Question bank — Potential flow — irrotational, inviscid; superposition of basic flows

False — it is a model. Real fluids have viscosity, but outside thin boundary layers the flow often behaves nearly inviscid, so the model captures lift and pressure well.

False — global circular paths are not the same as local spin. The free vortex has $\nabla \times \vec{v}=0$ everywhere except its singular centre, so it is irrotational.

True — Laplace's equation is linear, so any sum of solutions is a solution; this is exactly what makes superposition legal.

False — pressure comes from Bernoulli's equation, which is nonlinear in speed ($p+\frac{1}{2}\rho |\vec{v}{|}^2=$const). Add velocity fields first, then compute pressure once from the total speed.

False — it predicts zero net drag (d'Alembert's paradox) because there is no viscosity and no wake; it is trustworthy for lift and pressure, not drag.

True — with $u=\partial \psi /\partial y,v=-\partial \psi /\partial x$, the divergence becomes ${\psi }_{xy}-{\psi }_{yx}=0$ identically, so mass is conserved by construction.

True — the Cauchy–Riemann equations force $\nabla \varphi$ and $\nabla \psi$ to be perpendicular; geometrically $\nabla \varphi$ points along the flow while streamlines run with the flow, so equipotentials cut streamlines squarely, forming an orthogonal grid (see figure s02).

True — the stagnation streamline is the special streamline that passes through a stagnation point and splits the flow into inside/outside; since no fluid crosses any streamline, this separatrix acts exactly like a rigid boundary. That is how the cylinder and the Rankine nose "appear".

False — the vortex $\psi =-\frac{\Gamma }{2\pi }\ln r$ is constant on $r=a$, so the circle stays a streamline; it changes the speeds and hence the pressure, producing lift, not a new shape.

Only for irrotational flow, and even then $\varphi$ is single-valued globally only in a simply connected region; zero curl is the local precondition for a gradient, but a hole (vortex) can still make $\varphi$ multi-valued.

Wrong — a loop enclosing the centre has circulation $\Gamma \ne 0$. Irrotationality only guarantees zero circulation for loops that do not wrap the singular centre (the region isn't simply connected).

Reversed — $|\sin \theta |$ is largest at top/bottom ($\theta =\pm 90^{\circ }$), giving max speed $2U$ there; the stagnation points are at the front and back ($\theta =0,\pi$) where $\sin \theta =0$ (see figure s02).

As $r\to 0$, $u_r\to \infty$: the origin is a genuine singularity, not a real physical point. The model is valid only away from it.

A single source with uniform flow gives an open half-body (a rounded Rankine nose upstream that never closes downstream). You need a matching sink of equal strength to close the body into a Rankine oval.

Viscosity does not appear — the Kutta–Joukowski theorem is a purely potential-flow result. Viscosity's role is only to select the physical value of $\Gamma$ (via the Kutta condition), not to enter the lift expression.

Irrotational gives $\vec{v}=\nabla \varphi$ (in a simply connected region); incompressible gives $\nabla \cdot \vec{v}=0$; substituting yields ${\nabla }^2\varphi =0$, so all three velocity components are encoded in one harmonic scalar.

$\psi$ makes streamlines visible (its constant lines are the flow paths) and satisfies continuity automatically, whereas $\varphi$'s constant lines are equipotentials perpendicular to the flow.

Superposing a vortex onto the cylinder flow speeds one side and slows the other; Bernoulli then gives a pressure difference and a net sideways force — the Magnus effect, quantified by $L^{\prime }=\rho U\Gamma$.

Because Laplace's equation is linear, hard flows are just sums of a few catalogued simple flows (uniform, source, vortex, doublet), turning fluid dynamics into Lego assembly.

Bernoulli is quadratic in speed, so you must first add all velocity contributions into one total $\vec{v}$, then square its length once — squaring a sum is not the sum of squares.

$\nabla \varphi$ points in the direction of steepest potential rise, which is the flow direction $\vec{v}$; streamlines run along $\vec{v}$; so a curve of constant $\varphi$ (perpendicular to $\nabla \varphi$) must be perpendicular to the streamline — the Cauchy–Riemann equations state this algebraically.

Undefined/infinite — the centre is a singularity carrying all the circulation; everywhere else the vorticity is exactly zero.

The radial speed $m/(2\pi r)\to \infty$; the model breaks down at the singular point and is only meaningful for $r>0$.

With $m<0$ the source becomes a sink and its inflow now balances the stream on the upstream (left, $\theta =\pi$) side just as a source does; the on-axis balance $U-\frac{m}{2\pi x}=0$ still gives $x_s=m/(2\pi U)<0$, i.e. upstream. (The formula $x_s=-m/(2\pi U)$ is written for $\theta =\pi$ where $x=-|x|$.)

At the stagnation points $\theta =0,\pi$ where speed is zero — by Bernoulli, minimum speed means maximum pressure.

At the critical circulation ${\Gamma }_p=4\pi Ua$: the two stagnation points merge at the bottom ($\theta =-\frac{\pi }{2}$) there, and for $\Gamma >4\pi Ua$ a single stagnation point sits in the fluid below the cylinder.

Not necessarily — if circulation around the hole is nonzero, $\varphi$ increases by $\Gamma$ each loop, becoming multi-valued even though $\vec{v}$ stays single-valued.

Exactly zero (d'Alembert's paradox) — a consequence of no viscosity and fore-aft symmetric pressure, regardless of body shape.