Worked examples — Stream function, velocity potential

2.2.11 · D3 · Physics › Fluid Mechanics › Stream function, velocity potential

Yeh page parent topic ke liye worked-example workout hai. Hum theory dobara derive nahi karenge — balki hum har tarah ki situation ko dhundhte hain jo yeh topic throw kar sakta hai, aur har ek ke liye ek representative problem solve karte hain. Agar koi symbol unfamiliar lage, to parent note usse build karta hai; yahan hum use use karte hain aur hamesha batate hain kyun.

Shuru karne se pehle, do toolkits ki ek-line reminder jo hum baar baar use karenge:

Scenario matrix

Yeh hai cases ka poora landscape. Neeche ka har cell aage kisi example se cover hota hai.

Hum inhe order mein karte hain. Har numeric result neeche machine-check kiya gaya hai.

Ex 1 — Cell A: dono scalars, sab signs positive

Forecast: pehle guess karo — kya streamlines horizontal hongi ya vertical? Kya $\varphi =$const lines same set hongi ya perpendicular?

1. Divergence aur curl check karo. $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0+0=0$ aur $\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=0-0=0$. Yeh step kyun? Dono scalars tabhi legal hain jab unke existence conditions hold hon. Divergence-free ⇒ $\psi$ allowed; curl-free ⇒ $\varphi$ allowed. Dono zero hain, isliye dono exist karte hain.
2. $\varphi$ banao. $u={\varphi }_x=3$ se, $x$ mein integrate karo: $\varphi =3x$ (constant absorb kar liya). Yeh step kyun? $\varphi$ ko $u={\varphi }_x$ se define kiya gaya hai; derivative ko undo karne ka matlab hai integrate karna.
3. $\psi$ banao. $u={\psi }_y=3$ se, $y$ mein integrate karo: $\psi =3y$. Check karo $v=-{\psi }_x=0$ ✓. Yeh step kyun? $\psi$ cross index use karta hai ($u={\psi }_y$), isliye $y$ mein integrate karo, $x$ mein nahi.

Verify karo: $\psi =3y=$const ⇒ horizontal lines (streamlines flow ke saath). $\varphi =3x=$const ⇒ vertical lines. Yeh $90^{\circ }$ par cross karte hain. Recompute karo $u={\varphi }_x=3$ ✓, $v={\varphi }_y=0$ ✓.

Ex 2 — Cell B: negative velocity (sign flip)

Forecast: streamlines abhi bhi horizontal hain — lekin flux kis taraf count hoga? Positive ya negative?

1. $\psi$ banao. $u={\psi }_y=-2\Rightarrow \psi =-2y$. Yeh step kyun? Same recipe; $u$ ka sign seedha $\psi$ mein chala jaata hai.
2. Do streamlines ke beech Flux $={\psi }_2-{\psi }_1$ jahan $y_2=4,y_1=0$: $\psi (4)-\psi (0)=-8-0=-8$. Yeh step kyun? $\psi$ streamlines ke beech volume flow measure karta hai; difference flux hai (per unit depth).

Verify karo: magnitude $|-8|=|u|\cdot \Delta y=2\times 4=8$ ✓. Minus sign correctly report karta hai ki fluid high-$y$ se low-$y$ ki taraf cross karta hai jab standard orientation ke against dekha jaaye — $u$ mein sign flip flux sign ko flip karta hai, exactly jaise hona chahiye.

Ex 3 — Cell C: angle pe flow (dono components nonzero)

Forecast: streamlines tilt hongi. $v/u$ se aap kaunsa slope predict karte ho?

1. Streamline slope. $\frac{dy}{dx}=\frac{v}{u}=\frac{1}{2}$. Yeh step kyun? Velocity streamlines ke tangent hoti hai, isliye slope $v/u$ hai.
2. $\psi$ banao. $u={\psi }_y=2\Rightarrow \psi =2y+g(x)$. Phir $v=-{\psi }_x=-g^{\prime }(x)=1\Rightarrow g(x)=-x$. Toh $\psi =2y-x$. Yeh step kyun? Jab dono components live hoon, ek integrate karo, phir integration "constant" $g(x)$ ko pin karne ke liye doosra equation use karo.
3. $\varphi$ banao. $u={\varphi }_x=2\Rightarrow \varphi =2x+h(y)$; $v={\varphi }_y=h^{\prime }(y)=1\Rightarrow h=y$. Toh $\varphi =2x+y$.

Verify karo: streamline $\psi =2y-x=c\Rightarrow y=\frac{1}{2}x+\frac{c}{2}$, slope $\frac{1}{2}$ ✓ step 1 se match karta hai. Orthogonality: $\nabla \varphi =(2,1)$, $\nabla \psi =(-v,u)=(-1,2)$; dot product $=2(-1)+1(2)=0$ ✓ — perpendicular grid.

