A point source of volume strength sits on the axis of symmetry at

PHYSICS · Multi-part · Fermi University Physics · Fluids & Continuum Physics · Kinematics, conservation laws and ideal flow · Kinematics, conservation laws and ideal flow

Fluid Dynamics, Ch Kinematics in three dimensions and Cauchy's equation of motion

Problem

A point source of volume strength Q=0.060 m3 s−1Q=0.060\,\mathrm{m^3\,s^{-1}} sits on the axis of symmetry at z=−0.05 mz=-0.05\,\mathrm{m}, and a point sink of equal strength sits at z=+0.05 mz=+0.05\,\mathrm{m}, in an otherwise still, incompressible fluid. For a point source of strength QQ at height z0z_0 on the axis, the Stokes stream function is Ψsource(r,z)=−Q4πz−z0r2+(z−z0)2,\Psi_{\text{source}}(r,z)=-\frac{Q}{4\pi}\frac{z-z_0}{\sqrt{r^2+(z-z_0)^2}}, and for a sink of the same strength at z0z_0 it is Ψsink(r,z)=+Q4πz−z0r2+(z−z0)2,\Psi_{\text{sink}}(r,z)=+\frac{Q}{4\pi}\frac{z-z_0}{\sqrt{r^2+(z-z_0)^2}}, where rr is the distance from the axis. The two stream functions add, because each velocity field satisfies ∇⋅u=0\nabla\cdot\mathbf u=0 and Ψ\Psi depends linearly on u\mathbf u. ((a)) Write the combined stream function Ψ=Ψsource+Ψsink\Psi=\Psi_{\text{source}}+\Psi_{\text{sink}}. ((b)) Evaluate Ψ\Psi on the axis (r=0r=0) at z=0z=0 (the midpoint), and at (r,z)=(0.05 m, 0)(r,z)=(0.05\,\mathrm m,\,0). ((c)) Use the flux rule Δ(flux)=2πΔΨ\Delta(\text{flux})=2\pi\Delta\Psi to find the volume flux crossing the circle of radius 0.05 m0.05\,\mathrm m centred on the axis in the plane z=0z=0.

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