Visual walkthrough — Potential flow — irrotational, inviscid; superposition of basic flows

Step 1 — What a "flow picture" even is

WHAT. Before any algebra, fix the picture. A 2D flow lives on a flat sheet. At every point we draw a little arrow: which way the fluid moves there, and how fast (arrow length). Collect all arrows and you get a velocity field, written $\vec{v}=(u,v)$ where $u$ is the sideways (rightward, $+x$) speed and $v$ is the up ($+y$) speed.

WHY. Everything below is just stacking arrow-fields. If you can read an arrow diagram, you can follow the whole derivation.

PICTURE. Two ways to describe the same point:

• Cartesian $(x,y)$ — grid coordinates.
• Polar $(r,\theta )$ — distance $r$ from the origin, and angle $\theta$ measured counter-clockwise from the $+x$ axis. We need polar because the flows we add are circular / radial, and circles are ugly in $x,y$ but simple in $r,\theta$.

Step 2 — The straight river (uniform flow)

WHAT. The simplest flow: every arrow points right, all the same length $U$. So $u=U$ everywhere, $v=0$. In symbols the stream function is $\psi =Uy$.

WHY the stream function $\psi$? Recall from the parent: streamlines are the curves the fluid actually travels along, and they are exactly the level curves $\psi =\text{const}$. For the river, $\psi =Uy=\text{const}$ means $y=\text{const}$ — flat horizontal lines. Perfect: horizontal river, horizontal streamlines.

Term by term: $\underset{\text{streamline label}}{\underbrace{\psi }}=\underset{\begin{array}{c}\text{river speed}\\ \text{(a chosen number)}\end{array}}{\underbrace{U}}\cdot \underset{\begin{array}{c}\text{height of}\\ \text{the point}\end{array}}{\underbrace{y}}$

In polar (we will need it later), $y=r\sin \theta$, so ${\psi }_{\text{uniform}}=Ur\sin \theta .$

PICTURE. A family of equally spaced horizontal lines, all arrows the same length pointing right.

Step 3 — The magic dipole (the doublet)

WHAT. A doublet is what you get by placing a tiny source (fluid squirting out) infinitesimally close to a sink (fluid sucked in) and letting them merge, keeping their combined "strength" $\kappa$ finite. Its stream function is ${\psi }_{\text{doublet}}=-\frac{\kappa }{2\pi }\frac{\sin \theta }{r}.$

WHY do we reach for THIS flow and not, say, another source? We want something whose streamlines curl back on themselves — loops — because a solid round body must be surrounded by flow that goes around it, front to back. A lone source only pushes outward forever (no loops). A doublet's streamlines are closed circles, exactly the shape that can wrap a cylinder. That is why the doublet is the right Lego brick.

Term by term: ${\psi }_{\text{doublet}}=-\underset{\begin{array}{c}\text{strength}\\ \text{(how strong}\\ \text{the dipole is)}\end{array}}{\underbrace{\frac{\kappa }{2\pi }}}\cdot \underset{\begin{array}{c}\text{shape: strong near}\\ \text{origin, tilts with }\theta \end{array}}{\underbrace{\frac{\sin \theta }{r}}}$

The $\frac{1}{r}$ says the doublet's influence fades as you go out — near the origin it dominates, far away it dies. That "local push, global fade" is exactly what lets the river win far away and the doublet win up close.

PICTURE. Nested closed loops hugging the origin, arrows sweeping around them.

Step 4 — Stacking them (superposition)

WHAT. Laplace's equation is linear, so we may simply add the two stream functions: $\psi =\underset{\text{river}}{\underbrace{Ur\sin \theta }}-\underset{\text{doublet}}{\underbrace{\frac{\kappa }{2\pi }\frac{\sin \theta }{r}}}.$

WHY it's legal. Both $\psi$'s satisfy ${\nabla }^2\psi =0$. The sum of two solutions of a linear equation is again a solution, so the combined field is a genuine potential flow — no cheating.

The key algebra — factor out $\sin \theta$: $\psi =U\sin \theta \left(r-\frac{\kappa }{2\pi Ur}\right).$

Watch what we did: we pulled the common $\sin \theta$ to the front. This isolates a bracket that depends only on $r$. That bracket is about to hand us a circle.

PICTURE. Left: river arrows. Middle: doublet loops. Right (added): the arrows start to split and go around a blob near the origin.

Step 5 — A circle appears from nothing

WHAT. Ask: is there a streamline with $\psi =0$ that is a closed curve? Set $\psi =0$: $U\sin \theta \left(r-\frac{\kappa }{2\pi Ur}\right)=0.$

A product is zero when either factor is zero:

• $\sin \theta =0$ → the $x$-axis (the flow line coming straight in and out).
• $r-\frac{\kappa }{2\pi Ur}=0$ → $r^2=\frac{\kappa }{2\pi U}$, i.e. a fixed radius independent of $\theta$.

