Powertrain for aircraft in runways
OBJECTIVE - 1`to`Search and list out the total weight of various types of aircrafts.
OBJECTIVE - 2`to`Is there any difference between ground speed and air speed?
OBJECTIVE - 3`to`Why is it not recommended to use aircraft engine power to move it on the ground at Airport?
OBJECTIVE - 4`to`How an aircraft is pushed to runway when its ready to take off?
OBJECTIVE - 5`to`Learn about take off power, tyre design, rolling resistance, tyre pressure, brake forces when landing.
OBJECTIVE - 6`to`With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle.
OBJECTIVE - 7`to`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. Make all required assumptions. Prepare a table of assumed parameters. Draw a block diagram of powertrain.
OBJECTIVE - 1`to`Search and list out the total weight of various types of aircrafts.
The different terms represent different weights or masses. For each flight, the weights are taken into account for several reasons.
- Manufactre's Empty Weight (MEW) `to`Also called Manufacturer's Weight Empty (MWE) or Licensed Empty Weight. It is the weight of the aircraft "as built" and includes the weight of the structure, power plant, furnishings, installations, systems and other equipment that are considered an integral part of an aircraft. This excludes any baggage, passengers, or usable fuel.
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Zero Fuel Weight (ZFW)`to`This is the total weight of the airplane and all its contents (including unusable fuel), but excluding the total weight of the usable fuel on board. As a flight progresses and fuel is consumed, the total weight of the airplane reduces, but the ZFW remains constant. Maximum zero fuel weight (MZFW) is the maximum weight allowed before usable fuel and other specified usable agents (engine injection fluid, and other consumable propulsion agents) are loaded.
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Operating Empty Weight (OEW) `to`(Roughly equivalent to basic empty weight on light aircraft). It is the basic weight of an aircraft including the crew, all fluids necessary for operation such as engine oil, engine coolant, water, unusable fuel and all operator items and equipment required for flight but excluding usable fuel and the payload.
- Payload `to`It is the carrying capacity of an aircraft. It includes cargo, people, extra fuel. In the case of a commercial airliner, it may refer only to revenue-generating cargo or paying passengers.
- Maximum Takeoff Weight (MTOW) `to`This is the maximum weight at which the pilot of the aircraft is allowed to attempt to take off.
- Regulated Takeoff Weight (RTOW) `to`Depending on different factors (e.g. flap setting, altitude, air temperature, length of runway), RTOW or maximum permissible takeoff weight varies for each takeoff. It can never be higher than MTOW.
- Maximum Landing Weight (MLW) `to`This maximum weight at which an aircraft is permitted to land. The following image depicts takeoff weight components.
- Maximum Ramp Weight (MRW)`to`also called maximum taxi weight (MTW). It is the maximum weight authorized for maneuvering (taxiing or towing) an aircraft on the ground.
- Aircraft Gross Weight `to`It is the total aircraft weight at any moment during the flight or ground operation. This decreases during flight due to fuel and oil consumption.
Types of Aircraft `to`
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OBJECTIVE - 2`to`Is there any difference between ground speed and air speed?
Ground speed `to` It is when you are measuring the horizontal velocity of the plane relative to the ground. It’s the speed that is shown on the speedometer of a car. If your car suddenly started levitating for God knows what reason, the speedometer reading would not take into account your vertical velocity, so a plane could be moving vertically but still have zero ground speed. It is the velocity you have in any direction perpendicular to the vertical axis.
Air speed `to` is the velocity of the plane relative to the winds around it. It is like having two cars, one travelling at 100 km/h and the other at 60 km/h, both towards the right. These speeds are the ground speeds of the car. The air speed is wanting to know the difference in velocity between the two cars, which is 40 km/h, or -40 km/h depending on which car you are looking at. If the 100 km/h car represents the plane, and the 60 one the winds, then the difference is 40km/h. So you are essentially measuring how fast the plane is compared to the winds. And in the example above, if the car was travelling at 60 km/h to the left instead, the speed difference would be 160 km/h. It is the same thing for planes: when winds blow towards the plane, the plane’s air speed is bigger than the ground speed, because you are adding to the ground speed, and when winds blow in the same direction as the plane, the air speed is less than the ground speed.
In more technical terms, "ground speed is the velocity of the plane relative to a reference point on the ground. Wind speed is the velocity of wind with respect to that same reference point, and air speed is the velocity of the plane relative to that wind. Thus the air speed is defined as the vector difference between the ground speed and the wind speed. Also note that for whatever reason, the three quantities have both direction and magnitude; they are vectors despite their names".
