Exercises — Blasius solution — exact laminar boundary layer solution

Wo group local Reynolds number hai $R{e}_x=U_{\infty }x/\nu$. Blasius (laminar) roughly $R{e}_x\approx 5\times 10^5$ tak valid hai; uske upar hum Turbulent Boundary Layer ki taraf jaate hain. Humne kya kiya: yaad kiya ki inertia-vs-viscosity ratio hi ek maatra knob hai. Kyun: Blasius ODE tab nikla jab $x$ aur $U_{\infty }$ saare $\eta$ mein absorb ho gaye — jo cheez "scale jaanti hai" wo sirf $R{e}_x$ hai.

$f^{\prime }(\eta )=u/U_{\infty }$, similarity height $\eta$ par velocity ratio hai (yaad karo $\eta =y\sqrt{U_{\infty }/(\nu x)}$, stretched wall-distance hai). Wall par $\eta =0$ par hume $f^{\prime }(0)=0$ milta hai (no-slip: fluid plate se chipka hua hai). Door $f^{\prime }(\infty )=1$ (free stream $U_{\infty }$ se match karta hai). Kyun: $u=U_{\infty }{f}^{\prime }$ seedha $u=\partial \psi /\partial y$ (stream-function definition) se aaya, isliye $f^{\prime }$ literally "full speed ka kitna fraction" hai.

$\delta =5.0\sqrt{\frac{\nu x}{U_{\infty }}}$, isliye $\delta \propto \sqrt{x}$ (distance ke square-root ki tarah badhta hai) aur $\delta \propto 1/\sqrt{U_{\infty }}$ (tez flow ⇒ patli layer). Square-root kyun: inertia $U_{\infty }^2/x$ ko viscosity $\nu U_{\infty }/{\delta }^2$ ke saath balance karne par ${\delta }^2\propto \nu x/U_{\infty }$ force hota hai.

Step 1 — $R{e}_x$: $R{e}_x=\frac{0.8\times 0.10}{1.0\times 10^{-6}}=8.0\times 10^4$. Pehle kyun? Ye $<5\times 10^5$ hai, isliye laminar Blasius legal hai. Step 2 — $\delta$: $\delta =\frac{5.0\times 0.10}{\sqrt{8.0\times 10^4}}=\frac{0.5}{282.8}=1.77\times 10^{-3}\text{m}\approx 1.8\text{mm}$. Kaisa dikhta hai: plate se chipki hui "slow water" ki 2 mm se kam ki ek sheet.

Step 1 — $C_f$: $C_f=\frac{0.664}{\sqrt{8.0\times 10^4}}=\frac{0.664}{282.8}=2.348\times 10^{-3}$. Step 2 — un-normalise: ${\tau }_w=C_f\cdot \frac{1}{2}\rho U_{\infty }^2=2.348\times 10^{-3}\times \frac{1}{2}\times 1000\times 0.8^2$ $=2.348\times 10^{-3}\times 320=0.751\text{Pa}$. Kyun: $\frac{1}{2}\rho U_{\infty }^2$ dynamic pressure hai — natural "stress scale" jise tum dimensionless $C_f$ se multiply karte ho. Dekho Skin Friction Drag.

Step 1 — $R{e}_L$: $R{e}_L=\frac{20\times 0.30}{1.5\times 10^{-5}}=4.0\times 10^5$ (laminar, limit se thoda neeche). Step 2 — $C_D$: $C_D=\frac{1.328}{\sqrt{4.0\times 10^5}}=\frac{1.328}{632.5}=2.100\times 10^{-3}$. Step 3 — force: $F_D=C_D\cdot \frac{1}{2}\rho U_{\infty }^2\cdot (bL)$ $=2.100\times 10^{-3}\times \frac{1}{2}\times 1.2\times 400\times (0.15\times 0.30)$ $=2.100\times 10^{-3}\times 240\times 0.045=0.0227\text{N}$. $C_f$ ki jagah $C_D$ kyun: $C_D$ poori plate par friction ka average hai; $C_f$ ek point par local hai.

Idea: fixed $U_{\infty },\nu$ par $\delta \propto \sqrt{x}$, isliye $\frac{{\delta }_2}{{\delta }_1}=\sqrt{\frac{x_2}{x_1}}$. Compute: $x_2/x_1=0.80/0.20=4$, isliye ${\delta }_2={\delta }_1\sqrt{4}=2{\delta }_1=4.0\text{mm}$. Kaisa dikhta hai: distance chaar guna, lekin thickness sirf double — square-root growth decelerate ho rahi hai (neche figure dekho).

