4. Incompressible flow over finite wings 4.1 Consider a vortex filament of strength Γ in the shape of a closed circular loop of radius R. Obtain an expression for the velocity induced at the center of the loop in terms of Γ and R. 4.2 Consider the same vortex filament as in problem 4.1. Consider also a straight line through the center of the loop, perpendicular to the plane of the loop. Let z be the distance along this line, measured from the center of the loop. Obtain an expression for the velocity at distance z on the line, as induced by the vortex filament. 4.3 The measured lift slope for the NACA23012 airfoil is 0.1080 per degree, and 0 1.3 .Lα = = −Consider a finite wing using this airfoil, with AR=8 and taper ratio=0.8. Assume δ=τ . Calculate the lift and induced drag coefficients for this wing at geometric angle of attack 7= . 4.4 The Piper Cherokee (a light , single-engine general aviation aircraft) has a wing area of 15.8

and span 9.8 m. Its maximum gross weight is 10.9 kN. The wing uses an NACA 65-415 airfoil, which has a lift slope of 0.1033 per degree and

2m0 3 .Lα = = − Assume τ=0.12. If the

airplane is cruising at 193km/h at standard sea-level conditions at its maximum weight and is in straight level flight, calculate the geometric angle of attack of the wing. 4.5 Consider the airplane and flight conditions as in the previous problem. The span efficiency factor e for the complete aircraft is generally much less than that for the finite wing alone. Assume e=0.64. Calculate the induced drag for the airplane in the previous problem. 4.6 Consider a finite wing with aspect ratio of 6. Assume an elliptic lift distribution. The lift slope for the airfoil section is 0.1/degree. Calculate and compare the lift slopes for (a ) a straight wing, and (b) a swept wing, with a half-chord line sweep of 45 degrees. 4.7 Repeat the previous problem but for a lower aspect ratio of 3. From a comparison of the results of these two problems, draw some conclusions about the effect of wing sweep on the lift slope, and how the magnitude of this effect is affected by the aspect ratio. 4.8 An untwisted wing with an elliptical plan form and an elliptical lift distribution has an aspect ratio of 6 and a span of 12 m. The wing loading is 900 N/m2 when flying at a speed of 150 km/h(41.67 m/s) at sea-level. Calculate

a) the induced drag force b) the power to compensate for this drag c) the induced angle, effective angle and absolute angle of attack.

4.9 A wing with an elliptical planform is flying through sea-level air at a speed of 45 m/s. The wing loading W/S = l000 N/m2. The wing is untwisted and has the same section from root to tip. The lift curve slope of the section m0 is 5.7. The span of the wing is 10 m, and the aspect ratio is 5. Find he sectional-lift and induced-drag coefficients Find also the effective, induced and absolute angles of attack. What is the power that is required to overcome the induced drag of the wing ? 4.10 Massachusetts Institute of Technology's Light Eagle is a human-powered aircraft that held Federation Aeronautique Internationale closed-course world distance record of 36.4 miles (58.58 km), established in 2 hr, 14 min on January 23, 1987. Some specifications f the Light Eagle are shown in the figure. With limited power that could be provided by a human being, the demand for the design to have the lowest possible drag was critical. Although a small viscous drag was achieved by careful design of the cabin and airfoil profiles, the induced drag was greatly reduced by utilizing a slender wing, whose aspect ratio is AR = b2/S = 39.4 based on the data shown in the figure. The average speed in the record-breaking flight was 7.29 m/s (16.3 mph) that, at sea- level, gives a dynamic pressure q = 32.58 N/m2. In a level flight, the gross weight (W= 1076.4 N) of the aircraft is balanced by the wing lift L. Thus, the corresponding lift coefficient is (with S = 30.66 m2)

08.1==∞ SqWCL

Assuming an elliptical lift distribution, find the wing-induced drag coefficient, the induced drag and the power required to overcome this drag. Assume a profile drag coefficient of

021.00 =DC and estimate the total power required to overcome the drag.

4.11 The bound vortex AB shown in the accompanying figure is in the presence of a sea- level stream of speed equal to 100 m/s. The total force on the vortex is 10,000 N. At the end of 1/5 s, the vortex formation in the fluid is shown in the figure. Find using Biot-Savarts law the induced downward velocity at the point E.

