Project-1: Powertrain for aircraft in runways

 

 

 

AIM:-

Part A:

1. Search and list out the total weight of various types of aircraft. 
2. Is there any difference between ground speed and airspeed?
3. Why is it not recommended to use aircraft engine power to move it on the ground at the Airport? 
4. How an aircraft is pushed to the runway when it's ready to take off?
5. Learn about take-off power, tyre design, rolling resistance, tyre pressure, brake forces when landing.        

Part B:

6. A. With necessary assumptions, calculate the force and power required to push/pull an aircraft by a towing vehicle. 
7. Develop the model for the calculated force and power using Simulink.
8. A. Design an electric powertrain with the type of motor, its power rating, and energy requirement to fulfil 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 the powertrain. 

(Hint: DC7 Block)

1. Also, Design the parameters in an excel sheet.

Part A:

 QUESTION 1 

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

a)Aircraft Gross Weight:

• The aircraft gross weight (also known as the all-up weight and abbreviated AUW) is the total aircraft weight at any moment during the flight or ground operation.
• An aircraft's gross weight will decrease during a flight due to fuel and oil consumption. An aircraft's gross weight may also vary during a flight due to payload dropping or in-flight refuelling. At the moment of releasing its brakes, the gross weight of an aircraft is equal to its takeoff weight. During the flight, an aircraft's gross weight is referred to as the en-route weight or in-flight weight.
• Aircraft gross weight limits are established during an aircraft's design and certification period and are laid down in the aircraft type certificate and manufacturer specification documents
• The absolute maximum weight capabilities of a given aircraft are referred to as the structural weight limits. The structural weight limits are based on the aircraft maximum structural capability and define the envelope for the CG charts (both maximum weight and CG limits).
• An aircraft's structural weight capability is typically a function of when the aircraft was manufactured, and in some cases, old aircraft can have their structural weight capability increased by structural modifications.

b)Manufacturer's Empty Weight (MEW):

• Also called Manufacturer's Weight Empty (MWE) or Licensed Empty Weight It is the weight of the aircraft "as-built and includes the structure, power plant, furnishings installations, systems, and systems other equipment that are considered an integral part of an aircraft.

1. c) Maximum Flight Weight (MFW):

• The maximum design taxi weight (also known as the maximum design ramp weight (MRW)) is the maximum weight certificated for aircraft manoeuvring on the ground (taxing or towing) as limited by aircraft strength and airworthiness requirements

1. d) Maximum Takeoff Weight (MTOW)

• Is the maximum certificated design weight when the brakes are released for takeoff and is the greatest weight for which compliance with the relevant structural and engineering. requirements have been demonstrated by the manufacturer.

1. e) Maximum Landing Weight(MLW):

• The maximum certificated design weight at which the aircraft meets the appropriate landing certification requirements. It generally depends on the landing gear strength of the landing impact loads on certain parts of the wing structure. The MIW must not exceed the MTOW.
• The maximum landing weight is typically designed for 10 feet per second (600 feet per minute) sink rate at touch down with no structural damage.

1. f) Maximum Zero-Fuel Weight(MZFW):

• The maximum certificated design weight of the aircraft less all usable fuel and other specified usable agents (engine injection fluid, and other consumable propulsion agents). It is the maximum weight permitted before usable fuel and other specified usable fluids are loaded in specified sections of the aeroplane.
• The MZFW is limited by strength and airworthiness requirements. At this weight, the subsequent addition of fuel will not result in the aircraft design strength being exceeded. The weight difference between the MTOW and the MZFW may be utilised only for the addition of fuel

1. g) Payload:

It is the carrying capacity of an aircraft. It includes cargo, people, extra fuel. In the case of a commercial airliner, it may refer only to revenue-generating cargo or paying passengers.

