Fluid kinematics — Eulerian vs Lagrangian description: Worked examples

2.2.9 · D3 · Physics › Fluid Mechanics › Fluid kinematics — Eulerian vs Lagrangian description

Is page ki har cheez ek boxed tool par tiki hui hai. Koi bhi symbol use karne se pehle ise plain words mein dobara state karte hain.

Formula Ek hi tool (words mein dobara)

$\frac{D ϕ}{D t} = local \frac{\partial ϕ}{\partial t} + convective u \frac{\partial ϕ}{\partial x} + v \frac{\partial ϕ}{\partial y} + w \frac{\partial ϕ}{\partial z}$

$ϕ$ = fluid ke saath carry hone wali koi bhi quantity (temperature, velocity component, salt concentration...).
$\frac{\partial ϕ}{\partial t}$ = "field khud kitni tezi se change hota hai ek frozen point par space mein" — tum wahan khade raho aur number ko tick karte dekho.
$u, v, w$ = fluid ke velocity components $x, y, z$ directions mein.
$\frac{\partial ϕ}{\partial x}$ = " $ϕ$ kitna change hota hai per metre agar main is instant $x$ mein sideways step loon" (space mein ek slope).
Toh $u \partial ϕ / \partial x$ = (metres per second) × (change per metre) = change per second kyunki main ek nayi jagah carry ho raha hoon.

$\frac{D}{D t}$ ka jawab hai: "moving particle actually kya feel karta hai?" — wahi ek sawaal jo Newton ke $F = ma$ ko zaroori lagta hai.

Humein convective piece ki zaroorat kyun hai, sirf $\partial / \partial t$ kyun nahi? Kyunki ek particle aisi field mein baith sakta hai jo time mein kabhi nahi badlati, phir bhi woh ek aisi region mein sweep ho sakta hai jahan field ki alag value ho. Neeche diya tool (Figure s01) dikhata hai kyun: wahi number ek trajectory ke saath grow kar sakta hai chahe har fixed dot par frozen ho.

Figure — Fluid kinematics — Eulerian vs Lagrangian description

Scenario matrix

Neeche har case class hai jo yeh topic throw kar sakta hai. Har worked example us cell ko naam deta hai jo woh cover karta hai.

Cell	Kya special hai	Covered by
A. Steady, positive convective	$\partial / \partial t = 0$ , particle speeds up	Ex 1
B. Unsteady + convective, dono add	local aur convective same sign	Ex 2
C. Opposite signs (woh ladte hain)	local up, convective down (ya vice versa)	Ex 3
D. Degenerate: uniform field	$\nabla ϕ = 0$ ⇒ convective $= 0$	Ex 4
E. Degenerate: zero velocity (rest point)	$u = v = w = 0$ ⇒ convective $= 0$	Ex 4
F. Full 2-D / 3-D, sab components	har $u \partial_{x} + v \partial_{y} + w \partial_{z}$ nonzero	Ex 5
G. Lagrangian → Eulerian conversion	label $a$ ko eliminate karo	Ex 6
H. Limiting behaviour	ek parameter ko → 0 ya → ∞ jaane do, trend check karo	Ex 7
I. Real-world word problem	English → field translate karo, phir tool apply karo	Ex 8
J. Exam twist: vector acceleration, sign trap	$ϕ = v$ , ek negative dhyaan se dekho	Ex 9

Prereqs Steady vs unsteady flow mein hain (iska matlab " $\partial / \partial t = 0$ ") aur Streamlines, pathlines and streaklines mein (particle ka path kya hota hai). Tool khud Continuity equation, Euler's equation of motion aur Navier-Stokes equations ko power karta hai aage.

Worked Examples

Worked example Ex 1 — Cell A: steady flow, particle phir bhi accelerate karta hai

Ek converging pipe mein steady 1-D flow hai $u (x) = 3 (1 + 2 x)$ m/s ( $x$ metres mein), $v = w = 0$ , koi time dependence nahi. Point $x = 1$ m par particle ki acceleration $a_{x}$ nikalo.

Forecast: flow "steady" hai — kya $a_{x}$ zero hoga? Pehle guess karo.

Local term. $\partial u / \partial t = 0$ . Yeh step kyun? "Steady" matlab field har fixed point par time mein frozen hai, isliye yeh term khatam ho jaati hai. (Dekho Steady vs unsteady flow.)
Convective term. $u \frac{\partial u}{\partial x}$ . Compute karo $\partial u / \partial x = 3 \cdot 2 = 6$ . Aur $u (1) = 3 (1 + 2) = 9$ . Yeh step kyun? Particle tezi fluid mein ride karta hai; jis rate se yeh speed up hota hai = (uski speed) × (speed ka spatial slope).
Add karo. $a_{x} = 0 + 9 \times 6 = 54$ m/s $^{2}$ . Yeh step kyun? $a_{x} = D u / D t$ ; yahan sirf convective bachta hai.

Verify: general formula $a_{x} = u \partial_{x} u = 3 (1 + 2 x) \cdot 6 = 18 (1 + 2 x)$ ; $x = 1$ par deta hai $18 \times 3 = 54$ ✓. Units: $(m/s) (1/s) = m/s^{2}$ ✓. Steady ≠ zero acceleration.

Worked example Ex 4 — Cells D & E: convective term ko kill karne ke do tarike

(D) Uniform field. $T (x, y, t) = 290 + 7 t$ (koi space dependence nahi), koi bhi velocity $u = v = 5$ . (E) Rest point. $T (x, t) = 300 + 0.5 x + 4 t$ lekin ek stagnation point par jahan $u = v = w = 0$ .

Forecast: dono cases mein $D T / D t = \partial T / \partial t$ . Kyun?

Case D: $\partial_{x} T = \partial_{y} T = 0$ , isliye convective $= u (0) + v (0) = 0$ . Isliye $D T / D t = 7$ K/s. Kyun? Koi spatial slope nahi matlab jahan bhi particle move kare, value identical hai — motion kuch nahi deta.
Case E: velocity zero hai, isliye convective $= 0 \cdot 0.5 = 0$ . Isliye $D T / D t = 4$ K/s. Kyun? Ek particle jo move nahi karta woh kaheen carry nahi ho sakta — yeh exactly ek fixed thermometer ki tarah behave karta hai.

Verify: D: $D T / D t = 7$ ✓ (local se match karta hai). E: $D T / D t = 4$ ✓. Do independent routes ( $\nabla ϕ = 0$ ya $v = 0$ ) dono convective term ko zero kar dete hain.

Worked example Ex 5 — Cell F: full 2-D, har component alive

Velocity field $u = x$ , $v = - y$ (ek "stagnation flow", steady). Point $(2, 3)$ par acceleration components $a_{x}, a_{y}$ nikalo.

Forecast: kaun sa term bachta hai jab flow steady ho lekin 2-D ho?

$ϕ = u$ set karo. $a_{x} = \partial_{t} u + u \partial_{x} u + v \partial_{y} u$ . Steady ⇒ $\partial_{t} u = 0$ . Yahan $\partial_{x} u = 1$ , $\partial_{y} u = 0$ . Kyun? Acceleration velocity ka hi material derivative hai ( $ϕ = v$ ).
$a_{x} = 0 + x \cdot 1 + (- y) \cdot 0 = x$ . $(2, 3)$ par: $a_{x} = 2$ m/s $^{2}$ .
$ϕ = v$ set karo. $a_{y} = u \partial_{x} v + v \partial_{y} v$ . Yahan $\partial_{x} v = 0$ , $\partial_{y} v = - 1$ .
$a_{y} = x \cdot 0 + (- y) (- 1) = y$ . $(2, 3)$ par: $a_{y} = 3$ m/s $^{2}$ . $+ y$ kyun? Do negatives ( $v < 0$ $y = 0$ ki taraf move kar raha hai, aur $v$ $y$ ke saath decrease ho raha hai) ek positive push ke liye multiply hote hain.

Verify: $a_{x} = x = 2$ , $a_{y} = y = 3$ ✓. Note karo dono sirf convective term se nonzero hain — steady, phir bhi har direction mein accelerating. Is force ke liye Euler's equation of motion dekho.

Worked example Ex 6 — Cell G: Lagrangian → Eulerian conversion

Particle paths diye gaye hain $x = a e^{2 t}$ (label $a$ = position at $t = 0$ ). Velocity ko Eulerian field $u (x, t)$ ke roop mein nikalo aur uski acceleration bhi.

Forecast: jawab mein koi $a$ nahi hona chahiye — Eulerian fields sirf current position jaante hain.

Lagrangian velocity. $u = \frac{\partial x}{\partial t}_{a} = 2 a e^{2 t}$ . Kyun? Path ko label fixed rakh ke differentiate karo — tum particle ke saath ride kar rahe ho.
Label eliminate karo. $x = a e^{2 t}$ se, $a = x e^{- 2 t}$ . Substitute karo: $u = 2 (x e^{- 2 t}) e^{2 t} = 2 x$ . Kyun? Ek Eulerian field nahi bata sakta kaun sa particle; $a$ ko $(x, t)$ se replace karne ke liye path invert karo.
Acceleration, Eulerian way. $a_{x} = \partial_{t} u + u \partial_{x} u = 0 + (2 x) (2) = 4 x$ . Cross-check kyun? Lagrangian directly: $a = \partial_{t}^{2} x ∣_{a} = 4 a e^{2 t} = 4 x$ — match karna chahiye.

Verify: $u (x, t) = 2 x$ ; dono routes dete hain $a_{x} = 4 x$ ✓. Field mein $a$ chodna classic error hai (parent ke mistake list mein dekho).

Worked example Ex 7 — Cell H: limiting behaviour

Nozzle flow $u (x) = u_{0} (1 + k x)$ , steady. $a_{x}$ ka study karo jab $k \to 0$ aur jab $k \to \infty$ .

Forecast: $k$ measure karta hai pipe kitni sharply narrow hoti hai. Extremes par acceleration ka kya hoga?

General: $a_{x} = u \partial_{x} u = u_{0} (1 + k x) \cdot u_{0} k = u_{0}^{2} k (1 + k x)$ .
Limit $k \to 0$ (uniform pipe): $a_{x} \to u_{0}^{2} \cdot 0 \cdot 1 = 0$ . Kyun? Koi narrowing nahi ⇒ koi spatial speed change nahi ⇒ accelerate karne ke liye kuch nahi. Cell D recover karta hai.
Limit $k \to \infty$ (violent narrowing): $a_{x} \sim u_{0}^{2} k^{2} x \to \infty$ . Kyun? Sharper convergence har metre mein bigger speed jump pack karta hai; convective term $k$ mein quadratically blow up karta hai.

Verify (numbers): lo $u_{0} = 2, x = 1$ . $k = 0.1$ par: $a_{x} = 4 (0.1) (1.1) = 0.44$ . $k = 1$ par: $a_{x} = 4 (1) (2) = 8$ . Ratio $8/0.44 \approx 18.2$ — fast grow karta hai, super-linear trend confirm karta hai ✓.

Worked example Ex 8 — Cell I: real-world word problem

Ek cyclist $6$ m/s ki speed se ek seedhi road par chalti hai. Air temperature har metre east ki taraf $0.2$ K badhti hai aur poori dopahar $3$ K/hour warm hoti ja rahi hai. Cyclist east ki taraf ja rahi hai. Cyclist ke thermometer ko warming rate kya dikhegi (K/s mein)?

Forecast: pehle English ko fields mein translate karo — phir sirf tool lagao.

Field banao. $\partial T / \partial x = 0.2$ K/m. Time rate: $3$ K/hour $= 3/3600$ K/s $= 8.33 \times 1 0^{- 4}$ K/s. Speed $u = 6$ m/s east. Kyun? Combine karne se pehle sab kuch SI (K, m, s) mein convert karo — units agree karni chahiye.
Local: $\partial T / \partial t = 8.33 \times 1 0^{- 4}$ K/s.
Convective: $u \partial_{x} T = 6 \times 0.2 = 1.2$ K/s.
Add karo: $D T / D t = 0.000833 + 1.2 = 1.2008$ K/s. Kyun? Cyclist hi moving parcel hai; dono terms apply hote hain.

Verify: convective $= 1.2$ completely dominate karta hai; dopahar ki slow warming ( $< 0.001$ K/s) ek ride mein negligible hai. $D T / D t \approx 1.20$ K/s ✓. Lesson: spatial gradient mein motion usually slow global drift ko beat karta hai.

Worked example Ex 9 — Cell J: exam twist, vector acceleration with a sign trap

Unsteady flow $u (x, t) = x - 2 t$ , $v = 0$ , $w = 0$ . $x = 1$ , $t = 1$ par $a_{x}$ nikalo. (Signs dhyaan se dekho.)

Forecast: ek local aur ek convective term hai, aur $u$ yahan negative hai — bahut se students ek sign drop kar dete hain.

Local: $\partial u / \partial t = - 2$ . Minus kyun? $u$ literally time ke saath decrease karta hai ( $- 2 t$ ) — genuine unsteadiness.
Convective: $u \partial_{x} u$ . Yahan $\partial_{x} u = 1$ , aur $(1, 1)$ par: $u = 1 - 2 = - 1$ . Point par $u$ compute kyun karo? Convective term actual local velocity use karta hai, jo $- 1$ hai ( $- x$ mein move kar raha hai), $+ 1$ nahi.
Combine karo: $a_{x} = - 2 + (- 1) (1) = - 3$ m/s $^{2}$ . Kyun? Dono terms yahan negative hain — particle decelerate/reverse karta hai.

Verify: $a_{x} = \partial_{t} u + u \partial_{x} u = - 2 + (- 1) (1) = - 3$ ✓. Signed $u = - 1$ plug karke trap avoid kiya, magnitude nahi.

Recall Self-test: cell ka naam do, phir solve karo

Ex 1 kaun sa cell cover karta hai, aur $a_{x} = 0$ kyun nahi hai? ::: Cell A (steady but convective); particle faster fluid mein move karta hai, $a_{x} = 54$ m/s $^{2}$ . Ex 3 mein, kya cheez $D T / D t$ ka sign $+ 4$ se $- 2$ kar deti hai? ::: Wind speed ko $u = 4$ karna convective $- 12$ ko local $+ 10$ se overcome karwa deta hai. Do independent tarike jinse convective term zero ho jaata hai (Ex 4)? ::: Uniform field ( $\nabla ϕ = 0$ ) YA zero velocity ( $v = 0$ ). Ex 6 mein answer mein $a$ kyun nahi hona chahiye? ::: Eulerian fields sirf current $(x, t)$ par depend karte hain; particle label $a$ ko path equation ke through eliminate karna padta hai. Ex 9 mein $(1, 1)$ par convective term mein $u$ ki kya value jaati hai? ::: Signed local velocity $u = - 1$ , $+ 1$ nahi.