Project-1: Powertrain for aircraft in runways

Aim

Part A:

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

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

Type	MTOW [kg]	MLW [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
Airbus A380-800[1][2][3]	575,000	394	3100	1930	Super	Super
Boeing 747-8F	447,700	346.091	3100	1800	Heavy	Heavy
Boeing 747-8	443,613	306.175	3100		Heavy	Heavy
Boeing 747-400ER	412,770	295.742	3090		Heavy	Heavy
Antonov An-124-100M	405,060	330	2520	900	Heavy	Heavy
Boeing 747-400	396,900	295.742	3018	2179	Heavy	Heavy
Lockheed C-5 Galaxy[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
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
Airbus A330-900	251,000	191	3100		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
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[17]	84,000	70.1			Medium	Large
Airbus A321-100[18]	83,000	77.8	2200	1540	Medium	Large
Boeing 737-800	79,000	65.32	2308	1634	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	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[18]	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[18]	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[19]	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[20]	37,500	32.8	2244	1304	Medium	Large
Bombardier CRJ900[21]	36,500	33.345	1778	1596	Medium	Large
Embraer 170[22]	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[23]	9,752	8.709	1353	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	1036	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

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

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

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

Commercial aircrafts are equipped with engines rated from 18,000 to nearly 100,000 lbs of thrust. Such thrust levels provide for safe takeoff, flight, and landing over a wide range of temperatures, altitudes, gross weights, and payload conditions. However, the exhaust wake from these engines can pose hazards in commercial airport environments. Operators and airport authorities must carefully consider these hazards and the resulting potential for injury to people and damage to or caused by baggage carts, service vehicles, airport infrastructure, and other airplanes.

When modern jet engines are operated at rated thrust levels, the exhaust wake can exceed 603 Km/hr from the engine exhaust nozzle. This exhaust flow field extends in a rapidly expanding cone, with portions of the flow field contacting and extending along the pavement surface as shown in the figure above. Exhaust velocity components are attenuated with increasing distance from the engine exhaust nozzle. However, an airflow of 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 240 km/h at 200 ft beyond the airplane and 80 to 161 km/h well beyond this point.

If the power from the engine's aircraft is high when on the ground, the thrust produced can lead to Foreign Object Damage or FOD which can damage the aircraft's structure and/or equipments and personnel. Debris from the ground wether on the taxiway or runway can get sucked into the engine and can not only damage the components inside the engine but also the stabilizers, elevators or fuselage.

Thrust from the engine can also blow any ground equipment away if in case they are not fastened and can also harm ground personnel either from the suction force of the engine blades or from the exhaust wake or due to flying debris or equipment hitting them. The figure above shows the hazardous zones of an engine when the aircraft is on the ground and this can also include the thrust required for taxiing.

Therefore due to the high thrust produced by the engine, it is not recommended to use the engine's power when it is on the ground if it is stationary or taxiing due to the exahust wake and the force produced by the intake blades.

  

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

The process of pushing an aircraft to the runway when its ready to take-off is called as Pushback. It is an airport procedure during which an aircraft is pushed backwards away from an airport gate towards the runway by external power. Pushbacks are carried out by special, low-profile vehicles called pushback tractors or tugs.

Although many aircraft are capable Of moving themselves backwards on the ground using reverse thrust (a procedure referred to as a powerback), 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.

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

  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 2h 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:

-An aircraft tyre are deisgned to withstand extreme heavy loads by absording shocks during and there by providing cushioning effect.

-The number of tires required for an aircraft depends on the total weight of aircraft as the weight needs to be distributed evenly.

-Aircratt tyre are designed to tacillitate stability in high crosswind condition,to chanalize the water away tor preventing skidding and helps to give a good braking effect.

-Aircraft tires are usually inflated by nitrogen air inorder to reduce expansion and contraction from extrerme changes in ambient pressure and tempurature.This type of inert gases will eliminate the possibilty of tire explosion.

-Aircraft tire design are set into an aircraft based on its application and also suffeicient clearance with the ground for allowing tyre growth which happens when centrifugal force is at high speed.

rolling resistance:

Tire rolling resistance is the energy that your vehicle needs to send to your tyres to maintain movement at a constant speed over the surface. In other words, it is the effort required to keep a tyre rolling. It is mainly caused by non-elastic effects; i.e not all the energy needed for deformation (or movement) of the wheel, roadbed, etc. is recovered when the pressure is removed. The main contributor to the Rolling Resistance is the process known as Hysteresis. Hysteresis is essentially the energy loss that occurs as a tire rolls through its footprint. The energy loss must be overcome by the vehicle's engine which results in wasted fuel.

