S = the wing area, item e, in square metres
The diagrams are in figure 4 and 5.
Angle of Attack
The angle of attack of an aerofoil is the angle α between the aerofoil and the relative wind. Ideally, the direction of the wind is measured in the free stream, i.e. far enough away that it is undisturbed by the aerofoil. the angle α is measured relative to some reference on the aerofoil, and common choices of reference include
- The direction that produces zero lift, or
- The chord line of the aerofoil.
The diagrams are in figure 6.
Lift Coefficient
The lift coefficient is a number that aerodynamicists use to model all of the complex dependencies of shape, inclination and some flow conditions on lift. It is incorporated in the lift equation to predict the lift force generated by a wing using this particular cross-section.
The lift coefficient Cl is equal to:
Where,
L = lift force, ρ = density, V = speed and A = Area
Noticing that the lift equation does not include terms for angle of attack.
The diagram is in figure 7.
Drag
For a solid object moving through a fluid or gas, drag is the sum of all the aerodynamic or hydrodynamic forces in the direction of external fluid flow. It therefore acts to oppose the motion of the object, and in a powered vehicle it is overcome by thrust.
Types of drag are generally divided into three categories. And they are : a) Parasitic drag, b) lift-induced drag and c) wave drag.
In this experiment only lift-induced drag was used which is only relevant when wings or a lifting body is present.
The diagram is in figure 8.
Stagnation Point
The stagnation point is a theoretically assigned point in a flow where the velocity is zero, where any streamline touches a solid surface at an angle. In the example in figure 9 there are two stagnation points just outside the the fluid.
Discussion
From the pressure coefficient plots it can be found the stagnation point () and the table is as follows:
Table: Stagnation Point
From the above table it can be seen that for the angle of attack 11 and 16 stagnation points were almost similar and these points occurred at tube 16 of hole 2. But in those points the angle of attack 11 and 16 showed two different characteristics. The lift on the wing is usually due to the pressure difference between the upper surface and lower surface of the wing. As the angle of attack (α) increases an unsymmetrical flow pattern of high velocity is produced, the boundary layers are close together which causes low pressure on the upper surface and high pressure on the lower surface, which causes an increase in the lift force .
Hence from the lift curve slope it can be seen that from angle of attack 1 to 11 the lift was because of low pressure on the upper surface and high pressure on the lower surface while during the angle of attack 16 due to high pressure on the upper surface and low pressure on the bottom there was no lift. And it showed the characteristic of a Formula One car.
During the experiment no flow separation point was found i.e. the pressure distribution on the upper surface never became constant.
The value of at peak suction (i.e. negative value) and at the trailing edge is in the table below:
Table: Value of Cp at peak suction and trailing edge
The pressure coefficient for angles of attack (-4, 1, 6, 11) increased at the trailing edge and so was the lift force. But at angle of attack 16 there was a dramatically break down in the flow pattern which caused sudden drop in the lift force and causing stall. And this whole point of dramatically change was the pressure on the upper surface was very high near the trailing edge
After studying the lift curve it had been found that the angle of attack for no lift and for stall point was 16 degree. The reason behind this matter was high pressure on the upper surface and low pressure on the bottom that caused a decrease in lift force i.e. no lift. The lift curve slope was determined by plotting the lift coefficient against the angle of attack. Theoretically the gradient had been found out as 0.1 while in practice it was 0.080 which was quite close to 0.1.
The difference between the static pressure (hs) in tube 31 and the atmospheric pressure (ha) in tube 1for all angle of attacks was 0.1056 m. The wind tunnel speed was calculated as 24.58 m/s. Hence the Reynolds number was found to be 1.38*10^5. Comparing this value with the standard one was very small. Due to a low speed wind tunnel the range varies from 10^3 to 10^6.
Conclusion
Concluding this experiment the graph showed that increasing the angle of attack increases the area of the graph and the lift coefficient but only up to certain stage. There was a linear part on the graph, which can be used to find out the lift curve slope and it flattened just before the stall point. And it also can be proved that for this aerofoil 16 degrees angle of attack is critical. This experiment had been conducted successfully about 80%accuracy.
The accuracy of results may have been affected due to the group separation during the experiment. Errors occurred while exchanging results not noting them down properly. The readings were taken manually off the manometer. This could have been prevented if a digital display was available. The manometer was kept at an angle this could have effected the flow of the fluid through the tubes. Therefore before starting the experiment the zero data should be set and the angle should be measured accurately. Another cause for inaccuracy was less angle of attack measured, whether a few more angle of attacks might give better knowledge of this experiment.
References
1. John J. Bertin Aerodynamics For Engineers, Second edition. Prentice – Hall International Editions, 1989.
2. Dale Crane Dictionary of Aeronautical Terms, Third edition. Aviation Supplies & Academics, Inc., 1997.
3. www.google.com
Appendix
How to calculate the Pressure Coefficients.
Cp = (h – hs) / (ha – hs)
Cp pressure coefficient, it has no unit (hs)= static pressure (m)
h = readings for the tapping (m) ha = atmospheric pressure (m)
EXAMPLE: -4 angle of attack, at tube 3 or hole 1
h = 5.9 (m) hs = 6.9 (m) ha = 2.5 (m)
Cp = (5.9 – 6.9) / (2.5 –6.9) = 0.227 (no units)
This method is used to calculate each and every Pressure Coefficient, as shown in the results table. This helped us plot the graph of Cp against x/c.
How to get x/c
x = distance of tapping (inc) c = chord length (inc)
EXAMPLE: -4 angle of attack at, tube 3 or hole 1
x = 0.05 (inc) c = 3.5 (inc)
Calculating x/c = 0.05/3.5 = 0.014
This method is used to calculate each distance per unit chord length in the result table.
The Lift Coefficient (Cl) is worked out on a graph (Cp) against x/c by counting the number of squares and multiplying it by the area.
Cl = Lift Coefficient (no units) A = Area (m2)
N = number of squares
EXAMPLE: -4 angle of attack graph Cp against x/c
Cl = Area x Number of squares
Cl = 0.0672 x 6.5
Cl = -0.4368 (no units)
Negative sign cause α < 0. All (Cl) values are calculated in the same way, see the results table.
Calculating Tunnel Speed
½ ρU2 = ρmg (hs – ha) sinφ
U = tunnel speed (m/s) φ = inclination of the manometer 25.6(deg)
ρm = density of the manometer fluid (1.23 kg/m3) ρ= density of air (830kg/m3)
hs = static pressure (m) (6.9 inc) ha = atmospheric pressure (m) (2.5 inc)
(hs – ha) should be in meters (6.9 – 2.5) = 4.4 (inc) x 0.024 = 0.1056 (m)
U2 = 2ρmg (hs – ha) sinφ / ρ
U2 = 2 x 830x 9.81( 0.1056) 0.4320 / 1.23
U = 24.58 (m/s)
Calculating Reynolds Number
RE = Reynolds number U = Tunnel Speed (24.85m/s)
vair = viscosity of air( 1.5x10-5 ) c = chord length (3.5inc)
Convert chord length in to meters = (0.084m)
RE = Uc / νair
RE = 24.58 x 0.084 / 1.5x10-5
RE = 1.38 x 105
Figures and Graphs
Figure 2: Aerofoil
Figure 3: Aerofoil
Figure 4: Lift
Figure 5: Lift
Figure 6: Angle of Attack
Figure 7: lift coefficient
Figure 8: Drag
Figure 9: Stagnation point
Figure 0: Wind Tunnel Model
Figure 1: Ames Research Centre Wind Tunnels