Experiment 7
Flow over a symmetrical airfoil NACA0015
Hrishav Das AE21B023
April 3, 2023
Contents
0 AIM:
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1 Experimental Setup:
2
2 Theory:
2
3 Procedure:
4
4 Observations and Calculations:
5
5 Conclusions:
6
6 Sources of Error:
8
List of Figures
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Experimental Setup . . . . . . . . . . . . . . . . . . . . . .
Pressure Ports on the Airfoil Surface . . . . . . . . . . . .
Pressure Distribution . . . . . . . . . . . . . . . . . . . . .
Continuous Pressure Distribution represented with colours
NACA 0015 Airfoil . . . . . . . . . . . . . . . . . . . . . .
Cp vs x Graph . . . . . . . . . . . . . . . . . . . . . . . .
Cl vs Alpha Graph . . . . . . . . . . . . . . . . . . . . . .
Cl and Cd Vs Alpha . . . . . . . . . . . . . . . . . . . . .
Cl and Cd Vs Alpha . . . . . . . . . . . . . . . . . . . . .
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0
AIM:
To analyze flow over a symmetrical airfoil NACA0015 and hence calculate Lift and Drag
Coefficients.
1
Experimental Setup:
• The experimental setup consists of a simple low speed wind tunnel.
• At the test section of such a tunnel, we keep a NACA0015 airfoil attached wall to
wall.
• We measure the static pressure along the surface of the airfoil using static pressure
ports.
Figure 1: Experimental Setup
2
Theory:
By taking measurements of static pressure on the airfoil surface, we get pressure distribution as a function of distance along the chord.
Every airfoil has an equation for its surface. The equation for NACA0015 airfoil for its
surface is : √
y = 5t 0.2969 x − 0.1260x − 0.3516x2 + 0.2843x3 − 0.1015x4
Where y is the distance from the mean chord line to the surface, x is the non dimensional
position along the chord length, and t is the maximum thickness as a fraction of the chord
length.
2
Figure 2: Pressure Ports on the Airfoil Surface
From the pressure readings, we can find out Cp at each of point of measurement:
Cp =
PLocal −PStat
PDynamic
In the figure given above, the x and y axis is with respect to the airfoil frame. The
force in the x and y direction can be calculated if we have a continuous distribution of
pressure by breaking the airfoil surface into discrete panels. We have the data regarding
the surface profile (in terms of an equation describing the surface). So in effect, we have
the orientations of each panel.
We can assume each panel to have Cp as the mean of the value calculated at its ends.
Once we have Cp at each panel, we can find the force acting on that panel and resolve
this in the direction parallel and perpendicular to the free stream.
Figure 3: Pressure Distribution
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Once we have computed the forces in the direction of the freestream and those perpendicular to the free stream, we can find the Cd and Cl respectively.
CL =
CD =
3
L
1
2
ρU∞
2
D
1
2
ρU∞
2
Procedure:
• Set up the wind tunnel and all the connections appropriately.
• Turn the airfoil by an certain degrees to simulate and angle of attack.
• Turn on the turbine of the wind tunnel and let the flow develop inside the wind
tunnel across the airfoil specimen kept inside.
• Take readings from the pressure ports for the flow across this airfoil for this given
angle of attack.
• Turn of the turbine.
• Turn the airfoil by the same amount in the opposite direction.
• Turn on the turbine and let the flow develop.
• Take readings from the pressure ports again. This would give us the readings for
the other side of the airfoil. (since the airfoil is symmetric)
• Tabulate these readings.
• Find out the L and D and accordingly the CL and CD .
• Plot CL and CD with respect to angle of attack and find its slope in the linear
region of the graph.
