Chapter 17 Unsteady flow of liquids in pipes
Introduction
We have been concerned to date with steady flow of water in pipes but it is inevitable that there will be applications where the rate of flow changes. The behaviour of the water is by no means simple and, in the right circumstances, can be destructive. Transient pressures of 40 bar can be obtained easily. Most people will meet the effects of such changes as “water hammer” but it is a wider field and it is of interest to mechanical engineers. We shall find that when the rate of flow changes slowly the water behaves in a relatively simple way but if a steady flow is interrupted by, say, the complete and sudden closure of a valve the result is much more complicated. In between we have the behaviour of the water when a valve is closed rapidly but not suddenly.
I suppose that it would be possible to construct a computer programme to model the behaviour of water in any pipe, under any head, for any initial condition of flow and any given method of closure of any valve but I do not think that many engineers would be very interested in using it. What is needed is a general understanding of the main characteristics of the behaviour of the water with unsteady flow and an awareness of when the effects of these transients are going to become important.
We shall have to build the physics for this system and we shall find that it would become much easier if the effects of friction in the pipe are ignored. Fortunately we already know from our work on pipes and bearings that water is one of the most mobile liquids that we encounter in engineering and the losses caused by viscosity are amongst the lowest that we have to deal with. It is probably good enough for our purposes to limit our interest to water and then to try to account for the fundamental behaviour of the water in a pipe. This can then be transferred to other applications.
Slow changes in flow
Let
me start with the most simple system of just a horizontal pipe connected to a
tank of water with a valve at the end of the pipe. It is show diagrammatically
in figure 17-1.
Clearly, if the valve is partly open, water will flow through the pipe and the flow will become steady when the friction loss in the pipe plus the loss in the valve is equal to the head above the pipe at inlet. I looked at this system in Chapter 8 by ignoring the loss in the valve and ascribing an area ratio to the valve. That is complicated enough but here we need to consider adding acceleration to the water in the pipe. It makes sense to consider finding a relationship between acceleration and the head that must be impressed on the water to produce the acceleration. In order to do so we make the usual simplifying decision to treat the flow as one-dimensional. However we must also decide what to do about quantifying the acceleration of the water because we do not know whether it is uniform along the length of the pipe. We shall find that the pressures produced in a pipe by accelerations caused by changes in the flow of water in quite ordinary pipe systems is enough to make the compressibility of water become a significant factor. In the first instance we can consider changes in velocity that take place slowly enough for compressibility to be insignificant. Then we can take the acceleration to be uniform along the pipe.
The mass of water in the pipe is and, if the acceleration of the water is taken to be uniform and equal to where is the velocity of flow, we can write that the accelerating force .
The accelerating force must be a pressure difference applied over the length of the pipe. Then or :- .
It is likely that we shall use this as a head and then .
Sudden opening of a valve
Let me try to use this expression. Suppose that in the simple system of figure 17-1 the free surface is 10 metres above the valve, that the pipe is of 100 metres in length, of diameter 50 mm and has a friction coefficient of 0.001. Now suppose that the valve is suddenly opened from being fully closed. The flow will obviously increase in some way until a steady condition is achieved when the head lost to friction in the pipe plus the kinetic energy head in the outflow is equal to 10 metres. We can find out how the flow in the pipe varies with time.
The steady state that will finally be reached will come when:-
where is the final steady velocity of the water.
If the valve is a gate valve or a plug valve it would be useful to ignore the loss in the valve and then the equation becomes :- and we can evaluate for the given figures to give =4.67 m/s
We can set up a differential equation to relate the change of velocity and time using the energy equation and the expression for the head required to produce an acceleration. We get :- where is the instantaneous velocity of the water in the pipe at time after closure.
This can be rearranged to give :-
This
can be integrated repeatedly using Mathcad to give a velocity versus time
graph. This is graph 17-1. In principle the water in the pipe will take an
infinite time to reach its final steady state. The dotted red line gives the
final velocity nd we can see that, after one minute there is still a few
percent to go.
In engineering practice the system that I have proposed is most unlikely to exist, as the tank will have to be filled and then the water in it is not still. This will affect the flow and some equilibrium condition will quickly be established.
I have only once seen really steady flow. It was a fountain in an ornamental fishpond. The fountain was a nozzle set with its axis vertical and fed by a long pipe from a large lake. The jet was like a glass trumpet with no observable motion until the top edge broke up into separate parts held together by surface tension that fell back into the fishpond.
Slow opening or closing of a valve
On the face of it a decision to open the valve slowly seems to be a small step to take from sudden opening yet it is almost intractable.
