2 Engineering Example 1

2.1 Two-dimensional fluid flow

Introduction

Functions of a complex variable find a very elegant application in the mathematical treatment of two-dimensional fluid flow.

Problem in words

Find the forces and moments due to fluid flowing past a cylinder.

Mathematical statement of the problem

Figure 10 shows a cross section of a cylinder (not necessarily circular), whose boundary is $C$ , placed in a steady non-viscous flow of an ideal fluid; the flow takes place in planes parallel to the $xy$ plane. The cylinder is out of the plane of the paper. The flow of the fluid exerts forces and turning moments upon the cylinder. Let $X,Y$ be the components, in the $x$ and $y$ directions respectively, of the force on the cylinder and let $M$ be the anticlockwise moment (on the cylinder) about the orgin.

Figure 10

Blasius’ theorem (which we shall not prove) states that

$X-iY=\frac{1}{2}i\rho {\oint }_C\left(\right.\frac{dw}{dz}{\left)\right.}^{2}dz$ and $M=\text{Re}\left\{\right.-\frac{1}{2}\rho {\oint }_{C}z\left(\right.\frac{dw}{dz}{\left)\right.}^{2}dz\left\}\right.$

where Re denotes the real part, $\rho$ is the (constant) density of the fluid and $w=u+iv$ is the complex potential (see Section 261) for the flow. Both $\rho$ and $\omega$ are presumed known.

Mathematical analysis

We shall find $X,Y$ and $M$ if the cylinder has a circular cross section and the boundary is specified by $\left|z\right|=a$ . Let the flow be a uniform stream with speed $U$ .

Now, using a standard result, the complex potential describing this situation is:

$w=U\left(\right.z+\frac{a^2}{z}\left)\right.$ so that $\frac{dw}{dz}=U\left(\right.1-\frac{a^2}{z^2}\left)\right.$  and   $\left(\right.\frac{dw}{dz}{\left)\right.}^2=U^2\left(\right.1-\frac{2a^2}{z^2}+\frac{a^4}{z^4}\left)\right..$

Using Key Point 1 with $z_0=0:$

$X-iY=\frac{1}{2}i\rho {\oint }_C\left(\right.\frac{dw}{dz}{\left)\right.}^{2}dz=\frac{1}{2}i\rho U^2\oint \left(\right.1-\frac{2a^2}{z^2}+\frac{a^4}{z^4}\left)\right.dz=0$ so $X=Y=0$ .Also, $z\left(\right.\frac{dw}{dz}{\left)\right.}^2=U^2\left(\right.z-\frac{2a^2}{z}+\frac{a^4}{z^3}\left)\right.$ . The only term to contribute to $M$ is $\frac{-2a^2{U}^2}{z}$ .

Again using Key Point 1, this leads to $-4\pi a^2{U}^{2}i$ and this has zero real part. Hence $M=0$ , also.

Interpretation

The implication is that no net force or moment acts on the cylinder. This is not so in practice. The discrepancy arises from neglecting the viscosity of the fluid.

Exercises

1. Obtain the integral ${\int \; <!--nolimits\; -->}_{C}zdz$ along the straight-line paths
   1. from $z=2+2i$ to $z=5+2i$
   2. from $z=5+2i$ to $z=5+5i$
   3. from $z=2+2i$ to $z=5+5i$
2. Find ${\int \; <!--nolimits\; -->}_C\left(z^2+z\right)dz$ where $C$ is the part of the unit circle going anti-clockwise from the point $z=1$ to the point $z=i$ .
3. Find ${\oint }_{C}f\left(z\right)dz$ where $C$ is the circle $\left|z-z_0\right|=r$ for the cases
   1. $f\left(z\right)=\frac{1}{z^2}$ .   $z_0=1$
   2. $f\left(z\right)=\frac{1}{{\left(z-1\right)}^2},z_0=1$
   3. $f\left(z\right)=\frac{1}{z-1-i},z_0=1+i$

Answer