Deriving the Heat Equation with Exponential Term: Can We Solve for exp(theta)?

[squaremeplz]

Homework Statement

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

Homework Equations

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)?

 

[lanedance]

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' ?

 

[squaremeplz]

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.

 

[squaremeplz]

theta is the nodimensialization of the temperature.