# Figure 1a. The ellipse and tap set up ```Fluid Mechanics
Tuesday Week 6
MECH3261- Flow Laboratory
Introduction
Fluid mechanics involves the study of fluids and the forces that act upon them. It is a field that relies heavily on
mathematical models and experimental data to describe fluid flow phenomena. It is for this reason that the
following experiment is conducted; fluid behaviour is modeled in the provided apparatus in order to provide a
greater insight into concepts that define fluid mechanics, including but not limited to the growth of boundary
layers, pressure changes with varying orientations of an object as well as changes in velocity at different points
in the flow. Indeed, the study of fluid mechanics has a great significance in the physical world, allowing for the
development of more effective and efficient real-life applications.
Experiment objectives
The experiment is divided into two parts. The first part involves the use of a wind tunnel, while the second part
employs hydrogen bubble apparatus. Each experiment respectively aims to:
1. Observe air flow and behaviour of boundary layers around an object a wind tunnel, and to analyse the effects
of force and pressure due to the changing angles of attack.
2. Calculate flow velocity and boundary layer thickness for an open channel flow, thus observing the
development of the boundary layer.
Apparatus
- Wind Tunnel (as supplied by the lab)
- The test subject - Elliptical Prism
- Compass
- Manometers (attached to the test subject)
- Hydrogen bubble apparatus consisting of an electrode
- Flow channel with weir
- Ruler
Method Part A- Wind tunnel experiment
1. Allow the wind tunnel to run at a desirable pressure
2. Secure the test subject at the open end of the wind tunnel at an attack angle of 0&deg;, i.e. with the plane of
the minor axis normal to the air flow.
3. Record the manometer readings (in our case, take a photo for each angle)
4. Rotate the test subject at increments of 30&deg; (Up to 90&deg;), whilst taking a photo of the manometer and
making necessary observations.
5. Calculate the relevant data.
Fluid Mechanics
Tuesday Week 6
Figure 1a. The ellipse and tap set up
Flow
y
y
x
Characteristic
length
ϴ
Note that:
- Characteristic length is calculated based on simple geometry when the elliptical block’s orientation is changed.
- We take the height of the ellipse to be 30cm or 0.3m
Figure 1b. Wind tunnel apparatus
A1
A2
Fluid Mechanics
Tuesday Week 6
Method Part B- Hydrogen bubble apparatus
1. Set up the hydrogen bubble flow bench and measured all distances as shown in Figure 2a.
2. Inserted electrode into the first slot (marked A).
3. Switched the machine on to allow for the formation of hydrogen bubbles by hydrolysis. The hydrogen
bubbles mapped the location and size of the boundary layer.
