1-14-2019
1

1.1
Introduction
Water distribution system (WDS) is a component of a Water Supply System (WSS). The WSS starts with the source, treatment, storage, transport and distribution as depicted in the figure 1.
It starts with water abstraction (where pumps are used to assist in sucking/abstracting water through induction of pressure), pipes that convey raw water from the source to the raw water storage and/or the treatment plant (where raw water is treated), and storage reservoir (where treated water is stored), and distribution pipes (which convey treated water to the consumers).
Figure 1: Layout of water supply system
Source Treatment Storage Transport Distribution
Well
Handpump
Boreholes
River
Sedimentation
Sand filtration
Chlorination
Reverse osmosis
Elevated reservoir
Ground reservoir
Underground reservoir
Bulk pipe
Secondary pipe
Distribution pipe
Pipe and taps
Standpost
Water distribution system (WDS) comprises several components. It consists of pipes of various diameters, valves, and other appurtenances required to ensure adequate functioning of the system.
Water engineers are specialized in the design of water supply and distribution systems. The primary aspects to be addressed by the design engineer are as follows:
 Reliable yield of source(s) of water supply.
 Route selection: length, longitudinal section, pressure, air valves, scour valves, non- return and control valves, working space, slopes, access, servitudes, excavation conditions including rock and shoring needs, water table, services, rivers, railways, roads, traffic diversions, reinstatement requirements, environmental impacts.
 Pipe selection: diameter, internal pressure including water hammer, external loadings, wall thickness, material, internal protection, friction factor, external
2

protection, jointing, anchorage, expansion of above ground pipelines, maintenance.
 Economics- present value discount analysis.
Storage t
S
Secondary storage
Source
Treatment
Primary storage
Pump
Main line Branch line
S u b
Figure 2: Layout of water supply system
The main contents of this chapter include: o Module 1: Water Demand: This module entails the estimation of water demand for various uses. Students will be initiated to forecast water demand using various methods. o Module 2: Single pipe design . Water runs through a pressure pipe which comprise valves and other appurtenances. Students will learn how to design pipe, design and positioning of valves at critical points, and thrust bl o c ks . o Module 3: Pump design . From the resource, water is pumped to a treatment plant making use of a pump. Students will learn how a pump works and how to choose the most appropriate pump making use of pump curves as supplied by the pump manufacturer. o M o d u l e 4: Storage reservoir: There are a number of ways to calculate the size of a service reservoir. Students will be initiated to the design principles and calculations. o Module 5: Water distribution system design .
Water flows from the service reservoir to the consumers via a gravity feed pipe. Water must be distributed to make it possible for all the consumers to get water according to their needs The reticulation system will be investigated and designed making use of the Hardy Cross method.
3

1.1
Introduction
1.1.1
Overview
Pumps can be used for many reasons. Pumps can be used to pump clear water, sewage water, storm water, petrol, diesel, milk etc. All these substances have different properties and to be able to pump these substances, pumps with different properties and characteristic curves are selected.
In any pumping system, the purpose of pump is to provide sufficient pressure to overcome the operating pressure of the system to move fluid at a required flow rate. The operating pressure of the system is a function of the flow through the system and the arrangement of the system in terms of the pipe length, fittings, pipe size, the change in liquid elevation, pressure on the liquid surface, etc. To achieve a required flow through a pumping system, we need to calculate what the operating pressure of the system will be to select a suitable pump.
Water will only flow if enough energy is available to initiate the flow process. This energy can be obtained from gravity (potential energy) or from mechanical energy given to a pump by a motor or other device, and transformed t o fluid energy.
F i gu r e 3 : Pu m pi n g a r r a n g e m e n t
Water is pumped from the reservoir into a receiving tank. This kind of arrangement is used to lift water from a reservoir, or river, into a water treatment works for treatment before the water goes into the supply network. The water level in the reservoir varies but the discharge level in the receiving tanks remains constant as the water is discharged from a point above the water level.
Pumps can be divided up into two broad categories. The one category is referred to as positive displacement pumps, like what we will use for a windmill. These pumps are limited in their use
4

and can be considered as specialised pumps for limited use. The roto-dynamic pumps on the other hand have wide application of work that it can do. With different impellors, it is capable to pump both clear water as well as sewage water. Because of this ability, and because of the effectiveness of these pumps, it became very popular and is widely used. For the purpose of this study, only roto-dynamic pumps will be investigated.
1.1.2
Notion of hydraulics a) Friction
When you move a solid on a hard surface, there is friction between the object and the surface. In the case of moving fluids such as water, there is even less friction but it can become significant for long pipes. Friction can be also be high for short pipes which have a high flow rate and small diameter as in the syringe example. In fluids, friction occurs between fluid layers that are traveling at different velocities within the pipe. There is a natural tendency for the fluid velocity to be higher in the centre of the pipe than near the wall of the pipe. Friction will also be high for viscous fluids and fluids with suspended particles.
Figure 4: Flow in a closed pipe
The cause of friction is the interaction of the fluid with the pipe wall, the rougher the pipe, the higher the friction. Another cause of friction are the fittings (elbows, tees, y's, etc.) required to get the fluid from point A to B. Each one has a particular effect on the fluid streamlines. Friction depends on: o average velocity of the fluid within the pipe o viscosity o pipe surface roughness
An increase in any one of these parameters will increase friction
The amount of energy required to overcome the total friction loss within the system has to be supplied by the pump if you want to achieve the required flow rate.
5

b) Energy and head in pump systems
Energy and head are two terms that are often used in pump systems. We use energy to describe the movement of liquids in pump systems because it is easier than any other method. There are four forms of energy in pump systems: pressure, elevation, friction and velocity .
Pressure is produced at the bottom of the reservoir because the liquid fills up the container completely and its weight produces a force that is distributed over a surface which is pressure.
This type of pressure is called static pressure.
 Pressure energy is the energy that builds up when liquid or gas particles are moved slightly closer to each other.
 Elevation energy is the energy that is available to a liquid when it is at a certain height.
If you let it discharge it can drive something useful like a turbine producing electricity. It is the weight of the object W times the distance d;
 Friction energy is the energy that is lost to the environment due to the movement of the liquid through pipes and fittings in the system. It is the force of friction F times the distance the liquid is displaced or the pipe length l ;
 Velocity energy is the energy that moving objects have.
The three forms of energy: elevation, pressure and velocity interact with each other in liquids.
The energy that the pump must supply is the friction energy plus the difference in height that the liquid must be raised to which is the elevation energy.
PUMP ENERGY = FRICTION ENERGY + ELEVATION ENERGY
Figure 5: Pumping system: energy and head
 Head is defined as energy divided by weight or the amount of energy used to displace a object divided by its weight.
 the friction head FH is the friction energy divided by the weight of liquid displaced
6

Figure 6: Pumping system: energy and head c) Static head
Head is “a body of water kept in reserve at a height. Because of its height and weight the fluid produces pressure at the low point and the higher the reservoir the higher the pressure.
Figure 7: Pumping system: static head
The amount of pressure at the bottom of a reservoir is independent of its shape, for the same liquid level, the pressure at the bottom will be the same.
7

Figure 8: Pumping system: pressure in a reservoir
When a pump is used to displace a liquid to a higher level it is usually located at the low point or close to it. The head of the reservoir, which is called static head, will produce pressure on the pump that will have to be overcome once the pump is started.
To distinguish between the pressure energy produced by the discharge tank and suction tank, the head on the discharge side is called the discharge static head and on the suction side the suction static head.
Figure 9: Pumping system: static head
Usually the liquid is displaced from a suction tank to a discharge tank. The static head is then the difference in height of the discharge tank fluid surface minus the suction tank fluid surface. Static head is sometimes called total static head to indicate that the pressure energy available on both sides of the pump has been considered.
8

