Free Streamline Theory

Consider a situation in which steady, incompressible, irrotational flow is partly bounded by rigid walls, and partly by free streamlines of unknown shape on which the pressure takes a known constant value. For instance, the free streamlines might correspond to the interface of a liquid with the atmosphere, in which case the constant pressure would correspond to that of the atmosphere. According to Bernoulli's theorem, (6.40), (neglecting gravity) the fluid speed is constant on a free streamline.

Let us define the new complex variable

$${\mit\Omega} =\ln\left(\frac{dz}{d\bar{F}}\right)= \ln\left(\frac{1}{v_x-{\rm i}\,v_y}\right)=\ln \left(\frac{1}{v}\right)+ {\rm i}\,\theta,$$ (6.89)

where $z=x +{\rm i}\,y$ , $\bar{F}=\skew{3}\bar{\phi}+{\rm i}\,\skew{3}\bar{\psi}$ , and $v$ and $\theta$ are the magnitude and direction (relative to the $x$ -axis) of the velocity vector $(v_x,\,v_y)$ . (See Section 6.5.) Here, $\bar{F}=-F$ , $\skew{3}\bar{\phi}=-\phi$ , and $\skew{3}\bar{\psi}=-\psi$ . It follows that the real part of ${\mit\Omega}$ is constant on each free streamline, whereas the imaginary part is constant on each straight segment of the rigid boundary. Thus, the whole boundary of the fluid is represented in the ${\mit\Omega}$ -plane by a polygon. We know, from the Schwarz-Christoffel theorem (see Section 6.8), that it is always possible to map the interior (or exterior) of a polygon in one complex plane onto the upper half of another complex plane. Hence, it is possible to find conformal relations between ${\mit\Omega}$ and a new complex variable $\lambda$ , and between $\bar{F}$ and $\lambda$ , such that the flow region is mapped onto the upper half of the $\lambda$ -plane in both cases. In this way, we can find a relation between ${\mit\Omega}$ and $\bar{F}$ , from which an expression for $\bar{F}$ in terms of $z$ follows via integration.

As an example of the application of free streamline theory, let us calculate the contraction ratio of a two-dimensional jet of liquid emerging from an orifice. Suppose that the orifice in question is a long thin slot in a plane wall of small thickness, and that the wall forms part of a large vessel containing liquid. The speed of the liquid on the free streamlines that emerge from the edges of the orifice is uniform, and equal to $V$ , say. This is also the speed of the interior of the jet far downstream of the orifice, where (neglecting the effects of gravity) the streamlines are straight and parallel. (See Figure 6.16.) Let the two streamlines bounding the flow, on which $\skew{3}\bar{\psi}=+\skew{3}\bar{\psi}_1$ and $\skew{3}\bar{\psi}=-\skew{3}\bar{\psi}_1$ , say--be $ABC$ and $A'B'C'$ , respectively, where $A$ , $A'$ , $C$ , and $C'$ all represent points at infinity. Figure 6.15 shows the corresponding straight-line boundaries in the $\bar{F}$ - and ${\mit\Omega}$ -planes, where ${\mit\Omega}$ is now defined in a more convenient manner as

$${\mit\Omega} =\ln\left(V\,\frac{dz}{d\bar{F}}\right)=\ln\left(\frac{V}{v}\right)+{\rm i}\,\theta.$$ (6.90)

Figure 6.15: Conformal transformations required for the determination of the flow from an orifice in a plane wall in two dimensions.

We, first, need to map the semi-infinite strip $A'B'C'CBA$ in the ${\mit\Omega}$ -plane to the upper half of the $\lambda$ -plane. The conformal transformation of a semi-infinite strip onto the upper half of another complex plane is achieved by the relation (6.81). Adapting this relation to our current needs, we take $K'=1$ , $z_0=-{\rm i}\,\pi/2$ , and $b=-c=1$ , so that

$$\lambda = {\rm i}\,\sinh{\mit\Omega}$$ (6.91)

In particular, this transformation maps the points $B'={\rm i}\,\pi/2$ and $B={\rm i}\,\pi/2$ in the ${\mit\Omega}$ -plane to the points $B=1$ and $B=-1$ , respectively, in the $\lambda$ -plane.

