Challenge

Project-1: Powertrain for aircraft in runways

I AN A STUDENT OF LIVE BATCH -16

Part A:

1. Search and list out the total weight of various types of aircrafts.

SOLUTION:

the total weight of various types of aircrafts are

Type Maximum take-off weight [kg] Maximum landing weight [tonnes] Take-off run [m] Landing run [m]
Antonov An-225 640,000 591.7 3500
Scaled Composites Model 351 Stratolaunch 589,670 3660
Airbus A380-800[1][2][3] 575,000 394 3100 1930
Boeing 747-8F 447,700 346.091 3100 1800
Boeing 747-8 443,613 306.175 3100
Boeing 747-400ER 412,770 295.742 3090
Antonov An-124-100M 405,060 330 2520 900
Boeing 747-400 396,900 295.742 3018 2179
Lockheed C-5 Galaxy[4][5][6] 381,000 288.417 2530 1494
Boeing 747-200[7] 377,840 285.700 3338 2109
Boeing 747-300[7] 377,840 260.320 3222 1905
Airbus A340-500[8] 371,950 240 3050 2010
Airbus A340-600[8] 367,400 256 3100 2100
Boeing 777F 347,800 260.816 2830
Boeing 777-300ER 351,800 251.29 3100
Boeing 777-200LR 347,450 223.168 3000
Boeing 747-100[7] 340,200 265.300
Airbus A350-1000 308,000 233.5
Boeing 777-300 299,370 237.683 3380
Boeing 777-200ER 297,550 213.00 3380 1550
Airbus A340-300[8][9] 276,700 190 3000 1926
McDonnell Douglas MD-11 273,300 185 2990 1890
Airbus A350-900 270,000 175 2670 1860
Ilyushin Il-96M 270,000 195.04 3115 2118
McDonnell Douglas DC-10 256,280 183 2990 1890
Boeing 787-9[10] 254,000 192.777 2900
Boeing 787-10[10] 254,000 201.849
Airbus A340-200[8][11] 253,500 181 2990
Airbus A330-900 251,000 191 3100
Ilyushin IL-96-300 250,000 175 2600 1980
Airbus A330-300[12][13] 242,000 185 2500 1750
Airbus A330-200[12][13] 242,000 180 2220 1750
Lockheed L-1011-500 231,300 166.92 2636
Boeing 787-8[10] 228,000 172.365 3300 1695
Lockheed L-1011-200 211,400
Ilyushin IL-86 208,000 175
Boeing 767-400ER 204,000 158.758 3414
Airbus A300-600R[14] 192,000 140 2385 1555
Boeing 767-300ER 187,000 136.08 2713 1676
Concorde 185,000 111.1 3440 2220
Airbus A300-600[14] 163,000 138 2324 1536
Boeing 767-300 159,000 136.078 2713 1676
Airbus A310-300[14] 157,000 124 2290 1490
Vickers VC10 152,000 151.9
Boeing 707-320B[15] 151,000 97.5
Boeing 707-320C[15] 151,000 112.1
Douglas DC-8-61 147,000
Airbus A310-200[14] 142,000 123 1860 1480
Airbus A400M 141,000 122 980 770
Douglas DC-8-32 140,000
Douglas DC-8-51 125,000
Boeing 757-300 124,000 101.6 2550 1750
Boeing 707-120B[15] 117,000 86.3
Boeing 757-200 116,000 89.9 2347 1555
Boeing 720B[16] 106,000 79.5
Boeing 720[16] 104,000 79.5
Tupolev Tu-154M 104,000 80
Tupolev Tu-204SM 104,000 87.5 2250
Convair 880 87,500
Boeing 737-900 85,000 66.36 2500 1704
Boeing 737-900ER 85,000 71.35 2804 1829
Boeing 727-200 Advanced[17] 84,000 70.1
Airbus A321-100[18] 83,000 77.8 2200 1540
Boeing 737-800 79,000 65.32 2308 1634
Boeing 727-200[17] 78,000 68.1
McDonnell-Douglas MD-83 73,000 63.28
Boeing 727-100[17] 72,500 62.4
Boeing 727-100C[17] 72,500 62.4
McDonnell-Douglas MD-90-30 71,000 64.41 2165 1520
de Havilland Comet 4 70,700
Boeing 737-700 70,000 58.06 1921 1415
Airbus A320-100[18] 68,000 66 1955 1490
Boeing 737-400 68,000 54.9 2540 1540
de Havilland Comet 3 68,000
Boeing 377 67,000
Boeing 737-600 66,000 54.66 1796 1340
Airbus A220-300 65,000 57.61 1890 1494
Hawker Siddeley Trident 2E 65,000
Airbus A319[18] 64,000 62.5 1850 1470
Boeing 737-300 63,000 51.7 1939 1396
Boeing 737-500 60,000 49.9 1832 1360
Airbus A220-100 59,000 50.80 1463 1356
Airbus A318[18] 59,000 57.5 1375 1340
Boeing 717-200HGW 55,000 47.174 1950
Douglas DC-7 55,000
de Havilland Comet 2 54,000
Boeing 717-200BGW 50,000 46.265 1950
de Havilland Comet 1 50,000
Douglas DC-6A 48,600
Douglas DC-6B 48,500
Embraer 190[19] 48,000 43 2056 1323
Caravelle III 46,000
Fokker 100 46,000 39.95 1621 1350
Douglas DC-6 44,000
Avro RJ-85 42,000 36.74
Handley Page Hermes 39,000
Embraer 175[20] 37,500 32.8 2244 1304
Bombardier CRJ900[21] 36,500 33.345 1778 1596
Embraer 170[22] 36,000 32.8 1644 1274
Bombardier CRJ700 33,000 30.39 1564 1478
Douglas DC-4 33,000
Vickers Viscount 800 30,400
Bombardier Q400 28,000 28.01 1219 1295
Bombardier CRJ200 23,000 21.319 1918 1479
ATR 72-600 22,800 22.35 1333 914
Saab 2000 22,800 21.5 1300
Embraer ERJ 145 22,000 19.3 2270 1380
ATR 42-500 18,600 18.3 1165 1126
Saab 340 13,150 12.930 1300 1030
Embraer 120 Brasilia 11,500 11.25 1560 1380
BAe Jetstream 41 10,890 10.570 1493 826
Learjet 75[23] 9,752 8.709 1353 811
Pilatus PC-24[24] 8,300 7.665 893 724
Embraer Phenom 300[25] 8,150 7.65 956 677
Beechcraft 1900D 7,765 7.605 1036 853
Cessna Citation CJ4[26] 7,761 7.103 1039 896
de Havilland Hercules 7,000
Embraer Phenom 100 4,800 4.43 975 741

