Streamlines, pathlines, streaklines: Worked examples

2.2.10 · D3 · Physics › Fluid Mechanics › Streamlines, pathlines, streaklines

Yeh parent topic ka ek practice child page hai. Shuru karne se pehle, teen tools ka ek reminder taaki koi bhi symbol bina samjhe use na ho:

Recall Teen curves ek saanth mein

Streamline — time freeze karo, arrows follow karo. ODE: $\frac{d x}{u} = \frac{d y}{v}$ jisme $t$ ek constant hai.
Pathline — ek particle ki poori journey. ODE: $\frac{d x}{d t} = u$ , $\frac{d y}{d t} = v$ , time mein integrate karo.
Streakline — smoke plume: woh saare particles jo ek point se guzre, abhi dekhe gaye.

Yahan $u$ = fluid ki horizontal (x-direction) speed, $v$ = vertical (y-direction) speed. Dono milake $v = (u, v)$ velocity field hai.

Neeche har symbol pehli baar aane par define kiya gaya hai. "Field" ka matlab bas yeh hai: mujhe ek point $(x, y)$ aur ek time $t$ do, main tumhe do speeds $u$ aur $v$ wapas dunga.

Scenario matrix

Is topic ko ek aisi machine samjho jisme alag-alag tarah ke flow daale ja sakte hain. Har row ek case class hai; aakhri column us worked example ka naam deta hai jo uspe fit baithta hai.

#	Case class	Kya khas baat hai	Example
C1	Steady, curved field	$\partial v / \partial t = 0$ → teeno curves milte hain	Ex 1
C2	Steady, straight-but-shearing field	streamlines curve karti hain chahe flow time ke saath kabhi change na ho	Ex 2
C3	Unsteady field, teeno alag hain	time explicitly $v$ mein aata hai	Ex 3
C4	Sign / quadrant bookkeeping	$u, v$ sign change karte hain; har quadrant mein arrows kis taraf point karte hain	Ex 4
C5	Zero / degenerate input (stagnation point)	$u = 0$ aur/ya $v = 0$ ; ODE mein division-by-zero	Ex 5
C6	Limiting behaviour	$t \to \infty$ ya bahut pehle se switch on hua source	Ex 6
C7	Real-world word problem (gusty wind mein smoke plume)	physical story ko $u, v$ mein translate karo	Ex 7
C8	Exam twist — reverse question	streamlines diye hain, field reconstruct karo aur steadiness test karo	Ex 8

Hum aathon cover karenge.

Worked example Ex 1 — Steady curved field (C1): teeno milte hain

Field: $v = (u, v) = (- y, x)$ . Streamline, pathline ( $(1, 0)$ se $t = 0$ par release), aur streakline (source at $(1, 0)$ ) nikalo. Dikhao ki woh same curve hain.

Forecast: padhne se pehle andaaza lagao — kya woh match karenge? (Is field mein koi $t$ nahi hai, toh...)

Step 1 — Streamline. $\frac{d x}{- y} = \frac{d y}{x} \Rightarrow x d x = - y d y \Rightarrow x^{2} + y^{2} = C$ solve karo. Yeh step kyun? "Tangent parallel to $v$ " se ratio ODE milta hai jisme $t$ frozen hota hai; yahan freeze karne ke liye koi $t$ hai hi nahi toh kuch change nahi hota. $(1, 0)$ se: $C = 1$ , yani unit circle.

Step 2 — Pathline. $\frac{d x}{d t} = - y, \frac{d y}{d t} = x$ solve karo. Pehle ko differentiate karo: $\overset{x}{¨} = - \overset{y}{˙} = - x$ , toh $x = cos t$ , $y = sin t$ ( $x (0) = 1, y (0) = 0$ use karke). Yeh step kyun? Ek particle actual field ko time ke saath follow karta hai; yeh coupled system harmonic-oscillator pair hai, jiska solution unit angular speed par trace hone wala circle hai.

Step 3 — Streakline. Time $τ$ par $(1, 0)$ se release hua particle $x = cos (t - τ), y = sin (t - τ)$ par hai. $t$ fix karo, $τ$ sweep karo: point $(cos (t - τ), sin (t - τ))$ poore unit circle par ghoomta hai. Yeh step kyun? Steady flow ka matlab hai ki har released particle usi circular path par chalta hai; un sab ka plume wahi circle hai.

Verify: teeno $x^{2} + y^{2} = 1$ dete hain. ✔ Yahi golden rule hai (C1: steady ⇒ identical).

Worked example Ex 2 — Steady shear (C2): "straight" flow se curved streamlines

Field: $v = (u, v) = (y, 0)$ — fluid purely horizontal move karta hai, lekin upar zyada fast (ek shear layer). Streamlines nikalo.

