Project-1: Powertrain for aircraft in runways

although many aircrafts moves in the back wars using reverse thrust as a result. the jet blast or the propaller wash may damage the terminals and other accesories.there is dangerous when engine close to ground because they damage the engine.

and finally this may cause the huge damage to the aircraft.thats why pushback trucks provided for the backword movement of aircrafts in airports.this push back truck lifts the front wheel of the aircraft.the role of push back truck is to move the aircraft away from the parking area.

when we use high power engine then damage can occur itself to the aircraft.also for truck driver it is easy to see outside and drive the vehicle compared to the pilot.also push back driver knows well about the airport compared to pilot.

the other important thing is it saves the huge fuel for such a small thing used.also this is very noisy operation when done on the ground.if this is not used then not only damage the aircraft but also huge risk of jet blast and surrounding buildings.

the usage of reverse thrust to push back from the gate/terminal is also potentially hazardous because it can cause debris to blow towards the airplane which creates a foreign object hazardous.this foreign object hazard is damgerous in many ways.

for instance ,the ground staff are at risk because the reverse thrust can created an air flow which can blow them away and damage their ground operation equipment.the passengers in the airplane are also at risk because the debris can damage the jet engines of the aircraft which can create a fire hazard that can create an explosion due to the flammability of the fuel.

the debris can also pose a risk to the potential passengers waiting at the terminal/gate because it can also break the glass windows.

due to the above possible dangers,the tug vehicle is used for reversing the airplane until it is a safe distance from the terminal/gate and the airport ground staff.by using the taxiing vehicle,the fuel can be saved,and it can save the lives of the ground staff and their equipment and people at the terminal/gate.

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

after all the passengers have boarded and the pilots have an authority for the flight control,then the aircraft begins the first stage of a flight procedure called push back.in this stage,the front of the airplane is connected to a tug vehicle which helps for push back that is reverse the airplane from the terminal/gate.when airplane pushes back from the terminal/gate then airplane is ready for 'taxiing'.

then taxiing is the next procedure before flight in which the aircrafts moves on the ground using its own power on ots wheels without the assistance of any tug or towing vehicles.the same propulsion power of the aircraft that the jet engines produce for thrust during flight is used for moving forwards on the ground.since the taxiing procedure occurs with the engine power of an aircraft,this procedure only occurs once an aircraft is at a safe distance from the terminal/gate , as well as from the ground staff after pushback.this is because there are several dangers that exist when an aircraft uses the trust from the jet engines for the movement on the ground.for this reason ,a pushback vehicle is used to guide the aircraft away from the potential danger zones before taxiing can take place.during this taxiing procedure,the aircraft will move along taxiways.it is paths on an airport ground which connect runways,terminal/gates,hangers and other facilities.this airplane moving along these taxiways at a relatively low speed of 20-30 knots (23-35 mph or 37-56 kmph) ,as per speed limits set by operators and aircraft manufactures for safety reasons.the taxiways includes roads,have markings on their surface also to show the direction in which to conduct the taxiing procedure in.also,the taxiways have a different sign boargs which are used to identify runways ,taxiways etc.these sign boards may be either operational guidance signs or mandatory instruction signs.taxiways also use various types of lights and are mainly used for the purpose of guiding an aircraft to the runway from a taxiway or vice versa when the visibility is low.

operation guidance signs:

the 'A' symbol is a location sign for a taxiway whereas the 'B' symbol shows the direction to the taxiway with the designation 'Bravo'.

the mondatory instruction signs:

the below symbol is used as the stop sign on taxiways.

the below symbol is taxiways light.

takeoff power-

it is the phase of flight in which an aircraft leaves the ground and becomes airborne.for this take off ,an aircrafts wings are designed with an airfoil shape which is able to generate a lift force relative to the air that is when the aircraft moves.this lift force is required to lift the aircraft off of the ground during takeoff and is also needed to flight to keep the aircraft in the air.the lift force generated because of the way in which air flows around the airfoil shape.when the airfoil moves relative to the air,a "down wash" is created which creates a pressure gradient between the top and bottom of the airfoil.the bottom of the airfoil experiences a higher pressure and top of the airfoil will experiences a lower pressure.due to this pressure gradient, a lift force is generated .the total amount of force that is generated depends on 3 factors:

