Blasius Theorem

Consider some flow pattern in the complex $z$ -plane that is specified by the complex velocity potential $F(z)$ . Let $C$ be some closed curve in the complex $z$ -plane. The fluid pressure on this curve is determined from Equation (6.41), which yields

$$P = p_0 -\frac{1}{2}\,\rho\left\vert\frac{dF}{dz}\right\vert^{\,2}.$$ (6.173)

Let us evaluate the resultant force (per unit length), and the resultant moment (per unit length), acting on the fluid within the curve as a consequence of this pressure distribution.

Figure 6.19: Force acting across a short section of a curve.

Consider a small element of the curve $C$ , lying between $x$ , $y$ and $x+dx$ , $y+dy$ , which is sufficiently short that it can be approximated as a straight-line. Let $P$ be the local fluid pressure on the outer (i.e., exterior to the curve) side of the element. As illustrated in Figure 6.19, the pressure force (per unit length) acting inward (i.e., toward the inside of the curve) across the element has a component $P\,dy$ in the minus $x$ -direction, and a component $P\,dx$ in the plus $y$ -direction. Thus, if $X$ and $Y$ are the components of the resultant force (per unit length) in the $x$ - and $y$ -directions, respectively, then

$$dX =-P\,dy,$$ (6.174)

$$dY =P\,dx.$$ (6.175)

The pressure force (per unit length) acting across the element also contributes to a moment (per unit length), $M$ , acting about the $z$ -axis, where

$$dM = x\,dY-y\,dX = P\,(x\,dx+y\,dy).$$ (6.176)

Thus, the $x$ - and $y$ -components of the resultant force (per unit length) acting on the of the fluid within the curve, as well as the resultant moment (per unit length) about the $z$ -axis, are given by

$$X =-\oint_C P\,dy,$$ (6.177)

$$Y =\oint_C P\,dx,$$ (6.178)

$$M =\oint_C P\,(x\,dx+y\,dy),$$ (6.179)

respectively, where the integrals are taken (counter-clockwise) around the curve $C$ . Finally, given that the pressure distribution on the curve takes the form (6.173), and that a constant pressure obviously yields zero force and zero moment, we find that

$$X =\frac{1}{2}\,\rho\oint_C \left\vert\frac{dF}{dz}\right\vert^{\,2}dy,$$ (6.180)

$$Y =-\frac{1}{2}\,\rho\oint_C \left\vert\frac{dF}{dz}\right\vert^{\,2}dx,$$ (6.181)

$$M =-\frac{1}{2}\,\rho\oint_C \left\vert\frac{dF}{dz}\right\vert^{\,2}(x\,dx+y\,dy).$$ (6.182)

Now, $z=x +{\rm i}\,y$ , and $\skew{2}\bar{z}=x-{\rm i}\,y$ , where $\bar{~}$ indicates a complex conjugate. Hence, $d\skew{2}\bar{z} = dx -{\rm i}\,dy$ , and ${\rm i}\,d\skew{2}\bar{z} = dy+{\rm i}\,dx$ . It follows that

$$X - {\rm i}\,Y = \frac{1}{2}\,{\rm i}\,\rho\oint_C \left\vert\frac{dF}{dz}\right\vert^{\,2}d\skew{2}\bar{z}.$$ (6.183)

However,

$$\left\vert\frac{dF}{dz}\right\vert^{\,2}d\skew{2}\bar{z} = \frac{... ...,\frac{d\bar{F}}{d\skew{2}\bar{z}}\,d\skew{2}\bar{z} = \frac{dF}{dz}\,d\bar{F},$$ (6.184)

where $dF = d\phi+{\rm i}\,d\psi$ and $d\bar{F}=d\phi-{\rm i}\,d\psi$ . Suppose that the curve $C$ corresponds to a streamline of the flow, in which case $\psi={\rm constant}$ on $C$ . Thus, $d\psi=0$ on $C$ , and so $d\bar{F}=dF$ . Hence, on $C$ ,

$$\frac{dF}{dz}\,d\bar{F} = \frac{dF}{dz}\,dF = \left(\frac{dF}{dz}\right)^{\,2}dz,$$ (6.185)

which implies that

$$X - {\rm i}\,Y = \frac{1}{2}\,{\rm i}\,\rho\oint_C \left(\frac{dF}{dz}\right)^{\,2}dz.$$ (6.186)

This result is known as the Blasius theorem, after Paul Blasius (1883-1970).

Now, $x\,dx+ y\,dy={\rm Re}(z\,d\skew{2}\bar{z})$ . Hence,

$$M = {\rm Re}\left(-\frac{1}{2}\,\rho \oint_C \left\vert\frac{dF}{dz}\right\vert^{\,2} z\,d\skew{2}\bar{z}\right),$$ (6.187)

or, making use of an analogous argument to that employed previously,

$$M = {\rm Re}\left[-\frac{1}{2}\,\rho \oint_C \left(\frac{dF}{dz}\right)^{\,2} z\,dz\right],$$ (6.188)

In fact, Equations (6.186) and (6.188) hold even when $\psi$ is not constant on the curve $C$ , as long as $C$ can be continuously deformed into a constant-$\psi$ curve without leaving the fluid or crossing over a singularity of $(dF/dz)^{\,2}$ .

Figure 6.20: Source in the presence of a rigid boundary.