Now the reason why ground speed is important is because it lets you know how quickly you could reach your destination, in the same way that the speed of a car lets you know that. The air speed is important because it is what allows the plane to fly. The Bernoulli principle states that if a fluid like air increases its speed, then its pressure will decrease. For a plane, the plane’s wings will affect the way air flows above and below them, which will cause a difference in pressure and thus cause a force, known as the lift force which is what makes the plane fly. The way the wings affect the air flow is determined by how quickly the plane is moving relative to the air, which is the air speed.
OBJECTIVE - 3`to`Why is it not recommended to use aircraft engine power to move it on the ground at Airport?
Although most aircraft are capable of moving themselves on the ground, it is not typically done due to the following reasons,
- The jet blast from the engines can damage the terminal building and nearby equipment. It can even throw away people standing nearby.
- Engines can blow sand and debris.
- Engines can suck in sand and debris which can cause serious damage to the engine.
- It is easier for the truck driver to see the surroundings better than the pilot sitting in the cockpit with limited windows.
- The pushback truck driver knows the airport better than the pilot.
- The plane engines are very loud. This noise can add up substantially when multiple planes are moving around the airport. It won't be a pleasant experience for people sitting inside the terminal, as most people don't even like the sound of powerful motor-bikes.
- It saves fuel.
- Modern planes are huge. It is difficult to navigate such gigantic structures through limited space. An A380 is eight stories high. The pilots won't be able to see anything directly in-front of them or anything behind the wings.
Actually Pushback is for moving the aircraft backwards from the gate, and towing or taxing is the procedure to move the aircraft. These machineries are called as aircraft tugs. Aircraft tugs are real modern marvels, moving aircraft weight hundreds of tons all day, every day. A true pushback tug aims to allow the operator to have excellent visibility of the attaching mechanism between airplane and tug.
while moving the aircraft in the reverse direction by applying the forward thrust, care msut be taen because reverse thrust is unsafe when stopped or at low speed. This is because the nose may lift, dropping the tail on to the Tarmac. The cause is the engines being at some considerable height from the ground and the aircraft centre of gravity being just ahead of the main (rear) wheel set. Additionally, once moving backwards, if the aircraft’s own brakes are used, the same can happen, i.e., the plane can sit on its tail.
OBJECTIVE - 4`to`How an aircraft is pushed to runway when its ready to take off?
Aircrafts are usually ‘pushed’ by ground vehicles known as tow vehicles such as tow trucks. It is used mainly to reverse an airplane from the terminal area, until it is in a position to move by itself. These tow trucks 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.
Aircrafts don’t have powered wheels, and hence cannot move in reverse (let alone forward by themselves). Imagine a tiny jet strapped to one of your toy cars; that’s essentially how airplanes work on the ground! They go where the engines push them towards. (Note: There is a mechanism known as reverse thrusters, which allow the engine air flow to be pushed towards the front, but that is highly risky maneuver, and is only used to reduce speed immediately after landing.)
Another reason to push back an aircraft to a safe area, is for safety reasons. The engines produce tremendous thrust from the exit, and an immense vacuum from the front; enough to suck in people, small vehicles, ground equipment etc.Hence 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).
OBJECTIVE - 5`to`Learn about take off power, tyre design, rolling resistance, tyre pressure, brake forces when landing.
Takeoff `to` is the phase of flight in which an aircraft goes through a transition from moving along the ground (taxiing) to flying in the air, usually starting on a runway.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 `to`Thrust of propulsion system pushing aircraft along runway.
D `to` Aerodynamic Drag of vehicle resisting the aircraft motion.
F `to` 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.
`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 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.
For Small commuter aircraft `to` `V_R = 1.1 × V_(STALL)` or `V_R = 1.05 × V_(MIN CONTROL)`, whichever is great.
`V_(STALL) = sqrt(W/(0.5*C_(Lmax)*rho*S))`, where S `to` displacement
Take off Field Length `to` 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*S_1`
Therefore, Takeoff power = force * displacement = `(m*(dv)/(dt))*2*S_1= (T-D-F)*S_1`
Rolling Resistance `to`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 runway
Braking Forces in Landing `to`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.
`m*(dv)/(dt)=-T-D-F`
The negative acceleration or deceleration value will be based on friction coefficient for maximum braking and the value of reverse thrust (if available).
Aircraft Tyre design `to` Aircraft wheels are an important component of a landing gear system. With tires mounted upon them, they support the entire weight of the aircraft during taxi, tkeoff, and landing. The typical aircraft wheel is lightweight, strong, and made from aluminium alloy. Some magnesium alloy wheels also exist. Early aircraft wheels were of single piece construction, much the same as the modern automobile wheel.
As aircraft tires were improved for the purpose they serve, they were made stiffer to better absorb the forces of landing without blowing out or separating from the rim. Stretching such a tire over a single piece wheel rim was not possible. A two-piece wheel was developed.
