Project-1: Powertrain for aircraft in runways

PROJECT 1:

POWERTRAIN FOR AIRCRAFT IN RUNWAYS

PART A

1:WIEGHTS OF DIFFERENT TYPES OF AIRCRAFTS

2:DIFFERENCE BETWEEN GROUND SPEED AND AIR SPEED

• Ground speed- The speed at which aircraft moving relative to ground
• Air speed - The speed at which aircraft moving relative to ground
• If the aircraft flying in the direction of wind,it experiences tailwind and ground speed is higher than airspeed
• If the aircraft flying against the wind,it experiences headwind and ground speed is less than airspeed
• The relationship between ground and air speed can be written as

Ground speed = Air speed + Wind speed

3:WHY AIRCRAFT ENGINE POWER NOT USED TO MOVE ON THE RUNWAY

• The aircraft engine consumes large amount of fuel to move them
• The fuel used in aircraft are jet fuel and AVGAS ,which are very expensive
• Also considering the large amount of carbon emissions and noise pollutions
• Aircraft engine is not recommended to move it in the runway

4:HOW AIRCRAFT MOVED TO RUNWAY

• The aircraft moved in the runway using towing vehicle
• The towing vehicle canmove a aircraft forward and backwardby attaching them to towing vehicle
• The following image shows the towing vehicle moving an aircraft
• 

5:LEARN ABOUT AIRCRAFT PARAMETERS

a)TAKE OFF POWER:

• The power required to lift the aircraft from the ground is called take off power
• The engine full power may used to take off lighter aircraft
• But the large aircraft only use less power to take off in order to improve engine life ,reduce maintanenece and reduce noise
• The forces acting on an aircraft while take off are Air drag ,rolling resistance and the propulsion power
• The force required to take off the aircraft can be written as

Ftake-off = Fthrust - Frr - Fad

• Where Fthrust-propulsion force of turbine
• Frr- rolling reistance force
• Fad - Aerodynamic drag force
• The take off power can be calculated as

Ptake-off = Ftake-off * V

• Where V-velocity of the aircraft

b)TYRE DESIGN:

• An aircraft tyre designed to withstand extremely heavy loads for shorter duration
• The number of tires required for aircraft increases with the weight of the aircraft
• Aircraft tire tread patterns are designed to facilitate stability in high crosswind conditions and braking effect
• Aircraft tires are usually inflated with nitrogen to minimize expansion and contraction from extreme changes in ambient temperature and pressure experienced during flight
• The below image shows the design of tyre used in aircraft
• 

c)ROLLING RESISTANCE:

• The force acting between the tyre and the track or runway is called rolling resistance
• The rolling resistance may vary according to the tire presure and runway conditions (climate)
• The rolling resistance can be calculated as

Frr = urr*m*g

• Where,Frr-rolling resistance force
• urr - coefficient of rolling resistance
• m-mass of the vehicle
• g-acceleration due to gravity

d)TYRE PRESSURE:

• Aircraft tires generally operate at high pressures, up to 200 psi (1,400 kPa)
• The main landing gear on the Concorde was typically inflated to 232 psi (16.0 bar)
• Its tail bumper gear tires were as high as 294 psi (20.3 bar).
• The aircraft tyres should properly inflated to prevent it from any severe damage or explosion
• Research says that five percent of tire pressure losses per day under typical operations
• At this rate of loss ,the tire needs frequent servicing

e)BRAKE FORCES:

• Aircraft brakes stop a moving aircraft by converting its kinetic energy to heat energy by means of friction between rotating and stationary discs located in brake assemblies in the wheels.
• An electrical signal is sent from the flight deck to hydraulic actuators near the main landing gear to apply brake
• Hydraulic fluid at 3,000 pound per square inch is used to force the brake unit against the wheel,to slow it down
• Nowadays electric brakes also implemented on aircraft,which reduces the cost and maintanence
• The braking force can be calculated as

Fbrake = 0.5* (m*V^2)/d

• Where m - mass of the aircraft
• V-velocity of the aircraft
• d-stoppping distance

PART B:

6A:CALCULATE FORCE AND POWER REQUIRED TO MOVE AN AIRCRAFT BY A TOWING VEHICLE

PARAMETERS:

