Project-1: Powertrain for aircraft in runways

OBJECTIVE:

• To list out the total weight of various types of aircraft.
• To find the difference between ground speed and airspeed.
• Why is it not recommended to use aircraft engine power to move it on the ground at the Airport?
• How an aircraft is pushed to the runway when it's ready to take off?
• To learn about take-off power, tyre design, rolling resistance, tyre pressure, and brake forces when landing.
• To calculate the force and power required to push/pull an aircraft by a towing vehicle.
• To design an electric powertrain with the type of motor, its power rating, and energy requirement to fulfil aircraft towing applications in Simulink. And estimate the duty cycle range to control the aircraft speed from zero to highest.

Total weight of various types of aircraft

MTOW = Maximum take-off weight, MLW = Maximum landing weight, TOR = Take-off run (SL, ISA+15°, MTOW), LR = Landing run (SL, ISA+15°, MLW), SL = Sea Level, ISA = International Standard Atmosphere, ICAO= International Civil Aviation Organisation, FAA = Federal Aviation Administration

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	575,000	394	3100	1930	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	381,000	288.417	2530	1494	Heavy	Heavy
Boeing 747-200	377,840	285.700	3338	2109	Heavy	Heavy
Boeing 747-300	377,840	260.320	3222	1905	Heavy	Heavy
Airbus A340-500	371,950	240	3050	2010	Heavy	Heavy
Airbus A340-600	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	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	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	254,000	192.777	2900		Heavy	Heavy
Boeing 787-10	254,000	201.849			Heavy	Heavy
Airbus A340-200	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	242,000	185	2500	1750	Heavy	Heavy
Airbus A330-200	242,000	180	2220	1750	Heavy	Heavy
Lockheed L-1011-500	231,300	166.92	2636		Heavy	Heavy
Boeing 787-8	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	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	163,000	138	2324	1536	Heavy	Heavy
Boeing 767-300	159,000	136.078	2713	1676	Heavy	Heavy
Airbus A310-300	157,000	124	2290	1490	Heavy	Heavy
Vickers VC10	152,000	151.9			Heavy	Heavy
Boeing 707-320B	151,000	97.5			Heavy	Heavy
Boeing 707-320C	151,000	112.1			Heavy	Heavy
Douglas DC-8-61	147,000				Heavy	Heavy
Airbus A310-200	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	117,000	86.3			Medium	Large
Boeing 757-200	116,000	89.9	2347	1555	Medium	Large
Boeing 720B	106,000	79.5			Medium	Large
Boeing 720	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	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
Embraer Phenom 100	4,800	4.43	975	741	Light	Small

Difference between ground speed and airspeed

In this figure, the three main components are wind speed, ground speed, and airspeed. The wind has components in all three primary directions (north-south, east-west, and up-down) and is dynamic in nature. Here, we are considering only velocities along the aircraft's flight path. 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

The aircraft moves relative to the reference point at the ground speed for a reference point picked on the ground. Ground speed is also a vector quantity so a comparison of the ground speed to the wind speed must be made according to rules for vector comparisons. Ground speed is simply the sum of airspeed and wind speed.

`"ground speed" = "airspeed" + "wind speed"`

Airspeed

The significant 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.

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

Most aircraft are capable of moving themselves backwards on the ground using reverse thrust, a procedure called a powerback, it is not typically done due to the following reasons:

• 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 pushback truck driver knows the airport better than the pilot.
• The plane engines are very loud. This noise can add up substantially when multiple planes are moving around the airport. It won't be a pleasant experience for people sitting inside the terminal.
• It saves an enormous amount of fuel as more fuel is needed due to traction.

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

The power for taxiing comes from the engines themselves. If it is a propeller-powered aircraft, the propeller provides thrust and pushes the aircraft forward. If it is a jet engine, the exhaust air that comes out of the engine creates a forward force due to Newton's third law and moves the aircraft. Aircraft have nose wheel steering, controlled by a cockpit control called a tiller. The tiller allows the nose wheel to rotate at quite extreme angles making tighter turns possible. It also reduces fuel consumption as you do not need much engine power during taxiing. If rudders were used, the power has to be increased to make sure sufficient airflows over the rudder surface to make it effective.

