pressure distribution around a symmetrical aerofoil
TRANSCRIPT
Abstract
The relative velocity of the wind tunnel was found to be 19.44ms-1. The critical angle or stall angle is α =14⁰ for this experiment, when this point is reached, the relative wind approaching the aerofoil gradually stops to flow over the upper surface of the aerofoil resulting in turbulence being created towards the trailing edge of the aerofoil. The Reynolds’s number for this experiment is Re = 1.152 x 10-5 and comparisons will be made with the theoretical values.
Introduction
Basic aerodynamics principles suggest there are four forces acting on a plane whilst in flight as illustrated in Fig 1 below.
Fig1.
(http://www.waybuilder.net/sweethaven/aviation/aerodynamics01/lessonmain.asp?num=0204)
The following experiment is primarily concerned with the Lift force and in particular the effect it has on a cross-section of a wing of an aircraft known as an aerofoil. Fig 2 shows the wing of an aircraft on the left and a cross-sectional part of the wing on the right, known as an aerofoil.
Fig 2.
(Oertel, H et al, 2004, p268)
The Camber line is an imaginary line connecting the leading edge and trailing edge of the chord.
Angle of attack α, is the angle between the chord line L of an aerofoil and the vector representing the relative motion between the aerofoil and the air
A = lift, W= drag R= Resultant force
“In its simplest sense an aerofoil section may be defined as that profile designed to obtain a desirable reaction from the air through which it moves” (Dingle & Tooley, 2005, p548). Dingle & Tooley therefore imply that the lift force for flight is generated by converting air resistance.
“The purpose of this experiment is to measure the static pressure distribution around a symmetrical aerofoil, find the normal force and hence determine the lift-curve slope.” (ENGINEERING LABORATORY, INSTRUCTION BOOKLET)
The method to be used is a simulated environment using a T8 wind tunnel and manometer to calculate the static pressure at various points on the aerofoil for varying angles of attack (angles of incidence). Relative wind is generated by the wind tunnel to simulate the motion of the aerofoil travelling through air. Wind tunnel testing has provided factual experimental information to assist in verifying and supporting theoretically derived data about aerofoil, wing and an entire aircraft itself.
Theory
To determine the pressure distribution around the symmetrical aerofoil, Pressure difference will be taken from 15 different points along the aerofoil. The points are divided into further subsections. Points 1-7 represent the upper surface (US) of the aerofoil; Points 8-13 represent the lower surface (LS). Points 0 and 14 represent the leading and trailing edges respectively. This is illustrated by fig 3.
Fig 3.
Below are the measurements of each point (hole number) taken from the leading edge 0. Measurements are given in inches.
The pressure distribution on an aerofoil is calculated by resolving the value for Pressure Coefficient using the formula below and it can be shown that this value differs for various angles of incidence.
C = P−P∞12ρV 2
The equation can be rearranged using Bernoulli’s equation as shown below
Bernoullis Equation:
P∞ + 12ρV 2+ ρgz=C (ρgz ~ 0, C = PTOTAL = PATM)
P∞ + 12ρV 2 = PATM
Therefore , 12ρV 2 = PATM - P∞
Cp = P−P∞Patm−P∞ substituting P = ρgh
Cp = h−hsha−hs
Where
h = pressure at points(tappings) 0, 1, 2, 3, 4.........13, 14
hs= Static Pressure in the tunnel working section
ha= atmospheric pressure reading
h, hs, ha values are obtained from the readings on the manometer. Therefore Cp of all the tappings is calculated for each attack angle (angle of incidence).
There are 5 angles of attack (angles of incidence) α = -4, +1, +6, +11, +16
The equation of state is used to obtain the value for density of air in the lab, this value is necessary in order to enable to calculation of the tunnel speed
Equation of State
ρair = PlabRTlab
Plab = Barometer Reading for Lab Pressure
Tlab = Laboratory Temperature
R = Gas constant for air
Tunnel Speed is given as:
½ ρairV2=ρm g(hs - ha ) sinѲ
Where
g = acceleration due to gravity
ρair = density of air in the lab
ρm = density of manometer fluid
Ѳ = manometer inclination to the horizontal
Lift Coefficient
CL = L
12ρV 2 c
is given by:
The lift coefficient for each angle is used to calculate the lift force for each attack angle.
