Project-1: Powertrain for aircraft in runways

EV BATCH - 17

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

ANS :

• The answer for this question is prepared in the presentation format.
• The image snippets are placed below and the presentation is attached for your reference.
• Below is the general classification of the aircraft.
• Weight is a major factor in airplane construction and operation, and it demands respect from all pilots and particular diligence by all maintenance personnel.
• Excessive weight reduces the efficiency of an aircraft and the available safety margin if an emergency condition should arise.
• The total weight is the maximum take-off weight.
• It is the maximum mass at which the aircraft is certified for take-off due to structural or other limits

• MTOW Maximum take-off weight, MLW Maximum landing weight, TOR Take-off run, LR Landing run

Maximum Takeoff Weight:

The maximum weight limits for an aircraft to begin its takeo roll is called the max takeoff weight.

Maximum Landing Weight:

The maximum landing weight is, as you might imagine, the maximum aircraft weight limit approved for an aircraft to land. Landing weight can cause structural damage.

• In this above diagram refers that we canexplain the examples of heavy large and small aircrafts

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

ANS :

Ground speed

• Ground speed is the horizontal speed of an aircraft relative to the ground.
• An aircraft heading vertically would have a ground speed of zero.
• Ground speed can be determined by the vector sum of the aircraft's true airspeed and the current wind speed and direction; a headwind subtracts from the ground speed, while a tailwind adds to it.
• Winds at other angles to the heading will have components of either headwind or tailwind as well as a crosswind component.

Wind Speed :

• For a reference point picked on the ground, the air moves relative to the reference point at the wind speed.
• A positive velocity is defined to be in the direction of the aircraft's motion.
• We are neglecting cross winds, which occur perpendicular to the flight path but parallel to the ground, and updrafts and downdrafts, which occur perpendicular to the ground.

Airspeed :

• The important quantity in the generation of lift is the relative veloci between the object and the air, which is called the airspeed.
• Airspeed cannot be directly measured from a ground position, but must be computed from the ground speed and the wind speed.
• Airspeed is vector difference between the ground speed and the wind speed.

Airspeed Ground Speed - Wind Speed :

• The aircraft usually operates on land as well as in air and so need to study their speed differences in-ground and in air
• The ground speed and the airspeed usually change based on the wind component.
• If we consider that there is no wind component then, ground speed = airspeed
• If the wind is blowing in the tail direction of the aircraft then (airspeed ground speed) so, airspeed = ground speed+ win component.
• If the wind is blowing in the head direction of the aircraft then (airspeed ground speed)so, airspeed ground speed - wind component.
• Also, gravity plays a major role in increasing the aircraft's speswhile it is descending
• Here wind speed is the vector difference between ground andairspeed. On a particular day, the wind speed may be zero is equal to the ground speed.
• If the measured airspeed is greater than the observed ground speed the wind speed becames positive.

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

ANS :

• 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.
• But the thrust produced resulting jet blast might cause damage to the terminal building or equipment.
• Engines close to the ground may also blow sand and debris forward and then suck them into the engine, causing damage to the engine. A pushback is therefore the preferred method to move the aircraft away from the gate. And also it can reduce the cost for fuel. Taxiing can save money.
• Effectively a loss of control can happen, especially during high power engine running. Damage can occur to the aircraft itself, other aircraft nearby or to airsidestructures.
• There is a risk of injury to ground support personnel who may be in relatively close proximity to the aircraft.
• It is easier for truck driver to see the surroundings better than the pilot sitting in the cockpit with limited window.
• The pushback driver knows the airport better than the pilot.It also saves fuel virgin Atlantic hadproposed a new system in which pushback truck will move the aircraft all the way to runway to save fuel and reduce impact of environment.
• The plane engines are very loud. When more than one plane will add up their sound, than it will not be pleasent for passangers and others near airport.

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4 . How an aircraft is pushed to the runway when it's ready to take off?

