1. Civil Engineering Department
Prof. Majed Abu-Zreig
Hydraulics and Hydrology – CE 352
Chapter 5
Water Pumps
10/30/2012 1
2. Definition
• Water pumps are devices designed to
convert mechanical energy to hydraulic
energy.
• They are used to move water from lower
points to higher points with a required
discharge and pressure head.
• turbines turn fluid energy into electrical or
mechanical
4. • This classification is based on the
way by which the water leaves the
rotating part of the pump.
• In radial-flow pump the water
leaves the impeller in radial
direction,
• while in the axial-flow pump the
water leaves the propeller in the
axial direction.
• In the mixed-flow pump the water
leaves the impeller in an inclined
direction having both radial and
axial components
5. In a centrifugal pump flow enters along
the axis and is expelled radially. (The
reverse is true for a turbine.)
An axial-flow pump is like a propeller;
the direction of the flow is unchanged
after passing through the device.
A mixed-flow device is a hybrid device,
used for intermediate heads.
17. Selection of A Pump
It has been seen that the efficiency of a pump depends on the discharge,
head, and power requirement of the pump. The approximate ranges of
application of each type of pump are indicated in the following Figure.
18. Centrifugal Pumps
• Demour’s centrifugal pump - 1730
• Theory
– conservation of angular momentum
– conversion of kinetic energy to potential energy
• Pump components
– rotating element - impeller
– encloses the rotating element and seals the pressurized
liquid inside – casing or housing
19. Centrifugal Pumps
• Broad range of applicable flows and heads
• Higher heads can be achieved by increasing the
diameter or the rotational speed of the impeller
Flow Expansion
Discharge
Casing
Suction Eye Impeller
Impeller
Vanes
20. Centrifugal Pump:
• Centrifugal pumps (radial-flow pumps) are the most
used pumps for hydraulic purposes. For this reason,
their hydraulics will be studied in the following
sections.
21.
22.
23. Main Parts of Centrifugal Pumps
1. Impeller:
• which 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.
24. Main Parts of Centrifugal Pumps
2. 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.
25. 3. Suction Pipe.
4. Delivery Pipe.
5. The Shaft: which is the bar by which the
power is transmitted from the motor drive to
the impeller.
6. The driving motor: which is responsible for
rotating the shaft. It can be mounted directly
on the pump, above it, or adjacent to it.
26. Mechanics
• Fluid enters at the eye of the impeller
and flows outward. As it does so it picks
up the tangential velocity of the impeller
vanes (or blades) which increases
linearly with radius (u = r ). At exit the
fluid is expelled nearly tangentially at
high velocity (with kinetic energy
subsequently converted to pressure
energy in the expanding volute). At the
same time fluid is sucked in through the
inlet to take its place and maintain
continuous flow.
27. Mechanics cont..
• The analysis makes use of angular dynamics:
• torque = rate of change of angular momentum
• power = torque ´ angular velocity
• The absolute velocity of the fluid is the vector
sum of: impeller velocity (tangential)+ velocity
relative to the impeller (parallel to the vanes)
• Write:
• u for the impeller velocity (u = wr )
• n for the fluid velocity relative to the impeller
• V = u + w for the absolute velocity
28. Pump Efficiency
Pump output Power Po QH t
p
Power input to the pump Pi Tw
Pi is the power input delivered from the motor to the
impeller of the pump.
Pi Qw (vto ro vti ri ) Q(vtouo vui )
Motor efficiency
: Pi
m Pm is the power input delivered to the motor.
Pm
Overall efficiency of the motor-pump system:
o p m Po
o
Pm
29. Pump Efficiency
• Example.
• A pump lifts water from a large tank at a rate of
30 L/s. If the input power is 10 kW and the
pump is operating at an efficiency of 40%, find:
• (a) the head developed across the pump;
• (b) the maximum height to which it can raise
water if the delivery pipe is vertical, with D=100
mm and friction factor f = 0.015.
• Answer: (a) 13.6 m; (b) 12.2 m
30. 5.4 Pump Characteristic Curves
• 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.
33. Constant- and Variable-Speed Pumps
• The speed of the pump is specified by the angular
speed of the impeller which is measured in
revolution per minutes (rpm).