2. Is there any difference between ground speed and air speed?

SOLUTION:

ground speed

  • Ground speed is the horizontal speed of an aircraft relative to the Earth’s surface.
  • It is vital for accurate navigation that the pilot has an estimate of the ground speed that will be achieved during each leg of a flight.
  • An aircraft diving vertically would have a ground speed of zero.
  • Information displayed to passengers through the entertainment system of airline aircraft usually gives the aircraft ground speed rather than airspeed.
  • Ground speed can be determined by the vector sum of the aircraft's true airspeed and the current wind speed and direction;

groundspeed= airspeed+windspeed

air speed

  • airspeed is also the speed at which air is flowing around the aircraft's wings
  • An airspeed indicator indicates the aircraft's speed relative to the air mass.
  • The air mass may be moving over the ground due to wind, and therefore some additional means to provide position over the ground is required.

airspeed = groundspeed-windspeed

hence,Ground speed is quite different from airspeed. When an aircraft is airborne the ground speed does not determine when the aircraft will stall, and it doesn't influence the aircraft performance such as rate of climb.

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

SOLUTION:

The term Aircraft Ground Running is usually used to describe the operation of some or all of the engines of an aircraft, whilst on the ground, for the purpose of functionally checking the operation of either engines or aircraft systems.

Although many aircrafts are capable of moving in backward using reverse thrust, as result the jet blast or prop wash might cause damage to the terminal building or equipment. Engine close to ground may also blow sand and debris forward and then suck them into the engine, causing damage to the engine.

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

SOLUTION:

aircraft is pushed to runway by two methods taxing and towing

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

An airplane uses taxiways to taxi from one place on an airport to another; for example, when moving from a hangar to the runway.

There are different ways that airplanes can be taxied. Taxiing is most commonly performed using an airplane’s own propulsion system.

2. Towing: Movement of most aircraft around the airport and hangars is usually accomplished by towing with a towing vehicle or 'tug'.

The aircraft is attached it to the tow vehicle by the use of a towbar or other specialized methods that connect to or raise the nose or landing gear off the ground.

Towing aircraft can be dangerous and damage to the aircraft, other objects, or injuries to towing team members is possible if the proper procedures are not followed.

Most towing accidents are preventable by the proper handling of the aircraft, maintaining situational awareness, and practicing patience. Towing accidents cannot only result in aircraft damage but can also delay or cancel flights or lead to other airport disruptions.