Since there is a difference in height between the suction and discharge flanges or connections of a pump by convention it was decided that the static head would be measured with respect to the suction flange elevation.
Figure 10: Pumping system: static head
In case of the static head on the pump with the discharge pipe end open to atmosphere, when the end is open to atmosphere then the static head is measured with respect to the pipe end.
Figure 11: Pumping system: static head
9

Sometimes the discharge pipe end is submerged as shown in figure below, then the static head will be the difference in elevation between the discharge tank fluid surface and suction tank fluid surface.
Figure 12: Pumping system: static head d) Flow rate
The flow rate will vary with the static head. If the pipe end elevation is high, the flow rate will be low.
Figure 13: Pumping system: Flow rate – positive static head
If the liquid surface of the suction tank is at the same elevation as the discharge end of the pipe then the static head will be zero and the flow rate will be limited by the friction in the system.
10

Figure 14: Pumping system: Flow rate – zero static head
The discharge pipe end is raised vertically until the flow stops, the pump cannot raise the fluid higher than this point and the discharge pressure is at its maximum.
Figure 15: Pumping system: Flow rate – low flow due to maximum head
If the discharge pipe end is lower than the liquid surface of the suction tank then the static head will be negative and the flow rate high
Figure 16: Pumping system: Flow rate – negative static head
11

Pumps are most often rated in terms of head and flow, and flow rate depends on friction. The flow rate will vary with the size and diameter of the discharge pipe. A system with a discharge pipe that is generously sized will have a high flow rate. The smaller the pipe, the less the flow. The pump you install is designed to produce a certain average flow for systems that have their pipes sized accordingly. The impeller size and its speed predispose the pump to supply the liquid at a certain flow rate. If you attempt to push that same flow through a small pipe the discharge pressure will increase and the flow will decrease. Similarly, if you try to empty a tank with a small tube, it will take a long time to drain
When the discharge pipe is long, the friction will be high and the flow rate low, and if the pipe is short the friction will be low and the flow rate high.
Figure 17: Pumping system: Flow rate – length of pipe and friction e) Total head
Total head and flow are the main criteria that are used to compare one pump with another or to select a centrifugal pump for an application. Total head is related to the discharge pressure of the pump. The amount of pressure that a pump can produce will depend on the density of the fluid.
Figure 18: Pumping system: Total head
12

Discharge static head plus friction head equals pump discharge head
Figure 19: Pumping system: Flow rate – Total head
1.2
Hydraulic Analysis of Pumps and Piping Systems
We begin our study by defining all the different terms used to describe the pump performance in the piping system.
 h s
(static suction head): it is the difference in elevation between the suction liquid level and the centerline of the pump impeller.
 h d
(static discharge head): it is the difference in elevation between the discharge liquid level and the centerline of the pump impeller.
 H stat
(static head): it is the difference (or sum) in elevation between the static discharge and the static suction heads: H stat
= h d
± h s
 H ms
(manometric suction head): it is the suction gage reading (if a manometer is installed just at the inlet of the pump, then Hms is the height to which the water will rise in the manometer).
 H md
(manometric discharge head): it is the discharge gage reading (if a manometer is installed just at the outlet of the pump, then Hmd is the height to which the water will rise in the manometer).
 H m
(manometric head): it is the increase of pressure head generated by the pump.
Manometric head of the pump is the head that the pump must effectively overcome so that the transported fluid can be pumped away Hm = h md
± h ms
 H t
(total dynamic head): it is the total head delivered by the pump
Pump can be placed in two possible position in reference to the water levels in the reservoirs.
13

Case 1
Figure 20: Pumping system: positions of a pump
Datum pump center line
H t

H m d
V d
2
 
2 g
( H m s

V s
2
2 g
)
H t

H stat
 h f d
  h m d
 h f s
  h m s

V d
2
2 g
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Case 2
V d
2
V s
2
H t

H m d

2 g

( H m s

2 g
)
In the above equations; we define:
 hf s
: is the friction losses in the suction pipe.
 hf d
: is the friction losses in the discharge (delivery) pipe.
 hm s
: is the minor losses in the suction pipe.
 hm d
: is the minor losses in the discharge (delivery) pipe
In the majority of cases a pump is selected for the particular pipeline system and flow rate value is set in advance.
 Head – energy imparted by pump to the pumped medium and attributed to unit of pumped medium mass. It is denoted by letter H and has dimension in meters. It should be clarified that the head is not geometrical characteristic and is not the height to which a pump can lift pumped medium.
 Power consumption (shaft power) – power consumed by pump during operation.
Power consumption differs from pump useful capacity consumed directly for imparting of energy to the pumped medium. Part of consumed power can be lost due to leakages, bearings friction, etc. Performance factor determines ratio between these quantities.
Calculation of these characteristic may vary for different types of pumps, which is associated with differences in their design and operating principles.
15

1.3.1
Friction losses
Frictional losses in pipes can be related to the pipe diameter, the mean flow velocity, the wall roughness factor, the density and the viscosity of the fluid. For the purposes of this study, we will use the Darcy Weisbach formula mainly for pump design. a) Darcy – Weisbach equation
Darcy Weisbach formulated a basic formula that can be used to calculate frictional losses in pipes.
The equation is written as follows: h f

R f
Q n f


8

L
2 gD
5
Q
2 

L
12 .
1 D
Q
2
5
or
Where:
 d is the diameter pipeline (m) h f
  L
D v
2
2 g
 L is the length of pipeline (m)
 λ is the friction factor ( not that some books consider it as f )
f can be calculated or obtained from Moody diagramme
λ is calculated using the equation provided below
 h f is the head loss due to friction
 v is the mean velocity of the flow (m/s)
 g is the gravitational acceleration (9.81m/s 2 )
 Q is the flow r a t e (m 3 /s}
Note : all the two equations above K
Darcy Weisbach coefficient p
and R f represent the first part of the equation; and n f
is the b) Colebrook – White equation
For flows in pipelines that do not follow the smooth or rough pipe flow Laws, Colebrook and
White have developed an equation:
1

 
2 log


2
Re
.
51

 k
3 .
7 D


or
1

 
2 log


5 .
1286
Re
0 .
89
 k
3 .
7 D


This equation can be written as follows: (Barr equation)
The friction coefficient f can be found using a modified version of the Colebrook White equation:
16

Where
 k = Roughness factor (m)
 Re = Reynolds number
 D = pipe diameter (m)
 f or λ = friction factor – very often λ = 4f
The pipe roughness factor k is a standard value obtained from standard tables and is based upon the material of the pipe, including any internal coatings, and the internal condition of the pipeline i.e. good, normal or poor.
Reynolds number is a dimensionless quantity associated with the smoothness of flow of a fluid and relating to the energy absorbed within the fluid as it moves. For any flow in pipe, Reynolds number can be calculated using the following formula:
Re = vD / u …………..(8)
Where
 u = Kinematic viscosity (m 2 /s)
  c) Hazen Williams

497
T

42 .

5

10
1
.