Ex 4 — Cell D: rotational flow ($\varphi$ ke liye degenerate)

Forecast: whirlpool alarm — kya potential $\varphi$ survive karega?

1. Curl test. $\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega -(-\omega )=2\omega =2\ne 0$. Yeh step kyun? $\varphi$ exist karta hai sirf tabhi jab curl zero ho. Yahan yeh $2\omega$ hai — flow spin karta hai — isliye koi single-valued $\varphi$ nahi. Cell D genuinely degenerate hai.
2. Divergence test. $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0+0=0$. Yeh step kyun? $\psi$ ko sirf incompressibility chahiye, jo hoti hai. Isliye $\psi$ abhi bhi exist karta hai.
3. $\psi$ banao. $u={\psi }_y=-\omega y\Rightarrow \psi =-\frac{\omega }{2}{y}^2+g(x)$; $v=-{\psi }_x=-g^{\prime }(x)=\omega x\Rightarrow g=-\frac{\omega }{2}{x}^2$. Toh $\psi =-\frac{\omega }{2}(x^2+y^2)$.

Verify karo: $\omega =1$ pe, $\psi =-\frac{1}{2}(x^2+y^2)$; streamlines $\psi =$const ⇒ circles $x^2+y^2=$const ✓ (spinning fluid circles mein move karta hai). Recompute karo $u={\psi }_y=-y$ ✓, $v=-{\psi }_x=x$ ✓. Lesson: har flow ka $\varphi$ nahi hota — yeh pehli common galti hai, jo ab concrete ho gayi.

Ex 5 — Cell E: kya $\psi$ allowed bhi hai? (divergence gate)

Forecast: kya divergence zero hoga?

1. Divergence. $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=1+1=2\ne 0$. Yeh step kyun? Flux property wala koi $\psi$ exist nahi karta jab tak divergence zero na ho. Yahan yeh $2$ hai — fluid har jagah create ho raha hai. $\psi$ illegal.
2. Ek component ka sign flip karo: try karo $u=x$, $v=-y$. Ab divergence $=1+(-1)=0$ ✓. Yeh step kyun? $v$ ko flip karna woh banata hai jo $x$ mein baahir jaata hai usse $y$ ke through andar kheench lo — ek valid stagnation-point flow.
3. Fixed field ke liye $\psi$ banao. $u={\psi }_y=x\Rightarrow \psi =xy+g(x)$; $v=-{\psi }_x=-(y+g^{\prime }(x))=-y\Rightarrow g^{\prime }=0$. Toh $\psi =xy$.

Verify karo: streamlines $\psi =xy=$const ⇒ hyperbolas — classic corner/stagnation flow. Recompute karo $u={\psi }_y=x$ ✓, $v=-{\psi }_x=-y$ ✓, divergence $0$ ✓. Moral: pehle hamesha divergence gate run karo.

Ex 6 — Cell F: polar source, singular origin, all quadrants

Pehle figure padhein: magenta arrows streamlines hain — yeh origin se seedhe baahir shoot karte hain jaise ek wheel ki spokes, har angle $\theta$ ke liye ek. Violet dashed circles equipotentials $\varphi =$const hain. Notice karo ki har spoke har circle ko perfect right angle pe pierce karta hai: yeh crossing pattern woh flow net hai jo aapko extract karna chahiye. Orange dot centre mein singular source point hai.

Forecast: $r\to 0$ par speed ka kya hoga? Kya streamlines circles hain ya spokes?