ANS :

1) Taxing:

• Taxiing is the movement of an aircraft on the ground, under its own power, in contrast to towing or push-back where the aircraft is moved by a tug.
• The aircraft usually moves on wheels, but the term also includes aircraft with skis or floats.
• An airplane uses taxiways to taxi from one place on an airport to another; for example, when moving from a hangar to the runway

2) Towing:

• Movement of large aircraft about the airport, flight line, and hangar is usually accomplished by towing with a tow tractor.
• In the case of small aircraft, some moving is accomplished by hand pushing on the correct areas of the aircraft.
• Aircraft may also be taxied about the flight line but usually only by certain qualified personnel.
• Towing aircraft can be a hazardous operation, causing damage to the aircraft and injury to personnel, if done recklessly or carelessly.
• This article outline the general procedure for towing aircraft.
• However, specific instructions for each model of aircraft are detailed in the manufacturer's maintenance instructions and are to be followed in all instances.

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

ANS :

1) Take off power:

• Take off power can be predicted using a simple measure of the acceleration of the aircraft along the runway based on force equilibrium.
• The forces involved will be.
• T is Thrust of propulsion system pushing aircraft along the runway.
• D is Aerodynamic Drag of vehicle resisting the aircraft motion.
• F is Rolling resistance friction due to the contact of wheels or skids on the ground.
• During the take-off run, the imbalance in these forces will produce an acceleration along the runway.
• m*(dv/dt)=T-D-F

where, dv/dt is the acceleration along the runway and m is the mass of the vehicle.

• The procedure for take-off will be that the vehicle will accelerate until it reaches a safe initial flying speed.

• The pilot can then rotate the vehicle to an attitude to produce a climb lift and it will ascend from the ground The determination of this safe flying speed or rotation speed V. is a critical factor in determining take-off performance Take-off rules vary slightly depending on the aircraft category.

For Small commuter aircraft

V=1.1*VSTALL

V=1.05*V control

VSTALL =√(W/0.5CtmaxρS)

where S-Runway length

Take off power= force*displacement

(m*dv/dt)*2*s1 = (T-D-F)*S1

2) Tyre design :

• Designing aircraft tires is a very intense undertaking The tyre is designed to withstand 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.
• Aircraft tyre generates more heat during landing Aircraft drags the wheel until they match the rotational velocity of wheels.
• Due to this heat generates. Aircraft's tread pattern is designed to facilitate stability in high crosswind conditions, to channel water away to prevent hydroplaning and for braking effect.

3) Construction of tyre :

• Radial Tyre:

a ) It has fibre strands of the ply fabric oriented at 90° to the direction of rotation and the tyre sidewall.

b) This restricts flexibility bidirectional and the flexibility of the sidewall while it strengthens the tyre to carry heavy loads.

c) Due to their unique construction most radial tyres are resistant to penetration ad cuts. Due to stiffer tread they distribute weight evenly.

d) Thus provide greater traction. Radial tyres have an advantage in term of reducing the heat better.

• Bias-ply Tyre:

a) It has the fabric bias oriented with and across the direction of rotation and the sidewall.

b) Since fabric can stretchon the bias, the tyre is flexible and can absorb loads.

c) Tyre strength is obtained by adding plies.

d) It is cut resistant is sidewall area.

e) These tyres do not offer as much smoothness in terms of riding quality.

4) Rolling Resistance:

• Rolling resistance is the force resisting the motion when a body (such as ball, tyre or wheel) rolls on a surface.
• Rolling resistance can be seen in the aircraft when the aircraft takes-off from the runway & while landing at the runway with the help of their wheels

Rolling resistance can be expressed as Fm g (N).

Where

Frr the Rolling resistance(N)

μrr coefficient of rolling resistance.

m is mass(kg)

g is acceleration due to gravity= 9.81m/s^2

5) 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 tyre-related events is to maintain proper tire inflation pressure.
• Failure to keep aircraft tires properly inflated can lead to very serious consequences.
• One relatively easy way to help avoid those consequences is to check tire pressures regularly and service them as necessary.
• Following these three actions is a good way to make sure you get the most value out of your tires and contribute to the safe operation of your aircraft.
• Aircraft tires are usually inflated with nitrogen to minimize expansion and contraction from extreme changes in ambient temperature and pressure experienced during flight.
• Dry nitrogen expands at the same rate as other dry atmospheric gases (normal air is about 80% nitrogen), but common compressed air sources may contain moisture, which increases the expansion rate with temperature, Also, the use of an inert gas for tire inflation eliminates the possibility of tire explosion.