Towing of aircraft

5. Learn about take off power, tyre design, rolling resistance, tyre pressure, brake forces when landing.

SOLUTION:

Take off power:

  • The amount of power that an engine is allowed to produce for a limited period of time for takeoff.
  • The use of takeoff power is usually limited to 5 min for reciprocating engines and up to 2½ min for gas turbine engines. This may not always be the case. Specifically, with respect to reciprocating engines, it is the brake horsepower developed under standard, sea-level conditions and under the maximum conditions of the crankshaft rotational speed and the engine manifold pressure approved for the normal take-off.
  • It is limited in continuous use to the period of time shown in the approved engine specifications.
  • With respect to gas turbine engines, it is the thrust developed under static conditions at a specified altitude and atmospheric conditions of the rotor shaft rotational speed and gas temperature approved for the normal takeoff.
  • It is limited in continuous use to the period of time shown in the approved engine specifications.

Tyre design:

Carcass plies are used to form the tire. They are sometimes called casing plies. An aircraft tyre is constructed for the purpose it serves.

  1. They support the weight of an aircraft while it is on the ground.
  2. It also provides the necessary traction for braking and stopping of an aeroplane.
  3. Tyre also helps to absorb the shock of landing and provide cushioning the roughness of takeoff, roll-out, and taxi operations.

Unlike an automobile or truck tyre, it does not have to carry a load for a long period of continuous operation. However, an aircraft tyre absorbs the high impact loads of landing, and also it’s operating at high speeds for a short time when required.

Some technical aspects of the tyre are:

1. Retreading – Retreading is methods of restoring a worn tyre by renewing the tread area or by renewing the tread area plus one or both sidewalls. Repairs are included in the tyre retreading process.

2. Load Rating – Load rating is the maximum permissible load at a specified inflation pressure.

3. Ply Rating – Ply Rating is used to identify the maximum recommended load rating and inflation pressure for a specified tyre. It is an index of tyre strength.

4. Speed Rating – The speed rating is the maximum takeoff speed to which the tyre has been tested.

5. Skid Depth – Skid depth is the distance between the tread surface and the deepest groove as measured in the mould.

Aircraft tyres may be Tube-type or Tubeless:

Tubeless tyres are more advantageous over tube-type. There is no longer the use of tube-type tyre in recent aviation. Nowadays all airliners are using tubeless tyres. Tubeless that are meant to be used without a tube has the word TUBELESS on the sidewall of the tyre.

Aircraft tyres may be Radial or Bias

Almost all airliners are using Radial tyre. Bias is an older design, and it’s mainly used for road vehicles. Radial tyres have the word RADIAL on the sidewall. Radial tires are more expensive than bias-ply tyres. Radial tires are in demand because of their lower life cycle cost and long term value.

Aircraft Tyre Standards

Each tyre manufacturer produces their own designs of the tyre with different tread patterns, wear patterns, wear limits etc, and in some cases, different names for the same part of the tyre. But all tyres must meet the approved standards before being accepted by the FAA or CAA.

All commercial aircraft tyres approved under FAA Requirement Technical Standard Order (TSO) C62. A TSO is a minimum performance standard for specified materials, parts, and appliances used on civil aircraft.

Rolling resistance:

Rolling Resistance Coefficient (μr) is defined by μr = F/G,

where F is the force necessary to pull the axle of a tire horizontally in the direction of travel, and

G is the vertical load on the tire which is assumed to roll on a flat horizontal surface.

the rolling friction of airplane wheels and tires due to various conditions of wheel loading, tire inflation pressure, and ground surface

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 inflated to 232 psi (16.0 bar), whilst its tail bumper gear tires were as high as 294 psi (20.3 bar).The high pressure and weight load on the Concorde tyres were a significant factor in the loss of Air France Flight 4590.
  • Aircraft tires are usually inflated with nitrogen to minimize expansion and contraction from extreme changes in ambient temperature and pressure experienced during flight.
  • 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.
  • The requirement that an inert gas, such as nitrogen, be used instead of air for inflation of tires on certain transport category airplanes was prompted by at least three cases in which the oxygen in air-filled tires had combined with volatile gases given off by a severely overheated tire and exploded upon reaching autoignition temperature.
  • The use of an inert gas for tire inflation eliminates the possibility of tire explosion.

Brake forces when landing:

Airplanes rely on a braking system to safely land on runways. At cruising altitude, most commercial airplanes fly at a speed of roughly 500 to 600 mph. When landing, however, they must reduce their speed. A typical 747, for instance, has a landing speed of about 160 to 170 mph. And upon touching the runway, airplanes must quickly brake until they come to a complete stop.

WING SPOILER: Many airplanes use wing spoilers to assist with braking when landing. Not to be confused with ailerons, spoilers are extendable flaps on the ends of an airplane’s wings. Pilots can raise the spoilers to decelerate the airplane as it approaches the runway. And even while on the runway, pilots will typically leave the wing spoilers raised. Raised wing spoilers create drag, which essentially slows down the airplane so that it’s able to brake more quickly.

