GRUNDFOS
RESEARCH AND TECHNOLOGY
The Centrifugal Pump
The Centrifugal Pump
5
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Mechanical, electronic, photographic or other reproduction or copying from this book or parts
of it are according to the present Danish copyright law not allowed without written permission
from or agreement with GRUNDFOS Management A/S.
GRUNDFOS Management A/S cannot be held responsible for the correctness of the information
given in the book. Usage of information is at your own responsibility.
6
Preface
In the Department of Structural and Fluid Mechanics
we are happy to present the first English edition of the
book: ’The Centrifugal Pump’. We have written the book
because we want to share our knowledge of pump hydraulics, pump design and the basic pump terms which
we use in our daily work.
’The Centrifugal Pump’ is primarily meant as an internal book and is aimed at technicians who work with
development and construction of pump components.
Furthermore, the book aims at our future colleagues,
students at universities and engineering colleges, who
can use the book as a reference and source of inspiration in their studies. Our intention has been to write
an introductory book that gives an overview of the hydraulic components in the pump and at the same time
enables technicians to see how changes in construction and operation influence the pump performance.
In chapter 1, we introduce the principle of the centrifugal pump as well as its hydraulic components, and we
list the different types of pumps produced by Grundfos.
Chapter 2 describes how to read and understand the
pump performance based on the curves for head, power, efficiency and NPSH.
In chapter 3 you can read about how to adjust the
pump’s performance when it is in operation in a system.
The theoretical basis for energy conversion in a centrifugal pump is introduced in chapter 4, and we go through
how affinity rules are used for scaling the performance
of pump impellers. In chapter 5, we describe the different types of losses which occur in the pump, and how
the losses affect flow, head and power consumption. In
the book’s last chapter, chapter 6, we go trough the test
types which Grundfos continuously carries out on both
assembled pumps and pump components to ensure
that the pump has the desired performance.
The entire department has been involved in the development of the book. Through a longer period of time we
have discussed the idea, the contents and the structure
and collected source material. The framework of the
Danish book was made after some intensive working
days at ‘Himmelbjerget’. The result of the department’s
engagement and effort through several years is the book
which you are holding.
We hope that you will find ‘The Centrifugal Pump’ useful, and that you will use it as a book of reference in you
daily work.
Enjoy!
Christian Brix Jacobsen
Department Head, Structural and Fluid Mechanics, R&T
7
Contents
Chapter 1. Introduction to Centrifugal Pumps................11
1.1
1.2
Principle of centrifugal pumps. ......................................12
The pump’s hydraulic components.............................13
1.2.1 Inlet flange and inlet.............................................14
1.2.2 Impeller..........................................................................15
1.2.3 Coupling and drive.................................................17
1.2.4 Impeller seal................................................................18
1.2.5 Cavities and axial bearing................................ 19
1.2.6 Volute casing, diffuser and
outlet flange. ..............................................................21
1.2.7 Return channel and outer sleeve. ................23
1.3 Pump types and systems................................................... 24
1.3.1 The UP pump..............................................................25
1.3.2 The TP pump...............................................................25
1.3.3 The NB pump..............................................................25
1.3.4 The MQ pump............................................................25
1.3.5 The SP pump.............................................................. 26
1.3.6 The CR pump. ............................................................ 26
1.3.7 The MTA pump......................................................... 26
1.3.8 The SE pump...............................................................27
1.3.9 The SEG pump. ..........................................................27
1.4 Summary ..............................................................................................27
Chapter 2. Performance curves ............................................29
2.1
2.2
2.3
2.4
2.5
8
Standard curves . ..................................................................... 30
Pressure...........................................................................................32
Absolute and relative pressure.......................................33
Head ............................................................................................. 34
Differential pressure across the pump .....................35
2.5.1 Total pressure difference ..................................35
2.5.2 Static pressure difference..................................35
2.5.3 Dynamic pressure difference. .........................35
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.5.4 Geodetic pressure difference......................... 36
Energy equation for an ideal flow................................37
Power............................................................................................... 38
2.7.1 Speed. ............................................................................. 38
Hydraulic power....................................................................... 38
Efficiency. ...................................................................................... 39
NPSH, Net Positive Suction Head.................................40
Axial thrust.................................................................................. 44
Radial thrust. .............................................................................. 44
Summary........................................................................................45
Chapter 3. Pumps operating in systems............................ 47
3.1
3.2
3.3
3.4
3.5
3.6
3.7
Single pump in a system.................................................... 49
Pumps operated in parallel.............................................. 50
Pumps operated in series. ..................................................51
Regulation of pumps. ............................................................51
3.4.1 Throttle regulation.................................................52
3.4.2 Regulation with bypass valve.........................52
3.4.3 Start/stop regulation. ..........................................53
3.4.4 Regulation of speed...............................................53
Annual energy consumption . ........................................ 56
Energy efficiency index (EEI).............................................57
Summary....................................................................................... 58
Chapter 4. Pump theory.......................................................... 59
4.1
4.2
4.3
Velocity triangles.....................................................................60
4.1.1 Inlet.................................................................................. 62
4.1.2 Outlet.............................................................................. 63
Euler’s pump equation........................................................ 64
Blade shape and pump curve..........................................66
4.4
4.5
4.6
4.7
4.8
4.9
Usage of Euler’s pump equation. ................................... 67
Affinity rules................................................................................ 68
4.5.1 Derivation of affinity rules............................... 70
Pre-rotation..................................................................................72
Slip......................................................................................................73
The pump’s specific speed.................................................74
Summary........................................................................................75
Chapter 5. Pump losses............................................................. 77
5.1
5.2
Loss types.......................................................................................78
Mechanical losses...................................................................80
5.2.1 Bearing loss and shaft seal loss....................80
5.3 Hydraulic losses........................................................................80
5.3.1 Flow friction................................................................81
5.3.2 Mixing loss at
cross-section expansion............................. 86
5.3.3 Mixing loss at
cross-section reduction.......................................87
5.3.4 Recirculation loss....................................................89
5.3.5 Incidence loss............................................................90
5.3.6 Disc friction................................................................. 91
5.3.7 Leakage.......................................................................... 92
5.4 Loss distribution as function of
specific speed............................................................................. 95
5.5 Summary....................................................................................... 95
Chapter 6. Pumps tests................................................. 97
6.1
6.2
Test types......................................................................................98
Measuring pump performance. ....................................99
6.2.1 Flow................................................................................100
6.2.2 Pressure........................................................ 100
6.2.3 Temperature............................................................ 101
6.2.4 Calculation of head............................................. 102
6.2.5 General calculation of head..................... 103
6.2.6 Power consumption. .......................................... 104
6.2.7 Rotational speed................................................... 104
6.3 Measurement of the pump’s NPSH.......................... 105
6.3.1 NPSH3% test by lowering the
inlet pressure............................................... 106
6.3.2 NPSH3% test by increasing the flow. ......... 107
6.3.3 Test beds. ................................................................... 107
6.3.4 Water quality. ......................................................... 108
6.3.5 Vapour pressure and density....................... 108
6.3.6 Reference plane..................................................... 108
6.3.7 Barometric pressure...........................................109
6.3.8 Calculation of NPSHA and determination
of NPSH3%. ..................................................................109
6.4 Measurement of force.......................................................109
6.4.1 Measuring system. .............................................. 110
6.4.2 Execution of force measurement. ............. 111
6.5 Uncertainty in measurement of performance. . 111
6.5.1 Standard demands for uncertainties...... 111
6.5.2 Overall uncertainty..............................................112
6.5.3 Test bed uncertainty...........................................112
6.6 Summary......................................................................................112
Appendix....................................................................................... 113
A. Units.........................................................................................................114
B. Control of test results.................................................................. 117
Bibliography ............................................................................................122
Standards...................................................................................................123
Index ............................................................................................................ 124
Substance values for water...........................................................131
List of Symbols........................................................................................132
9
10
Chapter 1
Introduction to
centrifugal pumps
1.1 Principle of the centrifugal pump
1.2 Hydraulic components
1.3 Pump types and systems
1.4 Summary
1. Introduction to Centrifugal Pumps
1. Introduction to Centrifugal Pumps
In this chapter, we introduce the components in the centrifugal pump and
a range of the pump types produced by Grundfos. This chapter provides the
reader with a basic understanding of the principles of the centrifugal pump
and pump terminology.
The centrifugal pump is the most used pump type in the world. The principle
is simple, well-described and thoroughly tested, and the pump is robust, effective and relatively inexpensive to produce. There is a wide range of variations based on the principle of the centrifugal pump and consisting of the
same basic hydraulic parts. The majority of pumps produced by Grundfos
are centrifugal pumps.
Outlet
1.1 Principle of the centrifugal pump
An increase in the fluid pressure from the pump inlet to its outlet is created when the pump is in operation. This pressure difference drives the fluid
through the system or plant.
The centrifugal pump creates an increase in pressure by transferring mechanical energy from the motor to the fluid through the rotating impeller.
The fluid flows from the inlet to the impeller centre and out along its blades.
The centrifugal force hereby increases the fluid velocity and consequently
also the kinetic energy is transformed to pressure. Figure 1.1 shows an example of the fluid path through the centrifugal pump.
Impeller
Inlet
Direction of rotation
Outlet
Impeller Impeller
blade
Inlet
Figure 1.1: Fluid path through
the centrifugal pump.
12
12
1.2 Hydraulic components
The principles of the hydraulic components are common for most centrifugal pumps. The hydraulic components are the parts in contact with the fluid.
Figure 1.2 shows the hydraulic components in a single-stage inline pump.
The subsequent sections describe the components from the inlet flange to
the outlet flange.
Figure 1.2: Hydraulic
components.
Motor
Coupling
Shaft
Shaft seal
Cavity above impeller
Cavity below impeller
Volute
Diffuser
Outlet flange
Inlet flange
Pump housing
13
Impeller
Impeller seal
Inlet
13
1. Introduction to Centrifugal Pumps
1.2.1 Inlet flange and inlet
The pump is connected to the piping system through its
inlet and outlet flanges. The design of the flanges depends
on the pump application. Some pump types have no inlet
flange because the inlet is not mounted on a pipe but submerged directly in the fluid.
The inlet guides the fluid to the impeller eye. The design of
the inlet depends on the pump type. The four most common types of inlets are inline, endsuction, doublesuction
and inlet for submersible pumps, see figure 1.3.
Inline pumps are constructed to be mounted on a straight
pipe – hence the name inline. The inlet section leads the
fluid into the impeller eye.
Impeller
Inlet
Impeller
Inlet
Endsuction pumps have a very short and straight inlet section because the impeller eye is placed in continuation of
the inlet flange.
The impeller in doublesuction pumps has two impeller eyes.
The inlet splits in two and leads the fluid from the inlet
flange to both impeller eyes. This design minimises the axial
force, see section 1.2.5.
In submersible pumps, the motor is often placed below the
hydraulic parts with the inlet placed in the mid section of
the pump, see figure 1.3. The design prevents hydraulic losses related to leading the fluid along the motor. In addition,
the motor is cooled due to submersion in the fluid.
Impeller
Inlet
Impeller
Inline pump
Endsuction pump
Doublesuction pump
Inlet
Submersible pump
Figure 1.3: Inlet for inline, endsuction, doublesuction and submersible pump.
14
14
The design of the inlet aims at creating a uniform velocity profile into the
impeller since this leads to the best performance. Figure 1.4 shows an example of
the velocity distribution at different cross-sections in the inlet.
1.2.2 Impeller
The blades of the rotating impeller transfer energy to the fluid there by
increasing pressure and velocity. The fluid is sucked into the impeller at the
impeller eye and flows through the impeller channels formed by the blades
between the shroud and hub, see figure 1.5.
The design of the impeller depends on the requirements for pressure, flow
and application. The impeller is the primary component determining the
pump performance. Pumps variants are often created only by modifying
the impeller.
Figure 1.4: Velocity distribution in inlet.
 Axial direction
Hub plate
Hub
Impeller channel
(blue area)
Trailing edge
 Radial direction
The impeller’s direction of
rotation
 Tangential direction
The impeller’s direction of rotation
Leading edge
Shroud plate
Impeller blade
Figure 1.5: The impeller components, definitions of directions and flow relatively to the impeller.
15
15
1. Introduction to Centrifugal Pumps
The impeller’s ability to increase pressure and create flow depends mainly
on whether the fluid runs radially or axially through the impeller,
see figure 1.6.
In a radial impeller, there is a significant difference between the inlet
diameter and the outlet diameter and also between the outlet diameter
and the outlet width, which is the channel height at the impeller exit. In
this construction, the centrifugal forces result in high pressure and low
flow. Relatively low pressure and high flow are, on the contrary, found in an
axial impeller with a no change in radial direction and large outlet width.
Semiaxial impellers are used when a trade-off between pressure rise and flow
is required.
The impeller has a number of impeller blades. The number mainly depends
on the desired performance and noise constraints as well as the amount and
size of solid particles in the fluid. Impellers with 5-10 channels has proven to
give the best efficiency and is used for fluid without solid particles. One, two
or three channel impellers are used for fluids with particles such as wastewater. The leading edge of such impellers is designed to minimise the risk
of particles blocking the impeller. One, two and three channel impellers can
handle particles of a certain size passing through the impeller. Figure 1.7
shows a one channel pump.
Radial impeller
Semiaxial impeller
Axial impeller
Figure 1.6: Radial, semiaxial and
axial impeller.
Figure 1.7: One channel pump.
Impellers without a shroud are called open impellers. Open impellers are
used where it is necessary to clean the impeller and where there is risk of
blocking. A vortex pump with an open impeller is used in waste water application. In this type of pump, the impeller creates a flow resembling the
vortex in a tornado, see figure 1.8. The vortex pump has a low efficiency
compared to pumps with a shroud and impeller seal.
After the basic shape of the impeller has been decided, the design of the
impeller is a question of finding a compromise between friction loss and loss
as a concequence of non uniform velocity profiles. Generally, uniform velocity
profiles can be achieved by extending the impeller blades but this results in
increased wall friction.
16
Figure 1.8: Vortex pump.
16
1.2.3 Coupling and drive
The impeller is usually driven by an electric motor. The coupling between motor
and hydraulics is a weak point because it is difficult to seal a rotating shaft. In
connection with the coupling, distinction is made between two types of pumps:
Dry-runner pumps and canned rotor type pump. The advantage of the dry-runner
pump compared to the canned rotor type pump is the use of standardized motors.
The disadvantage is the sealing between the motor and impeller.
Motor
Shaft seal
In the dry runner pump the motor and the fluid are separated either by a shaft
seal, a separation with long shaft or a magnetic coupling.
In a pump with a shaft seal, the fluid and the motor are separated by seal rings, see
figure 1.9. Mechanical shaft seals are maintenance-free and have a smaller leakage
than stuffing boxes with compressed packing material. The lifetime of mechanical
shaft seals depends on liquid, pressure and temperature.
If motor and fluid are separated by a long shaft, then the two parts will not get
in contact then the shaft seal can be left out, see figure 1.10. This solution has
limited mounting options because the motor must be placed higher than the
hydraulic parts and the fluid surface in the system. Furthermore the solution
results in a lower efficiency because of the leak flow through the clearance between the shaft and the pump housing and because of the friction between the
fluid and the shaft.
Exterior magnets on
the motor shaft
Inner magnets on
the impeller shaft
Rotor can
Motor
Long shaft
Water level
Motor
Motor shaft
Hydraulics
Motor cup
Exterior magnets
Inner magnets
Motor cup
Figure 1.9: Dry-runner with shaft seal.
Figure 1.10: Dry-runner with long shaft.
Rotor can
Impeller shaft
Figure 1.11: Dry-runner with magnet drive.
17
17
1. Introduction to Centrifugal Pumps
In pumps with a magnetic drive, the motor and the fluid are separated by
a non-magnetizable rotor can which eliminates the problem of sealing a
rotating shaft. On this type of pump, the impeller shaft has a line of fixed
magnets called the inner magnets. The motor shaft ends in a cup where the
outer magnets are mounted on the inside of the cup, see figure 1.11. The
rotor can is fixed in the pump housing between the impeller shaft and the
cup. The impeller shaft and the motor shaft rotate, and the two parts are
connected through the magnets. The main advantage of this design is that
the pump is hermitically sealed but the coupling is expensive to produce.
This type of sealing is therefore only used when it is required that the pump
is hermetically sealed.
In pumps with a rotor can, the rotor and impeller are separated from the
motor stator. As shown in figure 1.12, the rotor is surrounded by the fluid
which lubricates the bearings and cools the motor. The fluid around the rotor results in friction between rotor and rotor can which reduces the pump
efficiency.
1.2.4 Impeller seal
A leak flow will occur in the gap between the rotating impeller and stationary
pump housing when the pump is operating. The rate of leak flow depends
mainly on the design of the gap and the impeller pressure rise. The leak flow
returns to the impeller eye through the gap, see figure 1.13. Thus, the impeller has to pump both the leak flow and the fluid through the pump from the
inlet flange to the outlet flange. To minimise leak flow, an impeller seal is
mounted.
The impeller seal comes in various designs and material combinations. The
seal is typically turned directly in the pump housing or made as retrofitted
rings. Impeller seals can also be made with floating seal rings. Furthermore,
there are a range of sealings with rubber rings in particular well-suited for
handling fluids with abrasive particles such as sand.
18
Fluid
Rotor
Stator
Rotor can
Outlet
Impeller
Inlet
Bearings
Figure 1.12: Canned rotor type pump.
Outlet
Inlet
Leak flow
Gap
Impeller seal
Figure 1.13: Leak flow through the gap.
18
Achieving an optimal balance between leakage and friction is an essential
goal when designing an impeller seal. A small gap limits the leak flow but
increases the friction and risk of drag and noise. A small gap also increases
requirements to machining precision and assembling resulting in higher
production costs. To achieve optimal balance between leakage and friction,
the pump type and size must be taken into consideration.
1.2.5 Cavities and axial bearing
The volume of the cavities depends on the design of the impeller and the
pump housing, and they affect the flow around the impeller and the pump’s
ability to handle sand and air.
Cavity above impeller
Primary flow
Cavity below impeller
Secondary flow
The impeller rotation creates two types of flows in the cavities: Primary
flows and secondary flows. Primary flows are vorticies rotating with the
impeller in the cavities above and below the impeller, see figure 1.14.
Secondary flows are substantially weaker than the primary flows.
Primary and secondary flows influence the pressure distribution on the
outside of the impeller hub and shroud affecting the axial thrust. The axial
thrust is the sum of all forces in the axial direction arising due to the pressure condition in the pump. The main force contribution comes from the
rise in pressure caused by the impeller. The impeller eye is affected by the
inlet pressure while the outer surfaces of the hub and shroud are affected
by the outlet pressure, see figure 1.15. The end of the shaft is exposed to the
atmospheric pressure while the other end is affected by the system pressure. The pressure is increasing from the center of the shaft and outwards.
19
Figure 1.14: Primary and secondary flows
in the cavities.
19
1. Introduction to Centrifugal Pumps
The axial bearing absorbs the entire axial thrust and is therefore exposed to
the forces affecting the impeller.
Outlet pressure
Atmospheric pressure
Outlet pressure
The impeller must be axially balanced if it is not possible to absorb the entire
axial thrust in the axial bearing. There are several possibilities of reducing
the thrust on the shaft and thereby balance the axial bearing. All axial
balancing methods result in hydraulic losses.
One approach to balance the axial forces is to make small holes in the hub
plate, see figure 1.16. The leak flow through the holes influences the flow
in the cavities above the impeller and thereby reduces the axial force but it
results in leakage.
Axial thrust
Inlet pressure
Figure 1.15: Pressure forces which cause
axial thrust.
Axial balancing hole
Another approach to reduce the axial thrust is to combine balancing holes
with an impeller seal on the hub plate, see figure 1.17. This reduces the pressure in the cavity between the shaft and the impeller seal and a better balance can be achieved. The impeller seal causes extra friction but smaller
leak flow through the balancing holes compared to the solution without the
impeller seal.
A third method of balancing the axial forces is to mount blades on the back
of the impeller, see figure 1.18. Like the two previous solutions, this method
changes the velocities in the flow at the hub plate whereby the pressure
distribution is changed proportionally. However, the additional blades use
power without contributing to the pump performance. The construction
will therefore reduce the efficiency.
Figure 1.16: Axial thrust reduction using
balancing holes.
Impeller seal
Axial balancing hole
Figure 1.17: Axial thrust reduction using impeller seal and balancing holes.
20
20
A fourth method to balance the axial thrust is to mount fins on the pump
housing in the cavity below the impeller, see figure 1.19. In this case, the primary flow velocity in the cavity below the impeller is reduced whereby the
pressure increases on the shroud. This type of axial balancing increases disc
friction and leak loss because of the higher pressure.
Blades
Figure 1.18: Axial thrust reduction through
blades on the back of the hub plate.
1.2.6 Volute casing, diffuser and outlet flange
The volute casing collects the fluid from the impeller and leads into the
outlet flange. The volute casing converts the dynamic pressure rise in the
impeller to static pressure. The velocity is gradually reduced when the crosssectional area of the fluid flow is increased. This transformation is called
velocity diffusion. An example of diffusion is when the fluid velocity in a pipe
is reduced because of the transition from a small cross-sectional area to a
large cross-sectional area, see figure 1.20. Static pressure, dynamic pressure
and diffusion are elaborated in sections 2.2, 2.3 and 5.3.2.
Figure 1.20: Change of fluid velocity
in a pipe caused by change
in the cross-section area.
Fins
Figure 1.19: Axial thrust reduction using fins
in the pump housing.
Diffusion
Small cross-section:
High velocity, low static
pressure, high dynamic pressure
Large cross-section:
Low velocity, high static
pressure, low dynamic
pressure
21
21
1. Introduction to Centrifugal Pumps
The volute casing consists of three main components:
Ring diffusor, volute and outlet diffusor, see figure 1.21.
An energy conversion between velocity and pressure occurs in each of the three components.
The primary ring diffusor function is to guide the fluid
from the impeller to the volute. The cross-section area in
the ring diffussor is increased because of the increase in
diameter from the impeller to the volute. Blades can be
placed in the ring diffusor to increase the diffusion.
The primary task of the volute is to collect the fluid from
the ring diffusor and lead it to the diffusor. To have the
same pressure along the volute, the cross-section area in
the volute must be increased along the periphery from
the tongue towards the throat. The throat is the place
on the outside of the tongue where the smallest crosssection area in the outlet diffusor is found. The flow conditions in the volute can only be optimal at the design
point. At other flows, radial forces occur on the impeller
because of circumferential pressure variation in the volute. Radial forces must, like the axial forces, be absorbed
in the bearing, see figure 1.21.
The outlet diffusor connects the throat with the outlet flange. The diffusor increases the static pressure by
a gradual increase of the cross-section area from the
throat to the outlet flange.
The volute casing is designed to convert dynamic pressure to static pressure is achieved while the pressure
losses are minimised. The highest efficiency is obtained
by finding the right balance between changes in velocity
and wall friction. Focus is on the following parameters
when designing the volute casing: The volute diameter,
the cross-section geometry of the volute, design of the
tongue, the throat area and the radial positioning as well
as length, width and curvature of the diffusor.
Radial force vector
Outlet flange
Outlet diffusor
Ring diffusor
Throat
Volute
Tongue
Radial force vector
Figure 1.21:
The components of the
volute casing.
22
22
1.2.7 Return channel and outer sleeve
To increase the pressure rise over the pump, more impellers can be connected in series. The return channel leads the fluid from one impeller to the next,
see figure 1.22. An impeller and a return channel are either called a stage or
a chamber. The chambers in a multistage pump are altogether called the
chamber stack.
Besides leading the fluid from one impeller to the next, the return channel
has the same basic function as volute casing: To convert dynamic pressure
to static pressure. The return channel reduces unwanted rotation in the fluid
because such a rotation affects the performance of the subsequent impeller.
The rotation is controlled by guide vanes in the return channel.
Outer
sleeve
Annular
outlet
Chamber
stack
Chamber
In multistage inline pumps the fluid is lead from the top of the chamber
stack to the outlet in the channel formed by the outer part of the chamber
stack and the outer sleeve, see figure 1.22.
When designing a return channel, the same design considerations of impeller and volute casing apply. Contrary to volute casing, a return channel does
not create radial forces on the impeller because it is axis-symmetric.
Impeller
Guide vane
Impeller blade
Return channel
Figure 1.22: Hydraulic components in an
inline multistage pump.
23
23
1. Introduction to Centrifugal Pumps
1.3 Pump types and systems
This section describes a selection of the centrifugal pumps produced by
Grundfos. The pumps are divided in five overall groups: Circulation pumps,
pumps for pressure boosting and fluid transport, water supply pumps, industrial pumps and wastewater pumps. Many of the pump types can be
used in different applications.
Circulation pumps are primarily used for circulation of water in closed systems e.g. heating, cooling and airconditioning systems as well as domestic
hot water systems. The water in a domestic hot water system constantly
circulates in the pipes. This prevents a long wait for hot water when the tap
is opened.
Pumps for pressure boosting are used for increasing the pressure of cold water and as condensate pumps for steam boilers. The pumps are usually designed to handle fluids with small particles such as sand.
Water supply pumps can be installed in two ways: They can either be submerged in a well or they can be placed on the ground surface. The conditions
in the water supply system make heavy demands on robustness towards
ochre, lime and sand.
Industrial pumps can, as the name indicates, be used everywhere in the industry and this in a very broad section of systems which handle many different homogeneous and inhomogeneous fluids. Strict environmental and
safety requirements are enforced on pumps which must handle corrosive,
toxic or explosive fluids, e.g. that the pump is hermetically closed and corrosion resistant.
Wastewater pumps are used for pumping contaminated water in sewage
plants and industrial systems. The pumps are constructed making it possible
to pump fluids with a high content of solid particles.
24
24
1.3.1. The UP pump
Circulation pumps are used for heating, circulation of cold water, ventilation and aircondition systems in houses, office buildings, hotels, etc. Some
of the pumps are installed in heating systems at the end user. Others are
sold to OEM customers (Original Equipment Manufacturer) that integrate
the pumps into gas furnace systems. It is an inline pump with a canned rotor which only has static sealings. The pump is designed to minimise pipetransferred noise. Grundfos produces UP pumps with and without automatic regulation of the pump. With the automatic regulation of the pump, it is
possible to adjust the pressure and flow to the actual need and thereby save
energy.
1.3.2 The TP pump
The TP pump is used for circulation of hot or cold water mainly in heating,
cooling and airconditioning systems. It is an inline pump and contrary to the
smaller UP pump, the TP pump uses a standard motor and shaft seal.
Outlet
Hydraulic
Motor
Inlet
Figure 1.23: UP pumps.
Inlet
Outlet
Figure 1.24: TP pump.
1.3.3 The NB pump
The NB pump is for transportation of fluid in district heating plants, heat
supply, cooling and air conditioning systems, washdown systems and other
industrial systems. The pump is an endsuction pump, and it is found in many
variants with different types of shaft seals, impellers and housings which
can be combined depending on fluid type, temperature and pressure.
Outlet
Inlet
Figure 1.25: NB pump.
1.3.4 The MQ pump
The MQ pump is a complete miniature water supply unit. It is used for
water supply and transportation of fluid in private homes, holiday
houses, agriculture, and gardens. The pump control ensures that it starts
and stops automatically when the tap is opened. The control protects
the pump if errors occur or if it runs dry. The built-in pressure expansion
tank reduces the number of starts if there are leaks in the pipe system.
The MQ pump is self-priming, then it can clear a suction pipe from air
and thereby suck from a level which is lower than the one where
the pump is placed.
25
Outlet
Inlet
Figure 1.26: MQ pump.
25
1. Introduction to Centrifugal Pumps
1.3.5 The SP pump
The SP pump is a multi-stage submersible pump which is used for raw water supply, ground water lowering and pressure boosting. The SP pump can
also be used for pumping corrosive fluids such as sea water. The motor is
mounted under the chamber stack, and the inlet to the pump is placed between motor and chamber stack. The pump diameter is designed to the size
of a standard borehole. The SP pump is equipped with an integrated nonreturn valve to prevent that the pumped fluid flows back when the pump is
stopped. The non-return valve also helps prevent water hammer.
1.3.6 The CR pump
The CR pump is used in washers, cooling and air conditioning systems,
water treatment systems, fire extinction systems, boiler feed systems and
other industrial systems. The CR pump is a vertical inline multistage pump.
This pump type is also able to pump corrosive fluids because the hydraulic
parts are made of stainless steel or titanium.
Figure 1.27: SP pump.
Outlet
Non-return valve
Chamber stack
Inlet
Motor
Motor
Chamber stack
1.3.7 The MTA pump
The MTA pump is used on the non-filtered side of the machining process
to pump coolant and lubricant containing cuttings, fibers and abrasive
particles. The MTA pump is a dry-runner pump with a long shaft and no
shaft seal. The pump is designed to be mounted vertically in a tank.
The installation length, the part of the pump which is submerged
in the tank, is adjusted to the tank depth so that it is possible to
drain the tank of coolant and lubricant.
Inlet
Outlet
Figure 1.28: CR-pump.
Outlet
Mounting flange
Outlet channel
Shaft
Pump housing
Figure 1.29: MTA pump.
26
Inlet
26
1.3.8 The SE pump
The SE pump is used for pumping wastewater, water containing sludge and
solids. The pump is unique in the wastewater market because it can be installed submerged in a waste water pit as well as installed dry in a pipe system. The series of SE pumps contains both vortex pumps and single-channel
pumps. The single-channel pumps are characterised by a large free passage,
and the pump specification states the maximum diameter for solids passing
through the pump.
1.3.9 The SEG pump
The SEG pump is in particular suitable for pumping waste water from toilets. The SEG pump has a cutting system which cuts perishable solids into
smaller pieces which then can be lead through a tube with a relative small
diameter. Pumps with cutting systems are also called grinder pumps.
Motor
Outlet
Inlet
Figure 1.30: SE pump.
1.4 Summary
In this chapter, we have covered the principle of the centrifugal pump and
its hydraulic components. We have discussed some of the overall aspects
connected to design of the single components. Included in the chapter is
also a short description of some of the Grundfos pumps.
Motor
Inlet
Outlet
Figure 1.31: SEG pumps.
27
27
28
28
Chapter 2
Performance
curves
2.1 Standard curves
2.2 Pressure
H
[m]
2.3 Absolute and relative pressure
50
2.4 Head
40
2.5 Differential pressure across the
pump - description of differential
pressure
2.6 Energy equation for an ideal
flow
2.7 Power
2.8 Hydraulic power
2.9 EfficiencyHead
2.10 NPSH,
Net Positive Suction Head
30
2.11 Axial thrust
Efficiency
20
2.12 Radial thrust
10
2.13 Summary
0
0
η [%]
10
20
30
40
70
60
50
50
P2
[kW]
10
8
6
4
2
0
Power
NPSH
60
70
Q [m3/h]
40
30
20
10
0
NPSH
(m)
12
10
8
6
4
2
0
2. Performance curves
2. Performance curves
The pump performance is normally described by a set of curves. This chapter
explains how these curves are interpretated and the basis for the curves.
2.1 Standard curves
Performance curves are used by the customer to select pump matching his
requirements for a given application.
The data sheet contains information about the head (H) at different flows
(Q), see figure 2.1. The requirements for head and flow determine the overall
dimensions of the pump.
η [%]
H [m]
Head
50
40
70
30
60
Efficiency
50
20
40
30
20
10
10
0
0
10
20
30
40
50
Q [m /h]
0
Power
10
8
8
6
6
4
4
2
0
30
70
NPSH [m]
P2 [kW]
10
60
3
NPSH
2
0
Fígure 2.1: Typical performance curves for a
centrifugal pump. Head (H), power
consumption (P), efficiency (η) and NPSH are
shown as function of the flow.
30
In addition to head, the power consumption (P) is also to be found in the data
sheet. The power consumption is used for dimensioning of the installations
which must supply the pump with energy. The power consumption is like
the head shown as a function of the flow.
Information about the pump efficiency (η) and NPSH can also be found in
the data sheet. NPSH is an abbreviation for ’Net Positive Suction Head’. The
NPSH curve shows the need for inlet head, and which requirements the
specific system have to fullfill to avoid cavitation. The efficiency curve is
used for choosing the most efficient pump in the specified operating range.
Figure 2.1 shows an example of performance curves in a data sheet.
During design of a new pump, the desired performance curves are a vital
part of the design specifications. Similar curves for axial and radial thrust are
used for dimensioning the bearing system.
Motor
Controller
The performance curves describe the performance for the complete pump
unit, see figure 2.2. An adequate standard motor can be mounted on the
pump if a pump without motor is chosen. Performance curves can be
recalculated with the motor in question when it is chosen.
Coupling
For pumps sold both with and without a motor, only curves for the hydraulic
components are shown, i.e. without motor and controller. For integrated
products, the pump curves for the complete product are shown.
Hydraulics
Figure 2.2.: The performance curves are
stated for the pump itself or for the
complete unit consisting of pump, motor
and electronics.
31
31
2. Performance curves
2.2 Pressure
Pressure (p) is an expression of force per unit area and is split into static and
dynamic pressure. The sum of the two pressures is the total pressure:
p tot = pstat + pdyn
[Pa]
where
1 ⋅ ρ ⋅ V[Pa]
2
ptot = Total
[Pa]
pdyn =pressure
2
pstat = Static pressure [Pa]
pdyn = Dynamic
∆p tot = ∆pressure
pstat + ∆p[Pa]
dyn + ∆p geo
(2.1)
(2.2)
[Pa]
(2.5)
∆pstat = ispstat,
(2.6)
[aPapressure
]
Static pressure
measured
gauge, and the measurement
of
out − pstat,with
in
static pressure must always be done in static fluid or through a pressure tap
2
1 ⋅ ρ ⋅ V to2 the
1 flow
mounted
[Pa] see figure 2.3.
(2.7)
∆pperpendicular
⋅ ρ ⋅ Vdirection,
dyn =
out −
in
2
2
pstat
Q
Total pressure can be measured
through a pressure tap with the opening
2


facing the flow direction,
pressure can be found
 1 2.3. The

Qsee
1 dynamic
 figure
 [Pa]
⋅  4 −
Δp = 1 ⋅ ρ ⋅ 
(2.8)
4 
difference
2
measuringdyn
the pressure
total
pressure
and static pressure.
π   Dbetween
D in 
out
 4
Such a combined pressure measurement can be performed using a pitot tube.
Dynamic pressure is a function of the fluid velocity. The dynamic pressure can
(2.1)
[Pa]
p tot = pwith
stat + p dynfollowing
be calculated
measured
∆p geo = ∆z ⋅the
ρ ⋅ g [Pa] formula,where the velocity (V) is (2.9)
and the fluid density (ρ) is know:
p V2
[Pa]
pdyn+ = 1 +⋅ ρg ⋅⋅ Vz 2= Constant
ρ
22
m 2 
 s2 
 
