Introduction to Aerospace Engineering

Transcription

1 Introduction to Aerosace Engineering Lecture slides hallenge the future

2 Introduction to Aerosace Engineering Aerodynamics & Prof. H. Bijl ir. N. Timmer

3 &. Airfoils and finite wings Anderson 5.9 end of chater 5 excl. 5.9

4 Toics lecture & Pressure distributions and lift Finite wings Swet wings 3

5 Pressure coefficient Tyical examle Definition of ressure coefficient : - uer side q lower side -.0 Stagnation oint: t t - q > 4

6 Examle 5.6 The ressure on a oint on the wing of an airlane is 7.58x0 4 N/m. The airlane is flying with a velocity of 70 m/s at conditions associated with standard altitude of 000m. alculate the ressure coefficient at this oint on the wing 000 m: N/m ρ.0066 kg/m 3.50 q 5

7 Obtaining lift from ressure distribution θ leading edge V s trailing edge ds dy θ dx ds cos θ 6

8 Obtaining lift from ressure distribution Normal force er meter san: with ds cos θ dx N N TE LE c Write dimensionless force coefficient : dx 0 l l TE cosθds cosθds n c 0 u LE dx N ρv c u N q c n l x u d q 0 c q 0 c x d n ( ) x d l u c 0 7

9 TLsinα- Dcosα NLcosα + Dsinα L R N α T D V α angle of attack 8

10 Obtaining lift from normal force coefficient L N cos α T sinα c c cosα c l n t sinα L N T cosα q c q c q c sinα For small angle of attack α 5 o : cosα, sinα 0 d x ( ) ( ) l l c 0 u 9

11 Examle 5. onsider an airfoil with chord length c and the running distance x measured along the chord. The leading edge is located at x/c0 and the trailing edge at x/c. The ressure coefficient variations over the uer and lower surfaces are given as, u, u, l x x 300 for 0 0. c c x x for 0..0 c c x x 0.95 for 0.0 c c alculate the normal force coefficient. 0

12 omressibility correction of the ressure coefficient at a certain oint omressibility correction Aroximate theoretical correction (valid for 0 < M < 0.7) :,0 M M.0 Prandtl-Glauert Rule

13 omressibility correction for lift coefficient ( ) d ( x) c c l u o l l c 0 M M c 0 ( ) ( ) d x u o c d x ( ) ( ) l,0 l c 0 l l,0 M u o l M> α

14 Examle 5. onsider an NAA 44 airfoil at an angle of attack of 4 o. If the free-stream Mach number is 0.7, what is the lift coefficient? -From A. D for angle of attack of 4 o, c l 0.83.However, these data were obtained at low seeds. 3

15 ritical Mach number and critical ressure coefficient M0.3 M eak 0.44 M0.5 M eak 0.7 M0.6 M eak.00 ritical Mach number for the airfoil 4

16 ritical ressure coefficient,min thick -.5 thick medium -.0 medium thin -0.5 thin, cr f (M ) M.0 Thicker airfoil reaches critical ressure coefficient at a lower value of M 5

17 6 ritical ressure coefficient Definition of ressure coefficient : q q Dynamic ressure : ρ ρ ρ / V V V q We have found before : ρ a thus : M a V q For isentroic flow we found : 0 M + Dividing these equations we find :... 0 M + and also :

18 7 ritical ressure coefficient )M ( )M ( + + Substitution in -equation gives : + + )M ( )M ( M q + + )M ( )M ( M or :

19 8 ritical ressure coefficient The critical ressure coefficient is found when M is reached : + + )M ( M,cr c cr ( ) M M

20 The intersection of this curve with the airfoil P,min will give the value of the M cr secific for the airfoil under consideration. +, cr M cr Airfoil indeendent curve cr [( ) / ] M + [( ) / ] M cr min,0min M Airfoil deendent curve 9

21 Drag divergence 0

22 Drag divergence M< M <M cr M> M< M cr <M <M drag divergence M< shock wave M> M< M >M drag divergence Searated flow

23 Drag divergence Develoment of suercritical airfoils -,cr x/c

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25 Finite Wings 4

26 haracteristics of finite wings ti vortex Flow around the wing ti generates ti vortices low ressure high ressure 5

