Project-1: Powertrain for aircraft in runways

Powertrain for aircraft in runways

PART - A

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

AIRCRAFT	EMPTY WEIGHT	MAX FUEL WEIGHT	MAX CARGO WEIGHT	MAX GROSS TAKE-OFF WEIGHT
Antanov AN-225	285,000 kg	300,000 kg	190,000 kg	640,000 kg
Airbus A380-800	277,000 kg	254,000 kg	84,000 kg	575,000 kg
Boeing 747-800	220,100 kg	194,000 kg	134,000 kg	447,700 kg
Airbus A340-500	168,000 kg	175,000 kg	54,000 kg	372,000 kg
Boeing 777-300ER	168,000 kg	145,500 kg	67,100 kg	352,000 kg
Boeing 787-10	135,500 kg	101,500 kg	57,300 kg	254,000 kg
Boeing 737-900	44,700 kg	24,000 kg	20,200 kg	85,000 kg
Airbus A320-100	42,600 kg	22,100 kg	20,000 kg	68,000 kg
Embraer 190	28,800 kg	13,000 kg	13,100 kg	48,000 kg
Gulfstream G650	24,500 kg	21,900 kg	3,000 kg	45,200 kg
Bombardier CRJ900	21,850 kg	8,900 kg	2,760 kg	36,500 kg
Bombardier Q400	17,800 kg	5,300 kg	8,500 kg	28,000 kg
Learjet 75	6,300 kg	2,750 kg	1,300 kg	9,700 kg
Cessna Citation CJ4	4,700 kg	2,650 kg	1,000 kg	7,800 kg
Beechcraft King Air B100	3,200 kg	1,450 kg	1,900 kg	5,400 kg
Diamond DA50 RG	1,450 kg	155 lbs	560 kg	2,000 kg
Beechcraft Bonanza G36	1,150 kg	200 kg	385 kg	1,660 kg
Cessna 206H	990 kg	235 kg	520 kg	1,630 kg
Sirrus SR22	1,000 kg	220 kg	430 kg	1,630 kg
Cessna 172R	770 kg	170 kg	255 kg	1,100 kg

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

Ground Speed

The velocity of the object measured relative to the ground is called the ground speed. Again, this is a vector quantity.

Airspeed

The important quantity in the generation of lift is the relative velocity 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 the vector difference between the ground speed and the wind speed. 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.

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

Airplanes get towed or pushed by tractors and tugs when jet blast from their engines or prop wash from their propellers will create a hazard to nearby buildings, ground crew, other aircraft & ground handling equipment. Moving aircraft from hangers to aprons does not require pilots if being towed.

When aircraft go in for maintenance or are not required for flight by far the easiest way is to send a tow vehicle to the gate and have the maintenance team tow the aircraft to the hanger or parking areas.

By doing this it allows time freedom for the airline, the airport ground controllers, and the maintenance team as pilots are not required to start the engines.

When an aircraft is being towed, trained and authorized maintenance personnel will be on board the aircraft to release and control the brakes to aid the tow vehicle operator.

Some airplanes will need to have their APU (Auxiliary Power Unit) started by the maintenance personnel so that electrical and hydraulic systems can be operational to operate the braking systems and aircraft lights when towing at night.

Moving an airplane away from the gate with a Tug or Tractor is by far the safest way around today’s busy airports.

Jet blast and propwash from an aircraft can easily send loose items airborne, turn grains of dust into projectiles and create debris that can be easily ingested into the multi-million-dollar engines.

By allowing the driver and apron walkers to reverse every aircraft out of their parking spots it allows many pairs of eyes to constantly scan the area for potential collisions and negates the need for engines to be running while in the close confines of a busy apron.

Tugs and tractors make manoeuvring an aircraft a simple and safe operation and that is why it is very; very rare you will ever see a large aircraft reverse using the power of its own engines.

Reference: Why Do Airplanes Need to Be Towed & Pushed? – Pilot Teacher

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

Pushback 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 plane's nose gear with a tow bar, or could be a "wheel-lift" tug. These tugs cradle the nose gear, then lift it up before moving the plane. That gives the tug driver control over the plane's direction during pushback.

New taxi technologies are appearing, like pilot-controlled tugs, and electric motors mounted to the plane's landing gear. Both promise to save fuel, and reduce airport noise.

Depending on the size of aircraft to be towed the tow vehicle can be as small a quad and be as large as a bus.

The heavier the aircraft is to be moved, the heavier and more power the tow vehicle has to have.

