Project-1: Powertrain for aircraft in runways
Aim : To study and design of electric powertrain for aircraft towing application.
Explanation :
1. Total weight of various types of aircrafts.
Aircraft designs have come a long way since the Wright Brothers made their first flight near Kitty Hawk, North Carolina. Computer Technology is allowing manufacturers to develop planes that are more reliable, aerodynamic, and powerful than ever. When you make the comparisons and consider the strengths and limitations of each type, it is easier to decide which airplane you might like to fly or fly in on a temporary or permanent basis.
Fig.1.1.
Jumbo Passenger Jets
Boeing 747 8 Intercontinental - Max Take Off Weight:2,138,448 Kg - 4,714,422 lbs,
Max Payload:75,000 Kg - 165,345 lbs
Mid Size Passenger Jets
Airbus A350-1000 - Max Take Off Weight:316,000 Kg - 696,654 lbs
Max Payload:68,000 Kg - 149,913 lbs
Light Passenger Jets
Bombardier CRJ 1000 - Max Take Off Weight:202,381 Kg - 446,169 lbs
Max Payload:12,156 Kg - 26,799 lbs
Passenger Turboprops
Bombardier Q400Max - Take Off Weight:142,196 Kg - 313,485 lbs
Max Payload:8,669 Kg - 19,112 lbs
Cargo Airplanes
Boeing Dreamlifter - Max Take Off Weight:1,770,282 Kg - 3,902,764 lbs
Max Payload:113,400 Kg - 250,002 lbs
Very Light Jets
Eclipse Aerospace EA500 - Max Take Off Weight:2,612 Kg - 5,758 lbs
Max Payload:1,027 Kg - 2,264 lbs
Light / Medium / Heavy Business Jets
Embraer Praetor 500 / Dessault Falcon 6X / Boeing BBJ MAX 9 - Max Take Off Weight:17,040 Kg - 37,566 lbs / 35,135 Kg - 77,459 lbs / 88,314 Kg - 194,697 lbs
Max Payload:1,325 Kg - 2,921 lbs / 30,025 Kg - 66,193 lbs / 17,427 Kg - 38,420 lbs
Military Jets
The Dassault Rafale - Max Take Off Weight: 24,500 Kg - 54,013 lbs
Max Payload: 8,000 Kg - 17,637 lbs
Twin Turboprops
Cessna 441 Conquest II - Max Take Off Weight:4,467 Kg - 9,848 lbs
Max Payload:1,039 Kg - 2,291 lbs
Aerobatic
The Zivko Edge 540T - Max Take Off Weight:3527 Kg - 7776 lbs
Max Payload:322 Kg - 710 lbs
Amphibious
The Bombardier 415 MP - Max Take Off Weight:96,671 Kg - 213,121 lbs
Max Payload:2,903 Kg - 6,400 lbs
Military Turboprops
The Airbus / EADS CASA C-295M - Max Take Off Weight:23,200 Kg - 51,147 lbs
Max Payload:9,250 Kg - 20,393 lbs
2) Ground speed and Air speed
Ground speed is how fast an aircraft is travelling, relative to a fixed point on the ground. Ground speed is how fast an airplane's shadow would move across the land. If there's a strong wind pushing an aircraft, that's reflected in the ground speed. Airspeed, in contrast, is how fast an airplane is really flying strictly under its own power, which is calculated by subtracting the wind speed from the ground speed.
Fig.1.2.
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.
Airspeed = Ground Speed - 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.
Suppose, we had an airplane that could take off on a windless day at 100 mph (liftoff airspeed is 100 mph). We are at an airport with an east-west runway that is 1 mile long. The wind is blowing 20 mph towards the west and the airplane takes off going east. The wind is blowing towards the aircraft which we call a headwind. Since we have defined a positive velocity to be in the direction of the aircraft's motion, a headwind is a negative velocity. While the plane is sitting still on the runway, it has a ground speed of 0 and an airspeed of 20 mph:
Airspeed = Ground Speed (0) - Wind Speed (-20) = 20 mph
The airplane starts its take off roll and has a constant acceleration a. From Newton's second law of motion, the ground speed V at any time t is:
V = a * t
And the distance d down the runway at any time is:
d = 1/2 * a * t^2
For a fixed length runway, this specifies the time to be used in the velocity equation. Let's assume that at 5000 feet down the runway, the velocity is 80 mph. Then the airspeed is given by
Airspeed = Ground Speed (80) - Wind Speed (-20) = 100 mph
And the airplane begins to fly. Now another pilot, with exactly the same airplane decides to take off to the west. The wind is now in the same direction as the motion and this is called a tailwind. The sign on the wind speed is now positive, not negative as with the headwind. The acceleration along the ground is the same, so at 5000 feet down the runway, the ground speed is again 80 mph. The airspeed is then given by:
Airspeed = Ground Speed (80) - Wind Speed (20) = 60 mph
This airplane doesn't have enough airspeed to fly. It runs off the end of the runway!
