Project-1: Powertrain for aircraft in runways

Project 1 - Powertrain for aircraft in runways

1. Search and list out the total weight of various types of aircrafts.
2. Is there any difference between the ground speed and the air speed?

3. Why is it not recommended to use the aircraft's engine power to move it on the ground at the airport?

4. How is an aircraft pushed to runway when its ready to takeoff?

5. Learn about takeoff power, tire design, rolling resistance, tire pressure and brake forces when landing.

6. With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle.
7. Design an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfill aircraft towing application. Estimate the duty cycle range to control the aircraft speed from zero to highest. Make all the required assumptions. Prepare a table of assumed parameters. Draw a block diagram of powertrain.

ANS 1:

Weight is a force that exists due to the force of gravity of the Earth that acts on a body with a mass. Airplanes are designed with numerous components of different sizes and weights such as the wings, the engine, the fuselage, the tail etc. Since every component of the airplane has a weight, the total weight of the airplane is given by the summation of the weights of each component. For instance:

`w = m*g`

`w_T = W_(fusel) + w_(eng) + w_(tail) + w_(payload) + .....`

So, the total weight of an aircraft that an engineer is designing is calculated by the summation of all the individual weights of each component of the aircraft. The weight of an aircraft is a very important factor to consider when designing an aircraft. This is because engineers need to know the maximum weight the aircraft can carry while travelling in the air with the maximum fuel capacity. In this way, engineers set limits and boundaries for an aircraft so that it operates safely without overloading the aircraft's structure or any issues that would damge the aircraft while in the air. These limits extend to the weights of the passengers as well as the baggage that each passenger is permitted to travel with (hand luggage & check in luggage). Other weights to consider are the weight that the aircraft fuel has and how that weight will decrease during the course of the flight as the fuel is being used up.

There are various different weights that are taken in to consideration when designing an aircraft and they are mentioned in the specifications of different aircraft. These various weights can be divided in to design weight limits and authorised weight limits. Design weight limits are the limits that engineers set during the design state of aircrafts and are the maximum permissible weights that the aircraft can carry. Authorised weight limits are those that are set by authorities in aircraft flight manuals (AFMs). The authorized weight limits may be lesser than or equal to the design weight limits but they can never exceed them during aircraft operation e.g. taxing, take off, landing, en-route.

The various nomenclature of weights pertaining to aircrafts are mentioned below:

Manufacturer's Empty Weight (MEW)

This is the weight of an aircraft as designed when it has been built by a manufacturer and this includes the structural weight, the furnishings, installations and other systems that are considered essential for the operation of the aircraft. This weight excludes the weight of passengers, payload, fuel etc.

Zero Fuel Weight (ZFW)

This is the total weight of an aircraft and all of its contents including the weight of the passengers and the payload and unusable fuel but it does not include the weight of the usable fuel. As an aircraft is en route to its destination, the fuel gets consumed and so the weight of the aircraft actually reduces over the journey however, the zero fuel weight remains constant.

Maximum Zero Fuel Weight (MZFW)

This is the maximum weight that an aircraft is allowed to carry before the usable fuel and other specified usable agents are loaded on to the aircraft in to specified regions of the aircraft. These agents include e.g. engine injection fluid, and other consumable propulsion agents.

Operating Empty Weight (OEW)

This weight is the basic weight of an aircraft which includes the weight of the aircraft crew, necessary fluids for operation e.g. engine coolant, water and all operator items and equipment required for the flight operation. This weight excludes the weight of the payload and the weight of the usable fuel.

Payload

The payload is the carrying capacity of any vehicle and is inclusive of cargo, the passengers, and extra fuel. In the case of commercial airliners, the payload is the weight of an aircraft that is revenue generating e.g. paying passengers, cargo.

Maximum Takeoff Weight (MTOW)

This weight is the maximum weight that the pilot of an aircraft is permitted to take off with. In practice, the maximum takeoff weight is maintained at a lesser value than the actual weight due to aircraft performance, environmental conditions, landing weight requirements etc.

Regulated Takeoff Weight (RTOW)

Depending on a variety of factors such as altitude, ambient air temperature, airfield characteristics, the regulated takeoff weight varies for different flights. It is referred to as regulated because it takes in to consideration numerous secondary factors which are applicable for certain flight conditions. The regulated takeoff weight must never exceed maximum takeoff weight.

Maximum Landing Weight (MLW)

This weight is the maximum weight at which an aircraft is permitted to land on to a runway or airfield with. The maximum landing weight must not exceed the maximum takeoff weight. In other words, an aircraft is permitted to land with a weight that is lesser than the weight at which it went in to takeoff.

Maximum Ramp Weight (MRW)

This weight is also known as the Maximum Taxi Weight (MTW) and it is the maximum weight that is permitted for maneuvering (taxiing/towing) an aircraft on the ground and is limited by the aircraft strength and air worthiness requirements. This weight is inclusive of the weights of the taxi and the run up fuel needed for the auxillary power unit (APU) and engines. This weight is greater than the maximum takeoff weight because the fuel weight included in the MRW will be consumed during taxiing/towing before takeoff. The difference between the MRW and MTOW is referred to as the maximum taxi fuel allowance and depends on the size of the aircraft, the number of engines, the APU fuel consumption etc.

Aircraft Gross Weight

This weight is also known as the 'all up weight' (AUW) and it is the total weight of an aircraft at any moment during the flight or ground operations. This weight will decrease during the flight operation due to fuel and oil consumption. This weight may also vary during the flight operation due to in - flight refuelling or payload dropping. When the aircraft releases its brakes, the aircraft gross weight is equal to the maximum takeoff weight.