1) the air speed

2) the angle of attack

3) the shape of the airfoil.

the angle of attack is the angle between the reference line of an airfoil and the direction of the velocity of the wind.if angle of attack is higher then down wash is greater and hence the lift force generated is greater.the greater the velocity of the air flowing over the airfoil will also increase the lift force.also when changing the shape of the airfoil then this will affects the lift force.the shape of the airfoil can be altered by the use of slats and flaps.the slats and flaps increase the chamber of the airfoil and allows it to generate more lift force at lower speeds.when slats and flaps activate then they extend and alter the shape of the airfoil in such a way that they increase the area of the wings which increases the lift force.

to control the lift force that the aircraft experiences,the other components which are present aside from the slats and flaps that can be adjusted.

the first component are the ailerons located on the wings.the ailerons have flaps which can move up then generated less lift force and when down to generate more lift force.

takeoff performance-

this performance of an aircraft is obtained by measuring its acceleration on the runway and observing the forces which acting on it . by using newtons second law of motion,the resultant force ,F acting on a body is given by:

F= m*a

therefore,Ft-Fd-Fr= m*a

so,

a = (Ft-Fd-Fr)/m

where,

Ft- thrust force

Fd- Drag force

Fr- Rolling Resistance force

a- acceleration

m- mass

this acceleration is continue untill aircraft achieving a safe flying speed refferred to as the rotation velocity (Vr).

at this speed,the pilot can rotate the aircraft which will produce enough lift to takeoff from the ground.this rotation velocity that is takeoff speed depends on the various factors e.g. air density (depends on air temperature and field elevation),aircraft gross weight and the aircraft configuration (slats and flaps position).typical takeoff speeds for commercial jetliners range from 240-285 kmph or 149-177 mph.this speed is lower in smaller aircrafts than light aircraft for example Cessna 150 having a takeoff speed of 100 kmph.

aircraft can be operated in the transport category and use a system of 'V-speeds' the three of them are V1,V2,and Vr.

V1 is the decision speed (critical engine speed) and it is the speed at which any decision to abort takeoff must be made.if the aircraft speed greater than V1 then it must continue to takeoff unless there is a reason to believe that takeoff will be unsuccessful.for this instance ,if an engine failure is detected before the V1 then takeoff should be aborted.

Vr is the rotational speed as mentioned before and it is greater than V1.

V2 is the takeoff safety speed ,which is the speed at which the aircraft must acchieve at the 35 feet height at the end of the runway distance.

the rotational speed is depends on the aircraft gross weight and flap settings but it also depends on the environmental conditions.the rotational speed is 1.1 times greater than the stall speed.the stall speed is the lowest speed at which an aircraft can be flown before the airflow begins to separate from the wings and the angle of attack becomes too large.since we know that the lift force must be equal to the weight of the aircraft for a flight to be maintained at constant level,the stall speed is as

Vstall= √(2*W)/(Clmax.ρ.A)

Where,

Vstall= stall speed

W= Weight of the aircraft

Clmax= Coefficient of lift at stall speed

ρ = Air density

A = Wing area

the rotational speed is given by,

Vr= 1.1*Vstall

Tire Design=

Aircraft tires are designed in such a way that they can withstand extremely high loads for short durations of time.the number of tires in aircraft is proportional to its weight that for heavier aircraft requires more tires.if number of tires are less for a given weight of an aircraft,then the weight cannot be distributed evenly enough amongst the tires.the tread patterns on aircraft tires are designed in such a way that they can provide stability to flights during crosswind conditions.the tread design also heps to direct water off of the tires so as to prevent hydropatting and they also aid in the braking action.

