Challenge

Project-1: Powertrain for aircraft in runways

POWERTRAIN OF AIRCRAFT

1.Weights of Airplanes

MTOW = Maximum take-off weight, MLW = Maximum landing weight, TOR = Take-off run (SL, ISA+15°, MTOW), LR = Landing run (SL, ISA+15°, MLW)

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 A380-800 575,000 394 3100 1930 Super 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 381,000 288.417 2530 1494 Heavy Heavy
Boeing 747-200 377,840 285.700 3338 2109 Heavy Heavy
Boeing 747-300 377,840 260.320 3222 1905 Heavy Heavy
Airbus A340-500 371,950 240 3050 2010 Heavy Heavy
Airbus A340-600 367,400 256 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 747-100 340,200 265.300 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 A340-300 276,700 190 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 787-9 254,000 192.777 2900 Heavy Heavy
Boeing 787-10 254,000 201.849 Heavy Heavy
Airbus A340-200 253,500 181 2990 Heavy Heavy
Ilyushin IL-96-300 250,000 175 2600 1980 Heavy Heavy
Airbus A330-300 242,000 185 2500 1750 Heavy Heavy
Airbus A330-200 242,000 180 2220 1750 Heavy Heavy
Lockheed L-1011-500 231,300 166.92 2636 Heavy Heavy
Boeing 787-8 228,000 172.365 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 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 A300-600 163,000 138 2324 1536 Heavy Heavy
Boeing 767-300 159,000 136.078 2713 1676 Heavy Heavy
Airbus A310-300 157,000 124 2290 1490 Heavy Heavy
Vickers VC10 152,000 151.9 Heavy Heavy
Boeing 707-320B 151,000 97.5 Heavy Heavy
Boeing 707-320C 151,000 112.1 Heavy Heavy
Douglas DC-8-61 147,000 Heavy Heavy
Airbus A310-200 142,000 123 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 117,000 86.3 Medium Large
Boeing 757-200 116,000 89.9 2347 1555 Medium Large
Boeing 720B 106,000 79.5 Medium Large
Boeing 720 104,000 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 84,000 70.1 Medium Large
Airbus A321-100 83,000 77.8 2200 1540 Medium Large
Boeing 737-800 79,000 65.32 2308 1634 Medium Large
Boeing 727-200 78,000 68.1 Medium Large
McDonnell-Douglas MD-83 73,000 63.28 Medium Large
Boeing 727-100 72,500 62.4 Medium Large
Boeing 727-100C 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 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 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 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 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 37,500 32.8 2244 1304 Medium Large
Bombardier CRJ900 36,500 33.345 1778 1596 Medium Large
Embraer 170 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 9,752 8.709 1353 811 Medium Small
Pilatus PC-24[ 8,300 7.665 893 724 Medium Small
Embraer Phenom 300 8,150 7.65 956 677 Medium Small
Beechcraft 1900D 7,765 7.605 1036 853 Medium Small
Cessna Citation CJ4 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

2. Air Speed & Ground Speed

Air speed is simply the speed at which an aircraft is moving relative to the air it is ying in. It’s also the speed at which the air is owing around the aircraft’s wings.


Ground speed, on the other hand, is the aircraft’s speed relative to the ground. Ground speed is simply the sum of airspeed and wind speed.


One thing that should be noted here is that ground speed is horizontal rather than vertical speed – an aircraft climbing completely vertically would have a ground speed of zero. In other words, while airspeed is what determines whether there is enough airow around an aircraft to make it y, ground speed is what determines how fast an aircraft will get to its destination

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

Most aircraft are capable of moving themselves backwards on the ground using reverse thrust, a procedure called as a powerback, it is not typically done due to the following reasons:

The jet blast from the engines can damage the terminal building and nearby equipment. It can even throw away people standing nearby. Engines can blow sand and debris. Engines can suck in sand and debris which can cause serious damage to the engine. It is easier for the truck driver to see the surroundings better than the pilot sitting in the cockpit with limited windows. The pushback truck driver knows the airport better than the pilot. The plane engines are very loud. This noise can add up substantially when multiple planes are moving around the airport. It won't be a pleasant experience for people sitting inside the terminal, as most people don't even like the sound of powerful motor-bikes. It saves fuel. Virgin Atlantic had proposed a new system in which the pushback trucks will push the plane all the way to the runway to save fuel and reduce impact on nearby environment.

