CodeCogs

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

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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.

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, i.e.

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