CENTRIFUGAL PUMP TEST
RIG
INTRODUCTION TO CENTRIFUGAL
PUMPS
The
Hydraulic Machines which converts the Mechanical energy into Hydraulic energy
are called Pumps. The
Hydraulic energy is in the form of Pressure energy. If the mechanical energy is
converted into Pressure energy by means of centrifugal force acting on the Fluid, the Hydraulic machines is called
Centrifugal Pumps.
The
centrifugal fan with consists of a blower driven by FHP Motor with a series of
curved redial vanes. Air is drowning in near the hub, called the blower eye,
and is whirled round at high speed by the vanes on the blower as the blower
rotates at high rotational speed. The static pressure of the air increases from
the eye of the blower to the tip of the blower in order to provide the centrifugal
force on the air. As the air leaves the blower tip it is passed through
diffuser passage which converted most of the kinetic energy of the air into an
increase in enthalpy and hence the pressure of the air is further increased. The
blower may be double sided, having an eye either side of the unit, so that air
is drowning in both side. The advantages of this type is that the blower is
subjected to approximately equal forces in an axial direction. In practice
nearly half the total practice is achieved in blower and the remaining half the
diffuser. A pressure ratio of around 4: 1 can be achieved with the centrifugal
fan.
The
Centrifugal pump acts as a reversed of an inward radial flow reaction turbine.
This means that the flow in centrifugal pumps is in the radial outward
directions. The centrifugal pump works on the principle of forced vortex flow
which means that when a certain mass of liquid is rotated by an external
torque, the rise in pressure head of the rotating liquid takes place. The rise
in pressure head at any point of the rotating liquid is proportional to the
square of tangential velocity of the liquid at that point. Thus at the outlet
of the impeller, whose radius is more, the rise in pressure head will be more
and the liquid will be discharged at the outlet with a high pressure head. Due
to this high pressure head, the liquid can be lifted to a high level.
Main
Parts of Centrifugal Pump: 
 Impeller
 Casing
Impeller:  The Rotating Part of centrifugal Pump is called “Impeller”.
It consists of a series of backward curved vanes. The Impeller is mounted on a
shaft which is connected to the shaft of an an electric motor.
Casing:  The Casing of a Centrifugal Pump is similar to the casing of
a reaction turbine. It is an airtight Passage surrounding the impeller and is
designed in such a way that the kinetic energy of the water discharged at the
outlet of the Impeller is converted into pressure Energy before the Water
Leaves the casing and enters the delivery pipe. Centrifugal pump is a rotodynamic
machine, which develop dynamic pressure of liquid by virtue of rotation for
pumping of liquid to a higher height. In centrifugal pump, liquid in the
impeller of a pump is made to rotate by external force, so that it is through
away from the center of rotation. As constant supply of liquid is made
available at the center liquid can be pumped to higher level.
The ‘UNICOOL’
unit consists of a centrifugal pump driven by a Dimmer Control motor. Input to
motor is measured on energy meter. A measuring tank is provided to measure the
discharge. Suction vacuum and discharge pressure is measure by gauges. A Ball
valve on discharge pipe varies the head. Thus, performance of pump can be
estimated at various speed and heads.
SPECIFICATIONS: 
Centrifugal
pump 25 x 25 mm. size, base mounted,
Motor: 
1 H.P. Dimmer Control motor directly coupled to pump.
Measuring
tank  x  x mm. height, fitted with drain valve.
Sump tank
600 x 900 x 600 mm heights.
Ball
valve to control the head.
Pressure
gauge to measure discharge pressure.
Vacuum
gauge to measure suction vacuum.
Energy
meter to measure input the motor.
EXPERIMENTAL PROCEDURE: 
 Fill up sufficient water in the sump tank.
 Open the priming nipple plug (At the top of pump) and fill up water up to the nipple and then tight the plug.
 Shut off the discharge valve.
 Start the pump. As discharge valve is closed, no discharge will be observed, but discharge pressure will be indicated. This is called ‘Shut off head’ of the pump.
 Slowly open the discharge valve, so that small discharge is observed.
 Note down discharge head, suction vacuum and time required for 10 ltrs. Of water level rise in measuring tank and 10 revolutions of energy meter disc.
 Note down the observations at different valve openings.
 Repeat the procedure for any different speed.
OBSERVATIONS: 
S.
NO.

