# Heat equation differentiation

1. Nov 22, 2009

### squaremeplz

1. The problem statement, all variables and given/known data

I'm having trouble deriving the following equation

$$\frac {\partial^2 {\theta}}{\partial {x'^2}} = -y^2*exp(\theta)$$

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

$$\frac {\partial^2 {\theta}}{\partial {x'^2}} + y*exp(\theta) = 0$$

and solve for exp(theta)?

Last edited: Nov 22, 2009
2. Nov 22, 2009

### 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' ?

3. Nov 23, 2009

### squaremeplz

ok, so ignore the first post please

the equation is

$$\frac { d^2 \theta }{d x'^2 } = -y *exp(\theta)$$ 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:$$\frac {\Delta T}{T} << 1$$

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 $$\theta = \frac {E}{RT^2_a} *(T - T_a)$$

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

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

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 $$\theta = f(y,x')$$ in case of high activation energy E. RT<<E

the book gives the following result.

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

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
4. Nov 23, 2009

### squaremeplz

theta is the nodimensialization of the temperature.

Last edited: Nov 23, 2009