|| Powertrain Of An Aircraft's Towing Vehicle ||

Last, But not the least reason as every so often is fuel saving. Example, A Boeing 737-800 burns approximately 300 kilos of fuel per hour with engines running at idle power. Now, let's say that we just landed and are stuck in inbound taxiing traffic for thirty minutes. We could cut engine number two and we could save 75 kilograms of fuel that equals two petrol tanks of a standard size car. Now, let's use Alaska Airlines as an example. They have 160 Boeing 737s in their fleet. Now, let's say each have an average taxi time of fifteen minutes with one engine shut down equals 37.5 kilograms of fuel times 160 planes equals six tons. But, each plane performs at least two flights per day, so the total fuel saved would be twelve tons. That's a lot of fuel for one day, multiply that by 365 days per year,

and we are saving a LOT of money.

In aviation, pushback is an airport procedure during which an aircraft is pushed backwards away from an airport gate by an external power. Pushbacks are carried out by special, low-profile vehicles called pushback tractors or tugs.

Image 1: Pushback of an airplane by a pushback tractor or tugs

Although, many aircraft are capable of moving themselves backwards on the ground using reverse thrust (a procedure referred to as a powerback), the resulting jetblast 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.

PROCEDURE:

Pushbacks at busy aerodromes are usually subject to ground control clearance to facilitate ground movement on taxiways. Once clearance is obtained, the pilot will communicate with the tractor driver (or a ground handler walking alongside the aircraft in some cases) to start the pushback. To communicate, a headset may be connected near the nose gear.

Since the pilots cannot see what is behind the aircraft, steering is done by the pushback tractor driver and not by the pilots. Depending on the aircraft type and airline procedure, a bypass pin may be temporarily installed into the nose gear to disconnect it from the aircraft's normal steering mechanism.

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.

Airbus A380, pushback is shown in the Link

Link: https://upload.wikimedia.org/wikipedia/commons/6/6f/Airbus-A380-Pushback-001.ogv

5.

• Take-Off Power:

Takeoff is the phase of flight in which an aerospace vehicle leaves the ground and becomes airborne.

For aircraft that take off horizontally, this usually involves starting with a transition from moving along the ground on a runway. For balloons, helicopters and some specialized fixed-wing aircraft (VTOL aircraft such as the Harrier), no runway is needed. Takeoff is the opposite of landing.

The total energy of an aircraft flying in the atmosphere can be calculated using equation.

E = ½ m v² + mgh

Taking an example of Boeing 737-300,

A Boeing 737-300 has a maximum takeoff weight of 5.65 × 10⁴ kg, a cruise altitude of h = 10,195 m, and cruise speed of 221 m/sec. Inserting these numbers into the above equation, we obtain 7.03 GJ for the energy at cruise conditions. [3] However, the engines mounted onto the wings of the plane are required to provide additional energy per time, power, in order to keep the aircraft flying at a constant altitude and speed. This extra power needed to overcome the drag acting on the airplane and is calculated using

P = (dh/dt) * m * g

where (dh/dt) is the decent speed required to keep the plane's speed constant if the engines fail. Assuming that a Boeing 737-300 has a similar glide ratio to a Boeing 747 of 17:1, a Boeing 737-300 has a descent speed of approximately (dh/dt)= (221m/sec)/17 = 13 m/sec. The power required to keep a Boeing 737-300 flying at a constant altitude and speed is 7.2 × 10⁶ watts. The rate of fuel burn necessary for the engines to produce enough power to keep the airplane flying can be calculated from the total power requirement. Assuming the efficiency of the engines adding thrust to the aircraft is roughly 30% and the specific gravity of jet fuel is about 0.82 kg per liter, a Boeing 737-300 burns approximately 0.665 liters per second. A Boeing 737-300 has a standard fuel capacity of 2.0104 × 10⁴ liters. Assuming that takeoff, climb, and landing burns approximately one third of the total fuel, and at cruise conditions with a fuel burn rate of 0.665 liters per second, a Boeing 737-300 can fly for approximately (2/3) × 8.4 hours= 5.6 hours at cruise conditions before running out of jet fuel.

• Tyre Design:

An aircraft tire or tyre is designed to withstand extremely heavy loads for short durations. The number of tires required for aircraft increases with the weight of the aircraft, as the weight of the airplane needs to be distributed more evenly. Aircraft tire tread patterns are designed to facilitate stability in high crosswind conditions, to channel water away to prevent hydroplaning, and for braking effect.

Aircraft tires also include fusible plugs (which are assembled on the inside of the wheels), designed to melt at a certain temperature. Tires often overheat if maximum braking is applied during an aborted takeoff or an emergency landing. The fuses provide a safer failure mode that prevents tire explosions by deflating in a controlled manner, thus minimizing damage to aircraft and objects in the surrounding environment.

Diagram: Construction of an aircraft tire

Some technical aspects which should be taken care while designing the tyre:

Retreading – Retreading is methods of restoring a worn tyre 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 tyre. It is an index of tyre strength.

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

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

• Rolling resistance:

Rolling resistance, sometimes called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface. It is mainly caused by non-elastic effects; that is, not all the energy needed for deformation of the wheel, roadbed, etc., is recovered when the pressure is removed.

