Viscous flow — Poiseuille flow, velocity profile in pipe: Question bank

2.2.17 · D5 · Physics › Fluid Mechanics › Viscous flow — Poiseuille flow, velocity profile in pipe

Shuru karne se pehle, teen symbols jo baar baar aayenge, seedhe shabdon mein — taaki kuch bhi mystery na rahe:

$R$ = pipe radius (centre se wall tak ki distance); $r$ = centre line se kisi point ki distance, toh $0 \leq r \leq R$ .
$Δ P = P_{1} - P_{2}$ = pipe ke along pressure drop — yaani push.
$Q$ = volume flow rate (har second mein kitne cubic metres fluid kisi cross-section se guzarti hai).
$η$ = viscosity (fluid ki internal stickiness); $L$ = pipe ki length.

Do master results jinhe hum test karte hain: $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2}), Q = \frac{π R ^{4} Δ P}{8 η L} .$

True or false — justify

Pipe ke bilkul centre pe fluid pe sabse zyada shear stress hota hai.

False. Shear stress hai

τ = η d v / d r

; gradient

d v / d r = - \frac{Δ P}{2 η L} r

— yeh centre pe zero hota hai aur wall pe sabse bada, toh wall — centre nahi — sabse zyada shear feel karta hai.

Fixed pressure drop par pipe ki length double karne se flow rate aadha ho jaata hai.

True.

Q \propto 1/ L

, toh double length matlab aadha flow — ek lambi pipe mein zyada wall area hota hai jo zyada distance tak drag karta hai, toh same push se kam fluid move hota hai.

Agar pressure drop double kar do, toh centre pe maximum speed double ho jaati hai.

True.

v_{m a x} = \frac{Δ P R ^{2}}{4 η L}

— yeh

Δ P

mein linear hai, toh double push se double centre speed milti hai (jab tak flow laminar rahega).

Mean speed, centre speed ke barabar hoti hai.

False. Parabola ke liye average exactly peak ka aadha hota hai:

\overset{v}{ˉ} = \frac{1}{2} v_{m a x}

, kyunki walls ke paas slow fluid average ko neeche kheench leta hai.

Poiseuille's law ek tez-bahti nadi par bhi utna hi kaam karta hai.

False. Iske liye laminar flow chahiye (low Reynolds number); nadi turbulent hoti hai aur bilkul alag, non-linear relationship follow karti hai — dekho Reynolds number and turbulence.

Fluid ko double viscous karo, baaki sab fixed rakho, toh flow rate aadhi ho jaayegi.

True.

Q \propto 1/ η

, toh stickiness double karne se flow aadhi ho jaati hai — zyada internal friction same push ko resist karta hai.

Pipe wall pe fluid stationary hota hai kyunki wall material ke saath friction use rok deta hai.

Effect mein True, cause mein subtle. Yeh no-slip condition hai: wall ko touch karne wale fluid molecules ka bulk velocity zero hota hai, aur yahi

v (R) = 0

ko pin karta hai aur poora profile fix karta hai.

Agar fluid non-viscous ho (

η = 0

), toh velocity profile phir bhi parabolic rahegi.

False. Zero viscosity mein koi shear stress nahi, koi wall drag nahi, aur wall fluid ke rukne ki koi wajah nahi — uniform "plug" flow milega, parabola nahi. Parabola hi viscosity ki pehchaan hai.

Fixed flow rate

Q

ke liye required pressure drop viscosity se independent hota hai.

False.

Δ P = \frac{8 η L Q}{π R ^{4}}

se, required push directly

η

ke proportional hai: ek stickier fluid ko same

Q

force karne ke liye zyada pressure chahiye.

Spot the error

"Kyunki

Q = v \times area

aur area

\propto R^{2}

, flow rate

R^{2}

ke hisaab se scale karta hai."

Error yeh hai: velocity khud bhi

R^{2}

ke hisaab se badhti hai (centre speed

\propto R^{2}

), toh

Q \propto R^{2} \cdot R^{2} = R^{4}

. Bada cross-section aur flatter gradient — dono effects milke multiply hote hain.

