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Chapter 23 Reciprocating and other positive displacement
pumps
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Introduction
The reciprocating pump
The force pump
The village pump
The semi-rotary pump
The powered reciprocating pump
Dynamic effects
Multi-throw pumps
The air vessel
Effect of pipe friction on the reciprocating pump
Revisiting the pumps figures 23-1 and 23-2
The sliding vane pump
The gear pump
The hydraulic ram
Worked examples on reciprocating pumps
© Ivor Bittle
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Chapter 23 Reciprocating and other positive displacement pumps
Introduction
I pondered on whether to include a chapter on these pumps in this book. Pumps are
everywhere, both positive displacement pumps and centrifugal pumps and I intend to
deal with centrifugal pumps in a later chapter. Reciprocating pumps have a long history
that has seen the early, large, slow-moving pumps superseded by centrifugal pumps but
also a re-emergence as small pumps for special applications.
Figure 23-2
Figure 23-1
Figure 23-1 is a photograph of an all-plastic reciprocating pump used in a dental irrigator
for cleaning teeth and in another guise for removing ear wax. Figure 23-2 shows some of
the parts of the three-cylinder swash plate pump used in a domestic jet washer. These are
just two of many applications and frequently these pumps are used for metering.
In Victorian England (1837-1901) coal was the universal fuel and the universal engine
was the steam engine. The confidence and sheer joie-de-vivre in that period was staggering
and never likely to be repeated. Public buildings, railways, roads, shipping, manufacture
and every other activity developed at an unprecedented rate. Pumping, especially of
water, was necessary in all sorts of applications, not least in water supply and later in
sewage handling and pumps developed along with everything else. High-pressure water
was used to carry energy round factories to drive capstans, presses, cranes and so on and
used in such iconic structures as the Tower Bridge to raise and lower the bascules and to
lock the free ends of the bascules in the lowered position.
Large steam engines are generally slow speed devices and reciprocating pumps were the
norm. These two were well matched. Reciprocating pumps were still being made in 1920
and I suppose that for special applications like priming centrifugal pumps on fire engines
still are, even if they are now operated by sophisticated control systems. The pump is a
device that can be analysed with some confidence and some of the practical details are
interesting.
These pumps have influenced other positive displacement pumps that have followed so
they are worth a few pages and it is necessary to understand reciprocating pumps in
order to appreciate the reasons for designs for the pumps that are used very widely today.
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The force pump
The force pump is the starting
point. There are so many
different ways that force
pumps can be designed that I
found it difficult to choose one
to show diagrammatically. In
the end I sought to show the
principal features of the design.
Figure 23-3 shows how the
force pump works. A piston or
plunger is essential, and it must
work either in a cylinder or in a
chamber of suitable shape with
some means of making a
durable and effective seal
Figure 23-3
between the piston and the
cylinder or between the plunger and the chamber. I will use a piston and cylinder for
analysis but it could just as easily be a plunger and chamber.
All sorts of configurations are used. Two valves are needed. Both are non-return valves
and I have shown mushroom valves because they are easily drawn and easily understood
but earlier pumps would have used leather-faced, flap valves and modern pumps might
use flap valves in plastic. The pump is fitted in a pipe so that liquid flows into the
cylinder through the lower valve and out of the cylinder through the upper valve.
Movement of the piston to the right draws liquid into the cylinder through the lower
valve and movement to the left pushes liquid upwards through the upper valve. At the
change in direction from moving right to moving left the lower valve closes and the
upper one opens and vice versa. It could not be more simple.
Who knows when the first reciprocating pump was made or what materials were used to
make it? We must look for reasons to want to lift water and four important applications
come to mind. One is to lift water out of the ground for drinking, another is for
irrigation where water is lifted from a stream on to the land and the third is the necessity
to lift water out of ships hulls and the last is for mine drainage. There must be many
others.
Probably making manually operated pumps for the removal of bilge water was the most
pressing because wooden hulls always leaked and water in a hull can move with the
rolling of the ship and impair the stability. There was adequate skill on board most ships
to maintain pumps and bilge pumps made from wood, iron and leather must have been
reliable. In an autobiography of the life of a slave from Africa in the latter half of the
18th century it is recorded that he made many voyages as a seaman. Several times he
records that the pumps were needed in emergency but there is no hint that the pumps
were unreliable. Similarly village pumps for drinking water took on such permanence that
they too must have worked for long periods without serious maintenance. In the heyday
of transport of goods by barge and lighter it is recorded that a woman made a living
pumping bilge water using a pump that she carried from boat to boat. Her pump could
not have been made mainly from iron for her to be able to carry it so we must presume
that it was mainly wooden. Our modern use of reciprocating pumps has a continuous
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line of development from these early pumps as manufacturing methods and new alloys
and plastics have emerged.
The village pump.
It is worth looking at the village pump. It
would have supplied drinking water from
a well and it was hand operated. The
valves were flap valves faced with
leather.1 Two arrangements became the
most common and they are shown in
figure 23-4. In both cases the pump
mechanism is above the free surface in
the well and water can only be lifted out
of the well if low pressure can be created
in the cylinder during the upstroke of the
pump. When the pump is started for the
first time it will contain no water only air.
Then the pump must act as an air pump.
Air can leak past the piston much more
easily than water and the pump will have
to be operated, by hand, very vigorously,
Figure 23-4
to get water into the lower part of the
cylinder. The deeper that the well is the
more trouble it will be to start the pump. Suppose that the pump is primed and
functioning. It is clear that on the up-stroke water will flow up the suction pipe and into
the cylinder passing through the open suction valve. When the piston reverses the
suction valve closes and the water cannot return to the well. On the down-stroke the
outlet valve opens and water passes to the spout.2
Water is not much use as a lubricant and sealing this pump could not have been very
reliable especially as the flap valves were probably faced with leather. The second pump
is a typical force pump but now with valves in the piston. It has the advantage that, if the
piston seal is not very good, water can be introduced above the piston to improve the
seal to get the pump primed.
The semi-rotary pump
Whilst many village pumps have been preserved
in various states of order the only direct
descendent that is still used widely is the handoperated semi-rotary pump. It is shown in figure
23-5. These pumps are typically used to bail open
boats and to pump fuel from drums into boats,
tractors, aeroplanes, and so on in remote sites.
The handle is moved backwards and forwards
and liquid is pumped during both strokes. It is
Figure 23-5
We think of using rubber for seal but whilst rubber was known from about 1750 it was not until 1840
with the invention of vulcanising that rubber became sufficiently durable to be used in pumps.
2 One might expect that a foot valve would be fitted at the lower end of the suction pipe to keep the
suction pipe full and the pump primed but this was not the case. Either the foot valve needed too much
attention or it was best not to risk frost damage if the suction pipe and the pump were always full.
1
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simple and ingenious. It is really two pumps in one
body.
Figure 23-6 shows the arrangement of the pump.
The body of the pump is designed to create a
cylindrical space in which a "wing" fitted to a
spindle fits snugly between the opposite flat faces
and the inner cylindrical surface of the pump body.
This wing can be made to oscillate backwards and
forwards by operating a handle on a spindle in a
boss in one end plate. Flap valves are fitted to the
two ports in the wing.
Figure 23-6
A splitter, in black, having an inverted V shape,
over the suction port is part of the body of the
pump and it is machined to make a seal with the boss of the wing. This splitter has two
ports with flap valves.
This arrangement creates three chambers A, B and C separated by the flap valves.
Chamber A is of constant volume wherever the wing may be but B and C can change in
volume as the wing is rotated. If the pump as shown is full of liquid, movement of the
handle to make the wing turn anti-clockwise will force liquid from chamber B into
chamber A and this will force liquid out through the delivery port because the volume of
chamber A is constant. At the same time the volume of C increases and liquid is sucked
into C. Reversal of the motion of the wing will make the right hand side deliver liquid
and suction to take place through the flap valve in B.
The powered reciprocating pump
The pump shown diagrammatically in figure 23-3 has the layout required for a
mechanically driven pump. The new
arrangement is shown in figure 23-7.
It is the obvious design and very
common. It would be installed in a
pipe system that can be typified by