The "rolling resistance coefficient" is defined by the following equation:

F=μrr⋅N">F=μrr⋅NF=μrr⋅N

where,

F">FF is the rolling resistance force

μrr">μrrμrris the dimensionless rolling resistance coefficient or coefficient of rolling friction (CRF), and

N">NNis the normal force, the force perpendicular to the surface on which the wheel is rolling or the normal force (equal to W)

• Tyre Pressure:

Aircraft tires generally operate at high pressures, up to 200 psi (14 bar; 1,400 kPa) for airliners, and even higher for business jets. The main landing gear on the concrode was typically inflated to 232 psi (16.0 bar), whilst its tail bumper gear tires were as high as 294 psi (20.3 bar). The high pressure and weight load on the Concorde tyres were a significant factor in the loss of Airplanes.

• Braking Forces Acting During Landing:

In the initial landing phase, when the object is approaching the runway, there are aerodynamic forces and moments acting on it as well as the gravitational force. In this situation, the object is the aircraft with its own motion laws. But as soon as the object’s wheels touch the runway, there appears a new force acting on this object while other forces are changing. The wheel force of friction starts to act on the object; the aerodynamic forces start to decrease. The wheel friction force depends on the magnitude of aerodynamic lift (in respect to the weight of the object). The more time the object moves on the runway, the smaller the aerodynamic lift and the greater the wheel friction force. The aircraft in its kinematic behavior turns into a ground transport vehicle. The object’s longitudinal motion in the aircraft phase is described in accordance with Newton's law:

6.

The force to move an air plane on an absolute flat runway with no wind at constant low speed is equal to the rolling resistance between the plane and the tug.

The rolling resistance for the air plane can be calculated as:

Fr=μrr⋅mr⋅g">Fr=μrr⋅mr⋅gFr=μrr⋅mr⋅g

where

F_r = Force (N)

c = rolling resistance coefficient

m_r = rolling mass (kg)

g = acceleration of gravity (9.81 m/s²)

Force required to pull an air plane at constant low speed

Assuming,

The mass of the AirPlane as 7000kg

Which is a very light jets on down to virtually all piston engined aircraft and single engined turboprops.

Rolling Resistance Coefficent = 0.02

Then,

F_r = 0.02*7000*9.81

F_r = 1374 N

This is the force that can be delivered by a pulling truck.

The friction force for the puller can be calculated to

F_f = μ W

= μ m_f a_g

where,

F_f = frictional force (N)

μ = friction coefficient

W = Weight (N)

m_f = friction mass (kg)

taking, value of μ = 0.9

then,

m_f = (1374N) / (0.9 (9.81 m/s²))

m_f = 156 kg

With a more realistic bending angle 60^o - the minimum vertical force or weight can be calculated as

W = tan(60^o) * 1374

W= 2380 N

Then, the mass of the puller could be,

m_f1 = W/g

=2380/9.81

= 243 kg

Taking, radius of the wheel as 0.25m

Now, Calculating Torque (T) = F*R

then, T = 1374*0.25

T = 344 N-m

So,

Power(P) = F*V

Taking V = 6 kph (including load)

i.e., V = 1.66 m/s

then,

P = 1374*1.66

P = 2.3 Kw

7.

Fig. 1. Basic Powertrain of an EV

By, the powertrain diagram. We can say, that output of the converter/ controller is the input of the motor

According to the previous Solution, We got the power consumption of the towing vehicle as 2.3 Kw

We know that,

By Electric Power Equation,

By the equation for electric power:

Vo⋅Io=η⋅Vi⋅Ii">Vo⋅Io=η⋅Vi⋅IiVo⋅Io=η⋅Vi⋅Ii

where,

V_o is the output voltage

I_o is the output current

η is the power efficiency (ranging from 0 to 1)

V_i is the input voltage

I_i is the input current

Here,

Considering, DC input voltage,

Vi">ViVi = 36V

and, Pi=2.5">Pi=2.5Pi=2.5KW

then,

Pi=Vi⋅Ii">Pi=Vi⋅IiPi=Vi⋅Ii

250036=Ii">250036=Ii250036=Ii

Ii=70Amps.">Ii=70Amps.Ii=70Amps.

So, Efficiency of the Buck Converter will be,

ηb=PoPi">ηb=PoPiηb=PoPi

ηb=23002500">ηb=23002500ηb=23002500

ηb=0.92">ηb=0.92ηb=0.92

Now, From the Equation

Vo⋅70=0.92⋅36⋅70">Vo⋅70=0.92⋅36⋅70Vo⋅70=0.92⋅36⋅70 (Ii=Io">Ii=IoIi=Io)

Vo=33.12V">Vo=33.12VVo=33.12V

Now, For Duty Cycle,

We know That,

Duty cycle (D">DD) = VoVi">VoViVoVi

So,

D=3336=0.91">D=3336=0.91D=3336=0.91

So, Duty cycle will be operated in the range of

"0.1<D<0.91"

Energy required to fulfill aircraft towing application is given by,

=power(P)⋅TotalTime(Tt)">=power(P)⋅TotalTime(Tt)=power(P)⋅TotalTime(Tt)

Let us Assume, Total Time taken = 10 min. = 600 sec.

So,

Assumed Parameters are taken from(Link): https://www.lektro.com/products/ap8360/