Engineering Example 3

Power transmitted through fluid-filled pipes is the basis of hydraulic braking systems and other hydraulic control systems. Suppose that power associated with a piston motion at one end of a pipeline is transmitted by a fluid of density $ρ$ moving with positive velocity $V$ along a cylindrical pipeline of constant cross-sectional area $A$ . Assuming that the loss of power is mainly attributable to friction and that the friction coefficient $f$ can be taken to be a constant, then the power transmitted, $P$ is given by

$P = ρ g A (h V - c V^{3})$ ,

where $g$ is the acceleration due to gravity and $h$ is the head which is the height of the fluid above some reference level ( $=$ the potential energy per unit weight of the fluid). The constant $c = \frac{4 f l}{2 g d}$ where $l$ is the length of the pipe and $d$ is the diameter of the pipe. The power transmission efficiency is the ratio of power output to power input.

Problem in words

Assuming that the head of the fluid, $h$ , is a constant find the value of the fluid velocity, $V$ , which gives a maximum value for the output power $P$ . Given that the input power is $P_{i} = ρ g A V h$ , find the maximum power transmission efficiency obtainable.

Mathematical statement of the problem

We are given that $P = ρ g A (h V - c V^{3})$ and we want to find its maximum value and hence maximum efficiency.

To find stationary points for $P$ we solve $\frac{d P}{d V} = 0.$

To classify the stationary points we can differentiate again to find the value of $\frac{d^{2} P}{d V^{2}}$ at each stationary point and if this is negative then we have found a local maximum point. The maximum efficiency is given by the ratio $P ∕ P_{i}$ at this value of $V$ and where $P_{i} = ρ g A V h .$ Finally we should check that this is the only maximum in the range of $P$ that is of interest.

Mathematical analysis

$P = ρ g A (h V - c V^{3})$

$\frac{d P}{d V} = ρ g A (h - 3 c V^{2})$

$\frac{d P}{d V} = 0$ gives $ρ g A (h - 3 c V^{2}) = 0$

$\Rightarrow V^{2} = \frac{h}{3 c} \Rightarrow V = \pm \sqrt{\frac{h}{3 c}}$ and as $V$ is positive $\Rightarrow V = \sqrt{\frac{h}{3 c}} .$

To show this is a maximum we differentiate $\frac{d P}{d V}$ again giving $\frac{d^{2} P}{d V^{2}} = ρ g A (- 6 c V)$ . Clearly this is negative, or zero if $V = 0$ . Thus $V = \sqrt{\frac{h}{3 c}}$ gives a local maximum value for $P$ .

We note that $P = 0$ when $E = ρ g A (h V - c V^{3}) = 0$ , i.e. when $h V - c V^{3} = 0$ , so $V = 0$ or $V = \sqrt{\frac{h}{C}}$ . So the maximum at $V = \sqrt{\frac{h}{3 C}}$ is the only max in this range between $0$ and $V = \sqrt{\frac{h}{C}}$ .

The efficiency $E$ , is given by (input power/output power), so here

E = \frac{ρ g A (h V - c V^{3})}{ρ g A V h} = 1 - \frac{c V^{2}}{h}

At $V = \sqrt{\frac{h}{3 c}}$ then $V^{2} = \frac{h}{3 c}$ and therefore $E = 1 - \frac{c \frac{h}{3 c}}{c} = 1 - \frac{1}{3} = \frac{2}{3}$ or $66 \frac{2}{3} %$ .

Interpretation

Maximum power transmitted through the fluid when the velocity $V = \sqrt{\frac{h}{3 c}}$ and the maximum efficiency is $66 \frac{2}{3}$ %. Note that this result is independent of the friction and the maximum efficiency is independent of the velocity and (static) pressure in the pipe.

Figure 19 :

{ Graphs of transmitted power as a function of fluid velocity for two values of the head}

Figure 19 shows the maxima in the power transmission for two different values of the head in an oil filled pipe (oil density $1100 kg m^{- 3}$ ) of inner diameter 0.01 m and coefficient of friction 0.01 and pipe length 1 m.

5 Engineering Example 3

5.1 Fluid power transmission