Reynolds number Re = ρvL - μ — laminar vs turbulent criterion: Question bank

2.2.19 · D5 · Physics › Fluid Mechanics › Reynolds number Re = ρvL - μ — laminar vs turbulent criterio

True or false — justify

Koi item true ho ya false, reason wahi matter karta hai. Pehle reason bolo, phir reveal karo.

Fluid ki viscosity badhane se, baaki sab fixed rakhne par, flow zyada turbulent ho jaati hai.

False.

μ

denominator mein hai, isliye bada

μ

ek chhota

R e

deta hai — viscosity woh peacemaker hai jo disturbances ko damp karti hai aur flow ko laminar ki taraf dhakelta hai.

Reynolds number newtons ki units carry karta hai kyunki humne ise forces se banaya hai.

False. Yeh do forces ka ratio hai jinki units same hain, isliye units cancel ho jaati hain —

R e

dimensionless hai. Dekho Dimensional analysis.

Agar do flows (ek toy pipe aur ek oil pipeline) ka same

R e

ho, toh woh same regime mein behave karte hain.

True. Kyunki

R e

dimensionless hai, ise match karne ka matlab hai ki inertia-to-viscosity balance identical hai — yahi ek universal number ka poora point hai.

Pipe flow ke liye sahi characteristic length

L

pipe ka radius hai.

False. Standard convention diameter

D

use karta hai. Radius use karne se

R e

aadha ho jaata hai aur regime galat predict ho sakta hai.

R e_{c} \approx 2000

ek fundamental constant of nature hai jaise

c

g

False. Critical

R e

geometry-dependent aur fuzzy hai — flat plates, spheres aur pipes sab alag hote hain, aur bahut smooth pipes

R e \sim 1 0^{5}

tak laminar rehti hain. Yeh ek practical guideline hai.

Flow speed ko double karne se Reynolds number double ho jaata hai.

True.

R e = ρ v L / μ

mein

v

ke saath linear hai (inertial force mein

v^{2}

tha; ek extra

v

viscous

v

ke saath cancel ho jaata hai), isliye

R e

directly

v

ke proportion mein scale hota hai.

Pipe diameter ko double karne se inertial force

ρ v^{2} L^{2}

double ho jaati hai.

False. Inertial force

\propto L^{2}

hai, isliye

L

double karne se yeh chaar guna ho jaati hai. Lekin

R e \propto L

hai (

L^{2}

nahi) kyunki viscous force bhi

\propto L

hai, aur ek

L

cancel ho jaata hai.

Bernoulli's principle bahut high-

R e

flow mein solid wall ke paas ek achha model hai.

False. Bernoulli's principle inviscid (ideal) flow assume karta hai — walls se door

R e \to \infty

ka idealisation. Wall ke bilkul paas viscosity hamesha ek patli boundary layer mein dominate karti hai, chahe

R e

kitna bhi bada ho.

Poiseuille's law ko pipe mein

R e = 8000

par safely apply kiya ja sakta hai.

False. Poiseuille's law laminar flow assume karta hai (

R e < 2000

R e = 8000

par flow turbulent hai, isliye parabolic velocity profile aur law ka pressure-flow relation break down kar jaata hai.

Kinematic viscosity

ν = μ / ρ

formula mein

μ

aur

ρ

dono ko replace kar sakta hai.

True.

R e = ρ v L / μ = v L / (μ / ρ) = v L / ν

. Density aur viscosity ko

ν

mein bundle karna exact hai aur aksar zyada convenient hota hai.

Spot the error

Har statement mein ek flaw hai. Use identify karo.

"Honey thick aur complicated hai, isliye yeh zaroor turbulently flow karta hoga."

Error yeh hai ki "thick" ko "chaotic" se equate kiya ja raha hai. Honey ka bada

μ

R e

ko tiny banata hai (aksar

\sim 1

), isliye yeh deeply laminar hai aur smooth ribbons mein girta hai.

R e = ρ v L^{2} / μ

kyunki inertial force

ρ v^{2} L^{2}

hai."

Student viscous force

μv L

se divide karna bhool gaya. Ratio hai

ρ v^{2} L^{2} / (μv L) = ρ v L / μ

; ek factor

v

ka aur ek

L

ka cancel ho jaata hai.

"Kyunki

R e

dimensionless hai, iska value plug kiye gaye units par depend nahi karta."

Value unit-independent hai tabhi jab aap har quantity mein consistent units use karo (sab SI, maano). cm ko m ke saath ya poise ko Pa·s ke saath mix karne se galat number aata hai chahe true

R e

ek pure number ho.

"Velocity gradient

d v / d y \approx v / L

hai jahan

L

pipe ki length hai."

Yahan

L

transverse length hai jiske across speed

v

0

tak girta hai (roughly diameter), pipe ki downstream length nahi. Dekho Viscosity and Newton's law of viscosity.

R e = 3000

par ek flow definitely laminar hai kyunki yeh 4000 se kam hai."

2000 < 3000 < 4000

transitional band hai — flow unstable aur intermittent hai, cleanly laminar nahi. Sirf

R e < 2000

reliably laminar hota hai.

"Stokes' law drag

F = 6 π μ r v

ek fast cannonball ke liye air mein valid hai."