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

Taxiing is the movement of an aircraft on the ground, under its own power, in contrast to towing or push-back where the aircraft is moved by a tug. The aircraft usually moves on wheels, but the term also includes aircraft with skis or oats (for water-based travel).

The power for taxiing comes from the engines itself. If it is a propeller powered aircraft, the propeller provides thrust and pulls the aircraft, like it would in air. If it is a jet engine, the exhaust air that comes out of the engine creates a forward force due to Newton's third law and moves the airplane.
Airliners have nose wheel steering, controlled by a cockpit control called a tiller. The tiller allows nose wheel to rotate at quite extreme angles making tighter turns possible. It also reduces the fuel consumption as you do not need much engine power during taxiing. If rudders were used, the power has to be increased to make sure sucient air ows over the rudder surface to make it effective.
Once on the take off roll and the airplane gains sucient speed i.e. enough air ows over the rudders, pilots go back to using the rudder pedals to maintain the centre line.

Some large airplanes with multi wheel main gears have a tendency to scrub the tyres on the tarmac when nose wheel is steered. To counter this problem, an automatic main gear steering is inbuilt in the aircraft. This makes the main gears turn the opposite direction to that of the nose wheel, greatly decreasing the wear on the tyres.
Small general aviation aircraft uses the rudder for turns during taxi. They don't have a separate control for nose wheel. In such airplanes, to make tighter turns, you could use differential braking. That is, if you want to make a tight turn for example to the left, you would push the upper part of your left rudder pedal which would apply left braking. This allows the airplane to make a sharper turn to the left.

5. Take off power, tyre design, rolling resistance, tyre pressure, brake forces when landing.

Power:

Take off is the phase of ight in which an aerospace vehicle leaves the ground and becomes airborne and the power required to fullll this task is Take off power. For aircraft that take off horizontally, this usually involves starting with a transition from moving along the ground on a runway. It is needed becuase for a given aircraft, the takeoff speed is usually dependent on the aircraft weight; the heavier the weight, the greater the speed needed. So speed is achieved by running on runways. The speeds needed for takeoff are relative to the motion of the air (indicated airspeed). A headwind will reduce the ground speed needed for takeoff, as there is a greater ow of air over the wings. Typical takeoff air speeds for jetliners are in the range of 240–285 km/h (130–154 kn; 149–177 mph). Light aircraft, such as a Cessna 150, take off at around 100 km/h (54 kn; 62 mph). Ultralights have even lower takeoff speeds. After this the nose is raised to a nominal 5°–15° nose up pitch attitude to increase lift from the wings and affect liftoff.
So, power required to do all this task is accounted into take off power. Most takeoffs use "derated" thrust to save engine wear. For each takeoff, performance is calculated, the necessary power setting is determined and the thrust setting is made. When using this method during takeoff, it is always possible to increase to full power if the situation requires. The captain always makes the nal decision whether to use full thrust or derate.

Tyres :

When it comes to safety, tyres are one of the most important components of aircraft. They help to absorb the shock of landing and provide cushioning. It also provides the necessary traction for braking and stopping of an aircraft. And all this is for short times. An aircraft tyre is designed to withstand extremely heavy loads while landing, take off, taxing and parking. The number of wheels required for aircraft increases with the weight of the aircraft, as the weight of the aeroplane needs to be distributed more evenly.
Tyres are radial type and tubeless. Aircraft tyres are conductive too as they are manufactured with tread rubber with conducting compounds to permit earthing of static charges.
Carcass plies are used to form the tire. They are sometimes called casing plies. An aircraft tyre is constructed for the purpose it serves.
They support the weight of an aircraft while it is on the ground. It also provides the necessary traction for braking and stopping of an aeroplane. Tyre also helps to absorb the shock of landing and provide cushioning the roughness of takeoff, roll-out, and taxi operations.

