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

2.2.20 · D4 · Physics › Fluid Mechanics › Boundary layer — Prandtl's concept, growth along flat plate

Yahan use ki gayi har formula parent note mein bani hui hai Boundary Layer parent. Jin teen par hum sabse zyada rely karte hain:

Formula Is page ke liye Toolkit

Pehle, do material properties jo batate hain ki fluid "kitna sticky" aur "kitna heavy" hai:

$μ$ = dynamic viscosity, fluid kitni strongly shearing resist karta hai (dekho Viscosity and Newton's law of viscosity), units $Pa\cdots = kg / (m\cdots)$ .
$ρ$ = density, mass per unit volume, units $kg/m^{3}$ .

Do velocities poore mein aati hain, toh inhe abhi pin down karo:

$U$ = free-stream speed, plate se door fluid ki steady speed (m/s).
$u$ = local flow speed inside the boundary layer, wall se height $y$ par fluid ki speed (m/s). Wall par $u = 0$ (no-slip); bahut door $u \to U$ .

Material properties se:

Local Reynolds number: $R e_{x} = \frac{U x}{ν}$ jahan $ν = \frac{μ}{ρ}$ kinematic viscosity hai (units $\frac{Pa\cdots}{kg/m ^{3}} = m^{2} / s$ ).
Laminar layer thickness (Blasius): $δ (x) = \frac{5.0 x}{R e _{x}} = 5.0 \frac{ν x}{U}$ .
Wall shear (order of magnitude): $τ_{w} \sim μ \frac{U}{δ}$ .

Yahan $x$ = leading edge (plate ki agali tip) se distance, $δ$ = thickness jahan $u = 0.99 U$ .

Yeh bhi dekho: Reynolds number, Skin friction drag, Laminar vs Turbulent flow.

Level 1 — Recognition

L1.1 Boundary-layer formula $δ = 5.0 ν x / U$ mein, batao ki har symbol ka kya matlab hai aur har ka SI unit kya hai.

Recall Solution

$δ$ = boundary-layer thickness (wall se distance jahan local speed $u$ , $0.99 U$ tak pahunchti hai), unit m.
$ν$ = kinematic viscosity $= μ / ρ$ , unit $m^{2} / s$ .
$x$ = leading edge se downstream distance, unit m.
$U$ = free-stream speed, unit m/s.
Constant $5.0$ dimensionless hai (exact Blasius solution se). Sanity check: root ke andar, $\frac{( m ^{2} / s ) ( m )}{m/s} = m^{2}$ , aur $m^{2} = m$ . ✔

L1.2 True ya false, har ek ke saath ek line ka reasoning: (a) Leading edge par $δ = 0$ . (b) Faster free stream layer ko thicker banata hai. (c) Layer thickness $x$ ke saath linearly grow karti hai.

Recall Solution

(a) True. $x = 0$ par, $δ = 5.0 ν \cdot 0/ U = 0$ . Koi bhi fluid abhi drag nahi hua. (b) False. $δ \propto 1/ U$ , toh faster stream → thinner layer (viscosity ke sideways diffuse hone ke liye kam time). (c) False. $δ \propto x$ , growth jo slow hoti jaati hai (neeche figure dekho).

Figure s01 ko walk through karo. Horizontal axis leading edge se $x$ distance hai (metres mein, $0$ se $2$ tak); vertical axis $δ$ millimetres mein hai, air ke liye compute kiya gaya $U = 15 m/s$ , $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ par. $δ = 0$ par moti grey line plate surface hai. Top par blue arrows uniform free stream $U$ hain jo left se aa raha hai. Red curve $δ (x) = 5 ν x / U$ hai: notice karo ki yeh origin se vertically (steep) leading edge par nikalta hai, phir bend over ho jaata hai aur flatten ho jaata hai — yeh bend-over hi $x$ law ka visual signature hai. (Formally, $δ = 5 ν x / U$ ko square karne se $x \propto δ^{2}$ milta hai, yaani $x$ , $δ$ mein ek parabola hai; $x$ horizontal plot karne par curve woh parabola sideways turn hua hai. Plate ke neeche faint mirror curve sirf us sideways parabola ko complete karta hai taaki tum shape pehchan sako.)

Figure — Boundary layer — Prandtl's concept, growth along flat plate

Level 2 — Application

L2.1 Air ( $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ ) flat plate par $U = 30 m/s$ se flow karta hai. $x = 0.20 m$ par $R e_{x}$ aur $δ$ nikalo.

