Project-1: Powertrain for aircraft in runways

Part A:
  
1. Search and list out the total weight of various types of aircraft. 
a)Aircraft Gross Weight:
• The aircraft gross weight (also known as the all-up weight and abbreviated AUW) 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 the flight, an aircraft's gross weight is referred to as
the en-route weight or in-flight weight.
• Aircraft gross weight limits are established during an aircraft's design and certification
period and are laid down in the aircraft type certificate and manufacturer specification
documents
• The absolute maximum weight capabilities of a given aircraft are referred to as the
structural weight limits. The structural weight limits are based on the aircraft maximum
structural capability and define the envelope for the CG charts (both maximum weight
and CG limits).
• An aircraft's structural weight capability is typically a function of when the aircraft was
manufactured, and in some cases, old aircraft can have their structural weight capability
increased by structural modifications.
b)Manufacturer's Empty Weight (MEW):
• Also called Manufacturer's Weight Empty (MWE) or Licensed Empty Weight It is the
weight of the aircraft "as-built and includes the structure, power plant, furnishings
installations, systems, and systems other equipment that are considered an integral part
of an aircraft.
c) Maximum Flight Weight (MFW):
• The maximum design taxi weight (also known as the maximum design ramp weight
(MRW)) is the maximum weight certificated for aircraft manoeuvring on the ground
(taxing or towing) as limited by aircraft strength and airworthiness requirements
d) Maximum Takeoff Weight (MTOW)
• Is the maximum certificated design weight when the brakes are released for takeoff
and is the greatest weight for which compliance with the relevant structural and
engineering. requirements have been demonstrated by the manufacturer.
e) Maximum Landing Weight(MLW):
• The maximum certificated design weight at which the aircraft meets the appropriate
landing certification requirements. It generally depends on the landing gear strength of
the landing impact loads on certain parts of the wing structure. The MIW must not
exceed the MTOW.
• The maximum landing weight is typically designed for 10 feet per second (600 feet per
minute) sink rate at touch down with no structural damage.

f) Maximum Zero-Fuel Weight(MZFW):
• 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 aeroplane.
• The MZFW 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 MTOW and the MZFW may be utilised only for the
addition of fuel
g) Payload:
It is the carrying capacity of an aircraft. It includes cargo, people, extra fuel. In the case
of a commercial airliner, it may refer only to revenue-generating cargo or paying
passengers.
The following image depicts Takeoff Weight Components:

h Types of aircraft
There are several ways to identify aircraft by type. the primary distinction is between
those that are lighter than air and those that are heavier than air.

2. Is there any difference between ground speed and airspeed?

 difference between ground speed and airspeed
 One of the most confusing concepts for young scientists is the relative
velocity between objects. Aerodynamic forces are generated by an object
moving through a fluid (liquid or gas).
 A fixed object in a static fluid does not generate aerodynamic forces. Hot air
balloons "lift" because of buoyancy forces and some aircraft like the Harrier use
thrust to "lift" the vehicle, but these are not examples of aerodynamic lift.
 To generate life, an object must move through the air, or air must move past the
object. Aerodynamic lift depends on the square of the velocity between the object
and the air. Now things get confusing because not only can the object be moved
through the air, but the air itself can move.
 To properly define the relative velocity, it is necessary to pick a fixed reference
point and measure velocities relative to the fixed point. The reference point is
fixed to the ground in this slide, but it could just as easily be fixed to the aircraft
itself.
 It is important to understand the relationships of wind speed to ground speed and
airspeed.
Wind Speed
For a reference point picked on the ground, the air moves relative to the reference
point at the wind speed. Notice that 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. A positive velocity is defined to
be in the direction of the aircraft's motion. We are neglecting crosswinds, which occur
perpendicular to the flight path but parallel to the ground, and updrafts and downdrafts,
which occur perpendicular to the ground.