1. $\varphi$ banao. $u_r=\frac{\partial \varphi }{\partial r}=\frac{m}{2\pi r}\Rightarrow \varphi =\frac{m}{2\pi }\ln r$. Yeh step kyun? Polar coordinates mein $\nabla \varphi$ ka radial component $\partial \varphi /\partial r$ hota hai; use integrate karo.
2. $\psi$ banao. Polar Cauchy–Riemann relation radial velocity ko $\psi$ ke angular derivative se tie karta hai: $u_r=\frac{1}{r}\frac{\partial \psi }{\partial \theta }$. Ise $\frac{m}{2\pi r}$ ke barabar set karne se milta hai $\frac{\partial \psi }{\partial \theta }=\frac{m}{2\pi }\Rightarrow \psi =\frac{m}{2\pi }\theta$. Yeh step kyun? Hum $\psi$ ki polar definition use karte hain (uska $\theta$-derivative $u_r$ deta hai), na ki generic "$\psi$ streamline ke saath constant hai." Yahan har streamline fixed $\theta$ ki ray hai, isliye $\psi$ actually rays ko unke angle se label karta hai — yeh ray se ray badlta hai, aur sirf ek given spoke ke saath constant hota hai.
3. All-quadrant behaviour (figure se). $\psi =\frac{m}{2\pi }\theta$ smoothly badhta hai quadrant I ($0\le \theta <\frac{\pi }{2}$), II, III, IV mein jaise aap spokes ko anticlockwise sweep karte ho; $\varphi$ $\theta$ ko ignore karta hai isliye uske equipotential circles $r=$const sab chaar quadrants mein valid hain. Singular origin: jaise $r\to 0$, $u_r=\frac{m}{2\pi r}\to \infty$ aur $\varphi =\frac{m}{2\pi }\ln r\to -\infty$ — source point ek genuine singularity hai, domain se excluded hai.
4. Branch-cut ($\psi$ ki multi-valuedness). Kyunki $\theta$ khud $2\pi$ se jump karta hai har baar jab aap ek chosen ray (jaise negative-$x$ axis) cross karte ho, $\psi =\frac{m}{2\pi }\theta$ us ray ke across $\frac{m}{2\pi }\cdot 2\pi =m$ se jump karta hai. Isliye $\psi$ origin ke aas-paas ek loop mein single-valued nahi hai — yeh branch cut tak defined hai, ek line jis par hum agree karte hain na cross karne ki taaki $\theta$ (aur isliye $\psi$) baaki plane pe single-valued rahe. Yeh step kyun? Cut ko ignore karne se aap "prove" kar sakte ho ki $\psi$ ek poore loop ke baad khud par return karta hai, jo neeche flux $=m$ result ke saath contradict karta hai.
5. Ek poore loop pe Flux. $Q=\psi (\theta =2\pi )-\psi (0)=\frac{m}{2\pi }\cdot 2\pi =m$. Yeh step kyun? Streamlines ke beech flux $\psi$ ka difference hai; ek poore circle pe woh difference exactly source strength $m$ hai — aur yeh nonzero hai precisely isliye kyunki $\psi$ cut ke across $m$ se jump karta hai.

Verify karo: $m=4\pi$ ke saath: $\varphi =2\ln r$, $\psi =2\theta$, aur $Q=4\pi$. Check karo $u_r=\partial \varphi /\partial r=2/r=\frac{4\pi }{2\pi r}$ ✓. Circles (equipotentials) radial spokes (streamlines) se $90^{\circ }$ par milti hain — orthogonality har quadrant mein hold karta hai, aur cut ke across $\psi$-jump of $m$ total outflow se match karta hai.

Ex 7 — Cell G: irrotational vortex, limiting behaviour

Pehle figure padhein: ab magenta circles streamlines hain — fluid round aur round swirl karta hai — aur violet dashed spokes equipotentials hain. Yeh source ka exact mirror image hai: jo wahan streamlines the woh yahan equipotentials hain, aur vice versa. Circles pe chhote arrows swirl sense dikhate hain; centre mein orange core woh jagah hai jahan speed blow up karti hai.

Forecast: yeh swirls karta hai — surely iske paas vorticity hogi? (Surprise dekho.)

1. $\varphi$ banao. Yahan $u_{\theta }=\frac{1}{r}\frac{\partial \varphi }{\partial \theta }=\frac{\Gamma }{2\pi r}\Rightarrow \varphi =\frac{\Gamma }{2\pi }\theta$. Yeh step kyun? Ab angular velocity component nonzero hai, isliye $\varphi$ $\theta$ par depend karta hai.
2. $\psi$ banao. $u_{\theta }=-\frac{\partial \psi }{\partial r}=\frac{\Gamma }{2\pi r}\Rightarrow \psi =-\frac{\Gamma }{2\pi }\ln r$. Yeh step kyun? Swirl component ke liye polar $\psi$ relation mein minus hota hai, $v=-{\psi }_x$ ko mirror karta hai.
3. Vorticity — twist (shown, not asserted). Polar coordinates mein purely azimuthal field ki out-of-plane vorticity hai ${\omega }_z=\frac{1}{r}\frac{\partial (r{u}_{\theta })}{\partial r}-\frac{1}{r}\frac{\partial u_r}{\partial \theta }.$ $u_r=0$ aur $r{u}_{\theta }=r\cdot \frac{\Gamma }{2\pi r}=\frac{\Gamma }{2\pi }$ ($r$ mein ek constant) ke saath, milta hai $\frac{\partial }{\partial r}\left(\frac{\Gamma }{2\pi }\right)=0$, isliye ${\omega }_z=0$ har $r>0$ ke liye. Yeh har jagah irrotational hai sirf origin ko chhod ke, jahan saari circulation concentrated hai. Yeh Ex 4 ke solid-body rotation ke opposite hai, jo har jagah rotational tha. Yeh step kyun? Yeh subtlety expose karta hai: swirling flow abhi bhi irrotational ho sakta hai — aur ab hum ne ise polar curl formula ke saath earn kiya hai.
4. Branch-cut ($\varphi$ ki multi-valuedness). Yahan $\varphi =\frac{\Gamma }{2\pi }\theta$ hai jo $\theta$ carry karta hai: yeh chosen cut ke across $\frac{\Gamma }{2\pi }\cdot 2\pi =\Gamma$ se jump karta hai. Isliye potential multi-valued hai, branch cut tak defined hai, exactly source case mein $\psi$ ko mirror karta hai. Woh jump hai origin ke around circulation $\Gamma$.
5. Limits (figure se). Jaise $r\to \infty$, circles bade ho jaate hain aur $u_{\theta }\to 0$ (fluid dur mein barely move karta hai). Jaise $r\to 0^{+}$, $u_{\theta }\to \infty$ — orange mein marked singular core.