The diagram below shows the various weights considered for aircraft operations:

The image below shows the various airliners listed by their maximum takeoff weight (MTOW):

Type MTOW [kg] MLW [tons] TOR [m] LR [m] ICAO category FAA category Antonov An-225 640,000 591.7 3500 Heavy Super Scaled Composites Model 351 Stratolaunch 589,670 3660 Heavy Super Airbus A380-800[1][2] 575,000 394 3100 1930 Heavy Super Boeing 747-8F 447,700 346.091 3100 1800 Heavy Heavy Boeing 747-8 443,613 306.175 3100 Heavy Heavy Boeing 747-400ER 412,770 295.742 3090 Heavy Heavy Antonov An-124-100M 405,060 330 2520 900 Heavy Heavy Boeing 747-400 396,900 295.742 3018 2179 Heavy Heavy Lockheed C-5 Galaxy[3][4][5] 381,000 288.417 2530 1494 Heavy Heavy Boeing 747-200[6] 377,840 285.700 3338 2109 Heavy Heavy Boeing 747-300[6] 377,840 260.320 3222 1905 Heavy Heavy Airbus A340-500[7] 371,950 240 3050 2010 Heavy Heavy Airbus A340-600[7] 367,400 256 3100 2100 Heavy Heavy Boeing 777F 347,800 260.816 2830 Heavy Heavy Boeing 777-300ER 351,800 251.29 3100 Heavy Heavy Boeing 777-200LR 347,450 223.168 3000 Heavy Heavy Boeing 747-100[6] 340,200 265.300 Heavy Heavy Airbus A350-1000 308,000 233.5 Heavy Heavy Boeing 777-300 299,370 237.683 3380 Heavy Heavy Boeing 777-200ER 297,550 213.00 3380 1550 Heavy Heavy Airbus A340-300[7][8] 276,700 190 3000 1926 Heavy Heavy McDonnell Douglas MD-11 273,300 185 2990 1890 Heavy Heavy Airbus A350-900 270,000 175 2670 1860 Heavy Heavy Ilyushin Il-96M 270,000 195.04 3115 2118 Heavy Heavy McDonnell Douglas DC-10 256,280 183 2990 1890 Heavy Heavy Boeing 787-9[9] 254,000 192.777 2900 Heavy Heavy Boeing 787-10[9] 254,000 201.849 Heavy Heavy Airbus A340-200[7][10] 253,500 181 2990 Heavy Heavy Ilyushin IL-96-300 250,000 175 2600 1980 Heavy Heavy Airbus A330-300[11][12] 242,000 185 2500 1750 Heavy Heavy Airbus A330-200[11][12] 242,000 180 2220 1750 Heavy Heavy Lockheed L-1011-500 231,300 166.92 2636 Heavy Heavy Boeing 787-8[9] 228,000 172.365 3300 1695 Heavy Heavy Lockheed L-1011-200 211,400 Heavy Heavy Ilyushin IL-86 208,000 175 Heavy Heavy Boeing 767-400ER 204,000 158.758 3414 Heavy Heavy Airbus A300-600R[13] 192,000 140 2385 1555 Heavy Heavy Boeing 767-300ER 187,000 136.08 2713 1676 Heavy Heavy Concorde 185,000 111.1 3440 2220 Heavy Heavy Airbus A300-600[13] 163,000 138 2324 1536 Heavy Heavy Boeing 767-300 159,000 136.078 2713 1676 Heavy Heavy Airbus A310-300[13] 157,000 124 2290 1490 Heavy Heavy Vickers VC10 152,000 151.9 Heavy Heavy Boeing 707-320B[14] 151,000 97.5 Heavy Heavy Boeing 707-320C[14] 151,000 112.1 Heavy Heavy Douglas DC-8-61 147,000 Heavy Heavy Airbus A310-200[13] 142,000 123 1860 1480 Heavy Heavy Airbus A400M 141,000 122 980 770 Heavy Heavy Douglas DC-8-32 140,000 Heavy Heavy Douglas DC-8-51 125,000 Medium Large Boeing 757-300 124,000 101.6 2550 1750 Medium Large Boeing 707-120B[14] 117,000 86.3 Medium Large Boeing 757-200 116,000 89.9 2347 1555 Medium Large Boeing 720B[15] 106,000 79.5 Medium Large Boeing 720[15] 104,000 79.5 Medium Large Tupolev Tu-154M 104,000 80 Medium Large Tupolev Tu-204SM 104,000 87.5 2250 Medium Large Convair 880 87,500 Medium Large Boeing 737-900 85,000 66.36 2500 1704 Medium Large Boeing 737-900ER 85,000 71.35 2804 1829 Medium Large Boeing 727-200 Advanced[16] 84,000 70.1 Medium Large Airbus A321-100[17] 83,000 77.8 2200 1540 Medium Large Boeing 737-800 79,000 65.32 2308 1634 Medium Large Boeing 727-200[16] 78,000 68.1 Medium Large McDonnell-Douglas MD-83 73,000 63.28 Medium Large Boeing 727-100[16] 72,500 62.4 Medium Large Boeing 727-100C[16] 72,500 62.4 Medium Large McDonnell-Douglas MD-90-30 71,000 64.41 2165 1520 Medium Large de Havilland Comet 4 70,700 Medium Large Boeing 737-700 70,000 58.06 1921 1415 Medium Large Airbus A320-100[17] 68,000 66 1955 1490 Medium Large Boeing 737-400 68,000 54.9 2540 1540 Medium Large de Havilland Comet 3 68,000 Medium Large Boeing 377 67,000 Medium Large Boeing 737-600 66,000 54.66 1796 1340 Medium Large Airbus A220-300 65,000 57.61 1890 1494 Medium Large Hawker Siddeley Trident 2E 65,000 Medium Large Airbus A319[17] 64,000 62.5 1850 1470 Medium Large Boeing 737-300 63,000 51.7 1939 1396 Medium Large Boeing 737-500 60,000 49.9 1832 1360 Medium Large Airbus A220-100 59,000 50.80 1463 1356 Medium Large Airbus A318[17] 59,000 57.5 1375 1340 Medium Large Boeing 717-200HGW 55,000 47.174 1950 Medium Large Douglas DC-7 55,000 Medium Large de Havilland Comet 2 54,000 Medium Large Boeing 717-200BGW 50,000 46.265 1950 Medium Large de Havilland Comet 1 50,000 Medium Large Douglas DC-6A 48,600 Medium Large Douglas DC-6B 48,500 Medium Large Embraer 190[18] 48,000 43 2056 1323 Medium Large Caravelle III 46,000 Medium Large Fokker 100 46,000 39.95 1621 1350 Medium Large Douglas DC-6 44,000 Medium Large Avro RJ-85 42,000 36.74 Medium Large Handley Page Hermes 39,000 Medium Large Embraer 175[19] 37,500 32.8 2244 1304 Medium Large Bombardier CRJ900[20] 36,500 33.345 1778 1596 Medium Large Embraer 170[21] 36,000 32.8 1644 1274 Medium Large Bombardier CRJ700 33,000 30.39 1564 1478 Medium Large Douglas DC-4 33,000 Medium Large Vickers Viscount 800 30,400 Medium Large Bombardier Q400 28,000 28.01 1219 1295 Medium Large Bombardier CRJ200 23,000 21.319 1918 1479 Medium Large ATR 72-600 22,800 22.35 1333 914 Medium Large Saab 2000 22,800 21.5 1300 Medium Large Embraer ERJ 145 22,000 19.3 2270 1380 Medium Large ATR 42-500 18,600 18.3 1165 1126 Medium Small Saab 340 13,150 12.930 1300 1030 Medium Small Embraer 120 Brasilia 11,500 11.25 1560 1380 Medium Small BAe Jetstream 41 10,890 10.570 1493 826 Medium Small Learjet 75[22] 9,752 8.709 1353 811 Medium Small Pilatus PC-24[23] 8,300 7.665 893 724 Medium Small Embraer Phenom 300[24] 8,150 7.65 956 677 Medium Small Beechcraft 1900D 7,765 7.605 1036 853 Medium Small Cessna Citation CJ4[25] 7,761 7.103 1039 896 Medium Small de Havilland Hercules 7,000 Medium Small Embraer Phenom 100 4,800 4.43 975 741 Light Small

