Project-1: Powertrain for aircraft in runways

 

Given that 

Part A:

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

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

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

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

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

Part B:

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

    B. Develop the model for the calculated force and power using Simulink.

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. 

(Hint :DC7 Block)

    B. Also, Design the parameters in excel sheet.

 

 

Aim : 

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

2.Differeniate between ground speed and air speed.

3.Find out why  it is not  recommended to use aircraft engine power to move it on the ground at Airport

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

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

6.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. 

 

PART A -

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

The different terms represent different weights or masses. For each flight, the weights are taken into account for several reasons.

(MEW)Manufacturer's Weight Empty - also known as Manufacturer's Weight Empty 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.

(ZFW) Zero-fuel weight

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.

(OEW) Operating empty weight

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

It is the carrying capacity of an aircraft. It includes cargo, people, and extra fuel. In the case of a commercial airliner, it may refer only to revenue-generating cargo or paying passengers.

Maximum take-off weight (MTOW)

This is the maximum weight at which the pilot of the aircraft is allowed to attempt to take off.

Regulated takeoff weight (RTOW)

Depending on different factors (eg, 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)

The maximum weight at which an aircraft is permitted to land. The following image depicts takeoff weight components:

(MRW) 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.

 

 

 

 

It is the total aircraft weight at any moment during the flight or ground operation. This decreases during the flight due to fuel and oil consumption

 a) Crew weight (Wc)

    The crew comprises the people necessary to operate the airplane in flight. e.g. Pilot, Co-pilot, Airhostess, etc.

 b) Payload weight (Wp)

    The payload is what the airplane is mentioned to transport passengers, baggage, freight, etc. (Military use the payload includes bombs, rockets, and other disposable ordnance).

  c) Fuel weight (W1)

    This is the weight of the fuel in the fuel tanks. Since fuel is consumed during the flight. is a variable, decreasing with time during the flight.

  d) Empty weight (We)

    This is the weight of everything else-the structure engines (with all accessory equipment), electronic equipment landing gear, fixed equipment, and anything else that is not crew, payload, or fuel.

  e) Gross weight (Wo)

    The sum of these weights is the total weight of the airplane. Gross weight or total weight varies through the flight because fuel is being consumed. The design take-off gross weight is the weight of the airplane at the instant it begins its mission. It includes the weight of the fuel.

 

Different types of aircraft with their weight:

 

QUES.2  Is there any difference between ground speed and airspeed?

ground speed

Ground speed is the airplane's speed relative to the surface of the Earth.

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 it's horizontal rather than vertical speed - an aircraft climbing completely vertically would have a ground speed of zero.

airspeed

airspeed (at least true airspeed) - is its speed relative to the air it is flying in.

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.

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.

 

 

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

Engine power is used all the time to move aircraft on the ground. I think you mean in the ramp area. if, so, again, engine power s used all the time in the ramp. They do get pushed back because only aircraft with engines on the tail can power back due to FOD Sometimes if it's a tight area, a tug will be used.

Aircraft can move using their engine power all the time on the airport apron and taxiways under their power to go towards runways and after landing taxi to their assigned gate. But the thrust produced resulting in a jet blast might cause damage to the terminal building or equipment. 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. And also it can reduce the cost of fuel. Taxing can save money.

QUES. 4 How an aircraft is pushed to the runway when it's ready to take off?

Airplanes have engines, and they can use them to move forward along the ground. So they push themselves to the runway.

Most of the time, a small vehicle is indeed needed to push the larger commercial airplanes backward away from the terminal building because even if a plane can revere its thrust, it's not always safe to hit the buildings with that reverse thrust. So, the little trucks push the bugger planes back away from the buildings, but after that, the planes can apply a small amount of engine power to roll forward just as all airplanes can. Aircraft push themselves on the runaway

Taxiing:

Taxiing (rarely spelled taxying) is the movement of an aircraft on the ground, under its own power, in contrast to towing or push-back where the aircraft is moved by a tug. The aircraft usually moves on wheels, but the term also includes aircraft with skis or floats (for water-based travel).

