Experiment Multi Pump Test Rig

UNIVERSITI KUALA LUMPUR MALAYSIAN INSTITUTE OF CHEMICAL & BIO-ENGINEERING TECHNOLOGY

FLUID MECHANICS

CLB 11003

TITLE : Experiment 6 : Multi Pump Test Rig

Lecturer’s Name : En.Eddyazuan

Name / Section 1)SURENDRAN BALAKRISHNAN 2)MUHAMMAD AKMAL HAKIN BIN RAMLAN 3)AHMAD IKHRAM ROSLAN Due Date : 13 APRIL 2016

ID Number 55201113445 55201113557 55201113682

OBJECTIVES 

Determine the operating characteristic of different pumps in a contained unit.



Understand the types of pumps in principle and design, and the selection of the appropriate pump for a particular application for optimal operation.

SUMMARY The objective of this experiment is to determine the operating characteristic of different pumps in a contained unit. In addition, this experiment was conducted to understand the types of pumps in principle and design and the selection of the appropriate pump for a particular application for optimal operation. The results for this experiment were obtained for pump 1, pump 2 and pump 3 according to different types of characteristics for each of the pump. This experiment is divided into four parts. First experiment is rotational speed vs volumetric flow rate, which is for a performance curve for a centrifugal pump. The second experiment is other performance curve for a centrifugal pump. The third experiment is rotational speed vs output pressure, which is performance curve for a positive displacement pump. Finally, the last experiment is other performance curve for a positive displacement pump. For each part of experiment, the respective graphs were plotted for different types of characteristics. In the discussion, the characteristics curves for each part of experiment was plotted according the pump 1, pump 2 and pump 3. In the each characteristics curves for pump 1, pump 2 and pump 3, the relationships between each characteristics have been discussed. In short, as a conclusion, students were able to determine the operating characteristics of different

pumps in a contained unit. Besides, students understood the types of pumps in principle and design and the selection of the appropriate pump for a particular application for optimal operation. Thus, the objectives of this experiment were achieved.

RESULTS Data Collected for Experiment 1: Table 1 : Rotational Speed and Flow rate for P1 Speed (RPM)

Flow rate (%)

2800

59.3

2600

57.0

2400

53.9

2200

49.1

2000

44.6

1800

40.1

1600

35.7

1400

30.8

1200

26.2

1000

21.7

800

17.3

600

12.8

Volume of Q was calculated using formula : a) Q 

q 113.56  60  100 1000

b) Q 

q 113.56  60  100 1000

q = Flow rate (%)

q = Flow rate (%)

When q is 59.4

When q is 57.0

59.3 113.56  60  100 1000 3 m Q  4.04 hr

Q

57.0 113.56  60  100 1000 3 m Q  3.88 hr

Q

c) Q 

q 113.56  60  100 1000

d) Q 

q 113.56  60  100 1000

q = Flow rate (%)

q = Flow rate (%)

When q is 53.9

When q is 49.1

49.1 113.56  60  100 1000 3 m Q  3.35 hr

53.9 113.56  60  100 1000 3 m Q  3.67 hr

Q

Q

e) Q 

q 113.56  60  100 1000

f)

Q

q 113.56  60  100 1000

q = Flow rate (%)

q = Flow rate (%)

When q is 44.6

When q is 40.1

44.6 113.56  60  100 1000 3 m Q  3.04 hr

Q

g) Q 

q 113.56  60  100 1000

40.1 113.56  60  100 1000 3 m Q  2.73 hr

Q

h) Q 

q 113.56  60  100 1000

q = Flow rate (%)

q = Flow rate (%)

When q is 35.7

When q is 30.8

35.7 113.56  60  100 1000 3 m Q  2.43 hr

Q

30.8 113.56  60  100 1000 3 m Q  2.10 hr

Q

i)

Q

q 113.56  60  100 1000

j)

Q

q 113.56  60  100 1000

q = Flow rate (%)

q = Flow rate (%)

When q is 26.2

When q is 21.7

26.2 113.56  60  100 1000 3 m Q  1.79 hr

21.7 113.56  60  100 1000 3 m Q  1.48 hr

Q

Q

k) Q 

q 113.56  60  100 1000

l)

Q

q 113.56  60  100 1000

q = Flow rate (%)

q = Flow rate (%)

When q is 17.3

When q is 12.8

17.3 113.56  60  100 1000 3 m Q  1.18 hr

Q

12.8 113.56  60  100 1000 3 m Q  0.87 hr

Q

Volume Flow, Q (m3/hr)

Flow rate (%)

Rotational Speed , N (RPM)

59.3

4.04

2800

57.0

3.88

2600

53.9

3.67

2400

49.1

3.35

2200

44.6

3.04

2000

40.1

2.73

1800

35.7

2.43

1600

30.8

2.10

1400

26.2

1.79

1200

21.7

1.48

1000

17.3

1.18

800

12.8

0.87

600

Rotational Speed (N) vs Volume Flow rate (Q) Rotatinal Speed, N (RPM)

3000 y = 667.63x - 0.2371 R² = 0.997

2500 2000 1500 1000 500 0 0

0.5

1

1.5

2

2.5

3

3.5

Volume Flow Rate, Q ( m3/hr)

Figure 1: Rotational Speed (N) vs Volume Flow rate (Q)

4

4.5

Data Collected for Experiment 2 : Table 2: Flow rate, Speed, Differential Pressure and Power for P1 Flow rate %

Speed RPM

Diff. Pressure %

Power kW

60

2800

16.4

0.53

50

2799

30.4

0.50

40

2807

42.9

0.48

30

2822

55.2

0.46

20

2836

61.5

0.42

10

2851

65.4

0.40

Volume of Flow rate , Q wac calculate using formula :

a) Q 

q 113.56  60  100 1000

b) Q 

q = Flow rate (%) When q is 60

q = Flow rate (%) When q is 50

50 113.56  60  100 1000 3 m Q  3.41 hr

60 113.56  60  100 1000 3 m Q  4.09 hr

Q

Q

c) Q 

q 113.56  60  100 1000

d) Q 

q = Flow rate (%) When q is 40

30 113.56  60  100 1000 3 m Q  2.04 hr

40 113.56  60  100 1000 3 m Q  2.73 hr

q 113.56  60  100 1000

q = Flow rate (%) When q is 20

20 113.56  60  100 1000 3 m Q  1.36 hr

Q

q 113.56  60  100 1000

q = Flow rate (%) When q is 30

Q

Q

e) Q 

q 113.56  60  100 1000

f)

Q

q 113.56  60  100 1000

q = Flow rate (%) When q is 10

10 113.56  60  100 1000 3 m Q  0.68 hr

Q

PMi is calculated as below :-

a).

c).

