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      <text>12.2.3.3 fluid power transmission example</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4><br>

<h5>Fluid power transmission</h5><br>

<h6>Introduction</h6>
<p>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 \(\rho\) 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 \[ P=\rho g A(hV -cV^3)\], 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 \(c=\dfrac{4fl}{2gd}\) 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.</p><br>

<h6>Problem in words</h6>
<p>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 \(P_i= \rho gAVh\), find the maximum power transmission efficiency
    obtainable.
</p><br>

<h6>Mathematical statement of the problem</h6>
<p>We are given that \(P=\rho gA(hV-cV^3)\) and we want to find its maximum value and hence maximum efficiency.</p>
<p>To find stationary points for \(P\) we solve \(\dfrac{ \mathrm{d} P}{ \mathrm{d} V}= 0.\)</p>
<p>To classify the stationary points we can differentiate again to find the value of \(\dfrac{ \mathrm{d}^2P}{ \mathrm{d} V^2}\) 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 \(P_i = \rho gAVh.\) Finally we should check that this is the only maximum in the range of \(P\) that is of interest.</p><br>

<h6>Mathematical analysis</h6>
<p>\[ P=\rho gA(hV-cV^3)\]\[ \dfrac{ \mathrm{d} P}{ \mathrm{d} V}=\rho gA(h-3cV^2)\]\[ \dfrac{ \mathrm{d} P}{ \mathrm{d} V}=0\;\; \mbox{ gives }\;\; \rho gA(h-3cV^2)=0\]\[ \Rightarrow\quad V^2=\dfrac{h}{3c}\quad \Rightarrow\quad V=\pm\ \sqrt{\dfrac{h}{3c}}\] and as \(V\) is positive \[\Rightarrow\quad
    V=\sqrt{\dfrac{h}{3c}}.\]
</p>
<p>To show this is a maximum we differentiate \(\dfrac{ \mathrm{d} P}{ \mathrm{d} V}\) again giving \(\dfrac{ \mathrm{d}^2P}{ \mathrm{d} V^2} =\rho gA(-6cV)\). Clearly this is negative, or zero if \(V=0\). Thus \(V =\sqrt{\dfrac{h}{3c}}\) gives a local maximum value for \(P\).</p>
<p>We note that \(P=0\) when \(E=\rho gA(hV-cV^3)=0\), i.e. when \(hV-cV^3=0\), so \(V=0\) or \(V=\sqrt{\dfrac{h}{C}}\). So the maximum at \(V=\sqrt{\dfrac{h}{3C}}\) is the only max in this range between \(0\) and \(V=\sqrt{\dfrac{h}{C}}\).</p>
<p>The efficiency \(E\), is given by (input power/output power), so here \[E=\frac{\rho gA(hV-cV^3)}{\rho gAVh}=1-\dfrac{cV^2}{h}\]</p>
<p>At \(V =\sqrt{\dfrac{h}{3c}}\) then \(V^2=\dfrac{h}{3c}\) and therefore \(E = 1-\dfrac{c\dfrac{h}{3c}}{c}=1-\dfrac{1}{3}=\dfrac{2}{3}\) or \(66\frac{2}{3}\%\).</p><br>

<h6>Interpretation</h6>
<p>Maximum power transmitted through the fluid when the velocity \(V=\sqrt{\dfrac{h}{3c}}\) 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.</p><br>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/12-2-diagram-power.svg" alt="Graphs of transmitted power as a function of fluid velocity for two values of the head." width="284" height="223" class="img-responsive atto_image_button_text-bottom" style="font-size: 0.9375rem;"></figure>

<p>The figure above shows the maxima in the power transmission for two different values of the head in an oil filled pipe (oil density \(1100 \text{kg m}^{-3}\)) of inner diameter \(0.01 \text{m}\) and coefficient of friction \(0.01\) and pipe length \(1
    \text{m}\).
</p>]]></text>
<file name="12-2-diagram-power.svg" path="/" encoding="base64">[encoded data, 58160 characters]</file>
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