Fluid Mechanics for Mechanical Engineers/Boundary Layer Approximation

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Boundary Layer
Boundary layer is a region where viscous force is relatively high compared to  inertia, force due to pressure gradient, gravitational or electromagnetic force. Viscous force appears when there is a velocity gradient in the flow. Velocity gradients occurs usually next to the walls, where the fluid takes the velocity of the wall (no-slip condition) and in the mixing regions where flow has high velocity gradients, such as (jet flows).

When we look to the flow around an airfoil in a wind tunnel, we can clearly see the how the velocity gradient occurs on the airfoil.

Boundary Layer Approximation
Prandtl (1905) had the following hypotheses: Now consider the figure shown below:
 * For small values of viscosity, viscous forces are only important close to the solid boundaries (within boundary layer) where no-slip condition has to be satisfied. And everywhere else they can be neglected.
 * Thickness of boundary layer approaches to zero as the viscosity becomes smaller.

As it is shown in the figure we are using a body-fitted coordinate system which means x direction is always tangent to the surface of the body and y direction is always normal to the surface of the body. Note that edge velocity is shown by $$ u(x) $$ and the boundary layer thickness is represented by $$ \delta(x) $$ which means they are varying along x direction. Also let $$ \overline{\delta} $$ be the average boundary layer thickness along the surface of length $$ L $$.

A measure for $$\overline{\delta}$$ can be obtained by an order of magnitude analysis using momentum equation in x direction.


 * $$ u{\partial u \over\partial x} + v{\partial u \over\partial y} = -{1\over\rho} {\partial p \over\partial x} + {\nu} \left({\partial^2 u \over\partial x^2 } + {\partial^2 u \over \partial y^2}\right) $$

Note that Cartesian form of the equation is only valid when $$ \overline{\delta} << R $$, where $$ R $$ is the local radius of curvature of the body. Let $$ u_\infty $$ be the characteristic magnitude of $$ u $$. In other words $$ u_c = u_\infty $$. $$ L $$ is the distance along which $$ u $$ changes appreciably, and it can be the length of the body. An order of magnitude analysis shows that $$ {\partial u \over \partial x} \sim {\frac{u_\infty}{L}} $$. Hence the first advection term in the momentum equation can be written as:


 * $$ u{\partial u \over \partial x} \sim \frac{u_\infty^2}{L} $$
 * The second advection term is then:
 * $$ v{\partial u \over \partial y} \sim v_c\frac{u_\infty}{\overline{\delta}} $$

where $$v_c$$ is the characteristic velocity normal to the wall in the boundary layer, which occurs due to fluid displacement caused by the walls. Doing the same analysis for the viscous terms in momentum equation would give us:


 * $$ \nu{\partial^2 u \over \partial y^2} \sim \nu \frac{u_\infty}{\overline{\delta}^2} $$ and $$ \nu {\partial^2 u \over \partial x^2} \sim \nu \frac{u_\infty}{L^2} $$

for which $$ \nu \frac{u_\infty}{L^2}<< \nu \frac{u_\infty}{\overline{\delta}^2} $$.

Within the boundary layer inertia (advection terms) and the viscous forces should be in the same order. In other words


 * $$ \frac{u_\infty^2}{L} \sim \nu \frac{u_\infty}{\overline{\delta}^2} $$
 * $$ \overline{\delta} \sim \sqrt{\frac{\nu L}{u_\infty}} $$
 * $$ \frac{\overline{\delta}}{L} \sim \sqrt{\frac{\nu}{u_\infty L}}=\frac{1}{\sqrt{Re}} $$

As the final equation shows, the higher the $ Re $ the thinner would be the boundary layer. Now, we can simplify the conservation equations within the boundary layer. The basic idea is to assume that the gradients across the boundary layer is much larger than the gradients along layer in the main flow direction.


 * $$ {\partial \over \partial x} << {\partial \over \partial y} \;, {\partial^2 \over \partial x^2} << {\partial^2 \over \partial y^2} $$

Looking at continuity equation one can find out an estimate for $$ v_c $$ by order magnitude analysis.


