Project-1: Powertrain for aircraft in runways
Part A:
1. Search and list out the total weight of various types of aircrafts.
Below is a table of some of the most common aircraft flying around our skies today:
| Aircraft | Empty Weight | Max Fuel Weight | Max Cargo Weight | Max Gross Takeoff Weight |
|---|---|---|---|---|
| Antanov AN-225 | 628,000 lbs 285,000 kg |
661,400 lbs 300,000 kg |
417,000 lbs 190,000 kg |
1,411,000 lbs 640,000 kg |
| Airbus A380-800 | 611,000 lbs 277,000 kg |
560,000 lbs 254,000 kg |
185,000 lbs 84,000 kg |
1,268,000 lbs 575,000 kg |
| Boeing 747-800 | 485,300 lbs 220,100 kg |
427,400 lbs 194,000 kg |
295,000 lbs 134,000 kg |
987,000 lbs 447,700 kg |
| Airbus A340-500 | 370,000 lbs 168,000 kg |
386,300 lbs 175,000 kg |
119,000 lbs 54,000 kg |
820,000 lbs 372,000 kg |
| Boeing 777-300ER | 370,000 lbs 168,000 kg | 320,800 lbs 145,500 kg | 148,000 lbs 67,100 kg |
776,000 lbs 352,000 kg |
| Boeing 787-10 | 299,000 lbs 135,500 kg | 223,500 lbs 101,500 kg |
126,300 lbs 57,300 kg | 560,000 lbs 254,000 kg |
| Boeing 737-900 | 98,500 lbs 44,700 kg |
53,100 lbs 24,000 kg |
44,600 lbs 20,200 kg |
187,000 lbs 85,000 kg |
| Airbus A320-100 | 93,900 lbs 42,600 kg |
48,700 lbs 22,100 kg |
44,000 lbs 20,000 kg |
150,000 lbs 68,000 kg |
| Embraer 190 | 63,500 lbs 28,800 kg |
28,600 lbs 13,000 kg |
28,800 lbs 13,100 kg |
106,000 lbs 48,000 kg |
| Gulfstream G650 | 54,000 lbs 24,500 kg | 48,200 lbs 21,900 kg | 6,500 lbs 3,000 kg |
100,000 lbs 45,200 kg |
| Bombardier CRJ900 | 48,160 lbs 21,850 kg | 19,600 lbs 8,900 kg | 6,075 lbs 2,760 kg | 80,500 lbs 36,500 kg |
| Bombardier Q400 | 39,300 lbs 17,800 kg |
11,700 lbs 5,300 kg |
18,700 lbs 8,500 kg |
62,000 lbs 28,000 kg |
| Learjet 75 | 13,900 lbs 6,300 kg | 6,060 lbs 2,750 kg |
2,900 lbs 1,300 kg |
21,500 lbs 9,700 kg |
| Cessna Citation CJ4 | 10,300 lbs 4,700 kg | 5,800 lbs 2,650 kg |
2,200 lbs 1,000 kg | 17,200 lbs 7,800 kg |
| Beechcraft King Air B100 | 7,100 lbs 3,200 kg | 3,200 lbs 1,450 kg |
4,150 lbs 1,900 kg |
12,000 lbs 5,400 kg |
| Diamond DA50 RG | 3,200 lbs 1,450 kg |
340 lbs 155 lbs |
1,250 lbs 560 kg |
4,400 lbs 2,000 kg |
| Beechcraft Bonanza G36 | 2,500 lbs 1,150 kg | 450 lbs 200 kg |
850 lbs 385 kg |
3,660 lbs 1,660 kg |
| Cessna 206H | 2,200 lbs 990 kg |
520 lbs 235 kg |
1,150 lbs 520 kg |
3,600 lbs 1,630 kg |
| Sirrus SR22 | 2,250 lbs 1,000 kg | 485 lbs 220 kg |
950 lbs 430 kg |
3,600 lbs 1,630 kg |
| Cessna 172R | 1,700 lbs 770 kg |
380 lbs 170 kg |
560 lbs 255 kg |
2,425 lbs 1,100 kg |
2. Is there any difference between ground speed and air speed?
There are two different types of speed when talking about airplanes – ground speed and airspeed.
While ground speed is the airplane’s speed relative to the surface of the Earth, airspeed – at least true airspeed – is its speed relative to the air it is flying in.
