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

2.2.28 · D3 · Physics › Fluid Mechanics › Potential flow — irrotational, inviscid; superposition of ba

Numbers touch karne se pehle, vocabulary ko pictures se pin karte hain.

Definition Char building-block flows, har ek ke saath uska

$ψ$ (yahan define kiya, poori matrix mein use hoga) Neeche har symbol re-anchor kiya gaya hai taaki tumhe parent note recall karne ki zaroorat na pade.

Uniform flow (speed $U$ ki ek steady wind rightward blow karti hai): $ψ_{uni} = U y = U r sin θ$ . Streamlines seedhi horizontal lines hain.
Source / sink, strength $m$ = source strength = fluid ka volume jo per second per unit depth pump out hota hai (units $m^{2} / s$ ). Ek hose imagine karo jo saari directions mein evenly spray karta ho. Iska stream function hai $ψ_{source} = \frac{m}{2 π} θ$ , aur iska outward speed hai $u_{r} = \frac{m}{2 π r}$ (Ex 1 mein derive kiya). $m > 0$ = source (outflow), $m < 0$ = sink (inflow, plughole ki tarah).
Free vortex, strength $Γ$ = circulation = fluid kitna swirl karta hai, ek loop ke around tangential speed add karke nikala jaata hai (units $m^{2} / s$ ). Chai stirring karna imagine karo. Iska stream function hai $ψ_{vortex} = - \frac{Γ}{2 π} ln r$ , aur iska tangential (anticlockwise) speed hai $u_{θ} = \frac{Γ}{2 π r}$ (Ex 4b mein derive kiya). $Γ > 0$ = anticlockwise, $Γ < 0$ = clockwise. Dekho Vorticity and circulation.
Doublet, strength $κ$ = doublet strength = ek source aur ek equal sink infinitely close saath squash kiye hue ("fluid yahan appear hota hai, ek bal ki door ghaib ho jaata hai"). Iska stream function hai $ψ_{doublet} = - \frac{κ}{2 π} \frac{sin θ}{r}$ . Yahi woh brick hai jise hum uniform wind pe superpose karte hain ek solid cylinder conjure karne ke liye (Ex 6, Ex 8).

Hum $ψ$ se velocity read karenge parent ke rule ka use karke, jo yahan restate kiya gaya hai taaki tumse galti se sign flip na ho:

The scenario matrix

Is topic ka har problem inhi cells mein se ek hai. Neeche ke worked examples cell(s) ke saath tagged hain jo woh cover karte hain, taaki milke poori table bhar jaaye. (Saare strength symbols $m, Γ, κ$ upar define kiye gaye hain.)

Cell	Kya vary karta hai	Example(s)
A. Source strength $m$ ka sign	$m > 0$ (source, outflow) vs $m < 0$ (sink, inflow)	Ex 1, Ex 2
B. Field point ka quadrant	$- x$ axis par stagnation ( $θ = π$ ) vs off-axis koi point	Ex 3
C. Circulation $Γ$ ka sign	$Γ > 0$ (anticlockwise) vs $Γ < 0$ (clockwise) → lift kis direction mein	Ex 5
D. Degenerate cancellation	equal source+sink annihilate, AUR vortex+uniform jahan $u_{θ}$ wind cancel kare	Ex 4, Ex 4b
E. Limiting behaviour	$r \to \infty$ (far field → uniform) aur $r \to 0$ (near field → singularity blow up)	Ex 6
F. Real-world word problem	spinning ball / drain whirlpool actual units ke saath	Ex 7
G. Exam twist	cylinder with circulation: do stagnation points kahan baithe hain, aur kab merge hote hain?	Ex 8

Worked example

Ex 1 — Cell A (source, $m > 0$ ): ek given radius par speed Ek 2-D source of strength $m = 12 m^{2} / s$ (volume flow per unit depth) origin par baitha hai. $r = 3 m$ par radial speed $u_{r}$ nikalo, aur uski direction batao.

