Project-1: Powertrain for aircraft in runways

Question 1:

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.

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.

Airline	MTOW	MLW
Antonov An-225	640,000	591.7
Scaled Composites Model 351 Stratolaunch	589,670
Airbus A380-800[1][2][3]	575,000	394
Boeing 747-8F	447,700	346.091
Boeing 747-8	443,613	306.175
Boeing 747-400ER	412,770	295.742
Antonov An-124-100M	405,060	330
Boeing 747-400	396,900	295.742
Lockheed C-5 Galaxy[4][5][6]	381,000	288.417
Boeing 747-200[7]	377,840	285.700
Boeing 747-300[7]	377,840	260.320
Airbus A340-500[8]	371,950	240
Airbus A340-600[8]	367,400	256
Boeing 777F	347,800	260.816
Airbus A330-900	352,000	251.5
Boeing 777-300ER	351,800	251.29
Boeing 777-200LR	347,450	223.168
Boeing 747-100[7]	340,200	265.300
Airbus A350-1000	308,000	233.5
Boeing 777-300	299,370	237.683
Boeing 777-200ER	297,550	213.00
Airbus A340-300[8][9]	276,700	190
McDonnell Douglas MD-11	273,300	185
Airbus A350-900	270,000	175
Ilyushin Il-96M	270,000	195.04
McDonnell Douglas DC-10	256,280	183
Boeing 787-9[10]	254,000	192.777
Boeing 787-10[10]	254,000	201.849
Airbus A340-200[8][11]	253,500	181
Ilyushin IL-96-300	250,000	175
Airbus A330-300[12][13]	242,000	185
Airbus A330-200[12][13]	242,000	180
Lockheed L-1011-500	231,300	166.92
Boeing 787-8[10]	228,000	172.365
Lockheed L-1011-200	211,400
Ilyushin IL-86	208,000	175
Boeing 767-400ER	204,000	158.758
Airbus A300-600R[14]	192,000	140
Boeing 767-300ER	187,000	136.08
Concorde	185,000	111.1
Airbus A300-600[14]	163,000	138
Boeing 767-300	159,000	136.078
Airbus A310-300[14]	157,000	124
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
Airbus A400M	141,000	122
Douglas DC-8-32	140,000
Douglas DC-8-51	125,000
Boeing 757-300	124,000	101.6
Boeing 707-120B[15]	117,000	86.3
Boeing 757-200	116,000	89.9
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
Convair 880	87,500
Boeing 737-900	85,000	66.36
Boeing 737-900ER	85,000	71.35
Boeing 727-200 Advanced[17]	84,000	70.1
Airbus A321-100[18]	83,000	77.8
Boeing 737-800	79,000	65.32
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
de Havilland Comet 4	70,700
Boeing 737-700	70,000	58.06
Airbus A320-100[18]	68,000	66
Boeing 737-400	68,000	54.9
de Havilland Comet 3	68,000
Boeing 377	67,000
Boeing 737-600	66,000	54.66
Airbus A220-300	65,000	57.61
Hawker Siddeley Trident 2E	65,000
Airbus A319[18]	64,000	62.5
Boeing 737-300	63,000	51.7
Boeing 737-500	60,000	49.9
Airbus A220-100	59,000	50.80
Airbus A318[18]	59,000	57.5
Boeing 717-200HGW	55,000	47.174
Douglas DC-7	55,000
de Havilland Comet 2	54,000
Boeing 717-200BGW	50,000	46.265
de Havilland Comet 1	50,000
Douglas DC-6A	48,600
Douglas DC-6B	48,500
Embraer 190[19]	48,000	43
Caravelle III	46,000
Fokker 100	46,000	39.95
Douglas DC-6	44,000
Avro RJ-85	42,000	36.74
Handley Page Hermes	39,000
Embraer 175[20]	37,500	32.8
Bombardier CRJ900[21]	36,500	33.345
Embraer 170[22]	36,000	32.8

Question 2:

There is a difference between ground speed and airspeed to understand this we should take into account wind speed too.

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 cross winds. 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.

For example, taking wind speed as 20 km/hr and the runway is the east-west runway the wind blowing towards the west and the flight runs at 80 km/hr the wind speed acting in the west and the flight takes off east than the Airspeed=Groundspeed(80)- Wind Speed(-20) =100 km/hr

Hence there is a difference between ground speed and airspeed.

Question 3:

The risks which may arise from engine ground running relate to the potential for loss of control of the aircraft by those persons occupying the pilot seats in the flight deck. Unwanted consequences from such static running are mainly related to those arising from the unintended movement of the aircraft during engine running - effectively a loss of control - especially during high power engine running.

Damage can occur to the aircraft itself, other aircraft nearby, or to airside structures. In addition, there is a risk of injury to ground support personnel who may be in relatively close proximity to the aircraft. When modern jet engines are operated at rated thrust levels, the exhaust wake can exceed 375 mi/h (325 km or 603 km/h) immediately aft of the engine exhaust nozzle. This exhaust flow field extends aft in a rapidly expanding cone, with portions of the flow field contacting and extending aft along the pavement surface. Exhaust velocity components are attenuated with increasing distance from the engine exhaust nozzle.

However, an airflow of 300 mi/h (260 km or 483 km/h) can still be present at the empennage, and significant people and equipment hazards will persist hundreds of feet beyond this area. At full power, the exhaust wake speed can typically be 150 mi/h (130 km or 240 km/h) at 200 ft (61 m) beyond the airplane and 50 to 100 mi/h (43 to 88 km or 80 to 161 km/h) well beyond this point

Question 4:

In aviation, pushback is an airport procedure during which an aircraft is pushed backward away from an airport gate by an external power. Pushbacks are carried out by special, low-profile vehicles called pushback tractors or tugs.

Although many aircraft are capable of moving themselves backward on the ground using reverse thrust (a procedure referred to as a power back), the resulting jet blast or prop wash might cause damage to the terminal building or equipment. Engines close to the ground may also blow sand and debris forward and then suck them into the engine, causing damage to the engine. A pushback is therefore the preferred method to move the aircraft away from the gate.

Pushbacks at busy aerodromes are usually subject to ground control clearance to facilitate ground movement on taxiways. Once clearance is obtained, the pilot will communicate with the tractor driver (or a ground handler walking alongside the aircraft in some cases) to start the pushback. To communicate, a headset may be connected near the nose gear. Since the pilots cannot see what is behind the aircraft, steering is done by the pushback tractor driver and not by the pilots. Depending on the aircraft type and airline procedure, a bypass pin may be temporarily installed into the nose gear to disconnect it from the aircraft's normal steering mechanism.

Once the pushback is completed, the towbar is disconnected, and any bypass pin removed. The ground handler will show the bypass pin to the pilots to make it clear that it has been removed. The pushback is then complete, and the aircraft can taxi forward under its own power.

Question 6:

The rolling resistance for the airplane can be calculated as

F, = cm. ag

where F, = rolling resistance (N)

c = rolling resistance coefficient

m = rolling mass (kg)

ag = acceleration of gravity (9.81 m/s2)

If the mass of a plane is 40000 kg and the rolling resistance co-efficient between the aircraft wheel and the runway is 0.02

The force required to overcome the rolling resistance can be calculated as

Fr = 0.02 ' (40000 kg) 4(9.81 m/s2) = 7848 N

The friction for the puller can be calculated to Fr = p W =pmras

where Fr = frictional force (N)

p = friction coefficient W = Weight (N)

W = Weight (N)

m1 = friction mass (kg)

The friction coefficient between a pulling truck's rubber wheels and a dry runway is 0.9.

Since pulling trucks are connected to air crafts nose wheels close to runway we simplify the calculation of required truck mass

m1= (7848 N) / (0.9 (9.81 m/s2)) = 889 kg

Taking radius of the wheel R as 0.3m

The torque is given by T=F+R T= 7848 * 0.3

= 2354 N.m

The power for pulling/push the aircraft can be calculated as P=F-V

V= velocity of pull vehicle taken as 5 km/hr

V=1.388 m/s

P=7848*1.388 =10893

W =10.89 kW

Question 7:

Taking the previous example power value 10 kW for this problem

Power =10 kW

The motor power should be greater than or equal to 10kW assuming it is 10 kW.

The energy required can be calculated as E=P*t

assuming taxi time of 5 mins.

E=10"(5'60) kJ

= 3000kJ

From power equation Vo*Io=rlVi*li where,

Vo is the output voltage 10 is the output current q is the power efficiency (ranging from 0 to 1)

Vi is the input voltage

Ii is the input current

Consider DC input voltage as 48V and Power as 12kW

P= Vi'Ii

12000=48*1i

Ii=2.50A

The efficiency is calculated as ri=10000/12000 =0.83

From the above power equation Vo'lo=r1Vitli Ve250=0.83*48*250 Vo=40V

The Vout =Vin/1-D

40=48/(1-D)

D= 0.2 the duty cycle has in the range of (0,0.2)