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CHNG2801/CHNG9201 FLUID MECHANICS
LABORATORY EXPERIMENT
Air Flow
Preparation and Prework
Running the Equipment
TAGL: March 2024
Air flow rig AR demo V2:
https://view.vuforia.com/command/view-experience?url=https%3A%2F%2Funiversity-of-
sydney-dev.es.thingworx.com%2FExperienceService%2Fcontent%2Fprojects
%2Fair_flow_rig_v3%2Findex.html%3FexpId%3D1
Safety:
You must wear safety glasses or spectacles, laboratory coat, pants or jeans and
closed, covered footwear while in the laboratory area at all times. Failure to follow this
requirement will result in your ejection from the laboratory for the rest of the laboratory
session. You will not be given the opportunity to repeat the experiment on another day.
This practice is intended to encourage the development of good safety discipline in
industry as well as for your direct protection.
Overall Learning Objectives
Students should be able to:
1.
Identify pipework and equipment used to make fluids move (fans and pumps). What do
these pieces of equipment look like, and how do they work? (in laboratory)
2.
Calculate flow rate using the Pitot tube, and the orifice plate flow meter to measure the
flowrate in a duct. (in laboratory)
3.
Construct fan and system curves from pressure and flow rate measurements. (in
laboratory)
1
4.
Operate chemical engineering equipment safely and extract relevant measurements and
data. (in laboratory)
5.
Synthesise findings with relevant fluid mechanics theory and report results concisely.
(after laboratory)
INTRODUCTION
The aims of this experiment are (a) to compare two different flow measurement techniques,
and (b) to measure the pump and system curves for the fan and the other equipment,
respectively.
We need to compare flow measurement devices, specifically Pitot tubes (and Pitot tube
traverses) with orifice plate flow measurements.
We need to create a set of pump and system curves. The pressure decrease through the
system is the system curve (the pressure difference required by the system at a given flow
rate), which starts at atmospheric pressure. This pressure drop must equal the pressure
increase across the pump, which is the pump or fan curve). Each measurement is an operating
point where a pump curve intersects a system curve. The system curve may be changed by
altering a valve position, while a pump curve may be changed by altering the pump speed.
EQUIPMENT
Figure 1 shows a Piping and Instrumentation Diagram for the equipment.
Figure 1:
Piping and Instrumentation Diagram for the air flow experiment.
DP
DP
Negative point for Pitot tube, static pressure: negative side of digital manometer
Positive point for Pitot tube, impact pressure: positive side of digital manometer
Negative point for orifice plate flow meter: negative side
of digital manometer
Positive point
for orifice plate flow meter: in the atmosphere
Pitot tube traverse
From atmosphere
Iris valve
Orifice plate flow meter
Fan
2
PREWORK CALCULATIONS (TO BE DONE BEFORE THE LABORATORY FOR
THE EXPERIMENTS STARTING SOON)
These calculations must
be shown to the laboratory tutor on the day of the experiment and must
be completed correctly before commencing. The next section 6 of these notes gives some
examples that might be helpful.
(a)
The following pressure differences (between the impact and static tappings) were recorded
at equispaced points in a 114 mm x 127 mm rectangular duct through which air (density,
1.2 kg m
-3
; viscosity, 1.8 x 10
-5
kg m
-1
s
-1
) is flowing. Calculate the volumetric flow rate
through the duct. The pressure differences (in Pa) are as shown in the following Table 1.
Equation (1) in the next section will be important. Use a Pitot-tube coefficient (
C
) of 1
(unity).
Table 1: Pressure differences recorded by the digital pressure meter (in Pa) in each section of
the Pitot tube traverse for the air flow experiment. 75
100
125
100
100
125
150
125
75
150
175
100
50
75
125
75
Each rectangle in the grid above represents an equal area, so averaging the point velocities
directly to give an average velocity is reasonable.