The following image depicts Takeoff Weight Components:

h Types of aircraft

There are several ways to identify aircraft by type. the primary distinction is between those that are lighter than air and those that are heavier than air   

Type	MTOW [kg]	MLW [tonnes]	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
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[4][5][6]	381,000	288.417	2530	1494	Heavy	Heavy
Boeing 747-200[7]	377,840	285.700	3338	2109	Heavy	Heavy
Boeing 747-300[7]	377,840	260.320	3222	1905	Heavy	Heavy
Airbus A340-500[8]	371,950	240	3050	2010	Heavy	Heavy
Airbus A340-600[8]	367,400	256	3100	2100	Heavy	Heavy
Boeing 777F	347,800	260.816	2830		Heavy	Heavy
Airbus A330-900	352,000	251.5	3100		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[7]	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[8][9]	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[10]	254,000	192.777	2900		Heavy	Heavy
Boeing 787-10[10]	254,000	201.849			Heavy	Heavy
Airbus A340-200[8][11]	253,500	181	2990		Heavy	Heavy
Ilyushin IL-96-300	250,000	175	2600	1980	Heavy	Heavy
Airbus A330-300[12][13]	242,000	185	2500	1750	Heavy	Heavy
Airbus A330-200[12][13]	242,000	180	2220	1750	Heavy	Heavy
Lockheed L-1011-500	231,300	166.92	2636		Heavy	Heavy
Boeing 787-8[10]	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[14]	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[14]	163,000	138	2324	1536	Heavy	Heavy
Boeing 767-300	159,000	136.078	2713	1676	Heavy	Heavy
Airbus A310-300[14]	157,000	124	2290	1490	Heavy	Heavy
Vickers VC10	152,000	151.9			Heavy	Heavy
Boeing 707-320B[15]	151,000	97.5			Heavy	Heavy
Boeing 707-320C[15]	151,000	112.1			Heavy	Heavy
Douglas DC-8-61	147,000				Heavy	Heavy
Airbus A310-200[14]	142,000	123	1860	1480	Heavy	Heavy
Airbus A400M	141,000	122	980	770	Heavy	Heavy
Douglas DC-8-32	140,000				Heavy	Heavy
Embraer Phenom 100	4,800	4.43	975	741	Light	Small
Douglas DC-8-51	125,000				Medium	Large
Boeing 757-300	124,000	101.6	2550	1750	Medium	Large
Boeing 707-120B[15]	117,000	86.3			Medium	Large
Boeing 757-200	116,000	89.9	2347	1555	Medium	Large
Boeing 720B[16]	106,000	79.5			Medium	Large
Boeing 720[16]	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[18]	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
Airbus A380-800	575,000	394	3100	1930	Super	Super

Question 2

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

 difference between ground speed and airspeed

• One of the most confusing concepts for young scientists is the relative velocity between objects. Aerodynamic forces are generated by an object moving through a fluid (liquid or gas).
• A fixed object in a static fluid does not generate aerodynamic forces. Hot air balloons "lift" because of buoyancy forces and some aircraft like the Harrier use thrust to "lift" the vehicle, but these are not examples of aerodynamic lift.
• To generate life, an object must move through the air, or air must move past the object. Aerodynamic lift depends on the square of the velocity between the object and the air. Now things get confusing because not only can the object be moved through the air, but the air itself can move.
• To properly define the relative velocity, it is necessary to pick a fixed reference point and measure velocities relative to the fixed point. The reference point is fixed to the ground in this slide, but it could just as easily be fixed to the aircraft itself.
• It is important to understand the relationships of wind speed to ground speed and airspeed.
• 

Wind Speed

For a reference point picked on the ground, the air moves relative to the reference point at the wind speed. Notice that the wind speed is a vector quantity and has both a magnitude and a direction. Direction is important. A 20 mph wind from the west is different from a 20 mph wind from the east. The wind has components in all three primary directions (north-south, east-west, and up-down). In this figure, we are considering only velocities along the aircraft's flight path. A positive velocity is defined to be in the direction of the aircraft's motion. We are neglecting crosswinds, which occur perpendicular to the flight path but parallel to the ground, and updrafts and downdrafts, which occur perpendicular to the ground.

Ground Speed

For a reference point picked on the ground, the aircraft moves relative to the reference point at the ground speed. Ground speed is also a vector quantity so a comparison of the ground speed to the wind speed must be done according to rules for vector comparisons.

Airspeed

The important quantity in the generation of lift is the relative velocity between the object and the air, which is called the airspeed. Airspeed cannot be directly measured from a ground position but must be computed from the ground speed and the wind speed. Airspeed is the vector difference between the ground speed and the wind speed.

Airspeed = Ground Speed - Wind Speed

On a perfectly still day, the airspeed is equal to the ground speed. But if the wind is blowing in the same direction that the aircraft is moving, the airspeed will be less than the ground speed.

