AA241A Sample Final Exam Solutions
This is an open note, open book exam to be completed in 3 hours. The maximum score for each problem is indicated in parentheses.
Many of the questions on this exam deal with modifications to a reference airplane design described below: Fuselage length = 80 ft
Fuselage maximum diameter = 10 ft
Distance from cruise c.g. to tail a.c. = 40 ft
Horizontal tail area = 281 ft2
Taper ratio = .25 Reference
Wing Area = 1000 ft2
T-tail with aft engines
AR = 8.
Sweep = 30deg.
t/c = .10 at both root and tip
Cruise Weight = 100,000 lbs
Cruise Altitude = 35,000 ft
Cruise Mach = 0.80
Total CDp = .020
Total e = .85
"Peaky" airfoil sections
(density = 7.382 x 10-4 lb s2/ft4 kinematic viscosity = 4.0575 x 10-4 ft2/s)
1. The company that builds this airplane decides to replace the wing with a new design of aspect ratio 10 rather than 8. Assume that the following parameters are held constant: fuselage geometry, taper ratio, reference wing area, sweep, t/c, airfoil type, tail geometry, weight, and cruise conditions. (10)
a. Find the L/D of the original design.
b. Find the L/D of the new design.
a. V = M a = 973.6 * 0.8 = 778.8 ft/sec
First we compute CL = 2W / S
rV2 = 2*100000 / (1000*.0007382*778.8^2) = .447.
Now CD = CDp + CDi + CDc = .020 + CL2 /
p AR e + CDc = .020 + .447^2 / (pi*8*.85) + CDc = 0.02935 + CDc
Computation of CDc:
Given CLnormal = .447/cos2(30) = .596 and t/cnormal = .10 / cos(30) = .1155
We find from plot in notes that Mccnormal = 0.680. So, Mcc = .68 / cos(30) = .785
So M/Mcc = 1.019. From plot of CDc in notes: CDc/cos3(30) = .00174, so CDc = .00113.
So CD = .00113 + .02935 = 0.03048 and L/D = CL/CD = 14.67.
b. When the wing span changes, the following parameters change: CDi due to changes in aspect ratio, lift-dependent viscous drag, and fuselage interference; CDp due to changes in Reynolds number, and exposed wing area. This requires computation of wing CDp for both designs. The difference is subtracted from the CDp given above. Details are left to the student for now.
2. If a new airfoil were designed for the reference airplane, which of the following airfoils would be most appropriate? The actual Cp distribution is shown at the cruise Mach number. (4)
Cp at the crest well above Cp* Rather conventional peaky section -- best choice
Very thin and too far below MDiv No lift.
3. Rather than the 10% peaky section assumed for the reference airplane, estimate the maximum thickness of a supercritical airfoil if it is used on the reference airplane at the specified cruising conditions. Assume a new supercritical airfoil is used with a section Mcc that is .06 higher than that of a peaky section at the same t/c and CL. (5)
From problem 1 CLnormal = .596. If the airplane is cruising at Mdiv = .8, then Mcc is given by:
MDiv = Mcc [ 1.02 +.08 ( 1 - Cos 30 ) ] . So Mcc = .8/(1.02+.08*(1-cos(30))) = 0.776
The sections see the normal component of this: Mccnormal = .776 cos(30) = .672.
For peaky sections, the plot in the notes yields t/cnormal = .123 for this value of CLnormal and Mccnormal .
Supercritical sections will behave as though Mccnormal were .06 lower or Mccnormal = .612.
This leads to t/cnormal = .196 or t/c = .170
4. What is wrong with the wing design with the characteristics shown below? Results are computed for the isolated wing in inviscid flow at low speeds. (Choose all that apply.) (5)
a. Cl near the root is too high.
b. The tip Cl's are too high.
c. The lift distribution is not close enough to elliptical (note the value of e)
d. There is not enough lift near the tip.
e. The two curves should not intersect.