3
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(b) This rectangular duct is followed by a circular duct of 140 mm diameter in which there is
an orifice plate flow meter. The orifice plate hole diameter is 108 mm and the pressure drop
between the upstream tapping and the throat (the hole) is 216 Pa. Given a discharge
coefficient of 0.6 for the orifice plate flow meter, calculate the volumetric flow rate as
measured by this flow meter. Equation (2) in the following section will be important.
= 0.25756 M^3/s
Note: Useful information for these calculations (above) is given in the equations in the next
section.
EXPECTATIONS IN THE LABORATORY
Suppose that we have a 140 mm diameter pipe, with an orifice plate flow meter having an
orifice diameter of 108 mm.
With a Pitot tube, in general, and here in this laboratory, the simple application of the Pitot
tube equation is helpful: u
=
C
pt
√
2
∆ P
ρ
(1)
Given that C
pt
≈ 1 and ρ
≈ 1 kg m
-3
, the pressure difference across a manometer for a velocity
of 5 ms
-1
through a test section is about (5 ms
-1
)
2
/2 = 12 Pa, and at 10 ms
-1
, the pressure
difference is (10 ms
-1
)
2
/2 = 50 Pa. In the experiment, we will use ρ
≈ 1.2 kg m
-3
, but here we
use ρ
≈ 1 kg m
-3 for convenience.
What should we expect in the orifice plate flow meter? Should the pressure differences be the
same as those from the Pitot tube?
Yes and no.
Yes, the general form of the equation is similar for all devices, and all equations for these
types of meter come from Bernoulli’s equation. For the Pitot tube, see equation (1) above.
For the orifice plate flow meter, equations (2) and (3) apply:
Q
=
C
D
A
t
√
2
∆ P
ρ
[
1
−
(
A
t
A
p
)
2
]
(2)
u
av
=
Q
A
p
(3)
4
Where C
D
is a discharge coefficient (about 0.6 for the orifice plate flow meter), u
av
is the
average pipe velocity, A
t
is the throat cross-sectional area, and A
p
is the pipe cross-sectional
area.
And no:
The orifice plate flow meter is located in a 140 mm diameter pipe, so A
p
= π/4 (0.14 m)
2
=
0.0154 m
2
.
For the orifice plate meter, A
t
= π/4 (0.108 m)
2
= 0.00916 m
2
, and rearranging equation (2),
for the orifice plate flow meter, gives
∆ P
=
(
Q
C
D
A
t
)
2
ρ
2
[
1
−
(
A
t
A
p
)
2
]
(4)
∆ P
=
(
u
av
A
p
C
D
A
t
)
2
ρ
2
[
1
−
(
A
t
A
p
)
2
]
(5)
Hence, for an average air velocity of 5 ms
-1
through the orifice plate flow meter, the pressure
difference is:
∆ P
=
(
5
×
0.0154
0.61
×
0.00916
)
2
1
2
[
1
−
(
0.00916
0.0154
)
2
]
= 61 Pa
For the Pitot tube, the pressure difference at a velocity of 5 ms
-1
is 12 Pa.
For an average air velocity of 10 ms
-1
through the orifice plate flow meter, the pressure
difference is:
∆ P
=
(
10
×
0.0154
0.61
×
0.00916
)
2
1
2
[
1
−
(
0.00916
0.0154
)
2
]
= 245 Pa
For the Pitot tube, the pressure difference at a velocity of 10 ms
-1
is 50 Pa.
For the orifice plate flow meter, the relationship between the velocity and the pressure
difference is affected by the pipe and throat cross-sectional areas and the discharge
coefficients.
TESTING PROCEDURE
Orifice Plate: To measure the pressure difference between the upstream location and the
throat in the orifice plate flowmeter, connect the plastic tube after the orifice plate to the
negative side of the digital manometer. The upstream pressure is just atmospheric pressure,
so the positive side of the manometer should just be open to the atmosphere.
5
Pitot Tube: To measure the pressure difference for the Pitot tube, connect the bottom of the
Pitot tube (connected to the impact pressure point) to the positive side of the digital
manometer with a plastic tube, and also use a plastic tube to connect the side of the Pitot tube
(connected to the static pressure point) to the negative side of the digital manometer.
