1 / 21

Lecture 27: Lift

Lecture 27: Lift. Many biological devices (Biofoils) are used to create Lift. How do these work?. chord section analysis…. (normal to U). total force. (normal to wing). lift. angle of attack = a. (parallel to U). drag. wing velocity = U. First, some definitions…. wing section,c

chesna
Download Presentation

Lecture 27: Lift

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 27: Lift Many biological devices (Biofoils) are used to create Lift. How do these work?

  2. chord section analysis…. (normal to U) total force (normal to wing) lift angle of attack = a (parallel to U) drag wing velocity = U First, some definitions… wing section,c (chord) winglength, R wingarea, S

  3. Two ways to derive lift: 1) mass deflection total force a Surface area, S U air deflecteddownward by wing Pressure always acts normal to the surface of an object. Therefore, this mass deflection force acts roughly perpendicular to surface of biofoil.

  4. 1) Massdeflection total force Lift and drag are defined ascomponents perpendicularand parallel to direction of motion. lift a drag Surface area, S U air deflecteddownward by wing

  5. dimensionless scaling parameters amplitude · length2 Reynolds number = frequency · viscosity forward velocity reduced frequency = length · angular velocity RoboFly

  6. CL Fs CD 1 0 0 3 . 5 90o 8 0 3 . 0 6 0 2 . 5 4 0 2 . 0 total force coefficient CT total force orientation q (degs) 2 0 1 . 5 0 1 . 0 - 2 0 0 . 5 - 4 0 - 6 0 0 . 0 - 9 0 9 1 8 2 7 3 6 4 5 5 4 6 3 7 2 8 1 9 0 - 9 0 9 1 8 2 7 3 6 4 5 5 4 6 3 7 2 8 1 9 0 angle of attack a (degs) angle of attack a (degs) total force a q

  7. 4 CT 3 CT cos a CT a 2 CT = 3.5 sin a CT sin a 1 0 0 1 5 3 0 4 5 6 0 7 5 9 0 angle of attack (a) 3 4 CD = CTsin a CL = CTcos a 3 2 CL CD 2 1 1 viscous drag { 0 0 0 1 5 3 0 4 5 6 0 7 5 9 0 0 1 5 3 0 4 5 6 0 7 5 9 0 angle of attack (a) angle of attack (a)

  8. total force lift a drag Surface area, S U

  9. 3 highestlift:drag ratio 2 a=45 3 . 5 a=22.5 lift coefficient 3 . 0 1 CD 2 . 5 2 . 0 a=-9 a=90 force coefficients CL 1 . 5 0 1 . 0 a=-9 0 . 5 0 . 0 - 0 . 5 1 - - 9 0 9 1 8 2 7 3 6 4 5 5 4 6 3 7 2 8 1 9 0 0 1 2 3 4 angle of attack (degs) drag coefficient Polar plot of lift and drag:

  10. fluid travels fasterover to of biofoil Flow is tangentialat trailing edge Flow separatesat leading edge Law of continuity applies to streamline 2. Circulation U

  11. Difference in velocityacross surface is equivalentto net circular flow around biofoil = Circulation, G mathematically: Kutta-Joukowski Theorem: combine withpreviousdefinition: R=biofoil lengthc= biofoil width (lift per unit span) U

  12. G=0 +G Required byKelvin’s Law startingvortex -G boundvortex Consider 2D biofoil starting from rest: G=0

  13. Consider 3D biofoil starting from rest: Helmholtz’ Law requires that a vortex filament cannot end abruptly: boundvortex Downward flowthrough center of vortex ring startingvortex tipvortex Circulation, G, is constantalong vortex ring

  14. How is structure of vortex ring related to lift on biofoil? forward velocity, U R Circulation, G Area = A Ring momentum =mass flux through ring= GrA Force = d/dt (GrA) = Gr d/dt(A) = Gr R U Force/R = GrU = Kutta-Joukwski Therefore, elongation of vortex ring is manifestation of force on biofoil.

  15. 1. velocity, u(x,y) ux uy u(x,y) 3. circulation, G Duy Dux Dx Dy w = G = S w x y Three important descriptors of fluid motion: 2. vorticity, w(x,y)

  16. Fstroke = r G A /t Fslap = m U / t where m is bolus of accelerated water, moving atvelocity, u impulse (F x t) = mass x velocity • Momentum • of vortex ring • r G A G = circulation A

More Related