An airplane uses taxiways to taxi from one place on an airport to another; for example, when moving from a hangar to the runway. The term "taxiing" is not used for the accelerating run along a runway prior to takeoff, or the decelerating run immediately after landing, which is called the takeoff roll and landing rollout, respectively.

 

Towing:

Movement of large aircraft about the airport, flight line, and hangar is usually accomplished by towing with a tow tractor (sometimes called a “tug”). In the case of small aircraft, some moving is accomplished by hand pushing on the correct areas of the aircraft. Aircraft may also be taxied about the flight line but usually only by certain qualified personnel.

Towing aircraft can be a hazardous operation, causing damage to the aircraft and injury to personnel, if done recklessly or carelessly. This article outline the general procedure for towing aircraft. However, specific instructions for each model of aircraft are detailed in the manufacturer’s maintenance instructions and are to be followed in all instances.

 

 

Before the aircraft to be towed is moved, a qualified person must be in the flight deck to operate the brakes in case the tow bar fails or becomes unhooked. The aircraft can then be stopped, preventing possible damage.

Some types of tow bars available for general use can be used for many types of towing operations. [Figure 2] These bars are designed with sufficient tensile strength to pull most aircraft, but are not intended to be subjected to torsional or twisting loads. Many have small wheels that permit them to be drawn behind the towing vehicle going to or from an aircraft. When the bar is attached to the aircraft, inspect all the engaging devices for damage or malfunction before moving the aircraft.

 

 

Some tow bars are designed for towing various types of aircraft. However, other special types can be used on a particular aircraft only. Such bars are usually designed and built by the aircraft manufacturer.

 When towing the aircraft, the towing vehicle speed must be reasonable, and all persons involved in the operation must be alert. When the aircraft is stopped, do not rely upon the brakes of the towing vehicle alone to stop the aircraft. The person in the flight deck must coordinate the use of the aircraft brakes with those of the towing vehicle. A typical smaller aircraft tow tractor (or tug) is shown in Figure 3. 

The attachment of the tow bar varies on different types of aircraft. Aircraft equipped with tail wheels are generally towed forward by attaching the tow bar to the main landing gear. In most cases, it is permissible to tow the aircraft in reverse by attaching the tow bar to the tail wheel axle. Any time an aircraft equipped with a tail wheel is towed, the tail wheel must be unlocked or the tail wheel locking mechanism may damage or break. Aircraft equipped with tricycle landing gear are generally towed forward by attaching a tow bar to the axle of the nosewheel. They may also be towed forward or backward by attaching a towing bridle or specially designed towing bar to the towing lugs on the main landing gear. When an aircraft is towed in this manner, a steering bar is attached to the nosewheel to steer the aircraft.

 

 

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

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 the 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 the 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.

 

1. Rotation Velocity.

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. is a critical factor in determining take-off performance.

For safety reasons, VR is usually determined as being 1.2*VSTALL  or 1.1*Vmin control whichever is greater.

Stall speed, VSTALL is the lowest speed that the aircraft can be flown before the airflow starts to separate from the wings as the angle of attack becomes too great, it can be calculated based on knowledge of the aircraft take-off configuration and hence the maximum achievable lift coefficient CL(max),

To maintain level flight the lift produced must equal the weight, at the point just before stall this leads to the following balance,

                          W = L = CL(max)* ½*rho*VSTALL2*S

hence stall speed can be calculated as,

                          VSTALL = [W/( CL(max)* ½* rho*S)]1/2

Minimum control speed, VMINCONTROL 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.

 

*Tyre design of an aircraft:

-An aircraft tyre are deisgned to withstand extreme heavy loads by absording shocks during and there by providing cushioning effect.

-The number of tires required for an aircraft depends on the total weight of aircraft as the weight needs to be distributed evenly.

-Aircraft tyre are designed to facillitate stability in high crosswind condition,to chanalize the water away for preventing skidding and helps to give a good braking effect.