3) Why is it not recommended to use aircraft engine power to move it on the ground at Airport?
If this was your car, you’d put the gear into drive or first gear and press the accelerator. This would inject fuel into the engine and provide power to turn the wheels and off you’d go. However, aircraft wheels aren’t connected to the engines in any way. When the pilots squeeze the thrust levers forwards, the thrust generated from the engines pushes the aircraft forwards. The wheels merely turn under this forward power allowing the aircraft to move.
They use high-torque motors and the aircraft APU to allow the aircraft to move itself without starting its engines or requiring a tug to be connected and disconnected.
4) How an aircraft is pushed to runway when its ready to take off?
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.
Pushback is an airport procedure that results in an aircraft being pushed backwards away from an airport gate by external power, which is usually provided by pushback tractors or aviation tugs as shown in Fig.1.3.. But, These tow trucks / Push backs will only help move the aircraft to an area where the pilots can safely switch on the engines, not the runway! (Ground vehicles are rarely allowed to move on a runway (save for emergencies and inspection), and the aircraft is supposed to move to the runway by itself).
Then it is a common procedure to start the engines once it is cleared away from the terminal area and lined up on a taxiway (roads that lead to the runway).
Fig.1.3.
5. Take off power, Tyre design, Rolling resistance, Tyre pressure, Brake forces when landing.
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 high ‘take-off power’ setting (although this is not necessarily full power) and the aircraft attitude is pitched up to maintain a specific speed (usually in the region of V2 + 15kts). Once the ‘safe height’ is reached the engine thrust can therefore be reduced to a more appropriate (i.e. efficient) setting and the aircraft flight path can be changed to a more appropriate (i.e. efficient) flight path.
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.
Tyre design - Aircraft tires may be tube-type or tubeless. They support the weight of the aircraft while it is on the ground and provide the necessary traction for braking and stopping. The tires also help absorb the shock of landing and cushion the roughness of takeoff, rollout, and taxi operations. Aircraft tires must be carefully maintained to perform as required. They accept a variety of static and dynamic stresses and must do so dependably in a wide range of operating conditions.
Fig.1.4.
The typical airliner tire can handle a 38-ton load. It can meet the ground 500 times before needing a re-tread, a refresh it can take on seven times in its life.
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.
Fig.1.5.
Aircraft tires are classified in various ways including by: type, ply rating, whether they are tube-type or tubeless, and whether they are bias ply tires or radials. Identifying a tire by its dimensions is also used.
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.
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.
Today’s tyres are conductive. Aircraft tyres are manufactured with tread rubber with conducting compounds to permit earthing of static charges.
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.
Rolling resistance -
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)
Tyre pressure - Aircraft tires operate at such high extremes of pressure, load, and speed, their care and service and is critically important. The most important action an operator can take to prevent tire-related events is to maintain proper tire inflation pressure. Failure to keep aircraft tires properly inflated can lead to very serious consequences. The most serious of these is the structural failure of the tire.
Fig.1.6. Tire condition monitoring . (The Pressure values are in PSI)
Brake forces when landing - Too strong a braking produces skid, which is an undesirable condition, while weak breaking results in a long landing run. The shortest possible (optimal) landing run without any skidding requires a variable braking force to match the frictional force at all times. The maximum allowable braking force at any instant is a random variable as the ground-induced excitation and the frictional force between the wheel and the tyre are random in nature.
To be effective, braking action at any speed depends upon sufficient friction existing between the tyres and the runway surface achieved through freely rotating wheels.
Fundamental to this are:
- Achieving the maximum possible weight on braked landing gear assemblies
- The condition of the tyre tread
- The tyre inflation pressure
- The condition of the runway surface
Braking effectiveness will also be affected by the degree of brake wear, which must be within AMM (Aircraft Maintenance Manual) specified limits, and the manually or automatically applied brake pressure as modified by system protections. These ensure adequate wheel rotation exists and is maintained before the commanded brake pressure is transmitted to the brake units and takes effect. When brake temperature indication is available on the flight deck, it must be within prescribed limits before a take off roll is commenced so that effective braking is available if a take off is rejected. System faults or inappropriate use of brakes during a long taxi out can raise brake temperatures into cautionary ranges where a delay for take off may be required.