Rolling Resistance:

Rolling resistance, is the force resisting the motion when a body rolls on a surface. It is experienced during take off and landinf of aircraft.
Factors that contribute to rolling resistance are the amount of deformation of the wheels, the deformation of the roadbed surface, and movement below the surface. Additional contributing factors include wheel diameter, load on wheel, surface adhesion, sliding, and relative micro-sliding between the surfaces of contact.
We also have a constant for calculation of rolling resistance force known as coecient of rolling resistance depnds upon typre pressure and velocity.
c = 0.005 + (1 / p) (0.01 + 0.0095 (v / 100)2)
Where,
c = coecient of rolling resistance
p = tyre pressure in bar
v = velocity of vehicle in Km/H.

Aircraft tyre pressure:


Aircraft tires generally operate at high pressures, up to 200 psi (14 bar; 1,400 kPa) for airliners, and even higher for business jets. The main landing gear on the Concorde was typically inated to 232 psi (16.0 bar), whilst its tail bumper gear tires were as high as 294 psi (20.3 bar). Tests of airliner aircraft tires have shown that they are able to sustain pressures of maximum 800 psi (55 bar) before bursting. During the tests the tires have to be lled with water, to prevent the test room being blown apart by the energy that would be released by a gas when the tire bursts. Aircraft tires are usually inated with nitrogen to minimize expansion and contraction from extreme changes in ambient temperature and pressure experienced during ight. Dry nitrogen expands at the same rate as other dry atmospheric gases (normal air is about 80% nitrogen), but common compressed air sources may contain moisture, which increases the expansion rate with temperature.

Brake forces when landing:


The three primary methods of achieving deceleration on the runway during the landing roll or a rejected take off are engine power, brakes and mechanical spoilers. Effective and co-ordinated use of whichever of these are available will result in a stopping distance appropriate to the available runway ahead, or the desired runway exit point if sooner. The key to success is the properly co-ordinated use of all the available methods.

Engine Power: In this the focus is to select all thrust/power levers to the ground idle position promptly, and if available, continue the action through to the selection of reverse thrust or reverse propeller pitch. Selection at these relatively high speeds must be symmetrical because otherwise, directional control may be prejudiced.

Wheel Brakes: It depends upon sucient friction existing between the tyres and the runway surface achieved through freely rotating wheels. These ensure adequate wheel rotation exists and is maintained before the commanded brake pressure is transmitted to the brake units and takes effect. An electrical signal is sent from the ight deck to hydraulic actuators near the main landing gear. Hydraulic uid pressure is used to force the brake unit against the wheel, thus slowing it down. But this resulted in less eciency due to increase in weight. So we switched to electrical braking, i.e. when the pilots press on the brake pedals, an electrical signal is sent to the brake unit on the wheel. Here, electrically powered actuators are used to press the carbon brake disc against the wheel, slowing it down. Here, when the pilots press on the brake pedals, it sends an electronic signal to the Left and Right Brake System Control Units (BSCUs). These then send signals to the four Electronic Brake Actuator Controllers (EBAC) which control the rate of braking on the wheels. Each wheel has four Electric Brake Actuators (EBA), a kind of piston which presses against the carbon brake discs.

Mechanical Spoilers : This is done by increasing aerodynamic drag on the moving aircraft. This can be achieved by raising upper wing surface panels called ground spoilers or by operation of a tail cone ‘clamshell’ type air brake. Both systems can also be used in the air on some aircraft types as in-ight air brakes, but the extent of their operation may be less in the airborne case than in the ground (weight on wheels) case.

6. With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle.

Assumptions :

Weight of the Aircraft (Airbus A-380) = 575000kg
Rolling Resistance Coecient of an aircraft tire(μrr) = 0.004
Wight of the Towing Tractor = 50000kg
Rolling Resistance Coecient of Towing Vehicle(μrr) = 0.001
Then,
Rolling Resistance Force of Tow Tractor = μrr × m. g = 0.001 × 50000 × 9.81 = 0.49KN

Rolling Resistance Force of aircraft = μrr × m. g = 0.004 × 575000 × 9.81 = 22.56KN

Total Rolling Resistance Force = Rolling Resistance Force of aircraft + Towing Tractor = 0.49 + 22.56
Total Rolling Resistance Force = 23.05KN

Air Drag Force :The drag force is the force that occures due to the Air Strike effect on the vehicle when it moves. The vehicle moves through a air medium of particular dencsity value with certain speed this causes the air strike against the vehicle body and it causesDrag Force on the vehicle.
The Aerodynamic Drag is given by the following expression

fad = 0.5* ρ *A *V^2 *Cd
Where,
ρ = Density of Air Medium
A = Frontal Area Of the Vehicle which strike the Air
V = Velocity with which the vehicle moves
Cd = Coecient of Dragwhich varies different shapes of the frontal area.

