Potential flow — irrotational, inviscid; superposition of basic flows: Worked examples

Before touching numbers, let us pin the vocabulary to pictures.

Definition The four building-block flows, each with its

$ψ$ (defined here, used all over the matrix) Every symbol below is re-anchored so you never need to recall the parent note.

Uniform flow (a steady wind of speed $U$ blowing rightward): $ψ_{uni} = U y = U r sin θ$ . Streamlines are straight horizontal lines.
Source / sink, strength $m$ = source strength = the volume of fluid pumped out per second per unit depth (units $m^{2} / s$ ). Picture a hose spraying evenly in all directions. Its stream function is $ψ_{source} = \frac{m}{2 π} θ$ , and its outward speed is $u_{r} = \frac{m}{2 π r}$ (derived in Ex 1). $m > 0$ = source (outflow), $m < 0$ = sink (inflow, like a plughole).
Free vortex, strength $Γ$ = circulation = how much the fluid swirls, found by adding up tangential speed all the way round a loop (units $m^{2} / s$ ). Picture stirring tea. Its stream function is $ψ_{vortex} = - \frac{Γ}{2 π} ln r$ , and its tangential (anticlockwise) speed is $u_{θ} = \frac{Γ}{2 π r}$ (derived in Ex 4b). $Γ > 0$ = anticlockwise, $Γ < 0$ = clockwise. See Vorticity and circulation.
Doublet, strength $κ$ = doublet strength = a source and an equal sink squashed infinitely close together ("fluid appears here, disappears a hair's breadth away"). Its stream function is $ψ_{doublet} = - \frac{κ}{2 π} \frac{sin θ}{r}$ . It is the brick we superpose onto a uniform wind to conjure a solid cylinder (Ex 6, Ex 8).

We will read velocity off $ψ$ using the parent's rule, restated so you never flip a sign by accident:

The scenario matrix

Every problem in this topic is one of these cells. The worked examples below are tagged with the cell(s) they cover, so together they fill the table. (All strength symbols $m, Γ, κ$ were defined just above.)

Cell	What varies	Example(s)
A. Sign of source strength $m$	$m > 0$ (source, outflow) vs $m < 0$ (sink, inflow)	Ex 1, Ex 2
B. Quadrant of the field point	stagnation on $- x$ axis ( $θ = π$ ) vs a point off-axis	Ex 3
C. Sign of circulation $Γ$	$Γ > 0$ (anticlockwise) vs $Γ < 0$ (clockwise) → which way lift points	Ex 5
D. Degenerate cancellation	equal source+sink annihilate, AND vortex+uniform where $u_{θ}$ cancels the wind	Ex 4, Ex 4b
E. Limiting behaviour	$r \to \infty$ (far field → uniform) and $r \to 0$ (near field → singularity blows up)	Ex 6
F. Real-world word problem	spinning ball / drain whirlpool with actual units	Ex 7
G. Exam twist	cylinder with circulation: where do the two stagnation points sit, and when do they merge?	Ex 8

Worked example

Ex 1 — Cell A (source, $m > 0$ ): speed at a given radius A 2-D source of strength $m = 12 m^{2} / s$ (volume flow per unit depth) sits at the origin. Find the radial speed $u_{r}$ at $r = 3 m$ , and state its direction.

Forecast: Guess before computing — as you walk away from a source, does the fluid speed up or slow down?

Derive the source velocity from mass conservation. All $m$ of the outflow must cross every circle of radius $r$ ; the circle's circumference is $2 π r$ , so speed = flow ÷ length = $u_{r} = \frac{m}{2 π r}$ . Why this step? Bigger circle, same flow spread thinner → slower. We derive rather than recall, honouring the no-memorization promise.
Plug in. $u_{r} = \frac{12}{2 π ( 3 )} = \frac{12}{6 π} = \frac{2}{π} \approx 0.637 m/s$ . Why this step? Just arithmetic once the formula is trusted.
Direction. $m > 0 \Rightarrow$ outward (away from origin), and $u_{θ} = 0$ (a pure source has no swirl).

Verify: Units: $(m^{2} / s) / m = m/s$ ✓. Sanity: your forecast should have been "slows down" — indeed $u_{r} \propto 1/ r$ . At $r = 6 m$ it would halve to $\approx 0.318$ . ✓

Worked example

Ex 3 — Cell B (off-axis field point): total velocity of uniform + source Combine a rightward uniform wind $U = 2 m/s$ with a source $m = 8 π m^{2} / s$ at the origin (a Rankine half-body). Find the velocity vector at the point $(x, y) = (3, 4)$ .

