Determining Flow Velocity.docx

TABLE OF CONTENTS

TITLE

PAGE NUMBER

Summary

2

Objective of Experiment

2

Theory

3

Equipment

8

Procedure

9

Data and Results

11

Sample Calculation

12

Analysis and Discussion

17

Conclusion

19

Reference

19

1

Summary -

As an overview for this experiment, this summary will encapsulate the major portion of the report and basically give an idea of what the experiment is all about.

-

This experiment is done in two stages. The first stage has an objective of understanding the concept behind fluid flow velocity and learning how to measure it by using a Pitot tube.

-

The second stage is to find and determine the discharge coefficients, CD for an orifice plate and a small nozzle.

1. Objective As previously mentioned, this experiment is divided into two stages with two distinct objectives. -

For the first stage, the objective is to learn the method of measuring air flow velocity using the Pitot tube as well as understanding the working principle of a Pitot tube and the importance of the Bernoulli equation in determining the air flow velocity.

-

For the second stage, the objective is to determine the discharge coefficient, CD when an orifice plate and a small nozzle is placed in the Test tube.

2

2. Theory

Experiment 1 A pitot tube is used to explore the developing boundary layer in the entry length of a pipe which has air drawn through it. With pitot tube, the velocity distribution profiles can be determined at a number of cross-sections at different locations along a pipe. With pitot tube, air flow velocities in the pipe can be obtained by first measuring the pressure difference of the moving air in the pipe at two points, where one of the points is at static velocity. The Bernoulli equation is then applied to calculate the velocity from the pressure difference.

v

2p



or

2 gh'

Δp : The pressure difference between the pitot tube and the wall pressure tapping m manometer). h’ : The pressure difference expressed as a 'head' of the fluid being measured (air) ρ : The air density at the atmospheric pressure and temperature of that day.(kg/m3) g : gravitational acceleration constant (9.81 m/s2)

When fluid flows past a stationary solid wall, the shear stress set up close to this boundary due to the relative motion between the fluid and the wall leads to the development of a flow boundary layer. The boundary layer may be either laminar or turbulent in nature depending on the flow Reynolds number. The growth of this boundary layer can be revealed by studying the velocity profiles at selected cross-sections, the core region still outside the boundary layer showing up as an area of more or less uniform velocity. If velocity profiles for cross-sections different distances from the pipe entrance are compared, the rate of growth of the boundary layer along the pipe length can be determined. Once the boundary layer has grown to the point where it fills the whole pipe cross-section this is termed "fully developed pipe flow".

3

Reynolds Number

The Reynolds number is a measure of the way in which a moving fluid encounters an obstacle. It's proportional to the fluid's density, the size of the obstacle, and the fluid's speed, and inversely proportional to the fluid's viscosity (viscosity is the measure of a fluid's "thickness"--for example, honey has a much larger viscosity than water does).

v

: fluid velocity

d

: obstacle size

A small Reynolds number refers to a flow in which the fluid has a low density so that it responds easily to forces, encounters a small obstacle, moves slowly, or has a large viscosity to keep it organized. In such a situation, the fluid is able to get around the obstacle smoothly in what is known as "laminar flow." You can describe such laminar flow as dominated by the fluid's viscosity--its tendency to move smoothly together as a cohesive material.

A large Reynolds number refers to a flow in which the fluid has a large density so that it doesn't respond easily to forces, encounters a large obstacle, moves rapidly, or has too small a viscosity to keep it organized. In such a situation, the fluid can't get around the obstacle without breaking up into turbulent swirls and eddies. You can describe such turbulent flow as dominated by the fluid's inertia--the tendency of each portion of fluid to follow a path determined by its own momentum.

The transition from laminar to turbulent flow, critcal flow, occurs at a particular range of Reynolds number (usually around 2500). Below this range, the flow is normally laminar; above it, the flow is normally turbulent.