32
Phyd
P1
[⋅ 100 % ]
ptot
pdyn
pstat
ptot
Figure 2.3: This is how static pressure pstat ,
total pressure ptot and dynamic pressure
pdyn are measured.
d
d
(2.10)
(2.2)
[Pa]
p tot= =p ∆p+statp + ∆p[Pa
(2.5)
dyn] + ∆p geo
where ∆
pabs
(2.3)
rel
bar
V = Velocity [m/s]
∆pstat = pstat,
(2.6)
− pstat, in [Pa]
r = Density
[kg/m
Δp tot 3] out
(2.4)
[m]
H=
ρ ⋅ 1g
2
2
1
] pressure and vice versa.
(2.7)
∆ppressure
⋅ ρ ⋅ V be transformed
− ⋅ ρ ⋅ Vin to[Pa
dyn =
Dynamic
static
Flow
2 can out
2
(2.11)
]
Pa pipe
= Hwhere
⋅ g ⋅ ρthe
⋅ Qpipe
= ∆diameter
p tot ⋅ Q is[W
through hyd
increased converts dynamic pressure
2 The flow through a pipe is called a pipe flow, and
to static pressure,
2.4.
Phydsee figure
 Q   1
1 
1 where
(2.12)
[⋅ 100
= =pipe
%]diameter is increasing
 the
the partη
the
is called a diffusor.
hyd
Δof
pdyn
(2.8)
P22 ⋅ ρ ⋅  π  ⋅  D 4 − D 4  [Pa]
out
in 

 4
η tot =
pstat
ptot
p
Figure 2.4: Example of conversion of
dynamic pressure to static pressure in
a diffusor.
(2.13)
32
p tot = pstat + pdyn
[Pa]
(2.1)
2.3 Absolute and relative pressure
1 ⋅ρ⋅V
Pressure
in2 two
or relative
[Pa]different ways: absolute pressure (2.2)
pdynis =defined
2
pressure. Absolute pressure refers to the absolute zero, and absolute
pressure
thus
only
a positive
[Pa] Relative pressure refers
∆pcan
∆pstat
+ be
∆pdyn
+ ∆pgeo number.
(2.5) to the
tot =
pressure of the surroundings. A positive relative pressure means that the
∆pisstatabove
(2.6)
]
= pstat,
− pstat, in [Pa
pressure
the
pressure,
and a negative relative
pressure
out barometric
means that the pressure is below the barometric pressure.
(2.7)
∆pdyn = 1 ⋅ ρ ⋅ Vout 2 − 1 ⋅ ρ ⋅ Vin2 [Pa]
2
2
The absolute and relative definition is also known from temperature
measurement where the absolute
temperature is measured in Kelvin [K] and
2
  1 in Celsius
 isQmeasured
1 [°C]. The temperature measured
the relative temperature
1
 ⋅ 
 [Pa]
−
Δpdyn =
⋅ρ⋅
(2.8)
4
4 



2
π
in Kelvin is always positive andrefers
Dout toD inthe
 absolute zero. In contrast, the
 4
temperature in Celsius refers to water’s freezing point at 273.15K and can
therefore be negative.
The barometric
measured
as absolute pressure. The (2.9)
barometric
]
∆p geo = ∆pressure
z ⋅ ρ ⋅ g is[Pa
pressure is affected by the weather and altitude. The conversion from relative
2
pressurepto absolute
pressure is done by
the current barometric pressure
madding

V2
g
z
Constant
+
+
⋅
=
(2.10)

2 
to the measured
relative
pressure:
ρ
2
s 
pabs = prel + pbar
[Pa]
(2.3)
Δp tot pressure is measured by means of three different types of
In practise, static
(2.4)
[m]
H=
ρ⋅ g
pressure gauges:
• An absolute pressure gauge, such as a barometer, measures pressure
(2.11)
Phyd = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q [W]
relative
to absolute zero.
• An standard
Phydpressure gauge measures the pressure relative to the
(2.12)
[⋅ 100 % ]
=
η
hyd
atmospherich
P2 pressure. This type of pressure gauge is the most
commonly used.
Phyd pressure gauge measures the pressure difference
• A differential
[⋅ 100 % ]
(2.13)
η tot =
between the two
P1 pressure taps independent of the barometric pressure.
P1 > P2 > Phyd
[W]
η tot = η control ⋅ η motor ⋅ η hyd
33
NPSH A =
(2.14)
[ ⋅ 100 % ]
( p abs,tot,in − pvapour )
ρ ⋅g
[m]
(2.15)
(2.16)
33
p tot = pstat + pdyn
[Pa]
(2.1)
2. Performance curves
pdyn = 1 ⋅ ρ ⋅ V 2 [Pa]
2
(2.2)
∆p tot = ∆pstat + ∆pdyn + ∆pgeo [Pa]
2.4 Head
(2.5)
The different performance curves are introduced on the following pages.
∆pstat = pstat, out − pstat, in [Pa]
(2.6)
Δp tot
H=
ρ⋅ g
[m]
(2.4)
34
30
20
10
0
0
[m]
[m]
or
10
20
30
40
50
60
70
Q [m3/h]
Figure 2.5: A typical QH curve for a centrifugal
pump; a small flow gives a high head and a
large flow gives a low head.
H [m]
12
10
8
6
Water at 20oC
998.2 kg /m3
1 bar = 10.2 m
1 bar
(2.11)
Phyd = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q [W]
The QH curve
will ideally be exactly the same if the test in figure 2.6 is made with
a fluid having aPdensity different from water. Hence, a QH curve is independent
hyd
(2.12)
[⋅can
]
100be
%explained
η hyd = fluid. It
of the pumped
based on the theory in chapter
4 where it
P2
is proven that Q and H depend on the geometry and speed but not on the density
of the pumpedPfluid.
hyd
[⋅ 100 % ]
(2.13)
η tot =
P1
The pressure increase across a pump can also be measured in meter water column
(2.14) with
[W] is a pressure unit which must not be confused
> Phydcolumn
1 > P2water
[mWC].PMeter
the head
in [m].
As seen
in the
of physical properties of water,(2.15)
the change
η tot
= η control
⋅ η motor
⋅ ηtable
hyd [ ⋅ 100 % ]
in density is significant at higher temperatures. Thus, conversion from pressure
to head is essential.
( p abs,tot,in − pvapour )
(2.16)
[m]
NPSH A =
ρ ⋅g
NPSH A > NPSH R = NPSH3% + 0.5
NPSH A > NPSH R = NPSH3% . SA
50
40
10.2 m
A QH curve or pump curve shows the head (H) as a function of the flow (Q). The
1 ⋅ ρ ⋅ V 2 − 1 ⋅ ρ ⋅ V 2 [Pa]
(2.7) stated
flow (Q)∆ispdyn
the=rate
the pump. The flow is generally
out going through
in
2 of fluid
2
3
in cubic metre per hour [m /h] but at insertion into formulas cubic metre per
second [m3/s] is used. Figure 2.5
2 shows a typical QH curve.
 Q   1
1 
 ⋅ 
 [Pa]
−
Δpdyn = 1 ⋅ ρ ⋅ 
(2.8)
4



π
The QH curve for 2a given pump
can
Dbe
D in 4  using the setup shown in
out determined

 4
figure 2.6.
The pump is started and runs with constant speed. Q equals 0 and H reaches
its highest value when the valve is completely closed. The valve is gradually
opened∆and
decreases.
H is the height of the fluid column
(2.9) in the
]
p geoas=Q∆increases
z ⋅ ρ ⋅ g H[Pa
open pipe after the pump. The QH curve is a series of coherent values of Q and H
2
represented
m2.5.

p by
V 2the curve shown in figure
+
+ g ⋅ z = Constant  2 
(2.10)
ρ
2
s
 
In most cases the differential pressure across the pump Dptot is measured and
] following formula:
pabs
prel + pbar by[Pa
(2.3)
the head
H is=calculated
the
H [m]
4
2
0
0
1.0
1.5
2.0
Q [m3/h]
Figure 2.6: The QH curve can be determined
in an installation with an open pibe after
the pump. H is exactly the height of the fluid
column in the open pipe. measured from
inlet level.
(2.17)
(2.17a)
34
2.5 Differential pressure across the pump - description of differential pressure
(2.1)
p tot = pstat + pdyn [Pa]
2.5.1 Total pressure difference
The total pressure difference across the pump is calculated on the basis of
three contributions:
(2.2)
pdyn = 1 ⋅ ρ ⋅ V 2 [Pa]
2
∆p tot = ∆pstat + ∆pdyn + ∆pgeo
[Pa]
(2.5)
where ∆pstat = pstat, out − pstat, in [Pa]
(2.6)
Δptot = Total pressure difference across the pump [Pa]
2
2
1 ⋅ ρ ⋅ V difference
1 ⋅ ρ ⋅ Vacross
Δpstat = ∆Static
the
[Pa
] pump [Pa]
(2.7)
pdyn =pressure
out −
in
2
2
Δpdyn = Dynamic pressure difference across the pump [Pa]
Δpgeo = Geodetic pressure difference between the pressure sensors [Pa]
2
(2.1)
p tot = pstat1 + pdyn Q[Pa]  1
1 
 [Pa]
⋅  4 −
Δpdynpressure
=
⋅ ρdifference
⋅
(2.8)
4
2.5.2 Static

π 
2
Dout
D in 

 4  can be measured directly with a differential
The static pressure difference
2
pressure
sensor,
(2.2) outlet
[Pa] sensor can be placed at the inlet and
pdyn
= 1 ⋅ or
ρ ⋅aVpressure
2
of the pump. In this case, the static pressure difference can be found by the
following
expression:
[Pa]
∆
= stat
∆p+statpdyn
+ ∆p[Pa
(2.5)
(2.1)
p
dyn]+ ∆p geo
tottot= p
(2.9)
∆p
p
geo = ∆z ⋅ ρ ⋅ g [Pa]
∆pstat =2 pstat, out − pstat, in [Pa] 2
(2.6)
m 
p V
+
+ g ⋅ z = Constant  2 
(2.10)
(2.2)
pρdyn =211⋅ ρ ⋅ V 2 2 [Pa]1
s 
2
2
[
(2.7)
∆pdyn = ⋅ ρ ⋅ Vout − ⋅ ρ ⋅ Vin
Pa]
2
2
2.5.3 Dynamic
pressure
difference
[Pa
pabs
(2.3)
[Pa]
∆
p tot= =prel
∆p+statpbar
+ ∆p
(2.5)
dyn] + ∆p geo
The dynamic pressure difference
between the inlet and outlet of the pump
2
 [Pa
is found∆p
by the
following
(2.6)
=tot
p1stat, out − pQformula:
1]
1 
Δp
 [Pa]
Δp=stat
⋅ [ρm⋅] stat, in ⋅  4 −
(2.8)
(2.4)
H
dyn =
4 


2
π
D in 
ρ⋅ g
 Dout
4


(2.7)
∆pdyn = 1 ⋅ ρ ⋅ Vout 2 − 1 ⋅ ρ ⋅ Vin2 [Pa]
(2.11)
Phyd = H2⋅ g ⋅ ρ ⋅ Q = 2∆p tot ⋅ Q [W]
2
Phyd
]  1

η
1
[
∆hyd
p geo= = P∆1z ⋅ [ρ⋅ 100
⋅gQ%
Pa
 ]⋅ 
−
Δpdyn = 2 ⋅ ρ ⋅
 π   Dout4
2
D in 4
 4 
2
2
m 
p V Phyd
%]
= Constant
η tot+ = + g [⋅ ⋅z100
 s2 
ρ
2 P1
 
35
]]
P
> P+ pbar [W[Pa
p > =P2prel
∆1abs
p geo = ∆z hyd
⋅ ρ ⋅ g [Pa]
η tot = η control ⋅ η motor ⋅ η hyd




[Pa]
(2.12)
(2.9)
(2.8)
(2.13)
(2.10)
(2.14)
(2.3)
(2.9)
[ ⋅ 100 % ]
(2.15)
35
2
[Pa]
pdyn = 1 ⋅ ρ ⋅ Vcurves
2. Performance
2
∆p
= ∆pstatpdyn
+ ∆p[Pa
p
dyn]+ ∆p geo
tottot= pstat +
(2.2)
[Pa]
(2.5)
(2.1)
In practise, the dynamic pressure and the flow velocity before and after the
∆pstat = pstat, out − pstat, in [Pa]
(2.6)
pump are not
measured during test of pumps. Instead, the dynamic pressure
difference
and pipe diameter of the(2.2)
inlet and
pdyncan
= 1be
⋅ ρcalculated
⋅ V 2 2 [Pa]1if the flow
2
1
2
[
]
(2.7)
∆
p
=
⋅
ρ
⋅
V
−
⋅
ρ
⋅
V
Pa
dyn
out
in
outlet of the pump
2 are known:
2
∆p tot = ∆pstat + ∆pdyn + ∆pgeo [Pa]
(2.5)
2
∆pstat = p1stat, out − pQstat, in [Pa
(2.6)
1]
1 
 [Pa]
⋅ 
−
Δpdyn =
⋅ρ⋅
(2.8)
4 
 π   Dout4
2
D in 

4


(2.7)
∆pdyn = 1 ⋅ ρ ⋅ Vout 2 − 1 ⋅ ρ ⋅ Vin2 [Pa]
2
2
The formula shows that the dynamic pressure difference is zero if the pipe
diameters are identical before
2 and after the pump.
 Q   1
1 
(2.9)
∆
= ∆1z ⋅⋅ ρρ ⋅ ⋅g [Pa
 ]⋅ 

Δp
pgeo
(2.8)
dyn =
 D 4 − D 4  [Pa]


2
π
2.5.4 Geodetic pressure difference
4   out 2 in 
2
m  inlet and outlet can be measured
p Vpressure
The geodetic
difference between
+
+ g ⋅ z = Constant  2 
(2.10)
ρ
2 way:
in the following
s 
]]
p p = =prel∆z+⋅ pρbar⋅ g[Pa
[Pa
∆abs
geo
(2.3)
(2.9)
Δp
m 2 
p = V 2 tot [m]
(2.4)
H
+ ρ ⋅ g+ g ⋅ z = Constant  2 
(2.10)
where ρ
2
s
 
Δz is theP difference
inρvertical
position
between
the gauge connected
(2.11) to the
[W]
=
H
⋅
g
⋅
⋅
Q
=
∆
p
⋅
Q
hyd
tot
[
]
p
=
p
+
p
Pa
(2.3)
outlet pipe
connected to the inlet pipe.
abs andrelthe gauge
bar
Phyd
(2.12)
[⋅ 100 % ]
=p
η hyd Δ
tot
The geodetic
P
(2.4)Hence,
2
[m] difference is only relevant if Δz is not zero.
H = pressure
the position ρ
of⋅ gthe measuring taps on the pipe is of no importance for the
hyd geodetic pressure difference.
calculation
of Pthe
[⋅ 100 % ]
(2.13)
ηhyd
(2.11)
P
tot = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q [W ]
P1
The geodetic pressure
difference is zero when a differential pressure gauge
Phyd
(2.14)
[W]% ]
P1hyd> =
P2 >
Phyd[ ⋅ 100
(2.12)
η
is used for
measuring
the
static pressure difference.
P2
(2.15)
η tot = η control ⋅ η motor ⋅ η hyd [ ⋅ 100 % ]
Phyd
[⋅ 100 % ]
(2.13)
η tot =
P1 ( p abs,tot,in − pvapour )
(2.16)
[m]
NPSH A =
ρ ⋅g
(2.14)
P1 > P2 > Phyd [W ]
36
NPSH
NPSH R = NPSH
0.5 % ][m] or
η tot = Aη >
⋅ η hyd3% [ ⋅+100
control ⋅ η motor
[m]
NPSH A > NPSH R = NPSH3% . SA
( p abs,tot,in − pvapour )
[m]
NPSH =
(2.17)
(2.15)
(2.17a)
(2.16)
36
∆pdyn = 1 ⋅ ρ ⋅ Vout 2 − 1 ⋅ ρ ⋅ Vin2
2
2
[Pa]
(2.7)
2
 Q   1
1 
 ⋅ 

Δpdyn = 1 ⋅ ρ ⋅ 
(2.8)
 D 4 − D 4  [Pa]


2
π
2.6 Energy equation
for
an
ideal
flow
out
in 

 4
The energy equation for an ideal flow describes that the sum of pressure
energy, velocity energy and potential energy is constant. Named after
the Swiss physicist Daniel Bernoulli, the equation is known as Bernoulli’s
(2.9)
∆p geo = ∆z ⋅ ρ ⋅ g [Pa]
equation:
p V2
+
+ g ⋅ z = Constant
ρ
2
pabs = prel + pbar
m 2 
 s2 
 
(2.10)
[Pa]
(2.3)
Bernoulli’s equation is valid if the following conditions are met:
Δp tot
(2.4)
[m]
H=
ρ⋅ g
1. Stationary
flow – no changes over time
2.P Incompressible
flow – true for most liquids
(2.11)
hyd = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q [W ]
3. Loss-free flow – ignores friction loss
Phyd flow – no supply of mechanical energy
4. Work-free
(2.12)
[⋅ 100 % ]
η hyd =
P2
Formula (2.10) applies along a stream line or the trajectory of a fluid particle.
Phyd the flow through a diffusor can be described by formula (2.10),
For
[⋅ 100 % ]
(2.13)
=
η totexample,
P1 flow through an impeller since mechancial energy is added.
but not the
(2.14)
P > P > Phyd [W ]
In1most2 applications,
not all the conditions for the energy equation are met. In
spite
the equation can be used for making a rough (2.15)
calculation.
η tot =ofηthis,
control ⋅ η motor ⋅ η hyd [ ⋅ 100 % ]
NPSH A =
( p abs,tot,in − pvapour )
ρ ⋅g
[m]
NPSH A > NPSH R = NPSH3% + 0.5
NPSH A > NPSH R = NPSH3% . SA
37
(2.16)
[m]
[m]
or
(2.17)
(2.17a)
NPSH A =
(pbar + ρ ⋅ g ⋅ Hgeo − ∆ p loss , suction pipe ) − pvapour
[m]
ρ ⋅g
NPSH A =
3500 Pa
7375 Pa
101300 Pa
− 3m −
−
992.2kg m3 ⋅ 9.81m s 2
992.2kg m3 ⋅ 9.81m s 2 992.2kg m3 ⋅ 9.81m s 2
NPSH A = 6.3m
(2.18)
37
2. Performance curves
2.7 Power
The power curves show the
flow, see
p tot = pstat + pdyn [Pa]energy transfer rate as a function of(2.1)
figure 2.7. The power is given in Watt [W]. Distinction is made between
three kinds of power, see figure 2.8:
• Supplied power
from external electricity source to the motor and
(2.2)
pdyn = 1 ⋅ ρ ⋅ V 2 [Pa]
2
controller (P1)
• Shaft power transferred from the motor to the shaft (P2)
∆p tot = ∆pstat + ∆pdyn + ∆pgeo [Pa]
(2.5)
• Hydraulic power transferred from the impeller to the fluid (Phyd)
∆pstat = pstat, out − pstat, in [Pa]
(2.6)
The power consumption depends on the fluid density. The power curves
3
2
1 ⋅ ρ ⋅on
are generally
with
which
[
(2.7)
∆pdyn =based
Vouta2 standard
− 1 ⋅ ρ ⋅ Vfluid
Pa] a density of 1000 kg/m
in
2
2
corresponds to water at 4°C. Hence, power measured on fluids with another
density must be converted.
2
 Q   1
1 
1


 [Pa]
⋅  4 − for4integrated
Δpdynsheet,
=
⋅ρ⋅
In the data
products, (2.8)
while P2 is
2 P1 isnormally
π   Dstated
D in 
out
4


typically stated for pumps sold with a standard motor.
2.7.1 Speed
Flow, head and power consumption vary with the pump speed, see section 3.4.4.
(2.9)
∆p geo = ∆z ⋅ ρ ⋅ g [Pa]
Pump curves
can only be compared if they are stated with the same speed. The
curves can be converted
to the same speed
by the formulas in section 3.4.4.
m 2 
p V2
+
+ g ⋅ z = Constant  2 
(2.10)
ρ
2
s 
2.8 Hydraulic
pabs = ppower
(2.3)
rel + pbar [Pa]
The hydraulic power Phyd is the power transferred from the pump to the
fluid. As seen
Δpfrom the following formula, the hydraulic power is calculated
(2.4)
[m]
H =flow, tot
based on
ρ ⋅ ghead and density:
Phyd = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q
[W]
P [W]
P1
P2
Q [m3/h]
Figure 2.7: P1 and P2 power curves.
P1
P2
Phyd
Figure 2.8: Power transfer in a pump unit.
(2.11)
Phyd
An independent
curve
for %
the
(2.12)in data
[⋅ 100
] hydraulic power is usually not shown
η hyd =
P
2
sheets but is part of the calculation of the pump efficiency.
η tot =
Phyd
P1
[⋅ 100 % ]
(2.13)
[W]
(2.14)
P1 > P2 > Phyd
38
η tot = η control ⋅ η motor ⋅ η hyd
[ ⋅ 100 % ]
(2.15)
38
⋅
∆p geo
geo = ∆z ⋅ ρ g [Pa
2]
 Q   1
1
 ⋅ 
−
Δpdyn =22 1 ⋅ ρ ⋅ 
  Dout4 m 22Din 4
p + V +2 g ⋅ z =πConstant
 4 
 s 22 
ρ
2
 
(2.9)
(2.9)




[Pa]
(2.8)
(2.10)
2.9 Efficiency
[Pa]
pabs = prel
+ pbar
(2.3)
(2.3)
rel
The totalabsefficiency
(ηbartot) is the ratio between hydraulic power and supplied
[Paefficiency
]
p geo =2.9
∆z shows
⋅ ρ ⋅ g the
power. ∆Figure
curves for the pump (ηhyd(2.9)
) and for a
Δp tot
tot
(2.4)
(2.4)
[
]
H
=
m
complete pump unit with motor and controller (ηtot).
ρ⋅ g
m 2 
p V2
+
+ g ⋅ z = Constant  2 
(2.10)
(2.11)
Pρhyd
⋅ g ⋅ ρ ⋅ refers
Q = ∆to
p tot
Q s [W
The hydraulic
efficiency
P2 ⋅, whereas
(2.11) to P1:
hyd = 2H
tot
 ] the total efficiency refers
pabs = pPrel
hyd + pbar [Pa]
hyd
[⋅ 100 % ]
η hyd
hyd =
P22
Δp
H = Ptot [m]
hyd
g [ ⋅ 100 % ]
ρ ⋅hyd
η tot
tot =
P11
Phyd = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q [W]
[W]
P11 > P22 > Phyd
hyd
Phyd
= control [⋅⋅η100
% ] hyd [ ⋅ 100 % ]
η
ηhyd
tot = η control
motor ⋅ η hyd
tot
motor
P
2
η[%]
ηhyd
ηtot
Q[m3/h]
(2.3)
(2.12)
(2.12)
Figure 2.9: Efficiency curves for the pump
(ηhyd) and complete pump unit with motor
and controller (ηtot).
(2.4)
(2.13)
(2.13)
(2.11)
(2.14)
(2.14)
(2.12)
(2.15)
(2.15)
The efficiency is always below 100% since the supplied power is always
Phyd
( p abs,tot,in
) to losses in controller, motor and pump
abs,tot,in power
vapour
larger than
the
hydraulic
due
[⋅ 100 −%p]vapour
(2.13)
(2.16)
η tot =
[m]
NPSH
(2.16)
A =
A
P1 total ρefficiency
⋅g
components. The
for the entire pump unit (controller, motor
and hydraulics) is the product of the individual efficiencies:
(2.17)
(2.14)
P1 > PA2 >
> NPSH
Phyd R[W
(2.17)
NPSH
=]NPSH3% + 0.5 [m] or
A
R
3%
(2.17a)
. S
(2.17a)
NPSH RR = NPSH
NPSH
(2.15)
η tot = AAη >
⋅ η hyd3%
3% [ ⋅ 100
A % ][m]
control ⋅ η motor
A
+ ρ ⋅ g ⋅ Hgeo
) − pvapour
− ∆ p loss
where NPSH = ((p
pbar
loss ,, suction
suction pipe
pipe
bar
vapour [m]
geo )
abs,tot,in − pvapour
(2.18)
(2.18)
A =
A
(2.16)
[
]
NPSH
m
. ρ ⋅g
A
hcontrol = Controller
efficiency
ρ ⋅ g [ 100%]
hmotor = Motor efficiency [ . 100%]
7375 Pa
Pa + 0.5− 3m
[m−] or 35003 Pa (2.17)2 −
NPSH
NPSHAAA => NPSH 101300
R = NPSH
3%
3
2
3
2
3 ⋅ 9.81m s 2
992.2kg
m
9.81m
s
992.2kg
m
992.2kg
m33 ⋅ 9.81m s 22
⋅
(2.17a)
The flow
where> the
pump
has the highest
efficiency
is called the
optimum
. S
[
]
NPSH
=
NPSH
m
NPSH
R
A
3%
A
point orNPSH
the best
efficiency point (QBEP).
A = 6.3m
A
(pbar + ρ ⋅ g ⋅ Hgeo − ∆ p loss , suction pipe ) − pvapour
[m] (2.18)
ρ ⋅g 2
pvapour
pstat,in + pbar + 0.5 . ρ . V112
vapour
(2.19)
[m] (2.19)
+ H geo
= stat,in bar
geo − H loss,
loss, pipe
pipe−
ρ
⋅
⋅
g
g
ρ
3500 Pa
7375 Pa
101300 Pa
=
− 3m −
−
992.2kg m3 ⋅ 9.81m s 2
992.2kg m3 ⋅ 9.81m s 2 992.2kg m3 ⋅ 9.81m s 2
47400 Pa
-27900 Pa + 101000 Pa + 500 Pa
= 6.3m
+ 3m − 1m −
3
2
3
2
3
2
=
973 kg m ⋅ 9.81m s
973 kg m3 ⋅ 9.81m s 2
NPSH A =
NPSH AA
NPSH A
NPSH AA
NPSH
A
39
NPSH AA = 4.7m
pvapour
p
+ p + 0.5 . ρ . V12
[m]
+ H geo − H loss, pipe−
NPSH A = stat,in bar
(2.19)
39
2
2
2
 Q  
2. Performance
curves
 ⋅ 
Δp = 1 ⋅ ρ ⋅ 
2
dyn
π 
 4
1
1
D 4 − D 4
in
 out