 * $$ {\partial u \over \partial x} + {\partial v \over \partial y} = 0 $$

Since $$ {\partial u \over \partial x} \sim \frac{u_\infty}{L} $$, so  $${\partial v \over \partial y} \sim \frac{v_c}{\overline{\delta}}$$

We know that $$ u >> v $$ but at the same time we know that $$ {\partial \over \partial x} << {\partial \over \partial y} $$. Hence we can say that both terms can have the same order of magnitude.In addition to that the continuity equation dictates also that both partial derivative terms should have the same value with opposite sign. Hence,


 * $$ \frac{u_\infty}{L} \sim \frac{v_c}{\overline{\delta}} $$
 * $$ v_c \sim \overline{\delta}{\frac{u_\infty}{L}}$$ since $$ \frac{\overline{\delta}}{L} \sim \frac{1}{\sqrt{Re}}$$
 * $$ v_c \sim u_\infty{\frac{1}{\sqrt{Re}}} $$
 * Finally, we should find an estimation for the pressure gradient. Experimental data at high $ Re $ numbers show that force due to pressure gradient is in the order of inertia.
 * $$ {\partial P \over \partial x} \sim \rho u{\partial u \over \partial x}\sim \rho{\frac{u_\infty^2}{L}} $$
 * Moreover, it is also true that in the boundary layer pressure difference w.r.t. to ambient pressure scales with the dynamic pressure
 * $$ {\partial P \over \partial x} \sim \frac{P - P_\infty}{L} \sim \rho{\frac{u_\infty^2}{L}} \longrightarrow P - P_\infty \sim \rho u_\infty^2 $$

Hence the order of magnitude for each term in the governing equations are as follows:

$$ \underbrace_{U_\infty/L} + \underbrace_{U_\infty/L} = 0 $$

$$ \underbrace{u{\partial u \over\partial x}}_{U_\infty^2/L} + \underbrace{v{\partial u \over\partial y}}_{U_\infty^2/L} = \underbrace{-{1\over\rho} {\partial p \over\partial x}}_{U_\infty^2/L} + \underbrace{{\nu} {\partial^2 u \over\partial x^2 }}_{\nu U_\infty/L^2} + \underbrace{{\nu} {\partial^2 u \over\partial y^2 }}_{\nu U_\infty/\overline{\delta}^2} $$

Both equations can be non-dimensionalized by $$ U_\infty/L $$ and $$ U_\infty^2/L  $$, respectively. Denoting the non-dimensional parameters with a *, the non-dimensional form of governing equations becomes:

$$ \underbrace_{1} + \underbrace_{1} = 0 $$

$$ \underbrace{u^*{\partial u^* \over\partial x^*}}_{1} + \underbrace{v^*{\partial u^* \over\partial y^*}}_{1} = \underbrace{-{1\over \rho^*} {\partial p^* \over\partial x^*}}_{1} + \underbrace{{\nu^*} {\partial^2 u^* \over\partial {x^*}^2 }}_{\nu / U_\infty L=1/Re_L} + \underbrace{{\nu^*} {\partial^2 u^* \over\partial {y^{*}}^2 }}_{\nu L / U_\infty \overline{\delta}^2= (L/\overline{\delta})^2 /Re_L=1}  $$

Note that, since $$ \frac{\overline{\delta}}{L} \sim \frac{1}{\sqrt{Re_L}}$$, the second viscous term $$ \left(\frac{L}{\overline{\delta}}\right)^2 \sim Re_L$$ and therefore its order of magnitude becomes 1.

For $$ Re_L \rightarrow \infty $$ and $$ \rho^*=\frac{\rho}{\rho} =1 $$ and $$ \nu^*=\frac{\nu}{\nu} =1 $$ our non-dimensional conservation equations would simplify to


 * $$ u^* \frac{\partial u^*}{\partial x^*} + v^* \frac{\partial u^*}{\partial y^*} = - \frac{\partial P^*}{\partial x^*} + \frac{\partial^2 u^*}{\partial {y^*}^2} $$


 * $$ - \frac{\partial P^*}{\partial y^*} = 0 $$


 * $$ \frac{\partial u^*}{\partial x^*} + \frac{\partial v^*}{\partial y^*} = 0 $$

Those are non-dimensional boundary layer equations. Going back to the dimensional variables


 * $$ u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = - \frac{1}{\rho} \frac{\partial P}{\partial x} + \nu \frac{\partial^2 u}{\partial y^2} $$


 * $$ - \frac{\partial P}{\partial y} = 0 $$


 * $$ {\partial u \over \partial x} + {\partial v \over \partial y} = 0 $$

These equations are parabolic although the original Navier-Stokes equation is elliptic. Note that the second equation only says that the pressure is not dependent to y direction and it is only a function of $$ x $$. Therefore we have 3 unknowns and 2 equations. Moreover, pressure at boundary layer is equal to the pressure at the edge which can be found from solving Euler equations for the outer portion of the fluid. In other words, pressure gradient along $$x$$ direction can be found by using Euler equations along the edge of the boundary layer.