Idea: $C_f=0.664/\sqrt{R{e}_x}\propto 1/\sqrt{x}$. $C_f$ ko half karne ka matlab $\sqrt{x}$ ko $2$ se multiply karna hai, yaani $x$ ko $4$ se. Compute: $x=4\times 0.10=0.40\text{m}$. Kyun: $C_f\propto x^{-1/2}$, isliye $C_f(4x)=C_f(x)/2$. Friction downstream fade hoti hai — profile flatten hoti hai, wall slope girti hai.

Step 1 — $\eta$ invert karo: $\eta =y\sqrt{U_{\infty }/(\nu x)}$ se, $y=\eta \sqrt{\nu x/U_{\infty }}$ milta hai. Step 2 — plug in: $\sqrt{\nu x/U_{\infty }}=\sqrt{\frac{1.5\times 10^{-5}\times 0.25}{15}}=\sqrt{2.5\times 10^{-7}}=5.0\times 10^{-4}\text{m}$. Step 3: $y=1.84\times 5.0\times 10^{-4}=9.2\times 10^{-4}\text{m}=0.92\text{mm}$. Ye kyun kaam karta hai: $\eta$ "clever height" hai. Ek fixed shape point (half-speed) hamesha fixed $\eta$ par hota hai; ek baar physical $y$ mein convert karo.

Step 1 — unit width per total drag: wall shear ko plate ke along integrate karo: $\frac{F_D}{b}={\int }_0^L{\tau }_{w}dx=0.332\mu U_{\infty }\sqrt{\frac{U_{\infty }}{\nu }}{\int }_0^L{x}^{-1/2}dx.$ Step 2 — integral karo: ${\int }_0^L{x}^{-1/2}dx=2\sqrt{L}$, isliye $\frac{F_D}{b}=0.332\mu U_{\infty }\sqrt{\frac{U_{\infty }}{\nu }}\cdot 2\sqrt{L}=0.664\mu U_{\infty }\sqrt{\frac{U_{\infty }L}{\nu }}.$ Step 3 — $C_D$ mein normalise karo: $C_D=\frac{F_D/b}{\frac{1}{2}\rho U_{\infty }^{2}L}$. $\mu =\rho \nu$ use karke: $C_D=\frac{0.664\rho \nu U_{\infty }\sqrt{U_{\infty }L/\nu }}{\frac{1}{2}\rho U_{\infty }^{2}L}=\frac{1.328}{\sqrt{U_{\infty }L/\nu }}=\frac{1.328}{\sqrt{R{e}_L}}.$ Step 4 — compare: $C_f(L)=0.664/\sqrt{R{e}_L}$, isliye $C_D(L)=2C_f(L)$. ✓ Physical kyun: leading edge ke paas ($x\to 0$) wall slope — aur isliye ${\tau }_w$ — bahut bada hota hai ($x^{-1/2}$ ki tarah blow up karta hai). Us bade upstream friction ko small trailing-edge friction ke saath average karne par ek mean milta hai jo exactly trailing-edge local value ka double hai. Integral $\int x^{-1/2}$ front se dominated hai.

Wall par: $\eta =0$ dene par $\eta f^{\prime }=0$, aur BC $f(0)=0$ doosra term bhi kill kar deta hai, isliye $v(x,0)=\frac{1}{2}\sqrt{\nu U_{\infty }/x}(0-0)=0$. ✓ (Koi fluid solid plate cross nahi karta — impermeability.) Wall ke upar: kyunki layer downstream thick hoti hai, iske andar $\partial u/\partial x<0$ hai (fixed $y$ par $x$ badhne ke saath fluid decelerate hota hai). Continuity $\partial u/\partial x+\partial v/\partial y=0$ phir $\partial v/\partial y>0$ force karta hai, isliye $v$ $0$ se upar chadhta hai — ek gentle outflow. Ye L4 synthesis hai continuity + Stream Function and Vorticity ka: stream function $\psi$ ko precisely is liye choose kiya gaya tha ki continuity automatically hold ho, aur phir bhi ye non-zero $v$ predict karta hai. Jo mistake ye kill karta hai: "flow purely parallel hai" — galat; $v\ne 0$ layer mein momentum carry karta hai.