Forecast: Compute karne se pehle guess karo — jab tum source se door jaate ho, toh fluid speed up hota hai ya slow down?

Source velocity mass conservation se derive karo. Saara $m$ outflow har radius $r$ ki circle cross karta hai; circle ki circumference $2 π r$ hai, toh speed = flow ÷ length = $u_{r} = \frac{m}{2 π r}$ . Yeh step kyun? Badi circle, same flow zyada phaila → slower. Hum derive karte hain recall karne ki jagah, no-memorization promise honor karte hue.
Plug in karo. $u_{r} = \frac{12}{2 π ( 3 )} = \frac{12}{6 π} = \frac{2}{π} \approx 0.637 m/s$ . Yeh step kyun? Jab formula trusted ho toh sirf arithmetic hai.
Direction. $m > 0 \Rightarrow$ outward (origin se door), aur $u_{θ} = 0$ (ek pure source mein koi swirl nahin).

Verify: Units: $(m^{2} / s) / m = m/s$ ✓. Sanity: tumhara forecast "slows down" hona chahiye tha — waakai $u_{r} \propto 1/ r$ . $r = 6 m$ par yeh half ho kar $\approx 0.318$ ho jaayega. ✓

Worked example

Ex 2 — Cell A (sink, $m < 0$ ): same formula, flipped sign Origin par ek sink of strength $m = - 20 m^{2} / s$ baitha hai (plughole soch lo). $r = 2 m$ par $u_{r}$ aur uski direction nikalo.

Forecast: Plughole ke liye, fluid andar jaata hai ya bahar?

Same formula. $u_{r} = \frac{m}{2 π r} = \frac{- 20}{2 π ( 2 )} = \frac{- 20}{4 π} = - \frac{5}{π} \approx - 1.592 m/s$ . Yeh step kyun? Hum sink ke liye koi nayi formula invent nahin karte. Sink sirf negative $m$ wala source hai; answer mein minus sign hi physics hai.
Sign interpret karo. $u_{r} < 0$ matlab outward speed negative hai → fluid andar move kar raha hai. Plughole ke liye sahi hai.

Verify: Magnitude $5/ π \approx 1.592 m/s$ positive aur finite hai; sirf direction $m$ ke sign ke saath badi. Cell A ka poora content yahi hai: ek formula source aur sink dono cover karta hai; sign meaning carry karta hai. ✓

Worked example

Ex 3 — Cell B (off-axis field point): uniform + source ki total velocity Ek rightward uniform wind $U = 2 m/s$ aur origin par ek source $m = 8 π m^{2} / s$ combine karo (ek Rankine half-body). Point $(x, y) = (3, 4)$ par velocity vector nikalo.

Figure — har arrow ka matlab (ek 3-4-5 triangle scale ke saath draw kiya gaya): Origin par coral dot (source) hai. Dashed line origin se field point $(3, 4)$ tak jaati hai; yeh ek 3-4-5 right triangle ki hypotenuse hai, isliye isko " $r = 5$ " label kiya gaya hai. Point $(3, 4)$ par teen arrows tip-to-tail start hote hain: mint arrow uniform wind hai (pure rightward, length $2$ ); coral arrow us point par source ka outward push hai (chhota, origin se door point karta hua, components $0.48$ right aur $0.64$ upar); lavender arrow unka vector sum hai — woh total velocity jo hum neeche compute karte hain. Algebra se pehle arrows padho: picture pehle se hi bata deti hai ki answer right-aur-thoda-upar lean karega.

Forecast: Source ke right aur upar wale ek point par, source out-aur-upar push karta hai jabki wind right push karti hai. Kya combined vector zyada "upar" point karega ya zyada "right"?

