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

2.2.18 · D2 · Physics › Fluid Mechanics › Navier-Stokes equations — derivation from Newton's second la

Step 1 — Humara object KYA hai, aur "velocity field" kya hota hai?

KYA. River ko arrows ke ek grid ki tarah socho. Space mein har point par ek chota arrow hai jo batata hai ki wahan ka paani kis direction mein ja raha hai aur kitni tezi se (arrow ki length). Arrows ka yeh collection velocity field kehlata hai, jise $\vec{u}(\vec{r},t)$ likha jaata hai.

Is notation ko symbol by symbol decode karte hain:

• $\vec{u}$ — chota arrow: ek velocity (direction + size dono hain). Upar ka tiny arrow matlab "yeh ek vector hai, sirf ek number nahi."
• $\vec{r}$ — kahan tum khade ho (ek position). Field ki value depend karti hai tum kahan dekhte ho par.
• $t$ — kab dekhte ho (time). Field pal pal badal sakta hai.

KYUN. Ek fluid ek particle nahi hota — yeh ek continuum hai. Hum har molecule ke liye $\vec{F}=m\vec{a}$ nahi likh sakte. Isliye hum fluid ka ek chota cube kaat lete hain (volume $dV$, mass $dm=\rho dV$, jahan $\rho$ = mass per unit volume, yaani density) aur hum Newton's law usi ek cube par apply karenge.

PICTURE.

Step 2 — Cube ka acceleration KYA hai? (material derivative)

KYA. Acceleration matlab hai "velocity cube ke liye khud kitni tezi se badal rahi hai." Yahan ek trap hai: cube move karta hai. Jaise-jaise yeh drift karta hai, woh ek aisi jagah pahunchta hai jahan paani ka arrow alag hota hai. Isliye uski velocity badlti hai do alag-alag wajahon se.

KYUN do wajahein. Ek patti ko ek sankri hoti channel mein daakhil hote dekho:

1. Poori river waqt ke saath speed up ho sakti hai → tumhare fixed point par arrow bada hota hai. Ise local (unsteady) part kaho.
2. Chahe river kabhi bhi time mein na bade, patti carry hoti hai chaudi slow paani se sankri fast paani mein → uski speed khud badh jaati hai. Ise convective (carried) part kaho.

PICTURE.

KAISE (chain rule, scratch se). Cube position $\vec{r}(t)=(x(t),y(t),z(t))$ par hai jo time ke saath badlti hai. Uski velocity hai $\vec{u}(x(t),y(t),z(t),t)$. Is "function-inside-a-function" ko chain rule se differentiate karte hain:

$\frac{d\vec{u}}{dt}=\underset{\text{time changes}}{\underbrace{\frac{\partial \vec{u}}{\partial t}}}+\underset{\text{position changes}}{\underbrace{\frac{\partial \vec{u}}{\partial x}\frac{dx}{dt}+\frac{\partial \vec{u}}{\partial y}\frac{dy}{dt}+\frac{\partial \vec{u}}{\partial z}\frac{dz}{dt}}}$

Ab key substitution: cube ki $x$-position badlne ki rate wohi uski $x$-velocity hai, isliye $\frac{dx}{dt}=u_x$, aur usi tarah $\frac{dy}{dt}=u_y$, $\frac{dz}{dt}=u_z$. Inhe plug karo aur teen position terms ek compact object mein bundle ho jaate hain:

$\boxed{\frac{D\vec{u}}{Dt}=\frac{\partial \vec{u}}{\partial t}+(\vec{u}\cdot \nabla )\vec{u}}$

Step 3 — Cube ko kaun push karta hai? Force #1: gravity

KYA. Mass ka har tukda gravity se neeche khicha jaata hai. Humare cube par pull hai uski mass times $\vec{g}$ (neeche gravitational acceleration, lagbhag $9.8{\text{m/s}}^2$).

$d{\vec{F}}_{\text{body}}=\underset{\text{density}}{\underbrace{\rho }}\underset{\text{down pull per kg}}{\underbrace{\vec{g}}}\underset{\text{cube volume}}{\underbrace{dV}}$

KYUN ise body force kehte hain. Yeh cube ke poore body par ek saath kaam karta hai, uski surface par nahi. Pressure aur viscosity se compare karo, jo sirf cube ke faces par kaam karte hain.

PICTURE.

Step 4 — Force #2: pressure ko difference chahiye, sirf value nahi

KYA. Pressure $P$ paani ka cube ke har face par andar ki taraf squeeze karna hai. Left face ko right push milti hai; right face ko left push milti hai.

KYUN sirf differences matter karte hain. Agar dono taraf se squeeze same ho, toh dono push cancel ho jaate hain — zero net force. Cube ko move karne ke liye, ek face par pressure dusre opposite face ke pressure se zyada hona chahiye. Isliye force depend karta hai ki pressure cube ke across kaise badlta hai.

KAISE (derivation). Left $x$-face (area $dydz$) par push hai $P(x)dydz$ jo $+x$ direction mein point karta hai. Right face par push hai $P(x+dx)dydz$ jo $-x$ mein point karta hai. Net $x$-force:

$d{F}_x=[P(x)-P(x+dx)]dydz=-\underset{\text{slope of }P}{\underbrace{\frac{\partial P}{\partial x}}}\underset{dV}{\underbrace{dxdydz}}$

Bracket mein tiny step $dx$ ke upar $P$ mein badlaav ka minus hai — woh hai derivative $\partial P/\partial x$. Teeno directions ke liye karte hain:

$\boxed{d{\vec{F}}_{\text{pressure}}=-\nabla PdV}$

Minus sign kehta hai: force high pressure se low pressure ki taraf point karta hai (pressure landscape par downhill). Dekho Hydrostatic pressure.

