Project-1: Powertrain for aircraft in runways

EV LIVE BATCH -16

Part A:

1. Search and list out the total weight of various types of aircrafts??

SOLUTION:

• Weight is a mojor factor in airplane construction and operation , and it demands respect from all pilots and particular diligence by all maintenance personal
• Excessive weight reduces the efficency of an aircraft and athe available safety margin if an emergency condition condition should arise

AIRCRAFT TOTAL WEIGHT (TAKE-OFF WEIGHT)

1. Average sized airplane

eg-Boeing 737-800 80,000kg

2. fighter jet

eg - boeing f-18 Hornet 113,646kg

3. cargo airplanes

eg- airbus A300 600f 165,900kg

4. Military Helicopter

eg- boeing RAH-66 comanche 7,791kg

5. mid -size helicopter

eg- bell 212 5,5080kg

6. turboprop aircraft

eg- grand caravan ex 3,994kg

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

SOLUTION:

• The airplanes always try to take off and land into the wind
• this requires a lower ground speed,which means the plane can take off or land in the shortest distance travelled along the ground
• since runway have a fixed length , you want to get airborne as fast as possible on take-off and stopped as soon as possible on landing
• now mechanical and electronic devices provide the information that is radioded to the cockpit
• it is important to understand the relationships of wind speed to ground speed and airspeed

AIR SPEED:

AIR SPEED = GROUND SPEED - WIND SPEED

• air speed is the vector difference between the ground speed and the wind speed
• The realtive velocity between the object and the object and the air is called air speed.
• airspeed cannot be directly measured from a ground position
• but must be computed from the ground speed and the wind speed
• speed relative to air
• measured quantity
• using pitot tube

GROUND SPEED:

• the speed on which the aircraft moves relative to the point on the ground
• ground speed is a vector quantity
• speed relative to ground
• calculated quantity (Air speed+ wind velocity)

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

SOLUTION:

• Aircraft can move using their own engine power all the time on the airport apron and taxiways under their own power to go towards runways and after landing taxi to their assigned gate
• Engine close to the ground may also blow sand debris forward and then suck them into engine,causing damage to the engine
• A pushback is therfore the prefered method to move the aircraft away from the gate
• and also it can reduce the cost of fuel
• taxing can save money

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

SOLUTION :

• pushbacks is when an aircraft is pushed backwards away from the airport gate by vehicles called tugs or tractors.
• closer to departure , an aircraft tug will park right in front of the nose wheel.

• the tug might be directly attached to the planes nose gear with a tow bear bar, or could be a wheel -lift tug.
• that gives the tug driver control over the planes direction during pushback
• like pilot -controlled tugs and electric motors mounted to the planes landing gear
• both promise to save fuel and reduce airport noise

SOLUTION:

5. Learn about take off power, tyre design, rolling resistance, tyre pressure, brake forces when landing??

SOLUTION:

TAKE OFF :

• A normal take -off is one which is the airplane is headed into the wind,or the wind is very light
• Also,the take -off surface is firm and of sufficient length to permit the airplane to gradually accelerate to normal lift-off and climb -out speed,and there are no obstruction along the take -off path.
• There are four forces which helps the aircraft to take off:
• wings keep an aircraft up in the air ,but the four forces are what make this happen
• they push a plane up,down,forward or slow it down.

THRUST :

• it is a force that moves an aircraft in the direction of the motion .
• it is created with a propeller,jet engine or rocket
• air is pulled out in an opposite direction
• one example is a household fan

T=pshaft/v

DRAG:

• it is a force that moves an aircraft in the direction of the motion
• it trends to slow an object
• Drag is caused by friction and difference in air pressure
• an example is putting your hand out of a moving car window and feeling it pull back

D=CD.0.5.p.v^2.s

WEIGHT:

it is the force caused by gravity

W=mg

LIFT:

• it is the force that holds an airplane in the air
• the wings create most of the lift used by airplanes

L=CL.0.5.p.v.s

THEREFORE TAKE-OFF POWER=TOTAL FORCE*VELOCITY

• For smaller aircrafts like turboprop we use full power for take-off.
• because they dont have take-off power settings
• but in aircraft like jet plane they have take off power setting so they have to take the full power,

ROLLING RESISTANCE :

• The friction between aircraft and runway will be proportional to the normal force exerted by aircraft on the runway
• F= M(W-L)
• The normal force will be the difference between weight (W) of aircraft and lift(L),the friction coefficient (m) will be typically of a magnitude of 0.02 for a standard tarmac runway

TYRE DESIGN"

• Designing aircraft tires is a very intense undertaking
• the tyre is designed to withstand heavy loads while heavy loads while landing,take -off taxing and parking
• the number of wheels required for aircraft increase with the weight of the aeroplane needs to be distributed more evenly
• Aircraft drags the wheel until they match the rotational velcoity of the wheels
• due to this heat generates

TWO TYPE OF TIRES ARE USED IN AIRCRAFT APPLICATION : Bias-ply tries and radical tries.