Har piece ko $x, y$ mein likho. Uniform: $u_{1} = U = 2, v_{1} = 0$ . Source: yeh radially outward point karta hai, toh iske Cartesian components hain $u_{2} = \frac{m}{2 π r} \frac{x}{r} = \frac{m}{2 π} \frac{x}{r ^{2}}$ , $v_{2} = \frac{m}{2 π} \frac{y}{r ^{2}}$ . Yeh step kyun? Superposition velocity fields add karta hai, isliye add karne se pehle hume dono ko same coordinates mein express karna hoga. Factor $x / r$ (= $cos θ$ ) simply hai "outward direction ka kitna hissa $x$ ke saath point karta hai" — figure mein dashed radius line se padh lo.
$r^{2}$ compute karo. $r^{2} = 3^{2} + 4^{2} = 25$ , toh $r = 5$ (dashed line par " $r = 5$ " label). Yeh step kyun? Hume source ke $1/ r$ decay ko scale karne ke liye distance chahiye.
Source components. $\frac{m}{2 π} = \frac{8 π}{2 π} = 4$ . Toh $u_{2} = 4 \cdot \frac{3}{25} = \frac{12}{25} = 0.48$ , $v_{2} = 4 \cdot \frac{4}{25} = \frac{16}{25} = 0.64$ — coral arrow ke components.
Add karo. $u = 2 + 0.48 = 2.48 m/s$ , $v = 0 + 0.64 = 0.64 m/s$ — lavender arrow. Yeh step kyun? Yahi actual superposition hai — legal hai kyunki Laplace's equation linear hai (dekho 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 wind ke $2$ se upar (source locally energy add karta hai). Forecast check: $u > v$ , toh arrow zyada right point karta hai upar se — is modest radius par wind dominate karti hai. ✓

Worked example

Ex 4 — Cell D (degenerate cancellation, source vs wind): Ex 3 ke flow ka stagnation point Same flow jaise Ex 3 mein (uniform $U = 2$ , source $m = 8 π$ ). Stagnation point nikalo — woh ek jagah jahan fluid bilkul still ho.

Forecast: Source har jagah outward push karta hai; wind right blow karti hai. Yeh exactly kahan cancel ho sakte hain — source ke upstream (left) mein ya downstream (right) mein?

$v = 0$ kahan ho sakta hai? $x$ -axis par ( $y = 0$ ) source ka koi vertical push nahin hai, toh $v = 0$ automatically ho jaata hai. Wahan dekho. Yeh step kyun? Stagnation point ke liye dono $u = 0$ aur $v = 0$ chahiye; $x$ -axis $v$ ko free mein kill kar deta hai, kaam aadha ho jaata hai.
Negative $x$ -axis par ( $θ = π$ , toh $x < 0$ ), source left push karta hai (wind ke against). $u = 0$ set karo: $U + \frac{m}{2 π x} = 0$ . Negative axis kyun? Sirf upstream par source ka push wind ka oppose karta hai; downstream pe dono add ho jaate hain aur kabhi cancel nahin hote.
Solve karo. $x_{s} = - \frac{m}{2 π U} = - \frac{8 π}{2 π \cdot 2} = - \frac{8 π}{4 π} = - 2 m$ .

Verify: $x = - 2$ par: source outward speed $= \frac{m}{2 π r} = \frac{8 π}{2 π \cdot 2} = 2 m/s$ left point karta hua, yani $u_{source} = - 2$ ; wind $u = + 2$ ; sum $= 0$ ✓. Yeh pehla "degenerate" case hai — ek source velocity ek point par ek wind velocity ko annihilate karti hai. ✓

Worked example

Ex 4b — Cell D (baaki do degeneracies): source+sink annihilation, aur vortex+wind swirl-cancellation Matrix ke do quick degenerate scenarios. (a) Ek source $m = + 6 m^{2} / s$ aur ek sink $m = - 6 m^{2} / s$ same point par placed hain. Kaunsa flow bachta hai? (b) Ek free vortex $Γ = 4 π m^{2} / s$ origin par centred ek rightward wind $U = 2 m/s$ mein baitha hai. $+ y$ -axis par vortex swirl leftward point karta hai (wind opposing). Kis radius $r$ par vortex ki tangential speed wind ki leftward component ko exactly cancel karti hai?

