Chapter Five Fluid Flow Pumping of Liquids

CHPTER FIVE
Pumping of Liquids
5.1 Introduction
Pupmps are devices for supplying energy or head to a flowing liquid in order to
overcome head losses due to friction and also if necessary, to raise liquid to a higher
level.
For the pumping of liquids or gases from one vessel to another or through long
pipes, some form of mechanical pump is usually employed. The energy required by
the pump will depend on the height through which the fluid is raised, the pressure
required at delivery point, the length and diameter of the pipe, the rate of flow, together
with the physical properties of the fluid, particularly its viscosity and density. The
pumping of liquids such as sulphuric acid or petroleum products from bulk store to
process buildings, or the pumping of fluids round reaction units and through heat
exchangers, are typical illustrations of the use of pumps in the process industries. On the
one hand, it may be necessary to inject reactants or catalyst into a reactor at a low, but
accurately controlled rate, and on the other to pump cooling water to a power station or
refinery at a very high rate. The fluid may be a gas or liquid of low viscosity, or it may
be a highly viscous liquid, possibly with non-Newtonian characteristics. It may be clean,
or it may contain suspended particles and be very corrosive. All these factors influence
the choice of pump.
Because of the wide variety of requirements, many different types are in use
including centrifugal, piston, gear, screw, and peristaltic pumps, though in the chemical
and petroleum industries the centrifugal type is by far the most important.
5.2 The Total Head (Δh)
The head imparted to a flowing liquid by a pump is known as the total head (Δh).
If a pump is placed between points  and  in a pipeline, the head for steady flow are
related by: -
.Pd

.Ps
zd

zs
Pump

Datum line
Suction side Discharge side

Figure (1) Typical pumping system.

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Chapter Five Fluid Flow Pumping of Liquids
η Ws ⎛ P u 22 ⎞ ⎛ P u2 ⎞
Δh = =⎜ 2 + + z 2 ⎟ − ⎜ 1 + 1 + z1 ⎟ − hF
g ⎝ ρ g 2α 2 g ⎠ ⎝ ρ g 2α 1 g ⎠

ΔP Δ u 2
⇒ Δh = + + Δz + hF
ρ g 2α g
5.3 System Heads
The important heads to consider in a pumping system are: -
1- Suction head
2- Discharge head
3- Total head
4- Net positive suction head (NPSH)
The following definitions are given in reference to typical pumping system shown
in preceding Figure, where the datum line is the centerline of the pump
1- Suction head (hs)
Ps
hs = z s + − ( hF ) s
ρg
2- Discharge head (hd)
Pd
hd = z d + + ( hF ) d
ρg
3- Total head (Δh)
The total head (Δh), which is required to impart to the flowing liquid is the
difference between the discharge and suction heads. Thus,
Δh = hd − hs
Pd − Ps
⇒ Δh = ( z d − z s ) + ( ) + [(hF ) d + (hF ) s ]
ρg
where,
⎛L Le ⎞ u d2
( hF ) d = 4 f d ⎜ + ∑ ⎟
⎝d d ⎠d 2 g
⎛L Le ⎞ u s2
( hF ) s = 4 f s ⎜ + ∑ ⎟
⎝d d ⎠s 2 g

The suction head (hs) decreases and the discharge head (hd) increases with
increasing liquid flow rate because of the increasing value of the friction head loss terms
(hF)s and (hF)d. Thus the total; head (Δh) which the pump is required to impart to the
flowing liquid increases with increasing the liquid pumping rate.
Note:
If the liquid level on the suction side is below the centerline of the pump, zs is
negative.
4- Net positive suction head (NPSH)
Available net positive suction head
⎛ Ps − Pv ⎞
NPSH = z s + ⎜ ⎟ − ( hF ) s
⎝ ρg ⎠
This equation gives the head available to get the liquid through the suction piping.

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Chapter Five Fluid Flow Pumping of Liquids
Pv is the vapor pressure of the liquid being pumped at the particular temperature in
question.
The available net positive suction head (NPSH) can also be written as:
Pv
NPSH = hs −
ρg
The available net positive suction head (NPSH) in a system should always be
positive i.e. the suction head always be capable of overcoming the vapor pressure (Pv)
since the frictional head loss (hF)s increases with increasing pumping rate.
At the boiling temperature of the liquid Ps and Pv are equal and the available
NPSH becomes [zs-(hF)s]. In this case no suction lift is possible since zs must be
positive. If the term (Ps-Pv) is sufficiently large, liquid can b lifted from below the
centerline of the pump. In this case zs is negative.
From energy consideration it is immaterial whether the suction pressure is below
atmospheric pressure or well above it, as long as the fluid remains liquid. However, if
the suction pressure is only slightly greater than the vapor pressure, some liquid may
flash to vapor inside the pump, a process called “Cavitation”, which greatly reduces the
pump capacity and severe erosion.
If the suction pressure is actually less than the vapor pressure, there will be
vaporization in the suction line, and no liquid can be drawn into the pump.
To avoid cavitation, the pressure at the pump inlet must exceed the vapor pressure
by a certain value, called the “ net positive suction head (NPSH)”. The required values
of NPSH is about 2-3 m H2o for small pump; but it increases with pump capacity and
values up to 15 m H2o are recommended for very large pump.
5.4 Power Requirement
The power requirement to the pump drive from an external source is denoted by
(P). It is calculated from Ws by:
QΔP QΔh ρ g m& Δhg
P = m& Ws = = =
η η η
The mechanical efficiency (η) decreases as the liquid viscosity and hence the
frictional losses increase. The mechanical efficiency is also decreased by power losses in
gear, Bering, seals, etc.
These losses are not proportional to pump size. Relatively large pumps tend to
have the best efficiency whilst small pumps usually have low efficiencies. Furthermore
high-speed pumps tend to be more efficient than low-speed pumps. In general, high
efficiency pumps have high NPSH requirements.
5.5 Types of Pumps
Pumps can be classified into: -
1- Centrifugal pumps.
2- Positive displacement pumps.
1- Centrifugal pumps
This type depends on giving the liquid a high kinetic energy, which is then
converted as efficiently as possible into pressure energy. It used for liquid with very
wide ranging properties and suspensions with high solid content including, for example,

