# Cylindrical continuity eq using cartesian substitution.

1. Jul 26, 2013

### psyclone

I've have two questions, but if my assumption is incorrect for the first, it will also be incorrect for the second. (in-terms of physics.)
For a two dimensional cylinder, using cylindrical co-ordinates (as follows), taking v(subscript-r) => velocity normal to cylinder surface & v(subscript-phi) => velocity parallel to surface (please refer to diagram for more detail).
$$\frac{1}{r}\frac{\partial v_{\phi}}{\partial \phi}+\frac{1}{r} \frac{\partial (r v_{r})}{\partial r}=0$$
Using the the product rule, can we separate out the second term into two terms, creating a continuity equation consisting of three terms?
$$\frac{\partial v_{\phi}}{\partial \phi}+ v_{r}\frac{\partial r}{\partial r} + r \frac{\partial v_{r}}{\partial r}=0$$
If that's valid. Can we make a substitution by using the cartesian form of the continuity equation, to change the normal velocity (v(subscript-r)) to the parallel velocity (v(subscript-phi))?
$$\frac{\partial u }{\partial x}+\frac{\partial v }{\partial y}=0 ==> \int d v =-\int_{0}^{\delta}\frac{\partial u } {\partial x}dy$$
To yield,
$$\frac{\partial v_{\phi}}{\partial \phi}-\frac{\partial u}{\partial y}\frac{\partial y}{\partial x}\delta + r \frac{\partial v_{r}}{\partial r} =0$$
Your help will be much appreciated. Ref: diagram Int. Comm. Heat Mass Transfer, Vol. 28, No. 8, pp 1127, 2001.

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