5
6
This empirical formula applies to the transition region, and is given by: h f
Where
H
L c d m

R f
Q n f
=
=
=
=
=

10 .
68 L
C
1 .
852 hw
D
4 .
87
Q
1 .
852 elevation height difference (m) length of pipe (m) roughness coefficient (see table 2) pipe diameter (m) constant (usually 1.85} d) Manning equation h f

R f
Q n f

10 .
29 N
/ 3
2 L
Q 2
D 16
Where:
N = Manning factor (m -1.3
s)
L = Pipe length (m)
D = pipe length (m)
Q = pipe flow (m 3 /s)
For the purpose of this study, Darcy Weisbach, Hazen Williams and Manning formula will be used in conjunction with the Hardy Cross method to perform Hydraulic pipe analyses for pipe network calculations and to calculate water flow for reservoir design.
17

When to use which formula
 Hazen-Williams formula: smooth pipes (sand roughness = 0.03 mm)’
 Manning's formula: rough pipes (sand roughness= 0.3 mm)
Which one is the best formula?
Example:
Find the head loss due to the flow of 24 l/s of water through 1,600 m of 160 mm diameter plastic pipe. If the temperature of the water 10
º
C.
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………….
18

1.3.2
Minor losses
Minor losses occur at all obstacles that changes the smooth flow of fluid in the pipe, like pipe inlets, contractions, expansions, bends, valves, branches and exits. For the purposes of this study, minor losses will be given as equivalent length of pipe. When doing a pipe design, minor losses will be available as equivalent length and will be given in a tabulated form in the catalogues of the manufacturers of the equipment.
Minonr losses hm = K 𝑣 2
2𝑔
or h m
= ξ 𝑣
2𝑔
2
Sometimes K is written as
ξ to symbolize the loss coefficient due to pipe appurtenances including bend,
T, Y, elbow etc.). Value of K
L
can be found in table and for the purpose of this subject they will be always given.
Type of Component or Fitting
Minor Loss Coefficient
- ξ -
Tee, Flanged, Dividing Line Flow
Tee, Threaded, Dividing Line Flow
Tee, Flanged, Dividing Branched Flow
Tee, Threaded , Dividing Branch Flow
Union, Threaded
Elbow, Flanged Regular 90 o
Elbow, Threaded Regular 90 o
Elbow, Threaded Regular 45 o
Elbow, Flanged Long Radius 90 o
Elbow, Threaded Long Radius 90 o
Elbow, Flanged Long Radius 45 o
Return Bend, Flanged 180 o
Return Bend, Threaded 180 o
Globe Valve, Fully Open
Angle Valve, Fully Open
Gate Valve, Fully Open
Gate Valve, 1/4 Closed
Gate Valve, 1/2 Closed
Gate Valve, 3/4 Closed
Swing Check Valve, Forward Flow
Ball Valve, Fully Open
Ball Valve, 1/3 Closed
Ball Valve, 2/3 Closed
Diaphragm Valve, Open
Diaphragm Valve, Half Open
Diaphragm Valve, 1/4 Open
Water meter
2
0.15
0.26
2.1
0.2
0.2
1.5
10
1.5
0.4
0.2
0.7
0.2
0.9
1.0
2.0
0.08
0.3
17
2
0.05
5.5
200
2.3
4.3
21
7
19

Example:
What will be the minor losses in a pipe of 1,600 m long coupled with two ball valves (1/3 and 2/3 closed), a swing check valve, a bulk water meter and 6 threaded regular elbows (90ᴼ) when the flow velocity is 2m/s?
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………….
1.3.3
Design headloss
There are a number of reasons as to why head loss occurs in pipeline design. To make sure that nothing is left out the designer should draw up a checklist. He should also take note that as pipes become older, it will deteriorate due to the quality of water conveyed, growth of fungi and the type of pipe material used (Van Breda, 2003).
Examples of such effects are:
• Corrosion of painted steel and cast-iron
• Physical damage due to transport of sand and debris
• Increasing roughness of cement and cement mortar linings in soft waters due to leaching of mortar and exposure of sand aggregate.
• Tuberculation of cast iron pipes
• Black slime growth.
The checklist should make provision for:
• Irregular laying of pipelines, Bends, Valves and off-takes
1.3.4
Pump curves (system and characteristics) a) System characteristics curve
The pump characteristic curves indicate the capability of the pump. Before we are starting to select a pump we have to calculate the system requirements. For this the system head curve should be calculated. The total head, H t
that the pump delivers includes the elevation head and the head losses incurred in the system. The friction loss and other minor losses in the pipeline depend on the velocity of the water in the pipe, and hence the total head loss can be related to the discharge rate.
 The system curve is the graphical representation of the relationship between discharge and headloss in a system of pipes;
 The system curve is completely independent of the pump characteristics
 The basic shape of the system curve is parabolic
 The system curve will start at zero flow and zero head if there is no static lift, otherwise the curve will be vertically offset from the zero head value
 The intersection between the system and pump curves is called the operating point
For a given pipeline system (including a pump or a group of pumps), a unique system headcapacity (H-Q) curve can be plotted. This curve is usually referred to as a system characteristic curve or simply system curve. It is a graphic representation of the system head and is developed by plotting the total head, over a range of flow rates starting from zero to the maximum expected value of Q. The task of the pump as a machine is to impart energy to a fluid. In steady state, the head H of the pump is equal to the head Hs of a system. The necessary head is equal to
20

H t
= H stat
+
h
L
The mean velocity is calculated from the continuity equation v = Q/A. As the flow rate is not changing on the pipe length, then h
L
= kQ 2 . where k is expressing the flow resistance. The system head curve could be expressed by the next way H = H st
+ kQ 2
This is parabola starting from the point H = H st
+kQ 2 as illustrated in figure below.
70
60
50
H (m)
40
30
20
10
0 3 6 9 12 15 18 b) Pump characteristics curve
Pump manufacturers provide information on the performance of their pumps in the form of curves, commonly called pump characteristic curves (or simply pump curves).
In pump curves the following information may be given:
the discharge on the x-axis,
the head on the left y-axis,
the pump power input on the right y-axis,
the pump efficiency as a percentage,
the speed of the pump (rpm = revolutions/min).
the NPSH of the pump.
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70
60
50
40
30
20
10
NPSH
Pump Curve ef fi ci ency
2
0
6
4
80%
70%
60%
50%
40%
0 100 200 300 400
Pump characteristics curve c) Duty point
The pump characteristic curves are very important to help select the required pump for the specified conditions. If the system curve is plotted on the pump curves in we may produce the following Figure:
The operating point of a pump is the intersection point of the pump characteristic curve and system head curve. This matching point indicates the actual working conditions, and therefore the proper pump that satisfy all required performance characteristic is selected. The intersection of the pump characteristic curve with the pipeline system curve is the only point common to both curves and determines the duty point to which the pump adjusts itself automatically (see figure above).
22

70
60
50
H (m)
40
30
20
10
Pump Curve efficienc y
System Curve
6
4
2
0
70%
60%
50%
40%
0 3 6 9 12 15 18
Q (m /hr)
To alter the duty point of a system to achieve a certain flow rate and/or pressure, the position or slope of the pump characteristic or the system curve must be altered. This can be done making use of one or more of the following possibilities:
 Change pump characteristic by: - change of pump speed
 Change of impeller diameter Pumps and/or impellers in series Pumps in parallel
 Change pipeline system curve.
To draw the system curve, the static head of the system and the energy loss as the fluid runs through the pipe is needed. To calculate the energy losses of the fluid running through the pipe, the Darcy Weisbach formula is used.
A corresponding variable for the power input, efficiency of the pump and NPSH
R
value is assigned to each duty point (see figure below). In the design of the operating data of a centrifugal pump, care should be taken to ensure that the pump works as close as possible to the point of best efficiency.
Pump performance curve
23

