Cylindrical continuity eq using cartesian substitution.

[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.

 

[Navin A S]

Your assumption is incorrect. The equation you have written is not a continuity equation and cannot be used to substitute for the normal velocity. The equation you have written is a conservation of momentum equation, which is used to describe the motion of a two-dimensional cylinder in cylindrical coordinates. The continuity equation describes the flow of mass or energy in a given area, while the conservation of momentum equation describes the motion of a particle in that same area. Therefore, they are two different equations and cannot be used interchangeably.