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Heat equation differentiation

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm having trouble deriving the following equation

    [tex] \frac {\partial^2 {\theta}}{\partial {x'^2}} = -y^2*exp(\theta) [/tex]

    and y = x/x'

    my main problem is the exponent



    2. Relevant equations



    3. The attempt at a solution

    Normally i would use the equation

    (x')'' + k^2*x' = 0

    x' = c1 * cos(kx') + c2 * sin (kx')

    can I rearrange the equation as

    [tex] \frac {\partial^2 {\theta}}{\partial {x'^2}} + y*exp(\theta) = 0 [/tex]

    and solve for exp(theta)?
     
    Last edited: Nov 22, 2009
  2. jcsd
  3. Nov 22, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    what are you trying to derive? do you mean solve the differntial equation

    and what does y = x/x' represent? what is x differentiated w.r.t in the equation to give x' ?
     
  4. Nov 23, 2009 #3
    ok, so ignore the first post please

    the equation is

    [tex] \frac { d^2 \theta }{d x'^2 } = -y *exp(\theta) [/tex] eq. 1

    first off, this is a steady state model. meaning, we consider the pre-explosion temperature to be small in comparison with the absolute temperature of the walls:[tex] \frac {\Delta T}{T} << 1 [/tex]

    2nd, the reaction rate only depends on the deperature in accordance with exp(-E/RT)

    3rd we regad the thermal conductivity of the walls as being infinitely large.

    x' = x/r is the nondimensionalization of x, r is the half length (i.e radius for cylinder), not the derivative, for -L < x < L we have -1 < x' < 1. x' drops unit (i.e m, cm, ..)

    theta is the nondimensionalization of temperature [tex] \theta = \frac {E}{RT^2_a} *(T - T_a) [/tex]

    y (although i used a different variable) is known as the frank kamenetskii parameter

    [tex] y = \frac {Q}{d}*\frac {E}{R*T^2_a}*r^2*z* exp(\frac {-E}{RT_a}) [/tex]

    E: activation energy
    T_a: ambient temperature
    Q: heat released
    z: frequency of particle collision
    r: radius or half width (depending on geometry)
    R: gas constant
    d: thermal conductivity

    all uniform except Q, i think..

    the book solves the differential equation 1, analytically, for a function [tex] \theta = f(y,x') [/tex] in case of high activation energy E. RT<<E

    the book gives the following result.

    [tex] exp(\theta) = \frac {a}{cosh^2(b \frac{+}{-} \sqrt \frac{a*y}{2} * x')} [/tex]

    im just trying to figure out what steps I need to take in order to arrive at the last solution.
     
    Last edited: Nov 23, 2009
  5. Nov 23, 2009 #4
    theta is the nodimensialization of the temperature.
     
    Last edited: Nov 23, 2009
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