Example 1:
For the following pump, determine the required pipes diameter to pump 60 L/s and also calculate the needed power. Assume water temperature is 15ºC.
𝑣
2
Minor losses = 10
2𝑔
Pipe length = 10 km
Roughness = 0.15 mm
h s
= 20 m
Q (l/s)
Ht
70
31
60
35
50
38
40
40.6
η p
40 53 60 60
To get 60 l/s from the pump Hs + H
L
must be > 35 m
30
42.5
57
20
43.7
50
10
44.7
35
0
45
-
Iteration 1 :
Assume the diameter = 300 mm, then:
A =0.070 m 2 and V = 0.85 m/s
Re = 2.25 x 10 5 and Ks /D = 0.0005 and 𝞴 = 0.019 h f
= 0.019 x 10,000 x (0.85) 2 / (0.3 x 19.62)
= 23.32 m h m
= 10 𝑣
2
2𝑔
= 10 x (0.85) 2 / 2g ……………= 0.37 h s
+ h f
+ h m
= 43.69 > 35 mm
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Iteration 2 :
Assume the diameter = 350 mm
A =0.0962 m 2 and V = 0.624 m/s
Re = 1.93 x 10 5 and Ks /D = 0.00043 and 𝞴 = 0.0185 h f
= 0.0185 x 10,000 x (0.624) 2 / (0.35 x 19.62) = 10.48 m h m
= 10 𝑣 2
2𝑔
= 10 x (0.624) 2 / 2g = 0.20 m h s
+ h f
+ h m
= 30.68 < 35 mm
The pump would deliver approximately 70l/s through the 350 mm pipe and to regulate the flow to 60 l/s an additional head loss of 4.32 m by valve closure would be required.
Power required = 𝜌 𝑔𝐻𝑄 𝜂
=
1000 𝑥 9.81 𝑥 (
0.53
60
1000
)𝑥 35
= 38869.8 W or 38.87KW
Example 2 :
A pump was designed to satisfy the following system
Q (m 3 /hr) h f
(m)
3
12
Atmospheric pressure = 10.3 m
Vapour pressure = 0.25 m
Pipe diameter is 50mm
h d
= 13 m
Suction pat h
L
= 24 𝑣
2
2𝑔
Check whether the pump is suitable or not.
6
20
9
38
25

a) Draw the system curve and check the operation point
Hstat = hd + hs ………………..= 13 + 7 = 20
There is an operation point at Q = 9 m 3 /hr and H = 58 m
 NPSHr = 4.1, then check NPSHa
 V = Q/A .......................….
9/3600 𝜋
𝑥 0.05²
4
= 1.27 𝑚/𝑠
 Then hm = 24 𝑥 1.27²
2𝑔
= 2.0 m
 NPHSa = ± H s
– h fs
– Σ h ms
+
𝑃𝑎𝑡𝑚 𝛾
 = -7 – 2 + 10.3 – 0.25
−
𝑃𝑣𝑎𝑝 𝛾
 We can say that NPSHa = 1.05 <4.1
Therefore, the pump is not suitable, cavitation will occur
26

Example 3 :
A centrifugal pump running at 1000 rev/min gave the following characteristics
Dischage (m 3 /s)
Total head (m)
Efficiency 𝜂 (%)
0
22.5
0
0.075
22.2
46
The pump is used in the following system:
 Pipe length = 15 m
0.150
21.6
70
0.225
19.5
76
0.3
14.1
76
0.375
0
74
 Pipe diameter = 300 mm
 Static head = 15 m
 Darcy friction factor = 0.006
 Minor losses = 6 m equivalent pipe length
Determine
3.1 The quantity that the pump will discharge in this system (10)
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
3.2 The corresponding head (7)
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
3.3 The power required to drive the pump (6)
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
3.4 The diameter of the main (6)
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
Use h f
= 𝜆 𝐿
12.1𝑑 5
2
1.3.5
Pump speed
In order to assist in pump selection, it is deemed convenient to use the numerical quantity, specific speed (ns) as an aid to classification. Specific speed (ns) may be defined as the speed (in rpm) at which a given impeller would operate if reduced (or increased) proportionally in size so as to deliver a capacity of 1 m 3 /s through a head of 1m. In other words, specific speed ns of a pump is the required speed of one of the present pumps which are geometrically similar in all parts, which delivers a flow rate of 1 m 3 /s at a head of 1 m. It is non-rotational, speedless, particularly dimensionless number which is directly proportional to the product of the pump’ best efficiency point (BEP) speed (N) and function of its flow (Q) and inversely proportional to a function of the pump’s BEP head (H). The BEP is the point of pumping operation at which the peak hydraulic efficiency is exhibited. The BEP, flow rate and head would be those values at this point.
Specific speed ns is calculated from the expression
𝑵𝑸
½
𝑯 ¾
Where
 Ns = specific speed of a pump …….. (unitless)
 Q
 H
= flow (m 3 /s)
= total manometric head (m)
 N = rotational speed of pump (rpm)
𝑵𝒔 𝑯
¾
𝑸 ½
27

Not that the resulting units in the equation above is not relatable in practices because of being inconsequential. The importance of the pure number, with or without labels, is that of a comparative tool. Lower absolute numbers indicate a low specific speed; higher numbers mean high Ns. Once determined, the number itself is not needed as a basis for further mathematical operations and therefore essentially requires no units. Moreover, the number relates to the physical geometry of impeller style. The specific speed increases with n and Q and decreases with
H.
Example of specific speed of certain pumps
Pump type N s range ( Q - l/s, H-m)
Centrifugal
Mixed flow up to 2600
2600 to 5000
Axial flow 5000 to 10 000
Example:
Determine the speed of a pump required to deliver 2.4 m 3 /s of water when the total manometric head is 24.6 m.
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
Example:
Assume ideal pump speed Ns = 2,500 and a given total dynamic head of 27.40 m and a flow rate of 15140 l/min. Determine the rotational speed of the impeller.
 Re-arrange the equation Ns = 𝑵𝑸
½
𝑯 ¾
………to N = 𝑵𝒔 𝑯
¾
𝑸 ½
 Calculate N = 𝟐𝟓𝟎𝟎 𝒙 𝟐𝟕.𝟒𝟎
¾
𝟐𝟓𝟐.𝟑 ½
1.3.6
Power required
= 1885 rpm
The shaft power P (in kW) required by a pump is given by equation:
P = 𝑸 𝒙 𝑯 𝒙 𝒈 𝒙 𝝆 𝛈
Where
P = Power (kW)
ρ = Density (Kg/m 3 ) = 1000 kg/m 3 for water
Q = Discharge (m 3 /s)
H = Operating head (m water)
g = Acceleration due to gravity (9,81 m/s 2 )
η = Pump efficiency
28

To obtain the input power, divide P by the motor efficiency and to obtain the kVA demand, divide the input power by the power factor which usually varies between 0, 82 and 0, 95 depending on motor size and whether power factor correction equipment has been installed.
It is frequently impossible to ensure that pumps will only operate at the specified duty point.
It is therefore essential to check the power absorbed under the most adverse conditions likely to be encountered to ensure that the motors do not overload and cause the switchgear to trip.
In order to prevent overload tripping it is recommended that the installed power of the motor should exceed the calculated absorbed power by at least the margins given below:
 50% for pumps requiring up to 2kW
 40% for pumps requiring from 2 to 5 kW
 30% for pumps requiring from 5 to 10 kW
 20% for pumps requiring from 10 to 30 kW
 15% for pumps requiring from 0 to 100 kW
 10% for pumps requiring over 100 kW
Example:
What is the power required for a pump to convey 2.4 m 3 /s when the total operating head is estimated to be 24.6 m. Consider the expected efficiency to be equivalent to70%?
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………….
1.3.7
Net Positive Suction Head (NPSH) and Cavitation
NPSH (Net Positive Suction Head) is the international dimension for the calculation of the supply conditions. Net Positive Suction Head (NPSH) It is another important pump parameter; it exists in two forms. NPSH available (NPSH
A
) is the resulting pressure value of the liquid entering the pump. The pressure drop inside a pump /∆H between the suction flange and the impeller vane entrance is usually determined by lowering the pressure H and manometer at the suction flange until the Q-H pump curve starts to depart from the characteristic curve.
The NPSH is in the end a dimension of the evaporation hazard in the pump inlet socket and is influenced by the vapour pressure and the pumped liquid.
Where we take the datum through the centerline of the pump impeller inlet (eye). This difference is called the Net Positive Suction Head (NPSH), so that
𝑁𝑃𝑆𝐻 =
𝑃𝑠 𝛾
+
𝑉𝑠 2
2𝑔
−
𝑃𝑣𝑎𝑝𝑜𝑢𝑟 𝛾
There are two values of NPSH of interest: The NPSH of the pump is called NPSH required, and that of the system is called NPSH a(vailable .
 NPSH
A
: The Net Positive Suction Head Available at the pump impeller inlet.
 NPSH
R
: The Net Positive Suction Head Required by the pump to operate without experiencing damaging cavitation and a dramatic reduction in pumping production.
The NPSH
R
– is the resulting pressure drop produced as the liquid passes through the pump.
29

