Exercise 1 C) As mentioned in the lectures the Wright brothers were the ﬁrst to achieve powered ﬂight with a heavier-than-air vehicle, whilst keeping control of the vehicle as well.

Exercise 2 B) In the NACA "XYZZ" code the Y indicates the position of the maximum camber, so not the thickness of the airfoil. Instead, ZZ indicates the maximum airfoil thickness as a percentage of the chord length and X does indeed show the maximum camber as a percentage of chord length.

Exercise 3 A) The lift-drag polar is a ﬁgure plotting the drag coefﬁcient of an airfoil on the x-axis and the lift coefﬁcient of the airfoil on the y-axis. Hence answer A is correct.

Exercise 4 B) Modern civil aircraft ﬂy at high altitude in order to reduce their drag (which depends on the air density). Hence the fact that air density is low at high altitude is not a reason to ﬂy there, instead we do so because we want low drag. In order to generate sufﬁcient lift in these low air density conditions, they need to have a high airspeed. The reduction of travel times is an additional advantage which originates from this reason.

Exercise 5 C) In a beam the shear forces are carried by the web plate. The two girders (attached to the web plate) transfer compressive and tensile forces, which are required for moment equilibrium.

Exercise 6 B) In a composite the ﬁbres carry the main loads only in the primary direction of loading (which is also the direction the ﬁbres are laid in), and the function of the resin is to protect the ﬁbres and transfer loads from ﬁbre to ﬁbre.

Exercise 7 A) By deﬁnition, for an anisotropic material the material properties vary with direction. Hence the material properties depend on the direction the material is tested in.

Exercise 8 B & C) In order to reduce drag at transonic and supersonic speeds, one can (as explained in the lectures) think of reducing the airfoil thickness and increasing the wing sweep angle. Increasing the wing aspect ratio won’t help (think of supersonic planes, which usually have short wings), reducing the tip chord of the wing has little eﬀect either.

Answers to Test Module A

1

AE1110x - Introduction to Aeronautical Engineering

Exercise 9 B) From ﬁgure 1 we can ﬁnd that, for an angle of attack of 7 the lift coeﬃcient is C L = 0.8. Given the airspeed (which we need to convert to m/s, so 300 km/h = 300000m / 3600 s = 83.3 m/s), wing area and air density, we can then compute that: ◦

L =

1 1 C L ρSV 2 = 0.8 0.92 29 83.32 = 74111.1 N = 74 kN 2 2

·

·

· ·

Exercise 10 A) The production process for metals in their liquid state is called Cast ing. B) The production process for metal sheets when loaded beyond their yield stress is called Forming. C) The assembly of multiple components is called Joining. D) Positioning yarns or rovings around a cylindrical mandrel is called (Filament) Winding.

Exercise 11 B) To assess what material could best be used, we need to investigate the speciﬁc properties of the three materials given. The relevant property for tensile applications is the (speciﬁc) material strength, hence we divide the strength by the density to obtain: Strength 1260 = = 161.5 MPa dm3 /kg 7.8 Density Strength 280 Speciﬁc strength aluminium = = = 103.7 MPa dm3 /kg 2.7 Density Strength 410 Speciﬁc strength composite = = = 186.4 MPa dm3 /kg 2.2 Density Speciﬁc strength steel =

· · ·

From these numbers it is concluded that the Aluminium alloy is the worst option to pick in this case. Note that this is a case where we consider one-directional tension only (one of the main advantages of aluminium is its isotropic behaviour).

2

Answers to Test Module A

AE1110x - Introduction to Aeronautical Engineering

Exercise 12 A) We are given that the air density at cruise altitude for this Gulfstream IV is 0.42 kg/m 3 . This we can directly substitute in the equation we know for the air density of the troposphere, the lower temperature gradient layer of the International Standard Atmosphere. This allows us to calculate the air temperature at cruise altitude: g −1 aR

−

T T T 0.42 = 1.225 T 288.15 0.342857 = T 288.15 ρcruise = ρ0

cruise 0

cruise

·

287.00

1

−

5.25684803−1

cruise

4.25684803

cruise

0.342857 =

9.80665

−

0.0065

−

288.15

√

T cruise 4.25684803 = 0.342857 = 0.77766193 288.15 T cruise = 0.77766193 288.15 = 224.0833 K

·

Now we can use Toussaint’s equation to determine the aircraft altitude: T cruise = T 0 + a (hcruise

−h ) 0

224.0833 = 288.15 + 0.0065(hcruise 0) 224.0833 288.15 = 9856.4 m = 32337.3 ft hcruise = 0.0065