Some large aeroplanes with multi-wheel main gears have a tendency to scrub the tyres on the tarmac when the nose wheel is steered. To counter this problem, an automatic main gear steering is inbuilt into the aircraft. This makes the main gears turn in the opposite direction to that of the nose wheel, greatly decreasing the wear on the tyres.
Small general aviation aircraft use the rudder for turns during taxiing. They don't have a separate control for the nose wheel. In such aeroplanes, to make tighter turns, you could use differential braking. That is, if you want to make a tight turn for example to the left, you would push the upper part of your left rudder pedal which would apply left braking. This allows the aeroplane to make a sharper turn to the left.

Take off power, tyre design, rolling resistance, tyre pressure, and brake forces when landing.

Take off power:

Take-off is the phase of flight in which an aerospace vehicle leaves the ground and becomes airborne and the power required to fulfil this task is take-off power. For aircraft that take off horizontally, this usually involves starting with a transition from moving along the ground on a runway. It is needed because for a given aircraft, the takeoff speed is usually dependent on the aircraft's weight; the heavier the weight, the greater the speed needed. So speed is achieved by running on runways. 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 ow of air over the wings. Typical takeoff air speeds for jetliners are in the range of 240–285 km/h (130–154 knots; 149–177 mph). Light aircraft, such as a Cessna 150, take off at around 100 km/h (54 knots; 62 mph). Ultralights have even lower takeoff speeds. After this, the nose is raised to a nominal 5°–15° nose up pitch altitude to increase lift from the wings and affect liftoff.

So, the power required to do all these tasks is accounted into take-off power. Most takeoffs use "derated" thrust to save engine wear. For each takeoff, performance is calculated, the necessary power setting is determined and the thrust setting is made. When using this method during takeoff, it is always possible to increase to full power if the situation requires it. The captain always makes the final decision whether to use full thrust or derate.

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 braking and stopping an aircraft. And all this is for short time. 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. Tyres are radial type and tubeless. Aircraft tyres are conductive too as they are manufactured with tread rubber with conducting compounds to permit earthing of static charges.

Designing aircraft tyres requires sizing, weight requirement, regulatory requirements including foreign regulation in many cases etc.

A cutaway of a typical bias-ply tire is shown in the above figure. Bias-ply tires are popular choices for aircraft tires because of their durability and retreadability. The tire construction consists of the following components:

• Tread: The tread is made of rubber mixed with other additives to obtain the desired level of toughness, durability, and resistance to wear. The tread pattern is designed to aircraft operational requirements, with the ribbed tread design used widely due to its good traction under varying runway conditions.
• Sidewall: The sidewall is a protective layer of rubber that covers the outer casing ply. It extends from the tread edge to the bead area.
• Tread Reinforcing fabric: One or more layers of fabric that strengthens and stabilizes the tread area for high-speed operation. It also serves as a reference for the buffing process when tires are retreaded.
• Casing Plies: Alternate layers of rubber-coated fabric (running at opposite angles to one another) provide the strength of the tire.
• Beads: Hoops of high tensile strength steel wire that anchor the casing plies and provide a firm mounting surface on the wheel. The outer edge of the bead that fits against the wheel flange is called the bead heel. The inner bead edge is called the bead toe.
• Apex Strip: A wedge of rubber affixed to the top of the bead bundle.
• Casing Ply Turnups: The casing plies are anchored by wrapping them around the wire beads, thus forming the ply turnups.
• Chafer: A protective layer of rubber and/or fabric located between the casing plies and wheel to minimize chafing.
• Inner liner: In tubeless tires, the liner is a layer of low permeability rubber that acts as a built-in tube and restricts gas from diffusing into the casing plies. In tube-type tires, a thinner liner is used to prevent tube chafing against the inside ply.
• Undertread: Undertread is reinforcing plies of rubber-coated fabric placed under the buff line cushion to protect casing plies and strengthen and stabilize the tread area. They are considered an integral part of the casing construction.

