Hydraulic Machines III
Dr. A. Oyieke
November 10, 2020
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Dr. A. Oyieke
Hydraulic Machines III
CENTRIFUGAL PUMPS
Construction
The main parts of a centrifugal pump are as shown:
(a) Cut away section of CP
(b) Closed CP
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Dr. A. Oyieke
Hydraulic Machines III
1) The Impeller - The rotating part where kinetic energy is
transferred by the vanes to the liquid while the liquid is
flowing outwards. A properly designed impeller; optimizes
flow , minimizes turbulence and maximizes efficiency .
F The impeller of a centrifugal pump can be of three basic types:
(a) Open
(b) Semi-open
(c) Closed impeller
Figure: Types of impellers
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Dr. A. Oyieke
Hydraulic Machines III
a) Open impeller - have their vanes free on both sides. Are
structurally weak and typically used in small-diameter pumps
handling suspended solids.
b) Semi-open impeller - The vanes are free on one side and
enclosed on the other. The shroud adds mechanical strength.
They also offer higher efficiencies than open impellers. They
can be used in medium-diameter pumps and with liquids
containing small amounts of suspended solids.
c) Closed impeller - The vanes are located between two discs, all
in a single casting. They are used in large pumps with high
efficiencies and low required Net Positive Suction Head. The
centrifugal pumps with closed impeller are the most widely
used pumps handling clear liquids. The closed impeller is a
more complicated in design and expensive.
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Dr. A. Oyieke
Hydraulic Machines III
2) Casing - The outer stationary part in which the impeller is
enclosed. The casing contains the liquid and acts as a pressure
containment vessel that directs the flow of liquid in and out
of the centrifugal pump through a single discharge pipe.
Water is collected around the impeller and eventually ends in .
The volute is curved funnel that has a constantly increasing
cross sectional area as it approaches the discharge port. The
volute receives the fluid being pumped by the impeller, slows
down its rate of flow. Therefore, according to Bernoulli’s
principle, the volute converts kinetic energy into pressure by
reducing speed while increasing pressure.
The concentric shape has a uniform cross sectional area
towards the exit/outlet
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Dr. A. Oyieke
Hydraulic Machines III
3) Inlet - most pumps have a single inlet on one side of the the
pump. However, low pressure, high volume pumps sometimes
have symmetrical flow with an inlet on both sides of the pump
and a common outlet.
4) Whirlpool chamber - space between the impeller and the
volute in which the water rotates as a free vortex while also
flowing outwards
5) Diffusers - diverting channels or guiding vanes fitted in the
whirlpool chamber to limit the amount of energy sapping
turbulence.Reduces the velocity of water and aid in conversion
of dynamic to static energy
Multi-stage pumps: Several impeller-volute assemblies
compounded with a common shaft where the fluid passes from a
diffuser outlet directly to the impeller inlet of the next stage
Can have pressure heads of several hundreds of metres
Often referred to as turbine pumps because they look like
turbines with reverse flow direction.
Dr. A. Oyieke
Hydraulic Machines III
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Single vs Multi-stage pumps
(a) Single stage CP
(b) Multi-stage CP
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Dr. A. Oyieke
Hydraulic Machines III
Theory of centrifugal pumps
The function of a centrifugal pump is to increase the
pressureof the liquid being pumped
The pressure increase is dependent on;
the size and shape of the vanes of the impeller
the rotational speed of the impeller
the losses that occur while the liquid is passing through the
pump
To determine the ideal pressure increase over the impeller, we
analyse the velocity vectors at the inlet and exit of the
impeller
Assume that the flow is without any shock this means the
relative velocity of the liquid enters the rotor vanes
tangentially
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Dr. A. Oyieke
Hydraulic Machines III
Velocity Triangles and Euler heads
γ
w2
u2
V2
f2
β
Vr1
α
u1
β
Vr2
r1
V1 = f1
r2
Figure: Velocity triangle of a centrifugal pump
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Dr. A. Oyieke
Hydraulic Machines III
From the velocity diagram;
Let subscripts 1 and 2 denote inlet and outlet(exit) velocities.
