Visual walkthrough — Viscous flow — Poiseuille flow, velocity profile in pipe

2.2.17 · D2 · Physics › Fluid Mechanics › Viscous flow — Poiseuille flow, velocity profile in pipe

Humein pehle se sirf ek idea chahiye: fluids chipchipe hote hain. Usi chipchipahaat ko viscosity kehte hain (dekho Viscosity and Newton's law of viscosity). Baaki sab yahan se hi develop karenge.

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Step 1 — Pipe aur uske andar ek chupi hui cylinder ko picture karo

KYA. Ek seedha horizontal pipe draw karo jiska radius $R$ hai (fluid jitni door tak ja sakta hai) aur length $L$ (kitni door hum dekh rahe hain). Ab socho ek chhoti invisible fluid tube uske andar hai, same centre-line share karte hue, kisi radius $r$ ke saath jahan $0\le r\le R$. Yeh inner tube hamari free body hai — fluid ka woh chunk jis par hum physics karenge.

KYU. Physics mein, jab tum forces directly nahi dekh sakte, tum ek tukda isolate karte ho aur maangte ho ke uski forces balance karein. Cylinder yahan natural piece hai kyunki pipe ki symmetry ka matlab hai ke sab kuch sirf iss par depend karta hai ke tum centre se kitni door ho — distance $r$.

PICTURE. Baahri grey wall pipe hai. Lavender inner cylinder hamara chosen chunk hai. Do cheezein usse push aur pull karengi: pressure uske do flat ends par, aur friction uski curved side par.

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Step 2 — Push: do flat ends par pressure

KYA. Fluid pressure $P_1$ par enter karta hai aur kam pressure $P_2$ par exit karta hai. Difference $\Delta P=P_1-P_2$ hi hamare cylinder ko aage dhakelta hai. Pressure force per unit area hai, toh force nikaalny ke liye hum area se multiply karte hain jis par woh act karta hai — flat circular end, area $\pi r^2$.

$$F_{\text{push}}=\underset{\begin{array}{c}\text{pressure drop}\\ \text{(shove per area)}\end{array}}{\underbrace{\Delta P}}\times \underset{\begin{array}{c}\text{cylinder ke end}\\ \text{ka area}\end{array}}{\underbrace{\pi r^2}}$$

YEH kyun aur kuch aur kyun nahi? Pressure se area multiply kyun karte hain? Kyunki pressure akela kuch nahi hila sakta — ek pinhole par huge pressure ek tiny force hai. Force hi matter ko accelerate karta hai, aur force $=$ pressure $\times$ area. Hum difference $P_1-P_2$ use karte hain kyunki peechwala end forward push karta hai jabki agla end backward push karta hai; sirf net maayine rakhta hai.

PICTURE. Do coral arrows: ek bada jo back face ko forward push karta hai ($P_1$), ek chhota jo front face ko backward push karta hai ($P_2$). Unka net forward drive hai.

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Step 3 — Brake: curved side par viscous friction

KYA. Hamare cylinder ke bahar wala fluid (radius $r$ aur wall ke beech ka shell) slower chalti hai, kyunki woh stationary wall ke zyada karib hai. Toh woh hamare cylinder ki curved skin par backward drag karti hai. Yeh drag per unit area shear stress $\tau$ hai, aur Newton's law of viscosity kehta hai ke woh viscosity times equals hai ke bahar jaane par speed kitni jaldi change hoti hai:

$$\tau =\underset{\begin{array}{c}\text{viscosity}\\ \text{(stickiness)}\end{array}}{\underbrace{\eta }}\underset{\begin{array}{c}\text{velocity gradient:}\\ v\text{ kitni jaldi }r\text{ ke saath change hoti hai}\end{array}}{\underbrace{\frac{dv}{dr}}}$$

Stress ko force mein convert karne ke liye hum phir area se multiply karte hain — lekin is baar cylinder ki curved side ka area, jo circumference $2\pi r$ times length $L$ hai:

$$F_{\text{visc}}=\tau \times \underset{\begin{array}{c}\text{curved side area}\\ =\text{(circumference)}\times L\end{array}}{\underbrace{2\pi rL}}=\eta \frac{dv}{dr}(2\pi rL)$$

Speed $v$ ki jagah gradient $dv/dr$ kyun? Ek layer jo tezi se chal rahi hai apne ek aisi neighbour ke saath jo utni hi tezi se chal rahi hai, koi friction feel nahi karti — woh slide nahi karte. Friction neighbours ke beech ke difference se aata hai, yaani speed ek chhote step mein $r$ ke saath kitni change hoti hai. Woh "change per step" exactly derivative $dv/dr$ measure karta hai. Hume yahan precisely calculus chahiye kyunki speed har radius par alag hai.