In chapter 8 I looked at valves and showed that when a valve is used to control the flow through a pipe it does so by introducing a loss at the valve. Then the relationship between the rate of flow and the valve position is much dependent on the configuration of the pipe and its supply system. There can be no independent valve characteristic that relates valve position and flow.
When I dealt with sudden opening I had to ignore the loss in the valve. Now it cannot be ignored because it is an important factor in determining the flow even if it is transient. So it is likely that a working engineer faced with slow opening would keep in mind graph 17-1 and design accordingly.
Sudden closing and slow closing of a valve
Sudden closing and slow closing of a valve is fundamentally more complicated than slow opening but it is not intractable. We can go a long way towards understanding and predicting the behaviour of the water in the pipe.
We have to build a mental picture of this behaviour using physics and then apply it as rigorously as circumstances warrant. So let me build that physics.
Behaviour of water in a pipe following sudden closure of a valve
Figure
17-2 shows a facility to test a horizontal pipe. The test pipe is U shaped for
convenience with a large radius sweep. In effect the roof tank and the small
closed tank, joined together by a pipe, replace a tank of very small
cross-section but very tall.
At the outlet end of the test pipe there is a plug cock that can be turned very quickly to give what amounts to sudden closure. A pressure transducer is fitted close to the plug cock.[1]
At this point we cannot analyse just any sudden closure. We must start with the case of a system with a high value of the static head and a low flow that would, of course, be the result of having the valve partially open. With some care it is possible to get a pressure-time trace from the pressure transducer near to the valve for the events following sudden closure of this partly open valve. It would look like figure 17-3.
This graph is a trace of the pressure against time as recorded by the transducer, processed to give readings above and below the pressure corresponding to atmospheric pressure, against time. Before closure the loss of head to friction would be small compared with the static head
The
most obvious features are the cyclic nature of the trace and that the periodic
time is constant. The ramps on the tops to each half wave are a little puzzling
but can be explained. If it were not for these ramps this trace would look like
a square wave. In fact the real trace is a square wave combined with the
effects of friction loss to make it into a decaying wave.
When one seeks an explanation it makes sense to start by ignoring friction during the events that produce the square wave and then take friction into account later. Such a decision leads us to the position where there is no mechanism for attenuation of the wave but does let us get to grips with the physical explanation of the cycle of events.
Figure 17-4 shows the square wave. It takes equal half waves that are alternately positive and negative relative to the pressure head at the instant of closure. I have shown the negative half waves as going below atmospheric pressure just to introduce the idea of pressures that can be less than atmospheric and, indeed, go down to the vapour pressure of the water.
In order to explain this trace we must look at events in the pipe adjacent to the valve just after the instant of closure. In figure 17-5 I am endeavouring to represent a wave that is moving through a pipe full of water that is moving with one-dimensional flow at uniform speed and pressure . I cannot usefully draw a closed plug cock but we have to imagine it to be just to the right of the arbitrary section so that between the closed plug cock and the wave there is water that is at rest at pressure . I use to indicate “a finite increase in” as distinct from to mean “a small increase in”.
So figure 17-5 represents a moving wave front going from right to left advancing into water moving at from left to right and bringing the water to rest as it advances. At the wave front kinetic energy is given up in exchange for pressure energy and that energy is stored as strain energy in the water. The wave front moves at constant speed and it can never be in equilibrium.
Now we must look to see what happens to this wave front.
We
have a system to analyse and we need to know the starting condition. Figure
17-6 shows both the system and the starting condition. The system is a
horizontal pipe connected to a supply tank. At the inlet end the pipe is
connected to a supply tank and, at the outlet end, there is a plug valve. Now
we must suppose that water flows in this pipe at a steady velocity with uniform pressure .
This defines the system and the steady flow condition before the valve is
suddenly closed.
We know that a pressure wave will be propagated at some uniform velocity towards the inlet end and that the pressure will rise by at the wave front. We can represent the activity in the pipe at some instant after closure. It is figure 17-7. The wave is part way along the pipe and moving right to left. Between the valve and the wave the water will be at rest at pressure . Between the inlet and the wave the water continues to flow at the same velocity at the original pressure . This can go on until the pressure wave reaches the open end.
Figure 17-9 shows the state of the water in the pipe at the instant that the pressure wave reaches the open end. All the water is at rest at pressure . However, at the wave front there is only a pressure from left to right and a pressure from right to left. This condition is unstable and we must decide what happens.
The first thing to recognise is that the wave front can only occur in an elastic medium and in this case the elastic medium is the water. So, during the passage of the wave over the length of the pipe, the kinetic energy once possessed by the water that is now at rest, is stored as strain energy in the water. (The pipe is also elastic and will store strain energy but we will take that into account later.)
