Exercises — Potential flow — irrotational, inviscid; superposition of basic flows

$U$ ::: the speed of the uniform "straight river" flow, pointing along the $+x$ axis. $r,\theta$ ::: polar coordinates — $r$ is distance from the origin, $\theta$ is the angle measured anticlockwise from the $+x$ axis. $m$ ::: source strength = volume of fluid pumped out per second per unit depth ($m>0$ pushes out, $m<0$ sucks in). $\Gamma$ ::: circulation of a vortex = how much "swirl" it carries (units of speed × length). $\kappa$ ::: doublet strength (a source and sink squeezed infinitely close). $\varphi$ ::: velocity potential; $\vec{v}=\nabla \varphi$. $\psi$ ::: stream function; $\psi =$const traces a streamline the fluid glides along. $u_r,u_{\theta }$ ::: the two velocity pieces in polar form — $u_r$ points outward, $u_{\theta }$ points anticlockwise.

A source spits fluid straight outward in every direction equally. So there is no swirl ($u_{\theta }=0$) and only outward motion. Mass conservation: all the fluid $m$ pumped out must cross every circle of radius $r$. The circle has circumference $2\pi r$, so $u_r(2\pi r)=m\Rightarrow \boxed{u_r=\frac{m}{2\pi r}}.$ Integrating $u_r=\partial \varphi /\partial r$ gives $\varphi =\frac{m}{2\pi }\ln r$. The stream function is $\psi =\frac{m}{2\pi }\theta$.

$u_{\theta }=\frac{\Gamma }{2\pi r}.$ The flow is irrotational everywhere except the exact centre $r=0$, where $u_{\theta }\to \infty$ — a singular point. Away from the centre a tiny fluid cross floats around the circle without spinning about its own axis (that is what "irrotational" means: zero local spin, not zero path curvature).

Uniform flow is the same everywhere, so position is irrelevant: $u=U=3\text{ m/s}$, $v=0$.

$u_r=\frac{m}{2\pi r}=\frac{12\pi }{2\pi \cdot 3}=\frac{12\pi }{6\pi }=2\text{m/s}.$

Top, $\theta =90^{\circ }$: $u_{\theta }=-2(4)\sin 90^{\circ }=-8\text{ m/s}$. Speed = $|-8|=8\text{ m/s}$ (the minus just means it points clockwise/backward there). Front, $\theta =180^{\circ }$: $u_{\theta }=-2(4)\sin 180^{\circ }=0$. This is a stagnation point — the fluid stops dead as it hits the nose. Note the top speed $8=2U$ is exactly twice the free-stream speed. See the figure below.

On the $x$-axis ahead of the source, $\theta =\pi$ and the total $x$-velocity is $u=U-\frac{m}{2\pi x}$. The source pushes fluid leftward here, opposing the stream. They cancel where $u=0$: $x_s=-\frac{m}{2\pi U}=-\frac{8\pi }{2\pi \cdot 2}=-2\text{m}.$

Streamline test: a streamline is a curve the fluid never crosses — that means the velocity component pointing across it (the radial one, $u_r$) must be zero on it. At $r=a$: $u_r=U\cos \theta (1-\frac{a^2}{a^2})=U\cos \theta \cdot 0=0\text{for every }\theta .$ So no fluid ever pierces the circle — it acts exactly like a solid wall. ✓ Stagnation points: both $u_r=0$ (already true on the surface) and $u_{\theta }=0$. On $r=a$, $u_{\theta }=-2U\sin \theta$, which vanishes at $\sin \theta =0\Rightarrow \theta =0$ (rear) and $\theta =\pi$ (front). Those are the two stagnation points, front and back.

On the surface the speed is $v=|u_{\theta }|=2U|\sin \theta |$, so $v^2=4U^2{\sin }^2\theta$. Bernoulli between far upstream ($p_{\infty }$, speed $U$) and the surface: $p_{\infty }+\frac{1}{2}\rho U^2=p+\frac{1}{2}\rho (4U^2{\sin }^2\theta ).$ Solve for $p-p_{\infty }$ and divide by $\frac{1}{2}\rho U^2$: $C_p=1-4{\sin }^2\theta .$ At $\theta =90^{\circ }$: $C_p=1-4(1)=-3$. Negative means suction — pressure drops far below ambient at the fast-moving shoulder. See the pressure figure.

Replace $\theta \to -\theta$ (top vs bottom): ${\sin }^2$ is unchanged, so the pressure at the front ($\theta =180^{\circ }$) equals that at the rear ($\theta =0^{\circ }$), and left–right the pattern is a mirror image. The pressure pushing backward on the front is exactly matched by pressure pushing forward on the rear. They cancel: net drag $=0$. This is d'Alembert's paradox — inviscid theory predicts zero drag because it has no wake and no friction, contradicting reality. (It still predicts lift correctly once a vortex is added.)

