Lab_Experiment_8_1
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California State University, Long Beach *
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Dec 6, 2023
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Uploaded by kalvarez22
Lab Experiment #8:
Centrifugal pump
CE 336: Fluids Mechanic Laboratory
Instructor: Andres Acosta, MS, PE
Submitted by:
Steven Ramirez
Eddie Rangel
Selena Nguyen
Shakir Shihabudeen
Justin Nguyen
Date of performing lab experiment: 11/30/2022
Date of submitting lab report: 12/19/2022
Introduction
1
This experiment examines the efficiency of the pump and the arrangement of the pumps singly
and parallel. The pump's overall head is calculated by reading the pressure head of its input and
output gauges. Calculating the system's efficiency can be done using its total head. To derive a
total head equation, Bernoulli's equation will be combined with other assumptions. To evaluate
the performance of a centrifugal pump, use its total head found.
Purpose of Study
The objective of this experiment is to determine the characteristics of a centrifugal pump. We
will operate a single pump as well as two pumps in parallel. At a fixed motor speed, the flow
rates through the pumps will be varied and corresponding efficiency of the pump will be
determined. We will compute pump efficiency at various other operating speeds.
Theory
Centrifugal pumps are radial flow rotodynamic machines. Fluid enters the rotor or impeller and
is thrown outwards by the centrifugal action. Because of the high speed of rotation, liquid
acquires a high kinetic energy and is thrown by centrifugal action.
Figure 1. Centrifugal pump
The total head a pump provides
H
=
P
2
−
P
1
?
+(
z
2
−
z
1
)+
V
2
2
−
V
2
1
2
g
On the experimental apparatus, diameters of the inlet and outlet pipes are similar, so, the velocity
head difference is considered insignificant. Therefore,
H
=
P
2
−
P
1
?
+(
z
2
−
z
1
)+
0
Pressure heads can be directly read in meters of H
2
O from the inlet and teh outlet gauge of the
test apparatus. The equation can be written as,
H
=(
h
2
−
h
1
)+(
z
2
−
z
1
)=(
h
2
−
h
1
)+
h
d
Datum is taken through the center of the pump impeller and each position is given a datum head
correction
factor.
Following variables are used to measure pump characteristics or performance,
2
Qt
=
V
/
t
Wo
=
γ QtH
Power
W
(
¿¿
o
/
W
i
)
×
100%
(
¿¿
out
/
Power
¿
)
×
100%
=
¿
η
p
=
¿
Where Q
t
is the flow rate through the pump, W
o
is the power output and W
i
is the power input.
W
i
is directly read the value from the inverter, explained in the procedure.
Equipment
●
A stopwatch (phone)
●
F1-27 Centrifugal pump test accessory
●
Hydraulic bench F1-10
Equipment Diagram:
Figure 2. Equipment Setup
Figure 2.2 Single Pump Setup
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3
Figure 2.3 Parallel Pump Setup
Experimental Setup & Procedures:
-
Experimental Set-Up
The F1-27 pump will be used to as the single pump for this experiment
1)
The inlet of the single pump should be connected to the sump drain valve of the F1-
10.The sump drain valve should be left fully open.
2)
Place the discharge manifold on the operating channel of the hydraulic bench.
3)
The outlet of the single pump should be connected to the discharge manifold.
The F1-27 pump will also be used in parallel with the pump in the hydraulic bench.
1)
The inlet of the F1-27 pump should be connected to the sump drain valve of the F1-10
and the sump drain valve should be fully open.
2)
The discharge manifold will be on the operating channel of the bench.
3)
The outlet of the F1-27 will be connected to the end of the Tee connector. One end of the
Tee connector will be connected to the outlet of the hydraulic bench. The other outlet will
be connected to the discharge manifold.
-
Procedure for Single Pump Operation
1)
The sump drain valve on the hydraulic bench should be open.
2)
The discharge control valve on the F1-27 should be kept closed.
3)
The motor should be turned on and be set to 50.00 hz.