Positive point for orifice plate flow meter: in the atmosphere
Pitot tube traverse points (distances from near wall):
2 mm
12 mm
19 mm
36 mm
69 mm
86 mm
These distances represent (almost) equal cross-sectional areas, so the point velocities can be
averaged to give the average velocity.
1. Flow meters: Set the flow to a high value by fully opening up the Iris valve (indicator at
the blue line) after the entry of the duct and setting the variable speed drive (v.s.d.) on
the fan to 50 Hz.
2. Flow meters: With the Iris valve fully opened (indicator at the blue line), carry out a Pitot
Tube traverse of the circular pipe using the points given above (2 mm, 12 mm, etc).
Without altering the flow, measure the pressure at the downstream tapping of the orifice
plate flow meter.
3. Flow meters: Again, with the Iris valve fully opened (indicator at the blue line), repeat the
experiment using two lower flow rates (e.g. 40 Hz, and 30 Hz on the v.s.d.), so that you
carry out measurements for three flow rates. 4. Fan and system curves (valve setting 1): Start with the Iris valve fully opened (indicator at
the blue line), set the variable speed drive (v.s.d.) on the fan to 30 Hz. Measure the static
pressure before the fan. Also, measure the pressure downstream of the orifice plate flow
meter, allowing us to calculate the flow rate. The static pressure will be less than
atmospheric pressure, due to the pressure drop along the pipe. This combination of the
static pressure and the flow rate from the flow meter is an operating point on both a
system curve (for the system with the Iris valve fully open) and on a fan curve (for the
fan at a frequency or speed of 30 Hz). The pressure drop down the pipe is equal to the
pressure at the inlet to the fan.
5. Fan and system curves (valve settings 2, 3 and 4): Change the Iris valve setting (indicator
at the blue line) to the numbers 4, 3, and the bottom line, and repeat the static pressure
measurements before the fan for settings of the variable speed drive (v.s.d.) on the fan at
6
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30 Hz, 40 Hz, and 50 Hz. You will have four valve settings and three fan speeds, so you
will have 12 operating points)
6. Calculate flow rates and pressure differences (increases and decreases) on the spot. If your
results are unexpected, measure them again, checking that your instrumentation is set up
correctly.
7
RESULTS AND DISCUSSION
Typical Data and Analysis
Use C
D
= 0.6 for the orifice plate flow meter. Circular duct diameter
:
105 mm
Orifice plate hole diameter
:
71 mm
Comparison between Pitot tube traverses and orifice plate flow meters
The raw data for the comparison between Pitot tube traverses and orifice plate flow meters
are shown in Table 2.
Table 2:
Raw data for the pressure differences across the Pitot tube and the orifice plate flow
meters, iris valve fully open
Fan
frequency
Pitot tube
Orifice
plate
flow
meter Point 1,
2 mm
Point 2,
12 mm
Point 3,
19 mm
Point 4,
36 mm
Point 5,
69 mm
Point 6,
86 mm
(Hz)
(Pa)
(Pa)
(Pa)
(Pa)
(Pa)
(Pa)
(Pa)
20
26
31
33
34
35
29
304
30
60
69
72
76
80
70
690
40
105
125
129
135
137
118
1200
50
170
195
201
202
209
186
1815
Processing the data and doing the calculations
Sample calculation for the Pitot tube
Row 1, 20 Hz fan frequency, iris valve fully open, column 1, point 1, 2 mm from wall, Δ
P
=
26 Pa on the Pitot tube.
Equation (1) becomes the following equation, with C
pt
≈ 1.29 at 20
o
C, including the density
of 1.2 kg m
-3
at 20
o
C:
u
=
C
pt
√
∆ P
(1)
Given that C
pt
≈ 1.29 at 20
o
C, includes the density of 1.2 kg m
-3
at 20
o
C.