Verify karo: $\Gamma =4\pi$ ke saath: $\varphi =2\theta$, $\psi =-2\ln r$. Origin enclosing kisi bhi loop ke around Circulation $=\oint u_{\theta }rd\theta =\frac{\Gamma }{2\pi }\cdot 2\pi =\Gamma =4\pi$ ✓, cut ke across $\varphi$-jump se match karta hai. Note karo equipotentials $\varphi =$const radial spokes hain aur streamlines $\psi =$const circles hain — exactly swapped vs source (Ex 6). Yeh source↔vortex swap ek favourite flow-net duality hai.

Ex 8 — Cell H: real-world flux word problem

Forecast: $\psi$ use karne se pehle "speed × width" se number guess karo.

1. Speed identify karo. $u={\psi }_y=5\text{m/s}$, $v=-{\psi }_x=0$. Yeh step kyun? Actual velocity recover karne ke liye $\psi$ differentiate karo — yahi fluid kar raha hai.
2. Flux $={\psi }_2-{\psi }_1$. $Q=\psi (3)-\psi (1)=15-5=10{\text{m}}^2/\text{s}$ (per unit depth). Yeh step kyun? $\psi$ ka yahi to point hai: do streamline values ke beech difference hai hi unke beech volumetric flow — koi integration nahi chahiye.

Verify karo (units + physics): $Q=u\times \Delta y=5\frac{\text{m}}{\text{s}}\times 2\text{m}=10\frac{{\text{m}}^2}{\text{s}}$ ✓. Units area/time hain (2D "flow rate per unit depth") — consistent. Channel depth, maano $2\text{m}$ se multiply karo to get true volume rate $20{\text{m}}^3/\text{s}$.

Ex 9 — Cell I: exam twist — doosra scalar recover karo aur orthogonality prove karo

Forecast: kya yahan $\varphi$ exist bhi karta hai? $\psi =x^2-y^2$ kaunsi velocity hide karta hai?

1. Velocity extract karo. $u={\psi }_y=-2y$, $v=-{\psi }_x=-2x$. Yeh step kyun? $\psi$ hamesha velocity encode karta hai cross-with-minus recipe ke through.
2. Dono conditions check karo. Divergence $={\partial }_x(-2y)+{\partial }_y(-2x)=0$ ✓ ($\psi$ yeh guarantee karta hai). Curl $={\partial }_{x}v-{\partial }_{y}u=-2-(-2)=0$ ✓ ⇒ $\varphi$ exist karta hai. Yeh step kyun? $\varphi$ dhundhne se pehle aapko uska right earn karna hoga.
3. $\varphi$ banao. $u={\varphi }_x=-2y\Rightarrow \varphi =-2xy+h(y)$; $v={\varphi }_y=-2x+h^{\prime }(y)=-2x\Rightarrow h^{\prime }=0$. Toh $\varphi =-2xy$. Yeh step kyun? Ek C–R equation integrate karo, bacha hua doosre se fix karo.
4. $(2,1)$ par Orthogonality. $\nabla \varphi =({\varphi }_x,{\varphi }_y)=(-2y,-2x)=(-2,-4)$; $\nabla \psi =({\psi }_x,{\psi }_y)=(2x,-2y)=(4,-2)$. Dot product $=(-2)(4)+(-4)(-2)=-8+8=0$ ✓. Yeh step kyun? Zero dot product ⇒ $\varphi$-line aur $\psi$-line wahan $90^{\circ }$ par cross karte hain.

Verify karo: $\varphi$ se recompute karo: $u={\varphi }_x=-2y$ ✓, $v={\varphi }_y=-2x$ ✓ — step 1 se match karta hai. Dono scalars Laplace satisfy karte hain: ${\nabla }^2\psi =2-2=0$, ${\nabla }^2\varphi =0+0=0$ ✓. Matrix ka har cell ab cover ho gaya.

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