At the instant that the wave front reaches the open end of the pipe it is in the condition shown in figure 17-10 where the water in the tank is not moving yet there is a large difference in pressure between the two sides of the wave. This cannot persist and the prevailing process of exchanging kinetic energy for strain energy is reversed so that the water flows out of the pipe from right to left and a wave front in which the pressure drops by to moves from left to right. In the process the strain energy changes back to kinetic energy but now in the opposite direction and the water flows out of the pipe into the supply tank.
Figure
17-11 shows the wave front on its way towards the valve leaving water to the
left moving at towards the open end. The water to the right
of the wave front is still stationary at a pressure of .
The next interesting point is when this wave front reaches the valve when the pipe is full of water all moving away from a closed valve!
This
instant is shown in figure 17-12 and the pressure is again throughout. We have to decide what now
happens at the valve because the situation shown in figure 17-12 cannot be in
equilibrium. In fact a new wave front is generated at the valve and in this
wave front the pressure drops by and the water between the valve and the wave
front comes to rest again. We can see the result in figure 17-13.
The
pressure at the valve drops by and I have shown the resulting pressure in
the pipe as being less than atmospheric but greater than zero absolute. We
shall see that typical conditions in a pipe is more likely to bring the
pressure to zero absolute when this diagram would not be correct.
The next changeover point will be when the wave reaches the open end. Then the water in the pipe will be at rest from end to end at pressure . This is again not in equilibrium and yet another wave front is propagated towards the valve causing the velocity of the water to change back to the original value and the pressure to rise to . This is shown in figure 17-14. When that wave reaches the valve the cycle is complete and will start again.
This
gives an explanation of the square wave that we postulated for the events
following sudden closure of a valve at the end of a pipe. We can use it to draw
the diagrams for the pressure variations with time at, say, the mid point of
the pipe and at the open end.
I have drawn the pressure-time traces for this square wave for pressures measured at these three points. The trace for the mid point is self-explanatory and the trace for the open end merely notes that there will be pressure spikes as the wave front reaches the open end and reverses.
In order to make this explanation of the events in a pipe after sudden closure we ignored friction as it suited the explanation. We still need an explanation of the ramps on the top of real traces but I will leave that until later in this text.
All this makes sense but it lacks any physical reality because we cannot put values to the magnitudes of the pressure changes or to the speed of the waves. This can be done using our normal physics.
Speed of propagation of a wave front
We need to relate the rise in pressure, the initial velocity of the water, the frequency of the cycle and the physical properties of the water and the pipe. Clearly the frequency depends on the length of the pipe and on the speed of propagation of the wave front. So we must have a value for the speed of the wave. We also need to find an expression for . The analysis is unusual because we have a finite rise in pressure producing changes in density and volume of the water that are both relatively very small. I shall use both and as appropriate.
Consider some instant at some time after the valve is suddenly closed. The wave will be at some point in the pipe that will be from the valve where is the speed of propagation of the wave.
Figure
17-16 shows a section of a pipe that is adjacent to a plug valve. It shows the
position of a wave front at time after the valve has been closed suddenly. The
water is flowing towards the wave front at velocity and at pressure and density .
The wave front moves at velocity where is the speed of propagation of the wave front
in stationary water. Between the wave front and the valve the water is at rest
at pressure and at .
During time the wave front moves and the area of cross section of the pipe
over this length changes to .
The mass of fluid that has come to rest in time is made up of the mass that was in the pipe when the valve was closed plus that which flowed in during time .
So we can say that :-
from which we get :-
.
The density term in these equations is always a nuisance so we must find a way to eliminate it in favour of pressure.
Density is mass per unit volume and the volume of a liquid is related to the pressure by the elastic properties of the liquid. We use a bulk modulus that is defined by :-
where is the change in a volume of liquid resulting from a change in pressure of the liquid of . is the constant of proportionality and is called the bulk modulus. Clearly has the units of pressure and, for water it has a value of about 2.05 GN/m2.
For the mass of water that is at rest between the valve and the wave front the change in density resulting from the rise in pressure will be from to where is the volume before the wave front and is the volume after the passing of the wave front. From this :- and because .
Rearranging this we get . Then we get the crucial switch
Now we can eliminate the density terms from
This looks very formidable but it will yield if we now apply momentum to the compression. The mass of water that has been brought to rest has given up momentum as a result of a force equal to acting on it for a time . So :-
or,
This simple step has given us an expression for in terms of and measurable quantities. It is important. Here we shall need to use it in the form .
This expression can be combined with to give a very useful outcome but it is tedious.
