1. What is wrong with the fuselage design shown above? Cruise is at M = .8.
a. Nose fineness ratio is too small
b. Tail fineness ratio too small
c. More seats on left than on right
d. Seats extend into tapering tail section
e. Not all rows contain the same number of seats
f. List one other item: Kinks at tailcone junction, straight sides on tailcone, 4 seats against wall
2. An SST fuselage has a length of 300 ft, a diameter of 13 ft, and a wetted area of 10,000 sq ft.Estimate the change in fuselage drag at Mach 2.0 and 60,000 ft if the length is reduced by 10 ft but the cross-section is unchanged. Assume a form factor of 1.0 and ignore the change in Reynolds number. Also assume about the same wave drag as a minimum drag body and ignore upsweep. At 60,000 ft, r = .0002256 sl/ft3 speed of sound = 968.1 ft/sec, n = .001316 ft2/sec
Two main effects: wetted area reduction saves skin friction drag and fineness ratio reduction increases wave drag.
Skin friction: Dfriction = q Cf k Swetted
Here Re = V l / n = M a l / n = 2.0 * 968.1 * 300 / .001316 = 441,383,000.
At M = 0 we would have Cf = .458/(log Re) 2.58 = .001743. At Mach 2.0 the Cf is reduced by about 22% (from plot in notes) so Cf = .00136.
With this large fineness ratio k is about 1.0, so Dfriction/q = 13.6 sq ft.
Wave drag: D = q CDpSx. If we assume a minimum drag body: CDp= 9.8 * D2 / L2 = 0.0184. So Dwave/q = .0184 * p* 6.52 = 2.4425 sq ft.
Repeating the calculation for a 290 ft long body yields Dfriction= 13.147 sq ft and Dwave = 2.614 sq ft. The total change in drag area is .2815 sq ft or 1.75% or 118.6 lbs.
3. We have designed a wonderful new airfoil section. 2-D wind tunnel tests of this 14% section show good characteristics at a Mach number of 0.7, and a design lift coefficient of 0.5. We would like to use this section on the wing of an airplane designed to fly at Mach = 0.80. What will be the wing sweep and approximate airplane lift coefficient, CL? (Note that this new airfoil section is much better than conventional sections, so it will not help to use the data on Mcc vs. CL and t/c for peaky sections in the notes.)
If the 2D section is designed to operate at Mach 0.7 and the wing is to be flown at Mach 0.8, we need the normal component of the flow to be Mach 0.7, so cos L= .7/.8 and L = 29deg .
The 2D section is to operate at CLeffective= 0.5 so the CL is about 0.5 cos2L= 0.383.
4. What is the maximum allowable streamwise t/c for an aircraft flying at M = 0.8 if it has a sweep of 35º, a conventional peaky-type section, and cruises at an altitude where CL = 0.375?
If the aircraft is flying at MDiv then Mdiv = 0.8 = Mcc (1.02 +.08*(1-cosL)). So Mcc = .773
Based on simple sweep theory the normal Mach number is then Mcc cosL = .633
The effective normal lift coefficient is .375 / cos2L = 0.559.
From the plot in the notes for peaky sections this requires t/c / cosL = .177so t/c = 14.5%
5. Which of the following statements about the various components of drag are true?
a. The vortex drag would be equal to the lift-dependent drag if the flow were inviscid
(If the lift-dependent drag does not include drag at zero lift then it may include only part of the vortex drag. We gave credit for either answer here, though.)
b. The lift-dependent viscous drag is smaller for high AR wings
No, it is changed very little with AR. It is approximately CDi_viscous= k CDpCL2
c. The zero lift drag is mostly due to skin friction --- Yes.
d. The presence of zero-lift twist drag makes the vortex drag higher than the ideal CL2 / p AR
Not necessarily. A twisted wing can still have elliptic loading and CDi= CL2 / p AR
e. The skin friction drag of a wing is typically 20% or more higher than a flat plate with the same wetted area and Reynolds number.
Yes. The combined effects of form factor and roughness typically cause an increment of 20% or more compared with the smooth, flat plate value.
6. Airplanes with higher aspect ratios tend to have lower values of CDi at a given C L. Why is this? (Select all contributing factors.)
a. From elliptic wing results we expect CDi to be inversely proportional to AR in inviscid flow.
b. A higher aspect ratio wing has lower lift-dependent viscous drag coefficient at a given CL.
No it is about the same and perhaps a bit higher due to lower Reynolds numbers.
c. Higher aspect ratio wings tend to bend more, increasing wing twist and reducing vortex drag at a given CL.
d. The fuselage interference effect is reduced for a wing of given area.
For a given fuselage, the factor s increases (less interference) with the span.
e. The tip vortices will be weaker for a high aspect ratio wing at a given CL.
7. The drag coefficient is expressed by the equation:
What measures will most likely reduce the magnitude of the second term for conventional aircraft:
a. Reduce fuselage diameter: Increases the factor s.
b. Eliminate any wing twist: This may hurt, not help.
c. Make the lift distribution closer to elliptical: Yes. This generally reduces vortex drag.
d. Increase the wing span while holding area constant: Yes this increases AR. Some people wondered if the lower e with higher AR would overwhelm the higher AR. The answer is clear from the small relative change in e. Doubling AR might reduce e by 10-20%, nothing close to 50%.
e. Reduce the wing t/c: This actually helps a little since the lift-dependent viscous drag increases with CDp and larger t/c means somewhat larger CDp. The effect is small though so we gave credit either way.
8. Which of the pressure distributions shown below corresponds to an airfoil with a much lower thickness than the maximum allowable for flight at MDiv at this CL? A, B, C, D, or E
Although the shape of the Cp distributions labeled A and C are similar A is the only section with Cpmin << Cp*. This represents a section with much lower thickness than would be allowed for flight at MDiv at the conditions shown.
