Title of experiment: CENTRIFUGAL PUMP

INTRODUCTION

Centrifugal pump consist of an impeller rotating within a spiral casing. The fluid enters the pump axially through the suction pipe via the eye of the impeller, it is discharge from impeller around the entire circumference both into the ring stationary diffuser vanes (and through them into the volute casing) or directly into the casing. The casing collect the fluid, decelerates it – thus converting some of the kinetic energy into pressure energy – and finally discharges the fluid through the delivery flange.

OBJECTIVE

The objective of this experiment are

to obtain the characteristics performance of the centrifugal pump

Student should be able to understand the principles and application of centrifugal pump

THEORY

Assuming steady flow, pump basically increase the Bernoulli head of flow between point 1 (suction) and point 2 (delivery), neglecting viscous work and heat transfer, this donate by head, H. The head of a pump is the mechanical work transfer red by the pump to the medium pumped under the local gravity conditions. The head, H tells us the increment of mechanical energy, E between inlet and outlet.

Head could be defined as:

H = (ZD- Zs) + (PD – Ps) + (VD2 – Vs2)

ρg 2g

Where

ZD – Zs : Difference in the height of the inlet and outlet cross section on the pump.

: Difference in the pressure head of the medium pumped between inlet and outlet

: Difference in the speed of the medium pumped between inlet and outlet.

Usually VD and Vs are about the same ZD – Zs is no more than a meter or so.

Therefore, the net pump head, H equals to change in pressure head:

H = PD – Ps ……………… (2)ρg

The power required to drive the pump is brake power, Pmech Pmech = T.ω

= T 2πN …………… (3) 60

Where T = torque (Nm)

Hydraulic power output, Phyr: Phydr = ρgQH ………… (4)

The pump efficiency, η is given by the ratio between the power output by a pump and the power drawn from the shaft, ie

η = P hydr ……………….. (5)

P mech

Pump Characteristic Curves

P (and thus the efficiency η) , as well as the parameter NSPH req depend on the flow rate Q. The relationship between these of the performance data is displayed in characteristic curves. The operating behavior of each centrifugal pump is characteristic by these characteristic curves.

System characteristic curve

The system characteristic curve is given by the pressure losses in the pipes at a specific flow rate. For increasing speed (= flow rate), these exist points in a graph of head H against flow rate Q using which the characteristic curve can be drawn. The operating point of a pump is positioned, as per figure 1, where the head of the pump and the system are the same that is at the point where the system characteristic curve and pump characteristic curves cross.

Centrifugal Pump

One of the most common radial-flow turbo machines is the centrifugal pump. This type

of pump has two main components: an impeller attached to a rotating shaft, and a stationary

casing, housing, or volute enclosing the impeller. The impeller consists of a number of blades 1

usually curved 2, also sometimes called vanes, arranged in a regular pattern around the shaft. A

sketch showing the essential features of a centrifugal pump is shown in figure below. As the

impeller rotates, fluid is sucked in through the eye of the casing and flows radially outward.

Energy is added to the fluid by the rotating blades, and both pressure and absolute velocity are

increased as the fluid flows from the eye to the periphery of the blades. For the simplest type of

centrifugal pump, the fluid discharges directly into a volute-shaped casing. The casing shape is

designed to reduce the velocity as the fluid leaves the impeller, and this decrease in kinetic

energy is converted into an increase in pressure. The volute-shaped casing, with its increasing

area in the direction of flow, is used to produce an essentially uniform velocity distribution as the

fluid moves around the casing into the discharge opening. For large centrifugal pumps, a

different design is often used in which diffuser guide vanes surround the impeller. The diffuser

vanes decelerate the flow as the fluid is directed into the pump casing. This type of centrifugal

pump is referred to as a diffuser pump.

EQUIPMENT

Equipment used for this experiment are:

1. Basic module water pumps, (GUNT Hamburg), Germany.2. Self Priming centrifugal pump.3. Drive and Brake unit.4. Tachometer

Procedure

Series of measurement at various speed must be performed on the pumps. The pump must be running at constant speed and the system must be more or less at the steady state. Steps to run the experiment are:

1. Turn On all main switches and check to ensure the apparatus is ready.2. The belt guard (1) must be in place and the direction of the rotation indicator in clock

wise direction is illuminated.3. The pump can only start with back pressure. The ball valve 2 for flow rate regulation

must be closed.4. Move the potentiometer 3 to start the motor. Fully open the ball valve.5. Set the speed to 1000 rpm and observe whether water is pumped back to the tank 46. Let the pump runs for some time and reach its operating temperature.

Experiment At constant Speed, 2900 rpm7. Ensure the ball valve is fully opened.8. Increase the speed in 2900 rpm by turning the potentiometer.9. For this experiment, take the flow rate reading from magnetic inductive only 510. Flow rate , P1, P2 and torque are recorded11. A little the ball valve are closed to reduce the flow rate.