Let V = absolute velocity of water (velocity as would be
observed by a stationary observer )
V is resolved into two components/sets i.e
Relative velocity Vr
Peripheral velocity u
It can also be resolved into two components/sets which are
perpendicular to each other i.e;
Radial velocity f
Whirl velocity w
To simplify the theory, we can assume that there is no whirl
velocity at the entrance and thus the flow is purely radial. i.e
W1 = 0 and V1 − f1
Let α = angle of vane inlet in relation to the tangent at the
inlet radius r1 and β = the exit angle of the vanes in relation
to the tangent at the exit radius r2
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Dr. A. Oyieke
Hydraulic Machines III
⇒ Let the breadth(width) of the impeller
be b (i.e internal width/the width of
the liquid)
⇒ The volumetric flow rate Q relations
can be found as:
Q1 = Q2 = Q
Q = 2πr1 b1 f1 = 2πr2 b2 f2
(1)
Q = πD1 b1 f1 = πD2 b2 f2
(2)
b1
b2
r2
D2
D1
r2
The angular ω velocity of the impeller is given by;
ω=
2πN
60
(3)
The peripheral velocity of the vane u is given as u = ωr ,
therefore;
u1 =
ωD1
2πND1
ωD2
2πND2
=
. . . u2 =
=
2
60
2
60
Dr. A. Oyieke
Hydraulic Machines III
(4)
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Euler Head
⇒ Consider the whirl velocity of water which can be associated
with a tangential momentum M = mass flow rate ×
tangential velocity of water ;
M = ṁw = ρQw
(5)
⇒ Taking moment of the momentum about the centre of the
impeller, then at inlet;
Mr1 = ρQw1 r1 = 0
since w1 = 0
(6)
and at the outlet;
Mr2 = ρQw2 r2
(7)
⇒ For this change in momentum of the water, a torque T is
required, i.e
T = M2 r2 − M1 r1
(8)
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Dr. A. Oyieke
Hydraulic Machines III
⇒ Substituting equations 6 and 7 into 8 we get;
XX
T = ρQw2 r2 − X
ρQw
r1
1X
T = ρQw2 r2
since
w1 = 0
(9)
(10)
⇒ Hydraulic power Phyd - The power received by the water from
the impeller is given by;
Phyd = T ω = ρQw2 r2 ω
but r2 ω = u2
∴ Phyd = ρQw2 u2
(11)
⇒ Euler head HE is the power per unit of weight flow rate. The
weight flow rate of the fluid is ρgQ(N/s), thus;
P
ρQw2 u2
w2 u2
=
= HE =
ρgQ
ρgQ
g
(12)
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Dr. A. Oyieke
Hydraulic Machines III
⇒ The unit of HE is metres and can be considered as the total
head (static plus dynamic) which has been added to the water
by the impeller.
⇒ If Bernoulli equation is set up for the flow of the water
through the pump, thus expression is used for the total
pressure head added to the water .
⇒ However, not all of this total pressure head becomes available
due to losses in the pump.
⇒ This pressure head is known as the vertical head, ideal head,
theoretical head or Euler head
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Dr. A. Oyieke
Hydraulic Machines III
Worked examples
Example 1
The ideal pressure increase across a centrifugal pump is 16 m. The
radial velocity through the pump is constant at 3.1 m/s. The exit
angle of the impeller is 60o and the water enters the impeller
radially. Ignore friction and other losses and determine;
i. the peripheral velocity at the impeller exit.
ii. the velocity head (dynamic head) at the impeller exit.
iii. the entrance angles of the diffuser guide vanes.
iv. the static pressure at the exit of the impeller if static head at
inlet to the impeller is 1.5 m of liquid.
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Dr. A. Oyieke
Hydraulic Machines III
Solution
Step one: Draw the diagram showing all the velocities and their
components applicable
Figure: Velocity diagram
Ideal pressure head HE = 16 m
velocity is constant)
f1 = f2 = 3.1 m/s ( radial
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Dr. A. Oyieke
Hydraulic Machines III
Step two: Calculate the required parameters.
i. Peripheral velocity; U2
HE =
u2 w2
g
w 2 = u2 − x = u2 −
(13)
f2
tan 60
3.1
= u2 − 1.79
tan 60
Substituting equation 15 into 16
w2 = u2 −
HE =
u2 (u2 − 1.79)
= 16
9.81
u2 2 − 1.79u2 − 157 = 0(quadraticequation)
p
1.79 ± 1.792 + 4(157)
u2 =
= 13.46 m/s
2
(14)
(15)
(16)
(17)
(18)
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Dr. A. Oyieke
Hydraulic Machines III
2
2
ii. The velocity head (dynamic head) at the impeller exit v2g
w2 = u2 − 1.79 = 13.46 − 1.79 = 11.67m/s
q
p
v2 = w2 2 + f2 2 = 11.672 + 3.12 = 12.07m/s
∴
v2 2
12.072
=
= 7.431 m
2g
2 × 9.81
iii. The entrance angle of diffuser guide vanes; γ
γ = tan−1
f2
3.1
= tan−1
= 14.88o
w2
11.6
γ = sin−1
f2
3.1
= tan−1
= 14.88o
v2
12.07
or
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Dr. A. Oyieke
Hydraulic Machines III
iv. static pressure at exit of the impeller if the impeller static
head at inlet is 1.5 m of liquid; P2
Consider Bernoulli equation across the impeller
P1
v1 2
P2
v2 2
+
+ z1 + H E =
+
+ z2 + HL,imp
ρg
2g
ρg
2g
on the centre line Z1 = Z2 and losses are to be ignored
1.5 +
3.12
P2
+ 16 =
+ 7.4831
2 × 9.81
ρg
∴
P2
= 10.56m
ρg
P2 = 10.56 × 103 × 9.81 = 103.6 kPa
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Dr. A. Oyieke
Hydraulic Machines III
Example 2
A pressure gauge installed at the inlet to the centrifugal pump
indicates a static pressure of -35 kPa (a vacuum) and a pressure
gauge at the exit indicates a static pressure of 320 kPa. The
suction side has a pipe diameter of 100 mm and the discharge pipe
has a diameter of 75 mm. The flow rate is 13 litres/second.
