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

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

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

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

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

The head lost over the elemental length is given by:

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

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

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

Thus the head lost can also be written as:

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

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