Examples

Suppose we had an aeroplane that could take off on a windless day at 100 mph (liftoff airspeed is 100 mph). We are at an airport with an east-west runway that is 1 mile long. The wind is blowing 20 mph towards the west and the aeroplane takes off going east. The wind is blowing towards the aircraft which we call a headwind. Since we have defined a positive velocity to be in the direction of the aircraft's motion, a headwind is a negative velocity. While the plane is sitting still on the runway, it has a ground speed of 0 and an airspeed of 20 mph:

Airspeed = Ground Speed (0) - Wind Speed (-20) = 20 mph

The aeroplane starts its takeoff roll and has a constant acceleration a. From Newton's second law of motion, the ground speed V at any time t is:

V = a * t

and the distance d down the runway at any time is:

d = 1/2 * a * t^2

For a fixed-length runway, this specifies the time to be used in the velocity equation. Let's assume that at 5000 feet down the runway, the velocity is 80 mph. Then the airspeed is given by

Airspeed = Ground Speed (80) - Wind Speed (-20) = 100 mph

and the aeroplane begins to fly. Now another pilot, with the same aeroplane decides to take off to the west. The wind is now in the same direction as the motion and this is called a tailwind. The sign on the wind speed is now positive, not negative as with the headwind. The acceleration along the ground is the same, so at 5000 feet down the runway, the ground speed is again 80 mph. The airspeed is then given by:

Airspeed = Ground Speed (80) - Wind Speed (20) = 60 mph

 

 QUESTION 3 

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

Aircraft Engine Power is Not Recommended to be Used to Move It on The Ground at Airport

• Although many aircraft are capable of moving in backwards using reverse thrust, as a result, the jet blast or prop wash might cause damage to the terminal building or equipment.
• Engine close to the ground may also blow sand and debris forward and then suck them into the engine, causing damage to the engine. Therefore there are pushback trucks in the airport.
• A pushback truck lifts the front wheel of the aircraft, their role in most airports is to move the aircraft away from the parking spot onto the taxiways, which is called towing
• For movement on the ground, the aircraft uses its own engine power, which is called taxing

 

  Unwanted consequences mainly arise from the unintended movement of during engine is running 

 

Although most aircraft are capable of moving themselves on the ground. it is not typically done due to the following reasons.

• Effectively a loss of control can happen, especially during high power engine running. Damage can occur to the aircraft itself, other aircraft nearby or to airside structures.
• There is a risk of injury to ground support personnel who may be in relative proximity to the aircraft. 
• The jet blast from the engines can damage the terminal building and nearby equipment. It can even throw away people standing nearby.
• 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 plane engines are very loud. When more than one plane will add up their sound then it will not be pleasant for passengers and s near the airport.
• It also saves fuel. Virgin Atlantic had proposed a new system in which pushback trucks will move the aircraft all the way to the runway to save fuel and reduce the impact on the environment.
• Modern planes are huge. It is difficult to navigate such gigantic structures through limited space. An A380 is eight stories high. The pilots won't be able to see anything directly in front of them or anything behind the wings
• In addition, response to potential but unexpected occurrences must be covered by appropriate training and, where a rapid response may be necessary, memory actions.
• Finally, engine ground running, especially if engine operation above Ground Idle is to take place, is best carried out with both flight deck pilot seats occupied and with clearly defined roles for the person in charge and their assistant. In particular, this allows checklists to be carried out using the challenge and response method.

 QUESTION 4  

4. How an aircraft is pushed to the runway when it's ready to take off?

4.1 Phases of A Flight

                 Image courtesy of Federal Aviation Administration (FAA) 4.1 Taxi

1. Taxiing refers to the movement of an aircraft on the ground, under its own power. The aircraft moves on wheels. An aeroplane uses taxiways to taxi from one place on an airport to another; for example, when moving from a terminal to the runway.
2. The aircraft always move on the ground following the yellow lines, to avoid any collision with the surrounding buildings, vehicles or other aircraft. The taxiing motion has a speed limit. Before making a turn, the pilot reduces the speed further to prevent tire skids. Just like cars, there is a certain list of priorities during taxiing. The aircraft that are landing or taking off have higher priority. The other aircraft have to wait for these aircraft before they start or continue taxiing.
3. The thrust to propel the aircraft forward comes from its propellers or jet engines. Steering is achieved by turning a nose wheel or tail wheel/rudder; the pilot controlling the direction travelled with their feet. The use of engine thrust near terminals is restricted due to the possibility of jet blast damage. This is why the aircraft is pushed back from the buildings by a vehicle before they can start their own engines for taxiing.