WHY this is the whole trick. A fixed radius, same for every angle, is a circle. Name it $a$: $\boxed{a^2=\frac{\kappa }{2\pi U}}⟺\kappa =2\pi U{a}^2.$

So if we choose the doublet strength to be $\kappa =2\pi U{a}^2$, the streamline $\psi =0$ contains the circle $r=a$. Fluid never crosses a streamline — so the flow behaves exactly as if a solid cylinder of radius $a$ sat there. We conjured a solid body by picking one number.

Term by term in the boxed result: $a^2=\frac{\kappa }{2\pi U}:a(\text{radius we want}),\kappa (\text{doublet strength we tune}),U(\text{river speed}).$

PICTURE. The blob of Step 4 snaps into a perfect circle of radius $a$; streamlines outside bend smoothly around it.

Step 6 — Reading the speed on the surface

WHAT. With $\kappa =2\pi U{a}^2$, rewrite: $\psi =U\sin \theta \left(r-\frac{a^2}{r}\right).$ The fluid on the surface moves along the circle, so only the tangential speed $u_{\theta }$ matters there. In polar, the tangential velocity is recovered from $\psi$ by $u_{\theta }=-\frac{\partial \psi }{\partial r}.$

WHY that formula (and why the minus sign)? By definition $\psi$ is built so that fluid flows along its level lines; the tangential speed is how fast $\psi$ changes as you step radially outward. The sign convention $u_{\theta }=-\partial \psi /\partial r$ makes counter-clockwise positive, matching how we drew $\theta$. This is the polar twin of $u=\partial \psi /\partial y,v=-\partial \psi /\partial x$ from the parent.

Do the derivative: $u_{\theta }=-\frac{\partial }{\partial r}\left[U\sin \theta \left(r-\frac{a^2}{r}\right)\right]=-U\sin \theta \left(1+\frac{a^2}{r^2}\right).$ Now stand on the surface, $r=a$, so $a^2/r^2=1$: $\boxed{u_{\theta }{|}_{r=a}=-2U\sin \theta }$

Term by term: $u_{\theta }=-\underset{\begin{array}{c}\text{from }1+1\end{array}}{\underbrace{2}}\underset{\text{river speed}}{\underbrace{U}}\underset{\begin{array}{c}\text{position on}\\ \text{the circle}\end{array}}{\underbrace{\sin \theta }}$

PICTURE. Arrows tangent to the circle, longest at top/bottom, shrinking to zero at front/back.

Step 7 — Every case on the surface (do not skip any!)

WHAT. Walk $\theta$ all the way around and read $u_{\theta }=-2U\sin \theta$ at each landmark.

WHY cover all four. A reader must never hit a spot on the cylinder we didn't explain. The two stagnation points (front and back) are where the incoming river is brought fully to rest — the flow splits there. The two speed maxima ($|u_{\theta }|=2U$, twice the free-stream speed) sit at top and bottom. The sign flip between top and bottom just records that fluid goes over the top one way and under the bottom the other way — symmetric.

Degenerate check — $U\to 0$: if the river stops, $u_{\theta }=-2U\sin \theta \to 0$ everywhere. No stream, no flow around anything. Consistent. Likewise $a\to 0$ (or $\kappa \to 0$) removes the doublet and we're left with the bare river $\psi =Ur\sin \theta$. Every limit is well behaved.

PICTURE. The circle with its four landmark arrows labelled: two zero-length (stagnation) at $0,\pi$; two full-length at $\pm 90^{\circ }$.

Step 8 — From speed to pressure (one honest warning)

WHAT. Where the fluid is fast, the pressure is low — that is Bernoulli's equation: $p+\frac{1}{2}\rho v^2=\text{const}.$ Since $|u_{\theta }|$ is largest at top and bottom, pressure is lowest there and highest at the front and back stagnation points.

WHY the warning. Bernoulli is nonlinear in $v$ (there's a $v^2$). So you may not add pressures the way you added velocities. The recipe is strict: add the stream functions → get total velocity → plug that single total velocity into Bernoulli once. For the plain (non-spinning) cylinder the pressure is perfectly symmetric front-to-back, so the net push is zero — this is d'Alembert's paradox from the parent. Add a vortex and the top/bottom symmetry breaks, giving lift via the Kutta–Joukowski theorem — the seed of the Magnus effect.

PICTURE. Color map on the circle: hot (low pressure) top & bottom, cool (high pressure) front & back.

The one-picture summary

Read it left to right: river $\psi =Ur\sin \theta$ $+$ doublet $\psi =-\frac{\kappa }{2\pi }\frac{\sin \theta }{r}$ $=$ cylinder flow, where choosing $\kappa =2\pi U{a}^2$ turns the streamline $\psi =0$ into the circle $r=a$, and the surface speed reads $u_{\theta }=-2U\sin \theta$.

Flashcards