Dedicated tractors or tugs are used to hook onto the airplanes to allow for controlled tows.

By far the most common sight around the world’s airports is the Tug and a Towbar. The towbars are connected to the axle of the aircraft’s front landing gear and then to the front or rear of the Tug. Once connected, the driver of the Tug can easily manoeuvre the aircraft from the gate and out onto the taxiway.

Becoming more popular around the large airports is the Towbar less Tug.

These tractors are purposely designed to be able to move any aircraft without the need for any additional apparatus like a towbar.

These Tugs come in a range of sizes to suit all of the larger aircraft flying today and they are heavy, powerful, moving machines, and due to their speed at moving aircraft, they are a popular choice among ground crews and airport ground controllers.

They work by having the driver reverse up to the nose landing gear of the airplane.

The airplane’s front tires are then placed against a stop and a bar or locking arm closes in around the other side of the plane’s tires.

Once secured, the Tug will lift the entire landing gear tires off the ground.

This then allows the Towbar less Tug to easily move the airplane around once the pilot/s have released the brakes.

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

TAKE OFF AND LANDING

Take Off Performance

Take off performance 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 – Thrust of propulsion system pushing aircraft along runway.

D – Aerodynamic Drag of vehicle resisting the aircraft motion.

F – Rolling resistance friction due to the contact of wheels or skids on the ground.

During take-off run the imbalance in these forces will produce an acceleration along the runway.

`(dV)/dt = (T - D - F)/m`

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

Rotation Velocity, V_R

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 climb lift and it will ascend from the ground. The determination of this safe flying speed or rotation speed, V_R, is a critical factor in determining take-off performance.

Take-off rules vary slightly depending on the aircraft category. Small commuter aircraft should be considered as meeting FAR 23 rules, transport category aircraft should comply with FAR 25 rules.

Small commuter aircraft :

For safety reasons V_R is usually determined as being 1.1 × V_STALL or 1.05 × V_{MIN CONTROL}

which ever is greater. Stall speed, V_STALL, is the lowest speed that the aircraft can be flown before the airflow starts to separate from wings as the angle of attack becomes too great. The wing is assumed in this case to be in take-off configuration or "clean".

It can be calculated based on knowledge of the aircraft take-off configuration and hence the maximum achievable lift coefficient C_L(max). As shown in the previous section , to maintain level flight the lift produced must equal the weight, hence the stall speed can be calculated as,

`Vstall = sqrt(W/(1/2 * Clmax*rho*S))`

As well as rotation speed there are other safety considerations as shown in the following Figure.Minimum control speed, V_MC is a more complex calculation and requires knowledge of the stall characteristics of the tailplane and elevator.

For conventional aircraft there is only a small difference between V_R calculations based on stall speed or minimum control speed.

V₁- Abort decision speed. Below this speed the take-off can be safely aborted. After this there will not be sufficient runway length to allow the aircraft to decelerate to a stop.

V₂ – Safe climb speed. V₂ must be no less than 1.2 * V_stall. Below this speed aircraft cannot attain sufficient climb rate.

Transport Aircraft :

V_R must not be less than V₁
V_R must be greater than 1.05 * V_MC
V_R must be set so that aircraft achieves V₂ before reaching a height of 35ft above the runway surface.

Aircraft must climb at a minimum gradient to avoid obstacles at the end of the runway. With engine failure on multi-engined aircraft, this speed should still be achievable.

Thrust

The thrust of gas turbine or turbofan engines will be relatively constant during take-off. A good assumption is to use the manufacturer's values for maximum static thrust for take-off calculations.

The thrust of a propeller driven aircraft can be found from the given shaft horsepower data for the engine and the use of the equations using propeller efficiency given in the previous section.

`T = (Pshaft * eta) / V`

DragIt is critical to correctly estimate the propeller efficiency for the particular aircraft velocity along the runway. At V=0 the efficiency is 0 so the above equation makes no sense. At V=V_R the efficiency will be in the range 50% to 80% depending on the type of propeller system used and the thrust value at this point will be easy to obtain. In practice, the thrust obtained throughout the take-off roll is roughly constant so this end point value is a good approximation from V=0 to V=V_R

The resistance to motion due to the air viscosity will give a drag of

`D = Cd* 1/2*rho*V^2*S`

where C_D can be considered constant and calculated using the formula shown in the previous sectionρ.V2S

Although Drag Coefficient is constant, Drag will increase in proportion to the square of velocity.

