Experiment 3 Bernoulli's Theorem

UNIVERSITI TEKNOLOGI MARAFAKULTI KEJURUTERAAN KIMIA

PROCESS ENGINEERING LABORATORY(CPE453)

No. Title Allocated Marks (%) Marks

1 Abstract/Summary 5 2 Introduction 10 3 Aims 5 4 Theory 10 5 Apparatus 5 6 Methodology/Procedure 10 7 Results 10 8 Calculations 10 9 Discussion 20 10 Conclusion 5 11 Recommendations 5 12 Reference / Appendix 5

TOTAL MARKS 100

Remarks:

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---------------------------Date: ABSTRACT

NAME : MOHAMAD AZRUL SOFI BIN MOHD TAHARSTUDENT ID : 2013982991GROUP : 3EXPERIMENT : BERNOULLI’S THEOREM DEMONSTRATIONDATE PERFORMED : 14TH APRIL 2015SEMESTER : 3PROGRAMME/ CODE : EH2413A / CPE453SUBMIT TO : MISS NURUL LIYANA BINTI ROSLI

The purpose of this experiment is to investigate the validity of Bernoulli’s equation when

applied to a steady flow of water in tapered duct and the flow rate of steady flow rates are

measured too based on the Bernoulli’s theorem by relating the pressure, velocity and elevation in

a moving fluid. Meanwhile, the compressibility and viscosity (internal friction) of which are

negligible and the flow where the flow is in steady or laminar. FM 24 Bernoulli’s Apparatus is

used for this experiment. The experiment was conducted in order to find out the time taken to

collect 3L of water, the volumetric flow rates of the water, the pressure difference at all

manometer tube at different cross section. From the 0.003 m3 of water, the time to collect them is

recorded based on the different flow rate for each experiment.

The combination of venture meter with the manometer tube and hydraulic bench were

used in this experiment. The water is fed up during the experiment through a hose connector and

the flow rate can be adjusted at the flow regulator valve at the outlet of the test section. The flow

measurement can be showed by the venture and the discharge coefficient can be determined by

the results which the reading of each manometer tubes increase when the pressure difference

increase. From the reading of height can be calculated from the data by using the Bernoulli’s

equation to find the velocity of the moving fluid.

The reading of the pressure level and velocity of the tube A to F is recorded. From the

Bernoulli’s theorem, the relationship between the pressure and the velocity is where the pressure

is inversely proportional to velocity. Bernoulli’s principle tells us that the fluid flows more

quickly through the narrow area, the pressure decreases rather than increases. Thus, it proves the

validity of Bernoulli’s theorem.

INTRODUCTION

Figure 1: Apparatus used to investigate the validity of Bernoulli's equation

Figure above shows the apparatus used in this experiment when applied to the steady flow of

water in a tapered duct. The apparatus consists of a clear acrylic duct of varying circular cross

section known as Venturi. The duct has a series of wall tapping that allows measurement of the

static pressure distribution along the converging duct, while the total head tube is provided to

traverse along the centre line of the test section. These tapping are connected to a manometer

bank incorporating a manifold with air bleed valve. Pressurisation of the manometer is facilitated

by a hand pump.

The unit is connected to the hydraulic bench using flexible hoses. The hoses and the connections

are equipped with rapid action couplings. The flexible hose attached to the outlet pipe which

should be directed to the volumetric measuring tank on the hydraulics bench. A flow control

valve is incorporated downstream of the test section. Flow rate and pressure in the apparatus may

be varied independently by adjustment of the flow control valve and the bench supply control

valve. The unit consists of the followings:

1) Venturi

The venture meter is made of transparent acrylic with the following specifications:

Throat diameter : 16 mm

Upstream diameter : 26 mm

Designed flow rate : 20 L/mins

2) Manometer

There are eight manometer tubes: each length 320 mm for static pressure and total head

measuring along the venture meter.

The manometer tubes are connected to an air bleed screw for air release as well as tubes

pressurization.

3) Baseboard

The baseboard is epoxy coated and designed with four height adjustable stands to level

the venture meter.

4) Discharge valve

One discharge valve is installed at the venture discharge section for flow rate control.

5) Connections

Hose connections are installed at both inlet and outlet.