• Mass of the towing vehicle Mt=1500 kg
• Mass of the aircraft Ma=184612.1kg
• Rolling reistance coefficient urr = 0.02
• Air density p= 0.015
• Width of the vehicle W=2.5 m
• Height of the vehicle H=2 m
• Drag coefficient Cd=0.8
• Velocity of the vehicele =5.0kmph
• Acceleration time = 12 s

EQUATIONS:

• The rolling resistance force Frr can be calculated

Frr= urr*m*g

• Where,urr - coefficient of rolling resistance
• m-mass of the vehicle
• g-acceleration due to gravity

• The Aerodynamic Drag force Fad can be calculated as

Fad=0.5*p*A*Cd*v^2

• Where,p-Air density in kg/m3
• A-Frontal Area (A=W*H ,W&H width and height of vehicle)
• Cd-Drag coefficient
• v-Velocity in m/s (speed in kmph convert to m/s by dividing it by 3.6)
• The acceleration force can be calculated as

Fac = m*a

• Where ,m-mass of thevehiclein kg
• a-acceleration in m/s2
• The total traction force can be calculated as

Fte = Frr+Fad+Fac

• The total traction power (Pte)can be calculated as

Pte= Fte *v

• Where,Fte-total traction force
• v-velcoity of vehicle in m/s

SOLUTION:

• The total mass m can be calculated
• m = Mt+Ma
• m=1500+184612.1
• m=186112.1kg
• The rolling resistance force can be calculated as
• Frr= urr*m*g
• Substituting all known values in the equation
• Frr = 0.02*186112.1*9.8
• Frr=36477.9716 N
• The Aerodynamic Drag force Fad can be calculated as
• Fad=0.5*p*A*Cd*v^2
• The frontal area can be calculated as
• A=W*H
• A=2.5*2
• A=5 m^2
• The velocity in kmph converted to m/s as
• v=5/3.6 m/s
• Substituting all known values
• Fad = 0.5*0.015*5*0.8*(5/3.6)^2
• Fad = 0.03*1.92901235
• Fad= 0.05787037 N
• The acceleration force can be calculated as
• Fac = m*a
• The acceleration can be calculated as
• a=v/t
• a=(5/3.6)/12
• a=0.1157 m/s^2
• Substituting alll known values
• Fac = 186112.1*0.1157
• Fac = 21540.74 N
• The total traction force can be calculated as
• Fte = Frr+Fad+Fac
• Substituting all known values
• Fte = 36477.9716 +0.05787037 +21540.74
• Fte = 58018.7695 N
• The total traction power (Pte)can be calculated as
• Pte= Fte *v
• Substituting all known values
• Pte = 58018.7695 *(5/3.6)
• Pte = 80581.62 w
• Converting to kw by dividing 1000
• Pte = 80.581 kw

CONCLUSION:

• The force and power required by a towing vehicle to move a Boeing 747-400 of 184612.1kg are 58018.8 N and 80.5kw respectively

6B:DEVELOP A SIMULINK MODEL FOR CALCULATED FORCE AND POWER

SIMULINK MODEL:

• The simulink model for calculation of force and power is shown below
• 
• The rolling resistance force calculation
• 
• The aerodynamic drag force calculation
• 
• The acceleratio force calculation
• 
• The traction power calculation
• 

REPORT:

• The input parameters used in the model logged into MATLAB workpace
• 
• The rolling resistance calculation
• Frr=urr*m*g
• urr-coefficient of rolling resistance defined by a constant block
• m-total mass which is addition of mass and payload of the vehicle
• g-acceleration due to gravity constant 9.8m/s2
• These values multiplied by a product block to obtain Frr-rolling resistance force
• Created a subsystem of these blocks
• The output connected to a display block through goto and from blocks
• The Aerodynamic Drag force Fad can be calculated as
• Fad=0.5*p*A*Cd*v^2
• Where,p-Air density defined by a constant block
• A-Frontal Area obtained by multiplying width and height using a product block
• Cd-Drag coefficient is a constant block
• v-Velocity in kmph converted to m/s by multiplying a gain block of value 1/3.6
• The velocity in m/s squared using a square block
• These values multiplied by a product block to obtain air drag force Fad
• Created a subsystem of these blocks
• The output connected to a display block through goto and from blocks
• The acceleration force can be calculated as
• Fac = m*a
• m-mass obtained by adding mass of vehicle and payload by add block
• a-acceleration obtained by dividing velocity in m/s by time using a divide block
• Created a subsystem of these blocks
• The output connected to a display block through goto and from blocks
• The total traction force can be calculated as
• Fte = Frr+Fad+Fac
• Adding outputs from the subsystems using a add block
• The total traction power (Pte)can be calculated a
• Pte= Fte *v
• The total traction force multiplied by velocity in m/s by product block to obatin the traction power
• The output multiplied by a gain block of value 1/1000 to convert watts to kilo watts
• Created a subsystem out of it
• The output displayed through goto and from block in display block
• Run the model
• Outputs displayed in the display blocks