4. Measured and recorded the estimated boundary layer thickness δm (see Table 2.1).
5. Moved the electrode to the next slot and repeated steps 3 to 5 for the remaining slots.
6. Measured the depth of water upstream and above the crest of the weir as shown in Figure 2b.
A
B
D
C
57mm
127mm
212mm
302mm
Figure 2a. Apparatus set-up (top view)
6mm
Figure 2b. Depth of water in apparatus set up (side view)
3mm
320mm
Direction of flow
Fluid Mechanics
Tuesday Week 6
Results Part A- Wind tunnel experiment
Imperial (Inches)
Tap
Number
TAP 1
TAP 2
Pref
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
θ=0
1.6
2.5
2.5
2.6
2.6
2.9
2.6
3
3.1
3.2
3.4
3.4
3.45
2.6
2.4
1.9
1.6
1.6
2
2.4
3.1
3.3
3.25
3.2
2.9
2.9
2.85
(Let θ be the attack angle)
θ = 30
1.25
1.95
2.5
2.4
2.4
2.4
3.2
2.4
3.2
3.3
2.9
2.4
1.9
1.45
2.4
1.55
2.2
3.2
3.7
4
4.3
3.2
2.85
2.5
3.2
3.1
3.15
θ = 60
1.1
1.9
2.5
2.5
2.5
3.3
2.5
3.4
2.6
1.7
1.3
1.1
1.15
2.5
2.7
3.4
3.45
3.4
3.3
3.35
3.3
3.5
3.5
3.5
3.5
3.4
3.3
θ = 90
1.05
1.5
2.5
2.5
2.5
3.4
2.5
3.2
1.7
1.15
1.1
1.65
2.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
Metric (Meters)
Tap
Number
TAP 1
TAP 2
Pref
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
θ=0
0.04064
0.0635
0.0635
0.06604
0.06604
0.07366
0.06604
0.0762
0.07874
0.08128
0.08636
0.08636
0.08763
0.06604
0.06096
0.04826
0.04064
0.04064
0.0508
0.06096
0.07874
0.08382
0.08255
0.08128
0.07366
0.07366
0.07239
θ = 30
0.03175
0.04953
0.0635
0.06096
0.06096
0.06096
0.08128
0.06096
0.08128
0.08382
0.07366
0.06096
0.04826
0.03683
0.06096
0.03937
0.05588
0.08128
0.09398
0.1016
0.10922
0.08128
0.07239
0.0635
0.08128
0.07874
0.08001
θ = 60
0.02794
0.04826
0.0635
0.0635
0.0635
0.08382
0.0635
0.08636
0.06604
0.04318
0.03302
0.02794
0.02921
0.0635
0.06858
0.08636
0.08763
0.08636
0.08382
0.08509
0.08382
0.0889
0.0889
0.0889
0.0889
0.08636
0.08382
θ = 90
0.02667
0.0381
0.0635
0.0635
0.0635
0.08636
0.0635
0.08128
0.04318
0.02921
0.02794
0.04191
0.0635
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
0.0889
Fluid Mechanics
Tuesday Week 6
Table 1.2 values used in calculation process
Wind Tunnel Dimensions:
Section
Length (m)
Area (m^2)
Given values
Air density (ρ air)
Air dynamic viscosity (μ air)
Water density (ρ water)
Water kinematic Viscosity (ν water)
Tap1
0.6
0.36
Tap2
0.3
0.09
1.2kg / m^3
1.8 x 10^-5 Pa.s
998kg/m^3
1x10^-6 m^2/s
Calculating the velocity of the flow in the wind tunnel:
Assuming a steady, incompressible and frictionless flow, the velocity can be found by applying Bernoulli’s
equations between section 1 and 2 of the wind tunnel:
𝑃1
𝑝
𝑃1
𝑝
+
+
𝑉1&sup2;
2
𝑉1&sup2;
2
+ 𝑔𝑧1 =
=
𝑃2
𝑝
+
𝑃2
𝑝
𝑉2&sup2;
+
𝑉2&sup2;
=&gt;
2
2
+ 𝑔𝑧2
and z1 = z2 (no elevation difference)
𝑉2&sup2; − 𝑉1&sup2; =
2(𝑃1−𝑃2)
𝑝ρair
V1A1= V2A2
Since flow is continuous, (A1 corresponds to Tap1, A2 corresponds to Tap2)
=&gt; V1 = (V2A2)/A1
=&gt;
𝑉22 (1 −
Now
𝐴22
2(𝑃1−𝑃2)
𝐴1
𝑝ρair
2) =
P1-P2 = ρwaterg(h2-h1)
therefore,
𝑉22 =
2(𝑃1−𝑃2)
(𝐴12 )
𝑝ρair
(𝐴12 −𝐴22 )
Since flow is continuous, (A1 corresponds to Tap1)
Now letting h1= tap 1 value and h2 = tap 2 value, we can calculate the pressure difference and hence the velocity
at each angle of attack.
Angle of Attack (degrees)
Pressure different (P1-P2) (Pa)
0
223.8080868
30
174.0729564
60
90
198.9405216 111.9040434
Velocity 2 (m/s)
19.94695574
17.59156143
Characteristic length (m)
0.04
0.05
Reynolds Number
53191.88197
58638.53811
Table 2.2 Pressure difference and velocity of each angle of attack
18.80617022 14.10462767
0.0866
0.1
108574.2894 94030.85111
Fluid Mechanics
Tuesday Week 6
Reynolds number for each angle of attack was calculated using
𝑅𝑒 =
ρVD
μ
Where V is the velocity, D is characteristic length, ρ is the air density and μ is the air viscosity.