4.2 Take-Off

• Takeoff is the phase of flight in which an aircraft goes through a transition from moving along the ground (taxiing) to flying in the air, usually starting on a runway. Usually, the engines are run at full power during takeoff. Following the taxi motion, the aircraft stops at the starting line of the runway. Before takeoff, the engines, particularly piston engines, are routinely run up at high power to check for engine-related problems. This makes considerable noise. When the pilot releases the brakes, the aircraft starts accelerating rapidly until the necessary speed for take-off is achieved.

• The takeoff speed required varies with air density, aircraft weight, and aircraft configuration (flap and/or slat position, as applicable). Air density is affected by factors such as field elevation and air temperature.
• Operations with transport category aircraft employ the concept of the takeoff V-Speeds, V1 and V2. These speeds are determined not only by the above factors affecting takeoff performance but also by the length and slope of the runway. Below V1, in case of critical failures, the takeoff should be aborted; above V1 the pilot continues the takeoff and returns for landing. After the co-pilot calls V1, Then, V2 (the safe takeoff speed) is called. This speed must be maintained after an engine failure to meet per- performance targets for rate of climb and angle of climb.
• 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 flow of air over the wings. This is why the aircraft always take off against the wind. Sidewind is not preferred as it would disturb the stability of the aircraft. Typical takeoff air speeds for jetliners are in the 130–155 knot range (150–180 mph, 240–285 km/h). For a given aircraft, the takeoff speed is usually directly proportional to the aircraft weight; the heavier the weight, the greater the speed needed. Some aircraft have difficulty generating enough lift at the low speeds encountered during takeoff. These are therefore fitted with high-lift devices, often including slats and usually flaps, which increase the camber of the wing, making it more effective at low speed, thus creating more lift. These have to be deployed from the wing before performing any manoeuvre.
• At the beginning of the climb phase, the wheels are retracted into the aircraft and the undercarriage doors are closed. This operation is audible by the passengers as a noise coming from below the floor.

 4.3 Climb

• Following take-off, the aircraft has to climb to a certain altitude (typically 30,000 ft or 10 km) before it can cruise at this altitude safely and economically. A climb is carried out by increasing the lift of wings supporting the aircraft until their lifting force exceeds the weight of the aircraft. Once this occurs, the aircraft will climb to a higher altitude until the lifting force and weight are again in balance. The increase in lift may be accomplished by increasing the angle of attack of the wings, by increasing the thrust of the engines to increase speed (thereby increasing lift), by increasing the surface area or shape of the wing to produce greater lift, or by some combination of these techniques. In most cases, engine thrust and angle of attack are simultaneously increased to produce a climb.

• Because lift diminishes with decreasing air density, a climb, once initiated, will end by itself when the diminishing lift with increasing altitude drops to a point that equals the weight of the aircraft. At that point, the aircraft will return to level flight at a constant altitude. During the climb phase, it is normal that the engine noise diminishes. This is because the engines are operated at a lower power level after the take-off. It is also possible to hear a whirring noise or a change in the tone of the noise during a climb. This is the sound of the flaps that are retracting. A wing with a retracted flap produces less noise.
• 

 QUESTION 5 

5. Learn about take-off power, tyre design, rolling resistance, tyre pressure, brake forces when landing.    
6. a) TAKE-OFF POWER:

• Takeoff Power is the phase of flight in which an aerospace vehicle leaves the ground and becomes airborne On take-off, to position the aircraft to a safe height away from terrain and obstacles (ie. a flight path of maximum height and minimum ground distance desired), the engine thrust is set to a high 'take-off power setting and the aircraft attitude is pitched up by to maintain a specific speed by fulfilling the task is known as the take-off power.
• Take off performance can be predicted using a simple measure of the acceleration of the aircraft along the runway based on force equilibrium.

The forces involved will be.

where

T=Thrust of propulsion system pushing aircraft along the runway.

D=Aerodynamic Drag of vehicle resisting the aircraft motion

F=Rolling resistance friction due to the contact of wheels or skids on the ground.