 

 

 

 

-Aircraft tires are usually inflated by nitrogen air inorder to reduce expansion and contraction from extrerme changes in ambient pressure and tempurature.This type of inert gases will eliminate the possibilty of tire explosion.

-Aircraft tire design are set into an aircraft based on its application and also suffeicient clearance with the ground for allowing tyre growth which happens when centrifugal force is at high speed.

 

1.Rolling Resistance

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

The rolling resistance force is mainly experienced by the tire when the vehicle moves over any surface. The rolling resistance force is given by the expression:

                                  Frr = urr* (m*g)

where m*g is weight in Newton

                                   urr = coefficient of rolling resistance

The coefficient of rolling resistance depends on many factors like temperature, tire temperature, the material of the tire, the threaded pattern of the tire, and also the road roughness and liquid on the road. These factors determine the rolling resistance force.

 

Tyre pressure:

The pressure generally is very high in an aircraft time to reduce the rolling resistance and to withstand the weight, usually, it will be around 200 to 300psi of pressure and they change based on the type of the aircraft. Tests of airliner aircraft tires have shown that they can sustain the pressure of a maximum of 800 psi. Aircraft tires are usually inflated with nitrogen to minimize expansion and contraction from the extreme changes in the ambient temperature and pressure experienced during flight

Brake force while landing

Stopping a 200-tonne aircraft landing at 180 mph requires a lot of braking force. To do this either hydraulic brakes or electric brakes are used. Hydraulic brakes put on the weight of the aircraft. Nowadays designers use an electric brake system. The aim of brake force when landing is to minimize the distance.

The touch-down velocity should be approximately the stall speed of the aircraft in landing configuration. This will be achieved by increasing the drag and decelerating the aircraft to minimum flying speed. The deceleration on the landing roll from to 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

 

Where,   T - Thrust of propulsion system pushing aircraft along the 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.

                  m - Mass

So, the brake force = Mass*Deceleration

6 question

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

  

ans:

considering:

Total mass of the air craft (including payload) for air a380=575,000kg

mass of the push block = 1000kg(assumed)

 

Rolling resistance(Frr)::

coefficient of rolling resistance(crr) =0.002

rolling resistance  force is given by Frr=crr.m.g”>Frr=crr*m*g

for the air craft

                         Frr=0.002*575000*9.81

                         =Frr=11281.5N

                          =Frr=11.2815KN

 

For The Towing Vehicle

                       Frrt=0.002*1000*9.81

                          Frrt=19.62

The Towing resistance value is so less to the compared to the air craft because this is the heavy aircraft these can be neglected

 

AERODYNAMICS RESISITANCE (Far):

                                                            Drag coefficient (Cd)=0.025

                                                            Air density=1.125kg/m^3

                                                                Frontal area(A)=150m^2

                                                               Velocity(V)=14.5kmph or 4m/s

                                                                 Far = ½*cd*p*A*V2

                                                               Far = ½*0.025*1.125*150*4^2

                                                                        =33.75N

As the vehicle and plane are moving at very low speed Aerodynamics resistance can be neglected

 

Hill climb Resistance:

Runways are constructed flat so gradient resistance/hill climb force is neglected

 

 TOTAL RESISTANCE:

                                            Ft=11.2815KN

POWER REQUIRED:

                                  P=Ft*V

                                  =P=11.2815*4

                                       =45.26K

The minimum power required by the vehicle is 45.26KW to taxi  a heavy aircraft at4m/s

  B. Develop the model for the calculated force and power using Simulink.

 

in this we can design a model  shown in below we can understand

Simulink model:

1. constant blocks are used to feed the parameters such as mass,coefficient and other parameters
2. using product block the resistive forces as calculated
3. using sum block  totaal resistive forces also calculated
4. the outputs are visualised using display blocks

 

 

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. 

