MANUFACTURING METHODSABRAHAM AUDU Group A2 [email protected] 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.Abstract The relative velocity of the wind tunnel was found to be 19.waybuilder. known as an aerofoil. when this point is reached.152 x 10-5 and comparisons will be made with the theoretical values. Fig 2 shows the wing of an aircraft on the left and a cross-sectional part of the wing on the right. 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. Fig1.44ms-1. Introduction Basic aerodynamics principles suggest there are four forces acting on a plane whilst in flight as illustrated in Fig 1 below. The critical angle or stall angle is α =14: for this experiment. . (http://www. The Reynolds’s number for this experiment is Re = 1.net/sweethaven/aviation/aerodynamics01/lessonmain. ” (ENGINEERING LABORATORY. (Oertel. 2004.Fig 2. 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. Dingle & Tooley therefore imply that the lift force for flight is generated by converting air resistance. p548). “The purpose of this experiment is to measure the static pressure distribution around a symmetrical aerofoil. 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. 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). find the normal force and hence determine the lift-curve slope. . 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. H et al. Angle of attack α. wing and an entire aircraft itself. p268) The Camber line is an imaginary line connecting the leading edge and trailing edge of the chord. 2005. Measurements are given in inches.Theory To determine the pressure distribution around the symmetrical aerofoil. Points 8-13 represent the lower surface (LS). Fig 3. . Points 0 and 14 represent the leading and trailing edges respectively. This is illustrated by fig 3. Below are the measurements of each point (hole number) taken from the leading edge 0. Points 1-7 represent the upper surface (US) of the aerofoil. Pressure difference will be taken from 15 different points along the aerofoil. The points are divided into further subsections. 1.. +11. ha values are obtained from the readings on the manometer... 4.. 14 hs= Static Pressure in the tunnel working section ha= atmospheric pressure reading h. 2.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. +16 . C = PTOTAL = PATM) substituting P = ρgh Where h = pressure at points(tappings) 0. There are 5 angles of attack (angles of incidence) α = -4. 3.P∞ Cp = Cp = (ρgz ~ 0. hs. C= ⁄ The equation can be rearranged using Bernoulli’s equation as shown below Bernoullis Equation: P∞ + P∞ + = PATM = PATM .13. +1.... Therefore Cp of all the tappings is calculated for each attack angle (angle of incidence). +6.. this value is necessary in order to enable to calculation of the tunnel speed Equation of State ρair = 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 = ⁄ is given by: The lift coefficient for each angle is used to calculate the lift force for each attack angle.The equation of state is used to obtain the value for density of air in the lab. . The manometer is inclined at an angle of 19. an aerofoil is placed in the wind tunnel attached to a beam on one side of the test section. 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.Barometer . +6.T8 Wind tunnel (City University Aero Lab) .Symmetric Aerofoil of 3. In this experiment. The tunnel will be run at a constant velocity while the angle of attack will be taken at -4. +1.5 inches chord length WIND TUNNEL T8 AEROFOIL TEST SECTION MANOMETER Fig4.Multi-tube manometer .9 degrees from the horizontal. +11 and +16 respectively. .Experimental Arrangements Apparatus . The airfoil’s angle of attack can be adjusted by rotating the mount which is connected to the beam holding the aerofoil.Thermometer . 