these tires are designed with fusible plugs.A fusible plug is a threaded metal cylinder made up of bronze ,brass and gunmetal with a drilled hole that extends through its entire length.the holes are filled with a metal of low melting point,this will cause the deflation of the tires.in case of energancy landing or aborted takeoffs, the aircrafts wheels experience high friction with the runway surface and the use of the plugs will control the deflation of the tires in a safe manner.without using this ,there is a risk of an explosion which can damage the aircraft and other nearby objects.

they also help to absorb the shock of landing and provide cushioning.it also provides the necessary traction for braking and stopping of an aircraft.

tire pressure=

these tires are inflated with an inert gas and mostly used nitrogen.the inert gases are used for inflation because they are non-reactive and not flammable.when the ambient air temperature and pressure changes,the nitrogen gas helps to minimize the expansion and contraction of the aircraft tires.using oxygen for the inflation of aircraft tires is very dangerous since it is flammable.using nitrogen instead eliminates any risk of there being a tire explosion.

generally commercial airliners use tires that can withstand a pressure of 800 psi before they burst. A Boeing 777-300ER has 12 tires and for each tire is inflated to a pressure of 220psi with a diameter of 134 cm and a weight of 120 kg.the tire cost is $5000 and this is changed after 300 cycles whereas the brakes are changed after every 2000 cycles.

by using a calibrated gauge check the aircraft tires because aircraft tires operate at high temperatures and pressures and they do not retain their air perfectly.the tire pressure reduce by 5% in 24 hours.michelin, one of the 4 main aircraft tire manufactures,recommends that the air pressuree of the tires be checked before the first flight that day or before each flight if the aircraft isnt flown daily.

the tires can be checking before the flight rather than after is benefacial because the tires need to be cold when performing the pressure check.if we check the tire after the flight,then the tires is hot due to its recent usage and this makes it very difficult to determine what the proper pressure should be.michelin recommended that the tire pressure is checked no earlier than 3 hours before a flight after that it has last been in use.

during servicing if we perform the pressure checks then we need to consider the ambient temperature's effect on the tires.if ambient temperature changes by 3 degree then there is 1 % change in the tire pressure.to account for this,machelin recommended that the tires be serviced in a way that the tire pressure is overshoot to 105% of the operating pressure.so it is a guideline that the highest pressure in the aircraft maintenance manual (AMM) be targeted.

for aircraft applications ,there are two main types of tires are used:

1) Bias ply tire

2) Radial ply tire

1) bias ply tire are used for their long durability and hence long working life as well as for their retreadability.the construction is

Tread -

it is made of rubber and other additives that are used to obtain desired properties such as high durability,high toughness and high resistance to wear.the pattern of the thread has a ribbed design which helps to provide excellent traction in different runway conditions whether it is good or bad.

the below figure shows bias ply tire-

Sidewall- it is a protective layer of rubber that covers the outer casing ply and it extends from the tread edge to the bead area.

tread reinforcing ply-

it is one or more layers of a fabric which strenthens and stabilizes the tread area when the aircraft ground speed is high.it also helps as a reference for the buffing process during retreading of the tires.

Buff line Cushion-

it is made up of rubbery fabric that increases the adhesion between the tread reinforcing ply and the breakers.

Breakers-

these are the plies reinforcing of a rubber coated fabric which are placed under the buff line cushion and it helps to stabilize and strengthen the tread area.

casing plies-

theses are the alternate layers of rubber coated fabric which run at opposite angles to each other and it will provide the strength of the tires.

Wire beads-

these are the hoops of high tensile strength steel which anchor the casing plies and provide a firm mounting surface on the wheel.

Apex Strip-

this is a wedge of rubber fixed to the top of the bead bundle

these are the layers of rubberized fabric which help to anchor the bead wires to the casing and it also improve the durability of the tire.

Ply Turnups-

the casing plies are wrapped around the wire beads to anchor them which forms ply turnups.

Chafer-

it is a protective layer or rubber or fabric coated loacted between the casing plies and the wheel to keep chafing to a minimum.