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 comparison of
the ground speed to the wind speed must be done according to rules for vector
comparisons.
Airspeed
The important quantity in the generation of lift is the relative velocity between the
object and the air, which is called the airspeed. 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

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.
Examples
Suppose we had an aeroplane that could take off on a windless day at 100 mph (liftoff
airspeed is 100 mph). We are at an airport with an east-west runway that is 1 mile long.
The wind is blowing 20 mph towards the west and the aeroplane takes off going east.
The wind is blowing towards the aircraft which we call a headwind. Since we have
defined a positive velocity to be in the direction of the aircraft's motion, a headwind is a
negative velocity. While the plane is sitting still on the runway, it has a ground speed of
0 and an airspeed of 20 mph:

Airspeed = Ground Speed (0) - Wind Speed (-20) = 20 mph

The aeroplane starts its takeoff roll and has a constant acceleration a. From Newton's
second law of motion, the ground speed V at any time t is:

V = a * t
and the distance d down the runway at any time is:
d = 1/2 * a * t^2

For a fixed-length runway, this specifies the time to be used in the velocity equation.
Let's assume that at 5000 feet down the runway, the velocity is 80 mph. Then the
airspeed is given by

Airspeed = Ground Speed (80) - Wind Speed (-20) = 100 mph

and the aeroplane begins to fly. Now another pilot, with the same aeroplane decides to
take off to the west. The wind is now in the same direction as the motion and this is
called a tailwind. The sign on the wind speed is now positive, not negative as with the
headwind. The acceleration along the ground is the same, so at 5000 feet down the
runway, the ground speed is again 80 mph. The airspeed is then given by:
Airspeed = Ground Speed (80) - Wind Speed (20) = 60 mph

3. Why is it not recommended to use aircraft engine power to move it on the

ground at the Airport?

Aircraft Engine Power is Not Recommended to be Used to Move It on The Ground
at Airport
•Although many aircraft are capable of moving in backwards using reverse thrust, as a
result, the jet blast or prop wash might cause damage to the terminal building or
equipment.
• Engine close to the ground may also blow sand and debris forward and then suck
them into the engine, causing damage to the engine. Therefore there are pushback
trucks in the airport.
• A pushback truck lifts the front wheel of the aircraft, their role in most airports is to
move the aircraft away from the parking spot onto the taxiways, which is called towing
• For movement on the ground, the aircraft uses its own engine power, which is called

Unwanted consequences mainly arise from the unintended movement of during engine
is running

Although most aircraft are capable of moving themselves on the ground. it is not
typically done due to the following reasons.
• 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 relative proximity
to the aircraft. 
• The jet blast from the engines can damage the terminal building and nearby
equipment. It can even throw away people standing nearby.
• Engines can suck in sand and debris which can cause serious damage to the engine.

• It is easier for the truck driver to see the surroundings better than the pilot sitting in the
cockpit with limited windows.

• The plane engines are very loud. When more than one plane will add up their sound
then it will not be pleasant for passengers and s near the airport.
• It also saves fuel. Virgin Atlantic had proposed a new system in which pushback trucks
will move the aircraft all the way to the runway to save fuel and reduce the impact on
the environment.
• Modern planes are huge. It is difficult to navigate such gigantic structures through
limited space. An A380 is eight stories high. The pilots won't be able to see anything
directly in front of them or anything behind the wings
• In addition, response to potential but unexpected occurrences must be covered by
appropriate training and, where a rapid response may be necessary, memory actions.
• Finally, engine ground running, especially if engine operation above Ground Idle is to
take place, is best carried out with both flight deck pilot seats occupied and with clearly
defined roles for the person in charge and their assistant. In particular, this allows
checklists to be carried out using the challenge and response method.

4. How an aircraft is pushed to runway when its ready to take off?
4.1 Phases of

Image courtesy of Federal Aviation Administration (FAA) 4.1 Taxi
a. Taxiing refers to the movement of an aircraft on the ground, under its own power.
The aircraft moves on wheels. An aeroplane uses taxiways to taxi from one place
on an airport to another; for example, when moving from a terminal to the
runway.
b. The aircraft always move on the ground following the yellow lines, to avoid any
collision with the surrounding buildings, vehicles or other aircraft. The taxiing
motion has a speed limit. Before making a turn, the pilot reduces the speed
further to prevent tire skids. Just like cars, there is a certain list of priorities during
taxiing. The aircraft that are landing or taking off have higher priority. The other
aircraft have to wait for these aircraft before they start or continue taxiing.
c. The thrust to propel the aircraft forward comes from its propellers or jet engines.
Steering is achieved by turning a nose wheel or tail wheel/rudder; the pilot
controlling the direction travelled with their feet. The use of engine thrust near
terminals is restricted due to the possibility of jet blast damage. This is why the
aircraft is pushed back from the buildings by a vehicle before they can start their
own engines for taxiing.
4.2 Take-Off
 Takeoff is the phase of flight in which an aircraft goes through a transition from
moving along the ground (taxiing) to flying in the air, usually starting on a runway.
Usually, the engines are run at full power during takeoff. Following the taxi
motion, the aircraft stops at the starting line of the runway. Before takeoff, the
engines, particularly piston engines, are routinely run up at high power to check
for engine-related problems. This makes considerable noise. When the pilot
releases the brakes, the aircraft starts accelerating rapidly until the necessary
speed for take-off is achieved.