Relation to Rolling Resistance and Tire Pressure :

Braking Forces during Landing :

Following braking devices are used to decelerate the aircraft -

1. Spoilers.
2. Wheel Brakes (Including Anti-Skid systems & Auto-Brake systems)
3. Reverse Thrust

6) Spoilers :

• The mechanical deflection of parts of the wing upper surface by increasing aerodynamic drag on the moving aircraft.
• This can be achieved by raising upper wing surface panels called ground spoilers or by operation of a tail cone 'clamshell type air brake and also by increasing the effective downward load on the landing gear and thereby increasing the efficiency of wheel braking .

7) Wheel Brakes :

• To be effective, braking action at any speed depends upon sufficient friction existing between the tires and the runway surface achieved through freely rotating wheels.

• Anti skid system :

  When landing on slippery runways, there's a chance that the wheels may start to skid as the brakes are applied. To stop this from happening and to maintain maximum effective braking, each wheel has anti-skid protection.

  Using a variety of sources to determine the aircraft speed, the brake units know how fast the wheels should be spinning. If that speed drops significantly, it's because the current brake pressure on that wheel is too great and the wheel is just skidding over the surface.

  In this situation, the anti-skid system automatically reduces the braking on that wheel to a point where the skid stops before reapplying the pressure.

8) Reverse Thrust :

• Thrust reversers provide a deceleration force that is independent of runway condition.
• Thrust reverser efficiency is higher at high airspeed therefore.
• Treversers must be selected as early as possible after touchdown.
• Thrust reversers should be returned after reverse idle at low airspeed to prevent engine stall or foreign object damage.

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

ANS :

• Considering : Total mass of ther aircraft (including payload) for airbus a380=575,000kg.

Mass of the push back: 1000kg (assumed)

• Rolling Resistance (Frr):

  Coefficient of rolling resistance (Crr) = 0.002 (on tarmac)

  Rolling resistance force is given by F, Cm-g For the aircraft: F= 0.002-575000 9.81 F, 11281.5N F,, = 11.2815kN

  For the Towing Vehicle: F= 0.002 1000-9.81 F = 19.62N

  As the towing vehicle resistance is so less compared to the Aircraft because this is a heavy aircraft these can be neglected

• Aerodynamic resistnace (Far):

  Drag coeffcient(Cd): 0.025

  Air Density (p) = 1.125kg/m^3

  Frontal Area (A): 150m^2.

  Velocity (v) = 14.5kmph or 4m/s

  Far=1/2*Cd*ρ*A*V^2

  Far= 1/2*0.025*1.125*150*4^2

  Far= 1/2*0.025*1.125*150*16

  Far=33.75N

  As the vehicle and plane is moving at very low speed aerodynamic resistance can be neglected

• Hill climbing Resistance :

  Runways are constructed flat so Gradient resistance/hill climb force is neglected.

• Total Resistance :

  Ft = 11.2815kN

• Power Required :

  P=Ft*v

  P=11.2815*4

  P=45.26KW

  The min power required by the tow vehicle is 45.26kw to taxi a heavy aircraft at 4m/s.

MATLAB MODEL :

clear all

close all

clc

ASSUMED VALUES

rr =[0.02,0.03,0.04,0.05, 0.06, 0.07];

u = 0.8

m = linspace (500000, 600000,10);

g = 9.81

rho = 1.25;

A = 158.3

cd = 0.02

v = 3;

r = 0.35

G = 20;

frr zeros(6,10);

fad zeros(6,10);

ffr zeros(6,10); fte zeros(6,10);

%FOR LOOP FOR FORCE CALCULATIONS for i = 1/6

for j-1:10

frr1 = urr(i).*m(j)*g;

frr(i,j)= frr1;

fad1 0.5 rho*A*cd v^2;

fad(i,j) = fad1;

ffr1 = u*m(j).*g:

ffr(i,j)= ffr1;

fte(i,j)= frr(i,j) + fad (i,j) + ffr(i,j);

end

end

display(fte)

P = (fte. v)/1000;

%TORQUE CALCULATION T = (fte. r)/G;