Next, we need to map the infinite strip $ABCA'B'C'$ in the $\bar{F}$ -plane to the upper half of the $\lambda$ -plane. The conformal transformation of a semi-infinite strip onto the upper half of another complex plane is achieved by the relation (6.88). Adapting this relation to our current needs, we take $-K'=\pi/(2\,\skew{3}\bar{\psi}_1)$ , $b=0$ , $z_0={\rm i}\,\skew{3}\bar{\psi}_1$ , so that

$$\lambda = {\rm i}\,\exp\left(-\frac{\pi}{2}\,\frac{\bar{F}}{\skew{3}\bar{\psi}_1}\right).$$ (6.92)

The flow field has now been mapped onto the upper half of the $\lambda$ -plane in two coincident ways, giving

$$\lambda = {\rm i}\,\exp\left(-\frac{\pi}{2}\,\frac{\bar{F}}{\skew... ...\rm i}}{2}\left(V\,\frac{dz}{d\bar{F}}-\frac{1}{V}\,\frac{d\bar{F}}{dz}\right).$$ (6.93)

Hence,

$$V\,\frac{dz}{d\bar{F}} = -{\rm i}\,\lambda+(1-\lambda^{\,2})^{1/2},$$ (6.94)

where, with cuts in the $z$ -plane at $AB$ and $A'B'$ , the correct branch of $(1-\lambda^{\,2})^{1/2}$ is that which is real and positive when $\skew{3}\bar{\psi}=0$ . Integration, with the aid of Equation (6.92), yields

$$\frac{\pi}{2}\,\frac{V}{\skew{3}\bar{\psi}_1}\,(z-z_1)= {\rm i}\,(\lambda-1)-(1-\lambda^{\,2})^{1/2}+\tanh^{-1}[(1-\lambda^{\,2})^{1/2}],$$ (6.95)

where $z_1$ is a constant. However, $\lambda=1$ at the point $B$ , where $z={\rm i}\,d$ ($2\,d$ being the width of the orifice), so $z_1={\rm i}\,d$ , and

$$\frac{\pi}{2}\,\frac{V}{\skew{3}\bar{\psi}_1}\,(z-{\rm i}\,d)= {\rm i}\,(\lambda-1)-(1-\lambda^{\,2})^{1/2}+\tanh^{-1}[(1-\lambda^{\,2})^{1/2}],$$ (6.96)

Finally, the required relationship between $z$ and $\bar{F}$ is obtained by eliminating $\lambda$ between Equations (6.92) and (6.96).

Figure 6.16: Free streamlines of a liquid jet emerging from an orifice in a plane wall in two dimensions.

On the free streamline $BC$ , we have

$$\skew{3}\bar{\psi}=\skew{3}\bar{\psi}_1, \skew{3}\bar{\phi}=V\,s, {\mit\Omega}={\rm i}\,\theta,$$ (6.97)

where $s$ denotes distance measured along the streamline from $B$ . It follows from Equations (6.91) and (6.92) that

$$\lambda=-\sin\theta = \exp\left(-\frac{\pi}{2}\,\frac{V}{\skew{3}\bar{\psi}_1}\,s\right).$$ (6.98)

Thus, making use of Equation (6.96), the equation of streamline $BC$ can be written, in parametric form, as

$$x =\frac{2}{\pi}\,\frac{\skew{3}\bar{\psi}_1}{V}\left[\tanh^{-1}(\cos\theta)-\cos\theta)\right],$$ (6.99)

$$y =d-\frac{2}{\pi}\,\frac{\skew{3}\bar{\psi}_1}{V}\left(1+\sin\theta\right).$$ (6.100)

