Post on 24-Dec-2015
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 1
The University of Jordan
Faculty of Engineering and Technology
Department of Civil Engineering _________________________________________
Hydraulics Laboratory
0931363
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 2
Objectives and Liabilities:
to study the performance of an axial flow pump.
Apparatus:
Water moves from a water tank through a calibrated nozzle. It then passes through the
pump and down to a fully adjustable delivery valve. It then returns to the water tank. The
delivery valve allows the user to gradually shut the downstream water flow for a range of
pump performance tests. Electronic transducers measure the pump inlet and outlet
Pressures and the pressure difference across the nozzle. A digital display shows all the
readings.
For quick and reliable tests, Equipment can supply the optional VDAS (Versatile Data
Acquisition System). VDAS gives accurate real-time data capture, monitoring and
Display, calculation and charting of all the important readings on a computer. The
computer is not supplied
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 3
Experimental produce :
1- Make sure all vales are open and turn on the motor in the control cabinet.
2- Turn the control knob to the required speed 2000rpm.
3- Set the outlet angle blade to 0 degree.
4- Set the rotor blade to 30 degree.
5- Adjust the lower left hand valve for changing the discharge pressure .
6- Record pressure, flow rate and torque for variance discharge pressure for the
given speed.
7- Turn the control knob to 2500rpm and repeat step 5 & 6.
8- Turn the control knob to 3000rpm and repeat step 5 & 6.
Theoretical of this experiment:
When the pump start, the fluid enters the suction nozzle and then into enter of the
impeller, ( suction eye ). As the impeller rotates, it spins the fluid sitting in the cavities
between the vanes outward and provides centrifugal acceleration. As the fluid leaves
the eye of the impeller, low pressure is developed, causing continuous flow into the
pump inlet.
Same in the axial flow pump, the head developed by a pump is determined by
measuring the pressures on the suction and discharge sides of the pump.
The velocities are computed by measuring the discharge and dividing it by the
respective pipe cross areas therefore, the net head delivered by the pump to the fluid
is:
(
) (
)
Where:
H: head developed by the pump in m.
P1,2: pressure head at the suction side and delivery side of the pump in Pa.
V1,2: velocity at the suction side and delivery side of the pump in
.
Z1,2: elevation at the suction side and delivery side of the pump in m.
ρ: the density of water =
g: gravity acceleration =
( )
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 4
Usually the intake pipe is larger than the discharge pipe, however in the current
apparatus the discharge and suction pipe are the same size, therefore the velocity
heads cancel out.
Also the assumption is made that both suction side and delivery side are on the same
elevation, resulting in neglecting the elevation head, the net total can be repressed:
Where:
H: head developed by the pump in m.
: Pressure head.
ρ: the density of water =
g: gravity acceleration =
( )
We will cross the equation with 10^5 to convert from bar to Pa.
The total power output, in watts, of the pump is equal to the production of the pump
total pressure and the volumetric flow rate:
Where:
Q: flow rate in m^3/sec
ρ: the density of water =
g: gravity acceleration =
( )
V: velocity at the suction side and delivery side of the pump in
.
The power input, in watts, from the dynamometer s given by:
(
)
Where:
τ: torque in
ω: angular velocity.
F: force ( measured load on motor ) in N
r: torque arm=0.178m.
N: impeller speed in rpm.
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 5
The total power output of the pump is equal to the production of the pump total
pressure and the volumetric flow rate:
ἠ
Calculation about this experiment:
For N = 2000rpm
Q
(
)
P1
(
)
P7
(
)
F
(KN) H (m)
Pout
(Watt)
Pin
(Watt) ἠ
0.012 0 14000 14 1.427115189 168 521.9012 32.19
0.009 2000 19000 14.5 1.732925586 153 540.5405 28.305
0.006 5000 23000 14.2 1.834862385 108 529.3569 20.40211
0.004 8100 26000 16 1.824668705 71.6 596.4585 12.00419
0
2
4
6
8
10
12
14
16
18
20
0 2 4 6 8 10 12 14 16
H
Q
h
Poly. (h)
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 6
For N = 2500rpm
Q
(
)
P1
(
)
P7
(
)
F
(KN) H (m)
Pout
(Watt)
Pin
(Watt) ἠ
0.015 -3000 19000 20.5 2.242609582 330 955.2656 34.54537
0.012 0 24000 20.2 2.44648318 288 941.2861 30.59644
0.008 3000 31000 20.5 2.854230377 224 955.2656 23.44898
0.005 9000 36000 22 2.752293578 135 1025.163 13.16864
0
5
10
15
20
25
30
35
0 2 4 6 8 10 12 14
ἠ
Q
ἠ
Poly. (ἠ)
0
0.5
1
1.5
2
2.5
3
0 2 4 6 8 10 12 14 16
Axi
s Ti
tle
Axis Title
H (m)
Poly. (H (m))
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 7
For N = 2500rpm
Q
(
)
P1
(
)
P7
(
)
F
(KN) H (m)
Pout
(Watt)
Pin
(Watt) ἠ
0.02 -11000 20000 28 3.160040775 620 1565.704 39.59881
0.015 -5000 28000 26.5 3.363914373 495 1481.827 33.40472
0.01 0 40000 28.3 4.077471967 400 1582.479 25.2768
0.005 15000 50000 32.5 3.567787971 175 1817.335 9.629487
0
5
10
15
20
25
30
35
40
0 2 4 6 8 10 12 14 16
ἠ
Q
ἠ
Poly. (ἠ)
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 5 10 15 20 25
H
Q
H (m)
Poly. (H (m))
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 8
The difference between it:
0
5
10
15
20
25
30
35
40
45
0 5 10 15 20 25
ἠ
Q
ἠ
Poly. (ἠ)
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 1 2 3 4 5 6 7
H
Q
N=2000rpm
N=2500rpm
N=3000rpm
Poly. (N=2000rpm)
Poly. (N=2500rpm)
Poly. (N=3000rpm)
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 9
Sample calculation of this experiment:
At Q=12m^3/sec & N=2000rpm
(
)
0
5
10
15
20
25
30
35
40
45
0 1 2 3 4 5 6 7
ἠ
Q
N=2000rpm
N=2500rpm
N=3000rpm
Poly. (N=2000rpm)
Poly. (N=2500rpm)
Poly. (N=3000rpm)
By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 10
ἠ
Conclusions: 1- The obtained plots of the performance relations was unexpected in there forms, there
was sever irregularity in the relation between head and 'Q', as the values of Q
decreased the H varied irregularly increasing and decreasing without a certain rhythm.
And the same problem was faced with the efficiency relations where they increased
for the taken flow rate. I think there was something wrong in the readings we
obtained.
2- The relation between the power output and the flow rate is direct according to the
plots, in theory it is assumed that the relation shall be inverse as H is supposed to
decrease uniformly with increasing Q.
3- Decreasing the flow rate increases the force of the motor linearly. I think this relation
can be explained by the following: higher flow rate has higher momentum force which
causes an opposite effect on the rotating propeller and so on the motor, as we didn't
change anything with the motor setting the only effect on its force can be caused by
the flow conditions (flow rate)