RADIAL TYRE:

• It has fibre stands of the ply fabric oriented at 90degree to the direction of rotation and the tyre sidewall.
• due to the their stiffer tread they distribute weight evenly
• thus provide greater traction
• radical tyres have an advantages in term of reducing the heat better

BIAS -PLY:

• it has the fabric bias oriented with and across the direction of rotation and the sidewall
• since fabric can stetch on the bias, the tyre is flexiable and can absorb loads
• tyre strength is obtained by adding plies

TYRE PRESSURE:

• Aircraft tyres operate at such high extremes of pressure,load and speed ,their care and service is critically important
• the most important action an operator can take to prevent can to prevent tyre-related events is to maintain proper tire inflation pressure
• following three actions is agood way to make sure you get the most value out of your tires and contributes to the safe operation of your aircraft

1. perform daily pressure checks with a calibrated guage
2. target the highest pressure recommended in the AMM or POH
3. compensate for ambient temperaturechanges.

BRAKES FORCES WHEN LANDING:

• stopping a 200-tonne aircraft landing at 180mph requires a lot of braking force
• to do this either hydraulic brakes or electric brakes or elctric brakes are used
• the aim of brake force when landing is to minimise the distance
• the touch down velocity should be approximetely the stall speed of the aircraft in landing configuration
• this can be solved average acceleration approach that was used to esminate the take off roll

dv\dt=-T-D-F \m

WHERE;

T= thrust of propulsion system pushing aircraft along runway

D=aerodynamics drag of vehicle resisting the aircraft motion

F=rolling resistance friction due to the correct of wheels or skids on the ground

m= mass

SO THE BRAKE FORCE=MASS* DECELERATION

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??

SOLUTION:

The force to move an airplane on an absolute flat runway no wind at constant low speed is equal to the rolling resistance between the plane and the tarmac.

The rolling resistance for the air plane can be calculated as ,

Frr=mrr.mg

where,

Frr=rolling resistance

mrr=coefficient of rolling resistance

mg=acceleration of gravity

EXAMPLE-- FORCE REQUIRED TO PULL AN AIRPLANE AT CONSTANT LOW SPEED ON TARMAC

If the mass of a plane is 40000kg and the rolling coefficient between the aircraft wheel and the tarmac is 0.02.

The force required to overcome the rolling resistance can be calculated as:

Frr=0.02(40000kg)(9.81m\s^2)

Frr=7848N

This force can be delivered by a pulling truck.Enough weight of truck is required to create friction force against the tarmac that equal the plane rolling resistance.

The friction force for the puller can be calculated to;

Ff=mw

=m.mf.ag

WHERE,

Ff=frictional force

m=friction coefficient

w=weight

mf=friction mass

Required mass for pulling truck:

The friction coefficient between a pulling rubber wheel and dry tarmac is 0.9

since pulling trucks are connected to air craft nose wheels close to tarmac we simply the calculation of required truck mass to:

mf =(7848N)\(0.9(9.81m\s2))

=889kg

power required for pulling truck with a velocity of 5m\s:

p=F*v

WHERE ,

F=total force acting on truck

V=velocity of truck

p=7848*5

p=39.24

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) B. Also, Design the parameters in excel sheet??

SOLUTION:

According to study 6 for towing an aircraft of 40000kg needed 39.42kw of power .

so we selecting a motor which can take this much power.

VEHICLE DRIVE: 58.3 HP\45.5Kw ,80VDC traction motor mated with Auburn \Dana 44 differnetial.

Auburn is a high performance series differential which deliver more traction and faster engagement for controlled power transfer & lightning fast reaction to changing load

power output = 45.5Kw

BATTERIES:Two 40 VDC industrial , 595 AMP -HR Batteries (6hr rate)wired in series.

power input= V*I

= 80*595= 47.6KW

Efficiency of motor=power output \power input

= 45.5\47.6

= 95.58percentage

Energy required for towering aircraft = required power* Time

Assume taxing time = 15min

Therefore energy required = 39.24 *0.25

= 9.81Kwh

Assume highest speed =25kmph

Duty cycle = Average power/peack power

=39.24/45.5

=86percentage

ASSUMED PARAMETERS:

S.NO PARAMETERS VALUES

01 weight of aircraft 40000kg

02 Rolling coefficient of aircraft 0.02

03 rolling coefficient of tractor 0.9

04 Highest speed 25Kmph

05 Taxing time 15min

06 Battery voltage 40v (2no's)

07 motor power 45.5 kw

08 terminal voltage 80v

BATTERY:

• it is stored the energy needed to the run the vehicle
• it provides the direct current (DC) output

CONTROLLER:

• between converter and motor we use a controller
• it is used to mange the power tranfer
• it control the frequency and magnitude of voltage supplied to electric motor in order to manage the speed and accelaration

CONVERTER;

• DC -DC converter with the DC current and the supply to the motor

ELECTRIC MOTOR:

• converts electrical energr to mechanical energy and deliver into wheels via a transmission
• many EVs use motor generators that can perform regeneration as well

DIFFERENTIAL :

• To give power to both wheels from motor a differential is used which also help in fast engagement in controlled power