Rolling resistance:

Rolling resistance sometimes called rolling friction or rolling drag is the force resisting the motion when a body (such as a ball, tire, or wheel) rolls on a surface. The cause of rolling resistance lies in the slippage between the wheel and the surface, which dissipates energy. Factors that contribute to rolling resistance are the amount of deformation of the wheels, the deformation of the roadbed surface, and movement below the surface. Additional contributing factors include wheel diameter, load on the wheel, surface adhesion, sliding, and relative micro-sliding between the surfaces of contact.

Rolling resistance can be given by the below formula,

`F_(rr) = mu_(rr)*m*g`

where, `F_(rr)="rolling resistance force"`

`mu_(rr)="rolling resistance coefficient"`

`m="mass in kg"`

`g="acceleration due to gravity"`

Tyre pressure:

Aircraft tires generally operate at high pressures, up to 200 psi (14 bar; 1400 kPa) for airliners, and even higher for business jets. Tests of airliner aircraft tires have shown that they are able to sustain pressures of a maximum of 800 psi( 55 bar; 5500 kPa ) before bursting.

Aircraft tires are usually inflated with nitrogen to minimize expansion and contraction from extreme changes in ambient temperature and pressure experienced during flight. Dry nitrogen expands at the same rate as other dry atmospheric gases (normal air is about 80% nitrogen), but common compressed air sources may contain moisture, which increases the expansion rate with temperature. The use of inert gas for tyre inflation will eliminate the possibility of a tyre explosion.

Brake forces when landing:

The three primary methods of achieving deceleration on the runway during the landing roll or a rejected take-off are engine power, brakes and mechanical spoilers. Effective and coordinated use of whichever of these are available will result in a stopping distance appropriate to the available runway ahead, or the desired runway exit point if sooner. The key to success is the properly coordinated use of all the available methods.

• Engine Power: In this, the focus is to select all thrust/power levers to the ground idle position promptly, and if available, continue the action through to the selection of reverse thrust or reverse propeller pitch. The selection at these relatively high speeds must be symmetrical because otherwise, directional control may be prejudiced.
• Wheel Brakes: It depends upon sufficient friction existing between the tyres and the runway surface achieved through freely rotating wheels. These ensure adequate wheel rotation exists and is maintained before the commanded brake pressure is transmitted to the brake units and takes effect. An electrical signal is sent from the right deck to hydraulic actuators near the main landing gear. Hydraulic fluid pressure is used to force the brake unit against the wheel, thus slowing it down. But this results in less efficiency due to an increase in weight. So electrical braking is prefered, i.e. when the pilots press on the brake pedals, an electrical signal is sent to the brake unit on the wheel. Here, electrically powered actuators are used to press the carbon brake disc against the wheel, slowing it down. Here, when the pilots press on the brake pedals, it sends an electronic signal to the Left and Right Brake System Control Units (BSCUs). These then send signals to the four Electronic Brake Actuator Controllers (EBAC) which control the rate of braking on the wheels. Each wheel has four Electric Brake Actuators (EBA), a kind of piston which presses against the carbon brake discs.
• Mechanical Spoilers: This is done by increasing aerodynamic drag on the moving aircraft. This can be achieved by raising upper wing surface panels called ground spoilers or by operation of a tail cone ‘clamshell’ type air brake. Both systems can also be used in the air on some aircraft types as in-flight air brakes, but the extent of their operation may be less in the airborne case than in the ground (weight on wheels) case.

