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Next: Exercises Up: Incompressible Boundary Layers Previous: Criterion for Boundary Layer

Approximate Solutions of Boundary Layer Equations

The boundary layer equations, (8.110)-(8.113), take the form

$\displaystyle \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}$ $\displaystyle = 0,$ (8.139)
$\displaystyle u\,\frac{\partial u}{\partial x} + v\,\frac{\partial u}{\partial y} - U\,\frac{dU}{dx}$ $\displaystyle = \nu\,\frac{\partial^{\,2} u}{\partial y^{\,2}},$ (8.140)

subject to the boundary conditions

$\displaystyle u(x,\infty)$ $\displaystyle = U(x),$ (8.141)
$\displaystyle u(x,0)$ $\displaystyle = 0,$ (8.142)
$\displaystyle v(x,0)$ $\displaystyle =0.$ (8.143)

Furthermore, it follows from Equations (8.140), (8.142), and (8.143) that

$\displaystyle \nu\left.\frac{\partial^{\,2} u}{\partial y^{\,2}}\right\vert _{y=0} = -U\,\frac{dU}{dx}.$ (8.144)

The previous expression can be thought of as an alternative form of Equation (8.143). As we saw in Section 8.4, the boundary layer equations can be solved exactly when $ U(x)$ takes the special form $ U_0\,x^{\,m}$ . However, in the general case, we must resort to approximation methods.

Following Pohlhausen (Schlichting 1987), let us assume that

$\displaystyle \frac{u(x,y)}{U(x)} = f(\eta),$ (8.145)

where $ \eta = y/\delta(x)$ , and $ \partial/\partial x\ll 1/\delta$ . In particular, suppose that

$\displaystyle f(\eta)=\left\{\begin{array}{lcl} a+b\,\eta+c\,\eta^{\,2}+d\,\eta...
...\,4}&\mbox{\hspace{1cm}}&0\leq\eta\leq 1\\ [0.5ex] 1&&\eta>1\end{array}\right.,$ (8.146)

where $ a$ , $ b$ , $ c$ , $ d$ , and $ e$ are constants. This expression automatically satisfies the boundary condition (8.141). Moreover, the boundary conditions (8.142) and (8.144) imply that $ a=0$ , and

$\displaystyle f''(0) = -{\mit\Lambda}(x),$ (8.147)

where $ '= d/d\eta$ , and

$\displaystyle {\mit\Lambda} = \frac{\delta^{\,2}}{\nu}\,\frac{dU}{dx}.$ (8.148)

Finally, let us assume that $ f$ , $ f'$ , and $ f''$ are continuous at $ \eta=1$ : that is,

$\displaystyle f(1)$ $\displaystyle = 1,$ (8.149)
$\displaystyle f'(1)$ $\displaystyle = 0,$ (8.150)
$\displaystyle f''(1)$ $\displaystyle =0.$ (8.151)

These constraints corresponds to the reasonable requirements that the velocity, vorticity, and viscous stress tensor, respectively, be continuous across the layer. Given that $ a=0$ , Equations (8.146), (8.147), and (8.149)-(8.151) yield

$\displaystyle f(\eta) = F(\eta)+{\mit\Lambda}\,G(\eta)$ (8.152)

for $ 0\leq \eta\leq 1$ , where

$\displaystyle F(\eta)$ $\displaystyle =1 - (1-\eta)^{\,3}\,(1+\eta),$ (8.153)
$\displaystyle G(\eta)$ $\displaystyle = \frac{1}{6}\,\eta\,(1-\eta)^{\,3}.$ (8.154)

Thus, the tangential velocity profile across the layer is a function of a single parameter, $ {\mit\Lambda}$ , which is termed the Pohlhausen parameter. The behavior of this profile is illustrated in Figure 8.13. Note that, under normal circumstances, the Pohlhausen parameter must lie in the range $ -12\leq {\mit\Lambda}\leq 12$ . For $ {\mit\Lambda}> 12$ , the profile is such that $ f(\eta)>1$ for some $ \eta<1$ , which is not possible in a steady-state solution. On the other hand, for $ {\mit\Lambda}<-12$ , the profile is such that $ f'(0)<0$ , which implies flow reversal close to the wall. As we have seen, flow reversal is indicative of separation. Indeed, the separation point, $ f'(0)=0$ , corresponds to $ {\mit \Lambda }=-12$ . Expression (8.152) is only an approximation, because it satisfies some, but not all, of the boundary conditions satisfied by the true velocity profile. For instance, differentiation of Equation (8.140) with respect to $ y$ reveals that $ (\partial^{\,3} u/\partial y^{\,3})_{y=0}\propto f'''(0)=0$ , which is not the case for expression (8.152).