The first is the required NPSH, denoted (NPSH)
R
, that must be maintained or exceeded so that cavitation will not occur and usually determined experimentally and provided by the manufacturer. NPSH
R
is a value that expresses the minimum absolute pressure that must be acting on a liquid as it enters the pump impeller to avoid excessive cavitation and degradation of pump performance. The NPSH of the pump is determined by measurements carried out on the suction and delivery side of the pump. This value is to be read from the pump characteristic curve and is indicated in meter (m).
This value is to be read from the pump characteristic curve and is indicated in meter (m).
The NPSH
A
– represents the head that actually occurs for the particular piping system. NPSH
A
is a value that expresses the absolute pressure acting on a liquid as it enters the pump. It is a measure of the pressure that stands between the liquid in its current state and the formation of vapour bubbles (boiling). It is defined by the system within which the pump operates; it must be calculated. NPSHR is a characteristic of the pump itself; it is determined from tests conducted by the pump manufacturer. This value can be determined experimentally, or calculated if the system parameters are known. The NPSH is a dimension of the evaporation hazard in the pump inlet socket and is influenced by the vapour pressure and the pumped liquid.
The NPSHa(vailable) should be greater than the NPSHr(equired) for the pump to operate without cavitation.
(NPSH)
A
> (NPSH)
R
For safety reasons another 0.5 m should be integrated into the calculation, i.e.:
(NPSH)a > (NPSH)r + 0.5m a) Determination of (NPSH)
A
P
 atm
 h
S

P

S

V
S
2
2 g
  h
L
P

S 
V
S
2
2 g

P
 atm  h
S
  h
L
P

S

V
S
2
2 g

P
Vapor


P
 atm
 h
S
  h
L

P
Vapor

( NPSH )
A

P
 atm  h
S
  h
L

P
Vapor

( NPSH )
A
  h s
 h f s
  h m s

P atm


P vapor
 ɤ is the specific gravity of the liquid and ɤ = ρg ………ρ is the density of the liquid
30

Example:
A pump is designed to convey 24l/s of water through a pipe of 200 mm diameter. The friction and minor losses at the suction are 24 m and 2.4m respectively when the maximum suction head is 10.4 m. Determine the NPSHa of the pump.
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………….
Example
A pump is designed to convey 24l/s of water through a pipe of 200 mm diameter. The friction and minor losses at the suction are 24 m and 2.4m respectively when the maximum suction head is 10.4 m. Determine the NPSH
A
of the pump.
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………. b) Cavitation
A discussion of Net Positive Suction Head (NPSH) in essence is a discussion of cavitation. As fluid moves through the pump inlet its pressure is reduced. If the decrease in pressure results in the liquid equalling or going below its vapour pressure, then a portion of the liquid boils or vaporises.
In other words the pressure increases in the impeller channel and causes an implosion of the vapour bubbles. Vapour bubbles form at the intake vane channel. As the now gas-entrained liquid enters the high pressure section of the pump these “bubbles” rapidly collapse releasing large amounts of damaging energy to pump internals. This process is known as cavitation . The reason is in most of the cases a too low static suction head. The result is pitting corrosion at the impeller, pressure drops and unsteady running of the pump. Finally, cavitation
Cavitation produces a crackling sound in the pump and causes damage to the pumped product.
Generally spoken is cavitation the formation and collapse of vapour bubbles in the liquid.
Cavitation may occur in pipes, valves and in pumps. First the static pressure in the pump falls below the vapour pressure associated to the temperature of a fluid at the impeller intake vane channel.
The greater the margin between NPSH
A
and NPSH
R
, the lesser the possibility of cavitation.
Therefore, pump suction designs which maximize NPSHA are superior to those that subject the suction liquid to pressure reductions which approach the liquid vapour pressure. c) Thoma’s cavitation constant
The cavitation constant: is the ratio of (NPSH)
R
Thoma’s cavitation constant ( σ )
to the total dynamic head (Ht) is known as the
 
( NPSH )
R
H t
Note: If the cavitation constant is given, we can find the maximum allowable elevation of the pump inlet (eye) above the surface of the supply (suction) reservoir.
31

Example
A Pump has a cavitation constant = 0.12, this pump was instructed on well using UPVC pipe of
10m length and 200mm diameter, there are elbow (ke=1) and valve (ke=4.5) in the system. The flow is 35l/s and the total Dynamic Head H t
= 25m (from pump curve)
= 0.0167. Calculate the maximum suction head considering atmospheric pressure head = 9.69 m and vapour pressure head = 0.2 m. Consider the atmospheric pressure head equal to 9.69 m and vapour pressure head as 0.2 m.
 NPSHr = σ x H t
…………0.12 x 25 = 3
 NPSHa = ± H s
– h fs
– Σ h ms
+ 𝑃𝑎𝑡𝑚 𝛾
−
𝑃𝑣𝑎𝑝 𝛾
 Vs = Q/A……………(0.035) / ( 𝜋 /4 x 0.2
2 ) …… = 1.11 m/s
 Then h valve
= 1.11
2 /(2 x 9.81) = 0.063 m and h elbow
= 4.5 [(1.11
2 /(2 x 9.81)] = 0.283 m
 Then hfs = 𝜆 𝐿𝑉 2
2𝑔𝐷
0.0167
𝑥 10 𝑥 1.111
2
= 0.053
2 𝑥 9.81 𝑥 0.2
 3 = H s
– 0.053 – (0.283 + 0.063) +
9.69
1
 Hs = - 6.088 m
+
0.2
1
 The pump suction is below the centreline d) Suction specific speed
In addition to pump specific speed, there exists two values of suction specific speed depending on the form of the NPSH used.
A suction specific speed available ( N
SSA
) can also be calculated using the equation N
SSA
=
𝑵𝑸
½
𝑵𝑷𝑺𝑯𝒂 ¾
A suction specific speed required (N
SSR
) can be obtained N
SSR
= 𝑵𝑸
½
𝑵𝑷𝑺𝑯𝒓 ¾
Generally, the larger the numerical value of N
SSR
, the more favourable the pump suction capabilities are. Normal pump designs exhibit N
SSR
values ranging from 6,000 to 12,000. Greater values are not uncommon. N
SSA
and N
SSR
are just numbers that describe the system condition available to the pump’s suction side. N
SSR
must exceed N
SSA
in order to preclude liquid cavitation.
The difference between these two is known as margin. Ideally N
SSR
≫N
SSA
While it is well known that pumps have been determined largely by experience, it is fairly well known among designers that cavitation occurs beyond Nss = 10,000 (based on cold water).
Special pumps designs can accommodate suction specific speeds of 12,000. Occasionally, cavitation can be experienced even when the value of NSS is well below the 12,000 limit.
NSSA can be used to determine or select the optimum rotational speed. Much like NSS number of
2,500 through extensive pump industry experience, it has been determined that the optimum suction conditions exist at suction specific speeds led than 8,500. Hence, the probable upper limit of rotational speed that a pump should be subjected to in order to achieve the optimum suction conditions can be calculated using the equation 𝟖𝟓𝟎𝟎 𝐱 𝑵𝑷𝑺𝑯𝒂
¾
𝑸 ½
32