ANS 2:

An aircraft has two different terms for the velocity at which it is moving with. The ground speed of an aircraft is a measure of how fast the aircraft is moving relative to a fixed point on the ground. It can also be thought of how fast the shadow of an aircraft is moving across land. The ground speed is a horizontal speed meaning that if the aircraft was moving vertically upwards or downwards, it would have no ground speed. The air speed however is the speed at which the aircraft is moving with its own power relative to the air it is flying in. The air speed is calculated by subtracting the wind speed from the ground speed. It can also be thought of the speed of the air flowing around the wings of an aircraft. This means that if there was no effect of wind resistance, the ground speed would be equal to the air speed. If an aircraft in the air is moving in the same direction as the wind blows, it will experience tail wind meaning that its ground speed is higher than its air speed. However, if the aircraft moves in the opposite direction to that of which the wind blows, it will experience head wind meaning that its ground speed is lower than its air speed.

Consider the example below:

Suppose an aircraft was travelling in the air with an air speed of 500 miles per hour (mph) and has to travel a distance of 2000 miles. If there was no wind at all, the air speed would be equal to the ground speed i.e. the ground speed would also be 500mph. This would mean that that the aircraft would arrive at its destination in 4 hours.

However, if there was a wind blowing with a speed of 100mph in the opposite direction to that of which the aircraft is moving, the aircraft would experience headwind. So, we know that the ground speed would be lesser than the air speed in this condition.

`V_(ground) = V_(air) + V_(w)`

If we were to consider the direction of the aircraft's velocity as positive, then the wind velocity would be negative. This would rearrange the equation to:

`V_(ground) = V_(air) - V_w`

`V_(ground) = 500 - 100`

`V_(ground) = 400mph`

So, we can see that the ground speed of the aircraft is 400mph even though the air speed of the aircraft is 500mph. Due to this, the time taken for the aircraft to reach its destination will extend with it now taking 5 hours.

Consider the opposite effect:

If the wind was blowing with the same speed of 100mph in the same direction as that of which the aircraft is moving, the aircraft would experience tailwind now. In this condition, the ground speed of the flight will be greater than the air speed.

As the wind velocity and the air speed are in the same direction, we can consider both as being in the positive direction.

`V_(ground) = V_(air) + V_(w)`

`V_(ground) = 500 + 100`

`V_(ground) = 600mph`

So, we can see that the ground speed is 100mph greater than the air speed and this will mean that the aircraft will arrive at its destination in 3 hours and 20 minutes.

ANS 3:

Unlike ground based vehicles, airplanes do not have a direct connection between the engine power and the wheels. The engines of an aircraft produce a thrust by sucking in air from the front and accelerating it backwards through a jet engine or propeller. This thrust action is what allows the aircraft to move forward. Whether an aircraft is travelling in the air or on the ground, the way in which it uses the propulsive power from the jet engines is the same whether it has to move forwards or backwards. This means to say that an aircraft uses its jet engines for propulsion in the air and on the ground.

"Pushback" is the procedure in which a tug vehicle is towed to the front of the airplane and are used to help reverse the airplane to tug it to a safe distance from the terminal/gate before takeoff. Tug vehicles are used for this purpose because even though they have a small size compared to the airplane, they have high torque and mass. In spite of using the tug vehicles for the pushback procedure, the engines of an airplane are more than capable of performing this procedure on their own. This can be done by using a 'reverse thrust' wherein the blades of the propeller are twisted in order for them to be able to take in air from behind the aircraft as opposed to the front and accelerate it forward. In this way, the airplane can use this propulsive power to move backwards however, it is very rarely used.