D=During take-off runs the imbalance in these forces will produce an acceleration along the runway.

dv/dt =Acceleration along with the runway

m =Mass of the vehicle.

whereas 

Power =(T-D-F)*V

1. b) TYRE DESIGN:

• 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 the braking and stopping of an aircraft.
• 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.
• A Boeing 737NG & 737MAX uses 6 wheels, Boeing 787 uses 10 wheels, Boeing 777 uses 14 wheels and Airbus A380 uses 22 wheels. Aircraft tyres work under extreme conditions, carrying up to 340 tons and accelerating at over 250 km/hour at takeoff, in addition to enduring varied environmental stress when in flight and taxiing.
• 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 the braking and stopping of an aeroplane.
3. Tyre also helps absorb the shock of landing and provides cushioning for 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.
• 

Let’s go through some technical aspects of the tyre.

• Retreading = Retreading is restoring a worn tyre by renewing the tread area or renewing the tread area plus one or both sidewalls. Repairs are included in the tyre retreading process.
• Load Rating = Load rating is the maximum permissible load at a specified inflation pressure.
• 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.
• Speed Rating = The speed rating is the maximum takeoff speed to which the tyre has been tested.
• Skid Depth = Skid depth is the distance between the tread surface and the deepest groove as measured in the mould.

Aircraft tyres must have an approved speed and load rating and have sufficient clearance when retracted through landing gear to allow for tyre growth. Tyre growth is the increase in the size of the tyre due to centrifugal forces at high speed.

• 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 tyres 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 tyres. 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.

1. c) ROLLING RESISTANCE:

Rolling resistance is the force resisting the motion when a body (such as a ball, tire, or wheel) rolls on a surface.

Rolling resistance can be seen in the aircraft when the aircraft Takes off at the runway with the help of from the runway and while landing its wheels.

The rolling resistance can be expressed

Fr =cW

where

         F= rolling resistance or rolling friction

         c=rolling resistance coefficient - dimensionless (coefficient of rolling friction-CRF)

         W=normal force or weight of the body

W = mag

where

            m=mass of the body (kg, lb)

            ag =acceleration of gravity (9.81 m/s2.32.174 ft/s²)

The rolling resistance can alternatively be expressed as

Fr=cl W/r

where

          cl=rolling resistance coefficient- dimension length (coefficient of rolling friction) (mm, in)

          r=radius of the wheel (mm, in)

d)TYRE PRESSURE:

1. Aircraft tires generally operate at high pressures, up to 200 psi(14 bar; 1.400 kPa) for airliners, and even higher for business jets. 
2. 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).
3. Tests of airliner aircraft tires have shown that they can sustain pressures of the maximum of 800 psi (55 bar, 5.500 kPa) before bursting. During the tests, the tires have to be filled with water, to prevent the test room from being blown apart by the energy that would be released by a gas when the tire bursts Aircraft tires are usually inflated with nitrogen to minimize
4. expansion and contraction from extreme changes in ambient temperature and pressure experienced during flight. Dry nitrogen  Laye expands at the same rate as other dry atmospheric gases (normal air is about 80% nitrogen), but common compressed air sources may contain moisture, rate with temperature. which increases the expansion
5. The requirement that an inert gas, such as nitrogen, be used instead of air for inflation of tires on certain transport category aeroplanes was prompted by at least three cases in which the oxygen in air-filled tires combined with volatile gases given off by a severely overheated tire and exploded upon reaching auto-ignition temperature The use of inert gas for tire inflation will eliminate the possibility of a tire explosion

1. e) BRAKE FORCES WHEN LANDING:

1. All commercial aircraft have automatic brake systems. Just like a car, these systems also have an Anti-Lock braking system (ABS) that prevents the tires from sliding down a wet runway.
2. The advantage of the automatic brake system is that the aircraft brakes comfortably at a constant deceleration. Passengers experienced a constant force that presses them into their seats.
3. The brake system automatically applies the wheel brakes just enough to create a constant deceleration independent of the air resistance, spoilers and reversers. It is only in the last phase of brakes by pressing on both brake pedals. This deactivates the automatic brake system.
4. Immediately after landing, the aircraft is still moving fast enough that aerodynamic devices are much more efficient than the brakes. The brake units are powered by the hydraulics system.
5. An electrical signal is sent from the flight deck to hydraulic actuators near the main landing gear. Here, hydraulic fluid at 3.000 pounds per square inch is used to force the brake unit against the wheel, thus slowing it down.