4)
The FUNC/DATA key can be used to through different settings and the up/down
key can be used to change the motor’s speed.
5)
The discharge control valve can open fully. Take measurements of the inlet and
outlet pressure head gauges.
6)
To get the flow rate, either use the table that is in the lab manual, if the water is
above 1.4 L/s. If it is below, you must find how long it takes to get 5L of water.
7)
Slightly close the discharge control valve and get the measurements of the inlet +
outlet pressure head gauges.
8)
Once the valve is fully closed, get the data for no flow rate.
9)
Repeat steps 5-8, for 45 Hz, 40 Hz, 35 Hz, and 30 Hz.
-
Procedure for Parallel Pump Operation
4
1)
Keep the control valve on the hydraulic bench closed and the sump valve open.
Keep the discharge valve closed as well.
2)
Turn on the hydraulic bench and open the control valve fully.
3)
The motor should be turned on as well and set to 50 Hz.
4)
Fully open the discharge valve
5)
Change the settings by having the valve fully open to fully closed. Record the
inlet and outlet pressure head gauges. The outlet head can be read from the F1-27
gauge and the outlet head can be read from the discharge control manifold.
6)
For flow rates below 1.4 L/s, record how long it takes to get 5L of water in the
tank
7)
For flow rates over 1.4 L/s, get the flow rate from the table from the lab manual.
To do this remove the sealing ball and weight from the tank. Let the water level in
the tank stabilize before getting the reading value from the tank.
Calculations:
Table 1: Reference flow rate
Scale Reading
(liters)
Flow rate
(L/s)
Scale Readings
(liters)
Flow rate
(L/s)
0
1.41
20
2.00
1
1.44
21
2.02
2
1.48
22
2.05
3
1.51
23
2.07
4
1.54
24
2.09
5
1.57
25
2.12
6
1.60
26
2.14
7
1.63
27
2.16
8
1.66
28
2.19
9
1.69
29
2.21
10
1.72
30
2.23
11
1.75
31
2.25
12
1.78
32
2.27
5
13
1.81
33
2.29
14
1.84
34
2.31
15
1.86
35
2.33
16
1.89
36
2.35
17
1.92
37
2.37
18
1.94
38
2.39
19
1.97
39
2.40
40
2.42
1.
Motor speed (n) is measured directly in Hertz from an inverter.
2.
The volumetric flow rate (
Q
v
)
by collecting 5 L of water and finding the time to
collect that amount of water.
Q
v
=
V
t
=
0.005
m
3
6.24
s
=
0.00801
m
3
/
s
3.
Inlet head and outlet head are read from inlet and outlet gauges.
4.
For the Inlet and Outlet head correction factors
h
di
=
0.02
m
h
do
=
0.17
m
5.
Pump power input
W
i
is read directly from the inverter by using the FUNC/DATA
key.
6.
Total head
H
d
=(
h
do
−
h
di
)+(
h
o
−
h
i
)
H
d
=(
0.17
−
0.02
)+(
5.3
−
0
)=
5.45
m
7.
Pump power output
w
o
=
ρg H
d
Q
v
=
γ H
d
Q
v
=(
9810
N
/
m
3
)(
5.45
m
)(
0.00175
m
3
/
s
)=
93.56
W
8.
Overall turbine efficiency
η
p
=
W
o
W
i
×
100%
=
93.56
730
×
100%
=
12.82%
Table 2: Single Centrifugal pump experiment. Motor speed: 50 Hz.