[Pa]
(2.8)
2.10 NPSH, Net Positive Suction Head
NPSH is a term describing conditions related to cavitation, which is
undesired
and= harmful.
(2.9)
∆p geo
∆z ⋅ ρ ⋅ g [Pa]
2
2
Cavitation
creation of vapourm
bubbles
in areas where the pressure

p isVthe
g
z
Constant
+
+
⋅
=
(2.10)


locally drops
The extent of cavitation
depends
2
ρ
2to the fluid vapour pressure.
s 
on how low the pressure is in the pump. Cavitation generally lowers the
head and
[Pavibration.
]
pabscauses
= prel noise
+ pbar and
(2.3)
Cavitation first
Δp totoccurs at the point in the pump where the pressure is
(2.4)
[m]
H=
lowest, which
is most often at the blade edge at the impeller inlet, see
ρ⋅ g
figure 2.10.
(2.11)
Phyd = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q [W]
The NPSH value
Phydis absolute and always positive. NPSH is stated in meter [m]
[⋅ 1002.11.
% ]Hence, it is not necessary to take the(2.12)
like theηhead,
density of
hyd = see figure
P2
different fluids into account because NPSH is stated in meters [m].
Phyd
[between
⋅ 100 % ]
(2.13)
η tot =
Distinction is made
two different NPSH values: NPSHR and NPSHA.
P
Figure 2.10: Cavitation.
NPSH [m]
1
(2.14)
[Available
P1 > P2 for
> PNPSH
W]
NPSHA stands
and is an expression of how close
the fluid
hyd
in the suction pipe is to vapourisation. NPSHA is defined as:
(2.15)
η tot = η control ⋅ η motor ⋅ η hyd [ ⋅ 100 % ]
NPSH A =
( p abs,tot,in − pvapour )
ρ ⋅g
[m]
Figure 2.11: NPSH curve.
(2.16)
Q[m3/h]
where
(2.17)
NPSH A > NPSH R = NPSH3% + 0.5 [m] or
pvapour = The vapour pressure of the fluid at the present temperature
[Pa].
(2.17a)
[
m]
NPSH A > NPSH R = NPSH3% . SA
The vapour pressure is found in the table ”Physical properties of
water” in the back of the book.
(p + ρ ⋅ g ⋅ Hgeo − ∆ p loss , suction pipe ) − pvapour
= bar pressure
pabs,tot,in NPSH
= TheAabsolute
at the inlet flange [Pa]. [m] (2.18)
ρ ⋅g
NPSH A =
3500 Pa
7375 Pa
101300 Pa
− 3m −
−
3
2
3
2
992.2kg m ⋅ 9.81m s
992.2kg m ⋅ 9.81m s 992.2kg m3 ⋅ 9.81m s 2
NPSH A = 6.3m
40
NPSH =
pstat,in + pbar + 0.5 . ρ . V12
+H
−H
−
pvapour
[m]
(2.19)
40
ρ
+
2
+ g ⋅ z = Constant
pabs = prel + pbar
(2.10)
 s2 
 
[Pa]
(2.3)
Δp tot
(2.4)
[m]
H=
g NPSH Required and is an expression of the lowest NPSH
NPSHR standsρ ⋅for
value required
operating
conditions. The absolute
pressure
(2.11)
Phyd = H for
⋅ g acceptable
⋅ ρ ⋅ Q = ∆p
tot ⋅ Q [W ]
pabs,tot,in can be calculated from a given value of NPSHR and the fluid vapour
Phyd
pressure by inserting
NPSH in the formula (2.16) instead of NPSHA.
(2.12)
[⋅ 100 %R ]
η hyd =
P2
To determine if a pump can safely be installed in the system, NPSHA and
Phyd
NPSHR η
should
be found
within the
[⋅ 100for
% ]the largest flow and temperature(2.13)
tot =
P
operating range.
1
(2.14)
P1 > P2 > Phyd [W ]
A minimum
safety margin of 0.5 m is recommended. Depending on the
application,
higher
safety
level
may%be
(2.15) noise
η tot = aη control
⋅ η motor
⋅ η hyd
] required. For example,
[ ⋅ 100
sensitive applications or in high energy pumps like boiler feed pumps,
European Association
of Pump
Manufacturers
indicate a safety factor SA of
( p abs,tot,in
)
− pvapour
(2.16)
[
]
=
NPSH
m
A
1.2-2.0 times the NPSH3%
ρ .⋅ g
NPSH A > NPSH R = NPSH3% + 0.5
NPSH A > NPSH R = NPSH3% . SA
[m]
[m]
or
(2.17)
(2.17a)
⋅ g ⋅ Hgeo −can
(pbar +inρsystems
∆ pbe
− pprevented
loss , reduced
suction pipe ) or
vapour
The riskNPSH
of cavitation
[m] by:(2.18)
A =
⋅
ρ
g
• Lowering the pump compared to the water level - open systems.
• Increasing the system pressure - closed systems.
3500 Pa
7375 Pa
101300 Pa
• Shortening
to reduce− the
3m −friction loss. 3
NPSH A =the suction line
−
3
2
2
992.2kg
m
9.81m
s
992.2kg
m
⋅
s
992.2kg
m3 ⋅ 9.81m s 2
⋅
9.81m
• Increasing the suction line’s cross-section area to reduce the fluid
velocity
thereby reduce friction.
NPSHand
A = 6.3m
• Avoiding pressure drops coming from bends and other obstacles in
the suction line.
pvapour
p
+ pbar + 0.5 . ρ . V12
• Lowering
temperature
to reduce
[m] (2.19)
+ vapour
H geo − Hpressure.
NPSH Afluid
= stat,in
loss, pipe−
ρ ⋅g
ρ ⋅g
The two following examples show how NPSH is calculated.
47400 Pa
-27900 Pa + 101000 Pa + 500 Pa
NPSH A =
+ 3m − 1m −
3
2
973 kg m ⋅ 9.81m s
973 kg m3 ⋅ 9.81m s 2
NPSH A = 4.7m
41
41
pabs = prel + pbar
Δp tot
[Pa]
]
∆pPerformance
Pa]curves
H ∆=z ⋅ ρ ⋅ g [[m
geo =
2.
ρ⋅ g
(2.3)
(2.9)
(2.4)
2
2
p VP
⋅ ρ ⋅ Q = ∆pmtot ⋅ Q [W]
hyd
+
+ g= ⋅Hz ⋅=gConstant
(2.10) (2.11)

2 
ρ
2 2.1 Pump drawing from
 s a well
Example
Phyd
[⋅]100 % ]
η hyd +=p
pabs = p
(2.3) (2.12)
bar
P2 [Pa
A pump rel
must draw
water from a reservoir where the water level is 3 meters
below the pump. To calculate the NPSHA value, it is necessary to know the
P
Δp tot
(2.4) (2.13)
[mhyd
] [⋅ 100 % ]
H = ηloss
friction
tot = in the inlet pipe, the water temperature and the barometric
ρ⋅ g
P1
pressure, see figure 2.12.
(2.11) (2.14)
Phyd = H
P1⋅ >g P⋅ 2ρ >⋅ Q
Phyd= ∆[pWtot] ⋅ Q [W]
Water temperature = 40°C
(2.15)
ηPhyd = η control ⋅ =η101.3
η hyd [ ⋅ 100 % ]
Barometric
(2.12)
[⋅ 100 % ] motor ⋅ kPa
η hyd = tot pressure
2
PressurePloss
in the suction line at the present flow = 3.5 kPa.
( p abs,tot,in of
− p40°C,
At a water temperature
vapour ) the vapour pressure is 7.37 kPa and ρ is
(2.16)
[m]
PNPSH
hyd 3 A =
[⋅ 100
% ] ρare
(2.13)
η tot =
⋅ g found in the table ”Physical properties
992.2kg/m
. The
values
of water”
P1
in the back
of the book.
(2.17)
NPSH > NPSH = NPSH3% + 0.5 [m] or
(2.14)
P1 > P2 > PhydA [W ] R
(2.17a) as:
. Sin formula
[m] (2.16) can be written
NPSH
For thisNPSH
system,
NPSH
expression
R =ANPSH
A >the
3%
A
(2.15)
η tot = η control ⋅ η motor ⋅ η hyd [ ⋅ 100 % ]
Reference plane
∆ploss, suction pipe
H<0
pbar
Figure 2.12: Sketch of a system where
water is pumped from a well.
(pbar + ρ ⋅ g ⋅ Hgeo − ∆ p loss , suction pipe ) − pvapour
[m] (2.18)
NPSH
A =
( p abs,tot,in
− pvapour )
(2.16)
[m] ρ ⋅ g
NPSH A =
ρ ⋅g
7375 Pa
Paposition in relation to3500
Hgeo is the
water
level’s101300
vertical
the Pa
pump. Hgeo
− 3m −
− can
A = = NPSH
3+ 0.5
2]
3
2
[
NPSH A NPSH
> NPSH
m
or 992.2kg m(2.17)
R992.2kg 3%
m
9.81m
s
⋅
s
992.2kg
m3 ⋅ 9.81m s 2
⋅
9.81m
either be above or below the pump and is stated in meter [m]. The water
. S
[the
m] pump. Thus, H (2.17a)
NPSHin
> NPSH
R = NPSH
A NPSH
level
this
system
is placed
below
is negative, Hgeo =
3%
A
geo
A = 6.3m
-3m.
(p + ρ ⋅ g ⋅ Hgeo − ∆ p loss , suction pipe ) − pvapour
[m] (2.18)
NPSH A = bar
pvapour
pstat,in
+ pis:
ρ
g0.5 . ρ . V12
bar⋅+
The system
NPSH
value
[m] (2.19)
+ H geo − H loss, pipe−
NPSH A = A
ρ ⋅g
ρ ⋅g
3500 Pa
7375 Pa
101300 Pa
NPSH A =
− 3m −
−
3
2
3
2
2
992.2kg-27900
m ⋅ 9.81m
s
m3 ⋅ 9.81m
+ 500 Pa m ⋅ 9.81m s 992.2kg 47400
Pa s
Pa + 101000
Pa992.2kg
3
m
1
m
NPSH A =
+
−
−
973 kg m3 ⋅ 9.81m s 2
973 kg m3 ⋅ 9.81m s 2
NPSH A = 6.3m
NPSH A = 4.7m
The pump chosen for the system2 in question must have a NPSHR value lower
pvapour
pstat,in + pbar + 0.5 . ρ . V1
] (2.19)
+ H geo
−
H loss,
NPSH6.3
than
of −0.5
m.pipe
Hence,
the[m
pump
must have a
A =m minus the safety margin
ρ ⋅g
ρ ⋅g
NPSHR value lower than 6.3-0.5 = 5.8 m at the present flow.
NPSH A =
42
47400 Pa
-27900 Pa + 101000 Pa + 500 Pa
+ 3m − 1m −
973 kg m3 ⋅ 9.81m s 2
973 kg m3 ⋅ 9.81m s 2
NPSH A = 4.7m
42
tot
(2.4)
[m]
H=
Δp tot ρ ⋅ g
(2.4)
[m]
H=
ρ⋅ g
(2.11)
Phyd = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q [W]
(2.11)
Phyd = H ⋅ g ⋅ ρ ⋅ Q = ∆p tot ⋅ Q [W]
Phyd
(2.12)
[⋅ 100 % ]
η hyd =
Phyd
2
Example
in
(2.12)
[⋅ P100
%a] closed system
η hyd = 2.2 Pump
P2
Phyd
System
[there
] free water surface to refer to. This
⋅ 100 is
% no
(2.13)
=
η tot system,
In a closed
example
Phyd
P1
[⋅ 100
% ] sensor’s placement above the reference
(2.13)
η tot = how the
pstat, in
shows
pressure
plane can
P1
be usedPto
find the absolute
2.13.
(2.14)
[W] pressure in the suction line, see figure
1 > P2 > Phyd
(2.14)
P1 > P2 > Phyd [W ]
Hgeo>0
(2.15)
η tot = η control ⋅ η motor ⋅ η hyd [ ⋅ 100 % ]
The relative
static pressure on the suction side is measured to be pstat,in =
(2.15)
η tot = η control ⋅ η motor ⋅ η hyd [ ⋅ 100 % ]
Reference plane
-27.9
kPa2. Hence, there is negative pressure in the system at the pressure
( p abs,tot,in − pvapour )
gauge. NPSH
The pressure
gauge is placed
in
(2.16)
[m] above the pump. The difference
A =
(
)
−
p
p
ρ
⋅
g
abs,tot,in
vapour
Figure 2.13: Sketch of a closed system.
height
the pressure[m
gauge
and the impeller eye(2.16)
Hgeo is therefore a
]
=
NPSH Abetween
ρ
⋅
g
positive value of +3m. The velocity in the tube where the measurement
of
(2.17)
NPSH A > NPSH R = NPSH3% + 0.5 [m] or
pressure is made results in a dynamic pressure contribution
of 500 Pa.
(2.17) (2.17a)
NPSH ANPSH
> NPSH>RNPSH
= NPSH
+ 0.5 . [m
[m]
NPSH
SA] or
R =3%
A
3%
(2.17a)
.
[m]
NPSH A > NPSH
R = NPSH
Barometric
pressure
= 1013%kPa SA
+ ρ ⋅ g ⋅ Hgeo − ∆point
(pbarmeasurement
p loss , suction
− pvapour
pipe)) and
Pipe loss
between
(pstat,in
pump
to
[mis
] calculated
NPSH
(2.18)
A =
ρ
⋅
⋅
g
+
(
)
p
∆
p
⋅
ρ
g
−
−
p
H
loss
,
suction
pipe
bar
vapour
geo
HNPSH
= 1m.
[m] (2.18)
loss,pipeA =
ρ ⋅g
System temperature = 80°C
3500 Pa
7375 Pa
101300 Pa
VapourNPSH
pressure
pvapour = 47.4
kPa and2 density
− 3m − is ρ = 973 3kg/m3, values
− are
A =
3
2
3500 Pa m ⋅ 9.81m s 992.2kg
7375 Pa m3 ⋅ 9.81m s 2
101300
992.2kgPam ⋅ 9.81m s
992.2kg
found
properties
NPSH Ain=the table ”Physical
− 3m −of water”. 3
−
3
2
2
992.2kg
m ⋅ 9.81m s
992.2kg m ⋅ 9.81m s 992.2kg m3 ⋅ 9.81m s 2
NPSH
= 6.3m
A
For
thisA system,
NPSH
= 6.3m formula 2.16 expresses the NPSHA as follows:
pvapour
p
+ p + 0.5 . ρ . V12
[m] (2.19)
+ H geo − H loss, pipe−
NPSH A = stat,in bar
pvapour ρ ⋅ g
pstat,in + pbar + 0.5 . ρρ⋅.gV12
[m] (2.19)
+ H geo − H loss, pipe−
NPSH A =
ρ ⋅g
ρ ⋅g
Inserting the values gives:
47400 Pa
-27900 Pa + 101000 Pa + 500 Pa
NPSH A =
+ 3m − 1m −
500 Pa s 2
47400
-27900 Pa +973
101000
Pa3 +
kg m
973
kg Pam3 ⋅ 9.81m s 2
⋅ 9.81m
NPSH A =
+ 3m − 1m −
973 kg m3 ⋅ 9.81m s 2
973 kg m3 ⋅ 9.81m s 2
NPSH A = 4.7m
NPSH A = 4.7m
Despite the negative system pressure, a NPSHA value of more than 4m is
available at the present flow.
43
43
2. Performance curves
2.11 Axial thrust
Axial thrust is the sum of forces acting on the shaft in axial direction,
see figure 2.14. Axial thrust is mainly caused by forces from the pressure
difference between the impeller’s hub plate and shroud plate, see section
1.2.5.
The size and direction of the axial thrust can be used to specify the size of
the bearings and the design of the motor. Pumps with up-thrust require
locked bearings. In addition to the axial thrust, consideration must be taken
to forces from the system pressure acting on the shaft. Figure 2.15 shows an
example of an axial thrust curve.
The axial thrust is related to the head and therefore it scales with the speed
ratio squared, see sections 3.4.4 and 4.5.
Figure 2.14: Axial thrust
work in the bearing’s
direction.
Force [N]
500
400
300
200
100
0
10
20
30
40
50
60
70
Q [m3/h]
Figure 2.15: Example of a axial thrust curve
for a TP65-410 pump.
Figure 2.16: Radial thrust
work perpendicular on
the bearing.
2.12 Radial thrust
Radial thrust is the sum of forces acting on the shaft in radial direction
as shown in figure 2.16. Hydraulic radial thrust is a result of the pressure
difference in a volute casing. Size and direction vary with the flow. The
forces are minimum in the design point, see figure 2.17. To size the bearings
correctly, it is important to know the size of the radial thrust.
Force [N]
100
80
60
40
20
0
10
20
30
40
50
60
70
Q [m3/h]
Figure 2.17: Example of a radial thrust curve
for a TP65-410 pump.
44
44
2.13 Summary
Chapter 2 explains the terms used to describe a pump’s performance
and shows curves for head, power, efficiency, NPSH and thrust impacts.
Furthermore, the two terms head and NPSH are clarified with calculation
examples.
45
45
Chapter 3
Pumps operating in systems
Tank on roof
3.1 Single pump in a system
3.2 Pumps operated in parallel
3.3 Pumps operating in series
3.4 Regulation of pumps
3.5 Annual energy consumption
3.6 Energy efficiency index (EEI)
3.7 Summary
�
Buffer tank
Hloss, pipe friction
Hoperation
Qoperation
3. Pumps operating in systems
3. Pumps operating in systems
This chapter explains how pumps operate in a system and how they can be
regulated. The chapter also explains the energy index for small circulation
pumps.
A pump is always connected to a system where it must circulate or lift
fluid. The energy added to the fluid by the pump is partly lost as friction in
the pipe system or used to increase the head.
Implementing a pump into a system results in a common operating point.
If several pumps are combined in the same application, the pump curve for
the system can be found by adding up the pumps’ curves either serial or
parallel. Regulated pumps adjust to the system by changing the rotational
speed. The regulation of speed is especially used in heating systems where
the need for heat depends on the ambient temperature, and in water supply systems where the demand for water varies with the consumer opening
and closing the tap.
48
48
3.1 Single pump in a system
A system characteristic is described by a parabola due to
an increase in friction loss related to the flow squared. The
system characteristic is described by a steep parabola if
the resistance in the system is high. The parabola flattens
when the resistance decreases. Changing the settings of
the valves in the system changes the characteristics.
The operating point is found where the curve of the
pump and the system characteristic intersect.
In closed systems, see figure 3.1, there is no head when
the system is not operationg. In this case the system characteristic goes through (Q,H) = (0,0) as shown in figure
3.2.
In systems where water is to be moved from one level to
another, see figure 3.3, there is a constant pressure difference between the two reservoirs, corresponding to the
height difference. This causes an additional head which
the pump must overcome. In this case the system characteristics goes through (0,Hz) instead of (0,0), see figure 3.4.
Elevated tank
Boiler
Valve
Hoperation
Qoperation
Hz
Buffer tank
Heat Exchanger
Hoperation
Qoperation
Figure 3.1: Example of a closed system.
Figure 3.3: Example of an open system
with positive geodetic lift.
H
H
Hmax
Figure 3.2: The system characteristics of a closed system resembles a
parabola starting at point (0.0).
Hmax
Figure 3.4: The system characteristics of an open system resembles a
parabola passing through (0,Hz).
Hoperation
Hoperation
Hloss,friktion
Hloss,friktion
Hz
Qoperation
49
Q
Qoperation
Q
49
3. Pumps operating in systems
3.2 Pumps operated in parallel
In systems with large variations in flow and a request for constant pressure,
two or more pumps can be connected in parallel. This is often seen in larger
supply systems or larger circulation systems such as central heating systems
or district heating installations.
H
Hmax
Hoperation, a= Hoperation, b
Parallel-connected pumps are also used when regulation is required or if an
auxiliary pump or standby pump is needed. When operating the pumps, it
is possible to regulate between one or more pumps at the same time. A nonreturn valve is therefore always mounted on the discharge line to prevent
backflow through the pump not operating.
Parallel-connected pumps can also be double pumps, where the pump
casings are casted in the same unit, and where the non-return valves are
build-in as one or more valves to prevent backflow through the pumps. The
characteristics of a parallel-connected system is found by adding the single
characteristics for each pump horisontally, see figure 3.5.
Qoperation, a = Qoperation, b
Qmax Q
Qoperation, a + Qoperation, b = Qsystem
Hoperation, a
Qoperation, a
Qsystem
Qoperation, b
Pumps connected in parallel are e.g. used in pressure booster systems, for water supply and for water supply in larger buildings.
Major operational advantages can be achieved in a pressure booster system
by connecting two or more pumps in parallel instead of installing one big
pump. The total pump output is usually only necessarry in a limited period.
A single large pump will in this case typically operate at lower efficiency.
Hoperation, b
Figure 3.5: Parallel-connected pumps.
By letting a number of smaller pumps take care of the operation, the system
can be controlled to minimize the number of pumps operating and these
pumps will operate at the best efficiency point. To operate at the most
optimal point, one of the parallel-connected pumps must have variable
speed control.
50
50
3.3 Pumps operated in series
Centrifugal pumps are rarely connected in serial, but a multi-stage pump
can be considered as a serial connection of single-stage pumps. However,
single stages in multistage pumps can not be uncoupled.
If one of the pumps in a serial connection is not operating, it causes a considerable resistance to the system. To avoid this, a bypass with a non-return valve
could be build-in, see figure 3.6. The head at a given flow for a serial-connected
pump is found by adding the single heads vertically, as shown in figure 3.6.
H
Hoperation, b
Hmax,total
Hmax,a
Hoperation,tot= Hoperation,a+Hoperation,b
Hoperation, a
Qoperation,a= Qoperation,b
3.4 Regulation of pumps
It is not always possible to find a pump that matches the requested performance exactly. A number of methods makes it possible to regulate the pump
performance and thereby achieve the requested performance. The most
common methods are:
1.
2.
3.
4.
Throttle regulation, also known as expansion regulation
Bypass regulation through a bypass valve
Start/stop regulation
Regulation of speed
Qmax
Q
Hoperation,b
Qoperation, a= Qoperation, b
Hoperation,a
Figure 3.6: Pumps connected in series.
There are also a number of other regulation methods e.g. control of preswirl rotation, adjustment of blades, trimming the impeller and cavitation
control which are not introduced further in this book.
51
51
3. Pumps operating in systems
3.4.1 Throttle regulation
Installing a throttle valve in serial with the pump it can
change the system characteristic, see figure 3.7. The resistance in the entire system can be regulated by changing the
valve settings and thereby adjusting the flow as needed.
A lower power consumption can sometimes be achieved
by installing a throttle valve. However, it depends on the
power curve and thus the specific speed of the pump.
Regulation by means of a throttle valve is best suited for
pumps with a relatve high pressure compared to flow (lownq pumps described in section 4.6), see figure 3.8.
3.4.2 Regulation with bypass valve
A bypass valve is a regulation valve installed
parallel to the pump, see figure 3.9. The bypass valve
guide part of the flow back to the suction line and concequently reduces the head. With a bypass valve, the
pump delivers a specific flow even though the system is
completely cut off. Like the throttle valve, it is possible
to reduce the power consumption in some case. Bypasss
regulation is an advantage for pumps with low head
compared to flow (high nq pumps), see figure 3.10.
Figure 3.7: Principle
sketch of a throttle
regulation.
Figure 3.9: The bypass valve
leads a part of the flow
back to the suction line
and thereby reduces
the flow into the system.
Valve
H
Hloss, throttling
H
Hloss,system
H
H
b
a
b
Q
b
η
b
Q
a
System Bypass
flow
flow
Q
P
b
a
Q
b
P1 P
2
Pa
Q
Q
η
a
Figure 3.8: The system characteristic is changed through throttle
regulation. The curves to the left show throttling of a low nq
pump and the curves to the right show throttling of a high nq
pump. The operating point is moved from a to b in both cases.
52
System
a
a
Pb
Q
a
H
Hloss,system
Q
a
Pb P
a
η
H
b
P
b
Q-Qbypass
Q
Bypass
flow
Q
b
Pb
a
Q bypass
b
P
a
System
flow
Bypass valve
a
Q
P
Pa
System
From an overall perspective neither regulation with
throttle valve nor bypass valve are an energy efficient solution and should be avoided.
η
b
b
a
Q
Q
Figure 3.10: The system characteristic is changed through bypass
regulation. To the left the consequence of a low-nq pump is
shown and to the right the concequences of a high nq pump is
shown. The operating point is moved from a to b in both cases.
52
3.4.3 Start/stop regulation
In systems with varying pump requirements, it can be an advantage to use
a number of smaller parallel-connected pumps instead of one larger pump.
The pumps can then be started and stopped depending on the load and a
better adjustment to the requirements can be achieved.
3.4.4 Speed control
When the pump speed is regulated, the QH, power and NPSH curves are
changed. The conversion in speed is made by means of the affinity equations. These are futher
described in section 4.5:
n
Q BB = Q AA ⋅ n BBB
(3.1)
(3.1)
QB = Q A ⋅ n
(3.1)
n BAAA
QB = Q A ⋅
(3.1)
nA
2
 n 22
HBB = H AA ⋅⋅  nBBB 2
(3.2)
(3.2)
HB = H A  n 
(3.2)
 nBAA 
HB = H A ⋅  A 
(3.2)
3
A 3
 n
3

n
n BB 
(3.3)
(3.3)
P
(3.3)
= P
PBBB =
PAAA ⋅⋅  n B 3
 n BAAA 
(3.3)
PB = PA ⋅  
2
 nA 
2
n  2
(3.4)
NPSH BB = NPSH AA ⋅  n BBB  2
(3.4)
(3.4)
NPSH B = NPSH A ⋅  n 
 n BAA 
(3.4)
NPSH B = NPSH A ⋅  A 
nA 

Index A in the equations describes the initial values, and index B describes the
PL,avg
= 0.06 ⋅ P100%
+ 0.15 ⋅ P75%
+ 0.35 ⋅ P50%
+ 0.44 ⋅ P25%
(3.5)
(3.5)
75%
50%
25%
PL,avg
= 0.06 ⋅ P100%
(3.5)
modified
values.
L,avg
100% + 0.15 ⋅ P75% + 0.35 ⋅ P50% + 0.44 ⋅ P25%
PL,avg = 0.06 ⋅ P100% + 0.15 ⋅ P75% + 0.35 ⋅ P50% + 0.44 ⋅ P25%
(3.5)
P
The equationsPL,avg
provide coherent points on an affinity parabola in the QH
L,avg
[[−−]] is shown in figure 3.11.
EEI = L,avg
(3.6)
(3.6)
= P parabola
EEI affinity
(3.6)
graph. The
Ref
PPL,avg
Ref
Ref
[−]
EEI =
(3.6)
PRef
Different regulation curves can be created based on the relation between
the pump curve and the speed. The most common regulation methods are
proportional-pressure control and constant-pressure control.
53
H
n = 100%
n = 80%
Coherent
points
Affinity
parabola
n = 50%
Q
Figure 3.11: Affinity parabola in the QH graph.
53
3. Pumps operating in systems
Proportional-pressure control
Proportional-pressure control strives to keep the pump head proportional
to the flow. This is done by changing the speed in relation to the current
flow. Regulation can be performed up to a maximum speed, from that point
the curve will follow this speed. The proportional curve is an approximative
system characteristic as described in section 3.1 where the needed flow and
head can be delivered at varying needs.
Proportional pressure regulation is used in closed systems such as heating
systems. The differential pressure, e.g. above radiator valves, is kept almost
constant despite changes in the heat consumption. The result is a low energy consumption by the pump and a small risk of noise from valves.
Figure 3.12 shows different proportional-pressure regulation curves.
Constant-pressure control
A constant differential pressure, independent of flow, can be kept by
means of constant-pressure control. In the QH diagram the pump curve for
constant-pressure control is a horisontal line, see figure 3.13. Constant-pressure control is an advantage in many water supply systems where changes
in the consumption at a tapping point must not affect the pressure at other
tapping points in the system.
54
54
H
H
Q
Q
P2
P2
Q
Q
η
η
Q
Q
n
n
Q
Figure 3.12: Example of proportional-pressure control.
55
Q
Figure 3.13: Example of constant-pressure control.
55
3. Pumps operating in systems
3.5. Annual energy consumption
Like energy labelling of refrigerators and freezers, a corresponding labelling
for pumps exists. This energy label applies for small circulation pumps and
makes it easy for consumers to choose a pump which minimises the power
consumption. The power consumption of a single pump is small but because
the worldwide number of installed pumps is very large, the accumulated energy consumption is big. The lowest energy consumption is achieved with
speed regulation of pumps.
The energy label is based on a number of tests showing the annual runtime
and flow of a typical
n circulation pump. The tests result in a load profile defined
QB = Q A ⋅ B
(3.1)
by a nominal operating
point (Q100% ) and a corresponding distribution of the
nA
operating time.
H
H max
max { Q . H } ~ P hyd,max
Q
P1
P100%
2
n 
HB = operating
H A ⋅  B  point is the point on the pump curve where the
(3.2)product
The nominal
nA 

of Q and H is the highest. The same flow point also refers to P100%, see figure
3
3.14. Figure 3.15 shows
 n B the time distribution for each flow point.
(3.3)
PB = PA ⋅  
 nA 
The representative power consumption is found by reading the power
2
consumption at the different
 n B operating points and multiplying this with
 
(3.4)
NPSH B = NPSH
A⋅ 
the time expressed
in percent.
n A 
PL,avg = 0.06 ⋅ P100% + 0.15 ⋅ P75% + 0.35 ⋅ P50% + 0.44 ⋅ P25%
EEI =
PL,avg
PRef
[−]
Phyd,max
(3.5)
(3.6)
Q100%
Figure 3.14: Load curve.
Flow %
Time %
100
6
75
15
H
50
35
P100%
25
44
Q
H
H max
P75%
3.15: Load profile.
PFigure
50%
P25%
Q25%
Q50%
Q75%
Q100%
H
Q
Q100%
Q75%
Q50%
Q25%
56
56
nB
(3.1)
nA
3.6 Energy efficiency index (EEI)
2
In 2003 a study of anmajor
part of the circulation pumps on the market was