This is because using the reverse thrust to push back from the terminal/gate consumes a lot of engine power and is also a very noisy operation when done on the ground. The usage of reverse thrust to push back from the terminal/gate is also potentially hazardous. This is because the usage of the reverse thrust can cause debris to blow toward the airplane which creates a foreign object hazard. This foreign object hazard is dangerous in many ways. For instance, the ground staff are at risk, because the reverse thrust can create an air flow which can blow them away and damage their ground operations equipment. The passengers in the airplane are also at risk because the debris can damage the jet engines of the aircraft which can create a fire hazard that can create an explosion due to the flammability of the fuel. The debris can also pose a risk to the potential passengers waiting at the terminal/gate because it can break through the glass windows. Due to the possible dangers, the tug vehicle is used for reversing the airplane until it is a safe distance from the terminal/gate and the airport ground staff. By using the taxiing vehicle, the fuel can be saved, and it can save the lives of the ground staff and their equipment and people at the terminal/gate.

In addition, some airplanes e.g. A380s or 747s, are very huge in size and their pilots would have great difficulty in reversing their airplanes from the terminal/gate. The person who maneuvers the tug vehicle has better visibility of the ground and will be more aware of the space around the airplane. Because of this, it is better that for the push back from the terminal/gate, a tug vehicle is used. From here, the airplane will use its own engine power to propel itself along the runway and this continues on to takeoff as well.

ANS 4:

Once all the passengers have boarded and the pilots have authorization from the flight control, the aircraft begins the first stage of a flight procedure which is called 'push back'. In this stage, the front of the airplane is connected to a tug vehicle which helps to push back i.e. reverse the airplane from the terminal/gate. The airplane pushes back from the terminal/gate and once this is done, it is ready for "taxiing".

The image below shows an airplane with a tug vehicle connected to the front side of it used for the push back procedure.

Taxiing is the next procedure before flight in which the aircraft moves on the ground using its own power on its wheels without the assistance of any tug or towing vehicles. The same propulsive power that the jet engines produce for thrust during flight is used for moving forwards on the ground. Since the taxiing procedure occurs with the engine power of an aircraft, this procedure only occurs once an aircraft is at a safe distance from the terminal/gate, as well as from the ground staff after pushback. This is because there are several dangers that exist when an aircraft uses the thrust from the jet engines for movement on the ground. It is for this reason that a pushback vehicle is used to guide the aircraft away from the potential danger zones before taxiing can take place. During the taxiing procedure, the aircraft will move along taxiways.

The diagram below shows two aircrafts in their taxiing procedure. The aircraft on the top right however has the right of way.

Taxiways are paths on an airport ground which connect runways, terminals/gates, hangars and other facilities. The airplane moves along these taxiways at a relatively low speed of 20 - 30 knots (23 - 35 mph or 37 - 56kmph) because of speed limits for safety reasons set by operators and aircraft manufacturers. The taxiways like roads, have markings on their surface to show the direction in which to conduct the taxiing procedure in. Also, the taxiways have sign boards which are used to identify runways, taxiways etc. These sign boards may be either operational guidance signs or mandatory instruction signs. Taxiways also use various types of lights and are mainly used for the purpose of guiding an aircraft to the runway from a taxiway or vice versa when the visibility is low.

Operational guidance signs:

The 'A' symbol is a location sign for a taxiway whereas the 'B' symbol refers to the direction to the taxiway with designation 'Bravo'.

Mandatory instruction signs:

This symbol is used as the stop sign on taxiways.

Taxiway lights:

ANS 5:

Takeoff Power

Takeoff is the phase of flight in which an aircraft leaves the ground and becomes airborne. To be able to takeoff, an aircraft's wings are designed with an airfoil shape which is able to generate a lift force when moved relative to the air (when the aircraft moves). This lift force is needed to lift an aircraft off of the ground during takeoff and is also needed in flight to keep the aircraft in the air. The lift force is generated because of the way in which air flows around the airfoil shape. When the airfoil moves relative to the air, a "down wash" is created which creates a pressure gradient between the top and the bottom of the airfoil. The bottom of the airfoil will experience a higher pressure whereas the top of the airfoil will experience lower pressure. Due to this pressure gradient, a lift force is generated. The amount of force that is generated depends on 3 factors:

• the air speed
• the angle of attack
• the shape of the airfoil

The angle between the reference line of an airfoil and the direction of the velocity of the wind is referred to as the "angle of attack". The higher the angle of attack, the greater the down wash and hence the greater the lift force that is generated. The greater the velocity of the air flowing over the airfoil will also increase the lift force. Similarly, changing the shape of the airfoil also affects the lift force. The shape of the airfoil can be altered by the use of slats and flaps. The slats and flaps increase the camber of the airfoil and allows it to generate more lift force at lower speeds. When the slats and flaps are activated, they extend and alter the shape of the airfoil in such a way that they increase the area of the wings which increases the lift force.

To control the lift force that the aircraft experiences, there are other components aside from the slats and flaps that can be adjusted. One component are the ailerons located on the wings. The ailerons have flaps also which can move up to generate less lift force and down to generate more. Other components are found at the tail of the aircraft and are called the rudders and the elevators. Adjustment of the elevators allow for the control of the vertical forces whereas adjustment of the rudder allows for the horizontal forces to be controlled.

During takeoff, the lift force generated needs to be greater than the weight of the aircraft. All the three aforementioned techniques of the airfoil are utilised to generate sufficient lift force that makes takeoff successful. First, the speed of the aircraft needs to increase by increasing the thrust force from the jet engines. This will generate some lift force but to increase it further, the pilots activate the slats and flaps. When the aircraft has accelerated to the required speed for takeoff, the pilots then activate the elevators upwards which produces a tail force tilting the aircraft upwards. In doing so, the angle of attack of the airfoil is increased, which produces a sudden increase in the lift force that makes the aircraft takeoff. An angle of attack of 15 degrees is needed for takeoff and is maintained during ascension.

The diagram below shows how the angle of attack is measured with respect to the reference line e.g. chord line on the airfoil.