PMi  Power (kW )  1000

PMi  Power (kW )  1000

when power  0.53

when power  0.50

PMi  0.53kW 

1000W 1kW

b).

1000W 1kW

PMi  530W

PMi  500W

PMi  Power (kW )  1000

PMi  Power (kW )  1000

when power  0.48

when power  0.46

PMi  0.48kW 

1000W 1kW

d).

PMi  0.46kW 

1000W 1kW

PMi  4800W

PMi  460W

PMi  Power (kW )  1000

PMi  Power (kW )  1000 when power  0.40

when power  0.42

e).

PMi  0.50kW 

PMi  0.42kW  PMi  420W

1000W 1kW

f).

PMi  0.40kW  PMi  400W

1000W 1kW

i.

Motor Input Power (PMI) Vs. Volume Flow rate (Q)

Volume Flow rate,

Flow rate (%)

Power (kW)

Q (m3/hr)

Motor Input Power, PMi, (W)

60

4.09

0.53

530

50

3.41

0.50

500

40

2.73

0.48

480

30

2.04

0.46

460

20

1.36

0.42

420

10

0.68

0.40

400

Motor Input Power (PMi) vs Vol Flow Rate (Q)

Motor Input Power , PMi (W)

600 y = 38.089x + 374.16 R² = 0.9898

500 400

300 200 100 0

0

0.5

1

1.5

2

2.5

3

3.5

Volume Flow Rate , Q (m3/hr)

Figure 1: Motor Input Power vs Volume Flow Rate

4

4.5

ii.

Pump Total Head (H) Vs. Volume Flow rate (Q)

Pump Total Head is calculated by using formula as below :4  DP   3  10.2  10  H  Z c 2  Z c1      w g  100    where

H  Pump Total Head , m Z c 2  Outlet Dis tan ce From Datum ( water )  860mm  0.86m Z c1  Inlet Dis tan ce From Datum ( water )  180mm  0.18m DP  Differential Pr essure,%

 w  Density water  1000 g  Gravity  9.81

kg m3

m s2

i). Differential Pressure, % = 16.4 when DP  16.4   2  3bar  10.2  10 4 N / m   16.4   bar  H  (0.86m  0.18m)     kg m   100   1000 3  9.81 2   m s       4 N / m2  16.4   3bar  10.2  10  bar H  0.68m    2  kg m 1N .s  100   1000 3  9.81 2   kg.m  m s  H  5.80m

ii). Differential Pressure, % = 30.4 when DP  30.4   2  3bar  10.2  10 4 N / m  30 . 4    bar  H  (0.86m  0.18m)    kg m   100   1000 3  9.81 2   m s       4 N / m2  30.4   3bar  10.2  10  bar H  0.68m    2  kg m 1N .s  100   1000 3  9.81 2   kg.m  m s  H  10.16m iii). Differential Pressure, % = 42.9 when DP  42.9   4 N / m2    42.9   3bar  10.2  10 bar  H  (0.86m  0.18m)    kg m   100   1000 3  9.81 2   m s       4 N / m2 3 bar  10 . 2  10  42.9    bar H  0.68m    2  100 kg m 1 N . s    1000 3  9.81 2   kg.m  m s  H  14.06m iv). Differential Pressure, % = 55.2 when DP  55.2   2  3bar  10.2  10 4 N / m   42.9   bar  H  (0.86m  0.18m)    kg m   100   1000 3  9.81 2   m s       4 N / m2  42.9   3bar  10.2  10  bar H  0.68m    2  kg m 1N .s  100   1000 3  9.81 2   kg.m  m s  H  14.06m

v). Differential Pressure, % = 61.5 when DP  61.5   2  3bar  10.2  10 4 N / m  61 . 5    bar  H  (0.86m  0.18m)    kg m   100   1000 3  9.81 2   m s       4 N / m2  61.5   3bar  10.2  10  bar H  0.68m    2  kg m 1N .s  100   1000 3  9.81 2   kg.m  m s  H  19.86m vi). Differential Pressure, % = 65.4 when DP  65.4   4 N / m2    65.4   3bar  10.2  10 bar  H  (0.86m  0.18m)    kg m   100   1000 3  9.81 2   m s       4 N / m2 3 bar  10 . 2  10  65.4    bar H  0.68m    2  100 kg m 1 N . s    1000 3  9.81 2   kg.m  m s  H  21.08m

Zc2-Zc1

Diff. Pressure, D

Pump Total Head, H

(m /hr)

(m)

(%)

(m)

4.09

0.68

16.4

5.80

3.41

0.68

30.4

10.16

2.73

0.68

42.9

14.06

2.04

0.68

55.2

17.90

1.36

0.68

61.5

19.86

0.68

0.68

65.4

21.08

Volume Flow rate, Q 3

Pump Total Head (H) vs Vol Flow Rate (Q) 25

Pump Total Head, H (m)

20

15

10 y = -4.5772x + 25.727 R² = 0.9611 5

0 0

0.5

1

1.5

2

2.5

Volume Flow Rate, Q

3

3.5

(m3/hr)

Figure 2 : Pump Total Head (H) Vs Volumetric Flow Rate (Q)

4

4.5

iii.

Pump Power Output (Po) Vs. Volume Flow rate (Q)

Pump Power Output was obtained by calculate using formula as below :-

Po   w gHQ  

1hr 3600s

where Po  Pump Power Output , W

 w  Density water  1000 m s2 H  Pump Total Head , m g  Gravity  9.81

m3 Q  Volume Flow rate, hr

kg m3

1.

when H  5.80m, Q  4.09 Po Po Po Po

2.

m3 hr

 kg m m 3  1hr     1000 3  9.81 2  5.80m  4.09 hr m s   3600s kg.m 2 1N .s 2  64.64 3  kg.m s N .m  64.64 s  64.64W

when H  10.16m, Q  3.41

m3 hr

 kg m m 3  1hr Po  1000 3  9.81 2  10.16m  3.41   hr  3600s m s  2 2 kg.m 1N .s Po  94.41 3  kg.m s N .m Po  94.41 s Po  94.41W 3.

when H  14.06m, Q  2.73 Po Po Po Po

m3 hr

 kg m m 3  1hr   1000 3  9.81 2  14.06m  2.73 hr  3600s m s  kg.m 2 1N .s 2  104.60 3  kg.m s N .m  104.60 s  104.60W

4.

