Blasius solution — exact laminar boundary layer solution

2.2.22 · Physics › Fluid Mechanics

WHY do we even need this?

WHY hum poore Navier–Stokes equations solve nahi kar sakte? Kyunki woh brutal nonlinear PDEs hain. Prandtl ki insight (1904): ek thin layer ke andar, equations bahut zyada simplify ho jaate hain. Blasius (1908, unka PhD) ne phir us simplified set ko flat plate ke liye solve kiya.

HOW: deriving the boundary-layer equations

Start karo 2D steady incompressible flow se, koi pressure gradient nahi (flat plate, $\partial p/\partial x=0$ bahar):

Continuity: $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

x-momentum (Navier–Stokes): $u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}\right)$

Prandtl boundary-layer equation: $u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{{\partial }^{2}u}{\partial y^2}$

boundary conditions ke saath: $u(x,0)=0,v(x,0)=0,u(x,\infty )=U_{\infty }$

HOW: the similarity transform (the heart of Blasius)

Similarity variable aur ek dimensionless stream function $f$ define karo: $\boxed{\eta =y\sqrt{\frac{U_{\infty }}{\nu x}}},\psi =\sqrt{\nu x{U}_{\infty }}f(\eta )$

Kyunki $u=\partial \psi /\partial y$ aur $v=-\partial \psi /\partial x$ (yeh automatically continuity satisfy karta hai — WHY hum stream function use karte hain), hum compute karte hain:

$u=U_{\infty }{f}^{\prime }(\eta )$

Yeh step kyun? $\partial \psi /\partial y=\sqrt{\nu x{U}_{\infty }}{f}^{\prime }(\eta )\cdot \partial \eta /\partial y=\sqrt{\nu x{U}_{\infty }}{f}^{\prime }\cdot \sqrt{U_{\infty }/(\nu x)}=U_{\infty }{f}^{\prime }$.

$v=\frac{1}{2}\sqrt{\frac{\nu U_{\infty }}{x}}(\eta f^{\prime }-f)$

$u,v$ aur unke derivatives ko Prandtl equation mein substitute karne par, saare $x$ aur $U_{\infty }$ cancel ho jaate hain, aur ek pure ODE bachti hai:

Yeh ek 3rd-order nonlinear ODE hai jiska koi closed form nahi hai — numerically solve kiya jaata hai. Key numerical result: $f^{\prime \prime }(0)=0.332$

HOW: extracting physical quantities

Yeh step kyun ($0.664$): $C_f=\frac{0.332\mu U_{\infty }\sqrt{U_{\infty }/\nu x}}{\frac{1}{2}\rho U_{\infty }^2}=\frac{2(0.332)}{\sqrt{U_{\infty }x/\nu }}=\frac{0.664}{\sqrt{R{e}_x}}$.

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Connections

• Prandtl Boundary Layer Theory — woh simplification jo Blasius solve karta hai.
• Navier–Stokes Equations — parent equations.
• Reynolds Number — laminar vs turbulent determine karta hai.
• Stream Function and Vorticity — $\psi$ trick.
• Skin Friction Drag — ${\tau }_w$ ka application.
• Similarity Solutions in PDEs — mathematical method.
• Turbulent Boundary Layer — $x^{4/5}$ contrast case.

Concept Map