Multiply top and bottom by to give :-
Divide top and bottom by
Now we have to extract
Rearranging gives :-
and then
Now will always be small when compared with 1 and will also be minute. Using this we get . It is important to realise that and are of the same order of magnitude so we have to decide whether to take the expansion of the pipe into account. We must put some figures to this equation to find some speeds.
We have for water and . Then .
This figure for the speed of the wave front is enormous when compared with the normally encountered speeds of water in practical pipe systems. Effectively the speed of the wave front is 1,432 m/s regardless.
However
we cannot just forget the expansion of the pipe under stress. We make use of
the elementary theories of hoop stress. Figure 17-17 represents one half of a
short length of a pipe of diameter and wall thickness .
It is in equilibrium under three concentrated forces that are the resultants of
three distributed forces. The downward force is due to the internal pressure and the two upward forces are due to the
stress acting in the wall of the pipe.
The pressure rise produces a downward force of acting on a total area . So the hoop stress = . This stress produces a hoop strain, that is the circumferential extension/the circumference = where is Young’s modulus for the material of the pipe.
We need the value of . The increase in length of the circumference is so the increase in area of the pipe is the circumference times the increase in radius which is equal to and . Then .
We had and we can now substitute for to give :-
or .
In the final form this becomes .
This expression does not contain so the speed of the wave in not dependent on but is dependent on the elastic properties of both the water and the metal of the pipe.
W e can quantify the effect of the elasticity of the pipe. Suppose that the pipe were to be of 32 mm diameter and of 1 mm wall thickness and made of steel. This would be a thin walled pipe and if there is any effect this would make it large. Given that E for steel is the speed of the wave front for this pipe :-
So the velocity of the wave front is primarily dependent on the elasticity of the water and for regular pipes a figure of 1350 m/s can be used.
We are now in a position to predict the pressure-time diagram for the pressure in a pipe, at the valve, after sudden closure. We know the speed of the wave front. If we know the length of the pipe we can predict the time taken for the wave front to traverse the pipe and if we know the velocity of flow before the valve is closed we can predict the pressure rise following sudden closure.
Figure
17-18 is figure 17-4 with the new information added. The duration of a crest is
the time for a wave front to go from end to end and back, that is, and the pressure rise is .
What we now need to know is what magnitude the pressure rise will have for any initial value of
This can be the result of using . I have plotted it for water and a range of from 0 to 10 m/s.
The
commonly used figure for the speed for economical flow of water in pipes is 3
m/s. This corresponds to a pressure rise in a wave front generated by sudden
closure of about 40 bar. This is the pressure at a depth of 400 metres in
water.
It is a large pressure when compared with the normal pressures encountered in, say, buildings and it must be treated with respect.
Effect of friction on the pressure/time
trace
We are now in a position to explain the ramps on the tops of the wave in figure 17-3.
In figure 17-19 I have shown the steady state flow of water through a horizontal pipe. The pressure at inlet will be as before, but now, at velocity , there will be a friction loss in the pipe and the pressure at the valve will be . Now we have to think out what happens when the valve is suddenly closed. We know that the pressure at the valve will rise instantly by and a wave front of this magnitude will be propagated along the pipe towards the inlet end bringing to rest the water between the wave front and the valve.
When we dealt with this before we ignored friction and the wave advanced into water all at the same pressure. Now, as the wave advances it meets water at an ever-increasing pressure. This must affect what goes on at the wave front.
In figure 17-20 I have drawn a wave front in the pipe just after the valve has been closed. The wave will have moved a short distance through the pipe and brought the water between the wave front and the valve to rest. In figure 17-20 I have shown the pressures on the two sides of the wave. Pressure must equal the pressure in the pipe at the position of the wave plus the rise in pressure caused by the destruction of the momentum of the water brought to rest from velocity . As the wave moves along the pipe that pressure will rise to a maximum value of when it started at . We made the following statement in deriving the expression for the rise in pressure through the wave. The mass of water that has been brought to rest in the short time has given up momentum as a result of a force equal to acting on it for a time . So :-
or,
.
The
continual change in upstream pressure has disturbed the original equality
because are all unchanged yet the pressure rise
through the wave is not equal to .
The water solves this problem by propagating a wave of continuously increasing
magnitude towards the valve increasing the pressure in the water at rest and
allowing a small additional compression of the water and a small increase in
diameter of the pipe to increase very slightly the volume of water at rest over
that which would be at rest if there were to be no friction. The reverse wave
will travel at the normal speed and, as a consequence in the time that the
main wave front takes to travel from end to end of the pipe and back the
pressure at the valve will rise by the original pressure loss. The net effect
of the returning pressure wave is to increase the pressure at the valve at a
rate that is one half of the rate that the pressure ahead of the wave front is
increasing due to the existing pipe friction.