B


⋅
H
H
=
(3.2)
B TheA purpose
conducted.
 n  was to create a frame of reference for a representa A
tive power consumption for a specific pump. The result is the curve shown
3
in figure 3.16. Based
 n on
 the study the magnitude of a representative power
(3.3)
PB = PA ⋅  B 
consumption of an
pump at a given Phyd,max can be read from the
A 
 n average
curve.
2
n 
(3.4)
NPSH B = NPSH A ⋅  B 
The energy index is defined
 n A as
 the relation between the representative
power (PL,avg) for the pump and the reference curve. The energy index can
be interpreted as an expression of how much energy a specific pump uses
PL,avg = 0.06 ⋅ P100% + 0.15 ⋅ P75% + 0.35 ⋅ P50% + 0.44 ⋅ P25%
(3.5)
compared
to average pumps on the market in 2003.
QB = Q A ⋅
EEI =
PL,avg
PRef
[−]
(3.6)
If the pump index is no more than 0.40, it can be labelled energy class A. If the
pump has an index between 0.40 and 0.60, it is labelled energy class B. The
scale continues to class G, see figure 3.17.
Reference power [W]
4000
3000
2000
1000
0
1
10
100
1000
10000
Hydraulic power [W]
Figure 3.16: Reference power as function
of Phyd,max.
Klasse
A
EEI
B 0.40 < EEI
C 0.60 < EEI
D 0.80 < EEI
E 1.00 < EEI
F 1.20 < EEI
G 1.40 < EEI
0.40
0.60
0.80
1.00
1.20
1.40
Figure 3.17: Energy classes.
Speed regulated pumps minimize the energy consumption by adjusting the
pump to the required performance. For calculation of the energy index, a reference control curve corresponding to a system characteristic for a heating
system is used, see figure 3.18. The pump performance is regulated through
the speed and it intersects the reference control curve instead of following
the maximum curve at full speed. The result is a lower power consumption
in the regulated flow points and thereby a better energy index.
H
Hmax
Q 0% ,
H100%
2
Q100% , H100%
n25%
n50%
n75%
n100%
Q
Figure 3.18: Reference control curve.
Q100%
Q75%
Q50%
Q25%
57
57
3. Pumps operating in systems
3.7 Summary
In chapter 3 we have studied the correlation between pump and system
from a single circulation pump to water supply systems with several parallel
coupled multi-stage pumps.
We have described the most common regulation methods from an energy
efficient view point and introduced the energy index term.
58
58
Chapter 4
Pump theory
4.1 Velocity triangles
4.2 Euler’s pump equation
4.3 Blade form and pump curve
C2
4.4 Usage of Euler’s pump equation
4.5 Affinity rules
C2m
U2
C2u
4.6 Inlet rotation
β2
4.7 Slip
2
W1
4.8 Specific speed of a pump
4.9 Summary
W2
α2
C1m
α1
β1
1
U1
r1
r2
4. Pump theory
4. Pump theory
The purpose of this chapter is to describe the theoretical foundation of energy conversion in a centrifugal pump. Despite advanced calculation methods which have seen the light of day in the last couple of years, there is still
much to be learned by evaluating the pump’s performance based on fundamental and simple models.
When the pump operates, energy is added to the shaft in the form of mechanical energy. In the impeller it is converted to internal (static pressure)
and kinetic energy (velocity). The process is described through Euler’s pump
equation which is covered in this chapter. By means of velocity triangles for
the flow in the impeller in- and outlet, the pump equation can be interpreted
and a theoretical loss-free head and power consumption can be calculated.
Velocity triangles can also be used for prediction of the pump’s performance
in connection with changes of e.g. speed, impeller diameter and width.
4.1 Velocity triangles
For fluid flowing through an impeller it is possible to determine the absolute
velocity (C) as the sum of the relative velocity (W) with respect to the impeller, i.e. the tangential velocity of the impeller (U). These velocity vectors
are added through vector addition, forming velocity triangles at the in- and
outlet of the impeller. The relative and absolute velocity are the same in the
stationary part of the pump.
The flow in the impeller can be described by means of velocity triangles,
which state the direction and magnitude of the flow. The flow is three-dimensional and in order to describe it completely, it is necessary to make two
plane illustrations. The first one is the meridional plane which is an axial
cut through the pump’s centre axis, where the blade edge is mapped into
the plane, as shown in figure 4.1. Here index 1 represents the inlet and index
2 represents the outlet. As the tangential velocity is perpendicular to this
plane, only absolute velocities are present in the figure. The plane shown in
figure 4.1 contains the meridional velocity, Cm, which runs along the channel
and is the vector sum of the axial velocity, Ca, and the radial velocity, Cr.
60
2
Cm
Cr
1
Ca
Figure 4.1: Meridional cut.
60
C2
Figure 4.2a: Velocity triangles
positioned at the impeller inlet
and outlet.
W2
C2m
U2
U2
β2
2
C1m
α1
1
W2
C2
C2U
C2m
β2
U2
W1
β1
C2
C2m
β2
W2
α2
C2U
2
α2
α2
C2U
1
W1
U1
ω
Wβ1 1
r1
r2
U1
β1
α1
α1
C1m
C1m
Figure
4.2b: Velocity triangles
U1
The second plane is defined by the meridional velocity and the tangential
velocity.
An example of velocity triangles is shown in figure 4.2. Here U describes the
impeller’s tangential velocity while the absolute velocity C is the fluid’s velocity
compared to the surroundings. The relative velocity W is the fluid velocity compared to the rotating impeller. The angles α and β describe the fluid’s relative
and absolute flow angles respectively compared to the tangential direction.
Velocity triangles can be illustrated in two different ways and both ways are
shown in figure 4.2a and b. As seen from the figure the same vectors are repeated. Figure 4.2a shows the vectors compared to the blade, whereas figure
4.2b shows the vectors forming a triangle.
W1
W1
U1
U1
C1
C1m
C1
C1m C1U
C1U
By drawing the velocity triangles at inlet and outlet, the performance curves
of the pump can be calculated by means of Euler’s pump equation which
will be described in section 4.2.
61
61
4. Pump theory
4.1.1 Inlet
Usually it is assumed that the flow at the impeller inlet is non-rotational.
This means that α1=90°. The triangle is drawn as shown in figure 4.2 position
1, and C1m is calculated from the flow and the ring area in the inlet.
The ring area can be calculated in different ways depending on impeller type
(radial impeller or semi-axial impeller), see figure 4.3. For a radial impeller
this is:
A1 = 2 π ⋅ r1 ⋅ b1 [m2]
(4.1)
where
 r1 , hub + r1 , shroud 
 ⋅ b1 inlet
(4.2)
[m2edge
] [m]
π ⋅ 
r1 = TheAradial
of the impeller’s
1 = 2 ⋅position
2


b1 = The blade’s height at the inlet [m]
Q
m
C 1 m = impeller [ s ]
(4.3)
A
1
= 2 π ⋅ r ⋅ bimpeller
[m2] this is:
(4.1)
and forAa semi-axial
1
62
1
b2
Blade
r2
1
U1 = 2 ⋅ π ⋅ r1⋅ r1n, hub=+rr11⋅, shroud
ω [ m
(4.4)
s]
2
60

(4.2)
[
]
m
A1 = 2 ⋅ π ⋅ 
b
⋅
 1
A1 = 2 π ⋅Cr1 ⋅ b1 [m22]
(4.1)

1m
[ ]
(4.5)
tan β1 =
Q impeller
Umust
m
1 r
r
+
The entire
flow
pass
through
this
ring
area.
C
is
then
calculated

2
1
,
hub
1
,
shroud
C
(4.3)
1m
A111 m==2 π⋅ π⋅ rA⋅1 ⋅ b1 [[ms ]]
(4.1)
 ⋅ b1
(4.2)
[m2]

1
from:
2

(4.6)
A2 = 2 π ⋅ r2 ⋅ b2 [m2]
r1n, hub + r1 , shroud  m

2
U
22⋅Q
r1 ⋅ ω [⋅ sb1]
π
⋅ r⋅1⋅
=m
(4.4)
(4.2)
[
]
m
A
π
=
⋅
impeller
11 =
 60 [ ]

C 1m =
(4.3)
A1 r2 , hub + rs22 , shroud  
(4.7)
 ⋅ b2 [m2]
A 2 = 2 ⋅QπC⋅1 m
[[m
(4.5)
tan β1 = impeller
2 ]]
m


n
=
C
(4.3)
U
U11 m= 2 ⋅ πvelocity
r1 ⋅ U= sequals
r1 ⋅ ω the
⋅A
[ sproduct
]
The tangential
of radius and angular(4.4)
frequency:
1 60 1
Q impeller m 2
(4.8)
C22m=
(4.6)
A
= =2 π⋅=π⋅CA
s[m]r1]]⋅ ω
U
⋅rr1221m⋅⋅ bn2 [ [=
(4.4)
[ms ]
(4.5)
tan
1 β2
1
60
U
2
1
A1 = 2 π ⋅ r1 ⋅ b1 n [m ]
(4.1)
where U = 2 ⋅ πC⋅1rm r⋅
+ r rshroud
 [m
]
(4.9)
2
s
2 β =
2 2 , hub[ =22,]
2⋅ω
(4.5)
tan
(4.7)
A
(4.6)
ω = Angular
[s
A22 ==221frequency
π⋅ π⋅Ur⋅21⋅ rb160
[
m
+-12]r1], shroud ⋅ b2 [m ] 2
2, hub


 ⋅ b1
(4.2)
[
m]
A1 = 2 ⋅ π ⋅ 
-1
n = RotationalCspeed [min2 ]

m
2m 
2
W
(4.10)
[ [s+m]r2,]shroud  ⋅ b [m2]
(4.6)
A22==2 πQ impeller
⋅ r2 ⋅ rb
2 , hub
(4.7)
A
⋅Qπβimpeller
⋅2  2 [m
]
(4.8)
C22m==2sin
2
s[m

]
=
C
A
2s has been drawn, see figure 4.4, based(4.3)
1 mvelocity
2 triangle
When the
on α1, C1m
A1 r
+ r2 , shroud  m
2 , hub
2
C
n
m
and U1,A
angle
Without inlet
rotation
2m
(4.7)
[m calculated.
]
b2 ] be
= 2relative
⋅ ππ −
⋅  flow
(4.11)
C
⋅U
Uthe
r [⋅ ω]β⋅1 [can
(4.9)
22 ==2 Q
2 ⋅ r2 ⋅ n m=
[β=s2 ]2r1 ⋅2ωs  [mss ]
(4.8)
C)212U
= =2 ⋅becomes:
πimpeller
⋅ r1tan
⋅ 60
(4.4)
(C1 = C1mU
this
m
A 2 60
Q
C2(impeller
C
n[m
m
TC
r2⋅1⋅rmC
r[1 =]⋅ C]r1U⋅)ω [Nm
(4.12)
(4.8)
=21 ⋅=⋅ π
W
(4.10)
(4.5)
[ms ]
tan
2⋅ U −
U22=2m==βm
(4.9)
U2212 60 s 2
sin βA
n
m
= 22πC
⋅π
⋅ r2⋅ C
⋅b2 mm=2 r2 ⋅ m
ω [s]
U2 =
(4.9)
2⋅mrT
(4.6)
A
[
]
m
60
[
W
(4.13)
=
ω
⋅
2 P=
2
2
(4.11)
C
[
]
U
−
W
(4.10)
[
]
2U
2
22
s
s
β
sin β2 tan
= m . ω . (2r2 . C2U − r1 . C1U )
C  r2 , hubm+ r2 , shroud 
W22 ==2=⋅ 2πm m
(4.10)
. ω[ .sr2]. C2U −
(4.7)
 ⋅[.Nm
A
b. [m2]
T = =msin
⋅ (βr2−⋅⋅C2CU(2m
− r1 ⋅ C1U[)mω
(4.12)
(4.11)
C
] r1 2 C]1U )
U
b1
r1
b2
W2
r2, shroud
C2
βb2
1
r1, shroud
Blade
C2m
U2
rα2, hub
2
r1, hubC2U
Figure 4.3: Radial impeller at the top,
semi-axial impeller at the bottom.
W1
β1
α1
C1m
U1
Figure 4.4: Velocity triangle at inlet.
W1
C1
62
63
 r1 , hub + r1 , shroud 
 ⋅ b1
(4.2)
[m2]
(4.1)
A1 = 2 π⋅ π⋅ r⋅1 ⋅ b1 [m2]
2


+ r21 , shroud 
A1 =
= 2 πQ⋅impeller
r1 ⋅ br11, hub[m
(4.1)
 ⋅ b1
(4.2)
[m2]
A
1 =2 ⋅ π ⋅ 
[ms2]]
C
(4.3)

1m
A1

r
r
+
 1 , hub 1 , shroud 
 ⋅m
(4.2)
[m2]
A = 2 Q⋅ πimpeller
b1
⋅  n m
[
C111 m= =
(4.3)
2]
U
2
r
r
⋅
π
⋅
⋅
=
⋅
ω
(4.4)
[
]
2
s
1
1]

s
A
=
2
π
⋅
r
⋅
b
[
m
(4.1)
A1 1 60
1
1
4.1.2 Outlet
Q Cimpeller
m
As withCthe
inlet,1 mthe
velocity
triangle at the outlet is drawn as(4.3)
shown in
1 m β=
[[+srr121]]⋅, shroud
(4.5)
tan
U
212⋅=π
r⋅1⋅ r1n, hub=
⋅A
ω [⋅m
1=
s]
(4.2)
[m2] area is calculated(4.4)
A
π
=
⋅
1
A11 position
= 2 π ⋅Ur11⋅2.b60
]
(4.1)
 b1
figure 4.2
a2radial
impeller,
outlet
as:
1For[m

 m
n
C
U
rr122]⋅ ω [ s ]
⋅π
π⋅⋅rr11m⋅⋅br =
(4.4)
(4.5)
tan
=Q
+m
A11 ==β22
[m
(4.1)
11, hub[
1 , shroud
(4.6)
A
2 [m ]
Ur⋅211⋅ b60
]
=12π⋅ π⋅impeller
C
(4.3)
 ⋅ b1
(4.2)
[m2]
A21 m==2
s
2
1
CA

1m
r
r
+


1 , hub[
1 , shroud
(4.5)
tan
(4.2)
[m2]
+ r22,]]shroud   ⋅m
A1 ==β212π=Q⋅ π⋅Ur⋅ ⋅ rb
b1
2n
, hub[
2
(4.6)
A
m
2 semi
21⋅ impeller
2 =m
impeller
(4.7)

[
]
m
A
2
b
⋅
=
⋅
π
⋅
2
2
r
r
=
⋅
π
⋅
⋅
ω
(4.4)
[
]
and forU
a
axial
it
is:
2
2
[
C11 m =

(4.3)
1
1]
s

s
A1 60 2

2
Q
m
+mr ]
(4.6)
A
r12m⋅ rb

, 2hub[
2
=2π=⋅ π⋅Cimpeller
C 21 m==β2
(4.3)
(4.7)
 ⋅ [bm
⋅ 2n
[ sr2,]⋅shroud
A
⋅A
(4.5)
tan
2= 2
2 ] [m ]
1Q
m
impeller
U
r
⋅
π
⋅
=
ω
(4.4)
1
1
1
1
(4.8)
C2 m = U 1 60[ s ]2
 s
A 2 r2n
r
+

, hub
2 , shroud
m
2
(4.7)
 ⋅ [b2sfor
U
r ⋅
⋅⋅ππC⋅ ⋅in
= r12 ⋅ ωway
(4.4)
] [m ]
A12 ==22Q
C2m is calculated
same
nm
m ] the inlet:
(4.5)
tan
(4.6)
A
⋅ ⋅r121rm⋅ the
b60
2 ]r ⋅ ω as
2 [ [m
]
(4.8)
C
[
=β=221 π⋅=πimpeller
⋅
=
U222m=
(4.9)
s
U
s
2
2
CA121m 60
[ ]
(4.5)
tan β1 Q
= impellerr m
2 , hub
n[ [s+m]r22,]shroud  m [m2]
Ur⋅ 1⋅ b
(4.8)
C
(4.6)
⋅
(4.7)
A
b
⋅
π
[
]
⋅⋅ π
⋅
⋅
=
⋅
ω
U222m===22πC
r
r
(4.9)
2
2
2
m
s
2
2

W 2 = 2 mA 2 60
(4.10)
[ s ]2

(4.6)
A2 = 2sin
π ⋅βr22 ⋅ rb2n [+mr2]
[ms ] [m2]
⋅ π ⋅ r2 ⋅2 , hub =2 , rshroud
U2 = 2 Q
(4.9)
2 ⋅ ω
C
(4.7)

2π
m ⋅  60m
A
2
b
⋅
=
⋅
impeller
2
2

 C2 m[ sU]2is calculated
W
The tangential
from the following:
(4.10)
(4.8)
C2 2m== velocity
m
(4.11)
C2U = sin
[ s ]
U 2βA−22 r2 , hub + r2 , shroud
2
(4.7)
 ⋅ b2 [m ]
A 2 = 2 C⋅ 2πm ⋅ tan βm2
n[ m
W
=2 Q
(4.10)
]2 r ⋅ m
 [m ]
C
2=
s
impeller
⋅
π
⋅
⋅
=
ω
U
r
(4.9)
s
2
2
2
2
m
[s] [s]
(4.8)
C2 m ==sin
(4.11)
C
U β−2 60
[Nm]
T 2U= mQ⋅ (2rA2 ⋅2Ctan
(4.12)
2U −βr21 ⋅ C1U )
m
[ sdesign
] mphase,
(4.8) value
C2 m = C impeller
Cthe
In the beginning
same
2n
mm
A−
22m
C
[ms ]β2 is assumed to have the(4.11)
=2 ⋅Uπ
⋅ of
=] r2 [⋅ ω
U2U
(4.9)
2r2 ⋅ [
2=
s ] [Nm
W
=
(4.10)
2
60
s
] can then be calculated from:
T = msin
⋅ (βr2 ⋅ Ctan
(4.12)
as the blade
angle.
The
21 ⋅ C1U ) velocity
2U −βrrelative
2
n
m
W]r ⋅ ω [ ]
(4.13)
U2 P=2 2 =⋅ π ⋅Tr2⋅ ⋅ω [=
(4.9)
s
2
60
C
.
.
.
.
m
2
m
(rr12]⋅ CC12UU)m
− r1[Nm
C1U])
T =2 =m=⋅ (r2m
⋅ C2CUω2 m−
(4.12)
W
(4.10)
[
(4.11)
C2UP= sin
U 2β−T ⋅ ω s[W] [ s ]
(4.13)
. ( ωβ. 2r2 . C2U − ω . r1 . C1U )
2 tan
2 =
=
C2 m m
m
W 2 = = m . ω[. (sr ]. C − r . C )
(4.10)
. ( U2 . 2C2U 2−Um
= βm
U11 . C11UU)
2T ⋅ C
and C2UCas:P= sin
2 m [W]
(4.13)
=
ω
(4.11)
[
]
U
−
2
T 2U= m=⋅ (2r2m
⋅ C2. U(.ω
−.rr12.⋅.CC12UU)−s ω[.Nm
(4.12)
. ]
= Q .tan
ρ
(U
β2 .C2U − U1.r.1C1CU1)U )
.
.
= m .Cω2 m (.r2 C2Um
− r1 . C1U )
(4.11)
C2U = =U 2 −m ( U2 C2U −
[ U]1 C1U )
. ( ωβ. 2r2 . C2U −s ω
. r1 . C1U )
tan
=
m
[
]
Nm
.⋅ C]]12UU )− U1 . C1U )
T =P m
(4.12)
=
Ur2[1[W
2Q
2Q
U. (−
(4.14)
P
∆⋅ p(rtot
W
(4.13)
T⋅ C.⋅ ρ
ω
hyd2= =
= m . ( U2 . C2U − U1 . C1U )
.
.
.
.
HerebyTthe
velocity
triangle
at
the
outlet
has
been
determined
and
=
m
ω
(
r
C
−
r
C
)
]
= m ⋅ (r2 ⋅ C. 2U. − r12.⋅ C12UU) 1[Nm
(4.12)can now
=
Q ⋅ ρQ ( U2[WC]2U − U1 . C11UU )
∆
p
tot
(4.14)
P
=
∆
p
be drawn,
figure
tot
. C]2U − ω . r1 . C1U )
m
Hhyd=
(4.15)
(4.5.
ω]. r[2W
(4.13)
P2see
T ⋅ .[ω
ρ=⋅ g m
. (ωU.2(.r2C.2CU 2−U −Ur11 . CC11UU))
=
(4.14)
PhydP2=∆p
∆tot
p totm
W]]
(4.13)
=
T⋅Q
ω [[W
H
=
(4.15)
.ω
.
.
..[m
.
.
.
.
(]⋅U
C
−
U
C
)
(4.16)
Phyd =ρ=Q⋅ g⋅ Q
H .⋅ρ
ρ
g
=
m
⋅
H
⋅
g
m
(
r
ω
r
C
)
[
]
W
1 1U)1U
= m ω . (2r22 . C222UUU − r11. C
1U
∆p
. 1.U )
tot m .. ( U2. . C
1 .C
[m
H= =
(4.15)
( ω] r2 . 2CU2U−−Uω
r C )
ρ=
⋅ pg⋅Pm
(4.16)
=
Q
H
⋅
ρ
⋅
g
=
m
⋅
H
⋅ .g1C 1[)UW]
[
]
W
(4.14)
Phyd
∆
⋅
Q
(4.17)
=
P
.
.
.
Q
ρ
(
U
C
−
U
hyd hyd = tot
2
2
2
U
1
1
U
= m . ( U2 . C2U − U1 . C1U )
⋅ (U2 ⋅ C2U − U. 1 ⋅ C1U )
m ⋅ H ⋅ g =. m
=Qtot⋅ Q
ρρ. ( ⋅Ug2 .=
Cm−⋅ HU ⋅ gC [)W]
(4.16)
P P =∆p
(4.17)
=pP2H ⋅⋅[Qm
hyd
]
H
=
[W]2U 1 1U
(4.14)
(4.15)
Phyd
=
∆
hyd ρ (⋅U
2tot⋅ C2U − U1 ⋅ C1U )
g
H=
m ⋅ H ⋅ g = mg⋅ (U2 ⋅ C2U − U1 ⋅ C1U )
W2
β2
U2
C2
C2m
α2
C2U
Figure 4.5: Velocity triangle at outlet.
W1
63
 r1 , hub + r1 , shroud 
 ⋅ b1
(4.2)
[m2]
A1 = 2 ⋅ π ⋅ 
2
2

A1 = 2 π ⋅ r1 ⋅ b1 [m ]
(4.1)
Q impeller
m
4. Pump
C 1 m =theory
(4.3)
 r1 , hub[+sr1], shroud 
 ⋅ b1
(4.2)
[m2]
A1 = 2 ⋅ π A⋅ 1
22


A1 ==22⋅ππ⋅⋅rr1 ⋅⋅b1n =
[mr ]⋅ ω [ms ]
(4.1)
U
(4.4)
1
1
Q impeller
60 m1
4.2 Euler’s
pump
equation
[
]
C 1m =
(4.3)
r1 , shroud 
 r1 , hubis+sthe
2 equation in connection with
CA
1
Euler’s tan
pump
equation
most
important
1⋅ m

(4.2)
[
]
m
A
2
b
π
=
⋅
⋅
[ 2]
(4.5)
1 β1 =

 1
U 1equation
 n
 derived
m
pump design.
The
can
be
in
many
different
ways.
The metU1 = 2 ⋅ π ⋅ r1 ⋅
= r1 ⋅ ω
(4.4)
[
]
s
60
Q
m
hod described
includes a control volume which limits the impeller,
the
impeller
= π here
C 1 m= 2
(4.3)
(4.6)
A
⋅CrA2 ⋅ b2 [[ms2]]
2
moment
of
momentum
equation
which
describes
flow
forces
and
velocity
1m
1
[ ]
(4.5)
tan β1 =
triangles at inletUand
outlet.
1
m
n
U1 = 2 ⋅ π ⋅ r1⋅r2 , hub=+ r21, ⋅shroud
ω  [s] 2
(4.4)
(4.7)
 ⋅ b2 [m ]
A 2 = 2 ⋅ π ⋅  60
2
A2 =
2 π ⋅Cr2 is
⋅ ban
m2 ]
 limited volume which is used (4.6)
2 [
A control
volume
imaginary
for setting
[ ]The equilibrium equation can be set (4.5)
tan β1 = 1 m
up equilibrium
equations.
up for torU1
Q impeller
m
+ ]r2 , shroud 
 r2 , hub
[
(4.8)
=
C
2
m
2
s
ques, energy
which
 ⋅ b2 [m
] are of interest. The(4.7)moment
A 2 = 2 and
⋅ πA⋅ 2other flow quantities
22one
 equilibrium equation, linking(4.6)
A2 = 2 π equation
⋅ r2 ⋅ b2 [m
]
of momentum
is
such
mass flow
n
m]
[
=
⋅
π
⋅
⋅
=
⋅
ω
U
2
r
r
(4.9)
s
2
2
2
and velocitiesQwith
impeller
60m diameter. A control volume between 1 and 2, as
[ s+ ]r2, shroud
(4.8)
= impeller
C2 mfigure
 r2is, hub
shown in
4.6,
used for
(4.7)
m2]
A 2 = 2 ⋅ πA⋅ 2 often
⋅ b2 an[impeller.

2
C

W 2 = 2 m  n[m
(4.10)
s =] r ⋅ ω [m ]
⋅ πβ⋅2 r2 ⋅we are
U2 = 2sin
s in is a torque balance. The(4.9)
2
The balance
which
torque (T)
60m interested
Q impeller
(4.8)
[
]
=
C
m
2
s
from the drive shaft
to the torque originating from the fluid’s
A 2 C2corresponds
m]
m
(4.11)
C2U = CU
[
m
22m−impeller
flow through
the
with
mass
flow m=rQ:
s
W2 =
(4.10)
tann[βs2 ]
m
⋅ πβ⋅2 r2 ⋅
= r2 ⋅ ω [ s ]
U2 = 2sin
(4.9)
60
64
T = m ⋅ (r2 ⋅ C2CU − r1 ⋅ C1U )m [Nm]
(4.12)
(4.11)
C2U = CU22m− 2 mm
[s]
W2 =
(4.10)
tan[βs2 ]
sin βthe
By multiplying
torque by the angular velocity, an expression for the
2
(4.13)
P = T ⋅ ω [W] At the same time, radius multiplied
shaft power
T = 2m ⋅(P(r22) ⋅ Cis2CUfound.
− r1 ⋅ C1U )m [Nm]
(4.12) by the
m . ω2 m. (the
r2 . Ctangential
(4.11) in:
U 2 −equals
2[
U−
angularCvelocity
2U = =
s r]1 . C1U ) velocity, r2w = U2. This results
tan
β
2
= m . ( ω . r2 . C2U − ω . r1 . C1U )
W]
(4.13)
P2 =
T ⋅.ω( U .[C
C1U ])
T = m=⋅ (r2m
⋅ C2U −2r1 ⋅ C2U1U−) U1 [.Nm
(4.12)
.
.
.
.
=
m. ρω
. ( U(2r2. CC22UU−−Ur11 .CC11UU))
= Q
= m . ( ω . r2 . C2U − ω . r1 . C1U )
W] − U . C )
(4.13)
P = m
T ⋅ ω [C
(4.14)
2]U
1
1U
Phyd2= =
∆p tot ⋅ .Q( U2 [. W
.
.
.
.
=
m
ω
(
r
C
−
r
C
)
= Q . ρ . ( U22. C22UU− U11 . C11UU )
∆p
=tot m . ( ω . r2 . C2U − ω . r1 . C1U )
[m] equation, the hydraulic power added (4.15)
H =to the
According
to the fluid
ρ=⋅ g menergy
C2]U − U1 . C1U )
(4.14)
Phyd = ∆p tot ⋅ .Q( U2 [. W
can be written as the increase in pressure Δptot across the impeller multi= Q . ρ . ( U2 . C2U − U1 . C1U )
plied byPhyd
the=∆flow
(4.16)
pQtot⋅ HQ:⋅ ρ ⋅ g = m ⋅ H ⋅ g [W]
[
m]
H=
(4.15)
ρ⋅g
(4.14)
Phyd = ∆p tot ⋅ Q [W]
(4.17)
Phyd = P2
(4.16)
Phydm=∆⋅ pH
Q ⋅⋅ gH=⋅ m
ρ ⋅⋅ (gU= ⋅m
⋅
H
⋅
g
[
]
W
C − U1 ⋅ C1U )
tot
[
m] 2 2U
H=
(4.15)
ρ (⋅Ug ⋅ C − U ⋅ C )
2
2U
1
1U
Hhyd= = P2
(4.17)
P
g
ω
r1
1
U1 = r1ω
2
r2
U2 = r2ω
2
1
Figure 4.6: Control volume for an impeller.
64
ω− (rr2⋅ CC2U) − r1[Nm
C1U])
T =P m=
⋅ (r m
(4.12)
m]
Un
1[W]
1U
(4.13)
T⋅rC⋅ 2⋅ω
[
⋅=π 2⋅m
⋅
ω
U2 =2 2 =
r
(4.9)
s.
2 . (ω. r =
2
.
.
60
2 C2U − ω r1 C1U )
.
.
.
.
= m ω (r2 C2U − r1 C1U )
= m . ( U2 . C2U − U1 . C1U )
. C − ω . r1 . C1U )
C m
(4.13)
P2 =
T ⋅ .ω( ω . r[2W
. ]2U
.
W 2 = = 2 mQ . ρ . ([Um
(4.10)
2 ]C2U − U1 C1U )
s
.
.
.
.
.
= βm
(
U
C
−
U
C
)
sin
ω 2(r2 2CU 2U − r11 C11UU )
2
. (.ω
Q.ρ
( U. .. C − U . . C. )
(4.14)
Phyd = =
∆p totm⋅ C
Q2 m r2[2W]22UU mω1 r1 1CU1U )
The head
as:
(4.11)
C2Uis=defined
[ sU] . C )
U
−
. ( U2 . C2U −
= 2 mtan
1
1U
β2
(4.14)
Phyd =∆p
∆p tot .⋅ Q. [W
. ]
.
H = =tot Q ρ[m(]U2 C2U − U1 C1U )
(4.15)
T = mρ ⋅⋅ (gr2 ⋅ C2U − r1 ⋅ C1U ) [Nm]
(4.12)
∆ptot
H
(4.15)
(4.14)
Phyd= =ρ∆⋅ pgtot ⋅ [Qm] [W]
and thePhyd
expression
(4.16) to:
= Q ⋅ H ⋅for
ρ ⋅ hydraulic
g = m ⋅ H power
⋅ g [Wcan
] therefore be transcribed
(4.13)
P2 ∆p
= T ⋅ ω [W]
[
]
m
H
(4.16)
(4.15)
Phyd= =ρ=Q⋅tot
⋅
H
⋅
ρ
⋅
g
=
m
⋅
H
⋅
g
[
]
W
(4.17)
Phyd =gP2m . ω . (r2 . C2U − r1 . C1U )
. ⋅ 2CU − ω
. C1U ))
= ⋅ gm=. m
( ω⋅.(rU
2 C
− .Ur1free,
If the flowm
is⋅ H
to
loss
the hydraulic and(4.17)
mechanical
2 be2 U
1 ⋅ C1Uthen
=assumed
(4.16)
PhydPhyd
= =Q
⋅P2m
H ⋅. (ρU⋅ g. C= m−⋅UH .⋅Cg )[W]
2
2
U
1
1
U
U ⋅ C − U1 ⋅ C1U )
power can be (equated:
H =⋅ =H ⋅2gQ=. 2ρUm
⋅ (U. 2C⋅ C2−U U
− U. 1C⋅ C)1U )
m
. (gU
2
2U
1
1U
(4.17)
Phyd =
P
(U22⋅ C2U − U1 ⋅ C1U )
H=
g
2 ⋅C 2)
Hp⋅Utot
g22 =⋅−Q
m
(4.14)
Phydm=⋅ ∆
U12⋅ ([UW2 ]⋅ C2U −WU
C22 − C12
1 1− W12U
[m] (4.18)
H =
+
+
2⋅g
2⋅g
(U2 ⋅22C⋅2gU −2 U1 ⋅ C1U )
=p U − U
H∆
W 2 − W22
C22Dynamic
− C12 head
head2as consequence
[m] (4.15)
(4.18)
H == Statictot
+ Static 1head as consequence
+
[
m1g]
of the velocity
ofρ
the⋅ centrifugal
g 2 ⋅ g force
2 ⋅ g change
2⋅g
the impeller
This is the equation known asthrough
Euler’s
equation, and it expresses the impel2
2
2
Static head
as consequence
Static head2as consequence
U
−
U
W
−
W
C 2Dynamic
− C 2 head outlet.
2
1
1
2
ler’s head
at
tangential
and
absolute
velocities
of
the
velocity
change
[m] (4.16)
(4.18)
Hhyd= of= the
+ in2inlet1 and
P
Qcentrifugal
⋅ H ⋅ ρforce
⋅ g = +m ⋅through
H ⋅ gthe impeller
[
]
W
2 ⋅ g are applied2 ⋅to
g the velocity
2 ⋅ gtriangles, Euler’s pump
If the cosine relations
headof
as consequence
Dynamic head
Staticbe
head
as consequence
equation can
written
as theStatic
sum
the three contributions:
of the velocity change
of the
force
(4.17)
=centrifugal
Phyd
P2
through the impeller
− Uthe
m ⋅ H ⋅ g = m ⋅ (U2 ⋅ C2U of
1 ⋅ Ccentrifugal
1U )
• Static head as consequence
force
(
⋅
−
⋅
)
U
C
U
C
2 consequence
2U
1
1U of the velocity change through the impeller
• Static head
H = as
g
• Dynamic head
U22 − U12
2⋅g
H =
+
Static head as consequence
of the centrifugal force
W12 − W22
2⋅g
+
Static head as consequence
of the velocity change
through the impeller
C22 − C12
2⋅g
[m]
(4.18)
Dynamic head
If there is no flow through the impeller and it is assumed that there is no
inlet rotation, then the head is only determined by the tangential velocity
based on (4.17) where C2U = U2:
H0 =
65
U 22
g
[m]
(4.19)
65
4. Pump theory
When designing a pump, it is often assumed that there is no inlet rotation
meaning that C1U equeals zero.
H =
U 2 ⋅ C 2U
g
[m]
(4.20)
⋅ ⋅v
[N]
F =m
4.3 Blade shape and pump curve
⋅ ⋅ v = ρ ⋅ A ⋅ v 2 [N]
∆I = m
β2
[N] β
∆I = F
1
β2
β1
(4.22)
β2
β2
β1
2
2
U
U2
−
⋅Q
H=
g
π ⋅ D2 ⋅ b2 ⋅ g ⋅ tan( β2 )
�2�>90 o
(4.21)
�2�>90 o�2�>90 o
β1
[m]
β2
β1
β1
β2 = 90o β2 = 90βo2 = 90o
U2 2
U 2 (4.23)
(4.21)
− ⋅b  
⋅ Q [m]
H = 2 D
Q B = Q A g⋅  B2π ⋅BD2⋅ b2 ⋅ g ⋅ tan( β2 )
 DA ⋅ bA  
2

