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

$U$ ::: uniform "seedhi nadi" flow ki speed, $+x$ axis ki taraf point karti hai. $r,\theta$ ::: polar coordinates — $r$ origin se doori hai, $\theta$ woh angle hai jo $+x$ axis se anticlockwise measure hota hai. $m$ ::: source strength = fluid ka volume jo per second per unit depth pump hota hai ($m>0$ bahar dhakelta hai, $m<0$ andar kheenchta hai). $\Gamma$ ::: vortex ki circulation = yeh kitna "swirl" carry karta hai (units hain speed × length). $\kappa$ ::: doublet strength (ek source aur sink jo infinitely paas squeeze hote hain). $\varphi$ ::: velocity potential; $\vec{v}=\nabla \varphi$. $\psi$ ::: stream function; $\psi =$const ek streamline trace karta hai jis par fluid glide karta hai. $u_r,u_{\theta }$ ::: polar form mein velocity ke do pieces — $u_r$ bahar ki taraf point karta hai, $u_{\theta }$ anticlockwise point karta hai.

Ek source fluid ko har direction mein equally seedha bahar phenkata hai. Toh koi bhi swirl nahi hai ($u_{\theta }=0$) aur sirf outward motion hai. Mass conservation: jo $m$ fluid pump hota hai woh radius $r$ ke har circle ko cross karna chahiye. Circle ki circumference $2\pi r$ hai, toh $u_r(2\pi r)=m\Rightarrow \boxed{u_r=\frac{m}{2\pi r}}.$ $u_r=\partial \varphi /\partial r$ ko integrate karne par $\varphi =\frac{m}{2\pi }\ln r$ milta hai. Stream function hai $\psi =\frac{m}{2\pi }\theta$.

$u_{\theta }=\frac{\Gamma }{2\pi r}.$ Flow bilkul centre $r=0$ ke siwa har jagah irrotational hai, jahan $u_{\theta }\to \infty$ — ek singular point. Centre se door ek choti fluid cross circle ke around float karti hai bina apne axis par spin kiye (yahi "irrotational" ka matlab hai: zero local spin, zero path curvature nahi).

Uniform flow har jagah same hota hai, toh position irrelevant hai: $u=U=3\text{ m/s}$, $v=0$.

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

Top, $\theta =90^{\circ }$: $u_{\theta }=-2(4)\sin 90^{\circ }=-8\text{ m/s}$. Speed = $|-8|=8\text{ m/s}$ (minus ka matlab sirf yeh hai ki woh wahan clockwise/backward point karti hai). Front, $\theta =180^{\circ }$: $u_{\theta }=-2(4)\sin 180^{\circ }=0$. Yeh ek stagnation point hai — fluid naak se takraate waqt bilkul ruk jaata hai. Note karo ki top speed $8=2U$ exactly free-stream speed se do guna hai. Neeche figure dekho.

Source ke aage $x$-axis par, $\theta =\pi$ hai aur total $x$-velocity hai $u=U-\frac{m}{2\pi x}$. Source fluid ko yahan leftward dhakelta hai, stream ke against. Woh tab cancel hote hain jab $u=0$: $x_s=-\frac{m}{2\pi U}=-\frac{8\pi }{2\pi \cdot 2}=-2\text{m}.$

Streamline test: streamline woh curve hai jise fluid kabhi cross nahi karta — matlab velocity component jo iske across point kare (radial wala, $u_r$) uska us par zero hona zaroori hai. $r=a$ par: $u_r=U\cos \theta (1-\frac{a^2}{a^2})=U\cos \theta \cdot 0=0\text{for every }\theta .$ Toh koi fluid kabhi circle ko pierce nahi karta — yeh bilkul solid wall jaisa behave karta hai. ✓ Stagnation points: dono $u_r=0$ (surface par pehle se true) aur $u_{\theta }=0$. $r=a$ par, $u_{\theta }=-2U\sin \theta$, jo $\sin \theta =0\Rightarrow \theta =0$ (rear) aur $\theta =\pi$ (front) par vanish karta hai. Yeh dono stagnation points hain, front aur back.

Surface par speed hai $v=|u_{\theta }|=2U|\sin \theta |$, toh $v^2=4U^2{\sin }^2\theta$. Far upstream ($p_{\infty }$, speed $U$) aur surface ke beech Bernoulli: $p_{\infty }+\frac{1}{2}\rho U^2=p+\frac{1}{2}\rho (4U^2{\sin }^2\theta ).$ $p-p_{\infty }$ solve karo aur $\frac{1}{2}\rho U^2$ se divide karo: $C_p=1-4{\sin }^2\theta .$ $\theta =90^{\circ }$ par: $C_p=1-4(1)=-3$. Negative matlab suction — fast-moving shoulder par pressure ambient se bahut neeche gir jaata hai. Pressure figure dekho.

