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

2.2.18 · D3 · Physics › Fluid Mechanics › Navier-Stokes equations — derivation from Newton's second la

Yeh page parent Navier–Stokes note ka drill ground hai. Wahan humne equation ko $F = m a$ se banaya tha. Yahan hum ise har us situation mein use karte hain jo exam mein aa sakti hai — steady aur unsteady, viscous aur inviscid, rest mein aur gravity-driven, plus ek word problem aur ek exam twist.

Shuru karne se pehle, ek reminder us master sentence ka jisme hum hamesha plug karte hain:

Har symbol, taaki koi line one se hi na bhatak jaaye:

Definition Is page ka alphabet

$ρ$ (rho) — fluid ki density, har cubic metre mein kitna mass packed hai, units $kg / m^{3}$ .
$u$ — velocity field: space aur time ke har point par yeh batata hai ki wahan fluid kis direction mein aur kitni tez move kar raha hai. $u_{x}$ is ka $x$ -axis ke along component hai.
$P$ — pressure, kisi bhi surface par inward push per unit area, units $Pa = N / m^{2}$ .
$μ$ (mu) — dynamic viscosity, fluid kitna "sticky/thick" hai, units $Pa \cdot s$ (dekho Viscosity & Newton's law of viscosity).
$g$ — gravity as a vector (iska ek direction hai, seedha neeche), magnitude likha jaata hai $g = 9.8 m/s^{2}$ . Jab bhi hum koi axis choose karte hain hume $g$ ko us par project karna padta hai: axis $x$ ke along component hai $g_{x} = g \cdot \overset{x}{^}$ . Toh $g$ (arrow) aur $g$ (number) alag objects hain — hamesha check karo ki formula kaunsa maang raha hai.
$\nabla$ (nabla) — "har direction mein slope" operator; $\nabla P$ woh vector hai jo fastest pressure rise ki taraf point karta hai. $\nabla^{2}$ ise do baar apply karta hai (curvature).
$\frac{D}{D t}$ — Material derivative, acceleration jo chalte hue fluid ke saath measure ki jaati hai: $\frac{D u}{D t} = \frac{\partial u}{\partial t} + (u \cdot \nabla) u$ . Yeh explicitly Ex 5 (pure convection) aur Ex 6 (pure local term) mein aata hai.

The scenario matrix

Har problem jo tum meet kar sakte ho woh in cells mein se ek hai (ya blend). Neeche ke examples par cell label laga hai.

Cell	Case class	Kya switch on / off hai	Example
C1	At rest ( $u = 0$ )	sirf pressure + gravity	Ex 1
C2	Steady viscous, pressure-driven	$\partial_{t} = 0$ , $μ \neq = 0$ , gradient drives	Ex 2
C3	Steady viscous, gravity-driven	$μ \neq = 0$ , $ρ g$ drives (falling film)	Ex 3
C4	Inviscid limit ( $μ = 0$ )	viscous term drop karo → Euler	Ex 4
C5	Convective term dominant	$(u \cdot \nabla) u \neq = 0$ , narrowing pipe	Ex 5
C6	Unsteady (transient start-up)	$\partial_{t} \neq = 0$	Ex 6
C7	Sign / direction check	$\partial_{x} P$ ka sign kis direction mein push karta hai?	Ex 7
C8	Degenerate / limiting numbers	$μ \to 0$ , $h \to 0$ , $G \to 0$	Ex 8
C9	Real-world word problem	Poiseuille flow rate mein numbers lagao	Ex 9
C10	Exam twist (dimensionless)	NS terms se Reynolds number banao	Ex 10
C11	All terms live at once	unsteady + convective + viscous + free surface	Ex 11

Ex 1 — Fluid at rest (Cell C1)

Worked example Paani ka ek tank, 3 m gehraai. Talne par gauge pressure nikalo. (

$ρ = 1000 kg/m^{3}$ , $g = 9.8$ .) Forecast: padhne se pehle number guess karo. Zyada gehraai → upar paani ka zyada weight → zyada pressure. Roughly $1 0^{4}$ Pa per metre. 3 m ke liye kya milega?

