3 Engineering Example 1

3.1 Water wheel efficiency

Introduction

A water wheel is constructed with symmetrical curved vanes of angle of curvature $\theta$ . Assuming that friction can be taken as negligible, the efficiency, $\eta$ , i.e. the ratio of output power to input power, is calculated as

$\eta =\frac{2\left(V-v\right)\left(1+cos\theta \right)v}{V^2}$

where $V$ is the velocity of the jet of water as it strikes the vane, $v$ is the velocity of the vane in the direction of the jet and $\theta$ is constant. Find the ratio, $v∕V$ , which gives maximum efficiency and find the maximum efficiency.

Mathematical statement of the problem

We need to express the efficiency in terms of a single variable so that we can find the maximum value.

$\text{Efficiency}=\frac{2\left(V-v\right)\left(1+cos\theta \right)v}{V^2}=2\left(1-\frac{v}{V}\right)\frac{v}{V}\left(1+cos\theta \right).$

Let $\eta =$ Efficiency and $x=\frac{v}{V}$ then $\eta =2x\left(1-x\right)\left(1+cos\theta \right).$

We must find the value of $x$ which maximises $\eta$ and we must find the maximum value of $\eta$ . To do this we differentiate $\eta$ with respect to $x$ and solve $\frac{d\eta }{dx}=0$ in order to find the stationary points.

Mathematical analysis

Now $\eta =2x\left(1-x\right)\left(1+cos\theta \right)=\left(2x-2x^2\right)\left(1+cos\theta \right)$

So $\frac{d\eta }{dx}=\left(2-4x\right)\left(1+cos\theta \right)$

Now $\frac{d\eta }{dx}=0\Rightarrow 2-4x=0\Rightarrow x=\frac{1}{2}$ and the value of $\eta$ when $x=\frac{1}{2}$ is

$\eta =2\left(\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1+cos\theta \right)=\frac{1}{2}\left(1+cos\theta \right).$

This is clearly a maximum not a minimum, but to check we calculate $\frac{d^2\eta }{d{x}^2}=-4\left(1+cos\theta \right)$ which is negative which provides confirmation.

Interpretation

Maximum efficiency occurs when $\frac{v}{V}=\frac{1}{2}$ and the maximum efficiency is given by

$\eta =\frac{1}{2}\left(1+cos\theta \right).$