Project-1: Powertrain for aircraft in runways

AIM :

• a) To calculate the force and power required for to push/pull aircr1aft by towing vehicle with necessary assumptions.

b) To develop the model for the calculated force and power using Simulink.

• a) To design an Electric Powertrain with type of motor,its power rating and enery requirement to fulfill aircraft towing application and draw a block diagram of powertrain.

b) To design the parameters in excel sheet for Electric Powertrain.

Study 1:

A) We have assumed the Towing Vehicle as KALMAR tow tractor TBL 190 and the aircraft considered is AIRBUS A319 TBL 190 is designed for pushback and maintenance towing from MD-80/A319 upto B777/A340.

KALMAR tow tractor TBL 190:

Mass of Tow Vehicle = 50,000 kg

Inertia of 5% = 2500 kg

Mass of Airplane = 371950 kg

Inertia of 5% = 18597.5 kg

Coefficient of Rolling Resistance for Tow Truck = 0.0035

Coefficient of Rolling Resistance for Airplane = 0.004

• The forces to be taken into consideration while calculating th traction power required are

1. `F_(rr)=mu_(rr)* m*g`

Where,

`F_(rr)`=Rolling resistance force at N

`mu_(rr)`=the coefficeint of Rolling Resistance

= total mass of vehicle in Kg

g(g=9.81)=acceleration the due to gravity in `m/(s^2)`

2. `F_(ad)=(1/2)*rho*A*C_d*v^2`

Where,

`rho=`Air density

A= frontol area of vehicle in `m^2`

`C_d`=Drag Coefficient

`v^2`=Velocity in m/s

3. `F_h=m*g*sin(psi)`

Where,

`F_h`=Hill climbing force in N

`psi=`hill climbing angle in radian

4. `F_a=m*a`

Where,

a = acceleration of the vehicle in `m^2`

• The summation of the all forces multiplied by the vehicle will give the required power.

• Rolling Resistance Force is given by,

`F_(rr)`=`F_(rr)=mu_(rr)* m*g`

Therefore, For Tow Truck Frr is,

`F_(rr1)` = `0.0035*(50000+2500)*9.81`

`F_(rr1)`= 1802.59 N

`F_(rr1)`= 1.80 kN

For Airplane `F_(rr)`is,

`F_(rr2)` = `0.004 *(371950+18597.5)*9.81`

`F_(rr2)` = 15325.08 N

`F_(rr2)` = 15.33 kN

• The Total Rolling Resistance Force is given by,

`F_(rr1) + F_(rr2)`

= 1.80 + 15.33

= 17.13 kN

• Aerodynamic Drag Force

Velocity of Tow Truck and Airplane = 15 km/hr = 1`5 * 5/18`= 4.167 m/sec

Air Density = 1.25 `(kg)/m^3`

Aerodynamic Drag Coefficient for tow truck (`C_d`) = 1.15

Aerodynamic Drag Coefficient for Airplane (`C_d`) = 0.025

• Frontal Area of Tow truck = 1.9 `m^2`
• Frontal Area of Airplane = 25 `m^2`

Therefore, Aerodynamic Drag Force is given by,

For Tow Truck,

Fad1 = 0.5 * Air Density * Frontal Area * Drag Coefficient * Velocity^2

Fad1 =`0.5 * 1.25 * 1.9 * 1.15 * 4.167^2`

Fad1 = 23.71 N

For Airplane,

Fad2 = 0.5 * air density * Frontal Area * Drag Coefficient * Velocity^2

Fad2 = `0.5 * 1.25 * 25 * 0.025 * 4.167^2`

Fad2 = 6.782 N

Therefore, Total Aerodynamic Drag Force is given by,

Fad1 + Fad2

= 23.71 + 6.782

= 30.49 N = 0.030 kN.

• Total Tractive Force is given by,

Ft = Frr + Fad

Ft = 17.13 + 0.030

Ft = 17.16 kN

• Power Required by Motor to tow airplane is given by,

Power = Tractive Force * Velocity

Power =`17.16 * 4.167`

Power = 71.49 kW

Calculation of all parameters in Matlab:

Subsystem of Rolling Resistance:

Subsystem of Aerodynamic Resistance:

Study 2 :

What is Powertrain?

• As the name suggests, the Powertrain provides power to the vehicle. Powertrain refers to the set of components that generates the power required to move the vehicle and deliver it to the wheels.

Assumption for Powertrain Design-

Mass of Tow Truck = 50000 kg

Mass of Airplane = 371950 kg

Inertia = 5% of Mass

Air Density = 1.25 kg/m^3

Velocity = 4.167 m/sec

Frontal Area of tow truck = 1.9 m^2

Frontal area of Airplane = 25 m^2

Coefficient of Rolling Resistance of tow truck = 0.0035

Coefficient of Rolling Resistance of Airplane = 0.004

Coefficient of Aerodynamic Drag of tow truck = 1.15

Coefficient of Aerodynamic Drag of Airplane = 0.025

Radius of wheel of Tow Truck = 0.50 m

Operating Time of vehicle = 15 min = 15*60 = 900 sec = 0.25 hr

Tractive Force = 17.16 kN (from Excel Calculator)

• Power Required to tug Airplane = 71.49 kW (from Excel Calculator)

Torque is given by,

Torque = Force * Radius of wheel/Gear Ratio

Where, Gear Ratio for delivering high Power = 11

Torque = 17.16 * 0.5/11

Torque = 0.78 kN-m

• Energy required to towing application is,

Energy = Power * Time

Energy = 71.49 * 0.25

Energy = 17.87 kWh

For Towing operation, we can use following types of machines

• 3 phase induction Motor
• BLDC Motor
• Permanent Magnet Synchronous Motor
• Switched Reluctance Motor

• From this we choose here 3 phase Induction Motor of standard rating of 90 kW.
• If we check that available Induction motor of the rating is 90 kw, 120 HP, 415V, 4 Pole, 3 Phase motor having 1500RPM.
• Let us consider the efficiency of 3 phase induction motor is 95%, then input side power must be,

Efficiency = Output/Input

Input = Output/Efficiency

Input = 90/95%

• Input side Mechanical Power for Induction motor = 94.74 kW
• The Motor output power is the input power of the towing vehicle so to obtain maximum speed, the duty cycle needed is calculated as below,

Duty Cycle = Output Power/Input Power

Duty Cycle = 71.49/90 = 0.7943 = 79.43%

• So, the duty cycle will be 79.43%
• The Duty Cycle range to control the airplane speed from zero to highest is from 0 to 79.43%

Block Diaram of EV Powertrain:

Excel Calculation for all design parameters:

CONCLUSION:

• Taxing & Pushback Aircrafts using the Towing power is undesirable.
• Towing Vehicles proes as a better alternative for that purpose.
• The power and energy requirements for a towing vehcile were calculated and a simple pre electric divetrain for the same is presented.