hi I'm Mike Crowley and to day at Fluid
Mechanics i'm going to explain water
hammer in pipes. Water hammer is a
special transient flow case. Transient
flow and the study of transition flow
which is called surge analysis is
concerned with dynamically changing
flow velocities in pipe. Water hammer
occurs when there is a sudden or rapid
change in the flow velocity. It's usually
associated with a valve slamming closed
or rapid closing of a valve. It can lead
to very high pressure transients which
can cause the pipe to fail often is
associated with a banging noise which
leads to the term water hammer. Basically
you have a long column of
water and you're rapidly stopping it.
It bangs against the valve and it causes a
banging noise. In this lesson I will
explain the theory behind water hammer
i'll show you how to calculate the
pressure transients that are induced
due to water hammer. I will explain this
shortly at Fluid Mechanics.
So let me explain what is happening and how
to calculate the induced pressures. So if I
draw a sketch of a tank, connected to a
pipe. This is a header tank. A pipe line connected
to it. And in this tank we have a head of
fluid and that is pushing the fluid
along the pipe. Its going to have an
initial velocity Ui and it's going
into a open tank, at this end here. So
this is our initial conditions, constant
velocity Ui initially along
the pipe into a tank. And there's a head
of fluid, which is pushing the flow along.
The pressure at the inlet to the pipe is...
The pressure equals.
Rho,
which is a density, Gravity times h
The head. Now knowing the pressure at the
inlet to the pipe and if you know the
other conditions along the pipe. You know
the length of the pipe, diameter of the
pipe, the viscosity of the fluid, it is
possible to calculate what the flow rate
is along the pipe. Now in this video I'm
not going to explain how to do that, but
it but it's not very difficult job to
calculate the velocity along the
pipe. So then what happens,
In the water hammer case, we have
a sudden closure of valve at the end of
the pipe, So that some instance in time
the end of the pipe is closed off.
I'm just going to show a blockage on the
end of the pipe, there to show that the
pipe has been closed
now into the instantanes you do that,
you still got flow coming into the
starts of the pipe.
But at this end of the pipe here the
flow has stopped, because it has got
nowhere to go. So what actually happens
is it sets up a pressure fronter or a wave
front which travels up the pipe and i will show
it at this position here. And this
pressure or way front travels up the
pipe at velocity C. And C is the velocity
the sonic velocity in the pipe. So on
this side of the way front here. The
velocity and U equals 0. And on
this side of the pipe, the velocity is
still the initial velocity. Now that is a
little bit theoretical, because it
assumes you had an
instantaneous closure valve. But no
matter how far you close it, it will take
some time to close the
valve. And in that case what happens
instead of just being a one plane in the
pipe the the change of velocity will
occur over a section of pipe, so this
is probably a bit more realistic and
basically what we're saying is that over
this length, here there will be a
pressure change, Delta P. Where on
this side the velocity is U, the
initial velocity. And on this side of the
wave front the velocity is zero.
So the velocity will be changing across
this wave front now the length of this
wave front from here to here, is to do
with how long it takes to close the
valve. So if the valve was closed
instantaneously it would be just be a plane but
if it takes a fraction of a second
basically it's how far that wave
front travels in the time. So
the time it takes to close the
valve, times the sonic velocity will
determine what the length of that wave
front is. Now across the wavefront the
velocity is going
from the initial velocity down to zero
velocity there's a change in momentum or
change in velocity across that wave front.
That wave front can only change momentum,
or the velocity can only change if
there's a force applied to the fluid,
okay. We've now got to look at Newton's
second law to work out what force is
applied to the fluid as it goes across
the wavefront. Newton's second law
is force equals mass times acceleration
now in our case we're not talking about
forces were talking about pressures and
we're not talking about mass and
acceleration. We're talk about
changes in momentum. So for us the the
force that's acting across that wave
front there is the the differential
pressure, DP across the wavefront acting
on the area of the pipe. So I will put down A
for the area of the pipe. So now we need
to look at what is the
momentum change across that wavefront
well the wavefront is traveling up the
pipe at velocity C so at any instance in
time we can actually work out how much
fluid is traveling through that wavefront
okay and the amount of fluid
that's traveling through that wavefront.
Is basically how fast it's going up the
pipe times the area of the of the area
wavefront times the density of the
fluid. So the mass flow rate part of
it is. The velocity of the wavefront C
times the area of the pipe A times
the density of the fluid rho
okay. So
the fluid that's actually
go through that wavefront in terms of
kilograms per second, going up through
the wavefront is C, A, rho. So in other other
words the velocity of the wavefront
that's the sonic
velocity of the wavefront, the area of
the pipe and the density of the fluid.
And that's the mass flow rate
going through that wavefront. And how
much is the velocity changing?
Well it's going from Ui down to zero. So
in other words it's going from the
initial velocity down to zero. So the
momentum change is Ui. So we can take
out A from both sides of that equation there.
so we've basically got delta p equals
C rho Ui. Or more generally we say
that the pressure for a sudden closure
of a valve is C rho U, okay. Now that
that equation there is called the
Joukowsky equation and it's a famous
equation, and that determines what the
maximum pressure rise you can get to
water hammer is.