The Rolling Resistance can be expressed by the generic equation:

Fr=c *m •g where; c is the rolling resistance coeficient(dimensionless) m is the mass of the body(kg or 1b) g is the acceleration of gravity(9.8m/sec2 or 32.174ft/sec2)

The Rolling Resistance Coefficient is greatly get influenced by the factors like wheel design, rolling surface, wheel dimensions, etc.

Some typical values of rolling resistance coefficient(c )for different vehicles is shown in the figure given below:

Rolling Resistance Goemclent
	c, (mm)
0.001 - 0.002	0.5	railroad steel wheels on steel rails
0.001		bicycle tire on wooden track low resistance tubeless tires bicycle lire on concrete bicycle tire on asphalt road
0.002 - 0.005
0.002
0.004
0.005		dirty tram rails
- 0.01		truck tire an asphalt
0.008		bicycle tire on rough paved road
0.01 - 0.015		ordinary car tires on concrete, new asphalt, coööles small new car tires on tar or asphalt car tires on gravel - rolled new car tires on coööles - large worn
o. 02
o. 02
o. 03
0.04 - 0.08		car tire on solid sand, gravel loose worn, soil medium nard
		car tire on loose sand

Part B:

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

The forces acting on an aircraft during the push and pull of the towing vehicle are Rolling Resistance, Air drag force.

Assumptions:

Let us assume an Aircraft of Antonov An-225 plane.

Mass of the Aircraft 'ml' = 247000 kg,

Mass of the Towing Vehicle m2 = 50000kg,

Coefficient of Rolling ground friction for Dry concrete or asphalt road 0.03,

Frontal Area of the Vehicle 35m ^A2,

Drag coefficient of an aircraft  0.6,

Drag coefficient of an Towing Vehicle Cd(Towing)= 0.15,

Density of air = 1.225 kgm ^A3,

Gravity g = 9.81ms^A2,

Speed of an aircraft = 20kmph = 20*5/18= 5.556 m/sec,

Rolling Resistance Force:

For the Aircraft:

Frraircraft=rrxmxg

Frraircraft=O.03x247000x9.81

Frraircraft=72692.1 N

For the Towing Vehicle:

We need to add the mass of the Aircraft as the Towing Vehicle is pushing/ pulling the aircraft as well therefore, the total rolling resistance force will be:

Total Rolling Resistance force

Frr=Frraircraft+Frr(vehicIe)

Err-72692.1+14715

Frr=87407.1 N

Aerodynamic Drag force:

Fad=12xpxAxv2x(Cd(Aircraft)+Cd(Towing))

Fad=4750.47N

Total Force of the Towing Vehicle,

F=Fad+Frr(vehicle)

F=4750.47+87407.l

F-92157.57 N Power Required,  92157.57x5.556

 512.027 KW

The Force and Power of the Towing vehicle carrying the Aircraft are 92.157KN and 512.027 KN.

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

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. 

(Hint :DC7 Block)

B. Also, Design the parameters in excel sheet.

DESCRIPTION:

The 200 HP DC motor is separately excited with a constant 150 V DC field voltage source. The armature voltage is provided by an IGBT converter controlled by two PI regulators. The converter is fed by a 515 V DC bus obtained by rectification Of a 380 V AC 50 Hz voltage source. In order 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 in order to obtain the electromagnetic torque needed to reach the desired speed The speed reference change rate follows acceleration and deceleration ramps in order 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 1 and 4 are opposite to those of IGBT devices 2 and 3. This generates the average armature voltage needed to obtain the desired armature current. In order to limit the amplitude of the current oscillations, a smoothing inductance is placed in series with the armature circuit.

OUTPUT:

SIMULATION:

Before starting the simulation, set the initial bus voltage to 515 V via the GUI block ('Initial States Setting' button and 'Cbus' variable).

Start the simulation. You can observe the motor armature voltage and current, the four IGBT pulses and the motor speed on the scope The current and speed references are also shown.