Part B:

 QUESTION 6 

6. A. With necessary assumptions, calculate the force and power required to push/pull an aircraft by a towing vehicle. 
7. Develop the model for the calculated force and power using Simulink.

The Force and Power required to push/pull an Aircraft by a Towing

Vehicle:

Let us Assume:

Weight of the Aircraft =500000kg

Rolling Resistance Coefficient of an aircraft tire =0.005

Wight of the Towing Tractor=50000kg

Rolling Resistance Coefficient of Towing Vehicle =0.002

The Velocity of the Aircraft while Towing=18KMPH=5m/s

The density of Air medium=1.225 kg/m^3

The Frontal area of the Aircraft=20 m^2

The Coefficient of Drag of the Aircraft = 0.26

Towing Tractor Gear ratio(G- for high torque output) = 11

Towing Tractor tyre radius(r) = 0.5 m

The forces that are acting on the aircraft are

1. i) Rolling Resistance
2. ii) Drag Force
3. i) Rolling Resistance for the Aircraft

Now. Here the Rolling resistance be

   Frr =c. mr. ag

Let.

     Frr =Rolling resistance

     c=coefficient of rolling resistance

     mr=mass

     ag =Acceleration of gravity

Now Substituting the values in the above formula

Resistance of the Aircraft gives us the Rolling

Fr(aircraft) = 0.005*(500000 kg)*(9.81 m/s^2)

Fr(aircraft)= 24.525 KN.

Therefore the Rolling resistance of the aircraft Fr(aircraft)= 24.525 KN.

Rolling Resistance for the Towing Vehicle

Now let us calculate the rolling resistance for the Towing Vehicle

   Frr =c. mr. ag

Fr(towing)=0.002(50000 kg)(9.81 m/s^2)

Fr(towing)=0.981 KN.

Therefore the Rolling resistance of the Towing vehicle Fr(towing)=0.981 KN.

1. ii) Drag Force

         

Where.

          D=Drag force

          C=Drag Coefficient

          V=Velocity of Towing Vehicle

          A=Frontal area

Now Substituting the assumed values in the above the Drag force of the Towing vehicle

Drag Force = 0.5* (0.26*1.225*20*5^2)

Drag Force = 0.079 KN

Therefore the Drag force is 0.079625 KN

Now to get the total force to push the aircraft by the towing vehicle

we add the

=Rolling resistance of Aircraft + Rolling resistance of Towing Vehicle+ Drag Force

= 24.525 KN +0,981 KN+ 0.079625 KN

=25.585625 KN

Therefore the Total force required for a towing vehicle to push the aircraft is=25.585625 KN=25585.625N

Power required to the towing vehicle to pull the aircraft

In this case,

Power=force*velocity

Power=25.585625*5.0

Power=127.928125 KW

Therefore the power required for the towing vehicle to pull the aircraft=127.928125 KW=127928.125 W

Simulink Model for the calculated Force and Power

 

 QUESTION 7 

7. A. Design an electric powertrain with the type of motor, its power rating, and energy required to fulfil 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 the powertrain. 

(Hint: DC7 Block)

1. Also, Design the parameters in an excel sheet
2. A) Design an Electric Powertrain with the type of Motor:

Let us Assume:

Weight of the Aircraft =500000kg

Rolling Resistance Coefficient of an aircraft tire =0.005

Wight of the Towing Tractor=50000kg

Rolling Resistance Coefficient of Towing Vehicle =0.002

The Velocity of the Aircraft while Towing=18KMPH=5m/s

The density of Air medium=1.225 kg/m^3

The Frontal area of the Aircraft=20 'm^2

The Coefficient of Drag of the Aircraft =0.26

Towing Tractor Gear ratio(G-for high torque output)=11

Towing Tractor tyre radius(r)=0.5 m

Operational time of the Towing for one cycle=0.25 Hours (15min)

Power converters used Bidirectional DC-DC Boost Converter

We taking calculated values from the above equation

The total force required for a towing vehicle to push the aircraft is =25.585625 KN=25585.625N

Power required for the towing vehicle to pull the aircraft=127.928125KW=127928.125 W

Torque Required with the wheels of towing vehicle to push the aircraft

Torque=force *(radius/Gear ratio)

