Blasius solution — exact laminar boundary layer solution: Question bank

2.2.22 · D5 · Physics › Fluid Mechanics › Blasius solution — exact laminar boundary layer solution

Intuition Yeh page kis liye hai

Parent note ne tumhe machinery di thi. Yeh page un ideas par attack karta hai jo uske peeche hain — woh jagahein jahan ek confident student galat baat kehta hai kyunki koi formula "sahi lag raha tha". Neeche har item ek one-line reveal hai: answer cover karo, pehle ek guess commit karo, phir check karo. Agar tum in sabko justify kar sako, toh tum Blasius ko sach mein samajhte ho.

Shuru karne se pehle, ek quick glossary taaki koi bhi symbol unearned na rahe:

$U_{\infty}$ (kaho "U-infinity") = free-stream speed, plate se door jahan viscosity feel nahi hoti.
$u$ = plate ke saath fluid ki speed ( $x$ -direction mein); $v$ = plate se door fluid ki speed ( $y$ -direction mein).
$δ$ (delta) = boundary-layer thickness, woh height jahan $u$ $U_{\infty}$ ke $99%$ tak pahunch jaati hai.
$η$ (eta) = "clever height" $y U_{\infty} / (ν x)$ jo har profile ko identical bana deta hai.
$ψ$ (psi) = stream function; $u = \partial ψ / \partial y$ , $v = - \partial ψ / \partial x$ (yeh pairing continuity automatically satisfy karti hai). Blasius ke liye $ψ = ν x U_{\infty} f (η)$ .
$f$ = dimensionless stream-function shape; $f^{'} = u / U_{\infty}$ velocity ratio hai (prime ka matlab $d / d η$ ).
$ν$ (nu) = kinematic viscosity (fluid kitna "sticky-and-spread-y" hai); $μ = ρ ν$ = dynamic viscosity.
$R e_{x} = U_{\infty} x / ν$ = Reynolds Number, position $x$ par inertia aur viscosity ka ratio. Plate ki total length $L$ ke saath yeh $R e_{L} = U_{\infty} L / ν$ ban jaata hai, poori plate ke liye ek single fixed number.
$τ_{w}$ = wall shear stress, fluid ka plate par tangential drag force per unit area.
$C_{f} = \frac{τ _{w}}{\frac{1}{2} ρ U _{\infty}^{2}} = \frac{0.664}{R e _{x}}$ = local skin-friction coefficient — wall shear ko dynamic pressure $\frac{1}{2} ρ U_{\infty}^{2}$ se dimensionless banaya gaya position $x$ par.
$C_{D} = \frac{1.328}{R e _{L}}$ = poori plate ke ek side ke liye length-averaged drag coefficient (yeh $x = L$ par evaluate kiye gaye $C_{f}$ se twice nikalta hai).
$β \approx 1.72$ = "displacement constant": wall se door $f (η) \to η - β$ , aur yeh offset $β$ displacement thickness hai — slow fluid ki wajah se outer streamlines kitni door push hoti hain.

Pehle teen pictures apne dimag mein rakho

Definition

$v$ -formula kahaan se aata hai — $ψ \to f$ mapping Humne $v$ kabhi guess nahi kiya; yeh stream function $ψ = ν x U_{\infty} f (η)$ se $v = - \partial ψ / \partial x$ ke zariye follow hota hai. Kyunki $ψ$ $x$ par depend karta hai dono front factor $ν x U_{\infty} = (ν U_{\infty})^{1/2} x^{1/2}$ ke through aur $f$ ke andar $η (x)$ ke through, product rule use karo: $v = - \frac{\partial ψ}{\partial x} = - front factor differentiated (ν U_{\infty})^{1/2} \frac{1}{2} x^{- 1/2} f + f differentiated via η (ν U_{\infty})^{1/2} x^{1/2} f^{'} \frac{\partial η}{\partial x} .$ Doosre term mein $\partial η / \partial x = - η /2 x$ daalo: $v = - \frac{1}{2} \frac{ν U _{\infty}}{x} f + \frac{1}{2} \frac{ν U _{\infty}}{x} η f^{'} = \frac{1}{2} \frac{ν U _{\infty}}{x} (η f^{'} - f) .$ Toh mysterious $(η f^{'} - f)$ kuch nahi hai bas "front-factor derivative minus $η$ -chain derivative" — $ψ = ν x U_{\infty} f$ ka direct consequence, koi alag assumption nahi. (Check: wall par $η = 0, f = 0, f^{'} = 0$ , toh $v (x, 0) = 0$ ✓.)