Step 1. $u = 0$ har jagah set karo. Yeh step kyun? "At rest" exactly $u = 0$ hai; yeh poora left side aur poora viscous term kill kar deta hai (dono ko moving field chahiye). $0 = - \nabla P + ρ g$

Step 2. Vertical component lo. $z$ ko upar point karne do, toh $g = (0, 0, - g)$ , yaani iska $z$ -projection $g_{z} = - g$ hai. Yeh step kyun? $g$ ek vector hai; ek scalar equation padhne ke liye hum ise $z$ par project karte hain. Horizontally kuch bhi vary nahi karta, toh $\partial_{x} P = \partial_{y} P = 0$ aur $P$ sirf $z$ par depend karta hai — partial ek ordinary derivative ban jaata hai. $0 = - \frac{\partial P}{\partial z} - ρ g \Rightarrow \frac{d P}{d z} = - ρ g$

Step 3. Surface se neeche integrate karo. Yeh step kyun? $d P / d z$ constant hai, toh pressure depth $h$ ke saath linearly badhta hai. $P = P_{surface} + ρ g h$

Verify: $P_{gauge} = ρ g h = 1000 \times 9.8 \times 3 = 29400 Pa$ . Units: $\frac{kg}{m ^{3}} \cdot \frac{m}{s ^{2}} \cdot m = \frac{kg}{m \cdot s ^{2}} = Pa$ ✓. Yeh exactly Hydrostatic pressure hai.

Ex 2 — Steady viscous, pressure-driven (Cell C2)

Worked example

$y = 0$ aur $y = h = 2 mm$ par do fixed plates ke beech paani, pressure gradient $G = - \partial P / \partial x = 500 Pa/m$ se push ho raha hai, viscosity $μ = 1 0^{- 3} Pa\cdots$ . Centreline velocity nikalo. Forecast: profile ek parabola hai, dono walls par zero (no-slip), middle mein fastest. Max speed millimetres/s hogi ya metres/s?

Figure mein kya dekhna hai: do moti horizontal slate lines fixed plates hain. Lavender curve jo daayein bulge kar rahi hai woh velocity profile $u_{x} (y)$ hai — notice karo ki yeh dono plates par zero se pinned hai (no-slip) aur middle mein sabse moti. Har chhota coral arrow us height par local flow speed hai: walls par kuch nahi se badhte hain centre par sabse lambe tak, parabola trace karte hue. Butter arrow neeche pressure push $G = - \partial P / \partial x$ hai jo sab kuch drive karta hai, aur mint dot $y = h /2$ par $u_{m a x}$ mark karta hai. Picture hi derivation hai: constant curvature (Step 1) jo no-slip walls se forced hai (Step 2) sirf ek symmetric parabola ho sakti hai.

Step 1. Steady ( $\partial_{t} = 0$ ), flow sirf $x$ ke along, sirf $y$ par depend karta hai. Kyunki field ab single-variable function $u_{x} (y)$ tak reduce ho gayi hai, hum $\partial^{2} / \partial y^{2}$ se ordinary $d^{2} / d y^{2}$ par switch karte hain. Yeh step kyun? har dropped term literally zero hai — koi time change nahi, koi cross-flow nahi, toh convection $(u \cdot \nabla) u = 0$ . Saath hi, continuity $\partial u_{x} / \partial x = 0$ guarantee karta hai ki $u_{x}$ $x$ par depend nahi kar sakta — yahi woh hai jo hume sirf $u_{x} (y)$ likhne deta hai. $0 = G + μ \frac{d ^{2} u _{x}}{d y ^{2}}$

Step 2. ODE ko no-slip $u_{x} (0) = u_{x} (h) = 0$ ke saath solve karo. Yeh step kyun? Second derivative constant hai $\Rightarrow$ do baar integrate karo; walls constants ko pin karte hain. Yeh Poiseuille flow hai. $u_{x} (y) = \frac{G}{2 μ} y (h - y)$

Step 3. Max $y = h /2$ par. Yeh step kyun? Symmetric parabola midpoint par peak karti hai. $u_{m a x} = \frac{G h ^{2}}{8 μ}$

Verify: $u_{m a x} = \frac{500 \times ( 0.002 ) ^{2}}{8 \times 1 0 ^{- 3}} = \frac{500 \times 4 \times 1 0 ^{- 6}}{8 \times 1 0 ^{- 3}} = 0.25 m/s$ . Units: $\frac{( Pa/m ) \cdot m ^{2}}{Pa\cdots} = \frac{m}{s}$ ✓.

Ex 3 — Steady viscous, gravity-driven (Cell C3)

Worked example Tel ki ek thin film (

$μ = 0.1 Pa\cdots$ , $ρ = 900$ ) ek vertical wall se neeche flow karti hai, film thickness $δ = 1 mm$ , free surface par air hai (zero shear). Surface velocity nikalo. Forecast: yahan gravity driver hai, pressure gradient nahi. Free surface sabse fast hai (koi cheez ise drag nahi karti), wall ruki hui hai. Half-parabola.

