Project-1: Powertrain for aircraft in runways

Part A:

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

Ans Part A 1 :

When looking at different aircraft's specifications or in POHs/AFMs, there are several different weights of the aircraft mentioned (e.g. maximum take-off weight, maximum landing weight, operating empty weight).

The different terms represent different weights or masses¹. For each flight, the weights are taken into account for several reasons.

A brief description about these is below:

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 weight of the structure, power plant, furnishings, installations, systems and other equipment that are considered an integral part of an aircraft.

This excludes any baggage, passengers, or usable fuel².

Zero-fuel weight (ZFW)

This is the total weight of the airplane and all its contents (including unusable fuel), but excluding the total weight of the usable fuel on board.

As a flight progresses and fuel is consumed, the total weight of the airplane reduces, but the ZFW remains constant.

Maximum zero fuel weight (MZFW) is the maximum weight allowed before usable fuel and other specified usable agents (engine injection fluid, and other consumable propulsion agents) are loaded.

Operating empty weight (OEW) (Roughly equivalent to basic empty weight on light aircraft)

It is the basic weight of an aircraft including the crew, all fluids necessary for operation such as engine oil, engine coolant, water, unusable fuel and all operator items and equipment required for flight but excluding usable fuel and the payload.

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.

Maximum takeoff weight (MTOW)

This is the maximum weight at which the pilot of the aircraft is allowed to attempt to take off³.

Regulated takeoff weight (RTOW)

Depending on different factors (e.g. flap setting, altitude, air temperature, length of runway), RTOW or maximum permissible takeoff weight varies for each takeoff. It can never be higher than MTOW. More information is on this answer.

Maximum landing weight (MLW)

This maximum weight at which an aircraft is permitted to land³.

The following image depicts takeoff weight components.

^{Image Source}

Maximum ramp weight (MRW)
also called maximum taxi weight (MTW)

It is the maximum weight authorized for maneuvering (taxiing or towing) an aircraft on the ground.

Aircraft gross weight

It is the total aircraft weight at any moment during the flight or ground operation. This decreases during flight due to fuel and oil consumption.

This is a list of aircraft sorted by maximum takeoff weight.