Thus, the asymptotic semi-width of the jet is

$$b= \lim_{s\rightarrow\infty} y(s)= d-\frac{2}{\pi}\,\frac{\skew{3}\bar{\psi}_1}{V}.$$ (6.101)

Far from the orifice, which corresponds to $\lambda\rightarrow 0$ , Equation (6.94) yields

$$\frac{d\bar{F}}{dz} =\frac{1}{v_x-{\rm i}\,v_y} \simeq \frac{1}{V}.$$ (6.102)

In other words, as expected, the velocity profile becomes uniform across the jet far downstream of the orifice, which implies that $\skew{3}\bar{\psi}_1=b\,V$ . It follows, from Equation (6.101), that the contraction ratio of a two-dimensional liquid jet emerging from an orifice in a plane wall takes the value

$$\frac{b}{d}=\frac{\pi}{\pi+2}= 0.61.$$ (6.103)

Somewhat surprisingly, this value is very close to the observed contraction ratio of a three-dimensional jet emerging from a circular hole in a plane wall. (See Section 4.6.) According to Equations (6.99), (6.100), and (6.103), the equation of the free streamline $BC$ can be written

$$\frac{x}{d} =\frac{2}{\pi+2}\left[\tanh^{-1}(\cos\zeta)-\cos\zeta)\right],$$ (6.104)

$$\frac{y}{d} =\frac{\pi+2\,\sin\zeta}{\pi+2},$$ (6.105)

for $\pi/2\geq\zeta\geq 0$ . By symmetry, the equation of the free streamline $B'C'$ is obtained from the previous equation via the transformation $x\rightarrow x$ , $y\rightarrow -y$ . The streamlines $BC$ and $B'C'$ are plotted in Figure 6.16.

Figure: Contraction ratio of a liquid jet emerging from a two-dimensional orifice formed by a gap between two semi-infinite plane walls that subtend an angle $2\,\alpha$ .

We can easily extend the previous calculation to determine the contraction ratio of a liquid jet emerging from a two-dimensional orifice formed by a gap between two semi-infinite plane walls that subtend an angle $2\,\alpha$ . In this case, the free streamline $BC$ , on which $\skew{3}\bar{\psi}=+\skew{3}\bar{\psi}_1$ , corresponds to $v=V$ and $\theta=-\alpha$ , whereas the free streamline $B'C'$ , on which $\skew{3}\bar{\psi}=-\skew{3}\bar{\psi}_1$ , corresponds to $v=V$ and $\theta=+\alpha$ . Hence, the transformation (6.93) generalizes to give

$$\lambda = {\rm i}\,\exp\left(-\frac{\pi}{2}\,\frac{\bar{F}}{\skew... ...\psi}_1}\right)={\rm i}\,\sinh\left(\frac{\pi}{2\,\alpha}\,{\mit\Omega}\right),$$ (6.106)

which implies that

$$V\,\frac{dz}{d\bar{F}} = \left[-{\rm i}\,\lambda + (1-\lambda^{\,2})^{1/2}\right]^{\,2\,\alpha/\pi}.$$ (6.107)

Far from the orifice, which corresponds to $\lambda\rightarrow 0$ , the previous expression yields $d\bar{F}/dz\simeq 1/V$ . In other words, a long way downstream of the orifice, the velocity profile is again uniform across the jet. Hence, we can write $\skew{3}\bar{\psi}_1=b\,V$ , where $b$ is the semi-width of the jet far from the orifice. Combining the previous two equations, we obtain

$$-\frac{\pi}{2}\,\frac{dz}{b}=\frac{d\lambda}{\lambda} \left[-{\rm i}\,\lambda + (1-\lambda^{\,2})^{1/2}\right]^{\,2\,\alpha/\pi}.$$ (6.108)

