Challenge

Project-1: Powertrain for aircraft in runways

AAnshuman Tejaswi24 May 2021

AIRCRAFTS --

These are the vehicles used as a medium of aerial locomotion. These vehicles fly by taking the air suport which is used to balance the downwards gravity.

Aircrafts can be classified into various types based on the various modes of classification, which are;-

Weight and the weight Balancing have a very significant effect in the handlng of aircrafts. Take for example, if the aircraft becomes too heavy then it may not take off as sufficient lift force cannot be generated moreover, if somehow we are able to do take off then also it would be difficult for the ongoing operations to maintain the aerial stability of the aircraft and in some worse case it may lead to fatal conclusion. Therefore, it is very vital to do a careful study of weight balacing of aircrafts.

Calculation of aircraft weight and balance

During this calculation every part of an aircraft signifies its importance.

We have taken a simple demo of calculation of weight of airplane:-

We have an airplane with two rows of seats, 88 gallons of usable fuel, and two baggage areas, each of which includes the distance from the Datum (Arm). The location of the empty weight center of gravity (EWCG) is shown and also includes its Arm.
To calculate weight and balance,we sum the weight of our passengers, baggage, and fuel into the appropriate locations. Then multiplying the weight by the Arm to determine the Moment. When we have done so for every location, we will divide the Total Moment by the Total Weight and in this way the total weight is computed.

Here a list is created for weight of differnt types of aircrafts:-

Type	MTOW [kg]	MLW [tonnes]	TOR [m]	LR [m]
Antonov An-225	640,000	591.7	3500
Scaled Composites Model 351 Stratolaunch	589,670		3660
Airbus A380-800	575,000	394	3100	1930
Boeing 747-8F	447,700	346.091	3100	1800
Boeing 747-8	443,613	306.175	3100
Boeing 747-400ER	412,770	295.742	3090
Antonov An-124-100M	405,060	330	2520	900
Boeing 747-400	396,900	295.742	3018	2179
Lockheed C-5 Galaxy	381,000	288.417	2530	1494
Boeing 747-200	377,840	285.700	3338	2109
Boeing 747-300	377,840	260.320	3222	1905
Airbus A340-500	371,950	240	3050	2010
Airbus A340-600	367,400	256	3100	2100
Boeing 777F	347,800	260.816	2830
Airbus A330-900	352,000	251.5	3100
Boeing 777-300ER	351,800	251.29	3100
Boeing 777-200LR	347,450	223.168	3000
Boeing 747-100^[7]	340,200	265.300
Airbus A350-1000	308,000	233.5
Boeing 777-300	299,370	237.683	3380
Boeing 777-200ER	297,550	213.00	3380	1550
Airbus A340-300	276,700	190	3000	1926
McDonnell Douglas MD-11	273,300	185	2990	1890
Airbus A350-900	270,000	175	2670	1860
Ilyushin Il-96M	270,000	195.04	3115	2118
McDonnell Douglas DC-10	256,280	183	2990	1890
Boeing 787-9	254,000	192.777	2900
Boeing 787-10	254,000	201.849
Airbus A340-200	253,500	181	2990
Ilyushin IL-96-300	250,000	175	2600	1980
Airbus A330-300	242,000	185	2500	1750
Airbus A330-200	242,000	180	2220	1750
Lockheed L-1011-500	231,300	166.92	2636
Boeing 787-8	228,000	172.365	3300	1695
Lockheed L-1011-200	211,400
Ilyushin IL-86	208,000	175
Boeing 767-400ER	204,000	158.758	3414
Airbus A300-600R	192,000	140	2385	1555
Boeing 767-300ER	187,000	136.08	2713	1676
Concorde	185,000	111.1	3440	2220
Airbus A300-600	163,000	138	2324	1536
Boeing 767-300	159,000	136.078	2713	1676
Airbus A310-300	157,000	124	2290	1490
Vickers VC10	152,000	151.9
Boeing 707-320B	151,000	97.5
Boeing 707-320C	151,000	112.1
Douglas DC-8-61	147,000
Airbus A310-200	142,000	123	1860	1480
Airbus A400M	141,000	122	980	770
Douglas DC-8-32	140,000
Douglas DC-8-51	125,000
Boeing 757-300	124,000	101.6	2550	1750
Boeing 707-120B	117,000	86.3
Boeing 757-200	116,000	89.9	2347	1555
Boeing 720B	106,000	79.5
Boeing 720	104,000	79.5
Tupolev Tu-154M	104,000	80
Tupolev Tu-204SM	104,000	87.5	2250
Convair 880	87,500
Boeing 737-900	85,000	66.36	2500	1704
Boeing 737-900ER	85,000	71.35	2804	1829
Boeing 727-200 Advanced	84,000	70.1
Airbus A321-100	83,000	77.8	2200	1540
Boeing 737-800	79,000	65.32	2308	1634
Boeing 727-200	78,000	68.1
McDonnell-Douglas MD-83	73,000	63.28
Boeing 727-100	72,500	62.4
Boeing 727-100C	72,500	62.4
McDonnell-Douglas MD-90-30	71,000	64.41	2165	1520
de Havilland Comet 4	70,700
Boeing 737-700	70,000	58.06	1921	1415
Airbus A320-100	68,000	66	1955	1490
Boeing 737-400	68,000	54.9	2540	1540
de Havilland Comet 3	68,000
Boeing 377	67,000
Boeing 737-600	66,000	54.66	1796	1340
Airbus A220-300	65,000	57.61	1890	1494
Hawker Siddeley Trident 2E	65,000
Airbus A319	64,000	62.5	1850	1470
Boeing 737-300	63,000	51.7	1939	1396
Boeing 737-500	60,000	49.9	1832	1360
Airbus A220-100	59,000	50.80	1463	1356
Airbus A318	59,000	57.5	1375	1340
Boeing 717-200HGW	55,000	47.174	1950
Douglas DC-7	55,000
de Havilland Comet 2	54,000
Boeing 717-200BGW	50,000	46.265	1950
de Havilland Comet 1	50,000
Douglas DC-6A	48,600
Douglas DC-6B	48,500
Embraer 190	48,000	43	2056	1323
Caravelle III	46,000
Fokker 100	46,000	39.95	1621	1350
Douglas DC-6	44,000
Avro RJ-85	42,000	36.74
Handley Page Hermes	39,000
Embraer 175	37,500	32.8	2244	1304
Bombardier CRJ900	36,500	33.345	1778	1596
Embraer 170	36,000	32.8	1644	1274
Bombardier CRJ700	33,000	30.39	1564	1478
Douglas DC-4	33,000
Vickers Viscount 800	30,400
Bombardier Q400	28,000	28.01	1219	1295
Bombardier CRJ200	23,000	21.319	1918	1479
ATR 72-600	22,800	22.35	1333	914
Saab 2000	22,800	21.5	1300
Embraer ERJ 145	22,000	19.3	2270	1380
ATR 42-500	18,600	18.3	1165	1126
Saab 340	13,150	12.930	1300	1030
Embraer 120 Brasilia	11,500	11.25	1560	1380
BAe Jetstream 41	10,890	10.570	1493	826
Learjet 75	9,752	8.709	1353	811
Pilatus PC-24	8,300	7.665	893	724
Embraer Phenom 300	8,150	7.65	956	677
Beechcraft 1900D	7,765	7.605	1036	853
Cessna Citation CJ4	7,761	7.103	1039	896