figure 23-8. The pump draws water
from the sump and makes a static lift
h that comprises two other lifts, the
suction lift hs from the sump to the
pump and the delivery lift hd from
Figure 23-7
the pump to the header.
If the pump were to be run very slowly
there would be virtually no losses to
friction in the pipe-work and we could
draw a pressure/volume diagram for
the pump. It is shown in figure 23-9.
Unlike pressure/volume diagrams for
engines and compressors no scale is
required for volume because the
concept of a clearance volume has no
Figure 23-8
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meaning for pumps handling incompressible liquid. All we need is a length representing
the volume of water displaced by one stroke of the pump and a vertical scale with heads
shown relative to the atmospheric pressure as the datum.
We can then track the cycle of events. Starting
at a , where the pressure is atmospheric and
the piston stationary at the end of its travel,
motion of the piston will cause the suction
valve to open as the pressure drops below
atmospheric pressure at b by hs , the suction
lift. As the pump is running slowly liquid will
keep up with the piston at constant pressure
head as it moves to the other end of its stroke
at c . This is the suction stroke. The piston
now reverses and suction valve closes and the
pressure head rises by h  hs  hd when the
Figure 23-9
delivery valve opens at d . Liquid now flows
through the delivery pipe as the piston moves
back to its starting point at e . This is the delivery stroke. During delivery the pressure
head will be constant. At the next reversal the pressure will drop to the suction pressure
and the cycle start again.
During the 19th century large pumps driven by steam engines drained mines all over the
world and drained polders in Holland to reclaim land and they operated on this cycle.
In practice the delivery will be less than the swept volume because there will be a small
loss as the valves close. It is unavoidable no matter what type of valve is used.
This single cylinder pump will deliver intermittently and the delivery/time diagram is as
shown in figure 23-10. The rate of delivery
cannot be steady during the delivery stroke
because the piston is driven by a crank. It
rises from zero to a maximum near midstroke and then falls to zero and then there
is no flow during the suction stroke. If this
is the delivery diagram for the pump there
must be a velocity/time diagram for the
flow in the delivery pipe and, depending on
the area ratio between the pipe and the
piston, the velocity is always greater and
perhaps much greater than that of the
Figure 23-10
piston.
This will mean that for any speed greater than the very slowest speed there will be losses
to pipe friction during delivery and, if the delivery pipe is long there will be inertia forces
to contend with and quite possibly water hammer effects as well. We need to know more
about the dynamic effects of this intermittent flow.
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Dynamic effects
Most pumps run at a high enough speed for dynamic effects to be important. However
we cannot deal with all the dynamic effects at once so I will start with the inertia effects.
We can deal with the suction pipe and the delivery pipe separately and I will start with
the suction.
This will necessitate some analysis that would be
much simplified if we suppose that the piston
moves with simple harmonic motion. One might
argue that it is easy to compute the motion if the
ratio between the length of the connecting rod and
the crank throw are known but, in the event, other
uncertainties dwarf that of simplifying the piston
motion.
Figure 23-11
Suppose that the pump in figure 23-7 has a crank
radius of r , a piston diameter of d and that the
crank turns at  radians per second. For SHM and
figure 23-11 the displacement of the piston from its
mid-position is given by:displacement = r  cos t and from this by
differentiation :velocity = r  sin t and :acceleration =  2 r  cos t
These expressions are for the piston and we want them for water moving in the suction
pipe. If the water is continuous from the surface in the sump to the face of the piston we
can apply the continuity equation to show that the ratio of the displacement of the water
in the suction pipe to the displacement of the piston is equal to the ratio of areas, that is,
the ratio of the square of the diameter of the piston to the square of that of the pipe. Let
the diameter of the suction pipe be d s and then :velocity of water in suction pipe =
acceleration of water in suction pipe =
d2
  r  sin t and :d s2
d2 2
  r  cos t .
d s2
From Chapter 17 of this book we have an expression for the head needed to give an
l
acceleration a to a column of liquid in a pipe of length l . It is :- inertia head =  a .
g
From this and the expression for the displacement of the piston we get
l d2
l d2
inertia head =  2   2 r  cos t =  2   2 times displacement.
g ds
g ds
The inertia head bears a linear relationship with displacement. When t = 0, cos t  1
and when t =180º, cos t  1 and we can add the inertia head in the suction pipe to
our head/displacement diagram.
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We
can
modify
the
pressure/displacement diagram in
figure 23-9 to take account of the
inertia head to give figure 23-12.
Then the pressure head at the start
l d2
of suction will be hs   2   2
g ds
below atmospheric and, at the end
of the suction stroke will be
l d2
hs   2   2 below atmospheric.
g ds
The head will vary linearly between
these two values. The net result is
that the pressure during the suction
stroke varies as shown in red in the
diagram and the suction stroke
Figure 23-12
starts at b and finishes at c .
I have also shown in blue the zero of pressure head. The vapour pressure of water is
small, in the range of 0.01 of an atmosphere and it is near enough for this purpose to be
l d2
regarded as zero. Clearly it is possible for the sum of hs and  2   2 to exceed one
g ds
atmosphere and then the pump will be in trouble because the absolute pressure in the
pump will try to fall below zero and the water will separate from the piston face. Once
separation has occurred the water in the suction pipe will flow with constant acceleration
under a difference in pressure of one atmosphere. It will inevitably catch up with the
piston and then there will be water hammer. This must be avoided.
Now let us look at the delivery stroke. There will be inertia effects in the delivery pipe of
exactly the same character as those in the suction pipe and we can add the resulting
pressure
changes
to
the
pressure/displacement diagram.
There will be a similar effect on the
delivery stroke as a result of inertia of the
water in the delivery pipe and the whole
diagram can be drawn as in figure 23-13.
It is likely that the length of the delivery
pipe will be greater than that of the
suction pipe so the inertia effects even for
the same diameter delivery pipe will be
greater. It is possible for the value of
ld d 2 2
   to be so large that the suction
g d d2
valve opens and delivery takes place
directly from the suction pipe.
Figure 23-13
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These effects and those in the suction pipe must be avoided.
There are two ways to achieve this. One is to redesign the single piston pump so that
there are odd numbers of pumps working in parallel with their cranks spaced at equal
angles. This evens out the delivery. The other is to fit air vessels to both the suction and
delivery pipes near to their connections to the pump. I will deal with these two
separately.
Multi-throw pumps
Figure 23-14 shows what this is about. In the diagram I have shown the deliveries from a
three-throw pump with three cylinders
delivering to a common delivery pipe.
The three delivery strokes are shown in
colour and the sum of these deliveries is
shown in black. Clearly, with this delivery,
the flow in the delivery pipe will fluctuate
by only a fraction of the fluctuation for a
single pump having the same delivery and
five or seven would reduce the
fluctuation still more.
Three-throw
pumps
were
made
throughout the 19th century and were
Figure 23-14
driven by water wheels and steam
engines. Currently they find application in oil hydraulics and as pressure washers where it
is high pressure that is wanted and not a large through-put.
One should not overlook the effects of using flexible hose with reciprocating pumps. We
saw in chapter 17 that the magnitude of the pressure transients in water hammer
depended on the initial velocity and on the speed of propagation of a pressure wave
through the water in the pipe. That speed of propagation, in turn, depends on the
modulus of elasticity ( E ) of the material used to make the pipe and on the wall
thickness. Hoses can be made of rubber or plastic and these have a very low modulus of
elasticity when compared with commonly used metals and then the speed of propagation
of a wave is very low. Water hammer effects are much attenuated by flexible hoses and
are very widely used with or without armouring.
The air vessel
An air vessel shown diagrammatically in figure 23-15
is shaped like any other pressure vessel being
cylindrical with domed ends. It is fitted using a tee to
the suction or delivery pipes at the points that are
closest to the pump. The vessel has a valve at the top
to allow air into the vessel and a glass sight gauge to
allow the volume of air to be kept at what is deemed
to be the best level.
Such air vessels would be fitted to a single cylinder
pump that rotates at a high enough speed for the
pressure/displacement diagram, in the absence of air
© Ivor Bittle
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Figure 23-15
vessels, to be like that in figure 23-13 and to convert it to the diagram in figure 23-9. We
need to see how the new arrangement works and it is probably most simple to start with
an air vessel that is fitted to the delivery pipe. The pump delivers during the delivery
stroke and has no delivery during the suction stroke. The intention is for the air vessel to
absorb about half of the water delivered during the delivery stroke and to return it to the