Step 1 — axes choose karo aur $g$ project karo. $x$ ko wall ke saath seedha neeche point karne do (flow direction) aur $y$ wall se horizontally bahar measure karo. Kyunki $x$ neeche ki vertical ke saath aligned choose kiya gaya hai, gravity vector fully $x$ par project hota hai: $g_{x} = g \cdot \overset{x}{^} = + g$ aur $g_{y} = 0$ . Yeh step kyun? $g$ ek vector hai; $x$ -momentum equation sirf iska $x$ -component chahti hai. $x$ ko $g$ ke saath align karke hume clean $g_{x} = g$ milta hai, aur $y$ -equation sirf thin film ke across (constant) hydrostatic pressure deta hai, jise hum ignore karte hain. Steady, sirf $u_{x} (y)$ (continuity phir se $x$ -dependence forbid karta hai, toh $\partial^{2} \to d^{2}$ ): $0 = ρ g_{x} + μ \frac{d ^{2} u _{x}}{d y ^{2}} = ρ g + μ \frac{d ^{2} u _{x}}{d y ^{2}} \Rightarrow \frac{d ^{2} u _{x}}{d y ^{2}} = - \frac{ρ g}{μ}$

Step 2 — free-surface boundary condition. Wall par no-slip $u_{x} (0) = 0$ ; free surface par zero shear $\frac{d u _{x}}{d y}_{y = δ} = 0$ . Yeh step kyun? Air oil ko grip nahi kar sakti, toh viscous stress $μ d u / d y$ open top par vanish hona chahiye — yeh free-surface pressure/stress condition hai: ek bahut halke fluid ke saath interface par, tangential stress $\to 0$ aur surface pressure $=$ atmospheric.

Step 3. Integrate karo: $u_{x} (y) = \frac{ρ g}{μ} (δ y - \frac{y ^{2}}{2})$ . Yeh step kyun? Pehla integration + zero-shear slope constant fix karta hai; doosra + no-slip offset fix karta hai. $y = δ$ par surface speed: $u_{surf} = \frac{ρ g δ ^{2}}{2 μ}$

Verify: $u_{surf} = \frac{900 \times 9.8 \times ( 0.001 ) ^{2}}{2 \times 0.1} = \frac{900 \times 9.8 \times 1 0 ^{- 6}}{0.2} = 0.0441 m/s$ ✓.

Ex 4 — Inviscid limit → Euler (Cell C4)

Worked example Ek frictionless ideal fluid (

$μ = 0$ ) steadily flow karta hai. Dikhao ki NS Bernoulli-friendly Euler ban jaata hai, aur pressure drop evaluate karo jab speed $2$ se $4 m/s$ tak double ho jaati hai ( $ρ = 1000$ , horizontal, streamline ke along koi gravity term nahi). Forecast: koi stickiness nahi, toh sirf pressure aur gravity bachte hain. Tez flow → lower pressure. Kitna lower?

Step 1. $μ = 0$ set karo. Yeh step kyun? Poora viscous term mar jaata hai, Euler equation (inviscid flow) bachta hai: $ρ \frac{D u}{D t} = - \nabla P + ρ g$

Step 2. Steady, horizontal streamline ke along, integrate karo → Bernoulli: $P + \frac{1}{2} ρ u^{2} = const$ . Yeh step kyun? $ρ (u \cdot \nabla) u$ kinetic term mein integrate hota hai.

Step 3. $Δ P = - \frac{1}{2} ρ (u_{2}^{2} - u_{1}^{2})$ . Yeh step kyun? Total constant hai: extra speed ke liye pay karne ke liye pressure girna padega.

Verify: $Δ P = - \frac{1}{2} \times 1000 \times (4^{2} - 2^{2}) = - \frac{1}{2} \times 1000 \times 12 = - 6000 Pa$ . Pressure 6 kPa drop karta hai ✓.

Ex 5 — Convective term dominant (Cell C5)

Worked example Ek narrowing channel mein steady flow:

$x$ ke along velocity hai $u_{x} (x) = u_{0} (1 + α x)$ jahan $u_{0} = 1 m/s$ , $α = 2 m^{- 1}$ . Material derivative use karke $x = 0.5 m$ par fluid particle ka acceleration nikalo. Forecast: bhaale ki $\partial u / \partial t = 0$ ho (steady!), ek particle tez paani mein move karke speed up karta hai. Toh acceleration zero nahi hai. Positive hai ya negative?