 * $$ - \frac{dP}{dx} = u_e \frac{du_e}{dx} \longrightarrow P + \rho \frac{u_e^2}{2} = const. $$

The boundary conditions of the problem are as follows


 * $$ u(x, 0) = 0 $$
 * $$ v(x, 0) = 0 $$
 * $$ u(x, \infty) = u(x) \longrightarrow edge \; velocity $$
 * $$ u(x_0, y) = u_{in}(y) \longrightarrow initial \; profile $$

Solution of this problem would be in an iterative manner in which one should guess the thickness of the boundary layer and then solve the Euler equation to find the pressure distribution over length $$ L $$, then by finding the pressure distribution one can find the velocity distribution within boundary layer and by having velocity distribution one can find the thickness of the boundary layer to compare it with the guessed one. If they match we have reached to the final solution if not the new thickness should be taken as next guess and the procedure would repeat the same way as before.

Different Measures of Boundary Layer Thickness
There are 3 different definitions for boundary layer listed below:


 * 1.  $$ \delta_{99} $$  where $$ u = 0.99 u_e $$. Based on this definition the boundary layer is going to be defined where the velocity of the fluid is 0.99 times of the edge velocity (upstream velocity).


 * 2. Displacement thickness ($$ \delta^* $$):The name of this definition comes from the displacement of the streamlines at the presence of walls contacting with the fluid.


 * $$ u_\infty h = \int_{0}^{h + \delta^*} u dy $$
 * $$ u_\infty h = \int_{0}^{h} u dy + u_\infty \delta^* $$
 * $$ u_\infty \left ({h} - \delta^* \right) = \int_{0}^{h} u dy $$

Solving for $$ \delta^* $$


 * $$ \delta^* = \int_{0}^{h} \frac{u_\infty - u}{u_\infty} dy = \int_{0}^{h} \left (1 - \frac{u}{u_\infty}\right) dy $$

As $$ h \rightarrow \infty \; : \delta^* = \int_{0}^{\infty} \left(1 - \frac{u}{u_\infty}\right) dy $$

There are some practical usages of using $$ \delta^* $$
 * Design of ducts and wind tunnels
 * Assume frictionless flow and dimension the tunnel
 * Estimate $$ {dp \over dx} $$ along the tunnel
 * Calculate $$ \delta^* $$
 * Enlarge the duct by $$ \delta^* $$ to obtain the same mass flow rate


 * Finding $$ {dp \over dx} $$ at the edge of the boundary layer
 * Neglect boundary layer (frictionless flow assumption) and calculate $$ {dp \over dx} $$ over the body surface
 * Solve boundary layer by $$ {dp \over dx} $$
 * Calculate $$ \delta^* $$
 * Displace the body by $$ \delta^* $$
 * Calculate $$ {dp \over dx} $$ again by frictionless flow assumption
 * Repeat steps from 2 to 5 until a convergence criterion is satisfied


 * 3.  Momentum thickness  ($$ \theta $$):

It is defined such that $$ \rho u^2 \theta $$ is the momentum loss due to presence of boundary layer. Hence,


 * $$ \rho u_\infty^2 \theta=\rho u_\infty^2 h - \int_{0}^{h + \delta^*} \rho u^2 dy $$
 * where
 * $$ \int_{0}^{h + \delta^*} \rho u^2 dy = \int_{0}^{h} \rho u^2 dy + {\rho \delta^* u_\infty^2 } $$


 * $$ \rho u_\infty^2 \theta=\rho u_\infty^2 h - \int_{0}^{h} \rho u^2 dy - \rho \delta^* u_\infty^2 $$


 * $$ u_\infty^2 \theta=\int_{0}^{h} \left (u_\infty^2 - u^2 \right) dy - u_\infty^2 \int_{0}^{h} \left(1 - \frac{u}{u_\infty} \right) dy $$

From which


 * $$ \theta = \int_{0}^{h} \frac{u}{u_\infty} \left(1 - \frac{u}{u_\infty} \right) dy $$

For $$ h \rightarrow \infty $$


 * $$ \theta = \int_{0}^{\infty} \frac{u}{u_\infty} \left(1 - \frac{u}{u_\infty} \right) dy $$

Flat Plate Boundary Layer
Flat plate is a special case where the edge velocity is constant ($$ u_e(x) = u_\infty = const. $$). Hence, as a consequence of Euler equation $$ {dp \over dx} = 0 $$ along the flat plate. The boundary layer equations reduces to


 * $$ u{\partial u \over \partial x} + v{\partial u \over \partial y} = \nu{\partial^2 u \over \partial y^2} $$


 * $$ {\partial u \over \partial x} + {\partial v \over \partial y} = 0 $$

Boundary conditions are as follows:


 * $$ u(x > 0,y = 0 ) = 0 $$


 * $$ v(x > 0, y = 0) = 0 $$


 * $$ u(x > 0, y = \delta(x)) = u_\infty $$


 * $$ u(x = x_0, y) = u(y) \; \longrightarrow \; given \; as \;initial \; condition $$

Now we have two equations and two unknowns which means that our system of equations is closed and can be solved. One way to solve it is through using the similarity solution. We consider the non-dimensional velocity profile as a function of one variable ($$ \eta $$).