PICTURE. Alag alag radii par alag alag lengths ke arrows — beech mein lambe, wall ke paas chhote. Jahan arrow lengths sabse zyada differ karti hain, sliding (aur friction) sabse zyada hai.

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Step 4 — Forces balance karo (steady flow ka matlab hai net push nahi)

KYA. Hamaara cylinder constant speed par move kar raha hai — fluid time ke saath speed up ya slow down nahi ho raha (steady flow ka yahi matlab hai). Koi acceleration nahi matlab, Newton's second law se, total force zero hai. Toh forward push backward drag ko exactly cancel karna chahiye:

$$\underset{\text{forward push}}{\underbrace{\Delta P(\pi r^2)}}=\underset{\text{backward drag}}{\underbrace{-\eta \frac{dv}{dr}(2\pi rL)}}$$

Minus sign kyun? Bahar jaao ($r$ badhao) aur speed girti hai, toh $dv/dr$ ek negative number hai. Lekin friction physically backward kheenchti hai, ek real resisting force. Minus sign us negative gradient ko ek positive resisting magnitude mein flip karta hai, toh dono sides genuine forward-vs-backward balance hain. Ab gradient ke liye solve karo:

$$\frac{dv}{dr}=-\frac{\Delta P}{2\eta L}r$$

Isko padhna: speed girane ki rate $r$ badhne ke saath steeper hoti jaati hai. Centre ke paas speed muskil se change hoti hai; wall ke paas woh tezi se girta hai. Yahi ek fact parabola ka beej hai.

PICTURE. Ek balance beam: ek pan mein pressure force, doosre mein viscous force, level kyunki woh equal hain.

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Step 5 — Slopes add karo: speed paane ke liye integrate karo

KYA. Step 4 ne humein har radius par velocity ka slope diya. Actual speed $v(r)$ recover karne ke liye hume un saare slopes ko sum karna hoga — integration yahi hai: infinitely many tiny changes ko add karna. Hum apne radius $r$ se wall $R$ tak integrate karte hain, kyunki wall par hume answer pakka pata hai.

$${\int }_{v(r)}^{0}dv=-\frac{\Delta P}{2\eta L}{\int }_r^R{r}^{\prime }d{r}^{\prime }$$

$r$ se $R$ tak integrate kyun? No-slip condition ki wajah se: wall ko touch karne wala fluid nahi move kar sakta, toh $v(R)=0$. Woh jaana hua endpoint anchor hai jo otherwise-unknown constant ko pin karta hai. Integral karne par (right side $\frac{1}{2}(R^2-r^2)$ hai):

$$0-v(r)=-\frac{\Delta P}{2\eta L}\cdot \frac{R^2-r^2}{2}⟹\boxed{v(r)=\frac{\Delta P}{4\eta L}(R^2-r^2)}$$

Term by term: $\Delta P$ upar (zyada push $\to$ zyada fast), $\eta L$ neeche (zyada sticky ya lamba $\to$ slower), aur shape-factor $R^2-r^2$ — centre par sabse bada ($r=0$), wall par exactly zero ($r=R$).

PICTURE. Bhari hui parabola: ek bullet-nose profile, beech mein mota aur tez, dono walls par zero par pinned.

