Answer:
Step-by-step explanation:
Given:
Rate k = αμ ( β - μ )
Find:
Derive an equation for the concentration u(x,t)
Solution:
- From the law of conservation we have:
[tex]\frac{d}{dt}\int\limits^b_a {u (x,t)} \, dx = sig (a,t) - sig(b,t) + \int\limits^b_a {\alpha }u(\beta \ - u). dx[/tex]
- After dividing the above expression by A i.e cross sectional area of the rod:
[tex]0 = \int\limits^b_a ({\frac{du}{dt} + \frac{dsig}{dx} - \alpha u(\beta - u)).dx} \,[/tex]
- This is valid for any for any interval [ a, b ] , and the integral is 0:
[tex]0 = {\frac{du}{dt} + \frac{dsig}{dx} - \alpha u(\beta - u)[/tex]
- Now use Fick's law we will obtain a PDE only as a function of u:
[tex]\frac{du}{dt} = k*\frac{d^2u}{dx^2} + \alpha u (\beta - u)[/tex]
