Boundary layer — Prandtl's concept, growth along flat plate: Question bank

2.2.20 · D5 · Physics › Fluid Mechanics › Boundary layer — Prandtl's concept, growth along flat plate

Yaad rakhne wale symbols (sab parent note mein define hain):

$U$ = free-stream speed, wall se bahut door.
$δ$ = boundary-layer thickness, $u = 0.99 U$ par define ki gayi.
$μ$ = viscosity; $ν = μ / ρ$ = kinematic viscosity (momentum-diffusion coefficient, units $m^{2} / s$ ).
$R e_{x} = U x / ν$ = local Reynolds number.
$τ_{w} = μ (\partial u / \partial y) ∣_{y = 0}$ = wall shear stress.

True or false — justify karo

Kya boundary layer ek aisi region hai jahan viscosity zyada hoti hai?

False. Viscosity

μ

fluid mein har jagah same hoti hai; layer ke andar jo zyada hota hai woh velocity gradient

\partial u / \partial y

hota hai, jo viscous stress

μ \partial u / \partial y

ko wahan significant banata hai.

High Reynolds number par boundary layer moti hoti hai.

False.

δ \propto 1/ R e_{x}

, isliye zyada

R e_{x}

matlab patli layer — inertia dominate karta hai aur viscosity wall ke paas ek p얇i si sliver mein simatt jaati hai.

Boundary layer plate ke neeche distance

x

ke saath linearly badhti hai.

False. Laminar layer ke liye

δ \propto x

; kyunki momentum diffuse hota hai

time

ki tarah, layer moti hoti hai lekin ever-decreasing rate se.

Boundary layer ke edge par velocity exactly

U

hoti hai.

False. Velocity

U

ko asymptotically approach karti hai aur kabhi exactly equal nahi hoti; "edge" ek convention hai jo

u = 0.99 U

par rakha jaata hai.

Wall shear stress sabse zyada downstream mein hoti hai jahan layer fully developed ho jaati hai.

False.

τ_{w} \sim μU / δ \propto 1/ x

, isliye shear leading edge par sabse zyada hoti hai jahan

δ

sabse choti aur gradient sabse steep hota hai.

Free-stream speed double karne se boundary-layer thickness bhi double ho jaati hai.

False.

δ \propto 1/ U

, isliye

U

double karne par

δ

1/ 2 \approx 0.71

se multiply ho jaati hai — tez stream kam diffusion time deti hai, isliye patli layer banti hai.

Ek perfectly inviscid ("ideal") fluid mein bhi boundary layer hoti hai.

False.

μ = 0

hone par na no-slip condition hoti hai aur na viscous stress, isliye koi boundary layer nahi banti — yahi reason hai ki ideal theory zero drag deti hai (d'Alembert's paradox).

Boundary layer ke bahar fluid ko ideal maanna ek achha approximation hai.

True. Wall se door gradients gentle hote hain, isliye

μ \partial u / \partial y

genuinely negligible hota hai aur flow inviscid outer stream ki tarah behave karti hai — yahi Prandtl ka two-region split hai.

No-slip condition ka matlab hai ki wall par fluid speed free-stream speed ke barabar hoti hai.

False. No-slip ka matlab hai fluid velocity wall ki velocity ke barabar hoti hai — stationary plate ke liye woh zero hai,

U

nahi.

Leading edge par boundary-layer thickness zero hoti hai.

True.

x = 0

par pahunchne wale fluid ne plate ke upar koi time spend nahi kiya, isliye viscosity ko wall ka influence diffuse karne ka koi time nahi mila:

δ \to 0

Error dhundo

"Kyunki air ki viscosity sirf

\sim 1 0^{- 5} Pa\cdots

hai, har jagah viscous stress negligible hai."

Stress

μ \partial u / \partial y

hai, sirf

μ

nahi. Wall ke paas

\partial u / \partial y \sim U / δ

bahut bada hota hai, isliye ek chota

μ

ek huge gradient se multiply hokar layer mein negligible nahi reh jaata.

"Kyunki

δ \propto ν x / U

, zyada viscous oil patli boundary layer deta hai."

Direction galat hai:

δ \propto ν

, isliye bada

ν

(chipchipa fluid) moti layer deta hai — viscosity momentum ko wall se zyada door diffuse karti hai.

R e_{x} = U x / ν

isliye jaise main downstream jaata hoon

R e_{x}

girta hai aur layer patli hoti hai."

R e_{x}

x

ke saath badhta hai (

U

aur

ν

fixed hain,

x

grow kar raha hai), aur phir bhi

δ = x / R e_{x} \propto x

badhta rehta hai — tumhe

x

ko numerator mein aur

R e_{x}

ko sahi se combine karna hoga.

"High

R e

par viscous term khatam ho jaata hai, isliye real plate par drag zero hona chahiye."