Step 1. Material derivative likho aur local term drop karo. Yeh step kyun? Material derivative true particle acceleration hai; steady flow $\partial u / \partial t$ ko zero karta hai, sirf convection bachta hai: $\frac{D u _{x}}{D t} = = 0 \frac{\partial u _{x}}{\partial t} + (u \cdot \nabla) u_{x} = u_{x} \frac{d u _{x}}{d x}$

Step 2. $\frac{d u _{x}}{d x} = u_{0} α$ , toh $\frac{D u _{x}}{D t} = u_{x} (x) \cdot u_{0} α = u_{0} (1 + α x) \cdot u_{0} α = u_{0}^{2} α (1 + α x)$ . Yeh step kyun? Diye gaye profile par chain rule — yeh convective acceleration concrete form mein hai.

Verify: $\frac{D u _{x}}{D t} = 1^{2} \times 2 \times (1 + 2 \times 0.5) = 2 \times 2 = 4 m/s^{2}$ . Positive → narrows mein accelerate ho raha hai, leaf-in-a-river picture se match karta hai ✓.

Ex 6 — Unsteady start-up (Cell C6)

Worked example Stokes' first problem: ek fluid plate ke upar rest mein hai.

$t = 0$ par plate suddenly move karti hai. Plate ke paas equation hai $\frac{\partial u}{\partial t} = ν \frac{\partial ^{2} u}{\partial y ^{2}}$ jahan $ν = μ / ρ = 1 0^{- 6} m^{2} / s$ . $t = 4 s$ baad motion kitni gehraai tak "diffuse" ho gayi hai? Forecast: viscosity momentum ko heat ki tarah spread karti hai. Reach ki gayi depth $ν t$ ki tarah badhti hai — linearly nahi. Penetration depth $δ \sim ν t$ estimate karo.

Step 1. Sirf $\partial_{t}$ aur viscous term rakho. Yeh step kyun? Plate ke along koi pressure gradient nahi, koi gravity nahi; yahan flow dono $y$ aur $t$ par depend karta hai, toh hum genuinely partial derivatives rakhte hain (unlike Ex 2/3 jo ek variable tak collapse ho gaye the). Material derivative apne local piece $\partial u / \partial t$ tak reduce ho jaata hai kyunki $u$ ka $x$ -dependence nahi hai. Yeh diffusion equation hai.

Step 2. Dimensional balance: $\frac{u}{t} \sim ν \frac{u}{δ ^{2}} \Rightarrow δ \sim ν t$ . Yeh step kyun? Do surviving terms ki sizes match karne se growth law reveal hota hai.

Verify: $δ = ν t = 1 0^{- 6} \times 4 = 4 \times 1 0^{- 6} = 2 \times 1 0^{- 3} m = 2 mm$ ✓. Slow diffusion — isliye viscous startup sluggish hota hai.

Ex 7 — Sign / direction check (Cell C7)

Worked example Pressure

$+ x$ ke along badhta hai: $\partial P / \partial x = + 300 Pa/m$ . Pressure force per volume kis direction mein hai, aur ek fluid parcel ke saath kya hoga (dusre forces ignore karo)? Forecast: sign trap. Common instinct: "high pressure aage push karta hai." Careful — force hai $- \nabla P$ .

Step 1. Pressure force per volume $f_{x} = - \partial P / \partial x$ . Yeh step kyun? Humne parent mein derive kiya tha ki sirf difference matter karta hai, aur minus sign kehta hai fluid high se low pressure ki taraf push hota hai.

Step 2. $f_{x} = - (+ 300) = - 300 Pa/m = - 300 N/m^{3}$ . Yeh step kyun? Diya gaya gradient plug karo.

Verify: negative → force $- x$ direction mein point karta hai, yaani backward, $+ x$ mein move kar rahe parcel ko decelerate karta hai. Intuition se match karta hai: fluid rising pressure ke against climb karta hai ✓. Magnitude $∣ f_{x} ∣ = 300 N/m^{3}$ .

Ex 8 — Degenerate / limiting numbers (Cell C8)

Worked example Ex 2 se Poiseuille result lo

$u_{m a x} = \frac{G h ^{2}}{8 μ}$ . Teen limits examine karo: (a) $μ \to 0$ , (b) $h \to 0$ , (c) $G \to 0$ . Har ek physically kya matlab rakhta hai, aur $u_{m a x}$ evaluate karo agar $h$ half karke $1 mm$ kar diya jaaye (Ex 2 ke $G, μ$ rakhte hue)? Forecast: thinner gap → dramatically slower flow kyunki $h$ squared appear karta hai. $h$ halving speed 4 factor se cut karegi, 2 se nahi.

