Question 2 of 6: Plane potential flow from a logarithmic velocity potential

Question 2: Plane potential flow from a logarithmic velocity potential (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 $\phi(r,\theta) = -\Gamma \ln r$ with $\Gamma > 0$ constant. Because $\phi$ has the dimensions of velocity times length, $\Gamma$ carries units of m²/s — an area per unit time, that is, a volume flow rate per unit depth. The potential depends on $r$ alone; there is no $\theta$-dependence anywhere in the problem.

Find. The conjugate stream function $\psi$; a sketch of the two families $\phi = \text{const}$ and $\psi = \text{const}$; the radial velocity $V_r(r)$ together with an identification of the flow pattern; and the physical meaning of the grouping $2\pi\Gamma$.

Approach. Differentiate $\phi$ to get the velocity components, then integrate the Cauchy–Riemann pair $V_r = \partial\phi/\partial r = (1/r)\,\partial\psi/\partial\theta$ and $V_\theta = (1/r)\,\partial\phi/\partial\theta = -\partial\psi/\partial r$ to recover $\psi$; read the two curve families off the functional forms; and identify the pattern from the sign of $V_r$ and the net flux through a closed circuit.

1. Extract the velocity components from the potential. In plane polars $V_r = \partial\phi/\partial r$ and $V_\theta = (1/r)\,\partial\phi/\partial\theta$. With $\phi = -\Gamma\ln r$, $$V_r = \frac{\partial}{\partial r}\left(-\Gamma\ln r\right) = -\frac{\Gamma}{r}, \qquad V_\theta = \frac{1}{r}\frac{\partial}{\partial\theta}\left(-\Gamma\ln r\right) = 0$$ The flow is purely radial, and $V_r$ is negative for every $r$ because $\Gamma > 0$.
2. Integrate for the stream function (part a). The polar definitions of $\psi$ are $V_r = (1/r)\,\partial\psi/\partial\theta$ and $V_\theta = -\partial\psi/\partial r$. The first gives $$\frac{\partial\psi}{\partial\theta} = r V_r = -\Gamma \quad\Longrightarrow\quad \psi = -\Gamma\theta + f(r)$$ and the second closes it: $V_\theta = -f'(r) = 0$, so $f$ is a constant that may be set to zero without loss of generality. Hence $$\boxed{\psi(r,\theta) = -\Gamma\,\theta}$$ Both functions are harmonic, $\nabla^{2}\phi = \nabla^{2}\psi = 0$, as any admissible potential–stream-function pair must be.
3. Identify the two curve families (part b). Setting $\phi = -\Gamma\ln r$ constant forces $r$ constant, so the equipotentials are concentric circles centred on the origin. Setting $\psi = -\Gamma\theta$ constant forces $\theta$ constant, so the streamlines are radial rays emanating from the origin. Figure 2 shows both families; they intersect at right angles everywhere, which is the geometric signature of a plane potential flow and a useful sanity check on the algebra of step 2. Note also that the equipotential circles crowd together as $r$ decreases, correctly indicating that the speed rises towards the centre.
4. State the radial velocity and name the pattern (part c). From step 1, $$\boxed{V_r = -\frac{\Gamma}{r},\qquad V_\theta = 0}$$ The minus sign is the whole answer to “identify the flow pattern”: fluid moves towards the origin along every ray, with speed growing as $1/r$. This is a two-dimensional line sink of strength $\Gamma$ located at the origin — equivalently, a point sink in the plane, or a line sink normal to the plane if the third dimension is restored. Had the potential been $+\Gamma\ln r$ the same algebra would give a source. Continuity is satisfied everywhere except at the origin, since $$\frac{1}{r}\frac{\partial (rV_r)}{\partial r} = \frac{1}{r}\frac{\partial}{\partial r}(-\Gamma) = 0$$ and the flow is irrotational everywhere except at the origin, where $r = 0$ makes $V_r$ singular. The origin is therefore a singular point that must be excluded from the flow domain.
5. Interpret the grouping $2\pi\Gamma$ (part d). Compute the volume flow rate per unit depth crossing any circle of radius $r$ centred on the origin: $$Q' = \oint \mathbf{V}\cdot \hat{\mathbf{n}}\,ds = \int_{0}^{2\pi} V_r\,r\,d\theta = \int_{0}^{2\pi}\left(-\frac{\Gamma}{r}\right) r\,d\theta = -2\pi\Gamma$$ The radius cancels identically, so the same volume flow crosses every circle — exactly what continuity demands for a flow with no sources or sinks except at the origin. Therefore $$\boxed{2\pi\Gamma = |Q'| = \text{the sink strength: the volume flow rate per unit depth swallowed at the origin}}$$ with SI units of m³/s per metre of depth, i.e. m²/s. The same quantity is the jump in $\psi$ on going once around the origin, $\Delta\psi = -2\pi\Gamma$, which is why the stream function of a sink is multivalued while that of a uniform stream is not.