− −

−

−

B) We are given that the Gulfstream is in cruise ﬂight at an altitude where ρ = 0.42 kg/m3 , at a true airspeed of 460 kts. In this ﬂight condition we know that lift is equal to drag and thrust is equal to weight. First we need to convert all units to the desired ones. In this case this means only the airspeed is to be converted, so we compute V T AS = 460 kts 0.5144444 = 236.644 m/s. Now we can compute the lift coeﬃcient, according to:

·

L = W 1 C L ρSV 2 = mg 2 mg C L = 1 ρSV 2 2 30000 9.80665 = 1 = 0.283 2 0.42 88.3 236.644 2

·

·

·

·

Knowing the lift coeﬃcient, we can now compute the drag coeﬃcient using: C L2 C D = C D0 + πAe The only unknown in this equation is the aspect ratio. However, given the wing surface area and 2 .72 the wingspan we can compute A = b S = 23 = 6.361. Substituting gives: 88.3 C D = 0.015 + Answers to Test Module A

0.2832 = 0.020995 π 6.361 0.67

·

·

3

AE1110x - Introduction to Aeronautical Engineering

The computation of the drag coeﬃcient allows us to compute the drag and, since drag is equal to thrust in cruise, we can also compute the thrust! This gives: T = D 1 = C D ρSV 2 2 1 = 0.020995 0.42 88.3 236.6442 2 = 21801.448 N

·

·

·

·

Now we are nearly there, as the power available of an engine is given by multiplying thrust by airspeed. Given the airspeed and knowing the thrust, we can determine the total available power. Dividing this by the number of engines gives the available power per engine, so: T V 2 21801.448 236.644 = = 2.58 MW 2

P a1 engine =

·

·

C) We are asked to determine the mass ﬂow through one engine for the given ﬂight condition. One equation in which we encounter the mass ﬂow is the equation for thrust:

− V

T = m˙ V j

0

However, since we don’t know the exhaust velocity this equation is not useful. Another way to calculate the mass ﬂow comes from the following equation: m = ρV ˙ Ainlet The only parameter still to determine here is the inlet area, as both the density (given) and the airspeed (given) are known. The inlet area can be computed from the given diameter: 2 d inlet 1.11762 = π = 0.981 m2 Ainlet = π 4 4

This leads to the following mass ﬂow: ˙ m = ρV Ainlet = 0.42 236.644 0.981 = 97.50 kg/s

·

·

D) We are given that the total engine thrust is 20 kN, with the mass ﬂow per engine to be 90 kg/s. The equation which we need to use to calculate the jet exhaust speed is:

− V

T = m˙ V j

0

Since the thrust T , the mass ﬂow m˙ and the airspeed V 0 are known, we can calculate the jet exhaust speed. There are now two ways to come to the ﬁnal answer. First one could compute the thrust per engine (20 kN / 2 = 10 kN) and then compute the jet velocity using the mass ﬂow of a single engine:

−

T = m˙ V V 0 j 10000 T + V 0 = + 236.644 = 347.756 m/s V j = ˙ m˙ 90 4

Answers to Test Module A

AE1110x - Introduction to Aeronautical Engineering

Alternatively one could use the total thrust and compute the total mass ﬂow (2 times 90 kg/s makes 180 kg/s) and then compute the jet exhaust velocity:

−

T = m˙ V V 0 j T 20000 V + V 0 = + 236.644 = 347.756 m/s j = ˙ m˙ 180 E) It is known that the given equation for thermal eﬃciency can be rewritten to: ηth =

P 2 = 1+ m˙ V − V eng

1 2

2

j

2 0

V j V 0

Filling in the airspeed V 0 and the given jet exhaust speed one obtains: ηth =

Answers to Test Module A

2 1+

380 236.644

= 0.768 = 76.8%

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AE1110x - Introduction to Aeronautical Engineering

Exercise 13 A) Given that the helium-ﬁlled balloon is designed to keep the skycar in the air, we know that the total lift force of the balloon should equal the weight of the skycar. Adding the skycar’s mass and the back-up system mass we ﬁnd a total mass of 1178 kg. This allows us to equate the weight of the skycar to the balloons lift, as follows:

ρV g 1

gas − M M ai r

L = W

= mg

We know the air density (0.86 kg/m 3 ), the gravitational acceleration (g = 9.80665 m/s 2 ), the molar mass of the helium (M He = 4.0 g/mol) and the molar mass of air ( M ai r = 28.97 g/mol). Hence we can ﬁll in the equation and solve for the volume:

M ρV g 1 − = mg M 4.0 gas ai r

0.86 V 9.80665 1

· ·

= 1178 9.80665

− 28.97

·

7.26924V = 11552.2 V = 1589.2 m3 B) The temperature the helium gas is stored at can be computed using the equation of state (since we are given both the pressure and the density). To use this equation of state however, we ﬁrst have to determine the speciﬁc gas constant for the helium gas. Given the value of the universal gas constant and the molar mass of helium, we compute: R =

R

M H e 8.314 J/(mol K) = 4.0 g/mol 8.314 J/(mol K) = = 2078.5 J/(kg K) 0.004 kg/mol

· ·

·

Using the equation of state we can now compute that: p = ρRT p T = ρR 542000 = = 283.440 K = 10.29 C 0.92 2078.5 ◦

·

6

Answers to Test Module A

AE1110x - Introduction to Aeronautical Engineering

C) In this question we are to derive an equation for C mα for the skycar concept. For clarity, the given diagram is repeated below:

Figure 1: A schematic drawing of the skycar concept, with the forces in vertical directions, moments around the pitch axis and relevant lengths indicated.

As usual, the ﬁrst step to perform is to set up a moment equation for this vehicle. This moment equation is always set up around the centre of gravity. This gives (taking clockwise moments as positive): M cw = M ac 1 + M ac 2 + Lw 1 l w Lw 2 l w LH l H

· −

· −

·

The next step is to non-dimensionalise this equation. While doing so, use will be made of the following expressions: Lw 1 = Lw 2 = LH = M ac 1 = M ac 2 =

1 C L ρS1 V 2 2 w 1 1 C L ρS2 V 2 2 w 2 1 C LH ρSH V 2 2 1 C Ma c w ρS1 V 2 c¯ 2 1 C Ma c w ρS2 V 2 c¯ 2

Non-dimensionalising the equation means we divide everything by 1 2

M cw = ρSV 2 c¯

M cw = 1 2 ρSV c ¯ 2 C mto t =

1 2

M ac 1 M ac 2 Lw 1 l w + + 1 1 ρSV 2 c¯ 2 ρSV 2 c¯ 2 ρSV 2 c¯

1 2

C Ma c w ρS1 V 2 c¯

·

1 2

ρSV 2 c¯

+

1 2

C Ma c w ρS2 V 2 c¯ 1 2

C mac w S1 C mac w S2 + + S S

Using S 1 = S 2 = 12 S and V H =

−

·

ρSV 2 c¯ C Lw 1 S1 l w

SH l H S c ¯

S c¯

·

Lw 2 l w ρSV 2 c¯

·

1 2

+

−

1 2

ρSV 2 c¯ , so:

LH l H ρSV 2 c¯

−

·

1 2

C Lw 1 ρS1 V 2 l w

·

1 2

1 2

−

C Lw 2 ρS2 V 2 l w 1 2

ρSV 2 c¯ ρSV 2 c¯ C Lw 2 S2 l w C LH SH l H S c¯ S c¯

·

·

−

1 2

C LH ρSH V 2 l H 1 2

ρSV 2 c¯

·

−

gives:

1 l w C mto t = C mac w + C Lw 1 2 c¯

Answers to Test Module A

1 2

− 12 C L

w 2

l w c¯

− C L

H

V H

7

·

AE1110x - Introduction to Aeronautical Engineering

Having non-dimensionalised the moment equation, the next step is to take its derivative with respect to α. This gives the following: dC mac w 1 dC Lw 1 l w 1 dC Lw 2 dC mto t = + 2 dα c¯ 2 dα dα dα dC mac w 1 dC Lw 1 l w 1 dC Lw 2 dC mto t = + 2 dα c¯ 2 dα2 dα dα 1 l w 1 l w C mα = 0 + C Lαw C Lαw 0.90 2 c¯ 2 c¯ l w C mα = 0.05 C Lαw c¯ 0.87 C LαH V H

l w dC LH V H c¯ dα dα2 l w dC LH dαH V H dα c¯ dαH dα

−

−

−

− −

·

−

− C L · 0.87 · V H

·

·

αH

·

D) We are given that the derivative of the non-dimensionalised moment equation with respect to the angle of attack α is given by: C mα = 0.11 C Lαw

·

l w c¯

− 1.7 · C L · V H αH

This allows us to compute the minimum horizontal tail size for longitudinal static stability (normally you would use the ’real’ equation just derived above). It is known that the value of C mα should be smaller than zero, so:

· l cw¯ − 1.7 · C L · V H l w 1.7 · C L · V H > 0.11 · C L c¯ l 0.11 · C L c V H > 1.7 · C L 0.11 · 0.1 . V H > 1.7 · 0.095 SH · l H > 0.210654 0 > 0.11 C Lαw

αH

αH

αw

w αw ¯ αH

3 0 97

S c¯

SH > 0.210654

· Sl c¯ H

16.2 0.97 SH > 0.210654 6 2 SH > 0.552 m

·

·

So the horizontal tail surface area should be bigger than 0.552 m 2 to provide longitudinal static stability. E) Answers b & c are correct here. The ﬁrst answer is incorrect, since the tail surface also produces drag. Besides, in many conventional aircraft the tail actually often produces negative lift to keep the aircraft in balance. A larger horizontal tail surface area increases the static margin (makes the aircraft longitudinally more stable), hence the allowed range of centre of gravity positions increases. This means the pilot has to worry less about the centre of gravity position of his aircraft and can place cargo more freely. For a given centre of gravity position, this eﬀect also means the aircraft becomes more stable and hence easier to ﬂy. The agility decreases when stability increases.

8

Answers to Test Module A

AE1110x - Introduction to Aeronautical Engineering

Exercise 14 A) We are given the stall speeds of the aircraft in equivalent airspeed, so at the speed corresponding to a standard air density of ρ = 1.225kg/m3 . Using this air density, the given wing area, the given mass and the given airspeeds, we can compute the lift coeﬃcients. The only thing that needs to be done is convert the airspeeds to m/s, so: V 0% ﬂaps = 99kts = 99 0.5144444 = 50.930 m/s

· V 100% ﬂaps = 78kts = 78 · 0.5144444 = 40.127 m/s Now the lift coeﬃcients are found by setting lift equal to weight (as this is steady ﬂight), giving: 1 C L ρSV 2 = mg 2 mg C L = 1 ρSV 2 2 4762.72 9.80665 = 1.04 0% Flaps: C L = 1 1.225 28.2 50.9302 2 4762.72 9.80665 100% Flaps: C L = 1 = 1.68 2 1.225 28.2 40.127 2

·

·

·

·

· ·

· ·

B We already know the equivalent stall speed of the aircraft without ﬂaps (99 kts, or 50.930 m/s). Therefore the only thing we need to do is convert this equivalent airspeed to a true airspeed at an altitude of 30,000 ft. This means we must ﬁrst compute the air density at an altitude of 30,000 ft (or 30,000 0.3048 = 9144 m).

·

To compute the temperature at this altitude we use Toussaint’s equation: T 9144 = T 0 + a (h9144 = 288.15 +

−h ) 0

−0.0065 (9144 − 0) = 228.714 K

From this temperature we can compute the air pressure at this altitude, using the relation between temperature and pressure in a layer with a temperature lapse rate: g aR

−

T T T = p T 228.714

p 9144 = p 0

9144 0

g aR

−

p 9144

9144

0

0

p 9144 = 101325

·

288.15

−9.80665 0.0065 287.00

−

·

= 30082.833 Pa Now knowing both the temperature and the pressure, we can use the equation of state to ﬁnd the air density at this altitude: p = ρRT 30082.833 p = = 0.4582 kg/m3 ρ = RT 287.00 228.714

·

Answers to Test Module A

9

AE1110x - Introduction to Aeronautical Engineering

Now this air density and the air density at sea level are used to convert from the given equivalent airspeed to the true airspeed: V T AS

ρ = V ρ 1.225 0

EAS

=

0.4582

· 99 kts = 161.87 kts = 83.27 m/s

C On the given warm day the temperature is given to be 25 C (or 298.15 K) and the air pressure is 998 hPa, or 99,800 Pa. In that case we can use the equation of state to compute the air density at sea level: ◦

p = ρRT 99800 p = = 1.1663 kg/m3 ρ = RT 287.00 298.15

·

Now this ’true’ air density can be used to convert between equivalent airspeed and true airspeed, as follows: 0% Flaps: V T AS

100% Flaps: V T AS

ρ = V ρ 1.225 = · 99 kts = 101.46 kts 1.1663 ρ = V ρ 1.225 0

EAS

0

EAS

= Given the temperature of 25

◦

1.1663

· 78 kts = 79.94 kts = 41.12 m/s

(or 298.15 K) on this day, the speed of sound is known to be:

√ a = γRT = 1.4 · 287.00 · 298.15 = 346.12 m/s Given the true airspeed at which stall occurs with full ﬂaps (41.12 m/s) we compute: M s tall =

10

V T AS 41.12 = = 0.1188 346.12 a

Answers to Test Module A

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