Project-1: Powertrain for aircraft in runways

1.) The aircraft's 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 consumption. It may also vary during a flight due to payload dropping or in-flight refueling. At the moment of releasing its brakes, the gross weight of an aircraft is equal to its take-off weight. During flight, an aircraft's gross weight is referred to as the en-route weight or in-flight weight.

Maximum design taxi weight (MDTW): The maximum design taxi weight (also known as the maximum design ramp weight (MDRW)) is the maximum weight certificated for aircraft maneuvering on the ground (taxiing or towing) as limited by aircraft strength and airworthiness requirements.

Maximum design take-off weight (MDTOW): Is the maximum certificated design weight when the brakes are released for take-off and is the greatest weight for which compliance with the relevant structural and engineering requirements has been demonstrated by the manufacturer.

Maximum design landing weight (MDLW): The maximum certificated design weight at which the aircraft meets the appropriate landing certification requirements. It generally depends on the landing gear strength or the landing impact loads on certain parts of the wing structure. The MDLW must not exceed the MDTOW.

Maximum design zero-fuel weight (MDZFW): The maximum certificated design weight of the aircraft less all usable fuel and other specified usable agents (engine injection fluid, and other consumable propulsion agents). It is the maximum weight permitted before usable fuel and other specified usable fluids are loaded in specified sections of the airplane. The MDZFW is limited by strength and airworthiness requirements. At this weight, the subsequent addition of fuel will not result in the aircraft design strength being exceeded. The weight difference between the MDTOW and the MDZFW may be utilized only for the addition of fuel.

Minimum flight weight (MFW): Minimum certificated weight for flight as limited by aircraft strength and airworthiness requirements.

further small categories of weights:

a) Crew weight (Wc )-The crew comprises the people necessary to operate the airplane in flight.
e.g., Pilot, Co-pilot, Airhostess, etc.
b) Payload weight ( Wp )-The payload is what the airplane is mentioned to transport passengers, baggage, freight, etc. (Military use the payload includes bombs, rockets, and other disposable ordnance).
c) Fuel weight ( Wf )-This is the weight of the fuel in the fuel tanks. Since fuel is consumed during the course of the flight. is a variable, decreasing with time during the flight.
d) Empty weight ( We )-This is the weight of everything else-the the structure engines (with all accessory equipment), electronic equipment landing gear, fixed equipment, and anything else that is not crew, payload, or fuel.

2.) Ground speed is the horizontal speed of an aircraft relative to the ground. An aircraft heading vertically would have a ground speed of zero. Airspeed is the relative velocity between the object and the air, which is an important quantity in the generation of lift.

Airspeed = Ground Speed - Wind Speed i.e. Airspeed is the vector difference between the ground speed and the wind speed. On a perfectly still day, the airspeed is equal to the ground speed. But if the wind is blowing in the same direction that the aircraft is moving, the airspeed will be less than the ground speed.

On a perfectly still day, the airspeed is equal to the ground speed. But if the wind is blowing in the same direction that the aircraft is moving, the airspeed will be less than the ground speed.

3.) Aircraft spends a minimum of 20 minutes using its main engine to taxi from the airport terminal gate to the runway, where the aircraft takes-off and similarly to the terminal gate after landing. Taxiing is the movement of an aircraft on the ground, under its own power, in contrast to towing or push-back where the aircraft is moved by a tug. During taxi operation, aircraft burns a lot of fuel, generates large quantities of emissions, and the engine in the process of taxiing is exposed to damage due to Foreign Object Damage (FOD). This results in huge operational and maintenance costs as well as carbon emission tax which are all at the expense of the airline. The use of engine thrust near terminals is restricted due to the possibility of jet blast damage. Hence, it is not recommended to use aircraft engine power to move it on the ground at airports.

4.)Pushback is a procedure during which an aircraft is pushed backward away from an airport gate by an external power. Pushbacks are carried out by special, low-profile vehicles called pushback tractors or tugs.

• Although many aircraft are capable of moving themselves backward on the ground using reverse thrust (a procedure referred to as a power-back), the resulting jet blast or prop wash 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.
• Closer to departure, an aircraft tug will park right in front of the nose wheel.
• The tug might be directly attached to the plane's nose gear with a tow bar, or could be a "wheel-lift" tug. These tugs cradle the nose gear, then lift it up before moving the plane. That gives the tug driver control over the plane's direction during pushback.

pushback tug carrying a towbar

5.) Take-off power: For light aircraft, usually full power is used during take-off. Large transport category (airliner) 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 emergency cases, the power used can then be increased to increase the aircraft's performance.

Tire design: 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.

Design and Construction of Aircraft tire:

Carcass plies are used to form the tire. They are sometimes called casing plies. An aircraft tire is constructed for the purpose it serves.