6 0.8 0.7 0.7 0.9 1 Graph 2 shows Cp vs x/c at +1 degree angle of attack (angle of incidence).5 -2 -1.3 0.2 0.2 0. Pressure distribution at incidence = +1 Degrees -3 -2.1 0.4 0.5 0 0.5 Cp -1 -0.5 1 US LS x/c 0 0. Pressure distribution at incidence = -4 Degrees -3 -2.Results Graph 1 shows Cp vs x/c at -4 degree angle of attack (angle of incidence).5 1 US LS 0.9 x/c 1 .5 0 0 0.8 0.3 0.5 0.1 0.5 0.5 Cp -1 -0.4 0.6 0.5 -2 -1. Graph 3 shows Cp vs x/c at +6 degree angle of attack (angle of incidence).8 0.1 0.5 1 US LS x/c 0 0.4 0.5 0.5 Cp -1 -0.9 1 Graph 4 shows Cp vs x/c at +11 degree angle of attack (angle of incidence).7 0.6 0.7 0.5 -2 -1. Pressure distribution at incidence = +6 Degrees -3 -2 Cp -1 0 0 x/c 1 US LS 0.6 0.5 0.3 0.9 1 .1 0.2 0. Pressure distribution at incidence = +11 Degrees -3 -2.8 0.2 0.4 0.5 0 0.3 0. 5 0 0 0.8 0.Graph 5 shows Cp vs x/c at +16 degree angle of attack (angle of incidence).5 1 US LS 0.5 -2 -1. (CL) -5 . Lift Coefficient Vs Angle of Incidence 10 8 Lift Coeffiecient CL 6 4 2 0 -2 -4 -6 0 5 10 15 angle of incidence (α) 20 CL Poly.7 0.4 0.1 0.5 Cp -1 -0.3 0.2 0. Pressure distribution at incidence = +16 Degrees -3 -2.5 0.9 x/c 1 Graph 6 shows CL vs angle of attack (angle of incidence).6 0. Lift force of point(tapping) 1 Vs angle of incidence 100 80 Lift Force 60 40 20 0 -20 0 -40 -60 Angle of incidence α 5 10 15 20 L for point 1 -5 See appendix for calculations.Graph 7 shows CL Theory and experimental vs angle of attack (angle of incidence). . 20 Lift Coefficient Vs Angle of Incidence 15 10 Lift Coefficient 5 CL 0 -5 -5 0 5 10 15 20 CL Theory angle of incidence (α) -10 -15 Graph 8 shows Lift force of point 1 on aerofoil vs angle of attack (angle of incidence). 4. 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.L.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.W. & Carpenter P. and the Reynolds number in this experiment is 115214. When the stall point is reached. Graph 2 to graph 3 shows the change in pressure on the upper surface increases from angle of incidence of +1 to +6. 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. 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. Looking at the Lift coefficient Vs angle of incidence graph this value is α =14: for this experiment and the maximum lift coefficient is 7. In textbooks this point is also referred to as the critical angle or stall angle.000 to 400. This value is known as the stall point. (Houghton E. Also as the angle of attack increases the lift coefficient at stagnation shifts from the trailing edge to the lower surface causing a stall. . 2003. Typical pressure distributions on an aerofoil Fig 5.000.5. Graph 8 shows that a maximum lift force of about 85 N (Newtons) is reached at angle of incidence of α ≈14: for point 1 on the aerofoil. The theoretical Reynolds number is in a range from 120. which theoretically is between 10: and 15:. Graph 7 shows the theoretical lift curve slope to have a value of 0.031 at -4 angle of attack to 0. Both Random uncertainties as well as systematic uncertainties occurred.7 as shown in graph 5 indicating the stall point has been reached. which in turn will provide more accurate results. I would have used a computer or another instrument to offset the uncertainties. 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 extent until. Looking at graph 6 it can be seen the aerofoil has no lift at α ≈1:. and the decrease in pressure on top of the aerofoil creates an upward force. The values obtained have been rounded up or down above and below the ‘true value’ respectively. The theoretical lift curve slope for graph 6 can only be used upto the stall point. . The shape of the airfoil causes a low pressure area above the aerofoil according to Bernoulli's Equation. (iv) The stagnation point moves progressively further back on the lower surface and the increased pressure on the lower surface covers a greater proportion of the surface.75 however there is a rapid decrease in the value of the pressure distribution on the upper surface to -1. To better the experiment i would have taken and accounted for these errors in my calculations. Conclusion Relative wind passing over the top of the aerofoil produces aerodynamic.625 at + 16 angle of attack.1097 per degree however the experimental value is As the angle of incidence increases the value of Cp increases from 0. 