(Hint :DC7 Block)

 

see the below we can highlated given hint to take the hint and designed a circuit DC7 Block shown in below figure

 

Description:

The 200 Hp Dc motor is separately excited with a constant 150v.Dc field voltage source .The armature voltage is provided  by an IGBT converter controller by two  pI regulator .The converter is fed by a515V Dc bus obtained by rectification of a380V Ac 50 hz voltage source to limit the Dc  Bus voltage during dynamic  braking mode .a braking chopper has beemn added between the diode rectifier and  the DC7 BLOCK

The first regulator is a speed regulator followed by a current regulator  the speed regulator outputs the armature current references in (p.u.) used by the current  controller  to obtain the electromagnetic torque needed to reach the desired speed  the speed reference changes rate follows acceleration and declaration ramps to avoid sudden references changes that could cause armature over current                     de stabalize  the system current  regulates control the armature by computing duty ratios 5Khz pulses of four IGBT devices (pulse with modulation) for proper  system behaviour instantaneous pulse values of IGBT devices 1 And 4 are opposite IGBT devices 2And 3 generates average armature voltage needed to obtain the desired armature  current to limit amplitude current oscillations smoothing inductance is placed in series in armature circuit

       see the above  figure the  scope2

 

OUTPUT From motor:

here we can observe the scope 1

see the below figure we can understand easily how wave outforms  in scope1:

showing waveforms  Armature Voltage (V), Vdc(V)

                               Speed(rpm), IGBT(1&4)

                             Armature current ia(A), IGBT(2&3)

      see the above figure waveforms in scope: motor waveforms

 

OUTPUT From BATTERY:

 

here we shown in output wave form from battery  from showing wave form in scope we can easily understood 

figure showing SOC (%)

                        Current(A)

                       And Voltage(V)

          see the above figuare  we can understand waveforms easily

SIMULATION :

Before simulation set the initial bus voltage 515V via the GUI block (the  initial stages setting button and C bus variable).

2.start the simulation we can observe the motor armature voltage and current  the four IGBT pulses and motor speed on the scope the current and speed references are also shown clearly 

3.the  motor is coupled to a linear load  which means the mechanical torque of the proportional to the speed

4.the speed references is set 500 rpm at t=0 , observe  that motor speed  follow the references ramp accurately (+400 rpm/s) and reaches a steady state around t=1.3s

the armature current flows the current references very well  with a fast response  time and small ripples that the current ripple frequency is 5KHz

At t=2 seconds  the speed references drops  1184rpm the current references decreases to reduce the electromagnetic torque causes the motor declaration with the help of load torque

At t= 2.2 seconds the current reverse to produce a braking electro magnetic torque (dynamic braking mode)

at t=3.25 seconds  motor reaches 0 rpm  and load torque reverse and becomes negative current  produce accerlatiomn and electromagnetic field torque to allow motor to follow  negative speed ramp -400 rpm seconds at t=63 sec speed reaches to -1184 rpm and stabilized

 

see the below image we can understand

now we assume that 

tyre radius to be  = 28 inch = 0.7112

torque required at wheel = total force*radius

                                              = 9.95*0.7112

                                                =7.7076 KNM

Motor Torque =(Wheel torque )/(gear ratio* efficiency)

 

assuming the gear ratio is 8 and transmission efficiency 82%

                               =7.706/(8*0.82)

                               =1079 KNM

if motor was 90% efficiency then motor torque =1.198KNM

power required by motor will be   =29.84/(0.82*90)

                                                 =40.43KW

and total energy  consumed = 40.43*0.1667 [ time to be in 10 mins]

                              =6.77KW/h

choosing a battery  whose power rating is around 60 KW

there fore the duty cycle range to control the speed from zero to high speed is

                        d=req.power/rated power

                            =40.43/60

                             =0.67

this is the mean controller should provide a duty cycle ranging from  0 to 67% to control towing speed  parameters in excel is showing in below we can easily understand

CONCLUSION:

finally we have conclusion that what ever he asking in question we explained and some calculation part and running a matlab circuit design with exact outputs  and we kept in an excel sheet and we attached below what we did