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Flux Calculations in Cylindrical Coordinates #589

Answered by c-white
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A partial answer: The form of the rotational terms is somewhat consistent with that of the nonrotational curvilinear source terms in that an effort is made to exactly conserve angular momentum $R \rho v^\phi$ rather than azimuthal momentum $\rho v^\phi$, but not to do anything special for radial momentum $\rho v^R$.

The radial momentum equation in differential form is

$$\partial_t (\rho v^R) + \frac{1}{R} \partial_R (R (\rho v^R v^R + p)) + \frac{1}{R} \partial_\phi (\rho v^R v^\phi) + \partial_z (\rho v^R v^z) = \frac{1}{R} (\rho v^\phi v^\phi + p) + \Omega^2 R \rho + 2 \Omega \rho v^\phi.$$

In integral form, this is exactly

$$\partial_t \langle \rho v^R \rangle_{R\phi z}$$

$$+ \frac{1}{…

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