 The takeoff speed required varies with air density, aircraft weight, and aircraft
configuration (flap and/or slat position, as applicable). Air density is affected by
factors such as field elevation and air temperature.
 Operations with transport category aircraft employ the concept of the takeoff V-
Speeds, V1 and V2. These speeds are determined not only by the above factors
affecting takeoff performance but also by the length and slope of the runway.
Below V1, in case of critical failures, the takeoff should be aborted; above V1 the
pilot continues the takeoff and returns for landing. After the co-pilot calls V1,
Then, V2 (the safe takeoff speed) is called. This speed must be maintained after
an engine failure to meet per- performance targets for rate of climb and angle of
climb.
 The speeds needed for takeoff are relative to the motion of the air (indicated
airspeed). A headwind will reduce the ground speed needed for takeoff, as there
is a greater flow of air over the wings. This is why the aircraft always take off
against the wind. Sidewind is not preferred as it would disturb the stability of the
aircraft. Typical takeoff air speeds for jetliners are in the 130–155 knot range
(150–180 mph, 240–285 km/h). For a given aircraft, the takeoff speed is usually
directly proportional to the aircraft weight; the heavier the weight, the greater the
speed needed. Some aircraft have difficulty generating enough lift at the low
speeds encountered during takeoff. These are therefore fitted with high-lift
devices, often including slats and usually flaps, which increase the camber of the
wing, making it more effective at low speed, thus creating more lift. These have
to be deployed from the wing before performing any manoeuvre.
 At the beginning of the climb phase, the wheels are retracted into the aircraft and
the undercarriage doors are closed. This operation is audible by the passengers
as a noise coming from below the floor.
 4.3 Climb

 Following take-off, the aircraft has to climb to a certain altitude (typically 30,000 ft
or 10 km) before it can cruise at this altitude safely and economically. A climb is
carried out by increasing the lift of wings supporting the aircraft until their lifting
force exceeds the weight of the aircraft. Once this occurs, the aircraft will climb to
a higher altitude until the lifting force and weight are again in balance. The
increase in lift may be accomplished by increasing the angle of attack of the
wings, by increasing the thrust of the engines to increase speed (thereby
increasing lift), by increasing the surface area or shape of the wing to produce
greater lift, or by some combination of these techniques. In most cases, engine
thrust and angle of attack are simultaneously increased to produce a climb.

 Because lift diminishes with decreasing air density, a climb, once initiated, will
end by itself when the diminishing lift with increasing altitude drops to a point that
equals the weight of the aircraft. At that point, the aircraft will return to level flight
at a constant altitude. During the climb phase, it is normal that the engine noise
diminishes. This is because the engines are operated at a lower power level after
the take-off. It is also possible to hear a whirring noise or a change in the tone of
the noise during a climb. This is the sound of the flaps that are retracting. A wing
with a retracted flap produces less noise.

5. Learn about take-off power, tyre design, rolling resistance, tyre pressure, brake
forces when landing.    
a) TAKE-OFF POWER:
Takeoff Power is the phase of flight in which an aerospace vehicle leaves the ground
and becomes airborne On take-off, to position the aircraft to a safe height away from
terrain and obstacles (ie. a flight path of maximum height and minimum ground distance
desired), the engine thrust is set to a high 'take-off power setting and the aircraft attitude
is pitched up by to maintain a specific speed by fulfilling the task is known as the take-
off power.
• Take off performance can be predicted using a simple measure of the acceleration of
the aircraft along the runway based on force equilibrium.
The forces involved will be.

where
T=Thrust of propulsion system pushing aircraft along the runway.
D=Aerodynamic Drag of vehicle resisting the aircraft motion
F=Rolling resistance friction due to the contact of wheels or skids on the ground.

