

Pipe
Transients


Transient
Maths
Transient
analysis is an involved and complicated process, and care
must be taken to assess the geometry of the distribution system and
it's operating regieme to understand it fully. However here we
give a short introduction and show that with a couple of assumptions we
can get some useful information.

First as was stated in here the pressure transients move
around the system at the speed of sound of the fluid in the
pipes. If we had a large container of water the speed of sound
would be about 1,484 m/s. However the pipe walls are always
slightly elastic and deform as a pressure wave passes. This deformation
has the effect of slowing the wave so we are left with the following
expression for the wave speed in a pipe:
where c is the speed of sound, rho the density of water, K the bulk modulus of water (2.05e9
Pa), D the diameter of the
pipe, t the pipe wall
thickness and E the modulus
of elasticity of the pipe.

If we call the upstream or downstream
length of pipe before a reflecting surface (reservoir, valve, junction
etc.), l, then if the
operation that causes the transient (valve opening closing, pump
startup trip, etc.) occurs in a time shorter than 2l/c,
then
the
pressure
wave
will not have had time to pass along the length
and reflect back to the original site. This type of operation is
then termed a 'fast' operation and will produce the largest
transients. If the
operation takes longer than 2l/c to occur then the returning
pressure wave will interact with the wave generated by the operation
and will reduce the magnitude of the pressure wave. In this case
a more complicated analysis is required to determine the actual finall
pressure rise.

The pressure rise in a fast operation
is given by the Joukowski Equation:
where dH is the magnitude of the pressure
wave, g gravitational
acceleration and dU the
change in the flow velocity.

Example
If
we have a valve in the middle of a straight section of pipe of total
length 1000 m. The pipe is connected to a pump at the upstream
end and a reservoir at the downstream end and the flow rate is 4 l/s
and runs at a steady state head of 20m. If we assume that the pipe is
made of MDPE (E = 2e9 Pa) and has diameter and wall thickness of 100mm
and 10mm
respectively, we shall calculate the pressure rise of we close the
valve in 1 s:
Putting
the pipe properties into the equation for the wave speed in a pipe
gives:
We
now calculate whether the valve operation time is less than the
reflection time: Given the total length of the pipe is 1000m and hte
valve is in the middle l = 1000/2, therefore the time for a
reflection to return to the valve is:
which
is
less
than
the valve operation time of 1 s, which means the valve
operation is 'fast' and the magnitude of the transient can be
calculated from the Joukowski Equation. We are assuming the valve
completely closes during th operation so the flow rate drops from 4 l/s
to zero. The initial velocity in the pipe is U = Q/A,
where U
is the flow velocity, Q the
flow rate and A the cross
sectional area of the pipe.
so the change in velocity during the operation, dU, is 0.51 m/s, which we can put
into the Joukowski Equation:
the initial pressure in the pipe was 20m so upstream of the valve
closure the pressure change will be positive giving a maximum pressure
of 42.19 m downstream the pressure change is negative and the transient
produces a minimum pressure of 2.19 m. This valve operation
produces a negative transient and thus a risk of contaminant intrusion
if there exists a pathway into the distribution system.





Maintained
and
updated
by
Richard
Collins,
June
2010