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Calculation of air flow velocity The manometer tube liquid levels must be used to calculate pressure differences, h and pressure heads in all these experiments. Starting with the basic equation of hydrostatics: p = gh

(2)

we can follow this procedure through using the following definitions: Example: Manometer tubes

Liquid surface

1(static

2(stagnation

‘pressure’*)

‘pressure’)

X1

X2

readings (mm) Angle of inclination,  = 0 ‘pressure’ term is used since this reading is in mm of manometer fluid and not the pressure of unit Pa. Therefore the equivalent vertical separation of liquid levels in manometer tubes, h = (x1 - x2)cos

(3)

If k is the density of the kerosene in the manometer, the equivalent pressure difference p is: p = k gh = k g(x1 - x2) cos

(4)

The value for kerosene is k = 787 kg/m3 and g = 9.81 m/s2. If x1 and x2 are read in mm, then: p = 7.72(x1 - x2)cos [N/m2]

(5)

The p obtained is then used in second equation (1) to obtain the velocity. To use the first equation (1), convert this into a 'head' of air, h’. Assuming a value of 1.2 kg/m3 for this gives: h' 

 k ( x1  x 2 ) . . cos  air 1000

[N/m2]

(6) 5

Experiment 2 The orifice plate meter forms a jet, which expands to fill the whole pipe, some diameter distance downstream. The pressure difference between the two sides of the plate is related to the jet velocity, and therefore the discharge, by the energy equation:

where ; Q

=

discharge (volume/time)

Aj

=

jet cross-section area at minimum contraction (vena contracta)

Ao

=

orifice cross-section are

vj

=

jet velocity at minimum contraction (vena contracta)

Cc

=

coefficient of contraction of jet

Cv

=

coefficient of velocity of jet

g

=

gravitational acceleration (9.81 ms -2)

h

=

pressure difference 'head' of air across orifice (refer to equation (6)

of Exp. I)

These two coefficients are normally combined to give a single coefficient of discharge: CD = Cc.Cv Equation (1) now becomes

(2) If Q can be determined independently, then the discharge coefficient can be determined as follows:-

(3)

6

Values of Qi can be determined if the standard nozzle is fitted at the pipe inlet.

(4) If hi = the drop in pressure head across the inlet, the discharge = (ρk/ρair )* (xbefore nozzle –xafter nozzle): in which Ai = standard nozzle cross-section area (= πd2 /4) and C’D assumed to be 0.97. Values of hI are obtained from the manometer tube levels connected to the pipe inlet pressure tapping and open to the atmosphere.

Calculating the CD of orifice plate From equation (4), with the Qi obtained from standard nozzle where CD of standard nozzle is assumed to be 0.97, we can calculate the CD of orifice plate. Assuming that Qi across standard nozzle and Qo across orifice plate is the same, apply equation (3) CD 

Qo Ao 2 gho

……………………………(5) Where ho

=

(k/air)*(x across orifice)

Ao = cross section area of orifice plate hole

7

3. Equipment The equipment used for Experiment 1 and 2:

Figure 1 Experiment 1 apparatus

Figure 2 Experiment 2 Diagram 8

4. Procedure Experiment 1 1) Five mounting positions were provided for the Pitot tube assembly which were: 54 mm, 294 mm, 774 mm, 1574 mm, and 2534 mm from the pipe inlet. 2) The standard inlet nozzle was ensured that is fitted for the experiment and that the orifice plate is removed from the pipe break line. 3) The manometer is set such that the inclined position is at 0o. 4) The Pitot tube is mounted at position 1 (at 54 mm, nearest to the pipe inlet). It is noted that the connecting tube, the pressure tapping at the outer end of the assembly, is connected to a convenient manometer tube. The tip which is a L- shaped metal tube of the Pitot tube is checked to be facing the incoming flow. 5) It is noted also that a pipe wall static pressure tapping near to the position where the Pitot tube assembly is placed. The static pressure tapping is connected to a manometer tube. 6) The Pitot tube is positioned with the traverse position at 0 mm, where the tip is touching the bottom of the pipe. Both manometer tube levels at the wall static and the pitot tube is measured and recorded until the traverse position touching the top of the pipe. 7) The velocity traverse for the same air flow value is repeated at the next position with the Pitot tube assembly. The blanking plugs is checked to be placed at the holes that are not in use.

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Experiment 2 1) The orifice plate is inserted into position (taking care to observe the instruction as to ) in which the surface should face the approaching airflow. 2) All the static pressure tapping points are connected to manometer tubes ensuring that one manometer tube remains unconnected to record the air pressure and that one is attached to the first tapping point adjacent to the standard inlet nozzle which should be fitted. 3) The fan is turned on with low airflow (damper plate closed) and all the manometer tubes are measured, including any opening to the air (reading is taken after the fan is turned on). 4) The air flow is gradually increased by increasing the damper opening to 100 percent. At all openings, the reading is taken.

