A Numerical Solution to the Blasius Boundary Layer Equation

Table of Content

• Overview
• Derivation
• Implementation summary
• How to Compile

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Overview

A Blasius boundary layer (named after Paul Richard Heinrich Blasius) describes the steady two-dimensional laminar boundary layer that forms on a semi-infinite plate which is held parallel to a constant unidirectional flow.

Blasius obtained an exact solution for the Boundary Layer Equations by assuming a zero-pressure gradient.

This numerical solution considers the Blasius Exact Solution for laminar boundary layer flow over a flat plate.
The similarity variables are given as:

$$ \begin{align*} & \eta ={\dfrac {y}{\delta (x)}}=y{\sqrt {\dfrac {U}{\nu x}}},\\ \\ & \psi ={\sqrt {U \nu x}}f(\eta )\\ \end{align*} $$

      Where:
$\quad \quad \quad \quad \delta (x)$ is the boundary layer thickness,
$\quad \quad \quad \quad \psi$ is the stream function, and
$\quad \quad \quad \quad U$ denotes the freestream velocity, and $u$ the local velocity.

The Blasius similarity equation is given by: $$ff'' + 2f''' = 0$$ The Boundary conditions are given as:

$$ \begin{align*} & f(0) = 0,\\ \\ & f'(0) = 0,\\ \\ & f'(\infty) = 1. \end{align*} $$

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Derivation

The momentum equation for a hydrodynamic boundary layer over a flat plate is given as:

$$ u \dfrac {\partial u}{\partial x} + v \dfrac {\partial u}{\partial x} = \nu \dfrac {\partial ^2 u}{\partial y^2}\\ \tag{1} $$

The velocity distribution in the boundary layer can be obtained by solving the equation above. The following boundary conditions are satisfied:

• $(i)$   At $y = 0, \quad u = 0,$

• $(ii)$ At $y = 0, \quad v = 0,$

• $(iii)$ At $y = \infty , \quad u = U.$

The Blasius technique for an exact solution for the hydrodynamic layer lies in the conservation of the momentum equation and the continuity equation:

$$ \dfrac {\partial u}{\partial x} + \dfrac {\partial v}{\partial y}=0\\ \tag{2} $$

As $u\ge v,$

$$ \rho u\dfrac {\partial u}{\partial x}=\mu \dfrac {\partial ^2 u}{\partial y^2} $$

      since $\nu = \dfrac {\mu} {\rho}$.

Also, as $u \propto U$ and $\dfrac {\partial u}{\partial x} \propto \dfrac {U}{L},$ along a plate length $L,$ therefore,

$$\dfrac {\rho U^2}{L} = \mu \dfrac {U}{\delta ^2}$$

$$ \therefore \delta = \sqrt {\dfrac {\mu L}{\rho U}} = \sqrt {\dfrac {\nu L}{U}} = \sqrt {\dfrac {\nu x}{U}}\\ \tag{3} $$

The dimensionless velocity $\dfrac{u}{U}$ can be expressed at any location $x$ as a function of the dimensionless distance from the wall $\dfrac {y}{\delta}$.

$$ \dfrac {u}{U}=f \Biggl(\dfrac {y}{\delta} \Biggr)\\ \tag{4} $$

Substituting the value of $\delta$ from eqn. $(3)$ in eqn. $(4)$, we obtain,

$$ \dfrac {u}{U}=f\Biggl[\dfrac {y}{\sqrt {x}} \sqrt {\dfrac {U}{\nu}}\Biggr] = f(\eta)\\ \tag{5} $$

Where, $\quad \eta = y \sqrt {\dfrac {U}{\nu x}}$ denotes the stretching factor

The stream function, $\psi$, can be defined such that,

$$ \dfrac {\psi}{U} = \Biggl [\sqrt {\dfrac {\nu x}{U}} \Biggr] f(\eta)\\ \tag{6} $$

Or,

$$ \psi = \sqrt {\nu x U} f(\eta)\\ \tag{7} $$

The partial differential of the stream function with respect to $x$ gives the velocity in the $Y$-direction (generally taken as negative), and its partial differential with respect to $y$ gives the velocity in the $X$-direction:

$$ u = \dfrac {\partial \psi}{\partial y}; \quad \quad v = \dfrac {\partial \psi}{\partial x} $$

$$ \therefore \dfrac {\partial \psi}{\partial y} = \dfrac {\partial \psi}{\partial \eta} \times \dfrac {\partial \eta}{\partial y} = \dfrac {\partial}{\partial \eta} \Biggl[U\sqrt{\dfrac {\nu x}{U}}f(\eta) \Biggr] \times \dfrac {\partial}{\partial y} \Biggl[y \sqrt{\dfrac {U}{\nu x}} \Biggr] $$

Or,

$$ u=U\sqrt {\dfrac {\nu x}{U}} \dfrac {df}{d\eta}\Biggl [\sqrt {\dfrac {U}{\nu x}} \Biggr] = U\dfrac {df}{d\eta}\\ \tag {8} $$

Here $f$ is abbreviated as $f(\eta)$

$$ \therefore \dfrac {\partial u}{\partial x} = U\dfrac {\partial}{\partial x} \left [\dfrac {df}{d\eta} \right] = U\dfrac {\partial}{\partial \eta} \left [\dfrac {df}{d\eta} \right] \dfrac {\partial}{\partial x} =-U\dfrac{d^2 f}{d \eta ^2} \cdot \dfrac {1}{2 x} \cdot y \sqrt{\dfrac {U}{\nu x}} $$

Or,

$$ \dfrac {\partial u}{\partial x} = -U \dfrac {\eta}{2x} \dfrac {d^2 f}{d\eta ^2}\\ \tag{9} $$

Now,

$$ \dfrac {\partial u}{\partial y}={U}{\dfrac {\partial}{\partial y}}\Biggl (\dfrac {df}{d\eta}\Biggr) ={U}{\dfrac {\partial}{\partial \eta}}\Biggl(\dfrac {\partial f}{\partial \eta}\Biggr) \dfrac {\partial \eta}{\partial y} \cdot {U}{\sqrt {\dfrac {U}{\nu x}}} \cdot \dfrac {d^2 f}{d\eta ^2} \\ \tag{10} $$

Similarly,

$$ \dfrac {\partial ^2 u}{\partial y^2} = \dfrac {U^2}{\nu x} \cdot \dfrac {d^3 f}{d\eta ^3}\\ \tag{11} $$

Again,

$$ \begin{align*} & v = - \dfrac {\partial \psi}{\partial x} = - \dfrac {\partial}{\partial x} \Biggl [\sqrt {\dfrac {\nu x}{U}}f(\eta) \Biggr] = -U\sqrt {\dfrac {\nu}{U}} \cdot \dfrac {\partial}{\partial x}[\sqrt {x}f(\eta)] \\ \\ & =-U\sqrt {\dfrac {\nu}{U}} \cdot \Biggl[\sqrt {x}\dfrac {\partial f}{\partial x} + f \cdot \dfrac {\partial}{\partial x}(\sqrt {x})\Biggr]\\ \\ & =-U\sqrt {\dfrac {\nu}{U}} \cdot \Biggl[\sqrt {x}\dfrac {\partial f}{\partial \eta} \cdot \dfrac {\partial \eta}{\partial x} + \dfrac {f}{2\sqrt{x}}\Biggr] \\ \\ & =-U\sqrt {\dfrac {\nu}{U}} \cdot \Biggl[\sqrt {x}\dfrac {\partial f}{\partial \eta} \cdot \dfrac {\partial}{\partial x} \Biggl(y\sqrt {\dfrac {U}{\nu x}}\Biggr) + \dfrac {f}{2\sqrt {x}} \Biggr] \\ \\ & =-U\sqrt {\dfrac {\nu}{U}} \cdot \Biggl[\sqrt {x}\dfrac {\partial f}{\partial \eta} \cdot y\sqrt {\dfrac {U}{\nu}} \Biggl(-\dfrac {1}{2} \Biggr) \dfrac {1}{x\sqrt {x}} + \dfrac {f}{2\sqrt {f}} \Biggr] \\ \\ & =-\sqrt {U\nu} \dfrac {1}{2\sqrt {x}} \Biggl[-\dfrac {df}{d\eta}y \sqrt {\dfrac {U}{\nu x}} + f \Biggr] \\ \\ & =-\dfrac {1}{2} \sqrt {\dfrac {Ux}{x}} \Biggl[-\eta \dfrac {df}{d\eta} + f \Biggr] \\ \\ \end{align*} $$

$$ \begin{align} & v = \dfrac {1}{2} \sqrt {\dfrac {U\nu}{x}} \Biggl[ -\eta \dfrac {df}{d\eta} -f \Biggr] \tag{12} \end{align} $$