Po Po Po Po 5.

m3 hr

when H  17.90m, Q  2.04

 kg m m 3  1hr     1000 3  9.81 2  17.90m  2.04 hr m s   3600s kg.m 2 1N .s 2  99.51 3  kg.m s N .m  99.51 s  99.51W

when H  19.86m, Q  1.36

m3 hr

 kg m m 3  1hr  Po  1000 3  9.81 2  19.86m  1.36 hr  3600s m s  kg.m 2 1N .s 2 Po  73.60 3  kg.m s N .m Po  73.60 s Po  73.60W 6.

when H  21.08m, Q  0.68 Po Po Po Po

m3 hr

 kg m m 3  1hr   1000 3  9.81 2  21.08m  0.68 hr  3600s m s  kg.m 2 1N .s 2  39.06 3  kg.m s N .m  39.06 s  39.06W

Volume Flow rate, Q (m3/hr)

Pump Total Head, H (m)

Pump Power Output, Po (W)

4.09

5.80

64.64

3.41

10.16

94.41

2.73

14.06

104.60

2.04

17.90

99.51

1.36

19.86

73.60

0.68

21.08

39.06

Pump Power Output (P0) vs Vol Flow rate (Q)

Pump Power Output, P0 (W)

120 100

y = 8.1807x + 59.792 R² = 0.1736

80 60 40 20 0 0

0.5

1

1.5

2

2.5

3

3.5

Volume Flow Rate, Q (m3/hr)

Figure 3 : Pump Power Output vs Volume Flow Rate

4

4.5

iv.

Pump Power Input (Pi) Vs. Volume Flow rate (Q)

Pump Power Input, Pi was calculated by using formula below :-

Pi  PMi  Pp1 min  where Pi  PumpPowerInput , W PMi  Motor Input Power , W Pp1 min  Pump 1 Power at No Load (50 Hz )  70W

a).

c).

e).

when PMi  530

b).

when PMi  500

Pi  (530  70)W

Pi  (500  70)W

Pi  460W

Pi  430W

when PMi  480

d).

when PMi  460

Pi  (480  70)W

Pi  (460  70)W

Pi  410W

Pi  390W

when PMi  420

f).

when PMi  400

Pi  (420  70)W

Pi  (400  70)W

Pi  350W

Pi  330W

Volume Flow rate, Q

Motor Input Power, PMi

Pp1min

Pump Power Input, Pi

(m3/hr)

(W)

(W)

(W)

4.09

530

70

460

3.41

500

70

430

2.73

480

70

410

2.04

460

70

390

1.36

420

70

350

0.68

400

70

330

Pump Power Input (Pi) vs Vol Flow Rate (Q) 500 y = 38.089x + 304.16 R² = 0.9898

Pump Power Input, Pi (W)

450 400 350 300 250 200 150 100 50 0 0

0.5

1

1.5

2

2.5

3

Volume Flow rate, Q (m3/hr)

Figure 4 : Pump Power Input vs Volume Flow rate

3.5

4

4.5

v.

Pump Efficiency (ETA) Vs. Volume Flow rate (Q)

Pump Efficiency was obtained by calculation:-

a).

ETA 

Po  100% Pi

b).

when Po  64.64 W , Pi  460W ETA 

ETA 

64.64W  100% 460 W

Po  100% Pi

ETA 

d).

ETA 

when Po  73.60 W , Pi  350W ETA 

73.60W  100% 350 W

ETA  21.03%

Po  100% Pi

99.51W  100% 390W ETA  25.52%

104.60W  100% 410W

Po  100% Pi

ETA 

when Po  99.51W , Pi  390W ETA 

ETA  25.51% e).

94.41W  100% 430W

ETA  21.96%

when Po  104.60W , Pi  410W ETA 

Po  100% Pi

when Po  94.41W , Pi  430W

ETA  14.05% c).

ETA 

f).

ETA 

Po  100% Pi

when Po  39.06W , Pi  330W ETA 

39.06W  100% 330 W

ETA  11.84%

Volume Flow rate, Q

Pump Power Output, Po

Pump Power Input, Pi

Pump Efficiency, ETA

(m3/hr)

(W)

(W)

%

4.09

64.64

460

14.05

3.41

94.41

430

21.96

2.73

104.60

410

25.51

2.04

99.51

390

25.52

1.36

73.60

350

21.03

0.68

39.06

330

11.84

Pump Efficiency (ETA) vs Vol Flow rate (Q) 30

Pump Efficiancy , ETA

25 y = 0.5786x + 18.605 R² = 0.0163

20 15 10 5 0 0

0.5

1

1.5

2

2.5

3

3.5

4

Volume of Flow rate, Q ( m3/hr)

Figure 5 : Pump Efficiency (ETA) vs Volume of Flow Rate (Q)

4.5

vi) Overall Efficiency (ETAgr) Vs. Volume Flow rate (Q)

Overall Efficiency was obtained by calculate using formula at below :a).

ETAgr 

Po  100% PMi

b).

when Po  64.64W , PMi  530W ETAgr 

ETAgr 

64.64W  100% 530W

Po  100% PMi

ETAgr 

d).

ETAgr 

104.60 W  100% 480W

Po  100% PMi

ETAgr 

Po  100% PMi

when Po  99.51W , PMi  460W ETAgr 

99.51W  100% 460W

ETAgr  21.63%

ETAgr  21.80% e).

94.41W  100% 500W

ETAgr  18.88%

when Po  104.60 W , PMi  480W ETAgr 

Po  100% PMi

when Po  94.41W , PMi  500W

ETAgr  12.20% c).

ETAgr 

f).

ETAgr 

Po  100% PMi

when Po  73.60W , PMi  420W

when Po  39.06W , PMi  400W

73.60W  100% 420W ETAgr  17.52%

ETAgr 

ETAgr 

39.06W  100% 400W

ETAgr  9.77%

Volume Flow rate, Q (m3/hr) 4.09 3.41 2.73 2.04 1.36 0.68

Pump Power Output, Po (W) 64.64 94.41 104.60 99.51 73.60 39.06

Motor Input Power, PMi (W) 530 500 480 460 420 400

Overall Efficiency, ETAgr (%)

12.20 18.88 21.80 21.63 17.52 09.77

Overall Efficiency (ETAgr) vs Vol Flow rate (Q)

Overall Efficiency, ETAgr (%)

25 20

y = 0.6863x + 15.33 R² = 0.0311

15 10 5 0 0

0.5

1

1.5

2

2.5

3

3.5

4

Volume of Flow rate, Q ( m3/hr)

Figure 6 : Overall Efficiency (ETAgr) vs Volume of Flow rate, Q

4.5

Table 3 b: Rotational Speed and Flow rate for P3 Speed (RPM)

