# Shell balances in cylindrical coordinates

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1. Sep 10, 2015

### MexChemE

Hello, PF! I have some doubts about setting up shell balances in a cylindrical geometry. Consider a fluid flowing down a vertical pipe. In order to perform the momentum balance, we take a cylindrical (annular) shell of lenght L and width Δr. The analysis of such system can be found in chapter 2 of BSL's Transport Phenomena (section 2.3).

Now, in the book, the rate of momentum entering the shell in the axial direction through z = 0 is given by
$$(2\pi r\Delta r) \left. \phi_{zz} \right|_{z=0}$$
Where apparently $2\pi r\Delta r$ is the cross-sectional area of the annular shell. This confuses me; if I calculate the area by integrating over the annular cross-section, I get this
$$A = \int_r^{r+ \Delta r} \int_0^{2 \pi} r \ d \theta dr = \pi (2r \Delta r + (\Delta r)^2)$$
I don't understand why was the $(\Delta r)^2$ term left out from the area expression in the book.

Also, after setting up a shell balance in cartesian coordinates, we have to divide the entire balance equation by the entire volume of the shell (see section 2.2 of BSL). Apparently, this is not the case in cylindrical coordinates. Instead of dividing by 2πrΔrL, the authors divided by 2πΔrL; why did they leave out r?

Thanks in advance for any input!

2. Sep 10, 2015

### Staff: Mentor

They are looking at a thin annular region where the $(\Delta r)^2$ is going to be negligible. They are going to be dividing by 2πΔrL and letting Δr approach zero, so the $(\Delta r)^2$ contribution will drop out anyway.
Maybe they liked the form of the resulting equation better. It doesn't really matter. If they divided by r, the resulting equation would be equivalent.

Chet

3. Sep 11, 2015

### MexChemE

Got it. I did think that at first, but the limit had not been taken at that point, so the confusion arised.
I checked for myself and the equations ended up being the same. Personally, I prefer dividing by the whole volume because it makes more sense to me.