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Step 6 — Flow ko rings mein slice karo aur volume add karo

KYA. Hume $Q$ chahiye, fluid ka volume jo per second guzarta hai. Alag alag radii alag alag speeds par move karti hain, toh hum ek speed ko ek area se multiply nahi kar sakte. Iske bajaaye cross-section ko thin rings mein kaato: radius $r$ par ek ring, thickness $dr$, area $2\pi rdr$ (isko unroll karo — length $2\pi r$ aur width $dr$ ki ek strip). Us ring par har cheez almost same $v(r)$ par move karti hai.

$$Q={\int }_0^R\underset{\text{ring speed}}{\underbrace{v(r)}}\underset{\text{ring area}}{\underbrace{(2\pi rdr)}}={\int }_0^R\frac{\Delta P}{4\eta L}(R^2-r^2)2\pi rdr$$

Squares ki jagah rings kyun? Flow circularly symmetric hai — speed sirf centre se distance par depend karti hai. Rings woh shape hain jis par speed constant rehti hai, toh har ring cleanly contribute karti hai. Integral solve karne par:

$$Q=\frac{\pi \Delta P}{2\eta L}{\left[\frac{R^2{r}^2}{2}-\frac{r^4}{4}\right]}_0^R=\frac{\pi \Delta P}{2\eta L}\cdot \frac{R^4}{4}⟹\boxed{Q=\frac{\pi R^4\Delta P}{8\eta L}}$$

PICTURE. Disc cross-section coloured concentric rings mein toota hua, har ek apne speed arrow ke saath labelled — beech mein lambe, rim par tiny.

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Step 7 — Edge cases (reader ko kabhi guess mein mat chhoddo)

KYA. Har corner check karo taaki koi scenario surprise na kare.

• Exact centre par, $r=0$: $v=\frac{\Delta P{R}^2}{4\eta L}=v_{\max }$. Wahan slope $dv/dr=0$ hai — parabola ka flat top. Speed sabse zyada hai aur sabse dheere change ho rahi hai.
• Wall par, $r=R$: $v=0$. Yeh no-slip hai, boundary condition itself. Slope yahan steepest hai, toh friction wall par sabse tez hai.
• Koi pressure drop nahi, $\Delta P=0$: toh $v(r)=0$ har jagah aur $Q=0$. Bina push ke, sticky fluid bas baithti rehti hai — exactly wahi jo experience kehta hai.
• Zero viscosity, $\eta \to 0$: $v_{\max }\to \infty$. Ek frictionless fluid ko race karne ke liye koi push nahi chahiye — yahi wajah hai ke Bernoulli's principle ki ideal, non-viscous duniya mein yeh parabola kabhi nahi hoti; wahan ek flat profile hai.
• Bahut tez (turbulence): agar tum itna hard push karo ke smooth layers chaotic swirls mein toot jaayein. Tab Poiseuille ki poori picture collapse ho jaati hai; Reynolds number and turbulence batata hai kab.

PICTURE. Do profiles side by side: viscous parabola (walls par zero par pinned) versus ek frictionless fluid ka idealised flat "plug" profile.

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Ek-picture summary

Sab ek saath: cylinder par force balance $\to$ $v$ ka slope $\to$ parabola tak integrate karo $\to$ $Q$ paane ke liye rings sum karo.

Recall Feynman retelling — poori walkthrough plain words mein

Maine ek mota pipe draw kiya aur socha ek patli fluid tube uske andar chupi hai. Peeche wala pressure us tube ko aage dhakelta hai; bahar ki slower fluid usse peeche kheenchti hai jaise chipchipa honey. Maine kaha "tum speed up nahi ho rahe, toh yeh dono equal hone chahiye" — isse mujhe ek rule mila ke bahar jaane par speed kitni tezi se girta hai. Un saare chhote drops ko add karna (yaani integrate karna), aur yeh fact use karna ke wall se chipki fluid bilkul nahi move kar sakti, mujhe ek parabola mila: beech mein fastest bullet-nose, edges par zero. Phir yeh jaanne ke liye ke per second kitna fluid nikalta hai, maine circle ko thin rings mein kaata, har ring ki speed ko uske area se multiply kiya, aur unhe stack kiya. Nikal aaya $R^4$ — kyunki dono room aur middle-speed radius ke saath badhte hain. Wider pipe, BAHUT ZYADA flow. Stickier fluid, kam flow. Koi push nahi, koi flow nahi. Yahi poori kahani hai.

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Connections

• Viscosity and Newton's law of viscosity — Step 3 mein use kiya gaya $\tau =\eta dv/dr$.
• Reynolds number and turbulence — jab yeh poori derivation valid nahi rehti (Step 7).
• Bernoulli's principle — frictionless flat-profile counterpart.
• Equation of continuity — connected pipes mein $Q$ aur area kaise link hote hain.
• Blood flow and circulatory system — tumhari arteries mein $R^4$ law.