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For most objects moving through a fluid, the significant fluid force is drag.

LIFT. For most objects moving through a fluid, the significant fluid force is drag. However for some specially shaped objects the lift force is also important. LIFT. DRAG. TIME OUT. Newtonian Theory (1687) entire second book of Principia dedicated to fluid mechanics

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For most objects moving through a fluid, the significant fluid force is drag.

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  1. LIFT For most objects moving through a fluid, the significant fluid force is drag. However for some specially shaped objects the lift force is also important. LIFT DRAG

  2. TIME OUT

  3. Newtonian Theory (1687) entire second book of Principia dedicated to fluid mechanics - assumed particles of fluid lose momentum normal to plate but keep momentum parallel to plate. p due to random motion of molecules p  p  Aside

  4. Force normal to plate, N = Time rate of change of the normal component of momentum = (mass flow) x change in normal component of velocity = N = ( V A sin) x (V sin) N/A = p – p  = (V sin)2 (p - p ) / (1/2  V 2) = Cp = 2 sin2 Cp = 1.86 as M  Aside

  5. Lift force is the component of R that is perpendicular to free stream velocity, and drag is the component of R parallel to the free stream velocity. If planes height is not changing then: Lift = Weight Forces on airplane at level speed and constant height and speed.

  6. Note: FL is not parallel to N FL FD CL = FL/(1/2 V2Ap) CD = FD /(1/2  V2A) Force generated if we brought fluid directly approaching area to rest

  7. Chord of an airfoil is the straight line joining the leading and trailing edge. If mean line and chord do not coincide then airfoil is cambered. Ap refers to planform area, maximum projection of wing (independent of angle of attack, )

  8. CL = FL/(1/2  V2Ap) CD = FD/(1/2  V2A) Ap = planform area max. proj. of wing Ap and c are independent of  A

  9. Given: Kite in standard air, mass = 0.2 kg; CL = 2sin(); CL/CD = 4. Find  Area = 1 m2 5o 0.2kg (g) U= 10 m/s  = ?

  10. Kite in standard air, CL = 2sin(); CL/CD = 4 FL U FD mg CL = FL/(1/2 V2Ap) CD = FD /(1/2  V2A) CL/CD = FL/FD = 4 (Ap = A) T  = ?

  11. 9.144 FL U FD mg T Fy = FL – mg –Tsin() = 0 Fx = FD –Tcos() = 0 FL = CL A ½  U2 CL = 2sin(5o) = 0.548 FL = 33.7N FD = FL/4 = 8.43N  = ?

  12. FL U FD mg T  = ? tan () = Tsin()/Tcos() = (FL – mg)/FD  = tan-1{{(FL – mg)/FD} = 75.1o

  13. IMPORTANCE OF CAMBER flat plate bent plate airfoil For all cases angle of attack is 4o and aspect ratio is 6. Lift to drag ratios of about 20 are common for modern transport planes.

  14. Otto Lilienthal (1848-96) is universally recognized as as the first flying human. His wings were curved (camber – height = 1/12th of chord). On August 9th, 1896 Lilienthal suffered a fatal spinal injury, falling 10-15 meters from the sky. Otto Lilienthal on a biplane glider in 1893 Otto Lilienthal on a monoplane glider in 1893

  15. Lift = U

  16. ASIDE “In order for lift to be generated there must be a net circulation around the profile.” Pg 448 ~ our book 2-D potential flow Lper unit span = U  = C vds

  17. ASIDE INVISCID INCOMPRESSIBLE FLOW AROUND AN 2-D CYLINDER

  18. ASIDE p + (/2)U2 = p + (/2)(2U sin)2 ON SURFACE FREE STREAM (2U sin) comes from 2-D potential flow for a source, sink and free stream flows

  19. ASIDE Governing equation for incompressible, irrotational flow: •V = 0;  x V = 0; V = ; 2 = 0 Laplace’s Equation, 2 = 0, is a second order, linear partial differential equation. Solutions can be added! Turns out that a 2-D source, 2-D sink and free stream makes a flow like that over a 2-D ellipse.

  20. Potential flow solutions: Source + Sink + Uniform flow m/(Ua) =1 ASIDE

  21. ASIDE Doublet: 2(a)m = constant as a shrinks to zero

  22. ASIDE add circulation: u = ro  = C(v+u)ds = 2rou p + (/2)U2 = p + (/2)(2U sin + ro)2 Lift is integral of pressure around cylinder

  23. ASIDE  = C vds = ro2ro; Lper unit span = U

  24. Bernoulli’s Equation

  25. We have just shown that if there is a net circulation, then there is lift. SO WHAT IS WRONG WITH THIS PICTURE TAKEN FROM A POPULAR TEXTBOOK?

  26. Both figures claim lift, which figure’s streamlines are consistent with lift?

  27. U = 4 m/s R = 7.7 cm Re = 4 x 104  = 0 U = 4 m/s R = 7.7 cm Re = 4 x 104  = 4U/R

  28. Both develop lift, see streamlines pinched on top (faster speeds, lower pressure) and wider on bottom (lower speeds and higher pressure)

  29. A CONSEQUENCE OF CIRCULATION AROUND WING IS STARTING VORTEX Kelvin’s theorem showed that the circulation around any closed curve in an inviscid, isentropic fluid is zero. Consequently there must be circulation around the airfoil in which the magnitude is the same as and whose rotation is opposite to that of the starting vortex.

