Question 2 of 6: The potential flow φ = −(A/2π) ln r

Question 2: The potential flow φ = −(A/2π) ln r (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A two-dimensional, incompressible, irrotational flow whose velocity potential in plane polar coordinates is $\phi=-\dfrac{A}{2\pi}\ln r$, with A a positive constant and no dependence on the angular coordinate θ.

Find. The stream function, a sketch of the two orthogonal families of curves, the radial velocity component with an identification of the flow pattern, and the physical interpretation of the constant that multiplies the logarithm.

Approach. Differentiate the potential to get the velocity components, then integrate the Cauchy–Riemann relations between φ and ψ in polar form to recover the stream function, and finally interpret the constant by evaluating the volume flux through a circuit enclosing the origin.

1. Part (a) — write down the polar velocity components from the potential. For an irrotational plane flow the velocity is the gradient of φ: $$V_r=\frac{\partial\phi}{\partial r}=-\frac{A}{2\pi r},\qquad V_\theta=\frac{1}{r}\frac{\partial\phi}{\partial\theta}=0 .$$ The potential contains no θ, so the flow has no swirl at all.
2. Integrate the first Cauchy–Riemann relation for the stream function. In polar coordinates the stream function satisfies $V_r=\dfrac{1}{r}\dfrac{\partial\psi}{\partial\theta}$ and $V_\theta=-\dfrac{\partial\psi}{\partial r}$. Substituting V_r from step 1, $$\frac{\partial\psi}{\partial\theta}=rV_r=-\frac{A}{2\pi} \quad\Longrightarrow\quad \psi=-\frac{A}{2\pi}\,\theta+f(r).$$
3. Close the integration with the second relation. Because V_θ = 0 we need $-\,\mathrm{d}f/\mathrm{d}r=0$, so f is a constant that may be set to zero without changing any velocity. Hence $$\boxed{\psi=-\frac{A}{2\pi}\,\theta}$$ The pair is exactly the complex potential $w(z)=\phi+i\psi=-\dfrac{A}{2\pi}\ln z$, which is the standard textbook sink.
4. Part (b) — identify the two families of curves. Setting φ = constant forces ln r = constant, so the equipotentials are concentric circles centred on the origin. Setting ψ = constant forces θ = constant, so the streamlines are straight radial rays. Figure 2 above is the required sketch; the two families meet at right angles everywhere, as they must for any analytic complex potential, and the streamline spacing crowds together as the origin is approached because the same volume flux must pass through an ever shorter circumference.
5. Part (c) — interpret the sign of the radial velocity. From step 1, $$\boxed{V_r=-\frac{A}{2\pi r},\qquad V_\theta=0}$$ With A > 0 the radial velocity is negative at every radius, meaning the fluid moves toward the origin along every ray. The pattern is therefore a two-dimensional (line) sink of strength A located at the origin: purely radial, irrotational everywhere except at the singular point r = 0, with a speed that grows without bound as 1/r. Had the sign of the potential been reversed the same algebra would describe a source.
6. Part (d) — evaluate the volume flux to fix the meaning of the constant. Take any circle of radius r about the origin and integrate the outward normal velocity over it, per unit depth into the page: $$Q=\int_0^{2\pi}V_r\,r\,\mathrm{d}\theta=\int_0^{2\pi}\left(-\frac{A}{2\pi r}\right)r\,\mathrm{d}\theta=-A .$$ The radius cancels, so the same volume passes every circle. The constant A is therefore the volume flow rate per unit depth swallowed by the sink, with units of m² s⁻¹, and the grouping A/2π that multiplies the logarithm is simply that flux spread over the 2π radians of a full circle. Equivalently, the jump in ψ on going once around the origin is $\Delta\psi=-A$, and a jump in stream function is by definition the volume flux between the two streamlines.
7. Address the circulation, since the printed symbol names it. The circulation about any closed circuit C is $$\Gamma=\oint_C \mathbf{V}\cdot \mathrm{d}\mathbf{s}=\int_0^{2\pi}V_\theta\,r\,\mathrm{d}\theta=0,$$ because V_θ vanishes identically. A sink carries volume flux but no circulation, so if the symbol Γ in the question is read literally then 2πΓ = 0 for this flow — the physically meaningful constant is A. The source–sink and the point vortex are the two conjugate members of the logarithmic family: $-\tfrac{A}{2\pi}\ln z$ carries flux and no circulation, while $\tfrac{i\Gamma}{2\pi}\ln z$ carries circulation Γ and no flux.
8. Put a number on it. For a sink of strength A = 0.60 m²/s the speed at r = 0.50 m is $|V_r| = 0.60/(2\pi\times0.50) = 0.191$ m/s, and the flux across that circle is $0.191\times 2\pi\times0.50 = 0.60$ m²/s per metre of depth — the radius has cancelled, exactly as step 6 predicts.