Velocity viscous flow around rotating cylinder

I am struggling to find an equation of flow velocity at distance $$r$$ around rotating cylinder with radius $$R$$, angular velocity $$w$$ in stationary viscous fluid with some density $$ρ$$ and viscosity $$\mu$$.

I found "Hagen–Poiseuille equation" $$U = \frac{(P_{2} - P_{1}) * (R^2-r^2)}{4\mu L}$$

But that equation is for pipe with radius $$R$$ and flow radius $$r$$ and require to know pressure difference, but rotating cylinder rotate flow due to viscosity, but not pressure explicitly.

• You are looking for Taylor–Couette flow Commented Feb 10, 2022 at 23:05
• Thank you, @Mauricio. That is exact what I need. Commented Feb 11, 2022 at 12:14

We can obtain a closed-form solution for steady unidirectional flow in a cylindrical coordinate system driven only by the rotation of the cylinder. All components of velocity vanish except the azimuthal component $$u_\theta$$ which is a function only of the radial coordinate $$r$$. For incompressible, viscous flow the Navier-Stokes equations in cylindrical coordinates reduce to

$$\mu\left[\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u_\theta}{\partial r} \right)- \frac{u_\theta}{r^2} \right]= 0,$$

Replacing partial with ordinary derivatives (since $$u_\theta$$ is a function of $$r$$ alone), we obtain

$$\frac{1}{r} \frac{d}{d r}\left(r \frac{d u_\theta}{d r} \right)- \frac{u_\theta}{r^2} = \frac{d^2 u_\theta}{dr^2}+\frac{1}{r} \frac{d u_\theta}{dr} - \frac{u_\theta}{r^2}=0$$

Multiplying both sides by $$r^2$$ we have the usual form of a second-order, homogeneous Euler-type differential equation

$$r^2\frac{d^2 u_\theta}{dr^2}+r\frac{d u_\theta}{dr} - u_\theta=0$$

This type of ODE can be solved by assuming solutions of the for $$r^n$$. Upon substitution of that form we get $$n^2 = 1$$ and a general solution

$$u_\theta = Ar + Br^{-1}$$

The boundary conditions for an infinite domain where the fluid is quiescent far from the cylinder are $$u_\theta(R) = \omega R$$ and $$u_\theta(r) \to 0$$ as $$r \to \infty$$. Applying these conditions and solving for the constants $$A,B$$ we get $$A= 0$$ and $$B= \omega R^2$$. Hence, the velocity is

$$u_\theta = \frac{\omega R^2}{r}$$

• Shouldn't velocity depend on viscos or density of medium. For instance, for air or water velocity distributions are different. Your conclusion is rigth for Navier-Stokes equations and Couette flow, but I don't know how to apply that equations to incompressible viscos fluid with some constant density. Thank you @RRL for efforts. Commented Feb 12, 2022 at 20:59
• @VadimOstantin: This is the basic laminar flow solution for circular Couette flow. In general for simple laminar flow the dependence of velocity on the viscosity and density will arise when there is a pressure gradient driving the flow. That is not the case here. For example in pressure driven flow in a tube you get the Hagen-Poiseuille profile you show above. This is because you don’t lose the term $-\frac{1}{\rho}\nabla p$ which balances the viscous term $\mu \nabla^2 \mathbf{u}$.
– RRL
Commented Feb 12, 2022 at 21:55
• You also get the dependence on the viscosity and density parameters when inertial forces are not negligible. That arises in flow between concentric cylinders at higher Reynolds number when Taylor vortices appear and the flow is no longer unidirectional.
– RRL
Commented Feb 12, 2022 at 21:59