Flow rate (%)

Volume flow rate, Q (m3/hr)

1400

29.2

0.497

1300

27.0

0.460

1200

24.9

0.424

1100

22.5

0.383

1000

20.2

0.344

900

17.9

0.305

800

15.6

0.266

700

13.3

0.227

600

11.1

0.189

500

08.8

0.150

400

06.6

0.112

Volume Flow, Q was calculated by using formula :

Q=

When q = 29.2, Q=

29.2

× 28.39 ÷ 103× 60 100

= 0.497 m3/hr When q = 24.9, Q=

24.9 100

× 28.39 ÷ 103× 60

= 0.424 m3/hr

𝑞 100

× 28.39 ÷ 103× 60

When q = 27.0, Q=

27.0 100

× 28.39 ÷ 103× 60

= 0.460 m3/hr When q = 22.5, Q=

22.5 100

× 28.39 ÷ 103× 60

= 0.383 m3/hr

When q = 20.2, Q=

When q = 17.9,

20.2

× 28.39 ÷ 103× 60 100

Q=

= 0.344 m3/hr

× 28.39 ÷ 103× 60

When q = 13.3,

15.6

× 28.39 ÷ 103× 60 100

Q=

13.3 100

× 28.39 ÷ 103× 60

= 0.227 m3/hr

= 0.266 m3/hr When q = 11.1, Q=

100

= 0.305 m3/hr

When q = 15.6, Q=

17.9

When q = 8.8,

11.1

8.8

× 28.39 ÷ 103× 60 100

Q = 100 × 28.39 ÷ 103× 60

= 0.189 m3/hr

= 0.150 m3/hr

When q = 6.6, 6.6

Q = 100 × 28.39 ÷ 103× 60 = 0.112 m3/hr

Figure 2 :Graph of Rotational Speed (N) Vs Volume Flow Rate (Q) for pump 3

Rotational Speed (N) Vs Volume Flow Rate (Q) 1600 y = 2582x + 112.03 R² = 0.9999

Rotational Speed (N)

1400 1200 1000 800 600 400 200 0 0

0.1

0.2

0.3

0.4

Volume Flow Rate (Q), m3/hr

0.5

0.6

Data Collected for Experiment 4 Table 4 b: Pressure, Flow rate, Speed and Power for P3 Pressure

Flow rate

Speed

Power

%

%

RPM

kW

60

29.6

1400

0.56

55

30.0

1407

0.51

50

30.3

1413

0.49

45

30.7

1419

0.47

40

31.0

1426

0.44

35

31.3

1432

0.43

30

31.6

1438

0.40

25

32.0

1440

0.39

20

32.2

1447

0.37

10

33.9

1452

0.36

Motor Power Input,PMi W 560 510 490 470 440 430 400 390 370 360

Volume Flow rate, Q m3/hr 0.50 0.51 0.52 0.52 0.53 0.53 0.54 0.55 0.55 0.57

Pump Total Head,H m 137.43 126.00 114.57 103.15 91.72 80.30 68.87 57.45 46.02 32.75

Pump Power Output,P0 W 170.40 159.35 147.73 133.01 120.54 105.54 92.22 78.35 62.76 32.75

Pump Power Input, Pi W 510 460 440 420 390 380 350 340 320 310

Pump Efficiency (ETA)

Overall Efficiency (ETAgr)

Volumetric Efficiency (ETAV)

33.41 34.64 33.58 31.67 30.91 27.77 26.35 23.04 19.61 10.56

30.43 31.25 30.15 28.30 27.40 24.54 23.06 20.09 16.96 9.10

94.35 95.76 97.22 96.81 98.18 97.77 99.20 100.90 100.41 103.70

i.

Motor Input Power (PMi) vs Output Pressure for P3

PMi was calculated as below : a).

PMi  Power (kW )  1000

b).

when power  0.51

when power  0.56 PMi  0.56kW 

1000W 1kW

PMi  0.51kW 

PMi  Power (kW )  1000

d).

1000W 1kW

PMi  0.47kW 

PMi  Power (kW )  1000

f).

1000W 1kW

PMi  0.43kW 

PMi  Power (kW )  1000

h).

1000W 1kW

PMi  0.39kW 

PMi  Power (kW )  1000 when power  0.37 PMi  0.37kW  PMi  370W

1000W 1kW

PMi  390W

PMi  400W i).

PMi  Power (kW )  1000 when power  0.39

when power  0.40 PMi  0.40kW 

1000W 1kW

PMi  430W

PMi  440W g).

PMi  Power (kW )  1000 when power  0.43

when power  0.44 PMi  0.44kW 

1000W 1kW

PMi  470W

PMi  490W e).

PMi  Power (kW )  1000 when power  0.47

when power  0.49 PMi  0.49kW 

1000W 1kW

PMi  510W

PMi  560W c).

PMi  Power (kW )  1000

1000W 1kW

j).

PMi  Power (kW )  1000 when power  0.36 PMi  0.36kW  PMi  360W

1000W 1kW

Output Pressure % 60 55 50 45 40 35 30 25 20 10

Motor Power Input,PMi W 560 510 490 470 440 430 400 390 370 360

Table 4.1 : Output Pressure (Pr) , Motor Power Input (PMi) for P3

Motor Input Power vs Output Pressure

Motor Input Power, PMi (W)

600 y = 3.9654x + 295.28 R² = 0.9519

500 400 300 200 100 0 0

10

20

30

40

50

60

Output Pressure, Pr (%)

Figure 2 : Motor Input Power vs Output Pressure for P3

70

Volume Flow (Q) vs Output Pressure (Pr) for P3 Volume Flow (Q) was calculated as below : a).

Q

q  28.39  1000  60 100 29.6 Q  28.39  1000  60 100 m3 Q  0.50 hr

b).

Q

c).

Q

q  28.39  1000  60 100 30.3 Q  28.39  1000  60 100 m3 Q  0.52 hr

d).

Q

e).

Q

q  28.39  1000  60 100 31.0 Q  28.39  1000  60 100 m3 Q  0.53 hr

f).

Q

g).

Q

q  28.39  1000  60 100 31.6 Q  28.39  1000  60 100 m3 Q  0.54 hr

h).

Q

i).

Q

q  28.39  1000  60 100 32.2 Q  28.39  1000  60 100 m3 Q  0.55 hr

j).