The images below show the slat and flat on the airfoil. The first image shows their default position and the image below it shows their adjusted position for more lift.

Takeoff Performance

The takeoff performance of an aircraft can be predicted by measuring its acceleration on the runway and observing the forces acting on it. Using Newton's second law of motion, the resultant force, F acting on a body is given by:

`F = ma`

`F_t - F_d - F_r = ma`

`a = (F_t - F_d - F_r)/m`

where,

Thrust force - `F_t`

Drag Force - `F_d`

Rolling resistance force - `F_r`

acceleration - `a`

mass - `m`

This acceleration will continue until the aircraft achieves a safe flying speed referred to as the rotation velocity, Vr. At this speed, the pilot can rotate the aircraft to an attitude which will create enough lift to takeoff from the ground. The rotation velocity i.e. takeoff speed depends on various factors e.g. air density (affected by air temperature and field elevation), aircraft gross weight and the aircraft configuration (slats and flaps position). Typical takeoff speeds for commercial jetliners range from 240 - 285kmph or 149 - 177mph. For smaller aircrafts, this speed becomes lower with light aircrafts e.g. Cessna 150 having a takeoff speed of 100kmph.

Aircrafts that operate in the transport category use a system of "V-speeds" and there are 3 of them: V1, V2 and Vr.

V1 is referred to as the decision speed (critical engine speed) and it is the speed at which any decision to abort takeoff must be made. If the aircraft exceeds V1, it must continue on to takeoff unless there is a reason to believe that takeoff will be unsuccessful. For instance, if an engine failure is detected before V1, takeoff can and should be aborted. Vr is the rotational speed as mentioned before and it must be greater than V1. V2 is the takeoff safety speed which is the speed at which the aircraft must achieve at the 35 feet height at the end of the runway distance.

The rotational speed is a function of the aircraft gross weight and the flap settings but it also depends on the environmental conditions. The rotational speed is usually determined as being 1.1 times greater than the stall speed. The stall speed is the lowest speed at which an aircraft can be flown before the airflow begins to separate from the wings and the angle of attack becomes too large. Since we know that the lift force must be equal to the weight of the aircraft for a flight to be maintained at a constant level, the stall speed is calculated as shown below:

`V_(stall) = sqrt((2*W)/(C_(lmax)*rho*A)`

where,

Stall speed - `V_(stall)`

Weight of the aircraft - `W`

Coefficient of lift at stall speed - `C_(lmax)`

Air density - `rho`

Wing area - `A`

The rotational speed is given by,

`V_r = 1.1 * V_(stall)`

Tire Design

Aircraft tires are designed in such a way that they can withstand extremely high loads for short durations of time. The number of tires on an aircraft is proportional to its weight, meaning that the heavier the aircraft is, the more tires it will need. This is because for a given weight of an aircraft, if the number of tires is insufficient, the weight cannot be distributed evenly enough amongst the tires. The tread patterns on aircraft tires are designed in such a way that they can provide stability to flights during crosswind conditions. The tread design also helps to direct water off of the tires to prevent hydroplaning and they also aid in the braking action.

Aircraft tires are designed with fusible plugs. A fusible plug is a threaded metal cylinder made of bronze, brass and gunmetal with a drilled hole that extends through its entire length. This hole is filled with a metal of low melting point. When the temperature of the metal exceeds its melting point, this will cause the deflation of the tires. During emergency landings or aborted takeoffs, an aircraft’s wheels experience high friction with the runway surface and the use of the plugs will control the deflation of the tires in a safe manner. Without them, there is a risk of an explosion which can damage the aircraft and other nearby objects.

Tire Pressure

Aircraft tires are inflated with an inert gas and the most commonly used is Nitrogen. Inert gases are used for inflation because they are non – reactive and not flammable. When the ambient air temperature and pressure changes, the Nitrogen gas helps to minimize the expansion and contraction of the aircraft tires. Using oxygen for the inflation of aircraft tires is very dangerous since it is flammable. Using Nitrogen instead eliminates any risk of there being a tire explosion.

General commercial airliners use tires that can withstand a pressure of 800psi before they burst. A Boeing 777-300ER has 12 tires and each one is inflated to a pressure of 220psi and has a diameter of 134cm and a weight of 120kg. Each tire costs $5000 and has to be changed after 300 cycles whereas the brakes are changed every 2000 cycles.

It is very important to check that the aircraft tires have been properly inflated using a calibrated gauge. This is because aircraft tires operate at high temperatures and pressures and they do not retain their air perfectly. The tire pressure can reduce by 5% in 24 hours. Michelin, one of the 4 main aircraft tire manufacturers, recommends that the air pressure of the tires be checked before the first flight that day or before each flight if the aircraft isn’t flown daily.

It is better to check before flight rather than after because the tires need to be cold when performing the pressure check. If we were to perform the check after the flight, the tires would be very hot due to its recent usage and this makes it very difficult to determine what the proper pressure should be. Michelin recommends that the tire pressure is checked no earlier than 3 hours before a flight after it has last been in use.

When performing the pressure checks during servicing, we need to consider the ambient temperature’s effect on the tires. Every 3 degree change in the ambient temperature results in a 1% change in the tire pressure. To account for this, manufacturers such as Michelin recommend that the tires be serviced in a way that the tire pressure is overshot to 105% of the operating pressure. So it is a guideline that the highest pressure in the aircraft maintenance manual (AMM) be targeted.

There are two main types of tires used for aircraft applications:

• Bias ply tire
• Radial ply tire

Bias ply tires are used for their long durability and hence long working life as well as for their retreadability. Their construction is explained below:

Tread

The tread is made of rubber and other additives that are used to obtain desired properties such as high durability, high toughness and high resistance to wear. The pattern of the tread has a ribbed design which helps provide excellent traction in different runway conditions whether good or bad.