Forecast: flow poori tarah sideways hai ( $v = 0$ ). Kya aapko straight ya curved streamlines expect hain?

Step 1 ODE: $\frac{d x}{y} = \frac{d y}{0}$ . Yeh step kyun? $u = y, v = 0$ ko $\frac{d x}{u} = \frac{d y}{v}$ mein daalo.

Step 2 Denominator $0$ hai matlab numerator bhi $0$ hona chahiye: $d y = 0 \Rightarrow y = const$ . Yeh step kyun? $\frac{d y}{0}$ ka sensible matlab yahi hai ki "woh change in $y$ jo ratio ko finite rakhta hai woh zero hai." Physically $v = 0$ toh kuch bhi vertically move nahi karta.

Verify: streamlines horizontal lines $y = c$ hain. Straight! (Contrast: yahan pathline bhi straight line hogi kyunki field steady hai — Ex 1 ka rule phir se. C2 dikhata hai "straight arrows ⇒ straight streamlines," ek sanity anchor baad mein cheezein complicated hone se pehle.) ✔

Worked example Ex 3 — Unsteady field (C3): teeno curves alag ho jaati hain

Field: $v = (u, v) = (1, t)$ . Origin par particle/source ke liye nikalo (a) $t = 2$ par streamline, (b) $t = 0$ par release hua pathline, (c) $t = 2$ par observe ki gayi streakline.

Forecast: field mein literally $t$ hai. Guess karo: kya (a), (b), (c) same honge ya teen alag curves?

Step 1 — Streamline at $t = 2$ . $t = 2$ freeze karo: $\frac{d x}{1} = \frac{d y}{2} \Rightarrow d y = 2 d x \Rightarrow y = 2 x$ . Yeh step kyun? Streamlines ek snapshot hain; $t = 2$ par har arrow direction $(1, 2)$ mein point karta hai, toh snapshot lines ki slope $2$ hogi.

Step 2 — Pathline from $(0, 0)$ at $t = 0$ . $\frac{d x}{d t} = 1 \Rightarrow x = t$ ; $\frac{d y}{d t} = t \Rightarrow y = \frac{1}{2} t^{2}$ . $t = x$ eliminate karo: $y = \frac{1}{2} x^{2}$ . Yeh step kyun? Ek particle time mein integrate hota hai; vertical speed $t$ ke saath badhti hai, trail ko ek parabola mein bend karti hai.

Step 3 — Streakline at $t = 2$ . Origin se $τ$ par release hua particle: $x = t - τ, y = \frac{1}{2} (t^{2} - τ^{2})$ . $t = 2$ rakho: $x = 2 - τ, y = \frac{1}{2} (4 - τ^{2})$ . $τ = 2 - x$ eliminate karo: $y = \frac{1}{2} (4 - (2 - x)^{2}) = 2 x - \frac{1}{2} x^{2}$ . Yeh step kyun? Alag-alag particles alag $τ$ par gaye, har ek badhti vertical speed ka alag "history" carry karta hai; abhi dikha plume $y = 2 x - \frac{1}{2} x^{2}$ hai.

Verify: teen alag curves — line $y = 2 x$ , parabola $y = \frac{1}{2} x^{2}$ , parabola $y = 2 x - \frac{1}{2} x^{2}$ . Woh sirf origin par milte hain. ✔ (C3: unsteady ⇒ sab alag.)

Worked example Ex 4 — Signs & quadrants (C4): har jagah arrow directions padhna

Field: $v = (u, v) = (- y, x)$ phir se (rotational). Kuch bhi solve kiye bina, charon quadrants mein flow ki direction batao, aur rotation ka sense confirm karo.

Forecast: counter-clockwise ya clockwise? Pehle point $(1, 0)$ se guess karo.

Step 1 — Quadrant I $(x > 0, y > 0)$ : $u = - y < 0$ (leftward), $v = x > 0$ (upward) ⇒ arrow up-left point karta hai. Step 2 — Quadrant II $(x < 0, y > 0)$ : $u = - y < 0$ (left), $v = x < 0$ (down) ⇒ down-left. Step 3 — Quadrant III $(x < 0, y < 0)$ : $u = - y > 0$ (right), $v = x < 0$ (down) ⇒ down-right. Step 4 — Quadrant IV $(x > 0, y < 0)$ : $u = - y > 0$ (right), $v = x > 0$ (up) ⇒ up-right. Yeh steps kyun? Direction dhundhne ke liye kabhi calculus nahi chahiye; bas do signed components evaluate karo. $u, v$ ke signs se arrow fully fix ho jaata hai.