Step 1 — drag law assemble karo: L4.1 se, $F_D=0.664b\mu U_{\infty }\sqrt{U_{\infty }L/\nu }\propto U_{\infty }^{3/2}{L}^{1/2}$ (kyunki $\sqrt{U_{\infty }}\cdot U_{\infty }=U_{\infty }^{3/2}$ aur $\sqrt{L}$). Step 2 — ratio banao: $\frac{F_B}{F_A}={\left(\frac{U_B}{U_A}\right)}^{3/2}{\left(\frac{L_B}{L_A}\right)}^{1/2}=2^{3/2}\cdot 2^{1/2}=2^2=4.$ Step 3 — sanity: speed zyada matter karti hai (exponent $3/2$) length se (exponent $1/2$). Dono double karne par clean $4\times$ milta hai. ✓ Kaisa dikhta hai: drag speed ke saath steeply badhta hai — $U_{\infty }^{3/2}$ term growth dominate karta hai.

Step 1 — ODE se $f^{\prime \prime \prime }(0)$ padho: $f^{\prime \prime \prime }=-\frac{1}{2}f{f}^{\prime \prime }$. $\eta =0$ par, $f(0)=0$, isliye $f^{\prime \prime \prime }(0)=-\frac{1}{2}f(0)f^{\prime \prime }(0)=-\frac{1}{2}(0)(0.332)=0.$ Step 2 — interpret karo: $f^{\prime \prime \prime }$ velocity gradient ki curvature hai. $f^{\prime \prime \prime }(0)=0$ ka matlab hai ki profile ka inflection-free, zero-jerk start hai — shear $f^{\prime \prime }$ wall par momentarily flat hai. Wall ke paas velocity $f^{\prime }\approx f^{\prime \prime }(0)\eta =0.332\eta$ (linear) ki tarah behave karti hai aur next correction sirf ${\eta }^4$ order par hai (kyunki ${\eta }^2$ aur ${\eta }^3$ coefficients, $\propto f^{\prime \prime \prime }(0)$, vanish ho jaate hain). Step 3 — parabola kyun fail hoti hai: ek parabola $u/U_{\infty }=a\eta +b{\eta }^2$ ka ek non-zero ${\eta }^2$ term hota hai, yaani non-zero curvature $f^{\prime \prime }$ jo constant rehti hai, jo $f^{\prime \prime \prime }(0)=0$ aur true flattening ke against hai. Parabola sirf ek crude integral-method approximation hai; exact Blasius wall region remarkably high order tak linear hai. Ye mastery hai: ODE khud tumhe $f^{\prime \prime \prime }(0)=0$ free mein deta hai — koi numerics nahi chahiye — aur ye profile shape pin kar deta hai.

Jab $x\to 0^{+}$ (leading edge):

• $\delta =5.0\sqrt{\nu x/U_{\infty }}\to 0$ (layer ki edge par zero thickness hai). ✓ sensible.
• ${\tau }_w=0.332\mu U_{\infty }\sqrt{U_{\infty }/(\nu x)}\to \infty$ (wall shear diverge karta hai). Ye leading-edge singularity hai.
• $C_f=0.664/\sqrt{R{e}_x}\to \infty$ bhi (kyunki $R{e}_x\to 0$). Kaun si prediction break hoti hai: infinite ${\tau }_w$ unphysical hai — koi bhi wall infinite drag feel nahi karta. Theory kyun forbid karti hai: Prandtl reduction ne assume kiya tha ki $\delta \ll x$ (yahi allow kiya tha ${\partial }^{2}u/\partial x^2$ drop karne ko). $x\to 0$ par, $\delta$ aur $x$ comparable ho jaate hain, isliye wo founding approximation khatam ho jaati hai aur boundary-layer equations simply invalid ho jaate hain. True tip flow ko full Navier–Stokes Equations ki zaroorat hai (aur dekho Prandtl Boundary Layer Theory jahan approximation hold karti hai). Jab $x\to \infty$: $\delta \to \infty$ lekin slowly ($\propto \sqrt{x}$), jabki ${\tau }_w\to 0$ aur $C_f\to 0$ — friction kuch nahi ki taraf fade hoti hai. Lekin $R{e}_x=U_{\infty }x/\nu$ badhta rehta hai aur eventually $\sim 5\times 10^5$ exceed karta hai, isliye flow "infinity" se pehle hi turbulent transition kar leti hai. Blasius isliye ek finite laminar window hai, sabhi $x$ ke liye valid formula nahi — window ke baad Turbulent Boundary Layer scaling $\delta /x\approx 0.37R{e}_x^{-1/5}$ use karo.

$\delta \propto x^{1/2}{U}_{\infty }^{-1/2}$ · ${\tau }_w\propto x^{-1/2}{U}_{\infty }^{3/2}$ · $C_f\propto R{e}_x^{-1/2}$ · $F_D\propto L^{1/2}{U}_{\infty }^{3/2}$ · aur constants double hote hain: $0.332\to 0.664\to 1.328$.