Project-1: Powertrain for aircraft in runways

Question 1

list of Weight Of Different aircrat has been attach in tbe Exel file report (Project 1)

Question 2

Is there any difference between ground speed and air speed?

Airspeed and ground speed are completely different when it comes to determining how fast an airplane flies.

Ground speed is that speed how fast the airplane is travelling on the ground relative to fixed point.

Air speed is that speed ,under his own engine power how fast airplane is travelling

In general ,If there no wind Ground speed=Air speed.

but as there is wind present

ground spped - Wind speed = Air speed.

NASA Explains,

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.

This is because that Drag is require for the lift of the airplane ,and its only generated from the Wind coming From the Opposite Side.

Question 3

Why is it not recommended to use aircraft engine power to move it on the ground at Airport?

Air craft Engine Power are Very much High ,as they are used to produced thrust.

Its works on sucking the wind and throwing backwards direction ,this way it Moved forward.

But the Wind Generating due to enhine power are high as Hurricane wind .

This types of wind can Damage the Buildings near air ports ,small houses, and the ROOF of Airports Buildings which are Prone to small forces.

Risk factor is high While using the aircfrat engine power At airport..

Question 4

How an aircraft is pushed to runway when its ready to take off?

A truck Called pushback truck Is used to push and pull the Aircraft,pushing it with a towbar attached to the nose-wheel or you can say Front Single wheel.

A Pushback Vehicle is shown in the Image.

(this the new versions of push back late where used a TOW bar attchteach to airplane and Truck)

By Lookinf this picture you may be wondering that ,How a little truck and pull this heavy airplane?

actuclly This they dont have to lift the heavy weight of the airplane ,airplane itself is mounted on high pressure tyre, thney Only have to use the Firction thing here to get Reduced.

This truck are slow speed but produce very large amount of torque ,used to pull or push aircraft.

IMPOTANT Terms When the Aircraft is Landing.

Tafe Off Power

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.

tyre design

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.

brake forces.

From below example you can get Which forces are needed to calculate a brake power for airplane.

some assumptions

Landing weight = takeoff weight in case of aborted takeoff =20000 lbf

Landing distance = 1000 ft

Speed (assume aborted takeoff) = 120 kt (200 ft/sec)

We will use Newton’s second law to determine force due to deceleration:

Deceleration: from simple kinematics, acceleration = velocity^2 / distance

= (200 ft/s)^2 / 1000 ft = 40 ft / s^2

Mass of aircraft = (weight) / g = 20000/32.2 = 621.11 lbf*s^2 / ft

Force = mass * acceleration (621.11 lbf*s^2/ft) * (40 ft / s^2) = 24844.4 lbf

So in this example, the braking (deceleration) force is 24844.4 lbf

Notice that in the English foot-pound-second units the mass unit is
pound-second^2 per foot, called a slug in order for the units to work out.

Question 6

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

Airplane Value

Gross weight = 35,000 kg

Now

Total Tractive froce = Rolling resistence +Arodynamic drag

Frontal area A = Height of the airbus * Width of the airbus

A =15.80* 43.90= 693.62 m^2

Substituting the row and A values in drag force formula,

Assume that vehicle running at the speed of 20 kmph

V= (20 5/18) = 5.56 m/s

Fad = 0.5 *1.225* 693.62* 0.0265*(5.56)^2

Fad=347.58 N

Now find the total force Fte = Frr+Fad

Fte-39043.8+347.58

Fte=39391.38 N

Now Find the power require in KW

Power =all foreces *velocity

39391.38*5.56=217431.550 W =219.431 KW

Now find total torqure of truck =Total force * radius of the wheel R

39391.38*0.5715=

22512.17 Nm

Question 6 B

Simple model develop using simulin blocks

BLOCKS USED

constant blocks for input values

products blocks for multiplication

ADD blocks for ADDing.

DISPLAY for values.

Here the formulas used in above question are converted into simulink model.

Question 7

Now will calculate the the Energy

Energy = Power output*time (considering time as 30min=1800 sec)

219.431*(1800/3600)=109.71 kilo watt hour

NOw find the Duty Cycle for the truck

WE know D>C =power output/power input

(we dont know the input power of truck therfore assuming it,

Power =VOLTAGE*CURRENT=440V*600A=264000watts=264KW)

Duty cycle = 219.431/264=0.83

83% is the duty cycle of the truck.

Block DIagram of electic Vehicle Power Train(front wheel drive)

DEsign of power train in sumulink

Result of the model