Plane potential flow from a logarithmic velocity potential: 22-Mec-B6 Advanced Fluid Mechanics, May 2015, Q1

Question 1: 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 $\phi(r,\theta) = -\tfrac{A}{2\pi}\ln r$, with $A>0$ a constant of dimensions $\text{m}^{2}\,\text{s}^{-1}$ (area per unit time, i.e. a volume flow rate per unit depth). The potential depends on $r$ alone.

Find. The conjugate stream function $\psi$, a sketch of the two families of curves $\phi=\text{const}$ and $\psi=\text{const}$, the radial velocity component $V_r(r)$ with an identification of the flow pattern, and the physical interpretation of $A$.

Figure 1.1 — Streamlines (solid rays, drawn with the flow direction) and equipotentials (dashed circles) of $\phi=-\tfrac{A}{2\pi}\ln r$. The two families are mutually orthogonal everywhere, as they must be for a potential flow.

Approach. Apply the polar Cauchy–Riemann relations that link $\phi$ and $\psi$, integrate them for $\psi$, differentiate $\phi$ for the velocity components, and interpret $A$ from the volume flux across a closed circuit.

Part (a) — write the polar Cauchy–Riemann relations. For a plane incompressible potential flow the velocity components follow from either function, $$\begin{aligned} V_r&=\frac{\partial\phi}{\partial r}=\frac{1}{r}\frac{\partial\psi}{\partial\theta}\cr V_\theta&=\frac{1}{r}\frac{\partial\phi}{\partial\theta}=-\frac{\partial\psi}{\partial r} \end{aligned}$$ These are the statements that $\phi$ satisfies irrotationality and $\psi$ satisfies continuity; together they let either function be recovered from the other.
Differentiate the given potential. Because $\phi$ carries no $\theta$ dependence, $$\begin{aligned} \frac{\partial\phi}{\partial r}&=-\frac{A}{2\pi r}\cr \frac{\partial\phi}{\partial\theta}&=0 \end{aligned}$$
Integrate for the stream function. The first relation gives $\tfrac{1}{r}\tfrac{\partial\psi}{\partial\theta}=-\tfrac{A}{2\pi r}$, so $\tfrac{\partial\psi}{\partial\theta}=-\tfrac{A}{2\pi}$ and $\psi=-\tfrac{A}{2\pi}\theta+f(r)$. The second relation gives $\tfrac{\partial\psi}{\partial r}=-V_\theta=0$, hence $f(r)$ is a constant, which may be set to zero without changing any velocity: $$\boxed{\;\psi(r,\theta)=-\frac{A}{2\pi}\,\theta\;}$$ Equivalently, in complex-potential form $F(z)=\phi+i\psi=-\tfrac{A}{2\pi}\ln z$, whose real and imaginary parts reproduce both functions at once.
Part (b) — identify the two families of curves. Setting $\phi=\text{const}$ requires $\ln r=\text{const}$, i.e. $r=\text{const}$: the equipotentials are concentric circles centred on the origin. Setting $\psi=\text{const}$ requires $\theta=\text{const}$: the streamlines are radial rays emanating from the origin. Circles and rays intersect at right angles everywhere, which is the geometric signature of a potential flow, and Figure 1.1 above is the required sketch. The origin itself is a singular point at which the model breaks down.
Part (c) — evaluate the velocity components. From step 2, $$\boxed{\;V_r=\frac{\partial\phi}{\partial r}=-\frac{A}{2\pi r}\quad\text{and}\quad V_\theta=0\;}$$ With $A>0$ the radial velocity is negative at every radius, so fluid moves steadily inward along the rays and there is no swirl. The speed varies as $1/r$: at $r=1\ \text{m}$ it is $A/2\pi$, and it grows without bound as the origin is approached. This is a plane line sink of strength A located at the origin — the mirror image of the familiar line source, whose potential carries the opposite sign.
Part (d) — interpret the constant from the volume flux. The volume flow crossing any circle of radius $r$, per unit depth normal to the plane, is $$Q=\oint \mathbf{V}\cdot\mathbf{n}\,\mathrm{d}s=V_r\,(2\pi r) =\left(-\frac{A}{2\pi r}\right)(2\pi r)=-A .$$ The radius has cancelled: the same volume crosses every circle, as continuity demands for a flow with no sources or sinks other than the one at the origin. Hence $$\boxed{\;\begin{aligned} A&=\text{volume flow rate per unit depth drawn into the sink}\cr [A]&=\text{m}^{3}\,\text{s}^{-1}\,\text{m}^{-1}=\text{m}^{2}\,\text{s}^{-1} \end{aligned}\;}$$ The same conclusion follows from the stream function: going once around the origin changes $\theta$ by $2\pi$ and therefore changes $\psi$ by $-A$, and the change in $\psi$ between two streamlines is the volume flux between them. A practical reading of $A$ is the suction rate of a long slotted drain, a line of well points, or a porous collector pipe per metre of its length.