4.12 The mechanism of drag reduction by formation flight of birds migrating over long distances is examined based on a crude three-bird model, in which each bird of span b is approximated by a single horseshoe vortex of circulation Γ. The downwash is computed only at the midspan of each wing using the Biot-Savarts formula. For a bird flying alone, show that the velocity induced at the midspan by the two trailing line vortices is w0=-Γ/πb with the negative sign denoting a downwash. When the three birds fly abreast with small distances between neighbouring wing tips, the velocities induced by the two trailing vortices at each of the two gaps are effectively cancelled, so that the combined vortex system may be represented by a single horseshoe vortex of width 3b as shown in the sketch. Show that the downwash velocities at the midspan of birds L, M, and R are equal to 3w0/5 w0/3, and 3w0/5 respectively. The result indicates that the total induced drag would be decreased significantly by flying abreast. It also shows that the bird in the middle experiences the lowest drag and thus could be the weakest among the three birds.

4.13*(difficult) If the three birds are rearranged in a V-formation with the middle bird flying at a distance d ahead of the others, the combined vortex system may be represented by a broken vortex line of strength Γ as shown in the sketch. Defining r = d/b show that the midspan downwash at bird M is

022 )94

614

21

31( wrrwM

−− +−++=

and those at bird L and R are

022 )94

1214

41

53( wrrww LR

−− +++−==

If the total power required to compensate the induced drag is proportional to 3w0 when the three birds fly separately, that power would be proportional to the sum of the individual downwash velocities when they fly in a formation. Let e = 3w0/(wM + wR+wL) be the induced drag efficiency of the formation flight; then

22 943614

121

45231 −− +−++= rr

e

represents the ratio of the induced power for the group flight to that without a formation. The role played by the stagger distance d in the analysis of formation flight three birds is examined in the following:

a) As d decreases toward zero, r -> 0 and the result obtained in Problem 7.2 is recovered. In the present abreast flight, the induced drag of the middle bird is 5/9 times that of the bird on either side. The value of e is 1.956 and e-1 = 0.511, giving a 48.9% saving in induced power.

b) When d increases from zero to infinity, the induced drag is redistributed in such a way that the drag saving is shifting continuously from the middle bird to the other two, whereas the efficiency e decreases monotonously from 1.956 to 1.608.

c) Show that when d = 0.209b, the drag savings are evenly distributed and all birds experience the same downwash velocity 0.511w0. The efficiency is e = 1.852 and

the saving in induced power is 46%. A further increase in d will make the bird at the middle a leader, in the sense that it needs to provide more power than the followers.

4.14 Two identical wings having the same weight and flying at the same speed are arranged in tandem. Which wing requires the greater horsepower? Hint: Consider the velocities induced by one wing on the other. 4.15 Design an untwisted elliptical wing to fly at 200 km/h in sea-level air with the following specifications: lift = 45,000 N planform area = 30 m2 wing span <12 m The absolute angle of attack αa cannot exceed 12° to avoid stall. For the wing of minimum induced drag, specify its aspect ratio, wingspan, maximum chord, the design αa, and the power required to compensate induced drag. 4.16 Shown in the accompanying photograph is the Gossamer Condor, a human-powered aircraft designed and built by Paul MacCready (Burke, 1980). On August 23, 1977, it completed a required figure-eight course around two turning points 0.5 miles apart to win the Kremer Prize of £50,000. The aircraft parameters are listed below: Rectangular wing: area S = 1056 ft2, span b = 96 ft Weight without pilot: W = 84 lb Cruise speed: V = 15 ft/s at sea-level

Profile drag coefficient: CD0 = 0.025 (estimated) It is assumed that the effects of the canard (the front steering airfoil) have been absorbed in the above data.

a) In the winning flight, the total weight of pilot Bryan Allen and equipment was 140 lb. Based on elliptical wing analysis, compute the power that was required to fly the Gossamer Condor at the cruise speed, and then determine whether Allen was able to carry out the mission. Ground tests revealed that he could deliver 0.4 hp in a period comparable to that of the mission b) Can a 120-lb person whose power output is 0.3 hp fly the aircraft at the same cruise speed? If not, redesign the main wing without changing S, W, and V, assuming that

the value of CD0 remains the same as before.