Experiment At Various Speed.12. The speed is set to 1500 rpm with the ball valve fully opened.13. Flow rate P1, P2 and torque are record.14. The speed is increase into 1100 rpm ( 100 rpm increment)15. Repeat from 13 until the maximum speed 3000 rpm.

RESULT

Calculate and fill up Table 1 and 2

Speed = 2900 rpm

No.

Flow rateQ

(l/s)

SuctionPressure P1 (Bar)

DeliveryPressureP2 (bar)

TorqueT

(Nm)

HeadH

(m)

MechanicalPower

Pmech (W)

Hydraulic Power

Phydr (W)

Efficiencyη

1 2.884 -0.89 0.37 2.72 13.01 829.45 368.08 0.442 2.883 -0.89 0.37 2.70 13.01 823.35 367.95 0.453 2.880 -0.89 0.38 2.68 13.11 817.25 370.65 0.454 2.869 -0.89 0.39 2.66 13.22 811.15 372.08 0.465 2.871 -0.88 0.40 2.65 13.22 808.10 372.33 0.466 2.856 -0.88 0.43 2.65 13.53 808.10 379.07 0.477 2.816 -0.85 0.49 2.63 13.66 802.00 377.36 0.478 2.757 -0.82 0.56 2.59 14.07 789.80 380.54 0.489 2.607 -0.74 0.67 2.56 14.37 780.66 367.51 0.4710 1.375 -0.63 0.85 2.50 15.09 762.36 203.55 0.2711 1.172 -0.54 0.97 2.45 15.39 747.11 176.94 0.2412 1.149 -0.30 1.35 2.14 16.82 652.58 189.59 0.2913 0.287 -0.07 1.78 1.58 18.86 481.81 53.10 0.11

No. MechanicalSpeed

N (rpm)

FlowrateQ

(l/s)

SuctionPressure P1

(Bar)

DeliveryPressureP2, (bar)

HeadH

(m)1 1500 1.486 -0.30 0.06 3.722 1600 1.592 -0.33 0.07 4.133 1700 1.700 -0.36 0.09 4.654 1800 1.800 -0.40 0.11 5.275 1900 1.902 -0.44 0.13 5.896 2000 2.006 -0.48 0.15 6.507 2100 2.100 -0.52 0.17 7.128 2200 2.220 -0.57 0.19 7.859 2300 2.330 -0.61 0.22 8.5710 2400 2.430 -0.66 0.24 9.2911 2500 2.540 -0.71 0.27 10.1212 2600 2.640 -0.76 0.30 10.9513 2700 2.750 -0.81 0.33 11.7714 2800 2.842 -0.86 0.37 12.70

15 2900 2.962 -0.90 0.40 13.42CALCULATION: Experiment At Constant Speed, 2900 rpm

Calculation of Head H (m)

1 bar = 1.013 x 105 N/m2 ρ water = 1000 kg/m3 gravity = 9.81 ms-2

H = PD - Ps

ρg

1 ). H = 0.3748 x 105 + 0.9016 x 105 = 13.01 m

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2 ). H = 0.3748 x 105 + 0.9016 x 105 = 13.01 m

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3 ). H = 0.3849 x 105 + 0.9016 x 105 = 13.11 m

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4 ). H = 0.3951 x 105 + 0.9016 x 105 = 13.22 m

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5 ). H = 0.4052 x 105 + 0.8914 x 105 = 13.22 m

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6 ). H = 0.4356 x 105 + 0.8914 x 105 = 13.53 m

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* The others calculation from data number 7 to 13 is same like above

Calculation of mechanical power

Mechanical Power , P mech (W) = T.ω

= T (2 N )

60

Where T = torque (Nm) N = Revolution/ minute (rpm)