PUMP
SPEED
(RPM)
N

DISCHARGE
PRESSURE
(Kg/cm^{2})
Pd

SUCTION
VACUUM
(mm
of Hg) Ps

TIMES
FOR 10 Lt.
Water
level rise
(Sec)
t_{w}

TIMES
FOR 10 rev. of
Energy meter
(Sec)






CALCULATIONS: 
Discharge pressure P_{d} =
__________Kg/cm^{2}
For
water, 10 m height corresponds to 1 Kg/ cm^{2}
Discharge
head, h_{d} = P_{d }x 10 m of water.
Suction Head
Suction
vacuum, P_{s} = ________mm of Hg
p_{s =
}13.6
Suction
head, h_{s} =  x 
1000x1
Where,
sp
gravity of hg = 13.6 and Sp gravity of water = 1
Total Head, h_{t} =
h_{d} + h_{s} + h_{r}
Where, h_{r }=
3 mtr. Is the head loss due to Friction?
Discharge: 
Let time
for 10 Cm. Level rise be t_{w} sec.
.380 x
.380 x .010
Then,
discharge, Q =  m^{3} / sec
t_{w}
_{}
Output power (or water
power)
W.Q. h_{t}
WP =
 kw
1000
Where,
W =
Specific weight of water = 9810 N / m^{3}
Q =
Discharge m^{3} / sec.
h_{t} =
Total head, mtrs.
Electrical Input: 
Let time
required for 10 rev. of energy meter disc be t_{e} sec.
10 3600
Electrical
input power, Ip =  x 
t_{e} 1300
Where,
Energy Meter constant = 1300 Imp / kw / hr.
Taking
motor efficiency as 60% we have input shaft power
SP =
Elect. I.P x 0.60
Overall
efficiency of the pump: 
WP
ŋ_{o} =
 x 100%
SP
PRECAUTIONS
Priming is must before
starting the pump. Pump should never be run empty.
Use clean water in the sump
tank.
Operate all the Controls
Gently.
CALCULATION
OBSERVATION TABLE: 
Sr
No.

Pump
speed
(RPM)
N

Discharge
Pressure
(Kg/cm^{2})
Pd

Suction
Vacuum
(mm
of Hg) Ps

Times
for 10 lit water level rise (Sec) tw

Time
for 10 rev of Energy meter (Sec) te

1

2400

0.35

200

5.65

20.9

2

2400

0.47

160

6.37

22.12

3

2400

0.57

130

8.22

23.06

4

2400

0.65

95

12.68

26.72

CALCULATIONS: 
1) Discharge Pressure Pd = 0.47 Kg / cm^{2}
For
water, 10 m height corresponds to 1 kg / cm^{2}
Discharge
head, h_{d} = 0.47 x 10 = 4.7 m of water
2) Suction Head –
Suction
vacuum, Ps = 160 mm of Hg = 160x13.6
Suction
head, hs =  x 
1000x1
= 2.18 m
of water
Where, sp
gravity of hg = 13.6
Sp.
Gravity of water = 1
Total Head, h_{t} = h_{d} + h_{s} +
h_{r}
Where, h_{r} =
2 mtr. is the head loss due to friction
h_{t} =
4.7 + 2.18 + 3
= 9.88 m
of water
Discharge –
Let time
10 lts. Level rise be tw sec.
0.01
Then,
Discharge, Q = 
6.37
= .90 x
10 m^{3} / sec
Output power (or water power)
W.Q.h_{}
WP =
 kW
1000
Where, W = Specific weight of water = 9810 N / m^{3}
Q =
Discharge m^{3} / sec
Ht =
Total head, mtrs
9810 x
.90 x 10^{3} x 9.88
WP = 
1000
= 0.097
kW
Electrial Input –
Let time
required for 10rev. of energy meter disc be t_{e} Sec. = 10x 3600
Electrical
Input Power, IP =  x  = 22 1300
= 0.362
kW
Where.
Energy meter constant = 1300 rev / kw / hr
Taking
motor efficiency as 60% we have input shaft power
SP =
0.362 x 0.75
= 0.29 kW
Overall efficiency of the
pump
.097
ŋ_{o} =
 x 100%
= 0.29 = 22%
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