I have forgotten

• https://me.yahoo.com

# Branched Pipes

The application of the Bernoulli equation to Branched pipes
View other versions (5)

### Key Facts

Gyroscopic Couple: The rate of change of angular momentum ($\inline&space;\tau$) = $\inline&space;I\omega\Omega$ (In the limit).
• $\inline&space;I$ = Moment of Inertia.
• $\inline&space;\omega$ = Angular velocity
• $\inline&space;\Omega$ = Angular velocity of precession.

## Introduction

It is common for a pipeline to be branched and for the system to be feeding more than one reservoir. This page examines this situation.

Bernoulli's principle states that for an inviscid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. Bernoulli's principle can be applied to various types of fluid flow, resulting in what is loosely denoted as Bernoulli's equation.

## Branched Pipes

##### MISSING IMAGE!

Applying Bernoulli's equation to the whol system but neglecting both the entry head and the junction head.

$H_1=\frac{4fl_1{v_{1}}^{2}}{2d_1g}+\frac{4fl_2{v_{2}}^{2}}{2d_2g}+\frac{{v_{2}}^{2}}{2g}$
$H_2=\frac{4fl_1{v_{1}}^{2}}{2d_1g}+\frac{4fl_3{v_{3}}^{2}}{2d_3&space;g}+\frac{{v_{2}}^{2}}{2g}$

It most cases it is possible to neglect the last terms of $\inline&space;\displaystyle\frac{v^2}{2g}$.

Applying the continuity equation:
$a_1v_1=a_2v_2+a_3v_3$
or
${d_{1}}^{2}v_1={d_{2}}^{2}2v_2+{d_{3}}^{2}v_3$

Example:
##### Example - Example 1
Problem
Water is pumped from a river to two reservoirs $\inline&space;A$ and $\inline&space;B$. The water surface in reservoir $\inline&space;A$ is at the same hight as the river whilst that in reservoir $\inline&space;B$ is 20 ft. higher.

Pumping from the river takes place by means of a centrifugal pump, the equation relating flow $\inline&space;Q$ (in cubic ft./sec.) and $\inline&space;H$ ft. at a constant speed being given by $\inline&space;H=75-10\;Q^2$

From the river to a junction $\inline&space;J$ is a common pipe is used of 8 in. diameter and 500 ft. long. The branch $\inline&space;J$ to the reservoir $\inline&space;A$ is 5 in. in diameter and 200 ft. long. The branch from $\inline&space;J$ to reservoir $\inline&space;B$ is 6 in. in diameter and 200 ft. long.

Neglecting all losses other than pipe friction, calculate the discharge to $\inline&space;A$ and $\inline&space;B$. Take $\inline&space;f$ as 0.007 throughout.

Workings
Darcy's equation can be rewritten as follows:

$h_f=\frac{4flv^2}{2dg}=\frac{flQ^2}{10d^5}$

Applying Bernoulli at the river and reservoir $\inline&space;A$:

$H=\frac{0.007\times&space;5000\times&space;Q^2}{10\times&space;\left&space;(&space;\tfrac{8}{12}&space;\right&space;)^5}+\frac{0.007\times&space;2000\times&space;{Q_{A}}^{2}}{10\times&space;\left&space;(&space;\tfrac{5}{12}&space;\right&space;)^5}=75-10\;Q^2$

$\therefore&space;\;\;\;\;\;\;26.57Q^2+111.5{Q_{A}}^{2}=75-10Q^2$

Similarly:
$H-20=\frac{0.007\times&space;5000\times&space;Q^2}{10\times&space;\left&space;(&space;\tfrac{8}{12}&space;\right&space;)^5}+\frac{0.007\times&space;2000\times&space;{Q_{B}}^{2}}{10\times&space;\left&space;(&space;\tfrac{6}{12}&space;\right&space;)^5}=75-10\;Q^2-20$

$\therefore&space;\;\;\;\;\;\;26.57Q^2+44.8{Q_{A}}^{2}=55-10Q^2$

But by continuity:
$Q=Q_A+Q_B$

From equation (1)
$Q^2=2.05-3.05{Q_{A}}^{2}$

Subtracting equation (2) from (1)
$111.5{Q_{A}}^{2}-44.8{Q_{B}}^{2}=20$
$\therefore&space;\;\;\;\;\;\;{Q_{A}}^{2}=2.49{Q_{B}}^{2}-0.446$

Substituting into equation (3) squared with values for $\inline&space;Q$ from equation (4) gives:
$2.05-3.05{Q_{A}}^{2}={Q_{A}}^{2}+2.49{Q_{A}}^{2}-0.446+20A\;\sqrt{2.49{Q_{A}}^{2}-0.446}$

Rearranging and collecting terms:
$2.496-6.54{Q_{A}}^{2}=2Q_A\sqrt{2.49{Q_{A}}^{2}-0.446}$

Squaring gives:
$32.84{Q_{A}}^{4}-30.77{Q_{A}}^{2}+6.23=0$

Treating this as a quadratic in $\inline&space;{Q_{A}}^{2}$
${Q_{A}}^{2}=0.297$
And:
$Q_A=0.545\;&space;cusec.$

From equation (4)
$Q_B=0.540\;cusec.$
Solution
$Q_A=0.545\;&space;cusec.$
$Q_B=0.540\;cusec.$