Q

q  28.39  1000  60 100 30.0 Q  28.39  1000  60 100 m3 Q  0.51 hr

q  28.39  1000  60 100 30.7 Q  28.39  1000  60 100 m3 Q  0.52 hr

q  28.39  1000  60 100 31.3 Q  28.39  1000  60 100 m3 Q  0.53 hr q  28.39  1000  60 100 32.0 Q  28.39  1000  60 100 m3 Q  0.55 hr q  28.39  1000  60 100 33.9 Q  28.39  1000  60 100 m3 Q  0.57 hr

Output Pressure % 60 55 50 45 40 35 30 25 20 10

Volume Flow rate, Q m3/hr 0.50 0.51 0.52 0.52 0.53 0.53 0.54 0.55 0.55 0.57

Table 4.2 : Volume Flow (Q) , Output Pressure (Pr) for P3

Volume Flow vs Output Pressure 0.58

Volume Flow, Q (m3/hr)

0.57 0.56 0.55 0.54 0.53 0.52 0.51 y = -0.0013x + 0.5799 R² = 0.9773

0.5 0.49 0

10

20

30

40

50

Output Pressure, Pr (%)

Figure 3 : Volume Flow vs Output Pressure for P3

60

70

ii.

Pump Power Output (P0) vs Output Pressure (Pr) for P3

Pump Total Head is calculated by using formula as below :H  Z G 2

4  Pr   20  10.2  10  Z G1      oil g  100  

  

where H  Pump Total Head , m Z G 2  Outlet Dis tan ce From Datum (oil)  380mm  0.38m Z G1  Inlet Dis tan ce From Datum (oil)  64mm  0.064m DP  Differential Pr essure,%

 oil  Density oil  910 g  Gravity  9.81

kg m3

m s2

a). Output Pressure, % = 60 when Pr  60

  4  60   20  10.2  10 H  (0.38m  0.064m)     100   910 kg  9.81 m  m3 s2  H  137.43m c). Output Pressure, % = 50 when Pr  50

  4  50   20  10.2  10 H  (0.38m  0.064m)     100   910 kg  9.81 m  m3 s2  H  114.57m e). Output Pressure, % = 40

b). Output Pressure, % = 55 when Pr  55

    4  H  (0.38m  0.064m)   55    20  10.2  10   100   910 kg  9.81 m   m3 s2   H  126.00m

     

d). Output Pressure, % = 45 when Pr  45

    4  H  (0.38m  0.064m)   45    20  10.2  10   100   910 kg  9.81 m   m3 s2   H  103.15m f). Output Pressure, % = 35

     

when Pr  35

when Pr  40   4  40   20  10.2  10 H  (0.38m  0.064m)     100   910 kg  9.81 m  m3 s2  H  91.72m g). Output Pressure, % = 30 when Pr  30

    4  H  (0.38m  0.064m)   35    20  10.2  10   100   910 kg  9.81 m   m3 s2   H  80.30m

     

h). Output Pressure, % = 25 when Pr  25

  4  30   20  10.2  10 H  (0.38m  0.064m)     100   910 kg  9.81 m  m3 s2  H  68.87m i). Output Pressure, % = 20 when Pr  20

    4  H  (0.38m  0.064m)   25    20  10.2  10   100   910 kg  9.81 m   m3 s2   H  57.45m

     

j). Output Pressure, % = 10 when Pr  10

  4 20    20  10.2  10 H  (0.38m  0.064m)      100   910 kg  9.81 m  m3 s2  H  46.02m

    4 10   20  10.2  10  H  (0.38m  0.064m)       100   910 kg  9.81 m   m3 s2   H  23.17m

Pump Power Output was obtained by calculate using formula as below :-

Po   oil gHQ 

1hr 3600s

where Po  Pump Power Output ,W

 oil  Density oil  910

kg m3

m s2 H  Pump Total Head , m g  Gravity  9.81

m3 Q  Volume Flow rate, hr

     

1.

 kg m m 3  1hr  Po   910 3  9.81 2  137.43m  0.50 hr  3600s m s  kg.m 2 N .m Po  170.40 3  170.40  170.40W s s

2.

 kg m m 3  1hr Po   910 3  9.81 2  126.00m  0.51   hr  3600s m s  kg.m 2 N .m Po  159.35 3  159.35  159.35W s s

3.

 kg m m 3  1hr  Po   910 3  9.81 2  114.57m  0.52 hr  3600s m s  kg.m 2 N .m Po  147.73 3  147.73  147.73W s s

4.

 kg m m 3  1hr    Po   910 3  9.81 2  103.15m  0.52 hr m s   3600s kg.m 2 N .m Po  133.01 3  133.01  133.01W s s

5.

 kg m m 3  1hr  Po   910 3  9.81 2  91.72m  0.53 hr  3600s m s  kg.m 2 N .m Po  120.54 3  120.54  120.54W s s

6.

 kg m m 3  1hr    Po   910 3  9.81 2  80.30m  0.53 hr m s   3600s kg.m 2 N .m Po  105.54 3  105.54  105.54W s s

7.

 kg m m 3  1hr  Po   910 3  9.81 2  68.87m  0.54 hr  3600s m s  kg.m 2 N .m Po  92.22 3  92.22  92.22W s s

8.

 kg m m 3  1hr  Po   910 3  9.81 2  57.45m  0.55 hr  3600s m s  kg.m 2 N .m Po  78.35 3  78.35  78.35W s s

9.

 kg m m 3  1hr    Po   910 3  9.81 2  46.02m  0.55 hr m s   3600s kg.m 2 N .m Po  62.76 3  62.76  62.76W s s

10.

 kg m m 3  1hr  Po   910 3  9.81 2  23.7 m  0.57 hr  3600s m s  kg.m 2 N .m Po  32.75 3  32.75  32.75W s s

Pump Power Output Vs Output Pressure 200

Pump Power Output (Po)

180

y = 1.2258x + 58.205 R² = 0.4116

160 140 120 100 80 60 40

20 0 0

10

20

30

40

50

60

70

80

90

100

Output Pressure (Pr)

Figure 3 :Pump Power Output (P0) vs Output Pressure (Pr) for P3

iii.