Calculating the pressures for each angle of attack:
In order to find the forces we must find the pressures for each tap in each angle case. This is done by using the
pressure difference equation on the previous equation and subbing in the water level for the reading for the
height difference. The coordinates of the taps were also calculated using Solidworks. We also need to calculate
the area normal to each tap to find the force. The area was assumed to be the arc length from the two points on
either side of the tap multiplied by the height of the elliptical body. For example, the area assumed for tap 6
would be the arc length from point 7 to point 5 times by 30cm. The following results are calculated using excel
Table 2.3
TAP
1
2
3
4
Pressure for different angles (Pa)
0&deg;
30&deg;
60&deg;
646.5567 596.821565 621.68913
646.5567 596.821565 621.68913
721.1594 596.821565 820.6296516
646.5567 795.762086 621.68913
90&deg;
621.68913
621.68913
845.497217
621.68913
Coordinates
x
y
49.5
2.82
49
3.98
44
9.5
34
14.66
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
746.027
770.8945
795.7621
845.4972
845.4972
857.931
646.5567
596.8216
472.4837
397.881
397.881
497.3513
596.8216
770.8945
820.6297
808.1959
795.7621
721.1594
721.1594
708.7256
795.762086
422.748608
285.977
273.543217
410.314826
621.68913
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
870.364782
18
0
-18
-34
-44
-49
-49.5
-50
-49.5
-49
-44
-34
-18
0
18
34
44
49
49.5
50
596.821565
795.762086
820.629652
721.159391
596.821565
472.483739
360.579695
596.821565
385.447261
547.086434
795.762086
920.099912
994.702608
1069.3053
795.762086
708.725608
621.68913
795.762086
770.894521
783.328304
845.4972168
646.5566952
422.7486084
323.2783476
273.5432172
285.9769998
621.68913
671.4242604
845.4972168
857.9309994
845.4972168
820.6296516
833.0634342
820.6296516
870.364782
870.364782
870.364782
870.364782
845.4972168
820.6296516
18.66
20
18.66
14.66
9.5
3.98
2.82
0
-2.82
-3.98
-9.5
-14.66
-18.66
-20
-18.66
-14.66
-9.5
-3.98
-2.82
0
Area (m^2)
0.0001231
0.0002596
0.0005524
0.000827
0.0010325
0.0001692
0.0010325
0.000827
0.0005524
0.0002596
0.0001231
0.0001692
0.0001231
0.0002596
0.0005524
0.000827
0.0010325
0.00108
0.0010325
0.000827
0.0005524
0.0002596
0.0001231
0.0001692
Fluid Mechanics
Tuesday Week 6
Calculating the drag forces due to pressure for each angle of attack:
In knowing the pressures and normal areas, we can find the normal forces using F=P*A
Table 2.4
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
TOTAL
CD
Normal Forces for different angles (N)
0&deg;
30&deg;
60&deg;
0.079598 0.0734754 0.076536876
0.167837 0.15492683 0.161382116
0.398375 0.32968985 0.453323544
0.534697 0.65808853 0.514131663
0.770276 0.61622052 0.872979066
0.130435 0.13464295 0.109397393
0.821627 0.84730321 0.436489533
0.699219 0.59639273 0.267348465
0.467061 0.32968985 0.151107848
0.222707 0.12265041 0.074235773
0.079598 0.04439139 0.076536876
0.100982 0.10098221 0.113604985
0.058168 0.04745286 0.104090151
0.103285 0.14201626 0.22270732
0.219793 0.43958647 0.467060622
0.411305 0.76091486 0.678653796
0.616221 1.0270342 0.860141138
0.832566 1.15484973 0.886280024
0.847303 0.82162736 0.898654921
0.668371 0.5861101 0.719784329
0.439586 0.34342693 0.480797699
0.187203 0.20656911 0.225934963
0.088783 0.09490573 0.104090151
0.119916 0.13253915 0.138850537
9.064915 9.76548661 9.094119788
3.164302 3.50624776 1.649501944
90&deg;
0.07653688
0.16138212
0.46706062
0.51413166
0.82162736
0.07152906
0.29527233
0.22621793
0.22666177
0.16138212
0.10715163
0.14726572
0.10715163
0.22593496
0.4807977
0.71978433
0.89865492
0.93999396
0.89865492
0.71978433
0.4807977
0.22593496
0.10715163
0.14726572
9.22812595
2.5770198
The total forces are also calculated and by using the following equation as well as the values from previous
calculations, the drag coefficient CD was calculated for each angle case.