No
of
Obs
Moto
r
Spee
d
n
(Hz)
Volu
me
of
wate
r
Vol
Time
to
colle
ct
t
(sec)
Flow
rate
Q
v
(m
3
/s
ec)
Inlet
Head
hi
(m
of
wate
Inlet
head
corre
ction
h
di
(m)
Outl
et
Head
h
0
(m
of
Outl
et
head
corre
ction
h
d0
Pum
p
pow
er
Input
W
i
Total
Head
H
(m
of
wate
Pum
p
powe
r
outp
ut
Over
all
Pum
p
Effic
iency
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6
(m
3
)
r)
wate
r)
(m)
(Watts)
r)
W
0
(Watts)
η
p
(%)
1
50
0
0
0.00
1750
00
0
0.02
5.3
0.17
730
5.45
93.5
6
12.8
2
2
50
0
0
0.00
1600
00
0
0.02
5.5
0.17
720
5.65
88.6
8
12.3
2
3
50
0
0
0.00
1620
00
-0.3
0.02
5.7
0.17
720
6.15
97.7
4
13.5
7
4
50
0
0
0.00
1660
00
0
0.02
5.2
0.17
720
5.35
87.1
2
12.1
0
5
50
0
0
0.00
1630
00
-0.2
0.02
6.5
0.17
710
6.85
109.
53
15.4
3
6
50
0
0
0.00
1510
00
-0.6
0.02
8.7
0.17
680
9.45
139.
98
20.5
9
7
50
0
0
0.00
1410
00
-0.7
0.02
11.1
0.17
550
11.9
5
165.
29
30.0
5
8
50
0.00
5
6.24
0.00
0801
00
-0.7
0.02
11.5
0.17
530
12.3
5
97.0
4
18.3
1
9
50
0.00
5
16.9
0
0.00
0295
80
-0.8
0.02
19.2
0.17
430
20.1
5
58.4
7
13.6
0
10
50
0
0
0
-0.6
0.02
20.1
0.17
380
20.8
5
0
0
Table 3: Single Centrifugal pump experiment. Motor speed: 45 Hz.
No
of
Obs
Moto
r
Spee
d
n
Volu
me
of
wate
r
Time
to
colle
ct
t
Flow
rate
Q
v
(m
3
/s
ec)
Inlet
Head
hi
(m
of
Inlet
head
corre
ction
h
di
Outl
et
Head
h
0
(m
Outl
et
head
corre
ction
Pum
p
pow
er
Input
Total
Head
H
(m
of
Pum
p
powe
r
outp
Over
all
Pum
p
Effic
7
(Hz)
Vol
(m
3
)
(sec)
wate
r)
(m)
of
wate
r)
h
d0
(m)
W
i
(Watts)
wate
r)
ut
W
0
(Watts)
iency
η
p
(%)
1
45
0
0
0.00
1520
00
-0.6
0.02
4.7
0.17
580
5.45
81.2
7
14.0
1
2
45
0
0
0.00
1480
00
-0.7
0.02
6.2
0.17
560
7.05
102.
36
18.2
8
3
45
0
0
0.00
1440
00
-0.8
0.02
6.3
0.17
560
7.25
102.
42
18.2
9
4
45
0
0
0.00
0146
00
-0.7
0.02
5.0
0.17
570
5.85
8.38
1.47
5
45
0
0
0.00
1410
00
-0.6
0.02
8.0
0.17
540
8.75
121.
03
22.4
1
6
45
0.00
5
4.08
0.00
1225
49
-0.7
0.02
10.1
0.17
500
10.9
5
131.
64
26.3
3
7
45
0.00
5
7.06
0.00
0708
22
-0.7
0.02
14.0
0.17
420
14.8
5
103.
17
24.5
6
8
45
0.00
5
13.5
3
0.00
0369
55
-0.8
0.02
15.2
0.17
360
16.1
5
58.5
5
16.2
6
9
45
0.00
5
32.7
1
0.00
0152
86
-0.8
0.02
15.4
0.17
330
16.3
5
24.5
2
7.43
10
45
0
0
0
-0.8
0.02
16.0
0.17
320
16.9
5
0
0
Table 4: Single Centrifugal pump experiment. Motor speed: 40 Hz.
