POLITEKNIK PORT DICKSON

JABATAN KEJUTERAAN MEKANIKAL

J3008 – MEKANIK BENDALIR

Laporan Amali( Venturi Meter )

DTP 3 Seksyen 1

Nama Pensyarah : En Azmi Bin Hj Udi

Nama Pelajar :

Wan Mohd Asraf Bin Wan Md Ariffin 06DTP09F1019

Mohd Syuawari Nadhir Bin Khalid 06DTP09F1033

Mohd Razif Bin Mohd Akbar 06DTP09F1034

Mohd Nasier Bin Omar 06DTP09F1035

Muhammad Muzammil Bin Nordin 06DTP09F1036

Muhammad Hafidzi Bin Japri 06DTP09F1037

Mohamad Shaiful Bin Mohd Razmi 06DTP09F1039

Tarikh Ujikaji : 11 Ogos 2010

Tarikh Hantar : 29 September 2010

TITLE : Venturi Meter

OBJECTIVE : To find the coefficient of discharge Cd for the Venturi Meter.

APPARATUS : Venturi Meter Apparatus

THEORY :

A venturi meter is a tube with a constricted throat that increases velocity and decreases

pressure. They are used for measuring the flowrate of compressible and incompressible

fluids in pipeline.

The venturi is manufactured from transparent acrylic materials and follows the classic 21º-10º

convergent-divergent design which forms the basis of most engineering standards for venturi

flow meters The design of the plate conforming with the British Standard for flow measurement

BS1042 for venturies in all respects other than that of minimum size.

Consider the flow of an incompressible fluid through the convergent – divergent pipe shown in

the figure below. Assuming that there is not loss of energy along the pipe and that the velocity

and piezometric heads are constant across each of the sections considered, the Bernoulli’s

Theorem states that.

For this apparatus:

Z₁ = Z₂ and P = ρgh

Hence if Bernoulli`s theorem is obeyed:

V 12

2 g+h

1=¿V 22

2 g+h2¿

In which v1 and v2 are the velocity of flow through section (1) and (2).

The equation of continuity is,

Q=a1V 1=a2 V 2

In which Q denotes the volume flow or dischange rate. Substituting in equation (2) and the solving this equation for v2 leads to,

So that the discharge rate, from the equation (2) becomes:

Where ,

Q1 = practical flow rate

Q2 = theoretical flow rate

Cd = Q1/Q2

EXPERIMENTAL PROCEDURE:

1) Switch on the pump.

2) While this is in process, value of h1 and h2 are read from scale.Similar redaing may be

taken at series of reducing value of (h1-h2).

3) Use time watch to obtain the flow rate. The corresponding flow rate should be measured

for each value of (h1-h2).

4) Repeat step (2) and (3) for the other reading.

RESULT AND CALCULATION:

Time Vol. Flow rate Venturi

t m3 q1 LevelQ2

sec m3/s h1 h2 (h1-h2) Cd

54 10 ×10ˉ⁴ 1.85x10-4 0.061 0.012 0.049 2.13 ×10ˉ⁴ 0.87

50 10 ×10ˉ⁴ 2.00x10-4 0.081 0.022 0.059 2.33 ×10ˉ⁴ 0.86

47 10 ×10ˉ⁴ 2.13x10-4 0.102 0.034 0.068 2.51 ×10ˉ⁴ 0.85

45 10 ×10ˉ⁴ 2.22x10-4 0.121 0.045 0.075 2.75 ×10ˉ⁴ 0.84

40 10 ×10ˉ⁴ 2.50x10-4 0.140 0.052 0.088 2.86 ×10ˉ⁴ 0.87

34 10 ×10ˉ⁴ 2.94x10-4 0.160 0.062 0.098 3.01 ×10ˉ⁴ 0.94

A1 = πd2 / 4

= π (26)2 /4

= 530.9 mm2 m2 = 530.9 / 1000 x 1000

= 0.53 x 10-3

A2 = πd2 / 4

= π (16)2 /4

= 201.1 mm2 m2 = 201.1 / 1000 x 1000

= 0.2011 x 10-3

Mohd Razif Bin Mohd Akbar 06DTP09F1034

Q2 = a2 √ 2 g(h1−h2)1− (a2/a 1 )2

= 0.2011 x 10-3 2 ( 9.81 )(0.061−0.012)

1−¿¿

= 0.2011 x 10 -3 √ 0.9610.856

= 2.13 x 10 -4

Cd = Q1Q2

= 1.852 x 10 -4 / 2.13 x 10 -4

= 0.87

Conclusion :

i. Venturi meter readings at 0.05m tall and it is the Q₂ 2.13×10-⁴

ii. And the reading of the Cd 0.87

Mohd Syuawari Nadhir Bin Khalid 06DTP09F1033

Q2 = a2 √ 2 g(h1−h2)1− (a2/a 1 )2

= 0.2011 x 10-3 2 ( 9.81 )(0.059)