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

The concept of relative velocity is very important in order to generate considerable lift force to drive the aircrafts. The lift force is directly proportional to square of the relative velocity of aircraft and wind speed.
Now we will see seperately about wind ground and air speed.

For a reference point situated on the ground, the air moves wrt. reference point at the wind speed. We observed that the wind speed is a vector and has both a magnitude and a direction.

For a reference point picked on the ground, the aircraft moves relative to the reference point at the ground speed. Ground speed is also a vector quantity.

Airspeed is the relative velocity between aircraft and the wind speed.

Airspeed= Ground Speed - Windspeed

3. Importance of towing of Aircrafts at airports:-

(i) The primary advantage of using the toeing reduces the cost of fuel. The aeroplane fuels are costlier than the general gasoline being used in day to day vehicles, compels us to do an economic use of it.

(ii) The other tempting reason is that towing allowing us to park the giant aircrafts as way the situation demands. It become so burdensome and hurculean task to park or settle the huge aircrafts at desired locations. So, this demand is fulfilled by just towing them.

(iii) Planes move by pulling or pushing themselves through the air, rather than by applying engine power to spin their wheels, and thus have no forward or reverse gears.
So therefore these important needs , the engine power is not used to move aircrafts on ground.

The whole ground locomotion process is achieved by the help of towing tractors. These are especially designed high power tractors used for locomotion of huge aeroplanes.

When the plane is ready to take off, they are taken to the initial region of the runway to initiate the take-off process.

From there, the planes take a giant speed up to achieve the minimum take off speed, an finally take off is achieved.