∑Fx = FD and 𝐶𝐷 =
*Note that A is frontal area for each orientation of the ellipse.
𝐹𝐷
1
𝜌v2 𝐴
2
Fluid Mechanics
Tuesday Week 6
1200
1000
Pressure (Pa)
800
Angle of attack = 0&deg;
Angle of attack = 30&deg;
600
Angle of attack = 60&deg;
400
Angle of attack = 90&deg;
200
0
1 2 3 4 5 6 7 8 9 101112131415161718192021222324
Tap
Figure 1c. Pressure at each tap for varying angles of attack
1.4
1.2
Force (N)
1
0.8
Angle of attack = 0&deg;
Angle of attack = 30&deg;
0.6
Angle of attack = 60&deg;
Angle of attack = 90&deg;
0.4
0.2
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Tap
Figure 1d. Force distribution at each tap for different angles of attack
Fluid Mechanics
Tuesday Week 6
Results Part B- Hydrogen bubble apparatus
At the crest of the weir, the flow is defined as critical with Froude number = 1, and thus using the equation:
𝑉
𝐹𝑟 =
√𝑔𝑦
We can calculate the free stream velocity V2 = √(9.81 x 0.003) = 0.1716m/s
Using the continuity equation V1A1 = V2A2, we can find V1 (the upstream velocity). However, since the width of
channel remains constant at 0.32m, the equation can be reduced to:
V1 = V2 (
Y2
0.003
) = 0.1716 (
) = 0.0858m/s
Y1
0.006
The Reynolds number Rex can then be calculated for each section of the flow, using:
𝑅𝑒𝑥 =
𝑉𝑥
𝜈
Where x is the distance from the beginning of the boundary layer,
V is the upstream velocity = 0.0858m/s and
νs is the kinematic viscosity of water = 1 x 10-06 m2/s.
Based on the calculated Rex, we can go on to calculate the boundary layer thickness for both the laminar and
turbulent flow regimes using the relations:
1
𝛿𝑙𝑎𝑚𝑖𝑛𝑎𝑟 = 4.65𝑥𝑅𝑒𝑥 −2
𝛿𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 = 0.38𝑥𝑅𝑒𝑥 −1/5
x (m)
0.057
0.127
0.212
0.302
Rex
4890.6
10896.6
18189.6
25911.6
δm (m)
0.004
0.008
0.011
N/A
δl (m)
0.00379
0.005657
0.007309
0.008724
δt (m)
0.003961
0.007518
0.011328
0.015035
Table 2.1 A comparison of measured and theoretical boundary layers
Note that no boundary layer was observed at point D. This is later addressed in the discussion.
Fluid Mechanics
Tuesday Week 6
We can also calculate the rate of boundary layer growth at each section of the flow using the following
relation:
𝛿𝑖
𝑚𝑖 = tan
𝑥𝑖
xi (m)
57
70
85
90
mi
0.070291
0.114786
0.130139
N/A
δi (m)
4
8
11
N/A
Table 2.2 Boundary layer growth rate
0.016
Boundary layer thickness (m)
0.014
0.012
0.01
0.008
δ laminar
δ turbulent
0.006
δ measured
0.004
0.002
0
0.057
0.127
0.212
0.302
Figure 2c. Theoretical and measured boundary layer thickness at different points in the flow
Note that the graph discontinues for the measured boundary layer thickness, as no boundary was observed at
point D. Theoretical values for laminar and turbulent thicknesses at that point however were calculated, and so
the graph continues for that distance.