Verify: positive $x$ -axis $(1, 0)$ par shuru karo: $u = 0, v = 1$ → seedha upar. QI mein up-left, phir aage... yeh counter-clockwise circulate karta hai. Ex 1 ke circles $x = cos t, y = sin t$ se consistent. ✔

Worked example Ex 5 — Zero / degenerate input (C5): ek stagnation point

Field: $v = (u, v) = (x, - y)$ (ek "saddle"/stagnation flow, jaise fluid wall se takraake split ho). Streamlines nikalo aur origin interpret karo.

Forecast: origin par $u$ aur $v$ dono... kya hain? Kya wahan baitha hua particle kabhi move kar sakta hai?

Step 1 Origin par $x = 0, y = 0 \Rightarrow u = 0, v = 0$ . Dono zero ⇒ stagnation point (fluid wahan rest mein hai). ODE $\frac{d x}{u} = \frac{d y}{v}$ ban jaata hai $\frac{0}{0}$ — undefined, toh origin ek khas degenerate point hai, kisi bhi single streamline ka hissa nahi. Yeh step kyun? Zeros pehle check karo: ratio ODE mein division-by-zero stagnation point flag karta hai.

Step 2 Origin se door: $\frac{d x}{x} = \frac{d y}{- y} \Rightarrow ln ∣ x ∣ = - ln ∣ y ∣ + C \Rightarrow x y = k$ . Yeh step kyun? Separate aur integrate karo; $\int d x / x = ln ∣ x ∣$ natural tool hai kyunki field har coordinate mein linear hai.

Step 3 — Degenerate lines cover karo. Value $k = 0$ dono axes $x = 0$ aur $y = 0$ deta hai (incoming/outgoing streamlines jo stagnation point se milti hain). Yeh step kyun? $k = 0$ family $x y = k$ ka ek legitimate limiting case hai; iske bina origin ke paas ki picture incomplete hai.

Verify: streamlines hyperbolas $x y = k$ hain, plus $k = 0$ par axes; origin par $v = 0$ (stagnation). Ek point check karo: $(2, 1)$ par, $v = (2, - 1)$ ; hyperbola $x y = 2$ ki slope $\frac{d y}{d x} = - y / x = - 1/2$ hai, aur $\frac{v}{u} = - 1/2$ — match karta hai. ✔

Worked example Ex 6 — Limiting behaviour (C6): ek purani streakline kahan baithti hai?

Field: $v = (1, t)$ (Ex 3 jaisa hi), source at origin. Bahut purane release $τ \to - \infty$ ke corresponding streakline point ki location $t = 2$ par observe karo. Plume ki far tail kya karti hai?

Forecast: purana smoke sabse zyada travel kar chuka hai. Guess karo ki tail $+ x$ ya $- x$ ki taraf jaati hai, aur upar ya neeche.

Step 1 $t = 2$ par streakline point: $x = 2 - τ, y = \frac{1}{2} (4 - τ^{2})$ . Yeh step kyun? Ex 3 ka release formula reuse karo, limit lene ke liye $τ$ symbolic rakho.

Step 2 $τ \to - \infty$ let karo: $x = 2 - τ \to + \infty$ ; $y = \frac{1}{2} (4 - τ^{2}) \to - \infty$ (the $- τ^{2}$ dominate karta hai). Yeh step kyun? Limiting analysis: check karo ki $τ$ bada hone par kaunsa term jeetta hai. Yahan $- τ^{2}$ sab ko beat karta hai.

Step 3 Curve $y = 2 x - \frac{1}{2} x^{2}$ par, $x \to + \infty$ par $y \to - \infty$ — match karta hai. Yeh step kyun? Limit ko Ex 3 ke closed-form curve ke against cross-check karo.

Verify: sabse purana smoke bahut right aur bahut neeche hai; plume ki tail neeche curve karti hai. Ek concrete purani value $τ = - 4$ par: $x = 6, y = \frac{1}{2} (4 - 16) = - 6$ , aur indeed $2 (6) - \frac{1}{2} (36) = 12 - 18 = - 6$ . ✔ (C6.)

Worked example Ex 7 — Real-world word problem (C7): gusty wind mein smoke

Story: Origin par ek chimney continuously smoke release karti hai. Wind steadily east mein $2 m/s$ se chalti hai, lekin ek rising thermal linearly badhta hua upward gust add karta hai: vertical speed $= 0.5 t m/s$ ( $t$ seconds mein, $t = 0$ jab observation start hua). $t = 4 s$ par visible smoke plume ki shape nikalo.

Forecast: steady wind + growing updraft — plume straight line hai ya curve? Guess karo.

Step 1 — Field mein translate karo. $u = 2$ , $v = 0.5 t$ . Time aata hai ⇒ unsteady ⇒ yeh ek streakline problem hai (dye/smoke ⇒ streakline, Flow visualization techniques (dye, smoke, PIV) ke mutabik). Yeh step kyun? Ek fixed point se continuous injection streakline ki definition hai; compute karne se pehle curve type identify karo.