1 ). P mech = (2.72 x 2 x 2912) / 60 = 829.45 W

2 ). P mech = (2.70 x 2 x 2912) / 60 = 823.35 W

3 ). P mech = (2.68 x 2 x 2912) / 60 = 817.25 W

4 ). P mech = (2.66 x 2 x 2912) / 60 = 811.15 W

5 ). P mech = (2.65 x 2 x 2912) / 60 = 808.10 W

6 ). P mech = (2.65 x 2 x 2912) / 60 = 808.10 W

7 ). P mech = (2.63 x 2 x 2912) / 60 = 802.00 W

8 ). P mech = (2.59 x 2 x 2912) / 60 = 789.80 W

9 ). P mech = (2.56 x 2 x 2912) / 60 = 780.66 W

10 ). P mech = (2.50 x 2 x 2912) / 60 = 762.36 W

11 ). P mech = (2.45 x 2 x 2912) / 60 = 747.11 W

12 ). P mech = (2.14 x 2 x 2912) / 60 = 652.58 W

13 ). P mech = (1.58 x 2 x 2912) / 60 = 481.81 W

Calculation to get hydraulic power

Hydraulic Power ,Phydr (W) = ρgQH

Where ρ = density ρ water = 1000 kg/m3

g = gravity g = 9.81 ms-2

Q = flow rate 1 m3 =1000 liter

H = head

1 ). Phydr = 1000 x 9.81 x 2.884 x 10-3 x 13.01 = 368.08 W

2 ). Phydr = 1000 x 9.81 x 2.883 x 10-3 x 13.01 = 367.95 W

3 ). Phydr = 1000 x 9.81 x 2.882 x 10-3 x 13.11 = 370.65 W

4 ). Phydr = 1000 x 9.81 x 2.869 x 10-3 x 13.22 = 372.08 W

5 ). Phydr = 1000 x 9.81 x 2.871 x 10-3 x 13.22 = 372.33 W

6 ). Phydr = 1000 x 9.81 x 2.856 x 10-3 x 13.53 = 379.07 W

7 ). Phydr = 1000 x 9.81 x 2.816 x 10-3 x 13.66 = 377.36 W

8 ). Phydr = 1000 x 9.81 x 2.757 x 10-3 x 14.07 = 380.54 W

9 ). Phydr = 1000 x 9.81 x 2.607 x 10-3 x 14.37 = 367.51 W

10). Phydr = 1000 x 9.81 x 1.375 x 10-3 x 15.09 = 203.55 W

11). Phydr = 1000 x 9.81 x 1.172 x 10-3 x 15.39 = 176.94 W

12). Phydr = 1000 x 9.81 x 1.149 x 10-3 x 16.82 = 189.59 W

13). Phydr = 1000 x 9.81 x 0.287 x 10-3 x 18.86 = 53.10 W

Calculation of pump efficiency

Pump Efficiency, η = P hydr / P mech

1 ). η = 340.12 / 811.15 = 0.44

2 ). η = 344.06 / 814.20 = 0.45

3 ). η = 344.68 / 814.20 = 0.45

4 ). η = 347.77 / 820.30 = 0.46

5 ). η = 356.64 / 832.50 = 0.46

6 ). η = 365.54 / 841.65 = 0.47

7 ). η = 379.65 / 835.55 = 0.47

8 ). η = 395.37 / 823.35 = 0.48

9 ). η = 389.17 / 820.30 = 0.47

10). η = 370.37 / 802.00 = 0.27

11). η = 297.85 / 741.01 = 0.24

12). η = 297.85 / 741.01 = 0.29

13). η = 297.85 / 741.01 = 0.11

CALCULATION: Experiment At Various Speed

1 bar = 1.013 x 105 N/m2 ρ water = 1000 kg/m3 gravity = 9.81 ms-2

H = PD - Ps

ρg

1 ). H = 0.06078 x 105 + 0.3039 x 105 = 3.72 m

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2 ). H = 0.07091 x 105 + 0.3343 x 105 = 4.13 m

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3 ). H = 0.09117 x 105 + 0.3647 x 105 = 4.65 m

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4 ). H = 0.11143 x 105 + 0.4052 x 105 = 5.27 m

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5 ). H = 0.13169 x 105 + 0.4457 x 105 = 5.89 m

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6 ). H = 0.48624 x 105 + 0.1519 x 105 = 6.50 m

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7 ). H = 0.52676 x 105 + 0.1722 x 105 = 7.12 m

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8 ). H = 0.57741 x 105 + 0.1925 x 105 = 7.85 m

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9 ). H = 0.61793 x 105 + 0.2229 x 105 = 8.57 m

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10 ). H = 0.66858 x 105 + 0.2431 x 105 = 9.29 m

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11 ). H = 0.71923 x 105 + 0.2735 x 105 = 10.12 m

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12 ). H = 0.76988 x 105 + 0.3039 x 105 = 10.95 m

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13 ). H = 0.82053 x 105 + 0.3343 x 105 = 11.77 m

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14 ). H = 0.87118 x 105 + 0.3748 x 105 = 12.70 m

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15 ). H = 0.91170 x 105 + 0.4052 x 105 = 13.42 m

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Convert Flow rate from l/s to m3/s

