Project-1: Powertrain for aircraft in runways

Part A:

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

All kinds of “Weights” for an Aircraft [1]-[2]:

Prerequisites:

• Weight of a male person – 88 kg (avg)
• Weight of a female person – 70 kg (avg)
• Weight of children (2-12) years – 35 kg (avg)
• Weight of cabin crew – 75 kg (including 3kg of crew baggage) (avg)
• Weight of fuel in aircraft can be approximated as approx 12000 kgs for 4000 km by an narrow body aircraft (A-320).
• Payload – Passengers, Cargo, Mail, Baggage. (mainly)

Maximum Take-Off Weight (MTOW): It is the maximum permitted weight at the start of the take-off. While determining this limit, the manufacturer takes into consideration the strength of the material used for the manufacturing of the aircraft and standard performance variables such as runway temperature, altitude, atmospheric pressure, wind components, flaps, etc. (MZFW + Take-Off Fuel = MTOW)

List of maximum take-off weight of the selected 10 aircraft [2]-[3]:

 

Sr. no.	Name of the 10 military aircraft	weight (kg)	Sr. no.	Name of the e 10 commercial aircraft	weight (kg)
1	F-16 Fighting Falcon	21,772	1	Boeing 737-900	74,389
2	F-15 Eagle	36,700	2	Boeing 777- 300ER	351,534
3	Saab JAS 39 Gripen	16,500	3	Boeing 787-10	254,011
4	MiG 35	29,700	4	Airbus A320	78,000
5	J-10	18,400	5	Airbus A330-300	242,000
6	Sukhoi-Su-35	35,000	6	Airbus A350-1000	311,000
7	Dassault Rafale	24,500	7	Airbus A380-800	569,000
8	F-35 Ligthning II	31,751	8	Concorde	185,000
9	Eurofighter Typhoon	23,500	9	Bombardier C Series	67,585
10	F-22 Raptor	38,000	10	Tupolev Tu-114	180,000

 

Sr. no.	Name of the 10 Narrow Body Aircraft aircraft	weight (kg)	Sr. no.	Name of the e Wide Body aircraft	weight (kg)
1	Hawker Siddeley Trident	65,318	1	Boeing 747-8	447,696
2	Boeing 707	151,315	2	Boeing 767- 400ER	204,116
3	Boeing 727-200	95,100	3	Ilyushin II-86	208,000
4	Boeing 757-200	115,666	4	Ilyushin II-96	226,796
5	Ilyushin II-62M	165,500	5	Lockheed L-1011 Tristar	211,375
6	Tupolev Tu-154M	100,000	6	McDonnell Douglas DC-10	259,450
7	Tupolev Tu-134A	47,000	7	McDonnell Douglas MD-11	273,314
8	McDonnell Douglas MD-88	67,810	8	McDonnell Douglas MD-12	430,459
9	McDonnell Douglas MD-90-55	78,245	9	Airbus A300-600	170,500
10	Vickers VC10	151,956	10	Airbus A340	275,000

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

ground speed is the airplane’s speed relative to the surface of the Earth, while airspeed –is its speed relative to the air it is flying in [4].

Wind Speed

For a reference point picked on the ground, the air moves relative to the reference point at the wind speed. Notice that the wind speed is a vector quantity

Ground speed

As per reference point picked on the ground, the aircraft moves relative to the reference point at the ground speed. Ground speed is also a vector quantity 

Air speed

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.

Airspeed is the vector difference between the ground speed and the wind speed.

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

Taxiing refers to the movement of an airplane while on the runway. 

Taxiing is simply the process of moving an airplane while it’s on the runway. It occurs after an airplane has landed, and it occurs before an airplane takes off. Airplanes don’t actually fly while on the ground. As a result, the process by which they move on the runway isn’t known as flying; it’s known as taxiing.

Pushback is when the flight crew is ready to depart (doors closed and jetway removed) and they require to be towed out onto the tarmac to initiate engine startup. Once the aircraft is clear, we communicate up to the captain all clear to start engines.

Towing is a method used by maintenance crews to move an aircraft from point A to point B based on needs of the airline (in my particular experience) without starting the engines but using a tow tractor of sufficient horsepower (very important) to move the airplane. 

By using jet engines as an aircraft’s primary thrust for taxiing, airlines spend huge amounts of jet fuel while still on the ground, which could be much better used for flying. Many airlines plan to adopt electric taxiing (within the aircraft) to save fuel and reduce environmental impacts.