Example
A pump has an ideal suction speed NSS <8,500, with a NPSHa = 13.72 m and a flow rate Q of
227,100 l/min. Find the upper limit rotational speed N (i.e. real speed which should not be exceeded to avoid cavitation).
 Rearrange N
SSA
=
𝑁𝑄 ½
𝑁𝑃𝑆𝐻𝑎 ¾
𝑡𝑜 𝑜𝑏𝑡𝑎𝑖𝑛 N =
8500 x 𝑁𝑃𝑆𝐻𝑎
𝑄 ½
¾
 Then find N value as 8500 x 13.72
¾
3785 ½
= 986.54 rpm
Measures for the avoidance of cavitation
In general, cavitation can be prevented by:
 Reducing the pressure drop in the suction pipe by a larger suction pipe diameter, shorter suction pipe length and less valves or bends
 Increasing the static suction head and/or supply pressure, e.g. by an upstream impeller
(Inducer)
 Lowering the temperature of the pumped liquid
In practices, the operator and pump manufacturer should take some of the measures outlined below:
Measures by the operator –
 Reduction of geodetic suction lift or increase in suction head –
 Short suction line with largest possible cross section –
 Valves, bends, curves avoided where possible or the maximum radii used –
 The temperature of the fluid to be kept to a minimum –
 Application of a gas pressure to the surface of the liquid in closed suction or supply vessels.
Measures by pump manufacturers –
 Impellers with double curvature blades drawn well forward into the suction orifice -
Avoidance of short deflection radii at the blade cover –
 Reduction of the thickness of the impeller blades –
 Use of a smaller blade inlet angle -
 Reduction in speed –
 Fitting an inducer –
 Aligning the flow to the impeller by fitting a guide vane in the inlet connection
1.3.8
Multiple pumps operation
To install a pumping station that can be effectively operated over a large range of fluctuations in both discharge and pressure head, it may be advantageous to install several identical pumps at the station.
33

a) Parallel Operation
Pumping stations frequently contain several (two or more) pumps in a parallel arrangement.
Manifold
Q total
Q total
=Q
1
+Q
2
+Q
3
Pump
Pump
In this configuration any number of the pumps can be operated simultaneously. The objective being to deliver a range of discharges, i.e.; the discharge is increased but the pressure head remains the same as with a single pump. This is a common feature of sewage pumping stations where the inflow rate varies during the day. By automatic switching according to the level in the suction reservoir any number of the pumps can be brought into operation.
How to draw the pump curve for pumps in parallel?
The manufacturer gives the pump curve for a single pump operation only. If two or pumps are in operation, the pumps curve should be calculated and drawn using the single pump curve. For pumps in parallel, the curve of two pumps, for example, is produced by adding the discharges of the two pumps at the same head (assuming identical pumps).
Pumps parallel
Q

Q
1

Q
2

Q
3

  

Q n
 j

 n j

1
Q
H m

H m1

H m2

H m3
    
H mn
34

b) Series operation
The series configuration which is used whenever we need to increase the pressure head and keep the discharge approximately the same as that of a single pump. This configuration is the basis of multistage pumps; the discharge from the first pump (or stage) is delivered to the inlet of the second pump, and so on. The same discharge passes through each pump receiving a pressure boost in doing so
Pump Pump
Q
H total
=H
1
+H
2
+H
3
How to draw the pump curve for pumps in series?
The manufacturer gives the pump curve for a single pump operation only. For pumps in series, the curve of two pumps, for example, is produced by adding the heads of the two pumps at the same discharge. Note that, of course, all pumps in a series system must be operating simultaneously.
35

Example 1
The characteristic curve of a pump is to be plotted from the test results below.
Dischage (m 3 /s)
Total head (m)
Efficiency 𝜂 (%)
0
73
0
0.02
63
71
0.03
55
81
0.04
45
71
0.05
32
50
0.055
17
29
The pump is connected to a water pumping system which has a system-head relationship of H syst
32 + 5250Q 2 with H in m and Q in m 3 /s. Determine (i) the flow rate, (ii) the head and (iii) the efficiency if: a) A single pump is connected in the system b) Two identical pumps are connected in parallel in the system c) Two identical pumps are connected in series in the system
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………….
Example 2
The characteristics curves of a centrifugal pump is tabulated below. It is proposed to install two of these pumps in order to pump water through a 200 mm diameter water main of effective length 2000m. The Darcy-Weisbach friction factor can be taken as 0.018. Allow for minor losses totalling 40 V 2 /2g. The centreline of the pump is located 5 m above the constant water level of the “suction” (upstream) reservoir and 25 m below the delivery point of the receiving
(downstream) reservoir.
Characteristic curves of centrifugal pump.
Discharge Q (l/s)
Total Head H (m)
0
40
10
37
20
33
30
26
40
16 ŋ (%) - 65 80 70 40
Plot the necessary pump curves and determine the flow rate, corresponding head and power required: a) When a single pump is connected in the pipeline system. b) When running two of these identical pumps in parallel in the pipeline system. c) When running two of these identical pumps in series in the pipeline system.
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………….
Example 3
Two identical pumps having the tabulated characteristics are to be installed in a pumping station to deliver sewage to a settling tank through a 200 mm μPVC pipeline 2.5 km long. The static lift is
15 m . Allowing for minor head losses of 10 V 2 /2g and assuming an effective roughness of 0.15 mm calculate the discharge and power consumption if the pumps were to be connected: a) In parallel b) In series
Discharge ( l/s )
Total head ( m )
Overall efficiency ( % )
0
30
-
10
27.5
44
20
23.5
58
30
17.0
50
40
7.5
18
36

Example 4
A variable speed pump having the tabulated characteristics, at 1450 rev/min , is installed in a pumping station to handle variable inflows. Static lift = 15 m ; diameter of pipeline = 250 mm ; length = 2000 m ; k = 0.06 mm . Minor loss = 10.0 V 2 /2g .
Determine the total head of the pump and discharge in the pipeline at pump speeds of 1000 rev/min and 500 rev/min.
Pump characteristics at 1450 rev/min
Discharge ( l/s )
Total head ( m )
0
45.0
10
44.0
20
42.5
30
39.5
40
35.0
50
29.0
60
20.0
70
6.0
………………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………….
1.3.9
Law of similitudes
Pump manufacturers usually provide sets of characteristic curves for pump speeds (N) of around
2 900 rpm and 1450 rpm for two and four pole motors respectively, and for each speed the characteristic curves for various impeller diameters (D) are provided. The Civil Designer software program has a variety of pumps for different situations to choose from. Virtual simulations can be set in motion to analyse the effectiveness of the pump].
As a rule of thumb it can be calculated that the NPSH- required changes as the square of the speed or the impeller diameter changes. The efficiency usually increases with increased pump speed although there are no accurate methods available to calculate the change.
The actual performance characteristics curves of pumps have to be determined by experimental testing. Furthermore, pumps belonging to the same family, i.e.; being of the same design but manufactured in different sizes and, thus, constituting a series of geometrically similar machines, may also run at different speeds within practical limits. Each size and speed combination will produce a unique characteristics curve, so that for one family of pumps the number of characteristics curves needed to be determined is impossibly large.
The problem is solved by the application of dimensional analysis and by replacing the variables by dimensionless groups so obtained. These dimensionless groups provide the similarity
(affinity) laws governing the relationships between the variables within one family of geometrically similar pumps.
Thus, the similarity laws enable us to obtain a set of characteristic curves for a pump from the known test data of a geometrically similar pump. a) Affinity laws
The pump speed needs to be reduced in order to achieve the required flow at the TWL and the required speed can be calculated using the affinity laws:
37

