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Simple dimensional analysis

  • Thread starter chewwy
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  • #1
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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:

[itex]\frac{\partial \theta}{\partial t} = \lambda \frac{\partial^2 \theta }{\partial x^2 }[/itex]

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

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


The Attempt at a Solution



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

  • #2
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I can get to the fact that F must be a function of [itex](\frac{\theta_0}{kt})^a (\frac{x}{\sqrt{\lambda t}})^b[/itex]

but then i'm stuck...
 
  • #3
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this is almost certainly wrong but if [itex]\theta_0[/itex] has units of temperature and t has units seconds and k is of units [itex]Ks^{-1}[/itex] then [itex]\frac{\theta_0}{kt}[/itex] has units [itex]\frac{K}{Ks^{-1}s}[/itex] which all cancel so thats just a constant, then you have

[itex]F = C \zeta^b[/itex] with C a constant. Don't you?
 
  • #4
6
0
this is almost certainly wrong but if [itex]\theta_0[/itex] has units of temperature and t has units seconds and k is of units [itex]Ks^{-1}[/itex] then [itex]\frac{\theta_0}{kt}[/itex] has units [itex]\frac{K}{Ks^{-1}s}[/itex] which all cancel so thats just a constant, then you have

[itex]F = C \zeta^b[/itex] with C a constant. Don't you?
just because a quantity is dimensionless does not make it constant - it's a function of t!
 
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