Sidewall

This is a protective layer of rubber that covers the outer casing ply and it extends from the tread edge to the bead area.

Tread Reinforcing Ply

This consists of one or more layers of a fabric that strengthens and stabilizes the tread area when the aircraft ground speed is high. It also helps as a reference for the buffing process during retreading of the tires.

Buff Line Cushion

This is a cushion made of a rubbery fabric that increases the adhesion between the tread reinforcing ply and the breakers.

Breakers

These are reinforcing plies of rubber coated fabric that are placed under the buff line cushion and help to stabilize and strengthen the tread area.

Casing Plies

These are alternate layers of rubber coated fabric which run at opposite angles to each other and provide the strength of the tires.

Wire Beads

These are hoops of high tensile strength steel which anchor the casing plies and provide a firm mounting surface on the wheel.

Apex Strip

This is a wedge of rubber fixed to the top of the bead bundle.

Flippers

These are layers of rubberized fabric which help to anchor the bead wires to the casing. They also improve the durability of the tire.

Ply Turnups

The casing plies are wrapped around the wire beads to anchor them which forms ply turnups.

Chafer

This is a protective layer or rubber or fabric located between the casing plies and the wheel to keep chafing to a minimum.

Liner

In a tubeless tire, the liner is a layer of low permeability rubber that acts as a built in tube and prevents the diffusion of gas in to the casing plies. In tube tires, the liner layer is thinner to prevent chafing between the tube and the inside ply.

The image below shows a diagram of a bias ply tire.

Radial ply tires are used in aircrafts because they have a rigid belt that allows them to be used for increase landings and have lower rolling resistance than bias ply tires. They also weigh lesser than bias ply tires and have fewer components and are yet the same size. The different components of radial ply tires is mentioned below:

Overlay

This is a layer of reinforcing rubber coated fabric which is placed on top of the rigid belts and help them when aircrafts move with high ground speed.

Belt Plies

This is a composite structure which stiffens the tread area and increase the tire strength in the tread are for increased landings.

Casing Plies

These are the same as those found in bias ply tires in that they consist of layer of rubber coated fabric. But instead of running at opposite angles to each other, these run radially from the bead to the head.

Chippers

These are layers of rubber coated fabric which are applied at diagonal angles to improve the durability of the tires in the bead area.

The image below shows a diagram of a radial ply tire.

Rolling Resistance

The rolling resistance or rolling friction is a force that resists the motion of a body that rolls on a surface. In the case of aircrafts, the rolling resistance force exists between the aircraft tires and the runway surface it rolls upon. For a given body, its rolling resistance is mathematically expressed as the product of the coefficient of rolling resistance and the normal force, which is the force that is acting perpendicularly to the surface on which rolling action occurs.

`F_(rr) = mu_(rr) * N`

Where,

Rolling resistance force – `F_(rr)`

Coefficient of rolling resistance – `mu_(rr)`

Normal Force - `N`

From this, we know that the rolling resistance force is proportional to the normal force but also to the coefficient of rolling resistance which is dependent on various factors such as:

• Tire material
• Tire dimensions
• Tread thickness and shape
• Tire temperature
• Tire inflation pressure
• Roughness of the surface for rolling
• Presence/absence of liquids on the road

The above expression for the rolling resistance can only be considered when there are no other vertical forces acting on the rolling body. However during takeoff, we must consider the effect of the lift force as well. In this case, the normal force would be equal to the difference between the weight of the aircraft acting on the tire and the lift force acting on the wheel. This can be written mathematically as:

`F_(rr) = mu_(rr) * (W-L)`

In general, a standard tarmac runway is considered to have a coefficient of rolling resistance of 0.02.

Braking Forces

Aircrafts use brakes mainly during the landing procedure. During braking, the kinetic energy of the aircraft is converted to heat energy when friction occurs between the aircraft tires and the runway surface. Braking is a very crucial part of flight because there is only a limited distance on the runway that the aircraft can use to reduce its speed. Braking has to be done in such a way that the aircraft or its passengers or its equipment does not get damaged in the process. There are a few different systems that an aircraft uses to perform braking.

Working of Disc Brakes

The main brake that is used on an aircraft is the disc brake which work by manipulating the friction between the stationary disc and rotating disc found inside the disc brake. First, a command signal is sent to the disc brakes by the pilot pressing the brake pedal in the cockpit. This causes the actuators in the brake to move a piston to squeeze the stationary and rotary discs together. This creates a frictional force which is used to slow down the tires and hence the aircraft. As the friction in braking generates high amounts of heat energy, the disc brakes are designed as heat sinks to absorb this heat. Temperatures of the disc brake can reach 1800 degrees Celsius.

The disc brakes are found on a carrier assembly which consists of a torque tube that transmits braking torque to the landing gear. Placed between the carrier assembly and a backing plate, the discs are arranged in an alternating pattern of stators and rotors. Each rotor has grooves and notches that allow it to fit in with a wheel with corresponding geometry so that the rotor rotates with it. The stators are fitted to the torque tube and is stationary because the torque tube is connected to the axle and the landing gear structure both of which are stationary. Cylindrical spaces in the carrier assembly contain the pistons which move to cause the squeezing of the two discs bringing about braking.

Design of Brakes

The main design considerations of the disc brakes are the number of discs they will hold, the diameter of the discs, and their material. The brakes are also designed considering worst case scenarios. One such scenario is when takeoff is aborted at the decision speed, V1. If V1 was exceeded, then the aircraft cannot safely abort takeoff because it is at risk of not being able to stop before the end of the runway. In that scenario the brakes would need to be able to absorb more kinetic energy than in any other worst case scenario. Consider this scenario with an Airbus A380 which is the world’s biggest passenger plane. Its specifications are mentioned below:

• Maximum Takeoff Weight (MTOW) = 575000kg
• Maximum Decision Speed, V1 = 170 knots (195mph or 87 m/s)
• Number of Disc Brakes – 16

We need to calculate the kinetic energy of the aircraft ignoring the effects of air brakes, reverse thrust, headwinds and tailwinds.

