How Do You Solve the Steady-State Heat Equation for High Activation Energy?

[squaremeplz]

Homework Statement

the math in here is a bit over my head.

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.seperation of vars?

Homework Equations

The Attempt at a Solution

 

[NateD]

This equation is a partial differential equation. To solve it, you will need to use the method of separation of variables.The first step is to separate the two variables in the equation. This can be done by writing the equation as:\frac {d^2 \theta}{dx'^2} = -y * exp(\theta)The next step is to integrate both sides of the equation with respect to x'. On the left side, this results in the following equation:\int \frac {d^2 \theta}{dx'^2} dx' = \int -y * exp(\theta) dx'.On the right side, the integral can be solved using the chain rule and the properties of exponentials. This results in the following equation:\frac {d \theta}{dx'} = - \frac{y}{2} exp(\theta) + CAt this point, we can now integrate both sides of the equation with respect to x'. On the left side, this results in the following equation:\int \frac {d \theta}{dx'} dx' = \int - \frac{y}{2} exp(\theta) + C dx'On the right side, the integral can be solved using the chain rule and the properties of exponentials. This results in the following equation:\theta = - \frac{y}{4} exp(\theta)x'^2 + Cx' + DAt this point, we can now rearrange the equation to solve for exp(\theta). This results in the following equation:exp(\theta) = \frac {a}{cosh^2(b \frac{+}{-} \sqrt \frac{a*y}{2} * x')},where a, b, and c are constants.