The maximum pressure
rises is the sonic velocity, the
speed of sound in the fluid the density
of the fluid times the change in
speed of the fluid. So its initial speed
going down to zero. So let's try and
apply this equation to say a
50-mmr copper pipe. And say we
had a 50mm copper pipe with
an initial speed of 1m/s
and what we want to do, is find out when
we suddenly closed the valve how much
pressure rise we're going to get for a 15mm
copper pipe. Well let's just
put down some details first of all of this
copper pipe, so the diameter of the
copper pipe is 15mm and
the initial velocity U equals 1m/s
1m/s in a 15mm
pipe is actually equivalent to 8.7 l/min
Okay. So when we look at this equation and
we try to apply it,
If we were talking about
water in a copper pipe, that's what I'm
talking about here, we know the initial
velocity that's going to be 1m/s we know
the density of water that's normally a
1000 kg/m^3
the thing we're not sure about is,
what's the sonic velocity. And that's
what I'm going to talk about next.
So to find the sonic velocity in a fluid
you need to apply Hooke's
law to it.
If we assume that the pipe is
perfectly rigid and does not flex okay, you
can apply this equation which is Hooke's
law which basically says, the
specific speed is equal to the square
root of the bulk modulus, divided by
the density of the fluid. Now for water
let's just calculate that.
For water we got C equals the square root of.
The bulk modulus of water, is
2.19x10^9 Pa and the density is
a 1000, so if you calculate that you
get a speed of 1480 m/s. Now
that's assuming that the pipe is
perfectly rigid, but pipes aren't
perfectly rigid they actually flex and
that actually affects the stiffness
of the system. And as it gets less stiff
the sonic speed comes down. So there's a
modification you can do to this equation
to take into account the stiffness of
the pie.
Basically you modify Hooke's law
equation, so that C equals the square
root of, one on,
rho
k plus D on.
So what's this equation saying? Basically
what this equation is saying
is that the sonic speed is the
density. Same as there, one on K, that's
the bulk modulus + D. D is
the diameter of the pipe, E is the Youngs
modulus of the pipe material. And then
little e is the wall thickness,
Okay.
This part of the equation here
is taking into account the
stiffness of the actual pipe itself. If
the pipe was perfectly rigid then
effectively what that's saying is that
you have infinite young's modulus,
for the material and if
that number was infinitely large then
this term would would drop down to zero
okay, and if thats zero, if you put zero in there
you'll effectively come back to this
this original equation here. So basically
that's that's how its modified, so as
this becomes less stiff then this term
in the equation becomes more important
and it actually reduces the speed. So if
we actually now put in some numbers for
that. Now for a
standard 15 mm copper
pipe, I believe the wall thickness is
0.7mm and for copper E,
young's modulus is 120x10^9 Pa.
Okay, so if i put those numbers
into that equation,
will get
Okay and if you
work that out. You get C equals 1254 m/s.
So the velocity has come down
from, for a copper pipe from 1484 a
perfectly rigid copper pipe down to 1254
m/s. Actually copper pipe is very stiff
but it all depends on the pipe your
choosing.
So if you're talking about the pipes
that take water to your house.
The plastic pipes that
nowadays they use in the road.
Typically you'd find the wave speed in
one of those would be around about a
1000 m/s, but if you
took a very flexible pipe likes a garden
hose pipe
you know you could be talking in terms
of 100s m/s the
other thing to bear in mind about the
wave speed though is the bulk
modulus. Water in particular is very
stiff okay. So that's 2.19x10^9
now that's true as long as there's no
air in the in the water. But often you
get little small air bubbles in the
water and they can have quite a
significant effect on the bulk modulus.
and bring down to speed quite
significantly. So that can be quite
an important factor but anyway we'll
carry on with the calculation.
So we now want to work out what the
pressure rises due to this closure of
this 15mm pipe with a 1m/s
flow in it, and we close the
end of the valve.
We
have the numbers now to apply to the
Joukowsky equation so the pressure rise
looking at the Joukowsky equation is going
to be C which is 1254 times the density
of water which is normally a
1000 kg/m^3
times the velocity which in our
particular example is one and if we work
that out, that comes out
12.54x10^5. I'm going
to put in x10^5 because
1x10^5 is 1 bar.
ok, So that's equals 12.5 bar. So
that's the pressure rise you'd get
in that particular case, maximum. I happen
to know that the pressure rating of
a copper pipe, of this specification
is 58 bar. So the safety factor for that
particular case is 58 bar divided by 12.5bar. Which
equals 4.64. So the safety factor is 4.64
Another way of looking at that is
if we actually had a much higher
velocity. If the
initial velocity was 4.64m/s for we would
have actually then got 58 bar. Now for
copper pipe, that will be going some. So
normally for copper pipes when
you close the end of the valve you don't
have a problem from a burst point of
view. But just be a little bit careful
with that because the burst pressure is
not the only thing that's important when
you're designing a hydraulic system you have
all the fittings on the end of the
pipes, there highly likely
to be ripped off you go to
excessive pressures you have all the
bracketry on the walls, and things like
that. If you've got movement in the pipes
you might affect that. If you have
bends in pipes they can tend to flex. So
there are other things to take
into account. So in summary to calculate
the pressure rise due to a sudden closure and water hammer
what you need to know is the
initial flow conditions, an and the initial
flow velocity. You need to understand
and work out what the the sonic speed is and
I have shown you in the in the lesson how to
calculate that. And you need to know the
density of the
fluid.
From that you can apply the Joukowsky
equation and basically the maximum
pressure rise is the the product of the
velocity, the wave speed, the
density. If you have any questions on
this then please leave a comment on my
website blog and I will endeavor to
answer any questions there. I cannot answer
any general questions directly by email
but I will if you leave a question on
the blog try and answer it there. If you need
any more detailed advice particularly
need advice on surge analysis on a
consultancy type basis. Then
please contact me directly.
That's it today from fluid mechanics
thank you for listening.
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