 D   Geometric
HB= HA ⋅  B  
scaling
Figure 4.7 and 4.8
the connection
between the theoretical pump
 DAillustrate

(4.22)
 nB  
=
⋅
Q
Q

4
curve and the
B blade
AD n⋅shape
b  indicated at β2.
PB = PA ⋅  B4 AB  
 DAn⋅ bA2  Scaling of
Real pumpHcurves
are,B however, curved due to different losses, slip, inlet
=
H
 rotational speed
A⋅ 
B
rotation, etc., This is
A
nfurther
 discussed in chapter 5.
3 
 nB  
UPBB =CPm,B
C
(4.24)
A ⋅ n
=
= Au,B
 
UA C m,A Cu,A
66
β2
β1
β1
β2
β1
(4.21)
U 2 ⋅ C 2U
(4.20)
H =shapes
[m]on outlet
Figure 4.7: Blade
depending
angle
(4.22)
g  nB  
QB = QA ⋅   
 nA  
⋅ ⋅that
If it is assumed
of Eul(4.21)
[N2 ] is no inlet rotation (C1U =0), a combination
F =m
v nthere
B   Scaling of
er’s pumpHequation
and
equation
(4.6),
(4.8)
and
(4.11)
show
that
the


B = HA ⋅  (4.17)
nA   rotational speed
2
⋅
head varies
linearly
∆I =
m ⋅ v =with
ρ3 ⋅Athe
⋅ v flow,
(4.22)outlet
[Nand
] that the slope depends on the
angle β2: PB = PA ⋅  nB  
(4.23)
∆I = F [Nn]A 
(4.23)
 DB2 ⋅bB  
 2 
QU
B Q An⋅B ⋅ D2,B
B=
=  D ⋅b 
UA nA ⋅ DA2,A 2 A 

 D   Geometric
HB= HA ⋅  B  
β2
β2
(4.23)
β2 < 90o
β2 < 90βo 2 < 90o
H
0°
es
>9
blad
r b2
ept
H fo
w
s
d
war
For
H for b2 = 90°
H fo
rb
< 90
°
d-sw
ept
b
2
Back
war
lade
s
Q
Figure 4.8: Theoretical pump curves calculated based on formula (4.21).
(4.25)
66
4.4 Usage of Euler’s pump equation
There is a close connection between the impeller geometry, Euler’s pump
equation and the velocity triangles which can be used to predict the impact
of changing the impeller geometry on the head.
The individual part of Euler’s pump equation can be identified in the outlet
velocity triangle, see figure 4.9.
W2
C2m
C2
α2
β2
Figure 4.9: Euler’s pump equation and the
matching vectors on velocity triangle
H=
1
⋅ U2 ⋅ C2U
g
This can be used for making qualitative estimates of the effect of changing
impeller geometry or rotational speed.
67
67
4. Pump theory
In the following, the effect of reducing the outlet width b2 on the velocity
triangles is discussed. From e.g. (4.6) and (4.8), the velocity C2m can be seen
to be inversely proportional to b2. The size of C2m therefore increases when b2
decreases. U2 in equation (4.9) is seen to be independent of b2 and remains
constant. The blade angle b2 does not change when changing b2.
W2
C2
C2m
β2
The velocity triangle can be plotted in the new situation, as shown in figure
⋅ C decrease and that
4.10. The figure shows that the velocities C2U and CU
will
H = 2 2 2U
[m]
W2 will increase. The head will then decrease according
g to equation (4.21).
The power which is proportional to the flow multiplied by the head will
⋅ ⋅see
decrease correspondingly. The head at zero Fflow,
[N] (4.20), is
=m
v formula
2
proportional to U2 and is therefore not changed in this case. Figure 4.11
⋅the
shows a sketch of the pump curves before and∆after
I =m
⋅ vchange.
= ρ ⋅ A ⋅ v 2 [N]
68
(4.20)
H
(4.21)
(4.22)
(4.23)
 D 2 ⋅b  
Q B = Q A ⋅  B2 B  
 DA ⋅ bA  
2

 DB   Geometric
HB= HA ⋅   
 DA   scaling
 D 4 ⋅b  
PB = PA ⋅  B4 B  
 DA ⋅ bA  
(4.23)
[m]
(4.21)
W
Q
Cm
Figure 4.11: Change of head curve as
consequence of changed b2.
CU
CB
WB
WA
β2
UB
UB C m,B Cu,B
=
=
UA C m,A Cu,A
C2U,B
Figure 4.10: Velocity triangle at changed
outlet width b2.
[N]
I = Fis changed,
Similar analysis can be made when the blade ∆form
see section
U 2 ⋅ C 2U
2
4.3, and
4.5.(4.20)
H by
= scaling of both
[m] speed and geometry, seeUsection
U2
⋅Q
H= 2 −
g
g
π ⋅ D2 ⋅ b2 ⋅ g ⋅ tan( β2 )
4.5 Affinity rules
⋅ ⋅v
(4.21) changes
[N] affinity rules, the consequences of certain
F =m
By means
of the so-called
in the pump geometry and speed can be predicted with much
precision.
(4.22)
n  
⋅ ⋅ v = ρ ⋅ A ⋅ v 2 [N]
∆ I = are
m
⋅  B  (4.22)
Q B =the
Q Avelocity
The rules
all derived under the condition that
triangles
are
 nA  
geometrically
before and after the change. In the formulas
below,
2 (4.23)
N]
∆I = F [similar
Scaling of
nB  and
derived in section 4.5.1, index A refers to the original
geometry
index
to
=
⋅
H
H
A 
B
2
nA   rotationalB speed

U
U

2
2
the scaled
(4.21)
−
⋅ Q [m]
H = geometry.
3
U
g
π ⋅ D2 ⋅ b2 ⋅ g ⋅ tan( β2 )
 nB  
PB = PA ⋅  
 nA 
(4.23)
 D 2 ⋅b  
Q B = Q A ⋅  B2 B  
 DA ⋅ bA  
2

 DB   Geometric
C2m,B
C2U
U
(4.22)
n  
QB = QA ⋅  B  
 nA  
2
n   Scaling of
HB = HA ⋅  B  
nA   rotational speed
3
n  
PB = PA ⋅  B  
 nA 
C2,B
W2,B
UA
Cm,B
Cm,A
CU,B
CA
CU,A
(4.24)
68
Figure 4.12 shows an example of the changed head and power curves for a
pump where the impeller diameter is machined to different radii in order to
match different motor sizes at the same speed. The curves are shown based
on formula (4.26).
η [%]
H [m]
ø260 mm
ø247 mm
20
ø234 mm
ø221 mm
16
80
70
12
60
50
8
40
30
20
4
10
4
8
12
16
20
24 28
32
36
40 Q (m 3/h)
P2 [kW]
3
2,5
2
1,5
1
0,5
69
Figure 4.12: Examples of curves for
machined impellers at the same speed but
different radii.
69
2,B
=UQ2 A−
⋅    U2
B
(4.21)
⋅ Q [m]
HQ=
n
g  πA⋅D2⋅ b2 ⋅ g ⋅ tan( β2 )
2
nB   Scaling of W
C2
=
⋅
H
H
2
 
A 
4. PumpBtheory
nA   rotational speed
C2m
C2m,B
3
(4.22)β 2
 nB  
PBB = Q
PAA⋅   
Q
C2U,B
C2U
U
nA  

4.5.1 Derivation ofthe
affinity
rules
2
 Scaling
The affinity method
preciseofwhen adjusting the speed up and down
nisvery
(4.23)
⋅ 2 ⋅bB   
HB = HA D
rotational
BnA B
and when
using
geometrical
scaling speed
in all directions (3D scaling). The affiniQ B = Q A ⋅  2  
⋅
D
b
3 

A
A
ty rules can also be nused
 when wanting to change outlet width and outlet
PB =scaling).
PA D
⋅ 2B 
diameter (2D
 B n A  Geometric
HB= HA ⋅   
 DA   scaling
2
When the velocity
are(4.23)
similar, then the relation between the
 DBB4 ⋅triangles
bB  


=
⋅
Q
Q

⋅
P
P
B
A
2

B
A
4
corresponding sides
inthe velocity triangles is the same before and after
 DAA ⋅ bA  
a change of all components,
see figure 4.13. The velocities hereby relate to
2

 DB   Geometric
=
⋅
H
H
each other
   scaling
B as:
A
 DA  
UB C m,B
(4.24)
 D4 C
⋅bu,B 
⋅  =B4 C B  
PBU= P=A C
m,AD ⋅ bu,A 
A
 A A
W
UB nB ⋅ D2,B
(4.25)
= velocity is expressed by the speed n and the impeller’s outer
The tangential
U nA ⋅ D2,A
C expression
C
(4.24)
diameterUDAB 2. The
above for the relation between components
= m,B = u,B
m
UA after
C m,AtheCchange
before and
of the impeller diameter can be inserted:
u,A
C
C
UB
UA
=
U
nB ⋅ D2,B
(4.25)
nA ⋅ D2,A
CU
Q = A 2 ⋅ C2m= π ⋅ D2 ⋅ b2 ⋅ C2m
(4.26)
2
Q B π ⋅ D2,B ⋅ b2,B C2m,B π ⋅D2,B ⋅ b2,B ⋅ nB ⋅ D2,B  D2,B  b2,B nB
 ⋅
=
⋅
=
=
⋅
Q A π ⋅ D2,A ⋅ b2,A C2m,A π ⋅D2,A⋅ b2,A⋅ nA⋅ D2,A  D2,A  b2,A nA
Q = A 2 ⋅ C2m= π ⋅ D2 ⋅ b2 ⋅ C2m
(4.26)
2
70
Q πU⋅ D⋅ C⋅2bU,A
π ⋅D2,B ⋅ b2,B ⋅ nB ⋅ D2,B  D2,B  b2,B nB
C
2,B
H B== 2,A 2,B
 ⋅
⋅ 2m,B =
=
⋅ (4.27)
g
Q A π ⋅ D2,A ⋅ b2,A C2m,A π ⋅D2,A⋅ b2,A⋅ nA⋅ D2,A  D2,ACB b2,A nA
2
2
HB U2,B ⋅ C2U,B ⋅ g U2,B ⋅ C2U,BWB nB⋅ D2,BW
⋅ nB ⋅ D2,B D2,B  nB  C
A
 ⋅   A
=
=
=
= 
m,B
HA U2,A ⋅ C2U,A ⋅ g U2,A ⋅ C2U,A nA⋅ D2,A⋅βnA ⋅ DC2,A
D2,ACm,A
 nA 
2
U ⋅C
UB
UA
CU,B
CU,A
H = 2,A g 2U,A
(4.27)
(4.28)
P = Q ⋅ ρ ⋅ U 2 ⋅ C2 U
2
2
HB U2,B ⋅ C2U,B ⋅ g U2,B ⋅ C2U,B nB⋅ D2,B ⋅ nB ⋅ D2,B D2,B  nB 
 ⋅ 4 
=
=
=
=
3
U ⋅nC⋅ D ⋅ nQ⋅ D H D D   nbA2,B nB 
⋅ ρ⋅⋅CU22,B
H
g2U,BU2,AQ⋅ CB 2U,A
PBA QUB2,A
U,A ⋅⋅ C

=
=
⋅ 2,B A 2U,B2,A= A B ⋅2,A B = 2,A2,B
⋅ 
PA QA ⋅ ρ ⋅ U2,A ⋅ C2U,A Q A U2,A ⋅ C2U,A Q A HA  D2,A  b2,A nA 
Figure 4.13: Velocity triangle
at scaled pump.
70
UB
UB =
U
A=
UA
C m,B
 DA
=
C
m,B
C m,A
=
C m,A
⋅ bu,B

C
A
C
u,B
Cu,A
Cu,A
(4.24)
(4.24)
UB n
⋅ D2,BCu,B
CBm,B
(4.24)
(4.25)
UB == nB ⋅ D=2,B
(4.25)
UA = n
CAm,A
⋅ D2,ACu,A
UA nA ⋅ D2,A
Neglecting inlet rotation, the changes in flow, head and power consumption
UB nB ⋅ D2,B
(4.25)
can be expressed
as follows:
=
UA nA ⋅ D2,A
Flow:
Q = A 2 ⋅ C2m= π ⋅ D2 ⋅ b2 ⋅ C2m
(4.26)
Q = A 2 ⋅ C2m= π ⋅ D2 ⋅ b2 ⋅ C2m
(4.26)
2
Q B π ⋅ D2,B ⋅ b2,B C2m,B π ⋅D2,B ⋅ b2,B ⋅ nB ⋅ D2,B  D2,B 2 b2,B nB
Q = π ⋅ D2,B ⋅ b2,B ⋅ C2m,B = π ⋅D2,B ⋅ b2,B ⋅ nB ⋅ D2,B =  D2,B  ⋅ b2,B ⋅ nB
⋅ D2,A ⋅ b2,A ⋅ C2m,A = π ⋅D2,A⋅ b2,A⋅ nA⋅ D2,A =  D2,A  ⋅ b2,A ⋅ nA
Q BA = π
π ⋅ DC2 ⋅ b2 ⋅ C2m
Q
2m⋅=
π ⋅D2,A⋅ b2,A⋅ nA⋅ D2,A  D2,A  b2,A (4.26)
nA
Q A= Aπ2⋅⋅DC2,A
b2,A
2m,A
2
Q B π ⋅ D2,B ⋅ b2,B C2m,B π ⋅D2,B ⋅ b2,B ⋅ nB ⋅ D2,B  D2,B  b2,B nB
 ⋅
=
⋅
=
=
⋅
Head: Q A π ⋅ D2,A ⋅ b2,A C2m,A π ⋅D2,A⋅ b2,A⋅ nA⋅ D2,A  D2,A  b2,A nA
U2,A ⋅ C 2U,A
H = U ⋅gC
(4.27)
H = 2,A g 2U,A
(4.27)
2
2
HB U2,B ⋅ C2U,B ⋅ g U2,B ⋅ C2U,B nB⋅ D2,B ⋅ nB ⋅ D2,B D2,B 2 nB 2



=
=
=
=
⋅
⋅ U,A⋅ g U2,B ⋅ C2U,B nB⋅ D2,B ⋅ nB ⋅ D2,B D2,B  nB 
HB= UU
2,A
2,B ⋅ C22U,B
H

A = U 2,A ⋅gC2U,A ⋅ g = U2,A ⋅ C2U,A = nA⋅ D2,A ⋅ nA ⋅ D2,A = 
D2,A  ⋅ nA(4.27)
HA U2,A ⋅ C2U,A ⋅ g U2,A ⋅ C2U,A nA⋅ D2,A⋅ nA ⋅ D2,A D2,A  nA 
2
2
HB U2,B ⋅ C2U,B ⋅ g U2,B ⋅ C2U,B nB⋅ D2,B ⋅ nB ⋅ D2,B D2,B  nB 
 ⋅  
=
=
=
= 
ρ ⋅⋅ U
PHA= QU⋅ 2,A
C 2 ⋅ C⋅2gU U2,A ⋅ C2U,A nA⋅ D2,A⋅ nA ⋅ D2,A D2,A  nA(4.28)

(4.28)
⋅ C2 U
P = Q ⋅ ρ ⋅ U22:U,A
Power consumption
4
3
PB QB ⋅ ρ ⋅ U2,B ⋅ C2U,B Q B U2,B ⋅ C2U,B Q B HB  D2,B 4 b2,B nB 3
⋅
= Q ⋅ H =  D  b ⋅ n 
QB ⋅⋅ ρ
U2,B ⋅⋅ C
C2U,B = Q
PB = Q
B U 2,B ⋅ C2U,B
B
B
ρ⋅⋅⋅UU
⋅ nBA 
2,A
2U,A = Q A ⋅ U 2,A ⋅ C2U,A = Q A ⋅ HA = 
2,A 
2,A
A== Q A
 D2,B
 b2,B
(4.28)
⋅
ρ
⋅
PP
C
PA QA ⋅ ρ ⋅ U22,A ⋅ 2CU2U,A Q A U2,A ⋅ C2U,A Q A HA  D2,A  b2,A nA 
4
3
1
PB QB ⋅ ρQ⋅ U
Q B U2,B ⋅ C2U,B Q B HB  D2,B  b2,B nB 
2 ⋅ C2U,B
d12,B


=
=
⋅
=
⋅
=
⋅
nq = nd ⋅ Q d3 2
(4.29) 
Q A U2,A ⋅ C2U,A Q A HA  D2,A  b2,A
4 ⋅ C2U,A
2,A
nPAq = nQdA⋅⋅ ρH⋅ U
(4.29)nA 
d3
4
Hd
nq = nd ⋅
71
Qd
Hd
1
3
2
4
(4.29)
71
U1
4. Pump theory
4.6 Inlet rotation
Inlet rotation means that the fluid is rotating before it enters the impeller.
The fluid can rotate in two ways: either the same way as the impeller
(co-rotation) or against the impeller (counter-rotation). Inlet rotation occurs
as a consequence of a number of different factors, and a differentation
between desired and undesired inlet rotation is made. In some cases inlet
rotation can be used for correction of head and power consumption.
In multi-stage pumps the fluid still rotates when it flows out of the
guide vanes in the previous stage. The impeller itself can create an inlet
rotation because the fluid transfers the impeller’s rotation back into the inlet
through viscous effects. In practise, you can try to avoid that the impeller
itself creates inlet rotation by placing blades in the inlet. Figure 4.14 shows
how inlet rotation affects the velocity triangle in the pump inlet.
According to Euler’s pump equation, inlet rotation corresponds to C1U being
different from zero, see figure 4.14. A change of C1U and then also a change
in inlet rotation results in a change in head and hydraulic power. Co-rotation
results in smaller head and counter-rotation results in a larger head. It is
important to notice that this is not a loss mechanism.
72
W1
C1
W1
C1
W1
b1
U1
b1
b1
C1
a1
a1
a1
C1U
C1U
No inlet rotation
Counter rotation
Co-rotation
Figure 4.14: Inlet velocity triangle at constant
flow and different inlet rotation situations.
72
4.7 Slip
In the derivation of Euler’s pump equation it is assumed that the flow follows the blade. In reality this is, however, not the case because the flow
angle usually is smaller than the blade angle. This condition is called slip.
Nevertheless, there is close connection between the flow angle and blade
angle. An impeller has an endless number of blades which are extremely
thin, then the flow lines will have the same shape as the blades. When the
flow angle and blade angle are identical, then the flow is blade congruent,
see figure 4.15.
The flow will not follow the shape of the blades completely in a real impeller with a limited number of blades with finite thickness. The tangential
velocity out of the impeller as well as the head is reduced due to this.
Suction side
C2
Pressure side
C'2
�
U2
Figure 4.15:
ω
Blade congruent flow line: Dashed line.
Actual flow line: Solid line.
When designing impellers, you have to include the difference between flow
angle and blade angle. This is done by including empirical slip factors in the
calculation of the velocity triangles, see figure 4.16. Empirical slip factors
are not further discussed in this book.
It is important to emphasize that slip is not a loss mechanism but just an
expression of the flow not following the blade.
C2
C'2
C2m
W2
W'2
β'2 β2
U2
W'2
W2
β'2
C2
C'2
β2
U2
ω
73
Figure 4.16: Velocity triangles where ‘ indicates the velocity with slip.
73
Q A π ⋅ D2,A ⋅ b2,A
C2m,A
π ⋅D2,A⋅ b2,A⋅ nA⋅ D2,A
 D2,A 
b2,A
nA
4. Pump theory
H=
U2,A ⋅ C 2U,A
g
(4.27)
2
2
4.8 Specific
a pump
HB speed
U2,B ⋅of
C2U,B
⋅ g U2,B ⋅ C2U,B nB⋅ D2,B ⋅ nB ⋅ D2,B D2,B  nB 



= chapter 1, =pumps are =classified in many
=
As described
 ⋅ n ways for
HA inU2,A
⋅ C2U,A ⋅ g U2,A ⋅ C2U,A nA⋅ D2,A⋅ nA ⋅ D2,A Ddifferent
2,A 
 A
example by usage or flange size. Seen from a fluid mechanical point of view,
this is, however, not very practical because it makes it almost impossible to
compare Ppumps
designed and used differently.
(4.28)
= Q ⋅ ρwhich
⋅ U2 ⋅ Care
2U
4
3
A model number,
(n2,Bq),⋅ is
classify
n 
 b2,B pumps.
C2therefore
⋅ U2,Bspecific
⋅ C2U,B speed
Q B used
HB toD2,B
PB QB ⋅ ρthe
Q U
U,B

