Paper 1, Section I, D

Fluid Dynamics | Part IB, 2009

A steady velocity field u=(ur,uθ,uz)\mathbf{u}=\left(u_{r}, u_{\theta}, u_{z}\right) is given in cylindrical polar coordinates (r,θ,z)(r, \theta, z) by

ur=αr,uθ=γr(1eβr2),uz=2αzu_{r}=-\alpha r, \quad u_{\theta}=\frac{\gamma}{r}\left(1-e^{-\beta r^{2}}\right), \quad u_{z}=2 \alpha z

where α,β,γ\alpha, \beta, \gamma are positive constants.

Show that this represents a possible flow of an incompressible fluid, and find the vorticity ω\omega.

Show further that

curl(uω)=ν2ω\operatorname{curl}(\mathbf{u} \wedge \boldsymbol{\omega})=-\nu \nabla^{2} \boldsymbol{\omega}

for a constant ν\nu which should be calculated.

[The divergence and curl operators in cylindrical polars are given by

divu=1rr(rur)+1ruθθ+uzzcurlu=(1ruzθuθz,urzuzr,1rr(ruθ)1rurθ) and ,when ω=[0,0,ω(r,z)],2ω=[0,0,1rr(rωr)+2ωz2].]\begin{aligned} \operatorname{div} \mathbf{u} &=\frac{1}{r} \frac{\partial}{\partial r}\left(r u_{r}\right)+\frac{1}{r} \frac{\partial u_{\theta}}{\partial \theta}+\frac{\partial u_{z}}{\partial z} \\ \operatorname{curl} \mathbf{u} &=\left(\frac{1}{r} \frac{\partial u_{z}}{\partial \theta}-\frac{\partial u_{\theta}}{\partial z}, \frac{\partial u_{r}}{\partial z}-\frac{\partial u_{z}}{\partial r}, \frac{1}{r} \frac{\partial}{\partial r}\left(r u_{\theta}\right)-\frac{1}{r} \frac{\partial u_{r}}{\partial \theta}\right) \\ \text { and ,when } \boldsymbol{\omega} &=[0,0, \omega(r, z)], \\ \nabla^{2} \boldsymbol{\omega} &\left.=\left[0,0, \frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial \omega}{\partial r}\right)+\frac{\partial^{2} \omega}{\partial z^{2}}\right] .\right] \end{aligned}

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