  30. “Trailing vortices can be very strong and persistent, possibly being a hazard to other aircraft for 5 to 10 miles behind a large plane – air speeds of greater than 200 miles have been measured.” U = 30 cm/s Chord = 180 mm Re = 5 x 105 Floating tracer method Starting vortex

  31. Bernoulli’s Equation

  32. “The phenomenon of aerodynamic list is commonly explained by the velocity increase causing pressure to decrease (Bernoulli effect) over the top surface of the airfoil..” ~ YOUR BOOK PG 448 HINT: SAME THING THAT IS WRONG WITH THIS PICTURE?

  33. “In spite of popular support, Bernoulli’s Theorem is not responsible for the lift on an airplane wing.” “The phenomenon of aerodynamic list is commonly explained by the velocity increase causing pressure to decrease (Bernoulli effect) over the top surface of the airfoil..” ~ YOUR BOOK PG 448 Norman Smith: Physics Teacher, Nov. 1972, pg 451-455.

  34. Lift is a result of Newton’s 3rd law. Lift must accompany a deflection of air downward.

  35. BERNOULLI EQUATION GOOD FOR STREAM TUBES WHERE ENERGY IS NOT BEING ADDED OR SUBTRACTED Yet one can argue that B.E. is valid for outer stream tubes so book not wrong.

  36. BERNOULLI’S EQUATION X-MOMENTUM EQUATION: INVISCID and NO BODY FORCES: (Du[t,x,y,z]/dt) = - p/x (u/t) + u(u/x) + v(u/y) + w(u/z) = - p/x STEADY: u(u/x) + v(u/y) + w(u/z) = - p/x dx[u(u/x) + v(u/y) + w(u/z) = - p/x] aside

  37. BERNOULLI’S EQUATION CONSIDER FLOW ALONG A STREAMLINE: ds x V = 0 udz-wdx = 0; vdx-udy = 0 u(u/x)dx + v(u/y)dx + w(u/z)dx = - p/xdx u(u/x)dx + u(u/y)dy + u(u/z)dz = - p/x dx aside

  38. BERNOULLI’S EQUATION u(u/x)dx + u(u/y)dy + u(u/z)dz = - p/x dx u{(u/x)dx + (u/y)dy + (u/z)dz} = - (1/)p/x dx du udu - (1/) p/x dx ½ d(u2) = - (1/) p/x dx aside

  39. BERNOULLI’S EQUATION X-MOMENTUM EQUATION: ½ d(u2) = - (1/) p/x dx Y-MOMENTUM EQUATION: ½ d(v2) = - (1/) p/y dy X-MOMENTUM EQUATION: ½ d(w2) = - (1/) p/z dz u2 + v2 + w2 = V2 p/x dx + p/y dy + p/z dz = dp ½ d(V2) = - (1/) dp dp = -  V dV Euler equation aside

  40. aside BERNOULLI’S EQUATION ½ d(V2) = - (1/) dp  {½  d(V2) = - dp} INCOMPRESSIBLE: ½  (V22) - ½  (V12) = - (p2 – p1) p2 + ½  (V22) = p1 + ½  (V12) = constant along streamline If irrotational each streamline has same constant.

  41. aside BERNOULLI’S EQUATION Momentum equation and steady, no body forces, inviscid, incompressible along a streamline. p2 + ½  (V22) = p1 + ½  (V12) Kinetic Energy / unit volume If multiply by volume have balance between work done by pressure forces and change in kinetic energy. (interesting that for an incompressible, inviscid flow energy equation is redundant for the momentum equation)

  42. LIFT ‘Measurements’ CL = FL/(½  U2Ap) = f(shape, Re, Ma, Fr, /l) Except at very low Re w does not contribute directly to lift Lift = U = unsymmetrical flow field p2 + ½  (V22) = p1 + ½  (V12) = constant along streamline

  43. Juan Lopez, which curve is for the upper surface? = (p-p)/(1/2 U2) Calculated (dots) and measured (circles) pressure coefficients for airfoil at  = 7o.

  44. _________ pressure gradient = (p-p)/(1/2 U2) __________ pressure gradient Calculated (dots) and measured (circles) pressure coefficients for airfoil at  = 7o.

  45. favorable pressure gradient unfavorable pressure gradient = (p-p)/(1/2 U2) First person in 3rd row, which surface iis flow likely to separate? Calculated (dots) and measured (circles) pressure coefficients for airfoil at  = 7o.

  46. Stall results from flow separation over a major section of the upper surface of airfoil Rec = 9 x 106 CL = FL/( ½  V2Ap)

  47. Rec = 9 x 106 CL = FL/( ½  V2Ap) Because of the asymmetry of a cambered airfoil, the pressure distribution on the upper and lower surfaces are different. Must have camber to get lift at zero angle of attack.

  48. Rec = 9 x 106 CL = FL/( ½  V2Ap) Typical lift force is of the order unity. (dynamic pressure x planform area) Consequently, FL ~ ½  V2Ap Wing loading = FL/Ap = 1.5 lb/ft2 1903 Wright Flyer = 150 lb/ft2 Boeing 747 = 1 lb/ft2 bumble bee

  49. * * As angle of attack increases stagnation point moves downstream along bottom surface, causing an unfavorable pressure gradient at the nose*.

  50. Stream tube narrows then widens favorable unfavorable favorable to unfavorable causes lam. to turb. trans. stagnation point

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