Intuition Transform

$x$ ko kyun khatam kar deta hai — mechanism Agla figure ek construction diagram hai. Do stations $x_{1} < x_{2}$ par physical profiles $u (y)$ alag dikhte hain (right wali layer taller hai). Lekin height ko $y / ν x / U_{\infty}$ se rescale karo aur dono ek curve par collapse ho jaate hain. Yeh luck nahi hai — yeh $η$ choose karke engineer kiya gaya hai taaki har derivative mein $x$ -factors cancel ho jaayein.

Yahan cancellation words mein hai, term by term (prime $d / d η$ hai), upar derive kiye gaye do chain-rule facts use karke:

$u = U_{\infty} f^{'} (η)$ — koi bare $x$ survive nahi karta, sirf $η$ ke through.
$\frac{\partial u}{\partial x} = U_{\infty} f^{''} \frac{\partial η}{\partial x} = U_{\infty} f^{''} (- \frac{η}{2 x})$ — ek $1/ x$ laata hai.
$\frac{\partial u}{\partial y} = U_{\infty} f^{''} \frac{\partial η}{\partial y} = U_{\infty} f^{''} \frac{U _{\infty}}{ν x}$ , aur $\frac{\partial ^{2} u}{\partial y ^{2}} = U_{\infty} f^{'''} \frac{U _{\infty}}{ν x}$ — ek $1/ x$ laata hai.
$v = \frac{1}{2} \frac{ν U _{\infty}}{x} (η f^{'} - f)$ — ek $1/ x$ laata hai jo $\partial u / \partial y$ ke $1/ x$ ke saath pair karta hai. Jab tum yeh sab $u u_{x} + v u_{y} = ν u_{y y}$ mein daalte ho, har term exactly $U_{\infty}^{2} / x$ carry karta hai, toh divide karne par $x$ aur $U_{\infty}$ dono erase ho jaate hain — aur pure ODE $f^{'''} + \frac{1}{2} f f^{''} = 0$ bach jaata hai. Yahi wajah hai ki do variables mein ek PDE ek variable mein ODE ban jaata hai.

Intuition

$δ \propto x$ aur $τ_{w} \propto x^{- 1/2}$ kyun hain — scaling argument sketch Last figure parabolic $δ (x)$ curve ko badhte aur wall shear ko decay karte dikhata hai. Pehle, kyun $u \partial u / \partial x \sim U_{\infty}^{2} / x$ : fluid leading edge se travel ki gayi streamwise distance $x$ mein $0$ se lagbhag $U_{\infty}$ tak speed hota hai, toh along-plate gradient $\partial u / \partial x \sim U_{\infty} / x$ hai (change $U_{\infty}$ divided by distance $x$ ); typical speed $u \sim U_{\infty}$ se multiply karne par inertial term $u \partial u / \partial x \sim U_{\infty}^{2} / x$ milta hai. Ab is inertia $\sim U_{\infty}^{2} / x$ ko cross-stream viscous force $ν \partial^{2} u / \partial y^{2} \sim ν U_{\infty} / δ^{2}$ ke saath balance karo (velocity change $U_{\infty}$ small height $δ$ par, do baar differentiate). Equate karne par $δ^{2} \sim ν x / U_{\infty}$ milta hai, hence $δ \sim ν x / U_{\infty}$ . Wall slope $u / δ \sim U_{\infty} / δ$ hai, toh $τ_{w} \sim μ U_{\infty} / δ \propto μ U_{\infty} U_{\infty} / (ν x) \propto x^{- 1/2}$ . Dono proportionalities ek hi force balance se directly nikalti hain.