Determine;
i. the the power required for the pump if the overall efficiency of
the pump is 70%
ii. the Euler head if manometric efficiency is 75%
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Dr. A. Oyieke
Hydraulic Machines III
Solution 2
Pressure head increase across the pump
vs 2
vd 2
− Hms +
Htotal = Hmd +
2g
2g
Hmd =
Pmd
320 × 103
= 3
= 32.62m
ρg
10 × 9.81
Also;
Hms =
vd =
Q
2
πdd /4
vs =
−35 × 103
= −3.568m
103 × 9.81
=
0.013
= 2.942 m/s
π/4 × 0.0752
Q
0.013
=
= 1.655 m/s
2
π/4 × 0.12
πds /4
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Dr. A. Oyieke
Hydraulic Machines III
Solution 2 cont’d
1.6552
2.9432
− −3.568 +
= 36.49m
Htotal = 32.62 +
2 × 9.81
2 × 9.81
i. Pump power Ppump
Ppump =
ρgQHtotal
103 × 9.81 × 0.013 × 36.49
= 6.648w
=
ηoverall
0.7
ii. Euler head HE
ηmano =
Htotal
Htotal
36.49
⇒ HE =
=
= 48.65m
HE
ηmano
0.75
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Dr. A. Oyieke
Hydraulic Machines III
Example 3
A centrifugal pump discharges 0.1 m3 /s of water against a total
head of 22 m when rotating at 1440 rev/min. The impeller
diameter is 0.3 m and the flow area is constant throughout the
impeller at 0.022 m2 . The outlet of the impeller vanes makes an
angle of 23o with the tangent and loss in the impeller (HL,Imp ) is
estimated at 1.2 m. The suction and delivery pipes have diameters
of 200 mm. Assume radial flow. Determine;
i. the head increase from the inlet flange upto the exit of the
impeller.
ii. the loss in the diffuser (in metres of water)
iii. the pressure head increase in the diffuser.
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Dr. A. Oyieke
Hydraulic Machines III
Solution 3
Figure: Velocity diagram
u2 =
πND2
π × 1440 × 0.3
=
= 22.62m/s
60
60
Q
0.1
f2 =
=
= 4.545m/s
A
0.022
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Dr. A. Oyieke
Hydraulic Machines III
Solution 3 cont’d
f2
4.545
= 22.62 −
= 11.91m/s
o
tan 23
tan 23o
u2 w2
22.62 × 11.91)
HE =
=
= 27.47m
g
9.81
q
p
v2 = w2 2 + f2 2 = 11.912 + 4.5452 = 12.75m/s
w 2 = u2 −
i. Set up Bernoulli’s equation between the inlet flange and the
exit of the impeller;
vs 2
v2 2
+ hin + HE =
+ h2 + HL,imp
2g
2g
h2 − hin =
vs 2 v2 2
−
+ HE − HL,imp
2g
2g
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Dr. A. Oyieke
Hydraulic Machines III
Solution 3 cont’d
h2 − hin =
0.1
π/42 × 0.22
2
×
1
12.752
−
+ 27.47 − 1.2
2 × 9.81 2 × 9.81
h2 − hin = 0.516 − 8.286 + 27.47 − 1.2 = 18.50 m
ii. total pressure head increase across the pump = (energy
supplied to the water - loss in the impeller - loss in the
diffuser)
HE − hL,imp − hL,diff = 22
27.47 − 1.2 − hL,diff = 22
hL,diff = 4.27 m
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Dr. A. Oyieke
Hydraulic Machines III
Solution 3 cont’d
iii.
∆himp + ∆hdiff = ∆hpump
18.5 + ∆hdiff = 22
∆hdiff = 22 − 18.5 = 3.5 m
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Dr. A. Oyieke
Hydraulic Machines III