Step 2 — $τ$ par release hue particle ka Pathline origin se: $\frac{d x}{d t} = 2 \Rightarrow x = 2 (t - τ)$ ; $\frac{d y}{d t} = 0.5 t \Rightarrow y = 0.25 (t^{2} - τ^{2})$ . Yeh step kyun? Streakline ko individual pathlines se banao, har ek apne release time $τ$ par origin se start karta hai.

Step 3 — $t = 4$ par observe karo, $τ \in [?, 4]$ sweep karo. $x = 2 (4 - τ) = 8 - 2 τ$ , $y = 0.25 (16 - τ^{2})$ . $τ = (8 - x) /2 = 4 - \frac{x}{2}$ eliminate karo: $y = 0.25 (16 - (4 - \frac{x}{2})^{2}) = 0.25 (16 - 16 + 4 x - \frac{x ^{2}}{4}) = x - \frac{x ^{2}}{16} .$ Yeh step kyun? "Now" fix karo ( $t = 4$ ), release time vary hone do — yahi "abhi dikha sara smoke" ka matlab hai.

Verify: plume curve $y = x - \frac{x ^{2}}{16}$ hai (ek downward parabola). Units: $x$ metres mein, $y$ metres mein. Sabse fresh puff ( $τ = 4$ ) origin $(0, 0)$ par hai ✔; $τ = 0$ par release hua puff $x = 8, y = 0.25 (16) = 4$ par hai, aur $8 - \frac{64}{16} = 8 - 4 = 4$ ✔.

Worked example Ex 8 — Exam twist (C8): field reverse-engineer karo

Twist: Tumhe diya gaya hai ki streamlines circles $x^{2} + y^{2} = C$ hain aur flow steady hai. Ek velocity field propose karo, aur phir test karo: kya $v = (- y, x)$ akela answer hai? Kya $v = (- 2 y, 2 x)$ bhi valid hai?

Forecast: kya field unique hona zaroori hai, ya bahut saare fields ek hi streamline pattern share kar sakte hain?

Step 1 — Direction constraint. Streamlines $x^{2} + y^{2} = C$ ki slope $\frac{d y}{d x} = - \frac{x}{y}$ hai (implicitly differentiate karo). Humein $\frac{v}{u} = - \frac{x}{y}$ chahiye. Yeh step kyun? Streamline slope $v / u$ ke barabar honi chahiye; yahi ek cheez hai jo streamlines pin down karti hain — ek ratio, magnitudes nahi.

Step 2 — $v = (- y, x)$ test karo: $\frac{v}{u} = \frac{x}{- y} = - \frac{x}{y}$ ✔. $v = (- 2 y, 2 x)$ test karo: $\frac{v}{u} = \frac{2 x}{- 2 y} = - \frac{x}{y}$ ✔. Yeh step kyun? Dono ratio satisfy karte hain, toh dono same circles trace karte hain — field unique nahi hai; sirf direction fix hai, speed (aur ek position ki scalar function bhi) free hai.

Step 3 — Steadiness check. Kisi bhi field mein $t$ nahi hai, toh $\partial v / \partial t = 0$ ⇒ steady ⇒ golden rule se, har field ke liye pathlines aur streaklines bhi yahi circles hain. Yeh step kyun? Exam chahta hai ki tum notice karo ki "same streamlines" ≠ "same flow": dono fields alag speeds par circulate karte hain (ek doosre se double) phir bhi direction ke snapshot mein identical dikhte hain.

Verify: $(1, 0)$ par, field 1 deta hai $v = (0, 1)$ (speed $1$ ); field 2 deta hai $v = (0, 2)$ (speed $2$ ). Dono unit circle ke tangent, alag magnitudes. ✔ Streamlines sirf direction encode karti hain.

Recall Cover-check: kya humne har cell hit ki?

Steady curved (C1) ::: Ex 1 Steady shear, straight streamlines (C2) ::: Ex 2 Unsteady, teeno alag (C3) ::: Ex 3 Signs/quadrants (C4) ::: Ex 4 Zero/degenerate stagnation (C5) ::: Ex 5 Limiting behaviour (C6) ::: Ex 6 Real-world word problem (C7) ::: Ex 7 Exam reverse twist (C8) ::: Ex 8

Connections

Velocity field and material derivative
Steady vs unsteady flow
Continuity equation
Stream function ψ
Lagrangian vs Eulerian description
Flow visualization techniques (dye, smoke, PIV)

Worked examples — Streamlines, pathlines, streaklines

Scenario matrix

Connections

Concept Map