Viscous term sirf outer region mein shrink hota hai;얇i layer ke andar woh finite rehta hai aur Skin friction drag produce karta hai. Layer ko ignore karna hi exactly woh hai jisne d'Alembert ki galat zero-drag prediction di.

"Turbulent boundary layers bhi same

δ = 5 x / R e_{x}

law follow karte hain."

Nahi — woh Blasius result sirf laminar layer ke liye hai. Jab flow turbulent ho jaati hai toh mixing kahin zyada vigorous hoti hai aur

δ

tezi se badhta hai (roughly

\propto x^{4/5}

"Kyunki boundary layer patli hai, pressure iske across bahut badalta hai."

Ulta hota hai: layer patli hone ki wajah se pressure essentially iske across constant hota hai (outer stream dwara impose kiya gaya). Strong variation velocity mein hoti hai, pressure mein nahi.

Why questions

Prandtl ne flow ko do regions mein kyun split kiya instead of puri jagah Navier–Stokes equations solve karne ke?

Full equations intractable hain, lekin viscous term sirf얇i wall layer mein matter karta hai; splitting se tum viscosity ko wahan rakh sakte ho jahan zaroori hai aur bahar simple ideal-flow theory use kar sakte ho.

Ek tiny distance

δ

itna bada velocity gradient kyun create karta hai?

Fluid ko wall par

0

se lekar sirf

δ

ke across almost

U

tak jaana hota hai, isliye

\partial u / \partial y \approx U / δ

;

δ

ko chota karne se ratio, aur isliye gradient, bahut bada ho jaata hai.

Layer

x

ki tarah kyun badhti hai,

x

ki tarah nahi?

Viscosity momentum ke diffusion ki tarah kaam karti hai, aur koi bhi diffusion

\sim D t

distance spread karta hai;

D = ν

aur

t \sim x / U

δ \sim ν x / U

milta hai — yeh square-root law hai, linear nahi.

δ

set karne wali quantity kinematic viscosity

ν = μ / ρ

kyun hai,

μ

nahi?

ν

ke units

m^{2} / s

hain, exactly diffusion coefficient jaisi, isliye yeh directly measure karta hai ki momentum kitni tezi se sideways spread hota hai — woh physical process jo layer ko motaa karta hai.

Wall shear stress leading edge par maximum kyun hai?

Wahan

δ

sabse choti hai, isliye velocity gradient

U / δ

sabse steep hai, aur

τ_{w} \sim μU / δ

δ

ke inversely scale karta hai.

High Reynolds number "almost ideal outer flow" picture ko kyun justify karta hai?

High

R e_{x}

δ \propto 1/ R e_{x}

vanishingly얇a ho jaata hai, saara viscous action ek sliver mein confined ho jaata hai aur baaki field significant friction se free reh jaati hai.

Edge cases

x = 0

par exactly

δ

ka kya hota hai?

Woh zero hai — koi exposure time nahi matlab wall ke influence ka koi diffusion nahi, isliye layer kuch bhi nahi se shuru hoti hai.

Boundary-layer concept ek truly inviscid fluid (

ν = 0

) ke liye kya predict karta hai?

δ \sim ν x / U \to 0

: layer khatam ho jaati hai, har jagah ideal flow recover ho jaata hai aur d'Alembert ka (real fluids ke liye galat) zero-drag result milta hai.

U \to \infty

hone par, baaki sab fixed rakhte hue, kya hota hai?

δ \propto 1/ U \to 0

, isliye layer infinitely얇i ho jaati hai — flow ek ever-shrinking wall film ke bahar perfectly ideal dikhti hai.

Idealised laminar formula mein

x \to 0

par

τ_{w}

ka kya hota hai?

τ_{w} \propto 1/ x \to \infty

: model leading edge par unbounded shear deta hai, jo ek known idealisation hai jise real (finite-thickness, non-sharp) edges soften karte hain.

Agar plate bahut choti hai aur

R e_{x}

kabhi transition value cross nahi karta, kya turbulence possible hai?

Nahi —

R e_{x}

ko transition se neeche rakh kar layer poori length par laminar rahti hai, aur

5 x / R e_{x}

Blasius law poori jagah apply hota hai.

Agar plate khud fluid ke saath speed

U

par move kare (koi relative motion nahi)?

Toh wall aur fluid ek hi velocity share karte hain, no-slip koi gradient create nahi karta,

\partial u / \partial y = 0

, aur koi boundary layer nahi banti.

Recall Le jaane wala ek-line summary

Lagbhag har question ka trap $μ$ aur $μ \partial u / \partial y$ ko confuse karna hai, aur "zyada flow / higher $R e$ " ko "moti layer" se confuse karna hai. Yaad rakho: 얇i layer, huge gradient, $x$ growth, $1/ R e_{x}$ scaling.