pipe during the suction stroke to give a more or less steady flow in the delivery pipe. A
very large air vessel will indeed give virtually steady flow but a very small one would not
alter the fluctuations in flow in the delivery pipe by very much. Somewhere there is a best
practical ratio, between the size of the pump and the size of the air vessel or, to be more
specific, between the volume of air in the air vessel and the volume of water delivered
during one cycle. It is said that this ratio is between 1 and 3 but we can do better than
that if we make a few simplifications.
I have modified figure 23-8 by adding
air vessels to the suction and delivery
pipes to give figure 23-16. What we
want is some basis from which to
choose volumes for the air vessels.
I will start with the delivery and let the
volume of the empty air vessel be k
times the displacement of the pump.
Now what we are looking for is a
value of k that gets rid of inertia heads
Figure 23-16
and give us fairly steady flow in the
delivery pipe. Suppose that we have
steady flow and let us ignore friction in the delivery pipe. Let us start with the air vessel
full of air at the pressure corresponding to the delivery head hd . If the flow in the
delivery pipe is steady during delivery the water that flows from the pump to the tee will
go half to the delivery pipe and half to the air vessel. When the delivery stroke is
complete half of the water that has been delivered will be in the air vessel and the volume
of the air will have been reduced by that amount. We can calculate the rise in pressure of
the air using the isothermal relationship pV  constant but, as we must use absolute
pressure, I will use it in the form (ha  h) V  constant where ha  h is the absolute
head and is proportional to p .
Let the displacement volume, that is the product of the piston area and the stroke, be Vd .
At the start of the delivery stroke the head will be ha  hd and the volume of the air
V
k  Vd . At the end of delivery the volume of the air will be k  Vd  d . Then the new
2
k  Vd
k
value of ha  h will be :- (ha  hd ) 
= (ha  hd ) 
Vd
k  0.5
k  Vd 
2
k
We now need a graph of
versus k .
k  0.5
It is shown in figure 23-17 and we need to assess the implications. If we multiplied the
ordinates by ha  hd we would have a value for the absolute head in the air vessel at the
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end of delivery. No one can pretend that this is accurate because this system is in reality
very complicated but the trend of
the graph is so clear that no
matter how accurate a new model
may be it will not change in its
essential features. Indeed the
graph does not change much if
the volume entering the air vessel
is changed between 30% and 70%
of the displacement. We see that,
for an air vessel having a volume
equal to the displacement, the
absolute pressure would rise very
rapidly by the same order as the
delivery lift and impress an
acceleration on the water in the
Figure 23-17
delivery pipe. As the reason for
fitting the air vessel is to get rid of
the inertia effects of the water in the delivery pipe this will not do. The graph suggests
that we must use a volume for the air vessel greater than 3 times the displacement
volume and values of k between say 2 and 3 look to be suspect. The quoted practical
figure given above is k lies between 1 and 3 but we now know that it would be unwise
to use a figure lower than 3 especially as, in ordinary use, a significant part of the volume
of the air vessel is occupied by water.
Turning to the suction pipe and its air vessel we see that the presence of the air vessel
close to the intake to the pump
means that, during the suction stroke,
water will be drawn from the air
vessel as well as though the suction
pipe and that during the delivery
stroke water flows into the air vessel
ready for the next suction stroke. In
order to put a size to the air vessel we
could model it in the same way as for
the delivery vessel but now the air
pressure will drop during suction.
Suppose that at the start of the
suction stroke the pressure of the air
Figure 23-18
in the suction air vessel is ha  hs
where ha is the head corresponding to atmospheric pressure and the volume of air in the
air vessel is k  Vd where Vd is the displacement volume as before. If again, one half of
the flow into the pump comes from the air vessel the volume at the end of the suction
V
stroke will be k  Vd  d . The pressure at the end of suction will then be given by :2
k  Vd
k
(ha  hs ) 
or (ha  hs ) 
V
k  0.5
k  Vd  d
2
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I have added the graph for suction to that for delivery in figure 23-18. The blue line is the
absolute pressure in the air vessel at the end of suction. Again we cannot pretend that it
is accurate but the presence of the air vessel increases the absolute pressure at the end of
suction as k increases. We know that we want to avoid separation as a result of inertia
effects in the suction pipe and it seems that an air vessel having the same size as the one
for the delivery pipe could be satisfactory.
However we must recognise that whilst it would not matter if there were to be no water
in the delivery air vessel it would matter for the suction air vessel. It is probably best to
run both vessels with the water level visible in the sight gauges.
Effect of pipe friction on the reciprocating pump
At this point in the consideration of the reciprocating pump it becomes hard to decide
exactly what components are legitimately part of the pump. If the pump is purchased
without air vessels, the pressure/volume
diagram and inertia effects will depend on the
system in which the pump is used. The
system in figure 23-8 leads to the
pressure/volume diagram in figure 23-13
with its inertia effects and that diagram takes
no account of friction losses in the two pipes.
However, if air vessels that match the pump
are fitted, the pump and its air vessels
produce a steady flow at a pressure that
maintains that flow in whatever pipe system
is attached to it. It behaves like a unit and one
might expect to buy it in this form. The effect
of pipe friction is different for the pump with
or without air vessels.
Figure 23-19
f  l V
3 d 5
For pipes the friction loss is given by loss =
2
in S I units where the symbols have their usual meanings.
If the pump has air vessels and these produce steady flow the friction loss just adds
steady pressure heads to the static heads in the suction and delivery to give the
pressure/volume diagram in figure 23-19 and adds to the work needed to drive the
pump.
Revisiting the pumps figures 23-1 and 2
Now that we have an explanation of how pumps work we can revisit the two pumps in
the opening of this chapter and see them with new eyes.
© Ivor Bittle
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I have repeated figure 23-1 as 23-20 for convenience. This pump takes in water through
the hole on the top to the left. This intake
forms part of an adjustable by-pass valve
that sets the maximum pressure. The
delivery is the small pipe protruding from
the end of the cylinder. A hollow wand is
attached to a this pipe by a metre or so of
visco-elastic3, plastic tubing of 4 mm
outside diameter that is flexible and can be
pinched with the thumb nail. The pump
has no air vessels and there must be the
potential for inertia effects in the plastic
pipe. They are prevented from occurring
by the visco-elasticity of the plastic pipe.
Figure 23-1 brought down as fig 23-20
I have also brought down figure 23-2 as 23-21. It is from a domestic pressure washer. It
has three pistons that are driven by a swash plate that is shown on the right. The swash
plate is no more than a circular, hardened
steel plate that is set at an angle to the axis
of the pump mechanism by a pressing. A
piston is shown in the foreground complete
with its spring. If it were to be placed into
its cylinder and the swash plate and all the
thrust races that support it were to be
turned over, the hardened swash plate
would rest on the three hardened pistons
and when the whole lot is in its proper
housing, rotation of the swash plate would
Figure 23-2 brought down as fig 23-21
make the pistons go up and down under the
actions of the springs. Flap valves are fitted in a separate die casting and then we have a
very compact version of the three-throw reciprocating pump and it would have a delivery
time diagram like that in figure 23-14.
This still gives a pulsating flow but, once again, the pump delivers through a long
rubberised hose that damps out the pulsation. The pump will not be able to run at the
ordinary speed of alternating current motors and this pump is run through an epicyclic
gear box.
The sliding vane pump
Oil hydraulics sounds like a contradiction but it is the term used to mean the systems
where oil is used at high pressure to operate a whole range of machines that shift earth,
power tractors and harvesters on farms, lift, carry and stack goods, operate cranes and an
untold number of other machines. All this is possible because a relatively, very small
pump driven by a diesel engine or electric motor can produce a flow of oil at high
pressure that can operate equally small hydraulic motors to drive wheels or rams of all
sizes to power scoops and diggers and pulley systems. All these can be manually
Some pipes are flexible in that they can be squashed across a diameter. Rubber and some plastic pipes,
when released after being squashed, regain their shape slowly. It is as if the forces in the material have to
work against a friction force to recover their shape. I am not sure that visco-elastic is quite the right word
for use with what appears to be a solid but that is what we have.
3
© Ivor Bittle
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controlled with banks of lever-operated
valves that give astonishingly versatile
control to all these machines.
Oil hydraulics is a study in its own right but
the pump is an appropriate machine for this
chapter on positive displacement pumps
I have drawn the essential features of the
sliding vane pump in figure 23-22. The
rotor of the pump is simply a thick disc with
an integral shaft and, in this case, ten,