• Based on this speed, N , pumps can be divided into
two types:
• Constant-speed pumps
• Variable-speed pumps
34. Constant-speed pumps
• For this type, the angular speed , N , is constant.
• There is only one pump curve which represents the
performance of the pump
NPSH - m
6
4
NPSH 2
0
70 Pump Curve 80%
60 70%
Efficiency %
50 60%
H (m)
cy
cien
40 50%
effi
40%
30
20
10
0 100 200 300 400
Q (m3/hr)
35. Variable-speed pumps
• For this type, the angular
speed , N , is variable, i.e.;
pump can operate at
different speeds.
• The pump performance is
presented by several pump
curves, one for each speed
• Each curve is used to suit
certain operating
requirements of the system.
36. Similarity Laws:
Affinity laws
• 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.
37. • 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.
38. (a) 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:
2 3
Q2 N 2 H2 N 2 Pi 2 N 2
Q1 N1 H1 N1 Pi1 N1
where Q = discharge (m3/s, or l/s).
H = pump head (m).
N = pump rotational speed (rpm).
Pi = power input (HP, or kw).
39. • Therefore, if the pump
curve for speed N1 is
given, we can construct N1
the pump curve for the
speed N2 using previous
relationships. N2
Effect of speed change on pump
characteristic curves.
40. (b) 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:
3 2 5
Q2 D2 H2 D2 P 2 D2
i
Q1 D1 H1 D1 P1 D1
i
where D = impeller diameter (m, cm).
Note : D indicated the size of the pump
41. 5.5 Hydraulic Analysis of Pumps and
Piping Systems
• Pump can be placed in two possible position in
reference to the water levels in the reservoirs.
• We begin our study by defining all the
different terms used to describe the pump
performance in the piping system.
43. Ht Dynamic Head
•Ht (total dynamic head): it is the total head
delivered by the pump:
Vd2
H t H stat hl
2g
Vd2
H t H stat h f d hmd h f s hms
2g
In the above equations; we define:
hfs : is the friction losses in the suction pipe.
hfd : is the friction losses in the discharge (delivery) pipe.
hms : is the minor losses in the suction pipe.
hmd: is the minor losses in the discharge (delivery) pipe.
Vd : velcity of the dilevery pipe
44. Duty Point
In general the pump has to supply enough energy to:
• lift water through a certain height the static lift Hs;
• overcome losses dependent on the discharge, Q.
Thus the system head is s losses H = Hs + hf. :Losses are
proportional to Qm, so that the system characteristic is often
quadratic:
Ht =hs +kQm (m =2 or m=1.85)
45. System Curve
70 H p ( z2 z1 ) kQ m
60
50 hf kQ m
H (m) 40
30
H t Z 2 Z1 hL
20
Static head (z2-z1)=Hs
H t H stat hL
10
0 3 6 9 12 15 18
Q (m3 /hr)
46. Duty Point
System
Duty characteristic
point
Pump
characteristic
48. Selection of A 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.
49. • 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:
Matching the system and pump curves.
• The point of intersection is called the operating point.
• This matching point indicates the actual working conditions,
and therefore the proper pump that satisfy all required
performance characteristic is selected.
50. System Characteristic Curve
The total head, Ht , 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
For a given pipeline system (including a pump or a group of
pumps), a unique system head-capacity (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.
55. System Curve & Pump Curve cases
Pump selection
Pump Curve
System Curve
Pump Curve
System Curve
Pump Curve
System Curve
56. Example 2
For the following pump, determine the required pipes diameter to
pump 60 L/s and also calculate the needed power.