= is given
= B ⋅ units.
=Europe
⋅ the
= following
⋅  B  is
Specific speed
in
different
In
form
PA QA ⋅ ρ ⋅ U2,A ⋅ C2U,A Q A U2,A ⋅ C2U,A Q A HA  D2,A  b2,A nA 
commonly used:
nq = nd ⋅
Qd
Hd
1
3
2
4
(4.29)
Where
nd = rotational speed in the design point [min-1]
Qd = Flow at the design point [m3/s]
Hd = Head at the design point [m]
The expression for nq can be derived from equation (4.22) and (4.23) as the
speed which yields a head of 1 m at a flow of 1 m3/s.
The impeller and the shape of the pump curves can be predicted based on the
specific speed, see figur 4.17.
Pumps with low specific speed, so-called low nq pumps, have a radial outlet with large outlet diameter compared to inlet diameter. The head curves
are relatively flat, and the power curve has a positive slope in the entire flow
area.
On the contrary, pumps with high specific speed, so-called high nq pumps,
have an increasingly axial outlet, with small outlet diameter compared to the
width. Head curves are typically descending and have a tendency to create
saddle points. Performance curves decreases when flow increases. Different
pump sizes and pump types have different maximum efficiency.
74
74
Impeller shape
nq
Outlet velocity
triangle
15
d1
0
d2
30
U2
d2/d1 = 2.0 - 1.5
50
W2
d1
C2
U2
C2U
H
0
100
H %
Hd
Pd
C2
C2U
d2/d1 = 1.2 - 1.1
100
Pd
H
C2
W2
110
U2
W2
C2U
C2
U2
C2U
100 P
%
70 Pd
60
0
d2
155 Q/Qd
100
H %
Hd
U2
100 P
%
80 Pd
H
0
d1
165 Q/Qd
55
d2
90
110 P
%
100 Pd
Pd
100
d2/d1 = 1.5 - 1.3
W2
170 Q/Qd
70
C2U
d2
d1 = d2
100
100
C2
%
H
H %
Hd
d1
100 Pd
80
C2U
W2
130 P
Pd
100
C2
U2
d2/d1 = 3.5 - 2.0
d1
P
H %
Hd
d2
W2
Figure 4.17: Impeller shape, outlet velocity
triangle and performance curve as function
of specific speed nq.
Performance curves
100
H %
Hd
100
Pd
H
140 Q/Qd
100 P
%
65 Pd
45
0
100 130 Q/Qd
4.9 Summary
In this chapter we have described the basic physical conditions which are the
basis of any pump design. Euler’s pump equation has been desribed, and we
have shown examples of how the pump equation can be used to predict a
pump’s performance. Furthermore, we have derived the affinity equations
and shown how the affinity rules can be used for scaling pump performance.
Finally, we have introduced the concept of specific speed and shown how
different pumps can be differentiated on the basis of this.
75
75
Chapter 5
Pump losses
5.1 Loss types
5.2 Mechanical losses
5.3 Hydraulic losses
5.4 Loss distribution as function of specific speed
5.5 Summary
5. Pump losses
5. Pump losses
As described in chapter 4, Euler’s pump equation provides a simple, lossfree description of the impeller performance. In reality, because of a number
of mechanical and hydraulic losses in impeller and pump casing, the pump
performance is lower than predicted by the Euler pump equation. The losses
cause smaller head than the theoretical and higher power consumption, see
figures 5.1 and 5.2. The result is a reduction in efficiency. In this chapter we
describe the different types of losses and introduce some simple models for
calculating the magnitude of the losses. The models can also be used for
analysis of the test results, see appendix B.
H
Euler head
Recirculation losses
Leakage
Flow friction
Incidence
Pump curve
H
Q
Figure 5.1: Reduction of theoretical Euler
head due to losses.
5.1 Loss types
Distinction is made between two primary types of losses: mechanical losses
and hydraulic losses which can be divided into a number of subgroups. Table
5.1 shows how the different types of loss affect flow (Q), head (H) and power
consumption (P2).
Q
P
Shaft power P2
Mechanical losses
Disk friction
Hydraulic losses
Hydraulic power Phyd
P
Loss
Mechanical
losses
Hydraulic
losses
Smaller
flow (Q)
Lower head (H)
Higher power
consumption (P2)
Bearing
X
Shaft seal
X
Flow friction
X
Mixing
X
Recirculation
X
Incidence
X
Disk friction
Leakage
Q
X
X
Chart 5. 1: Losses in pumps and their influence on the pump curves.
Q
Figure 5.2: Increase in power consumption
due to losses.
Pump performance curves can be predicted by means of theoretical or empirical calculation models for each single type of loss. Accordance with the
actual performance curves depends on the models’ degree of detail and to
what extent they describe the actual pump type.
78
78
Figure 5.3 shows the components in the pump which cause mechanical and
hydraulic losses. It involves bearings, shaft seal, front and rear cavity seal, inlet, impeller and volute casing or return channel. Throughout the rest of the
chapter this figure is used for illustrating where each type of loss occurs.
Volute
Diffuser
Inner impeller surface
Outer impeller surface
Front cavity seal
Inlet
Bearings and shaft seal
Figure 5.3: Loss causing
components.
79
79
5. Pump losses
5.2 Mechanical losses
The pump coupling or drive consists of bearings, shaft seals, gear, depending
on pump type. These components all cause mechanical friction loss. The
following deals with losses in the bearings and shaft seals.
5.2.1 Bearing loss and shaft seal loss
Bearing and shaft seal losses - also called parasitic losses - are caused by
friction. They are often modelled as a constant which is added to the power
consumption. The size of the losses can, however, vary with pressure and
rotational speed.
The following model estimates the increased power demand due to losses
in bearings and shaft seal:
Ploss, mechanical = Ploss, bearing + Ploss, shaft seal = constant
(5.1)
where
V2
= ζ ⋅ H dyn, inpower
= ζ ⋅demand because of mechanical (5.2)
H loss, friktion
Ploss, mechanical
= Increased
loss [W]
2g
Ploss, bearing
= Power lost in bearings [W]
LV 2lost in shaft seal [W]
Ploss, shaftH
= Power
seal
(5.3)
loss, pipe = f
Dh 2 g
(5.4)
Dh = 4 A
5.3 Hydraulic O
losses
VD harise on the fluid path through the pump. The losses occur
Hydraulic
(5.5)
Re losses
=
ν
because of friction
or because the fluid must change direction and velocity
64 the pump. This is due to cross-section changes and the
on its path
(5.6)
flaminarthrough
=
passage throughRethe rotating impeller. The following sections describe the
individual hydraulic losses depending on how they arise.
Q
(10/3600) m3 s
Mean velocity: V =
=
= 3.45 m s
π
A
0.032 2 m2
4
Reynolds number: Re =
80
3.45m s ⋅ 0.032m
VD h
= 110500
=
ν
1 ⋅ 10 −6 m 2 s
Relative roughness: k/D =
0.15mm
= 0.0047
80
5.3.1 Flow friction
Flow friction occurs where the fluid is in contact with the rotating impeller surfaces and the interior surfaces in the pump casing. The flow friction
causes a pressure loss which reduces the head. The magnitude of the friction
loss depends on the roughness of the surface and the fluid velocity relative
to the surface.
Model
Flow friction occurs in all the hydraulic components which the fluid flows
through. The flow friction is typically calculated individually like a pipe friction loss, this means as a pressure loss coefficient multiplied with the dy(5.1)
Ploss, mechanical = Ploss, bearing + Ploss, shaft seal = constant
namic head into the component:
H loss, friktion = ζ ⋅ H dyn, in = ζ ⋅
V2
2g
(5.2)
LV 2
H loss, pipe = f
(5.3)
Dh 2 g
= Dimensionless
loss coefficient [-]
4A
head into the component [m]
(5.4)
D=h =Dynamic
O
= Flow velocity into the component [m/s]
VD h
(5.5)
Re =
ν
64
(5.6) 5.4.
flaminar
= thus grows quadratically with the flow velocity, see figure
The friction
loss
Re
where
ζ
Hdyn, in
V
Loss coefficients can be found e.g. in (MacDonald,
1997). Single components
Q
(10/3600) m3 s
such asMean
inlet and
outerVsleeve
affected
velocity:
=
=which are not directly
m s by the impeller
= 3.45
π
A
2
0.032 2 m
can typically be modelled with a 4constant
loss coefficient. Impeller, volute
housing and return channel will on the contrary typically have a variable loss
3.45m
s ⋅ 0.032m
VD hin the
coefficient.
When
the flow
impeller
is calculated,
the relative
= 110500
Reynolds
number:
Refriction
=
=
2
−6
ν
1
⋅
10
m
s
velocity must be used in equation (5.2).
Relative roughness: k/Dh =
81
LV 2
Hloss,friktion
V
Figure 5.4: Friction loss as function of
the flow velocity.
0.15mm
= 0.0047
32mm
2m ⋅ ( 3.45 m s) 2
81
5. Pump losses
Friction loss in pipes
Pipe friction is the loss of energy which occurs in a pipe with flowing fluid. At
the wall, the fluid velocity is zero whereas it attains a maximum value at the
pipe center. Due to these velocity differences across the pipe, see figure 5.5,
the fluid molecules rub against each other. This transforms kinetic energy to
heat energy which can be considered as lost.
V
To maintain a flow in the pipe, an amount of energy corresponding to the
energy which is lost must constantly be added. Energy is supplied by static
pressure difference from inlet to outlet. It is said that it is the pressure difference which drives the fluid through the pipe.
Figure 5.5: Velocity profile in pipe.
(5.1)
Ploss, mechanical = Ploss, bearing + Ploss, shaft seal = constant
The loss in the pipe depends on the fluid velocity, the hydraulic diameter
of the pipe, lenght and inner surface
V 2 roughness. The head loss is calculated
H loss, friktion = ζ ⋅ H dyn, in = ζ ⋅
(5.2)
2g
as:
H loss, pipe = f
LV 2
Dh 2 g
(5.3)
where
(5.4)
Dh = 4 A
O loss [m]
Hloss, pipe = Head
VD h coefficient [-]
f
= Friction
(5.5)
Re
=
L
= Pipeνlength [m]
64 velocity in the pipe [m/s]
V
=
Average
(5.1)
= Ploss, bearing + Ploss, shaft seal = constant
(5.6)
fPlaminar
=
loss, mechanical
Re diameter [m]
Dh
= Hydraulic
V2
= ζ ⋅ H dyn,
ζ ⋅ of the cross-sectional
H loss, friktion
(5.2)wetted
inQ =ratio
The hydraulic
diameter
is the
area to the
(10/3600)
m3 s
2g
Mean velocity: V =
=
= 3.45 m s
π
circumference. The hydraulic
is suitable for calculating the friction
A diameter
0.032 2 m2
LV 2
4
for cross-sections
H loss, pipe = fof arbitrary form.
(5.3)
Dh 2 g
3.45m s ⋅ 0.032m
VD h
= 110500
Reynolds
number: Re =
=
4
A
ν
(5.4)
Dh =
1 ⋅ 10 −6 m 2 s
O
VD h
0.15mm
where Relative
(5.5)
Re =
roughness: k/Dh =
= 0.0047
ν
32mm
A = The cross-section area of the pipe
[m2]
64
O = Thef wetted
circumference
of the pipe [m]
(5.6)
laminar =
2
Re
2m ⋅ ( 3.45 m s) 2
LV
= 0.031
= 1.2 m
Pipe loss: H loss, pipe = f
2
D h 2g
0.032m ⋅ 2 ⋅ 9.81 m s
Q
(10/3600) m3 s
=
= 3.45 m s
π 2 2 2
A
0.032 m
V
= ζ⋅H
= ζ 4⋅ 1
(5.7)
Mean velocity: V =
82
H
(5.8)
82
Equation (5.4) applies in general for all cross-sectional shapes. In cases where
(5.1) to the
= Ploss,
= constant
mechanical
bearing + Ploss, shaft
the pipePloss,
has
a circular
cross-section,
thesealhydraulic
diameter is equal
pipe diameter. The circular pipe is 2the cross-section type which has the
smallest
possible
interior surface V
compared to the cross-section(5.2)
area and
H loss,
friktion = ζ ⋅ H dyn, in = ζ ⋅
2g
therefore the smallest flow resistance.
LV 2
H loss, pipe = f
(5.3)
Dh 2 is
g not constant but depends on whether the flow is
The friction coefficient
laminar or turbulent.
This is described by the Reynold’s number, Re:
(5.4)
Dh = 4 A
O
VD h
(5.5)
Re =
ν
64
(5.6)
flaminar =
Re
where
83
n = Kinematic viscosity of the fluid [m2/s]
Q
(10/3600) m3 s
Mean velocity: V =
=
= 3.45 m s
π
The Reynold’s number is aAdimensionless
which expresses the re0.032 2 number
m2
4
lation between inertia and friction forces in the fluid, and it is therefore a
number that describes how turbulent
the
flow
The following guidelines
3.45m
s is.
⋅ 0.032m
VD h
= 110500
Reynolds number: Re =
=
−6
ν
apply for flows in pipes:
1 ⋅ 10 m 2 s
(5.1)
Ploss, mechanical = Ploss, bearing + Ploss, shaft seal = constant
0.15mm
Relative
roughness:: k/D
= 0.0047
Re < 2300
Laminar
flow
h =
2
V 32mm
=
ζ
⋅
=
ζ
⋅
H2300
H
(5.2)
<
Re
<
500
:
Transition
zone
loss, friktion
dyn, in
2g
Re > 5000
: Turbulent
flow.
2
2m ⋅ ( 3.45 m s) 2
2
= f LV = 0.031
= 1.2 m
Pipe loss: H loss,
LVpipe
2
D h 2g
H loss, pipe = f
0.032m ⋅ 2 ⋅ 9.81 m s (5.3)
D
2
g
h
Laminar flow only occurs
at relatively low velocities and describes a calm,
well-ordered4flow
without
eddies. The friction coefficient for laminar flow is
(5.4)
Dh = A
O the surface roughness
V2
independent of
Reynold’s
= ζ ⋅ H dyn,1 = ζ ⋅ 1 and is only a function of the(5.8)
H loss, expansion
VD
2 g with circular cross-section:(5.5)
h
number.ReThe
following
applies
for
pipes
=
2
ν

A 
ζ = 1 − 641 
(5.9)
(5.6)
flaminar = A2 
Re
2
 A 
V2
(5.10)
H loss, contraction = 1 − 0  ⋅ 0
AQ2  (10/3600)
2g
m3 s

Mean velocity: V =
=
= 3.45 m s
π
A
2
2
0.032
m
2
4V
H loss, contraction = ζ ⋅ H dyn,2 = ζ ⋅ 2
(5.11)
2g
3.45m s ⋅ 0.032m
VD h
= 110500
Reynolds number: Re =
=
2
νw 1 − w 1, kanal
1 ⋅ 10 −6 m 2 s
ws2
=ϕ
H loss, incidence = ϕ
(5.12)
(5.7)
83
5. Pump losses
Turbulent flow is an unstable flow with strong mixing. Due to eddy motion
most pipe flows are in practise turbulent. The friction coefficient for turbulent flow depends on the Reynold’s number and the pipe roughness.
Figure 5.6 shows a Moody chart which shows the friction coefficient f as
function of Reynold’s number and surface roughness for laminar and turbulent flows.
Figure 5.6: Moody chart:
Friction coefficient for laminar (circular
cross-section) and turbulent flow (arbitrary
cross-section). The red line refers to the
values in example 5.1.
0.1
0.09
0.08
0.05
inar
0.06
Lam
0.07
0.04
0.03
0.015
h
0.04
0.01
0.008
0.006
0.03
0.004
0.025
0.002
n zo
sitio
0.02
Relative roughness ( k/D )
0.02
64
Re
Tran
Friction coefficient
(f)
0.05
ne
0.001
0.0008
0.0006
Sm
oo
th
0.015
0.0004
pip
e
0.0002
0.0001
Turbulent
0.00005
0.01
0.009
0.000001
0.008
10
84
3
10
4
10
5
Reynolds number ( Re=V · Dh /ν )
10
6
10
0.000005
7
0.00001
10
8
84
Table 5.2 shows the roughness for different materials. The friction increases
in old pipes because of corrosion and sediments.
Materials
PVC
Ploss, mechanical
= Ploss, bearing + Ploss, shaft seal = constant
Pipe in aluminium, copper og brass
Table 5.2: Roughness for different
surfaces (Pumpeståbi, 2000).
Roughness k [mm]
0.01-0.05
(5.1)
0-0.003
0.01-0.05
V2
(5.2)
2g
Welded steel pipe, new
0.03-0.15
Ploss, mechanical
= Ploss, bearing
+ Ploss, shaft seal = constant 0.15-0.30 (5.1)
Welded
steel
deposition
LV 2 pipe with
H loss, pipe = f
(5.3)
Galvanised
0.1-0.2
Dh 2steel
g pipe, new
2
V
Galvanised
with
0.5-1.0
= ζ pipe
⋅ H dyn,
= ζ⋅
HAloss, friktionsteel
(5.2)
in deposition
4
2
g
(5.4)
Dh =
O
LV 2 of pipe loss
VD
Example
5.1:
Calculation
h
(5.5) (5.3)
pipe = f
Re = H loss,
Dh 2 in
g a 2 meter pipe with the diameter d=32 mm and a
Calculate
ν the pipe loss
64 m3/h. The pipe is made of galvanized steel with a roughness of
flow of Q=10
(5.6) (5.4)
flaminar =Dh = 4 A
Re Othe fluid is water at 20°C.
0.15 mm, and
VD h
(5.5)
Re =
ν
Q
(10/3600) m3 s
Mean velocity: V64=
=
= 3.45 m s
π
A
(5.6)
flaminar =
0.032 2 m2
Re
4
Steel pipe
H loss, friktion = ζ ⋅ H dyn, in = ζ ⋅
3.45m s ⋅ 0.032m
VD h
= 110500
Q = (10/3600) −6m3 s2
ν
1 ⋅ 10 m s= 3.45 m s
Mean velocity: V =
=
π
A
0.032 2 m2
4
0.15mm
Relative roughness: k/Dh =
= 0.0047
32mm
3.45m s ⋅ 0.032m
VD h
= 110500
Reynolds number: Re =
=
ν
1 ⋅ 10 −6 m 2 2s
2
2m ⋅ ( 3.45 m s)
(5.7)
f LVthe= 0.031
= 1.2when
m
Pipe loss:
From
the H
Moody
friction
coefficent (f) is 0.031
Re =
loss, pipe =chart,
2
0.15mm
D h 2g
0.032m
2
9.81
m
s
⋅
⋅
Relative
roughness:
k/D
=
=
0.0047
h
110500 and the relative roughness
k/D
=0.0047. By inserting the values in
h
32mm
the equation (5.3), the pipe loss can be calculated to:
V 12 2
2m ⋅ ( 3.45 m s(5.8)
)2
ζ ⋅H
H loss, expansion
H dyn,1 = =ζ f⋅ LV
= 0.031
= 1.2 m
Pipe =loss:
loss, pipe
2
g
2
D h 2g
0.032m ⋅ 2 ⋅ 9.81 m s
2

A 
ζ = 1 − 1 
(5.9)
A2 

V 12
(5.8)
H loss, expansion = ζ ⋅ H
2 dyn,1 2= ζ ⋅
2g
 A 
V
(5.10)
H loss, contraction = 1 − 0  ⋅ 0
2
2g
 A A 2 
ζ = 1 − 1 
(5.9)
Reynolds number: Re =
85
(5.7)
85
5. Pump losses
Ploss, mechanical = Ploss, bearing + Ploss, shaft seal = constant
(5.1)
V2
5.3.2 Mixing loss at cross-section expansion
H loss, friktion = ζ ⋅ H dyn, in = ζ ⋅
(5.2)
2g pressure energy at cross-section exVelocity energy is transformed to static
2
pansions in the pump,
the energy equation in formula (2.10). The converLVsee
H loss, pipe = f
(5.3)
sion is associated with
Dh 2agmixing loss.
4A
(5.4)
Dh = is
The reason
O that velocity differences occur when the cross-section exVD h 5.7. The figure shows a diffuser with a sudden expansion
pands, see figure
(5.5)
Re =
beacuse all water
ν particles no longer move at the same speed, friction occurs
64
between the molecules
in the fluid which results in a diskharge head loss.
(5.6)
flaminar =
Even though theRe
velocity profile after the cross-section expansion gradually
is evened out, see figure 5.7, a part of the velocity energy is turned into heat
Q
energy instead of static pressure
energy. m3 s
(10/3600)
Mean velocity: V =
=
= 3.45 m s
π
A
0.032 2 m2
4 in the pump: At the outlet of the imMixing loss occurs at different places
peller where the fluid flows intoVD
the volute
casing
or return channel as well
3.45m
s ⋅ 0.032m
h
= 110500
Reynolds
number: Re =
=
as in the
diffuser.
−6
ν
1 ⋅ 10 m 2 s
A2
A1
A2
A1
H loss, expansion = ζ ⋅ H dyn,1 = ζ ⋅
V 12
2g
A2
A1
When designing the hydraulic components,
0.15mm it is important to create small
Relative roughness: k/Dh =
= 0.0047
and smooth cross-section expansions
as possible.
32mm
2
Model
2m ⋅ ( 3.45 m s) 2
= 1.2 m
Pipe loss: H loss, pipe = f LV = 0.031
2
D h 2 g is a function
The loss at a cross-section expansion
dynamic
0.032mof⋅ 2the
m s head into
⋅ 9.81
the component.
V2
V1
Figure 5.7: Mixing loss at cross-section
expansion shown for a sudden expansion.
(5.7)
(5.8)
2

A 
where
ζ = 1 − 1 
A2into
V1 = Fluid velocity

 the component [m/s]
(5.9)
2
 A 
V2
The pressure
loss coefficient
the com(5.10)
H loss, contraction
= 1 − 0ζ depends
⋅ 0 on the area relation between
g evenly the area expansion happens.
2  as2
 asAwell
ponent’s inlet and outlet
how
H loss, contraction = ζ ⋅ H dyn,2 = ζ ⋅
86
ws2
V 22
2g
w 1 − w 1, kanal
(5.11)
2
86
Pipe loss: H loss, pipe = f
2m ⋅ ( 3.45 m s) 2
LV 2 = 0.031
= 1.2 m
2
D h 2g
0.032m ⋅ 2 ⋅ 9.81 m s
(5.7)
V 12
(5.8)
=
ζ
⋅
=
ζ
⋅
H
H
loss, expansion
For a sudden
expansion, as dyn,1
shown in figure
2 g 5.7, the following expression is used:

A 
ζ = 1 − 1 
A
2 

2
(5.9)
2
where
 A 
V2
(5.10)
H loss, contraction = 1 − 0  ⋅ 2 0
A1= Cross-section areaat inlet
A 2  [m ]2 g
A2= Cross-section area at outlet [m2]
V 22
=
ζ
⋅
=
ζ
⋅
H
H
(5.11)
dyn,2
loss,
contraction
The model gives a good estimate of2 gthe head loss at large expansion
ratios
(A1/A2 close to zero). In this case the loss coefficient
is
ζ
=
1
in
equation
(5.9)
2
2
w
−
w
w
1
1, kanal
which means
that= almost
is
H loss, incidence
ϕ s the
= ϕ entire dynamic head into the component
(5.12)
2 ⋅g
2⋅g
lost in a sharp-edged diffuser.
(5.13) with
For small
well
H loss,expansion
⋅ (Q − Qas
)2 +as
k 2 for other diffuser geometries
incidence = k 1 ratios
design
smooth area expansions, the loss coefficient ζ is found by table lookup
(MacDonalds)
Ploss, disk or
= by
kρ measurements.
U 32 D2 ( D2 + 5e )
k = 7.3 ⋅ 10
−4
 2ν ⋅ 10 6 


 U 2 D2 
m
(5.14)
(n3D52 )A
5.3.3 Mixing
loss
contraction
(Ploss, disk
)A =at( Pcross-section
)
(5.15)
loss, disk B
(n3D52 )B
Head loss at cross-section contraction occurs as a consequence of eddies being
created Q
inimpeller
the flow
it comes close to the geometry edges, see
figure 5.8.
(5.16)
= Q when
+ Q leakage
It is said that the flow ’separates’. The
for this is that the flow because
2 reason
2
2 ( D2 − Dgap )
H stat, pressure
gap = H stat,
impeller − ω
fl longer adheres in parallel to the(5.17)
of the local
gradients
no
surface but
8g
instead will follow curved streamlines. This means that the effective cross2
V 2flow
L V 2 + 1.0 isVreduced.
section H
area
which the
It is said that a(5.18)
contraction
+ fexperiences
stat, gap = 0.5
2g
s 2g
2g
is made. The contraction with the area A0 is marked on figure 5.8. The contraction
accelerates the flow and it must therefore subsequently decelerate again to
(5.19)
fill the cross-section.
occurs in this process. Head loss as a
2gH stat, gapA mixing loss
V=
consequence off Lcross-section
contraction
occurs typically at inlet to a pipe
s + 1.5
and at the impeller eye. The magnitude of the loss can be considerably reduced
= inlet
Q leakagethe
VA gapedges and thereby suppress separation. If the inlet is
by rounding
adequately rounded off, the loss is insignificant. Losses related to cross-section
contraction is typically of minor importance.
87
Contraction
A1
A0
V1
A2
V0
V2
Figure 5.8: Loss at cross-section contraction.
87
Relative roughness: k/Dh =
32mm
= 0.0047
Q
(10/3600) m3 s
2
Mean
velocity:
V
=
=
= ⋅ 3.45
2m
( 3.45mmss) 2
5. Pump
0.0312 m2
= 1.2 m
Pipe losses
loss: H loss, pipe = Af LV π =0.032
2
D h 2 g4
0.032m ⋅ 2 ⋅ 9.81 m s
3.45m s ⋅ 0.032m
VD h
= 110500
Model Reynolds number: Re = ν =2
1 ⋅ 10 −6 m 2 s
V1
(5.8)from V
= ζ ⋅ that the acceleration of the fluid
H loss,
H dyn,1
Based on
experience,
assumed
expansion = ζit⋅ is
1
2g
0.15mm
to V0 is Relative
loss-free,
whereas
the
subsequent
mixing
loss
depends
on
the
area
roughness:
k/Dh =
= 0.0047
2

32mm
A1  to the contraction
ratio now
A0 as well as the dynamic head in the
1
ζ =compared
−
(5.9)

A2 
contraction:
2
2m ⋅ ( 3.45 m s)
2LV 2
= 1.2 m
Pipe loss: H loss, pipe
 =Af0 
V =2 0.031
2
H loss, contraction = 1 − D h 2⋅ g 0
0.032m ⋅ 2 ⋅ 9.81 m s (5.10)
2g
 A2 
(5.7)
(5.7)
where
V0
A0/A2
VV22 2
=
ζ
⋅
=
ζ
⋅
H
H
(5.11)
dyn,2
loss,
contraction
(5.8)
= ζ⋅H
= ζ ⋅ 2 g1 [m/s]
H=loss,
expansion
Fluid
velocity
indyn,1
contraction
2g
= Area ratio 2[-] 2
2
w 1 − w 1, kanal

A1  w s
ζ loss,
= incidence
H
(5.12)
(5.9)
1 − = ϕ 2 ⋅ g = ϕ
A2 
2⋅g

The disadvantage
of this model is that it assumes knowledge of A0 which is
2
not directly measureable.
is therefore
 The
following
AQ
V2 2 alternative formulation (5.13)
0
H loss,
=
k
⋅
(
Q
−
(5.10)
⋅ ) 0+ k 2
design
loss, incidence
contraction =1 1 −

often used:
A
2
g
2 

Ploss, disk = kρ U D2 ( D2 + 5eV
) 2
H loss, contraction = ζ ⋅ H dyn,26 =m ζ ⋅ 2
(5.11)
2g
− 4  2ν ⋅ 10 

(5.14)
k = 7.3 ⋅ 10 
2
where
 U22D2  w − w
w
1
1, kanal
= ϕ outs of=the
ϕ component [m]
H loss, incidencehead
(5.12)
Hdyn,2 = Dynamic
2 ⋅ g (n3D52 )A 2 ⋅ g
(Ploss,
( Ploss,of
)B component
(5.15)
V2 = Fluid
velocity
[m/s]
disk )A = out
diskthe
(n3D52 )B
2
(5.13)
H loss, incidence = k 1 ⋅ (Q − Q design ) + k 2
(5.16)
Q impeller = Q + Q leakage
Figure 5.9 compares loss
coefficients at sudden cross-section expansions
Ploss, disk == H
kρ U 32 D2 (−D2ω+2 5(eD)22 −D2gap )
(5.17)
H
stat, gap
stat, impeller
fl the area ratio A /A between the
and –contractions
as
functionmof
inlet and
1
2
8g
6


ν
⋅
2
10
−
4
outlet. As
and thereby also the head
loss, is in
 2loss coefficient,

(5.14)
k =shown,
7.3 ⋅ 10 the
2
U2 D2LV 2than in expansions.
V
V

generalHsmaller
at
contractions
This
applies
in
particular
0.5
+f
+ 1.0
(5.18)
stat, gap =
2g
s 2g
2g
at large area ratios.
3 5
(n D2 )A
(Ploss, disk )A = ( Ploss, disk )B
(5.15)
(n3D52(5.19)
)B
The head loss
coefficient
for geometries with smooth area changes can be
2gH
stat, gap
V=
Q impeller
=L Q+ 1.5
+ QAs
found by
table flookup.
mentioned earlier, the pressure loss in (5.16)
a cross-secleakage
s
2
2
tion contraction can be reduced2 to
zero by rounding off the edges.
( Dalmost
2 − Dgap )
(5.17)
H
Qstat,
gap = H
stat,gap
impeller − ω fl
VA
leakage
3
2
Pressure loss coefficient ζ
A1
A2
A1
AR = A2 /A1
1,0
A2
AR = A1 /A2
0,8
0,6
0,4
0,2
0
0
0,2
0,4
0,6
0,8
1,0
Area ratio
Hloss,contraction = ζ . Hdyn,2
Hloss,expansion = ζ . Hdyn,1
Figure 5.9: Head loss coefficents at sudden
cross-section contractions and expansions.
8g
2
2
2
Hstat, gap = 0.5 V + f L V + 1.0 V
2g
s 2g
2g
88
2gH stat, gap
(5.19)
(5.18)
88
5.3.4 Recirculation loss
Recirculation zones in the hydraulic components typically occur at part
load when the flow is below the design flow. Figure 5.10 shows an example
of recirculation in the impeller. The recirculation zones reduce the effective cross-section area which the flow experiences. High velocity gradients
occurs in the flow between the main flow which has high velocity and
the eddies which have a velocity close to zero. The result is a considerable
mixing loss.
Recirculation zones can occur in inlet, impeller, return channel or volute
casing. The extent of the zones depends on geometry and operating point.
When designing hydraulic components, it is important to minimise the size
of the recirculation zones in the primary operating points.
Recirculation zones
Figure 5.10: Example of recirculation in
impeller.
Model
There are no simple models to describe if recirculation zones occur and if so to
which extent. Only by means of advanced laser based velocity measurements
or time consuming computer simulations, it is possible to map the recirculation
zones in details. Recirculation is therefore generally only identified indirectly
through a performance measurement which shows lower head and/or higher
power consumption at partial load than predicted.
When designing pumps, the starting point is usually the nominal operating
point. Normally reciculation does not occur here and the pump performance
can therefore be predicted fairly precisely. In cases where the flow is below
the nominal operating point, one often has to use rule of thumb to predict the
pump curves.
89
89
Dh 2 g
Dh = 4 A
O
5. Pump losses
VD h
Re =
ν
64
flaminar =
Re
(5.4)
(5.5)
(5.6)
Q
(10/3600) m3 s
Mean velocity:
= 3.45 m s
5.3.5 Incidence
loss V = A = π
0.032 2 m2
4 a difference between the flow angle and
Incidence loss occurs when there is
blade angle at the impeller or guide
leading
edges. This is typically the
3.45m
s ⋅ 0.032m
VD h vane
= 110500
=
=exists.
−6
case at Reynolds
part loadnumber:
or when Re
prerotation
ν
1 ⋅ 10 m 2 s
0.15mm
A recirculation
occurs on
side
of the=blade
when there is difference
Relativezone
roughness:
k/Done
0.0047
h =
32mm
between the flow angle and the blade angle, see figure 5.11. The recirculation
zone causes a flow contraction after the blade leading edge.2 The flow must
2
2m ⋅ ( 3.45 m s)
LVcontraction
once again
after
= f the
= 0.031 to fill the entire blade
= channel
1.2 m
Pipedecelerate
loss: H loss, pipe
2
D h 2g
0.032m ⋅ 2 ⋅ 9.81 m s
and mixing loss occurs.
Figure 5.11: Incidence loss at inlet to impeller
(5.7) or guide vanes.
2 occur at the volute tongue. The deAt off-design flow, incidence losses V
also
(5.8)
H loss, expansion = ζ ⋅ H dyn,1 = ζ ⋅ 1
signer must therefore make sure that
2 gflow angles and blade angles match
each other so the incidence
loss is minimised. Rounding blade edges and vo2

A1 
1
ζ
=
−
lute casing tongue
can reduce the incidence loss.
(5.9)

A2 

2
 A 
V2
Model
(5.10)
H loss, contraction = 1 − 0  ⋅ 0
The magnitude of the incidence
A 2  loss
2 gdepends on the difference between relative velocities before and after the blade leading edge and is calculated using
V 22
the following
model=(Pfleiderer
1990, p 224):
ζ ⋅ H dyn,2 =og
ζ ⋅Petermann,
H loss, contraction
(5.11)
2g
W1
2
where H loss, incidence = k 1 ⋅ (Q − Q design ) + k 2
(5.13)
ϕ = Emperical value which is set to 0.5-0.7 depending on the size of the recirculation
blade leading edge.
Ploss, diskzone
= kρafter
U 32 Dthe
2 ( D 2 + 5e )
ws= difference between relative
m velocities before and after the blade edge
6
− 4  2ν ⋅ 10 
using

see figure 5.12.
(5.14)
k = vector
7.3 ⋅ 10calculation,
 U 2 D2 
90
Q impeller = Q + Q leakage
(n3D52 )A
(n3D52 )B
l
(5.12)
2
(Ploss, disk )A = ( Ploss, disk )B
na
,ka
w 1 − w 1, kanal
w2
H loss, incidence = ϕ s = ϕ
2 ⋅g
2⋅g
β1
β´1
W
1
Figure 5.12: Nomenclature for incidence loss
model.
(5.15)
(5.16)
90
2
 A 
V2
H loss, contraction = 1 − 0  ⋅ 0
2g
 A2 
(5.10)
V 22
(5.11)
2g
Incidence loss is alternatively modelled as a parabola
with minimum at the
2
2
w 1 − w 1, kanal
w
s
best efficiency
point.
loss increases quadratically with
= ϕThe incidence
=ϕ
H loss, incidence
(5.12)the dif2 ⋅ g flow and 2the
⋅ g actual flow, see figure 5.13.
ference between the design
H loss, contraction = ζ ⋅ H dyn,2 = ζ ⋅
H loss, incidence = k 1 ⋅ (Q − Q design )2 + k 2
where
Qdesign
k1
k2
P=loss,Design
U 32[m
D23/s]
( D 2 + 5e )
disk = kρ
flow
m
5 6 
ν ⋅ 10
= Constant [s22/m
]

k = 7.3 ⋅ 10 − 4 
= Constant [m]
 U 2 D2 
(Ploss, disk )A = ( Ploss, disk )B
(n3D52 )A
(n3D52 )B
(5.13)
k2
Qdesign
(5.14)
Q
Figure 5.13: Incidence loss as function of
the flow.
(5.15)
(5.16)
Q
= Q + Q leakage
5.3.6 Diskimpeller
friction
2
2
2 ( Dconsumption
2 − Dgap )
Disk friction
is the
power
which occurs on(5.17)
the shroud
H stat, gap
= Hincreased
stat, impeller − ω fl
8g
and hub of the impeller because it rotates in a fluid-filled pump casing. The
2
2
fluid in the cavity between
impeller
and 2pump casing starts to rotate and
Hstat, gap = 0.5 V + f L V + 1.0 V
(5.18)
2
g
s
2
g
2 g The rotation velocity equals the
creates a primary vortex, see section 1.2.5.
impeller’s at the surface of the impeller, while it is zero at the surface of
the pump casing. The average velocity
(5.19) of the primary vortex is therefore as2gH stat, gap
V =be equal to one half of the rotational velocity.
sumed to
f Ls + 1.5
The centrifugal
Q leakage =force
VA gapcreates a secondary vortex movement because of the
difference in rotation velocity between the fluid at the surfaces of the impeller and the fluid at the pump casing, see figure 5.14. The secondary vortex increases the disk friction because it transfers energy from the impeller surface
to the surface of the pump casing.
The size of the disk friction depends primarily on the speed, the impeller diameter as well as the dimensions of the pump housing in particular the distance between impeller and pump casing. The impeller and pump housing
surface roughness has, furthermore, a decisive importance for the size of the
disk friction. The disk friction is also increased if there are rises or dents on
the outer surface of the impeller e.g. balancing blocks or balancing holes.
91
Hloss, incidence
e
Secondary
vortex
Figure 5.14: Disk friction on impeller.
91
0.15mm
Relative roughness: k/Dh = 2
= 0.0047
V 232mm
H loss, contraction = ζ ⋅ H dyn,2 = ζ ⋅
2g
5. Pump losses
(5.11)
2
2 2m ⋅ ( 3.45 m s )
LV 2w =−0.031
w 1, kanal
= 1.2 m
Pipe loss: H loss, pipe
w s=2 f D 2 g
1
2
= ϕh
H loss, incidence = ϕ
0.032m ⋅ 2 ⋅ 9.81 m s (5.12)
2 ⋅g
2⋅g
(5.7)
Model
Pfleiderer
Petermann (1990, p.
2 322) use the following model to deter(5.13)
H loss,and
incidence = k 1 ⋅ ( Q − Q design ) +
V 1k2 2
mine the
increased
power
consumption
caused by disk friction: (5.8)
=
ζ
⋅
=
ζ
⋅
H loss,
H
expansion
dyn,1
2g
3
2 2 D 2 ( D 2 + 5e )
Ploss, disk = kρ U