Experimental Arrangements
Apparatus- T8 Wind tunnel (City University Aero Lab)- Multi-tube manometer- Barometer- Thermometer- Symmetric Aerofoil of 3.5 inches chord length
WIND TUNNEL T8 AEROFOIL TEST SECTION
MANOMETERFig4.
The airfoil spans the tunnel test section and a multi-tube manometer is used to measure the pressure distribution of the wall tappings on the surface of the aerofoil for varying attack angles so that the corresponding values of lift coefficient can be calculated. In this experiment, an aerofoil is placed in the wind tunnel attached to a beam on one side of the test section. The airfoil’s angle of attack can be adjusted by rotating the mount which is connected to the beam holding the aerofoil. The tunnel will be run at a constant velocity while the angle of attack will be taken at -4, +1, +6, +11 and +16 respectively.
The manometer is inclined at an angle of 19.9 degrees from the horizontal.
Results
Graph 1 shows Cp vs x/c at -4 degree angle of attack (angle of incidence).
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
Pressure distribution at incidence = -4 Degrees
US LS
x/c
Cp
Graph 2 shows Cp vs x/c at +1 degree angle of attack (angle of incidence).
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
Pressure distribution at incidence = +1 Degrees
US LS
x/c
Cp
Graph 3 shows Cp vs x/c at +6 degree angle of attack (angle of incidence).
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-3-2.5
-2-1.5
-1-0.5
00.5
1
Pressure distribution at incidence = +6 Degrees
US LS
x/c
Cp
Graph 4 shows Cp vs x/c at +11 degree angle of attack (angle of incidence).
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
Pressure distribution at incidence = +11 Degrees
US LS
x/c
Cp
Graph 5 shows Cp vs x/c at +16 degree angle of attack (angle of incidence).
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
Pressure distribution at incidence = +16 Degrees
US LS
x/c
Cp
Graph 6 shows CL vs angle of attack (angle of incidence).
-5 0 5 10 15 20
-6
-4
-2
0
2
4
6
8
10
Lift Coefficient Vs Angle of Incidence
CLPolynomial (CL)
angle of incidence (α)
Lift C
oeffi
ecie
nt C
L
Graph 7 shows CL Theory and experimental vs angle of attack (angle of incidence).
-5 0 5 10 15 20
-15
-10
-5
0
5
10
15
20
Lift Coefficient Vs Angle of Incidence
CLCL Theory
angle of incidence (α)
Lift C
oeffi
cient
Graph 8 shows Lift force of point 1 on aerofoil vs angle of attack (angle of incidence).
-5 0 5 10 15 20
-60-40-20
020406080
100
Lift force of point(tapping) 1 Vs angle of in-cidence
L for point 1
Angle of incidence α
Lift F
orce
See appendix for calculations.
Discussion
From the results above it can be seen clearly that as the attack angle (angle of incidence) is increasing to a positive value the lift component increases up to a maximum value. Looking at the Lift coefficient Vs angle of incidence graph this value is α =14⁰ for this experiment and the maximum lift coefficient is 7.5. This value is known as the stall point. In textbooks this point is also referred to as the critical angle or stall angle. Graph 8 shows that a maximum lift force of about 85 N (Newtons) is reached at angle of incidence of α ≈14⁰ for point 1 on the aerofoil.
Typical pressure distributions on an aerofoil
Fig 5.
(Houghton E.L. & Carpenter P.W, 2003, p29)
Using Fig 5 for illustration purpose it can be clearly seen that as the angle of attack increases so does the pressure on the upper surface of the aerofoil. Also as the angle of attack increases the lift coefficient at stagnation shifts from the trailing edge to the lower surface causing a stall. When the stall point is reached, the relative wind approaching the aerofoil gradually stops to flow over the upper surface of the aerofoil resulting in turbulence being created towards the trailing edge of the aerofoil.