DISC BRAKES: In addition to wing spoilers, airplanes use disc brakes. Airplane disc brakes are similar to the braking system in automobiles. They consist of a pair of calipers that, when engaged, squeeze pads against the rotors of an airplane’s landing gear.Disc brakes are designed to remain static at all times. In other words, they don’t rotate with the wheels of an airplane’s landing gear. As the wheels turn, the disc brakes will remain static and stationary. They are a vital component of an airplane’s braking system because they are designed to apply pressure to the airplane’s wheels. Disc brakes will squeeze the wheels, thereby slowing down the speed at which they spin. In turn, this reduces the speed of the airplane so that it can come to a complete stop on the runway.

REVERSE THRUST: Finally, many jet airplanes use reverse thrust to assist with braking during landings. Reverse thrust is a feature in jet engine airplanes that, as the name suggests, involves the reversal of the engines’ thrust. When flying, the thrust is projected out the rear of an airplane’s engines. When landing, however, pilots may use the reverse thrust feature. Reverse thrust changes the direction of the engines’ thrust. Rather than projecting out the rear, the thrust will be projected out the front. This reversal of thrust provides deceleration that allows airplanes to slow down more quickly when landing.

Part B:

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

SOLUTION:

The given data is taken from the towing truck TMX-150-9 & Aircraft Airbus A319

towing truck TMX-150-9 Aircraft Airbus A319

The data is from towing truck TMX-150-9 and Aircraft Airbus A319.

Data given :

Mass of the truck — 8300kg

Mass of the aircraft — 75500kg

Length of the aircraft — 33.84m

Velocity of the truck — 25kmph

Height of the aircraft — 11.76m

Air density (rho) = 1.225kg/m^3

Drag coefficient(Cd) = 0.75

Rolling resistance coefficient = 0.013

Gravitational constant — 9.8Nm

CALCULATION:

Rolling Resistance force

(Frr)=μrr*m*g

m= mass of the truck + mass of the aircraft

m= 8300+75500 =83800kg

`Frr = 0.013*83800*9.8 =

Frr= 10678.12 N

Aerodynamic drag force (Fad)

`Fad =0.5*rho*Cd*A*V^2`

V = velocity of the truck(m/s) = 6.95m/s

Area(A) = Length of the aircraft*Height of the aircraft

A= 33.84*11.76

A= 397.95m

Fad=0.5*1225*0.75*397.95*6.95^2

Fad= 8830 N

Total force required to push or pull an aircraft by towing vehicle is

Ftotal = Frr+Fad

Ftotal= 10678.12+8830

Ftotal= 19508.12N

Total power = Ftotal*Velocity

Total power=219508.12*6.95

Total power= 135.581kW.

B. Develop the model for the calculated force and power using Simulink.

SOLUTION:

7. A. Design an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfill aircraft towing application in Simulink. 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.

SOLUTION:

Simulink model of electric powertrain of aircraft towing truck

replaced AC input supply by DC battery source in DC7-Four-Quadrant chopper 200HP Dc Drive and SOC current and voltage is ovserved here

plots:

Designing of an electric power train aircraft towing

The tyre specification of TMX-150-9 is 315/70 R15

Data:

Force required to tow the aircraft = 19508.12 N

Power required for towing the aircraft = 135.581kW

Tyre radius for the towing truck = Aspect ratio x section width + (Rim diameter/2) = 0.411m

Gear ratio= 7

Transmission efficiency(ηt) = 80%

Motor efficiency (ηm) — 92%

Calculations

  1. Torque required at the wheels (Tw)

Tw = Total force tire radius = 19508.12*0.411

Tw = 8017.83Nm

  1. Actual power required (Pac)

Pac = Power required/ (η t* η m)

Pac =35.581/ (0.8*0.92)

Pac= 184.21kW

  1. Motor Torque (Tm)

Tm=Tw/ (gear ratio* η t)

Tm=3017.83/ (0.8*7)

Tm=1431.75 Nm

  1. Actual motor torque (Tmac)

Tmac = Tm/ η m

Tmac= 1.431/0.92

Tmac=1.555kNm

  1. Battery capacity

Towing tractor takes 15 min for taxing

15 mins = 0.25hours

Battery capacity =Pac*hours = 184.21*.25 =

Battery capacity= 46.05 kWh

The battery must deliver the power output of 46.05 kWh

If we use 210kW battery we can run the tow truck easily.

  1. Duty Cycle

Duty Cycle= power required/power rated= (Pac/input power)

Duty Cycle = 184.21/210

Duty Cycle = 0.87

Block diagram of powertrain:

B. Also, Design the parameters in excel sheet.

SOLUTION:

https://docs.google.com/spreadsheets/d/1jndqoas5rGpjXANUEXQDAu_o0ez8iWPv/edit?usp=sharing&ouid=118176628456011014646&rtpof=true&sd=true

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