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Chapter Five Fluid Flow Pumping of Liquids
cement slurries, and may be constructed from a very wide range of corrosion resistant
materials. Process industries commonly use centrifugal pumps. The whole pump casing
may be constructed from plastics such as polypropylene or it may be fitted with a
corrosion resistant lining. Because it operates at high speed, it may be directly coupled
to an electric motor and it will give a high flow rate for its size. They are available in
sizes about 0.004 to 380 m3/min [1-100,000 gal/min] and for discharge pressures from a
few m H2o head to 5,000 kPa.
In this type of pump (Figure 2), the fluid is fed to the center of a rotating impeller
and is thrown outward by centrifugal action. As a result of the high speed of rotation the
liquid acquires a high kinetic energy and the pressure difference between the suction and
delivery sides arises from the interconversion of kinetic and pressure energy.

Figure (2) Section of centrifugal pump

The impeller (Figure 3) consists of a series
of curved vanes so shaped that the flow within
the pump is as smooth as possible. The greater
the number of vanes on the impeller, the greater
is the control over the direction of motion of the
liquid and hence the smaller are the losses due to
turbulence and circulation between the vanes. In
the open impeller, the vanes are fixed to a central
hub, whereas in the closed type the vanes are
held between two supporting plates and leakage Figure (3) Types of impeller
across the impeller is reduced. (a) for pumping suspensions (b) standard
The liquid enters the casing of the pump, closed impeller (c) double impeller
normally in an axial direction, and is picked up
by the vanes of the impeller. In the simple type of centrifugal pump, the liquid
discharges into a volute, a chamber of gradually increasing cross-section with a
tangential outlet. A volute type of pump is shown in Figure 4. In the turbine pump

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Chapter Five Fluid Flow Pumping of Liquids
(Figure 4(b)) the liquid flows from
the moving vanes of the impeller
through a series of fixed vanes
forming a diffusion ring. This
gives a more gradual change in
direction to the fluid and more
efficient conversion of kinetic
energy into pressure energy than is
obtained with the volute type.