Now, $\lambda=1$ corresponds to the point $B$ , at which $z={\rm i}\,d$ , where $d$ is the semi-width of the orifice. Thus, integration of the previous expression yields

$$\frac{z-{\rm i}\,d}{b} = \frac{2}{\pi}\int_\lambda^1 \frac{dk}{k}\left[-{\rm i}\,k + (1-k^{\,2})^{1/2}\right]^{\,2\,\alpha/\pi}.$$ (6.109)

Making the transformation $k=\sin\beta$ , we get

$$\frac{z-{\rm i}\,d}{b} = \frac{2}{\pi}\int_{\sin^{-1}(\lambda)}^{... ...p\left(-{\rm i}\,\frac{2\,\alpha}{\pi}\,\beta\right)\,\frac{d\beta}{\tan\beta}.$$ (6.110)

As before, we have $\lambda = \exp[-(\pi/2)\,(s/b)]$ on the free streamline $BC$ , where $s$ denotes distance measured along the streamline from $B$ . Far downstream of the orifice, $\lambda\rightarrow 0$ and ${\rm Im}(z)\rightarrow b$ . Hence, we obtain the following expression for the contraction ratio of the jet:

$$\frac{b}{d} =\left(1+\frac{2}{\pi}\int_0^{\pi/2}\frac{\sin[(2\,\alpha/\pi)\,\beta]}{\tan\beta}\,d\beta\right)^{-1}.$$ (6.111)

This contraction ratio is plotted as a function of $\alpha$ in Figure 6.17. It can be seen that the ratio takes the value $1$ when $\alpha=0$ , which corresponds to a two-dimensional jet emerging from a gap between two parallel planes. Not surprisingly, there is no contraction in this case, because the streamlines of the flow are already parallel before they emerge from the orifice. On the other hand, the ratio takes the value $1/2$ when $\alpha=\pi$ , which corresponds to the two-dimensional equivalent to the Borda mouthpiece discussed in Section 4.6. In this case, the contraction ratio is exactly the same as that of a three-dimensional Borda mouthpiece, which is not surprising, because there is nothing in the discussion of Section 4.6 that depends crucially on the shape of the mouthpiece cross-section.

Figure 6.18: Conformal transformations required for the determination of the flow past a flat plate with a cavity at ambient pressure.

As a second example of the use of free streamline theory, consider the flow past a flat plate normal to a liquid stream of infinite extent. The pressure in the cavity formed behind the plate is assumed to be equal to that of the undisturbed stream. Thus, the speed of the liquid on the free streamlines bounding the cavity is equal to $V$ , which is the uniform flow speed far upstream of the plate. Let us take $\skew{3}\bar{\psi}=0$ on the central streamline that divides at the stagnation point $O$ , and later becomes the two free streamlines. (See Figure 6.18.)

The correspondence between the various points on the streamline $\skew{3}\bar{\psi}=0$ in the $z$ -, $\bar{F}$ -, and ${\mit\Omega}$ -planes is specified in Figure 6.18. The flow occupies the whole of the $\bar{F}$ -plane, apart from a thin slit running along the positive section of the real axis. As previously, the procedure is to find transformations that map the flow regions in both the $\bar{F}$ - and ${\mit\Omega}$ -planes coincidentally onto the upper half of the $\lambda$ -plane. The semi-infinite strip in the ${\mit\Omega}$ -plane has the same width, position, and orientation as that shown in Figure 6.15. Thus, the appropriate relationship between ${\mit\Omega}$ and $\lambda$ is again

$$\lambda = {\rm i}\,\sinh {\mit\Omega},$$ (6.112)

with the locations of the points $B$ and $B'$ again being $\lambda=-1$ and $\lambda=+1$ , respectively. The appropriate relationship between $\bar{F}$ and $\lambda$ can be recognized by noting, first, that the flow occupies the upper half of the $\bar{F}^{\,1/2}$ -plane, and, second, that an inversion and change of sign are needed to bring corresponding points on the two real axes into coincidence. Thus, we obtain

$$\lambda =-\left(\frac{k\,V}{\bar{F}}\right)^{1/2},$$ (6.113)

where $k$ is some positive constant.