Recall Solution

Step 1 — Reynolds number. Pehle kyun? Yeh batata hai ki kaun sa formula apply hota hai aur $δ$ ke andar bhi appear karta hai. $R e_{x} = \frac{U x}{ν} = \frac{30 \times 0.20}{1.5 \times 1 0 ^{- 5}} = \frac{6}{1.5 \times 1 0 ^{- 5}} = 4.0 \times 1 0^{5} .$ Yeh usual flat-plate transition ( $\approx 5 \times 1 0^{5}$ ) se neeche hai, toh laminar Blasius valid hai. Step 2 — Thickness. Kyun? $R e_{x}$ known hone ke baad, $δ = 5.0 x / R e_{x}$ mein direct substitution hi ek remaining move hai — yeh dimensionless $R e_{x}$ ko physical length mein convert karta hai. $δ = \frac{5.0 x}{R e _{x}} = \frac{5.0 \times 0.20}{4.0 \times 1 0 ^{5}} = \frac{1.0}{632.46} = 1.58 \times 1 0^{- 3} m \approx 1.58 mm .$ 20 cm plate → 1.6 mm layer: genuinely thin. ✔

L2.2 Same air flow ke liye, kis distance $x$ par layer $δ = 3.0 mm$ thick hoti hai?

Recall Solution

Step 1 — $x$ isolate karo. Kyun? Hume $δ$ pata hai, $x$ chahiye. $δ = 5.0 ν x / U$ likho aur square karo: $δ^{2} = 25 \frac{ν x}{U} \Rightarrow x = \frac{δ ^{2} U}{25 ν} .$ Step 2 — Substitute karo ( $δ = 3.0 \times 1 0^{- 3} m$ ). Kyun? Teeno inputs ( $δ$ , $U$ , $ν$ ) known hain, toh rearranged formula mein plug karne se directly number milta hai. $x = \frac{( 3.0 \times 1 0 ^{- 3} ) ^{2} \times 30}{25 \times 1.5 \times 1 0 ^{- 5}} = \frac{9 \times 1 0 ^{- 6} \times 30}{3.75 \times 1 0 ^{- 4}} = \frac{2.7 \times 1 0 ^{- 4}}{3.75 \times 1 0 ^{- 4}} = 0.72 m .$ Check: $x = 0.72$ par, kya flow abhi bhi laminar hai? $R e_{x} = 30 \times 0.72/1.5 \times 1 0^{- 5} = 1.44 \times 1 0^{6}$ — transition se upar, toh reality mein turbulent hota. Hum $0.72 m$ laminar-formula value ke roop mein report karte hain.

Level 3 — Analysis

L3.1 Ek plate par, water steadily flow karta hai. $x_{2} = 1.6 m$ par $δ$ ko $x_{1} = 0.1 m$ par $δ$ se compare karo. Yeh kitne factor se thicker hai?

Recall Solution

Step 1 — Ratio use karo, constants ko khatam karo. Kyun? $U, ν$ aur $5.0$ dono stations par identical hain, toh ratio banao aur woh cancel ho jaate hain — $ν$ ke liye koi numbers nahi chahiye. Step 2 — Evaluate karo. Kyun? Sirf $x$ ratio mein bachta hai, toh answer $x$ -ratio ke single square root mein reduce ho jaata hai. $\frac{δ _{2}}{δ _{1}} = \frac{x _{2}}{x _{1}} = \frac{1.6}{0.1} = 16 = 4.$ Layer $x_{2}$ par 4 guna thicker hai, bhaavein distance 16 guna zyada hai — square root ise tame kar deta hai.

L3.2 Wall shear stress $τ_{w} \propto x^{- 1/2}$ scale karta hai. Agar $x = 0.25 m$ par $τ_{w} = 0.80 Pa$ hai, toh $x = 1.0 m$ par $τ_{w}$ nikalo.

Recall Solution

Step 1 — Scaling ratio set up karo. Kyun? Same fluid aur speed, sirf $x$ change hota hai, toh directly $τ_{w} \propto x^{- 1/2}$ use karo. $\frac{τ _{w, 2}}{τ _{w, 1}} = (\frac{x _{2}}{x _{1}})^{- 1/2} = (\frac{1.0}{0.25})^{- 1/2} = 4^{- 1/2} = \frac{1}{2} .$ Step 2 — Multiply karo. Kyun? Ratio woh factor hai jo dono stations ko connect karta hai, toh known $τ_{w, 1}$ times us factor se unknown $τ_{w, 2}$ milta hai. $τ_{w, 2} = 0.80 \times \frac{1}{2} = 0.40 Pa .$ Friction leading edge par sabse strong hoti hai (thinnest layer → steepest velocity gradient) aur 4 guna downstream jaane par half ho jaati hai.