Pump Power Input (Pi) vs Output Pressure (Pr) for P3

Pi was calculated was below : Pi = PMi - P3min

Pi = PMi - P3min

= PMi - 50W

= PMi - 50W

= 560 - 50

= 510 - 50

= 510 W

= 460 W

Pi = PMi - P3min

Pi = PMi - P3min

= PMi - 50W

= PMi - 50W

= 490 - 50

= 470 - 50

= 440 W

= 420 W

Pi = PMi - P3min

Pi = PMi - P3min

= PMi - 50W

= PMi - 50W

= 440 - 50

= 430 - 50

= 390 W

= 380 W

Pi = PMi - P3min

Pi = PMi - P3min

= PMi - 50W

= PMi - 50W

= 400 - 50

= 390 - 50

= 350 W

= 340 W

Pi = PMi - P3min

Pi = PMi - P3min

= PMi - 50W

= PMi - 50W

= 370 - 50

= 3600 - 50

= 320 W

= 310 W

Pump Power Input Vs Output Pressure 600 y = 3.9654x + 245.28 R² = 0.9519

Pump Power Input (Pi)

500 400 300 200 100

0 0

10

20

30

40

50

60

70

Output Pressure (Pr)

Figure 4 :Pump Power Input (Pi) vs Output Pressure (Pr) for P3

iv.

Pump Efficiency (ETA) vs Output Pressure (Pr) for P3

ETA was calculated was below :

1.

ETA 

Po  100% Pi

when Po  170.40 W , Pi  510W ETA 

70.40W  100% 510W

ETA  33.41%

2.

ETA 

Po  100% Pi

when Po  159.35W , Pi  460W ETA 

159.35W  100% 460W

ETA  34.64%

3.

ETA 

Po  100% Pi

4.

when Po  147.73W , Pi  440W ETA 

5.

ETA 

6.

7.

120.54W  100% 390W

ETA  30.91% P ETA  o  100% Pi

9.

8.

when Po  62.76 W , Pi  320W ETA 

62.76W  100% 320 W

ETA  19.61%

105.54W  100% 380W

ETA  27.77% P ETA  o  100% Pi when Po  78.35W , Pi  340W

92.22W  100% 350W

ETA  26.35% P ETA  o  100% Pi

ETA  31.67% P ETA  o  100% Pi ETA 

when Po  92.22 W , Pi  350W ETA 

133.01W  100% 420W

when Po  105.54 W , Pi  380W

when Po  120.54W , Pi  390W ETA 

Po  100% Pi

when Po  133.01W , Pi  420W

147.73W  100% 440W

ETA  33.58% P ETA  o  100% Pi

ETA 

ETA 

10.

78.35W  100% 340W

ETA  23.04% P ETA  o  100% Pi when Po  32.75W , Pi  310W ETA 

32.75W  100% 310 W

ETA  10.56%

Pump Efficiency Vs Output Pressure 40 y = 0.3612x + 14.55 R² = 0.9092

Pump Efficiency (ETA)

35 30

25 20 15 10 5

0 0

10

20

30

40

50

60

Output Pressure (Pr)

Figure 5 : Pump Efficiency (ETA) vs Output Pressure (Pr) for P3

v.

Overall Efficiency (ETAgr) vs Output Pressure (Pr) for P3

ETAgr was calculated as below :

1.

ETAgr 

Po  100% PMi

ETAgr 

170.40W  100% 560W

ETAgr  30.45%

2.

ETAgr 

Po  100% PMi

ETAgr 

159.35W  100% 510W

ETAgr  31.25%

70

3.

ETAgr 

Po  100% PMi

ETAgr 

147.73W  100% 490W

4.

ETAgr  30.15% 5.

ETAgr 

Po  100% PMi

ETAgr 

120.54W  100% 440W

ETAgr 

Po  100% PMi

6.

8.

92.22 W  100% 400W ETAgr  23.06%

ETAgr 

Po  100% PMi

ETAgr 

62.76 W  100% 370W

ETAgr  16.96%

ETAgr 

133.01W  100% 470W

ETAgr 

Po  100% PMi

ETAgr 

105.54W  100% 430W

ETAgr  24.54%

ETAgr 

9.

Po  100% PMi

ETAgr  28.3%

ETAgr  27.40% 7.

ETAgr 

ETAgr 

Po  100% PMi

ETAgr 

78.35W  100% 390W

ETAgr  20.09% 10.

ETAgr 

Po  100% PMi

ETAgr 

32.75W  100% 360W

ETAgr  9.10%

Overall Efficiency Vs Output Pressure

Overall Efficiency (ETAgr)

35 y = 0.351x + 11.758 R² = 0.93

30

25 20 15 10 5

0 0

10

20

30

40

50

60

70

Output Pressure (Pr)

Figure 6 : Overall Efficiency (ETAgr) vs Output Pressure (Pr) for P3

vi.

Volumetric Efficiency (ETAv) vs Output Pressure (Pr)for P3

Volumetric Efficiency (ETAV) was calculated as below :

1.

ETAv = =

𝑄

2.

𝑥 100

𝑉𝑖 𝑋 𝑥 𝑁 𝑥 60 0.50 6

6.309 𝑥 10ֿ

x100 𝑚3 𝑟𝑒𝑣 𝑥1400 𝑥 60

=

ETAv = =

6.309 𝑥 10ֿ

𝑄

4.

𝑥 100

𝑉𝑖 𝑋 𝑥 𝑁 𝑥60 0.52 6.309 𝑥 10ֿ

x100 𝑚3 𝑟𝑒𝑣 𝑥1413 𝑥 60

=

x100 𝑚3 𝑟𝑒𝑣 𝑥1407 𝑥 60

𝑄

𝑥 100

𝑉𝑖 𝑋 𝑥 𝑁 𝑥60 0.52 6

6.309 𝑥 10ֿ

𝑚3 𝑟𝑒𝑣 𝑥1419 𝑥

x100 60

= 96.81 𝑄

6.

𝑥 100

𝑉𝑖 𝑋𝑥 𝑁 𝑥60 0.53 6

6.309 𝑥 10ֿ

= 98.18

ETAv = =

= 97.22 ETAv =

𝑥 100

𝑉𝑖 𝑋𝑥 𝑁 𝑥60 0.51

= 95.76

6

5.

𝑄

6

= 94.35 3.

ETAv =

𝑚3 𝑟𝑒𝑣 𝑥1426 𝑥 60

3x100

ETAv = =

𝑄

𝑥 100

𝑉𝑖 𝑋 𝑥 𝑁 𝑥60 0.53 6

6.309 𝑥 10ֿ

= 97.77

𝑚3 𝑟𝑒𝑣 𝑥 1432 𝑥

x100 60

7.

𝑄

ETAv = =

8.

𝑥 100

𝑉𝑖 𝑋 𝑥 𝑁 𝑥60 0.54 6

6.309 𝑥 10ֿ

ETAv =

x100 𝑚3 𝑟𝑒𝑣 𝑥1438 𝑥 60

=

6.309 𝑥 10ֿ

=

x100 𝑚3 𝑟𝑒𝑣 𝑥1440 𝑥 60

= 100.90 𝑄

ETAv =

𝑥 100

𝑉𝑖 𝑋 𝑥 𝑁 𝑥60 0.55 6

= 99.20 10.