4.17 Air is streaming past a thin plate with chord length c=1m and at an angle of attack α. The lift force of the plate is 40 N per meter. Estimate the tip moment with respect to the leading edge

4.18 An airplane with weight 2000 kg is flying at a speed of 270 km/h at an altitude of 2000m.The wing area is 20 m2, the aspect ratio is 10 and the efficiency factor is 0.8. Estimate the induced drag of the airplane 4.19 Use a horseshoe vortex model to estimate the downwash at a position P of the swept wing in the figure as a function of the lift coefficient C

∞Vw /L and the aspect ratio AR.

4.20* A biplane consists of two plane plates, both with chord length c. The plates are placed according to the figure. Assume a lumped-vortex model in which a vortex is placed at the position of 25% of the chord measured from the leading edge for each plate. Estimate the lift force on both plates. (Hint: Use the condition that the flow should be tangential at a position of 75% of the chord for each plate)

4.21 A model of a rectangular wing has been tested in a wind tunnel. It is found that

09.0==αα d

dCC LL per degree, α0=−3 degrees and the efficiency e=0.87. Estimate the lift

force of a prototype with span of 14m and chord length 2 m when the speed is 50 m/s and the angle of attack is α=5 degrees.

4.22 Air is flowing with a velocity of 50 m/s past a plane plate with an angle of attack α=2 degrees. The span of the plate is b=8m and the area S=9.6 m2. Assume an elliptic chord distribution and calculate the induced drag force. Put ρ=1.2 kg/m3 and assume a profile efficiency is ξ=1. 4.23 Two identical airplanes are flying straight forward at the same altitude and the same speed. The distance between the wing tips is the same as the span for each airplane. There is a mutual interaction between the airplanes due to the up and down wash induced by the vortex system. Calculate the downwash at a position in the middle of the two tip vortices on one airplane from the plane itself and the up or downwash at the same position induced by the other airplane.

Answers

4.1 2

30

ˆ ˆ ˆ ˆ24 4

Rd Rr z z2R R R

π φ φ ππ π

′Γ × Γ= − = =∫V Γ

4.2 2

2 2 3/

ˆ2 ( )

R zz R

Γ=

+V 2

4.8 a) 970 N b) 40.4 kW c) –0.045 rad , 0.135rad , 0.18 rad 4.9 cl=CL=0.8062, cd=CD=0.0414, α0=0.14 rad, αi=−0.05rad, αa=0.19 rad. 46.2 kW=63hp 4.10 CDi=CL

2/πAR=0.0094, Di=9.39N, Pi=V Di=68.5 W=0.092 hp ,CD0=0.021 Profilmotstånd D0=CD0 ½ ρ V2 S=21.0 N Totalt motstånd=30.4 N. Ptot=221.6 W= 0.297hp

4.11 Lösning:

Biot-Savarts lag ger hastighetsbidrag från den bundna virveln ABΓ

dd

w ABABAB ⋅⋅

Γ=−−

⋅⋅Γ

=ππ 4

))60sin(60(sin4

b

C

ΓΑΒ

D

B

A

Bidrag

⋅Γ

=

=

b

w

AB

BC

π

d=b/4

b

från virveln BC

)5

1133(

((

24

2

+

−⋅⋅

Γ⋅

bb

AB

π

och AD

))

2()

4(

4/

)2

()4

3(

4/3(

24

2))

2()

2() 22222222

=+

++

⋅⋅

⋅⋅

Γ⋅=

++

+

−bb

bbb

bbbd

dbd

db AB

π

Bidrag från startvirveln

103)

)23()

2(

2

)23()

2(

2(

434 2222 bbb

b

bb

b

bw ABAB

CD⋅⋅⋅

Γ=

++

+⋅

⋅⋅

Γ=

ππ

Totalt nersvep vid E

smb

w ABAB /156.020/14.3/4.208.44.220

)103

15

11331( =⋅=⋅

⋅Γ

=⋅

+++⋅

Γ=

ππ

där ΓAB beräknas enligt

08.4)10020225.1/(10000 =⋅⋅=⋅⋅

=Γ∞ bV

LAB ρ

4.19 3bVπ ∞

Γ

4.22 ξ=1 ska användas i formeln 2lc πξα= Svar: 33.1 N

4.23 ,15b bπ π

Γ Γ−