True ya false — justify karo

Boundary layer ka ek sharp top edge hota hai jahan flow suddenly free-stream ban jaati hai.

False —

U_{\infty}

tak pahunchna smooth aur asymptotic hai (

f^{'} (\infty) = 1

sirf

η \to \infty

par reach hota hai);

δ

sirf ek arbitrary cutoff hai jo

u = 0.99 U_{\infty}

par define hai.

Agar free-stream speed

U_{\infty}

double karo, toh fixed

x

par boundary layer thinner ho jaati hai.

True —

δ \propto ν x / U_{\infty}

, toh zyada speed matlab zyada inertia slow zone ko downstream sweep kar deta hai thicken hone se pehle,

δ

1/ 2

se shrink kar deta hai.

Blasius velocity profile plate ke downstream move karne par shape change karti hai.

False — yahi toh ek similarity solution ka poora point hai:

η

ke against plot karne par profile ek universal curve hai; sirf physical

y

-scale stretch hoti hai.

Vertical velocity

v

boundary layer ke andar har jagah zero hai.

False — continuity

\partial u / \partial x + \partial v / \partial y = 0

ek small upward

v = - \partial ψ / \partial x = \frac{1}{2} ν U_{\infty} / x (η f^{'} - f) \neq = 0

force karta hai jab layer thick hoti hai; combination

(η f^{'} - f)

exactly wahi hai jo

- \partial ψ / \partial x

ψ = ν x U_{\infty} f

se produce karta hai. Sirf

v (x, 0) = 0

(wall par no through-flow).

Blasius equation

f^{'''} + \frac{1}{2} f f^{''} = 0

ka ek neat closed-form solution hai.

False — yeh nonlinear hai aur iska koi elementary closed form nahi hai; constant

f^{''} (0) = 0.332

numerical shooting method se aata hai.

Wall shear stress

τ_{w}

plate ke leading edge par sabse zyada hota hai.

True —

τ_{w} = 0.332 μ U_{\infty} U_{\infty} / (ν x) \propto 1/ x

, jo

x \to 0

(singular leading edge) par blow up karta hai aur downstream decay karta hai.

Kyunki

τ_{w} \to \infty

x = 0

, plate par total drag infinite hai.

False —

τ_{w} \propto x^{- 1/2}

integrable hai;

\int_{0}^{L} x^{- 1/2} d x = 2 L

finite hai, well-behaved

C_{D} = 1.328/ R e_{L}

deta hai.

Blasius result laminar aur turbulent dono boundary layers par equally apply hota hai.

False — yeh smooth laminar flow assume karta hai (

R e_{x} ≲ 5 \times 1 0^{5}

); ek Turbulent Boundary Layer momentum ko kaafi zyada efficiently mix karta hai aur

x^{1/2}

ki jagah

x^{4/5}

ki tarah badhta hai.

Plate length

L

double karne par total skin-friction drag double ho jaata hai.

False —

F_{D} \propto C_{D} \cdot L \propto L^{- 1/2} \cdot L = L^{1/2}

, toh drag sirf

L

ki tarah badhta hai; length double karne par drag

2 \approx 1.41

se multiply hota hai.

Pressure gradient

\partial p / \partial x

Blasius layer ke andar zero assume kiya jaata hai.

True (effectively) — Prandtl's scaling dikhata hai ki

y

-momentum equation

\partial p / \partial y \approx 0

par collapse ho jaati hai, toh pressure thin layer mein vary nahi karta; yeh uniform outer flow se simply impress hota hai, jahan flat plate ke liye

\partial p / \partial x = 0

hai. Hence andar bhi zero hai.

Error dhundo

"Layer downstream mein thick hoti hai, toh

δ

x

ke saath linearly badhta hai."

Error: growth

δ \propto x

hai, ek decelerating parabolic growth; ek straight line far downstream mein thickness ko bahut zyada over-predict karta hai.