Below, I will explain the two types of speed in more detail, as well as talk about the four types of airspeed that are commonly used.
Airspeed vs. Ground Speed
As mentioned above, true airspeed is simply the speed at which an aircraft is moving relative to the air it is flying in. As such, it’s also the speed at which the air is flowing around the aircraft’s wings.
Ground speed, on the other hand, is the aircraft’s speed relative to the ground. One thing that should be noted here is that it’s its horizontal rather than vertical speed – an aircraft climbing completely vertically would have a ground speed of zero.
In other words, while airspeed is what determines whether there is enough airflow around an aircraft to make it fly, ground speed is what determines how fast an aircraft will get to its destination.
Wind’s Effect on Ground Speed
The relationship between airspeed and ground speed is fairly simple. Ground speed is simply the sum of airspeed and wind speed.
If the aircraft is flying in the same direction as the wind is blowing, the aircraft experiences tailwind, and its ground speed is higher than its airspeed. On the other hand, if the wind is blowing against the direction the aircraft is traveling in, the aircraft experiences headwind, and its ground speed is lower than its airspeed.
To give you an actual example, imagine an aircraft that cruises at an airspeed of 500 miles per hour that has to cover a ground distance of 2,000 miles.
If there is no wind at all, then both the aircraft’s airspeed and ground speed would be the same 500 miles per hour, and the aircraft would reach its destination in four hours.
If there was a 100 miles per hour headwind – wind blowing against the aircraft’s direction of travel – the aircraft would still be traveling at an airspeed of 500 miles per hour. However, its ground speed would be just 400 miles per hour (100 miles per hour slower than its airspeed). And as such, it would take the aircraft five hours to reach its destinations.
Finally, if there was a 100 miles per hour tailwind – wind blowing in the same direction as the aircraft’s travel – the aircraft would still be traveling at an airspeed of 500 miles per hour, but its ground speed would be 100 miles faster. And, at 600 miles an hour, the aircraft would reach its destination in just three hours and twenty minutes.
The above is the reason why some flights go “out of their way” to avoid headwinds or catch tailwinds. And, why some flights might appear to be traveling at “supersonic speeds,” even though their airspeed – the speed that would actually matter in determining whether or not the flight truly is supersonic – is subsonic.
3. Why is it not recommended to use aircraft engine power to move it on the ground at Airport?
Airplanes use their jet engines for taxing just like for flying. And they have no additional engines to power the wheels.
The wheels of an airplane rotate freely as the plane depends on the thrust from its jet engines to propel it forwards or backwards.
However despite the engines being capable of powering the airplane's 'pushback' from the terminal/gate, they are rarely used for this purpose.
'Tug vehicles' are generally used to pushback the plane away from the terminal prior to take-off. These are specialized vehicles with high torque and heavy mass.
The reason for the use of tugs is that the alternative (i.e., pushback using the jet-engine's thrust-reverser) is a highly fuel-consuming and noisy process. It is also potentially hazardous for personnel working near the plane.
Once the plane is tugged away to a safe distance from the terminal, the plane begins using its own engine power for the rest of the taxing process.
We are assuming passenger commuter aircraft and large airliners. Aircraft 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.The only time they use external assistance is to pushback from the gate. Large turbofan aircraft need help to reverse via a pushback tractor. The engines are turned on after the pushback is complete and the aircraft can move forward under It’s own power.To be accurate, some aircraft can reverse using a variable pitch prop or thrust reversing on turbofans but generally reversing is done via pushback tractors. Blasting jet reversers blows debris In the crowded gate area creating foreign object damage hazard for adjacent aircraft, gate equipment or terminal waiting area glass windows.
Large jetliner passenger aircraft are pushed away from the passenger loading gate by a tug. This is a machine with wheels, some form of power such as an internal combustion engine, or a battery. When the large aircraft is being pushed backwards, there is someone on each wing tip. They are looking out to make sure nothing will touch the aircraft. The whole idea is to avoid expensive crunching noises.
I hold mechanic certificate, trust me, when you hear a crunching sound and a modern airliner is involved, it will be expensive beyond your wildest imagination.
An airliner might be pushed backwards off the ramp, and onto a taxi way. Then the engines can start, the aircraft can proceed forward under its own power. A twin engine jet might not start the second engine until it is close to takeoff runway, clearance is certain.
