Self-Similar Boundary Layers

The boundary layer equation, (8.35), takes the form of a nonlinear partial differential equation that is extremely difficult to solve exactly. However, considerable progress can be made if this equation is converted into an ordinary differential equation by demanding that its solutions be self-similar. Self-similar solutions are such that, at a given distance, $x$ , along the layer, the tangential flow profile, $v_x(x,y)$ , is a scaled version of some common profile: that is, $v_x(x,y)= U(x)\,F[y/\delta(x)]$ , where $\delta(x)$ is a scale-factor, and $F(z)$ a dimensionless function. It follows that $\psi(x,y)=-U(x)\,\delta(x)\,f[y/\delta(x)]$ , where $f'(z)=F(z)$ .

Let us search for a self-similar solution to Equation (8.35) of the general form

$$\psi(x,y) = -\left[\frac{2\,\nu\,U_0\,x^{\,m+1}}{m+1}\right]^{1/2... ...)=-U_0\,x^{\,m}\left[\frac{2\,\nu}{(m+1)\,U_0\,x^{\,m-1}}\right]^{1/2} f(\eta),$$ (8.41)

where

$$\eta =\left[\frac{(m+1)\,U_0\,x^{\,m-1}}{2\,\nu}\right]^{1/2}y.$$ (8.42)

This implies that $\delta(x)=[2\,\nu/(m+1)\,U_0\,x^{\,m-1}]^{1/2}$ , and $U(x)=U_0\,x^{\,m}$ . Here, $U_0$ and $m$ are constants. Moreover, $U_0\,x^{\,m}$ has dimensions of velocity, whereas $m$ , $\eta$ , and $f$ , are dimensionless. Transforming variables from $x$ , $y$ to $x$ , $\eta$ , we find that

$$\left.\frac{\partial}{\partial x}\right\vert _y = \left.\frac{\partial}{\partial x}\right\vert _\eta +\frac{m-1}{2}\,\frac{\eta}{x}\left.\frac{\partial}{\partial\eta}\right\vert _x,$$ (8.43)

$$\left.\frac{\partial}{\partial y}\right\vert _x =\left[\frac{(m+1)\,U_0\,x^{\,m-1}}{2\,\nu}\right]^{1/2}\left.\frac{\partial}{\partial\eta}\right\vert _x.$$ (8.44)

Hence,

$$\frac{\partial\psi}{\partial x} = -\left[\frac{\nu\,U_0\,x^{\,m-1}}{2\,(m+1)}\right]^{1/2}[(m+1)\,f+(m-1)\,\eta\,f'],$$ (8.45)

$$\frac{\partial\psi}{\partial y} = - U_0\,x^{\,m}\,f',$$ (8.46)

$$\frac{\partial^{\,2}\psi}{\partial y^{\,2}} =-\left[\frac{(m+1)\,U_0^{\,3}\,x^{\,3\,m-1}}{2\,\nu}\right]^{1/2}f'',$$ (8.47)

$$\frac{\partial^{\,2}\psi}{\partial x\,\partial y} = -\frac{U_0\,x^{\,m-1}}{2}\,[2\,m\,f'+(m-1)\,\eta\,f''],$$ (8.48)

$$\frac{\partial^{\,3}\psi}{\partial y^{\,3}} = -\frac{(m+1)\,U_0^{\,2}\,x^{\,2\,m-1}}{2\,\nu}\,f''',$$ (8.49)

where $'= d/d\eta$ . Thus, Equation (8.35) becomes

$$(m+1)\,f'''+ (m+1)\,f\,f'' -2m\,f'^{\,2} = -\frac{1}{U_0^{\,2}\,x^{\,2\,m-1}}\,\frac{dU^{\,2}}{dx}.$$ (8.50)

Because the left-hand side of the previous equation is a (non-constant) function of $\eta$ , while the right-hand side is a function of $x$ (and as $\eta$ and $x$ are independent variables), the equation can only be satisfied if its right-hand side takes a constant value. In fact, if

$$\frac{1}{U_0^{\,2}\,x^{\,2\,m-1}}\,\frac{dU^{\,2}}{dx}=2\,m$$ (8.51)

then

$$U(x) = U_0\,x^{\,m}$$ (8.52)

(which is consistent with our initial guess), and

$$f'''+f\,f'' + \beta\,(1-f'^{\,2}) = 0,$$ (8.53)

where

$$\beta = \frac{2\,m}{m+1}.$$ (8.54)

Expression (8.53) is known as the Falkner-Skan equation. The solutions to this equation that satisfy the physical boundary conditions (8.36)-(8.38) are such that

$$f(0) = f'(0) = 0,$$ (8.55)

and

$$f'(\infty) = 1,$$ (8.56)

$$f''(\infty) =0.$$ (8.57)

(The final condition corresponds to the requirement that the vorticity tend to zero at the edge of the layer.) Note, from Equations (8.39), (8.42), (8.47), (8.52), (8.55), and (8.56), that the normally integrated vorticity within the boundary layer is

$$\int_0^\infty \omega\,dy = -U(x).$$ (8.58)