RESULTS:

• The results of the above simulink model is shown below
• 

CONCLUSION:

• Developed the simulink model for the calculation of force and power
• The outputs displayed and verified

7A:ELECTRIC POWERTRAIN OF AIRCRAFT TOWING APLLICATION

POWER AND ENERGY CALCULATION:

• The power obtained for the aircraft towing vehicle application is 80.6kw
• Assuming towing vehicle travelled for 15 mins
• The energy can be calculated as
• E = P*t in kwh
• Where P-power required by the vehicle
• t-time in hours
• Substituting all known values in the energy equation
• E=80.6*0.25
• E = 20.15kwh
• The towing vehcle energy consumption will be 20kwh for15mins
• Assuming the motor parameters
• Motor input voltage =600v
• Motor input current =150A
• The input power can be calculated as
• Pin = V*I
• Pin = 600*150
• Pin = 90000w or 90kw
• The motor torque assumed to be 2000 Nm for pulling aircraft using towing vehicle

DUTY CYCLE:

• The duty cycle of the motor converter to achieve the required power can be calculated as
• Duty cycle = output power/input power
• Duty cycle = 80600/90000
• Duty cycle = 0.89 or 89%

SIMULINK MODEL:

• The electric powertrain model of towing vehicleforaircraft applicationis shown below
• 

REPORT:

• The speed reference to the DC7 converter provided by a step block
• 
• The battery mask block used to provide the input supply to the DC7 converter
• 
• The DC7 converter block parameters set as shown below
• 
• The torque input for the DC7 power converter is provided by a feedback from the motor speed
• 
• The gain value set to achieve the assumed motor torque of 2000Nm
• The output parameters from the DC7 block are motor ,controller and converter output
• The output parameters connected to a bus selector as shown below
• 
• The output 1 &2 are the duty cycles of IGBT in the converters
• The output 3 is the armature volatge of the motor
• The output 4 is the armature current of the motor
• The output 5 represents the actaual speed and the reference speed
• The outputs are connected to the scope block
• The power gui block added and changed simulation type to discrete
• The DC7 converter will act as four quadrant chopper circuit
• The switching operation of IGBT done in a way to achieve the required motor speed and torque

RESULT:

• The reference speed inputgiven to the DC7 converter is shown below
• 
• The final results of DC7 converter and motor are shown below
• 
• Graph 1:The duty cycles of IGBT 1&4
• Graph 2:The duty cycles of IGBT 2&3
• Graph 3:The armature voltage of the motor
• Graph 4:The armature current of the motor
• Graph 5:The actual speed and the reference speed
• By analying the graph ,the actual speed of the motor is equal to reference speed input

BLOCK DIAGRAM:

• The block diagram of the electric powertrain is shown below
• 
• BATTERY -DC source or input voltage to power converter ,controller input voltage also provided by the battery
• CONTROLLER-It provides contol signals or gate signal to the power converter ,controlling the duty cycle with repect to the reference value
• POWER CONVERTER- Four-quadrant chopper circuit ,which act as speed controller of motor
• MOTOR-DC motor or inductio motors
• GEAR-Motor connected to the axles through a gear system
• WHEELS-Axles connected to the wheels ,which moves the vehicle body

CONCLUSION :

• The simulink model of electric powertrain of towing vehicle application is designed
• The duty cycle and energy requirements of the motor calculated
• Block diagram of electric powertrain is designed

7B:DESIGN THE PARAMETERS IN EXCEL SHEET:

• The excel calculation for the force and power requirement of aircraft towing vehicle application is shown below
• 

CONCLUSION:

• Designed the parameters in the excel sheet and the outputs verified