Wake Downstream of a Flat Plate

As we saw in the previous section, if a flat plate of negligible thickness, and finite length, is placed in the path of a uniform high Reynolds number flow, directed parallel to the plate, then thin boundary layers form above and below the plate. Outside the layers, the flow is irrotational, and essentially inviscid. Inside the layers, the flow is modified by viscosity, and has non-zero vorticity. Downstream of the plate, the boundary layers are convected by the flow, and merge to form a thin wake. See Figure 51. Within the wake, the flow is modified by viscosity, and possesses finite vorticity. Outside the wake, the downstream flow remains irrotational, and effectively inviscid.

Since there is no solid surface embedded in the wake, acting to retard the flow, we would expect the action of viscosity to cause the velocity within the wake, a long distance downstream of the plate, to closely match that of the unperturbed flow. In other words, we expect the fluid velocity within the wake to take the form

$$v_x(x,y) = U_0 - u(x,y),$$ (752)

$$v_y(x,y) = v(x,y),$$ (753)

where

$$\vert u\vert\ll U_0.$$ (754)

Assuming that, within the wake,

$$\frac{\partial}{\partial x} \sim \frac{1}{x},$$ (755)

$$\frac{\partial}{\partial y} \sim \frac{1}{\delta},$$ (756)

where $\delta\ll x$ is the wake thickness, fluid continuity requires that

$$v\sim \frac{\delta}{x}\,u.$$ (757)

Now, the flow external to the boundary layers, and the wake, is both uniform and essentially inviscid. Hence, according to Bernoulli's theorem, the pressure in this region is also uniform--see Equation (691). However, as we saw in Section 7.3, there is no $y$-variation of the pressure across the boundary layers. It follows that the pressure is uniform within the layers. Thus, it is reasonable to assume that the pressure is also uniform within the wake, since the wake is formed via the convection of the boundary layers downstream of the plate. We conclude that

$$p(x,y)\simeq p_0$$ (758)

everywhere in the fluid, where $p_0$ is a constant.

The $x$-component of the fluid equation of motion is written

$$v_x\,\frac{\partial v_x}{\partial x} + v_y\,\frac{\partial v... ...x}{\partial x^2} + \frac{\partial^2 v_x}{\partial y^2}\right).$$ (759)

Making use of (752)-(758), the above expression reduces to

$$U_0\,\frac{\partial u}{\partial x} \simeq \nu\,\frac{\partial^2 u}{\partial y^2}.$$ (760)

The boundary condition

$$u(x,\pm\infty) = 0$$ (761)

ensures that the flow outside the wake remains unperturbed. Note that Equation (760) has the same mathematical form as a conventional diffusion equation, with $x$ playing the role of time, and $\nu/U_0$ playing the role of the diffusion coefficient. Hence, by analogy with the standard solution of the diffusion equation, we would expect $\delta\sim (\nu\,x/U_0)^{1/2}$.

As can easily be demonstrated, the self-similar solution to (760), subject to the boundary condition (761), is

$$u(x,y) = \frac{Q}{\sqrt{\pi}\,\delta}\,\exp\left(-\frac{y^{\,2}}{\delta^{\,2}}\right),$$ (762)

where

$$\delta(x) =2\left(\frac{\nu\,x}{U_0}\right)^{1/2},$$ (763)

and $Q$ is a constant. It follows that

$$\int_{-\infty}^\infty u\,dy = Q,$$ (764)

since, as is well-known, $\int_{-\infty}^\infty \exp(-t^2)\,dt = \sqrt{\pi}$. As expected, the width of the wake scales as $x^{1/2}$.

Figure 55: Tangential velocity profile across the wake of a flat plate of negligible thickness located at $y=0$. The profile is calculated for $Q/(U_0\,\delta )=0.5$.

The tangential velocity profile across the wake, which takes the form

$$\frac{v_x(x,y)}{U_0} = 1 - \frac{Q}{U_0\,\delta}\,\frac{1}{\sqrt{\pi}}\,\exp(-y^2/\delta^{\,2}),$$ (765)

is plotted in Figure 55. In addition, the vorticity profile across the wake, which is written

$$\frac{\omega(x,y)}{U_0/\delta}= -\frac{Q}{U_0\,\delta}\,\frac{2}{\sqrt{\pi}}\,\frac{y}{\delta}\,\exp(-y^2/\delta^{\,2})$$ (766)

is shown in Figure 56. It can be seen that the profiles pictured in Figures 55 and 56 are essentially smoothed out versions of the boundary layer profiles shown in Figures 52 and 53, respectively.

Figure 56: Vorticity profile across the boundary layers above and below a flat plate of negligible thickness located at $y=0$. The profile is calculated for $Q/(U_0\,\delta )=0.5$.

Suppose that the plate and a portion of its trailing wake are enclosed by a cuboid control volume of unit depth (in the $z$-direction) that extends from $x=-l$ to $x=+l$ and from $y=-h$ to $y=h$. See Figure 57. Here, $l\gg L$ and $h\gg \delta(l)$, where $L$ is the length of the plate, and $\delta(x)$ the width of the wake. Hence, the control volume extends well upstream and downstream of the plate. Moreover, the volume is much wider than the wake.