Liner-

In a tubeless tire,the liner is a layer of low permeability rubber that acts as a built in tube and prevents the diffusion of gas in to the casing plies.in tube tires,the liner layer is thinner to prevent chafing between the tube and the inside ply.

2) radial ply tires are used in aircrafts because they have a rigid belt that allows them to be used for increase landings and have lower rolling resistance than bias ply tires.they have a lesser weight than bias ply tires and have less components of same size.the different components of radial ply tires is shown below:

Overlay-

this is a layer of reinforcing rubber coated fabric which is placed on top of the rigid belts and help them when aircraft move with high ground speed.

Belt Plies-

this is a composite structure which stiffens the tread area and increase the tire strength in the tread area for increased landings.

Casing Plies-

they consist of layer of rubber coated fabric same as bias ply tires but instead of running at opposite angles to each other,these run radially from the bead to the head.

Chippers-

this rubber coated fabric layer applied at diagonal angles to improve the durability of the tires in the bead area.

rolling resistance-

it is a force that resists the motion of a body that rolls on a surface.in the case of aircrafts,the rolling resistance force exists between the aircraft tires and the runway surface it rolls upon.

for a given body,the rolling resistance is expressed mathematically as the product of the coefficient of rolling resistance and normal force,this force acting perpendicular to the surface on which rolling action occurs.

Frr = urr*N

where,

Frr= rolling resistance

urr = coefficient of rolling resistance

N = Normal Force

Braking Forces-

in aircrafts uses brakes mainly during landing procedure. during braking,kinetic energy of the aircraft conveted in to heat energy because friction occurs between the aircraft tires and the runway surface.the braking is a very crucial part of flight because there is a limited distance on the runway that the aircraft can use to reduce its speed.in the braking process ,braking has to be done in such a way that the aircraft or passengers or it equipment does not get damaged .there are different systems are used for braking process.

working of Disc brakes-

the main brake used in aircraft is the disc brake which work by manipulating the friction between the stationary disc and rotating disc found inside the disc brake.

first of all, a command signal is to the disc brakes by the pilot by pressing the brake pedal in the cockpit.this will causes the actuators in the brake to move a piston to squeeze the stationary and rotary discs together.which creates a frictional force and used to slow down the tires and hence the aircraft.as the friction in braking generates high amounts of heat energy,the disc brakes are designed as heat sinks to absorb this generated heat.this temperature of the disc brake is up to 1800 degrees celcius.

this brakes are found on a carrier assembly which consists of a torque tube that transmits braking torque to the landing gear and which is placed between the carrier assembly and a braking plate.the discs are arranged in an alternating pattern of stator and rotors.each rotor has grooves and notches that allow it to fit in with a wheel with a corresponding geometry so that rotor rotates with it.the stators are fitted to the torque tube and it is stationary because the torque tube is connected to the axle and landing gear structure both are stationary.the cylindrical spaces in the carrier assembly contains a pistons which move to cause the squeezing of the two discs bringing about braking.

Design of brakes-

the main design considerations of the disc brakes are the number of discs they will hold together,the disc diameter and their material.the brakes are also designed by considering the worst case situation.one of such scenario is when takeoff is aborted at the decision speed,V1.if V1 exceeded,then the aircraft cannot safely abort takeoff because it is a risk situation of not being able to stop before the end of the runway.in this case ,the brakes would need to be able to absorb more kinetic energy than in any other worst case condition.

In an Airbus A380 is the worlds biggest passenger plane,its specifications are mentioned below:

1) maximum takeoff weight (MTOW) = 575000 Kg

2) Maximum decision speed , V1= 170 knots (195 mph or 87 m/s)

3) Number of disc brakes = 16

so,we calculate the kinetic energy of the aircraft ignoring the effect of air brake reverse thrust headwinds and tailwinds.