Figure: Pohlhausen velocity profiles for $ {\mit \Lambda }=12$ (solid curve) and $ {\mit \Lambda }=-12$ (dashed curve).

It follows from Equations (8.115), (8.116), and (8.152)-(8.154) that

$\displaystyle \delta_1(x)$ $\displaystyle =\delta\int_0^1(1-f)\,d\eta = \delta\left(\frac{3}{10}-\frac{{\mit\Lambda}}{120}\right),$ (8.155)
$\displaystyle \delta_2(x)$ $\displaystyle =\delta\int_0^1 f\,(1-f)\,d\eta=\delta\left(\frac{37}{315}-\frac{{\mit\Lambda}}{945}-\frac{{\mit\Lambda}^{\,2}}{9072}\right).$ (8.156)


$\displaystyle \left.\frac{\partial u}{\partial y}\right\vert _{y=0} = \frac{U}{\delta}\,f'(0) = \frac{U}{\delta}\left(2 + \frac{{\mit\Lambda}}{6}\right).$ (8.157)

The von Kármán momentum integral, (8.117), can be rearranged to give

$\displaystyle \frac{U}{\nu}\,\delta_2\,\frac{d\delta_2}{dx} +\frac{\delta_2^{\,...
...ight) =\frac{\delta_2}{U}\left.\frac{\partial u}{\partial y}\right\vert _{y=0}.$ (8.158)


$\displaystyle \lambda(x) = \frac{\delta_2^{\,2}}{\nu}\,\frac{dU}{dx},$ (8.159)

we obtain

$\displaystyle U\,\frac{d}{dx}\!\left(\frac{\lambda}{dU/dx}\right) = 2\left[F_2(\lambda)-\lambda\left\{2+F_1(\lambda)\right\}\right] = F(\lambda),$ (8.160)


$\displaystyle \lambda$ $\displaystyle = \left(\frac{37}{315}-\frac{{\mit\Lambda}}{945}-\frac{{\mit\Lambda}^{\,2}}{9072}\right)^2{\mit\Lambda},$ (8.161)
$\displaystyle F_1(\lambda)$ $\displaystyle = \frac{\delta_1}{\delta_2} = \left(\frac{3}{10}-\frac{{\mit\Lamb...
...{315}-\frac{{\mit\Lambda}}{945}-\frac{{\mit\Lambda}^{\,2}}{9072}\right)\right.,$ (8.162)
$\displaystyle F_2(\lambda)$ $\displaystyle =\frac{\delta_2}{U}\left.\frac{\partial u}{\partial y}\right\vert...
...rac{37}{315}-\frac{{\mit\Lambda}}{945}-\frac{{\mit\Lambda}^{\,2}}{9072}\right),$ (8.163)
$\displaystyle F(\lambda)$ $\displaystyle = 2\left(\frac{37}{315}-\frac{{\mit\Lambda}}{945}-\frac{{\mit\Lam...
...{1}{120}\right){\mit\Lambda}^{\,2}+ \frac{2}{9072}\,{\mit\Lambda}^{\,3}\right].$ (8.164)

It is generally necessary to integrate Equation (8.158) from the stagnation point at the front of the obstacle, through the point of maximum tangential velocity, to the separation point on the back side of the obstacle. At the stagnation point we have $ U=0$ and $ dU/dx\neq 0$ , which implies that $ F(\lambda)=0$ . Furthermore, at the point of maximum tangential velocity we have $ dU/dx=0$ and $ U\neq 0$ , which implies that $ {\mit\Lambda}=\lambda=0$ . Finally, as we have already seen, $ {\mit \Lambda }=-12$ at the separation point, which implies, from Equation (8.161), that $ \lambda=-0.1567$ .

Figure: The function $ F(\lambda )$ (solid curve) and the linear function $ 0.47-6\,\lambda$ (dashed line).

