Project-1: Powertrain for aircraft in runways

TASK 1.

PART A.

Q1. Search and list out the total weight of various types of aircrafts.

ANS:

Aircraft Gross Weight:

As per federal aviation department,aircraft gross weight is defined as the total weight at any moment during flight or ground operation.
The aircraft gross weight dicreases during flight due and oil consumption.In some case gross weight (aircraft) may vary due to drop in payload or inflight refueling,where as on the other hand when the brakes are released the gross weight will be equal to take of weight.

During flight an aircraft gross weight will refered as inflight weight.

Types of aircraft weights:

1.Manufacures Empty weight:

It includes the weight of aircraft of the entire flight with its intergal working parts,but excludes baggage,passenger and fuels.

2.Zero Fuel Weight:

It includes total weight aeroplanes including unuseable fules but excludes total weight of useable fuel at board.

3.Payload:

It is load carrying capacity of an aircraft which includes passenger,cargo and fuels,but if it is in the case of cargo flight it determines the total load of logistics weight of the container which it needs to transport.

4.maximum Take Off Weight:

This is the maximum weight at which the aircraft would be made to takeoff.

5.Regulated Take Off Weight:

Depending on different factors (flap setting,altitude,lenght of runway),RTOW varies for each take off,can never be high for MOTW.

6.Maximum Landing Weight:

It is maximum weight at which an aircraft is permitted to land.

7.Maximum Ramp Weight:

it is the maximum weight used for manuevering (towing) aircraft on the ground.

List are the types of aircraft with its weight in excel file:

Q2.Is there any difference between ground speed and air speed?

ANS:

Ground Speed:

• They are the airplane speed which is related to the ground.
• They are the speed which determines how fast the airplane will reach the detination.
• They are the calculated by summing up the airspeed and windspeed.

Air Speed:

• It referes to the speed which is relative to air it is flying in.
• They determine whether is enough air flow around an aircraft to make it lift.
• They are independent of ground speed.

Here wind speed is the vector difference between ground and air speed.On a perfect still day the wind speed is zero and it is equal to the ground speed.

If the measured air speed is greater than the observed ground speed the wind speed becomes positive.

Q3.why is it not recommended to use aircraft engine power to move it on the ground at airport.

ANS:

Generally most of the airplane can move itself in reverse manner by making use of reverse thrust technology.But in most of the practical it is not possible because when airplanes start using reverse thrust it will consume lot of fuels as well as causes lot of noise and also hazardus for the people working near by.

The another for not using reverse thrust is that the jet blast can cause damage o terminal buildings or its equipments.Since the engine is alos close to the ground it will blow off the sand and debreis forward and sucks it to the engine and couses damage to it.To solve thid kind of issue we use a towing vehicle which helps to tow the vehicles in reverse direction and it eill be tugged until it reaches a safe place and from there onwards airplanes will move by itself for further take off.

Q4.How an aircraft is pushed to runways when its ready to take off.

ANS:

Airplanes have engines,and they can use them to move forward along the ground.So they push themselves to the runway.

It is true that most of thebtime,a small vwhicle is needed to push the larger commercial airplanes backwards away from the terminal building,because even if a plane can reverse it's thrust,it's not always safe to hit the buildings with that reverse thrust.So the little trucks push the bigger planes back away from the buildings,but after that,the planes can apply a small amount of engine power to roll forward just as all airplanes can.

Aircraft push themselves to the runway.

Q5.Learn about take off power,tyre design,rolling resistance,tyre pressure,brake forces when landing.

ANS:
TAKE OFF POWER:

The phase of flight in which an aerospace vehicle leaves the ground and becomes airborne.For aircraft travelling vertically,this is known as lift off.For aircraft that take off horizontally , this usually involves starting with a transition from moving along the ground on a runway.Take off performance can be predicated using a simple measure of the acceleration of the aircraft along the runway based on equilibrium.

The forces involved will be,

T - thrust of propulsion system pushing aircraft along the runways.

D - Aerodynamic drag of vehicle resisting the aircraft motion.

F - Rolling resistance friction due to the contact of wheels or skilds on the ground.

THRUST:

It is the force that moves an aircraft through the air.Also it is a mechanical force generated by the engines of the airplane.The engines does work on the gas and as the gas is accelerated to the rear,the engine is accelerated in the opposite direction.The acceleration of the engine mass produces a force on theaircraft.

TYRE DESIGN:

An aircraft tire is designed to withstand extremely heavy loads for short duration.The number of tires required for aircraft increases with the weight of the aircraft,as the weight of the airplane needs to be distibuted more evenly.Tire tread patterns are designed to facilities stability in high crosswind conditions,to channel water away to prevent hydroplaning and for braking effect.Tire includes fusible plugs,desiganted to melt at a certin temperature.The fuses provide a safer failure mode that prevents tire explosions by deflating in a controlled manner,thus minmizing damage to aircraft and objects in the surrounding enviroment.

ROOLING RESISTANCE:

The force resisting the motion when a body rolls on a surface.It is mainly caused by non-elastic effects.Two forms of this is hysterisis losses and permanent deformation of the object or the surface.slippage between the wheel and the surface also results in energy dissipation.Although some reserchers have included this in rolling resistance,sime suggest that this dissipation term should be treated separately from rolling resistance because it is due to the applied torque to the wheel and ground which is slip loss.

TYRE PRESSURE:

They usually have high pressures up to 200psi for airlines,and even higher for business jets.The main landing gear on the concorde was typically inflated to 232 psi,while its tail bumper gear tires were as high as 294 psi.They are usually inflated with nitrogen to minimize expansion and contraction from extreme changes in ambient temperature and pressur.Dry nitrogen expands at the same rate as other dey atmospheric gases but compressed air may contain moisture,which increases the expansion rate with temperature.

LANDING BRAKE FORCE:

The brake units are powered by the hydraulics system.An electrical signal is sent from the flight desk to hydraulic actuators near the main landing gear.Here,hydraulic fluid at 3000 pounds per sqauare inch is used to force the brake unit against the wheel tjhus slowing it down.

PART B.

Q6 (A).

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

Q (B). Develop the model for the calcualated force and power using simulink.

(A).ANS:

Considering:

Total mass of their aircraft (including payload ) for airbus a380 = 575,000 kg.

Mass of the push back = 1000 kg (assumed)

ROLLING RESIATNCE:

Coefficient of rolling resistance = 0.002

Rolling resistance force is given by

Frr=Crr.m.g

For the aircraft:

Frr= 0.002*575000*9.81

Frr=112815 N => 11.2815 kN

For the Towing vehicle:

Frr=0.002*1000*9.81

Frr=19.62 N

As the towing vehicle resistance is so less compared to the Aircraft because this is a heavy aircraft these can be neglected.

AERODYNAMIC RESISTANCE :

Drag coefficient (Cd) : 0.025

Air density (ρ)= 1.125 kg/m^3

Frontal Area (A): 150 m^2

Velocity (V): 14.5 kmph

(14.5/3.6)= 4 m/s

Fad = 1/2.ρ.A.Cd.V^2

Fad = 1/2*0.025*150*4^2

Fad = 33.75 N

As the vehicle and place is moving at very low speed Aerodynamic resistance can be neglected.

HILL CLIMBING RESISTANCE:

Runways are constructed flat so gradient resistance / Hill climb force is neglected.

Total Resistance:

Ft = 11.2815 kN

Power required:

P = Ft*v

P = 11.2815*4

P = 45.26 kW

The main power required by the tow vehicle is 45.26 kW to taxi a heavy air craft at 4 m/s

(B) .ANS:

simulink model :

constant blocks are used to feed the parameters such as mass , coefficients and other parameters.

Using product block the resistive forces are calculated.

Using sum block,total resistive forces can be calculated.

The outputs are visualised using display blocks.

Q7.(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 reuired assumptions.Prepare a table of assumed parameters.Draw a block diagram of powertrain.( Hint:DC7 block)

Q(B).Also,design the parameters in excel sheet.

(A).ANS:

We will be taking DC7 motor as an example:

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 Pl 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 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 armatureover-current and destabilize the system.the current regulator controls the armature current by computing the appropriate duty ratios of the 5hz pulses of the four IGBT devices 1 and 4 are opposite to those of IGBT devices 2 and 3.This generats 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.

SIMULATION:

Set the intial bus voltage to 515 V via GUI block.Start the simulation you can observe the motor armature voltage and current,the four IGBT pulses and the motor speed on the scope.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 500rpm at t=0 s. obesrve that the motor speed follows the reference ramp accurately (+400 rpm/s) and reaches steady sate 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 hz.At t=2 s , speed reference drops to 1184 rpm.The current reference decreases to reduce the electromegnetic 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.This causes the DC bus voltage to increase.At t=3.25 s,the motor reaches 0 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 reaches -1184 rpm and stabilizes around its reference.

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 ultracapacitors are selected to fullfill the required power deliver.

Motor selection:

Here we have selected a BLDC motor type that has a power starting 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

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 uoscale the battery armature voltage to the desired high voltage values so that the motor may spin at full power and required torque can be achieved using a bidirectional DC DC converter.Finally the speed control of the vehicle is done with switching techniques where the duty cycle adjustment can change the motor speed from the minimum to maximum.

(B).ANS

Design the parameters in excel sheet.

Frr=11281.5 N => 11.2815 kN