axial flow pump.pdf

By: Qusai Waleed Al-Qudah · · · · · · · · · · · · · · 1

The University of Jordan

Faculty of Engineering and Technology

Department of Civil Engineering _________________________________________

Hydraulics Laboratory

0931363


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


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 =

( )


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.



( )

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



( )

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.


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)


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))


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))


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)


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)


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)