1 l/s = 3600/1000 m3/s

For experiment at constant speed, 2900 rpm

No. Flow rate l/s Flow rate m3/hr

1 2.884 10.382

2 2.883 10.379

3 2.880 10.368

4 2.869 10.328

5 2.871 10.336

6 2.856 10.282

7 2.816 10.138

8 2.757 9.925

9 2.607 9.385

10 1.375 4.950

11 1.172 4.219

12 1.149 4.136

13 0.287 1.033

For experiment at various speed

No. Flow rate l/s Flow rate m3/hr

1 1.486 5.350

2 1.592 5.731

3 1.700 6.120

4 1.800 6.480

5 1.902 6.847

6 2.006 7.222

7 2.100 7.560

8 2.220 7.992

9 2.330 8.388

10 2.430 8.748

11 2.540 9.144

12 2.640 9.504

13 2.750 9.900

14 2.842 10.231

15 2.962 10.663

OBSERVATION

Pemerhatian yang dapat dibuat semasa ekperimen ini dijalankan adalah pada eksperimen

pertama iaitu pada halaju malar, apabila injap ditutup sedikit demi sedikit kadar alir akan

bertambah sehingga pada bacaan yang 5 kadar alir mula menurun, sehinggalah apabila injap

hampir ditutup sepenuh, kadr alir yang didapati cumalah 1 m3 /hr. Ini adalah disebabkan oleh

laluan yang semakin sempit di bahagian injap. Selepas itu kita dapat lihat apabila diplotkan data

ke dalam bentuk graf, didapati bahawa pada graf kelajuan malar dengan graf kelajuan berubah

terdapat satu persilangan diantara 2 graf ini. Persilangan graf mempunyai makna yang penting

iaitu dimana persilangan ini merupakan titik operasi terbaik bagi pam. Apabila ditunjukan ke

bawah didapati titi operasi terbaik adalah pada kadar alir 9.93 m3/hr. Dalam ekperimen ini juga

kita dapat melihat bahawa kebanyakkan bacaan yang diambil tidak mempunyai ketepatan yang

tepat disebabkan pengesan yang mengesan perubahan perubahan yang berlaku tidak berhenti

apabila mendapat bacaan. Bacaan naik turun menyebabkan bacaan diantara nilai tertinggi dengan

yang terendah dibahagi dua.

DISCUSION

-What suggestion to improve the experiment?

1. Gantikan injap biasa yang digunakan untuk melaras pengaliran air masuk kepada injap

yang mempunyai nilai sudut pusingan agar pusingan setiap sudut daripada bacaan

pertama sehingga bacaan yang ke 13 adalah sama

2. Tachometer yang digunakan semasa mendapatkan bacaan putaran ( rpm ) mestilah lebih

persis untuk mendapatkan bacaan yang lebih jitu.

3. Bacaan untuk mendapatkan kadar alir hendaklah dibaca dengan tepat sehingga stabil iaitu

diambil setiap 2 minit dan hendaklah dibahagi dua antara bacaan tertinggi dan terendah.

- Find the operating point of the pump. Mark on the graph

Pam beroperasi pada keadaan terbaik pada kadar alir 9.93 m3/hr dan Head 13.5 H(m)

- Describe the significant of the operating point to the pump

Titik operasi ini penting bagi menentukan kadar alir dan head yang paling sesuai dalam

sistem paip bagi pam empar yang berada di makmal kerana apabila kita mengetahui titik empar

dan kadar alir yang paling sesuai maka kita dapat pengaliran bendalir yang paling baik dalam

paip. Kita tahu bahawa di dalam paip terdapat terdapat beberapa factor lain yang dapat

mempengaruhi kadar alir seperti geseran antara bendalir dengan permukaaan paip, diameter paip

dan juga panjang paip.

CONCLUSION

Kesimpulan yang dapat dibuat diakhir ekperimen adalah objektif yang ditetapkan dalam

ekperimen ini telah tercapai di mana kami dapat tahu tentang ciri ciri dalam sebuah pam empar.

Dengan pelarasan head yang betul kita boleh mendapatkan kadar alir yang terbaik di dalam paip.

Dimana head adalah 13.5 m untuk mendapatkan kadar alir yang terbaik dalam paip iaitu 9.93

m3/hr.

REFERENCE

1) Bruce R. Munson & Donald F.Young (2002), “ Fundamental of Fluid Mechanic”, Department

of Mechanical Engineering, Iowa state Univesity.

2) John A Roberson, Clayton T Crowe,” Engineering Fluid Mechanics, Sixth Edition” , John

Wiley& Sons, Inc

3) Lab Sheet Mechanic Fluid

Observation can make during experiment this conducted is in first experiments that is on

constant velocity, when valve closed little by little rate of flow will increase until in reading 5

degrees flow started to decrease, until when valve almost closed fully, flow rate which found

only 1 m3 /hr. This is due to more cramped route in valve part. After that we can see when is

plotted data into graph form, found that in constant speed graph with speed graph changed found

a intersection between 2 this graph. Graph intersection has meaning importantly namely where

this intersection is best operation point for pump. When ditunjukan down found best operation

bridge is in value flow 9.93 m3/hr. In experiment this also we can see that the reading adopted

had no accuracy right because detectors detect change which occurred not stop upon obtaining

reading. Reading up and down cause reading between highest values with lowest bisected.