With Δ
P
= 26 Pa,
u
=
1.29
√
26
=
6.58
m s
−
1
(1)
Pitot tube: For the first row, 20 Hz fan frequency, iris valve fully open, the average velocity
for all the Pitot tube measurements is 7.21 m s
-1
. The cross-sectional area of the pipe is π/4
(0.105 m)
2
= 0.008659 m
2
, and this cross-sectional area applies for all the fan frequencies and
flow rates. Hence, the volumetric flow rate from the Pitot tube measurements for the first
row, 20 Hz fan frequency, iris valve fully open, is 7.21 m s
-1
× 0.008659 m
2
= 0.0624 m
3
s
-1
(
Q
Pitot
).
8
Sample calculation for the orifice plate flow meter
Row 1, 20 Hz fan frequency, iris valve fully open, Δ
P
= 304 Pa for the orifice plate flow
meter.
Equation (2) for the orifice plate flow meter follows:
Q
=
C
D
A
t
√
2
∆ P
ρ
[
1
−
(
A
t
A
p
)
2
]
(2)
Where C
D
is a discharge coefficient (about 0.6 for the orifice plate flow meter), A
t
is the
throat cross-sectional area, and A
p
is the pipe cross-sectional area (0.008659 m
2
, from above).
The throat cross-sectional area (
A
t
) is π/4 (0.071 m)
2
= 0.003959 m
2
.
Substituting the values into equation (2) gives:
Q
orifice
=
0.6
×
0.003959
√
2
×
304
1.2
[
1
−
(
0.003959
0.008659
)
2
]
=
0.0601
m
3
s
−
1
(2)
The comparisons between the volumetric flow rates from the Pitot tube traverses and the
orifice plate flow meters are shown in Table 3.
Table 3:
Volumetric flow rate comparison between the Pitot tube and the orifice plate flow
meters, iris valve fully open
Pitot tube
Fan
freq.
Point
1, 2
mm
Point
2,
12
mm
Point
3, 19
mm
Point
4, 36
mm
Point
5, 69
mm
Point
6, 86
mm
Pitot
tube
average
velocity
Q
Pitot
,
Pitot
tube
volu-
metric
flow rate
Q
orifice
,
orifice
plate
volu-
metric
flow rate
(Hz)
(m s
-1
)
(m s
-1
)
(m s
-1
)
(m s
-1
)
(m s
-1
)
(m s
-1
)
(m s
-1
)
(m
3
s
-1
)
(m
3
s
-1
)
20
6.58
7.18
7.41
7.52
7.63
6.95
7.21
0.0624
0.0601
30
10.0
10.7
10.9
11.2
11.5
10.8
10.9
0.0941
0.0906
40
13.2
14.4
14.7
15.0
15.1
14.0
14.4
0.125
0.119
50
16.8
18.0
18.3
18.3
18.6
17.6
17.9
0.155
0.147
Comparison between Pitot tube traverses and orifice plate flow meters
The raw data for the pump and system curves are shown in Tables 4 and 5.
9
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Table 4:
Raw data for fan and system curves, the pressure differences across the orifice plate
flow meter (for calculating the volumetric flow rates)
Valve position
Fan
frequency
Fully open
4/5 open
3/5 open
2/5 open
1/5 open
(Hz)
(Pa)
(Pa)
(Pa)
(Pa)
(Pa)
20
293
274
225
143
38
30
663
612
505
326
92
40
1173
1079
891
578
158
50
1813
1670
1378
890
261
Table 5:
Raw data for fan and system curves, the pressure at the inlet of the fan and the outlet
of the measurement tube (below atmospheric pressure)
Valve position
Fan
frequency
Fully open
4/5 open
3/5 open
2/5 open
1/5 open
(Hz)
(Pa)
(Pa)
(Pa)
(Pa)
(Pa)
20
314
323
330
340
340
30
714
726
743
766
770
40
1265
1285
1316
1359
1360
50
1960
1990
2035
2100
2110
The raw data in Table 5 may be used directly, since the pressure below atmospheric pressure
at the inlet to the fan and the outlet of the tube gives both the pressure drop across the tube
and the pressure increase across the fan, for the following reasons. The pressure drops along
the tube from atmospheric pressure to the pressure at the inlet to the fan, so this pressure is
the pressure required for this volumetric flow rate on the system curve. The pressure
increases across the fan from the pressure at the inlet to the fan to atmospheric pressure, so
this pressure is the pressure increase across the fan at this volumetric flow rate on the fan or
pump curve.