Figure (4) Radial flow pumps

(a) with volute (b) with diffuser vanes
2- Positive Displacement Pumps
In this type, the volume of liquid delivered is directly related to the displacement
of the piston and therefore, increases directly with speed and is not appreciably
influenced by the pressure. It used for high pressure and constant rates this type can be
classified into: -
2.1-Reciprocating Pumps, such as
a- The Piston Pump
This pump may be single-acting, with the liquid admitted only to the portion of
the cylinder in front of the piston or double-acting, in which case the feed is
admitted to both sides of the piston. The majority of pumps are of the single-
acting type typically giving a low flow rate of say 0.02 m3/s at a high pressure of
up to 100 Mpa.
b- The Plunger (or Ram) Pump
This pump is the same in principle as the piston type but differs in that the
gland is at one end of the cylinder making its replacement easier than with the
standard piston type. The piston or ram pump may be used for injections of
small quantities of inhibitors to polymerization units or of corrosion inhibitors to
high-pressure systems, and also for boiler feed water applications.
c- The Diaphragm Pump
The diaphragm pump has been developed for handling corrosive liquids and
those containing suspensions of abrasive solids. It is in two sections separated
by a diaphragm of rubber, leather, or plastics material. In one section a plunger
or piston operates in a cylinder in which a non-corrosive fluid is displaced. The
particularly simple and inexpensive pump results, capable of operating up to 0.2
Mpa.
d- The Metering (or Dosing) Pump
Metering pumps are driven by constant speed electric motors. They are used
where a constant and accurately controlled rate of delivery of a liquid is
required, and they will maintain this constant rate irrespective of changes in the
pressure against which they operate. The pumps are usually of the plunger type
for low throughput and high-pressure applications; for large volumes and lower
pressures a diaphragm is used. In either case, the rate of delivery is controlled by
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Chapter Five Fluid Flow Pumping of Liquids
adjusting the stroke of the piston element, and this can be done whilst the pump
is in operation. A single-motor driver may operate several individual pumps and
in this way give control of the actual flows and of the flow ratio of several
streams at the same time. The output may be controlled from zero to maximum
flow rate, either manually on the pump or remotely. These pumps may be used
for the dosing of works effluents and water supplies, and the feeding of
reactants, catalysts, or inhibitors to reactors at controlled rates, and although a
simple method for controlling flow rate is provided, high precision standards of
construction are required.
2.2-Rotary Pumps, such as
a- The Gear Pump
Gear and lobe pumps operate on the principle of using mechanical means to
transfer small elements or "packages" of fluid from the low pressure (inlet) side
to the high pressure (delivery) side. There is a wide range of designs available
for achieving this end. The general characteristics of the pumps are similar to
those of reciprocating piston pumps, but the delivery is more even because the
fluid stream is broken down into so much smaller elements. The pumps are
capable of delivering to a high pressure, and the pumping rate is approximately
proportional to the speed of the pump and is not greatly influenced by the
pressure against which it is delivering. Again, it is necessary to provide a
pressure relief system to ensure that the safe operating pressure is not exceeded.
b- The Cam Pump
A rotating cam is mounted eccentrically in a cylindrical casing and a very
small clearance is maintained between the outer edge of the cam and the casing.
As the cam rotates it expels liquid from the space ahead of it and sucks in liquid
behind it. The delivery and suction sides of the pump are separated by a sliding
valve, which rides on the cam. The characteristics again are similar to those of
the gear pump.
c- The Vane Pump
The rotor of the vane pump is mounted off centre in a cylindrical casing. It
carries rectangular vanes in a series of slots arranged at intervals round the
curved surface of the rotor. The vanes are thrown outwards by centrifugal action
and the fluid is carried in the spaces bounded by adjacent vanes, the rotor, and
the casing. Most of the wear is on the vanes and these can readily be replaced.
d- The Flexible Vane Pump
The pumps described above will not handle liquids containing solid particles in
suspension, and the flexible vane pumps has been developed to overcome this
disadvantage. In this case, the rotor (Figure 8.10) is an integral elasomer
moulding of a hub with flexible vanes which rotates in a cylindrical casing
containing a crescent-shaped block, as in the case of the internal gear pump.
e- The Flow Inducer or Peristaltic Pump
This is a special form of pump in which a length of silicone rubber or other
elastic tubing, typically of 3 to 25 mm diameter, is compressed in stages by
means of a rotor as shown in Figure 8.11. The tubing is fitted to a curved track
mounted concentrically with a rotor carrying three rollers. As the rollers rotate,
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Chapter Five Fluid Flow Pumping of Liquids
they flatten the tube against the track at the points of contact. These "flats" move
the fluid by positive displacement, and the flow can be precisely controlled by
the speed of the motor. These pumps have been particularly useful for biological
fluids where all forms of contact must be avoided. They are being increasingly
used and are suitable for pumping emulsions, creams, and similar fluids in
laboratories and small plants where the freedom from glands, avoidance of
aeration, and corrosion resistance are valuable, if not essential. Recent
developments^ have produced thick-wall, reinforced moulded tubes which give
a pumping performance of up to 0.02 m3/s at 1 MN/m2. The control is such that
these pumps may conveniently be used as metering pumps for dosage processes.
f- The Mono pump
Another example of a positive acting rotary pump is the single screw-extruder
pump typified by the Mono pump, in which a specially shaped helical metal
rotor revolves eccentrically within a double-helix, resilient rubber stator of twice
the pitch length of the metal rotor. A continuous forming cavity is created as the
rotor turns — the cavity progressing towards the discharge, advancing in front of
a continuously forming seal line and thus carrying the pumped material with it.
The Mono pump gives a uniform flow and is quiet in operation. It will pump
against high pressures; the higher the required pressure, the longer are the stator
and the rotor and the greater the number of turns. The pump can handle
corrosive and gritty liquids and is extensively used for feeding slurries to filter
presses. It must never be run dry. The Mono Merlin Wide Throut pump is used
for highly viscous liquids.
g- The Screw pumps
A most important class of pump for dealing with highly viscous material is
represented by the screw extruder used in the polymer industry. The screw pump
is of more general application and will be considered first. The fluid is sheared
in the channel between the screw and the wall of the barrel. The mechanism that
generates the pressure can be visualized in terms of a model consisting of an
open channel covered by a moving plane surface. If a detailed analysis of the
flow in a screw pump is to be carried out, then it is also necessary to consider
the small but finite leakage flow that can occur between the flight and the wall.
With the large pressure generation in a polymer extruder, commonly 100 bar
(107 N/m2), the flow through this gap, which is typically about 2 per cent of the
barrel internal diameter, can be significant. The pressure drop over a single pitch
length may be of the order of 10 bar (106 N/m2), and this will force fluid
through the gap. Once in this region the viscous fluid is subject to a high rate of
shear (the rotation speed of the screw is often about 2 Hz), and an appreciable
part of the total viscous heat generation occurs in this region of an extruder.