Airplanes

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

Type	MTOW [kg]	MLW [tonnes]	TOR [m]	LR [m]	ICAO category	FAA category
Antonov An-225	640,000	591.7	3,500		Heavy	Super
Scaled Composites Model 351 Stratolaunch	589,670		3,660		Heavy	Super
Airbus A380-800[1][2][3]	575,000	394	3,100	1,930	Heavy	Super
Boeing 747-8F	447,700	346.091	3,100	1,800	Heavy	Heavy
Boeing 747-8	443,613	306.175	3,100		Heavy	Heavy
Boeing 747-400ER	412,770	295.742	3,090		Heavy	Heavy
Antonov An-124-100M	405,060	330	2,520	900	Heavy	Heavy
Boeing 747-400	396,900	295.742	3,018	2,179	Heavy	Heavy
Lockheed C-5 Galaxy[4][5][6]	381,000	288.417	2,530	1,494	Heavy	Heavy
Boeing 747-200[7]	377,840	285.7	3,338	2,109	Heavy	Heavy
Boeing 747-300[7]	377,840	260.32	3,222	1,905	Heavy	Heavy
Airbus A340-500[8]	371,950	240	3,050	2,010	Heavy	Heavy
Airbus A340-600[8]	367,400	256	3,100	2,100	Heavy	Heavy
Boeing 777F	347,800	260.816	2,830		Heavy	Heavy
Boeing 777-300ER	351,800	251.29	3,100		Heavy	Heavy
Boeing 777-200LR	347,450	223.168	3,000		Heavy	Heavy
Boeing 747-100[7]	340,200	265.3			Heavy	Heavy
Airbus A350-1000	308,000	233.5			Heavy	Heavy
Boeing 777-300	299,370	237.683	3,380		Heavy	Heavy
Boeing 777-200ER	297,550	213	3,380	1,550	Heavy	Heavy
Airbus A340-300[8][9]	276,700	190	3000	1926	Heavy	Heavy
McDonnell Douglas MD-11	273,300	185	2,990	1,890	Heavy	Heavy
Airbus A350-900	270,000	175	2,670	1,860	Heavy	Heavy
Ilyushin Il-96M	270,000	195.04	3,115	2,118	Heavy	Heavy
McDonnell Douglas DC-10	256,280	183	2,990	1,890	Heavy	Heavy
Boeing 787-9[10]	254,000	192.777	2,900		Heavy	Heavy
Boeing 787-10[10]	254,000	201.849			Heavy	Heavy
Airbus A340-200[8][11]	253,500	181	2990		Heavy	Heavy
Airbus A330-900	251,000	191	3,100		Heavy	Heavy
Ilyushin IL-96-300	250,000	175	2,600	1,980	Heavy	Heavy
Airbus A330-300[12][13]	242,000	185	2500	1750	Heavy	Heavy
Airbus A330-200[12][13]	242,000	180	2,220	1,750	Heavy	Heavy
Lockheed L-1011-500	231,300	166.92	2,636		Heavy	Heavy
Boeing 787-8[10]	228,000	172.365	3,300	1,695	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	3,414		Heavy	Heavy
Airbus A300-600R[14]	192,000	140	2,385	1,555	Heavy	Heavy
Boeing 767-300ER	187,000	136.08	2,713	1,676	Heavy	Heavy
Concorde	185,000	111.1	3,440	2,220	Heavy	Heavy
Airbus A300-600[14]	163,000	138	2,324	1,536	Heavy	Heavy
Boeing 767-300	159,000	136.078	2,713	1,676	Heavy	Heavy
Airbus A310-300[14]	157,000	124	2,290	1,490	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	1,860	1,480	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	2,550	1,750	Medium	Large
Boeing 707-120B[15]	117,000	86.3			Medium	Large
Boeing 757-200	116,000	89.9	2,347	1,555	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	2,250		Medium	Large
Convair 880	87,500				Medium	Large
Boeing 737-900	85,000	66.36	2,500	1,704	Medium	Large
Boeing 737-900ER	85,000	71.35	2,804	1,829	Medium	Large
Boeing 727-200 Advanced[17]	84,000	70.1			Medium	Large
Airbus A321-100[18]	83,000	77.8	2,200	1,540	Medium	Large
Boeing 737-800	79,000	65.32	2,308	1,634	Medium	Large
Boeing 727-200[17]	78,000	68.1			Medium	Large
McDonnell-Douglas MD-83	73,000	63.28			Medium	Large
Boeing 727-100[17]	72,500	62.4			Medium	Large
Boeing 727-100C[17]	72,500	62.4			Medium	Large
McDonnell-Douglas MD-90-30	71,000	64.41	2,165	1,520	Medium	Large
de Havilland Comet 4	70,700				Medium	Large
Boeing 737-700	70,000	58.06	1,921	1,415	Medium	Large
Airbus A320-100[18]	68,000	66	1,955	1,490	Medium	Large
Boeing 737-400	68,000	54.9	2,540	1,540	Medium	Large
de Havilland Comet 3	68,000				Medium	Large
Boeing 377	67,000				Medium	Large
Boeing 737-600	66,000	54.66	1,796	1,340	Medium	Large
Airbus A220-300	65,000	57.61	1,890	1,494	Medium	Large
Hawker Siddeley Trident 2E	65,000				Medium	Large
Airbus A319[18]	64,000	62.5	1,850	1,470	Medium	Large
Boeing 737-300	63,000	51.7	1,939	1,396	Medium	Large
Boeing 737-500	60,000	49.9	1,832	1,360	Medium	Large
Airbus A220-100	59,000	50.8	1,463	1,356	Medium	Large
Airbus A318[18]	59,000	57.5	1,375	1,340	Medium	Large
Boeing 717-200HGW	55,000	47.174	1,950		Medium	Large
Douglas DC-7	55,000				Medium	Large
de Havilland Comet 2	54,000				Medium	Large
Boeing 717-200BGW	50,000	46.265	1,950		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[19]	48,000	43	2,056	1,323	Medium	Large
Caravelle III	46,000				Medium	Large
Fokker 100	46,000	39.95	1,621	1,350	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[20]	37,500	32.8	2,244	1,304	Medium	Large
Bombardier CRJ900[21]	36,500	33.345	1,778	1,596	Medium	Large
Embraer 170[22]	36,000	32.8	1,644	1,274	Medium	Large
Bombardier CRJ700	33,000	30.39	1,564	1,478	Medium	Large
Douglas DC-4	33,000				Medium	Large
Vickers Viscount 800	30,400				Medium	Large
Bombardier Q400	28,000	28.01	1,219	1,295	Medium	Large
Bombardier CRJ200	23,000	21.319	1,918	1,479	Medium	Large
ATR 72-600	22,800	22.35	1,333	914	Medium	Large
Saab 2000	22,800	21.5	1,300		Medium	Large
Embraer ERJ 145	22,000	19.3	2,270	1,380	Medium	Large
ATR 42-500	18,600	18.3	1,165	1,126	Medium	Small
Saab 340	13,150	12.930	1,300	1,030	Medium	Small
Embraer 120 Brasilia	11,500	11.25	1,560	1,380	Medium	Small
BAe Jetstream 41	10,890	10.57	1,493	826	Medium	Small
Learjet 75[23]	9,752	8.709	1,353	811	Medium	Small
Pilatus PC-24[24]	8,300	7.665	893	724	Medium	Small
Embraer Phenom 300[25]	8,150	7.65	956	677	Medium	Small
Beechcraft 1900D	7,765	7.605	1,036	853	Medium	Small
Cessna Citation CJ4[26]	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