Minor losses 10 v2/2g
Pipe length 10 km
roughness = 0.15 mm
hs = 20 m
Q 70 60 50 40 30 20 10 0
L/s
Ht 31 35 38 40.6 42.5 43.7 44.7 45
P 40 53 60 60 57 50 35 -
57. To get 60 L/s from the pump hs + hL must be < 35 m
Assume the diameter = 300mm
Then:
A 0.070m 2 , V 0.85m / s
Re 2.25 10 5 , K S / D 0.0005, f 0.019
0.019 10000 0.85
2
hf 23.32m
0.3 19.62
10 V 2 10 0.85
2
hm 0.37m
2g 2g
hs h f hm 43.69m 35m
58. Assume the diameter = 350mm
Then:
A 0.0962m 2 ,V 0.624m / s
Re 1.93 10 5 , K S / D 0.00043, f 0.0185
h f 10.48m,
10 V 2 10 0.624
2
hm 0.2m
2g 2g
hs h f hm 30.68m 35m
QH t 1000 9.81 1000 35
60
Pi 38869.8W 38.87kW
p 0.53
59. Example 3
A pump was designed to satisfy the following system
Q (m3/hr) 3 6 9
hf (m) 12 20 38
Check whether the pump is suitable or not
atm. pressure head 10.3 m
Vapour pressure head 0.25m
Pipe diameter is 50mm
hd 13m
24 V 2
suction Part hL
2g
60.
61. 1- Draw the system curve and check the operation point
HSTAT h d hS 13 7 20m
62. There are an operation point at:
Q = 9 m3/hr H =58m
NPSHR =4.1
Then Check NPSHA
Q 9 / 3600
V 1.27m/s
π
A 0.05
2
4
24 1.27
2
hL 2.0m
2g
Patm PVapor
(NP SH)A h S h f S h mS
γ air γ Vapor
(NP SH)A 7 2 10.3 0.25
(NP SH)A 1.05 4.1
pump is not suitable, the cavitation will occur
63. Multiple-Pump 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.
Pumps in Parallel Pumps in Series
64. (a) Parallel Operation
• Pumping stations frequently contain several (two or
more) pumps in a parallel arrangement.
Manifold
Qtotal
Qtotal =Q1+Q2+Q3
Pump Pump Pump
Q1 Q2 Q3
65. • 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.
66. 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).
67. Pumps in series & Parallel
j n
Pumps in Parallel: Q Q1 Q 2 Q3 Q n Q
j1
H m H m1 H m2 H m3 H mn
68.
69. (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
71. 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
72. H
3H1
Three pumps
in series
H1
2H1 Two pumps
in series
H1
H1
Single pump
H1
Q
Q1
74. Cavitation of Pumps and NPSH
• In general, cavitation occurs when the liquid pressure
at a given location is reduced to the vapor pressure of
the liquid.
• For a piping system that includes a pump, cavitation
occurs when the absolute pressure at the inlet falls
below the vapor pressure of the water.
• This phenomenon may occur at the inlet to a pump
and on the impeller blades, particularly if the pump is
mounted above the level in the suction reservoir.
75. • Under this condition, vapor bubbles form (water
starts to boil) at the impeller inlet and when these
bubbles are carried into a zone of higher pressure,
they collapse abruptly and hit the vanes of the
impeller (near the tips of the impeller vanes). causing:
• Damage to the pump (pump impeller)
• Violet vibrations (and noise).
• Reduce pump capacity.
• Reduce pump efficiency
76. How we avoid Cavitation ??
• To avoid cavitation, the pressure head at the inlet should not fall
below a certain minimum which is influenced by the further
reduction in pressure within the pump impeller.
• To accomplish this, we use the difference between the total head
at the inlet P V , and the water vapor pressure head
2
s s
Pvapor 2g
77. 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
2 Pvapor
Ps Vs
NPSH
2g
There are two values of NPSH of interest. 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.
The second value for NPSH of concern is the available NPSH, denoted
(NPSH)A , which represents the head that actually occurs for the particular
piping system. This value can be determined experimentally, or calculated if
the system parameters are known.
78. How we avoid Cavitation ??
• For proper pump operation (no cavitation) :
(NPSH)A > (NPSH)R
79. Determination of datum
(NPSH)A hs
applying the energy equation between
point (1) and (2), datum at pump
center line
Patm PS VS2
hS hL
air 2g
PS VS2 Patm
hS hL
2 g air
PS VS2 PVapor Patm P
hS hL
Vapor
2 g Vapor air Vapor
Patm PVapor
( NPSH ) A hS hL
air Vapor
80. P Pvapor
( NPSH ) A hs h f s hm s atm
Note that (+) is used if hs is above the pump centerline (datum).
at T 20 o
Patm 10.14 kN / m 2
Vapor 2.335 kN / m
2
P
81. Thoma’s cavitation constant
The cavitation constant: is the ratio of (NPSH)R to
the total dynamic head (Ht) is known as the Thoma’s
cavitation constant ( )
( NPSH )R
Ht
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.
