Why is F a function of the similarity variable in dimensional analysis?

[chewwy]

Just read this, and got a bit confused when trying to do it...

Homework Statement

We have a diffusion equation situation with a semi-infinite rod, so:

\frac{\partial \theta}{\partial t} = \lambda \frac{\partial^2 \theta }{\partial x^2 }

at infinity the rod is at some fixed temperature \theta_0, whilst at x=0, the temperature increases proportionally to time. write \theta(x,t)=\theta_0 + ktF.

explain with the help of dimensional analysis why F is a function only of the similarity variable \zeta = \frac{x}{\sqrt{\lambda t}}, and is independent of \theta_0 and k.

The Attempt at a Solution

Ok, right... so F must be dimensionless. but we have five variables here - \theta_0 , x, t, \lambda , k. how do we show \theta_0 and k aren't involved?

 

[chewwy]

I can get to the fact that F must be a function of (\frac{\theta_0}{kt})^a (\frac{x}{\sqrt{\lambda t}})^b

but then I'm stuck...

 

[latentcorpse]

this is almost certainly wrong but if \theta_0 has units of temperature and t has units seconds and k is of units Ks^{-1} then \frac{\theta_0}{kt} has units \frac{K}{Ks^{-1}s} which all cancel so that's just a constant, then you have

F = C \zeta^b with C a constant. Don't you?

 

[chewwy]

this is almost certainly wrong but if \theta_0 has units of temperature and t has units seconds and k is of units Ks^{-1} then \frac{\theta_0}{kt} has units \frac{K}{Ks^{-1}s} which all cancel so that's just a constant, then you have

F = C \zeta^b with C a constant. Don't you?

just because a quantity is dimensionless does not make it constant - it's a function of t!