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Test Module A - Answers

Delft University of Technology

Prof. dr. ir. Jacco Hoekstra Ir. Jos Sinke Tim van Leeuwen

B. Tome - CC - BY

AE1110x - Introduction to Aeronautical Engineering

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

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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 CL = 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 CL 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 Casting. 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 Density 7.8 Strength 280 Specific strength aluminium = = = 103.7 MPa · dm3 /kg Density 2.7 410 Strength = = 186.4 MPa · dm3 /kg Specific strength composite = Density 2.2 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).

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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/m3 . 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 ρcruise Tcruise aR −1 = ρ0 T0   −9.80665 Tcruise −0.0065 · 287.00 −1 0.42 = 1.225 288.15   Tcruise 5.25684803−1 0.342857 = 288.15   Tcruise 4.25684803 0.342857 = 288.15 √ Tcruise 4.25684803 = 0.342857 = 0.77766193 288.15 Tcruise = 0.77766193 · 288.15 = 224.0833 K

Now we can use Toussaint’s equation to determine the aircraft altitude: Tcruise = T0 + a (hcruise − h0 ) 224.0833 = 288.15 + −0.0065 (hcruise − 0) 224.0833 − 288.15 hcruise = = 9856.4 m = 32337.3 ft −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 VT AS = 460 kts · 0.5144444 = 236.644 m/s. Now we can compute the lift coefficient, according to: L=W 1 CL ρSV 2 = mg 2 mg CL = 1 2 2 ρSV 30000 · 9.80665 = 1 = 0.283 2 2 · 0.42 · 88.3 · 236.644 Knowing the lift coefficient, we can now compute the drag coefficient using: CL2 πAe The only unknown in this equation is the aspect ratio. However, given the wing surface area and 2 2 the wingspan we can compute A = bS = 23.7 88.3 = 6.361. Substituting gives: CD = CD 0 +

CD = 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 = CD 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

Pa1 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: T = m˙ Vj − V0




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: Ainlet = π

2 dinlet 1.11762 =π = 0.981 m2 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: T = m˙ Vj − V0




Since the thrust T , the mass flow m˙ and the airspeed V0 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˙ Vj − V0 T 10000 Vj = + V0 = + 236.644 = 347.756 m/s ˙ 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˙ Vj − V0 T 20000 Vj = + V0 = + 236.644 = 347.756 m/s ˙ m˙ 180 E) It is known that the given equation for thermal efficiency can be rewritten to: ηth =

Peng 1 ˙ 2m



2

Vj −

V02

=

2 1+

Vj V0

Filling in the airspeed V0 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: L=W   Mgas ρV g 1 − = mg Mair We know the air density (0.86 kg/m3 ), the gravitational acceleration (g = 9.80665 m/s2 ), the molar mass of the helium (MHe = 4.0 g/mol) and the molar mass of air (Mair = 28.97 g/mol). Hence we can fill in the equation and solve for the volume:   Mgas = mg ρV g 1 − Mair   4.0 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 MH e 8.314 J/(mol · K) = 4.0 g/mol 8.314 J/(mol · K) = = 2078.5 J/(kg · K) 0.004 kg/mol

R=

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

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Answers to Test Module A

AE1110x - Introduction to Aeronautical Engineering

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

Lw1

LH

Lw2

Mac

Mac

c.g.

lw

lw

W

lH

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): Mcw = Mac1 + Mac2 + Lw1 · lw − Lw2 · lw − LH · lH The next step is to non-dimensionalise this equation. While doing so, use will be made of the following expressions: Lw1 = Lw2 = LH = Mac1 = Mac2 =

1 CL ρS1 V 2 2 w1 1 CL ρS2 V 2 2 w2 1 CL ρSH V 2 2 H 1 CM ρS1 V 2 c¯ 2 acw 1 CM ρS2 V 2 c¯ 2 acw

Non-dimensionalising the equation means we divide everything by 12 ρSV 2 c¯, so: Mcw 1 2¯ 2 ρSV c

