Experiment 3 Bernoulli's Theorem
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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
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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
Average Time (s) 14.66 s
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
Average Time (s) 8.6 s
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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 0.3132 – 0.21453
= 0.09867 m/s
Calculation: Result 2
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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 0.5943 – 0.38536
= 0.20894 m/s
Calculation: Result 3
V iB=√ 2 g (h¿−hi )1000
A: V iB=√ 2 X 9.81 (220−194 )1000
= 0.7142 m/s
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
Difference = ViB - ViC
= 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
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