Anti Skid Units are fitted to the braking systems of all modern transport aircraft. They modulate applied brake system hydraulic pressure before it is transmitted to the actuators in the brake units so as to obtain optimum braking based upon wheel rotational speed data received at the Unit. A minimum wheel rotational speed must be detected before any brake application will be achieved to prevent tyre destruction resulting from a locked wheel and to guard against the risk of quaplanning on wet or icy runway surfaces.
Autobrake Systems provide pre-selectable rates of deceleration which usually vary between 3 and 6 knots per second constant deceleration rate. Selection of ‘Low’ autobrake on an aircraft equipped with thrust reversers will usually have the effect of delaying brake application to allow the thrust reversers to work efficiently in reducing the initial high ground speed. Maximum manual braking through the toe brakes can produce deceleration rates of up to 10 kts per second subject to the operation of anti skid units.
Modern landing gear assemblies on fixed wing transport aircraft are fitted with carbon brakes, although steel brakes may still be encountered on older aircraft types. The application techniques for the two types differ slightly. Caution is required if a current aircraft type rating includes aircraft types or type variants which have both brake types and both types are likely to be encountered. If this applies to either pilot, the subject should be included in pre-departure and approach briefings.
6. With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle.
Consider,
mass of a plane is 40000 kg
Drag Coefficient Cd = 0.25
Frontal Area = 2 m^2
Air Density = 1.225 Kg / m^3
Max speed v = 4.5 m/s
Rolling Resistance of aircraft (RRa)
RRa=CrrN
RR of aircraft =(40000*0.006*9.81) , RR coefficient of aircraft tire is assumed as 0.006
Rolling Resistance of aircraft = 2354 N
Rolling Resistance of Tug(RRt)
RRt=CrrN
RR of aircraft =(40000*0.001*9.81) , RR coefficient of tow vehicle is assumed as 0.001
Rolling Resistance of Tug= 392 N
Drag Force while Taxing(D)
D = 0.5 X C X Density X Frontal Area X (Velocity)^2
D =(0.5 x 0.25 X 1.225 X 2 x 4.5 x 4.5 )/2
D = 3.100 N
Total Force required by Tug to push the Aircraft = RRa + RRt + D
Total Force = 2750 N ........(1)
Assuming, Tug wheel dia = 0.4445 m ,
Torque Required by Tug to push the Aircraft = Force x Radius
= 2750 x 0.4445 = 1222 Nm........ (2)
Hence, Power required by Tug to push the Aircraft = Force X Velocity
= 2750 X 4.5
Power required by Tug to push the Aircraft = 12.4 KW .........(3)
7) Design an Electric powertrain with 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. Block diagram of powertrain.
The Basic Powertrain Circuit of one of the Electric Towing vehicle is as shown in the fig.1.7.
Fig.1.7.
Motor Type
Assume,
motor loss = 10%
Transmission loss = 20%
Gear Ratio : 1:10
Ftom Equtaion (2)
Motor Torque Required = 1222 / 10 = 122 Nm
Considering the losses,
Motor Torque Required = 122 + 122(0.1 +0.2) = 158.6 N-m
Power Required = 12.4 KW , From eq (3)
Assume Motor Efficiency = 90% & Designed run at 80 % full capacity
Motor Power = 12.4 / (0.9*0.8) = 17.22 KW or 23 HP
Now we have to choose a motor,
Model: HPM-20KW -- High Power BLDC Motor
1. Voltages: 72V
2. Rated power: 20-25KW
3. Peak power: 50KW
4. Speed: 4000rpm
5. Rated torque: 80 Nm
6. Peak torque: 160 Nm
7. Efficiency: >90%
8. Dimensions: 30x30x25cm
9. Weight: 39kg(20KW),27kg(15KW)
10: Cooling: liquid cooling
Duty Cycle = Required power / Rated Power = 17.22 / 50 = 0.34,
Hence Duty Cycle range is from 0 - 0.34 %
Energy Required from Battery,
Energy = Power X time
Energy = 17.22 X 15 X 60 ( Say, 15 min Taxing Time )
Energy = 15498 KJ = 15.498 MJ
Energy = 4.31 KWh
Consider the extra energy of 0.85KWh required to move to and fro from the parking place to aircraft.
Hence Energy = 4.31 + 0.85 = 5.16 KWh
To consider battery capacity for 20 trips, 5.16 X 20 = 103 KWh
Conclusion
1) Motor Required
- HPM-20KW -- High Power BLDC Motor
- Motor Torque - 160 N-m (Peak)
- Motor Power - 50KW (Peak)
2) Duty Cycle range is from 0 - 0.34
3) Battery Capacity = 103 KWh