D=During take-off runs the imbalance in these forces will produce an acceleration
along the runway.
dv/dt =Acceleration along with the runway
m =Mass of the vehicle.
whereas 
Power =(T-D-F)*V
b) TYRE DESIGN:
o When it comes to safety tyres 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 the braking and stopping of an aircraft.
o An aircraft tyre 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 aeroplane needs to
be distributed more evenly.
o A Boeing 737NG & 737MAX uses 6 wheels, Boeing 787 uses 10 wheels, Boeing
777 uses 14 wheels and Airbus A380 uses 22 wheels. Aircraft tyres work under
extreme conditions, carrying up to 340 tons and accelerating at over 250 km/hour
at takeoff, in addition to enduring varied environmental stress when in flight and
taxiing.
o Carcass plies are used to form the tire. They are sometimes called casing plies.
An aircraft tyre is constructed for the purpose it serves.
a. They support the weight of an aircraft while it is on the ground.
b. It also provides the necessary traction for the braking and stopping of an
aeroplane.
c. Tyre also helps absorb the shock of landing and provides cushioning for the
roughness of takeoff, roll-out, and taxi operations.
o Unlike an automobile or truck tyre, 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.

Let’s go through some technical aspects of the tyre.
 Retreading = Retreading is restoring a worn tyre by renewing the tread area or
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.
Aircraft tyres must have an approved speed and load rating and have sufficient
clearance when retracted through landing gear to allow for tyre growth. Tyre growth is
the increase in the size of the tyre due to centrifugal forces at high speed.
 Aircraft tyres may be Tube-type or Tubeless
Tubeless tyres are more advantageous over tube-type. There is no longer the use of
tube-type tyres in recent aviation. Nowadays all airliners are using tubeless tyres.

Tubeless that are meant to be used without a tube has the word TUBELESS on the
sidewall of the tyre.
 Aircraft tyres may be Radial or Bias
Almost all airliners are using Radial tyres. Bias is an older design, and it’s mainly used
for road vehicles. Radial tyres have the word RADIAL on the sidewall. Radial tires are
more expensive than bias-ply tyres. Radial tires are in demand because of their lower
life cycle cost and long term value.
c) ROLLING RESISTANCE:
Rolling resistance is the force resisting the motion when a body (such as a ball, tire, or
wheel) rolls on a surface.
Rolling resistance can be seen in the aircraft when the aircraft Takes off at the runway
with the help of from the runway and while landing its wheels.
The rolling resistance can be expressed
Fr =cW
where
         F= rolling resistance or rolling friction
         c=rolling resistance coefficient - dimensionless (coefficient of rolling friction-CRF)
         W=normal force or weight of the body
W = mag
where
            m=mass of the body (kg, lb)
            ag =acceleration of gravity (9.81 m/s2.32.174 ft/s²)
The rolling resistance can alternatively be expressed as
Fr=cl W/r
where
          cl=rolling resistance coefficient- dimension length (coefficient of rolling friction)
(mm, in)

          r=radius of the wheel (mm, in)
d)TYRE PRESSURE:
a. Aircraft tires generally operate at high pressures, up to 200 psi(14 bar; 1.400
kPa) for airliners, and even higher for business jets. 
b. The main landing gear on the Concorde was typically inflated to 232 psi (16.0
bar), whilst its tail bumper gear tires were as high as 294 psi (20.3 bar).
c. Tests of airliner aircraft tires have shown that they can sustain pressures of the
maximum of 800 psi (55 bar, 5.500 kPa) before bursting. During the tests, the
tires have to be filled with water, to prevent the test room from being blown apart
by the energy that would be released by a gas when the tire bursts Aircraft tires
are usually inflated with nitrogen to minimize
d. expansion and contraction from extreme changes in ambient temperature and
pressure experienced during flight. Dry nitrogen  Laye expands at the same rate
as other dry atmospheric gases (normal air is about 80% nitrogen), but common
compressed air sources may contain moisture, rate with temperature. which
increases the expansion
e. The requirement that an inert gas, such as nitrogen, be used instead of air for
inflation of tires on certain transport category aeroplanes was prompted by at
least three cases in which the oxygen in air-filled tires combined with volatile
gases given off by a severely overheated tire and exploded upon reaching auto-
ignition temperature The use of inert gas for tire inflation will eliminate the
possibility of a tire explosion
e) BRAKE FORCES WHEN LANDING:
i. All commercial aircraft have automatic brake systems. Just like a car, these
systems also have an Anti-Lock braking system (ABS) that prevents the tires
from sliding down a wet runway.
ii. The advantage of the automatic brake system is that the aircraft brakes
comfortably at a constant deceleration. Passengers experienced a constant force
that presses them into their seats.
iii. The brake system automatically applies the wheel brakes just enough to create a
constant deceleration independent of the air resistance, spoilers and reversers. It
is only in the last phase of brakes by pressing on both brake pedals. This
deactivates the automatic brake system.
iv. Immediately after landing, the aircraft is still moving fast enough that
aerodynamic devices are much more efficient than the brakes. The brake units
are powered by the hydraulics system.
v. An electrical signal is sent from the flight deck to hydraulic actuators near the
main landing gear. Here, hydraulic fluid at 3.000 pounds per square inch is used
to force the brake unit against the wheel, thus slowing it down.

