Project-1: Powertrain for aircraft in runways

The amount of power that an engine is allowed to produce for a limited period of time for takeoff. The use of takeoff power is usually limited to 5 min for reciprocating engines and up to 2½ min for gas turbine engines. This may not always be the case. Specifically,

with respect to reciprocating engines, it is the brake horsepower developed under standard,sealevel conditions and under the maximum conditions of the crankshaft rotational speed and the engine manifold pressure approved for the normal takeoff.

It is limited in continuous use to the period of time shown in the approved engine specifications. With respect to gas turbine engines,it is the thrust developed under static conditions at a specified altitude and atmospheric conditions of the rotor shaft rotational speed

and gas temperature approved for the normal takeoff. It is limited in continuous use to the period of time shown in the approved

engine specifications.

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 braking and stopping of an aircraft.

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.

Rolling Resistance :

Rolling resistance is the energy that your vehicle needs to send to your tyres to maintain
movement at consistent speed over a surface. In other words, it's the effort required to keep a tyre rolling. The main
contributor to the rolling resistance is the process known as hysteresis. Hysteresis is essentially the energy loss that
occurs as tyre rolls through its footprint.

Frr = `mu`rr mg
Where, Fry = the rolling resistance force
`mu`r = rolling resistance coefficient
m = mass of the vehicle
g = acceleration due to gravity

Tyre Pressure :

The tyre pressure of an aircraft depends on number of tyres and weight of the aircraft under cold
pressure. It is important to maintain the desired inflation pressure of the tyre or else it may result in damage of tyre.
For aircrafts the pressure is 200-300 psi depends on the weight and number of tyres. Nitrogen is used for tyre
inflation.

Braking Force While Landing :

Following braking techniques are used to decelerate the aircraft i.e.,

• Aircraft disc brakes in the landing gear is used to brake the wheels while touching the ground. These brakes are operated hydraulically or pneumatically.
• Thrust reversers allows the thrust from the engines to be used to slow the aircraft.
• Air brakes dedicated flight control surfaces that work by increasing the drag.
• Large drogue parachutes are used by several former and current military and civilian aircraft.

Part B:

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

Assuming,

Mass of the aircraft = `m_1`= 640000 kgs
Rolling resistance of aircraft = `rho_r` = 0.023
Acceleration due to gravity = g = 9.81 m/s2
Air density = `rho` = 1.25 kg/m3
Drag coefficient of the aircraft = `C_a` = 0.05
Frontal Area of aircraft = `A_1` = 250 m2
Mass of the tug = `m_2` = 2500 kgs
Rolling resistance of tug= `mu_(rt)` = 0.01
Drag coefficient of the tug = `C_t` = 0.02
Frontal area of the Tug = `A_2` = 20 m2
Speed of the tug vehicle = v = 15 kmph =15*5/18 =4.167 m/s

Where `mu_(rr) = mu_r+ mu_(rt)`
`M = m_1+ m_2`
`A = A_1+A_2`
`C_d= C_a + C_t`

Calculations:

Rolling Resistance Force = `F_(rr)=mu_(rr) *M * g`

= (0.023+0.01) *(640000+2500)*9.81

=0.033 *642500*9.81

= 207.996 KN

Aerodynamic Drag Force = `F_(ad) = 0.5*C_a*A*v^2*rho`

= `0.5* (0.05+0.02)* (250+20)* (4.167)^2 * 1.25`

= `0.5*0.07*270* (4.167)^2*1.25`

= 205.078 N

Total Tractive Force = `F_(te) = F_(rr) + F_(ad)`

= `207.996 + 0.205078`

= `208.195 KN`

`Total Power = P = F_(te)

=208.195K*4167

= 867.54 K`

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

Rolling Resistance Force :

Aero Dynamical Drag Force :

Overall :

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.

(Hint :DC7 Block)

Process :

Assuming,

Mass of the aircraft = `m_1`= 48000 kgs
Rolling resistance of aircraft = `rho_r` = 0.023
Acceleration due to gravity = g = 9.81 m/s2
Air density = `rho` = 1.25 kg/m3
Drag coefficient of the aircraft = `C_a` = 0.05
Frontal Area of aircraft = `A_1` = 250 `m^2`
Mass of the tug = `m_2` = 2500 kgs
Rolling resistance of tug= `mu_(rt)` = 0.01
Drag coefficient of the tug = `C_t` = 0.02
Frontal area of the Tug = `A_2` = 20 m2
Speed of the tug vehicle = v = 15 kmph =15*5/18 =4.167 m/s

Where `mu_(rr) = mu_r+ mu_(rt)`
`M = m_1+ m_2`
`A = A_1+A_2`
`C_d= C_a + C_t`

Calculations:

Rolling Resistance Force = `F_(rr)=mu_(rr) *M * g`

= (0.023+0.01) *(48000+2500)*9.81

= 0.033 *50500*9.81

= 16.348 KN

Aerodynamic Drag Force = `F_(ad) = 0.5*C_a*A*v^2*rho`

= `0.5* (0.05+0.02)* (250+20)* (4.167)^2 * 1.25`

= `0.5*0.07*270* (4.167)^2*1.25`

= 205.078 N

Total Tractive Force = `F_(te) = F_(rr) + F_(ad)`

= `16.348 + 0.205078`

= `16.5534 KN`

`Total Power = P = F_(te)*v

= 16.5534K*4167

= `68.97 K`

The Torque-speed relation is T=24.7+0.005lw^2

Taking the Reb 90 electric motor with 80KW power,400-800VT=300Nm, 107 hp

Duty cycle = Output Power/Input Power

= 68.97/80 = 86.21%

Parameters	SI unit
Tractive Force	16.55 KN
Power Output	6897 KW
Power Input	80 KW
Duty Cycle	0.86
Motor Voltage	400 V
Battery Pack	550V
Mass	50500 kgs
Rolling resistance coefficient	0.033
Air density	1.25 kg/m^3
Aero drag coefficient	0.07
Frontal Area	270 m^2
Velocity	4.16 m/s

Result :

Block Diagram :

B. Also, Design the parameters in excel sheet.