6) Hydraulic Bench

Sump tank : 120 L

Volumetric tank : 100 L

Centrifugal pump : 0.37 kW, 50 L/mins

AIMS

For this experiment, one of the aims is to investigate the validity of Bernoulli’s equation

when applied to the steady flow of the water in a tapered duct. Then it is also to measure the flow

rates and both static and total pressure heads in a rigid convergent/divergent tube of known

geometry for a range of steady flow rates.

THEORY

Figure 2: Pipe of varying cross section

(Giles R.V 1994) Bernoulli’s law indicates that, if an inviscid fluid is flowing along a pipe of

varying cross section, then the pressure is lower at constrictions where the velocity is higher, and

higher where the pipe opens out and the fluid stagnates. The well-known Bernoulli’s equation is

derived under the following assumptions:

1) The fluid is incompressible (density constant)

2) The flow is steady; ∂∂ t

=0

3) The flow is frictionless

4) Flow along a streamline

Then, it is expressed with the following equation:

Pρg

+ v2

2 g+z=h¿=constant

Where (in SI units);

P = fluid static pressure at the cross section in N/m2

ρ = density of the flowing fluid in kg/m3

g = acceleration due to gravity in m/s2 (its value is 9.81 m/s2)

v = mean velocity of fluid flow at the cross section in m/s

z = elevation head of the center of the cross section with respect to a datum

h = total head (stagnation) in m

The terms on the left-hand side of the above equation represent the pressure head (h), velocity

head (hv) and elevation head (z) respectively. The sum of these terms is known as the total head

(h*). According to the Bernoulli’s Theorem of fluid flow through a pipe, the total head (h*) at

any cross section is constant. In a real flow results will deviate from the theoretical ones.

In our experimental setup, the centreline of all the cross sections we are considering lie on the

same horizontal plane (which we may choose as the datum, z = 0 and thus, all the ‘z’ values are

zeros so that the above equation reduces to :

Pρg

+ v2

2 g=h¿=constant

This represents the total head at a cross section.

For the experiments, the pressure head is denoted as hi and the total head as h*I, where I

represents the cross sections at different tapping points.

Other form of Bernoulli’s equation

If the tube is horizontal the difference in height can be disregarded z1 = z2. Hence,

Pρg

+ v2

2 g= P

ρg+ v2

2 g

Total pressure head

Water at constant head from a tank is allowed to flow through a horizontal pipe line of varying

cross section. The pressure heads h1, h2 and h* are measured from a probe with an end hole facing

into the flow such that it brings the flow to rest locally at the probe end. Thus,

h¿=h+ v2

gwhere ;h1

¿=h2¿

Volume flow rate

Q̇=A5[ 2 g (h1−h5 )

1−( A5

A1)

2 ]where ; A=πd1

2

4

Velocity measurement

The velocity of the flow is measured by measuring the volume of the flow, V, over a time period,

t. Thus gives the rate of volume flow (m3/s), which in turn gives the velocity of the flow through

a defined area.

v= Q̇A

Continuity equation

For an incompressible fluid, conservation of mass requires that volume is also conserved.

A1 V 1=A2 V 2

APPARATUS

1) Manometer Tubes

2) Test Section

3) Water Inlet

4) Unions

5) Air Bleed Screw

6) Discharge Valve

7) Gland Nut

8) Hypodermic Probe

9) Adjustable Feet

PROCEDURE

Part A: General set up

1) The Bernoulli’s equation apparatus is set up on the hydraulic bench so that the base is

horizontal.

2) The test-section is ensured to have the 14-tappered section converging in the direction of

the flow.

3) The rig outflow tube is positioned above the volumetric tank.

4) The rig inlet is connected to the bench flow supply, where the bench valve and apparatus

flow control valve are closed and pump is started.

5) Gradually, the bench valve is opened to fill the test rig with water. Air bubble is

removed from the tube to allow an accurate reading.

6) In order to bleed air pressure tapping point and manometers both the bench valve and rig

flow control valve are closed. The air bleed screw is opened and the cap from the water

adjacent air valve is removed.

7) A length of small-bore tubing from the air valve is connected to the volumetric tank.

8) The bench valve is opened and allowed to flow through the manometer to purge all air

from them.

9) Then, the air bleed screw is tightened where the bench valve and test rig flow control

valve are partly opened.

10) Next, the air bleed is opened slightly to allow air to enter the top of the manometers. The

screw is re-tightened when the manometer reach a convenient height.