Figure — what each arrow means (a 3-4-5 triangle drawn to scale): The origin holds the coral dot (the source). The dashed line runs from origin to the field point $(3, 4)$ ; it is the hypotenuse of a 3-4-5 right triangle, so it is labelled " $r = 5$ ". At the point $(3, 4)$ three arrows start tip-to-tail: the mint arrow is the uniform wind (pure rightward, length $2$ ); the coral arrow is the source's outward push at that point (small, pointing away from the origin, components $0.48$ right and $0.64$ up); the lavender arrow is their vector sum — the total velocity we compute below. Read the arrows before the algebra: the picture already tells you the answer leans right-and-slightly-up.

Forecast: At a point up and to the right of the source, the source pushes out-and-up while the wind pushes right. Will the combined vector point more "up" or more "right"?

Write each piece in $x, y$ . Uniform: $u_{1} = U = 2, v_{1} = 0$ . Source: it points radially outward, so its Cartesian components are $u_{2} = \frac{m}{2 π r} \frac{x}{r} = \frac{m}{2 π} \frac{x}{r ^{2}}$ , $v_{2} = \frac{m}{2 π} \frac{y}{r ^{2}}$ . Why this step? Superposition adds velocity fields, so we must express both in the same coordinates before adding. The factor $x / r$ (= $cos θ$ ) is just "how much of the outward direction points along $x$ " — read it off the dashed radius line in the figure.
Compute $r^{2}$ . $r^{2} = 3^{2} + 4^{2} = 25$ , so $r = 5$ (the " $r = 5$ " label on the dashed line). Why this step? We need the distance to scale the source's $1/ r$ decay.
Source components. $\frac{m}{2 π} = \frac{8 π}{2 π} = 4$ . So $u_{2} = 4 \cdot \frac{3}{25} = \frac{12}{25} = 0.48$ , $v_{2} = 4 \cdot \frac{4}{25} = \frac{16}{25} = 0.64$ — the coral arrow's components.
Add. $u = 2 + 0.48 = 2.48 m/s$ , $v = 0 + 0.64 = 0.64 m/s$ — the lavender arrow. Why this step? This is the actual superposition — legal because Laplace's equation is linear (see Laplace's equation).

Verify: Speed $= 2.4 8^{2} + 0.6 4^{2} = 6.1504 + 0.4096 = 6.56 \approx 2.561 m/s$ , safely above the wind's $2$ (the source adds energy locally). Forecast check: $u > v$ , so the arrow points more right than up — the wind dominates at this modest radius. ✓

Worked example

Ex 4 — Cell D (degenerate cancellation, source vs wind): the stagnation point of Ex 3's flow Same flow as Ex 3 (uniform $U = 2$ , source $m = 8 π$ ). Find the stagnation point — the one place where the fluid is perfectly still.

Forecast: The source pushes outward everywhere; the wind blows right. Where can these exactly cancel — upstream (left) or downstream (right) of the source?

Where can $v = 0$ ? On the $x$ -axis ( $y = 0$ ) the source has no vertical push, so $v = 0$ automatically. Look there. Why this step? A stagnation point needs both $u = 0$ and $v = 0$ ; the $x$ -axis kills $v$ for free, halving the work.
On the negative $x$ -axis ( $θ = π$ , so $x < 0$ ), the source pushes left (against the wind). Set $u = 0$ : $U + \frac{m}{2 π x} = 0$ . Why the negative axis? Only upstream does the source's push oppose the wind; downstream they'd add and never cancel.
Solve. $x_{s} = - \frac{m}{2 π U} = - \frac{8 π}{2 π \cdot 2} = - \frac{8 π}{4 π} = - 2 m$ .

Verify: At $x = - 2$ : source outward speed $= \frac{m}{2 π r} = \frac{8 π}{2 π \cdot 2} = 2 m/s$ pointing left, i.e. $u_{source} = - 2$ ; wind $u = + 2$ ; sum $= 0$ ✓. This is the first "degenerate" case — a source velocity annihilating a wind velocity at one point. ✓

Worked example

Ex 4b — Cell D (the two other degeneracies): source+sink annihilation, and vortex+wind swirl-cancellation Two quick degenerate scenarios promised by the matrix. (a) A source $m = + 6 m^{2} / s$ and a sink $m = - 6 m^{2} / s$ are placed at the same point. What flow remains? (b) A free vortex $Γ = 4 π m^{2} / s$ centred at the origin sits in a rightward wind $U = 2 m/s$ . On the $+ y$ -axis the vortex swirl points leftward (opposing the wind). At what radius $r$ does the vortex's tangential speed exactly cancel the wind's leftward component there?

