Project-1: Powertrain for aircraft in runways

Part A

Q1)

• The aircraft gross weight is the total aircraft weight at any moment during the flight or ground operation.
• An aircraft’s gross weight will decrease during a flight due to fuel and oil consumption. An aircraft’s gross weight may also vary during a flight due to payload dropping or in-flight refuelling.
• At the moment of releasing its brakes, the gross weight of an aircraft is equal to its takeoff weight. During flight, an aircraft’s gross weight is referred to as the en-route weight or in-flight weight.

TYPE	MAXIMUM TAKE OFF WEIGHT [KG]
Antonov An-225	640,000
Boeing 747-8F	447,700
Boeing 747-8	443,613
Boeing747-400ER	412,770
Boeing 747-400	396,900
Boeing 747-200	377,840
Boeing 747-300	377,840
Airbus A340-500	371,950
Airbus A340-600	367,400
McDonnell Douglas DC-10	256,280
McDonnell Douglas MD-11	273,300
Lockhead L-1011-500	231,300
Lockhead L-1011-200	211,400
Vickers VC10	152,000

Maximum takeoff weight (MTOW):- The maximum take off weight is the maximum weight authorised at brake release for takeoff, or at the start of the takeoff roll.

The maximum takeoff weight is always less than the maximum taxi/ramp weight to aloow for fuel burned during taxi by the engines and the APU.

Q2)

There is a difference between ground speed and air speed to understand this we should take in to account wind speed too.

Wind Speed:- For a reference point picked on the ground, the air moves relative to the reference point at the wind speed. 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.

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 comparision of the ground speed to the wind speed must be done according to rules for vector comparisions.

Airspeed:- The important quantity in the generation of lift is the relative velocity between the object and the air, which is called the air speed. 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.

For example taking wind speed as 20 km/hr and runway is east-west runway the wind blowing towards west and the flight runs at 80 km/hr the wind speed acting in the west and the flight takes off east then the

Airspeed=Groundspeed(80)- Wind Speed(-20) =100 km/hr

If the planes runs towards west along the wind blowing direction then

the Airspeed=Groundspeed(80) - Wind Speed(20)=60km/hr

Hence there is a difference between ground speed and air speed.

Q3)

• 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 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.
• Exhaust velocity components are attenuated with increasing distance from the engine exhaust nozzle. 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.

Q4)

• In aviation pushback is an airport procedure during which an aircraft is pushed backwards away from an airport gate by applying external power.
• Although many aircraft are capable of moving themselves backwards on the ground using reverse thrust, the resulting jetblast or propwash 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 aerodomes are usually subject to ground control clearance is obtained, the pilot will communicate with the tractor driver to start the pushback.
• Once the pushback is completed the towbar is disconnected, and any bypass pin removed. The ground handler will show the bypass pin to the pilots to make it clear that it has been removed.
• The pushback is then complete, and the aircraft can taxi forward under its own power.

Q5)

TAKE-OFF POWER :-

• For light aircraft, usually full power is used during take-off. Large transport category aircraft may use a reduced power for take-off, where less than full power is applied in order to prolong engine life, reduce maintenance costs and reduce noise emissions.
• In some cases the power used can then be increased to increase the aircraft’s performance. Before take-off, the engines, particularly piston engines, are routinely run up at high power to check for engine-related problems.
• The aircraft is permitted to accelerate to rotation speed. The term rotation is used because the aircraft pivots around the axis of its main landing gear while still on the ground, usually because of gentle manipulation of the flight controls to make or facilitate this change in aircraft attitude.
• For most aircraft attempting a take-off without a pitch-up would require cruise speeds while still on the runway.

TYRE DESIGN:-

Based on the type of plied used, the tyres can be:- (a) Bias ply tyres (b) Radial ply tyres (c) Tube-type tyres (d) Tubeless tyres.

• Bias ply tyres:-These tyres feature a casing which is constructed of alternate layers of the rubber-coated ply cords which extend around the beads and are at alternate angles substantially less than 90 degree to the center line of the tread.
• Radial ply tyres:-These tyres feature a flexible casing which is constructed of rubber- coated ply cords which extend around the beads and are substantially at 90 degree to the centreline of the tread. The casing is stabilized by an essentially in extensible circumferential belt.
• Tube-type tyres:-These require tubes for inflation retention.
• Tubeless tyres:-These do not require tubes. They are constructed with an inner liner.

ROLLING RESISTANCE:-

The resistance offered by the tyres against motion of vehicle/aircraft is called as rolling resistance, the rolling resistance is produced during take-off/landing in the runway.

TYRE PRESSURE:-

Aircraft tyres 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, but common compressed air sources may contain moisture, which increases the expansion rate with temperature. The main landing gear on the Concorde was typically inflated to 232 psi, whilst its tail bumper gear tires were as high as 294 psi. The high pressure and weight load on the Concorde tyres were a significant factor in the loss of Air France flight 4590.

BRAKES FORCES WHEN LANDING:-

Aircraft have highly-engineered, multi-disc brakes made of a carbon- steel material. Manufactures include Honeywell and Goodrich, part of Collins Aerospace. In general, the brakes are composed of discs that rotate with the wheel and separate discs that are fixed. When pressure is applied-and below you see pressure applied from the right fixed disc on all of the discs-the entire assembly compresses to create the braking action. Carbon brakes are standard today, they cool faster than steel brakes and can weight up to 700 pounds less. That represents a significant weight for airlines.

Auto Brakes:-Wheel brakes are usually selected to engage automatically after landing. These are called ‘autobrakes’ that all airliners have several levels of autobrake ‘severity’. The crew will decide which is the most appropriate to use for every landing depending upon factors like the length of runway, if it is wet or dry, or if the pilots want to exit the runway at a particular point. For a short wet runway you might select autobrake 3, or MED on the Airbus. For a long dry runway you might select autobrake 1 or LOW on the Airbus.

PART B

Q6) A)

Mass of the aircraft (m_a) = 40000kg; Coefficient of rolling resistance (μ_r) = 0.03; Acceleration due to gravity (g) = 9.81ms^-2; Frontal area of an aircraft (A) = 85m²; Final velocity of the aircraft (v_f) = 3m/s; Initial velocity of the aircraft (u_i) = 0m/s; Coefficient of drag of an aircraft (c_d) = 0.15; Density of air (ρ) = 1.225 kg/m³; Acceleration time of the aircraft (t_a) = 10seconds,

Rolling resistance:-

Frr= μ_r mg; where, Frr= rolling resistance force; μ_r = coefficient of rolling resistance; m= total mass of the vehicle; g= acceleration due to gravity;

Frr= μ_r mg= 0.03*40000*9.81= 11,772 N

Acceleration force:-

Acceleration force is defined as the defined as the rate of change of velocity with respect to time.F= m*a; where, m= mass; a= acceleration,

F=m*a= m*(vf-vi)/t= 40000*(3-0)/10= 4000*3= 12000N

Aerodynamic drag force:-

It is represented by the formula Fd=1/*roh*A*cd*v^2; where, roh=density of air; A= frontal area; cd= drag coefficient; v^2= velocity,

Fd=1/*roh*A*cd*v^2= (1/2)*1.225*85*0.15*3^2= 70.284

Towing vehicle calculations:-

Mass of the Towing vehicle (m_a) = 8000kg; Coefficient of rolling resistance (μ_r) = 0.013; Acceleration due to gravity (g) = 9.81ms^-2; Frontal area of the Towing vehicle (A) = 8m²; Final velocity of the Towing vehicle (v_f) = 3m/s; Initial velocity of the Towing vehicle (u_i) = 0m/s; Coefficient of drag of the towing vehicle (c_d) = 0.60; Density of air (ρ) = 1.225 kg/m; Acceleration time of the towing vehicle (t_a) = 10seconds

Rolling resistance:- Frr= μ_r mg= 0.013*8000*9.81= 1020.24N

Acceleration force:- F=m*a= F=m*(vf-vi)/t= 8000*(3-0)/10= 8000*3/10= 800*3=2400N

Aerodynamic drag force:- F_d = ½ x ρ x A x c_{d x} v^2 = ½*(1.225*8*0.60*3^2)=26.46 N

Total tractive force:- F_{T =} F_rr (a) + F(a) + F_d (a) + F_rr (t) + F(t) + F_d (t); 11772+12000+70.284+1020.24+2400+2646= 27288.984N= 27.288984 KN

Power required to push or pull the aircraft P=Ft*V = 27288.984 * 3= 81866.952 W= 81.866952 KW

Q6) B)

Explanation of the model:-

• First we consider all 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 and acceleration force.
• Similarly, 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 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 block.

Q7) A)

AIRCRAFT PARAMETERS:-

PARAMETERS	VALUES	UNITS
MASS OF THE AIRCRAFT	40000	Kg
COEFFICIENT OF R.R	0.03	-
FRONTAL AREA	85	M^2
FINAL VELOCITY	3	m/s
INITIAL VELOCITY	0	m/s
COEFFICIENT OF DRAG	0.15	-
TOWING VEHICLE ACCELERATION TIME	10	sec

TOWING VEHICLE PARAMETERS:-

PARAMETERS	VALUES	UNITS
MASS OF TOWING VEHICLE	8000	KG
COEFFICIENT OF R.R	0.013	-
FRONTAL AREA	8	M^2
FINAL VELOCITY	3	m/s
INITIAL VELOCITY	0	m/s
COEFFICIENT OF DRAG	0.60	-
ACCELERATION	10	SEC
GEAR RATIO	10	-

The total power required to push or pull the aircraft by the towing vehicle is 81.866952 KW and the total tractive force is 27288.984N. to calculate the torque required by the towing vehicle the following parameters are assumed as shown above.

T = Fr* r/G = 27288.984*0.4/10 = 1091.56 Nm

The torque required for towing the vehicle to push or pull the aircraft is 1091.56 Nm

MOTOR:-

The motor is selected with a power rating of 97.022 KW and torque of 1091.56Nm.

MOTOR SPECIFICATION:-

Output power = 110 KW, 150 hp

Full load speed = 740 rpm

Full load efficiency = 94.5%

Full load current = 210 A

Full load torque = 1420 Nm

Rated voltage = 400V/50Hz

DUTY CYCLE:-

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

Pi = V * I =400*280 = 112000 W = 112KW

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

Duty cycle = Po/Pi = 97.02/112 = 0.866 = 86%

ENERGY :-

Assuming battery usage for 5 minutes is 300 sec

E = Po * t = 970.02 *300= 29106 kJ

We have to convert KJ to KWH we divide it by 3600.

E = 29106 / 3600 = 8.08 KWH

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

RESULT :-

• In Q1 types of aircraft with its maximum take-off weight is been mentioned.
• In Q2 detailed information of the difference between ground speed and air speed is been explained.
• In Q3 detailed information is provided why it is not recommended to use aircraft engine power to move it on the ground at airport.
• In Q4 how an aircraft is pushed to runway when it is ready to take-off is explained.
• Detailed information is provide in Q5 about rolling resistance, tyre design, tyre pressure etc.
• In Q6 & Q7 problem is solved by applying the formulas and the simulation model is also been attached with the Q6 and for Q7 excel calculation is also done both the values are same which we have calculated in book and in excel sheet.