A 
m
ζ = 1 − 1 
(5.9)
 2ν ⋅ 10 6 
A

(5.14)
k = 7.3 ⋅ 102− 4 
  U2 D2  2
A0
V0 2
(5.10)
H
1
=
−
⋅
where loss, contraction 

A 2 (n3D52 )2Ag

(Ploss, disk )diameter
)B
(5.15)
= ( Ploss, disk[m]
D2 = Impeller
A
(n3D52 )B
2
e = Axial distance to wall at the Vperiphery
of the impeller [m], see figure
H contraction = ζ ⋅ H dyn,2 = ζ ⋅ 2
(5.11)
(5.16)
=
Q
+
Q
5.14 Qloss,
2
g
impeller
leakage
U2 = Peripheral velocity [m/s] 2 ( D2 −D2 ) 2
2
gap
(5.17)
H stat, gap = H stat, impeller
w 2 −2 ω fl νw=10
-6 1, kanal
2
1−w
ν = Kinematic
8 g [m /s] for water at 20°C.(5.12)
H loss, incidenceviscosity
= ϕ s [m= /s],
ϕ
2⋅g
k = Emperical value 2 ⋅ g
2
2
V
L
V2
V
Hstat, gap = equals
0.5
+ ffor smooth
+2 1.0surfaces
m = Exponent
1/6
and between 1/7(5.18)
to 1/9
2g ) + k 2 2 g
(5.13)
H loss, incidence = k21 g⋅ (Q − Qs design
for rough surfaces
(5.19)
Ploss,are
=
kρ U 2 Dthe
disk
2 ( Ddesign
2 + 5e ) of the impeller, calculated disk friction
2gH
stat, gapto
If changes
made
V=
m
6
L +− 41.5
 2toν ⋅estimate
Ploss,disk,A can be fscaled
the disk friction Ploss,disk,B at another impel10 
s 

(5.14)
k = 7.3 ⋅ 10
ler diameter or speed:
 U 2 D2 
Q leakage = VA gap
(n3D52 )A
(Ploss, disk )A = ( Ploss, disk )B
(5.15)
(n3D52 )B
3
The scaling
(5.16)
Q equation
= Q + can
Q only be used for relative small design changes.
impeller
leakage
H stat, gap = H stat, impeller − ω 2fl
( D22 −D2gap )
8g
(5.17)
2
2
2
Hstat, gap = 0.5 V + f L V + 1.0 V
(5.18)
2g
s 2g
2g
5.3.7 Leakage
Leakage loss occurs because of smaller
(5.19) circulation through gaps between
2gH stat, gap
V = and fixed parts of the pump. Leakage loss results in a loss in efthe rotating
+ 1.5flow in the impeller is increased compared to the flow
f Lsthe
ficiency because
throughQthe
entire pump:
leakage = VA gap
92
92
 2ν ⋅ 10 6 

(5.14)
k = 7.3 ⋅ 10 − 4 
U DQ  (10/3600) m3 s
Mean velocity: V =2 2 =
= 3.45 m s
A 3 5π 0.032 2 m2
(n D24)A
(Ploss, disk )A = ( Ploss, disk )B
(5.15)
(n3D52 )B
3.45m s ⋅ 0.032m
VD h
= 110500
Reynolds number: Re =
=
ν
1 ⋅ 10 −6 m 2 s
(5.16)
Q impeller = Q + Q leakage
( D22 −D2gap )
(5.17)
H stat, gap =roughness:
H stat, impeller −k/D
ωh2fl = 0.15mm
= 0.0047
where Relative
8g
32mm
Qimpeller = Flow through impeller [m3/s], Q = Flow through pump [m3/s] , Qleakage
V 2 + f L V 2 2+ 1.0 V 2
= Leakage
[m3/s]
Hstat,flow
gap = 0.5
2m ⋅ ( 3.45 m s) 2 (5.18)
s 2LV
g = 0.031
2g
f
=
= 1.2 m
Pipe loss: H loss,2 g
pipe
2
D h 2g
0.032m ⋅ 2 ⋅ 9.81 m s
Leakage occurs many different places in the pump and depends on the pump
(5.19)
type. Figure 5.15
shows
where leakage
typically occurs. The pressure differ2gH stat,
gap
V=
2
V
L
ences in the pump
which
drives
the
leakage
flow as shown in figure 5.16.
1
f s +=1.5
(5.8)
ζ ⋅ H dyn,1 = ζ ⋅
H loss, expansion
2g
= VA 2gap the impeller and the casing at impeller eye and
Q leakagebetween
The leakage

A 
= 1 relief
− 1 are typically of the same size. The leakage flow
(5.9)
throughζaxial
between
A
2 

guidevane and shaft in multi-stage
pumps
are
less
important
because
both
2
2


A
V
Q
leakage,1
0
0
pressure
difference =and1 −
gap area⋅ are smaller.
(5.10)
H loss,
contraction


A
2
g
2 

To minimise the leakage flow, it is important
to make the gaps as small as
V2
ζ ⋅ H dyn,2 =
ζ⋅ 2
H loss,
possible.
When
the=pressure
difference
across the gap is large, (5.11)
it is in parcontraction
2g
ticular important that the gaps are small.
2
w 1 − w 1, kanal
w2
H loss, incidence = ϕ s = ϕ
(5.12)
Qleakage,2
Model
2 ⋅g
2⋅g
The leakage can be calculated by combining two different expressions for
2
(5.13)
H loss, incidence
k 1 ⋅ (Q
− Q design
+leakage,1
k 2 The head difference generated
the difference
in =head
across
the) Qgap:
by
the impeller, equation (5.17) and the head loss for the flow through the gap
3
equation
(5.18).
Both expressions
are necessary to calculate the leak flow.
Ploss,
disk = kρ U 2 D2 ( D 2 + 5e )
m
6
− 4  2ν ⋅ 10 

 the leakage between impeller eye(5.14)
=
⋅
k
7
.
3
10
In the following an example
and pump
 U D of
 2 2 
housing is shown. First the difference in head across the gap generated by
5 Qleakage,2
(n3D
the impeller is calculated. The
head
2 )A difference across the gap depends on
(Ploss, disk )A = ( Ploss, disk )B
(5.15)
3 5
the static head above the impeller
(n D2 )B and of the flow behaviour in the cavity
Qleakage,1impeller and pump casing:
between
Qleakage,1
(5.16)
Q impeller = Q + Q leakage
H stat, gap = H stat, impeller − ω 2fl
( D22 −D2gap )
8g
2
2
2
Hstat, gap = 0.5 V + f L V + 1.0 V
2g
s 2g
2g
93
2gH stat, gap
(5.19)
(5.17)
Figure 5.15: Types of leakage
(5.7)
Qleakage,1
Leakage between impeller eye and
pump casing.
Qleakage,2
Leakage above blades in an open
impeller
Qleakage,1
Qleakage,3
Leakage between guidevanes and shaft in a
multi-stage pump
Qleakage,4
Qleakage,3
Qleakage,1
Leakage as a result of balancing holes
(5.18)
93
2 2g
2m ⋅ ( 3.45 m s) 2
= 1.2 m
Pipe loss: H loss, pipe = f LV = 0.031
2
D h 2g
2
0.032m
⋅ 2 ⋅ 9.81 m s
2
w
−
w
w
1
1, kanal
= ϕ s =ϕ
H loss, incidence
(5.12)
5. Pump
losses
2 ⋅g
2⋅g
V2
= ζ )⋅2 + 1k
H
H dyn,1
expansion==kζ⋅ ⋅( Q
Hloss,
−
Q
loss, incidence
1
design
2 g2
(5.7)
(5.8)
(5.13)
where
2
 Rotational
A1  3 velocity of the fluid in the cavity between impeller
ωfl
=
1
ζ
=
−
Ploss, disk
(5.9)
 = kρ U 2 D2 ( D2 + 5e )
A2pump

 and
casing
m [rad/s]
6
 2ν ⋅ 10  2
2
Dgap
[m]
  A 0 of the
(5.14)
⋅ 10 − 4diameter
k = =7.3 Inner
V0gap
U
D
(5.10)
H
1
=
−
⋅
2 2 head
 static

Hstat, impellerloss,=contraction
Impeller
rise
[m]
2g
 A2 
(n3D52 )A
2
(
)
(
)
(5.15)
P
=
P
disk A
loss,
disk B the gap
The headloss,
difference
across
5 Vcan also be calculated as the head loss of
n3D
ζ ⋅2 )B 2
H loss, contraction = ζ ⋅ H dyn,2 (=
(5.11)
the flow through the gap, see figure2 g
5.17. The head loss is the sum of the fol(5.16)
= Q of
+Q
lowing Q
three
losses:
contraction when
the fluid
impellertypes
leakage Loss due to sudden
2
2
w 12 − w21, kanal
w
s
runs into
the gap,=friction
loss
loss due
H loss,
ϕ
= ϕ between
( D −D fluid
) and wall, and mixing
(5.12)
(5.17)
H incidence
2 ⋅ g − ω 2fl 2 2 ⋅gap
g
gap = H stat, impeller
to suddenstat,expansion
of the outlet
of8the
g gap.
2 )2 + k
2
H loss, incidence = kV1 2⋅ (Q − QL design
Hstat, gap = 0.5
+ f V + 1.0 2 V
2g
s 2g
2g
3
where Ploss, disk = kρ U 2 D2 ( D2 + 5e )
m
(5.19)
f = Friction coefficient
2gH stat,− 4gap 2[-]
ν ⋅ 10 6 
V
=


=
⋅
k
7
.
3
10
L = Gap length [m]
f Ls + 1.5
 U 2 D2 
s = Gap width [m]
(n3D52 )A
Q velocity
gap [m/s]
leakage = VA
V = Fluid
in
gap
(Ploss,
)
(
)
=
P
disk A
loss, disk B
5
(n3D
[m2 )2B]
Agap = Cross-section area of gap
Low pressure
High pressure
Figure 5.16: The leakage is drived by the
pressure difference across the impeller.
(5.13)
(5.18)
(5.14)
(5.15)
(5.16)
Q impeller = Q + Q leakage
The friction coefficient can be set to 0.025 or alternatively be found more
( D22 −5.6.
D2gap )
precisely
in gap
a Moody
see 2figure
(5.17)
H stat,
= H stat,chart,
impeller − ω fl
Figure 5.17: Pressure difference across the
gap through the friction loss consideration.
8g
2
2 equation
2 (5.18) and inserting H
By isolating the velocity
V in the
from
stat,gap
Hstat, gap = 0.5 V + f L V + 1.0 V
(5.18)
2
g
s
2
g
2
g
equation (5.17), the leakage can be calculated:
V=
2gH stat, gap
f Ls + 1.5
L
(5.19)
s
Q leakage = VA gap
Dspalte
D2
94
94
5.4 Loss distribution as function of specific speed
The ratio between the described mechanical and hydraulic losses depends
on the specific speed nq, which describes the shape of the impeller, see section 4.6. Figure 5.18 shows how the losses are distributed at the design point
(Ludwig et al., 2002).
Flow friction and mixing loss are significant for all specific speeds and are the
dominant loss type for higher specific speeds (semi-axial and axial impellers).
For pumps with low nq (radial impellers) leakage and disk friction on the hub
and shroud of the impeller will in general result in considerable losses.
At off-design operation, incidence and recirculation losses will occur.
η [%]
Mechanical loss
Leakage loss
Disk friction
100
95
90
Flow friction and mixing losses
85
Hydraulic efficiency
80
75
70
65
60
55
10
15
20
30
40
50
60 70 80 90
Figure 5.18: Loss distribution in a centrifugal
pump as function of specific speed nq
(Ludwig et al., 2002).
nq [min -1]
5.5 Summary
In this chapter we have described the individual mechanical and hydraulic
loss types which can occur in a pump and how these
� � � � � ���losses affect flow, head
��
and power consumption. For each loss type we �have
made a simple physical
� � �� ���� � � ��� �
description as well as shown in which hydraulic
the loss typi� � ��components
�
cally occurs. Furthermore, we have introduced some simple models which
can be used for estimating the magnitude of the losses. At the end of the
chapter we have shown how the losses are distributed depending on the
specific speeds.
95
95
Chapter 6
Pump tests
6.1 Test types
6.2 Measuring pump performance
6.3 Measurement of the pump’s NPSH
Hloss,friction,2
6.4 Measurement of force
H
U22
2.g
6.5 Uncertainty in measurement of performance
6.6 Summary
U'22
2.g
z'M2
Hloss,friction,1
U'12
2.g
pM2
U12
2.g
p'2
r.g
p2
r.g
pM1
p'1
r.g
z'M1 p1
r.g
H1
H2
H'2
H'1
z'2
z'1
z1
z2
S'1
S1
S2
S'2
6. Pump tests
6. Pump tests
This chapter describes the types of tests Grundfos continuosly performs on
pumps and their hydraulic components. The tests are made on prototypes in
development projects and for maintenance and final inspection of produced
pumps.
6.1 Test types
For characterisation of a pump or one of its hydraulic parts, flow, head, power consumption, NPSH and force impact are measured. When testing a complete pump, i.e. motor and hydraulic parts together, the motor characteristic
must be available to be able to compute the performance of the hydraulic
part of the pump. For comparison of tests, it is important that the tests are
done identically. Even small differences in mounting of the pump in the test
bench can result in significant differences in the measured values and there
is a risk of drawing wrong conclusions from the test comparison.
Flow, head, power consumption, NPSH and forces are all integral performance parameters. For validation of computer models and failure finding,
detailed flow field measurements are needed. Here the velocities and pressures are measured in a number of discrete points inside the pump using
e.g. LDA (Laser Doppler Anemometry) and PIV (Particle Image Velocimetry)
for velocity, see figure 6.1 and for pressure, pitot tubes and pressures transducers that can measure fast fluctuations.
The following describes how to measure the integral performance parameters, i.e. flow, head, power consumption, NSPH and forces. For characterisation of motors see the Motor compendium (Motor Engineering, R&T). For
flow field measurements consult the specialist literature, e.g. (Albrecht,
2002).
98
Figure 6.1: Velocity field in impeller measured
with PIV.
98
6.2 Measuring pump performance
Pump performance is usually described by curves of measured head and
power consumption versus measured flow, see figure 6.2. From these measured curves, an efficiency curve can be calculated. The measured pump performance is used in development projects for verification of computer models and to show that the pump meets the specification.
During production, the performance curves are measured to be sure they
correspond to the catalogue curves within standard tolerances.
Flow, head and power consumption are measured during operation in a test
bench that can imitate the system characteristics the pump can be exposed
to. By varying the flow resistance in the test bench, a number of corresponding values of flow, differential pressure, power consumption and rotational
speed can be measured to create the performance curves. Power consumption can be measured indirectly if a motor characteristic that contains corresponding values for rotational speed, electrical power, and shaft power is
available. Pump performance depends on rotational speed and therefore it
must be measured.
During development, the test is done in a number of operating points from
shut-off, i.e. no flow to maximum flow and in reversal from maximum flow
to shut-off. To resolve the performance curves adequately, 10 - 15 operating
points are usually enough.
H
50
40
30
20
10
0
0
10
20
30
40
50
60
70
Q
P2
8
4
0
Q
Figure 6.2: Measured head and power curve
as function of the flow.
Maintenance tests and final inspection tests are made as in house inspection tests or as certificate tests to provide the customer with documentation
of the pump performance. Here 2 - 5 predefined operating points are usally
sufficient. The flow is set and the corresponding head, electrical power consumption and possibly rotational speed are measured. The electrical power
consumption is measured because the complete product performance is
wanted.
99
99
6. Pump tests
Grundfos builds test benches according to in-house standards where
GS241A0540 is the most significant. The test itself is in accordance with the
international standard ISO 9906.
6.2.1 Flow
To measure the flow, Grundfos uses magnetic inductive flowmeters. These
are integrated in the test bench according to the in-house standard. Other
flow measuring techniques based on orifice, vortex meters, and turbine
wheels exist.
6.2.2 Pressure
Grundfos states pump performance in head and not pressure since head is independent of the pumped fluid, see section 2.4. Head is calculated from total
pressure measured up and down stream of the pump and density of the fluid.
Valve
Pipe contraction
Pipe expansion
Pipe bend
4 x D
2 x D
2 x D 2 x D
Figure 6.3: Pressure measurement outlet
before and after the pump. Pipe diameter, D,
is the pipe’s internal diameter.
The total pressure is the sum of the static and dynamic pressure. The static
pressure is measured with a pressure transducer, and the dynamic pressure
is calculated from pipe diameters at the pressure outlets and flow. If the
pressure transducers up and down stream of the pump are not located at
the same height above ground, the geodetic pressure enters the expression
for total pressure.
To achieve a good pressure measurement, the velocity profile must be uniform and non-rotating. The pump, pipe bends and valves affect the flow
causing a nonuniform and rotating velocity profile in the pipe. The pressure
taps must therefore be placed at a minimum distance to pump, pipe bends
and other components in the pipe system, see figure 6.3.
The pressure taps before the pump must be placed two pipe diameters upstream the pump, and at least four pipe diameters downstream pipe bends
and valves, see figure 6.3. The pressure tap after the pump must be placed
two pipe diameters after the pump, and at least two pipe diameters before
any flow disturbances such as bends and valves.
100
100
The pressure taps are designed so that the velocity in the pipe affects the
static pressure measurement the least possible. To balance a possible bias in
the velocity profile, each pressure tab has four measuring holes so that the
measured pressure will be an average, see figure 6.4.
Pressure gauge
Venting
Dz
The measuring holes are drilled perpendicular in the pipe wall making them
perpendicular to the flow. The measuring holes are small and have sharp
edges to minimise the creation of vorticies in and around the holes, see
figure 6.5.
It is important that the pressure taps and the connection to the pressure
transducer are completely vented before the pressure measurement is
made. Air in the tube between the pressure tap and transducer causes errors in the pressure measurement.
The pressure transducer measures the pressure at the end of the pressure
tube. The measurements are corrected for difference in height Δz between
the center of the pressure tap and the transducer to know the pressure at
the pressure tap itself, see figure 6.4. Corrections for difference in height are
also made between the pressure taps on the pump’s inlet and outlet side.
If the pump is mounted in a well with free surface, the difference in height
between fluid surface and the pressure tap on the pump’s outlet side must
be corrected, see section 6.2.4.
+
Figure 6.4: Pressure taps which average over
four measuring holes.
Figure 6.5: Draft of pressure tap.
6.2.3 Temperature
The temperature of the fluid must be known to determine its density. The
density is used for conversion between pressure and head and is found by
table look up, see the chart ”Physical properties of water” at the back of the
book.
101
101
6. Pump tests
Figure 6.6: Draft of pump test on
a piping.
H2
H loss,friction, 2
H 2'
H
H'
1
H loss,friction,1
H1
S1'
6.2.4 Calculation of head
The head can be calculated when flow, pressure, fluid
type, temperature and geometric sizes such as pipe
diameter, distances and heights are known. The total
head from flange to flange is defined by the following
equation:
H = H2 − H1
(6.1)
(6.2)
H = ( H2' + H loss, friction,2 ) − ( H '1 − Hloss, friction,1 )
Figure 6.6 shows where the measurements are made.
The pressure outlets and the matching heads are marked
p − p1 are
U 2 − U12 found in the powith a (H’ ).=The
(6.3)
z2 −pressure
z1 + 2outlets
+ 2 thus
ρ⋅ g
2 ⋅ g for the total head
sitions S’1Geodetic
and S’
and the
expression
pressure
2
H = H 2 − H 1 Static pressure Dynamic pressure
(6.1)
is therefore:
H = ( H2' + H loss, friction,2 ) − ( H '1 − Hloss, friction,1 )
S1
S2
S'
2
where Hloss,friction,1 and Hloss,friction,2 are the pipe friction losses between pressure outlet and pump flanges.
The size of the friction loss depends on the flow velocity,
the pipe diameter, the distance from the pump flange to
the pressure outlet and the pipe’s surface roughness. Calculation of pipe friction loss is described in section 5.3.1.
If the pipe friction loss between the pressure outlets and
the flanges is smaller than 0.5% of the pump head, it is
normally not necesarry to take this into consideration in
the calculations. See ISO 9906 section 8.2.4 for further
explanation.
(6.2)
  p' p − p UU

'2 2 − U 2
H = zz2'2+−zρ1 M⋅+2g + 2z 'M 2 1+ + 2 2 + H1 loss, friction, 2 −(6.3)
ρ ⋅ g 2 ⋅ g2 ⋅ g

 pressure

Geodetic
Static pressure
102

z 1'

Dynamic pressure

 p'
 U '2
+ M1 + z'M1  + 1 − H loss, friction,1 (6.4)
ρ⋅g

 2⋅g

102
S'
1
Figure 6.7: Pump test where the pipes
are at an angle compared to horizontal.
S1
Total head
S1
S2
H'
1
U 12
2.g
p1
r.g
p'
2
r.g
p2
r.g
H1
2
U'
2
2.g
H2
H'
2
z'
2
H = H2 − H1
S'
1
S'
2
2
z'
2
Because the manometer only measures the static
pressure, the dynamic pressure must also be taken into acz'
1
count. The dynamic pressure depends on the pipe diamz1
eter and can be different on eachz2side of the pump.
S'
S1
S'
1 illustrates
2
2 a pump test in
Figure 6.8
the basic Sversion
of
a pipe. The total head which is defined by the pressures
p1 and p2 and the velocities U1 and U2 in the inlet and
outlet flanges S1 and S2 can be calculated by means of
the following equation:
103
z'
M2
z1
6.2.5 General calculation of head
In practise a pump test is not always made
on a horiH loss,friction,2
zontal pipe, see figure 6.7.H This results
in a difference in
U 22
2.g pumppM2
2
height between the centers of the
in- and
outlet,
U'
2
2.g
otal head z’1 and z’2 , and the centers of the inlet and outlet flanges,
z'
M2
H loss,friction,1
z112 and z2 respectively.
The manometer can, furthermore,
U'
tatic head
p'
2
2
.g placed withUa
1
2be
p2
difference in height
compared
to the
r.g
2.g
r.g
pipe centre.pM1These differences in height must be taken
p1'
z'
p1 inH 1the calculation
H2
M1
consideration
of head. H'
.g
rinto
r.g
z'
M1
pM2
z'
1
z2
z1
U
2.g
2
2
H'
1
z'
2
S'
1
pM1
p1'
r.g
z'
1
H
H loss,friction,1
2
1
Static head U'
2.g
z'
M1
S'
2
H loss,friction,2
Figure 6.8: General draft of a
pump test.
z'
M2
S2
z2
S1
S2
(6.1)
S'
2
H
H2' −+ H
friction,2 ) − ( H '1 − Hloss, friction,1 )
H == (H
Hloss,
2
1
(6.2)
(6.1)
2
H = ( H2' + H loss, friction,2
p − )p−1 ( H 'U
−loss,
U12friction,1 )
1−H
H = z 2 − z1 + 2
+ 2
ρ⋅ g
2⋅g
Geodetic pressure
(6.2)
(6.3)
H = z 2 − z1 +
Geodetic pressure
2
pressure
p − p Dynamic
U2 −U
1
+ 2
ρ⋅ g
2⋅g
Static pressure
2
1
Static pressure
(6.3)
Dynamic pressure
Using the measured sizes in S’1 and S’2, the general
expression for the

 p'total head
 is:
U '2
H = z 2' +  M 2 + z 'M 2  + 2 + H loss, friction, 2 −
ρ⋅g
g
2
⋅

 

  p'M 2

 U2'2
H = z 2' +  p' + z 'M 2 + U '2 + H loss, friction, 2 −
ρ⋅g
2⋅g
z 1' + ρM⋅1g + z'M1 + 1 − H loss, friction,1 (6.4)
2⋅g






 p'M1
 U1'2
z 1' + ρ ⋅ g + z'M1  + 2 ⋅ g − H loss, friction,1 (6.4)




NPSH A =
pva
pstat,in + pbar + 0 .5 ⋅ ρ ⋅ V12
+ z geo− H loss, friction, −
ρ⋅g
ρ⋅
NPSH A =
pva
pstat,in + pbar + 0 .5 ⋅ ρ ⋅ V12
+ z geo− H loss, friction,103
−
ρ⋅g
ρ⋅
6. Pump tests
6.2.6 Power consumption
Distinction is made between measurement of the shaft power P2 and added
electric power P1. The shaft power can best be determined as the product of
measured angular velocity w and the torque on the shaft which is measured
by means of a torque measuring device. The shaft power can alternatively
be measured on the basis of P1. However, this implies that the motor characteristic is known. In this case, it is important to be aware that the motor
characteristic changes over time because of bearing wear and due to changes in temperature and voltage.
The power consumption depends on the fluid density. The measured power
consumption is therefore usually corrected so that it applies to a standard
fluid with a density of 1000 kg/m3 which corresponds to water at 4°C. Head
and flow are independent of the density of the pumped fluid.
6.2.7 Rotational speed
The rotational speed is typically measured by using an optic counter or magnetically with a coil around the motor. The rotational speed can alternatively be measured by means of the motor characteristic and measured P1. This
method is, however, more uncertain because it is indirect and because the
motor characteristic, as mentioned above, changes over time.
The pump performance is often given for a constant rotational speed. By
means of affinity equations, described in section 4.5, the performance can
be converted to another speed. The flow, head and power consumption are
hereby changed but the efficiency is not changed considerably if the scaling
of the speed is smaller than ± 20 %.
104
104
6.3 Measurement of the pump’s NPSH
The NPSH test measures the lowest absolute pressure at the inlet before
cavitation occurs for a given flow and a specific fluid with vapour pressure
pvapour , see section 2.10 and formula (2.16).
A typical sign of incipient cavitation is a higher noise level than usual. If the
cavitation increases, it affects the pump head and flow which both typically
decrease. Increased cavitation can also be seen as a drop in flow at constant
head. Erosion damage can occur on the hydraulic part at cavitation.
The following pages introduce the NPSH3% test which gives information
about cavitation’s influence on the pump’s hydraulic performance. The test
gives no information about the pump’s noise and erosion damage caused
by cavitation.
In practise it is thus not an actual ascertainment of cavitation but a chosen
(3%) reduction of the pump’s head which is used for determination of NPSHR
- hence the name NPSH3%.
To perform a NPSH3% test a reference QH curve where the inlet pressure is
sufficient enough to avoid cavitation has to be measured first. The 3% curve
is drawn on the basis of the reference curve where the head is 3% lower.
Grundfos uses two procedures to perform an NPSH3% test. One is to gradually lower the inlet pressure and keep the flow constant. The other is to gradually increase the flow while the inlet pressure is kept constant.
105
105
6. Pump tests
6.3.1 NPSH3% test by lowering the inlet pressure
When the NPSH3% curve is flat, this type of NPSH3% test is the best suited.
The NPSH3% test is made by keeping the flow fixed while the inlet pressure
pstat,in and thereby NPSHA is gradually lowered until the head is reduced with
more than 3%. The resulting NPSHA value for the last measuring point before
the head drops below the 3% curve then states a value for NPSH3% at the
given flow.
The NPSH3% curve is made by repeating the measurement for a number of
different flows. Figure 6.9 shows the measuring data for an NPSH3% test
where the inlet pressure is gradually lowered and the flow is kept fixed. It is
these NPSH values which are stated as the pump’s NPSH curve.
Procedure for an NPSH3% test where the inlet pressure is gradually lowered:
1. A QH test is made and used as reference curve
2. The 3% curve is calculated so that the head is 3% lower than the
reference curve.
3. Selection of 5-10 flow points
4. The test stand is set for the seleted flow point starting with the
largest flow
5. The valve which regulates the counter-pressure is kept fixed
6. The inlet pressure is gradually lowered and flow, head
and inlet pressure are measured
7. The measurements continue until the head drops below the 3% curve
8. Point 4 to 7 is repeated for each flow point
H
Q
Figure 6.9: NPSHA measurement by lowering
the inlet
pressure.
Referencekurve
3% kurve
Reference
curve
Målt
løftehøjde
3% curve
Measured head
106
106
6.3.2 NPSH3% test by increasing the flow
For NPSH3% test where the NPSH3% curve is steep, this procedure is preferable. This type of NPSH3% test is also well suited for cases where it is difficult
to change the inlet pressure e.g. an open test stand.
H
The NPSH3% test is made by keeping a constant inlet pressure, constant water level or constant setting of the regulation valve before the pump. Then
the flow can be increased from shutoff until the head can be measured below the 3% curve, see figure 6.10. The NPSH3% curve is made by repeating the
measurements for different inlet pressures.
Procedure for NPSH3% test where the flow is gradually increased
1. A QH test is made and used as reference curve
2. The 3% curve is calculated so that the head is 3% lower than the
reference curve.
3. Selection of 5-10 inlet pressures
4. The test stand is set for the wanted inlet pressure
5. The flow is increased from the shutoff and flow, head and inlet
pressure are measured
6. The measurements continue until the head is below the 3% curve
7. Point 4-6 is repeated for each flow point
Q
Figure 6.10: NSPHA measurement by
increasing
flow.
Referencekurve
3% kurve
Reference curve
Målt løftehøjde
3% curve
Measured head
Vacuum pump
Shower
6.3.3 Test beds
When a closed test bed is used for testing pumps in practise, then the inlet pressure can be regulated by adjusting the system pressure. The system
pressure is lowered by pumping water out of the circuit. The system pressure can, furthermore, be reduced with a throttle valve or a vacuum pump,
see figure 6.11.
Flow valve
Flow meter
Baffle plate
Heating/
cooling coil
Test pump
Throttle valve
Pressure
control
pump
Figure 6.11: Draft of closed
test bed for NPSH measurement.
107
107
6. Pump tests
In an open test bed, see figure 6.12, it is possible to adjust the inlet pressure
in two ways: Either the water level in the well can be changed, or a valve can
be inserted before the pump. The flow can be controlled by changing the
pump’s counter-pressure by means of a valve mounted after the pump.
Adjustable water level
for flow valve
and flow meter
Pump
6.3.4 Water quality
If there is dissolved air in the water, this affects the pump performance which
can be mistaken for cavitation. Therefore you must make sure that the air content in the water is below an acceptable level before the NPSH test is made. In
practise this can be done by extracting air out of the water for several hours.
The process is called degasification.
In a closed test bed the water can be degased by lowering the pressure in
the tank and shower the water hard down towards a plate, see figure 6.11,
forcing the air bubbles out of the fluid. When a certain air volume is gathered in the tank, a part of the air is removed with a vacuum pump and the
procedure is repeated at an even lower system pressure.
Throttle valve
Figure 6.12: Drafts of open test beds for
NPSH measurement.
6.3.5 Vapour pressure and density
The vapour pressure and the density for water depend on the temperature
and can be found by table look-up in ”Physical properties of water” in the
back of the book. The fluid temperature is therefore measured during the
execution of an NPSH test.
6.3.6 Reference plane
NPSH is an absolute size which is defined relative to a reference plane. In
this case reference is made to the center of the circle on the impeller shroud
which goes through the front edge of the blades, see figure 6.13.
Reference plan
108
Figure 6.13: Reference planes at
NPSH measurement.
108
  p'