No
of
Obs
Moto
r
Spee
d
Volu
me
of
wate
Time
to
colle
ct
Flow
rate
Q
v
(m
3
/s
Inlet
Head
hi
(m
Inlet
head
corre
ction
Outl
et
Head
h
0
Outl
et
head
corre
Pum
p
pow
er
Total
Head
H
(m
Pum
p
powe
r
Over
all
Pum
p
8
n
(Hz)
r
Vol
(m
3
)
t
(sec)
ec)
of
wate
r)
h
di
(m)
(m
of
wate
r)
ction
h
d0
(m)
Input
W
i
(Watts)
of
wate
r)
outp
ut
W
0
(Watts)
Effic
iency
η
p
(%)
1
40
0
0
0.00
1410
00
0
0.02
4.0
0.17
430
4.15
57.4
0
13.3
5
2
40
0.01
6.25
0.00
1600
00
0
0.02
4.0
0.17
430
4.15
65.1
4
15.1
5
3
40
0.01
6.84
0.00
1461
99
0
0.02
5.0
0.17
410
5.15
73.8
6
18.0
2
4
40
0.01
7.56
0.00
1322
75
0
0.02
5.5
0.17
400
5.65
73.3
2
18.3
3
5
40
0.01
9.09
0.00
1100
11
0
0.02
7.0
0.17
380
7.15
77.1
6
20.3
1
6
40
0.00
5
6.05
0.00
0826
45
0
0.02
9.0
0.17
340
9.15
74.1
8
21.8
2
7
40
0.00
6
12.3
5
0.00
0485
83
0
0.02
12.0
0.17
290
12.1
5
57.9
1
19.9
7
8
40
0.00
6
25.7
7
0.00
0232
83
0
0.02
12.5
0.17
260
12.6
5
28.8
9
11.11
9
40
0.00
6
37.9
8
0.00
0157
98
0
0.02
13.0
0.17
250
13.1
5
20.3
8
8.15
10
40
0
0
0
0
0.02
13.5
0.17
240
13.6
5
0
0
Table 5: Single Centrifugal pump experiment. Motor speed: 35 Hz.
No
of
Obs
Moto
r
Spee
Volu
me
of
Time
to
colle
Flow
rate
Q
v
Inlet
Head
hi
Inlet
head
corre
Outl
et
Head
Outl
et
head
Pum
p
pow
Total
Head
H
Pum
p
powe
Over
all
Pum
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9
d
n
(Hz)
wate
r
Vol
(m
3
)
ct
t
(sec)
(m
3
/s
ec)
(m
of
wate
r)
ction
h
di
(m)
h
0
(m
of
wate
r)
corre
ction
h
d0
(m)
er
Input
W
i
(Watts)
(m
of
wate
r)
r
outp
ut
W
0
(Watts)
p
Effic
iency
η
p
(%)
1
35
0.01
7
0.00
1410
00
0
0.02
2.5
0.17
310
2.65
36.6
6
11.8
2
2
35
0.01
7.49
0.00
1335
11
0
0.02
2.5
0.17
310
2.65
34.7
1
11.2
0
3
35
0.01
7.87
0.00
1270
65
0
0.02
4.0
0.17
300
4.15
51.7
3
17.2
4
4
35
0.01
10.8
7
0.00
0919
96
0
0.02
5.0
0.17
280
5.15
46.4
8
16.6
0
5
35
0.00
5
6.72
0.00
0744
05
0
0.02
7.0
0.17
250
7.15
52.1
9
20.8
8
6
35
0.00
6
10.1
8
0.00
0589
39
0
0.02
8.5
0.17
240
8.65
50.0
1
20.8
4
7
35
0.00
6
12.4
9
0.00
0480
38
0
0.02
9.0
0.17
230
9.15
43.1
2
18.7
5
8
35
0.00
6
15.7
0
0.00
0382
17
0
0.02
9.5
0.17
220
9.65
36.1
8
16.4
4
9
35
0.00
6
32.2
8
0.00
0185
87
0
0.02
9.5
0.17
200
9.65
17.6
0
8.80
10
35
0
0
0
0
0.02
10.0
0.17
190
10.1
5
0
0
Table 6: Single Centrifugal pump experiment. Motor speed: 30 Hz.