Rolling Resistance

The friction between aircraft and runway will be proportional to the normal force exerted by the aircraft on the runway.

`F = mu*(W-L)`

The normal force will be the difference between Weight of aircraft and Lift, the friction coefficient will be typically of a magnitude of 0.02 for a standard tarmac runway.

Average acceleration and distance to rotation

The rate of change of velocity can be predicted at any point on the take-off roll by substituting results for T, D and F into the initial equation for dV/dt. The subsequent velocity at any point can be found by integrating this resulting equation and the distance traveled found by then integrating the velocity.

Typically acceleration will be dominated by the drag component as thrust, weight and friction are relatively constant during this period. This leads to the result shown where acceleration is inversely proportional to velocity squared.

Due to the quadratic nature of acceleration change, an average value, (`(dV)/dt(avg) = bara`

, can be used for the take-off run. This average acceleration can be found at the point where,

`V = (Vr)/sqrt2`
VR=This average acceleration can be used to simplify calculations and the take-off run can be calculated as an equivalent constant acceleration over the complete period of time (t_R) taken to get from 0 to V_R. For a constant acceleration take-off calculation,

`Vr = bara*Tr` and distance travelled `S = 1/2*bara*Tr^2`

Rearranging leads to a relatively simple calculation to predict distance to rotation point.tR2

`S1 = 1/2 * (Vr^2) /2bara`

s2=From the rotation point, the end of the runway can be defined by the requirement to clear a 35ft obstacle at the end. During rotation it can be assumed that any residual excess thrust is absorbed in overcoming the lift induced drag as the aircraft begins to climb. Acceleration reduces and a constant flight speed during this climb phase can be assumed. The distance along the ground from rotation point to obstacle clearance point with thus be,Obstacle Clearance Distance

`S2 = 35/tan(theta)*ft`

This distance estimate will require knowledge of the climb gradient which can be calculated by using the methods in the following section on Climb and Descent. ft

Take-Off (Balanced) Field Length

The required length of runway will be the sum of the distance required to get to rotation speed and the extra length required to clear a 50ft obstacle or the extra length required to allow for rapid braking if the pilot decides to abort take-off at the decision speed V₁.

This length will typically be considerably longer than the distance required to achieve rotation (flying) speed. A rough approximation is that runway total length is around 2 x s₁

Distance to V₁ can be calculated in a manner similar to that shown for V_R. The calculation of braking distance will require knowledge of the maximum braking friction coefficient that can be generated by the aircraft. This information should be available from manufacturer's data. Braking distance calculations should also be done without any assumption of reverse thrust from engines as during a take-off abort, engine power may not be available.

Landing

The landing run can be calculated in a similar fashion to the take off distance. The aim is again to minimise the distance.

The touch down velocity should be approximately the stall speed of the aircraft in landing configuration. This will be achieved by a pitch manoeuvre during the flare portion of the approach which which increase drag an decelerate the aircraft to minimum flying speed.

The deceleration on the landing roll from V_TD to V₀ will be accomplished by braking and reverse thrust. This can be solved by the average acceleration approach that was used to estimate the take-off roll.

`(dV)/dt = (T - D - F)/m`

The negative acceleration or deceleration value will be based on friction coefficient for maximum braking and the value of reverse thrust (if available).

Reference: Resources : Aerodynamics for Students (aerodynamics4students.com)

Design and Construction of Aircraft tyre

Carcass plies are used to form the tire. They are sometimes called casing plies. An aircraft tyre is constructed for the purpose it serves.

1. They support the weight of an aircraft while it is on the ground.
2. It also provides the necessary traction for braking and stopping of an aeroplane.
3. Tyre also helps to absorb the shock of landing and provide cushioning the roughness of take-off, roll-out, and taxi operations.

Unlike an automobile or truck tyre, it does not have to carry a load for a long period of continuous operation. However, an aircraft tyre absorbs the high impact loads of landing, and also it’s operating at hig speeds for a short time when required.

Let’s go through some technical aspects of the tyre.

Retreading – Retreading is methods of restoring a worn tyre by renewing the tread area or by renewing the tread area plus one or both sidewalls. Repairs are included in the tyre retreading process.

Load Rating – Load rating is the maximum permissible load at a specified inflation pressure.

Ply Rating – Ply Rating is used to identify the maximum recommended load rating and inflation pressure for a specified tyre. It is an index of tyre strength.

Speed Rating – The speed rating is the maximum takeoff speed to which the tyre has been tested.