equally-spaced axial slots in it. Vanes or
blades having the same axial length as the
rotor fit snugly in these slots. I have shown
an end view of the rotor with the blades in
Figure 23-22
place. The diagram shows a section through
the body of the pump. The body has a central bearing for the shaft of the rotor and an
eccentric cylindrical hole to the depth of the rotor thickness is bored in it so that the
rotor just touches at the top as in my diagram to create a seal. The eccentricity of the
bore in the body creates a crescent-shaped space between the rotor and the body that I
have cross-hatched in blue and in red as is appropriate. When the rotor rotates the blades
are thrown out to keep in contact with the inner cylindrical surface of the body. The
body incorporates two bosses to make pipe connections and these bosses are blended
internally with two ports. These ports have a profile as I have shown and an axial width
of about one third of the axial thickness of the rotor.
I have drawn the diagram with the rotor with two blades
aligned with the centre line through the bosses. Then the
top two blades are symmetrically spaced about the
vertical centre line and the ports terminate about half
way across these blades. The two top blades seal the high
pressure side from the low pressure. The only route from
the low pressure side to the high pressure side is through
the space between the rotor and the body at the bottom
that forms a rectangular duct. The blades sweep through
this space. The length of that duct is determined by the
Figure 23-23
distance between the two lips of the ports. The pumping
action takes place in this short duct. In figure 23-22 I have shown the rotor in a position
where the blades A and B overlap the lips of the two ports and the space in the duct
between the two blades A and B is filled with oil at low pressure and sealed. If the rotor
now rotates by a few degrees the volume of oil between the blades A and B is suddenly
exposed to high pressure oil and this is the start of delivery. Blade B now has high
pressure on one side and low pressure on the other and as it sweeps through the space it
will perform a suction stroke behind it and a delivery stroke ahead of it. These strokes
will end when blade C reaches the inlet lip and low pressure oil again fills the space in the
duct and is trapped. I have drawn a scrap of diagram 23-22 with the rotor at mid-stroke
in figure 23-23. It is clear what is happening.
The sliding vane pump acts like a reciprocating pump in making suction and delivery
strokes but they are made simultaneously and, from the outside, the pump appears to
© Ivor Bittle
14
produce a continuous flow with no inertia effects. It is very well suited to running at
much higher speeds than a conventional reciprocating pump and, for closed, oilhydraulic systems, the oil that it pumps can be designed to have good lubrication
properties even when hot and to operate at high pressure.
The blades are only subjected to a large pressure difference as they go through the 36º of
the duct and have no net pressure forces on them elsewhere. Pressures of 200 bar are not
uncommon and if, say, the area of the blade exposed to this pressure is 3 square
centimetres the force will be about 6,000 Newtons. This is a large force and, coupled
with high speed, accounts for the small size of the pump required to produce high
pressure flow.
In oil hydraulics pumps may need to rotate but produce no flow. The pumps are fitted
with pressure relief valves joining the two ports.
The pump as shown in figure 23-22 could be used as a hydraulic motor and such motors
drive the massive wheels of earth movers.
The gear pump
The essentials of the gear pump are shown in figure 23-24. The moving parts are two
gear wheels. These gear wheels have teeth with an involute profile and, if they are well
made and mounted properly, at all positions of the wheels there will be line contact
between at least two pairs of teeth. There will be a force acting between the teeth at the
line or lines of contact and this makes an effective seal between the wheels. The wheels
are mounted in a body that is accurately
machined so that the tips of the teeth fit
snugly in the body and form seals at the
tips. The side plates fit closely on the
sides of the wheels to limit leakage. One
wheel, on its shaft, is wholly enclosed in
the body and the other has an extended
shaft that is driven from the outside. So
one wheel drives the other and the
several seals ensure that the liquid
passes through the pump as is intended
with a minimum of leakage.
Figure 23-24
Figure 23-24 shows the wheels at a
typical position and that the suction and
delivery sides of the pump are separated by the seals at the tips of two teeth and by the
line contact between two teeth. As the pump rotates liquid is carried continuously from
the suction side to the delivery side and the rise in pressure of the liquid takes place at the
instant that the tip of a tooth separates from the circular arc at the point of tangency. At
all times at least two of the teeth have high pressure on one side and low pressure on the
other.
The engagement between the teeth is more complex. As I have drawn the diagram the
teeth are only in contact on one line and two teeth are exposed to high pressure over
some part of their surfaces. As the wheels turn it is essential that there should always be
one pair of teeth in contact to make a seal. I have checked carefully and the shape of the
© Ivor Bittle
15
gears shown in figure 23-24 satisfy this requirement and during the change from one
tooth to the next, two pairs come into contact for a very short time.
The hydraulic ram
The hydraulic ram has been available for use for a very long time and some of these
pumps have run without attention for more than a century. A typical use is in animal
farming. Frequently valleys have a stream in grassland with areas fenced off and used in
rotation to make the best use of the grazing. Then the animals may not have access to
the stream. Water for the animals can be provided in troughs that are fed by the mains
water supply with a float valve to control the flow. Where mains water is not available a
hydraulic ram can lift water from the stream to the level of the trough even if that is
several metres above the stream. It
makes use of the energy in the water in
the stream. In order to do so a pipe of
uniform bore is laid alongside a stream
and the water from the stream allowed
to flow into the pipe from a forebay
constructed on the bed of the stream in
such a way that plant debris and other
debris cannot enter the pipe, clean water
will flow through the pipe to be
returned to the stream at the lower level
of the down stream end. If this flow
Fig 23-25
were to be stopped suddenly there
would be water hammer effects such
as those described in chapter 17 of this
book but, if the pipe is fitted with an
air vessel at its end, instead of the
pressure transients due to water
hammer, there would be inertia effects
in the water in the pipe and the water
would flow into the air vessel causing a
rise in pressure in the air in the vessel
and bring the flow to zero in a finite
time and in an orderly way. It is this
mechanism that is utilised in the design
of a hydraulic ram. What is required is
a delivery pipe, that would be
connected to the bottom of the air
vessel, and some arrangement of
valves to stop and start the flow in the
pipe.
The device is called the hydraulic ram
and it will run continuously pumping
water at a low but steady rate for
decades without attention. Figure 2322 shows a collection of old hydraulic
rams.
Figure 23-26
© Ivor Bittle
16
Figure 23-23 shows a common arrangement of the sort of hydraulic ram shown in figure
23-22 although I am sure that this layout has evolved over the years to be a bit more eyecatching. We see that two valves are needed to make the ram operate cyclically. In my
diagram the pump is built on a
chamber shaped like a cuboid. The
pipe running beside the stream is
called the driver pipe or just the
driver and it is connected to one end
of the chamber.
I have drawn the arrangement of the
whole installation diagrammatically
in figure 23-24. The ram is installed
at the surface level of the down
stream end of the driver and the
intake to the driver is at some
suitable point upstream.
Figure 23-27
The head hD that ultimately drives the ram depends on the length of the pipe and the
slope of the surface of the stream. In operation water flows through the driver and
through the gravity-operated valve to the stream until the gravity-operated valve closes.
Then inertia effects cause the pressure in the chamber to rise and open the non-return
valve to deliver water either directly to the upper reservoir or to the air vessel. When the
inertia effect dies away the non-return valve closes and the gravity-operated valve opens
for a new cycle to start.
There is much more to this simple device than is suggested in this short description of its
mode of operation. We can explore it.
It is clear that the important parts of this mechanism are the driver, which is aptly named,
and the gravity-operated valve and the only time that the delivery system has an effect is
during the short period of delivery when it creates a virtually steady pressure above the
non-return valve.
We clearly need to know more about the orders of magnitude of the velocity of the water
in the driver and the way that the inertia force in it changes with time.
We need to examine the driver pipe in operation. It is just a pipe of uniform diameter
that follows the gradient of the stream. We could use what we know about pipe friction
to get an idea of the magnitudes of the velocity that are possible for different gradients,
lengths and diameters
Suppose that the gradient is expressed as a fall per unit length, say s . For typical
gradients of natural streams s ranges from 0.01 to 0.04 that is from 1 metre in 100
metres to 4 metres in 100 metres. Whatever the gradient may be, when the flow is steady,
the fall in a given length is equal to the loss of head to friction whether it is in a stream or
in a pipe.
© Ivor Bittle
17
The Darcy expression for steady
flow in a pipe gives us loss
4  f  l c2
=
where l and d are the