Forecast: (a) Two equal-and-opposite pumps at one spot — is there any net fluid appearing anywhere? (b) The vortex speed dies as $1/ r$ ; is there one special radius where it matches the wind?

(a) Re-anchor the source stream function. From the definitions box, a source of strength $m$ has $ψ_{source} = \frac{m}{2 π} θ$ — this simply says "the amount of flow you have swept past grows in proportion to the angle $θ$ you have swung through," which is exactly what a hose spraying evenly must give. So source: $ψ_{+} = \frac{6}{2 π} θ$ ; sink: $ψ_{-} = \frac{- 6}{2 π} θ$ . Why use $ψ$ and not velocities here? Because we want to know whether any flow survives everywhere at once. Adding the two stream functions and getting a single constant instantly proves "no streamlines, no flow" — a global statement that one velocity vector at one point could never establish. Sum: $ψ = \frac{6}{2 π} θ - \frac{6}{2 π} θ = 0$ everywhere. Why is this legal? Stream functions of co-located flows add for the same reason velocities do — Laplace's equation is linear.
(b) Re-anchor and use the vortex tangential speed. From the definitions box, a free vortex has $ψ_{vortex} = - \frac{Γ}{2 π} ln r$ . Apply our reading rule $u_{θ} = - \frac{\partial ψ}{\partial r}$ : since $\frac{d}{d r} ln r = \frac{1}{r}$ , we get $u_{θ} = - (- \frac{Γ}{2 π} \cdot \frac{1}{r}) = \frac{Γ}{2 π r}$ — derived here, not recalled. On the $+ y$ -axis, anticlockwise means pointing in the $- x$ direction, so its $x$ -component is $- \frac{Γ}{2 π r}$ . The wind adds $+ U$ . Set the $x$ -component to zero: $U - \frac{Γ}{2 π r} = 0$ . Why this step? Cancellation needs the two horizontal pushes equal and opposite; only the horizontal parts matter on this axis.
Solve. $r = \frac{Γ}{2 π U} = \frac{4 π}{2 π ( 2 )} = 1 m$ .

Verify: (a) $ψ \equiv 0$ ⇒ no streamlines, no flow ✓ — matches intuition that a co-located source+sink of equal strength cancel. (b) At $r = 1$ : vortex $x$ -speed $= - \frac{4 π}{2 π ( 1 )} = - 2$ , wind $+ 2$ , sum $0$ ✓. Both remaining Cell-D degeneracies are now on the table. ✓

Worked example

Ex 5 — Cell C (sign of $Γ$ ): direction of Magnus lift A cylinder in a rightward stream $U$ , air density $ρ = 1.23 kg/m^{3}$ , carries circulation. Case (i): $Γ = + 6 m^{2} / s$ (anticlockwise). Case (ii): $Γ = - 6 m^{2} / s$ (clockwise). With $U = 10 m/s$ , find the lift per unit span $L^{'}$ and its direction for each.

Forecast: A ball spinning so its top surface moves with the wind — does it climb or dive?

State the formula. $L^{'} = ρ U Γ$ . Why this step? This is the standard sign convention above: anticlockwise $Γ > 0$ ⇒ upward lift.
Case (i) $Γ = + 6$ . $L^{'} = (1.23) (10) (6) = + 73.8 N/m$ → force upward.
Case (ii) $Γ = - 6$ . $L^{'} = (1.23) (10) (- 6) = - 73.8 N/m$ → force downward.

Verify: Units $kg/m^{3} \cdot m/s \cdot m^{2} / s = kg / (m \cdot s^{2}) = N/m$ ✓ (force per length). Flipping the sign of $Γ$ flips the lift direction, exactly as the Magnus effect demands — the whole content of Cell C. ✓

Worked example

Ex 6 — Cell E (limits $r \to \infty$ and $r \to 0$ ): building the cylinder and taking its limits Non-spinning cylinder of radius $a = 1 m$ in a stream $U = 5 m/s$ . (a) Derive the cylinder stream function and surface speed from uniform + doublet. (b) What does the flow look like far away ( $r ≫ a$ )? (c) What is the maximum surface speed and where? (d) What happens to the doublet alone as $r \to 0$ ?

Figure — what to look at: The butter-yellow disc is the solid cylinder ( $r = a$ ). The lavender curves are streamlines: far out they are nearly straight (the wind barely notices the obstacle), close in they part to flow around the disc. The two coral dots on the left and right of the disc are the front and rear stagnation points (fluid momentarily at rest). The two mint squares at top and bottom mark where streamlines squeeze tightest → fastest flow.

Forecast: Far from a small cylinder, should the wind notice it at all?

