Heat diffusion with arrhenius 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 well??

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.

 

[Jhero]

Thank you for your question about the substitutions in the given problem. it is important to understand the reasoning behind any equations and substitutions used in a problem.

In this case, the substitution of \theta = \frac{E}{RT_a^2}(T-T_a) is used to make the equation dimensionless. By substituting this expression, the units of temperature and energy (T and E) are canceled out, leaving \theta as a dimensionless quantity. This is important because the heat equation is a partial differential equation, and solving for dimensionless quantities makes the problem more manageable.

Similarly, the substitution x = L*b is used to make the spatial variable x dimensionless. This allows for easier manipulation of the equation and solving for the unknown variables.

I hope this explanation helps to clarify the reasoning behind the substitutions. It is important to always understand the underlying principles and logic behind any equations or substitutions used in scientific problems. Good luck with your solution!