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Tapered Pipe

Pressure head loss within a tapered pipe due to friction

Overview

23287/uniform_run_off_and_tapered_pipes_004.png
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Let the velocity and diameter at the sections 1 and 2 be as shown on the diagram. Consider the head lost over a short length dx which is st a distance x from position 1, and let the velocity and diameter associated with this short length be v and d.

The head lost over the element
=\frac{4fv^2dx}{2dg}

But,
\frac{\displaystyle\frac{1}{2}(d_1-d)}{x}=\tan\theta
Note that \theta is the half angle of the taper.


Or,
x=\frac{d_1-d}{2\tan\theta}
\therefore \;\;\;\;\;\;dx=-\frac{d(d)}{2\tan\theta}
Using the continuity equation:

{d_{1}}^{2}v_1=d^2v
Or,
v=\left ( \frac{d}{d_2} \right )^2v_1

The head lost over the elemental length is given by:

h_{fdx}=\frac{4f}{2dg}\left ( \frac{d_1}{d} \right )^4{v_{1}}^{2}\times- \frac{d(d)}{2\tan \theta}
Thus the head lost over the total length l is:

h_f=-\frac{f{v_{1}}^{2}{d_{1}}^{4}}{g\tan\theta}\int_{d_1}^{d_2}\frac{d(d)}{d^{\;5}}
h_f=\frac{f{v_{1}}^{2}{d_{1}}^{4}}{4g\tan\theta}\left [ \frac{1}{{d_{2}}^{4}} -\frac{1}{{d_{1}}^{4}}\right ]
h_f=\frac{f{v_{1}}^{2}}{4g\tan\theta}\left [ \left (  \frac{d_1}{d_{2} \right )^4} -1\right ]

Putting:
\tan\theta=\frac{1}{2}\times \frac{d_1-d_2}{l}
h_f=\frac{f{v_{1}}^{2}\times 2l}{4g(d_1-d_2)}\left ( \frac{{d_{1}}^{4}-{d_{2}}^{4}}{{d_{2}}^{4}} \right )
=\frac{f{v_{1}}^{2}\times l}{2gd_1d_2}\times \frac{({d_{1}}^{2}-{d_{2}}^{2})({d_{1}}^{2}+{d_{2}}^{2})}{{d_{2}}^{4}}

Thus the head lost can also be written as:

h_f=\frac{f{v_{1}^{2}}\times l}{2g{d_{2}}^{4}}\times (d_1+d_2)({d_{1}}^{2}+{d_{2}}^{2})