We can draw a plot of the pressure distribution when the wave is say half way between the valve and the open end. It is figure 17-21.
The important area of this diagram in figure 17-22 is the area depicting these returning waves that add to become a distributed wave. I have enlarged this small area to give figure 17-22. We can see that the net effect of the procession of returning waves generated by moving into a region of rising pressure resulting from the gradient produced by friction is the distributed wave shown in green and the effect of that wave at the valve is to give a steady rise in pressure.
We
can move on the show the pressure distribution when the wave front reaches the
tank. It is figure 17-33 and we can see that the pressure at the valve has
risen by one half of the pressure drop due to friction.
At the instant shown in figure 17-23 the wave front reverses just as it did in figure 17-9. As the wave front progresses towards the valve the distributed wave goes on travelling unchanged towards the valve and we can see a part way position in figure 17-24.
In
figure 17-24 the wave front is half way back. The pressure at the valve has
risen by ¾ of the pressure drop due to friction. There is now a very important
new line in blue. As the wave front moves towards the valve the water flows out
of the pipe and into the tank and in doing so reverses the pressure gradient
due to friction. It is this pressure gradient that is shown in blue.
Clearly the arrival of the wave front at the valve will have established this reverse pressure gradient from end to end with some loss due to friction. It will also have raised the pressure at the valve by the original pressure drop due to friction. When this is transferred to figure 17-18 it gives the ramps on the top of the square wave shown originally in figure 17-3.
We can also see that the reversal of the friction gradient gives the upside down ramps on the negative going waves in 1figure 17-3. We now know enough to understand the pressure-time trace of figure 17-3 including the attenuation of the wave as a result of the steady loss of energy to friction in the water in the pipe with no energy input from the tank.
This is an analysis that I think can be called physics although others might see it as an application of physics to engineering. However it is viewed, an engineer involved with some application involving unsteady flow will, in the first instance, need to understand this behaviour even if it is difficult to quantify. Let me give an example.
Suppose that you were in charge of a water turbine that ran under a high head and the supply pipe was 1000 metres long. If the machine were to be delivering its full output, when the velocity in the pipe might be 5 metres/sec, a sudden reduction of the load of 25% would have to be met by a rapid reduction in the output of the machine. Let us now suppose that the machine was a Pelton wheel where the water emerges into the machine to interact with the moving blades as a jet at atmospheric pressure. The jet is created in an adjustable nozzle. Closing the nozzle is not an option because that will produce a rapid, very large, increase in pressure in the pipe and the whole gamut of unsteady flow pressure waves running back and forth in the pipe with a transit time of about 1.4 seconds. Furthermore the rise in pressure will increase the velocity of the jet which is the opposite to what is required. One of two solutions must be used; the jet must be broken up by making it swirl or the valve must close as required to reduce the load and a by pass valve must open to release surplus water to waste. That sounds simple but there is now a problem of what to do with the bypass water. It is my understanding that the jet d’eau in Geneva is a replacement for a fountain in the lake that was the by pass flow from a Pelton wheel in a factory. It would have been very attractive with its continual random fluctuation of height. The importance of this example lies in the fact that only a qualitative knowledge of unsteady flow is needed to devise a relatively simple mechanical solution. Computers would not help even if the full analysis of the unsteady flow would be quite interesting academically.
The concepts of rapid and slow closure
We know that the speed of a wave front does not vary much from about 1400 m/s. It follows that, as a wave will take a time of to travel from the valve to the inlet end of the pipe and back, this time is quite predictable. It follows that we can categorise the times taken to close a valve and the character of the subsequent events by whether the valve has closed fully before the first returning wave reaches the valve. We call a closure where the closure time < a rapid closure
I know from electronic measurement of pressure transients that even small changes in valve setting produce wave fronts with measurable magnitudes of pressure. Provided that the time of closure is less than we can say that the original velocity of flow where is a small change in and we can say that the total rise in pressure at the vale . If we propose some relationship between velocity at the valve and time we can draw a diagram showing the change of pressure at the valve.
I have drawn the diagram for a closure in less than in figure 17-25. It is probably impossible to predict the graph of pressure at the valve-time during closure but, for the sake of illustration, I have drawn a smooth curve in blue. I have divided the rise in pressure, that will equal where is the original velocity of flow, into seven equal increments of pressure . The diagram has been drawn on the supposition that each rise in pressure takes place at the instant when the pressure has risen by . The seven rises in pressure are shown as red up-going arrows. Once the valve is fully closed the pressure at the valve remains constant with time until the first returning wave front reaches the valve. Then the pressure drops by shown by a red down-going arrow. Ultimately the seven returning wave fronts reduce the pressure at the valve by to give pressure of . I have drawn a blue line through the mid points of the down-going red arrows to give pressure-time graph up to the instant when the pressure reaches when this pressure prevails until the instant .