Torque= 25.585 *(0.5/11)

Torque=11629 KN-m

Therefore the Torque required the towing vehicle to push the aircraft =1.1629 KN-m

1. i) Energy Requirements:

The Energy requirement for an EV similar to the Towing vehicle to Push/Pull the aircraft is calculated on the Basis of Time

Let

        Energy =Power x Time taken by the towing vehicle to push-pull the aircraft

        Energy=127.928 x 0.25

        Energy =31.982 KWH

Therefore power required for one single towing is=31.982 KWh

Therefore, the battery must store the energy 31.982 KWH and deliver the energy throughout the run time for the motor to produce the 1.1629 KN-m of torque. Thus the storage of energy in batteries with high-performance Ultracapacitors are selected to fulfil the required power delivery.

Motor Selection:

Here we have selected a Permanent-Magnet Synchronous Motor (PMSM) type that has a power rating of 127.928 kw and with the gear ratio and the transmission, it should be able to deliver 1.1629 KN-m of torque instantaneously.

Motor Parameters:

Type=PMSM

Model No.=APEV80-12(16)

Phase=3

Application Electric bus,Truck,passenger,car

AC Voltage=200-750V

Pole=12

Direct Current Voltage (VDC)=540V

Rated current (A)=255

Peak Current (A)=650

Rated Power (kW)-150

Peak Power (kW) =220

Rated Speed (RPM)=1650

Peak Speed (RPM)-3.300

Rated Torque (Nm)-868

Peak Torque (Nm) =3.000

BEMF (VAC)/rpm=660/2401

Insulation Grade-H Protection

Class IP67

Cooling Water

Weight (Kg)=260

The efficiency of motor=90 % (Assumed Parameter)

Duty Cycle:

Here in this case the Duty cycle of the electric power train is calculated by the output power by the input power By this, we get the efficiency of the duty cycle.

Duty Cycle = Output Power /Input Power

We know that.

Output Power=127.92 kW

Input Power =Rated Motor Power / Motor Efficiency-

Input Power =150/0.9

Input Power=166.666kW

Then 

Duty Cycle =127.92/166.666

Duty Cycle =0.76

Duty Cycle =76%

Therefore the Efficiency of the Duty cycle that we get for this electric power train is=0 to 76%

BLOCK DIAGRAM OF THE POWERTRAIN

 

Simulink Model of Aircraft Towing Application Using DC7 Block:

• The example shown the DC7 four-quadrant chopper DC drive during speed regulation
• The 200 HP DC motor is separately excited with a constant 150 VDC field voltage source. The armature voltage is provided by an IGBT converter controlled by two Pl regulators
• The converter is fed by a 515 V DC bus obtained by rectifying a 380 V AC 50 Hz voltage source. To limit the DC bus voltage during dynamic braking mode, a braking chopper has been added between the diode rectifier and the DC7 block
• The first regulator is a speed regulator, followed by a current regulator. The speed regulator outputs the armature current reference (in .p.u) used by the current controller to obtain the electromagnetic torque needed to reach the desired speed.
• The speed reference change rate follows acceleration and deceleration ramps to avoid sudden reference changes that could cause armature over-current and destabilize the system
• The current regulator controls the armature current by computing the appropriate duty ratios of the 5 kHz pulses of the four IGBT devices (Pulse Width Modulation). For proper system behaviour, the instantaneous pulse values of IGBT devices land, 4 are opposite to those of IGBT devices 2 and 3. This generates the average armature voltage needed to obtain the desired armature current
• To limit the amplitude of the current oscillations, a smoothing inductance is placed in scenes with the armature circuit.

Output Plot:

1. Also, Design the parameters in an excel sheet

+

CONCLUSION:

• The reference Aircraft taken here is the Heavy Aircraft build and so our designed towing tractor is a heavy-duty operational vehicle.
• The power requirements are calculated for the particular load to be pushed/pulled and to achieve this power at the motor output, the motor type and its voltage at the terminals are discussed.
• Then the power converters are used to upscale the battery armature voltage to the desired high voltage value so that the motor may spin at full power and required torque can be achieved using a Bidirectional DC-DC Converter.
• Finally, the speed control of the vehicle is done with switching techniques where the Duty Cycle adjustment can change the motor speed from the minimum to maximum