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Dimension analysis

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img410.imageshack.us/img410/8495/88748860vk8.png [Broken]

    2. Relevant equations



    3. The attempt at a solution
    I don't know how to deal with the q1 part. q2 = u* =u x sqrt(... so that is clear.

    Normally I just substitute the expression for u in terms of u* (= f* )and t in terms of t* in [tex] \frac{ du }{dt } [/tex] but how should I deal with the differentiation of the q1 -term?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 7, 2009 #2

    Tom Mattson

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    You have to use the chain rule for that.
     
  4. Feb 8, 2009 #3
    To be honest I don't how to begin. I have this:

    [tex] u= \frac{u_0 \sqrt{q}f^*}{x} [/tex]

    [tex] t = \frac{x^2}{qk} [/tex]


    so [tex] \frac{du}{dt} = ? [/tex]
     
  5. Feb 8, 2009 #4

    Tom Mattson

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    OK, follow these steps.

    1.) As directed, assume that [itex]q_2=f^*(q_1)[/itex]. So [itex]f^*[/itex] is a function of [itex]q_1=x^2/kt[/itex].

    2.) Solve [itex]q_2=u\sqrt{kt}/u_0[/itex] for [itex]u[/itex]. Identify [itex]q_2[/itex] with [itex]f^*[/itex] at this point.

    By now you should have the following result.

    [tex]u(x,t)=\frac{u_0}{\sqrt{kt}}f^*\left(\frac{x^2}{kt}\right)[/tex]

    Now you can take partial derivatives with respect to [itex]x[/itex] and [itex]t[/itex] until your heart's content. And as I said before, you'll need the chain rule to do it.

    Try that and let us know if you get stuck again.
     
  6. Feb 9, 2009 #5
    Tom your approach does work I found out after 2 full pages of writing. But what I don't understand is where my approach is going wrong. You see I just used the usual method in dimensional analysis. Writing u in terms of u* and t in terms of t* and calculating the derative.

    If you calculate it you'll see that u* = f* but then things get nasty because there is no t*. So what I did was rewrite the expression for q1 in terms of t. Why is that approach wrong?
     
  7. Feb 9, 2009 #6

    Tom Mattson

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    The problem there is that q_1 is a function of x and t. If you're going to plug u(x,t) into the heat equation then you have to see all of those hidden x's and t's, because you are taking partial derivatives of u.

    That q under the square root sign should be q_1. Substitute the expression for q_1 in there and you'll get what I got.

    [tex]u= \frac{u_0 \sqrt{q_1}f^*\left(q_1\right)}{x}[/tex]

    [tex]u= \frac{u_0 \sqrt{\frac{x^2}{kt}}f^*\left(\frac{x^2}{kt}\right)}{x}[/tex]

    [tex]u= \frac{u_0}{\sqrt{kt}}f^*\left(\frac{x^2}{kt}\right)}[/tex]
     
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