The theoretical Reynolds number is in a range from 120,000 to 400,000, and the Reynolds number in this experiment is 115214.4. The transition from graph 1 to graph 2 shows as the angle of incidence moves from -4 to +1 the pressure on the upper surface is proportional to
the pressure on the lower surface. Graph 2 to graph 3 shows the change in pressure on the upper surface increases from angle of incidence of +1 to +6.
The Pressure distribution on the upper surface is seen to approach maximum in graph 4 with a value of -2.75 however there is a rapid decrease in the value of the pressure distribution on the upper surface to -1.7 as shown in graph 5 indicating the stall point has been reached. The theoretical lift curve slope for graph 6 can only be used upto the stall point, which theoretically is between 10⁰ and 15⁰. Graph 7 shows the theoretical lift curve slope to have a value of 0.1097 per degree however the experimental value is
As the angle of incidence increases the value of Cp increases from 0.031 at -4 angle of attack to 0.625 at + 16 angle of attack. Looking at graph 6 it can be seen the aerofoil has no lift at α ≈1⁰. Both Random uncertainties as well as systematic uncertainties occurred. To better the experiment i would have taken and accounted for these errors in my calculations. The values obtained have been rounded up or down above and below the ‘true value’ respectively. I would have used a computer or another instrument to offset the uncertainties, which in turn will provide more accurate results.
Conclusion
Relative wind passing over the top of the aerofoil produces aerodynamic. The shape of the airfoil causes a low pressure area above the aerofoil according to Bernoulli's Equation, and the decrease in pressure on top of the aerofoil creates an upward force. It is apparent from the experiment that an aerofoil has lift only if the pressure is higher on the lower surface than the upper surface. It can be concluded that:
(i) At low angle of incidence the lift is caused by the difference between the pressure on the upper surface and lower surface(ii) At higher incidences the lift is partly due to pressure reduction on the upper surface(iii) The pressure reduction on the upper surface increases both in intensity and extentuntil, at large angle of incidence.(iv) The stagnation point moves progressively further back on the lower surface andthe increased pressure on the lower surface covers a greater proportion of thesurface.
References
Dingle,L & Tooley,M (2005) Aircraft Engineering Principles. Butterworth-Heineman
Fundamentals of Aerodynamics (2008) Forces acting on an aircraft Avaliable at:http://www.waybuilder.net/sweethaven/aviation/aerodynamics01/lessonmain.asp?num=0204 (Accessed: 18 November 2008)
Houghton E.L & Carpenter P.W (2003) Aerodynamics for Engineering Students Butterworth-Heineman
Oertel, H, Prandtl,L, Böhle,M, & Mayes, K (2004) Prandtl's Essentials of Fluid Mechanics. Springer
Appendix
RAW DATA FROM EXPERIMENT
Attack Angle, α -4 +1 +6 +11 +16Tube No. Pressure Point on aerofoil Pressure Value (inches)1 Atmospheric pressure 4.2 4.2 4.2 4.2 4.22 0 6.8 4.9 5.9 10.6 8.43 1 6.8 9.3 11.7 15.5 10.54 8 12.4 8.4 6.1 4.9 5.05 2 5.3 7.8 11.4 15.9 12.76 9 11.4 9.1 7.5 6.1 5.97 3 7.7 9.2 10.5 11.7 10.38 10 10.7 9.5 8.6 7.6 7.59 4 8.3 9.1 10.0 10.2 10.210 11 10.2 9.3 8.6 7.9 7.811 5 8.3 8.9 9.5 9.5 10.112 12 9.4 8.8 8.6 8.2 8.413 6 8.1 8.5 8.2 8.7 10.414 13 8.4 8.4 8.1 8.0 8.515 7 7.8 7.9 8.0 8.1 9.916 14 7.7 7.5 7.7 7.9 9.431 Reference Pressure 7.4 7.4 7.4 7.4 7.4