Part B: Bernoulli’s Theorem Demonstration

1) General Start-up procedure in part A is performed.

2) All the manometer tubes are properly connected to the corresponding pressure taps and

free from air bubble.

3) The discharge valves are adjusted to a high measurable flow rate.

4) Water flow rates are measured by using volumetric method after the level are stabilizes.

5) Slide the hypodermic tube (total head measuring) connected to manometer #H gently, so

that its end reached the cross section of venturi tube at #A and the reading are taken.

6) The step at 5 are repeated for other sections (#B,#C,#D,#E and #F)

7) Steps 3 to 6 are repeated with three other decreasing flow rates by regulating the venturi

discharge valve.

8) The velocity, ViB is calculated using the Bernoulli’s equation where;

V iB=√2 g ( hB−hi )

9) The velocity, ViC is calculated using the continuity equation where; V iC=Qav

Ai

10) The differences between two velocities are calculated.

RESULTS

Volume (L) 3 L

Average Time (s) 26.35 s

Cross

Section

Using Bernoulli’s EquationUsing Continuity

EquationDifference

h* =

hH

(mm)

hi

(mm)

V iB=√2 g (h¿−hi )(m/s)

Ai =πD i2

4

(m2)

V iC=Qav

Ai

(m/s)

ViB - ViC

(m/s)

A 166 161 0.3132 5.3093 ×10−4 0.21453 0.09867

B 166 159 0.3706 3.6644 × 10−4 0.31083 0.05977

C 165 145 0.6264 2.0106 ×10−4 0.56650 0.05990

D 164 153 0.4646 3.1416 ×10−4 0.36255 0.10205

E 163 155 0.3706 3.8013 ×10−4 0.29963 0.07097

F 162 157 0.3132 5.3093 ×10−4 0.21453 0.09867

Table 1

Volume (L) 3 L


Cross Using Bernoulli’s Equation Using Continuity Difference

Section

equation

h* = hH

(mm)

hi

(mm)

V iB=√2 g (h¿−hi )(m/s)

Ai =πD i2

4

(m2)

V iC=Qav

Ai

(m/s)

VIb – VIc

(m/s)

A 178 166 0.4852 5.3093 ×10−40.38536 0.09984

B 180 159 0.6419 3.6644 × 10−40.55835 0.08355

C 177 105 1.1885 2.0106 ×10−41.01766 0.17089

D 174 138 0.8404 3.1416 ×10−40.65126 0.18914

E 173 145 0.7412 3.8013 ×10−40.53824 0.20296

F 172 154 0.5943 5.3093 ×10−40.38536 0.20894

Table 2

Volume (L) 3 L


Cross

Section

Using Bernoulli’s EquationUsing Continuity

equation

Differenc

e

h* = hH

(mm)

hi

(mm)

V iB=√2 g (h¿−hi )(m/s)

Ai =πD i2

4

(m2)

V iC=Qav

Ai

(m/s)

ViB - ViC

(m/s)