1−¿¿

= 0.2011 x 10 -3 √ 1.1580.856

= 2.33 x 10 -4

Cd = Q1Q2

= 2.00 x 10 -4 / 2.33 x 10 -4

= 0.86

Conclusion :

i. Venturi meter readings at 0.06m tall and it is the Q₂ 2.36×10-⁴

ii. And the reading of the Cd 0.86

Wan Mohd Asraf Bin Wan Md Ariffin 06DTP09F1019

Q2 = a2 √ 2 g(h1−h2)1− (a2/a 1 )2

= 0.2011 x 10-3 2 ( 9.81 )(0.068)

1−¿¿

= 0.2011 x 10 -3 √ 1.330.856

= 2.51 x 10 -4

Cd = Q1Q2

= 2.128 x 10 -4 / 2.51 x 10 -4

= 0.85

Conclusion :

i. Venturi meter readings at 0.066m tall and it is the Q₂ 2.36×10-⁴

ii. And the reading of the Cd 0.85

Mohd Nasier Bin Omar 06DTP09F1035

Q2 = a2 √ 2 g(h1−h2)1− (a2/a 1 )2

= 0.2011 x 10-3 2 ( 9.81 )(0.075)

1−¿¿

= 0.2011 x 10 -3 √ 1.470.856

= 2.64 x 10 -4

Cd = Q1Q2

= 2.22 x 10 -4 / 2.64 x 10 -4

= 0.84

Conclusion :

i. Venturi meter readings at 0.076m tall and it is the Q₂ 2.65×10-⁴

ii. And the reading of the Cd 0.84

Muhammad Muzammil Bin Nordin 06DTP09F1036

Q2 = a2 √ 2 g(h1−h2)1− (a2/a 1 )2

= 0.2011 x 10-3 2 ( 9.81 )(0.088)

1−¿¿

= 0.2011 x 10 -3 √ 1.7270.856

= 2.86 x 10 -4

Cd = Q1Q2

= 2.50 x 10 -4 / 2.86 x 10 -4

= 0.87

Conclusion :

i. Venturi meter readings at 0.088m tall and it is the Q₂ 2.86×10-⁴

ii. And the reading of the Cd 0.87

Muhammad Hafidzi Bin Japri 06DTP09F1037

Q2 = a2 √ 2 g(h1−h2)1− (a2/a 1 )2

= 0.2011 x 10-3 2 ( 9.81 )(0.098)

1−¿¿

= 0.2011 x 10 -3 √ 1.9230.856

= 3.01 x 10 -4

Cd = Q1Q2

= 2.94 x 10 -4 / 3.01 x 10 -4

= 0.94

Conclusion :

i. Venturi meter readings at 0.078m tall and it is the Q₂ 3.01×10-⁴

ii. And the reading of the Cd 0.94

Mohamad Shaiful Bin Mohd Razmi 06DTP09F1039

3) Discuss about the two graph obtained from the experiment.

i) (h1-h2) against Qp

m= y 2− y 1x 2−x1

¿ 0. 098−0.0 68

2.941× 10−4−2.128 ×10−4

¿ 0. 03

8.13 ×10−6

= 369.00

y = mx + c

0.098 = (369.00)(2.941 x10-4) + C

0.098 – 0.109 = C

-0.011 = C

Conclusion

- The position of a balanced top and bottom and the graph is a straight line graph

ii) Cd against Q1

m= y 2− y 1x 2−x1

¿ 0.94−0.84

2.941× 10−4−2. 22×10−4

¿ 0. 10

7.21×10−5

= 1386.963

y = mx + c

0.94 = (1386.963)(2.941 x10-4) + C

0.94 – 0.408 = C

0.532= C

Conclusion

The position of a balanced top and bottom and the graph is a straight line graph

CALCULATIONS

4) Using the value of Cd which you have obtained by experiment, determine the throat

diameter of the venture meter which would measure a flow of 0.4 m³ /s in a pipe of 0.6

diameter with a differential head of 0.8 m.

Qs = cd A1 √ 2 g(h)m2

0.4 = ( 0.94 ) (0.3) √ 2(9.81)(0.8)m2−1

0.4 = 0.282 √ 1.57m2−1

1.418 = √ 1.57m2−1

2.012 = 1.57

m2−1

m2 – 1 = 15.72.012

= 7.803

m2 = 8.803

m = 2.97

m = A 1A 2

2.97 = 0.3

2.97

= 0.1

A2 = πd22

4

0.1 = πd22

4

0.4 = πd22

d22 = 0.129

d2 = 0.4