Helicopters

MTOW = Maximum take-off weight

Type	MTOW [pounds]	MTOW [kg]	Power output [kW]	Power to weight [W/kg]	Rotor diameter [m]
Mil Mi-12[27]	231,485	105,000	19,200	182	2 x 35
Mil Mi-26[28]	123,459	56,000	17,000	343	1 x 32
Sikorsky CH-53K	88,000	39,916	16,200	405	1 x 24
Sikorsky CH-53E[29]	73,500	33,300	9,810	294	1 x 24
Boeing CH-47D/F Chinook[30]	50,000	22,680	7,058	311	2 x 18.3
AgustaWestland AW101[31]	34,392	14,600	4,698	321	1 x 18.59
Sikorsky S-92[32]	26,500	12,020	3,758	312	1 x 17.17
Eurocopter EC225 Super Puma	24,692	11,200	3,552	317	1 x 16.2
Boeing Vertol CH-46 Sea Knight[33]	24300	11,000	2,800	254	2 x 15.24
NHIndustries NH90	23,370	10,600	3,324	314	1 x 16.3
Eurocopter AS532 Cougar	19,840	9,000	2,370	263	1 x 15.6
Airbus Helicopters H160	13,340	6,050	1,910	316	1 x 13.4
Bell 412EP	1,1900	5,397	1,864	345	1 x 14.0
Eurocopter EC145 C-2	7,903	3,585	1,100	307	1 x 11.0
Eurocopter EC135 P2+/T2+	6,415	2,910	944	325	1 x 10.2
Eurocopter EC635 P2	6,393	2,900	1,218	420	1 x 10.2
AeroVelo Atlas (human powered)	282	128	1.1	8.5	4 x 20.2

Convertiplanes

MTOW = Maximum take-off weight

​

Type	MTOW [kg]	Power output [kW]	Power to weight [W/kg]	Rotor diameter [m]
Bell Boeing V-22 Osprey	27,445	6,150	224	2 x 11.58
Bell-Agusta BA609	7,600	1,447	190	2 x 7.92
Bell XV-15	6,009	1,156	192	2 x 7.62

Read this from left to right, from detailed to generalized. Boxes on the right sum up the separate items which they run across.

The maximum landing weight is what the manufacturer has certified for landings. Generally, this is MTOW minus some fuel.

Note that every manufacturer has his own standard and might draw the lines between the different weight groups differently. Say, for one the engine quadrant is part of the propulsion system, whereas the other sees it as part of the control system.

he aircraft has:

• Basic Empty Weight (BEW) is the weight of the aircraft "as built" and includes the weight of the structure, power plant, furnishings, installations, systems and other equipment that are considered an integral part of an aircraft before additional operator items are added for operation.
• Dry Operating Wieght (DOW) BEW + Weight of Crew (Pilot + Cabin including their bags) + Pantry
• Operating Weight (OW) DOW + Takeoff fuel (i.e. Ramp Fuel - Taxi fuel)
• Maximum Zero Fuel Weight (MZFW) DOW + Payload (anything put on the aircraft that generates revenue to the company, e.g. passenger, baggage, cargo, mail and fret)
• Maximum Taxi Weight (MTW) MZFW + Ramp fuel
• Maximum Takeoff Weight (MTOW) MZFW + Takeoff fuel, or MTW - Taxi fuel

Maximum Landing Weight (MLW) MTOW - Trip Fuel

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

Ans Part A 2 :

Fig A : Air speed of aircraft.