%PLOTTING

figure

plot(m, fte)

xlabel('mass of aircraft (kg)')

ylabel('traction force (N)')

title('Variation of Traction force with mass of aircraft')

legend("Icy concrete/asphalt 0.02", "Dry concrete/asphalt 0.03", "Firm turf 0.04", "Hard turf 0.05","

figure

plot(urr, fte)

xlabel('coefficent of rolling resistance')

ylabel('traction force (N)')

title('Variation of Traction force with rolling resistance coefficent')

legend("mass 1", "mass 2", "mass 3", "mass 4","mass 5", "mass 6", "mass 7", "mass 8","mass 9", "mass 10")

OUTPUT :

SIMULATION MODEL :

Constant blocks are used to feed the parameters such as mass, coefficients and other parameters

• Using product block the resistive forces are calculated.
• Using sum block, total resistive forces can be calculated.
• The outputs are visualised using display blocks.

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7a) Design an electric powertrain with the type of motor, it's power rating, and energy requirement to fulfill aircraft towing application. 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 the powertrain.

ANS :

Estimation of Duty Cycle:

Based on the calculation done in study 6

The Power Required for towing of the aircraft is 45.26kW.

Motor Considred for the drive train is REB 90

Considering 60kW as continuous power, so the voltage will be 400V

Battery power is considred to be 550v .

Duty cycle is given by Vo/Vb

where Vo is motor required voltage

Vb is battery voltage.

d= 400 550 ⇒ 0.7272

As there will be losses due to the resistance of FET and Inductance.

Correction factor is considred to be 90% As there is 10% loss, the voltafge demanded by the motor should be 10% more than required d 0.7272 (1+0.1) 0.799≈ 0.8

so the voltage demand should be around 440V from the battery.

Assumed parameters

Mass of aircraft 575000 kg

Mass of the push back track 1000 kg

coefficient of rolling resistance 0.002

Drag coefficient 0.025

Air density 1.125 kg/m^3

Frontal Area 150 m^2

Velocity 4 m/s

uphill gradient 0 deg

Battery voltage 550 v

correction Factor 0.8

Based on the calculations done in 6A.

Force required to push/pull the aircraft =19.508 KN

Power required =135.478 kw

Let us assume,

The Radius of the Tyre for the Tug - 25inches =0.635m

Gear ratio=8

Transmission Effeciency (nt) =87%

Motor Efficiency (nm)=95%

then,Torque required for the wheels (Tw)=Total Force *Radius in meters

=19.508 *0.635

=12.387KNm

Motor Torque-Wheel= torque/(Gear ratio'nt)

=12.387/(8 0.87)

=1.778KNm

For nm =95%. The required motor Torque (Tm)= 1.778/0.95 1.871 kNm

Similarly. The total power required by the motor considering all the efficiencies (Pt) =135.478/(0.87 0.95)=163.917Kw

Let us assume the towing operation was done for a 15 minutes (0.25Hrs) duration.

Then, Total Energy consumed = 163.917* 0.25 =40.98 KWh

If we consider the battery capacity=220 kw is used by the towing vehicle then. The duty cycle range to control the speed from O to max speed (3m/s) is given as

Duty Cycle (d)= (Required Power)/ (Rated power) = 163.917/220=0.75

Therefore the controller will be able to provide a duty cycle ranging from 0 to 75% in order to control the towing speed.

Block diagram of electric power train :

Implementation of Four quadrant dc chooper in simulink model :

For the iinput singnal we are using step bolck and its input vale is 500 and and having 0.77 dyty cylce ratio. as shown below.

RESULTS :

MOTOR OUPUT :

From the above module is simluated and motor output parameters of armature voltage and speed and armature current are shown below.