The pressure difference data in Table 4 must be converted to volumetric flow rates as shown
in the sample calculation applying equation (2) to the flow meter comparison.
Taking the pressure difference in the first row, first column, fan frequency 20 Hz, fully open
iris valve, Δ
P
= 293 Pa for the orifice plate flow meter, and using equation (2), gives:
Q
orifice
=
0.6
×
0.003959
√
2
×
293
1.2
[
1
−
(
0.003959
0.008659
)
2
]
=
0.0590
m
3
s
−
1
(2)
The remaining data in Table 4 have been calculated as above and placed into Table 6.
10
When plotting the fan (pump) and system curves, each point is a point on both fan and system
curves. The x-coordinate (abscissa) comes from a row, column location in Table 6, for the
volumetric flow rate. The y-coordinate (ordinate) comes from a row, column location in
Table 4, for both the pressure drop across the tube (system curve) and the pressure increase
across the fan (fan curve). The resulting fan and system curves are shown in Figure 2.
Table 6:
Calculated volumetric flow rates corresponding to the pressure differences across
the orifice plate flow meter (from Table 4) for the fan and system curves
Valve position
Fan
frequency
Fully open
4/5 open
3/5 open
2/5 open
1/5 open
(Hz)
(m
3
s
-1
)
(m
3
s
-1
)
(m
3
s
-1
)
(m
3
s
-1
)
(m
3
s
-1
)
20
0.0590
0.0571
0.0517
0.0412
0.0213
30
0.0888
0.0853
0.0775
0.0623
0.0331
40
0.1181
0.1133
0.1029
0.0829
0.0433
50
0.1468
0.1409
0.1280
0.1029
0.0557
0.0000
0.0500
0.1000
0.1500
0
1000
2000
3000
Fully open
4/5 open
3/5 open
2/5 open
1/5 open
20 Hz
30 Hz
40 Hz
50 Hz
Air flow rate, m3/s
Pressure increase or decrease, Pa
Figure 2:
Fan (pump) and system curves for the air flow experiment.
11
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Discussion
1.
You must prepare a concise report to your manager summarising the key findings.
Your support calculations and raw data should be included in the Appendices that
should be attached to your report.
2.
Calculate and compare the flows given by the two methods of measurement. 3.
Calculate the Reynolds Number for each flow rate and comment on the likely state of
flow (laminar or turbulent). Do not plot Reynolds number against flow rate because they
are directly proportional by definition!
4.
Discuss the uncertainties in the measurements and how these uncertainties propagate
through to affect the results. The appendix has notes on how to do this. What effects
do the uncertainties in all the input numbers (diameters, pressure differences,
densities) for the calculations have on the final results (e.g. flow rates)?
5.
The fan/pump and system curves show that there are two ways to control the flow rate
through the equipment (below the maximum flow rate). One way is to reduce the fan
frequency using the variable speed drive. The other way is to close the valve. Given
that the power required to pump the fluid (air) is the product of the pressure rise
across the pump (fan), in Pa, and the volumetric flow rate, in m
3
s
-1
, what is the lowest
energy cost way to achieve a volumetric flow rate that is half the maximum
volumetric flow rate? At this halfway value of the maximum volumetric flow rate,
what is the difference in power requirements between reducing the volumetric flow
rate by (a) closing the valve, or (b) reducing the fan/pump speed?
REFERENCES
Coulson, J M, Richardson, J F, Backhurst, J R, Harker, J H, (1999) Chemical Engineering.
Volume 1: Fluid Flow, Heat Transfer and Mass Transfer
, 6th Edition, Pergamon, Oxford.