Forecast: (a) Do equal-and-opposite pumps ek jagah — kya koi net fluid kahin appear kar raha hai? (b) Vortex speed $1/ r$ se die hoti hai; kya ek special radius hai jahan yeh wind match kare?

(a) Source stream function re-anchor karo. Definitions box se, strength $m$ wale source ka $ψ_{source} = \frac{m}{2 π} θ$ hota hai — yeh simply kehta hai "jitna flow tumne sweep kiya hai woh angle $θ$ ke saath proportionate grow karta hai jitna tumne ghuma," jo bilkul wahi hai jo evenly spraying hose dena chahiye. Toh source: $ψ_{+} = \frac{6}{2 π} θ$ ; sink: $ψ_{-} = \frac{- 6}{2 π} θ$ . Velocities ki jagah $ψ$ kyun use karein? Kyunki hum jaanna chahte hain ki koi bhi flow har jagah bachta hai ya nahin. Do stream functions add karke ek single constant mila toh instantly prove hota hai "no streamlines, no flow" — ek global statement jo ek jagah ek velocity vector kabhi establish nahin kar sakta. Sum: $ψ = \frac{6}{2 π} θ - \frac{6}{2 π} θ = 0$ har jagah. Yeh legal kyun hai? Co-located flows ke stream functions same reason se add hote hain jaise velocities — Laplace's equation linear hai.
(b) Re-anchor karo aur vortex tangential speed use karo. Definitions box se, ek free vortex ka $ψ_{vortex} = - \frac{Γ}{2 π} ln r$ hota hai. Hamara reading rule $u_{θ} = - \frac{\partial ψ}{\partial r}$ apply karo: kyunki $\frac{d}{d r} ln r = \frac{1}{r}$ , hum paate hain $u_{θ} = - (- \frac{Γ}{2 π} \cdot \frac{1}{r}) = \frac{Γ}{2 π r}$ — yahan derive kiya, recall nahin kiya. $+ y$ -axis par, anticlockwise matlab $- x$ direction mein point karna hai, toh iska $x$ -component hai $- \frac{Γ}{2 π r}$ . Wind $+ U$ add karta hai. $x$ -component zero set karo: $U - \frac{Γ}{2 π r} = 0$ . Yeh step kyun? Cancellation ke liye dono horizontal pushes equal aur opposite chahiye; sirf horizontal parts is axis par matter karte hain.
Solve karo. $r = \frac{Γ}{2 π U} = \frac{4 π}{2 π ( 2 )} = 1 m$ .

Verify: (a) $ψ \equiv 0$ ⇒ no streamlines, no flow ✓ — intuition se match karta hai ki equal strength wala co-located source+sink cancel ho jaata hai. (b) $r = 1$ par: vortex $x$ -speed $= - \frac{4 π}{2 π ( 1 )} = - 2$ , wind $+ 2$ , sum $0$ ✓. Baaki dono Cell-D degeneracies ab table par hain. ✓

Worked example

Ex 5 — Cell C ( $Γ$ ka sign): Magnus lift ki direction Ek rightward stream $U$ mein ek cylinder, air density $ρ = 1.23 kg/m^{3}$ , circulation carry karta hai. Case (i): $Γ = + 6 m^{2} / s$ (anticlockwise). Case (ii): $Γ = - 6 m^{2} / s$ (clockwise). $U = 10 m/s$ ke saath, har ek ke liye lift per unit span $L^{'}$ aur uski direction nikalo.

Forecast: Ek ball jo itni spin kare ki uski top surface wind ke saath move kare — kya woh climb karega ya dive?