Figure 57: Control volume surrounding a flat plate and its trailing wake.

Let us apply the integral form of the fluid equation of continuity to the control volume. For a steady-state, this reduces to (see Section 2.9)

$$\oint_S \rho\,{\bf v}\cdot d{\bf S}=0,$$ (767)

where $S$ is the bounding surface of the control volume. The normal fluid velocity is $-U_0$ at $x=-l$, $U_0-u(y)$ at $x=l$, and $v(x)$ at $y=\pm h$, as indicated in the figure. Hence, (767) yields

$$-\int_{-h}^h\rho\,U_0\,dy + \int_{-h}^{h}\rho\,[U_0-u(y)]\,dy+2\int_{-l}^{l}\rho\,v(x)\,dx = 0,$$ (768)

or

$$\int_{-h}^h u(y)\,dy = 2\,\int_{-l}^l v(x)\,dx.$$ (769)

However, given that $u\rightarrow 0$ for $\vert y\vert\gg\delta$, and since $h\gg \delta$, it is a good approximation to replace the limits of integration on the left-hand side of the above expression by $\pm \infty$. Thus, from Equation (764),

$$\int_{-h}^h\,u(y)\,dy=2\,\int_{-l}^lv(x)\,dx \simeq Q,$$ (770)

where $Q$ is independent of $x$. Note that the slight retardation of the flow inside the wake, due to the presence of the plate, which is parameterized by $Q$, necessitates a small lateral outflow, $v(x)$, in the region of the fluid external to the wake.

Let us now apply the integral form of the $x$-component of the fluid equation of motion to the control volume. For a steady-state, this reduces to (see Section 2.11)

$$\int_S \rho\,v_x\,{\bf v}\cdot d{\bf S} = F_x + \int_S \sigma_{xj}\,dS_j,$$ (771)

where $F_x$ is the net $x$-directed force exerted on the fluid within the control volume by the plate. It follows, from Newton's third law of motion, that $F_x=-D$, where $D$ is the viscous drag force per unit width (in the $z$-direction) acting on the plate in the $x$-direction. Now, in an incompressible fluid (see Section 2.6),

$$\sigma_{ij} = -p\,\delta_{ij} + \rho\,\nu\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right).$$ (772)

Hence, we obtain

$$-\int_{-h}^h\rho\,U_0^{\,2}\,dy+\int_{-h}^h\rho\left[U_0-u(y)\right]^{\,2}\,dy$$

$$+ 2\int_{-l}^l \rho\,U_0\,v(x)\,dx = -D -2\,\rho\,\nu\,\frac{d}{dl}\!\int_{-h}^h u(y)\, dy,$$ (773)

since the pressure within the fluid is essentially uniform, the tangential fluid velocity at $y=\pm h$ is $U_0$, and $v$ is assumed to be negligible at $x=\pm l$. Making use of Equation (770), as well as the fact that $Q$ is independent of $l$, we get

$$D = \rho\,U_0\,Q.$$ (774)

Here, we have neglected any terms that are second-order in the small quantity $u$. A comparison with Equation (748) reveals that

$$Q = 1.33\,(\nu\,U_0\,L)^{1/2},$$ (775)

or

$$\frac{Q}{U_0\,\delta} = 0.664\left(\frac{L}{x}\right)^{1/2}.$$ (776)

Hence, from (765) and (766), the velocity and vorticity profiles across the layer are

$$\frac{v_x(x,y)}{U_0} = 1 -0.375\left(\frac{L}{x}\right)^{1/2}\exp(-y^2/\delta^{\,2}),$$ (777)

and

$$\frac{\omega(x,y)}{U_0/\delta} = -0.749 \left(\frac{L}{x}\right)^{1/2}\frac{y}{\delta}\,\exp(-y^2/\delta^{\,2}),$$ (778)

where $\delta(x)= 2\,(\nu\,x/U_0)^{1/2}$. Finally, since the above analysis is premised on the assumption that $\vert 1-v_x/U_0\vert=\vert u\vert/U_0\ll 1$, it is clear that the previous three expressions are only valid when $x\gg L$ (i.e., well downstream of the plate).

The above analysis only holds when the flow within the wake is non-turbulent. Let us assume, by analogy with the discussion in the previous section, that this is the case as long as the Reynolds number of the wake, $U_0\,\delta(x)/\nu$, remains less than some critical value that is approximately $600$. Since the Reynolds number of the wake can be written $2\,{\rm Re}^{1/2}\,(x/L)^{1/2}$, where ${\rm Re} = U_0\,L/\nu$ is the Reynolds number of the external flow, we deduce that the wake becomes turbulent when $x/L \stackrel {_{\normalsize >}}{_{\normalsize\sim}}9\times 10^4/{\rm Re}$. Hence, the wake is always turbulent sufficiently far downstream of the plate. Our analysis, which effectively assumes that the wake is non-turbulent in some region, immediately downstream of the plate, whose extent (in $x$) is large compared with $L$, is thus only valid when $1\ll {\rm Re}\ll 9\times 10^4$.