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Given the Reynolds' Equation in cartesian coordinates:

$$ \frac{\partial}{\partial x}(h^3\frac{\partial p}{\partial x}) + \frac{\partial}{\partial z}(h^3\frac{\partial p}{\partial z})=6\mu r \omega \frac{\partial h}{\partial x} $$

how do I change it to obtain the Reynolds equation in cylindrical coordinates, given the following coordinate transformation: $$ x=r \cos (\theta)\\ z=-r \sin (\theta) $$

I would like to know how to find the differential operators and how to substitute them in the equation. The equation to obtain is $$ \frac{\partial}{\partial r}(rh^3\frac{\partial p}{\partial r}) + \frac{1}{r}\frac{\partial}{\partial \theta}(h^3\frac{\partial p}{\partial \theta})=6\mu r \omega \frac{\partial h}{\partial \theta} $$

Edit:

So I followed the suggestion and find that: $$ \frac{\partial}{\partial x} = \cos(\theta)\frac{\partial}{\partial r}-\frac{1}{r}\sin(\theta)\frac{\partial}{\partial \theta}\\ \frac{\partial}{\partial z} = -\sin(\theta)\frac{\partial}{\partial r}-\frac{1}{r}\cos(\theta)\frac{\partial}{\partial \theta} $$

Substituting this operators on the original equation, $$ \left( \cos(\theta)\frac{\partial}{\partial r}-\frac{1}{r}\sin(\theta)\frac{\partial}{\partial \theta} \right) \left(h^3 \left(\cos(\theta)\frac{\partial}{\partial r}-\frac{1}{r}\sin(\theta)\frac{\partial}{\partial \theta}\right) p\right) + \left(-\sin(\theta)\frac{\partial}{\partial r}-\frac{1}{r}\cos(\theta)\frac{\partial}{\partial \theta} \right) \left(h^3 \left( -\sin(\theta)\frac{\partial}{\partial r}-\frac{1}{r}\cos(\theta)\frac{\partial}{\partial \theta} \right) p \right)=6\mu r \omega \left( \cos(\theta)\frac{\partial}{\partial r}-\frac{1}{r}\sin(\theta)\frac{\partial}{\partial \theta} \right)h $$

Evaluating the expressions on the left-hand side of the equation, using the trigonometric identity $\sin^2(\theta)+\cos^2(\theta)=1$, the product rule in some derivatives and canceling out a few terms, I get to:

$$ \frac{\partial}{\partial r}\left(h^3\frac{\partial p}{\partial r}\right)+ \frac{1}{r}h^3\frac{\partial p}{\partial r} + \frac{1}{r^2}\frac{\partial}{\partial \theta}\left( h^3 \frac{\partial p}{\partial \theta} \right) = 6\mu \omega r \left( \cos \theta\frac{\partial h}{\partial r} -\frac{1}{r}\sin \theta\frac{\partial h}{\partial \theta} \right) $$

If I multiply this equation by $r$, and use the chain rule, I get what I want on the left-hand side: $$ \frac{\partial}{\partial r}\left(rh^3 \frac{\partial p}{\partial r} \right) + \frac{1}{r}\frac{\partial}{\partial \theta}\left( h^3 \frac{\partial p}{\partial \theta} \right) = 6\mu \omega r \left( \cos \theta\frac{\partial h}{\partial r} -\frac{1}{r}\sin \theta\frac{\partial h}{\partial \theta} \right) $$

The right-hand side of this equation, however, is wrong. How do I get from $6\mu \omega r \left( \cos \theta\frac{\partial h}{\partial r} -\frac{1}{r}\sin \theta\frac{\partial h}{\partial \theta} \right)$ to $6\mu r \omega \frac{\partial h}{\partial \theta}$ ?

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I didn't have time to solve the problem completely but I can sketch the idea. This is literally an application of the chain rule. It's probably easier to go the other way, writing $\partial/\partial r$ and $\partial / \partial \theta$ in terms of $\partial / \partial x$ and $\partial / \partial z$.

$$ \frac \partial {\partial r} = \frac {\partial x}{\partial r} \frac \partial {\partial x} + \frac {\partial z}{\partial r} \frac \partial {\partial z} = \cos \theta \frac \partial {\partial x} -\sin \theta \frac \partial {\partial y} = \frac x r \frac \partial {\partial x} +\frac z r \frac \partial {\partial z} $$

$$ \frac \partial {\partial \theta } = \frac {\partial x}{\partial \theta} \frac \partial {\partial x} + \frac {\partial z}{\partial \theta} \frac \partial {\partial z} = - r\sin\theta \frac \partial {\partial x} -r\cos\theta \frac \partial {\partial y} = z \frac \partial {\partial x} -x\frac \partial {\partial z} $$

If you wish, you can solve these for $\partial / \partial x$ and $\partial / \partial z$ in terms of $\partial/\partial r$ and $\partial / \partial \theta$.

Then substitute these derivatives into your equation! Good luck.

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  • $\begingroup$ Hey, just did that. I'm getting to right track, just edit the document. The left-hand side of the equation is the final form I want. The only problem now is the right-hand side of the equation. Any ideas? $\endgroup$ – Thales Mar 4 '17 at 15:32

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