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Tapered Pipe

Pressure head loss within a tapered pipe due to friction

Overview

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 $\inline dx$ which is st a distance $\inline x$ from position 1, and let the velocity and diameter associated with this short length be $\inline v$ and $\inline d$ .

The head lost over the element

$=\frac{4fv^2dx}{2dg}$

But,

$\frac{\displaystyle\frac{1}{2}(d_1-d)}{x}=\tan\theta$

Note that $\inline \theta$ 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 $\inline l$ 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:

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

Last Modified: 23 Nov 11 @ 12:57 Page Rendered: 2022-03-14 17:39:35