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

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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.

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

(1)

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

Using an equivalent length of pipe be L, then

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

(2)

i.e.

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

(3)

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

(4)

Let

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

(5)

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

(6)

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

(7)

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

(8)

The final pipe velocity v\_s is when

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

(9)

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

(10)

Substituting in the equation for dt above we get

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

(11)

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

(12)

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

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

(13)

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

(14)

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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

Rigid Column Theory