First affinity law – Flow is proportional to the shaft speed, i.e.
𝑄1
𝑄2
=
𝑁1
𝑁2
Where
 Q = Flow through the pipe (m 3 /sec)
 N = Shaft speed (rpm)
Second affinity law – Head is proportional to the square of the shaft speed, i.e.,
𝐻1
𝐻2
=
(𝑁1)²
(𝑁2)²
Where o Q = discharge (m3/s, or l/s). o H = pump head (m). o N = pump rotational speed (rpm). o Pi = power input (HP, or kw).
 Change in pump speed (constant size)
If a pump delivers a discharge Q1 at a head H1 when running at speed N1, the corresponding values when the same pump is running at speed N2 are given by the similarity (affinity) laws:
Q
Q
2
1

N
N
2
1
H
H
2
1


N
N
2
1

2
P i 1
N
2
3
P i 2 

N
1

 Change in pump size (constant speed)
A change in pump size and therefore, impeller diameter (D), results in a new set of characteristic curves using the following similarity (affinity) laws:
Q
Q
2
1


D
D
2
1

3
H
H
2
1


D
D
2
1

2
P i 2
5
P i 1


D
2
D
1

Where D = impeller diameter (m, cm).
Example 1
Determine the total head of the pump and discharge in the pipeline at pump speeds of 1000 rev/min and 500 rev/min. Pump characteristics at 1450 rev/min.
Discharge (l/s)
Total head (m)
0
45
10
44
20
42.5
30
39.5
40
35
50
29
60
20
70
6
38

The system curve is computed giving the following data:
Discharge (l/s)
System head (m)
20
16.40
40
20.08
60
26.12
80
34.23
N (rev/min
1450
1000
500
Q
2
= 20 x (1000/1450)
= 13.79 l/s
H
2
= 42.5 (1000/1450) 2
= 20.2 m
Q
1
N
1
Q
2
N
2
Q
2
H
2
0
45
0
21.4
0
5.35
20
42.5
13.79
20.2
6.9
5.05
40
35
27.59
16.65
13.90
4.16
60
20
41.38
9.50
20.70
2.40
Operating conditions o At N = 1450 rev/min: Q = 55l/d and H m
= 25.0 m o At N = 1000 rev/min: Q = 33l/s and H m
= 18.5 m o At N = 500 rev/min there is no discharge produced
For this pump, at the maximum head of 10.39 m and a flow of 2500 m 3 /hr (0.694m
3 /s) the pump efficiency is 84%. Therefore, the power requirement is:
P = 0.694 𝑥 10.39 𝑥 9.81 𝑥 1000
0.84
P = 84210 W = 84. 21 k W
39

Hence, we can say that to overcome the required head of 10.39 m, we need a variable speed pump with 84.21 W.
Example 2
A centrifugal pump running at 1000 rpm gave the following relation between head and discharge:
Discharge (m
3
/min)
0 4.5
9.0
13.5
18.0
22.5
Head (m) 22.5
22.2
21.6
19.5
14.1
0
The pump is connected to a 300 mm suction and delivery pipe the total length of which is 69 m and the discharge to atmosphere is 15 m above sump level. The entrance loss is equivalent to an additional 6m of pipe and friction factor is assumed as 0.024. o Calculate the discharge in m 3 /min. o If it is required to adjust the flow by regulating the pump speed, estimate the speed to reduce the flow to one-half
Solution
1) System curve:
The head required from pump = static + friction + velocity head
H t

H stat
 h f d
  h m d
 h f s
  h m s

V d
2
2 g
H stat
= 15 m
Friction losses (including equivalent entrance losses) =
 h fs
 h ms
  h fd
 h md

8
 f
2
L Q
2 g D
5
(consider f as 𝞴 )

8

0 .
024


2 g
( 69
( 0 .
3 )
5

6 )
Q
2
……………….. = 61.21Q
2 where Q in m 3 /s
V d
Velocity head in delivery pipe =
2 g
2

2
1 g 


Q
A 

 2

10 .
2 Q
2
Thus:
 Ht = 15 + 71.41Q
2 …… where Q in m 3 /s
 H t
= 15 + 19.83 x 10 -3 Q 2 …..where Q in m 3 /min
From this equation and the figures given in the problem the following table is compiled:
Discharge (m 3 /min) 0 4.5 9.0 13.5 18.0 22.5
Head available (m) 22.5 22.2 21.6 19.5 14.1 0
Head required (m) 15.0 15.4 16.6 18.6 21.4 25.0
40

Pump and Sytem Curves
18
16
14
12
10
8
6
4
2
0
28
26
24
22
20
0
Pump Curve
System Curve
2 4 6 8 10 12 14 16 18 20 22 24
Discharge, Q (m
3
/min)
From the figure above, the operating point is:
QA = 14 m 3 /min
HA = 19 m
At reduced speed: For half flow (Q = 7 m 3 /min) there will be a new operating point B at which:
QB = 7 m 3 /min
HB = 16 m
1.4
Pump selection
As pointed out in the introduction, pumping of water is in essence, adding energy to make the water flow to where it is needed. The civil engineer will analyse the proposed pipeline to establish all the energy needs, energy build up and energy losses that will occur as the water flows through the pipe. This information will help the designer to place valves needed to control the flow of the water, to make the pipe safe by preventing water hammer, to dissipate surplus energy build-up in the pipeline and to clean the pipe preventing pipe corrosion and to supply enough energy to let the water flow.
To be able to select a pump, the designer must have a thorough understanding of pumps, how it works, the various parts needed in a pump, the pump curves, the fluids and where it comes from, the system curve and how to draw the system curve, how to find the duty point and use the duty point to read off all the necessary detail from the pump curves.
Pumps can be used for many reasons. Pumps can be used to pump clear water, sewage water, storm water, petrol, diesel, milk etc. All these substances have different properties and to be able to pump these substances, pumps with different properties and characteristic curves are selected.
When you buy a pump you don’t specify the maximum total head that the pump can deliver since this occurs at zero flow. You instead specify the total head that occurs at your required flow rate.
This head will depend on the maximum height you need to reach with respect to the suction tank fluid surface and the friction loss in your system.
41

It has been seen that the efficiency of a pump depends on the discharge, head, and power requirement of the pump.
In selecting a particular pump for a given system:
 The design conditions are specified and a pump is selected for the range of applications.
 A system characteristic curve (H-Q) is then prepared.
 The H-Q curve is then matched to the pump characteristics chart which is provided by the manufacturer.
 The matching point (operating point) indicates the actual working conditions.
1.5
Centrifugal pumps
Pumps can be divided up into two broad categories. The one category is referred to as positive displacement pumps, like what we will use for a windmill. These pumps are limited in their use and can be considered as specialised pumps for limited use. The roto-dynamic pumps on the other hand have wide application of work that it can do. With different impellors, it is capable to pump both clear water as well as sewage water. Because of this ability, and because of the effectiveness of these pumps, it became very popular and is widely used. For the purpose of this study, only roto-dynamic pumps will be investigated. Note that a centrifugal pump can be either submersible
(wet) or dry.
1.5.1
Description
The rota-dynamic pump is so called because of the fact that the pressure head created is largely attributable to centrifugal action. A centrifugal pump consists basically of an impeller (see figure
2.2), which consists of a number of vanes, curved backwards from the direction of motion, and rotates in a casing. Water enters at the centre or eye of the impeller with high velocity and is forced outwards in a radial direction.
In the casing, part of the lunatic energy in the fluid is converted into pressure as the flow progresses into the discharge pipe. The efficiency of the conversion depends on the type and shape of the casing.
Pump components
 rotating element - impeller
 encloses the rotating element and seals the pressurized liquid inside – casing or housing
42