BATTERY OUTPUT :

From this garph we are plots results of battery peromance parmeters are soc(safe operation condition)and battry current and baterry volatge charerstics are shown below

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7B) Also,Design the parametrs in excel sheet

ANS :

EXCELL SHEET :

TimeSpeed in mphSpeed (m/s)Energy CalculationBrake EnergyTotal Brake Energy (kj)

0	0	0	0	FALSE	115646.4464
1	0	0	0	FALSE	115646.4464
2	0	0	119.906857	FALSE	115646.4464
3	1	0.44704	0	FALSE	115646.4464
4	1	0.44704	359.7205709	FALSE	115646.4464
5	2	0.89408	0	FALSE	115646.4464
6	2	0.89408	599.5342848	FALSE	115646.4464
7	3	1.34112	839.3479987	FALSE	115646.4464
8	4	1.78816	1079.161713	FALSE	115646.4464
9	5	2.2352	1318.975427	FALSE	115646.4464
10	6	2.68224	1558.78914	FALSE	115646.4464
11	7	3.12928	1798.602854	FALSE	115646.4464
12	8	3.57632	2038.416568	FALSE	115646.4464
13	9	4.02336	17266.5874	FALSE	115646.4464
14	15	6.7056	16307.33255	FALSE	115646.4464
15	19	8.49376	4676.367421	FALSE	115646.4464
16	20	8.9408	33094.29252	FALSE	115646.4464
17	26	11.62304	19784.6314	FALSE	115646.4464
18	29	12.96416	-25899.8811	25899.8811	115646.4464
19	25	11.176	47842.83593	FALSE	115620.5465
20	32	14.30528	69066.34961	FALSE	115620.5465
21	40	17.8816	-18705.46969	18705.46969	115620.5465
22	38	16.98752	28417.9251	FALSE	115601.8411
23	41	18.32864	74701.97189	FALSE	115601.8411
24	48	21.45792	-129379.4987	129379.4987	115601.8411
25	35	15.6464	457564.5662	FALSE	115472.4616
26	71	31.73984	-565600.6443	565600.6443	115472.4616
27	18	8.04672	36091.96394	FALSE	114906.8609
28	25	11.176	287776.4567	FALSE	114906.8609
29	55	24.5872	68946.44275	FALSE	114906.8609
30	60	26.8224	29257.2731	FALSE	114906.8609
31	62	27.71648	126621.6409	FALSE	114906.8609
32	70	31.2928	16906.86683	FALSE	114906.8609
33	71	31.73984	-97843.99528	97843.99528	114906.8609
34	65	29.0576	-314755.4995	314755.4995	114809.0169
35	40	17.8816	50960.41421	FALSE	114494.2614
36	45	20.1168	69066.34961	FALSE	114494.2614
37	51	22.79904	494375.9712	FALSE	114494.2614
38	82	36.65728	164991.8352	FALSE	114494.2614
39	90	40.2336	110913.8427	FALSE	114494.2614
40	95	42.4688	-67267.74675	67267.74675	114494.2614
41	92	41.12768	-86332.93701	86332.93701	114426.9937
42	88	39.33952	798100.0399	FALSE	114340.6608
43	120	53.6448	2974169.68	FALSE	114340.6608
44	198	88.51392	-2827283.78	2827283.78	114340.6608
45	125	55.88	8918072.486	FALSE	111513.377
46	300	134.112	8393479.987	FALSE	111513.377
47	400	178.816	-14388822.84	14388822.84	111513.377
48	200	89.408	92328279.86	FALSE	97124.55414
49	900	402.336	-97124554.14	97124554.14	97124.55414
50	0	0	0	FALSE	0
51	0	0	0	FALSE	0
52	0	0	0	FALSE	0
53	0	0	0	FALSE	0
54	0	0	0	FALSE	0
55	0	0	0	FALSE	0
56	0	0	0	FALSE	0
57	0	0	0	FALSE	0
58	0	0	0	FALSE	0
59	0	0	0	FALSE	0
60	0	0	0	FALSE	0
61	0	0	0	FALSE	0
62	0	0	0	FALSE	0
63	0	0	0	FALSE	0
64	0	0	0	FALSE	0
65	0	0	0	FALSE	0
66	0	0	0	FALSE	0
67	0	0	0	FALSE	0
68	0	0	0	FALSE	0

CONCLUSION :

With necessary assumptions, calculated the force and power required to push / pull an aircraft by a towing vehicle. Develop the model for the calculated force and power using Simulink. Designed an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfill aircraft towing application in Simulink. Estimated the duty cycle range to control the aircraft speed from zero to highest. plotted necessory required graphs.

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