No
of
Moto
r
Volu
me
Time
to
Flow
rate
Inlet
Head
Inlet
head
Outl
et
Outl
et
Pum
p
Total
Head
10
Obs
Spee
d
n
(Hz)
of
wate
r
Vol
(m
3
)
colle
ct
t
(sec)
Q
v
(m
3
/s
ec)
hi
(m
of
wate
r)
corre
ction
h
di
(m)
Head
h
0
(m
of
wate
r)
head
corre
ction
h
d0
(m)
pow
er
Input
W
i
(Watts)
H
(m
of
wate
r)
1
30
0.01
9
0.00
1410
00
0
0.02
2.0
0.17
230
2.15
29.7
4
12.9
3
2
30
0.01
8.90
0.00
1123
60
0
0.02
2.0
0.17
220
2.15
23.7
0
10.7
7
3
30
0.01
9.25
0.00
1081
08
0
0.02
3.0
0.17
220
3.15
33.4
1
15.1
9
4
30
0.01
10.3
4
0.00
0967
12
0
0.02
3.0
0.17
220
3.15
29.8
9
13.5
8
5
30
0.00
5
5.72
0.00
0874
13
0
0.02
4.0
0.17
200
4.15
35.5
9
17.7
9
6
30
0.00
5
9.49
0.00
0526
87
0
0.02
5.5
0.17
190
5.65
29.2
0
15.3
7
7
30
0.00
6
17.1
2
0.00
0350
47
0
0.02
6.5
0.17
170
6.65
22.8
6
13.4
5
8
30
0.00
6
24.3
8
0.00
0246
10
0
0.02
6.5
0.17
170
6.65
16.0
5
9.44
9
30
0.00
6
47.6
7
0.00
0125
87
0
0.02
7.0
0.17
160
7.15
8.83
5.52
10
30
0
0
0
0
0.02
7.0
0.17
150
7.15
0
0
Table 7: Parallel pump configuration (pump speed 50 Hz)
No
of
Moto
r
Volu
me
Time
to
Flow
rate
Inlet
Head
Inlet
head
Outl
et
Outl
et
Pum
p
Total
Head
11
Obs
Spee
d
n
(Hz)
of
wate
r
Vol
(m
3
)
colle
ct
t
(sec)
Q
v
(m
3
/s
ec)
hi
(m
of
wate
r)
corre
ction
h
di
(m)
Head
h
0
(m
of
wate
r)
head
corre
ction
h
d0
(m)
pow
er
Input
W
i
(Watts)
H
(m
of
wate
r)
1
50
0
0
0.00
1860
00
0
0.02
7.0
0.17
230
7.15
130.
46
56.7
2
2
50
0
0
0.00
1720
00
0
0.02
8.0
0.17
220
8.15
137.
52
62.5
1
3
50
0
0
0.00
1630
00
0
0.02
10.0
0.17
220
10.1
5
162.
30
73.7
7
4
50
0
0
0.00
1410
00
0
0.02
14.0
0.17
220
14.1
5
195.
72
88.9
7
5
50
0.00
5
5.55
0.00
0900
90
0
0.02
17.0
0.17
200
17.1
5
151.
57
75.7
8
6
50
0.00
5
8.31
0.00
0601
68
0
0.02
19.0
0.17
190
19.1
5
113.
03
59.4
9
7
50
0.00
5
9.23
0.00
0541
71
0
0.02
20.0
0.17
170
20.1
5
107.
08
62.9
9
8
50
0.00
5
12.7
0
0.00
0393
70
0
0.02
21.0
0.17
170
21.1
5
81.6
9
48.0
5
9
50
0.00
5
46.5
8
0.00
0107
34
0
0.02
23.0
0.17
160
23.1
5
24.3
8
15.2
4
10
50
0
0
0
0
0.02
24.0
0.17
150
24.1
5
0
0
9.
Graph of measured total head
H
d
for different flow rate
Q
v
for rotational speed 50
Hz.
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12
Figure 1: Total head vs Flow rate for 50 Hz speed
10. Plot the total head
H
d
for different flow rate
Q
v
for rotational 45 Hz, 40 Hz, 35
Hz, 30 Hz.