Figure 4: Continuous Pressure Distribution represented with colours
4
4
Observations and Calculations:
For α = 8◦ and U = 15m/s
Port Number
Location (mm)
Pu (mm H2O)
Pl (mm H2O)
Cpu
Cpl
1
0
-15.8
-24.0
-0.18
-0.76
2
3
-42.7
-2.9
-2.09
0.74
3
5
-37.8
-7.8
-1.74
0.39
4
7
-36.5
-10.0
-1.65
0.23
5
9
-33.7
-12.8
-1.45
0.04
6
22
-30.4
-14.9
-1.22
-0.11
7
29
-23.3
-14.9
-0.71
-0.11
8
36
-19.8
-13.7
-0.46
-0.03
9
43
-20.1
-15.3
-0.48
-0.14
10
50
-17.5
-14.2
-0.30
-0.06
Equation of Airfoil Surface :
√
y = 5(0.15) 0.2969 x − 0.1260x − 0.3516x2 + 0.2843x3 − 0.1015x4
Figure 5: NACA 0015 Airfoil
Chord Length (c) = 65mm
Port Number
Location (mm)
x (Location/c)
y (non dimensional)
y * c (mm)
1
0
0
0
0
2
3
0.05
0.043
2.79
3
5
0.08
0.053
3.45
4
7
0.12
0.06
3.91
5
9
0.14
0.065
4.24
6
22
0.34
0.074
4.85
7
29
0.45
0.07
4.55
8
36
0.55
0.061
4.00
9
43
0.66
0.05
3.27
10
50
0.77
0.037
2.40
Since we have 10 pressure ports, we can divide each of the top and bottom surfaces
into 9 panels. And we can assume the Cp acting on each of these panels is the mean of
the Cp at its ends. We find the angle (θ) each panel makes with the x axis along with
the length of each such panel. These values are tabulated below.
Panel Number
Panel Length (b)
Mean Cpu
Mean Cpl
θ Upper (degree)
1
0.0630
-1.14
-0.01
42.93
2
0.0324
-1.92
0.57
18.15
3
0.0316
-1.70
0.31
12.94
4
0.0312
-1.55
0.14
9.52
5
0.2002
-1.33
-0.04
2.68
6
0.1078
-0.96
-0.11
-2.46
7
0.1080
-0.59
-0.07
-4.49
• We find Cl by summing (Cpl − Cpu) ∗ b ∗ cos(θ) for each panel pair.
• We find Cd by summing (Cpl − Cpu) ∗ b ∗ sin(θ) for each panel pair.
We end up with:
Cl = 0.7213
Cd = 0.0919
5
8
0.1083
-0.47
-0.09
-5.94
9
0.1085
-0.39
-0.10
-7.08
For U = 15m/s
Angle of Attack (Degrees)
2
4
6
8
10
12
Cl
0.2686
0.4670
0.6264
0.7213
0.7557
0.5590
Cd
0.0094
0.0330
0.0628
0.0919
0.1339
0.1189
Cl
0.1760
0.3960
0.5244
0.7393
0.8922
1.0829
Cd
0.0340
0.0776
0.0551
0.1034
0.1575
0.3493
For U = 20m/s
Angle of Attack (Degrees)
2
4
6
8
10
12
5
Conclusions:
For U = 15m/s and AOA = 8◦
Figure 6: Cp vs x Graph
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From the above graph we can see that the pressure above the airfoil is lesser than the
pressure below the airfoil. This makes sense as this pressure difference is what drives the
lift force.
Cl vs Alpha Graph
Figure 7: Cl vs Alpha Graph
From the graph we can clearly see that for U=15m/s the Cl starts to reduce after
a certain angle, indicating stall (flow separation). Whereas for U=20m/s the Cl keeps
increasing for the given alpha values we have observed. Stall for higher Reynold’s Number
flow occurs at a greater alpha than that for a lower Reynold’s Number flow. (here
Reynold’s Number is closely associated with U)
Cl and Cd in the same plot
U = 15m/s
Figure 8: Cl and Cd Vs Alpha
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U = 20m/s
Figure 9: Cl and Cd Vs Alpha
In the Cl vs Alpha Graphs, let us analyze the slope of the graph in the linear region.
We see that near the origin at the bottom left part of the graph, Cl changes by approximately 0.1 for 1 degree change in alpha. 1 degree translates to π/180 radians. So the
slope we get would be 0.1 * 180/π = 5.7325. Note that the theoretical value of this slope
should be 2π. 5.7325 is pretty close to 2π. So the experimental value holds good within
acceptable limits of deviations from theory.
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Sources of Error:
• The ports might not be connected the correct way.
• The Data Acquisition System might be faulty.
• All our relations and equations applied are valid for Infinite Aspect Ratio Airfoil.
So the Airfoil must be mounted wall to wall. If there are any gaps, it might lead to
3D flow effect like tip vortices being generated leading to different conclusions.
• Readings might have been noted down when zero errors have not been removed
from the Data Acquisition System.
• Limited number of pressure ports might lead to incomplete information about
what’s changing by how much especially near the trailing edge.
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