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5. Data and Results Experiment 1 Data Sheet for Velocity Measurement Using Pitot Tube

Traverse Position (mm)

0 10 20 30 40 50 60 70 80

Traverse Position (mm)

0 10 20 30 40 50 60 185 80

Pitot Tube at 54 mm Static 'Pressure' Reading: 116(mm) Stagnation 'Pressure' Reading (mm) 100 98 98 98 96 96 96 96 98

x (mm)

p 2 (N/m )

velocity (m/s)

20 22 22 22 22 22 22 22 22

154.40 167.26 159.60 147.09 130.10 109.17 84.92 58.09 29.49

16.04 16.70 16.31 15.66 14.73 13.49 11.90 9.84 7.01

Pitot Tube at 774 mm Static 'Pressure' Reading: 120(mm)

Pitot Tube at 294 mm Static 'Pressure' Reading: 118(mm) Stagnation 'Pressure' Reading (mm) 110 100 100 100 100 100 100 100 100

x (mm)

p (N/m )

Velocity (m/s)

14 24 24 24 24 24 24 24 24

108.08 182.47 174.11 160.46 141.93 119.10 92.64 63.37 32.17

13.42 17.44 17.03 16.35 15.38 14.09 12.43 10.28 7.32

2

Pitot Tube at 1574 mm Static 'Pressure' Reading: 124(mm)

Stagnation 'Pressure' Reading (mm)

x (mm)

p 2 (N/m )

velocity (m/s)

Stagnation 'Pressure' Reading (mm)

x (mm)

p (N/m )

Velocity (m/s)

112 112 106 106 104 103 102 102 102

17 22 28 28 30 31 32 32 32

131.24 167.26 203.12 187.20 177.42 153.83 123.52 84.49 42.90

14.79 16.70 18.40 17.66 17.20 16.01 14.35 11.87 8.46

120 110 108 106 105 104 104 104 104

13 23 25 27 28 29 29 29 29

100.36 174.86 181.36 180.51 165.59 143.91 111.94 76.57 38.88

12.93 17.07 17.39 17.35 16.61 15.49 13.66 11.30 8.05

2

11

Pitot Tube at 2534 mm Static 'Pressure' Reading: 126(mm)

Traverse Position (mm)

0 10 20 30 40 50 60 70 80

Stagnation 'Pressure' Reading (mm) 124 118 114 112 110 109 109 109 109

x

(mm)

14 20 24 26 28 29 29 29 29

p (N/m )

velocity (m/s)

108.08 152.05 174.11 173.83 165.59 143.91 111.94 76.57 38.88

13.42 15.92 17.03 17.02 16.61 15.49 13.66 11.30 8.05

2

Sample Calculation

1) Pitot tube at 2534 mm Traverse Position

= 70 mm

Static Pressure

= 126 mm

Δx

= (Static Pressure) – (Stagnation Pressure) = 126 – 110 = 16 mm

2) Θ = 0o Δp

= ρk g (Δx) cos Θ = (7.72kg

) (16

m) cos (0)

= 124 N/m2

3) v

=√ =√ = 14.35 m/s

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Graph 1

Velocity vs Traverse Position 25

Velocity(m/s)

20

15

Pitot tube at 54 mm Pitot tube at 294 mm Pitot tube at 774 mm

10

Pitot tube at 1574 mm Pitot tube at 2534 mm

5

0 0

10

20

30

40

50

60

70

80

Traverse Positions (mm)

13

Experiment 2 Table 5.1 Static ‘Pressure’ Readings when using Standard Nozzle (80 mm) Damper Openings (% Openings) 0%

25%

Points

50%

75%

100%

mm of kerosene

Room “pressure” After nozzle

96

96

94

94

94

100

104

104

104

104

54mm

100

104

104

104

104

294mm

100

104

104

104

104

774mm

100

104

104

104

104

Before Orifice

100

104

104

104

104

After Orifice

120

186

200

204

208

1574mm

126

170

178

184

186

2534mm

110

154

162

166

168

Damper Openings (% Openings) hi (m) ho (m) 3 Qi (m /s) CD Re

0% 0 8.95772358 0 0 0

25% 3.839024 49.90732 0.04232 0.26903 46277.91

50% 5.118699 60.14472 0.048866 0.282978 46277.13

75% 0 74.22114 0 0 0

100% 0 76.78049 0 0 0

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Table 5.2 Static ‘Pressure’ Readings when using Small Nozzle (50 mm) Damper Openings (% Openings) 0%