"Pipe ke andar pressure walls pe sabse zyada aur centre pe sabse kam hota hai."

Error yeh hai: pressure pipe ke along vary karta hai (inlet pe high, outlet pe low), across pipe nahi. Kisi given cross-section pe pressure essentially uniform hota hai; velocity hai jo radius ke across vary karti hai.

"Kyunki velocity baahir ki taraf decrease hoti hai,

d v / d r

positive hai."

Error yeh hai: jaise

r

badhta hai (baahir jaate hue)

v

kam hoti hai, toh slope

d v / d r

negative hai. Wahi minus sign hai kyun derivation mein

\frac{d v}{d r} = - \frac{Δ P}{2 η L} r

carry hota hai.

"Ek thicker (zyada viscous) fluid bhaari hota hai, toh gravity use zyada tez push karti hai."

Error yeh hai ki do cheezein confuse ho rahi hain: viscosity internal friction hai, density/weight nahi — aur Poiseuille flow yahan horizontal hai toh gravity koi driving work karti hi nahi. Zyada

η

matlab slower flow.

"Derivation mein coaxial cylinder ke flat circular ends par shear force act karta hai."

Error yeh hai: pressure flat ends par act karta hai (area

π r^{2}

); viscous shear curved side surface par act karta hai (area

2 π r L

). Kaun sa force kaun se area par act karta hai — yeh mix up karne se poora force balance toot jaata hai.

"Profile ko integrate karte waqt hum

v (0) = 0

se shuru kar sakte hain kyunki centre ek natural reference hai."

Error yeh hai: centre woh jagah hai jahan speed maximum hoti hai, zero nahi. Correct boundary condition hai

v (R) = 0

wall par — yahi physical no-slip fact hai jo integration constant fix karta hai.

"Agar radius aadha kar dein, toh flow quarter ho jaayegi."

Error yeh hai:

Q \propto R^{4}

, toh

R

aadha karne se

Q

(1/2)^{4} = 1/16

se multiply hota hai — flow solvahwa hissa ho jaata hai, quarter nahi.

Why questions

Flow rate

R^{4}

par kyun depend karta hai,

R^{2}

par kyun nahi?

Kyunki wider pipe do baar help karta hai: zyada cross-sectional area deta hai (

\propto R^{2}

) aur gentler gradient ke saath higher centre speed allow karta hai (

v_{m a x} \propto R^{2}

). Dono effects multiply hokar

R^{4}

dete hain.

Velocity profile parabola kyun hoti hai, seedha cone kyun nahi?

Pressure force (

\propto r

— area ke upar act karta hua) ko viscous drag ke against balance karne se milta hai

d v / d r \propto r

;

r

mein linear term ko integrate karne par

r^{2}

term aata hai, aur

v \propto (R^{2} - r^{2})

ek parabola hai.

Steady flow ke liye pressure drop hona zaroori kyun hai?

Viscosity continuously momentum ko wall friction mein drain karti rehti hai; fluid ko constant speed par chalate rehne ke liye (steady flow mein koi acceleration nahi), woh loss continuously ek sustained push

Δ P

se compensate karna padta hai.

Mean speed exactly aadhi kyun hoti hai maximum ki, do-tihai kyun nahi?

Kyunki parabola

v_{m a x} (1 - r^{2} / R^{2})

ko disc pe average karte waqt (outer rings ko zyada weightage dete hue, kyunki ring area

r

ke saath badhta hai) exactly

\frac{1}{2} v_{m a x}

nikalta hai — us specific profile ki yeh geometric property hai.

Wall fluid sabse zyada viscous stress kyun feel karta hai jabki woh sabse slow move karta hai?

Stress gradient

d v / d r

par depend karta hai, speed par nahi. Wall ke paas speed chhoti distance mein sabse steeply change hoti hai, toh wahan ki layers sabse zyada rub karti hain — wall maximum-gradient region hai.

Real pipe mein flow describe karne ke liye simply Bernoulli's principle kyun use nahi kar sakte?

Bernoulli's principle assume karta hai koi viscosity nahi, toh woh streamline ke along mechanical energy conserve karta hai. Real pipe flow friction se energy lose karta hai, aur wahi pressure drop produce hota hai jise Poiseuille's law account karta hai.