Step 1 (a). $μ \to 0$ : $u_{m a x} \to \infty$ . Yeh step kyun? Koi friction nahi matlab push resist karne ke liye kuch nahi — unphysical, humein bata raha hai real fluids hamesha kuch $μ$ rakhte hain.

Step 2 (b). $h \to 0$ : $u_{m a x} \to 0$ . Yeh step kyun? Gap band kar do aur dono walls fluid ko har jagah rok dete hain — flow vanish ho jaata hai.

Step 3 (c). $G \to 0$ : $u_{m a x} \to 0$ . Yeh step kyun? Koi driving gradient nahi, koi flow nahi — Cell C1 (at rest) ke saath consistent.

Verify: $h$ half karke $0.001 m$ karo: $u_{m a x} = \frac{500 \times ( 0.001 ) ^{2}}{8 \times 1 0 ^{- 3}} = \frac{500 \times 1 0 ^{- 6}}{8 \times 1 0 ^{- 3}} = 0.0625 m/s$ . Yeh $0.25/4$ hai — exactly Ex 2 ke answer ka quarter ✓.

Ex 9 — Real-world word problem (Cell C9)

Worked example Blood (

$μ = 3.5 \times 1 0^{- 3} Pa\cdots$ maano) ek capillary-jaisi round tube se flow karta hai, radius $R = 1 mm$ , length $L = 0.1 m$ , pressure difference $Δ P = 200 Pa$ . Volume flow rate $Q$ nikalo. Forecast: round-pipe Poiseuille law hai $Q = \frac{π R ^{4} Δ P}{8 μL}$ . Notice karo $R^{4}$ — flow radius ke liye extremely sensitive hai. $m^{3} / s$ mein ek tiny number expect karo.

Step 1. Cylindrical coordinates mein same NS reduction jaisa Ex 2 mein hai $u (r) = \frac{Δ P}{4 μL} (R^{2} - r^{2})$ deta hai. Yeh step kyun? Round geometry flat-plate parabola ko ek paraboloid se replace karti hai, phir bhi $μ \nabla^{2} u = - \partial_{x} P$ se.

Step 2. Speed ko cross-section par integrate karo: $Q = \int_{0}^{R} u (r) 2 π r d r = \frac{π R ^{4} Δ P}{8 μL}$ . Yeh step kyun? Flow rate speed ko area par sum karna hai.

Verify: $Q = \frac{π ( 1 0 ^{- 3} ) ^{4} \times 200}{8 \times 3.5 \times 1 0 ^{- 3} \times 0.1} = \frac{π \times 1 0 ^{- 12} \times 200}{2.8 \times 1 0 ^{- 3}} \approx 2.244 \times 1 0^{- 7} m^{3} / s$ ✓ (lagbhag $0.22 mL/s$ ). Dekho Poiseuille flow.

Ex 10 — Exam twist: build the Reynolds number (Cell C10)

Worked example NS mein convective term aur viscous term ki sizes compare karke, woh dimensionless group derive karo jo decide karta hai "inertia vs friction." Water in a pipe ke liye evaluate karo:

$U = 1 m/s$ , $L = 0.05 m$ , $ν = 1 0^{- 6} m^{2} / s$ . Forecast: jo ratio milega woh Reynolds number hai $R e = U L / ν$ . Bada $R e$ = inertia wins (turbulent-prone); chhota $R e$ = viscosity wins (smooth).

Step 1 — convective (inertia) term ki size nikalo. Har derivative ko (typical change)/(typical length) se replace karo: scale $U$ ki velocity jo distance $L$ par change ho rahi hai. Yeh step kyun? Hume sirf order of magnitude chahiye, toh exact derivatives unnecessary hain. $ρ (u \cdot \nabla) u \sim ρ U \cdot \frac{U}{L} = \frac{ρ U ^{2}}{L}$

Step 2 — viscous term ki size nikalo. Viscous term second derivative hai, toh isme do length divisions hain: $μ \nabla^{2} u \sim μ U / L^{2}$ . Yeh step kyun? Same order-of-magnitude accounting, $\nabla^{2}$ par apply kiya (curvature $\Rightarrow$ $L$ se do baar divide karo). $μ \nabla^{2} u \sim μ \frac{U}{L ^{2}}$