C_{D} = C_{f}

, kyunki dono skin-friction coefficients hain."

Error:

C_{f} = 0.664/ R e_{x}

local hai (ek point);

C_{D} = 1.328/ R e_{L}

length-averaged value hai — aur yeh

x = L

par evaluate kiye gaye local

C_{f}

se twice ke barabar hota hai.

"Hum

\partial^{2} u / \partial y^{2}

drop kar sakte hain kyunki layer thin hai."

Error: ulta hai — hum streamwise

\partial^{2} u / \partial x^{2}

drop karte hain; cross-stream

\partial^{2} u / \partial y^{2}

dominant viscous term hai kyunki thin layer mein variation fast hoti hai.

"Kyunki

f^{'} (\infty) = 1

hai, wall condition bhi

f^{'} (0) = 1

honi chahiye."

Error: wall no-slip enforce karta hai, toh

f^{'} (0) = 0

(zero velocity); free stream

f^{'} = 1

reach karta hai.

η = y U_{\infty} / (ν x)

, toh fixed

y

par similarity variable downstream jaane par badhta hai."

Error:

η \propto 1/ x

, toh fixed physical height

y

par

η

ki value downstream shrink hoti hai — ek fixed point universal curve par neeche baith jaata hai jab layer uske aage swell hoti hai.

R e_{x}

mein Reynolds number plate length use karta hai, toh yeh ek single fixed number hai."

Error:

R e_{x} = U_{\infty} x / ν

leading edge se local distance

x

use karta hai, toh yeh plate ke saath badhta hai;

R e_{L}

(total length

L

ke saath) fixed wala hai.

"Wall shear ko

μ

chahiye, toh yeh sirf fluid par depend karta hai, flow par nahi."

Error:

τ_{w} = 0.332 μ U_{\infty} U_{\infty} / (ν x)

U_{\infty}^{3/2}

aur

x^{- 1/2}

ke saath bhi scale karta hai — yeh flow state ki property bhi bahut zyada hai.

Why questions

Hum

u

aur

v

ko alag solve karne ki jagah stream function

ψ

kyun use karte hain?

Kyunki

u = \partial ψ / \partial y, v = - \partial ψ / \partial x

define karne se continuity automatically satisfy hoti hai, do unknowns ek mein collapse ho jaate hain aur ek equation sirf momentum balance mein reh jaati hai.

Similarity transform ek PDE (

x, y

mein) ko ODE (

η

mein) kyun reduce kar deta hai?

Kyunki jab

u = U_{\infty} f^{'} (η)

substitute kiya jaata hai, har derivative ek power of

x

laati hai (

\partial η / \partial x = - η /2 x

aur

\partial η / \partial y = U_{\infty} / ν x

ke zariye); momentum equation mein har term same factor

U_{\infty}^{2} / x

carry karta hai, toh divide karne par

x

aur

U_{\infty}

dono cancel ho jaate hain aur

f^{'''} + \frac{1}{2} f f^{''} = 0

bach jaata hai — akele

η

mein ODE.

δ = 5.0 ν x / U_{\infty}

mein coefficient

5.0

fundamental constant kyun nahi hai?

Yeh "edge" ki arbitrary

99%

definition se tied hai;

f^{'} = 0.99

η \approx 5.0

par hota hai, lekin

95%

99.9%

definition alag coefficient deti.

Vertical velocity

v

non-zero kyun honi chahiye jab bhi plate horizontal ho?

Jab layer downstream mein thick hoti hai, mass conservation fluid ko upar squeeze karta hai; layer ke andar

\partial u / \partial x < 0

\partial v / \partial y > 0

force karta hai, toh fluid gently wall se door drift karta hai.

Blasius Navier–Stokes Equations ke un terms ko kyun ignore kar sakta hai jo Prandtl ne drop kiye?

Prandtl Boundary Layer Theory dikhata hai ki thin layer ke andar streamwise diffusion aur

y

-momentum equation retained balance of inertia against cross-stream viscous shear ke comparison mein negligibly small hain.