43

 Impeller
 is the rotating part of the centrifugal pump.
 It consists of a series of backwards curved vanes (blades).
 The impeller is driven by a shaft which is connected to the shaft of an electric motor
 Casing
 Which is an air-tight passage surrounding the impeller
 designed to direct the liquid to the impeller and lead it away
 Volute casing. It is of spiral type in which the area of the flow increases gradually.
 Suction Pipe.
 Delivery Pipe.
 The Shaft: which is the bar by which the power is transmitted from the motor drive to the impeller.
 The driving motor: which is responsible for rotating the shaft. It can be mounted directly on the pump, above it, or adjacent to it.
1.6
Guidelines for selection of suitable pump
Pump specifications
Pumps are designed by mechanical engineers. The civil engineer will design a pipe system, and then look for a pump to supply the needs for the system. How to design the pipe system will be discussed in chapter 3.
The general principles for selecting a pump can be summarized as follows:
 Determine the flow rate
 Determine the static head: This a matter of taking measurements of the height between the suction tank fluid surface and the discharge pipe end height or the discharge tank fluid surface elevation.
 Determine the friction head: the friction head depends on the flow rate, the pipe size and the pipe length.
 Calculate the total head - is the sum of the static head (remember that the static head can be positive or negative) and the friction head
 Select the pump - based on the pump manufacturer’s catalogue information using the total head and flow required as well as suitability to the application.
The first step to find the correct pump is to find the duty point for the pump and the system.
The duty point is the point where the pump curve and the system curve cross and at this point maximum efficiency will occur. To establish where this point is, one has to go about as follows:
 Draw a system curve (H system = H stat
+ H f
+ 𝑣
2
2𝑔
) over the pump curve (H-Q curve)
 The H-Q curve is provided by the pump manufacturer. For each pump, the curve will depend on the impeller, the pump casing, the revolutions per minute of the impellor and the material the pump is made of. Where the two curves cross, is the duty point of the pump. At this point, the designer can read off the delivery height, the flow capacity and the efficiency of the pump. This will enable the designer to see if the pump fits the needs of the system.
44

 Some pumps will have multiple H-Q curves for different impellers and the designer will be able to find a suitable pump by merely checking which impeller will give the highest efficiency to supply to the need. Once decided on a specific pump and impeller, the designer will look at the NPSH and the power needed to operate the pump. This can all be read from the pump curves supplied b y the pump manufacturer.
45

Self-study examples
Example 1
A centrifugal pump has a 100 mm diameter suction pipe and a 75 mm diameter delivery pipe.
When discharging 15 l/s of water, the inlet water mercury manometer with one limb exposed to the atmosphere recorded a vacuum deflection of 198 mm; the mercury level on the suction side was 100 mm below the pipe centerline. The delivery pressure gauge, 200 mm above the pump inlet, recorded a pressure of 0.95 bar. The measured in put power was 3.2 kW. Calculate the pump efficiency. (See figure below)
Solution
Manometric head = rise in total head
Hm
 p
2
 g

V
2
2
2 g
 z


 p
2
 g

0 .
95 * 10 .
198
 p
1
 g

V
1
2
2 g
9 .
65 m of

 1bar water
V
2
V
1

3 .
39 m / s ;

1 .
91 m / s ;
Then Hm

9.69

0.588

0.2
(-2.793

0.186)

13.09m
Efficiency (

 
)

V
2
2
2 g
V
1
2
2 g output power input power
3 .
2

0 .
588 m

0 .
186 m

3 .
2

0 .
015

13 .
09

10.198m
of water
 gQHm ( watts )
3200 ( watts )

0 .
602 ( 60 .
2 percent )
46

Example 2 (Pipeline selection in pumping system design)
As existing pump, having the tabulated characteristics is to be used to pump raw sewage to a treatment plant through a static lift of 20 m. An uPVC pipeline 10 km long is to be used. Allowing for minor losses totaling 10 𝑣 2
2𝑔
and taking an effective roughness of 0.15 mm because of sliming, and water temperature of 15
º
C, select a suitable commercially available pipe size to achieve a discharge of 60 l/s. Calculate this power consumption..
Discharge (l/s) 0 10 20 30 40 50
Total head (m) 45 44.7 43.7 42.5 40.6 38
60
35
70
31
Overall efficiency
35 50 57 60 60 53 40
(per cent)
Solution
At 60 l/s, total head = 35.0 m, therefore the sum of the static lift and pipeline losses must not exceed 35.0 m.
Try 300 mm diameter: A=0.0707 m 2 V=0.85 m/s
Re = 2.25*100000 ; k/D = 0.0005  =0.019= f
Friction head loss 
0 .
019

10000

0 .
3

19 .
62
0 .
85
2
Hs + h f
= 43.32 (> 35) pipe diameter too small

23 .
32 m
Try 350 mm diameter: A=0.0962 m 2 V=0.624 m/s;
Re = 1.93*100000 ; k/D = 0.00043  =0.0185
H f
=10.48m; hm

10

19
0
.
.
624
612
Hs + h f
+ hm =30.68 (< 35 m) O.K
2

0 .
2 m
47

The pump would deliver approximately 70 l/s through the 350 mm pipe and to regulate the flow to 60 l/s an additional head loss of 4.32 m by valve closure would be required.
Power consumption P
Example 3:

1000

9 .
81

0 .
06

0 .
55

1000
35

38 .
85 kW
Calculate the pipe friction loss of a 200 mm pipe with a flow rate of 12.6 l/s for water at 12
º
C and a pipe length of 900m. Consider the roughness is 0.00015m.
Example 4
A centrifugal pump has the following characteristics:
Discharge Q ( l/s )
Head H ( m )
Efficiency ( % )
0
16.0
0
6
15.0
44
12
12.7
60
18
9.5
63
24
5.6
53
30
1.0
10
It is used to lift water between two reservoirs, with free surfaces separated by 4.5 m , through 600 m of 125 mm diameter pipe with friction factor 0.004. The pipeline includes four standard elbows and two gate valves, and has square entrances into each reservoir; all connections are flanged.
Determine: a) The rate of flow delivered by the pump. b) The power input required.
Example 5
An existing pump, having the tabulated characteristics, is to be used to pump raw sewage to a treatment plant through a static lift of 20 m . A μPVC pipeline 10 km long is to be used. Allowing for minor losses totaling 10 𝑣
2
2𝑔 and taking an effective roughness of 0.15 mm because of sliming, select a suitable commercially available pipe size to achieve a discharge of 60 l/s . Calculate the power consumption.
Discharge ( l/s )
Total head ( m )
Overall efficiency ( % ) -
0
45
10
44.7
35
20
43.7
50
30
42.5
57
40
40.6
60
50
38.0
60
60
35.0
53
70
31.0
40
Example 5
Calculate the steady discharge and the power consumption of water pumped between two tanks in a system. The suction and the delivery pipe diameter is 200 mm and the total length is 2000 m. The friction coefficient is 0.0096. The losses in valves, bends plus the velocity heads amount to 6.2 𝑣
2
2𝑔
𝑎𝑛𝑑 ℎ𝑓 = 𝜆𝐿𝑣
2 𝑑2𝑔
. The static lift is 10 m. The pump characteristics are given as follows:
Discharge ( l/s )
Total head ( m )
Efficiency ( % )
0
25
-
10
23.2
45
20
20.8
65
30
16.5
71
40
12.4
65
50
7.3
45
48

Hint:
Hm = Hstat +HL
= 10 + 𝜆𝐿𝑣 𝑑2𝑔
2
+6.2 𝑣 2
2𝑔
= 10 + 5.209V
2
Make V factor of Q from V = Q/A equation
Hm = 10 + 5277.821Q
2
Apply values of Q provided in the table to obtain Hm
Plot the graph and read values as:
H = 15.5 m
Q = 0.0325 m 3 /s 𝜂 = 70%
Calculate Power as = 7.06 KW
49