=

Mcw = 1 2¯ 2 ρSV c Cmtot =

Mac1 Mac2 Lw · lw Lw · lw LH · lH + 1 + 1 1 2 − 1 2 2 − 1 1 2 2 2¯ ¯ ¯ ¯ ¯ 2 ρSV c 2 ρSV c 2 ρSV c 2 ρSV c 2 ρSV c 1 1 2 2 1 1 2¯ 2¯ 2 CLw1 ρS1 V · lw 2 CLw2 ρS2 V · lw 2 CMacw ρS1 V c 2 CMacw ρS2 V c + + − 1 1 1 1 2¯ 2¯ 2¯ 2¯ 2 ρSV c 2 ρSV c 2 ρSV c 2 ρSV c

−

1 2 2 CLH ρSH V · lH 1 2¯ 2 ρSV c

CLw2 S2 · lw Cmacw S1 Cmacw S2 CLw1 S1 · lw CL SH · lH + + − − H S S S c¯ S c¯ S c¯

Using S1 = S2 = 12 S and VH =

SH · lH S c¯

gives:

1 lw 1 lw Cmtot = Cmacw + CLw1 − CLw2 − CLH VH 2 c¯ 2 c¯

Answers to Test Module A

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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: dCmacw dCLH dCmtot 1 dCLw1 lw 1 dCLw2 lw = + − − VH dα dα 2 dα c¯ 2 dα c¯ dα dCmacw dCLH dαH dCmtot 1 dCLw1 lw 1 dCLw2 dα2 lw = + − − VH dα dα 2 dα c¯ 2 dα2 dα c¯ dαH dα 1 lw 1 lw − CLαw · 0.90 − CLαH · 0.87 · VH Cmα = 0 + CLαw 2 c¯ 2 c¯ Cmα = 0.05 · CLαw lcw¯ − 0.87 · CLαH · VH

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

lw − 1.7 · CLαH · VH c¯

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 Cmα should be smaller than zero, so: 0 > 0.11 · CLαw 1.7 · CLαH · VH > 0.11 · CLαw

lw − 1.7 · CLαH · VH c¯ lw c¯

VH >

0.11 · CLαw lcw¯ 1.7 · CLαH

VH >

3 0.11 · 0.1 0.97 1.7 · 0.095

SH · lH > 0.210654 S c¯ S c¯ lH 16.2 · 0.97 SH > 0.210654 · 6 2 SH > 0.552 m

SH > 0.210654 ·

So the horizontal tail surface area should be bigger than 0.552 m2 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.

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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: V0% flaps = 99kts = 99 · 0.5144444 = 50.930 m/s V100% 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 CL ρSV 2 = mg 2 mg CL = 1 2 2 ρSV 4762.72 · 9.80665 = 1.04 0% Flaps: CL = 1 2 2 · 1.225 · 28.2 · 50.930 4762.72 · 9.80665 100% Flaps: CL = 1 = 1.68 2 2 · 1.225 · 28.2 · 40.127

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: T9144 = T0 + a (h9144 − h0 ) = 288.15 + −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 T9144 aR T0   −g T9144 aR = p0 T0   −9.80665 228.714 −0.0065 · 287.00 = 101325 · 288.15

p9144 = p0 p9144 p9144



= 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 p 30082.833 ρ= = = 0.4582 kg/m3 RT 287.00 · 228.714 Answers to Test Module A

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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: r ρ0 VT AS = VEAS ρ r 1.225 · 99 kts = 161.87 kts = 83.27 m/s = 0.4582 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 p 99800 ρ= = = 1.1663 kg/m3 RT 287.00 · 298.15 Now this ’true’ air density can be used to convert between equivalent airspeed and true airspeed, as follows: r ρ0 0% Flaps: VT AS = VEAS ρ r 1.225 = · 99 kts = 101.46 kts 1.1663 r ρ0 100% Flaps: VT AS = VEAS ρ r 1.225 · 78 kts = 79.94 kts = 41.12 m/s = 1.1663 Given the temperature of 25 ◦ (or 298.15 K) on this day, the speed of sound is known to be: p √ 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: Mstall =

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VT AS 41.12 = = 0.1188 a 346.12

Answers to Test Module A