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Chapter Five Fluid Flow Pumping of Liquids
5.6 The advantages and disadvantages of the centrifugal pump
The main advantages are:
(1) It is simple in construction and can, therefore, be made in a wide range of
materials.
(2) There is a complete absence of valves.
(3) It operates at high speed (up to 100 Hz) and, therefore, can be coupled directly to
an electric motor. In general, the higher the speed the smaller the pump and motor
for a given duty.
(4) It gives a steady delivery.
(5) Maintenance costs are lower than for any other type of pump.
(6) No damage is done to the pump if the delivery line becomes blocked, provided it
is not ran in this condition for a prolonged period.
(7) It is much smaller than other pumps of equal capacity. It can, therefore, be made
into a sealed unit with the driving motor, and immersed in the suction tank.
(8) Liquids containing high proportions of suspended solids are readily handled.
The main disadvantages are:
(1) The single-stage pump will not develop a high pressure. Multistage pumps will
develop greater heads but they are very much more expensive and cannot readily
be made in corrosion-resistant material because of their greater complexity. It is
generally better to use very high speeds in order to reduce the number of stages
required.
(2) It operates at a high efficiency over only a limited range of conditions: this applies
especially to turbine pumps.
(3) It is not usually self-priming.
(4) If a non-return valve is not incorporated in the delivery or suction line, the liquid
will run back into the suction tank as soon as the pump stops.
(5) Very viscous liquids cannot be handled efficiently.
5.7 Priming The Pump
The theoretical head developed by a centrifugal pump depends on the impeller
speed, the radius of the impeller, and the velocity of the fluid leaving the impeller. If
these factors are constant, the developed head is the same for fluids of all densities and
is the same for liquids and gases. A centrifugal pump trying to operate on air, then can
neither draw liquid upward from an initially empty suction line nor force liquid a full
discharge line. Air can be displaced by priming the pump.
For example, if a pump develops a head of 100 ft and is full of water, the increase
in pressure is [100 ft (62.3lb/ft3) (ft2 / 144 in2)] = 43 psi (2.9 atm). If full of air the
pressure increase is about 0.05 psi (0.0035 atm).

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Chapter Five Fluid Flow Pumping of Liquids
5.8 Operating Characteristics
The operating characteristics of a pump are conveniently shown by plotting the
head (h), power (P), efficiency (η), and
sometimes required NPSH against the
η
flow (or capacity) (Q) as shown in Figure
(5). Theses are known as characteristic h
curves of the pump. It is important to
note that the efficiency reaches a
maximum and then falls, whilst the head Duty point
at first falls slowly with Q but eventually
falls off rapidly. The optimum conditions P
for operation are shown as the duty point,
i.e. the point where the head curve cuts
the ordinate through the point of
maximum efficiency.
Characteristic curves have a variety
of shapes depending on the geometry of
the impeller and pump casing. Pump Q
manufactures normally supply the curves Figure (5) Radial flow pump characteristics
only for operation with water.
In a particular system, a centrifugal
pump can only operate at one point on the Pump Operating
Δh against Q curve and that is the point
point
where the Δh against Q curve of the pump
intersect with the Δh against Q curve of Δh
the system as shown in Figure.
The system total head at a particular
liquid flow rate
⎛ Pd − Ps ⎞
Δh = ( z d − z s ) + ⎜ ⎟ + [(hF ) d + (hF ) s ]
⎝ ρg ⎠ System
where,
⎡L Le ⎤ u d2
( hF ) d = 4 f d ⎢ ∑ ⎥ Q
⎣d d ⎦d 2 g
Figure (6) System and pump total head
⎡ L Le ⎤ us2
(hF ) s = 4 f s ⎢ ∑ ⎥ against capacity
⎣d d ⎦s 2 g
For the same pipe type and diameter for suction and discharge lines: -
ΔP ⎡⎛ L Le ⎞ ⎛ L Le ⎞ ⎤ u 2
Δh = Δz + +4f⎢ + ∑
⎜ ⎟ + ⎜ +∑ ⎟ ⎥
ρg ⎣⎝ d d ⎠d ⎝ d d ⎠s ⎦ 2g
Q
but u =
(π / 4 d 2 )
ΔP 4 f ⎡⎛ L Le ⎞ ⎤⎛ ⎞
2
Le ⎞ ⎛ L Q
⇒ Δh = Δz + + ⎢⎜ + ∑ ⎟ + ⎜ + ∑ ⎟ ⎥⎜ ⎟
ρ g 2 g ⎣⎝ d d ⎠d ⎝ d d ⎠ s ⎦⎝ (π / 4 d 2 ) ⎠

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Chapter Five Fluid Flow Pumping of Liquids
Example -5.1-
A petroleum product is pumped at a rate of 2.525 x 10-3 m3/s from a reservoir under
atmospheric pressure to 1.83 m height. If the pump 1.32 m height from the reservoir, the
discharge line diameter is 4 cm and the pressure drop along its length 3.45 kPa. The
gauge pressure reading at the end of the discharge line 345 kPa. The pressure drop along
suction line is 3.45 kPa and pump efficiency η=0.6 calculate:-
(i) The total head of the system Δh. (ii) The power required for pump. (iii) The NPSH
Take that: the density of this petroleum product ρ=879 kg/m3, the dynamic viscosity
μ=6.47 x 10-4 Pa.s, and the vapor pressure Pv= 24.15 kPa.
Solution: Pd
(i) zd
⎛ Pd − Ps ⎞ Δu 2
Δh = ( z d − z s ) + ⎜ ⎟ + [(hF ) d + (hF ) s ] +
⎝ ρg ⎠ 2g α
1.83 m
us = 0 1.32 m
ud = (2.525 x 10-3 m3/s)/(π/4 0.042)
= 2 m/s Ps
Red = (879 x 2 x 0.04)/ 6.47 x 10-4
= 1.087 x 105
The pressure drop in suction line 3.45kPa
⇒ (hF)s = 3.45 x 103/(879 x 9.81)
= 0.4 m
And in discharge line is also 3.45 kPa ⇒ (hF)d = 0.4 m
The kinetic energy term = 22/(2 x 9.81) =0.2 m
The pressure at discharge point = gauge + atmospheric pressure = 345 + 101.325
= 446.325 kPa
The difference in pressure head between discharge and suction points is
(446.325 – 101.325) x 103 /(879 x 9.81) = 40 m
Δz = 1.83 m