5. Discussion on Take-off Power:-

On take-off, in order to position the aircraft to a safe height away from terrain and obstacles (i.e. a flight path of maximum height and minimum ground distance desired), the engine thrust is set to a very high ‘take-off power’ setting (although this is not necessarily full power) and the aircraft attitude is pitched up to maintain a specific speed. Once the ‘safe height’ is reached the engine thrust can therefore be reduced to a more appropriate and efficient setting and the aircraft flight path can be changed to a more appropriate flight path.

Discussion on tyre design of aircrafts:-

The tyres play a crucial in the whole mechanical mobility process of aircrafts. The tyres of aircrafts are designed more metaculously to withstand a very high pressure and temperature gradients. It also signifies its importance in the braking process as more advanced and highly engineered applies an efficient braking. The number of wheels for any aircrafts depends on the amount of weight it has to carry.

A simple schematic of aircrafts tyres are given below:-

Aircraft tires generally operate at high pressures, up to 200 psi (14 bar; 1,400 kPa) for airliners, and even higher for business jets.

Each of the twelve Boeing 777-300ER main tires is inflated to 220 psi (15 bar; 1,500 kPa), weighs 120 kg (260 lb), has a diameter of 134 cm (53 in) and is changed every 300 cycles while the brakes are changed every 2000 cycles.Each tire is worth about $5,000. Aircraft tires generally operate at high pressures, up to 200 psi (14 bar; 1,400 Kpa) for airliners, and even higher for business jets. The main landing gear on the Concorde was typically inflated to 232 psi (16.0 bar), whilst its tail bumper gear tires were as high as 294 psi (20.3 bar). The high pressure and weight load on the Concorde tyres were a significant factor in the loss of Air France Flight 4590.

Braking system of aircrafts on landing:-

The braking is a very important aspect in the engineering of aircrafts. The very high speed of aircrafts need to be stumbled to zero through a very careful process. With these things keeping in mind , three types of braking system are included in aircrafts which are:-

(i) Ground spoilers (ii) Disc brakes (iii) Reverse thrust

Ground spoilers

After touching down, the spoilers take a fully vertical shape by standing straight and out of the wings. In many cases, an additional ground spoiler, which is closer to the fuselage, is used as well. The main application of an aircraft spoiler is to cut the lift and provide some sort of a drag. The rest is done by disc brakes and reverse thrusters.

Disc Brakes

These are the brakes applied by the wheels just as common in general cars and bikes. But here comes a crucial methodology ,that at which extend it should be applied. Applying it more than a certain extent can skid the aircrafts ehich can result into overturning of planes.

Reverse Thrust

In this process the turbofans are just compelled to run in reverse direction with the power of engine. This technique is generaly not applied in small or lite aircrafts but it become a very crucial method in heavy crafts , such as Boeing, due to requirement of high braking force.

So by cummulative mixing of these threee type of technique proper braking is achieved in aircrafts.

Part B:-

The towing of aeroplane is done by specialy designed high torque tractors to move the aeroplanes at their respective places.
We are going to calculate the force which will be applied by the tractors for a general aeorplane to tow it.

Assumptions:-

(i) The road is horizontal, i.e gradeability is zero.

(ii) Coefficient of friction remains constant during the motion for tractor as well as aeroplane.

(iii) The aeroplane will be moved very slowly and which constant velocity,i.e acceleration=0

Let for our general case, BOEING 377 be our test vehicle

For finding out the values of different parameters , we have gone through various documentations and finally these are the sources from which informatopn have been gathered:-

Weight of aeroplane,M1= 67000 Kg
Weight of tractor,M2=54000 Kg
Friction coeff. between aeroplane and road,`mu1`=0.018
friction coeff, between tractor and road `mu2`= 0.

Friction fprce between road and aeroplane,F1 = `mu1*M1*g`

=`0.018*67000*9.8`

Therefore, F1= 11,818.80 N

Friction force between road and tractior,F2= `mu2*M2*g`

= `0.02*54000*9.8`

Therefore, F2= 10,584 N

Now, for air drag force:-

We know ,

`F3= 0.5*rho*A.v^2.Cd`

For general porposes, thses locomotives run at a average speed of 3 m/s.

F3= `0.5*21*1.225*0.15*3^2=17.364 N`

F4= `0.5*6*1.225*0.44*3^2=14.553 N`

`Total force required, F= F1+F2+F3+F4`

`F= 22,434.717 N`

Power= force*velocity

`22434.717*3`

P= 67304.151 Watt

6. b

A simulink model is framed in simulink by using constant , add,product and display blocks. The same equations are used to frame the model as are ued above.

7 Design of a simulink model for Electric powertrain of the towing tractor

To develop a heavy duty electric tractor we simply move from running DC motor on a regulated dc upply.The H-bridge controller is used b to regulate the the running of dc motor.