Formula state karo. $L^{'} = ρ U Γ$ . Yeh step kyun? Upar wala standard sign convention: 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). $Γ$ ka sign flip karne se lift direction flip ho jaati hai, exactly jaise Magnus effect demand karta hai — Cell C ka poora content. ✓

Worked example

Ex 6 — Cell E (limits $r \to \infty$ aur $r \to 0$ ): cylinder banana aur uske limits lena Non-spinning cylinder of radius $a = 1 m$ , stream $U = 5 m/s$ . (a) Uniform + doublet se cylinder stream function aur surface speed derive karo. (b) Door se flow kaisa dikhta hai ( $r ≫ a$ )? (c) Maximum surface speed kya hai aur kahan? (d) Akele doublet ka $r \to 0$ par kya hota hai?

Figure — kya dekhna hai: Butter-yellow disc solid cylinder hai ( $r = a$ ). Lavender curves streamlines hain: door pe almost straight hain (wind obstacle barely notice karti hai), paas mein disc ke around part ho jaati hain. Disc ke left aur right par do coral dots front aur rear stagnation points hain (fluid momentarily rest par). Top aur bottom par do mint squares mark hain jahan streamlines sabse tight squeeze hoti hain → fastest flow.

Forecast: Ek chhote cylinder se door, kya wind use notice karegi?

Uniform + doublet superpose karo (formula ka WHY). Uniform flow ka $ψ_{uni} = U y = U r sin θ$ hai. Doublet (upar define kiya) ka $ψ_{dou} = - \frac{κ}{2 π} \frac{sin θ}{r}$ hai. Add karo: $ψ = U r sin θ - \frac{κ}{2 π} \frac{s i n θ}{r} = U sin θ (r - \frac{κ}{2 π U} \frac{1}{r}) .$ Yeh step kyun? Hum chahte hain ki ek circle streamline ban jaaye. $a^{2} = \frac{κ}{2 π U}$ choose karo; tab $ψ = U sin θ (r - \frac{a ^{2}}{r})$ , aur $r = a$ par bracket $a - a = 0$ hai, toh $ψ = 0$ saare $θ$ ke liye. Constant- $ψ$ curve ek streamline hai, toh woh circle ek solid wall ki tarah act karta hai — ek cylinder! Yahi WHY hai quoted $ψ = U sin θ (r - a^{2} / r)$ ke peechhe.
Surface speed ( $- 2 U sin θ$ ka WHY). Surface par sirf $u_{θ}$ bachta hai (wall $u_{r}$ block karta hai). $u_{θ} = - \frac{\partial ψ}{\partial r} = - U sin θ (1 + \frac{a ^{2}}{r ^{2}})$ use karke, $r = a$ set karo: $u_{θ}_{r = a} = - U sin θ (1 + 1) = - 2 U sin θ$ . Yeh step kyun? Streamlines ke across $ψ$ ki derivative hi speed hai (hamara top rule); $r = a$ par evaluate karne se surface value milti hai, koi guessing nahin.
Far field $r \to \infty$ . $ψ = U sin θ (r - \frac{a ^{2}}{r})$ mein $a^{2} / r$ term $\to 0$ ho jaata hai, bacha $ψ \to U r sin θ = U y$ — pure uniform flow. Figure mein outer streamlines already almost straight hain. Yeh step kyun? Doublet $1/ r$ ki tarah decay karta hai; ek bounded obstacle sirf locally ek unbounded stream disturb kar sakta hai.
Max surface speed. $∣ u_{θ} ∣ = 2 U ∣ sin θ ∣$ sabse bada hota hai jab $∣ sin θ ∣ = 1$ , yani $θ = \pm 9 0^{\circ}$ (top aur bottom, figure mein mint squares). Value $= 2 U = 2 (5) = 10 m/s$ . Yeh step kyun? Streamlines cylinder ke "shoulders" par sabse zyada squeeze hoti hain → wahan fastest flow.
Near field $r \to 0$ (doublet alone). Doublet ki speed $κ / (2 π r^{2})$ ki tarah scale karta hai, jo $r \to 0$ par $\to \infty$ ho jaata hai. Yeh ek singularity hai — mathematically infinite, physically solid cylinder ke andar chhupaa hua hai (butter-yellow disc), toh koi real fluid kabhi wahan nahin pahunchta.