As an example of the use of the Blasius theorem, consider again the situation, discussed in Section 6.6, in which a line source of strength $Q$ is located at $(0$ , $a)$ , and there is a rigid boundary at $y=0$ . As we have seen, the complex velocity in the region $y>0$ takes the form

$$\frac{dF}{dz} = -\frac{Q}{\pi}\,\frac{z}{z^{\,2}+a^{\,2}}.$$ (6.189)

Suppose that we evaluate the Blasius integral, (6.188), about the contour $C$ shown in Figure 6.20. This contour runs along the boundary, and is completed by a semi-circle in the upper half of the $z$ -plane. As is easily demonstrated, in the limit in which the radius of the semi-circle tends to infinity, the contribution of the curved section of the contour to the overall integral becomes negligible. In this case, only the straight section of the contour contributes to the integral. Note that the straight section corresponds to a streamline (because it is coincident with a rigid boundary). In other words, the contour $C$ corresponds to a streamline at all constituent points that make a finite contribution to the Blasius integral, which ensures that $C$ is a valid contour for the application of the Blasius theorem. In fact, the Blasius integral specifies the net force (per unit length) exerted on the whole fluid by the boundary. Observe, however, that the contour $C$ can be deformed into the contour $C'$ , which takes the form of a small circle surrounding the source, without passing over a singularity of $(dF/dz)^{\,2}$ . (See Figure 6.20.) Hence, we can evaluate the Blasius integral around $C'$ without changing its value. Thus,

$$X - {\rm i}\,Y = \frac{1}{2}\,{\rm i}\,\rho\oint_{C'}\left(\frac{... ...ft(\frac{Q}{\pi}\right)^2\oint_{C'}\frac{z^{\,2}}{(z^{\,2}+a^{\,2})^{\,2}}\,dz,$$ (6.190)

or

$$X-{\rm i}\,Y = \frac{1}{8}\,{\rm i}\,\rho\left(\frac{Q}{\pi}\righ... ...}+\frac{2}{(z+{\rm i}\,a)\,(z-{\rm i}\,a)} + \frac{1}{(z+{\rm i}\,a)}\right]dz.$$ (6.191)

Writing $z={\rm i}\,a+\epsilon\,{\rm e}^{\,{\rm i}\,\theta}$ , $dz = {\rm i}\,\epsilon\,{\rm e}^{\,{\rm i}\,\theta}\,d\theta$ , and taking the limit $\epsilon\rightarrow 0$ , we find that

$$X - {\rm i}\,Y = \frac{{\rm i}\,\rho\,Q^{\,2}}{4\pi\,a}.$$ (6.192)

In other words, the boundary exerts a force (per unit length) ${\bf F} = -(\rho\,Q^{\,2}/4\pi\,a)\,{\bf e}_y$ on the fluid. Hence, the fluid exerts an equal and opposite force $-{\bf F} = (\rho\,Q^{\,2}/4\pi\,a)\,{\bf e}_y$ on the boundary. Of course, this result is consistent with Equation (6.47). Incidentally, it is easily demonstrated from Equation (6.188) that there is zero moment (about the $z$ -axis) exerted on the boundary by the fluid, and vice versa.

Consider a line source of strength $Q$ placed (at the origin) in a uniformly flowing fluid whose velocity is ${\bf V} = V_0\,(\cos\theta_0$ , $\sin\theta_0)$ . From Section 6.4, the complex velocity potential of the net flow is

$$F(z)=-\frac{Q}{2\pi}\,\ln z - V_0\,z\,{\rm e}^{-{\rm i}\,\theta_0}.$$ (6.193)

The net force (per unit length) acting on the source (which is calculated by performing the Blasius integral around a large loop that follows streamlines, and then shrinking the loop to a small circle centered on the source) is (see Exercise 1)

$${\bf F} =- \rho\,Q\,{\bf V}.$$ (6.194)

This force acts in the opposite direction to the flow. Thus, an external force $-{\bf F}$ , acting in the same direction as the flow, must be applied to the source in order for it to remain stationary. In fact, the previous result is valid even in a non-uniformly flowing fluid, as long as ${\bf V}$ is interpreted as the fluid velocity at the location of the source (excluding the velocity field of the source itself).

Finally, consider a vortex filament of intensity ${\mit\Gamma}$ placed at the origin in a uniformly flowing fluid whose velocity is ${\bf V} = V_0\,(\cos\theta_0$ , $\sin\theta_0)$ . From Section 6.4, the complex velocity potential of the net flow is

$$F(z)={\rm i}\,\frac{{\mit\Gamma}}{2\pi}\,\ln z - V_0\,z\,{\rm e}^{-{\rm i}\,\theta_0}.$$ (6.195)

The net force (per unit length) acting on the filament (which is calculated by performing the Blasius integral around a small circle centered on the filament) is (see Exercise 2)

$${\bf F} =\rho\,{\mit\Gamma}\,{\bf V}\times{\bf e}_z.$$ (6.196)

This force is directed at right-angles to the direction of the flow (in the sense obtained by rotating ${\bf V}$ through $90^{\circ}$ in the opposite direction to the filament's direction of rotation). Again, the previous result is valid even in a non-uniformly flowing fluid, as long as ${\bf V}$ is interpreted as the fluid velocity at the location of the filament (excluding the velocity field of the filament itself).