25%

50%

Points

75%

100%

mm of kerosene

Room “pressure” After nozzle

98

94

64

94

94

110

130

134

136

134

54mm

106

118

120

120

120

294mm

106

118

120

120

120

774mm

106

118

120

120

120

Before Orifice

104

114

114

116

116

After Orifice

124

186

196

200

202

1574mm

120

170

180

184

184

2534mm

116

158

166

168

170

Damper Openings (% Openings) hi (m) ho (m) 3 Qi (m /s) CD Re

0% 3.83902439 7.67804878 0.01650014 0.65853598 28923.6939

25% 23.03415 37.11057 0.040417 0.764203 70848.29

50% 28.15285 52.46667 0.044683 0.710544 78325.73

75% 29.43252 49.90732 0.045687 0.744909 80086.73

100% 30.7122 49.90723 0.046669 0.760931 81808.56

15

Graph 2- Standard Nozzle

mm Kerosene vd Tapping position

Kerosene Height (mm)

250 200 0

150

25 100

50 75

50

100

0 1

2

3

4

5

6

7

8

9

Tapping Position

Graph 3- Small Nozzle

mm Kerosene vd Tapping position

Kerosene Height (mm)

250 200 0

150

25 100

50 75

50

100

0 1

2

3

4

5

6

7

8

9

Tapping Position

16

6. Analysis and Discussion

Discussion – Experiment 1 -

From this experiment, we study how the Pitot tube is used to measure two different types of pressure, which is stagnation pressure and static pressure. These two vales used as the variables to calculate and determine the fluid flow velocity.

-

The results show that, as the traverse position increases in millimeter, the maximum value of velocity in the middle of the pipe decreases to a minimum value at the traverse position of 70 mm due to the fact that friction is present between molecules of the pipe wall and molecules of the fluid and causes an attraction.

-

The decrease in velocity of the fluid molecules causes a momentum transfer between these molecules and causes much slower fluid layer causes the faster fluid layer to be impeded instantaneously. So that we can conclude that the velocity of the fluid gains back stream velocity at the center of the pipe.

-

A chain reaction occurs forming a boundary layer. This shows that the incident occurring at different flow velocities along the traverse distance of the pipe. The velocity of the fluid is highest at the center of pipe. The fluid velocity will also increase if there is less restrictions to the flow. The maximum value of fluid velocity can be seen clearly in Graph 1, which shows the highest value of velocity in the middle region.

-

Error analysis of this experiment while holding the pitot tube it may fluctuate or move to opposite directing may cause error to get the reading. The may be leak or loosing causing the air flow out some where else which we cannot detect it due to disturbance sound of motor applied. This also may cause the error of reading

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Discussion – Experiment 2 -

From the data the value of discharge coefficient for both standard nozzle and small nozzle differs from each other. This fact that Reynold’s number is proportional to the velocity of fluid. Hence, the two nozzles have different values of velocity and flow rate of the fluid.

-

During the air flow passing through the orifice plate, the orifice area is smaller than the pipe and pressure will drop. This because pressure is proportional to the contact area. Thus, the air flow velocity will increase when it passes through the orifice plate due to velocity being proportional to area. This will produce a large amount of pressure drop.

-

The discharge coefficient, CD of the orifice will increase as the damper openings is increase. This due to the flow is approaching an ideal because the flow velocity is also increasing. The higher the rate of air flow, the less the effect due to obstructions to the flow.

-

In experiment 2, when different damper openings are used, the discharge coefficient, CD obtained from the orifice and small nozzle will be different. Pressure will drop when there are obstructions occurring along the pipe as in expectation. Then, the kerosene reading increases

-

From analysis of the experiment, the airflow is near to the orifice plate then the manometer level will increase. From the graph of discharge coefficient versus Reynold’s number (Graph 4 and 5), we can see that the increase in Reynold’s number will increase in discharge coefficient.

18

7. Conclusion Experiment 1

As a conclusion, we can state that the Bernoulli equation can be used to derive the velocity profile of the flowing fluid and to determine the forming of boundary layers of the flow of the fluid. Even with some certain errors in the reading, the graph of velocity profile against the transverses positions was made successfully. This shows that the boundary layer is in existence as there was an increase and decrease of the velocity profile of the air flow.

Experiment 2 From the results obtained, the coefficient of discharge along with the Reynold’s number were determined successfully. The Reynold’s number obtained was rather high. Hence, we concluded that the flow was laminar even when the damper was in 0 percent opening. Theoretically, the pressure when the fluid passes through the orifice opening would be much lower, in our experiment we obtained a much higher pressure reading due to the effect of boundary layers impeding the velocity profile of the fluid flow.

8. References 8.1. Measurements Laboratory Lab Manual 8.2. Experimental Method for Engineers, 7th Edition, JP.Holman

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