Exercise 1 B) With the term ’multidisciplinary’ we mean that knowledge on many topics (aerodynamics, materials and structures, propulsion, mechanics, etc.) is needed to design and build an aircraft.

Exercise 2 The various aircraft designs with their associated disciplines are shown below.

Figure 1: ’Ideal’ aircraft designs with their respective disciplines.

Exercise 3 On December 17th, 1903, Orville Wright managed to ﬂy the Wright Flyer for 120 feet, or 36.5 metres. As of November 2013 (when Singapore Airlines cancelled its 15,345 kilometre non-stop service from Singapore to Newark), the longest non-stop ﬂight is the 13,804 kilometre long service from Sydney to Dallas.

Exercise 4 A& C) Modern aircraft (such as the Boeing 787 and the Airbus A350) are characterised by the use of modern materials (such as composites) and the ever increasing size of the engines.

Exercise 5 A , B & D ) Thicker wings oﬀer a number of advantages. Most importantly, thick airfoils are capable of generating more lift (for the same angle of attack) than (very) thin wings. A second advantage is that it is easier to make them structurally stiﬀ enough, due to their increased thickness. A ﬁnal advantage is that all this volume can be used to store fuel.

Exercise 6 In July 1936 the Hindenburg set a record time for a double crossing of the Atlantic ocean of 5 days, 19 hours and 51 minutes, or about 140 hours. Compare that to current aircraft, which can ﬂy back and forth in under 16 hours.

Answers to Exercises Lecture 1 - Introduction & Ballooning

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AE1110x - Introduction to Aeronautical Engineering

Exercise 7 D) The cylinder shape of airships is relatively aerodynamic, and though they do ﬂy at low altitude, this is not the reason for their high drag. The biggest problem is the enormous frontal area of airships combined with their long length, which means their volume/displacement of air (causing drag) is enormous.

Exercise 8 In this lecture series (or actually, in Aeronautical Engineering in general) we use the following equation of state: p = ρ · R · T This we do because it allows us to ’erase’ the number of moles n and the volume V (which we in general do not know) out of the equation of state.

Exercise 9 The only parameters we need to convert are the temperature diﬀerence (∆T = 120 − 15 = 105) and the outside air temperature in Kelvin (T = 288.15 K). Now the lift of this hot air balloon is found using:

∆T L = ρV g T + ∆T

L = 1.225 · 2500 · 9.81

105 288.15 + 105

= mg

m · 9.81 = 8023 .727 m = 818kg

So this means the total mass of the balloon (including payload) can be 818 kilograms.

Exercise 10 Given that the molar mass of nitrogen gas is 28 grams per mole and knowing that the molar mass of air is 28.97 grams per mole, we can compute that the lift of the nitrogen balloon is:

L = ρV g 1 −

28 28.97

= ρV g · 0.0334829

∆T This latter number should be equal to the T +∆ T factor in the lift equation for a hot air balloons. Solving this equation (using T = 288.15K) leads to a ∆T of 9.982 degrees. Adding this to the 15 degrees Celsius room temperature gives a balloon temperature of 24.98 degrees Celsius.

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Answers to Exercises Lecture 1 - Introduction & Ballooning

AE1110x - Introduction to Aeronautical Engineering

Exercise 11 Using the given air pressure and outside air temperature (which is 283.15 K), we compute the air density to be: 101325 p = = 1.24686kg/m3 ρ = RT 287.00 · 283.15 This gives us:

∆T L = mg = ρV g T + ∆T

780 · 9.81 = 1.24686 · 1700 · 9.81

∆T T + ∆T

∆T T + ∆T

= 0.368

Using this ratio and T = 283.15 K, we ﬁnd ∆T = 165 K , so T + ∆T = 175 C. ◦

Exercise 12 In this case the volume of the balloon (which is a sphere with a radius of 7 metres and then elongated by a factor 23/14 in length) is: V =

4 23 4 23 = π73 · = 2360m3 πR 3 · 3 14 3 14

This gives again

∆T L = mg = ρV g T + ∆T

780 · 9.81 = 1.24686 · 2360 · 9.81

∆T T + ∆T

∆T T + ∆T

= 0.265

Using this ratio and T = 283.15 K, we ﬁnd ∆T = 102 K , so T + ∆T = 112 C. ◦

Exercise 13 Given the radius of the balloon, we can compute the volume of the balloon: V =

4 4 πR 3 = π · 0.163 = 0.01716m3 3 3

Now the lift of one balloon is given by (using a molar mass of 4.003 g/mol for helium gas and 28.97 g/mol for air):

L = ρV g 1 −

M gas M air

= 1.225 · 0.01716 · 9.81 1 −

4.003 28.97

= 0.1777N

Given that we want to lift a 75 kilogram human (W = mg = 735.75N ) we can compute we need 735.75N #balloons = = 4141 ≈ 4.1 · 103 0.1777N 3 We thus need roughly 4.1 · 10 balloons. Answers to Exercises Lecture 1 - Introduction & Ballooning

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