A 194 220 0.7142 5.309×10−4 0.657 0.0572

B 179 223 0.9291 3.664 × 10−4 0.952 - 0.0229

C 56 220 1.7938 2.011×10−4 1.734 0.0598

D 133 214 1.2606 3.142 ×10−4 1.110 0.1506

E 152 212 1.0850 3.801 ×10−4 0.917 0.1680

F 171 210 0.8747 5.310 ×10−4 0.657 0.2177

Table 3

CALCULATIONS

Area, A = π D2

4

Area A = π(26 X 10−3 )2

4

= 5.310 ×10−4 m2

Area B = π(21.6 X 10−3 )2

4

= 3.664 × 10−4 m2

Area C = π(16 X 10−3 )2

4

= 2.011×10−4 m2

Area D = π(20 X 10−3 )2

4

= 3.142 ×10−4m2

Area E = π(22 X 10−3 )2

4

= 3.801 ×10−4m2

Area F = π(26 X 10−3 )2

4

= 5.310 ×10−4m2

Calculation: Result 1

V iB=√ 2 g (h¿−hi )1000

A: V iB=√ 2 X 9.81 (166−161 )1000

= 0.3132 m/s

V iC=Qav

Ai

V iC=1.139 ×10−4

5.3093 ×10−4

= 0.21453 m/s

Difference = ViB - ViC

= 0.3132 – 0.21453

= 0.09867 m/s

B: V iB=√ 2 X 9.81 (166−159 )1000

= 0.3706 m/s

V iC=1.139 ×10−4

3.6644 ×10−4

= 0.31083 m/s


= 0.3706 – 0.31083

= 0.05977 m/s

C: V iB=√ 2 X 9.81 (165−145 )1000

= 0.6264 m/s

V iC=1.139 ×10−4

2.0106 ×10−4

= 0.5665 m/s


= 0.6264 – 0.5665

= 0.0599 m/s

D: V iB=√ 2 X 9.81 (164−153 )1000

= 0.4646 m/s

V iC=1.139 ×10−4

3.1416 ×10−4

= 0.36255 m/s


= 0.4646 – 0.36255

= 0.10205 m/s

E: V iB=√ 2 X 9.81 (163−156 )1000

= 0.3706 m/s

V iC=1.139 ×10−4

3.8013 ×10−4

= 0.29963 m/s


= 0.3706 – 0.29963

= 0.07097 m/s

F: V iB=√ 2 X 9.81 (162−157 )1000

= 0.3132 m/s

V iC=1.139 ×10−4

5.3093 ×10−4

= 0.21453 m


= 0.3132 – 0.21453

= 0.09867 m/s


V iB=√ 2g (h¿−hi )1000

A: V iB=√ 2 X 9.81 (178−166 )1000

= 0.4852 m/s

V iC=Qav

Ai

V iC=2.046 ×10−4

5.3093 ×10−4

= 0.38536 m/s


= 0.4852 – 0.38536

= 0.09984 m/s

B: V iB=√ 2 X 9.81 (180−159 )1000

= 0.6419 m/s

V iC=2.046 ×10−4

3.6644 ×10−4

= 0.55835 m/s


= 0.6419 – 0.55835

= 0.08355 m/s

C: V iB=√ 2 X 9.81 (177−105 )1000

= 1.1885 m/s

V iC=2.046 ×10−4

2.0106 ×10−4

= 1.0176 m/s


= 1.1885 – 1.0176

= 0.17089 m/s

D: V iB=√ 2 X 9.81 (174−138 )1000

= 0.8404 m/s

V iC=2.046 ×10−4

3.1416 ×10−4

= 0.65126 m/s


= 0.8404 – 0.65126

= 0.18914 m/s

E: V iB=√ 2 X 9.81 (173−145 )1000

= 0.7412 m/s

V iC=2.046 ×10−4

3.8013 ×10−4

= 0.53824 m/s


= 0.7412 – 0.53824

= 0.20296 m/s

F: V iB=√ 2 X 9.81 (172−154 )1000

= 0.5943 m/s

V iC=2.046 ×10−4

5.3093 ×10−4

= 0.38536 m/s


= 0.5943 – 0.38536

= 0.20894 m/s


V iB=√ 2 g (h¿−hi )1000

A: V iB=√ 2 X 9.81 (220−194 )1000

= 0.7142 m/s


= 0.7142 – 0.6570

= 0.0572 m/s

V iC=Qav

Ai

V iC=3 .488 ×10−4

5.309 ×10−4

= 0.6570 m

B: V iB=√ 2 X 9.81 (223−179 )1000

= 0.9291 m/s

V iC=3 .488 ×10−4

3.664 ×10−4

= 0.9520 m/s


= 0.9291 – 0.9520

= - 0.0229 m/s

C: V iB=√ 2 X 9.81 (220−56 )1000

= 1.7938 m/s

V iC=3 .488 ×10−4

2.011× 10−4

= 1.7340 m/s


= 1.7938 – 1.7340

= 0.0598

D: V iB=√ 2 X 9.81 (214−133 )1000

= 1.2606 m/s

V iC=3 .488 ×10−4

3.142× 10−4

= 1.110 m/s


= 1.2606 – 1.110

= 0.1506

E: V iB=√ 2 X 9.81 (212−152 )1000

= 1.0850 m/s

V iC=3 .488 ×10−4

3.801× 10−4

= 0.917 m/s


= 1.0850 – 0.9170

= 0.168 m/s

F: V iB=√ 2 X 9.81 (210−171 )1000

= 0.8747 m/s

V iC=3 .488 ×10−4

5.309 ×10−4

= 0.657 m/s


= 0.8747 – 0.6570

= 0.2177

DISCUSSION

As we know, this experiment was being conducted based on the objective to investigate

the validity of the Bernoulli’s equation when applied to the steady flow of water in a tapered

duct. Since the volume passing through at the given length of pipe during a given period of time

will be the same, the value of pressure will decrease. From the Bernoulli’s principle, it states that

the slower the rate of flow, the higher the pressure and the fastest the rate flow the lower the

pressure.