There are many other books on fluid mechanics, fluids engineering in the library. 13
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Refrigerant 134a enters a well-insulated nozzle at 200 Ibf/in.?, 140°F, with a velocity of 120 ft/s and exits at 50 Ibf/in.? with a velocity
of 1500 ft/s.
For steady-state operation, and neglecting potential energy effects, determine the temperature, in °F, and the quality of the refrigerant
at the exit.
T2 =
i
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X2 =
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The force, F, of the wind blowing against a building is given by F = CopV²A/2, where Vis the wind speed, p the density of the air, A the
cross-sectional area of the building, and CD is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient,
x, y, and z in the expression CD = M*LYT.
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Wind Farms
Consider the arrangement of three wind turbines in the following schematic in which wind
turbine C is in the wakes of turbines A and B.
Given the following:
- Uo = 12 m/s
A
-XẠC = 500 m
-XBC = 200 m
- z = 60 m
- Zo = 0.3 m
U.
-r, = 20 m
B
- CT = 0.88
Compute the total velocity deficit, udef(C) and the velocity at wind turbine C, namely Vc.
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As shown in the figure, a 300-ft tank contains 25 lb of H2O initially at 30 lbf/in?. The tank is connected to a large steam line carrying
steam at 200 Ibf/in?, 450°F. Steam flows into the tank through a valve until the tank pressure reaches p2 = 120 lbf/in? and the
temperature is 400°F, at which time the valve is closed.
Steam at
200 lbf/in.².
450°F
Tank
Valve
(1)
Initially:
30 lbfin.?, mı = 25 lb
|(2)
Finally:
Ibfin.?, 400°F.
P2
Determine the amount of mass that enters the tank, in Ib, and the heat transfer to the tank from its…
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Problem 4.137
3 of 4
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Replace the three forces acting on the plate by a wrench. Suppose
that FA = {450i}N, FB = {-350k} N, and Fc = {300j} N.
(Figure 1)
Part B
eТext
Determine the couple moment of the wrench.
Express your answer to three significant figures and include the appropriate units. Enter positive value if the sense of direction of the couple moment is the
same as that of the resultant force and negative value if the sense of direction of the couple moment is opposite to that of the resultant force.
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The spring of modulus k = 203 N/m is compressed a distance of 300 mm and suddenly released with the system at rest. Determine the
absolute velocities of both masses (positive if to the right, negative if to the left) when the spring is unstretched. Neglect friction.
k
2 kg
= 203 N/m
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Answers:
2kg mass,
v=
5.6kg mass,
v =
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Newton’s 2nd Law Lab (Modeling friendly lab)
Go to the PhET simulation Forces & Motion. https://phet.colorado.edu/sims/html/forcesandmotionbasics/latest/forcesandmotionbasics_en.html
Select “Acceleration”
Click to show Forces, Sum of Forces, Values, Mass, and Acceleration.
There are two experiments for this activity – make sure you include both.
Experiment #1: Acceleration vs. Force
In this lab you will determine the relationship between acceleration and net force.
Choose a mass at the beginning, and keep it constant for this entire experiment.
Set the friction to zero. This will make your Applied Force equal to the net force.
Record data for five different values of Applied Force.
Graph Acceleration vs. Net Force.
Graph this in Google sheets(you want a line graph - it should only have one line).
Make sure that Applied Force information is used as the x value
Make sure that Acceleration information is used as the y value
Add a trendline – see what fits best –…
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As shown in the figure, Refrigerant 22 enters the compressor of an air conditioning unit operating at steady state at 40°F, 80 lbf/in?
and is compressed to 160°F, 200 Ib;/in?. The refrigerant exiting the compressor enters a condenser where energy transfer to air as a
separate stream occurs, and the refrigerant exits as a liquid at 200 lb;/in?, 90°F. Air enters the condenser at 70°F, 14.7 Ib;/in? with a
volumetric flow rate of 1000 ft°/min and exits at 110°F. Neglect stray heat transfer and kinetic and…
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