⇒ Δh = 40 m + 1.83 m + 0.2 m + 0.4 m + 0.4 m = 42.83 m

(ii)
QΔP QΔh ρ g
P= = = [(2.525 x 10-3 m3/s)(42.83 m)(879 kg/m3)(9.81 m/s2)]/0.6
η η
⇒ P = 1.555 kW
(iii)
⎛ Ps − Pv ⎞
NPSH = z s + ⎜ ⎟ − ( hF ) s
⎝ ρg ⎠
= (- 1.32) + (1.01325 x 105 - 24150)/ (879 x 9.81) – 0.4 m
= 7.23 m

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Chapter Five Fluid Flow Pumping of Liquids
Example -5.2-
It is required to pump cooling water from storage pond to a condenser in a process
plant situated 10 m above the level of the pond. 200 m of 74.2 mm i.d. pipe is available
and the pump has the characteristics given below. The head loss in the condenser is
equivalent to 16 velocity heads based on the flow in the 74.2 mm pipe. If the friction
factor Φ = 0.003, estimate the rate of flow and the power to be supplied to the pump
assuming η = 0.5
Q (m3/s) 0.0028 0.0039 0.005 0.0056 0.0059
Δh (m) 23.2 21.3 18.9 15.2 11.0
Solution:
Δh = Δz +
ΔP
+
Δu 2
[
(h ) + (hF )s + (hF )condenser
ρ g 2g α F d
]
L u2
( hF ) d + s = 4 f = 4(0.006)(200/0.0742)(u2/2g) = 3.3 u2
d 2g
u2
(hF ) condenser = 16 = 0.815 u2
2g
u = Q/A = 321.26 Q
⇒ Δh = 10 + (0.815 + 3.3)(321.26 Q)2 = 10 + 2.2 x 105 Q2
To draw the system curve
Q (m3/s) 0.003 0.004 0.005 0.006
Δh (m) 11.98 13.52 15.5 17.92
24

22
Δh (m)

20
From Figure
18
Q = 0.0054 m3/s
Δh = 16.4 m 16

14

12

10
0.003 0.004 0.005 0.006 0.007
Q (m3/s)

QΔh ρ g
Power required for pump = = (0.0054)(16.4)(1000)(9.81)/0.5
η
= 17.375 kW

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Chapter Five Fluid Flow Pumping of Liquids
Example -5.3-
A centrifugal pump used to take water from reservoir to another through 800 m length
and 0.15 m i.d. if the difference in two tank is 8 m, calculate the flow rate of the water
and the power required, assume f =0.004.
Q (m3/h) 0 23 46 69 92 115
Δh (m) 17 16 13.5 10.5 6.6 2.0
η 0 0.495 0.61 0.63 0.53 0.1
Solution:
Δh = Δz +
ΔP
+
Δu 2
[
(h ) + (hF )s
ρ g 2g α F d
]
u = Q/A = 56.59 Q
L u2
( hF ) d + s = 4 f = 4(0.004)(800/0.15)(56.59 Q(h/3600 s) )2/2g
d 2g
= 1.0747 x 10-3 Q2 ----------------------- (Q in m3/h)
⇒ Δh = 8 + 1.0747 x 10-3 Q2
To draw the system curve
Q (m3/h) 0 20 40 60 80
Δh (m) 8.0 8.43 9.72 11.87 14.88
20 1
18 0.9
16 0.8
14 0.7
Δh (m)

12 0.6
From Figure 10 0.5
Q = 60 m3/h

η
8 0.4
Δh = 11.8 m 6 0.3
η = 0.64 4 0.2
2 0.1
0 0
0 20 40 60 80 100 120
Q (m3/s)
QΔh ρ g
Power required for pump = = (60)(1 h/3600 s)(11.8)(1000)(9.81)/0.64
η
= 3.014 kW
Example -5.4-
A pump take brine solution at a tank and transport it to another in a process plant
situated 12 m above the level in the first tank. 250 m of 100 mm i.d. pipe is available
sp.gr. of brine is 1.2 and μ = 1.2 cp. The absolute roughness of pipe is 0.04 mm and f =
0.0065. Calculate (i) the rate of flow for the pump (ii) the power required for pump if η
= 0.65. (iii) if the vapor pressure of water over the brine solution at 86°F is 0.6 psia,
calculate the NPSH available, if suction line length is 30 m.
Q (m3/s) 0.0056 0.0076 0.01 0.012 0.013
Δh (m) 25 24 22 17 13

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Chapter Five Fluid Flow Pumping of Liquids
Solution:
ΔP Δu 2
(i) Δh = Δz + + (h )
ρ g 2g α F d +s
u = Q/A = 127.33 Q
L u2
( hF ) d + s = 4 f = 4(0.0065)(250/0.1)(127.33 Q)2/2g
d 2g
= 53.707 x 103 Q2
⇒ Δh = 12 + 53.707 x 103 Q2
To draw the system curve
Q (m3/h) 0.005 0.007 0.009 0.011 0.013
Δh (m) 13.34 14.63 16.35 18.5 21.08
26

