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VinnyCee
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Homework Statement
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]
Homework Equations
http://en.wikipedia.org/wiki/Nondimensionalization"
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]
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