d
2g
length and diameter of the pipe, c is
the velocity of the water and f is
the friction coefficient. If we
rearrange this we can write :-
fall 4  f c 2


 s.
l
d 2g
From this we can relate c and d
for different values of s .
I have done this in figure 23-25. In
order to draw these graphs I put
f  0.008 and re-arranged to give :-
s
c
Figure 23-28
4  0.008 c 2
c2
and

 0.00163 
d
2g
d
sd
.
0.00163
The graphs show that for the
relatively large gradient of 0.1 the order of the maximum velocity of flow is say 1.7
metre/sec but for natural gradients it is about half a metre per second. Generally we can
say that these velocities are low and that inertia heads will also be low but not so low that
we cannot get a useful lift of water from the stream.
These figures tell us that, for ordinary bed slopes, we need to look carefully at the way in
which heads and pipe lengths are related to delivery. I am going to explore simple models
of the system to find out how the variables are related.
Figure 23-27 shows the hydraulic ram in a complete system. The driver is a long pipe
with a total fall of hD . This fall will depend on the gradient of the stream and the length
of the pipe. The delivery pipe could be any length but the static head between say the
water level in the air vessel and the free surface of the top reservoir is hd .
Let us start by supposing that the driver can actually generate sufficient head for the nonreturn valve to open and consider the instant when the non-return valve has just opened
because the gravity-operated valve has just closed. We then have water flowing through
the driver, through the non-return valve and to the junction between the air vessel and
the delivery pipe. There some water goes through the delivery pipe and some into the air
vessel. Now I want to make two simplifications and suppose that all the flow goes into
the air vessel and that the pressure in the air vessel is equal to the pressure corresponding
to the static had hD . Then, if we just ignore the small difference in level between the free
surface in the air vessel and the datum for hd , we can say that the water in the driver is
acted on be a head hd  hD to reduce its velocity. Then we can write that, at the instant
© Ivor Bittle
18
dc g  (hd  hD )
l dc

