Project-1: Powertrain for aircraft in runways

Airspeed = Ground Speed - Wind Speed.

Ans 3: Aircraft can move using their own engine power all the time on the airport and taxiways under their own power to go towards runways and after landing taxi to their assigned gate. But the thrust produced resulting jet blast 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. And also it can reduce the cost for fuel. Taxiing can save money.

• Effectively a loss of control can happen, especially during high power engine running. Damage can occur to the aircraft itself, other aircraft nearby or to airside structures.
• There is a risk of injury to ground support personnel who may be in relatively close proximity to the aircraft.
• It is easier for truck driver to see the surroundings better than the pilot sitting in the cockpit with limited window. The pushback driver knows the airport better than the pilot.

Ans 4 : There are two methods are used to push an aircraft into runway when it is ready to take-off. such as:-

• Taxiing
• Towing

(A) Taxiing:-

• It is defined as the movement of an aircraft to the ground by it's own power.
• An aircraft moves through wheels on ground the term also used for the air-craft that has skies or floating equipments.
• An airplane uses taxiways to move from one place to another place on the same airport.
• Refer image Ans 4 (A)
• (B) Towing:-

• • In case of towing or pushback, an aircraft is moved by a tug.
  • Movement of large aircraft about the airport, flight line & hangar is usually accomplished by towing with a tow tractor called as tug.
  • A pushback truch lifts the front wheel of the aircraft, their role in most airports is to move the aircraft away from the parking spot on the taxiways.
  • Here the truck is removed & when all platform staffs are out of the way, then the engines are started one after another.
  • Refer image Ans 4 (B)

Ans 5 : (A) Take off Power : Takeoff power in an aircraft refers to the amount of engine power that is applied during the takeoff phase of flight to achieve sufficient speed for lift-off. It is the power setting that ensures the aircraft accelerates to the speed needed for the wings to generate enough lift to become airborne.

This power level is typically higher than the normal cruising power to overcome the aircraft's weight, drag, and other resistances during takeoff.

(B) Tire design : When it comes to safety tires 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 of an aircraft.

An aircraft tire 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 airplane needs to be distributed more evenly.

A 737NG & MAX uses 6 wheels, Boeing 787 uses 10 wheels, Boeing 777 uses 14 wheels, Airbus A380 uses 22 wheels. Aircraft tires work under extreme conditions, carrying up to 340 tons and accelerating at over 250 km/hr. at takeoff, in addition to enduring varied environmental stress when in flight and taxing.

Due to this heat generates. Aircraft's tread pattern is designed to facilitate stability in high crosswind conditions, to channel water away to prevent hydroplaning and for braking effect.

Refer Tire image for reference

• Radial Tire

a ) It has fiber strands of the ply fabric oriented at 90° to the direction of rotation and the tire sidewall.

b) This restricts flexibility bidirectional and the flexibility of the sidewall while it strengthens the tire to carry heavy loads.

c) Due to their unique construction most radial tires are resistant to penetration ad cuts. Due to stiffer tread they distribute weight evenly.

d) Thus provide greater traction. Radial tires have an advantage in term of reducing the heat better.

• Bias ply tire

a) It has the fabric bias oriented with and across the direction of rotation and the sidewall.

b) Since fabric can stretch on the bias, the tire is flexible and can absorb loads.

c) Tire strength is obtained by adding plies.

d) It is cut resistant is sidewall area.

e) These tires do not offer as much smoothness in terms of riding quality.

(C) Rolling Resistance :

• Rolling resistance is the force resisting the motion when a body (such as ball, tyre or wheel) rolls on a surface.
• Rolling resistance can be seen in the aircraft when the aircraft takes-off from the runway & while landing at the runway with the help of their wheels
• Rolling resistance force Frr = μrr * m * g

• μrr Typical Values for Aircraft

  SURFACE BRAKES OFF BRAKES ON

  Dry concrete/ 0.03-0.05 0.3-0.5

  asphalt

  Wet concrete/ 0.05 0.15-0.3

  asphalt

  Icy concrete/ 0.02 0.06-0.10

  asphalt

  Hard turf 0.05 0.4

  Firm dirt 0.04 0.3

  Soft turf 0.07 0.2

  Wet grass 0.08 0.2

• Where,

  m = total mass of the aircraft (Actual mass + Payload) in kg

  g = Gravitational constant, that is 9.8 m/s^2.

(D) Tire Pressure : The pressure generally is very high in an aircraft tire to reduce the rolling resistance and to withstand the weight. usually the tire pressure will be around 190 to 320 psi based on the type of aircraft. Aircraft tires are usually inflated with nitrogen to minimize expansion and contraction from extreme changes in ambient temperature and pressure experienced during flight.

Refer Image RR vs TP for rolling resistance variation w.r.t tire pressure

(E) Brake forces while landing : Following braking devices are used to decelerate the aircraft -

1. Spoilers.
2. Wheel Brakes (Including Anti-Skid systems & Auto-Brake systems)
3. Reverse Thrust

1. Spoilers :

• The mechanical deflection of parts of the wing upper surface 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 and also by increasing the effective downward load on the landing gear and thereby increasing the efficiency of wheel braking

2. Wheel Brake :

• To be effective, braking action at any speed depends upon sufficient friction existing between the tires and the runway surface achieved through freely rotating wheels.

• Anti skid system :

  When landing on slippery runways, there's a chance that the wheels may start to skid as the brakes are applied. To stop this from happening and to maintain maximum effective braking, each wheel has anti-skid protection.

  Using a variety of sources to determine the aircraft speed, the brake units know how fast the wheels should be spinning. If that speed drops significantly, it's because the current brake pressure on that wheel is too great and the wheel is just skidding over the surface.

  In this situation, the anti-skid system automatically reduces the braking on that wheel to a point where the skid stops before reapplying the pressure.

3. Reverse Thrust :

• Thrust reversers provide a deceleration force that is independent of runway condition.
• Thrust reverser efficiency is higher at high airspeed therefore.
• reversers must be selected as early as possible after touchdown.
• Thrust reversers should be returned after reverse idle at low airspeed to prevent engine stall or foreign object damage.

Ans 6 :The forces required to push/ pull an aircraft by a towing are rolling resistance force and aerodynamic drag force.

Calculation:

let us assume that,

Mass of Towing Vehicle= 40,000kg

Mass of aircraft = 155,000kg

Total mass (m)= 40,000+155,000 =195,000kg

Frontal Area of the aircraft (A) =18 m²

Speed of the vehicle (2)=25Kmph

converting it into m/sec

u=25*(1000/3600)=6.9445m/sec

Drag coefficient (Cd) =0.75

Air density (P)=1.225Kg/m³

Calculating Rolling resistance Force (Frr) =Crr*m*g

=0.01*195.000*9.8

= 19.11ON

Calculating Aerodynamic Drag Force (Fd)=0.5*p Cd*A* v^2

=0.5 *1.225 *18* 0.75* (6.9445)2

=398.769N

Therefore the total Force required to push/pull the aircrcraft by the towing vehicle (F) Frr+Fd

=19,110+398.769

=19,508.769 N

Calculating the power (p) =Force*Velocity

= 19508.769* 6.9445

=135.478kW

Refer tb.slx file for MATLAB model

Ans 7 :

Based upon calculations done in Ques 6

Force required to push/pull the aircraft = 19,508.769 N

Power required = 135.478 kW

Radius of tire for the tug = 25 inches = 0.635 m

Gear ratio = 8

Transmission efficiency (nt) = 87%

Motor efficiency = (nm) = 95 %

Torque required at wheel = Total force required * radius of tire = 19508.769*0.635 = 12,388.06 Nm

Required motor torque = Torque at wheel/nt*gear ratio = 12,388.06/0.87*8 = 1,779.89 Nm

For nm = 95 % , required motor torque(Tm) = 1779.89/0.95 = 1,873.56 Nm

Total power required by motor considering all efficiencies = 135.478/0.87*0.95 = 163.91 kW

Motor used here is 200 hp DC motor with nominal values as : speed = 1184 rpm , nominal voltage = 440 V

Considering 250kW as continous power, so the voltage will be 400V, Battery power is considered to be 550V we know that duty cycles given by

d = Va/Vb = 400/550 = 0.7272.

There will also be loses due to the resistance of FET and inductance. Correction factor is considered to be 90% that makes us consider that there is 10% loss So the voltage which is demanded by the motor should be 10% more than the required voltage

d = 0.7272*(1+0.1) = 0.799 ~ 0.8,

so here the voltage demand is 440V from the battery.

For your reference input signal image attached

For you reference powertrain output signals are attached.

Simulink model (quessevenweek12) file is attached for your reference.