The required relation between $\bar{F}$ and ${\mit\Omega}$ is given by

$$\lambda =- \left(\frac{k\,V}{\bar{F}}\right)^{1/2}= {\rm i}\,\sin... ...\rm i}}{2}\left(V\,\frac{dz}{d\bar{F}}-\frac{1}{V}\,\frac{d\bar{F}}{dz}\right).$$ (6.114)

Hence,

$$\frac{1}{V}\,\frac{d\bar{F}}{dz} =-{\rm i}\left(\frac{k\,V}{\bar{F}}\right)^{1/2}+\left(1-\frac{k\,V}{\bar{F}}\right)^{1/2},$$ (6.115)

where the relevant branch of $(1-k\,V/\bar{F})^{1/2}$ is that which is positive on $AO$ . Integration, making use of the fact that $\bar{F}=0$ when $z=0$ , yields

$$\frac{z}{k}=\int_0^{\bar{F}/k\,V}\frac{du}{-{\rm i}\,u^{\,-1/2}+(... ...{1/2}}=\int_0^{\bar{F}/k\,V}\left[\,{\rm i}\,u^{\,-1/2}+(1-1/u)^{1/2}\right]du,$$ (6.116)

or

$$\frac{z}{k} = \left(\frac{\bar{F}}{k\,V}\right)^{1/2}\left(\frac{... ...ight]+2\,{\rm i}\left(\frac{\bar{F}}{k\,V}\right)^{1/2}+{\rm i}\,\frac{\pi}{2}.$$ (6.117)

We can now evaluate the constant $k$ making use of the fact that at the point $B$ , where $\lambda=-1$ ,

$$z={\rm i}\,b, \frac{\bar{F}}{k\,V}= 1.$$ (6.118)

Here, $b$ is the semi-width of the plate. Thus, we obtain

$$k=\frac{2\,b}{\pi+4}.$$ (6.119)

On the free streamline $BC$ ,

$$\bar{F}=\skew{3}\bar{\phi}=V\,(k+s), {\mit\Omega}={\rm i}\,\theta,$$ (6.120)

where $s$ represents distance along the streamline measured from $B$ . Thus, on this streamline,

$$\lambda=-\sin\theta =-\left(\frac{k}{k+s}\right)^{1/2}.$$ (6.121)

Hence, according to Equation (6.117), the equation of the streamline $BC$ can be written, in parametric form, as

$$x =(k+s)^{1/2}\,s^{1/2} -k\ln\left[\left(1+\frac{s}{k}\right)^{1/2}+\left(\frac{s}{k}\right)^{1/2}\right],$$ (6.122)

$$y = 2\,(k+s)^{1/2}\,k^{1/2}+\frac{\pi}{2}\,k.$$ (6.123)

By symmetry, the equation of the free streamline $B'C'$ is obtained from the previous equation via the transformation $x\rightarrow x$ , $y\rightarrow -y$ . We deduce that the cavity downstream of the plate extends to infinity, and that its boundary asymptotes to the parabola

$$y^{\,2} = 4\,k\,x = \left(\frac{8\,b}{\pi+4}\right)x.$$ (6.124)

From Bernoulli's theorem, the net force per unit length exerted by the fluid on the plate, which is obviously normal to the plate, and, therefore, directed parallel to the $x$ -axis, is

$$D =\int_{-b}^b(p-p_0)_{x=0}\,dy=\int_{-b}^b\left(\frac{1}{2}\,\rho\,V^{\,2}-\frac{1}{2}\,\rho\,v^{\,2}\right)dy$$ (6.125)

$$=\rho\,V^{\,2}\,b-\rho\int_0^{k\,V}\,\frac{\partial\skew{3}\bar{\phi}}{\partial y}\,d\skew{3}\bar{\phi}.$$ (6.126)