Figure 2: Total head vs Flow rate for 45 Hz, 40 Hz, 35 Hz and 30 Hz
11. Plot measured efficiency for different flow rate for rotational 50 Hz, 45 Hz, 40 Hz, 35
Hz, 30 Hz.
13
Figure 3: Measured efficiency vs Flow rate for 50 Hz
Figure 4: Measured efficiency vs Flow rate for 45 Hz
14
Figure 5: Measured efficiency vs Flow rate for 40 Hz
Figure 6: Measured efficiency vs Flow rate for 35 Hz
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15
Figure 7: Measured efficiency vs Flow rate for 30 Hz
12. Plot measured power output vs flow rate.
Figure 8: Power output vs Flow rate for 50 Hz
16
Figure 9: Power output vs Flow rate for 45 Hz
Figure 10: Power output vs Flow rate for 40 Hz
17
Figure 11: Power output vs Flow rate for 35 Hz
Figure 12: Power output vs Flow rate for 30 Hz
13. Graph of head
H
d
for different flow rates using a dataset for different settings of sump
drain valve opening.
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18
Figure 13: Total head vs flow rate for parallel pump
Questions and Discussion:
1.
In terms of pump head, efficiency and rotation speed, discuss shape of the pump
characteristics curves for a single pump or parallel pump settings.
a.
For a single pump, pump head, efficiency , and rotation speed have distinct
characteristics when related to flow rate. According to Figure 1, pump heads seem
to decrease as flow rate increases. This is a disproportionate linear relationship to
flow rate. According to Figures 2-6, efficiency has a parabolic relationship to flow
rate. This non-linear relationship shows how this pump runs more efficiently at a
certain flow rate that is not the highest. According to Figures 7-11, power also has
a nonlinear relationship to flow rate. They are closely related to the efficiency
graphs, which means there is a flowrate that produces the most power that isn't the
fastest flow rate. This information is useful because it means that the pump can
save energy by just running at the most efficient flow rate and not the highest to
produce the largest amount of water.
2.
How the head and flow rates are affected when the pumps are in series or in parallel?
a.
When in series the total head increases while flow rate remains the same. In
parallel, the total head will decrease and flow rates will increase.
Exercise 1
1.
Based on the Head vs Flow rate graph, discuss how the pump characteristics curve
changed with the rotational speed.
a.
In both series and parallel head vs flow rate graphs, the total head decreased in a
linear manner as the flow rate increased. It can be concluded that the total head is
disproportionately related to the flow rate.
2.
Based on efficiency vs Flow rate graph, discuss how the pump characteristics curve
changed with the rotational speed.
a.
Based on Figures 2-6, there is a non-linear relationship between efficiency of the
pump and the flowrate. There seems to be a negative parabolic relationship
19
between the two. This means there is a flow rate, that is not the fastest, that is the
best for operating the pump.
3.
Based on the power vs Flow rate graph, discuss how the pump characteristics curve
changed with the rotational speed.
a.
Like the efficiency graphs, there is a non-linear relationship between power
output and flow rate. There is a negative parabolic relationship between the two.
This means there is a flow rate, that is not the fastest, that produces the most
amount of power. This is important to understand when trying to save the most
amount of operation cost.
4.
Based on the Head vs Flow, efficiency vs Flow rate, and power vs flow rate, what is the
optimum operating point at each speed tested.
a.
The most optimum operating point seems to be around 0.0014
m
3
sec
across all
different types of graphs.
5.
Discuss using the plot of head H_d vs Q_v (the dataset for different settings of sump
drain valve opening) the effect of inlet suction head on the performance of the pump.
a.
According to Figure 13, there seems to be a negative linear relationship in the plot
between Total head and flow rate for different settings of sump drain valve
opening. This means that the effect of inlet suction on the head is
disproportionately related to the performance of the pump.
Exercise 2
1.
Discuss the pump characteristics curve for two pumps in parallel settings.
a.
