Calculation min. operating time butterflyvalve ?!?

[mrctchlr]

Who can help me?

Look at the following formulas

v = (354 * Q)/ DN^2 then Q = (v * DN^2) / 354


t = (5157 * Q ) / (DN^2 * dP)

Added those formulas together i get;

t = (14.56 * v)/ dP

For liquids v is aprrox. 5m/s

So the valve size is not a factor in closing time, only pressure?!? So the operating time to prevent waterhammer is equal for every size? Were do i go wrong?

Many thanks in forward!

 

[ione]

The change in velocity to cause water hammer has to be made within the time period t, which is defined as

t = 2 L / a
where:

t = critical period, sec
L = length of the pipe, m
a = speed of the pressure wave, m/s

If the condition above is realized, that is the valve is closed/opened within the time t, this will be considered as instantaneous and the water hammer H will be evaluated using the following formula

H = a v / g
where:

H = water hammer, m water column
a = speed of pressure wave, m/s
v = change in flow velocity, m/s
g = gravity (9.8 m/s^2)


The speed of the pressure wave (a) depends on the fluid.

You stated that “For liquids v is aprrox. 5m/s”, but to get this value for any given flow rate you have only one pipe size.

 

[ione]

To integrate my previous post , I’d like to add that pressure wave has different speeds depending not only on fluid, but also on pipe material and size.

To be more precise the speed of pressure wave (a) in ft/s can be written as

a = 4600/SQRT[1+k*(OD/s-2)/E]

where

k = Fluid bulk modulus, psi

E = Pipe modulus of elasticity, psi

OD = pipe outer diameter, inch
s = pipe thickness, inch