𝑥 100

𝑉𝑖 𝑋 𝑥 𝑁 𝑥60 0.55 6

6.309 𝑥 10ֿ

𝑚3 𝑟𝑒𝑣 𝑥1447𝑥

ETAv =

x100

=

𝑄

𝑥 100

𝑉𝑖 𝑋 𝑥 𝑁 𝑥60 0.57 6

60

6.309 𝑥 10ֿ

= 100.41

x100 𝑚3 𝑟𝑒𝑣 𝑥1452 𝑥 60

= 103.70

Volumetric Efficiency Vs Output Pressure 102

Volumetric Efficiency (ETAv)

9.

𝑄

101 100 99 98 97 96 y = -0.1486x + 103.79 R² = 0.9229

95 94 0

10

20

30

40

50

60

Output Pressure (Pr)

Figure 7 : Volumetric Efficiency (ETAv) vs Output Pressure (Pr)for P3

70

DISCUSSION The main objective of this experiment is to determine the operating characteristic of different pumps in a contained unit. Besides that, it also helps to understand the types of pumps in principle and design, and the selection of the appropriate pump for a particular application for optimal operation. In experiment 1, the reading that was recorded in the table shows that when the speed is decrease the reading of flowrate also decreases. Then, the graph of Rotational Speed (N) vs. Volume Flow rate (Q) is plotted, a straight line graph is produced. At speed = 2800 rpm, the volume flowrate is 59.3% and when at the lowest speed = 600 rpm, the flowrate is lower where its 12.8 %. Based on the theory, it can be said that when the rotational speed is increased, the volume flow is also increased. The objective is achieved.

Rotatinal Speed, N (RPM)

Rotational Speed (N) vs Volume Flow rate (Q) 3000 y = 667.63x - 0.2371 R² = 0.997

2500 2000 1500 1000 500 0 0

1

2

3

4

5

Volume Flow Rate, Q ( m3/hr)

In experiment 3, the readings for flow rate when there is a decrease in the speed is recorded. The formula of volumetric flow rate,

 q   113.56 x 60      Q =  100  x  1000 

is used to determine the volume flow (Q). From the table, it is known that once the values of speed decreases, the values of flow rate and volume flow rate are also decreasing. A graph of Rotational speed (N) vs. Volume Flow Rate (Q) is plotted and it shows a straight line graph

which means that the speed is directly proportionally to the volume flow rate as said by the theory.

Overall Efficiency, ETAgr (%)

Overall Efficiency (ETAgr) vs Vol Flow rate (Q) 25

y = 0.6863x + 15.33 R² = 0.0311

20 15 10 5 0 0

1

2

3

4

5

Volume of Flow rate, Q ( m3/hr)

Rotational Speed (N)

Rotational Speed (N) Vs Volume Flow Rate (Q) 1500 y = 2582x + 112.03 R² = 0.9999

1000 500 0 0

0.1

0.2

0.3

0.4

Volume Flow Rate (Q),

0.5

0.6

m3/hr

In experiment 2, the readings for flow rate, differential pressure, power and speed are recorded from the speed and output flow rate are maximum. When the output flow rate is decreased, the table shows that the values differential pressure and speed increase when the power is decreased. A range of graph is plotted. The graph for Motor Input Power (PMI) vs. Volume Flow rate (Q)) shows an increasing curve.

Motor Input Power , PMi (W)

Motor Input Power (PMi) vs Vol Flow Rate (Q) 600 500 400

y = 38.089x + 374.16 R² = 0.9898

300 200 100 0 0

1

2

3

4

5

Volume Flow Rate , Q (m3/hr)

The graphs for Pump Power Output (Po) vs. Volume Flow Rate and Pump Power Input (Pi) vs. Volume Flow Rate (Q) also shows increasing curve, which shows a directly proportional graph to volumetric flow rate. The Pump Efficiency (ETA) vs. Volume Flow Rate (Q) and Pump Total Head (H) vs. Volume Flow Rate (Q) graph shows a constant decrease.

Pump Total Head (H) vs Vol Flow Rate (Q) Pump Total Head, H (m)

25 20 15 10 y = -4.5772x + 25.727 R² = 0.9611

5 0 0

1

2

3

Volume Flow Rate, Q

4

(m3/hr)

5

Pump Power Output, P0 (W)

Pump Power Output (P0) vs Vol Flow rate (Q) 120 100

y = 8.1807x + 59.792 R² = 0.1736

80

60 40 20 0 0

1

2

3

4

5

Volume Flow Rate, Q (m3/hr)

Pump Power Input, Pi (W)

Pump Power Input (Pi) vs Vol Flow Rate (Q) 500 450 400 350 300 250 200 150 100 50 0

y = 38.089x + 304.16 R² = 0.9898

0

1

2

3

Volume Flow rate, Q (m3/hr)

4

5

Pump Efficiency (ETA) vs Vol Flow rate (Q) Pump Efficiancy , ETA

30 y = 0.5786x + 18.605 R² = 0.0163

25

20 15 10 5 0 0

1

2

3

4

5

Volume of Flow rate, Q ( m3/hr)

Overall Efficiency, ETAgr (%)

Overall Efficiency (ETAgr) vs Vol Flow rate (Q) 25 y = 0.6863x + 15.33 R² = 0.0311

20 15 10 5 0 0

1

2

3

4

5

Volume of Flow rate, Q ( m3/hr)

The last section of this experiment is experiment 4. In this experiment, the readings for flow rate, differential pressure, power and speed are recorded from the speed and output flow rate are maximum. When the pump head (pressure) is decreased, the table shows that the values of volume flow rate increased and the power is decreased. Pump Efficiency, ETA and Overall Efficiency (ETAgr) decreases when pressure is decreased. Volumetric Efficiency, % ETAVA decreases when pressure is decreased. A range of graph is plotted. The graphs for Motor Input Power (PMi) Vs Output Pressure (Pr) and Pump Power Input (Pi) Vs Output Pressure (Pr) show increasing curves. While, Pump Power Output (Po) Vs Output Pressure (Pr) gives a straight line graph. The Volume Flow (Q) Vs Output Pressure (Pr)decreases, Pump Efficiency (ETA) Vs Output Pressure (Pr)and Overall Efficiency (ETAgr) Vs Output Pressure (Pr) shows an increaese. The graph of

Volumetric Efficiency (ETAv) Vs y≈100.