Two pumps in a parallel setting mean that the sum of both single flow rates will
be the total head of the system. Overall, the flowrate of the system will be higher
than that of a single pump. When the overall flow rate of the system increases, the
total head will decrease.
2.
Discuss if you get two different flow rates when two pumps are operating in parallel.
a.
There will not be two different flow rates when two pumps are operating in
parallel. The sum of the two individual flow rates will be the one flow rate for the
whole system.
3.
Are there any differences between observed and theoretical heads? Provide reasons for
any differences
a.
There is a slight difference in observed head and theoretical head. Between each
different flow rate the heads differ by around .15 m of water. Possible reasons for
any differences may include energy losses in the system.
Conclusion:
In this experiment we were to observe the characteristics of both a single and parallel centrifugal
pump. In the experiment, we ran the pump at different flow rates at different motor speeds.
Overall the results of the experiment showed us that in a single pump, having a constant flow
rate that doesn't flow too fast, will result in having an efficient pump.This can be seen from
figures 2-6, as the flow rate remained at a constant speed, the pump was working most
efficiently. This also is the same for the power produced by the pump. From figures 8-12, we see
that as the flow rate remains at a constant speed, the pump produces the most amount of power.
So for both pump efficiency and power, it is related to the flow rate. It also showed us that if the
flow rate increases, the total head will also decrease disproportionately. For a parallel pump,
20
generally the flow rate will be higher than a single pump. So as the parallel pump increases in
flow rate, the total head will also decrease as well.Overall, a big takeaway from this experiment,
shows us that flow rate can affect how both pumps work.
References
Sultana, Rebeka. “Lab #8 Centrifugal pump.”
Fluid Mechanics Laboratory Student Manual.
2017. Print.
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Related Questions
5.19 Consider the piping system shown in the following diagram. The pump operates
at 1750 rpm.
El 1475'
L= 8000'; D = 12";f=0.026
El 1200'
Pump
106
Turbomachinery: Concepts, Applications, and Design
The
pump curve is given by h,=350-24Q² where h, is given in ft. and Q in cfs.
What would the head and flow rate delivered be under these conditions?
Ans: Q=160 cfs; h,=288 ft
5.20 Solve the previous problem for two identical pumps connected in (a) series and (b)
parallel.
Ans: a) Q=2.83 cfs, h,=315 ft; b) Q=2.58 cfs; h,=310 ft
%D
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QUESTION 1:
A centrifugal pump consists out of an impeller, diffuser and casing and is powered by 300
kW (shaft power) and rotates at 96.3 rad/s. The flow rate generated by the machine is
54.7 Kilo-litres/min. The impeller blade inlet angle and width is 23° and 180 mm
respectively. The blades occupy 12% of the impeller area and the water is assumed to
enter the impeller radially. Suction and delivery pipes are equal in size.
Determine:
1.1.
The impeller inlet (i) diameter and inlet blade velocity.
(Hint - Substitution: Vn in terms of Di and U, in terms of D; both sub into Tan of Blade inlet angle,
only unknown is then solving for D?)
ANS: 300 mm; 14.42 m/s
If the pump has a mechanical efficiency of 80%, calculate the blade velocity at
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guide vane angle. (Hint - Substitution: Write Vwo in terms of U. sub in Euler equation only
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ANS: Vwo = 11.54…
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D = 9.0 inches operating at a speed of N = 1750 rpm collected on a pump test stand using water at STP. For
this pump test there are no changes in elevation or velocity between the suction inlet and pump discharge. (a)
Plot the actual pump head and pump efficiency versus flowrate. (b) Use a curve fit to obtain a pump curve,
H₂ = H0 - C₂ Q². (c) Use a curve fit to obtain a 2nd order polynomial fit for the efficiency curve. (d) Calculate
the best efficiency point.
Q (gpm)
20
40
60
80
100
120
140
Ap (psi)
40.2
40.1
38.1
36.2
33.5
30.1
25.8
Wm (hp)
1.58
2.27
2.67
2.95
3.19
3.49
4.00
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Example 4:
If the diameter at section 1 is D₁ = 0.5 m, and
at section 2 the diameter D₂ = 0.2 m.