5. To fix the wing above, how would you change the taper and twist? (Choose all that apply.) (5)
a. Increase taper ratio -- to permit higher tip loads without higher tip Cl
b. Decrease taper ratio
c. Leave taper ratio the same
d. Increase washout
e. Decrease washout -- to increase tip loading
6. The parameters:
Area, AR, t/c, Sweep, Taper Ratio, and Washout are important in the initial wing design and have many different effects.Which of the above parameters leads to each of the following results when its value is increased
and the others are fixed? (Choose best answer for each part. (5)
a. This adds extra drag, lowers stall speed and lowers
D CDc at given altitude and Mach number
Increased Area
b. This reduces fuel volume, and increases the lift near the tip.
Increased Taper Ratio
c. This lowers stall speed a bit, increases fuel volume, and reduces Mcc.
Increased t/c
d. This improves the climb rate at a given (low) speed.
Span or AR
e. This reduces the induced drag when the tip lift is larger than optimal.
Washout
7. In problem #1 we considered what would happen to the reference airplane drag if the wing aspect ratio were changed. Assume that the original tail was sized according to the statistical sizing method we used in AA241 and compute the tail size required if the wing aspect ratio is changed from 8 to 10. Assume that the distance from cruise c.g. to tail a.c. is not changed and the c.g. range (in % MAC) is constant. (Consider just the horizontal tail) (5)
In the original case span = sqrt(AR Sw) = 89.44 ft
Croot = 2*Sw / (b (1+
l)) = 17.89 ft
MAC = 2/3(Croot+Ctip - Croot*Ctip/(Croot+Ctip)) = 2/3 Croot ( 1 +
l -
l / (1+
l)) = 12.52 ft
For the new wing with AR = 10 we find MAC = 11.2 ft
The tail volume coefficient is given by Vh = Sh/Sw * lh / c
From the data in problem 1 we find Vh = 281/1000 * 40/12.52 = 0.898
From the statistical sizing method: Vh = f (wf2 lf / Sw c ).
The correlation parameter wf2 lf / Sw c is, for this case = 102 * 80 / (1000 *12.52) = .639
For the new case the correlation parameter = 102 * 80 / (1000 *11.2) = .714
From the correlation plot we expect Vh = 0.30*3.2 = 0.96
The required tail area is then: Sh = Vh Sw c / lh ~ .96 * 1000 * 11.2 / 40 = 268.8 sq ft.
8. According to statistical tail sizing method we used in problem set #8, the required horizontal tail size is just set by the fuselage and wing parameters. This is not the best approach. Use the scissors curve shown below to estimate the required horizontal tail size if the c.g. range were 0 and the wing was positioned to satisfy the stability and control requirements with the smallest possible tail for the reference airplane discussed earlier.(5)
If the c.g. range were 0 we could get by with Vh = .54 if the scissors curve above applied to our airplane. This would mean Sh = Vh Sw c / lh ~ .54 * 1000 * 12.52 / 40 = 169 sq ft.
9. Estimate the clean wing FAR stall speed for the reference airplane if the outer wing section Clmax is 1.2 at low speed, sea level conditions. Assume flight at sea level and ignore Mach effects. (5)
If the outer panel has Clmax = 1.2, then from the relation between overall wing CL and outer panel Clmax shown in the plot in the notes, with sweep = 30deg. , CLmax = 1.2 * 82 = .984.
The only correction to be made is the correction for VSmin associated with the 1kt/sec FAA flight test. This adds about .2 to the 1-g CLmax so CLmax = 1.184.
The associated FAA Stall Speed is given by Vs = sqrt (2*100000 / (1000*.002377*1.184)) = 266.6 ft/sec = 158 kts.
10. Two wings for an SST are shown below. (5)
Use the expressions for minimum drag of supersonic wings and assume the following. The wings have the same area, carry the same lift, and fly at the same altitude. The drag of the two wings is the same at Mach 1.6. Which has lower drag at Mach 2.4 and why?
Wing B must have lower drag at M=2.4.
Since the wings have the same area, they have roughly the same skin friction drag. At Mach 1.6 the sum of volume wave drag, vortex drag, and lift-dependent viscous drag is equal. Since the length of wing B is longer, its reduced wave drag is offset by a higher vortex drag. Now at the higher Mach numbers and higher q, the volume-dependent drag will become a larger fraction of the total, so wing B will have lower drag.
If, however, we increase altitude as we increase M so that q is held fixed, we expect that the lift-dependent wave drag will increase while the other terms remain relatively unchanged. In this case, wing B, with its greater length will have less drag.