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28 Ti vortices ross section of ti vortex R V θ Induced velocity V V local flow vector w The trailing ti vortex causes downwash, w 7

29 Induced drag α i L L geometric angle of attack α V α i α 8

30 Induced drag α i D i Lsin α i L geometric angle of attack α induced drag, D i V α i α α eff 9

31 Induced drag α i D i Lsin α i L geometric angle of attack α induced drag, D i V α i α Since α small : sinα i α i D i L α i α eff 30

32 Induced drag FRONT VIEW Examle : ellitical lift distribution ellitical lift distribution downwash, w Ellitical lift distribution results in constant downwash and therefore constant induced angle of attack From incomressible flow theory : where : A α i L πa b (Asect ratio) Thus : S D i L πa 3

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35 Effect of asect ratio on induced drag High asect ratio : Low Induced Drag DG 00 DG 500 Low asect ratio : High Induced Drag Mirage

36 San efficiency factor D i L π A e san efficiency factor (Oswald factor) Ellitical loading Non-ellitical loading : e ; minimum induced drag : e < ; higher induced drag The total drag of the wing can now be written as : + D D π A Profile drag Induced drag L e 35

37 Examle 5.9 onsider a flying wing with wing area of 06 m, an asect ratio of 0, a san effectiveness factor of 0.95, and an NAA 44 airfoil. The weight of the airlane is 7.5x0 5 N. If the density altitude is 3 km and the flight velocity is 00 m/s, calculate the total drag on the aircraft. 36

38 Lift curve sloe geometric angle of attack α α eff α i V The effect of a finite wing is to reduce the wing lift curve sloe 37

39 Lift curve sloe The induced angle of attack reduces The local effective angle of attack : α eff α α i For a wing of a general lan form we may write : α α i i L π A e 57.3 L π Ae Angle is in radians! e is san effectiveness factor (theoretically different from e but in ractice more or less the same) For angle in degrees 38

40 Lift curve sloe L infinite wing (effective angle of attack) a 0 d l d α a a 0 finite wing (geometric angle of attack) d a dα L α From thin airfoil theory we find : a 0 π 39

41 Lift curve sloe Effective local angle of attack α eff α α i For an arbitrary loading distribution wing we find : d L L α i a0 πae d( α αi ) ( α α ) const, L a0 i + L a0α a + 0 πae const + a + 0 πae Integrate : a L L 0 α + Ae π const 40

42 Lift curve sloe Differentiating this equation results in : d dα Examle : d Infinite wing : A then L π dα L + a0 a0 πae Finite wings : A (Fokker 50) then A5 then d L dα d L dα π π π π 4

43 a d L a0 dα a0 + π Ae a 0 d l dα d L a0 dα 57.3a + π Ae 0 a 0 er degree 4

44 Examle 5. onsider a wing with an asect ratio of 0 and NAA 30 airfoil section. Assume Re 5 x 0 6. the san efficiency factor is ee If the wing is at 4 o angle of attack, calculate L and D. 43

45 Swet wings M V M cr 0.7 Λ40 44

46 Swet wings M V M cr 0.7 Λ40 V cos Λ V M cr for swet wing 0.7 cos Λ 0.9 By sweeing the wings of subsonic aircraft, the drag divergence is delayed to higher Mach numbers 45

47 Swet wings Effect of wing thickness and sweeback angle on minimum wing drag coefficient A B D,min t/c9% t/c6% t/c4% A B.0 M.0 M 46

48 Total drag D + D rofile L π Ae + + D D D D rofile f ressure w 47

49 Flas δ W L L ρv S Thus : V W ρs L and : V stall ρ W S L max The landing seed is decreased when the maximum lift coefficient is increased 48

50 Flas δ f 60 L flas extended δ δ f 5 flas retracted δ f 0 α 49

51 Flas Tyical examle 50

52 Flas and slats flas slats 5

53 Flas 5

54 Flas 53

55 Flas 54