In conventional parlance, this force constitutes a drag, because it is directed parallel to the undisturbed flow. (See Section 5.8.) Now, $\bar{F}/(k\,V)=\skew{3}\bar{\phi}/(k\,V)<1$ on $OB$ , so

$$\frac{1}{V}\,\frac{d\bar{F}}{dz} = -{\rm i}\,\gamma^{\,-1/2}+{\rm i}\,\left(\frac{1}{\gamma}-1\right)^{1/2},$$ (6.127)

where $\gamma=\skew{3}\bar{\phi}/(k\,V)$ . However, $d\bar{F}/dz=v_x-{\rm i}\,v_y=-{\rm i}\,\partial \skew{3}\bar{\phi}/\partial y$ . Hence,

$$D= \rho\,V^{\,2}\,b-\rho\, V^{\,2}\,k\int_0^1\left[\frac{1}{\gamma^{1/2}}-\left(\frac{1}{\gamma}-1\right)^{1/2}\right]d\gamma,$$ (6.128)

which yields

$$D= \left(\frac{2\pi}{\pi+4}\right)\rho\,V^{\,2}\,b.$$ (6.129)

Finally, if we define the drag coefficient, in the conventional manner (see Section 8.8), then we obtain

$$C_D=\frac{f}{\rho\,V^{\,2}\,b} = \frac{2\pi}{\pi+4}= 0.88.$$ (6.130)

We can extend the previous calculation to determine the flow past a flat plate inclined at an angle $\alpha$ to a liquid stream of infinite extent. Let the plate lie in the plane $x=0$ . The flow field then occupies a semi-infinite strip in the ${\mit\Omega}$ -plane that has the same width, position, and orientation as that in the previous calculation. Consequently, the appropriate relation between ${\mit\Omega}$ and $\lambda$ is again

$$\lambda={\rm i}\,\sinh{\mit\Omega} = \frac{{\rm i}}{2}\left(V\,\frac{dz}{d\bar{F}}-\frac{1}{V}\,\frac{d\bar{F}}{dz}\right).$$ (6.131)

It follows that

$$\frac{1}{V}\,\frac{d\bar{F}}{dz}= {\rm i}\,\lambda+\sqrt{1-\lambda^{\,2}}.$$ (6.132)

As before, the two edges of the plate, $B$ and $B'$ , correspond to $\lambda=-1$ and $\lambda=+1$ , respectively. Furthermore, the stagnation point, $O$ , on the front surface of the plate corresponds to $\vert\lambda\vert\rightarrow\infty$ . (See Figure 6.18.) However, the latter point is no longer necessarily located at the plate's midpoint.

The transformation (6.113) generalizes to give

$$\bar{F} = \frac{k\,V}{(\lambda-\cos\alpha)^2}.$$ (6.133)

At the points $A$ , $C$ , and $C'$ , $\vert\bar{F}\vert\rightarrow \infty$ and, hence, $\lambda\rightarrow \cos\alpha$ . (See Figure 6.18.) It follows from Equation (6.132) that $d\bar{F}/dz\rightarrow V\,(\sin\alpha+{\rm i}\,\cos\alpha)$ , which implies that $v_x\rightarrow V\,\sin\alpha$ and $v_y\rightarrow -V\,\cos\alpha$ . In other words, far upstream and downstream of the plate, the fluid velocity vector subtends an angle $\alpha$ with the plane $x=0$ .