Early two piece aircraft wheels were essentially one-piece wheels wth a removable rim to allow mounting access for the tire.These are still found on older aircraft. Later, wheels with two nearly symmetrical halves were developed. Nearly all modern aircraft wheels are of this two piece construction
two piece aircraft wheel found on modern airliner `to` The typical modern two piece aircraft wheel is cast or forged from aluminium or magnesium alloy. The halves are bolted together and contain a groove at the mating surface for an o-ring, which seals the rim sice most modern aircraft utilize tubeless tires. The bead seat area of a wheel is where the tire actually contacts the wheel. It is the critical area that accepts the significat tensile loads from the tire during landing. To strengthen this area during manufacturing, the bead seat area is typically rooled to pre stress it with a compressive stres load. Modern tyres are much more rigid, due to the load bearing requirements, which results in the wheels having to be of two piece construction.
Aircraft Tyre Pressure `to` Aircraft tires generally operate at high pressures, up to 200 psi (14 bar or 1,400 kPa) for airliners, and even higher for business jets. The main landing gear 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 tyres were a significant factor.
Tests of airliner aircraft tires have shown that they are able to sustain pressures of maximum 800 psi (55 bar; 5,500 kPa) before bursting. During the tests the tires have to be filled with water, to prevent the test room being blown apart by the energy that would be released by a gas when the tire bursts.
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.
The requirement that an inert gas, such as nitrogen, be used instead of air for inflation of tires on certain transport category airplanes was prompted by at least three cases in which the oxygen in air-filled tires had combined with volatile gases given off by a severely overheated tire and exploded upon reaching autoignition temperature. The use of an inert gas for tire inflation eliminates the possibility of tire explosion
OBJECTIVE - 6`to`With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle?.
What is a Traction Force?`implies`the force propelling the vehicle forward, transmitted to the ground through the drive wheels. The force propelling the vehicle forward, the tractive effort, has to accomplish
the following:
• overcome the rolling resistance;
• overcome the aerodynamic drag;
• provide the force needed to overcome the component of the vehicle’s weight acting
down the slope
• accelerate the vehicle, if the velocity is not constant.
1.Rolling Resistance Force `(F_(rr))to`due to the friction of the vehicle tyre on the road. Friction in bearings and the gearing system also play their part. The rolling resistance is approximately constant, and hardly depends on vehicle speed. It is proportional to vehicle weight. The equation is:
`Funderset( rr) ( )=underset( )(mu) underset( rr) ( )mg`
where `underset( ) (mu )underset( rr) ( )to`coefficient of rolling resistance
2.Aerodynamic Drag Force (`Funderset(rr)()`)`to`This part of the force is due to the friction of the vehicle body moving through the air. It is a function of the frontal area, shape, protrusions such as side mirrors, ducts and air passages, spoilers, and many other factors. The formula for this component is:
`Funderset(ad)()=(1/2)(rhoACunderset(D)()V^2)`
where `rhoto`density of air, `Ato`frontal area, `C_Dto`coffecient of drag, `Vto`velocity.
3.Hill Climbing Force (`F_(hc)`)`to`The force needed to drive the vehicle up a slope. It is simply the component of the vehicle weight that acts along the slope.The equation is:-
`Funderset(HC)()=mgsin(psi)`
`Funderset(TE)()=Funderset(rr)()+Funderset(ad)()+Funderset(hc)()+Funderset(la)()+Funderset(omegaa)()`
Here i have neglected the hill climbing force, acceleration force.
I have defined an array for rolling resistance coefficent and an array for mass of the aircraft. So for each rolling resistance coefficents and each mass of aircraft, the traction force of the aircraft, power and torque produced is calculated.