• They support the weight of an aircraft while it is on the ground.
• It also provides the necessary traction for braking and stopping of an airplane.
• Tyre also helps to absorb the shock of landing and provide cushioning the roughness of take-off, roll-out, and taxi operations.

Unlike an automobile or truck tire, it does not have to carry a load for a long period of continuous operation. However, an aircraft tyre absorbs the high impact loads of landing and also, it’s operating at high speeds for a short time when required.

Tube type or tubeless tire: Tires that are made to be used without a tube inserted inside have an inner liner specifically designed to hold air. Tube-type tires do not contain this inner liner since the tube holds the air from leaking out of the tire.

Bias-ply or radial ply: The direction of the plies used in construction of the tire, either bias or radial. Traditional aircraft tires are bias-ply tires. The plies are wrapped to form the tire and give it strength. The angle of the plies in relation to the direction of rotation of the tire varies between 30° and 60°. In this manner, the plies have the bias of the fabric from which they are constructed facing the direction of rotation and across the tire. Hence, they are called bias tires. The result is flexibility as the sidewall can flex with the fabric plies laid on the bias.

The plies in radial tires are laid at a 90° angle to the direction of rotation of the tire. This configuration puts the non-stretchable fiber of the plies perpendicular to the sidewall and direction of rotation. This creates strength in the tire allowing it to carry high loads with less deformation.

Retreading – Retreading is methods of restoring a worn tire by renewing the tread area or by renewing the tread area plus one or both sidewalls. Repairs are included in the tyre retreading process.

Load Rating – Load rating is the maximum permissible load at a specified inflation pressure.

Ply Rating – Ply Rating is used to identify the maximum recommended load rating and inflation pressure for a specified tire. It is an index of tire strength.

Speed Rating – The speed rating is the maximum takeoff speed to which the tire has been tested.

Skid Depth – Skid depth is the distance between the tread surface and the deepest groove as measured in the mold.

• Aircraft tires must have an approved speed and load rating and have sufficient clearance when retracted through landing gear to allow for tire growth. Tire growth is the increase in the size of the tire due to centrifugal forces at high speed.

Rolling Resistance: Rolling resistance is the force resisting the motion when a body (such as a ball, tire, or wheel) rolls on a surface. It is mainly caused by non-elastic effects; that is, not all the energy needed for deformation (or movement) of the wheel etc., is recovered when the pressure is removed. Another cause of rolling resistance lies in the slippage between the wheel and the surface, which dissipates energy; the deformation of the roadbed surface, and movement below the surface. Additional contributing factors include wheel diameter, load on wheel, surface adhesion, sliding, and relative micro-sliding between the surfaces of contact.

The rolling resistance can be expressed as

Fr = c W

where

Fr = rolling resistance or rolling friction (N, lbf)

c = rolling resistance coefficient - dimensionless (coefficient of rolling friction - CRF)

W = weight of the body

Tyre pressure: The aircraft tires are properly inflated in order to maximizing tire utilization and minimizing risks to safe operations.

Although the car and aircraft tires are dimensionally similar, they differ significantly in rated inflation pressure, load, and speed. Additionally, the passenger car and truck tires are able to operate continuously because they reach thermal equilibrium while aircraft tires do not. An aircraft tire which taxis continuously at rated load and 40 mph will continue to heat up until the tire fails. The failure to reach thermal equilibrium is a consequence of the high deflections (the difference between the unloaded and loaded tire section height) at which aircraft tires operate.

• High inflation pressure and high deflection rates – 2.5 times greater than a car tire – allow relatively small tires to support the high loads of an aircraft.
• If the tire operates underinflated or over deflected/overloaded, the nylon cords which form the structure of the tire go in and out of compression as the tire rotates. This weakens the cords until eventually, the cords break. If enough cords break, the entire structure will eventually fail. This particular break damaged the inner liner, allowing air to migrate through the tire to the atmosphere.
• Another serious consequence of under-inflation is the thrown tread. Over deflection increases shear between components in the tire as it rolls in and out of contact and deforms. This results in a more rapid build‐up of heat within the tire than would be evident if the tire were properly inflated. At some point the excessive heat will cause the rubber to revert, or reverse cure. This reverted rubber is like grease with no strength or adhesion to contain the structure. This can allow the tire to decompose, throwing the entire tread or pieces of it. In addition to destroying the tire, thrown treads often result in expensive aircraft damage.
• Because of the high pressures and extreme temperatures at which aircraft tires operate, they do not hold air perfectly. In fact, an aircraft tire can lose up to 5 percent of its pressure in 24 hours and still be perfectly serviceable.

Brake forces: The aircraft braking coefficient is dependent upon the surface friction between the tires on the aircraft wheels and the pavement surface. Less friction means less aircraft braking coefficient and less aircraft braking response. The degree of surface friction for a specific aircraft at a given moment is directly proportional to the braking action, subject only to the activation of wheel lock-up and anti-skid protection systems.

The braking force coefficient rises to a 'peak' value as the wheel progresses towards the locked wheel condition and falls to a lower value when the wheel locks.

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

Moving any aircraft requires a certain amount of drawbar pull (UP), which is measured in terms of pound-force (Lbf.) The unknown part is that the amount of DBP needed to pull the same aircraft will vary based on circumstances Likewise, the amount of DAP delivered by an aircraft tug will also vary based on operating conditions.

The calculations look like this:

Tractive Effort Required = (Wa+ Wt) x (Gr+ Br+ Rr)

Drawbar Pull Delivered - Wd x COF where,

Wa - maximum takeoff weight of the aircraft (in lbs.)

Wt- the weight of the aircraft tup (in lbs)

Gr = the steepest slope or grade the aircraft tug wit have to traverse (a percentage, treated as a decimal)

Br- breakaway coefficient (assumed 40 based on SAE AIR 1363 standards, treated as 0.40)

Rr - rolling resistance percentage, treated as a decimal)

Wd- total weight on drive axle(s) of aircraft tug

COF - coefficient of friction (the amount of tractive effort transferred to the ground, which varies by weather conditions)

The COF used in the calculation should be based on the worst conditions an aircraft tug will encounter because it can vary significantly. For example, the COF will be 0.9 on dry brushed concrete, however, it will be 0.66 on standard wet concrete and as low as 0.5 on snow or Ice, even with chains on the aircraft tug tires.

While the exact numbers will vary in each scenario, the drawbar pull delivered must always be higher than the drawbar pull required for an aircraft tug to be recommended in a given application

Considering this as our case:

If the mass of a plane is 40000 kg and the rolling coefficient between the aircraft wheel and the tarmac is 0.02- the force required to overcome the rolling resistance can be calculated as

Fr= 0.02 (40000 kg) (9.81 m/s^2)= 7848 N

Air density= 1.25 kg/m^3

Taking other parameters affecting the total tractive efforts the slope,

Drag force=0.625*A*Cd*V^2, wind, etc. As a typical value of 70% extra for this.

We get the effective tractive force as

Feff= 7848*1.7= 13341.6 N

This force can be covered by a puling truck. Enough Weight for the pullers is required to create friction forces against the tarmac that equals the plane rolling resistance.

The friction force for the pulley can be calculated to

Ff= µ*W

F= frictional force ( N,lb)

µ= Friction coefficient (N, Ib)

mf= friction mass (kg)

The friction coefficient between a pulling truck rubber wheels and dry tarmac 0.9. Since pulling trucks are connected to air crafts nose wheels close to tarmac, we simplify the calculation of required truck mass to

mf= 13341.6N / (0.9*(9.81m/s^2))

= 1511.12kg

Tyre size(d) = 0.6858m

Limit aircraft towing speed to a maximum of 5 MPH

Omega= v/r= 2.2352*2/0.6858= 6.518 rad/sec

TORQUE= r x F=(0.6858/2)*13341.6= 4,574.83 Nm

Power= T X w= 4,574.83 Nm*6.518rad/sec= 29.81 KW

Acc energy required to increase the vel from 0 to 5 MPH = 0.5*m*(Vf^2 – Vi^2) = 0.5*40000*(2.2352^2 – 0)= 99,922.3J
Acc power in 30 second= 3.33 kW (then steady speed maintained after 30 sec)

Tractive power required:

Limit aircraft towing speed to a maximum of 5 MPH= 2.2352 m/sec

Preq= Feff*velocity= 13341.6 N* 2.2352 m/sec= 29.82 KW

Total power req= 33.15KW

7.

Parameters:

Air density= 1.25 Kg/m^3
Limit aircraft towing speed to a maximum of 5 MPH

Omega= v/r= 2.2352*2/0.6858= 0.6518 rad/sec

Taxi time= 5 min

Drag coefficient= 0.0412

Rolling coefficient between the aircraft wheel and the tarmac is 0.02

Mass of the plane= 40000kg
Feff= 7848*1.7= 13341.6 N

Here we have obtained 29.82 KW.

Its power is around 30KW but motor with 4574.83Nm torque is not designed generally.

So will have to derate the motor for around 650KW

T= kt*Ia

For rated torque= 5728 for rated arm current= 2015A

Kt= T/Ia= 2.84

E= kt*w= 2.84*6.518= 18.51V

Treq= 4574.83Nm= Kt*Ia

Ia= Treq/Kt= 1610.85A

V= E+Ia*Ra= 18.51+1610.85*0.08= 147.378V

Vin= 600V

Duty cycle required at steady state= Vo/Vin= 0.24= 24%

Assumption: we have assumed the running continuously along a straight line but it is generally not the case and there are many motors set with gear system to propel the aircraft and also the drag is not constant and the ambient condition may vary with the weather condition. All these are neglected for ease of calculation.

BLOCK DIAGRAMS: