Heat diffusion with source

[squaremeplz]

Homework Statement

a material occupies -L < x < L and has uniform ambient temperature T_a. A chemical reaction begins within the body leading to the 1-d heat equation:

$$pc \frac {\partial{T}}{\partial{t}} = k \frac {\partial^2{T}}{\partial{x^2}} + pQAe^\frac{-E}{RT}$$

with BC and IC

$$T(+/- L, t) = T_a$$ and $$T(x,0) = T_a$$

Homework Equations

The Attempt at a Solution

The book gives the following substitutions without any justification

$$\theta = \frac{E}{RT_a^2}(T-T_a)$$ and $$x = L*b$$

and arrives at the following linear equation

$$L^2 \frac {pc}{k} { \frac {\partial{\theta}}{\partial{t}} = \frac {\partial^2{\theta}}{\partial{b^2}} + z*e^\frac{\theta}{1+y\theta}$$

then it asks to give the equations for z and y.

However, before I begin to solve this problem.. I would love to understand where the substitutions

$$\theta = \frac{E}{RT_a^2}(T-T_a)$$ and $$x = L*b$$

came from. I presume that x = L*b is used to make x dimensionless.. but what about theta?

Im pretty sure I know how to solve the problem (seperation of variables on the last eq.) , but I really don't understand the substitutions. It seems that whoever wrote the book solved the problem first and then realized the substitutions were a good fit. Any information (about the substitutions) is appreciated.

 

[mighty2000]

Thank you for your question. The substitutions \theta = \frac{E}{RT_a^2}(T-T_a) and x = L*b are commonly used in solving problems involving the 1-D heat equation. Let me explain the reasoning behind these substitutions.

Firstly, let's look at the term e^\frac{-E}{RT} in the heat equation. This term is known as the Arrhenius factor and it represents the effect of temperature on the rate of a chemical reaction. As you can see, it is dependent on the ratio \frac{E}{RT}. Now, in order to make this term dimensionless, we can divide it by T_a, the ambient temperature. This gives us \frac{E}{RT_a}. If we further divide by T_a, we get \frac{E}{RT_a^2}. This is the coefficient that is used in the substitution for \theta.

Now, let's look at the term x = L*b. As you correctly noted, this is used to make x dimensionless. However, it is not just any arbitrary substitution. In fact, b is known as the dimensionless variable and it is defined as \frac{x}{L}. This means that when x = L, b = 1 and when x = -L, b = -1. This substitution is commonly used in solving problems involving the heat equation because it simplifies the mathematics and makes the problem easier to solve.

I hope this explanation helps you understand the reasoning behind these substitutions. If you have any further questions, please do not hesitate to ask. Good luck with solving the problem!