On the front surface of the plate, $B'OB$ , $\vert\lambda\vert\geq 1$ and $d\bar{F}/dz=-{\rm i}\,V\,\partial\skew{3}\bar{\phi}/\partial y$ . Hence, from Equation (6.132),

$$\frac{1}{V}\,\frac{\partial\skew{3}\bar{\phi}}{\partial y}=-\lambda\pm \sqrt{\lambda^{\,2}-1}=\frac{-1}{\lambda\pm\sqrt{\lambda^{\,2}-1}},$$ (6.134)

where the upper/lower signs correspond to $\lambda>0$ and $\lambda<0$ , respectively. (The signs are chosen to ensure that $\partial\skew{3}\bar{\phi}/\partial y\rightarrow 0$ as $\vert\lambda\vert\rightarrow\infty$ .) Moreover, $\bar{F}=\skew{3}\bar{\phi}$ on $B'OB$ , so Equation (6.133) implies that

$$\skew{3}\bar{\phi} = \frac{k\,V}{(\lambda-\cos\alpha)^2}.$$ (6.135)

It follows that

$$\frac{d y}{d\lambda} = \frac{\partial y}{\partial\skew{3}\bar{\ph... ...2\,k}{(\lambda-\cos\alpha)^3}\,\left(\lambda\pm\sqrt{\lambda^{\,2}-1}\,\right).$$ (6.136)

Writing

$$\lambda = \frac{1-\cos\alpha\,\cos\omega}{\cos\alpha-\cos\omega},$$ (6.137)

the points $B$ , $O$ , and $B'$ correspond to $\omega=0$ , $\omega=\alpha$ , and $\omega=\pi$ , respectively. Furthermore,

$$\frac{d\lambda}{(\lambda-\cos\alpha)^3} =-\left(\frac{\cos\alpha-\cos\omega}{\sin^4\alpha}\right)\sin\omega\,d\omega,$$ (6.138)

$$\pm\sqrt{\lambda^{\,2}-1} =\frac{\sin\alpha\,\sin\omega}{\cos\alpha-\cos\omega}.$$ (6.139)

Hence, Equations (6.136)-(6.139) yield

$$dy = -\frac{2\,k}{\sin^4\alpha}\,(1-\cos\alpha\,\cos\omega+\sin\alpha\,\sin\omega)\,\sin\omega\,d\omega,$$ (6.140)

which can be integrated to give

$$y = \frac{k}{\sin^4\alpha}\left[2\,\cos\omega+\cos\alpha\,\sin^2\... ...pha\,\sin\omega\,\cos\omega+\sin\alpha\left(\frac{\pi}{2}-\omega\right)\right],$$ (6.141)

where we have chosen the constant of integration so as to ensure that the midpoint of the plate lies at $y=0$ . We expect $y=b$ when $\omega=0$ , where $b$ is the semi-width of the plate. We, thus, deduce, from the previous expression, that

$$k =\frac{2\,b\,\sin^4\alpha}{\pi\,\sin\alpha+4}.$$ (6.142)

In particular, the stagnation point, $O$ , at which $\omega=\alpha$ , lies a distance

$$y(\alpha)= \frac{2\,b}{\pi\,\sin\alpha+4}\left[2\,\cos\alpha\,(1+\sin^2\alpha)+\sin\alpha\left(\frac{\pi}{2}-\alpha\right)\right]$$ (6.143)

from the plate's midpoint.

Suppose that

$$\lambda =\cos\alpha+\zeta,$$ (6.144)

where $\vert\zeta\vert\ll 1$ . It follows, from Equation (6.133), that

$$\bar{F}=\frac{k\,V}{\zeta^{\,2}},$$ (6.145)

and, from Equation (6.132), that

$$\frac{1}{V}\,\frac{d\bar{F}}{dz'}\simeq 1+\frac{{\rm i}\,\zeta}{\sin\alpha},$$ (6.146)

where $z'=(\sin\alpha+{\rm i}\,\cos\alpha)\,z=x'+{\rm i}\,y'$ . Here, $x'$ and $y'$ are Cartesian coordinates oriented such that the unperturbed flow is parallel to the $x'$ -axis. The previous two expressions can be combined to give

$$\frac{z'}{k}\simeq \zeta^{\,-2}-\frac{2\,{\rm i}}{\zeta\,\sin\alpha}.$$ (6.147)