. This expression tells us that the
 . Then
dt
l
g dt
retardation of the water flowing in the driver and into the air vessel is constant and this
means that the flow will decay uniformly until the weight of the non-return valve causes
it to close. This closure means that the water is flowing in the driver at the instant of
closure and a small water-hammer wave is propagated towards the intake end of the
driver only to be reflected and produce a pressure drop on its arrival in the chamber.
This pressure drop will open the gravity-operated valve. So the delivery will take place at
a steadily reducing rate until it is stopped by the closure of the non-return valve.
of closure, hd  hD 
We need some idea of numbers for the period of delivery. Suppose that the driver is 100
metres long, and hd  hD is 10 metres. The deceleration is then 0.98 metres/sec/sec. For
values of s less than 0.04 and noting the graphs in figure 23-25 this will mean that
delivery takes less than 1 second. This tells us that the ram will operate at frequencies that
are about 30 cycles/minute.
The gravity-operated valve can be set to open at any desired velocity in the driver. If it is
set low the delivery per cycle will be small but the frequency high and vice versa for a
higher velocity. We need to get an idea of how the velocity in the driver increases with
time so that we can make a better decision of how to set the valve. Let us look at the
behaviour of the water in the driver.
Once the gravity-operated valve is open the water will start to flow through the driver,
into the chamber and out to the stream through the gravity-operated valve. It will
accelerate under the head hD but the flow will be resisted by the fluid friction in the
driver that will ultimately set a maximum to the velocity of the flow. At some lower
velocity the drag on the mushroom of the valve will overcome the weight and the valve
will close. We could learn more if we put values to the velocity/time relationship using
typical figures.
Suppose that the driver has a length l and a diameter d and that it is laid with a slope s .
The head causing the increase in flow is clearly l  s and this head produces the
acceleration and overcomes friction in the pipe. We can write :4  f  l c 2 l dc
and this can be re-written in the form :ls 