Higher viscosity

ν

boundary layer ko thinner ki jagah thicker kyun banata hai?

Zyada viscosity matlab wall ka "no-slip drag" flow mein aur door tak pahunchta hai, toh slow zone

δ \propto ν

wider spread hoti hai inertia usse overcome karne se pehle.

f^{''} (0) = 0.332

poore solution mein sabse important number kyun hai?

Yeh velocity profile ka wall slope hai, toh yeh directly wall shear set karta hai (

τ_{w} \propto f^{''} (0)

), skin-friction coefficient

C_{f} = 0.664/ R e_{x}

, aur hence drag

C_{D} = 1.328/ R e_{L}

Edge cases

Leading edge

x = 0

par

δ

aur

τ_{w}

ka exactly kya hota hai?

δ \to 0

(abhi koi layer form nahi hui) jabki

τ_{w} \to \infty

— ek mathematical singularity jo thin-layer approximation resolve nahi kar sakti, lekin jo finite drag mein integrate hoti hai.

Inviscid limit

ν \to 0

mein Blasius profile kya predict karta hai?

δ \propto ν \to 0

: layer wall par collapse ho jaati hai, har jagah slip, inviscid free-stream recover ho jaata hai — viscosity hi layer ki sole cause hai, iske saath consistent.

Kya Blasius bahut low Reynolds number (jaise

R e_{x} \sim 1

) par valid hai?

Nahi — derivation

δ ≪ x

require karti hai, yaani large

R e_{x}

;

R e_{x} \sim 1

par layer thin nahi hai aur dropped Navier–Stokes Equations terms waapis aa jaate hain.

Blasius ki validity ki upper Reynolds limit kya set karta hai?

R e_{x} \approx 5 \times 1 0^{5}

ke paas Turbulent Boundary Layer mein transition; uske baad smooth laminar assumption fail ho jaati hai aur

δ / x \approx 0.37 R e_{x}^{- 1/5}

use karte hain.

Wall se door

f

(not

f^{'}

) ki value kya hai, aur iska kya matlab hai?

Large

η

ke liye

f (η) \to η - β

jahan

β \approx 1.72

; constant offset

β

displacement thickness hai — slow fluid ki wajah se streamlines ka effective outward shift.

Wall par (

η = 0

f, f^{'}, f^{''}

mein se kaun zero hain aur kaun nahi?

f (0) = 0

(no through-flow) aur

f^{'} (0) = 0

(no-slip) zero hain, lekin

f^{''} (0) = 0.332 \neq = 0

— profile finite slope ke saath wall se nikalta hai, finite skin friction deta hai.

η \to \infty

par

f^{''} (η)

aur

f^{'''} (η)

ka kya hota hai, aur iska physically kya matlab hai?

Dono vanish ho jaate hain:

f^{'}

constant

1

par flat ho chuka hai, toh iska slope

f^{''} \to 0

(wall se door koi shear nahi) aur

f^{'''} \to 0

. Physically wahan flow uniform hai — velocity gradient nahi matlab viscous stress nahi, aur

v

apni small constant entrainment value

\frac{1}{2} ν U_{\infty} / x β

par level off ho chuka hai.

Zero pressure-gradient agar adverse (

\partial p / \partial x > 0

) ho jaaye toh is solution ka kya hoga?

Blasius ab apply nahi hoga — adverse gradient near-wall fluid ko decelerate karta hai aur separation drive kar sakta hai; flat-plate case ki jagah aapko more general Falkner–Skan family ki zaroorat padegi.

Recall Ek-line self test

Agar tum yeh answer kar sako, toh page land hua: Blasius profile "har $x$ par same" kyun hai jab bhi boundary layer visibly downstream mein thick hoti hai? Answer ::: Shape stretched variable $η$ mein identical hai; sirf physical conversion $y = η ν x / U_{\infty}$ $x$ ke saath badhti hai, toh same curve ek ever-taller $y$ -axis par draw hoti hai.