Quantity	Result
Stream function	$\psi=-\dfrac{A}{2\pi}\theta$ (arbitrary constant dropped)
Equipotential lines	Concentric circles $r=\text{const}$
Streamlines	Radial rays $\theta=\text{const}$, directed inward
Radial velocity	$V_r=-\dfrac{A}{2\pi r}$
Tangential velocity	$V_\theta=0$
Flow pattern	Plane line sink at the origin (no circulation)
Meaning of A	Volume flow rate per unit depth into the sink, $\text{m}^{2}\,\text{s}^{-1}$

Topic. Elementary plane potential flows: the line source and sink, and the pairing of velocity potential with stream function.

Key relations. Polar Cauchy–Riemann relations $V_r=\partial\phi/\partial r=(1/r)\partial\psi/\partial\theta$ and $V_\theta=(1/r)\partial\phi/\partial\theta=-\partial\psi/\partial r$; complex potential $F(z)=\phi+i\psi$; source/sink pair $F=\pm\tfrac{m}{2\pi}\ln z$; volume flux between two streamlines $Q=\psi_2-\psi_1$.

Why this works. A plane flow that is both incompressible and irrotational admits two scalar descriptions. Continuity, $\nabla\!\cdot\!\mathbf{V}=0$, guarantees a stream function whose level curves are the streamlines; irrotationality, $\nabla\times\mathbf{V}=0$, guarantees a velocity potential whose gradient is the velocity. The two are harmonic conjugates, so knowing one determines the other up to an additive constant, and their level curves must cross at right angles because $\nabla\phi\cdot\nabla\psi=V_rV_\theta-V_\theta V_r=0$. The logarithm is the fundamental solution of Laplace’s equation in two dimensions, which is exactly why it represents a point-like injection or withdrawal of fluid. Its strength is fixed not by the algebra but by the flux it carries, and because the flux is radius-independent the constant $A$ is a genuine property of the singularity rather than of the circle chosen to measure it. In practice such a solution is never used alone: superposed with a uniform stream it gives the Rankine half-body, and paired with an equal source at a finite spacing it gives the Rankine oval and, in the limit, the doublet that models flow past a cylinder.

Common pitfall. Two errors dominate. The first is a sign slip: a source has $\phi=+\tfrac{A}{2\pi}\ln r$, so the leading minus sign here makes the flow a sink, and answering “source” loses the identification mark in part (c). The second is quoting $A$ as a velocity or as a total discharge in $\text{m}^{3}\,\text{s}^{-1}$; it is a discharge per unit depth, and the units $\text{m}^{2}\,\text{s}^{-1}$ are worth checking dimensionally before writing the answer. A third, subtler slip is forgetting the $1/r$ factor in the polar form of the Cauchy–Riemann relations, which produces a stream function proportional to $r\theta$ and a flow that does not satisfy continuity.

Source. F. M. White, Fluid Mechanics, 8th ed., §4.4 and §8.2–8.3; P. K. Kundu, I. M. Cohen & D. R. Dowling, Fluid Mechanics, 6th ed., Ch. 6 (complex potential, source and vortex).

Question 1 of 6: Plane potential flow from a logarithmic velocity potential

Question 1: Plane potential flow from a logarithmic velocity potential (20 marks)