Superpose uniform + doublet (the WHY of the formula). Uniform flow has $ψ_{uni} = U y = U r sin θ$ . The doublet (defined at the top) has $ψ_{dou} = - \frac{κ}{2 π} \frac{sin θ}{r}$ . Add them: $ψ = U r sin θ - \frac{κ}{2 π} \frac{s i n θ}{r} = U sin θ (r - \frac{κ}{2 π U} \frac{1}{r}) .$ Why this step? We want a circle to become a streamline. Choose $a^{2} = \frac{κ}{2 π U}$ ; then $ψ = U sin θ (r - \frac{a ^{2}}{r})$ , and at $r = a$ the bracket is $a - a = 0$ , so $ψ = 0$ for all $θ$ . A constant- $ψ$ curve is a streamline, so that circle acts as a solid wall — a cylinder! This is the WHY behind the quoted $ψ = U sin θ (r - a^{2} / r)$ .
Surface speed (WHY of $- 2 U sin θ$ ). On the surface only $u_{θ}$ survives (the wall blocks $u_{r}$ ). Using $u_{θ} = - \frac{\partial ψ}{\partial r} = - U sin θ (1 + \frac{a ^{2}}{r ^{2}})$ , set $r = a$ : $u_{θ}_{r = a} = - U sin θ (1 + 1) = - 2 U sin θ$ . Why this step? The derivative of $ψ$ across streamlines is the speed (our top rule); evaluating at $r = a$ gives the surface value, no guessing.
Far field $r \to \infty$ . In $ψ = U sin θ (r - \frac{a ^{2}}{r})$ the $a^{2} / r$ term $\to 0$ , leaving $ψ \to U r sin θ = U y$ — pure uniform flow. In the figure the outer streamlines are already almost straight. Why this step? The doublet decays like $1/ r$ ; a bounded obstacle can only locally disturb an unbounded stream.
Max surface speed. $∣ u_{θ} ∣ = 2 U ∣ sin θ ∣$ is largest when $∣ sin θ ∣ = 1$ , i.e. $θ = \pm 9 0^{\circ}$ (top and bottom, the mint squares in the figure). Value $= 2 U = 2 (5) = 10 m/s$ . Why this step? Streamlines are squeezed most tightly over the "shoulders" of the cylinder → fastest flow there.
Near field $r \to 0$ (doublet alone). The doublet's speed scales like $κ / (2 π r^{2})$ , which $\to \infty$ as $r \to 0$ . It is a singularity — mathematically infinite, physically hidden inside the solid cylinder (the butter-yellow disc), so no real fluid ever reaches it.

Verify: At $θ = 0$ and $θ = 18 0^{\circ}$ , $u_{θ} = - 2 U sin θ = 0$ — the front and rear stagnation points (the coral dots in the figure), as expected for a symmetric non-spinning cylinder ✓. Max $10 m/s = 2 U$ ✓. Both ends of the limit spectrum (Cell E) are covered: gentle uniform flow at $\infty$ , blow-up at $0$ . ✓

Worked example

Ex 7 — Cell F (real-world word problem): the bathtub whirlpool Water drains through a small hole. Far from the hole the water rotates as a free vortex: $u_{θ} = \frac{Γ}{2 π r}$ . You measure a tangential speed of $0.30 m/s$ at radius $r = 0.10 m$ . (a) Find the circulation $Γ$ . (b) Predict the speed at $r = 0.02 m$ .

Forecast: Closer to the drain, does the water swirl faster or slower?

Solve for $Γ$ . From $u_{θ} = \frac{Γ}{2 π r}$ (derived in Ex 4b step 2): $Γ = 2 π r u_{θ} = 2 π (0.10) (0.30) = 0.06 π \approx 0.1885 m^{2} / s$ . Why this step? Circulation $Γ$ is the fixed strength of the vortex; measuring speed at one radius pins it down everywhere.
Predict at $r = 0.02$ . $u_{θ} = \frac{Γ}{2 π r} = \frac{0.06 π}{2 π ( 0.02 )} = \frac{0.06}{0.04} = 1.5 m/s$ . Why this step? Same $Γ$ , smaller $r$ → the $1/ r$ law makes the swirl race.

Verify: Ratio check: radius shrank by factor $5$ ( $0.10 \to 0.02$ ), so speed should grow by $5$ : $0.30 \times 5 = 1.5$ ✓. Note the free vortex is irrotational everywhere except the singular centre — the local element does not spin even though the whole path is circular (see the parent's mistake box). ✓

Worked example

Ex 8 — Cell G (exam twist): where do the stagnation points of a spinning cylinder go, and when do they merge? Cylinder radius $a$ , stream $U$ , circulation $Γ$ . (a) Derive the surface tangential speed by adding a vortex to the cylinder flow of Ex 6, then find the stagnation angles $θ_{s}$ . (b) For $U = 4 m/s$ , $a = 0.5 m$ , $Γ = - 4 π m^{2} / s$ , locate them. (c) At what $Γ$ do the two stagnation points merge — and what happens beyond that?