Furthermore, from Equations (8.40), (8.47), and (8.52), the $x$ -component of the viscous force per unit area acting on the surface of the obstacle is

$$\left.\sigma_{xy}\right\vert _{y=0} =\frac{1}{2}\,\rho\,U^{\,2}\left(\frac{\nu}{U\,x}\right)^{1/2}(m+1)^{1/2}\,\sqrt{2}\,f''(0).$$ (8.59)

It is convenient to parameterize this quantity in terms of a skin friction coefficient,

$$c_f = \frac{\left.\sigma_{xy}\right\vert _{y=0}}{(1/2)\,\rho\,U^{\,2}}.$$ (8.60)

It follows that

$$c_f(x) = \frac{(m+1)^{1/2}\,\sqrt{2}\,f''(0)}{[{\rm Re}(x)]^{1/2}},$$ (8.61)

where

$${\rm Re}(x) = \frac{U(x)\,x}{\nu}$$ (8.62)

is the effective Reynolds number of the flow on the outer edge of the layer at position $x$ . Hence, $c_f(x)\propto x^{\,-(m+1)/2}$ . Finally, according to Equation (8.41), the width of the boundary layer is approximately

$$\frac{\delta(x)}{x} \simeq \frac{1}{[{\rm Re}(x)]^{\,1/2}},$$ (8.63)

which implies that $\delta(x)\propto x^{\,-(m-1)/2}$ .

Figure: $f''(0)$ calculated as a function of $\beta$ for solutions of the Falkner-Skan equation.

If $m>0$ then the external tangential velocity profile, $U(x)=U_0\,x^{\,m}$ , corresponds to that of irrotational inviscid flow incident, in a symmetric fashion, on a semi-infinite wedge whose apex subtends an angle $\alpha\,\pi$ , where $\alpha=2\,m/(m+1)$ . (See Section 5.10, and Figure 5.10.) In this case, $U(x)$ can be interpreted as the tangential velocity a distance $x$ along the surface of the wedge from its apex (in the direction of the flow). By analogy, if $m=0$ then the external velocity profile corresponds to that of irrotational inviscid flow parallel to a semi-infinite flat plate (which can be thought of as a wedge whose apex subtends zero angle). In this case, $U(x)$ can be interpreted as the tangential velocity a distance $x$ along the surface of the plate from its leading edge (in the direction of the flow). (See Section 8.5.) Finally, if $m<0$ then the external velocity profile is that of symmetric irrotational inviscid flow over the back surface of a semi-infinite wedge whose apex subtends an angle $(1-\alpha')\,\pi$ , where $\alpha'=-m/(1+m)$ . (See Section 5.11, and Figure 5.11.) In this case, $U(x)$ can be interpreted as the tangential velocity a distance $x$ along the surface of the wedge from its apex (in the direction of the flow).

Unfortunately, the Falkner-Skan equation, (8.53), possesses no general analytic solutions. However, this equation is relatively straightforward to solve via numerical methods. Figure 8.2 shows $f''(0)$ , calculated numerically as a function of $\beta=2\,m/(m+1)$ , for the solutions of Equation (8.53) that satisfy the boundary conditions (8.55)-(8.57). In addition, Figure 8.3 shows $f'(\eta)$ versus $\eta$ , calculated numerically for various different values of $m$ . Because $\beta\rightarrow 2$ as $m\rightarrow \infty$ , solutions of the Falkner-Skan equation with $\beta>2$ have no physical significance. For $0<\beta<2$ , it can be seen, from Figures 8.2 and 8.3, that there is a single solution branch characterized by $f'(\eta)>0$ and $f''(0)>0$ . This branch is termed the forward flow branch, because it is such that the tangential velocity, $v_x(\eta)\propto f'(\eta)$ , is in the same direction as the external tangential velocity [i.e., $v_x(\infty)$ ] across the whole layer (i.e., $0<\eta<\infty$ ). The forward flow branch is characterized by a positive skin friction coefficient, $c_f\propto f''(0)$ . It can also be seen that for $\beta<0$ there exists a second solution branch, which is termed the reversed flow branch, because it is such that the tangential velocity is in the opposite direction to the external tangential velocity in the region of the layer immediately adjacent to the surface of the obstacle (which corresponds to $\eta=0$ ). The reversed flow branch is characterized by a negative skin friction coefficient. The reversed flow solutions are probably unphysical, because reversed flow close to the wall is generally associated with a phenomenon known as boundary layer separation (see Section 8.10) that invalidates the boundary layer orderings. It can be seen that the two solution branches merge together at $\beta=\beta_\ast=-0.1989$ , which corresponds to $m=m_\ast = -0.0905$ . Moreover, there are no solutions to the Falkner-Skan equation with $\beta<\beta_\ast$ or $m<m_\ast$ . The disappearance of solutions when $m$ becomes too negative (i.e., when the deceleration of the external flow becomes too large) is also related to boundary layer separation.

Figure: Solutions of the Falkner-Skan equation. In order from the left to the right, the various solid curves correspond to forward flow solutions calculated with $m=4$ , $1$ , $1/3$ , $1/9$ , 0 , $-0.05$ , and $-0.0904$ , respectively. The dashed curve shows a reversed flow solution calculated with $m=-0.05$ .