82. Head analysis in the suction side of the
pump
hp
Hs
hc
hfs
V2/2g
H’s
83. Example 1
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 35m3
and The total Dynamic Head Ht = 25m (from pump curve)
f=0.0167
Calculate the maximum suction head
atm. pressure head 9.69 m
Vapour pressure head 0.2m
84. σ 0.12
NPSH R σ H t 0.12 25 3
Patm P
(NPSH)A hS h f S hmS
Vapor
γair γVapor
Q 0.035
VS 1.11 m/s
A π 0.22
4
VS2 1.112 VS2 1.112
he 0.063 hV 4.5 4.5 0.283m
2g 2g 2g 2g
L V2 10 1.112
h fS f 0.0167 0.053m
D 2g 0.2 2 g
Patm P
(NPSH)A hS h f S hmS
Vapor
γair γVapor
3 hS 0.053 0.283 0.063 9.69 0.2
hS 6.088m
85. Specific Speed
• Pump types may be more explicitly defined by the
parameter called specific speed (Ns) expressed by:
N Q
Ns 3
H 4
Where: Q = discharge (m3/s, or l/s).
H = pump total head (m).
N = rotational speed (rpm).
86. • This expression is derived from dynamical similarity
considerations and may be interpreted as the speed in
rev/min at which a geometrically scaled model would have
to operate to deliver unit discharge (1 l/s) when generating
unit head (1 m).
• The given table shows the range of Ns values for the turbo-
hydraulic pumps:
Pump type Ns range (Q - l/s, H-m)
centrifugal up to 2600
mixed flow 2600 to 5000
axial flow 5000 to 10 000
89. Example 5
• A centrifugal pump running at 1000 rpm gave the following
relation between head and discharge:
Discharge (m3/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 f is assumed as
0.024.
1. Calculate the discharge in m3 per minute.
2. If it is required to adjust the flow by regulating the pump
speed, estimate the speed to reduce the flow to one-half
90. 1) System curve:
• The head required from pump =
static + friction + velocity head
Vd2
H t H stat h f d hmd h f s hm s
2g
• Hstat = 15 m
• Friction losses (including equivalent entrance losses) =
8 f LQ 2
h fs hms h fd hmd 2 g D5
8 0.024 (69 6) 2
Q
g (0.3)
2 5
61.21Q 2 where Q in m3/s
91. 2
V 1 Q
2
• Velocity head in delivery pipe = d
10.2Q
2
2g 2g A
3
where Q in m /s
Thus:
• H t 15 71.41Q 2 where Q in m3/s
or
• H t 15 19.83 10 3 Q 2 where Q in m3/min
• From this equation and the figures given in the problem the
following table is compiled:
Discharge (m3/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
93. From the previous Figure, The operating point is:
• QA = 14 m3/min
• HA = 19 m
• At reduced speed: For half flow (Q = 7 m3/min) there
will be a new operating point B at which:
• QB = 7 m3/min
• HB = 16 m
• HomeWork
How to estimate the new speed ?????
94. Pump and Sytem Curves
28
26
24
22
20
A
18
Head,Ht (m)
16
14 B
12
10
8
Pump Curve
6 System Curve
4 A
2 B
0
0 2 4 6 8 10 12 14 16 18 20 22 24
Discharge, Q (m3 /min)
95. 2
Q2 N 2 H2 N 2
Q1 N1 H1 N1
2
H Q
Q
HB B
16 2
H 2 Q 0.327Q 2
7
This curve intersects the original curve for N1 = 1000 rpm
at C where Qc= 8.2 m3/ hr and Hc= 21.9 m, then
QB N 2 7 N2
N2 = 855rpm
QC N1 8.2 1000
96. Pump and Sytem Curves
28
26
24
22
C
20
A
18
Head,Ht (m)
16
14 B
12 Pump Curve
10 System Curve
8 A
6
B
4
C
2
0
0 2 4 6 8 10 12 14 16 18 20 22 24
Discharge, Q (m3 /min)