Calculation:
Let us assume,

The Velocity of the Aircraft while Towing = 30KMPH = 8.33m/s
Density of Air medium(ρ) = 1.225
The Frontal area of the Aircraft(A) = 20m^2
The Coecient of Drag of the Aircraft = 0.026

Air drag force = 0.5*rho*A*V^2*Cd = 0.022KN
The Air Drag Force of the Aircraft = 0.022KN

Therefore the total force that the motor needs to overcome = Rolling Resistance of the aircraft + Drag Force acting on the Aircraft,

Total forces acting = 23.05 + 0.022
The Total Force that the Motor needs to overcome in order to push/pull the Aircraft = 23.07KN

The Power required to push/pull the airplane :


The power required by the motor is given by the velocity with which the towing operation takes place.
The average speed of push/pull = 20 kmph = 5.5m/s
Power required by the motor = Total Force*Velocity = 23.07*5.5

Power Required by The Motor = 126.8kw

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 required assumptions. Prepare a table of assumed parameters. Draw a block diagram of powertrain.

Assumptions :

1. Weight of the Aircraft 575000 kg
2. Weight of the Towing Tractor 50000 kg
3. Rolling Resistance Coecient of Aircraft 0.004
4. Rolling Resistance Coecient of Towing Tractor 0.001
5. The velocity of the Aircraft while Towing 8.33 m/s
6. Density of Air medium(rho) 1.225 kg/m^3
7. The Frontal area of the Aircraft 20 m^2
8. The Coecient of Drag of the Aircraft 0.026
9. Towing Tractor Gear ratio(G- for high torque output) 11
10. Towing Tractor tyre radus(r) 0.635 m
11. Operational time of the Towing for one cycle 0.33 Hour
12. The Battery Armature Voltage 350 V
13. The Battery Armature Current 350 A
14. The Motor Terminal Voltage for full power output 630 V @ motor terminals
15. Power converters used Bidirectional DC-DC Boost Converter
16. The maximum duty ratio for acieving full power 70%

Energy :

Assuming the operational time of the Towing Truck is around 20 minutes for one cycle
20 minutes = 0.33 hour

Energy(kwh) = power(kw)*Time(h) = 162.8*0.33
Energy(kwh) = 41.8 kwh
Therefore the battery must be able to store the energy 41.8 kwh and should deliver the energy throughout the run time in order for the motor to produce the 1.32 KNm of torque instantaneously. Thus the storage of energy in batteries with high performance Ultracapacitors are selected to fulll the required power deliver.


Motor Selection:

Here PMSM motor type is selected that has a power rating of 126.8kw and with the gear ratio and the transmission it should be able to deliver 1.32KNm of torque instantaneously.
The Battery Armature Voltage (V) = 350 V
The Battery Armature Current (A) = 350 A

Power Converter:
A boost converter (step-up converter) is a DC-to-DC power converter that steps up voltage (while stepping down current) from its input (supply) to its output (load).


The Electric Vehicle uses power from the battery source to drive the motor and then further to drive the wheels in order to move the vehicle. The speed and torque generated from the motor is very important to drive the vehicle as per the driver's desire. If we connect the battery directly to the motor the motor will spin with a speed and torque characteristics which will be based on the full power of the battery voltage. We do not require full power in all the driving conditions so we need to adopt a power controller between the bettery and the motor so that the motor speed may be controlled by the driver to drive the vehicle as desired.

For controlling the vehicle power we use the soft switching operation technique here. The amount of power can be controlled by changing the Duty Cycle value. Considering the losses we take 70% of the duty cycle to deliver the maximum power of the motor.
Motor Terminal Voltage = Armature Voltage * Duty Cycle = 900*0.7
With 70% Duty Cycle the voltage at motor terminals = 630V, Which gives the full power with all the gearings instantaneously. Thus by varying the Duty Cycle value the speed of the vehicle is controlled.

Block Diagram of Powertrain :

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