`E_k = ½ * m*v^2`

`E_k = ½ * 575000 * 87^2`

`Ek = 2.2GJ`

This is the amount of energy that is required by the combined efforts of all brakes to completely brake the aircraft.

To be able to generate this much frictional force, there are multiple discs per brake are needed and brakes need to be on most, if not all of them. An A380 has 22 tires that are distributed on 5 landing gear legs. Two wheels support the nose on a leg, eight wheels split between 2 legs to support the left and right wings, and 12 body wheels split between 2 legs under the fuselage. Only 16 of these 22 wheels have brakes.

Each of the 16 brake assemblies has 5 rotors made of Honeywell Carbenix 4000 carbon – carbon composite. The brake assemblies within the wheels are made of 2014 T-6 aluminum alloy and are attached to the axle and landing gears which are made of 300M high strength alloy steel, titanium and aluminium components.

Materials

Until 1963, the most common material used in brake discs was steel. Later, Beryllium was used instead because of its better thermal properties however it was expensive to handle since its oxide was toxic. Eventually, the industry shifted towards carbon brakes. They are made of carbon fibers in a graphite matrix. They were used because they were lighter, more durable and were less thermally sensitive. They also had higher absorption of thermal energy and cooled faster than compared to steel brakes. Carbon has a higher specific heat of energy compared to steel meaning this can reduce the brake weight. Carbon has a higher thermal conductivity meaning the heat will be distributed uniformly throughout the disc. It has a lower thermal expansion, higher thermal shock resistance and a higher temperature limit than steel. In addition, carbon’s specific strength remains consistent under different temperatures unlike Steel and Beryllium can.

Brake Cooling

Brakes need to cool due to the high heat buildup and they do this in two main ways:

Passive cooling

This cooling occurs via natural conduction from the brake disc material to the surrounding components. Cooling occurs via radiation of heat that comes off the brakes as well as by natural convection as air flows past and through the brake assembly.

Active cooling

This is done using fans that force air through the brakes to cool them down.

ANS 6:

To calculate the force and hence the power required to push/pull an aircraft using a towing vehicle, we need to first consider the forces acting on the bodies in the system. There are 2 bodies, the aircraft and the towing vehicle and each vehicle experiences the rolling resistance, drag and acceleration forces.

Aircraft Calculations

The aircraft experiences a rolling resistance force when its tires move on the tarmac as well as a drag and acceleration force (initially).

Let us make the following assumptions:

Mass of airplane, `m_a = 60,000kg`

Coefficient of Rolling resistance between the aircraft tires and the tarmac, `mu_(rra) = 0.02`

Acceleration due to gravity, `g = 9.81 ms^-2`

Coefficient of Drag of aircraft, `C_(da) = 0.019`

Density of Air, `rho = 1.225kgm^-3`

Wing Span Area of aircraft wings, `A_a = 91.04m^2`

Final Velocity of aircraft, `v_a = 5ms^-1`

Initial velocity of aircraft, `u_a = 0ms^-1`

Acceleration time of aircraft, `t_a = 10s`

Acceleration of Aircraft, `a_a`

Rolling Resistance Force

`F_(rra) = mu_(rra)*(m_a*g)`

`F_(rra) = 0.02*(60000*9.81)`

`F_(rra) = 11772N`

The rolling resistance force between the aircraft tires and the tarmac is 11,772N.

Drag Force

`F_(da) = 1/2*C_(da)*rho*A_a*(v_a)^2`

`F_(da) = 1/2*0.019*1.225*91.04*5^2`

`F_(da) = 26.49N`

The drag force experienced by the aircraft when it moves at a constant velocity of 5m/s is 26.49N.

Acceleration Force

`F_(aa) = m_a * a_a`

`F_(aa) = m_a*((v_a-u_a)/t_a)`

`F_(aa) = 60000*((5-0)/10)`

`F_(aa) = 30000N`

The acceleration force for the aircraft when it accelerates from rest to a velocity of 5m/s is 30,000N.

Towing Vehicle Calculations

The towing vehicle will experience the same forces as the aircraft but needs to be able to exert a thrust force to move itself and the aircraft.

The following assumptions are made:

Mass of Towing Vehicle, `m_t = 10,000kg`

Coefficient of Rolling Resistance between Towing Vehicle wheels and the tarmac, `mu_(rrt) = 0.04`

Acceleration due to gravity, `g = 9.81 ms^-2`

Coefficient of Drag of Towing Vehicle, `C_(dt) = 0.82`

Density of Air, `rho = 1.225kgm^-3`

Frontal Area of Towing Vehicle, `A_t = 8m^2`

Final Velocity of Towing Vehicle, `v_t = 5ms^-1`

Initial Velocity of Towing Vehicle, `u_t = 0ms^-1`

Acceleration time of Towing Vehicle, `t_a = 10s`

Acceleration of Towing Vehicle, `a_t`

Rolling Resistance Force

`F_(rrt) = mu_(rrt) *(m_t*g)`

`F_(rrt) = 0.04* (10000*9.81)`

`F_(rrt) = 3924N`

The rolling resistance force between the towing vehicle's tires and the tarmac is 3,924N.

Drag Force

`F_(dt) = 1/2*(C_(dt))*rho*A_t*(v_t)^2`

`F_(dt) = 1/2*0.82*1.225*8*5^2`

`F_(dt) = 100.45N`

The drag force experienced by the towing vehicle when it moves at a constant velocity of 5m/s is 100.45N.

Acceleration Force

`F_(at) = m_t*a_t`

`F_(at) = 10000*((v_t - u_t)/t_a)`

`F_(at) = 10000*0.5`

`F_(at) = 5000N`

The acceleration force of the towing vehicle when it accelerates from rest to a velocity of 5m/s is 5,000N.