◦ Even when reversing the aircraft, jet reversers blows debris creating foreign object damage hazard for adjacent aircraft, gate equipment or terminal waiting area glass windows and can also damage the crowded gate area.

◦ This is accompanied with the high noise that the engine produces.

◦ The engine produces large thrust from the exit and vacuum from the front which may suck in objects which may damage the engine.

Thus, it is not recommended to use the aircraft engine power to move the aircraft on the ground.

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.But the thrust produced resulting 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. 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 Its 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.

The disadvantages of using aircraft engine power to move it on the ground at Airport is given below,

• Fuel consumption.
• Wear and tear on the engine.
• Effectively a loss of control can happen, especially during high power engine running. Damage can occur to the aircraft itself, other aircraft nearby or to air side structures.
• There is a risk of injury to ground support personnel who may be in relatively close proximity to the aircraft.
• It is easier for truck driver to see the surroundings better than the pilot sitting in the cockpit with limited window. The pushback driver knows the airport better than the pilot.
• It also saves fuel virgin Atlantic had proposed a new system in which pushback truck will move the aircraft all the way to runway to save fuel and reduce impact of environment.
• The plane engines are very loud. When more than one plane will add up their sound, than it will not be pleasant for passengers and others near airport.

                   

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

Pushback is when an aircraft is pushed backward 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.

Phases of take-off flight:

The take-off flight is generally divided into two phases namely

Take-off ground run

Transition and climb phases

1. i) Take-off ground run

Take-off ground run during the ground run the airplane starts from rest and accelerates to the take-off speed (VT0 or V1). The flaps and engine(s) are adjusted for their take-off settings. In the case of an airplane with a tricycle type of landing gear, all three wheels remain in contact with the ground till a speed of about 85% of the VT0 is reached. This speed is called ‘Nose wheel lift-off speed’. At this speed, the pilot pulls the stick back and increases the angle of attack of the airplane so as to attain a lift coefficient corresponding to take-off (CLT0). At this stage, the nose wheel is off the ground and the speed of the airplane continues to increase. As the speed exceeds the take-off speed the airplane gets airborne and the main landing gear wheels also leave the ground. When the airplane has a tailwheel type of landing gear, the angle of attack is high at the beginning of the take-off run. However, the tail wheel is lifted off the ground as soon as some speed is gained and the deflection of the elevator can rotate the airplane about the main wheels. This action reduces the angle of attack and consequently the drag of the airplane during most of the ground run. As the take-off speed is approached the tail wheel is lowered to get the incidence corresponding to CLT0. When VT0 is exceeded, the airplane gets airborne. The point at which all the wheels have left the ground is called ‘Unstick point’.

(ii) Transition and climb phases

During the transition phase, the airplane moves along a curved path and the pilot tries to attain a steady climb. As soon as the airplane attains an altitude equal to the screen height, the take-off flight is complete. For airplanes with a high thrust to weight ratio, the screen height may be attained during the transition phase itself.

                                    Phases of take-off flight

 

5. Learn about take off power, tyre design, rolling resistance, tyre pressure, brake forces when landing. 
6. Take off Power [5]:

Take off performance can be predicted using a simple measure of the acceleration of the aircraft along the runway based on force equilibrium [5].

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.

      (1)

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

he 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 a 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. Take-off rules vary slightly depending on the aircraft category.

For Small commuter aircraft

V= 1.1*VSTALL

or

V=1.05*V Control

Where S= Length of Runway

Take-off Power = Force* Displacement

(m.dvdt).2.S1→(T–D−F).S1

2. Tire Design:

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

• They support the weight of an aircraft while it is on the ground.
• It also provide 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 tyres, it does not have to carry a load for a long period of continuous operations. However, an aircraft tyre absorbs the high impact loads of landing and also its operated at high speed for a short period of time when required.

TIRE CONSTRUCTION:

The aircraft tire construction consists of the following components:

1. Tread: The tread is a special rubber compound which is resistant to abrasive wear, cutting, chunking, and heat. Around its circumference the tread has moulded grooves which help to remove water between the tread and the runway during wet conditions and increase the traction and the grip.
2. Inter-Tread Fabric:

1.  An aircraft tyre can include Inter-Tread fabric(ITF) also known as tread reinforcement layers. This is one or more layers of nylon fabric between the casing piles and the base of the tread. The ITF keeps the tread stable and free from distortion during high speeds when there is increased centrifugal force. The ITF also gives protection to the casing piles if debric cuts the tread.
2. For special high speed applications, the ITF can be moulded into the rubber of tread. As the tyre wears, the ITF is seen in the tread pattern.