$\theta \to -\theta$ replace karo (top vs bottom): ${\sin }^2$ unchanged rehta hai, toh front par pressure ($\theta =180^{\circ }$) rear ke barabar hai ($\theta =0^{\circ }$), aur left–right pattern mirror image hai. Front par backward push karne wala pressure exactly rear par forward push karne wale pressure se match karta hai. Woh cancel ho jaate hain: net drag $=0$. Yeh d'Alembert's paradox hai — inviscid theory zero drag predict karta hai kyunki usmein koi wake aur koi friction nahi hoti, jo reality se contradict karta hai. (Vortex add karne par yeh lift sahi predict karta hai.)

$L^{\prime }=\rho U\Gamma =1.2\times 10\times 20=240\text{N/m}.$ Vortex ek taraf flow ko speed up karta hai aur doosri taraf slow — Bernoulli se yeh pressure difference banata hai — yeh Magnus effect hai. Kutta–Joukowski poore surface integral ko tidy product $\rho U\Gamma$ mein pack kar deta hai.

Stagnation: $u_{\theta }=0$ ⇒ $\sin \theta =-\frac{\Gamma }{4\pi Ua}$. $\sin \theta =-\frac{20}{4\pi (10)(0.5)}=-\frac{20}{20\pi }=-\frac{1}{\pi }\approx -\mathrm{0.3183.}$ Kyunki $|\sin \theta |<1$, surface par do stagnation points hain: $\theta =\arcsin (-0.3183)\approx -18.55^{\circ }$ (yaani $341.45^{\circ }$) aur $\theta =180^{\circ }-(-18.55^{\circ })=198.55^{\circ }$. Dono lower half par baithe hain — vortex ne stagnation points ko neeche drag kiya hai, yeh lift ki signature hai.

Woh tab merge hote hain jab $\sin \theta =-1$, yaani bottom $\theta =-90^{\circ }$ par. Tab $1=\frac{{\Gamma }_c}{4\pi Ua}\Rightarrow {\Gamma }_c=4\pi Ua=4\pi (10)(0.5)=20\pi \approx 62.8{\text{m}}^2/\text{s}.$ Is critical value se aage stagnation point surface se uth kar fluid mein chala jaata hai.

Dividing streamline (body ki surface) stagnation point $\theta =\pi$ se guzarti hai, jahan $\psi =Ur\sin \pi +\frac{m}{2\pi }\pi =\frac{m}{2}$ hai. Toh body streamline $\psi =\frac{m}{2}$ hai: $Ur\sin \theta +\frac{m}{2\pi }\theta =\frac{m}{2}.$ Far downstream ($x\to +\infty$) streamlines horizontal ho jaati hain, toh $\theta \to 0$ jab $y=r\sin \theta$ finite rehta hai. Tab $Ur\sin \theta \to Uy$ aur $\frac{m}{2\pi }\theta \to 0$: $U{y}_{\max }=\frac{m}{2}\Rightarrow \boxed{y_{\max }=\frac{m}{2U}}.$ Numerically: $y_{\max }=\frac{8\pi }{2\cdot 2}=2\pi \approx 6.283\text{ m}$. Toh far downstream total width hai $2y_{\max }=4\pi$.

Bernoulli se surface pressure: $p=p_{\infty }+\frac{1}{2}\rho (U^2-v^2)$ jahan $v^2=(2U\sin \theta +\frac{\Gamma }{2\pi a}{)}^2$ hai. Lift, pressure force ka upward ($y$) component hai; surface par outward normal $(\cos \theta ,\sin \theta )$ hai, aur pressure andar dhakelta hai, toh span ke perpendicular $y$-force hai $L^{\prime }=-{\int }_0^{2\pi }p\sin \theta ad\theta .$ $v^2=4U^2{\sin }^2\theta +\frac{2U\Gamma }{\pi a}\sin \theta +\frac{{\Gamma }^2}{4{\pi }^2{a}^2}$ expand karo. Jab $\sin \theta$ se multiply karke $0$ se $2\pi$ integrate karo:

• $\int \sin \theta d\theta =0$ (constant $p_{\infty }$, $U^2$, aur ${\Gamma }^2$ terms ko khatam karta hai),
• $\int {\sin }^3\theta d\theta =0$ ($4U^2{\sin }^2\theta$ term ko khatam karta hai),
• sirf ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ bachta hai, cross term $\frac{2U\Gamma }{\pi a}\sin \theta$ se. $L^{\prime }=-(-\frac{1}{2}\rho )a{\int }_0^{2\pi }\frac{2U\Gamma }{\pi a}{\sin }^2\theta d\theta =\frac{1}{2}\rho \cdot a\cdot \frac{2U\Gamma }{\pi a}\cdot \pi =\rho U\Gamma .✓$ Yeh Kutta–Joukowski theorem scratch se kamaaya gaya hai — lift poori tarah vortex–stream cross term se aati hai, isliye $L^{\prime }\propto \Gamma$ hai.

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

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 $\Gamma =2.4\pi ,L^{\prime }=43.2\pi$.