Step 3 — ratio banao. Inertia ko viscosity se divide karo; common piece $U / L$ cancel ho jaata hai aur $μ / ρ = ν$ (kinematic viscosity) appear karta hai. Yeh step kyun? Do forces ka pure ratio dimensionless hona chahiye — woh number Reynolds number hai. $R e = \frac{ρ U ^{2} / L}{μU / L ^{2}} = \frac{ρ U L}{μ} = \frac{U L}{ν}$

Verify: $R e = \frac{1 \times 0.05}{1 0 ^{- 6}} = 50000$ . Bada → inertia dominate karta hai, flow likely turbulent ✓. Dimensionless check: $\frac{( m/s ) ( m )}{m ^{2} / s}$ = pure number ✓.

Ex 11 — All terms live at once (Cell C11)

Worked example Impulsively started pipe:

$t = 0$ par ek lamba round pipe (radius $R$ ) pressure gradient $G$ par switch hota hai. Parabola settle hone se pehle, saare NS terms matter kar sakte hain. Confirm karo kaunse survive karte hain, aur steady state ke baad mean speed nikalo agar $R = 1 mm$ , $G = 500 Pa/m$ , $μ = 1 0^{- 3} Pa\cdots$ . Forecast: start-up par flow unsteady hai ( $\partial_{t} \neq = 0$ ); agar pipe narrow bhi ho toh parcel accelerate karta hai ( $(u \cdot \nabla) u \neq = 0$ ); friction hamesha present hai ( $μ \nabla^{2} u$ ); aur pipe outlet air se marta hai (free-surface / atmospheric pressure condition). Final settled mean speed centreline ki half hai.

Step 1 — kaunse terms alive hain? Transient ke dauran, $x$ -momentum equation har term rakhti hai: $ρ (unsteady \frac{\partial u _{x}}{\partial t} + convective (u \cdot \nabla) u_{x}) = G + viscous μ \nabla^{2} u_{x}$ Yeh step kyun? Yeh ek hi example hai jahan teen left-block effects mein se koi bhi switch off nahi hai — yeh genuine "combined" cell hai jo matrix mein promise ki gayi thi. Straight uniform pipe mein continuity $\partial u_{x} / \partial x = 0$ convection kill karta hai; narrowing pipe mein yeh survive karta.

Step 2 — steady state tak settle karo. Jaise $t \to \infty$ , $\partial_{t} \to 0$ ; straight pipe mein convection off hai, toh hum steady Poiseuille $u (r) = \frac{G}{4 μ} (R^{2} - r^{2})$ recover karte hain centreline ke saath $u_{m a x} = \frac{G R ^{2}}{4 μ}$ . Yeh step kyun? Transient die away karta hai aur balance Cell C2/C9 tak reduce ho jaata hai — cells ke across ek healthy consistency check.

Step 3 — mean speed. Round pipe ke liye ek paraboloid ka area-average exactly peak ka half hota hai: $\overset{u}{ˉ} = \frac{1}{2} u_{m a x} = \frac{G R ^{2}}{8 μ}$ . Yeh step kyun? Flow rate ÷ area; paraboloid ka mean uska max ka half hai.

Verify: $\overset{u}{ˉ} = \frac{500 \times ( 0.001 ) ^{2}}{8 \times 1 0 ^{- 3}} = \frac{500 \times 1 0 ^{- 6}}{8 \times 1 0 ^{- 3}} = 0.0625 m/s$ ✓. (Atmospheric pressure par free-surface exit ke saath consistent jo downstream boundary set karta hai; continuity ne yahan convection ko zero fix kiya.)

Recall Tumhare liye kaunsa cell sabse mushkil tha?

Cell C7 sign trap (force hai $- \nabla P$ ) ::: fluid high se low pressure ki taraf push hota hai, toh positive $\partial_{x} P$ ek negative $x$ -force deta hai. Ex 5 mein steady flow ke bawajood acceleration kyun hai? ::: convective term $(u \cdot \nabla) u$ material derivative mein — parcel tez fluid mein move karta hai. Ex 8 mein $h$ half karne par $u_{m a x}$ kitne factor se change hota hai? ::: ek quarter, kyunki $u_{m a x} \propto h^{2}$ . Pipe mein convective term kab survive karta hai (Ex 11)? ::: sirf tab jab pipe narrow ho; continuity ise straight uniform pipe mein kill kar deta hai.