3. Casing: 

1. The primary strength of the tyres is in the casing piles. A casing pile is a layer of high modulus cord which is coated with rubber compound to make a fabric. This fabric is assembeld in layers.
2. For bias(cross-ply) aircraft tyres, the casing piles are installed around the bead coils to amke the sidewalls of the tyres. This gives the bias tyre its strength. Adjacent layers of casing ply are put at opposite bias angles. The dstrength and load capacity of the tyre is related to the number of piles and bias angles.
3. In a tyre for a millitary applications, the casing cord could be a red colour for easy visibility and indication of wear.

4. Breakers:A breaker is an added layer of fabric that gives more strength. The breaker layers are laid directly on the top ply across the width of the tread. They are a part of the tyre casing, and are not removed when the tyre is retreaded.
5. Beads:

1. The tyre beads hold the tyre on its wheel. The beads are made up of high tensile wire strands. Each wire strand is coated with rubber compound and is wound into a coil of specified diameter for the tyre. A rubber component known as the apex is applied to the top of the bead coil to make bead assembly.
2. The bead filler is a fabric component which helps to hold the bead coil assembly together. The bead filler also gives more strength to the bead assembly and more stiffness to the area of the tyre that is compressed with a high load against the wheel rim.

6. Chafers:

1. A chafer components prevents chafing damage when the tyre is against the rim. It also decreases the conduction of brake heat to the tyre from the rim.
2. The chafer component is a strong nylon material which is coated with rubber. The chafer assembly is attached to the first ply and pulled around the bead face to the clinch area.

7. Sidewall: The sidewall is a layer of special rubber formulated with anti-oxidants. This rubber layer provides protection against weathering to the casing piles and is resistant to contamination, cuts and flexing.
8. Inner Liner: The inner liner is a layer of special rubber compound around the inner face of the tubeless tyre. The inner liner is attached to the first casing ply and is a seal to prevent leakage inflation gas and moisture into the tyre casing.
9. Chined Tyres: Some types of nose wheel tyres include one(or two) chines moulded into the rubber at one of the shoulder buttresses. A chine is a rubber extension, (arounf the tyre circumference), that deflects water from rear fuselage mounted engines during aircraft movement on a wet runway.
10. Breaker Package(Belt Piles): The belt piles is a number of piles made up of nylon or aramid attached below the tread of a radial-ply tyre. The breaker package give stiffness to the tread area and helps to keep the circumference constant as the tyres expand when it is inflated.

The aircraft tyre(bias type which is most common) cross-section is shwon on the picture below:

The Classification of the tyres w.r.to sidewall markings are shown in the figure below:

 

Configuration identification of S/N

AEA code No.

3. Rolling Resistance

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

     (4)

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.

4. tyre pressure

 

Aircraft tires generally operate at high pressures, up to 200 psi (14 bar; 1,400 kPa) for airliners, and even higher for business jets. The main landing gear on the Concorde was typically inflated to 232 psi (16.0 bar), whilst its tail bumper gear tires were as high as 294 psi (20.3 bar).

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.

5. Brake Forces when Landing:

The brakes used in the aircraft while landing are as follows:

1. Aircraft Disc Brakes: It is used to brake the wheels while touching the ground. These brakes are operated hydraulically or pneumatically. In most modern aircraft they are activated by the top section of the rudder pedals(toe brakes). In some older aircraft the bottom section is used instaed(heel brakes). Levers are used in a few aircraft. Most aircraft are capable of differential braking.
2. Thrust Reversers: It is used on a jet aircraft to provide a significant way of increasing the rate of deacceleration during the initial stages of both a landing roll or a rejected takeoff from a high speed. Use of reverse thrust on low wing aircraft with mounted engines should be limited to the time when the aircraft is on an active runway.
3. Air Brakes: Air brakes or speed brakes are used to cause the aircraft to slow more rapidly when they were deployed. They produce a drag which required a high power setting, these settings are much more responsive and rapid thus making power and glide slope corrections on approach more rapidly with speed brakes deployed.
4. Large Drogue Parachutes: It is a parachute designed for deployment from a rapidly moving object. It is used for various purposes such as to decrease the speed, to provide control and stability, or as a pilot parachute to deploy a larger parachute.

Part B:

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

ASSUMPTIONS:

• FOR TOWING TRUCK:
• Friction Coefficient = 0.006
• Mass of the vehicle = 25,000 Kg
• Drag coefficient = 0.608
• Frontal Area = 120 m^2
• Velocity of the vehicle = 3 m/sec
• Wheel radius = 0.6096 m
• Gear Ratio = 10

• FOR AIRCRAFT:
• Friction Coefficient = 0.004
• Mass of the vehicle = 3,51,800 Kg
• Drag coefficient = 0.026
• Frontal Area = 6500 m^2
• Velocity of the vehicle = 3 m/sec

• Total Tractive Force = Rolling Resistance Force + Aerodynamic Drag Force + Hill Climbing Force + Acceleration Force
• For calculation of force in this case the Hill climbing Force and the Acceleration Force can be neglected.
• And the Rolling resistance force and the Aerodynamic drag force will be present for both the Aircraft and the Towing Truck.

• Calculating the forces:

• Rolling Resistance Force = mu * m * g
• For Towing Truck:
• Frr1 = 0.006 * 25000 * 9.81
• Frr1 = 1471.5 N
• For Aircraft:
• Frr2 = 0.004 * 351800 * 9.81
• Frr2 = 13,804.6 N

• Aerodynamic drag Force = 1/2*(rho * Cd * A * V^2)
• For Towing Truck:
• Fad1 = 1/2 * (1.225 * 0.608 * 120 * 9)
• Fad1 = 402.192 N
• For Aircraft:
• Fad2= 1/2 * (1.225 * 0.026 * 650 * 9)
• Fad2 = 93.16 N

• Total Force = Frr1 + Frr2 + Fad1 + Fad2
• Total Force = 1471.5 + 13804.6 + 402.192 + 93.16
• Total Force = 15,771.484 N

• Power required by the Towing truck to move the Aircraft is = Force * Velocity
• Power = 15771.484 * 3
• Power = 47314.45 Watts
• Power = 47.314 KW

• Let us assume that the Towing of the aircraft takes place for about 5 minutes so, the energy requirement is:
• Energy = Power * Time
• Energy = 47.314 * 5/60
• Energy = 3.94 KWH

• As we consider a gear ratio of 10 so, the speed of motor is 3*10 = 30m/sec
• Torque generated by the motor is:
• T = (F * r)/G
• T = (15771.484*0.6096)/10
• T = 961.43 N-m

• Speed of Motor is 30m/sec
• Omega = 30/0.6096
• Omega = 49.2 rad/sec
• N = (omega*60) / (2*pi)
• N = 470 RPM

• Power required by the Motor is:
• P = T*omega
• P = 961.43 * 49.2
• P = 47.302 KW

MATLAB MODEL TO COMPUTE THE SAME:

close all

clear all

clc

v = 3;

g = 9.81;

rho = 1.225; %(density of air)

t = 5; %(min)

% PARAMETERS OF TOWING TRUCK

m1 = 25000;

mu1 = 0.006;

cd1 = 0.608;

a1 = 120;

% PARAMETERS OF AIRCRAFT

m2 = 351800;

mu2 = 0.004;

cd2 = 0.026;

a2 = 650;

% ROLLING RESISTANCE FORCE = mu*m*g

% FOR TOWING TRUCK

Frr1 = mu1 * m1 * g;

% FOR AIRCRAFT

Frr2 = mu2 * m2 * g;

% AERODYNAMIC DRAG FORCE = 1/2*rho*Cd*A*V^2

% FOR TOWING TRUCK

Fad1 = 0.5 * rho * cd1 * a1 * v^2;

% FOR AIRCRAFT

Fad2 = 0.5 * rho * cd2 * a2 * v^2;

% TOTAL TRACTIVE FORCE

Ft = Frr1 + Frr2 + Fad1 + Fad2;

% POWER = FORCE * VELOCITY

P = (Ft * v)/1000;

% ENERGY = POWER*TIME

E = P*(t/60);

fprintf('The total force required by the towing truck is %f N \n',Ft);

fprintf('The total Power required by the towing truck is %f KW \n',P);

fprintf('The total energy requirement for towing the aircraft is %f KWH \n',E)

Results:

The total force required by the towing truck is 15771.485250 N

The total Power required by the towing truck is 47.314456 KW

The total energy requirement for towing the aircraft is 3.942871 KWH

1. 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)

The following are the assumed parameters:

1. Also, Design the parameters in excel sheet

The following are the assumed parameters:

AIRCRAFT PARAMETERS

 

PARAMETERS	VALUES	UNITS
Mass of the aircraft	48500	kg
Coefficient of rolling resistance	0.03	--
Frontal area of the aircraft	85	M2
Final velocity of the aircraft	3	m/s
Initial velocity of the aircraft	0	m/s
Coefficient of drag of an aircraft	0.15	--
Acceleration time of the Towing vehicle	10	Seconds

 

 

TOWING VEHICLE PARAMETERS

 

PARAMETERS	VALUES	UNITS
Mass of the Towing vehicle	8000	kg
Coefficient of rolling resistance	0.013	--
Frontal area of the Towing vehicle	8	M2
Final velocity of the Towing vehicle	3	m/s
Initial velocity of the Towing vehicle	0	m/s
Coefficient of drag of the Towing vehicle	0.60	--
Acceleration of the Towing vehicle	10	seconds
Gear ratio	10

The total power required to push or pull the aircraft by the towing vehicle is 97.022952KW and the total tractive force is 32340.984N. To calculate the torque required  by the Towing vehicle the following parameters are assumed as shown above:

T = F_{r *} r /G

T = 32340.984 * 0.4 / 10

T = 1293.63 Nm

The torque required for towing the vehicle to push or pull the aircraft is 1293.63 Nm

To design the electric power train of the towing vehicle a motor is selected based on its power rating and the battery required based on energy requirement and duty cycle range to control the aircraft speed from zero to highest. 

MOTOR  

 

The motor is selected with a power rating of 97.022KW and torque of 1293.63 Nm .  TECO IE2 electric induction motor is selected which has a  power output of 110 KW and torque at full load is 1420Nm that satisfies the power output and torque needed for the vehicle.

MOTOR SPECIFICATIONS

Output power – 110 KW, 150 hp

Full load speed – 740 rpm

Full load Efficiency – 94.5%

Full load current – 210A

Full load Torque – 1420 Nm

Rated voltage – 400V/50 Hz

DUTY CYCLE

The rated voltage is 400V and the current is assumed to be 250A. So, the input power of the motor is calculated as product of voltage and current.

P_{i =} V * I

P_{i =} 400 *280

P_{i =} 112000W

P_{i =} 112KW

To calculate the duty cycle we divide output power to input power

 𝛿 = P_o / P_i  

 𝛿 = 97.02 / 112

 𝛿 = 0.866

 𝛿 = 86%

ENERGY

To calculate the energy required from batteries or battery pack. The energy is obtained by multiplying the output power and time. Assuming the battery usage for 5 minutes is 300 seconds.

E = P_o * t

E = 97.02 * 300

E = 29106kJ

So, the battery pack provides 29106 KJ of energy to the vehicle. Since the battery are rated in kilowatt hour (KWH). We have to convert KJ to KWH we divide it by 3600.

E = 29106/3600

E= 8.08KWH

So, the battery pack should provide at least 8.08 KWH to the vehicle.

 

Simulink model:

EXCEL SHEET CALCULATIONS 

RESULTS

Required torque output of the motor	1293.63 Nm
Input power of the motor	112KW
Duty cycle range	86%
Energy required from the batteries	8.08KWH

Block Diagram of Powertrain for Towing Vehicle-                                                                                                                                                                                                                                              

Electric Vehicles, consist of a battery, connected to an electric motor (either AC or DC) via an inverter and a motor controller. As automotive batteries generally operate on DC current, while the gird operates AC, a converter is needed to transfer power to the battery.One of the key advantages of an EV is its much higher levels of energy efficiency. Typically an EV will achieve a system efficiency of 80% from battery to wheel (roughly 60% from grid to wheel), while with a conventional IC engine it’s more like 15-25%. This is largely due to the fact that an IC engine is a “heat engine” (i.e. it has to burn fuel and produce heat to operate) and much of all that heat is simply wasted.

Reference:

1. https://www.grc.nasa.gov/www/k-12/VirtualAero/BottleRocket/airplane/weight2.html
2. http://aviationavi.com/all-kinds-of-weight-for-an-aircraft/
3. Ariffin , A. Rostam , W. Shibani, “Study of Aircraft Thrust-to-Weight Ratio”
4. https://www.grc.nasa.gov/www/k-12/airplane/move.html#:~:text=Airspeed%20is%20the%20vector%20difference,speed%20and%20the%20wind%20speed.&text=On%20a%20perfectly%20still%20day,less%20than%20the%20ground%20speed
5. http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/perf/to/index.html
6. http://www.dept.aoe.vt.edu/~lutze/AOE3104/