So, the thrust force that needs to be produced by the towing vehicle to pull/push the aircraft will be equal to the summation of the resistive forces of both the towing vehicle and the aircraft when moving from rest to a constant velocity.

Assuming, no other losses of energy and that after the initial acceleration of both the towing vehicle and the aircraft, both travel at constant speed, we can find the thrust force:

`F_t = F_(rra) + F_(da)+ F_(aa) + F_(rrt) + F_(dt) + F_(at)`

`F_t = 11772+26.49+30000+ 3924+100.45+5000`

`F_t = 50822.94N`

`F_t = 50.82kN`

The total thrust force needed is 50.82kN.

The total power needed is given by:

`P = F_t * v`

`P = 50822.94*5`

`P = 254,114.70W`

`P = 254.11kW`

Total power needed by the towing vehicle to push/pull the aircraft from rest to a velocity of 5m/s is 254.11kW.

ANS 7:

So, we know that the power needed by the towing vehicle to push/pull the aircraft from rest to a velocity of 5m/s is 254.11kW.

The torque needed by the towing vehicle is given by:

`T = (F_t*r)/(G*eta)`

Let us make the following assumptions:

Radius of wheels of the towing vehicle, `r = 0.35m`

Gear Ratio of Transmission, `G = 10`

Efficiency of Gear system, `eta = 0.9`

`T = (50,882.94*0.35)/(10*0.9)`

`T = 1978.78Nm`

So, the torque required for the towing vehicle to to push/pull the aircraft is 1,978.78Nm.

We are required to design an electric powertrain for the towing vehicle in order for it to be able to perform the towing operation with aircrafts. We need to select a motor by considering the power rating and the energy requirement as well as the duty cycle needed to control the aircraft speed from rest to the highest speed. We also need a battery to supply the DC power to the power converter to convert it to AC power for the induction motor.

Motor

We need a motor that can generate a power rating of at least 254kW.

I have selected the 'TECO' IE2 efficiency 3-phase induction motor for the towing operation. This motor is capable of producing a power output of upto 315kW, and a torque of 2416Nm which is more than suitable for the towing operation requirements. The motor's efficiency is 95%. From the efficiency and the power needed for the towing operation, we can calculate the input power that the motor needs.

`eta = P_o/P_i`

`P_i = P_o/eta`

`P_i = 254.11/0.95`

`P_i = 267.48kW`

The input power needed by the induction motor is 267.48kW and this is more than the power needed for the towing vehicle for the towing operation. This is because we need to overcompensate the required power due to energy losses in the induction motor which is reflected in the induction motor's efficiency.

The rated voltage of the motor is 400V and the full load current is 439.1A.

Power Converter

We need a power converter that will convert DC power from a battery and supply as AC power to the induction motor. Therefore, we need to use an inverter and we need to determine the duty cycle it has to operate at to generate the required input power to the induction motor. We will assume that the inverter is 100% efficient.

We know that,

`V_o = V_(bat)*delta`

where,

Output Voltage - `V_o`

Battery Voltage - `V_(bat)`

Duty cycle - `delta`

We also know that, power is given by:

`P = V*I`

`P_(i) = V_i*I_i`

where,

Input Power - `P_i`

Input Voltage - `V_i`

Input Current - `I_i`

`V_i = P_i/I_i`

`V_i = (267.48*1000)/439.1`

`V_i = 609.16V`

The inverter needs an input voltage of 609.16V to produce the required 267.48kW input power in to the induction motor.

Similarly,

`P_(o) = V_(o)*I_(o)`

where,

Output Power - `P_o`

Output Voltage - `V_o`

Output Current - `I_o`

Assuming, that the full load current is the same for input and output i.e. `I_i = I_o` as we assume that the inverter is 100% efficient.

Then, we can rewrite both power equations as:

`P_i/V_i = P_o/V_o`

Since, we want to find the duty cycle which is a ratio of the output voltage to the input voltage,

`P_i/P_o = V_o/V_i`

So, the duty cycle can be determined from the ratio of the power input to the power output.

`delta = P_i/P_o`

`delta = 254.11/267.48`

`delta = 0.9 or 90%`

So, the duty cycle is found to be the same as the efficiency of the motor. The range of the duty cycle is from 0 to 90%. From the duty cycle, we can now find the output voltage of the power converter i.e. the average voltage.

`V_o = V_i*delta`

`V_o = 609.16*0.9`

`V_o = 548.24V`

Battery

We need to know which battery to select in order for it to supply the required power to the power converter and hence the battery.

We know that,

`E = P*t`

where,

Energy - `E`

Power input - `P`

Time - `t`

Assume that the time of transmission of energy from the battery to the power converter and hence the motor is equal to the time needed for the towing operation. Consider this time to be 2 minutes i.e. 120 seconds. Assume that the battery is 100% efficient with no losses during energy transmission.

`E = 267.48*120`

`E = 32097.60kJ`

So the battery needed for this operation should have an energy storage of at least 32,097.60 kJ.

But, batteries are rated in terms of kilowatt hours i.e. kWh. To convert kJ to kWh, we need to divide by 3600.

`E = 32097.60/3600`

`E = 8.92kWh`

So for the towing application, the battery needs to have a rating of atleast 8.92kWh.

Assumptions

Parameter	Value
Towing Vehicle Radius (m)	0.35
Gear Ratio	11
Gear Ratio Efficiency (%)	90
Motor Efficiency (%)	90
Rated Voltage (V)	400
Full Load Current (A)	439.1
Inverter Efficiency (%)	100
Input Voltage to Inverter(V)	609.16
Average voltage to motor (V)	548.24V
Duty Cycle (%)	90
Battery Efficiency (%)	100

Block Diagram

Conclusion

In conclusion, the required research on the various aspects of aircrafts is completed and a powertrain for a towing vehicle that can push/pull an aircraft has been designed consisting of a motor, power converter and a battery. The powertrain was designed based on assumed parameters and conditions.