24
From Figure
Q = 0.0114 m3/s 22
Δh = 18.9 m
20
(ii)
Δh (m)

Power required for pump = 18

QΔh ρ g
=(00.0114)(18.9)
η 16

(1200)(9.81)/0.65=3.9 kW 14

(iii) 12
⎛ Ps − Pv ⎞ 0.005 0.0075 0.01 0.0125 0.015
NPSH = z s + ⎜ ⎟ − ( hF ) s 3
Q (m /s)
⎝ ρg ⎠
u = Q/A = 0.0114/(π/4 0.12) =1.45 m/s
For datum line passes through the centerline of the pump (zs = 0)
Ls u 2
( hF ) s = 4 f = 4(0.0065)(30/0.1)(1.45 )2/2g = 0.84 m
d 2g

⇒ NPSH = (101.325 x 103 - 0.6psi 101.325 x 103Pa/14.7psi)/(1200 x 9.81) – 0.84

= 7.416 m

5.9 Centrifugal Pump Relations

The power (PE) required in an ideal centrifugal pump can be expected to be a
function of the liquid density (ρ), the impeller diameter (D), and the rotational speed of
the impeller (N). If the relationship is assumed to be given by the equation,
PE = c ρa Nb Dc ---------------------------------(1)
then it can be shown by dimensional analysis that
PE = c1 ρ N3 D5 ---------------------------------(2)
where, c1 is a constant which depends on the geometry of the system.

13-Ch.5 Dr. Salah S. Ibrahim

Chapter Five Fluid Flow Pumping of Liquids
The power (PE) is also proportional to the product of the volumetric flow rate (Q)
and the total head (Δh) developed by the pump.
PE = c2 Q Δh ---------------------------------(3)
where, c2 is a constant.
The volumetric flow rate (Q) and the total head (Δh) developed by the pump are: -
Q = c3 N D3 ---------------------------------(4)
2 2
Δh = c4 N D ---------------------------------(5)
where, c3 and c4 are constants.
Equation (5) could be written in the following form,
Δh3/2 = c43/2 N3 D3 ---------------------------------(6)
Combine equations (4) and (6) [ eq. (4) divided by eq. (6)] to give;
Q c3 1 QN 2
3 = 3 ⇒ 3 = const. ------------------------------(7)
N
Δh 2
c 42 Δh 2
N Q
or, 3 = const. = N s ---------------------------------(8)
Δh 4

When the rotational speed of the impeller N is (rpm), the volumetric flow rate Q
in (USgalpm) and the total head Δh developed by the pump is in (ft), the constant Ns in
equation (8) is known as the specific speed of the pump. The specific speed is used as
an index of pump types and always evaluated at the best efficiency point (bep) of the
pump. Specific speed vary in the range (400 – 10,000) depends on the impeller type, and
has the dimensions of (L/T2)3/4. [ British gal=1.2USgal, ft3=7.48USgal, m3=264USgal]
5.9.1 Homologous Centrifugal Pumps
Two different size pumps are said to be geometrically similar when the ratios of
corresponding dimensions in one pump are equal to those of the other pump.
Geometrically similar pumps are said to be homologous. A sets of equations known as
the affinity laws govern the performance of homologous centrifugal pumps at various
impeller speeds.
For the tow homologous pumps, equations (4), and (5) are given
Q1 ⎛ N 1 ⎞⎛ D1 ⎞
3

= ⎜ ⎟⎜ ⎟ ---------------------------------(9)
Q2 ⎝ N 2 ⎠⎝ D2 ⎠
Δh1 ⎛ N 1 ⎞ ⎛ D1 ⎞
2 2

=⎜ ⎟ ⎜ ⎟ ---------------------------------(10)
Δ h2 ⎝ N 2 ⎠ ⎝ D 2 ⎠

Similarly for the tow homologous pumps equation (2)can be written in the form;
PE1 ⎛ N 1 ⎞ ⎛ D1 ⎞
3 5

=⎜ ⎟ ⎜ ⎟ ---------------------------------(11)
PE 2 ⎝ N 2 ⎠ ⎝ D2 ⎠
And by analogy with equation (10),
NPSH 1 ⎛ N 1 ⎞ ⎛ D1 ⎞
2 2

=⎜ ⎟ ⎜ ⎟ ---------------------------------(12)
NPSH 2 ⎝ N 2 ⎠ ⎝ D2 ⎠
Equations (9), (10), (11), and (12) are the affinity law for homologous centrifugal
pumps.
14-Ch.5 Dr. Salah S. Ibrahim
Chapter Five Fluid Flow Pumping of Liquids
For a particular pump where the impeller of diameter D1, is replaced by an
impeller with a slightly different diameter D2 the following equations hold
Q1 ⎛ N 1 ⎞⎛ D1 ⎞
=⎜ ⎟⎜ ⎟ ---------------------------------(13)
Q2 ⎝ N 2 ⎠⎝ D2 ⎠
Δh1 ⎛ N 1 ⎞ ⎛ D1 ⎞
2 2