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 lift, 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. In this slide, the reference point is fixed to the ground, 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

Fig B : 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.

Examples

Suppose we had an airplane 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 airplane 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 airplane starts its take off 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 airplane begins to fly. Now another pilot, with exactly the same airplane 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

This airplane doesn't have enough airspeed to fly. It runs off the end of the runway!

Significance of Understanding Relative Velocity

The importance of the relative velocity explains why airplanes take off and land on different runways on different days. Airplanes always try to take off and land into the wind. This requires a lower ground speed to become airborne, which means the plane can take off or land in the shortest distance traveled along the ground. Since runways have a fixed length, you want to get airborne as fast as possible on takeoff and stopped as soon as possible on landing. In the old days, a large "wind sock" was hung near the runway for pilots to see which way the wind was blowing to adjust their takeoff and landing directions. Now mechanical or electronic devices provide the information that is radioed to the cockpit.

The relationship between airspeed, wind speed, and ground speed explains why wind tunnel testing is possible and how kites fly.

• In the wind tunnel, the ground speed is zero because the model is fixed to the walls of the tunnel. The airspeed is then the negative of the wind speed that is generated in the tunnel. Whether the object moves through the air, or the air moves over the object, the forces are the same.
• A kite usually has no ground speed because a kite is held on the end of a string. But the kite still has an airspeed that is equal to the wind speed. You can fly a kite only with the wind at your back.

there are two different types of speed when talking about airplanes – ground speed and airspeed.

While ground speed is the airplane’s speed relative to the surface of the Earth, airspeed – at least true airspeed – is its speed relative to the air it is flying in.

Airspeed vs. Ground Speed

As mentioned above, true airspeed is simply the speed at which an aircraft is moving relative to the air it is flying in. As such, it’s also the speed at which the air is flowing around the aircraft’s wings.

Ground speed, on the other hand, is the aircraft’s speed relative to the ground. One thing that should be noted here is that it’s its 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 airflow around an aircraft to make it fly, ground speed is what determines how fast an aircraft will get to its destination.

Wind’s Effect on Ground Speed

The relationship between airspeed and ground speed is fairly simple. Ground speed is simply the sum of airspeed and wind speed.

If the aircraft is flying in the same direction as the wind is blowing, the aircraft experiences tailwind, and its ground speed is higher than its airspeed. On the other hand, if the wind is blowing against the direction the aircraft is traveling in, the aircraft experiences headwind, and its ground speed is lower

Fig C : View of wing of aeroplane with wind force acting.

To give you an actual example, imagine an aircraft that cruises at an airspeed of 500 miles per hour that has to cover a ground distance of 2,000 miles.

If there is no wind at all, then both the aircraft’s airspeed and ground speed would be the same 500 miles per hour, and the aircraft would reach its destination in four hours.

If there was a 100 miles per hour headwind – wind blowing against the aircraft’s direction of travel – the aircraft would still be traveling at an airspeed of 500 miles per hour. However, its ground speed would be just 400 miles per hour (100 miles per hour slower than its airspeed). And as such, it would take the aircraft five hours to reach its destinations.

Finally, if there was a 100 miles per hour tailwind – wind blowing in the same direction as the aircraft’s travel – the aircraft would still be traveling at an airspeed of 500 miles per hour, but its ground speed would be 100 miles faster. And, at 600 miles an hour, the aircraft would reach its destination in just three hours and twenty minutes.

The above is the reason why some flights go “out of their way” to avoid headwinds or catch tailwinds. And, why some flights might appear to be traveling at “supersonic speeds,” even though their airspeed – the speed that would actually matter in determining whether or not the flight truly is supersonic – is subsonic.

The Three Types of Airspeed

However, pilots commonly use three different types of airspeed: indicated airspeed, calibrated airspeed, and true airspeed.

Indicated airspeed is an airspeed that is calculated directly off an aircraft’s pitot-static system. It’s the calculated off the aircraft’s dynamic pressure – the difference between its total pressure and static pressure.

