Tapered Pipe

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, $v=\left ( \frac{d_1}{d} \right )^2v_1$

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

Thus the head lost can also be written as:

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