=⎜ ⎟ ⎜ ⎟ ---------------------------------(14)
Δ h2 ⎝ N 2 ⎠ ⎝ D 2 ⎠
PE1 ⎛ N 1 ⎞ ⎛ D1 ⎞
3 3

=⎜ ⎟ ⎜ ⎟ ---------------------------------(15)
PE 2 ⎝ N 2 ⎠ ⎝ D2 ⎠
The characteristic performance curves are available for a centrifugal pump
operating at a given rotation speed, equations (13), (14), and (15) enable the
characteristic performance curves to be plotted for other operating speeds and for other
slightly impeller diameters.
Example -5.5-
A volute centrifugal pump with an impeller diameter of 0.02 m has the following
performance data when pumping water at the best efficiency point (bep). Impeller speed
N = 58.3 rev/s capacity Q = 0.012 m3/s, total head Δh = 70 m, required NPSH = 18 m,
and power = 12,000 W. Evaluate the performance data of an homologous pump with
twice the impeller diameter operating at half the impeller speed.
Solution:
Let subscripts 1 and 2 refer to the first and second pump respectively,
N1/N2 = 2, D1/D2 = 1/2
Ratio of capacities
Q1 ⎛ N 1 ⎞⎛ D1 ⎞
3

= ⎜ ⎟⎜ ⎟ = 2 (1/8) =1/4
Q2 ⎝ N 2 ⎠⎝ D2 ⎠

⇒ Capacity of the second pump Q2 = 4 Q1 = 4(0.012) = 0.048 m3/s

Ratio of total heads
Δh1 ⎛ N 1 ⎞ ⎛ D1 ⎞
2 2

=⎜ ⎟ ⎜ ⎟ = 4 (1/4) = 1
Δ h2 ⎝ N 2 ⎠ ⎝ D2 ⎠

⇒ Total head of the second pump Δh2 = Δh1= 70 m

Ratio of powers
PE1 ⎛ N 1 ⎞ ⎛ D1 ⎞
3 5

= ⎜ ⎟ ⎜ ⎟ = 8 (1/32) = ¼
PE 2 ⎝ N 2 ⎠ ⎝ D2 ⎠
P P 1
assume B1 = E1 =
PB 2 PE 2 4

⇒ Break power of the second pump PB2 = 4 PB1= 4(12,000) = 48,000 W

NPSH 1 ⎛ N 1 ⎞ ⎛ D1 ⎞
2 2

=⎜ ⎟ ⎜ ⎟ = 4 (1/4) = 1
NPSH 2 ⎝ N 2 ⎠ ⎝ D2 ⎠

⇒ NPSH of the second pump NPSH2 = NPSH1= 18 m

H.W.
Calculate the specific speed for these two pumps. Ans. Ns1 = Ns2 = 816.4

15-Ch.5 Dr. Salah S. Ibrahim

Chapter Five Fluid Flow Pumping of Liquids
Note: -
The break power PB can be defined as the actual power delivered to the pump by
prime mover. It is the sum of liquid power and friction power and is given by the
PE
equation, PB =
η
Example -5.6-
A centrifugal pump was manufactured to couple directly to a 15 hp electric motor
running at 1450 rpm delivering 50 liter/min against a total head 20 m. It is desired to
replace the motor by a diesel engine with 1,000 rpm speed and couple it directly to the
pump. Find the probable discharge and head developed by the pump. Also find the hp of
the engine that would be employed.
Solution:
With the same impeller D1 = D2,
then Q1/Q2 = N1/N2
⇒ Q2= 50 (1000 / 1450) = 34.5 liter/min
and Δh2 = Δh1 (N2/N1)2 = 20 (1000/1450)2 = 9.5 m
PE2 = PE1 (N2/N1)3 = 15 (1000/1450)3 = 4.9 hp

H.W.
1- Repeat example 5.6 with Q1 = 850 lit/min, Δh1 = 40 m, N = 1450 rpm, and Power
= 15 hp.
2- Calculate the pump efficiency (η) for pumping of water, and the specific speed for
these two pumps.
Ans. η = 0.497 ≈ 0.5, and Ns1 = Ns2 = 650

16-Ch.5 Dr. Salah S. Ibrahim

Chapter Five Fluid Flow Pumping of Liquids
5.10 Centrifugal Pumps in Series and in Parallel
5.10.1 Centrifugal Pumps in Parallel
Consider two centrifugal pumps in parallel. The total head for the pump
combination (ΔhT) is the same as the total head for each pump,
ΔhT = Δh1 = Δh2 ΔhT
QT = Q1 + Q2 Q1
The operating characteristics curves for QT
two pumps in parallel are: - Q2
Δh QT
Solution by trail and error
1- Draw Δh versus Q for the two pumps Pump-1
and the system.
2- Draw horizontal ΔhT line and determine Δh Pump-2
ΔhS
T
Q1, Q2, and QS.
QS
3- QT (Total) = Q1 + Q2 = QS (system).
4- If QT ≠ QS repeat steps 2, 3, and 4 until
QT = QS.
Another procedure for solution
1- The same as above.
2- Draw several horizontal lines (4 to 6)
for ΔhT and determine their QT. Q2 Q1 QT Q
3- Draw ΔhT versus QT.
4- The duty point is the intersection of ΔhT curve with ΔhS curve.
5.10.2 Centrifugal Pumps in Series
Consider two centrifugal pumps in series. The total head for the pump
combination (ΔhT) is the sum of the total heads for the two pumps,
ΔhT = Δh1 + Δh2
QT = Q1 = Q2 Δh1 Δh2
The operating characteristics curves for QT
two pumps in series are: -
Solution by trail and error Δh QT
ΔhS
1- Draw Δh versus Q for the two pumps ΔhT
QS
and the system. Pump-1
2- Draw vertical QT line and determine Pump-2
Δh1, Δh2, and ΔhS.
3- QT (Total) = Q1 + Q2 = QS (system).
4- If ΔhT ≠ ΔhS repeat steps 2, 3, and 4
until ΔhT = ΔhS. Δh2
Another procedure for solution Δh1
1- The same as above.
2- Draw several Vertical lines (4 to 6)
for QT and determine their ΔhT. QT Q
3- Draw ΔhT versus QT.
4- The duty point is the intersection of ΔhT curve with ΔhS curve.