On this graph I have added the limiting case for this construction to be valid when the time of closure is . If time of closure exceeds and the same construction is used to produce a pressure-time diagram at the valve the fact that wave fronts begin to return to the valve before the valve is closed adds a severe complication. I have started to draw a diagram for the case of time of closure greater than and it is quickly evident that the complexity makes it unlikely that a working engineer would see it as a viable way to look at this closure.
I
have used the same proportions for figure 17-26 as I used for figure 17-35. The
graph of pressure versus time is now much less steep and before the valve is
fully closed the first returning wave arrives to start reducing the pressure. A
way of constructing the graph after the first wave arrives is to imagine that
the pressure goes on rising as if the wave was absent and then subtract the
appropriate pressure drop = .
Then one must imagine the pressure to rise again as if the valve is still
producing a pressure rise at the same time as the returning waves are reducing
the pressure. This is shown as two steps one in blue and the other in red. This
can be repeated until the valve is fully closed. Then the imaginary rise ceases
and the blue line goes horizontal.
Figure 17-26 is for a closure time of about 1.5 times the value of and it is complex. The value of the closure time could be in minutes not milliseconds. If the closure time is long compared with we know that we can use and we know that when the closure time is of the same order as the pressure transients are complex and probably unpredictable. This whole range is called slow closure and it is all too complicated to be viable. The only conclusion that we can draw is that it is better to design to eliminate the effects of pressure transients than to try to try to predict the transients.
Water hammer
However difficult these transients may be there is another very serious problem to contend with. It is separation and its audible effect, water hammer.
I am referring to the pipe system in which where is the velocity at the start of closure when the pressure is . In figure 17-4 I stressed that the diagram had negative going half waves that only just went below atmospheric pressure. I did not point out how easy it is for the pressure to attempt to drop below zero absolute.
We have seen that for typical velocities of flow e.g. 3 m/s the rise in pressure following sudden or rapid closure the value of can be large.
For . .
This means that almost any system fed from a tank will have less than 428 m. (The typical height for a power station building is about 40 metres.) It is very common indeed for to exceed . This leads to separation and water hammer.
We need to understand what actually happens. We need to go back to figure 17-12 in which friction is ignored and the instant when the first returning wave front reaches the valve and the water in the pipe is flowing from the valve towards the tank at velocity . The pressure now attempts to fall by hundreds of feet of head when it can only fall to zero absolute. So the pressure falls to zero absolute and the water continues to flow towards the tank at an initial velocity .
Two things now happen. The first is that the water separates from the valve and starts to create a void that can contain nothing other that water vapour and continues to flow towards the valve against a pressure difference of and this causes the column of water of length to be retarded uniformly to rest and then to be accelerated towards the valve to just reach velocity when it returns to the valve. Whilst this phase is going on the void in the pipe between the valve and the surface of the flowing water increases to some maximum length and then collapses again as the water returns to the valve. The second thing is that a negative going wave of magnitude sets off towards the tank and, in the time taken for the whole column to come to rest and then return to the valve, the wave front makes many passes from end to end of the column and may even be attenuated to nothing because of friction. In trying to assimilate this it should be recalled that the speed of the water through the pipe would be about 3 m/s where the speed of the wave would be 1400 m/s.
We must decide how we are to quantify this cycle. The new feature is that we have a wave front producing the basic square wave for the first half of the cycle and then a half cycle where the dominant feature is the behaviour of the a column of water of length moving under the effect of a constant force.
So
let me start pressure-time diagram for a point just upstream of the valve
following sudden and complete closure from velocity in the system depicted in figure 17-1 that I
repeat here as figure 17-27.
In order to draw this diagram we need an expression for the interval of time between the start of separation and the return of the column to contact with the valve. The column separated from the valve with a velocity of and it moves under a retarding force that equals the pressure difference multiplied by the area. The pressure difference is given by and when this is multiplied by the area we get a force of . The mass involved is given by if the length of the void is small when compared with .
Then if we equate the impulse of force and the change in momentum per second we can write :- where is the time taken for the void to form and collapse and is the net change in velocity.
From this we get :-
When the column hits the closed valve there is a loud noise produced by the collision. Then the pressure rises by as before and the cycle starts again but this time the rise in pressure is from zero absolute and not from .
I have gathered all this together in figure 17-28.
At one time I had a rig for student use to demonstrate water hammer. It comprised a water tank on a roof from which a steel pipe of 1² diameter and 50 metres long ran down to the laboratory in straights and swept bends to terminate in a plug cock. There were tappings for pressure transducers at the valve and at the mid length.
I made many tests on this rig and only one has survived in my care. It is for the pressure-time diagram at the valve. I have copied it to give figure 17-29. The length of 50 metres gave a value of of about 0.075 seconds. The rig was of realistic proportions so this trace is not just of academic interest. Separation is real and potentially destructive
Other aspects of transient flow in pipes
Any sort of model of these pressure transients gives no idea of the details of the real system. For instance the pressure transients in the metal of the pipe carry on running backwards and forwards at the same time as those in the water. They are easily detectable using strain gauges and show how much faster the wave fronts move in the metal than they do in the water.
In real pipe-work there will elbows and tees and these will produce reflections that just add to the general complexity.
I gave the impression that the void was at the valve but, depending on the configuration of the pipe, especially the slope, the void could be anywhere. Even when it is at the valve the void will not have a simple plane face but will have a ramp if the pipe is horizontal and be hollow[2] if the pipe is vertical.
Implications of this chapter
I think that I have made a case for engineers to regard pressure transients other than slow ones to be virtually unpredictable and therefore best eliminated at the design stage. This raises the question of what methods can be used to deal with applications where it is essential to be able to reduce a flow quickly in, say, a pipe feeding a water turbine.
There are three ways to the best of my knowledge. They have one aspect in common. A pressure relief system is fitted close to the machine to separate the short pipe supplying the machine from the long pipe from the reservoir from which the water flows. Then the water from the pressure relief system either goes to a surge tank, to a air pressure vessel or is discharged through some device like a swirling nozzle to dissipate the energy. I will deal with this topic in connection with reciprocating pumps and with turbines but it is appropriate to refer to the basic methods at this point.
The hydraulic ram
There is a device that makes use of these pressure transients. It is called the hydraulic ram and it is used to pump water and I will deal with it in the Section 2 chapter 19.
Worked examples
Q1 A steel pipe is 1500 metres long. It has an inside diameter of 150 mm and a wall thickness of 6mm. It is designed to convey 0.08 of water. For steel and for water .
Calculate :-
(i) the velocity of propagation of a pressure wave through the water if the elasticity is ignored,
(ii) the velocity of the pressure wave taking the elasticity of the pipe into account,
(iii) the velocity of flow for the design condition,
(iv) the rise in pressure that would follow the sudden and complete closure of a valve at the end of the pipe,
(v) the hoop stress in the pipe that would be produced by this pressure. Ignore the longitudinal strain.
(vi) the time that would elapse before the first wave returns to the valve,
(vii) the time taken for the first cycle to be completed by ignoring friction in the pipe and taking the initial pressure in the pipe to be 2.5 bar absolute,
(viii) the minimum time in which the valve can be closed if separation is to be avoided.
Q2 Water flows at a steady velocity of 2.5 m/s through a steel pipe of 1.5 m diameter and 12 mm wall thickness. The pipe is 1,200 m long.
Calculate :-
(i) the velocity of propagation of a pressure wave in the pipe,
(ii) the value of ,
(iii) the rise in pressure at a valve at the delivery end of the pipe if the valve is closed in 2 seconds,
(iv) the stress in the pipe wall at this pressure, and,
(v) the shortest closure time that could be used if the stress in the wall is to be limited to . Suppose that the valve can be closed in such way that, during the start of closure the rate of rise = where is the time of closure, that is, the closure is “linear”.
Q3 Water flows steadily from a reservoir through a steel pipe of 300 mm diameter, of wall thickness 8 mm, and length 2,000 m. The level in the reservoir is 40 m above that of a valve at the delivery end of the pipe.
(i) Take to be equal to 0.005 and calculate the velocity of flow in the pipe.
(ii) Using the method in Q2(v) estimate the pressure rise at the valve when it is closed in 10 seconds.
(iii) Calculate the pressure rise at the valve when the valve is suddenly closed to ultimately reduce the flow to one half of the original value.
Q4 Suppose that in Q3 the valve were to be fully closed with no flow in the pipe and that the valve was then suddenly opened. We have an expression for the retardation (or acceleration) of the water in a long pipe for cases when the time for significant changes to occur is long compared with . It is where is the head causing acceleration. Use this, the Darcy expression and the steady flow equation to estimate the time taken for the velocity of flow to reach 2 m/s. Suppose to be independent of the velocity.
Q5 Plot a graph showing how the loss of energy to friction varies with time during the acceleration phase.