 
d
2 g g dt

4  f c2 

  g  t
d 2g 

It is clear that the length can be cancelled in this expression because the head driving the
flow and the resistance to motion both increase directly with the length of the driver. We
have already seen that the slope determines the maximum velocity for any pipe diameter
and it becomes clear that the slope is a dominant quantity for this device.
c  s 
So let me find a graph for velocity in the driver against elapsed time for some specified
pipe. I will put d = 0.025 m, f = 0.008 and s =0.02 the expression above becomes :-
 c   0.02  0.0652  c 2   9.81   t and this can be used find data for
a graph of c versus t .
© Ivor Bittle
19
In figure 23-25 I have plotted the velocity of flow in a driver from zero velocity for a
pipe of 25 mm diameter for slopes of
0.02 and 0.04 and for a pipe of 50 mm
diameter at a slope of 0.04. The plots
show the effect of a change in slope and
the effect of a change in diameter.
The delivery per cycle will increase with
the magnitude of the velocity in the driver
when the gravity-operated valve closes.
For the three drivers in figure 23-25 it is
clear that up to 1.5 seconds the increase
in velocity is proportional to the elapsed
time. This means that 1.5 seconds may be
taken to be the shortest desirable interval
between closure of the non-return valve
and opening of the gravity-operated
Figure 23-29
valve. For longer times the increase in
velocity with time starts to fall away. There must be some optimum between about 1.5
seconds and say 2.5 seconds. This optimum can be found by testing.
When I started this paragraph on the hydraulic ram I knew little beyond the mode of
operation. I have used very simple models to get a much better idea of what I can expect
should I try to make one. Any engineer can do the same. This hydraulic ram is essentially
a workable device that can be made by anyone and work. Getting it to work for the best
is about selecting the best length for the driver and setting the gravity-operated valve.
Anyone with an actual site where a hydraulic ram could be installed might usefully
consider the possibility of creating a fall in the stream by building a low weir or just
placing blocks like stepping stones in the stream to cause an obstruction without
preventing water life from moving upstream. Then the slope can be increased for a short
driver and, whilst the frequency of operation will be much higher the compact
arrangement might be seen to be acceptable even with a reduced flow.
The hydraulic ram is a device that has found a niche in places where electricity is not
available and for these applications manufactured designs are readily available and those
will be probably be in metal. Many have been working unattended for over 100 years.
However the device might make a comeback in developed countries but the rise in metal
theft might lead to a desire to make everything in plastic with no scrap value. There are
no significant water hammer effects and the maximum pressures are only those of
delivery so the use of plastic is an option. The plastic must be chosen carefully for its
ability to survive continuous fluctuation of stress. Plastic materials have not been around
for 100 years yet.
© Ivor Bittle
20
Worked examples on reciprocating pumps
Q1 A single-acting reciprocating pump has a piston of 150 mm diameter and a stroke of
300 mm. The suction pipe is 10 metres long with a diameter of 75 mm and the suction
lift is 3.5 metres. If the pump runs at 30 rpm calculate the pressure head in the cylinder at
the start of the suction stroke. If the vapour pressure is 0.024 bar estimate the maximum
speed of the pump before separation occurs.
Q2 (i) A reciprocating pump has a bore of 150 mm and a stroke of 300 mm. It is to be
used to raise water to a height of 20 metres when running at 40 rpm. Ignore losses and
calculate the maximum flow and the power required to produce this flow.
(ii) The pump is used without air vessels and discharges through a pipe of diameter 100
mm and of length 20 metres. Take the friction factor f for the pipes to be 0.008 and
find a value for the maximum loss of head to friction in the delivery pipe. Take the
motion of the piston to be simple harmonic motion.
(iii) Calculate the inertia head impressed on the water at the start of delivery.
(iv) If the suction lift is 3 metres and the suction pipe is also 100 mm in diameter with a
length of 5 metres calculate the inertia head at the start of suction.
(v) Draw a diagram showing the variation of pressure head in the cylinder during one
cycle.
Q3 The bore and stroke of a single acting reciprocating pump are 100 mm and 200 mm
respectively. The suction pipe is 80 mm in diameter and 4.5 metres long and the pump is
4 metres above the water in the sump. If separation occurs when the absolute pressure
falls below 0.25 bar estimate the speed at which separation will occur. Take the piston to
have simple harmonic motion.
Q4 A single acting reciprocating pump has a piston of 250 mm diameter and a stroke of
450 mm. The delivery pipe is 110 mm in diameter and 50 metres long. The speed of
rotation is 60 rpm. Take the pump to have simple harmonic motion and estimate the
difference in the power consumption attributable to the friction loss in the delivery pipe
between running with or without an air vessel of substantial volume. Take f to be 0.01.
© Ivor Bittle
21
Q1 The angular velocity,  , of the crank at 30 rpm =
30  2
  radians /sec.
60
2
l  dp 
The inertia head at the start of the stroke =      2 r where d p is the diameter of
g  ds 
the piston, d s is the diameter of the suction pipe and r is the crank throw.
2
10  150 
2

