Exercises Lecture 1-Answers Introduction & Ballooning

Prof. dr. ir. Jacco Hoekstra Tim van Leeuwen

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Univ

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J. Mrachina - CC - BY - NC - ND

AE1110x - Introduction to Aeronautical Engineering

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

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

Figure 1: Ideal aircraft designs with their respective disciplines.

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

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

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

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

Answers to Exercises Lecture 1 - Introduction & Ballooning 1

AE1110x - Introduction to Aeronautical Engineering

Exercise 7D) 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 ofairships combined with their long length, which means their volume/displacement of air (causingdrag) is enormous.

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

Exercise 9The 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:

L = V g

(T

T + T

)L = 1.225 2500 9.81

(105

288.15 + 105

)= mg

m 9.81 = 8023.727m = 818kg

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

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

L = V g

(1 28

28.97

)= V g 0.0334829

This latter number should be equal to the TT+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 15degrees Celsius room temperature gives a balloon temperature of 24.98 degrees Celsius.

2 Answers to Exercises Lecture 1 - Introduction & Ballooning

AE1110x - Introduction to Aeronautical Engineering

Exercise 11Using the given air pressure and outside air temperature (which is 283.15 K), we compute the airdensity to be:

=p

RT=

101325

287.00 283.15 = 1.24686kg/m3

This gives us:

L = mg = V g

(T

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 = 165K, so T + T = 175 C.

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

V =4

3piR3 23

14=

4

3pi73 23

14= 2360m3

This gives again

L = mg = V g

(T

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 = 102K, so T + T = 112 C.

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

V =4

3piR3 =

4

3pi 0.163 = 0.01716m3

Now the lift of one balloon is given by (using a molar mass of 4.003 g/mol for helium gas and28.97 g/mol for air):

L = V g

(1 Mgas

Mair

)= 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 weneed

#bal loons =735.75N

0.1777N= 4141 4.1 103

We thus need roughly 4.1 103 balloons.

Answers to Exercises Lecture 1 - Introduction & Ballooning 3