Inserting the values of $u, \dfrac {\partial u}{\partial x}, \dfrac {\partial u}{\partial y}$, $\dfrac {\partial ^2 u}{\partial x^2}$ and $v$ from eqns. $(8), (9),$ $(10),$ and $(11)$ in eqn $(1)$, we get

$$ -U\dfrac {df}{d\eta} \cdot U \dfrac {\eta}{zx} \cdot \dfrac {d^2 f}{d\eta ^2} + \dfrac {1}{2} \sqrt {\dfrac {U\nu}{x}} \Biggl[\eta \dfrac {df}{d\eta} -f \Biggr] \cdot U\sqrt {\dfrac {U}{\nu x}} \dfrac {d^2 f}{d\eta ^2} = \nu \dfrac {U^2}{\nu x} \cdot \dfrac {d^3 f}{d \eta ^3} $$

Or,

$$ -\dfrac {1}{2x} \cdot U^2 \dfrac {df}{d\eta} \cdot \eta \dfrac {d^2 f}{d\eta ^2} + \dfrac {1}{2x}U^2 \dfrac {d^2 f}{d\eta ^2} \Biggl[\eta \dfrac {df}{d\eta} - f \Biggr] =\dfrac {U^2}{x} \cdot \dfrac {d^3 f}{d\eta ^3} $$

Or,

$$ -\dfrac {1}{2x} U^2 \dfrac {d^2 f}{d\eta ^2} \Biggl[\eta \dfrac {df}{d\eta}- \eta \dfrac {df}{d\eta} + f \Biggr] = \dfrac {U^2}{x} \cdot \dfrac {d^3 f}{d\eta ^3} $$

Or,

$$ -\dfrac {1}{2x}U^2 \dfrac {d^2 f}{d\eta ^2} \cdot f = \dfrac {U^2}{x} \cdot \dfrac {d^3 f}{d\eta ^3} $$

Or,

$$ 2 \dfrac {d^3 f}{d\eta ^3} + f \dfrac {d^2 f}{d\eta ^2} = 0 $$

Or,

$$ 2f''' + f'' = 0\\ \tag{13} $$

The physical and transformed boundary conditions are:

Physical boundary conditions	Transformed boundary conditions
at $y = 0, \quad u = 0$	at $\eta = 0, \quad \dfrac {df}{d\eta} = f'= 0$
at $y = 0, \quad v = 0$	at $\eta = 0, \quad f = 0$
at $y = \infty, \quad u = U$	at $\eta = \infty, \quad \dfrac {df}{d\eta} = f'= 1$

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Implementation summary

The Blasius equation (eqn. $13$) is decomposed to a system of coupled 3 linear ODEs:

$$ f' = f_1 \\ $$

$$ f_1' = f_2 \\ $$

$$ f_2' = - \dfrac {f \cdot f_2}{2} $$

This system of equation is then solved numerically using the fourth-order Runge-Kutta (RK4) numerical scheme.

A random number is used as the initial guess for $f''(0)$ in this project. The guss is then adjusted until the boundary condition $f'(\infty) = 1$ is satisfied (with a tolerance of ${10}^{-12}$). The data is written to a file, which can be accessed and evaluated.

Running the solution

Manual compilation

The code in this project requires at least C++17 to run. Compile with flag -std=c++17.

Using the shell scripts (Linux/macOS)

You can run the run.sh script from the project root to run the solution, or with CMake and Ninja build using run_with_cmake.sh. The script requires CMakeversion $\ge$ 3.20 and a decent C++ compiler.

Both scripts compile and run the solution in the /solution/ directory in the working directory, but should be executed from the project root.

clean.sh can be used to clean the compiled files and results.