Dear Zegait, Some of the ...
Published by Prem Baboo, Researcher at www.researchGate.net
Dear Zegait,
Some of the useful equations of calculation of water hammer in a pipe.
A water hammer commonly occurs when a valve closes suddenly at an end of a pipeline system, and a pressure wave propagates in the pipe. It is also called hydraulic shock. This pressure wave can cause major problems, from noise and vibration to pipe collapse.
Hammer calculation
From Perry you get the most common "rule of thumb":
h(wh)=a*DV/gc
where
a=sqrt(1/(den/gc*(1/K+D/b*E)))
(velocity of wave propagation)
and
h(wh)= water hammer head
V=change in fluid velocity
gc: grav. constant
den: Fluid density
K: Bulk modulus
D= pipe inside diameter
b: Pipe wall thickness
E: Young modulus for pipe material
Other equation,
P= 0.8 x W x V
Where P surge pressure psi
W Fluid density lb/cu.ft
V Velocity change ft/sec
This equation gives < 5% deviation from the standard equation
Pressure transients in pipe lines caused by a shock wave when closing or opening a valve can be calculated as
dp = 0.070 dv l / t (1)
where
dp = increase in pressure (psi)
dv = change in flow velocity (ft/s)
t = valve closing time (s)
l = upstream pipe length (ft)
1 ft (foot) = 0.3048 m
1 ft/s = 0.3048 m/s
1 psi (lb/in2) = 6894.8 Pa (N/m2)
Example - Water Hammer generated when closing a Solenoid Valve
The pressure increase (water hammer) in a 100 ft water pipe where water flow velocity is reduced from 6 ft/s to 0 ft/s when a solenoid valve close in 0.1 s can be calculated as
dp = 0.070 ((6 ft/s) - (0 ft/s)) (100 ft) / (0.1 s)
= 420 (psi)
With a closing time of 1 s (solenoid valve with damper) - the pressure increase (water hammer) can be calculated as
dp = 0.070 ((6 ft/s) - (0 ft/s)) (100 ft) / (1 s)
= 42 (psi)
Note! - it is important to open and close valves slowly and use soft starters to start and stop pumps to avoid damaging water hammers in piping systems.
http://www.engineeringtoolbox.com/water-hammer-d_966.htm
Regards
Prem Baboo