 * $$ \frac{u}{u_\infty} = g(\eta) $$ where $$\eta = \frac{y}{\delta(x)}$$

Hence,

$$ \frac{u}{u_\infty} = g(\frac{y}{\delta(x)}) $$

Also based on our earlier findings ($$ \delta \sim \sqrt{\frac{\nu x}{u}} $$) we conclude that $$ \delta(x) = \delta(u_\infty, \nu, x) $$

Since the problem is two dimensional, it is easier to work with stream function ($$ \psi $$). It is a function whose isolines represents the streamlines and they are perpendicular to the velocity potential ($$ \phi $$). So


 * $$ u = \frac{\partial \psi}{\partial y}, v =- \frac{\partial \psi}{\partial x} $$

Note that, inserting those equalities into the continuity equation will result in a zero sum. In other words, stream function satisfies the continuity equation per definition.

At $$ x = const. $$ location $$ \psi = \int_{0}^{y} u dy $$ represents the flow rate. Using the similarity variable $$\eta$$ would give us


 * $$ \psi = \delta(x) u_\infty \int_{0}^{\eta} g(\eta) d\eta $$

Non-dimensional form or the similarity form of $$ \psi $$ function is


 * $$ \frac{\psi}{u_\infty \delta(x)} = f(\eta) = \int_{0}^{\eta} g(\eta) d\eta \longrightarrow g(\eta) = \frac{df}{d\eta} $$

After inserting the velocities written in terms of stream function into the momentum equation, momentum equations would become


 * $$ \frac{\partial \psi}{\partial y} \frac{\partial^2 \psi}{\partial x \partial y} - \frac{\partial \psi}{\partial x} \frac{\partial^2 \psi}{\partial y^2} = \nu{\frac{\partial^3 \psi}{\partial y^3}} $$

Each term can be written as a function of $$\eta$$. From here onwards, for simplicity $$\delta=\delta(x), g=g(\eta) $$ and $$f=f(\eta)$$ are used

Since $$ \psi = u_\infty \delta f $$


 * $$ \frac{\partial \psi}{\partial x} = u_\infty \frac{\partial}{\partial x} \left( \delta f \right) = u_\infty \left( \delta \frac{\partial f}{\partial x} + f \frac{\partial \delta}{\partial x}\right) $$

Note that to find $$ \frac{\partial f}{\partial x} $$ we should use chain rule in differentiation.


 * $$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial \eta} \frac{\partial \eta}{\partial x} $$ where $$ \eta = \frac{y}{\delta} $$
 * using the quotient rule of differentiation:
 * $$ \frac{\partial \eta}{\partial x} = \frac{\frac{\partial y}{\partial x} \delta - y \frac{\partial \delta}{\partial x}}{\delta^2} =-\frac{y\frac{\partial \delta}{\partial x}}{\delta^2} =-\frac{y}{\delta^2} \frac{d \delta}{d x} $$

Combining derived relations for $$ \frac{\partial f}{\partial x} $$ and $$ \frac{\partial \eta}{\partial x} $$ we would obtain


 * $$ \frac{\partial \psi}{\partial x} = u_\infty \frac{d\delta}{dx} \left(f - \eta f^' \right) $$

Similarly


 * $$ \frac{\partial \psi}{\partial y} = u_\infty \frac{\partial }{\partial y} \left(\delta f \right) = u_\infty \left(f \frac{\partial \delta}{\partial y} + \delta \frac{\partial f}{\partial y}\right) $$

where $$\frac{\partial \delta}{\partial y} = 0$$ and


 * $$ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial \eta} \frac{\partial \eta}{\partial y} $$

since $$ \frac{\partial \eta}{\partial y}=\frac{1}{\delta} $$

$$ \frac{\partial f}{\partial y} = \frac{1}{\delta} \frac{\partial f}{\partial \eta}=\frac{1}{\delta}f^' $$

Combining derived relations for $$ \frac{\partial \psi}{\partial y} $$ and we would obtain


 * $$ \frac{\partial \psi}{\partial y} = u_\infty \delta \frac{1}{\delta} \frac{\partial f}{\partial \eta} = u_\infty f^' $$