Output Pressure (Pr) gives a constant straight line graph at

Motor Input Power (PMi)

Motor Input Power Vs Output Pressure 660 y = 1.1167x + 541.94 R² = 0.9902

640 620 600 580 560 540 0

20

40

60

80

100

Output Pressure (Pr)

Volume Flow Rate Vs Output Pressure

Volume Flow Rate (Q)

1.32 1.315 1.31 1.305 1.3

y = -0.0002x + 1.3121 R² = 0.8062

1.295

1.29 0

20

40

60

Output Pressure (Pr)

80

100

Pump Power Output Vs Output Pressure Pump Power Output (Po)

120

y = 1.0973x + 2.6965 R² = 1

100

80 60 40 20 0 0

20

40

60

80

100

Output Pressure (Pr)

Pump Power Input Vs Output Pressure Pump Power Input (Pi)

580 y = 1.1167x + 471.94 R² = 0.9902

560 540 520 500 480 460 0

20

40

60

Output Pressure (Pr)

80

100

Pump Efficiency Vs Output Pressure Pump Efficiency (ETA)

20 y = 0.1864x + 1.3208 R² = 0.9983

15 10 5 0 0

20

40

60

80

100

Output Pressure (Pr)

Overall Efficiency (ETAgr)

Overall Efficiency Vs Output Pressure 20 y = 0.1668x + 1.0831 R² = 0.9987

15 10 5 0 0

20

40

60

80

100

Output Pressure (Pr)

Volumetric Efficiency (ETAv)

Volumetric Efficiency Vs Output Pressure 110.4 110.2 110 109.8 109.6

y = -0.0044x + 109.89 R² = 0.2115

109.4 109.2 0

20

40

60

Output Pressure (Pr)

80

100

Motor Input Power, PMi (W)

Motor Input Power vs Output Pressure 600 y = 3.9654x + 295.28 R² = 0.9519

500 400 300

200 100 0 0

20

40

60

80

Output Pressure, Pr (%)

The characteristic curves for the experiment 2 and 4 were plotted in one graph. For pump 1 : Motor Input

Pump Total

Overall Efficiency, ETAgr

Power, PMi,

Head, H

(W)

(m)

Pump Power Output, Po (W)

4.09

530

5.80

64.64

460

14.05

12.20

3.41

500

10.16

94.41

430

21.96

18.88

2.73

480

14.06

104.60

410

25.51

21.80

2.04

460

17.90

99.51

390

25.52

21.63

1.36

420

19.86

73.60

350

21.03

17.52

0.68

400

21.08

39.06

330

11.84

09.77

Volume Flow, Q (m3/hr)

Pump Power Input, Pi

Pump Efficiency, ETA

(%)

Characteristics VS Volume Flow rate for P1 600

Characteristics

500 400

Pmi H

300

P0 Pi

200

ETA 100

ETAgr

0 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Volume Flow rate, Q

From the graph plotted for pump 1, the pressure head (H) increases when the volume flow rate (Q) increases. In addition, the motor input power (Pmi), pump output (Po), pump input (Pi), pump efficiency (ETA), and overall pump efficiency (ETAgr) decreases as the Q increases.

For pump 3: Motor Power Input,PMi W 560 510 490 470 440 430 400 390 370 360

Volume Flow rate, Q 3 m /hr 0.50 0.51 0.52 0.52 0.53 0.53 0.54 0.55 0.55 0.57

Pump Total Head,H m 137.43 126.00 114.57 103.15 91.72 80.30 68.87 57.45 46.02 32.75

Pump Power Output,P0 W 170.40 159.35 147.73 133.01 120.54 105.54 92.22 78.35 62.76 32.75

Pump Power Input, Pi W 510 460 440 420 390 380 350 340 320 310

Pump Efficiency (ETA)

Overall Efficiency (ETAgr)

Volumetric Efficiency (ETAV)

33.41 34.64 33.58 31.67 30.91 27.77 26.35 23.04 19.61 10.56

30.43 31.25 30.15 28.30 27.40 24.54 23.06 20.09 16.96 9.10

94.35 95.76 97.22 96.81 98.18 97.77 99.20 100.90 100.41 103.70

Characteristics VS Output Pressure for P3 600

Characteristics

500 Pmi

400

Q P0

300

Pi

200

ETA ETAgr

100

ETAv 0 0

10

20

30

40

50

60

70

Output Pressure

From the graph plotted for pump 3, the volume efficiency (ETAv) and volume flowrate (Q) increases when the output pressure (Pr) increases. In addition, the motor input power (Pmi), pump output (Po), pump input (Pi), pump efficiency (ETA), and overall pump efficiency (ETAgr) decreases as the output pressure (Pr) increases.

CONCLUSION AND RECOMMENDATION The main objective of this experiment is to determine the operating characteristic of different pumps in a contained unit. Besides that, it also helps to understand the types of pumps in principle and design, and the selection of the appropriate pump for a particular application for optimal operation. This experiment allows the students to measure the operating characteristic of different pump in a contained unit. The principles of the pump are different from each other. Pump is a device use to move fluid such as liquid, gases by physical or mechanical action. The results show different types of curve and line graphs according to different pumps. The function, principle and design of each pump vary according to its type. Different pumps hold different

operating characteristics. From this experiment, it is proven that centrifugal pump, plunger pump and gear pump has different working principle due to the type of fluid in which the pump is used to move the fluid. The design of three pumps has a big difference as centrifugal pump and plunger pump need two motor to run the pump. While the gear pump only needs a motor. To ensure the experiment successfully, before conducting this experiment, it is necessary to do some check up towards the equipment to avoid any misuse and malfunction. Each valve should be properly open/closed according to the type of pump. Next, the pump should not be operating when there is no liquid in the pipeline to avoid serious damage to the equipment. Besides that, adjust the potentiometer to its minimum setting before switch off the pump. Lastly, make sure that HV2 is not completely closed when P2 is running.

REFERENCES 1) Kirby, B.J. (2010). Micro- and Nanoscale Fluid Mechanics: Transport in Microfluidic Devices.. Cambridge University Press . 2) Emulsions, Foams, and Suspensions: Fundamentals and Applications, Laurier L. Schramm, Publisher: Wiley VCH, 26 July 2005 3) Cameron Tropea, Alexander L. Yarin, John F. Foss, Springer handbook of experimental fluid mechanics Publisher: Springer, 9 October 2007 4) Falkovich, Gregory (2011), Fluid Mechanics (A short course for physicists), Cambridge University Press

5) Batchelor, George K. (1967), An Introduction to Fluid Dynamics, Cambridge University Press 6) White, Frank M. (2003), Fluid Mechanics, McGraw–Hill