To calculate the force and power required to push/pull an aircraft by a towing vehicle

Assumptions made

• Aircraft: Airbus A380-800
• Weight of the aircraft: 575 tonnes
• Weight of the towing vehicle: 20 tonnes
• Coefficient of rolling resistance of the aircraft: 0.005
• Coefficient of rolling resistance of the towing vehicle: 0.002
• Velocity of towing: 15kmph/4.1mps
• Density of air: 1.225
• Coefficient of drag of aircraft: 0.02
• Coefficient of drag of towing truck: 0.02
• Frontal area of aircraft: 410 m^2
• Frontal area of the towing truck: 40 m^2

The power requirement calculation is made using an Excel calculator:

The link for the excel sheet: https://docs.google.com/spreadsheets/d/1WCm3a-ASPBB-QXt2e3BEa0xa28pE8WZn/edit?usp=sharing&ouid=118244381142113766406&rtpof=true&sd=true

The tractive force required is

Fte = 28.69 KN

Power is evaluated from the total force at a given speed

Therefore, if the aircraft is to be moved at 15kmph/4.1mps then,

Power (P) = 28.69 * 4.1 = 117.629 KW

P = 117.629 KW

To design an electric powertrain with the type of motor, its power rating, and energy requirement to fulfil aircraft towing applications. To estimate the duty cycle range to control the aircraft speed from zero to highest. Prepare a table of assumed parameters. And to draw a block diagram of the powertrain.

Torque calculation

Since the total tractive force (Fte) is known the torque (T) can be calculated using the below equation,

`T = ((Fte*r)/(G*eta))`

where, `T` = Torque in N-m

`Fte` = Total tractive force in N

`G` = Gear ratio

`eta` = Gear efficiency

From the table,

Towing tractor tyre radius = 0.635 m

Gear ratio = 11

efficiency = 0.95

thus, T = ((28.69 *0.635)/(11*0.95))

T = 1729.6 Nm

Selection of motor

Selection of motor is done based on the derived power and torque values.

A TEFC (Totally enclosed fan cooled) class F motor is selected with the following specifications

There are three standards of efficiency in motors. They are

• IE1 - Standard efficiency motor
• IE2 - High efficiency motor
• IE3 - Premium efficiency motor

The motor which selected comes under the IE2 standard. Therefore the motor has high efficiency(`eta`) of 94.5 during full load with power output of 110 KW and full load torque of 1420 N-m.

The rated voltage of the motor - 400V

The rated current of the motor at full load - 210A

Calculating output power from power converter

To get the maximum required output from the motor the losses in the motor is compensated by the input power of the motor. Therefore the input power of the motor can be calculated by the following equation.

Efficiency of the selected motor during full load will be `eta = 0.945`

`P_i = (P_o/eta)`

`P_i = 117.629 /0.945`

`P_i = 124.47 kW`

Therefore power output from power converter will be 124.47 kW.

Calculating battery voltage output

The battery output voltage = Power converter output voltage (when `delta = 1`)

wkt,

`P = V*I`

where, `P` = Power in W`

`V` = Voltage in V

`I` = Current in A

therefore, `V_b= P_i/I_i`

`V_b = 124470/210`

`V_b = 592.71 V`

The battery voltage output given to the power converter will be 592.71 V.

Selection of battery

wkt,

`E = P*t`

where, `E` = Energy

`P` = Power in KW

`t` = Average time required for towing the aircraft from gate to taxiway

Assuming the time required for towing the aircraft from gate to taxiway to be 10 minutes

Since the voltage provided by the battery will be 463.38 V, the power of the battery will be

`P_b = V_b*I_b`

`P_b = 592.71 xx 210`

`P_b = 124.47 kW`

Therefore, `E = 124.47 xx 600`

`E = 74682 kJ`

`E = 20.745 kWh`

A 20.745 kWh battery is required to push or pull the aircraft for 10 minutes.

Duty cycle of the power converter

If the input current is maintained equal to the output current then,

`(P_o/V_o) = (P_b/V_b)`

`(P_o/P_i) = (V_o/V_i)`

`delta = (V_o/V_i)`

where, `delta` = duty cycle

`delta = 117.629/124.47`

`delta = 0.945`

Which is equivalent to the efficiency of the motor

Therefore the duty cycle ranges from `delta = 0 - 0.945`

Then, with the battery voltage input as 592.71, the output voltage from the power converter varies from 0 - 560.11 V.

Block diagram of the powertrain

The block diagram is drawn as shown below.