All the other conditions are the same as in
Exercise #3.
What power in kilowatts and in horsepower
must be supplied to the flow by the pump?
Assume h₁ = 3 m of water and a₁ = α₂ = 1.
Water is being pumped through a system.
D₁ = 0.5 m
Z₁ = 30 m
P₁ = 70 kPa gage
α₁ = 1.0
1
Pump
Sual
Z₂ = 40 m
P2 = 350 kPa gage
%₂ = 1.0
D₂ = 0.2 m
Head loss in pipe = 3 m
Water
Q = 0.5 m³/s
Properties: Water (10°C, 1 atm, Table A.5): y = 9810 N/m³
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Note : Please Solve as Per Question's Statement.
Need a Neat and Clear Solution.See Attached Picture For Detail .
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Problem 1. A centrifugal water pump is designed to operate at a speed of N = 1200 rpm at a volume flow rate
of Q = 35 L/s. The impeller has the following dimensions:
Parameter
Radius, r (cm)
Blade width, b (cm)
Blade angle, B (deg)
Inlet
10
1.0
30
Outlet
17.5
0.75
40
(a) Draw the inlet and outlet velocity diagrams to scale. (b) Determine the inlet and outlet absolute flow angles
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b) Calculate the cylinder output speed in in/s.Cylinder output speed (VS) = Piston side flow (Q) / Piston area (Ap)Cylinder output speed (VS) = (Displacement x Speed of revolution) / Piston area (Ap)Cylinder output speed (VS) = (0.2 in3/rev x 1725 RPM) / 1.77 in2Cylinder output speed (VS) = (0.2 in3/rev x 1725 RPM) / 1.77 in2Cylinder output speed (VS) = 196.02in/s = 196.02 / 60s = 3.20 in/s
Questions:
c) Knowing the output velocity (ram speed), calculate the rod side flow in GPM when the ram is extending.
d) Calculate the piston exit time in seconds.
e) Calculate the piston entry time in…
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HEAD IN FEET
50
0 0 0 0
40
20
10
O 81%
8.625"
30 (219mm).
O 51%
O 71%
○ 61%
O
Taco
L/SEC 5
9.25"
(235mm)
8.00"
(203mm)
7.375"
(187mm)
6.75"
(171mm)
0
10
15
125
Model 5009
FI & CI Series
20
25
CURVES BASED ON CLEAR WATER
WITH SPECIFIC GRAVITY OF 1.0
8 8
52
75%
77%
30
79%
1160 RPM
FEBRUARY 19. 2002
35 40 45
REQUIRED NPSH
olº
1.5HP
(1.1KW)
dot
17%
Curve no. 2140
Min. Imp. Dia. 6.75"
Size 6 x 5 x 9.0
50
55
60
75%
72%
250 375 500
625
FLOW IN GALLONS PER MINUTE
8%
2HP(1.5KW)
750
%09.
55%
(2.2KW)
3HP
7.5HP(5.6KW)
2015
do
35%
875
FEET
15
12
-9
6
SHP(3.7KW)
3
0
10
8
NPSH
1000
HEAD IN METERS
5 KPo
45
-36
27
18
9
-0
100
-80
60
-40
2 -20
8
HEAD IN KILOPASCALS
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###
Develop a conclusion of a piping design report for the choice of a pump, based on the calculated parameters of flow Q = 10 l/s, power of 10kw taking into account that the head of 60 mca adopted was higher than that calculated for safety, and that the pump was chosen through the WebCAPS catalogue of the company GRUNDFOS.
The pump chosen was the GRUNDFOS CR 32-3 A-F-A-E-HQQE
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a. Sketch the problem.
b.Draw lines identifying the control volume, or control mass.
c.Identify the states with numbers, letters, or descriptions such as “in” and “out”.
d.Write down the knowns and unknowns.
e.Identify what is being asked for.
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Can yu please complete question 4
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