### 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

$\phantom{\rule{2em}{0ex}}X-iY=\frac{1}{2}i\rho {\oint }_{C}\left(\right\frac{dw}{dz}{\left)\right}^{2}dz$ and $M=\text{Re}\left\{\right\-\frac{1}{2}\rho {\oint }_{C}z\left(\right\frac{dw}{dz}{\left)\right}^{2}\phantom{\rule{0.3em}{0ex}}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:

$w=U\left(\rightz+\frac{{a}^{2}}{z}\left)\right$ so that $\frac{dw}{dz}=U\left(\right1-\frac{{a}^{2}}{{z}^{2}}\left)\right$  and   $\left(\right\frac{dw}{dz}{\left)\right}^{2}={U}^{2}\left(\right1-\frac{2{a}^{2}}{{z}^{2}}+\frac{{a}^{4}}{{z}^{4}}\left)\right.$

Using Key Point 1 with ${z}_{0}=0:$

$\phantom{\rule{2em}{0ex}}X-iY=\frac{1}{2}i\rho {\oint }_{C}\left(\right\frac{dw}{dz}{\left)\right}^{2}\phantom{\rule{0.3em}{0ex}}dz=\frac{1}{2}i\rho {U}^{2}\oint \left(\right1-\frac{2{a}^{2}}{{z}^{2}}+\frac{{a}^{4}}{{z}^{4}}\left)\right\phantom{\rule{0.3em}{0ex}}dz=0$ so $X=Y=0$ .Also, $\phantom{\rule{1em}{0ex}}z\left(\right\frac{dw}{dz}{\left)\right}^{2}={U}^{2}\left(\rightz-\frac{2{a}^{2}}{z}+\frac{{a}^{4}}{{z}^{3}}\left)\right$ . The only term to contribute to $M$ is $\frac{-2{a}^{2}{U}^{2}}{z}$ .

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 ${\int }_{C}z\phantom{\rule{0.3em}{0ex}}dz$ 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 ${\int }_{C}\left({z}^{2}+z\right)\phantom{\rule{0.3em}{0ex}}dz$ where $C$ is the part of the unit circle going anti-clockwise from the point $z=1$ to the point $z=i$ .
3. Find ${\oint }_{C}f\left(z\right)\phantom{\rule{0.3em}{0ex}}dz$ 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. $f\left(z\right)=\frac{1}{{\left(z-1\right)}^{2}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{z}_{0}=1$
3. $f\left(z\right)=\frac{1}{z-1-i},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{z}_{0}=1+i$
1. Here $y$ is constant along the given path $z=x+2i$ so that $u=x$ and $v=2$ . Also $dy=0$ . Thus $\begin{array}{rcll}{\int }_{C}z\phantom{\rule{0.3em}{0ex}}dz& =& {\int }_{C}\left(udx-vdy\right)+i{\int }_{C}\left(vdx+udy\right)={\int }_{2}^{5}xdx+i{\int }_{2}^{5}2dx\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}& \text{}\\ & =& {\left[\frac{{x}^{2}}{2}\right]}_{2}^{5}+i{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}2x\right]}_{2}^{5}=\left(\frac{25}{2}-\frac{4}{2}\right)+i\left(10-4\right)=\frac{21}{2}+6i.& \text{}\end{array}$
2. Here $dx=0$ , $v=y,\phantom{\rule{1em}{0ex}}u=5$ . Thus $\begin{array}{rcll}{\int }_{C}z\phantom{\rule{0.3em}{0ex}}dz& =& {\int }_{2}^{5}\left(-y\right)dy+i{\int }_{2}^{5}5dy& \text{}\\ & =& {\left[-\frac{{y}^{2}}{2}\right]}_{2}^{5}+i{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}5y\right]}_{2}^{5}=\left(-\frac{25}{2}+\frac{4}{2}\right)+i\left(25-10\right)=-\frac{21}{2}+15i.\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}& \text{}\end{array}$
3. $z=x+ix,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}u=x,\phantom{\rule{1em}{0ex}}v=x,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}dz=\left(1+i\right)dx$ , so

$\phantom{\rule{2em}{0ex}}{\int }_{C}z\phantom{\rule{0.3em}{0ex}}dz={\int }_{C}\left(xdx-xdx\right)+i{\int }_{C}\left(xdx+xdx\right)=i{\int }_{C}2xdx=2i{\left[\frac{{x}^{2}}{2}\right]}_{2}^{5}=21i.$

Note that the result in c. is the sum of the results in a. and b.

1. ${\int }_{C}\left({z}^{2}+z\right)\phantom{\rule{0.3em}{0ex}}dz={\left[\frac{{z}^{3}}{3}+\frac{{z}^{2}}{2}\right]}_{1}^{i}=\left(\frac{1}{3}{i}^{3}+\frac{{i}^{2}}{2}\right)-\left(\frac{1}{3}+\frac{1}{2}\right)=-\frac{4}{3}-\frac{1}{3}i$ .
2. Using Key Point 1 we have
1. 0,
2. 0,
3. $2\pi i$ .

Note that in all cases the result is independent of $r$ .