    0.15  6.03 metres.
9.81  75 
The pressure at the start of the suction stroke = 3.5  6.03  9.53 metres (ie below
atmospheric pressure).
Then inertia head 
If the atmospheric pressure is taken to be 1 bar the vapour pressure of 0.024 bar = 0.976 105 0.976 105
0.976 bar gauge. This =

 9.95 metres.
  9.81
1000  9.81
From this the maximum value of the inertia head at the start of the stroke = 9.95-3.5 =
6.45 metres.
2
10  150 
2

    r from which   3.25 rads/sec or 31 rpm
9.81  75 
Q2 (1) The delivery per second will be equal to the swept volume times the number of
  0.152 
30
0.3   0.00265m3 / s .
strokes per second. So delivery per second 
4
60
For this delivery per second the minimum power consumption will be
  g  (delivery / sec)  lift  1000  9.81 0.00265  20  520 watts .
At this value 6.45 
(ii) The maximum speed of the piston is   r 
2    30
 0.15  0.47 m/s . The
60
2
 0.15 
maximum velocity of the water in the delivery pipe = 0.47  
  1.06 m/s . The
 0.1 
4  f  l c 2 4  0.008  20 1.06 2



 0.36 m
maximum loss to friction is the given by
dp
2g
0.1
2  9.81
2
l  dp 
     2 r where d p is the
(iii) The inertia head at the start of the stroke =
g  dd 
diameter of the piston, d d is the diameter of the delivery pipe and r is the crank throw.
20  0.15 

Inertia head =

9.81  0.1 
2
 30  2 

  0.15  6.75 metres.
 60 
2
(iv) By proportion to the pipe lengths the inertia head at the start of the suction stroke is
5
6.75 
 1.7 m .
20
(v) The diagram shows the required cycle to scale.
© Ivor Bittle
22
2
l  dp 
     2 r where d p is the
Q3 The inertia head at the start of the stroke =
g  ds 
diameter of the piston, d s is the diameter of the suction pipe and r is the crank throw.
4.5  100 

The inertia head is

9.81  80 
rotation in rpm.
2
 2  n 
2

  0.1  0.00078  n where n is the speed of
60


2
p
0.25 105
A pressure of 0.2 bar is equal to

 2.5 m of water. Atmospheric
 g 1000  9.81
pressure is equal to 1 bar = 10metres. The maximum value that the inertia head may have
is 10  4  2.5  3.5 m . Therefore 0.00078  n 2  3.5 , from which n  69 rpm .
Q4 When an air vessel is fitted the best outcome is for the flow during delivery to be
  0.252
 0.45  0.022 m3 . There are
steady. The delivery per revolution of the pump is
4
60 rpm so in 1 second the delivery is 0.022 m3. V  0.022 m3 / s .
0.022
 2.3 m/s
Then the steady velocity in the delivery pipe is
  0.112 / 4
4  f  l c 2 4  0.01 50 2.32
The loss to friction will be



 4.9 m .
d
2g
0.11
2  9.81
The power loss to friction will be   g V  loss  1000  9.81 0.022  2.3  496 W
© Ivor Bittle
23
When there is no air vessel the flow through the delivery pipe will be intermittent with
flow varying sinusoidally for one stroke followed by a stroke during which there is no
flow. The graph shows one cycle that takes place in one second. Necessarily the power
required to maintain this flow against pipe friction will also vary sinusoidally. Now power
is energy transfer per second so in this case, where one cycle takes place in one second,
the total energy transferred to the water by friction during one cycle is also the power
and takes place during the delivery
stroke. We can integrate to find that
energy transfer. The basis of any
calculation must be the graph of the
velocity of the piston versus time. For
the graph the maximum velocity of the
piston is given by the tangential velocity
of the crankpin =   r which =
2    r  2    0.225  1.414m / s .
Now we need the speed of the water in
the delivery pipe c at some instant of
time t .
2
 250 
c
    r  sin t  5.165  2  0.225  sin 2  t  7.3sin 2  t .
 110 
The instantaneous value of V at time t =
  d p2
  0.112
c
 7.3  sin 2  t 
 0.0694  sin 2  t
4
4
4  f  l c2

Energy loss/second at time t =  g  V 
= :dp
2g
4  0.01 50  7.3  sin 2  t 
3
1000  9.81 0.0694  sin 2  t  

 33, 600  sin 2  t 
0.11
2  9.81
3
This can be integrated energy lost per cycle as  33,600   sin 2  t  dt = 7.13kJ .
0.5
3
0
As this energy is lost in one second the mean power is 7.13 kW. The saving is 6.63 kW.
© Ivor Bittle
24