Figure s02 ko walk through karo. Horizontal axis phir se distance $x$ hai (m, $0.05$ se $2.0$ tak); vertical axis wall shear $τ_{w}$ pascals mein hai. Orange curve $τ_{w} \propto x^{- 1/2}$ law hai, normalize kiya gaya taaki $τ_{w} = 0.40 Pa$ at $x = 1.0 m$ . Do marked dots exactly is problem ke stations hain: red dot $(0.25, 0.80)$ par aur green dot $(1.0, 0.40)$ par — value half hoti hai jab $x$ quadruple hota hai. Notice karo ki curve left edge ki taraf kitni steeply climb karti hai: yeh leading-edge blow-up hai jo toolkit mein flag hua tha ( $τ_{w} \to \infty$ jab $x \to 0$ ), isliye plot $x = 0.05$ se start hota hai na ki $0$ se.

Level 4 — Synthesis

L4.1 Ek plate of length $L = 1.0 m$ aur width $b = 0.5 m$ air mein hai ( $ρ = 1.2 kg/m^{3}$ , $ν = 1.5 \times 1 0^{- 5} m^{2} / s$ ) $U = 15 m/s$ par. Plate ke ek side ke liye laminar drag-coefficient formula use karte hue, $C_{D} = \frac{1.328}{R e _{L}}, F_{D} = C_{D} \cdot \frac{1}{2} ρ U^{2} \cdot (b L),$ nikalo (a) $R e_{L}$ , (b) $C_{D}$ , (c) ek side par friction drag force $F_{D}$ .

Recall Solution

Step 1 — Length-based Reynolds number. Kyun? $C_{D}$ poori plate par shear integrate karta hai, toh yeh trailing edge par $R e$ par depend karta hai, $R e_{L}$ . $R e_{L} = \frac{U L}{ν} = \frac{15 \times 1.0}{1.5 \times 1 0 ^{- 5}} = 1.0 \times 1 0^{6} .$ Step 2 — Drag coefficient. Kyun? Formula $C_{D} = 1.328/ R e_{L}$ abhi-compute kiye $R e_{L}$ ko dimensionless friction number mein convert karta hai. $C_{D} = \frac{1.328}{1.0 \times 1 0 ^{6}} = \frac{1.328}{1000} = 1.328 \times 1 0^{- 3} .$ Step 3 — Force. Yeh form kyun? Dynamic pressure $\frac{1}{2} ρ U^{2}$ times area $b L$ times dimensionless $C_{D}$ se force milti hai. $q = \frac{1}{2} ρ U^{2} = \frac{1}{2} (1.2) (1 5^{2}) = 0.5 \times 1.2 \times 225 = 135 Pa .$ $A = b L = 0.5 \times 1.0 = 0.5 m^{2} .$ $F_{D} = C_{D} q A = 1.328 \times 1 0^{- 3} \times 135 \times 0.5 = 0.089640 N .$ Toh plate ka ek side skin-friction drag ka lagbhag $0.0896 N$ feel karta hai — small, jaise thin laminar layer se expect hota hai. Dekho Skin friction drag.

L4.2 Same plate. Trailing edge ( $x = L = 1.0 m$ ) par layer thickness $δ$ kya hai, aur yeh plate length ka kitna fraction hai?

Recall Solution

Step 1 — $R e_{L}$ reuse karo. Kyun? Trailing edge par $δ$ ke liye local $R e_{x}$ chahiye $x = L$ par evaluate kiya hua, jo exactly $R e_{L} = 1.0 \times 1 0^{6}$ hai jo humne L4.1 mein compute kiya — same $U$ , $ν$ , aur same $x = L$ . Koi nayi arithmetic nahi chahiye. Step 2. Kyun? $R e_{L}$ ko $δ = 5.0 L / R e_{L}$ mein substitute karne se whole-plate Reynolds number trailing-edge thickness mein convert hota hai. $δ (L) = \frac{5.0 L}{R e _{L}} = \frac{5.0 \times 1.0}{1000} = 5.0 \times 1 0^{- 3} m = 5.0 mm .$ Step 3 — Fraction. Kyun? $δ$ ko $L$ se divide karne se thickness plate ke fraction ke roop mein express hoti hai, woh quantity jo judge karti hai ki Prandtl ka thin-layer split valid hai ya nahi. $\frac{δ}{L} = \frac{5.0 \times 1 0 ^{- 3}}{1.0} = 0.5%.$ Length ka half percent: yahi tiny ratio exactly wajah hai ki Prandtl ka two-region split kaam karta hai.