The pressure, velocity and elevation are related to the Bernoulli’s theorem and it is valid

in a steady, incompressible flow where net frictional forces are negligible. The equation was

obtained when the Euler’s equation was integrated along the streamline for a constant density for

incompressible fluid. The constant of integration (called the Bernoulli’s constant)

varies from one streamline to another but remains constant along a

streamline in steady, frictionless, incompressible flow. Despite its simplicity,

it had been proven to be a very powerful tool for fluid mechanics. Bernoulli’s

equation states that the “sum of the kinetic energy (velocity head), the

pressure energy (static head) and Potential energy (elevation head) per unit

weight of the fluid at any point remains constant” provided the flow were

steady and frictionless and the fluid used is incompressible. The energy is

assumed as it is added to nor taken away by some external agency.

From the data recorded, we can conclude that the velocity of water decrease as the water

flow rate decrease. This is because the diameter of the tube is affected the differences in velocity.

The bigger the diameter of the tube, the differences in velocity will bigger too. The flow rate of

the differences flow of each tube are also different. From the data, in the flow rate of 1.139 x 10 -4

m3/s at the section F the difference in velocity is 0.09867 m/s. Meanwhile for the 2.046 x 10 -4

m3/s the difference in velocity is 0.20894 m/s. Thus, for the 3.488 x 10-4 m3/s it is 0.2177 m/s.

From the results, we knew that the wider the area for the fluid to flow to the narrower

area, the higher the velocity of flowing fluid. This shown in the results where the velocity of

water flow in the tapered duct increases as the duct are decreases. So we can analysis that, the

difference velocity increases as the pressure difference increases. There was must be a parallax

error and zero error while taking a reading. The observer must have not read the level of static

head properly. Thus, the eyes are not perpendicular to the water level on the manometer.

Lastly, it can be concluded that the Bernoulli’s equation was valid when applied to steady

flow of water in tapered duct and absolute velocity values increase along the same channel.

Although the experiment proof that the Bernoulli’s equation was valid for

both flow but the values obtain might be slightly differ from the actual value.

CONCLUSION

As for the conclusion, we can concluded that the Bernoulli’s equation is

valid for convergent and divergent flow as both of it does obey the equation.

For both flow, as the pressure difference increase, the time taken for 3L

water collected increases and the flow rates of the water will also increase.

The results show the reading of the manometer tubes increases when the

pressure difference increases. Thus, the velocity for the same channel

increases, the total head pressure will also increase for both convergent and

divergent flows. Bernoulli’s theorem has several applications in everyday lives. Example,

Bernoulli’s equation can be used to find unknown pressure of the streamlines. All the flow rates

and both static and total pressure head in a rigid convergent and divergent are managed to be

calculated. The experiment was successfully conducted.

RECOMMENDATIONS

1) The experiment is repeated several times to get an accurate result.

2) The eye of observes should be placed parallel to the scale of manometer to get an

accurate reading.

3) Make sure the bubbles in the manometer are completely removed by adjusting the bleed

screw.

4) The valve should be control slowly so that the pressure difference can be maintained.

5) The time keeper must be alert with the rising of water volume to avoid error and must be only a person who taking the time.

6) The leakage of water in the instrument must be avoided.

REFERENCES

1) Fluid Mechanics for Chemical Engineers, Prentice Hall

2) Lab Manual: Bernoulli’s Theorem demonstration Unit.

3) Bernoulli experiment, 27 August 2010 at

http://www.scribd.com/doc/23125607/Bernoulli- Experiment

4) Bernoulli Lab Report, 27 August

2010,athttp://www.scribd.com/doc/23106099/Bernoulli- Lab-Report

APPENDICES

http://www.scribd.com/doc/23125607/Bernoulli-

Documents

Experiment 3 Bernoulli's Theorem