Verify: $θ = 0$ aur $θ = 18 0^{\circ}$ par, $u_{θ} = - 2 U sin θ = 0$ — front aur rear stagnation points (figure mein coral dots), jaise ek symmetric non-spinning cylinder ke liye expected ✓. Max $10 m/s = 2 U$ ✓. Limit spectrum ke dono ends (Cell E) cover ho gaye: $\infty$ par gentle uniform flow, $0$ par blow-up. ✓

Worked example

Ex 7 — Cell F (real-world word problem): bathtub whirlpool Paani ek chhote hole se drain karta hai. Hole se door paani ek free vortex ki tarah rotate karta hai: $u_{θ} = \frac{Γ}{2 π r}$ . Tum radius $r = 0.10 m$ par tangential speed $0.30 m/s$ measure karte ho. (a) Circulation $Γ$ nikalo. (b) $r = 0.02 m$ par speed predict karo.

Forecast: Drain ke paas, kya paani zyada tezi se swirl karta hai ya slow?

$Γ$ ke liye solve karo. $u_{θ} = \frac{Γ}{2 π r}$ se (Ex 4b step 2 mein derive kiya): $Γ = 2 π r u_{θ} = 2 π (0.10) (0.30) = 0.06 π \approx 0.1885 m^{2} / s$ . Yeh step kyun? Circulation $Γ$ vortex ki fixed strength hai; ek radius par speed measure karna use har jagah pin kar deta hai.
$r = 0.02$ par predict karo. $u_{θ} = \frac{Γ}{2 π r} = \frac{0.06 π}{2 π ( 0.02 )} = \frac{0.06}{0.04} = 1.5 m/s$ . Yeh step kyun? Same $Γ$ , chhota $r$ → $1/ r$ law swirl race karata hai.

Verify: Ratio check: radius $5$ se shrink hua ( $0.10 \to 0.02$ ), toh speed $5$ se grow karni chahiye: $0.30 \times 5 = 1.5$ ✓. Note karo ki free vortex har jagah irrotational hai except singular centre par — local element spin nahin karta chahe poora path circular ho (dekho parent ka mistake box). ✓

Worked example

Ex 8 — Cell G (exam twist): spinning cylinder ke stagnation points kahan jaate hain, aur kab merge hote hain? Cylinder radius $a$ , stream $U$ , circulation $Γ$ . (a) Ex 6 ke cylinder flow mein ek vortex add karke surface tangential speed derive karo, phir stagnation angles $θ_{s}$ nikalo. (b) $U = 4 m/s$ , $a = 0.5 m$ , $Γ = - 4 π m^{2} / s$ ke liye unhe locate karo. (c) Kis $Γ$ par do stagnation points merge hote hain — aur uske baad kya hota hai?

Figure — kya dekhna hai: Butter-yellow disc cylinder hai; mint arrows left par rightward stream $U$ dikhate hain; lavender arc added clockwise ( $Γ < 0$ ) swirl dikhata hai. Do coral dots neeche compute kiye gaye stagnation points hain — dono cylinder ke upper half mein baithe hain kyunki clockwise swirl ne unhe apni plain-cylinder positions (front/back) se upar drag kar liya hai.

Forecast: Rightward wind mein clockwise swirl ( $Γ < 0$ ) add karo — do stagnation points cylinder par upar slide karenge ya neeche?