PICTURE.

Step 5 — Force #3: viscosity, aur WHY yeh second derivative kyun hai

KYA. Neighbouring fluid layers alag-alag speeds par chalte hue ek doosre ko ragadti hain — fast layer ke molecules slow layer ko aage kheechte hain aur vice-versa. Yeh sideways drag viscous friction hai. Dekho Viscosity & Newton's law of viscosity.

Newton's law of viscosity stress deta hai (force per area) jo ek layer doosri par lagaati hai: $\tau =\mu \frac{\partial u_x}{\partial y}$ jahan $\mu$ ("mu") fluid ki stickiness (viscosity) hai aur $\partial u_x/\partial y$ hai ki $y$ mein oopar jaate waqt flow speed kitni tezi se badlti hai.

KYUN second derivative. Yahan subtle jump hai. $\tau$ sirf ek face par pull hai. Cube ko top face ki pull minus bottom face ki pull feel hoti hai. Force ke liye cube ke across stress mein difference chahiye — woh stress ka derivative hai, jo already ek derivative tha. Derivative of a derivative = second derivative.

KAISE (derivation). Do $y$-faces se net $x$-force: $\left[\mu \frac{\partial u_x}{\partial y}{|}_{y+dy}-\mu \frac{\partial u_x}{\partial y}{|}_y\right]dxdz=\mu \underset{\text{difference of slopes}}{\underbrace{\frac{{\partial }^2{u}_x}{\partial y^2}}}dV$

Saare faces ke contributions add karo (aur incompressibility use karke extra pieces drop karo):

$\boxed{d{\vec{F}}_{\text{visc}}=\mu {\nabla }^2\vec{u}dV}$

PICTURE.

Step 6 — Sab kuch assemble karo: cube ke liye $\vec{F}=m\vec{a}$

KYA. Left par teen forces stack karo, right par mass × acceleration rakho.

$\underset{\text{gravity}}{\underbrace{\rho \vec{g}dV}}\underset{\text{pressure}}{\underbrace{-\nabla PdV}}+\underset{\text{viscous}}{\underbrace{\mu {\nabla }^2\vec{u}dV}}=\underset{m}{\underbrace{\rho dV}}\underset{\vec{a}}{\underbrace{\frac{D\vec{u}}{Dt}}}$

KYUN $dV$ se divide karte hain. Har term mein cube ka volume $dV$ hai. Use cancel karo, aur hume ek statement milti hai per unit volume — chahe cube kitna bhi tiny ho, yeh sach hai. Phir Step 2 se $\frac{D\vec{u}}{Dt}$ expand karo:

PICTURE.

Step 7 — Edge case A: viscosity switch off karo ($\mu =0$)

KYA. Ek idealized fluid jisme koi stickiness nahi. $\mu {\nabla }^2\vec{u}$ term delete karo.

$\rho \frac{D\vec{u}}{Dt}=-\nabla P+\rho \vec{g}$

KYUN important hai. Yeh Euler equation hai — ek result jis par hum pehle se trust karte hain. Ek sahi derivation ko is limit mein iske barabar collapse karna chahiye. Karta hai. ✓

PICTURE.

Step 8 — Edge case B: motion switch off karo ($\vec{u}=0$)

KYA. Fluid rest mein hai. Har velocity term disappear ho jaata hai ($\partial \vec{u}/\partial t=0$, convective $=0$, viscous $=0$).

$0=-\nabla P+\rho \vec{g}\Rightarrow \nabla P=\rho \vec{g}\Rightarrow \frac{dP}{dz}=-\rho g$

KYUN. Jab kuch nahi move kar raha, pressure ko weight ko exactly hold up karna hoga. Integrate karne par milta hai $P=P_0+\rho gh$ ($h$ neeche measure kiya) — simple Hydrostatic pressure. Humara sabse basic fluid fact seedha nikal aata hai. ✓

PICTURE.

Step 9 — Edge case C: plates ke beech steady flow (Poiseuille)

KYA. Paani $y=0$ aur $y=h$ par do fixed plates ke beech gap se push kiya jaata hai, ek pressure se drive hota hua jo $x$ ke saath steadily drop karta hai: $\partial P/\partial x=-G$ (jahan $G>0$). Steady, straight flow: sirf $u_x(y)$ bachta hai.

KYUN yeh cancellations. Steady ⟹ $\partial \vec{u}/\partial t=0$. Straight parallel flow ⟹ convective term $=0$. Sirf $x$-equation bachti hai:

$0=G+\mu \frac{d^2{u}_x}{d{y}^2}\Rightarrow \frac{d^2{u}_x}{d{y}^2}=-\frac{G}{\mu }$

Do baar integrate karo, phir no-slip apply karo ($u_x=0$ bilkul har wall par, kyunki fluid solid se chipak jaata hai):

$u_x(y)=\frac{G}{2\mu }y(h-y)$

Ek parabola: dono walls par zero, middle mein maximum $\frac{G{h}^2}{8\mu }$. Yeh parabola viscous pressure-driven flow ki pehchaan hai. Dekho Poiseuille flow.

PICTURE.

Ek-picture summary

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