Fluid Mechanics
Tuesday Week 6
Discussion
Part A
Most force vector plots roughly showed a consistent or obvious pattern however there were some adjacent taps
which showed large variations in pressure. When observing the force distribution over the ellipse, the profile at
θ = 60&deg; was most erratic out of the four angles of attack tested, showing small irregularities in magnitude of
normal forces and direction of forces toward the front of the ellipse (relative to the wind tunnel). There were
also some discrepancies with theoretical expectations. For example, the plots should have been symmetrical but
they weren’t due to experimental errors which could have included: the body was not a perfect ellipse, tapping
locations were not exact, nor symmetrical (in our calculations they were assumed symmetrical), and the leading
edge tap may not have been exactly centered.
Theoretically, increasing the angle of attack should increase the total drag force on the elliptical cylinder. In the
experimental results, the total drag force increased from α = 0&deg; to α = 60&deg;, however it then decrease when α =
90&deg; where it should have been at a maximum. Thus there were experimental errors which could have been due
to turbulence, or the manometer not functioning correctly. Further, the increasing drag force is due to the
increased projected area, which was observed in our experiment.
Several improvements could have been made to the experiment. Taking photos of the manometer levels and
simply reading off images creates large room for error. Perhaps an electronic manometer counter would have
significantly increased the accuracy of readings. Also, the arc lengths that were calculated were straight lines
from point to point and not a curve. This then affects the calculated area and hence the forces etc.
Overall the experiment gives a good understanding on the behaviour of fluid flow, and shows the effects on the
orientation of a certain shape within in a flow. This understanding leads to better knowledge on design and
hence better engineering practice.
Part B
Based on the calculation of the Reynolds number and assuming that for an Rex of greater than 5x10^5 the flow is
turbulent, we find that the flow throughout the open-channel in the experiment remains laminar. One would
therefore be inclined to use the relation for δlaminar to calculate the theoretical boundary layer thickness. In the
results however, both relations (laminar and turbulent) have been used to calculate boundary layer thickness as
tabulated in Table 2.1. It is interesting to note that although the flow is determined to be laminar, the measured
boundary layer thickness actually holds almost the exact values as the calculated δturbulent. This leads to the idea
that there has been some form of calculation error, which is almost expected with every experiment.
The calculations relied primarily on the measured depths of the flow channel, which in itself is open to all sorts
of human error, especially considering the fact that the depths dealt with were in millimeters and therefore
difficult to read off a regular ruler.
In calculating the upstream velocity, the cross-sectional area of the channel was assumed constant when in
reality, it was slightly less due to the placement of the plate in the middle of the channel. As no measurements
were taken of the width of the plate, its area was eliminated from the calculations thus affecting calculated
velocity, which in turn was used to calculate the Reynolds number at each point in the flow.
Fluid Mechanics
Tuesday Week 6
The measurements of the boundary layer were also inaccurate, as the thickness was estimated based on the
point where we determined where the hydrogen bubbles paralleled out to the plate. Not only was this difficult
to observe because of the poor visibility of the hydrogen bubbles, but the constant movement of the bubbles
and their eventual diffusion into the stream made that parallel point difficult to determine.
At the fourth point (point D), no boundary layer was observed. This was accounted for by the abrupt ending of
the plate, which did not allow for enough length over which a boundary layer could be determined.
The rate of boundary layer growth was found to increase as the distance across the plate increased, which is
expected at the boundary layer behaves in such a manner that it eventually broadens as the flow becomes
turbulent and therefore mixes with the free stream.
Conclusion
The experiments conducted allowed for an observation of flow and the behaviour of boundary layers around an
object in a wind tunnel, and the effects of angle of attack on the force and pressure on the object at different
points, as well as helping to model boundary layer growth over a plate in an open channel flow. Although both
experiments were open to many sources of error, they provide a valuable insight into the behaviour of fluids
and demonstrate the importance of mathematical modeling in fluid mechanics.
```