Vortex superpose karo (surface-speed formula ka WHY). Non-spinning cylinder ka surface speed $- 2 U sin θ$ hai (Ex 6). Ek free vortex ek uniform tangential speed add karta hai $u_{θ}^{vortex} = - \frac{Γ}{2 π r}$ , jo surface $r = a$ par $- \frac{Γ}{2 π a}$ ke barabar hai (har $θ$ par same, kyunki vortex ki speed sirf $r$ par depend karti hai). Add karo: $u_{θ}_{r = a} = - 2 U sin θ - \frac{Γ}{2 π a} .$ Yeh step kyun? Velocities ka superposition legal hai (linear Laplace); vortex circle ko streamline rehne deta hai (sirf swirl add karta hai, wall ke through koi radial flow nahin), toh cylinder survive karta hai.
Surface speed zero set karo. $- 2 U sin θ - \frac{Γ}{2 π a} = 0 \Rightarrow sin θ_{s} = - \frac{Γ}{4 π U a}$ . Yeh step kyun? Stagnation point woh hai jahan surface speed zero ho; $sin θ$ ke liye solve karne se angle(s) milte hain.
Plug in karo. $sin θ_{s} = - \frac{- 4 π}{4 π ( 4 ) ( 0.5 )} = \frac{4 π}{8 π} = \frac{1}{2}$ . Toh $θ_{s} = 3 0^{\circ}$ aur $θ_{s} = 15 0^{\circ}$ — figure mein do coral dots, dono upper half mein. Do answers kyun? $sin$ ek angle aur uske supplement par same value leta hai — do stagnation points, top ke baare mein symmetric.
Merge condition (part c). $sin θ_{s} = c$ ke do solutions coincide karte hain jab $∣ c ∣ = 1$ : $\frac{Γ}{4 π U a} = 1 \Rightarrow ∣Γ∣ = 4 π U a$ . Us value par do dots ek single point par collapse ho jaate hain top par ( $θ = 9 0^{\circ}$ , agar $Γ < 0$ ) ya bottom par ( $θ = - 9 0^{\circ}$ , agar $Γ > 0$ ). Uske baad ( $∣Γ∣ > 4 π U a$ ) $sin θ_{s}$ ko $1$ se zyada hona padega, jo impossible hai — toh surface par koi stagnation point nahin hai; woh cylinder ke upar (ya neeche) fluid mein lift off kar jaata hai. Yeh step kyun? $sin$ $\pm 1$ par capped hai; us cap ki algebra hi merge-then-detach behaviour ki physics hai jo examiners ko bahut pasand hai.

Verify: Merge value $∣Γ∣ = 4 π (4) (0.5) = 8 π \approx 25.13 m^{2} / s$ . Hamare test case mein $∣Γ∣ = 4 π \approx 12.57 < 8 π$ tha, toh do points exist karte hain — $3 0^{\circ}$ / $15 0^{\circ}$ answer ke consistent ✓. Forecast check: $Γ < 0$ (clockwise) ke saath, $sin θ_{s} > 0$ , toh dono points upper half mein baithe hain — swirl ne unhe upar drag kiya. ✓

Recall Quick self-test (guess karne ke baad reveal karo)

Source $m$ sign vs sink ::: $m > 0$ = source (outward), $m < 0$ = sink (inward); same formula $u_{r} = m / (2 π r)$ . Uniform+source ka stagnation point ::: $x_{s} = - m / (2 π U)$ , upstream ( $- x$ ) axis par jahan source push wind cancel kare. Ek jagah equal source+sink ::: Fields exactly cancel, $ψ \equiv 0$ , bilkul koi flow nahin. $Γ > 0$ (anticlockwise) ke liye rightward wind mein Magnus lift ki direction ::: Upward, $L^{'} = ρ U Γ > 0$ . Plain cylinder par max surface speed ::: $2 U$ , top aur bottom par ( $θ = \pm 9 0^{\circ}$ ). Free vortex speed law ::: $u_{θ} = Γ/ (2 π r)$ ; $r \to 0$ par faster. Spinning cylinder ke do stagnation points kab merge hote hain? ::: Jab $∣Γ∣ = 4 π U a$ (tab $sin θ_{s} = \pm 1$ ); uske baad woh fluid mein detach ho jaate hain.