Pressure Coefficient:
C = P−P∞12ρV 2
Bernoullis Equation:
P∞ + 12ρV 2+ ρgz=C (ρgz ~ 0, C = PTOTAL = PATM)
P∞ + 12ρV 2 = PATM
Therefore , 12ρV 2 = PATM - P∞
Cp = P−P∞Patm−P∞
substituting P = ρgh
Cp = h−hsha−hs
c = chord length = 3.5” =0.0889m
h = pressure at points0, 1, 2, 3, 4.........13, 14
Convert inches to metres: inches * 0.0254 = metres
hs= Static Pressure in the tunnel working section = reference pressure (tube 31) = 7.4inches = 0.188
ha= atmospheric pressure reading = atmospheric pressure (tube 1) = 4.2 inches = 0.1067m
hs, ha values are in meters
Cp is calculated for each point of the aerofoil and for each angle of attack (angle of incidence). There are 5 angles of attack (angle of incidence), α = -4, +1, +6, +11, +16
Foil number
x (distance from leading edge) x/c α =(-4) α =(+1) α =(+6) α = (+11) α = (+16)
0 0 0 0.1875 0.78125 0.46875 -1 -0.3125
10.05
0.014286 0.1875-
0.59375-
1.34375 -2.53125 -0.968752 0.18 0.051429 0.65625 -0.125 -1.25 -2.65625 -1.65625
30.53
0.151429-
0.09375 -0.5625-
0.96875 -1.34375 -0.90625
41.23
0.351429-
0.28125-
0.53125 -0.8125 -0.875 -0.875
51.75
0.5-
0.28125-
0.46875-
0.65625 -0.65625 -0.84375
62.45
0.7-
0.21875-
0.34375 -0.25 -0.40625 -0.9375
73.15
0.9 -0.125-
0.15625 -0.1875 -0.21875 -0.781258 0.09 0.025714 -1.5625 -0.3125 0.40625 0.78125 0.75
90.26
0.074286 -1.25-
0.53125-
0.03125 0.40625 0.46875
100.88
0.251429-
1.03125-
0.65625 -0.375 -0.0625 -0.03125
111.49
0.425714 -0.875-
0.59375 -0.375 -0.15625 -0.12512 2.1 0.6 -0.625 -0.4375 -0.375 -0.25 -0.3125
132.8
0.8 -0.3125 -0.3125-
0.21875 -0.1875 -0.34375
143.5
1-
0.09375-
0.03125-
0.09375 -0.15625 -0.625
Equation of State
ρair = PlabRTlab
Plab = Barometer Reading for Lab Pressure = 1010mb = 101kPa
Tlab = Laboratory Temperature = 22⁰C + 273 = 295K
R = Gas constant for air = 287 JKg-1K-1
Substituting values into the equation of state
ρair = 101000/ (295 * 287)
ρair = 1.193 kgm-3
Tunnel Speed is given as:
½ ρairV2=ρm g(hs - ha ) sinѲ
ρair = density of air in the lab = 1.193 kgm-3
ρm = density of manometer fluid = 830 kgm-3
Ѳ = manometer inclination to the horizontal = 19.9⁰
Therefore rearranging equation gives tunnel speed V =√ ( 2 ρmg(h s−ha)sinѲ❑ρair )
Substituting values into the equation for tunnel speed
V=√{( 2 * 830 kgm-3 *9.81ms-2 *(0.1880m -0.1067m)*sin19.9)/1.193}
V = 19.44ms-1
Lift Coefficient
CL = L
12ρV 2 c
is given by:
Using a variety of methods such as the trapezium rule etc the value of Lift coefficient CL can be calculated for each attack angle (angle of incidence) by getting an approximation for the area between the upper surface and lower surface on each graph. Theoretical value of CL is given as 2π per radian.
α CL CL Theory-4 -4.55 -12.5664
1 0.3553.14159
3
6 3.52518.8495
611 7.175 16 7.325
By rearranging the equation L (lift force) can be calculated
CL = L
12ρV 2 c
αL for point 1
-4 -51.28431 4.0013026 39.73124
11 80.8713816 82.56208
Reynolds number:
Re = Vc/ vair
V= tunnel speed = 19.44 ms-1
c= chord length = 0.0889
vair = Kinematic viscosity of air = 1.5 x 10-5 m2s-1
Substituting values into equation
Re = 19.44 ms-1 * 0.0889m/ 1.5 x 10-5 m2s-1
Re = 115214.4