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.The Pressure distribution on the upper surface is seen to approach maximum in graph 4 with a value of -2. at large angle of incidence. 3 7.9 6.0 8.2 8.7 7. K (2004) Prandtl's Essentials of Fluid Mechanics.3 7.6 15.1 11.5 8. & Mayes.4 7.8 8.5 5.L.9 9.2 4.5 7.1 7. Böhle.9 7.L & Carpenter P.2 10.7 7.1 7.5 8.7 7.4 10.5 10.1 7.4 10. Butterworth-Heineman Fundamentals of Aerodynamics (2008) Forces acting on an aircraft Avaliable at: http://www.2 4.1 8.0 7.2 7.4 8.9 5.0 12.2 10. Prandtl. Springer Appendix RAW DATA FROM EXPERIMENT Tube No.9 8.8 10.8 9.1 10.9 15.7 12.waybuilder.4 8.8 7.6 8.3 9.References Dingle.9 6.0 10.M.3 11.6 8.4 6.W (2003) Aerodynamics for Engineering Students Butterworth-Heineman Oertel.7 9.2 8.3 8.6 10.2 7. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 31 Attack Angle.4 7.1 8.L & Tooley.9 9.5 4. α Pressure Point on aerofoil Atmospheric pressure 0 1 8 2 9 3 10 4 11 5 12 6 13 7 14 Reference Pressure -4 +1 +6 Pressure Value (inches) 4.5 9.5 9.5 10.8 4.4 7.4 .1 5.7 5.2 6.4 11.4 8.4 +16 4.9 9.2 9.5 7.4 8.5 8. H.net/sweethaven/aviation/aerodynamics01/lessonmain.4 +11 4.3 8.6 8.M (2005) Aircraft Engineering Principles.8 11.9 10.4 8.4 9.asp?num=0204 (Accessed: 18 November 2008) Houghton E.2 8.7 9.7 8. 0889m h = pressure at points0. 3. 4.5” =0..P∞ Cp = Cp = (ρgz ~ 0.. +6..Pressure Coefficient: C= ⁄ Bernoullis Equation: P∞ + P∞ + = PATM = PATM . 14 Convert inches to metres: inches * 0. α = -4. +11.188 ha= atmospheric pressure reading = atmospheric pressure (tube 1) = 4.. 1.0254 = metres hs= Static Pressure in the tunnel working section = reference pressure (tube 31) = 7. 2.1067m hs.13..2 inches = 0.4inches = 0... +1. There are 5 angles of attack (angle of incidence).. ha values are in meters Cp is calculated for each point of the aerofoil and for each angle of attack (angle of incidence). C = PTOTAL = PATM) substituting P = ρgh c = chord length = 3. +16 . 96875 0.5625 -1.25 0.05 x/c 0 α =(-4) 0.074286 0.025714 0.014286 0.25 -0.1875 0.84375 -0.21875 0.65625 0.65625 0.625 -0.125 -1.78125 0.3125 0.23 0.45 0.0625 -0.15625 -0.5 2.40625 0.09375 0.875 -0.3125 0.875 -0.46875 0.375 -0.96875 -1.375 0.90625 -0.125 -1.5 1 α =(+1) α =(+6) 0.03125 0.21875 0.9375 -0.75 0.46875 -0.875 -0.03125 0.34375 -0.53 0.15625 α = (+16) -0.46875 0.18 0.25 1.75 0.8 0.1875 -0.25 -0.34375 -0.7 3.8125 0.1875 -0.1 0.26 0.051429 0.Foil number 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 x (distance from leading edge) 0 0.3125 0.251429 1.351429 1.59375 1.40625 -0.78125 0.5625 0.65625 -1.88 0.49 0.53125 -2.53125 -0.3125 -0.34375 -0.15625 -0.03125 -0.151429 1.4375 -0.15 0.3125 -0.34375 -0.28125 0.625 .09375 α = (+11) -1 -2.425714 2.65625 -0.65625 -0.59375 -0.125 -0.03125 -0.8 3.78125 0.9 0.65625 -0.1875 0.21875 -0.53125 0.40625 -0.09 0.09375 0.375 -0.6 2.28125 0. 81ms-2 *(0.1067m)*sin19.9)/1.193 kgm-3 Tunnel Speed is given as: ½ ρairV2=ρm g(hs .1880m -0.193} V = 19.ha ) sinѲ ρair = density of air in the lab = 1.Equation of State ρair = Plab = Barometer Reading for Lab Pressure = 1010mb = 101kPa Tlab = Laboratory Temperature R = Gas constant for air = 22:C + 273 = 295K = 287 JKg-1K-1 Substituting values into the equation of state ρair = 101000/ (295 * 287) ρair = 1.9: Therefore rearranging equation gives tunnel speed V = ( Substituting values into the equation for tunnel speed V=√{( 2 * 830 kgm-3 *9.44ms-1 Lift Coefficient CL = ⁄ ) is given by: .193 kgm-3 ρm = density of manometer fluid = 830 kgm-3 Ѳ = manometer inclination to the horizontal = 19. 0889 vair = Kinematic viscosity of air = 1. Theoretical value of CL is given as 2π per radian.56208 Reynolds number: Re = Vc/ vair V= tunnel speed = 19.141593 3.2843 4.525 18.4 -5 .55 -12.73124 80.5 x 10-5 m2s-1 Substituting values into equation Re = 19.325 By rearranging the equation L (lift force) can be calculated CL = ⁄ α -4 1 6 11 16 L for point 1 -51.44 ms-1 * 0.001302 39.175 7.84956 7.0889m/ 1.87138 82.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.5 x 10 m2s-1 Re = 115214.44 ms-1 c= chord length = 0.355 3.5664 0. α -4 1 6 11 16 CL CL Theory -4.
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