The motor is coupled to a linear load, which means that the mechanical torque of the load is proportional to the speed

The speed reference is set at 500 rpm at t O s. Observe that the motor speed follows the reference ramp accurately (+400 rpm/s) and reaches steady state around t = 1.3 s.

The armature current follows the current reference very well, with fast response time and small ripples Notice that the current ripple frequency is 5 kHz.

At t = 2 s, speed reference drops to -1184 rpm. The Current reference decreases to reduce the electromagnetic torque and causes the motor to decelerate with the help of the load torque.

At t = 2.2 s, the current reverses in order to produce a braking electromagnetic torque (dynamic braking mode). This causes the DC bus voltage to increase

At t = 3.25 s, the motor reaches O rpm and the load torque reverses and becomes negative. The negative current now produces an accelerating electromagnetic torque to allow the motor to follow the negative speed ramp (-400 rpm's). At t 6.3 s, the speed reaches -1184 rpm and stabilizes around its reference.

CONCLUSION:-

The parameters assumed for designing this vehicle is taken from online references

S.No. Assumed Parameters	Assumed Values
Weight of the Aircraft	575000 kg
2.    Weight of the Towing Tractor	50000 kg
3.    Rolling Resistance Coefficient of Aircraft	0.004
4.    Rolling Resistance Coefficient of Towing Tractor	0.001
5.    The velocity of the Aircraft while Towing	8.33 m,'s
6.      Density of Air medium(rho)	1.225 kg/mA3
7.    The Frontal area of the Aircraft	20 rrf2
8.     The Coefficient of Drag of the Aircraft	0.026
9.   Towing Tractor Gear ratio(G- for high torque output)	11
10. Towing Tractor tyre radius(r)	0.635 m
11. Operational time of the Towing for one cycle	0.33 Hour
12. The Battery Armature Voltage	350 v
13. The Battery Armature Current	350 A
14. The Motor Terminal Voltage for full power output	630 V @ motor terminals
15.   Power converters used	Bidirectional DC-DC Boost Converter
16. The maximum duty ratio for acieving full power	70%

Assuming the operational time of the Towing Truck is around 20 minuts for one cycle

20 minutes - 0.33 hour

Energy(kwh) = power(kw)*Time(h) = 162.8*0.33

Energy(kwh) = 53.7 kWh

Therefore the battery must be able to store the energy 41.8 kWh and should deliver the energy throughout the run time in order for the motor to produce the 1.32 KNm of torque instantaneously. Thus the storage of energy in batteries with high performance Ultracapacitors are selected to fulfill the required power deliver.

Motor Selection:

Here we have selected a BLDC Motor type that has a power rating of 126.8kw and with the gear ratio and the transmission it should be able to deliver 1.32KNm of torque instantaneously.

The Battery Armature Voltage (V)	350 v
The Battery Armature Current (A)	350 A

Power Converter:

A boost converter (step-up converter) is a DC-to-DC power converter that steps up voltage (while stepping down current) from its input (supply) to its output (load). It is a class of sui.rched.:mQd.zuu.eL-supply (SMPS) containing at least two semiconductors (a and a and at least one energy storage element: a inductor, or the two in combination. To reduce  filters made of capacitors (sometimes in combination with inductors) are normally added to such a converter's output (load-side filter) and input (supply-side filter).

Speed Control of the Vehicle:

The Electric Vehicle uses power from the battery source to drive the motor and then further to drive the wheels in order to move the vehicle. The speed and torque generated from the motor is very important to drive the vehicle as per the driver's desire. If we connect the battery directly to the motor the motor will spin with a speed and torque characteristics which will be based on the full power of the battery voltage. We does not require full power in all the driving conditions so we need to adopt a power controller between the bettery and the motor so that the motor speed may be controlled by the driver to drive the vehicle as desired. For controlling the vehicle power we use the soft switching operation technique here. The amount of power can be controlled by changing the Duty Cycle value. Considering the losses we take 70% of the duty cycle to deliver the maximum power of the motor.

Motor Terminal Voltage = Armature Voltage * Duty Cycle = 900*0.7

With 70% Duty Cycle the voltage at motor terminals = 630V, Which gives the full power with all the gearings instantaneously. Thus by varying the Duty Cycle value the speed of the vehicle is controlled

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 sin at full newer and required torque can be achieved using a Bidirectional DC-DC Converter. Finally, the speed control of the vehicle is done we