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# Incompressible Flow

An anlysis of Water Hammer based on the assumption that water is incompressible

## Overview

Using The Rigid Column Theory the compressibility of the fluid is ignored and it is assumed that pressure changes caused by opening or closing a valve are felt instantaneously through out the pipe. In effect the water column is a solid column, which can accelerate or decelerated as an entity.

## The Gradual Valve Closure

A valve is a device that regulates, directs or controls the flow of a fluid by opening, closing, or partially obstructing various passageways.

As the valve closes the pressure at the valve rises decelerating the water column.

Let $\inline&space;p$ be the pressure rise above the initial steady flow pressure, $\inline&space;\alpha$ the deceleration of the water column, $\inline&space;p$ the rise in pressure, $\inline&space;a$ the cross sectional area and $\inline&space;l$ the length of the pipe.

The increase in Pressure Head $\inline&space;H$ is given by:
$H=\frac{p}{w}=-\frac{l}{g}\times&space;\alpha=-\frac{l}{g}\left&space;(&space;\frac{dv}{dt}&space;\right&space;)&space;/)&space;&space;&space;&space;\begin{quote}&space;&space;By&space;Newton's&space;Second&space;Law:&space;&space;&space;&space;Pressure&space;at&space;Valve=Mass&space;of&space;water&space;multiplied&space;to&space;Deceleration&space;&space;&space;&space;\(\therefore\;\;\;pa=\frac{w\times&space;a\times&space;l}{g}\times\alpha$
$\therefore\;\;\;p=\frac{w\timesl}{g}$

i.e.
$H=\frac{p}{w}=-\frac{l}{g}\times\frac{dv}{dt}$
{quote}

## Valve Closed To Produce Constant Acceleration

Acceleration is the rate of change of velocity over time.

$T=\frac{w}{g}\times&space;\frac{l}{p}\times&space;v_0$

Let the time taken to close the valve be $\inline&space;T$ seconds.

Then:
$\alpha=\frac{v}{T}$
$\therefore\;\;\;\;p=\frac{wl}{g}\times\frac{v_0}{T}\times&space;H=+\frac{l\times&space;v_0}{gT}$

So for any given pipeline, if the maximum pressure rise is specified, the time $\inline&space;T$ taken to close the valve is directly proportional to the initial water velocity.

$T=\frac{w}{g&space;}\times&space;\frac{l}{p}}\times&space;v_0$

## Valve Closed To Produce Uniform Rate Of Pressure Increase

Pressure is the force per unit area applied in a direction perpendicular to the surface of an object.

The maximum pressure rise
$=&space;\frac{wl}{g}\times&space;\alpha_{max}=\frac{wl}{g}\times&space;KT$

Where $\inline&space;K$ is a constant and $\inline&space;T$ is the total time of valve closure.

At any instant during closure
$\alpha=-K&space;\times&space;t$
$\frac{dv}{dt}=&space;-K\times&space;t$
$\therefore\:\;\;v=&space;-K\times&space;t^2+C$
where $\inline&space;C$ is Constant.
The values of $\inline&space;K$ and $\inline&space;C$ can be found from the known initial and final conditions $\inline&space;v=v_0$ at $\inline&space;t=0$ and $\inline&space;v=0$ when $\inline&space;t=T$.

then:
$\alpha_{max}=K\times&space;T_{max}$
And the maximum pressure rise
$=\frac{wl}{g}\times\alpha_{max}=\frac{w\times&space;l}{g}\times&space;K\:T$

## The Valve Closes So That The Area Decreases Uniformly With Time

$\left(\frac{h_m}{h_0}&space;\right)&space;=&space;\frac{K}{2}\;+\:\sqrt[]{\frac{K^2}{4}\:+\:K}$

This equation again relates the maximum pressure rise directly to the time of valve closure. The difference between this and the first case is that the rate of increase in $\inline&space;p$ has been controlled. The equation gives for a known pipe, the maximum pressure rise given any closure time. Note: The suffix "o" refers to the initial conditions when the head behind the valve and is equal to the gross supply head if the friction in the pipe is neglected and $\inline&space;h$ is the Inertia head at a time $\inline&space;t$.

For Valve opening
$\frac{h_m}{h_0}&space;=&space;\frac{K}{2}&space;-&space;\sqrt[]{\frac{K^2}{4}\:+\:K}$

A nozzle is a device designed to control the direction or characteristics of a fluid flow as it exits an enclosed chamber or pipe via an orifice.

Consider the valve as a nozzle at the end of the pipe with a $\inline&space;C_d$ which is constant.

Initial flow
$Q_0=av_o&space;=&space;C_d&space;a_{v_{0}}\sqrt{2gh_0}$

At a time $\inline&space;t$ the discharge
$Q&space;=&space;a&space;v&space;=&space;C_d\:a_v\:\sqrt[]{2\,g\,(h_0\:-\:h)}$

$\therefore\;\;\;\;\frac{Q}{Q_0}\:=&space;\frac{v}{v_0}\:=&space;\frac{a_v}{a_{v_0}}}$

But:
$a_v&space;=&space;a_{v_0}}\times&space;left(1-\frac{t}{T})&space;\right)$

$\therefore\;\;\;\;\frac{v}{v_0}&space;=&space;\left(1-\frac{t}{T}&space;\right)\sqrt[]{1+\frac{h}{h_0}}$

$\therefore\;\;\;\;{v}&space;=&space;v_0\left(1-\frac{t}{T}&space;\right)\sqrt[]{1+\frac{h}{h_0}}$
$\therefore\;\;\;\frac{dv}{dt}=&space;-&space;\frac{v_0}{T}\sqrt[]{1+\frac{h}{h_0}}+&space;&space;v_0\left(1-\frac{t}{T}&space;\right)\frac{1}{2}\left(1+\frac{h}{h_0}&space;\right)}^{-&space;&space;\frac{1}{2}}\frac{dh}{dt}$

When $\inline&space;h$ is a maximum:
$\frac{dh}{dt}\&space;=\&space;0$

$\therefore\;\;\;\;\;\left\(\frac{dv}{dt}&space;\right)_{h\,=\,h_{max}}=-\frac{v_0}{T}\sqrt[]{1+\frac{h}{h_0}}$

$h_m&space;=&space;-&space;\left(\frac{dv}{dt}&space;\right)_{h=h_{max}}&space;=&space;\frac{l}{g}\frac{v_0}{T}\sqrt[]{1+\frac{h_m}{h_0}}$

$\therefore\;\;\;\frac{h_m}{h_0}=&space;\frac{l}{g}\frac{v_0}{Th_0}\sqrt[]{1+\frac{h_m}{h_0}}$

Putting:
$K&space;=&space;\left(&space;\frac{l}{g}\frac{v_0}{Th_0}&space;\right)^2$
$\therefore\;\;\;\left(\frac{h_m}{h_0}&space;\right)^2&space;=&space;K\left(1&space;+&space;\frac{h_m}{h_0}&space;\right)$
$\therefore\;\;\;\left(\frac{h_m}{h_0}&space;\right)^2&space;-&space;K\left(\frac{h_m}{h_0}&space;&space;&space;\right)-K&space;=&space;0$

$\left(\frac{h_m}{h_0}&space;\right)&space;=&space;\frac{K}{2}+\sqrt[]{\frac{K^2}{4}+K}$

This equation again relates the maximum pressure rise directly to the time of valve closure. The difference between this and the first case is that the rate of increase in $\inline&space;p$ has been controlled. The equation gives for a known pipe, the maximum pressure rise given any closure time.

Which must be the positive root for valve closure.

For valve opening.
$\frac{h_m}{h_0}&space;=&space;\frac{K}{2}&space;-&space;\sqrt[]{\frac{K^2}{4}\:+\:K}$

For pipe 1
$h_1=-\frac{1}{g}\left&space;(&space;l_1+\frac{a_1}{a_2}l_2&space;\right&space;)\;\frac{dv_1}{dt}$

Note: The above expression is the same as equation (1) with an "equivalent" pipe length of:
$l=&space;l_1+\frac{a_1}{a_2}l_2$

And for a number of pipe sections of $\inline&space;n$ different cross sectional areas:
$l=&space;l_1+\frac{a_1}{a_2}l_2+\frac{a_1}{a_3}l_3+.........\frac{a_1}{a_n}l_n$

These equations allow the calculation of an "overall length" which can be used to calculate pressure rises under differing valve opening regimes.

Applying Newton's Second Law for each pipe:

$wh_0+wh_2a_2-wh_0a_2=\frac{wa_2l_2}{g}(-\frac{dv_2}{dt})$
$\therefore\;\;\;\;\;h_2&space;=&space;-\frac{l_2}{g}\,\frac{dv_2}{dt}$

$\therefore\;\;\;\;\;w(h_0+h_1)a_1&space;w(h_0-h_2)\times&space;a_1$
$=\frac{wa_1l_1}{g}\times&space;\left&space;(-&space;\frac{dv_1}{dt}&space;\right&space;)$
$\therefore\;\;\;\;\;h_1-h_2&space;=&space;-\frac{l}{g}\times&space;\frac{dv_1}{dt}$
By continuity:
$a_1v_1&space;=&space;a_2v_2$
$\therefore\;\;\;\;\;a_1\times&space;\frac{dv_1}{dt}&space;=&space;a_2\frac{dv_2}{dt}$

Putting equation (9) into equation (7)
$h_2&space;=&space;-\frac{l_2}{g}\times&space;\frac{a_1}{a_2}\times&space;\frac{dv_1}{dt}$

Equation (10) can now be inserted into equation (8)
$\therefore\;\;\;\;\;h_1&space;-\left&space;(&space;-\frac{l_1}{g}\times&space;\frac{dv_1}{dt}&space;\right&space;)&space;&space;-\frac{l_2}{g}\times&space;\frac{a_1}{a_2}\times&space;\frac{dv_1}{dt}$
$=&space;-\:\frac{1}{g}\:\left(l_1\:+\:\frac{a_1}{a_2}\:l_2&space;\right)\:\frac{dv_1}{dt}$

Note: The above expression is the same as equation (1) with an "equivalent" pipe length of:
$l=&space;l_1+\frac{a_1}{a_2}l_2$

And for a number of pipe sections of $\inline&space;n$ different cross sectional areas:
$l=&space;l_1+\frac{a_1}{a_2}l_2+\frac{a_1}{a_3}l_3+.........\frac{a_1}{a_n}l_n$

These equations allow the calculation of an "overall length" which can be used to calculate pressure rises under differing valve opening regimes.

N.B. The following two examples are based on the assumption that water is not compressible.

Example:
[imperial]
##### Example - Example 1
Problem

Find the maximum increase in pressure head if the valve closes in 12 seconds.
• a) So that the velocity decreases uniformly with time.
• b) So that the pressure rises uniformly with time.
• c) So that the valve discharge area decreases uniformly with time.
• d) The last 1000 ft. of pipe is 8 ft. diam. and the velocity decreases uniformly with time.
Workings
Initial velocity $\inline&space;V_0$
$=&space;&space;\frac{1100}{\displaystyle\frac{\pi}{4}\times&space;100}&space;=&space;14.01\:ft./sec.$

• a) Velocity decreases uniformly with time.

Using equation (2) The Inertia Head
$h&space;=&space;&space;\frac{l}{g}\frac{v_0}{T}$
$=&space;\frac{3000}{32.2}\times&space;\frac{14.01}{12}&space;=&space;108.8\:ft.$

• b) Pressure rises uniformly with time.