$L^{\prime }=\rho U\Gamma =1.2\times 10\times 20=240\text{N/m}.$ The vortex speeds the flow on one side and slows it on the other; by Bernoulli that makes a pressure difference — this is the Magnus effect. Kutta–Joukowski packages the whole surface integral into the tidy product $\rho U\Gamma$.

Stagnation: $u_{\theta }=0$ ⇒ $\sin \theta =-\frac{\Gamma }{4\pi Ua}$. $\sin \theta =-\frac{20}{4\pi (10)(0.5)}=-\frac{20}{20\pi }=-\frac{1}{\pi }\approx -\mathrm{0.3183.}$ Since $|\sin \theta |<1$, there are two stagnation points on the surface: $\theta =\arcsin (-0.3183)\approx -18.55^{\circ }$ (i.e. $341.45^{\circ }$) and $\theta =180^{\circ }-(-18.55^{\circ })=198.55^{\circ }$. Both sit on the lower half — the vortex has dragged the stagnation points downward, the signature of lift.

They merge when $\sin \theta =-1$, i.e. at the bottom $\theta =-90^{\circ }$. Then $1=\frac{{\Gamma }_c}{4\pi Ua}\Rightarrow {\Gamma }_c=4\pi Ua=4\pi (10)(0.5)=20\pi \approx 62.8{\text{m}}^2/\text{s}.$ Beyond this critical value the stagnation point lifts off the surface into the fluid.

The dividing streamline (the body's surface) passes through the stagnation point at $\theta =\pi$, where $\psi =Ur\sin \pi +\frac{m}{2\pi }\pi =\frac{m}{2}$. So the body is the streamline $\psi =\frac{m}{2}$: $Ur\sin \theta +\frac{m}{2\pi }\theta =\frac{m}{2}.$ Far downstream ($x\to +\infty$) the streamlines become horizontal, so $\theta \to 0$ while $y=r\sin \theta$ stays finite. Then $Ur\sin \theta \to Uy$ and $\frac{m}{2\pi }\theta \to 0$: $U{y}_{\max }=\frac{m}{2}\Rightarrow \boxed{y_{\max }=\frac{m}{2U}}.$ Numerically: $y_{\max }=\frac{8\pi }{2\cdot 2}=2\pi \approx 6.283\text{ m}$. So the total width far downstream is $2y_{\max }=4\pi$.

Surface pressure from Bernoulli: $p=p_{\infty }+\frac{1}{2}\rho (U^2-v^2)$ with $v^2=(2U\sin \theta +\frac{\Gamma }{2\pi a}{)}^2$. Lift is the upward ($y$) component of pressure force; the outward normal on the surface is $(\cos \theta ,\sin \theta )$, and pressure pushes inward, so the $y$-force per span is $L^{\prime }=-{\int }_0^{2\pi }p\sin \theta ad\theta .$ Expand $v^2=4U^2{\sin }^2\theta +\frac{2U\Gamma }{\pi a}\sin \theta +\frac{{\Gamma }^2}{4{\pi }^2{a}^2}$. When multiplied by $\sin \theta$ and integrated over $0$ to $2\pi$:

• $\int \sin \theta d\theta =0$ (kills the constant $p_{\infty }$, $U^2$, and ${\Gamma }^2$ terms),
• $\int {\sin }^3\theta d\theta =0$ (kills the $4U^2{\sin }^2\theta$ term),
• only ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ survives, from the cross term $\frac{2U\Gamma }{\pi a}\sin \theta$. $L^{\prime }=-(-\frac{1}{2}\rho )a{\int }_0^{2\pi }\frac{2U\Gamma }{\pi a}{\sin }^2\theta d\theta =\frac{1}{2}\rho \cdot a\cdot \frac{2U\Gamma }{\pi a}\cdot \pi =\rho U\Gamma .✓$ This is the Kutta–Joukowski theorem earned from scratch — the lift comes entirely from the vortex–stream cross term, which is why $L^{\prime }\propto \Gamma$.

$\Gamma =2\pi a{V}_s=2\pi (0.2)(6)=2.4\pi \approx 7.540{\text{m}}^2/\text{s}.$ $L^{\prime }=\rho U\Gamma =1.2\times 15\times 2.4\pi =43.2\pi \approx 135.7\text{N/m}.$

L2.1 $u_r=2$ · L2.2 top $8$, front $0$ · L2.3 $x_s=-2$ · L3.2 $C_p(90^{\circ })=-3$ · L4.1 $L^{\prime }=240$ · L4.2 $\sin \theta =-1/\pi$ · L4.3 ${\Gamma }_c=20\pi$ · L5.1 $y_{\max }=2\pi$ · L5.3 $\Gamma =2.4\pi ,L^{\prime }=43.2\pi$.