Part B:

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

 B. Develop the model for the calculated force and power using Simulink.

The Force and Power required to push/pull an Aircraft by a Towing
Vehicle:
Let us Assume:
Weight of the Aircraft =500000kg
Rolling Resistance Coefficient of an aircraft tire =0.005
Wight of the Towing Tractor=50000kg
Rolling Resistance Coefficient of Towing Vehicle =0.002
The Velocity of the Aircraft while Towing=18KMPH=5m/s
The density of Air medium=1.225 kg/m^3
The Frontal area of the Aircraft=20 m^2
The Coefficient of Drag of the Aircraft = 0.26
Towing Tractor Gear ratio(G- for high torque output) = 11
Towing Tractor tyre radius(r) = 0.5 m
The forces that are acting on the aircraft are
i) Rolling Resistance
ii) Drag Force
i) Rolling Resistance for the Aircraft
Now. Here the Rolling resistance be
   Frr =c. mr. ag
Let.

     Frr =Rolling resistance
     c=coefficient of rolling resistance
     mr=mass
     ag =Acceleration of gravity
Now Substituting the values in the above formula
Resistance of the Aircraft gives us the Rolling
Fr(aircraft) = 0.005*(500000 kg)*(9.81 m/s^2)
Fr(aircraft)= 24.525 KN.
Therefore the Rolling resistance of the aircraft Fr(aircraft)= 24.525 KN.
Rolling Resistance for the Towing Vehicle
Now let us calculate the rolling resistance for the Towing Vehicle
   Frr =c. mr. ag
Fr(towing)=0.002(50000 kg)(9.81 m/s^2)
Fr(towing)=0.981 KN.
Therefore the Rolling resistance of the Towing vehicle Fr(towing)=0.981 KN.
ii) Drag Force

         
Where.
          D=Drag force
          C=Drag Coefficient
          V=Velocity of Towing Vehicle

          A=Frontal area
Now Substituting the assumed values in the above the Drag force of the Towing vehicle
Drag Force = 0.5* (0.26*1.225*20*5*5)
Drag Force = 0.079 KN
Therefore the Drag force is 0.079625 KN
Now to get the total force to push the aircraft by the towing vehicle
we add the
=Rolling resistance of Aircraft + Rolling resistance of Towing Vehicle+ Drag Force
= 24.525 KN +0,981 KN+ 0.079625 KN
=25.585625 KN
Therefore the Total force required for a towing vehicle to push the aircraft
is=25.585625 KN=25585.625N
Power required to the towing vehicle to pull the aircraft
In this case,
Power=force*velocity
Power=25.585625*5.0
Power=127.928125 KW
Therefore the power required for the towing vehicle to pull
the aircraft=127.928125 KW=127928.125 W
Simulink Model for the calculated Force and Power

7. A. Design an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfill aircraft towing application in Simulink. 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. 

A) Design an Electric Powertrain with the type of Motor:
Let us Assume:
Weight of the Aircraft =500000kg
Rolling Resistance Coefficient of an aircraft tire =0.005

Wight of the Towing Tractor=50000kg
Rolling Resistance Coefficient of Towing Vehicle =0.002
The Velocity of the Aircraft while Towing=18KMPH=5m/s
The density of Air medium=1.225 kg/m^3
The Frontal area of the Aircraft=20 'm^2
The Coefficient of Drag of the Aircraft =0.26
Towing Tractor Gear ratio(G-for high torque output)=11
Towing Tractor tyre radius(r)=0.5 m
Operational time of the Towing for one cycle=0.25 Hours (15min)
Power converters used Bidirectional DC-DC Boost Converter
We taking calculated values from the above equation
The total force required for a towing vehicle to push the aircraft is =25.585625
KN=25585.625N
Power required for the towing vehicle to pull the aircraft=127.928125KW=127928.125 W
Torque Required with the wheels of towing vehicle to push the aircraft
Torque=force *(radius/Gear ratio)
Torque= 25.585 *(0.5/11)
Torque=11629 KN-m
Therefore the Torque required the towing vehicle to push the aircraft =1.1629 KN-
m
i) Energy Requirements:
The Energy requirement for an EV similar to the Towing vehicle to Push/Pull the aircraft
is calculated on the Basin of Time
Let
        Energy =Power x Time taken by the towing vehicle to push-pull the aircraft

