) = 0

    (1/r^2)*d/dr(r^2*D*dCa/dr) + (Rxn) = 0

    So here I have the "Rxn" term common in both equations and I combined and got the following after some simplifications:

    d^2Ca/dr^2 + 2*dCa/dr = [k/(H*D)]* { r*d^2T/dr^2 + 2*dT/dr}

    But in this case it looks way more complicated and I'm not sure how to deal with this type of an equation, since I have 2 derivates on both sides of the equation one with respect to the Concentration Ca and one with respect to temperature T.

    :confused: not sure which way to proceed and how to proceed
  2. jcsd
  3. May 11, 2011 #2
    Let Ca(r)=(1/r)exp(f(r)) :

    Attached Files:

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