The dynamic pressure depends not only on the aircraft’s speed, but also on the density of the air it is flying in. As such, the higher the aircraft flies – and the lower the air density as a result – the bigger the difference between indicated and true airspeed is.

Calibrated airspeed is indicated airspeed adjusted for a variety of errors.

Just as an example, one of the things it’s adjusted for is the flap position. The reason for that is that at different flap positions, air flows differently around the pitot-static system and affects the indicated airspeed readings.

True airspeed is, as has been mentioned numerous times in this article, the actual speed at which an aircraft is moving relative to the air it is traveling in. It’s calibrated airspeed adjusted for the the exact conditions (altitude, air temperature, etc.) that the aircraft is flying in.

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

Ans 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 kn 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 (fig. 1). Exhaust velocity components are attenuated with increasing distance from the engine exhaust nozzle. However, an airflow of 300 mi/h (260 kn 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 kn or 240 km/h) at 200 ft (61 m) beyond the airplane and 50 to 100 mi/h (43 to 88 kn or 80 to 161 km/h) well beyond this point.

High engine thrust during maintenance activity can cause considerable damage to airplanes and other elements in the airport environment. An example of this problem occurred after an airplane arrived at its final destination with a log entry indicating the flight crew had experienced anomalous engine operation. Subsequent evaluation resulted in replacement of an engine control component, followed by an engine test and trim run to verify proper engine operation. The airplane was positioned on an asphalt pad adjacent to a taxiway, with the paved surface extending from the wingtips aft to the empennage. During the high- power portion of the test run, a 20- by 20-ft (6.1- by 6.1-m) piece of the asphalt immediately aft of the engine detached and was lifted from the pad surface. This 4-in (10.2-cm)-thick piece of asphalt drifted up and into the core area of the left engine exhaust wake, where it shattered into numerous smaller pieces. The pieces were driven aft at substantial velocity, striking the aft fuselage and left outboard portion of the horizontal tail. The maintenance crew was alerted to the ramp disintegration and terminated the engine run. Subsequent inspection found that the outboard 4 ft (1.2 m) of the left horizontal stabilizer was missing, as was the entire left elevator. Corrective action included replacing the stabilizer and left elevator and repairing holes in the fuselage.

Major disadvantage while use aircraft engine power to move it on ground at Airport are :
1. Fuel Consumption.
2. Wear and tear on the engine.
3. Unnecessary jet blast and noise.

⦁ Too Much Fuel is Wasted While Taxiing :
Taxiing, sometimes called “taxying”, refers to an aircraft’s movement while on the ground, under its own power. This term excludes the accelerating run along a runway before takeoff and the decelerating run immediately upon landing.
By using jet engines as an aircraft’s primary thrust for taxiing, airlines are expending huge amounts of jet fuel while still on the ground, which could be much better used for flying. With the Cost of Fuel being their single biggest expense, many airlines plan to adopt electric taxiing to save fuel and reduce environmental impacts.

⦁ Jet Efflux Hazard is defined as hazards associated with the blast force generated behind a jet engine. The hazard exists whenever a jet engine is running but is increased and extends over a greater area and at greater distance behind the engine, at high engine power settings when taxiing, before and during take-off, and during engine maintenance activity.

⦁ Operationally, jet efflux can be hazardous in three ways:

⦁ 1. In the ramp environment where it can damage other aircraft, blow over ground equipment (baggage carts, aircraft steps, vehicles, etc.), cause structural damage to buildings, or injure or kill passengers, crew and ground personnel who may be in the vicinity.

⦁ 2. On taxiways where other aircraft (and especially jet engines) may be damaged by Foreign Object Debris (FOD) and smaller or light aircraft blown over or subjected to loss of directional control.

⦁ 3. On surface manoeuvring areas that are either damaged or not designed to withstand the efflux from engines running at high power.

⦁ The Risk to Operational Safety:

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. 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.

Because of these some drawbacks it is not recommended to use engine power to move on ground at airport.

The extensive noise arises from the engine power also dangerous for humans and other living things around the vicinity.