A piston engine can run at idle power using only a small percentage of the fuel consumption used at operating power level. Turbine engines such as modern airliner turbo fan engine burn a lot of fuel once they are started. So, there is a lot of incentive to not run them until it is time to fly.
How to get off the gate? Three thugs, a tug, and a tow bar! No disrespect to ramp guys, they have a hard job, low pay, they are in the weather and the noise.
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.
where dV/dt is the acceleration along the runway and m is the mass of the vehicle.
Rotation Velocity, VR
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, VR, 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 VR is usually determined as being 1.1 × VSTALL or 1.05 × VMIN CONTROL
which ever is greater. Stall speed, VSTALL, 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 CL(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,
Minimum control speed, VMC 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 VR calculations based on stall speed or minimum control speed.
As well as rotation speed there are other safety considerations as shown in the following Figure.
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 * Vstall. Below this speed aircraft cannot attain sufficient climb rate.
Transport Aircraft :
VR must not be less than V1
VR must be greater than 1.05 * VMC
VR 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.
It 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=VR 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=VR
Drag
The resistance to motion due to the air viscosity will give a drag of
where CD can be considered constant and calculated using the formula shown in the previous section
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=μ(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, (
, can be used for the take-off run. This average acceleration can be found at the point where,
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 (tR) taken to get from 0 to VR. For a constant
Rearranging leads to a relatively simple calculation to predict distance to rotation point.
Obstacle Clearance Distance
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
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.
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 VR. 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 VTD 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.
The negative acceleration or deceleration value will be based on friction coefficient for maximum braking and the value of reverse thrust (if available).
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.
- They support the weight of an aircraft while it is on the ground.
- It also provides the necessary traction for braking and stopping of an aeroplane.
- Tyre also helps to absorb the shock of landing and provide cushioning the roughness of takeoff, 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 high 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.
Aircraft tyres may be Tube-type or Tubeless
Tubeless tyres are more advantageous over tube-type. There is no longer the use of tube-type tyre in recent aviation. Nowadays all airliners are using tubeless tyres. Tubeless that are meant to be used without a tube has the word TUBELESS on the sidewall of the tyre.
Aircraft tyres may be Radial or Bias
Almost all airliners are using Radial tyre. Bias is an older design, and it’s mainly used for road vehicles. Radial tyres have the word RADIAL on the sidewall. Radial tires are more expensive than bias-ply tyres. Radial tires are in demand because of their lower life cycle cost and long term value.
Aircraft Tyre Standards
Each tyre manufacturer produces their own designs of the tyre with different tread patterns, wear patterns, wear limits etc, and in some cases, different names for the same part of the tyre. But all tyres must meet the approved standards before being accepted by the FAA or CAA.
All commercial aircraft tyres approved under FAA Requirement Technical Standard Order (TSO) C62. A TSO is a minimum performance standard for specified materials, parts, and appliances used on civil aircraft.
PART B:
The tractive force or the pulling force required by the tow will be
Pulling force = Drag force+friction force+rolling resistance
Below data is taken
range of weight = 200 - 250000kg
Area range = 50-100m^2
Coefficent of rolling resistance : 0.02
Coefficient of friction :0.01
Below matlab script is used for calculating pulling force required:
clc
clear all
%Input
c=0.02; %rolling resistance coefficient
m=linspace (200,25000,2000); % weight
g=9.81; % acceleration of gravity
u=0.6; % friction coeefficient
A=linspace(50,100,2000); %area of planes
v=2; %velocity
Cd=0.8; %Drag coefficient
%calculation
F1=c*m*g; %rolling rsistance force
F2=u*g*m; %friction force
F3=0.5*A*1.2*Cd*(v)^2;
Ft=F1+F2+F3;
plot(m,Ft);
title('Tractive force vs Weight of planes')
xlabel('Weight of planes(kg)')
ylabel('Tractive Force')
axis([0 2000 0 20000]);
Results:
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.
Vechicle Drive : 2 hp, 36 vdc traction motor mated
Batteries: 6 Vdc,190 amp-hr
Vechicle speed,loaded: 3.6mph / 5.8 kph
Lift cradle capacity : 1500 lbs / 680kg
The matlab model is attached and the results obtained are shown below
The rsults for 1000 kg mass is 