Navier-Stokes equations — derivation from Newton's second law for fluid: Exercises

2.2.18 · D4 · Physics › Fluid Mechanics › Navier-Stokes equations — derivation from Newton's second la

Poore note mein, incompressible Navier–Stokes equation yeh hai:

$ρ (\frac{\partial u}{\partial t} + (u \cdot \nabla) u) = - \nabla P + μ \nabla^{2} u + ρ g, \nabla \cdot u = 0.$

Yahan $ρ$ density hai (mass per volume), $u$ velocity field hai, $P$ pressure hai, $μ$ dynamic viscosity hai (fluid kitna "chipchipa" hai), aur $g$ gravity hai. $\nabla$ ("nabla") gradient operator hai — yeh measure karta hai ki koi quantity space mein kitni tezi se change hoti hai. $\nabla^{2}$ Laplacian hai, jo ek doosri spatial derivative hai.

Level 1 — Recognition

L1.1 — Terms ko naam do

Upar di gayi full equation mein, identify karo ki chaar terms mein se har ek kaun si physical force represent karti hai, aur uski units (SI) batao.

Recall Solution

$ρ \frac{\partial u}{\partial t}$ — local (unsteady) inertia. Units $kg m^{- 2} s^{- 2} = N m^{- 3}$ (force per volume).
$ρ (u \cdot \nabla) u$ — convective inertia. Same units, $N m^{- 3}$ .
$- \nabla P$ — net pressure force per volume. $\frac{Pa}{m} = \frac{N / m ^{2}}{m} = N m^{- 3}$ . ✓
$μ \nabla^{2} u$ — net viscous force per volume. $[μ] = Pa\cdots$ , $[\nabla^{2} u] = s^{- 1} m^{- 1}$ , product $= N m^{- 3}$ . ✓
$ρ g$ — gravity (weight) per volume. $kg m^{- 3} \cdot m s^{- 2} = N m^{- 3}$ . ✓

Har term ek force per unit volume hai — isliye inhe add kiya ja sakta hai.

L1.2 — Acceleration dhundho

Ek patta ek steady river mein beh raha hai ( $\partial u / \partial t = 0$ ) jo narrow hoti ja rahi hai. Kya patta accelerate karta hai? Kaun sa term ise capture karta hai?

Recall Solution

Haan. Flow steady hai, isliye local term $\partial u / \partial t$ zero hai, lekin patta faster paani mein carry ho raha hai. Acceleration poori tarah convective term $(u \cdot \nabla) u$ se aata hai. Ek fixed observer apne point par constant speed dekhta hai; lekin particle phir bhi speed up karta hai. Dekho Material derivative.

Level 2 — Application

L2.1 — Material derivative compute karo

Ek 1-D steady flow mein $u_{x} (x) = 3 x$ hai (SI units mein, toh $u_{x}$ mein $m/s$ , $x$ mein m). $x = 2 m$ par ek fluid particle ka acceleration dhundho.

Recall Solution

Steady ⟹ $\partial u_{x} / \partial t = 0$ . 1-D mein convective term: $a = u_{x} \frac{\partial u _{x}}{\partial x} = (3 x) (3) = 9 x .$ $x = 2$ par: $a = 9 (2) = 18 m/s^{2}$ . Dhyaan do ki acceleration nonzero hai steady flow ke bawajood — convective term ne sara kaam kiya.

L2.2 — NS se hydrostatic pressure

Paani ki ek tank ( $ρ = 1000 kg/m^{3}$ ) rest mein hai, $u = 0$ . NS use karke, surface se $h = 3 m$ ki depth par pressure dhundho jahan surface pressure $P_{0} = 1.01 \times 1 0^{5} Pa$ hai. $g = 9.8 m/s^{2}$ lo.

Recall Solution

$u = 0$ ke saath, saare inertia aur viscous terms khatam ho jaate hain. NS collapse hoti hai $0 = - \nabla P + ρ g$ mein, yaani $\nabla P = ρ g$ . Depth $h$ ko neeche ki taraf lete hue: $\frac{d P}{d h} = ρ g \Rightarrow P = P_{0} + ρ g h .$ $P = 1.01 \times 1 0^{5} + (1000) (9.8) (3) = 1.01 \times 1 0^{5} + 2.94 \times 1 0^{4} = 1.304 \times 1 0^{5} Pa .$ Yahi bilkul Hydrostatic pressure hai — NS ise zero-velocity case ke roop mein apne andar rakhti hai.

L2.3 — Reynolds number

Paani ( $ρ = 1000 kg/m^{3}$ , $μ = 1.0 \times 1 0^{- 3} Pa\cdots$ ) diameter $L = 0.05 m$ ki ek pipe mein $U = 2 m/s$ ki speed se flow karta hai. Reynolds number $R e = ρ U L / μ$ compute karo aur batao kaun sa NS term dominate karta hai.

Recall Solution

$R e = \frac{ρ U L}{μ} = \frac{( 1000 ) ( 2 ) ( 0.05 )}{1.0 \times 1 0 ^{- 3}} = \frac{100}{1 0 ^{- 3}} = 1.0 \times 1 0^{5} .$ $R e$ (inertial force)/(viscous force) ka ratio hai. Yahan $R e ≫ 1$ hai, isliye inertia ( $ρ (u \cdot \nabla) u$ term) dominate karta hai aur flow turbulent hai. Dekho Reynolds number.

Level 3 — Analysis

L3.1 — Kaun se terms bachte hain?

NS ke har term ko kept ya dropped classify karo: (a) inviscid flow ke liye, (b) static fluid ke liye, (c) steady flow ke liye.

Recall Solution

Case	$\partial u / \partial t$	$(u \cdot \nabla) u$	$- \nabla P$	$μ \nabla^{2} u$	$ρ g$
(a) inviscid ( $μ = 0$ )	keep	keep	keep	drop	keep
(b) static ( $u = 0$ )	drop	drop	keep	drop	keep
(c) steady ( $\partial_{t} = 0$ )	drop	keep	keep	keep	keep

Case (a) se Euler equation (inviscid flow) milti hai; case (b) se hydrostatics milti hai.

L3.2 — Viscous term ka dimensional check

Dimensional analysis se dikhao ki $μ \nabla^{2} u$ ke units $- \nabla P$ ke barabar hain, jo confirm karta hai ki inhe add kiya ja sakta hai.

Recall Solution

$[μ] = Pa\cdots$ . $[\nabla^{2} u] = \frac{m/s}{m ^{2}} = s^{- 1} m^{- 1}$ . Product: $Pa\cdots \cdot s^{- 1} m^{- 1} = Pa/m = N m^{- 3}$ . $[- \nabla P] = Pa/m = N m^{- 3}$ . ✓ Identical — dono forces per volume hain.

L3.3 — Second derivative kyun?

Ek shear stress $τ = μ \partial u_{x} / \partial y$ hai. Ek blob ke bottom face par stress $τ (y)$ hai aur top face par $τ (y + d y)$ . Net viscous force per volume derive karo aur explain karo ki ek derivative kyun kaafi nahi hai.

Recall Solution

Do horizontal faces se net $x$ -force (har face ka area $d x d z$ ): $d F_{x} = [τ (y + d y) - τ (y)] d x d z = \frac{\partial τ}{\partial y} d y d x d z = μ \frac{\partial ^{2} u _{x}}{\partial y ^{2}} d V .$ Per volume: $μ \partial^{2} u_{x} / \partial y^{2}$ . Ek derivative kyun nahi? $τ$ khud ek stress hai, net force nahi. Net force ke liye stress ko blob ke across differ karna padta hai — yeh difference ek aur spatial derivative hai. Dekho Viscosity & Newton's law of viscosity.

Figure — Navier-Stokes equations — derivation from Newton's second law for fluid

Level 4 — Synthesis

L4.1 — Plane Poiseuille flow

$y = 0$ aur $y = h$ par plates ke beech steady flow $u_{x} (y)$ hai, pressure gradient $\partial P / \partial x = - G$ (jahan $G > 0$ ) se driven hai, $x$ mein gravity nahi. No-slip: $u_{x} (0) = u_{x} (h) = 0$ . $u_{x} (y)$ derive karo aur dhundho ki speed maximum kahan hai.

Recall Solution

Steady, unidirectional ⟹ inertia terms khatam, $x$ -momentum NS ban jaata hai $0 = - \frac{\partial P}{\partial x} + μ \frac{d ^{2} u _{x}}{d y ^{2}} = G + μ \frac{d ^{2} u _{x}}{d y ^{2}} \Rightarrow \frac{d ^{2} u _{x}}{d y ^{2}} = - \frac{G}{μ} .$ Do baar integrate karo: $u_{x} = - \frac{G}{2 μ} y^{2} + C_{1} y + C_{2} .$ $u_{x} (0) = 0$ lagao $\Rightarrow C_{2} = 0$ . $u_{x} (h) = 0$ lagao $\Rightarrow C_{1} = \frac{G h}{2 μ}$ . Toh $u_{x} (y) = \frac{G}{2 μ} y (h - y) .$ Maximum jahan $d u_{x} / d y = \frac{G}{2 μ} (h - 2 y) = 0 \Rightarrow y = h /2$ (centre par), jahan $u_{m a x} = \frac{G}{2 μ} \cdot \frac{h}{2} \cdot \frac{h}{2} = \frac{G h ^{2}}{8 μ} .$ Yeh parabola viscous pressure-driven flow ki pehchaan hai. Dekho Poiseuille flow.

L4.2 — Numeric Poiseuille

L4.1 ko $G = 200 Pa/m$ , $μ = 1.0 \times 1 0^{- 3} Pa\cdots$ , $h = 0.01 m$ ke saath use karke, $u_{m a x}$ aur $y = h /4$ par speed compute karo.

Recall Solution

$u_{m a x} = \frac{G h ^{2}}{8 μ} = \frac{( 200 ) ( 0.01 ) ^{2}}{8 ( 1 0 ^{- 3} )} = \frac{200 \cdot 1 0 ^{- 4}}{8 \times 1 0 ^{- 3}} = \frac{0.02}{0.008} = 2.5 m/s .$ $y = h /4 = 0.0025$ par: $u_{x} = \frac{G}{2 μ} y (h - y) = \frac{200}{2 \times 1 0 ^{- 3}} (0.0025) (0.0075) = (1 0^{5}) (1.875 \times 1 0^{- 5}) = 1.875 m/s .$

Level 5 — Mastery

L5.1 — Wall ke neeche gravity-driven film

Thickness $h$ ki ek thin film gravity ke under ek vertical wall se neeche behti hai. Steady, unidirectional flow $u_{z} (x)$ jahan $x$ wall se distance hai ( $x = 0$ wall, $x = h$ free surface). Koi pressure gradient nahi. Wall par no-slip $u_{z} (0) = 0$ ; free surface par zero shear $d u_{z} / d x ∣_{x = h} = 0$ . $u_{z} (x)$ derive karo aur maximum (surface) speed nikalo.

Recall Solution

Flow direction ( $z$ , neeche) ke along, NS deta hai $0 = μ \frac{d ^{2} u _{z}}{d x ^{2}} + ρ g$ , toh $\frac{d ^{2} u _{z}}{d x ^{2}} = - \frac{ρ g}{μ} .$ Integrate karo: $\frac{d u _{z}}{d x} = - \frac{ρ g}{μ} x + C_{1}$ . $x = h$ par zero-shear: $0 = - \frac{ρ g}{μ} h + C_{1} \Rightarrow C_{1} = \frac{ρ g h}{μ}$ . Phir integrate karo: $u_{z} = - \frac{ρ g}{2 μ} x^{2} + \frac{ρ g h}{μ} x + C_{2}$ . No-slip $u_{z} (0) = 0 \Rightarrow C_{2} = 0$ . Toh $u_{z} (x) = \frac{ρ g}{μ} (h x - \frac{x ^{2}}{2}) .$ Maximum at the free surface $x = h$ : $u_{m a x} = \frac{ρ g}{μ} (h^{2} - \frac{h ^{2}}{2}) = \frac{ρ g h ^{2}}{2 μ} .$

L5.2 — Numeric film

$ρ = 1000 kg/m^{3}$ , $g = 9.8 m/s^{2}$ , $μ = 0.1 Pa\cdots$ , $h = 0.002 m$ lo. Surface speed $u_{m a x}$ dhundho.

Recall Solution

$u_{m a x} = \frac{ρ g h ^{2}}{2 μ} = \frac{( 1000 ) ( 9.8 ) ( 0.002 ) ^{2}}{2 ( 0.1 )} = \frac{( 1000 ) ( 9.8 ) ( 4 \times 1 0 ^{- 6} )}{0.2} = \frac{0.0392}{0.2} = 0.196 m/s .$

L5.3 — Convective vs. viscous (order-of-magnitude)

L2.3 ke pipe flow ke liye ( $ρ = 1000$ , $U = 2$ , $L = 0.05$ , $μ = 1 0^{- 3}$ ), convective inertia term $ρ U^{2} / L$ aur viscous term $μU / L^{2}$ ka ratio estimate karo. Confirm karo ki yeh $R e$ ke barabar hai.

Recall Solution

$\frac{ρ U ^{2} / L}{μU / L ^{2}} = \frac{ρ U ^{2} L ^{2}}{L μU} = \frac{ρ U L}{μ} = R e .$ Numerically: $\frac{( 1000 ) ( 2 ^{2} ) /0.05}{( 1 0 ^{- 3} ) ( 2 ) /0.0 5 ^{2}} = \frac{80000}{0.8} = 1.0 \times 1 0^{5} = R e$ . ✓ Isliye $R e$ inertia-to-viscosity measure karta hai — yeh seedha do NS terms ko compare karne se nikalta hai.

Recall Master checklist

Har term ek force per volume hai ::: isliye wo add hote hain; check karo ki units $N m^{- 3}$ hain. Steady flow mein bhi acceleration ho sakta hai ::: convective term $(u \cdot \nabla) u$ ki wajah se. Viscous force ek second derivative hai ::: yeh blob ke across stress ka difference hai. Second-order ODE ::: ko do boundary conditions chahiye. Free surface ::: zero shear, no-slip nahi.