Figure — what to look at: The butter-yellow disc is the cylinder; the mint arrows on the left show the rightward stream $U$ ; the lavender arc shows the added clockwise ( $Γ < 0$ ) swirl. The two coral dots are the stagnation points computed below — both sit in the upper half of the cylinder because the clockwise swirl has dragged them up from their plain-cylinder positions (front/back).

Forecast: Adding a clockwise swirl ( $Γ < 0$ ) to a rightward wind — will the two stagnation points slide up or down the cylinder?

Superpose the vortex (WHY of the surface-speed formula). The non-spinning cylinder has surface speed $- 2 U sin θ$ (Ex 6). A free vortex adds a uniform tangential speed $u_{θ}^{vortex} = - \frac{Γ}{2 π r}$ , which at the surface $r = a$ equals $- \frac{Γ}{2 π a}$ (same at every $θ$ , since a vortex's speed depends only on $r$ ). Add: $u_{θ}_{r = a} = - 2 U sin θ - \frac{Γ}{2 π a} .$ Why this step? Superposition of velocities is legal (linear Laplace); the vortex leaves the circle a streamline (it only adds swirl, no radial flow through the wall), so the cylinder survives.
Set surface speed to zero. $- 2 U sin θ - \frac{Γ}{2 π a} = 0 \Rightarrow sin θ_{s} = - \frac{Γ}{4 π U a}$ . Why this step? A stagnation point is where surface speed vanishes; solving for $sin θ$ gives the angle(s).
Plug in. $sin θ_{s} = - \frac{- 4 π}{4 π ( 4 ) ( 0.5 )} = \frac{4 π}{8 π} = \frac{1}{2}$ . So $θ_{s} = 3 0^{\circ}$ and $θ_{s} = 15 0^{\circ}$ — the two coral dots in the figure, both in the upper half. Why two answers? $sin$ takes the same value at an angle and its supplement — two stagnation points, symmetric about the top.
Merge condition (part c). The two solutions of $sin θ_{s} = c$ coincide when $∣ c ∣ = 1$ : $\frac{Γ}{4 π U a} = 1 \Rightarrow ∣Γ∣ = 4 π U a$ . At that value the two dots collapse to a single point at the top ( $θ = 9 0^{\circ}$ , if $Γ < 0$ ) or bottom ( $θ = - 9 0^{\circ}$ , if $Γ > 0$ ). Beyond it ( $∣Γ∣ > 4 π U a$ ) $sin θ_{s}$ would have to exceed $1$ , which is impossible — so there is no stagnation point on the surface at all; it lifts off into the fluid above (or below) the cylinder. Why this step? $sin$ is capped at $\pm 1$ ; the algebra of that cap is the physics of the merge-then-detach behaviour beloved of examiners.

Verify: Merge value $∣Γ∣ = 4 π (4) (0.5) = 8 π \approx 25.13 m^{2} / s$ . Our test case used $∣Γ∣ = 4 π \approx 12.57 < 8 π$ , so two points exist — consistent with the $3 0^{\circ}$ / $15 0^{\circ}$ answer ✓. Forecast check: with $Γ < 0$ (clockwise), $sin θ_{s} > 0$ , so both points sit in the upper half — the swirl pulled them upward. ✓

Recall Quick self-test (reveal after guessing)

Source $m$ sign vs sink ::: $m > 0$ = source (outward), $m < 0$ = sink (inward); same formula $u_{r} = m / (2 π r)$ . Stagnation point of uniform+source ::: $x_{s} = - m / (2 π U)$ , on the upstream ( $- x$ ) axis where source push cancels wind. Equal source+sink at one point ::: Fields cancel exactly, $ψ \equiv 0$ , no flow at all. Direction of Magnus lift for $Γ > 0$ (anticlockwise) in rightward wind ::: Upward, $L^{'} = ρ U Γ > 0$ . Max surface speed on a plain cylinder ::: $2 U$ , at top and bottom ( $θ = \pm 9 0^{\circ}$ ). Free vortex speed law ::: $u_{θ} = Γ/ (2 π r)$ ; faster as $r \to 0$ . When do a spinning cylinder's two stagnation points merge? ::: When $∣Γ∣ = 4 π U a$ (then $sin θ_{s} = \pm 1$ ); beyond that they detach into the fluid.