Q6 Water flows steadily at a velocity of 3 m/s through a steel pipe 0f 50 mm diameter and 3.5 mm wall thickness. The pipe is 70 m long and the pressure at the downstream end of the pipe is 2 bar gauge.
(i) Calculate the velocity, , of propagation of a pressure wave in the water in the pipe given that :-
where is the density of water = , is the bulk modulus of elasticity of water = , is the diameter of the pipe in metres, is the thickness of the pipe wall in metres and is the modulus of elasticity of steel = .
(ii) If the valve is now closed suddenly calculate the rise in pressure at the valve and the time for which this pressure persists.
(iii) Take the vapour pressure of water at ambient temperatures to be zero and the atmospheric pressure to be 1 bar and estimate the further time that will elapse before the pressure at the valve rises above the vapour pressure.
Q7 A horizontal pipe that is 20 m long and of 48 mm diameter is connected to a reservoir containing water at a depth of 4 metres below the free surface.
Take the surface roughness of the pipe to be 0.045 mm, the effective bulk modulus of water in the steel pipe to be , ignore loss other than pipe friction and calculate :-
(i) the maximum flow through the pipe,
(ii) the time taken for the flow to reach 95% of this maximum after sudden opening of the valve at the end of the pipe,
(iii) the maximum pressure reached in the pipe as a result of suddenly closing the valve after this maximum pressure has been achieved.
Q8 Water flows at a stead velocity of 2.5 m/s through a pipe that is 100 m long, has a diameter of 200 mm and has a wall thickness of 5 mm.
Take the bulk modulus of water to be , the value of Young’s modulus for steel to be and calculate the pressure rise when :-
(i) the flow is reduced uniformly to zero in 5 seconds and,
(ii) a valve at the end of the pipe is closed in a very short time.
Solutions
Q1 (i) , and for a rigid pipe
(ii)
(iii)
(iv)
(v) Hoop stress . For comparison the yield stress of mild steel is so this stress is not minor.
(vi) Time , that is the time for the wave to traverse the pipe from end to end and back. time .
(vii)
During the second phase the pressure will drop to zero. Then the
water is retarded by a pressure difference of 2.5 bar from an initial velocity
of 4.53 m/s We need to find the time taken to bring the water to rest and then
to give it a velocity of 4.53 m/s towards the valve.
Force on the water
The mass of water in the pipe Using impulse of force = change of momentum :-
.
Total time for a cycle is
(viii) This needs some thought but one must conclude that the safe time is the one derived from slow closure and linear change in velocity, that is :-
.
Then . Of course the time would be longer because cannot be constant.
Q2 (i)
(ii)
(iii) This is a rapid closure because the closure time is less than . The pressure rise is given by
(iv) The hoop stress
(v)
The first returning wave will reach the valve after 2.5 seconds from the start
of closure. The maximum pressure will at the valve occur at this time. The
maximum pressure will be .
In this case it is limited to the pressure that produces a hoop stress of .
We know that at 24 bar the hop stress is 150 so 8 bar will produce 50 .
Therefore , that is,
Q3 (i) For steady flow :-
.
From this :-
It follows that .
Then and .
(ii) A value of is required.
We also need a value of . . From this the maximum value of
Us8ing “linear” closure as above and ignoring the rise in pressure produced by the wave advancing into water at higher pressure the rise in pressure due to valve closure = .
(iii)
Q4 Continuing from Q3, during the period of acceleration the head available is used to produce the kinetic energy at exit, to meet the friction loss and to accelerate the water. So, applying the energy equation for some instant of time seconds after opening the valve we have :-
or
Mathcad gives
Integrating longhand gives
Q5 Loss to friction
We also have
This can be solved by using Mathcad or longhand.
Q6 (i)
(ii) The pressure rise
This pressure will persist for a time .
(iii) The further time required is that to bring the water in the pipe to rest from 3 m/s and to give it a velocity of 3 m/s back towards the valve under a pressure difference of 2 bar atmospheric pressure, that is 3 bar.
Accelerating force .
Mass of water
Acceleration
Time
Q7 (i)
Applying the energy equation to 1 in the
free surface in the tank and to 2 at discharge from the pipe :-
.
Substituting
must come from the Moody diagram and we do not know the velocity of flow for the Reynolds number. We can calculate the relative roughness from and will probably lie between . If we make a first guess at =0.005 . .
This gives the same value of . So and
(ii) For accelerating flow the head causing acceleration
This can be handled in Mathcad but for examination purposes we ca proceed longhand:-
.
(iii) The maximum rise in pressure is given by where and is the equivalent bulk modulus.
Q8 For this pipe :-
So .
(i) Compared with 0.16 seconds 5 seconds is a long time and this can be treated as a slow closure. Then :-
(ii)