Hence for the above reasons it is not recommended to use engine power to move on the ground.

i) 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.

ii) There is a risk of injury to ground support personnel who may be in relatively close proximity to the aircraft.

iii) It is easier for truck driver to see the surroundings better than the pilot sitting in the cockpit with limited window. The pushback driver knows the airport better than the pilot.

iv) It also saves fuel. virgin Atlantic had proposed a new system in which pushback truck will move the aircraft all the way to runway to save fuel and reduce impact of environment.

v) The plane engines are very loud. When more than one plane will add up their sound, than it will not be pleasent for passangers and others near airport.

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

Ans 4 :

Part B:

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

Ans 6 :

Calculation of the force and power required to push/pull an aircraft by a towing vehicle With necessary assumptions are as follows.

To push or pull an aircraft by a towing vehicle, 3 type of forces will act on that vehicle. Those are -

(i) Rolling resistance force

(ii) Aerodynamic drag force and

(iii) Hill climbing force.

(i) Rolling resistance force(Frr) : Rolling resistance is the force resisting the motion when a body rolls on a surface. It is the friction force that is needed to be overcome to move a vehicle.

The kinetic energy of the wheels is partially converted into heat. The rolling resistance is proportional to the normal force perpendicular to the rolling wheel and is approximated by the rolling resistance coefficient urr. Therefore, Equation that can be used for the rolling resistance force of the entire system is-

Frr = urr * m * g

where,

urr= Rolling resistance co-efficient (It depends on the tire material, tire structure, tire temperature, road roughness, road material etc.)

m= Total mass of the vehicle (vehicle mass + payload) in kg

g = gravitational constant , that is 9.8 m/s^2

urr of Car for different condition-
CONDITION	urr
Tires on concrete or asphalt	0.013
Tires on rolled gravel	0.02
Tar macadam	0.025
Unpaved road	0.05
Field	0.1 - 0.35

Here in this case for towing vehicle for aircraft urr= 0.013

(ii) Aerodynamic drag force(Fad) : Aerodynamic drag force is the force which is required to keep a vehicle moving. There are two component of this force , one is shape drag (due to the shape of the vehicle) and another is skin friction (due to friction between air and skin of the vehicle). When the vehicle moves the velocity of air near boundary is higher than the air surrounding. Both the tug and the aircraft will experience some aerodynamic drag. For completeness, the aerodynamic drag force caused by the aircraft and towing vehicle can be given by Equation below-

Fad = (1/2) * ρ * V^2 * A * (Cd_t + Cd_a)

where, ρ = Air density, that is normally 1.25 kg/m^3

A = Frontal area of the vehicle in m^2

Cd_t = Air drag co-efficient of towing vehicle (It depends on frontal area, shape, protrusions, ducts, air passage etc.)

Cd_a = Air drag co-efficient of aircraft (It depends on frontal area, shape, protrusions, ducts, air passage etc.)

v = Velocity of the vehicle in m/s

With regard to the drag equation, the term A simply refers to some reference area. For airplanes, this is generally the planform area of the wing The designers design an aircraft in such a way that the reference area is optimal. So, for calculation this has been neglected.

Cd for different types of vehicle-
VEHICLE TYPE	Cd
Open Convertible	0.5 - 0.7
Van body	0.5 - 0.7
Ponton body	0.4 - 0.55
wedge- shaped body, head lamps and bumpers are integrated into the body, covered undeready optimized cooling air flow	0.3 - 0.4
Motor cycle	0.6 - 0.7
racing car	0.2 - 0.25

In this study, we want to calculate the force and power required to push and pull an aircraft by towing vehicle so we take one of the towing vehicle model and airbus model on the website and consider the data for our calculation.

DOUGLAS TOWBARLESS AIRCRAFT TRACTOR: Model: TBL 280Mk4

The Douglas Tugmaster type TBL- 280 Mk4 towbarless aircraft handling tractor has been designed primarily for pushback, inter-gate towing, and longer distance maintenance towing operations at higher speeds with the following aircraft:
Embraer: ERJ 190/195
McDonnell Douglas: DC9/MD80/MD90
Boeing: B717, B737, B727, B757, B767, B787, and B777.
Airbus: A318/A319/A320/A321, A310/A300, A330 and A340-200/300
Boeing: DC10/MD11
Lockheed: L 1011

Gross weight = 35,000 kg

Width = 3400 mm = 3.4 m

Height = 4400 mm = 4.4 m

Length (excluding coupler) = 8470 mm = 8.47 m

Tyre size = R 22.5 = 571.5 mm

= 0.5715 m

Maximum vehicle speed = 32 kmph

= (32*5/18)

Velocity of the tug = 8.89 m/s

The coefficient of friction depends on the objects that are causing friction. The value is usually between 0 and 1 but can be greater than 1. A value of 0 means there is no friction at all between the objects. A value of 1 means the frictional force is equal to the normal force. It is a misconception that the coefficient of friction is limited to values between zero and one. A coefficient of friction that is more than one just means that the frictional force is stronger than the normal force.