Skid Depth – Skid depth is the distance between the tread surface and the deepest groove as measured in the mould.

Aircraft tyres must have an approved speed and load rating and have sufficient clearance when retracted through landing gear to allow for tyre growth. Tyre growth is the increase in the size of the tyre due to centrifugal forces at high speed.

Reference: Aircraft Tyre | Design and Construction - AviationHunt

ROLLING RESISTANCE

The force to move an airplane on an absolute flat runway with 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

Fr = c mr ag (1)

where

Fr = rolling resistance (N, lbf)

c = rolling resistance coefficient

mr = rolling mass (kg, lbm)

ag = acceleration of gravity (9.81 m/s2)

Example - Force required to pull an airplane at constant low speed on tarmac

If the mass of a plane is 40000 kg 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

Fr = 0.02 (40000 kg) (9.81 m/s2)

= 7848 N

This force can be delivered by a pulling truck or - as occasionally seen - by a number of pulling people. Enough weight for the pullers are required to create friction forces against the tarmac that equals the plane rolling resistance.

The friction force for the puller can be calculated to

Ff = μ W

= μ mf ag (2)

where

Ff = frictional force (N, lb)

μ = friction coefficient

W = weight (N, lbf)

mf = friction mass (kg)

Reference: Pulling an Airplane (engineeringtoolbox.com)

TYRE PRESSURE

Part B:

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

Mototok: Electric tugs for aircrafts over 95 t

Towing vehicel for study:

SPACER 250

Tugs for aircraft above 95T.

Dimensions:

width	3988mm
length	3999mm
overall height	879mm
weight	13000kg
wheel radius (r)	0.5m
speed	4km/h

AIRBUS 350

Length	60.54m
Height	17.10m
weight	190tonnes
width	43.90m

TO CALCULATE POWER RQUIRED BY TOWING VEHICLE;

Force acting on vehicel ;

Ft = Frr + Fad;

Ft = tractive force;

Frr = rolling resistance force;

Fad = air drag force;

now,

ROLLING RESISTANCE;

`Frr = murr * m * g`

`murr =`co-efficient of rolling resistance = 0.02`

m= mass of vehicle + mass of aircraft;

m = 13000 + 190000;

m = 2,03,000 kg.

g = gravity = 9.81m/s2

calculating Total rolling resistance ;

`Frr = 0.02 * (203000)*9.81`

`Frr = 39828.6`;

AIR DRAG FORCE;

`Fad = 1/2 * rho * A * Cd * V^2`

`rho = 1.225 (kg)/(m^3)`

A = front area of aircraft

A = height * width;

A = ( 15.80 * 43.90 );

`A = 693.62 m^2`

Cd = drag co-efficient = 0.0265;

v = velocity of vehicle = 1.2m/s;

calculating Total AIR DRAG ;

`Fad = 1/2 * rho * A * Cd * V^2`

`Fad = 1/2 * 1.225 * 693.63 * 0.0265 * (1.2)^2`

`Fad = 16.2122` ;

CALCULATING TOTAL TRACTIVE FORCE = Frr + Fad ;

Ft = 39828.6 + 16.211

Ft = 39298.811

now,

calculating TOTAL POWER REQUIRED BY VEHICEL;

`Pout = Ft * V`

`Pout = 39298.811 * 1.2`

`Pout = 47158.5732` watts`

`Pout = 47.2 kw`;

THE POWER REQUIRED BY VEHICLE IS 47.2 KW.

calculating TOTAL TORQUE REQUIRED BY VEHICEL;

`T = Ft * r`

`T = 39298.811 * 0.5`

`T = 19649.4055 Nm`

THE TORQUE REQUIRED BY VEHICLE IS 19649.4055 Nm.

SIMULINK MODEL OF THE SAME CALCULATION IS SHOWN BELOW;

7) A. Design an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfil 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.

Power rating of required motor : 47.2KW

torque required : 19649Nm

Energy = Pout * time ;

E = 47.2 * (10) ;

E = 472 KWH ;

Duty cyle = power input / power output

assuming battery speicfication ;

voltage = 400v

current = 150A

POWER INPUT = 400 * 150

POWER INPUT = 60kW

duty cycle = 47.2/60

DUTY CYCLE = 0.7866

%DUTY CYCLE = 78%

Also, Design the parameters in excel sheet.

EXCEL CALCULATOR:

SIMULINK MODEL OF POWERTRAIN: