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Non-dimensionalization - Given heat balance EQ, convert to a dimensionless EQ

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Define a dimensionless radial coordinate as

    [tex]x\,=\,r\,\sqrt{\frac{2\,h}{b\,k}}[/tex]

    and introduce [itex]y\,=\,T\,-\,T_a[/itex], and thus show the elementary equation

    [tex]x^2\,\frac{d^2y}{dx}\,+\,x\,\frac{dy}{dx}\,-\,x^2\,y\,=\,0[/tex]

    describes the physical situation.

    It gives the "physical situation" equation in the previous problem:

    [tex]\frac{1}{r}\,\frac{d}{dr}\,\left(r\,\frac{dT}{dr}\right)\,-\,\left(\frac{2\,h}{b\,k}\right)\,\left(T\,-\,T_a\right)\,=\,0[/tex]



    2. Relevant equations

    Non-dimensionalization



    3. The attempt at a solution

    Re-expressing the constants

    [tex]z\,=\,\frac{2\,h}{b\,k}[/tex]

    solving the dimensionless radial coordinate for r

    [tex]r\,=\,\frac{x}{\sqrt{z}}[/tex]

    Now, I substitute these along with [itex]y\,=\,T\,-\,T_a[/tex] into the "physical situation" equation that I am supposed to non-dimensionalize

    [tex]\frac{\sqrt{z}}{x}\,\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,\left[\left(\frac{x}{\sqrt{z}}\right)\,\frac{dT}{d\left(\frac{x}{\sqrt{z}}\right)}\right]\,-\,z\,y\,=\,0[/tex]

    What do I do about the denominator in the second and fourth fractions?

    [tex]\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}[/tex]
     
  2. jcsd
  3. Jul 18, 2007 #2

    Dick

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    It looks like sqrt(z) is just a constant. So d(x/sqrt(z))=dx/sqrt(z). z should cancel out of the equation altogether.
     
  4. Jul 18, 2007 #3
    Yes, [itex]\sqrt{z}[/itex] is only a constant.

    So you are saying that

    [tex]\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,=\,\frac{d}{\frac{dx}{\sqrt{z}}}[/tex]

    Right? So, does that mean that

    [tex]\frac{d}{\frac{dx}{\sqrt{z}}}\,=\,\frac{d\sqrt{z}}{dx}\,=\,\frac{d}{dx}\sqrt{z}\,=\,0[/tex]

    since [itex]\sqrt{z}[/itex] is constant?
     
    Last edited: Jul 18, 2007
  5. Jul 18, 2007 #4

    Dick

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    Riggghhhhtttt.
     
  6. Jul 18, 2007 #5
    But when I apply that to the equation above

    [tex]\frac{\sqrt{z}}{x}\,\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,\left[\left(\frac{x}{\sqrt{z}}\right)\,\frac{dT}{d\left(\frac{x}{\sqrt{z}}\right)}\right]\,-\,z\,y\,=\,0[/tex]

    There is division by zero and it just zeros out the whole equation. I am supposed to show that it is of the form

    [tex]x^2\,\frac{d^2y}{dx}\,+\,x\,\frac{dy}{dx}\,-\,x^2\,y\,=\,0[/tex]
     
  7. Jul 18, 2007 #6

    Dick

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    Division by zero?????? z is nonzero. It cancels inside the brackets and gives a common factor of z for the two terms.
     
  8. Jul 18, 2007 #7
    [tex]\frac{\sqrt{z}}{x}\,\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,\left[\left(\frac{x}{\sqrt{z}}\right)\,\frac{dT}{d\left(\frac{x}{\sqrt{z}}\right)}\right]\,-\,z\,y\,=\,0[/tex]

    The [itex]\frac{\sqrt{z}}{x}[/itex] terms cancel each other out to one

    [tex]\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,\frac{dT}{d\left(\frac{x}{\sqrt{z}}\right)}\,-\,z\,y\,=\,0[/tex]

    Now what? Multiply it all by [itex]d\left(\frac{x}{\sqrt{z}}\right)[/itex]?

    If I do that, I get

    [tex]d\left(dT\right)\,-\,\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,z\,y\,=\,0[/tex]

    Then what?
     
  9. Jul 18, 2007 #8

    Dick

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    It (sqrt(z)) cancels inside the brackets and gives a common factor of z for the two terms on the left hand side of the equation. Which can be factored out. 0/z=0. That is not division by zero. Don't make me repeat this again.
     
    Last edited: Jul 18, 2007
  10. Jul 18, 2007 #9
    You are saying that z "cancels inside the brackets". Can you show how you arrive at that conclusion, because I don't see how z cancels.
     
    Last edited: Jul 18, 2007
  11. Jul 18, 2007 #10

    Dick

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    Ok. (x/sqrt(z))*dT/d(x/sqrt(z))=(x/sqrt(z))*sqrt(z)*(dT/dx)=x*(dT/dx). 1/sqrt(z) is in the denominator of the differential. Move it into the numerator as sqrt(z) and then cancel with the sqrt(z) in the denominator of the first factor. Differentials are a lot like fractions if that is what is bothering you.
     
    Last edited: Jul 18, 2007
  12. Jul 18, 2007 #11
    OIC, from the equality in the third post

    [tex]\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,=\,\frac{d}{\frac{dx}{\sqrt{z}}}\,=\,\sqrt{z}\,\frac{d}{dx}[/tex]

    If I apply that to the steady-state heat balance equation as you have shown, I get

    [tex]\frac{\sqrt{z}}{x}\,\sqrt{z}\,\frac{d}{dx}\left(x\,\frac{dT}{dx}\right)\,-\,z\,y\,=\,0[/tex]

    [tex]\frac{z}{x}\,\frac{d}{dx}\left(x\,\frac{dT}{dx}\right)\,-\,z\,y\,=\,0[/tex]

    [tex]\frac{z}{x}\left(\frac{dT}{dx}\,+\,x\,\frac{d^2T}{dx^2}\right)\,-\,z\,y\,=\,0[/tex]

    [tex]z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx}\,-\,z\,y\,=\,0[/tex]

    Which is the form I am looking to show if I consider [itex]z\,=\,x^2[/itex].

    [tex]x^2\,\frac{d^2y}{dx}\,+\,x\,\frac{dy}{dx}\,-\,x^2\,y\,=\,0[/tex]

    So, the [itex]x^2[/itex] term in the general equation is equivalent to z in the heat balance equation. Does this conclude the problem?

    Thank you for the help:smile:
     
    Last edited: Jul 19, 2007
  13. Jul 19, 2007 #12

    Dick

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    You cannot consider x^2=z. z is constant, x is not. Cancel z from this:

    [tex]z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx }\,-\,z\,y\,=\,0[/tex]

    and multiply the result by x^2.
     
  14. Jul 19, 2007 #13
    So, I divide both sides of

    [tex]z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx}\,-\,z\,y\,=\,0[/tex]

    by z to get

    [tex]\frac{d^2T}{dx^2}\,+\,\frac{1}{x}\,\frac{dT}{dx}\,-\,y\,=\,0[/tex]

    and then multiply everything by [itex]x^2[/itex] to get

    [tex]x^2\,\frac{d^2T}{dx^2}\,+\,x\,\frac{dT}{dx }\,-\,x^2\,y\,=\,0[/tex]

    Why do I multiply by [itex]x^2[/itex] though?
     
  15. Jul 19, 2007 #14

    Dick

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    Just because that was the form you said you wanted to show in the problem statement.
     
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