        Energy=127.928 x 0.25
        Energy =31.982 KWH
Therefore power required for one single towing is=31.982 KWh
Therefore, the battery must store the energy 31.982 KWH and deliver the energy
throughout the run time for the motor to produce the 1.1629 KN-m of torque. Thus the
storage of energy in batteries with high-performance Ultracapacitors are selected to fulfil
the required power delivery.
Motor Selection:
Here we have selected a Permanent-Magnet Synchronous Motor (PMSM) type that has
a power rating of 127.928 kw and with the gear ratio and the transmission, it should be
able to deliver 1.1629 KN-m of torque instantaneously.
Motor Parameters:
Type=PMSM
Model No.=APEV80-12(16)
Phase=3
Application Electric bus,Truck,passenger,car
AC Voltage=200-750V
Pole=12
Direct Current Voltage (VDC)=540V
Rated current (A)=255
Peak Current (A)=650
Rated Power (kW)-150
Peak Power (kW) =220
Rated Speed (RPM)=1650
Peak Speed (RPM)-3.300
Rated Torque (Nm)-868

Peak Torque (Nm) =3.000
BEMF (VAC)/rpm=660/2401
Insulation Grade-H Protection
Class IP67
Cooling Water
Weight (Kg)=260
The efficiency of motor=90 % (Assumed Parameter)
Duty Cycle:
Here in this case the Duty cycle of the electric power train is calculated by the output
power by the input power By this, we get the efficiency of the duty cycle.
Duty Cycle = Output Power /Input Power
We know that.
Output Power=127.92 kW
Input Power =Rated Motor Power / Motor Efficiency-
Input Power =150/0.9
Input Power=166.666kW
Then 
Duty Cycle =127.92/166.666
Duty Cycle =0.76
Duty Cycle =76%
Therefore the Efficiency of the Duty cycle that we get for this electric power train
is=0 to 76%
BLOCK DIAGRAM OF THE POWERTRAIN

• The example shown the DC7 four-quadrant chopper DC drive during speed regulation

• The 200 HP DC motor is separately excited with a constant 150 VDC field voltage
source. The armature voltage is provided by an IGBT converter controlled by two Pl
regulators
• The converter is fed by a 515 V DC bus obtained by rectifying a 380 V AC 50 Hz
voltage source. To limit the DC bus voltage during dynamic braking mode, a braking
chopper has been added between the diode rectifier and the DC7 block
• The first regulator is a speed regulator, followed by a current regulator. The speed
regulator outputs the armature current reference (in .p.u) used by the current controller
to obtain the electromagnetic torque needed to reach the desired speed.
• The speed reference change rate follows acceleration and deceleration ramps to avoid
sudden reference changes that could cause armature over-current and destabilize the
system
• The current regulator controls the armature current by computing the appropriate duty
ratios of the 5 kHz pulses of the four IGBT devices (Pulse Width Modulation). For proper
system behaviour, the instantaneous pulse values of IGBT devices land, 4 are opposite
to those of IGBT devices 2 and 3. This generates the average armature voltage needed
to obtain the desired armature current
• To limit the amplitude of the current oscillations, a smoothing inductance is placed in
scenes with the armature circuit.

B. Also, Design the parameters in an excel sheet

CONCLUSION:
• The reference Aircraft taken here is the Heavy Aircraft build and so our designed
towing tractor is a heavy-duty operational vehicle.
• The power requirements are calculated for the particular load to be pushed/pulled and
to achieve this power at the motor output, the motor type and its voltage at the terminals
are discussed.
• Then the power converters are used to upscale the battery armature voltage to the
desired high voltage value so that the motor may spin at full power and required torque
can be achieved using a Bidirectional DC-DC Converter.
• Finally, the speed control of the vehicle is done with switching techniques where the
Duty Cycle adjustment can change the motor speed from the minimum to maximum