Using equation (3)
$v=&space;-&space;\frac{K}{2}\:t^2\:+\:C$

When $\inline&space;T&space;=&space;0$ $\inline&space;v&space;=&space;v_0$ which equals 14.01 ft./sec.

When $\inline&space;t$ = 12sec. $\inline&space;v&space;=&space;0$
$\therefore\;\;\;\;\;0&space;=&space;1\frac{K}{2}\times&space;144+14.01$
$\therefore\;\;\;\;\;K&space;=&space;\frac{14.01}{72}$
Thus:
$\alpha=&space;\frac{14.01}{72}\times&space;t$
And:
$\alpha_{max}=\frac{14.01\times&space;12}{72}&space;=&space;\frac{14.01}{6}\:ft,/sec^2$
But from equation (4)
$h_{max}&space;=&space;\frac{l}{g}\;\alpha_{max}$
$=&space;\frac{3000}{32.2}\;X\;\frac{14.01}{6}&space;=&space;217.6\:ft.$

• c) Valve Discharge Area decreases Uniformly with time

Using equation (5)
$\frac{h_m}{h_0}&space;=&space;\frac{K}{2}&space;+&space;\sqrt[]{\frac{K^2}{4}&space;+&space;K}$
Where:
$K&space;=&space;\left(\frac{lv_0}{gTh_0}&space;\right)^2&space;=&space;\left(\frac{3000\times&space;14.01}{32.2\times&space;12\times&space;500}&space;\right)^2&space;&space;=&space;\:0.04731$

Thus from equation (4)
$\frac{h_m}{h_0}&space;=&space;\frac{0.04731}{2}&space;=&space;\sqrt[]{\frac{0.04731^2}{4}+0.04731}=0.2425$
Thus:
$h_m&space;=&space;0.2425&space;\times&space;500&space;&space;&space;=&space;&space;121.3\;&space;ft.$

• d) The last 1000 ft. of pipe is 8 ft. diam. and the velocity decreases uniformly with time.

Using equation (6) The initial velocity in the 8" pipe
$=&space;\left(\frac{10}{8}&space;\right)^2\times&space;14.01=$
$=&space;21.9\:ft/sec.$

And from equation (7) The Equivalent Length
$l&space;=&space;l_1&space;+&space;\frac{a_1}{a_2}\:l_2&space;=&space;1000+\left(\frac{8}{10}&space;\right)^2\times&space;2000&space;=$
$=2280\.&space;ft.$

From equation (2) The Pressure on the Valve
$=&space;\frac{l}{g}\;\frac{v_{1.0}}{T}&space;=&space;\frac{2280}{32.2}\times&space;\frac{21.9}{12}&space;=&space;&space;129.2\:&space;ft.$
And the Pressure rise at the junction
$h_1&space;=&space;\frac{l_2}{g}\times\frac{v_2}{T}$
$=\frac{2000}{32.2}\times&space;\frac{14.01}{12}&space;=&space;72.5\:ft.$

Notes:
• 1. Pipe friction is small compared to the pressure rise and has been neglected
• 2. The time of closure $\inline&space;T$ is greater than $\inline&space;\displaystyle\frac{L}{1000}$
Solution
• a) When Velocity decreases uniformly with time the Inertia Head
is $\inline&space;108.8\;ft.$
• b) When Pressure rises uniformly with time, $\inline&space;217.6\;ft.$
• c) When Valve Discharge Area decreases Uniformly with time, $\inline&space;121.3\;&space;ft.$
• d) When the last $\inline&space;1000&space;ft.$ of pipe is $\inline&space;8&space;ft.$ diam. and the velocity decreases uniformly with time, $\inline&space;72.5\;ft.$

Last Modified: 10 Jan 12 @ 17:08     Page Rendered: 2022-03-14 15:56:36