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Engineering › Fluid Mechanics › Pipes › Surge ›

Rigid Column Theory

The sudden opening of a valve at the end of a pipe.

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Contents

Introduction
Sudden Valve Opening
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Introduction

The Rigid Column Theory is a significant factor in the design, construction and operation of pipelines. Owing to the potential damage including that caused by the water hammer effect, it is particularly important to arrangements that use pipelines transporting hazardous or flammable fluids.

The Rigid Column Theory assumes that the walls of the pipe are non-elastic, and that the fluid flowing in it is incompressible. This section discusses the effect of the Rigid Column Theory, and the subsequent head and velocity of the fluid, on the sudden opening of a valve.

Sudden Valve Opening

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

Sudden opening of a valve using The Rigid Column Theory.

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Let

the pipe velocity at an instant $t$ secs after the valve is thrown open be $v$
$H$ is the head causing the flow, which equals the entry loss + pipe friction + velocity head + valve loss + acceleration head. That is,

$H = \frac{1}{2}\frac{v^2}{2g} + \frac{4flv^2}{2dg} + \frac{v^2}{2g} + h_L + \frac{L}{g}\;\frac{dv}{dt}$

where $h_L$ may given as $\left( k\:\displaystyle\frac{v^2}{2g} \right)$ or as an equivalent length of pipe.

Using an equivalent length of pipe be $L$ , then:

$H = \frac{v^2}{2g}\;1.5 + \frac{4flv^2}{2dg}+\frac{l}{g}\:\frac{dv}{dt}$

i.e.,

$H\;=\frac{v^2}{2g}\left(1.5+\frac{4fL}{d} \right)+\frac{l}{g}\;\frac{dv}{dt}$

$\therefore\;\;\;2gH=v^2\left(1.5+\frac{4fL}{d} \right)+2l\;\frac{dv}{dt}$

Let

$k=1.5+\frac{4fl}{d}$

$\therefore\;\;\;2gH=kv^2+2l\:\frac{dv}{dt}$

$\therefore\;\;\;\frac{dv}{dt}=\frac{2gH-kv^2}{2l}$

Or,

$dt=\frac{2l}{2gH-kv^2}\;dv$

The final pipe velocity $v_s$ is when:

$\frac{l}{g}\;\frac{dv}{dt}=0$

$\therefore\;\;\;v_s^2=\frac{2gH}{k}$

Substituting in the equation for $dt$ above we get:

$dt = \frac{2l}{k}\;\frac{dv}{v_s^2 - v^2}$

$= \frac{2l}{k}\:\frac{l}{2v_s}\left[\frac{1}{v_s+v} + \frac{1}{v_s-v} \right]dv$

Integrating over the time $t$ , when the velocity goes from 0 to $v$ , gives:

$t=\frac{l}{kv_s}\left[\int_{0}^{v} \frac{1}{v_s + v}+\int_{0}^{v}\frac{1}{v_s-v}\right]\;dv$

$\therefore\;\;\;t=\frac{l}{kv_s}\;\ln\frac{v_s+v}{v_s-v}$

NOTE: This equation will give the time taken $t$ for the pipe velocity to reach a given value or the velocity after a given time.

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