Level 5 — Mastery

L5.1 (Total drag ka speed ke saath scaling derive karo.) Ek fixed laminar plate ke liye, dikhao ki total friction force $F_{D}$ free-stream speed $U$ ke saath kaise scale karta hai. Phir predict karo ki $F_{D}$ kitne factor se grow karta hai agar $U$ triple ho jaaye.

Recall Solution

Step 1 — Pieces symbolically assemble karo. Symbolic kyun? Hume scaling law chahiye, toh letters rakho aur sirf $U$ ke powers track karo. $C_{D} = \frac{1.328}{R e _{L}} = 1.328 (U L / ν)^{- 1/2} \propto U^{- 1/2} .$ $q = \frac{1}{2} ρ U^{2} \propto U^{2} .$ $F_{D} = C_{D} q A \propto U^{- 1/2} \cdot U^{2} = U^{3/2} .$ Step 2 — Law state karo. Laminar skin-friction drag is tarah grow karta hai: $F_{D} \propto U^{3/2} .$ Step 3 — Speed triple karo. $F_{D}$ , $3^{3/2} = 33 \approx 5.196$ se multiply hota hai. Physical read: dynamic pressure ( $U^{2}$ ) drag ko push karta hai upar, par thinning layer ( $C_{D} \propto U^{- 1/2}$ ) $U$ ka half power wapas chheen leti hai — net $U^{3/2}$ , us $U^{2}$ se weaker jo tum naively guess karte.

L5.2 (Design problem.) Tumhe water ( $ν = 1.0 \times 1 0^{- 6} m^{2} / s$ ) mein $2.0 m$ plate par trailing-edge boundary layer exactly $δ = 4.0 mm$ chahiye $x = L = 2.0 m$ par, laminar rehte hue. Yeh free-stream speed $U$ kaunsa achieve karta hai? Phir check karo ki flow genuinely laminar hai wahan.

Recall Solution

Step 1 — $δ = 5.0 ν L / U$ ko $U$ ke liye solve karo. Kyun? $δ$ aur $x = L$ fixed targets hain; $U$ unknown dial hai. Square karo: $δ^{2} = 25 \frac{ν L}{U} \Rightarrow U = \frac{25 ν L}{δ ^{2}} .$ Step 2 — Substitute karo ( $δ = 4.0 \times 1 0^{- 3} m$ ). Kyun? Saare inputs known hain, toh rearranged formula $U$ directly deta hai. $U = \frac{25 \times ( 1.0 \times 1 0 ^{- 6} ) \times 2.0}{( 4.0 \times 1 0 ^{- 3} ) ^{2}} = \frac{5.0 \times 1 0 ^{- 5}}{1.6 \times 1 0 ^{- 5}} = 3.125 m/s .$ Step 3 — Laminarity check. Kyun? Blasius sirf transition se neeche valid hai, toh hume $R e_{L}$ ko $5 \times 1 0^{5}$ threshold ke against confirm karna hai design trust karne se pehle. $R e_{L} = \frac{U L}{ν} = \frac{3.125 \times 2.0}{1.0 \times 1 0 ^{- 6}} = 6.25 \times 1 0^{6} .$ Yeh transition $R e \approx 5 \times 1 0^{5}$ se bahut upar hai — toh reality mein flow $x = 2 m$ se pehle hi turbulent ho jaata. Laminar target is plate par physically achievable nahi hai; ise laminar rakhne ke liye tumhe shorter plate ya zyada $ν$ (thicker fluid) chahiye hoga. Honest verdict report karna hi mastery step hai. Dekho Laminar vs Turbulent flow.

Active Recall

Recall One-line answers (inhe cover karo)

$δ$ , $x$ ke saath kaise scale karta hai ::: $x$ (grow karta hai, par decelerating) $δ$ , $U$ ke saath kaise scale karta hai ::: $1/ U$ (faster stream → thinner layer) Target $δ$ ke liye $x$ find karne ke liye tumhe ::: relation ko square karna hoga, kyunki $x \propto δ^{2}$ Wall shear $τ_{w}$ , $x$ ke saath kaise scale karta hai ::: $x^{- 1/2}$ (leading edge par sabse bada) $τ_{w}$ ka kya hota hai jab $x \to 0$ ::: yeh blow up karta hai ( $\to \infty$ ); Blasius sharp tip par break down karta hai Laminar total drag speed ke saath kaise scale karta hai ::: $F_{D} \propto U^{3/2}$ Is page par do velocities, $U$ aur $u$ , hain ::: free-stream speed aur local in-layer speed Kisi bhi Blasius result par trust karne se pehle, check karo ki ::: $R e_{x} ≲ 5 \times 1 0^{5}$ (abhi bhi laminar)