 U '2
H = z 2' +  M 2 + z 'M 2  + 2 + H loss, friction, 2 −
ρ⋅g
g
2
⋅

 


 p'M1

U1'2

− H loss, friction,1 (6.4)
+ z'M1  +
6.3.7 Barometric
z 1' +pressure
ρ⋅g
2⋅g




In practise the inlet pressure is measured as a relative
pressure in relation to
the surroundings. It is therefore necesarry to know the barometric pressure
at the place and time where the test is made.
6.3.8 Calculation of NPSHA and determination of NPSH3%
NPSHA can be calculated by means of the following formula:
NPSH A =
H
pvapour
pstat,in + pbar + 0 .5 ⋅ ρ ⋅ V12
+ z geo− H loss, friction, −
(6.5)
ρ⋅g
ρ⋅g
pstat,in = The measured relative inlet pressure
pbar
= Barometric pressure
V1
= Inlet velocity
zgeo
= The pressure sensor’s height above the pump
Hloss,friction
= The pipe loss between pressure measurement and pump
pvapour = Vapour pressure (table look-up)
ρ
= Density (table look-up)
The NPSH3% value can be found by looking at how the head develops during
the test, see figure 6.14. An NPSH3% value is determined by the NPSHA value
which is calculated from the closest data point above the 3% curve.
•
•
•
•
•
•
•
•
•
•
•
•
Figure 6.14: Determination of NPSH3%.
Q
Referencekurve
3% kurve
Reference curve
Målt løftehøjde
3% curve
NPSH 3%
Measured head
NPSH A
NPSH3%
NPSHA
6.4 Measurement of force
Measurement of axial and radial forces on the impeller is the only reliable
way to get information about the forces’ sizes. This is because these forces
are very difficult to calculate precisely since this requires a full three-dimensional numerical simulation of the flow.
109
109
6. Pump tests
6.4.1 Measuring system
The force measurement is made by absorbing the forces on the rotating system
(impeller and shaft) through a measuring system.
Dynamometer
Axial bearing
The axial force can e.g. be measured by moving the axial bearing outside the motor and mount it on a dynamometer, see figure 6.15. The axial forces occuring
during operation are absorbed in the bearing and can thereby be measured with a
dynamometer.
Axial and radial forces can also be measured by mounting the shaft in a magnetic
bearing where it is fixed with magnetic forces. The shaft is fixed magnetical both in
the axial and radial direction. The mounting force is measured, and the magnetic
bearing provides information about both radial and axial forces, see figure 6.16.
Radial and axial force measurements with magnetic bearing are very fast, and
both the static and the dynamic forces can therefore be measured.
By measurement in the magnetic bearing, the pump hydraulic is mounted directly
on the magnetic bearing. It is important that the fixing flange geometry is a precise reflection in the pump geometry because small changes in the flow conditions in the cavities can cause considerable differences in the forces affecting on
the impeller.
Axial sensor
Axial magnetic bearing
Support
bearing
Radial sensor
Radial magnetic bearing
Radial sensor
Radial magnetic bearing
Radial sensor
Figure 6.15: Axial force measurement
through dynamometer on the bearing.
Axial sensor
Support bearing
Figure 6.16: Radial and axial force measurement with magnetic bearing.
110
110
6.4.2 Execution of force measurement
During force measurement the pump is mounted in a test bed, and the test
is made in the exact same way as a QH test. The force measurements are
made simultaneously with a QH test.
At the one end, the shaft is affected by the pressure in the pump, and in the
other end it is affected by the pressure outside the pump. Therefore the system pressure has influence on the size of the axial force.
If comparison between the different axial force measurement is wanted, it
is necesarry to convert the system pressure in the axial force measurements
to the same pressure. The force affecting the shaft end is calculated by multiplying the area of the shaft end with the pressure in the pump.
6.5 Uncertainty in measurement of performance
At any measurement there is an uncertainty. When testing a pump in a test bed,
the uncertainty is a combination of contributions from the measuring equipment, variations in the test bed and variations in the pump during the test.
6.5.1 Standard demands for uncertainties
Uncertainties on measuring equipment are in practise handled by specifying a set of measuring equipment which meet the demands in the standard
for hydrualic performance test, ISO09906.
ISO09906 also states an allowed uncertainty for the complete measuring
system. The complete measuring system includes the test beds’ pipe circuit,
measuring equipment and data collection. The uncertainty for the complete measuring system is larger than the sum of uncertainties on the single
measuring instrument because the complete uncertainty also contains variations in the pump during test which are not corrected for.
Variations occuring during test which the measurements can be corrected for are the fluid’s characteristic and the pump speed. The correction is
111
111
to convert the measuring results to a constant fluid temperature and a
constant speed.
To ensure a measuring result which is representative for the pump, the
test bed takes up more measurements and calculates an average value.
ISO09906 has an instruction of how the test makes a representative average value seen from a stability criteria. The stability criteria is a simplified
way to work with statistical normal distribution.
6.5.2 Overall uncertainty
The repetition precision on a test bed is in general better than the collected
precision. During development where very small differences in performance
are interesting, it is therefore a great advantage to make all tests on the
same test bed.
There can be significant difference in the measuring results between several
test beds. The differences correspond to the overall uncertainty.
6.5.3 Measurement of the test bed’s uncertainty
Grundfos has developed a method to estimate a test beds’ overall uncertainty. The method gives a value for the standard deviation on the QH curve
and a value for the standard deviation on the performance measurement.
The method is the same as the one used for geometric measuring instruments, e.g. slide gauge.
The method is outlined in the Grundfos standard GS 241A0540: Test benches and test equipment.
6.6 Summary
In this chapter we have introduced the hydraulic tests carried out on complete pumps and their hydraulic components. We have described which
sizes to measure and which problems can occur in connection with planning
and execution of a test. Furthermore, we have described data treatment,
e.g. head and NPSH value.
112
112
Appendix
Appendix A. Units
Appendix B. Check of test results
H
�
�
�
�
�
�
�
�
�
�
�
�
Q
A. Units
A. Units
Some of the SI system’s units
Basic
Basicunits
units
Unit for
Name
Unit
Length
meter
m
Mass
kilogram kg
Time
second
S
Temperature
Kelvin
K
Unit for
Name
Unit
Definition
Angle
radian
rad
One radian is the angle subtended at the centre of a
circle by an arc of circumference that is equal in length
to the radius of the circle..
Unit for
Name
Unit
Definition
Force
Newton
N
N = kg ⋅ m / s2
Pressure
Pascal
Pa
Pa = N/m2 = kg /(m ⋅ s2 )
Energy, work
Joule
J
J = N⋅m = W ⋅ s
W
W = J/ s = N ⋅ m/ s = Kg ⋅ m2/s3
Additional units
Derived units
Power
114
Watt
Impulse
kg ⋅ m/ s
Torque
N⋅m
Conversion of units
Length
m
in (inches)
ft (feet)
1
39.37
3.28
0.0254
1
0.0833
s
min
h (hour)
1
16.6667 . 10-3
0.277778 . 10-3
60
1
16.6667 . 10-3
3600
60
1
m3/h
l/s
gpm (US)
3600
1000
15852
0.277778 . 10
1
0.277778
4.4
10
3.6
1
15.852
0.2271
0.063
1
Time
Flow, volume flow
m3/s
1
-3
-3
0.000063
Mass flow
Speed
kg/s
kg/h
kg/s
kg/h
ft/s
1
3600
1
3600
3.28
1
0.277778 . 10-3
1
0.9119
0.3048
1.097
1
0.277778 . 10-3
115
A. Units
Rotational speed
RPM = revolution per minute
s-1
rad/s
1
16.67 . 10-3
0.105
60
1
6.28
9.55
0.1592
1
kPa
bar
mVs
1
0.01
0.102
100
1
10.197
9.807
98.07 . 10-3
1
Pressure
Work, energy
Temperature
K
1
o
C
J
kWh
t(oC) = T - 273.15K
1
0.277778 . 10-6
3.6 . 106
1
T(Kelvin) = 273.15 C + t 1
o
Kinematic viscosity
116
Dynamic viscosity
m /s
cSt
Pa . s
cP
1
106
1
103
10-6
1
10-3
1
2
B. Check of test results
B. Check of test results
When unexpected test results occur, it can be difficult to find out why. Is the
tested pump in reality not the one we thought? Is the test bed not measuring correctly? Is the test which we compare with not reliable? Have some
units been swaped during the data treatment?
Typical examples which deviate from what is expected is presented on the
following pages. Furthermore, some recommendations of where it is appropriate to start looking for reasons for the deviating test results are presented.
The test shows that the efficiency is below the catalogue curve.
Possible cause
What to examine
How to find the error
Power consumption is too high
and/or the head is too low
Decide whether it is the power
consumption or the head which
deviates
Use one of the three schemes
below, scheme 1 -3
117
B. Check of test results
Table 1: The test shows that the power consumption for a produced pump lies
above the catalogue value but the head is the same as the catalogue curve.
118
Possible cause
What to examine
How to find the error
The catalogue curve does not
reflect the 0-series testen.
Compare 0-series test with
catalog curve
If the catalogue curve and
0-series test do not correspond,
it can not be expected that the
pump performs according to
the catalogue curve.
The impeller diameter or outlet
width is bigger than on the
0-series
Make a scaling of the test where
the impeller diameter D2 is
reduced until the power matches
most of the curve. If the head
also matches the curve, then the
diameter on the produced pump
is probably too big. Repeat the
same procedure with the impeller
outlet width b2. Scaling of D2
and b2 is discussed in chapter 4.5
Make sure that the right impeller
is tested.
Mechanical drag is found
Listen to the pump. If it is noisy,
turn off the pump and rotate by
hand to identify any friction.
Look at the difference of the
two power curves. Is it constant,
there is probably drag.
Remove the mechanical drag
The motor efficiency is lower
than specified.
Separate motor and pump. Test
them separately. The pump can
be in a test bench with torque
meter or with a calibrated motor.
If the pump’s power consumption
is ok, the motor is the problem.
Find cause for motor error.
Measure the impeller’s outlet on
the 0-series pump.
Adjust impeller diameter and
outlet width in the production
Table 2: The test shows that the power consumption and head lies below
the catalogue curve.
Possible cause
What to examine
How to find the error
Curves have been made at
different speeds.
Find the speed for the catalogue
curve and the test.
Convert to the same speed and
compare again.
The catalogue curve does not
reflect the 0-series test.
Compare the 0-series test with
the catalogue curve.
If the catalogue curve and 0-series
test do not correspond, it can not
be expected that the pump
performs according to catalogue
curve.
The impeller’s outlet diameter or
outlet width is smaller than on
the 0-series test.
Make a scaling of the test where
the impeller diameter D2 is
increased until the power matches
over most of the curve. If the
head also matches over most of
the curve, then the diameter on
the produced pump is probably
too small. Repeat the same
procedure with the impeller’s
outlet width b2. Scaling of D2
and b2 is discussed in chapter 4.5
Measure the impeller outlet on
the 0-series pump. Adjust impeller
diameter and outlet width in
the production.
Curve 1
Impellere D2/D1: 99/100=0.99 Curve 1
Impellere D2/D1: 100/99=1.01010101010101 Curve 1
H[m]
100
83.3333
66.6667
50
33.3333
16.6667
0
0
20
40
60
80
100
120
Q [m³/h]
Curve 1
Impellere D2/D1: 99/100=0.99 Kurve 1
Impellere D2/D1: 100/99=1.01010101010101 Curve 1
P1 [kW]
34,2857
28,5714
22,8571
17,1429
11,4286
5,71429
0
0
20
40
60
80
100
120
Q [m³/h]
119
B. Check of test results
Table 3: The power consumption is as the catalogue curve but the head is too low.
120
Possible cause
What to examine
How to find the error
The catalogue curve does not
reflect the 0-series test.
Compare O-series test with
catalogue curve.
If the catalogue curve and 0-series
test do not correspond, it can not
be expected that the pump performs
according to the catalogue curve.
Increased hydraulic friction
Compare the QH curves at the same
speed. Is the difference developing
as a parabola with the flow, there
could be an increased friction loss.
Examine surface roughness and
inlet conditions.
Remove irregularities in the
surface. Reduce surface roughness.
Remove elements which block
the inlet.
Calculation of the head is not
done correctly.
Examine the information about pipe Repeat the calculation of the head.
diameter and the location of the
pressure transducers. Examine
whether the correct density has been
used for calculation of the head.
Error in the differential pressure
measurement.
Read the test bed’s calibration report.
Examine whether the pressure
outlets and the connections to the
pressure transducers have been bleed.
Examine that the pressure transducers
can measure in the pressure range
in question.
If it has been more than a year
since the pump has been
calibrated, it must be calibrated
now. Use the right pressure
transducers.
Cavitation
Examine whether there is
sufficient pressure at the pump’s
inlet. See section 2.10 and 6.3)
Increase the system pressure.
Table 3 (continued)
Possible cause
What to examine
How to find the error
Increased leak loss.
Compare QH curves and power
Replace the impeller seal.
curves. If the curve is a horizontal
Close all unwanted circuits.
displacement which decreases
when the head (the pressure
difference above the gap) falls,
there could be an increased leak l
oss. Leak loss is described in section
5.3.7. Measure the sealing diameter
on the rotating and fixed part.
Compare the results with the
specifications on the drawing.
Examine the pump for other
types of leak loss.
0-series
Pump with leakage
H [m ]
40
35
30
25
20
15
10
5
0
0
5
10
15
20
25
30
35
Q [m ^3/h]
15
20
25
30
35
Q [m ^3/h]
0-series
Pump with leakage
H [m ]
2200
2000
1800
1600
1400
1200
1000
800
500
0
5
10
121
Bibliography
European Association of Pump Manufacturers (1999), ”NPSH for rotordynamic pumps: a reference guide”, 1st edition.
R. Fox and A. McDonald (1998), ”Introduction to Fluid Mechanics”.
5. edition, John Wiley & Sons.
J. Gulich (2004), ”Kreiselpumpen. Handbuch für Entwicklung, Anlagenplanung
und Betrieb”. 2nd edition, Springer Verlag.
C. Pfleiderer and H. Petermann (1990), ”Strömungsmachinen”.
6. edition, Springer Verlag, Berlin.
A. Stepanoff (1957), ”Centrifugal and axial flow pumps: theory, design and
application”. 2nd edition, John Wiley & Sons.
H. Albrecht and others (2002), ”Laser Doppler and Phase Doppler Measurement Techniques”. Springer Verlag, Berlin.
H. Hansen and others (1997), ”Danvak. Varme- og klimateknik. Grundbog”.
2nd edition.
Pumpeståbi (2000). 3rd edition, Ingeniøren A/S.
Motor compendium. Department of Motor Engineering, R&T, Grundfos.
G. Ludwig, S. Meschkat and B. Stoffel (2002). ”Design Factors Affecting
Pump Efficiency”, 3rd International Conference on Energy Efficiency in Motor
Driven Systems, Treviso, Italy, September 18-20.
122
Standards
ISO 9906 Rotodynamic pumps – Hydraulic performance acceptance testGrades 1 and 2. The standard deals with hydraulic tests and contains
instructions of data treatment and making of test equipment.
ISO2548 has been replaced by ISO9906
ISO3555 has been replaced by ISO9906
ISO 5198 Pumps – Centrifugal-, mixed flow – and axial pumps – Hydraulic
function test – Precision class
GS 241A0540 Test benches and test equipment. Grundfos standards for
contruction and rebuilding of test benches and data loggers.
123
Index
A
Absolute flow angle............................................................................. 61
Absolute pressure. .................................................................................33
Absolute pressure sensor..................................................................33
Absolute temperature.........................................................................33
Absolute velocity . .................................................................................60
Affinity .......................................................................................................... 70
Affinity equations . .........................................................53, 104
Affinity laws . .............................................................................. 68
Air content................................................................................................ 108
Angular frequency . .............................................................................. 62
Angular velocity............................................................................. 64, 104
Annual energy consumption ........................................................ 56
Area relation . ........................................................................................... 86
Auxiliary pump ....................................................................................... 50
Axial bearing............................................................................................. 20
Axial forces.........................................................................................44, 110
Axial impeller. .......................................................................................... 16
Axial thrust. .........................................................................................19, 20
Axial thrust reduction........................................................................ 20
Axial velocity.............................................................................................60
124
Cavity. ............................................................................................................ 19
Centrifugal force ....................................................................................12
Centrifugal pump principle ............................................................12
Chamber. ......................................................................................................23
Circulation pumps .......................................................................... 24, 25
Closed system . ........................................................................................ 49
Constant-pressure control ..............................................................54
Contraction.................................................................................................87
Control. ......................................................................................................... 39
Control volume . ..................................................................................... 64
Corrosion .................................................................................................... 85
Cross section contraction. ................................................................87
Cross section expansion. .................................................................. 86
Cross section shape.............................................................................. 83
Cutting system ........................................................................................27
B
Balancing holes ...................................................................................... 20
Barometric pressure . ..................................................................33, 109
Bearing losses ..........................................................................................80
Bernoulli’s equation. ............................................................................37
Best point.................................................................................................... 39
Blade angle. .........................................................................................73, 90
Blade shape. ..............................................................................................66
Bypass regulation...................................................................52
D
Data sheet . ................................................................................................ 30
Degasification ....................................................................................... 108
Density........................................................................................................ 108
Detail measurements ........................................................................98
Differential pressure . ................................................................... 34, 35
Differential pressure sensor ...........................................................33
Diffusor...................................................................................................21, 86
Disc friction................................................................................................ 91
Double pump............................................................................................ 50
Double suction pump .........................................................................14
Down thrust ............................................................................................. 44
Dryrunner pump. .................................................................................... 17
Dynamic pressure...................................................................................32
Dynamic pressure difference .........................................................35
C
Cavitation .......................................................................................... 40, 105
E
Eddies..............................................................................................................87
Efficiency...................................................................................................... 39
Electrical motor........................................................................................17
Electrical power. ................................................................................... 104
End-suction pump..................................................................................14
Energy class. ...............................................................................................57
Energy efficiency index (EEI). ..........................................................57
Energy equation. .....................................................................................37
Energy labeling........................................................................................ 56
Equilibrium equations. ...................................................................... 64
Euler’s turbomachinery equation........................................64, 65
F
Final inspection test . ..........................................................................99
Flow angle . ................................................................................... 61, 73, 90
Flow forces . ............................................................................................... 64
Flow friction . .............................................................................................81
Flow meters ............................................................................................100
Fluid column. ............................................................................................ 34
Force measurements........................................................... 110
Friction.......................................................................................................... 19
Friction coefficient ............................................................................... 82
Friction loss..........................................................................................49, 81
G
Geodetic pressure difference ................................................. 35, 36
Grinder pump ...........................................................................................27
Guide vanes ...............................................................................................23
H
Head ........................................................................................31, 34, 100, 102
Head loss calculation........................................................... 85
High specific speed pumps .............................................................74
Hydraulic diameter. ............................................................................. 82
Hydraulic losses................................................................................78, 80
Hydraulic power .................................................................................... 38
I
Ideal flow .....................................................................................................37
Impact losses............................................................................................90
Impeller. ........................................................................................................15
Impeller blades...................................................................................15, 16
Impeller outlet heigth........................................................................ 70
Impeller shape..........................................................................................75
Industrial pumps ................................................................................... 24
Inlet...........................................................................................................14, 62
Inlet flange..................................................................................................14
Inline-pump................................................................................................14
L
Laminar flow............................................................................................. 83
Leakage......................................................................................................... 92
Leakage loss.........................................................................................19, 92
Load profile. ................................................................................................54
Loss distribution..................................................................................... 95
Loss types.....................................................................................................78
Low specific speed pumps................................................................74
M
Magnetic bearing. ............................................................................... 110
Magnetic drive..........................................................................................18
Maintenance test. .................................................................................99
Measuring holes................................................................................... 101
Mechanical losses. .................................................................................78
Meridional cut. ........................................................................................60
Meridional velocity...............................................................................60
meterWaterColumn............................................................................ 34
Mixing losses............................................................................................ 86
Momentum equation......................................................................... 64
125
Index
Moody chart.............................................................................................. 84
Motor. .............................................................................................................17
Motor characteristics..........................................................................98
N
Non-return valve.....................................................................................51
NPSH . ..................................................................................... 31, 40, 105, 109
NPSH3%-test................................................................................ 105
NPSHA (Available)....................................................................40
NPSHR (Required)......................................................................41
O
Open impeller. ......................................................................................... 16
Open system............................................................................................. 49
Operating point............................................................................... 48, 49
Optical counter. .................................................................................... 104
Outlet............................................................................................................. 63
Outlet diameter...................................................................................... 70
Outlet diffusor..........................................................................................22
Outlet flange..............................................................................................14
P
Parasitic losses.........................................................................................80
Pipe diameter........................................................................................... 36
Pipe friction. .............................................................................................. 82
Pipe friction losses.............................................................................. 102
Potential energy......................................................................................37
Power consumption. ...................................................................31, 104
Power curves............................................................................................. 38
Prerotation.......................................................................................... 62, 72
Pressure.........................................................................................................32
Pressure loss coefficient..............................................................81, 88
Pressure measurement. ....................................................................98
126
Pressure sensor. .......................................................................................33
Pressure taps................................................................. 100, 102
Pressure transducer................................................................. 100, 101
Primary eddy............................................................................................. 91
Primary flow.............................................................................................. 19
Proportional-pressure control.......................................................54
Pump curve.................................................................................................31
Pump efficiency....................................................................................... 39
Pump losses............................................................................................... 79
Pump performance. ............................................................................. 30
Pumps for pressure boosting........................................................ 24
Pumps in parallel................................................................................... 50
Pumps in series. .......................................................................................51
Pumps in series....................................................................... 51
Q
QH-curve...................................................................................................... 34
R
Radial forces............................................................................... 22, 44, 110
Radial impeller......................................................................................... 16
Radial velocity..........................................................................................60
Recirculation losses..............................................................................89
Recirculation zones..............................................................................89
Reference curve. ................................................................................... 105
Reference plane............................................................................. 36, 108
Regulering af omdrejningstal..................................................51, 53
Regulation of pumps............................................................................51
Relative flow angle............................................................................... 61
Relative pressure.....................................................................................33
Relative speed..........................................................................................60
Relative velocity......................................................................................60
Representative power consumption. ...................................... 56
Return channel.........................................................................................23
Reynolds’ number. ................................................................................ 83
Ring area...................................................................................................... 62
Ring diffusor...............................................................................................22
Rotational speed.................................................................................... 91
Rotor can.......................................................................................................18
Roughness......................................................................................81, 82, 85
S
Seal ..................................................................................................................18
Secondary eddy....................................................................................... 91
Secondary flow........................................................................................ 19
Self-priming................................................................................................25
Semi axial impeller............................................................................... 16
Separation. ..................................................................................................87
Sewage pumps........................................................................................ 24
Shaft bearing lossses . ........................................................................80
Shaft power............................................................................................. 104
Shaft seal......................................................................................................17
Shaft seal loss...........................................................................................80
Single channel pumpe. ............................................................... 16, 27
Slip factor.....................................................................................................73
Specific number............................................................................... 74, 95
Speed control.......................................................................................51, 53
Stage ...............................................................................................................23
Standard fluid. ......................................................................................... 38
Standby pump. ........................................................................................ 50
Start/stop regulation ....................................................................51, 53
Static pressure..........................................................................................32
Static pressure difference.................................................................35
Submersible pump. ...............................................................................14
Suction pipe...............................................................................................40
Surface roughness. ............................................................................... 91
System characteristics. ...................................................................... 49
System pressure................................................................................... 107
T
Tangential velocity...............................................................................60
Temperature........................................................................................... 101
Test bed uncertainty. .........................................................................112
Test results................................................................................................ 117
Test types....................................................................................................98
Test uncertainty. ................................................................................... 111
Throat. ............................................................................................................22
Throttle regulation..........................................................................51, 52
Throttle valve. ...........................................................................................52
Tongue. ..........................................................................................................22
Torque. .......................................................................................................... 64
Torque balance........................................................................................ 64
Torque meter.......................................................................................... 104
Total efficiency. ....................................................................................... 39
Total pressure. ..........................................................................................32
Total pressure difference. .................................................................35
Transition zone. ...................................................................................... 83
Turbulent flow...................................................................................83, 84
U
Up-thrust..................................................................................................... 44
V
Vapour bubbles.......................................................................................40
Vapour pressure............................................................................ 40, 108
Velocity diffusion....................................................................................21
Velocity measurements....................................................................98
Velocity profile. .....................................................................................100
127
Index
Velocity triangles.............................................................................60, 75
Volute. ............................................................................................................22
Volute casing.............................................................................................21
Vortex pump............................................................................................. 16
W
Water quality. ........................................................................................ 108
Water supply pumps........................................................................... 24
Wetrunner pump....................................................................................17
128
List of Symbols
Symbol
Definition
FLOW
Q
Flow, volume flow
Design flow
Qdesign
Flow through the impeller
Qimpeller
Leak flow
Qleak
m
Mass flow
Unit
[m3/s]
[m3/s]
[m3/s]
[m3/s]
[kg/s]
HEAD
H
Head
[m]
Hloss,{loss type}
Head loss in {loss type}
[m]
NPSH
Net Positive Suction Head [m]
NPSHA
NPSH Available
(Net Positive Suction Head available
in system)
[m]
NPSH Required
(The pump’s net positive suction
head system demands)
[m]
NPSHR, NPSH3%
GEOMETRIC DIMENSIONS
A
Cross-section area [m2]
b
Blade height [m]
b
Blade angle [o]
b’
Flow angle [o]
s
Gap width [m]
D, d
Diameter [m]
Hydraulic diameter [m]
Dh
k
Roughness [m]
L
Length (gap length, length of pipe) [m]
O
Perimeter [m]
r
Radius [m]
z
Height [m]
Dz
Difference in height [m]
PRESSURE
p
Pressure
∆p
Differential pressure
The fluid vapour pressure
psteam
Barometric pressure
pbar
Positive or negative pressure
pbeho
compared to pbar if the fluid is in a
closed container.
Pressure loss in {loss type}
Ploss,{loss type}
EFFICIENCIES
hhyd
hcontrol
hmotor
htot
Hydraulic efficiency
Control efficiency
Motor efficency
Total efficiency for control,
motor and hydraulics
[Pa]
[Pa]
[Pa]
[Pa]
[Pa]
[Pa]
[-]
[-]
[-]
[-]
Symbol
Definition
POWER
P
Power
Power added from the electricity
P1
supply network
Power added from motor
P2
Hydraulic power transferred to
Phyd
the fluid
Power loss in {loss type}
Ploss,{loss type}
SPEED
w
f
n
Angular frequency
Frequency
Speed
VELOCITIES
V
U
C
W
The fluid velocity
The impeller tangential velocity
The fluid absolute velocity
The fluid relative velocity
Unit
[W]
[W]
[W]
[W]
[W]
[1/s]
[Hz]
[1/min]
[m/s]
[m/s]
[m/s]
[m/s]
SPECIFIC NUMBERS
Re
Reynold’s number
Specific speed
nq
[-]
FLUID CHARACTERISTICS
[kg/m3]
r
The fluid density
n
Kinematic viscosity of the fluid
[m2/s]
MISCELLANEOUS f
Coefficient of friction
[-]
g
Gravitational acceleration
[m/s2]
z
Dimensionless pressure loss coefficient
[-]
General indices
Index
Definition
Examples
1, in
2, out
m
r
U
a
stat
dyn
geo
tot
abs
rel
Operation
At inlet, into the component
At outlet, out of the component
Meridional direction
Radial direction
Tangential direction
Axial direction
Static
Dynamic
Geodetic
Total
Absolute
Relative
Operation point
A1, Cin
A2, Cout
Cm
Wr
C1U
Ca
pstat
pdyn, Hdyn,in
pgeo
ptot
pstat,abs, ptot,abs,in
pstat,rel
Qoperation
Affinity rules
Physical properties for water
T
[°C]
pvapour
[105 Pa]
r
[kg/m3]
n
[10-6 m2/s]
0
0.00611
1000.0
1.792
4
0.00813
1000.0
1.568
1.307
10
0.01227
999.7
20
0.02337
998.2
1.004
25
0.03166
997.1
0.893
30
0.04241
995.7
0.801
40
0.07375
992.3
0.658
50
0.12335
988.1
0.554
60
0.19920
983.2
0.475
70
0.31162
977.8
0.413
80
0.47360
971.7
0.365
90
0.70109
965.2
0.326
100
1.01325
958.2
0.294
110
1.43266
950.8
0.268
120
1.98543
943.0
0.246
130
2.70132
934.7
0.228
140
3.61379
926.0
0.212
150
4.75997
916.9
0.199
160
6.18065
907.4
0.188
Valve
Stop valve
n  
QB = Q A⋅  B  
 nA  
2
nB   Scaling of
HB = HA⋅   
nA   rotational speed
3
 nB  
PB = PA ⋅   
 nA 



 Geometric
 scaling

4
 D ⋅b  
PB = PA ⋅  B4 B  
 DA ⋅ bA  
 D 2 ⋅b 
Q B = Q A⋅  B2 B 
 DA ⋅ bA 
2
 DB 
HB = HA ⋅  
 DA 
Pictograms
Pump
Pressure gauge
Heat exchanger
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