17-Ch.5 Dr. Salah S. Ibrahim

Chapter Five Fluid Flow Pumping of Liquids
Home Work
P.5.1
Show that for homologous pumps, the specific speed (NS) of them is not depended on
the impeller rotational speed (N) and its diameter (D).
P.5.2
Figure 1.5 diagrammatically represents the heads in a liquid flowing through a pipe.
Redraw this diagram with a pump placed between points 1 and 2.

P.5.3
Calculate the available net positive section head NPSH in a pumping system if the
liquid density ρ = 1200 kg/m3, the liquid dynamic viscosity μ = 0.4 Pa s, the mean
velocity u = 1 m/s , the static head on the suction side zs = 3 m, the inside pipe
diameter di = 0.0526 m, the gravitational acceleration g = 9.81 m/s2, and the equivalent
length on the suction side (∑Le)s = 5.0 m.
The liquid is at its normal boiling point. Neglect entrance and exit losses.
P.5.4
A centrifugal pump is used to pump a liquid in steady turbulent flow through a smooth
pipe from one tank to another. Develop an expression for the system total head Δh in
terms of the static heads on the discharge and suction sides zd and zs respectively, the
gas pressures above the tanks on the discharge and suction sides pd and Ps respectively,
the liquid density ρ, the liquid dynamic viscosity μ, the gravitational acceleration g, the
total equivalent lengths on the discharge and suction sides (∑Le)d and (∑Le)s
respectively, and the volumetric flow rate Q.
P.5.5
A system total head against mean velocity curve for a particular power law liquid in a
particular pipe system can be represented by the equation
Δh = (0.03)( 100n)(un) + 4.0 for u ≤1.5 m/s

18-Ch.5 Dr. Salah S. Ibrahim

Chapter Five Fluid Flow Pumping of Liquids
where, Δh is the total head in m, u is the mean velocity in m/s, and n is the power law
index.
A centrifugal pump operates in this particular system with a total head against mean
velocity curve represented by the equation
Δh = 8.0 - 0.2u - 1.0u2 for u ≤1.5 m/s
(This is a simplification since Δh is also affected by n).
(a) Determine the operating points for the pump for
(i) a Newtonian liquid
(ii) a shear thinning liquid with n = 0.9
(iii) a shear thinning liquid with n = 0.8.
(b) Comment on the effect of slight shear thinning on centrifugal pump operation.
P.5.6
A volute centrifugal pump has the following performance data at the best efficiency
point:
Volumetric flow rate Q = 0.015 m3/s
Total head Δh = 65 m
Required net positive suction head NPSH = 16 m
Liquid power PE = 14000 W
Impeller speed N = 58.4 rev/s
Impeller diameter D = 0.22 m
Evaluate the performance of a homologous pump which operates at an impeller speed
of 29.2 rev/s but which develops the same total head Δh and requires the same NPSH.
P.5.7
Two centrifugal pumps are connected in series in a given pumping system. Plot total
head Δh against capacity Q pump and system curves and determine the operating
points for
(a) only pump 1 running (b) only pump 2 running (c) both pumps running
on the basis of the following data:
operating data for pump 1
Δh1 m, 50.0 49.5 48.5 48.0 46.5 44.0 42.0 39.5 36.0 32.5 28.5
3
Q m /h, 0 25 50 75 100 125 150 175 200 225 250
operating data for pump 2
Δh2 m, 40.0 39.5 39.0 38.0 37.0 36.0 34.0 32.0 30.5 28.0 25.5
3
Q m /h, 0 25 50 75 100 125 150 175 200 225 250
data for system
Δhs m, 35.0 37.0 40.0 43.5 46.5 50.5 54.5 59.5 66.0 72.5 80.0
3
Q m /h, 0 25 50 75 100 125 150 175 200 225 250
P.5.8
Two centrifugal pumps are connected in parallel in a given pumping system. Plot total
head Ah against capacity Q pump and system curves for both pumps running on the
basis of the following data:
operating data for pump 1 operating data for pump 2
Δh m, 40.0 35.0 30.0 25.0 Δh m, 0.0 35.0 30.0 25 .0
3 3
Q1m /h, 169 209 239 265 Q2m /h 0 136 203 267
data for system
19-Ch.5 Dr. Salah S. Ibrahim
Chapter Five Fluid Flow Pumping of Liquids
Δh m, 20.0 25.0 30.0 35.0
Qsm3/h, 0 244 372 470

20-Ch.5 Dr. Salah S. Ibrahim