Stokes' law and terminal velocity ek low- $R e$ result hai (

R e ≪ 1

). Cannonball ka

R e

enormous hota hai, isliye uski drag inertial (pressure) drag hai, viscous Stokes drag nahi.

Why questions

Mechanism explain karo, sirf fact state mat karo.

Ek critical Reynolds number exist hi kyun karta hai?

Disturbances hamesha rehti hain; viscosity unhe damp karti hai jabki inertia unhe amplify karta hai. Ek threshold ratio par amplification damping ko overtake kar leti hai, aur uske upar disturbances turbulence mein grow ho jaati hain.

R e

forces ke difference ki jagah ratio kyun hai?

Ratio dimensionless aur scale-free hota hai, isliye yeh relative strength compare karta hai jo regime govern karta hai. Difference units rakhta hai aur absolute size par depend karta hai, isliye woh universal nahi hota.

Inertial force

ρ v^{2} L^{2}

kyun scale karti hai?

Area

L^{2}

se per second flowing mass

\overset{m}{˙} = ρ v L^{2}

hai; har unit mass velocity

v

carry karta hai, isliye momentum delivered per second (ek force)

\overset{m}{˙} v = ρ v^{2} L^{2}

hai.

Viscous force

μv L

kyun scale karti hai?

Shear stress

τ = μ d v / d y \approx μv / L

(speed thickness

L

ke across

v

se drop karti hai); stress ko area

L^{2}

se multiply karne par

μ (v / L) L^{2} = μv L

milta hai.

Drag coefficient

C_{d}

ek fixed constant kyun nahi hai aur

R e

par kyun depend karta hai?

[[Drag force and drag coefficient|

C_{d}

]] capture karta hai ki flow structure (attached vs separated, laminar vs turbulent wake) kaise change hoti hai, aur woh structure inertia-to-viscosity balance se set hoti hai — yaani

R e

se.

Ek bahut smooth pipe

R e \sim 1 0^{5}

tak laminar kyun reh sakti hai?

Roughness ke bina disturbances seed karne ke liye kuch nahi hai, isliye inertia ke paas amplify karne ke liye almost kuch nahi hota, aur viscosity flow ko usual

\sim 2000

threshold se kaafi aage tak ordered rakhti hai. Yeh dikhata hai ki

R e_{c}

empirical hai, fundamental nahi.

Edge cases

Formula ko uski limits tak push karo.

Fluid rest par ho (

v = 0

) toh

R e

kya hoga?

R e = 0

. Koi motion nahi toh koi inertial force nahi, isliye flow trivially laminar hai (aslmein motionless) — viscosity ke paas kuch fight karne ke liye nahi hai.

Jab viscosity

μ \to 0

(ideal/inviscid fluid) ho toh

R e

ka kya hoga?

R e \to \infty

. Yahi Bernoulli's principle ke peeche ka idealisation hai: inertia poori tarah dominate karti hai, aur flow ko laminar rakhne ka koi viscous mechanism nahi hota.

Jab characteristic size

L \to 0

ho (e.g. ek bacterium swimming) toh

R e

kya karta hai?

R e \to 0

. Micro-scale swimmers ek aise duniya mein rehte hain jahan viscosity completely dominate karti hai — inertia negligible hoti hai, isliye propulsion band hote hi motion ruk jaati hai.

Do pipes ek hi fluid ko same speed par carry kar rahe hain lekin ek ka diameter double hai — kiska

R e

bada hoga?

Wider pipe ka.

R e \propto L

hai, isliye diameter double karne se

R e

double ho jaata hai, bada pipe turbulence ke zyada prone ho jaata hai.

Agar aap

R e

fixed rakhte hue poora system scale up karein (bada

L

), toh compensate karne ke liye kya change karna hoga?

Speed ya kinematic viscosity adjust karni padegi: kyunki

R e = v L / ν

hai, bada

L

proportionally chhota

v

(ya bada

ν

) maangta hai

R e

constant rakhne ke liye — yahi scale-model testing ka basis hai.

Kya low

R e

par turbulence possible hai agar aap fluid ko violently shake karo?

Sustained turbulence nahi. Strong forcing momentary disorder create kar sakta hai, lekin low

R e

par viscosity disturbances ko quickly re-damp kar deti hai, isliye flow laminar mein wapas relax ho jaata hai.

Active Recall

Recall One-line self-check

Q: $R e$ ke bottom mein kaunsi force hai? → viscous ( $μ$ ).
Q: $R e$ ki units? → koi nahi, dimensionless.
Q: $R e = 3000$ ka regime? → transitional.
Q: Pipe ke liye $L$ ? → diameter $D$ .
Q: $μ \to 0$ se $R e \to$ ? → $\infty$ (inviscid limit).

Connections

Parent topic (Hinglish)
Viscosity and Newton's law of viscosity — viscous force ke peeche $τ = μ d v / d y$ .
Poiseuille's law — laminar-only regime.
Stokes' law and terminal velocity — low- $R e$ drag.
Bernoulli's principle — $R e \to \infty$ idealisation.
Dimensional analysis — dimensionless ratios universal kyun hote hain.
Drag force and drag coefficient — $R e$ ke function ke roop mein $C_{d}$ .