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

1. Jul 18, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

Define a dimensionless radial coordinate as

$$x\,=\,r\,\sqrt{\frac{2\,h}{b\,k}}$$

and introduce $y\,=\,T\,-\,T_a$, and thus show the elementary equation

$$x^2\,\frac{d^2y}{dx}\,+\,x\,\frac{dy}{dx}\,-\,x^2\,y\,=\,0$$

describes the physical situation.

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

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

2. Relevant equations

http://en.wikipedia.org/wiki/Nondimensionalization" [Broken]

3. The attempt at a solution

Re-expressing the constants

$$z\,=\,\frac{2\,h}{b\,k}$$

solving the dimensionless radial coordinate for r

$$r\,=\,\frac{x}{\sqrt{z}}$$

Now, I substitute these along with $y\,=\,T\,-\,T_a[/tex] into the "physical situation" equation that I am supposed to non-dimensionalize $$\frac{\sqrt{z}}{x}\,\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,\left[\left(\frac{x}{\sqrt{z}}\right)\,\frac{dT}{d\left(\frac{x}{\sqrt{z}}\right)}\right]\,-\,z\,y\,=\,0$$ What do I do about the denominator in the second and fourth fractions? $$\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}$$ Last edited by a moderator: May 3, 2017 2. Jul 18, 2007 ### Dick 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. 3. Jul 18, 2007 ### VinnyCee Yes, [itex]\sqrt{z}$ is only a constant.

So you are saying that

$$\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,=\,\frac{d}{\frac{dx}{\sqrt{z}}}$$

Right? So, does that mean that

$$\frac{d}{\frac{dx}{\sqrt{z}}}\,=\,\frac{d\sqrt{z}}{dx}\,=\,\frac{d}{dx}\sqrt{z}\,=\,0$$

since $\sqrt{z}$ is constant?

Last edited: Jul 18, 2007
4. Jul 18, 2007

### Dick

Riggghhhhtttt.

5. Jul 18, 2007

### VinnyCee

But when I apply that to the equation above

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

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

$$x^2\,\frac{d^2y}{dx}\,+\,x\,\frac{dy}{dx}\,-\,x^2\,y\,=\,0$$

6. Jul 18, 2007

### Dick

Division by zero?????? z is nonzero. It cancels inside the brackets and gives a common factor of z for the two terms.

7. Jul 18, 2007

### VinnyCee

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

The $\frac{\sqrt{z}}{x}$ terms cancel each other out to one

$$\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,\frac{dT}{d\left(\frac{x}{\sqrt{z}}\right)}\,-\,z\,y\,=\,0$$

Now what? Multiply it all by $d\left(\frac{x}{\sqrt{z}}\right)$?

If I do that, I get

$$d\left(dT\right)\,-\,\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,z\,y\,=\,0$$

Then what?

8. Jul 18, 2007

### Dick

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
9. Jul 18, 2007

### VinnyCee

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
10. Jul 18, 2007

### Dick

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
11. Jul 18, 2007

### VinnyCee

OIC, from the equality in the third post

$$\frac{d}{d\left(\frac{x}{\sqrt{z}}\right)}\,=\,\frac{d}{\frac{dx}{\sqrt{z}}}\,=\,\sqrt{z}\,\frac{d}{dx}$$

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

$$\frac{\sqrt{z}}{x}\,\sqrt{z}\,\frac{d}{dx}\left(x\,\frac{dT}{dx}\right)\,-\,z\,y\,=\,0$$

$$\frac{z}{x}\,\frac{d}{dx}\left(x\,\frac{dT}{dx}\right)\,-\,z\,y\,=\,0$$

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

$$z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx}\,-\,z\,y\,=\,0$$

Which is the form I am looking to show if I consider $z\,=\,x^2$.

$$x^2\,\frac{d^2y}{dx}\,+\,x\,\frac{dy}{dx}\,-\,x^2\,y\,=\,0$$

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

Thank you for the help

Last edited: Jul 19, 2007
12. Jul 19, 2007

### Dick

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

$$z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx }\,-\,z\,y\,=\,0$$

and multiply the result by x^2.

13. Jul 19, 2007

### VinnyCee

So, I divide both sides of

$$z\,\frac{d^2T}{dx^2}\,+\,\frac{z}{x}\,\frac{dT}{dx}\,-\,z\,y\,=\,0$$

by z to get

$$\frac{d^2T}{dx^2}\,+\,\frac{1}{x}\,\frac{dT}{dx}\,-\,y\,=\,0$$

and then multiply everything by $x^2$ to get

$$x^2\,\frac{d^2T}{dx^2}\,+\,x\,\frac{dT}{dx }\,-\,x^2\,y\,=\,0$$

Why do I multiply by $x^2$ though?

14. Jul 19, 2007

### Dick

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