Thodi si narrow artery blood flow mein itna bada drop kyun cause karti hai?

R^{4}

dependence ki wajah se: radius mein thodi si bhi kami

Q

ko dramatically shrink kar deti hai (20% narrowing flow ko roughly

0. 8^{4} \approx 41%

tak cut kar deta hai). Yeh Blood flow and circulatory system mein central concept hai.

τ = η d v / d r

speed

v

ki bajaye gradient ke proportional kyun hai?

Friction adjacent layers ke beech relative sliding se aata hai; agar do neighbouring layers same speed se move kar rahi hain toh koi rubbing nahi hogi, chahe woh speed kitni bhi badi ho. Sirf difference — gradient — shear create karta hai. Dekho Viscosity and Newton's law of viscosity.

Edge cases

Bilkul centre par (

r = 0

), shear stress kya hai, aur kya yeh physically sensible hai?

Zero.

τ = η d v / d r \propto r

, toh

r = 0

par koi shear nahi — centre layer apne neighbours ke saath profile ke flat top par essentially lockstep mein move karti hai, toh koi rubbing nahi hoti.

Agar

Δ P = 0

ho toh velocity profile kya ban jaata hai?

v (r) = 0

har jagah — koi push nahi matlab koi flow nahi. Parabola flat zero line mein collapse ho jaata hai; kuch bhi fluid ko friction ke against drive nahi karta.

Agar

η \to 0

(ideal, frictionless fluid) ho toh formula fixed

Δ P

ke liye kya predict karta hai?

Dono

v_{m a x}

aur

Q

infinity tak blow up ho jaate hain, jo signal hai ki model toot raha hai: zero viscosity mein koi drag nahi jo push ko balance kare, toh koi steady finite-speed solution exist nahi karta — poori non-viscous picture pe switch karna padega.

Agar tum

R

badhao aur

Δ P

ghatao taaki unka product fixed rahe, toh kya flow rate change hoti hai?

Haan, flow phir bhi badhti hai.

Q \propto R^{4} Δ P

, toh agar

R

badhta hai aur

Δ P

ghatta hai taaki

R \cdot Δ P

constant rahe,

R

ke extra teen powers dominate karte hain aur

Q

sharply increase hoti hai — radius hamesha jeet jaata hai.

Jab flow turbulent hone lagta hai toh parabolic profile ka kya hota hai?

Woh clean parabola nahi rehta: turbulence core ko flatten kar deta hai (ek "plug-like" middle) aur wall gradient ko steep kar deta hai. Critical Reynolds number cross hote hi, Poiseuille's parabola us flow ko bilkul describe nahi karta.

Normal Poiseuille flow mein kya

v (r)

pipe mein kahin bhi negative ho sakta hai?

Nahi. Kyunki

R^{2} - r^{2} \geq 0

sab

0 \leq r \leq R

ke liye, aur

Δ P / (4 η L) > 0

, speed har jagah non-negative hai — zero sirf wall par, andar har jagah positive.

Active Recall

Recall Ek-line trap summary (cover karo aur recall karo)

Shear stress maximum kahan hai? → Wall par (max gradient), centre par zero.
$Q$ vs $R$ ? → $Q \propto R^{4}$ — radius do baar help karta hai.
$Q$ vs $η$ aur $L$ ? → $Q \propto 1/ η$ aur $Q \propto 1/ L$ .
$\overset{v}{ˉ}$ vs $v_{m a x}$ ? → Exactly aadha.
Profile fix karne wali boundary condition? → No-slip, $v (R) = 0$ .
Law kab fail karta hai? → Turbulent, non-steady, non-Newtonian, ya short-pipe flow mein.

Connections

Parent: Poiseuille flow — poori derivations jinhe yeh traps target karti hain.
Viscosity and Newton's law of viscosity — kyun $τ$ gradient par depend karta hai.
Reynolds number and turbulence — jab parabola break down karta hai.
Bernoulli's principle — frictionless contrast case.
Blood flow and circulatory system — real arteries mein $R^{4}$ trap.