Using this relation, the relationships for $$ \frac{\partial^2 \psi}{\partial y^2} $$ and $$ \frac{\partial^3 \psi}{\partial y^3} $$ can be written as:


 * $$ \frac{\partial^2 \psi}{\partial y^2} =\frac{u_\infty}{\delta}f^{''} $$ ,
 * $$ \frac{\partial^3 \psi}{\partial y^3} =\frac{ u_\infty}{\delta^2}f^{'''} $$

$$ \frac{\partial^2 \psi}{\partial x \partial y} $$ should be derived also:

$$ \frac{\partial^2 \psi}{\partial x \partial y} = \frac{\partial }{\partial y} \frac{\partial \psi}{\partial x} $$

$$ \frac{\partial^2 \psi}{\partial x \partial y} = \frac{\partial }{\partial y} \left( u_\infty \frac{d\delta}{dx}\left(f - \eta f^'\right)\right) $$

$$ \frac{\partial^2 \psi}{\partial x \partial y} = u_\infty \frac{d\delta}{dx} \left( \frac{\partial f }{\partial y} - \frac{\partial \eta f^' }{\partial y} \right)= u_\infty \frac{d\delta}{dx} \left( \frac{\partial f }{\partial y} - \eta \frac{\partial f^' }{\partial y}- f^'\frac{\partial \eta   }{\partial y}  \right) $$

Using the equalities found already $$ \frac{\partial f}{\partial y} =\frac{1}{\delta}f^' $$ and $$ \frac{\partial \eta}{\partial y}=\frac{1}{\delta} $$ and the relation $$ \eta \frac{\partial f^' }{\partial y}=\eta \frac{\partial  f^' }{\partial \eta}\frac{\partial  \eta}{\partial y}=\frac{\eta }{\delta}f^{''} $$

$$ \frac{\partial^2 \psi}{\partial x \partial y} = - u_\infty \frac{d\delta}{dx} \frac{\eta}{\delta}f^{''} $$

Finally substituting partial derivatives of $$ \psi $$ into the momentum equation would give us


 * $$ -\left(\frac{u_\infty \delta}{\nu} \frac{d\delta}{dx}\right) f f^{} = f^{'} $$

It shows that if we are looking for a x-independent solution for the similarity variable $$ f $$ then


 * $$ \frac{u_\infty \delta}{\nu} \frac{d\delta}{dx} = const. $$

Note that if we choose that constant as $$ \frac{1}{2} $$ then $$ \delta = \sqrt{\frac{\nu x}{u_\infty}} $$

That way our equation would become


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

Noting that $$ g(\eta) = \frac{u}{u_\infty} $$, $$ f(\eta) = \int_{0}^{\eta} g(\eta) d\eta $$ and $$ \eta=\frac{y}{\delta}  $$

Then the boundary conditions are as follows


 * $$ f(0) = 0 \longrightarrow \; \text{since} \; \psi(0) = 0 $$
 * $$ f^'(0) = g(0)=0 \longrightarrow \; \text{since} \; \psi(0) = 0 $$ (since $$ u(0)=v(0)=0 $$ at the wall)

In order to solve the above differential equation, one need to know $$ f^{''}(0) $$, which is not known. However, instead

$$ f^'(\infty) = g(\infty) = \frac{u}{u_\infty} = 1 $$ is known.

Therefore, it is possible to solve this equation by shooting method. Briefly, with an initial guess for $$ f^{''}(0) $$ the equation will be solved and the condition $$ f^'(\infty) = 1 $$ can be  be checked. Depending on the result $$ f^{''}(0) $$

will be modified iteratively in the further steps till solution satisfies $$ f^'(\infty) = 1 $$.

The exact solution delivers:

$$ \delta_{99}=5 \delta =5\sqrt{\frac{\nu x}{u_\infty}}=\frac{5x}{\sqrt{Re_x}} $$

There are approximate solutions for the laminar velocity profile, such as:

$$\frac{u(x,y)}{u_\infty}=2\frac{y}{\delta(x)}-\frac{y^2}{\delta(x)^2}=2\eta-\eta^2$$

Utilizing this relation and the integral form of mass and momentum equations, one can show:

$$ \delta_{99}=\sqrt{30\frac{\nu x}{u_\infty}}=5.48 \delta =\frac{5.48x}{\sqrt{Re_x}} $$

For turbulent flow, the following power law can be used for the non-dimensional velocity profile:

$$\frac{u(x,y)}{u_\infty}=\left(\frac{y}{\delta(x)}\right)^{1/7}=\eta^{1/7}$$

$$ \delta_{99}=\frac{0.382x}{Re_x^{1/5}} $$