Now, $\zeta$ is real and negative on the free streamline $BC$ , and real and positive on the free streamline $B'C'$ . (See Figure 6.18.) We conclude that, in the small-$\vert\zeta\vert$ limit, the equations of these streamlines can be written, in parametric form, as

$$x' \simeq \frac{k}{\zeta^{\,2}},$$ (6.148)

$$y' \simeq -\frac{2\,k}{\zeta\,\sin\alpha}.$$ (6.149)

In other words, far downstream of the plate, the free streamlines, which form the boundaries of the cavity behind the plate, asymptote to the parabola

$$y'^{\,2} =\left(\frac{4\,k}{\sin^2\alpha}\right)x' = \left(\frac{8\,b\,\sin^2\alpha}{\pi\,\sin\alpha+4}\right)x'.$$ (6.150)

From Bernoulli's theorem, the net force per unit length exerted by the fluid on the plate, which is obviously normal to the plate, and, therefore, directed parallel to the $x$ -axis, is

$$f =\int_{-b}^b(p-p_0)_{x=0}\,dy=\frac{1}{2}\,\rho\,V^{\,2}\int_{-b}^b\left(1-\frac{v_y^{\,2}}{V^{\,2}}\right)_{x=0}dy$$

$$= \frac{1}{2}\,\rho\,V^{\,2}\int_{+1}^{\,-1}\left(1-\frac{v_y^{\,... ... y}{\partial\skew{3}\bar{\phi}}\,\frac{d\skew{3}\bar{\phi}}{d\lambda}\,d\lambda$$

$$=-\rho\,k\,V^{\,2}\int_{+1}^{\,-1}\left(\frac{V}{v_y}-\frac{v_y}{V}\right)\frac{d\lambda}{(\lambda-\cos\alpha)^3},$$ (6.151)

where use has been made of Equation (6.135). However, according to Equations (6.134) and (6.139),

$$\frac{V}{v_y}-\frac{v_y}{V}=V\,\frac{\partial y}{\partial\skew{3}... ...sqrt{\lambda^{\,2}-1}=-\frac{2\,\sin\alpha\,\sin\omega}{\cos\alpha-\cos\omega}.$$ (6.152)

Thus, making use of Equation (6.138), we obtain

$$f = \frac{2\,\rho\,k\,V^{\,2}}{\sin^3\alpha}\int_0^\pi\sin^2\omega\,d\omega = \frac{\pi\,\rho\,k\,V^{\,2}}{\sin^3\alpha}.$$ (6.153)

Hence, it follows from Equation (6.142) that

$$f =\left(\frac{2\pi\,\sin\alpha}{\pi\,\sin\alpha+4}\right)\rho\,V^{\,2}\,b.$$ (6.154)

The drag, $D$ , is the component of the force acting on the plate in the direction of the unperturbed flow: that is,

$$D = f\,\sin\alpha = \left(\frac{2\pi\,\sin^2\alpha}{\pi\,\sin\alpha+4}\right)\rho\,V^{\,2}\,b.$$ (6.155)

On the other hand, the lift, $L$ , is the component of the force acting perpendicular to the direction of the unperturbed flow. (See Section 5.8.) Thus,

$$L = f\,\cos\alpha = \left(\frac{2\pi\,\sin\alpha\,\cos\alpha}{\pi\,\sin\alpha+4}\right)\rho\,V^{\,2}\,b.$$ (6.156)

Finally, it can be seen, from a comparison between Equations (6.124), (6.129), (6.150), and (6.155), that there appears to be a connection between the drag force exerted on the plate, and the shape of the parabola to which the cavity formed behind the plate asymptotes. In fact,

$$\frac{D}{\rho\,V^{\,2}} =\pi\lim_{x\rightarrow\infty}\left(\frac{y^{\,2}}{4\,x}\right)_{\rm cavity~boundary},$$ (6.157)

where the unperturbed flow runs parallel to the $x$ -axis. It turns out that this relation is a general one for two-dimensional flow past a body of arbitrary shape with an attached cavity at ambient pressure (Batchelor 2000).