As was first pointed out by Walz (Schlichting 1987), and is illustrated in Figure 8.14, it is a fairly good approximation to replace $ F(\lambda )$ by the linear function $ 0.47-6\,\lambda$ for $ \lambda$ in the physically relevant range. The approximation is particularly accurate on the front side of the obstacle (where $ \lambda>0$ ). Making use of this approximation, Equations (8.159) and (8.160) reduce to the linear differential equation

$\displaystyle \frac{d}{dx}\!\left(\frac{U\,\delta_2^{\,2}}{\nu}\right) = 0.47 - 5\,\frac{dU}{dx}\,\frac{\delta_2^{\,2}}{\nu},$ (8.165)

which can be integrated to give

$\displaystyle \frac{\delta_2^{\,2}}{\nu} = \frac{0.47}{U^{\,6}}\int_0^x U^{\,5}(x')\,dx',$ (8.166)

assuming that the stagnation point corresponds to $ x=0$ . It follows that

$\displaystyle \lambda= \frac{0.47}{U^{\,6}}\,\frac{dU}{dx}\int_0^x U^{\,5}(x')\,dx'.$ (8.167)

Recall that the separation point corresponds to $ x=x_s$ , where $ \lambda(x_s)\equiv\lambda_s= -0.1567$ .

Suppose that $ U(x) = U_0$ , which corresponds to uniform flow over a flat plate. (See Section 8.5.) It follows from Equations (8.166) and (8.167) that

$\displaystyle \frac{\delta_2(x)}{x} = \frac{0.69}{{\rm Re}^{1/2}},$ (8.168)

where $ {\rm Re}=U_0\,x/\nu$ , and $ \lambda=0$ . Moreover, according to Equations (8.148) and (8.162), $ {\mit\Lambda}=0$ and $ \delta_1/\delta_2=2.55$ . Hence, the displacement width of the boundary layer becomes

$\displaystyle \frac{\delta_1(x)}{x} = \frac{1.75}{{\rm Re}^{1/2}}.$ (8.169)

This approximate result compares very favorably with the exact result, (8.73).

Figure: The function $ \lambda (\theta )$ for flow around a circular cylinder.

Suppose that $ x=a\,\theta$ and $ U(\theta)=2\,U_0\,\sin\theta$ , which corresponds to uniform transverse flow around a circular cylinder of radius $ a$ . (See Section 8.8.) Equation (8.167) yields

$\displaystyle \lambda(\theta) = 0.47\,\frac{\cos\theta}{\sin^6\theta}\int_0^\theta \sin^5 \theta'\,d\theta'.$ (8.170)

Figure 8.15 shows $ \lambda (\theta )$ determined from the previous formula. It can be seen that $ \lambda=\lambda_s=-0.1567$ when $ \theta=\theta_s\simeq 108^\circ$ . In other words, the separation point is located $ 108^\circ$ from the stagnation point at the front of the cylinder. This suggests that the low pressure wake behind the cylinder is almost as wide as the cylinder itself, and that the associated form drag is comparatively large.

Figure 8.16: Flow over the back surface of a semi-infinite wedge.
\epsfysize =1.5in

Suppose, finally, that $ U=U_0\,x^{\,m}$ . If $ m$ is negative then, as illustrated in Figure 8.16, this corresponds to uniform flow over the back surface of a semi-infinite wedge whose angle of dip is

$\displaystyle \theta = - \frac{m}{1+m}\,\frac{\pi}{2}.$ (8.171)

(See Section 8.4.) It follows from Equation (8.167) that

$\displaystyle \lambda = \frac{0.47\,m}{1+5\,m} = -\frac{0.47\,\theta}{\pi/2-4\,\theta}.$ (8.172)

We expect boundary layer separation on the back surface of the wedge when $ \lambda<\lambda_s=-0.1567$ . This corresponds to $ \theta>\theta_s$ , where

$\displaystyle \theta_s = \frac{\pi}{2}\,\frac{(-\lambda_s)}{0.47+4\,(-\lambda_s)} \simeq 13^\circ.$ (8.173)

Hence, boundary layer separation can be prevented by making the wedge's angle of dip sufficiently shallow: that is, by streamlining the wedge, which has the effect of reducing the deceleration of the flow on the wedge's back surface. The critical value of $ m$ (i.e., $ m_s=-0.0125$ ) at which separation occurs in our approximate solution is very similar to the critical value of $ m$ (i.e., $ m_\ast=-0.0905$ ) at which the exact self-similar solutions described in Section 8.4 can no longer be found. This suggests that the absence of self-similar solutions for $ m<m_\ast$ is related to boundary layer separation.

next up previous
Next: Exercises Up: Incompressible Boundary Layers Previous: Criterion for Boundary Layer
Richard Fitzpatrick 2016-03-31