%PROGRAM TO CALCULATE THE TRACTION FORCE, POWER & TORQUE OF AN AIRCRAFT
clear all
close all
clc
%ASSUMED VALUES
urr =[0.02,0.03,0.04,0.05,0.06,0.07]; %rolling resistance coefficent ARRAY
u = 0.8; %friction resistance coefficent
m = linspace(500000,600000,10); %array for aircraft mass in Kg
g = 9.81; %acceleration of gravity m/sec^2
rho = 1.25; %density of air in kg/m^3
A = 158.3; % frontal area array m^2
cd =0.02; %coefficent of drag
v = 3; %velocity m/sec
r = 0.35; %radius of towing vehicle wheels in m
G = 20; %gear ratio
frr = zeros(6,10); %array to store rolling resistance force
fad = zeros(6,10); %array to store aerodynamic force
ffr = zeros(6,10); %array to store frictional force
fte = zeros(6,10); %array to store total traction force
%FOR LOOP FOR FORCE CALCULATIONS
for i=1:6
for j=1:10
frr1 = urr(i).*m(j)*g; %rolling resistance force in N
frr(i,j) = frr1;
fad1 = 0.5*rho*A*cd*v^2; %aerodynamic force in N
fad(i,j) = fad1;
ffr1 = u*m(j).*g; %friction force in N
ffr(i,j)= ffr1;
fte(i,j) = frr(i,j) + fad(i,j) + ffr(i,j); %total traction force in N
end
end
display(fte)
%POWER CALCULATION
P = (fte .* v)/1000; %aircraft power calculation KW
%TORQUE CALCULATION
T = (fte .* r)/G; %torque calculation in Nm
%PLOTTING
figure
plot(m,fte) %plotting trcation force and masss of the aircraft
xlabel('mass of aircraft (kg)') %labelling x axis
ylabel('traction force (N)') %labelling y axis
title('Variation of Traction force with mass of aircraft') %tittle
legend("Icy concrete/asphalt - 0.02","Dry concrete/asphalt - 0.03","Firm turf - 0.04","Hard turf - 0.05","Wet concrete/asphalt - 0.06","Soft turf - 0.07")%legend
figure
plot(urr,fte) %plotting trcation force and masss of the aircraft
xlabel('coefficent of rolling resistance') %labelling x axis
ylabel('traction force (N)') %labelling y axis
title('Variation of Traction force with rolling resistance coefficent') %tittle
legend("mass 1","mass 2","mass 3","mass 4","mass 5","mass 6","mass 7","mass 8","mass 9","mass 10") %legend
OUTPUT `to`
OBJECTIVE - 7`to`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. Make all required assumptions. Prepare a table of assumed parameters. Draw a block diagram of powertrain.
Assumptions `to`
for doing this problem, i have considered the values realated to aircraft like -
aircraft weight = 31298 kg
towing tractor = 1100kg
coefficent of drag = 0.02
coefficent of rolling resistance = 0.03
frontal area = 158.3`m^2`
Rolling resistance coefficent = 0.03
Velocity = 10km/hr =3 m/sec
Power = 19000 pound-force = 84.516 KN = 84516.21 N
towing tractors wheel radius = 0.35m
gear ratio = 20
velocity = 10km/hr = 3m/sec
density of air = 1.25kg/m^3
acceleration of gravity = 9.81m/sec^2
i have used the advisor tool to design the motor ratings.
Type of Motor and Motor Power ratings`to`
motor specifications and rating
motor model - MC_AC30, Siemens AC 30 kW induction motor/controller
Maximum power - 250KW
Efficiency - 86%
Mass of motor - 58Kg
Voltage - 216V
Peak current - 280A
Minimum Voltage - 140V
Powertrain for Aircraft Towing Tractor `to`
Now the program to calculate the traction force required to tow the aircraft using the above deifned motor ratings is as shown below -
%PROGRAM TO CALCULATE THE TRACTION FORCE, POWER & TORQUE OF AN AIRCRAFT
clear all
close all
clc
%ASSUMED VALUES
urr = 0.03; %rolling resistance coefficent
u = 0.2; %friction resistance coefficent
ma = 31298; %mass of aircraft in kg
mt = 1100; %mass of towing tractor in Kg
g = 9.81; %acceleration of gravity m/sec^2
rho = 1.25; %density of air in kg/m^3
A = 158.3; % frontal area array m^2
cd =0.02; %coefficent of drag
v = 3; %velocity m/sec
r = 0.35; % radius of towing vehicle wheels in m
G = 20; %gear ratio
%FORCE CALCULATION
frr = urr*(ma+mt)*g; %rolling resistance force in N
fad = 0.5*rho*A*cd*v^2; %aerodynamic force in N
ffr = u*(ma+mt)*g; %friction force in N
%TRACTION FORCE CALCULATION
Traction_force_in_N = (frr + fad + ffr) %total traction force in N
%POWER CALCULATION
Power_KW = (Traction_force_in_N * v)/1000 %aircraft power calculation KW
output `to`
so the Output power required to pull the aircraft & the towing tractor will be 219.3522 KW
And the selcted motors power rating is 250KW, which becomes the input power required by the motor.
Hence the duty cycle range is calculated by finding the ratio of output to input power
i.e. `P_(OUT)/P_(IN) =219.3522/250 =0.8774`
so we can say the duty ycle range is from 0 to 87.77% .
Now, assume that the tractor takes 10 min or 600sec to taxing (i.e. tow the aircraft from gate to runway)
Hence the energy requirement to fulfill aircraft towing application (E) = `P_(OUT)` * t
i.e. E = 219.3522 * 600
E = 131.611MJ
Google drive files - https://drive.google.com/drive/folders/146wXuzKBEXmAXQ7fgS20cs826ny_AnZ7?usp=sharing