Coefficient of rolling resistance (μrr) = 0.02 (Truck tyre in asphalt concrete road)

AIRCRAFT MODEL:

We are taking an aircraft model as Airbus A310 as considered as pushback operation. The advanced technologies of A310 and some specifications are given below,

The A310 helped pioneer the use of composites in commercial jetliners, with its vertical stabiliser built with carbon-fibre materials that offer significant weight savings over conventional metallic structures. Other major applications of composites on the A310 include floor struts and panels, spoilers and main landing gear doors.

Standard or Higher gross weight = 164000 kg

Height = 15.80 m

Width = 43.90 m

Drag coefficient (Cd)= 0.0265

To find:

The Total Force and Power required to push/pull an aircraft by the towing vehicle.

Formula used:

Total force = Fte = Frr+Fad

Rolling resistance force = Frr= μrr⋅m⋅g

Drag force = Fad = 0.5⋅ρ⋅A⋅Cd⋅v^2

Solution:

Total force = Fte = Frr+Fad

In the first step, we want to find the rolling resistance force and air drag force for both towing vehicle and airbus to obtain the force and power value during the towing process,

To find rolling resistance force

Rolling resistance force = Frr= μrr⋅m⋅g

Here mass (m) = mass of the towing vehicle+mass of the airbus

(m) = 35000+164000

m = 199000kg

Substitute mass (m) value in rolling resistance formula

Rolling resistance force = Frr = 0.02 * 199000 * 9.81

Frr = 39043.8 N

To find the air drag force

Drag force = Fad = 0.5⋅ρ⋅A⋅Cd⋅v^2

According to the International Standard Atmosphere (ISA) values = 1.225 kg/m^3.

ρ = 1.225 kg/m^3

Frontal area A = Height of the airbus * Width of the airbus

A = 15.80 * 43.90 = 693.62 m^2

Substituting the ρ and A values in drag force formula,

Assume that vehicle running at the speed of 20 kmph

V = (20*5/18) = 5.56 m/s

Fad = 0.5 * 1.225 * 693.62 * 0.0265 * (5.56)^2

= 11.258 * (5.56)^2

Fad = 62.594 N

Now find the total force = Fte = Frr+Fad

Fte = 39043.8+62.594

Fte = 39106.394 N

To find the power required by towing vehicle

Therefore find the power value of towing vehicle by equating the total force value in power formula,

Power required for towing vehicle Ptow = Fte * v

= 39106.394 * 5.56

Ptow = 217431.550 W

By converting the unit of power we get,

Ptow = 217.431 KW

To find the torque of the vehicle

Torque (T) = Fte * R

= 39106.394 * 0.5715

T = 22349.30 N-m

Hence power and torque of towing vehicle during the speed of 20kmph is obtained and they are given below

Power = 217.431 KW

Torque = 22822.03 N-m

B. SIMULINK MODEL FOR CALCULATION OF FORCE AND POWER

We have assumed some data for calculation for push or pull an aircraft by a towing vehicle, 3 type of forces will act on that vehicle. Those are -

(i) Rolling resistance force

(ii) Aerodynamic drag force and

(iii) Hill climbing force.

Details of simulink model :

Observations :

calculated the force and power required to push / pull an aircraft by a towing also Developed the model for the calculated force and power using Simulink.

The values are assumed for mass of the aircraft, coefficient of rolling resistance, frontal area, drag coefficient and velocity and substituted in the respective formulas for rolling resistance force, drag coefficient force and acceleration force

• From the above figures for the creation of the Simulink model, firstly we consider all the forces acting on the aircraft such as rolling resistance force, aerodynamic drag force and acceleration force.
• The values are assumed for mass of the aircraft, coefficient of rolling resistance, frontal area, drag coefficient and velocity and substituted in the respective formulas for rolling resistance force, drag coefficient force and acceleration force.
• Similarly for the above figure the forces are calculated for the towing vehicle the values are assumed and substituted in the respective formulas for rolling resistance force, aerodynamic drag and acceleration force.Finally, all the forces are added to obtain the total tractive force and the power is obtained by multiplying the total tractive force with the velocity.
• The simulation is run and the both the force and the power values are displayed using the display block.

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.

Ans 7 :

Table for the necessary assumed parameters-

The following are the assumed parameters:

AIRCRAFT PARAMETERS

PARAMETERS	VALUES	UNITS
Mass of the aircraft	48500	kg
Coefficient of rolling resistance	0.03	--
Frontal area of the aircraft	85	M2
Final velocity of the aircraft	3	m/s
Initial velocity of the aircraft	0	m/s
Coefficient of drag of an aircraft	0.15	--
Acceleration time of the Towing vehicle	10	Seconds

TOWING VEHICLE PARAMETERS

PARAMETERS	VALUES	UNITS
Mass of the Towing vehicle	8000	kg
Coefficient of rolling resistance	0.013	--
Frontal area of the Towing vehicle	8	M2
Final velocity of the Towing vehicle	3	m/s
Initial velocity of the Towing vehicle	0	m/s
Coefficient of drag of the Towing vehicle	0.60	--
Acceleration of the Towing vehicle	10	seconds
Gear ratio	10

The total power required to push or pull the aircraft by the towing vehicle is 97.022952KW and the total tractive force is 32340.984N. To calculate the torque required by the Towing vehicle the following parameters are assumed as shown above:

T = F_{r *} r /G

T = 32340.984 * 0.4 / 10

T = 1293.63 Nm

The torque required for towing the vehicle to push or pull the aircraft is 1293.63 Nm

To design the electric power train of the towing vehicle a motor is selected based on its power rating and the battery required based on energy requirement and duty cycle range to control the aircraft speed from zero to highest.

MOTOR

The motor is selected with a power rating of 97.022KW and torque of 1293.63 Nm . TECO IE2 electric induction motor is selected which has a power output of 110 KW and torque at full load is 1420Nm that satisfies the power output and torque needed for the vehicle.

MOTOR SPECIFICATIONS

Output power – 110 KW, 150 hp

Full load speed – 740 rpm

Full load Efficiency – 94.5%

Full load current – 210A

Full load Torque – 1420 Nm

Rated voltage – 400V/50 Hz

DUTY CYCLE

The rated voltage is 400V and the current is assumed to be 250A. So, the input power of the motor is calculated as product of voltage and current.

P_{i =} V * I

P_{i =} 400 *280

P_{i =} 112000W

P_{i =} 112KW

To calculate the duty cycle we divide output power to input power

= P_o / P_i

= 97.02 / 112

= 0.866

= 86%

ENERGY

To calculate the energy required from batteries or battery pack. The energy is obtained by multiplying the output power and time. Assuming the battery usage for 5 minutes is 300 seconds.

E = P_o * t

E = 97.02 * 300

E = 29106kJ

So, the battery pack provides 29106 KJ of energy to the vehicle. Since the battery are rated in kilowatt hour (KWH). We have to convert KJ to KWH we divide it by 3600.

E = 29106/3600

E= 8.08KWH

So, the battery pack should provide at least 8.08 KWH to the vehicle.

TOWING VEHICLE SIMULATION POWER TRAIN BLOCK DIAGRAM

Fig A: Block diagram of power train

RESULTS

Required torque output of the motor	1293.63 Nm
Input power of the motor	112KW
Duty cycle range	86%
Energy required from the batteries	8.08KWH

Electric Vehicles, consist of a battery, connected to an electric motor (either AC or DC) via an inverter and a motor controller. As automotive batteries generally operate on DC current, while the gird operates AC, a converter is needed to transfer power to the battery.One of the key advantages of an EV is its much higher levels of energy efficiency. Typically an EV will achieve a system efficiency of 80% from battery to wheel (roughly 60% from grid to wheel), while with a conventional IC engine it’s more like 15-25%. This is largely due to the fact that an IC engine is a “heat engine” (i.e. it has to burn fuel and produce heat to operate) and much of all that heat is simply wasted.

Results: