Project-1: Powertrain for aircraft in runways

AIM:

1. List out the total weight of various types of aircrafts.
2. Difference between ground speed and air speed.
3. Why is it not recommended to use aircraft engine power to move it on the ground at Airport?
4. How an aircraft is pushed to runway when its ready to take off?
5. Learn about take-off power, tire design, rolling resistance, tire pressure, brake forces when landing.
6. With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle.
7. Design an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfil aircraft towing application. 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.

Procedure:

To design the electric powertrain of tug vehicle we will use the reverse design approach, i.e. first we will understand and determine the application requirements of the vehicle, for eg. maximum weight to be towed. Based on that we will calculate the tractive force, torque and power requirement of the vehicle. Based on the torque, power & speed requirement we will choose our motor. Then, we will calculate the energy required for successful application of the vehicle, which will be stored in the vehicle's traction battery.

1. Weight components of an airplane are as follows:

• Crew weight (Wc):

The crew comprises the people necessary to operate the aeroplane in flight. E.g., Pilot, Co-pilot, Airhostess etc.

• Payload weight (Wp):

The payload is what the aeroplane is mentioned to transport passengers, baggage, freight etc. (Military use the payload includes bombs, rockets and other disposable ordnance)

• Fuel weight (Wf):

This is the weight of the fuel in the fuel tanks. Since fuel is consumed during the course of flight, it is a variable, decreasing with time during the flight.

• Empty weight (We):

This is the weight of everything else the structure, engines (with all accessory equipment), electronic equipment landing gear, fixed equipment and anything else that is not crew, payload or fuel.

• Gross weight (W0):

The sum of all these weights is the total weight of the aeroplane. Gross weight or total weight varies through the flight because fuel is being consumed. The design take-off gross weight is the weight of the aeroplane at the instant it begins its mission. It includes the weight of the fuel.

Below is a list of some major airliners sorted by maximum take-off weight.

*MTOW = Maximum take-off weight & MLW = Maximum landing weight

2. Difference between ground speed and airspeed.

Difference between ground speed and airspeed of an aircraft can be understood by the concept of ‘Relative velocity’. Relative velocity is the velocity of an object with respect to an another body in motion or at rest.

Ground Speed

Ground speed is the speed of aeroplane relative to a fixed point on the ground. Ground speed is a vector quantity so a comparison of the ground speed to the wind speed must be done according to rules for vector comparisons.

Airspeed

Airspeed is the actual speed of aeroplane under its own power, which is responsible for generating lift. Airspeed cannot be directly measured from the ground position but must be computed from the ground speed and the wind speed. Airspeed is the vector difference between the ground speed and wind speed.

• Airspeed = Ground Speed - Wind Speed

If the wind is blowing in the direction of aircraft’s motion its value is taken as positive and if blowing in the opposite direction it is taken as negative.

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. Why is it not recommended to use aircraft engine power to move it on the ground at Airport?

Taxing is the procedure of moving aeroplanes on the ground. Aeroplanes have no additional engines to power the wheels. The wheels of an aeroplane rotate freely. Aeroplanes depend on the thrust of their jet engines to propel forward. But in critical scenarios like ‘pushback’, it is not advised to use engine power to move it on the ground. During pushback from the terminal/gate, Tug vehicles are used, which are high torque vehicles.

The reason for that is that the pushback using the jet engine’s power is a highly fuel-consuming & noisy process. It is also hazardous for personnel working near the plane. The thrust produced by the jet engines can damage nearby buildings and other aircraft.

4. How an aircraft is pushed to the runway when it is ready to take off?

When an aircraft is ready to take-off it is pushed from the gate/terminal to runway with the help of special vehicles called ‘Tug vehicles or tractor’. This process is called ‘pushback’. These tug vehicles are heavy vehicles with high torque. Extra weight is added to these vehicles to achieve better traction.

5. Take-off power, tire design, rolling resistance, tire pressure & brake forces when landing.

Take-off Power:

• Take-off is the phase of flight in which an aircraft leaves the ground and becomes airborne. Take-off power is always calculated before flight according to the airspeed required for lift based on the length of runway, payload, altitude and other environmental factors. Aircraft rarely use full engine power during take-off. Most of the time, a derated or reduced power is used for take-off. This ensures longer engine life which translates to lower maintenance cost and also savings in fuel cost.

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.

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 aeroplane.
• Tire 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 tire absorbs the high impact loads of landing and also it’s operating at high speeds for a short time when required.

Let’s go through some technical aspects of the tire.

• 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 tire 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 take-off 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 mould.

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.

Almost all airlines today use tubeless radial tires because of their lower life cycle cost & long term value. Today’s aircraft tires are also conductive. Aircraft tires are manufactured with tread rubber with conducting compounds to permit earthing of static charges.

Aircraft tires also have ‘Chines’, also called deflectors. Chine tire are used on the nose wheel of aircraft, specially fuselage-mounted jet engines. It diverts runway water away from the engine inlets.

Chines are circumferential protrusions that are moulded into the sidewall of nose tires that deflect water sideways to help reduce excess water ingestion into the engines. Tires may have chines on one or both sides, depending on the number of nose tires on the aircraft.

Rolling-resistance:

Rolling resistance is the resistive force that opposes the motion of a body rolling on a surface. In an aircraft, it comes into play during take-off, landing, and during ground movement.

The rolling resistance can be expressed as,

`F_r=μmg`

where

`F_r` = rolling resistance or rolling friction (N)

µ = rolling resistance coefficient

m = mass of body (kg)

g = acceleration due to gravity (m/s^2)

The value of µ depends on road conditions, roughness, tire pressure, tire temperature, tire design, etc.

Tire Pressure:

Inflation pressure of aircraft tire should be sufficient enough to withstand the weight of aircraft and provide enough cushion while landing. Tire pressure also affects the coefficient of rolling resistance.

Aircraft tires generally operate at high pressures, up to 200 psi (14 bar; 1,400 kPa) for airliners, and even higher for business jets. Aircraft tires 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 (normal air is about 80% nitrogen), but common compressed air sources may contain moisture, which increases the expansion rate with temperature.

Brake forces while landing:

During braking, the kinetic energy of the aircraft is converted to heat energy when friction occurs between the aircraft tires and runway. Braking is a crucial part of flight because there is only a limited distance on the runway that the aircraft can use to reduce its speed. Braking has to be done in such a way that the aircraft or its passengers or its equipment does not get damaged. There are different types of braking systems employed in an aircraft.

• Aircraft disc brakes: There are disc rotors in the landing gear, which are used to brake the wheels while touching the ground. These brakes are operated hydraulically or pneumatically.
• Thrust reversers: Aircraft use thrust reversers, that allow thrust from the engines to be used to slow the aircraft.
• Mechanical Spoilers: The mechanical deflection of parts of the wing upper surfaces and tail cone assembly can assist deceleration 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.

6. We have already learned that a ‘Tug vehicle/tractor’ is used to move an aircraft on the ground from gate to runway and from runway to gate/terminal. To determine the force and power required by the ‘Tug vehicle’ we need to know the ‘Gross Weight’, i.e., the total weight of airliner, its equipment, crew, passenger & cargo weight, as well as the weight of the tug vehicle itself.

For our calculations, we are considering the maximum take-off weight of the ‘Airbus A380’ which is the among the heaviest in industry. Also, we are assuming that the maximum speed of towing is about 18 kmph = 5 m/s. And let’s assume it takes 60 sec. to accelerate to that speed. These are the conditions for an extreme case.

And for Tug vehicle we are considering total weight of 54tonnes, which is typical for a large aircraft tractor.

To push/pull the aircraft on ground three forces will be into play, Rolling resistance force, Aerodynamic drag & acceleration force. Let’s calculate them:

Data:

• Gross weight of aircraft, `M_a = 575000` Kg
• Weight of tug, `M_t = 54000`
• Acc. due to gravity, g = 9.81 m/s^2
• Coeff. of rolling resistance, µ = 0.02 (assumed)
• Air density, `rho` = 1.225 Kg/m^3
• Drag coeff., `C_d` = 0.0265
• Wing area, A = 843 m^2
• Speed, v = 5 m/s
• Acceleration, a = (v-0)/t = 0.083 m/s^2

Rolling resistance force

`F_(rr) = mu(M_a+M_t)g` = 0.02*9.81*(575000 + 54000) = 123,409.8 N

Aerodynamic Drag

`F_d = 1/2rhoC_dAv^2` = 0.5*1.225*0.0265*843*5^2 = 342.07 N

Acceleration Force

`F_a = (M_a+M_t)a` = (575000+54000) * (5-0)/60 = 52416.67 N

Total Force, `F_(te) = F_(rr)+F_d+F_a` = 176.17 kN

and,

Total power = Total Force X Speed = 880.84 kW

Therefore, the tug vehicle needs to produce 880.84 kW of power to be able to push/pull an aircraft.

Motor selection

Before deciding on the motor for our application we need to determine the power & torque requirements. We have already calculated the power required. Now we will calculate the motor torque required from the total  force we calculated above. Also, we are assuming the transmission will be a single-speed transmission.

Tire rolling radius, r = 0.5 m

Gear Ratio, G = 10

Total Force, `F_(te)` = 176.17 kN

Motor torque, `T_m = F_(te)*r/G` = 8.8 kNm

Therefore, we need a motor which can produce 880.84 kW of power and 8.8 kNm of torque. For these specifications, a high torque and low-speed motor will be the ideal choice. And to achieve such a large amount of torque multiple motors will be employed in the tug vehicle.

Here, for simplicity, we are considering just one motor. We are assuming a motor which can produce 880.84 kW of power and 8.8 kNm of torque at rated voltage of 1200 V & 750 A of current.

Energy requirement

We are assuming it takes about 10min. for pushback procedure. So, let’s calculate the energy required for this operation.

Energy = Power X time = 880.84 X 10/60 = 146.8kWh

Therefore, the battery must be able to store 146.8 kWh of energy.

Duty Cycle

The motor will use power from the battery to run and then further drive the vehicle. But we cannot directly connect the battery to the motor, because we need to supply power according to the load requirement. For that purpose, a motor controller is used. The motor controller will control the speed of the motor by changing the duty cycle continuously according to the requirement.

Also, a bidirectional DC-DC converter will be employed to Boost up the battery voltage and also allow charging of battery during regenerative braking. Here, we are assuming the output voltage from DC-DC converter, which is the input voltage to the controller, to be 1500 V, using that we can calculate the maximum duty cycle for full speed of the motor as,

Duty cycle, `D = V_o/V_i` = 1200/1500 = 0.8

Therefore, duty cycle of 80% will be used for maximum speed.

List of parameters assumed for above calculations:

Block diagram of powertrain:

References:

• https://en.wikipedia.org/wiki/List_of_airliners_by_maximum_takeoff_weight
• https://www.aviationhunt.com/aircraft-tyre-construction/
• https://en.wikipedia.org/wiki/Drag_coefficient#Aircraft
• https://modernairliners.com/airbus-a380/airbus-a380-specs/
• https://www.aviationhunt.com/aircraft-tire-construction/