E = 1/2*m*v^2

= (1/2* 575000*(87)^2)

= 2.2 GJ

This is the amount of enegy that is required by the combined efforts of all brakes to completely brake the aircraft.

this generate frictional force,for that there are multiple disc per brake are needed.An A380 has 22 tires that are distributed on wheels support the nose on a leg,eight wheels split between 2 legs to support the left and right wings and 12 body wheels split between 2 legs under the fuselage.only 16 of these 22 wheels have brakes.

each of the 16 brake assemblies has 5 rotors of honeywell carbenix 4000 carbon -carbon composite.the brake assembly within the wheels are made up of 2014 T-6 aluminium alloy and are attached to the axle and landing gears which are made up of 300M high strength alloy steel,titanium and aluminium components.

Materials

up to 1963 ,steel material was used in brake discs.after that because of better thermal properties ,beryllium material is used but it was expensive to handle since its oxide was toxic.after that ,the industry was shifted to carbon brakes.they are made of carbon fibers in a graphite matrix.carbon material is used because they were lighter ,more durable and less thermally sensitive.they also had higher absorption of thermal energy and cooled faster than steel brakes.also carbon has a higher specific heat of energy compared to steel so this can reduce the brake weight.carbon has a higher thermal conductivity meaning the heat will be distributed uniformly throughout the disc.it has a lower thermal expansion,higher thermal shock resistance and a higher temperature limit than steel.in addition to this,carbon's specific strength remains consistent under different temperatures unlike steel and beryllium.

Brake cooling

cooling is required for cooling due to the high heat buildup and they do this in two ways:

Passive cooling-

this cooling occurs via natural conduction from the brake disc material to the surrounding components.cooling occurs through radiation of heat that comes off the brakes as well as by natural convection as air flows through the brake assembly.

Active cooling

this cooling is done using fans that force air through the brakes to cool them down.

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

Assumption :

Weight of the aircraft = 500000 kg

rolling resistance coefficient of an aircraft tire = 0.005

weight of the towing tractor = 50000 kg

rolling resistance coefficient of towing vehicle = 0.002

the velocity of the aircraft while towing = 18 kmph = 5 m/s

density of air medium = 1.225 kg/m^3

the frontal area of the aircraft = 20 m^2

the coefficieent of drag of the aircraft = 0.26

towing tractor gear ratio (G-for high torque output) =11

towing tractor tyre radius (r) =0.5 m

the two forces are acting on the aircraft

1) rolling resistance for the aircraft

in this case,rolling resistance be

Fr = c*mr*ag

where,

Fr = rolling resistance

c = coefficient of rolling resistance

mr= rolling mass

ag= Acceleration due to gravity.

now substituting the values in the above equation gives us the rolling resistance of the aircraft

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

therefore the rolling resistance of the aircraft Fr = 24.525 KN

Rolling resistance for the towing vehicle

Now let us calculate the rolling resistance for the towing vehicle

Fr = c*mr*ag

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

therefore the rolling resistance of the towing vehicle Fr= 0.981 KN

2) Drag force=

D = 1/2*Cp*A*V^2

where,

D = drag force

Cp= drag coefficient

V = velocity of towing vehicle

A = frontal area

now substituting the assumed values in the above equation gives us the drag force of the towing vehicle.

Drag force = 0.5*(0.26*1.225*20*5*5)

= 0.079 KN

therefore the drag force is 0.079625 KN

so ,the total force to push the aircraft by the towing vehicle

therefore , total force= rolling resistance of the aircraft + rolling resistance of towing vehicle + Drag force

= 24.525 KN + 0.981 KN +0.079625 KN

= 25.585625 KN = 25585.625 N

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

Power required to the towing vehicle to pull the aircraft .

in this case,Power = force * velocity

= 25.585625 * 5.0

= 127.928125 KW

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

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

In this model we use constant block,product block,sum block,square block,display block,divide block etc.

1) constant block-the constant block generates a real or complex constant value signal .use this block to provide a constant signal input.

2) product block-this block outputs the results of multiplying two inputs.

3) sum block-the sum block performs addition on its inputs.

4) square block-there is a math function block in simulink-> math operation,where you can configure it to perform various math operation including the square function.

5) display block-the display block shows the value of the input data. you can specify the frequency of the display .for numeric input data,you can also specify the format of display.

6) divide block-the divide block outputs the result of dividing its first input by its second.the inputs can be scalars,a scalar and a nonscalar,or two nonscalar ,or two nonscalars that have the same dimensions.this block supports only complex input values at division ports when all ports have the same single or double data type.

in this model ,aircraft rolling resistance is calculated as Fr=0.005*500000*9.81 ,the product of three constant block is calculated in product block.

then ,rolling resistance force of the towing vehicle = 0.002*50000*9.81 ,the product of three constant block is calculated in product block.

then ,Drag force = 0.5*Cp*A*V^2= 0.5*0.26*1.225*20*5*5. all four constant block can be multiply in product block.

then, in sum block all this forces are summed to get total force and is displayed in display block =2.559e+04=25.585625KN=25585.625N.

Then power = force *velocity and power is displayed in display block= 1.279e+05= 127.928125 KW=127928.125W.

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)

Here ,i assumed the rafele fighter aircraft.

Motor selection for the power train:

based on some of the parameters we selcected DC machine for the power train.

the power rating and energy requirement to fulfill aircraft towing application:

assumed parameters=

here i assumed the rafale fighter aircraft=

parameters assumption

mass of the aircraft (ma) 24500 kg

mass of the towing vehicle 10000 kg

rolling resistance (urr) 0.05

air density (ρ) 1.5 kg/m^3

frontal area (A) 10 m^2

aero drag coefficient (Cd) 0.6

velocity (V) 21.6 km/hr=6 m/sec

gravitational constant (g) 9.81 m/s^2

above model represents the simple power train model for aircraft on the runway.this model consist of converter,four quadrant chopper DC drive,motor,motor controller.the dc drive is mainly used as a controller.

this is the example of DC-7 four quadrant chopper dc drive for speed regulation.

the dc motor of 200 HP rating is separately excited with a constant 150 V DC field voltage source.the 380 V ,50 Hz AC voltage source is conveted to 515 V DC by rectification ,this armature voltage is provided by an IGBT converter controlled by two PI regulators.for limiting DC bus voltage during dynamic braking mode,a braking chopper added between the diode rectifier and DC7 block.

the speed regulator outputs the armature current reference used by the current controller in order to obtain the electromagnetic torque require to do the desired speed.this changes the rate follows acceleration and deceleration ramps in order to avoid sudden reference changes that could cause armature over-current and destabilize the system.

by using four IGBT devices of 5KHz pulses duty ratios controls the armature current by current regulator.the IGBT 1 & 4 output and IGBT 2 & 3 output are opposite for proper system behaviour that is for getting desired armature current.a smoothing inductance is used in series with the armature circuit to limit the amplitude of the current oscillations.

simulation- first of all,set the initial bus voltage to 515 V via the GUI block before starting the simulation.then start the simulation ,you can observe on scope various result like motor armature voltage,current,four IGBT pulses,the motor speed.mechanical torque is proportional to speed bcause of linear load coupled to motor.

the output response is shown in below figure.

output response=there are 5 output responses are shown in below figure.

first response shown is for duty cycles IGBT 1 & 4

second response shown is for duty cycles IGBT 2 & 3

third response for armature voltage.

fourth response for armature current

fifth response for motor speed.

B. Also, Design the parameters in excel sheet.

Conclusion=

1) Studied, Searched and listed the total weight of various types of aircrafts.

2) Explained difference between ground speed and air speed.

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

4) Explaned, How an aircraft is pushed to runway when its ready to take off?

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

6) the total force required for a towing vehicle to push the aircraft = 25.585625 KN =25585.625 N.=2.559e+04.

the power required for the towing vehicle to pull the aircraft = 127.928125 KW = 127928.125 W=1.279e+05.

BOTH ANSWERS ARE SAME IN CALCULATION AND SIMULATION.

7) A) Designed an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfill aircraft towing application in Simulink. Estimated the duty cycle range to control the aircraft speed from zero to highest. Prepared a table of assumed parameters. also Drawing a block diagram of powertrain.

B) Prepared a design parameters in excel sheet.