Design of an Electric Powertrain for Aircraft Towing application

OBJECTIVES:

• Design and simulate an electric powertrain capable of towing a Boeing 737
• Understand aircraft towing procedures and the reasons behind them 
• Learn about aircraft takeoff and other ground procedures to manuever the plane

 

1) Calculate the force and power required by a towing vehicle to push/pull an aircraft. Make all necessary assumptions.

 

         

 

Above are two figures that present the towing that is done to move the aircraft. The figure on the left represents the traditional towing method via a tow truck/tractor. The figure on the right represents an aircraft towing system (ATS) that was created by a company founded in 2003 called Mototok. The ATS' they create are considered nothing less of a revolutionary technology due to the following factors, but not limited to:

• Fully clean electric drive
• Extremely powerful electric motors – driven by high performance, maintenance-free batteries with high recycling capability
• Radio remotely controlled - only 1 person required for operation (as opposed to atleast 4 that are required for towing with a truck/tractor). The operator is able to move around to see every vantage point, ensuring high safety and reliability
• Extremely compact
• Little to no maintenance costs

 

Having chosen the Boeing 737-400 (68,000kg) as the aircraft to evaluate the following questions, the ATS chosen is the Mototok Twin 7500, with a max towing capacity of 75,000kg.

(Note - all numbers used are actual parameters of both Boeing 737-400 and Mototok Twin 7500, unless stated otherwise)

 

The ATS will need to account for the following to produce enough thrust force to tow the aircraft:

1. Rolling resistance force of aircraft
2. Rolling resistance force of ATS
3. Inertial force of aircraft
4. Inertial force of ATS
5. Drag force aircraft experiences
6. Drag force ATS experiences

 

Boeing 737-400:

Max ramp weight	68,000kg
Wingspan area	105.4`m^2`
Assumptions
Coefficient of rolling resistance	0.02
Air density	1.225 `kgm^-3`
Coefficient of drag	0.019
Towing time	20 seconds

 

Mototok Twin 7500:

Mass	2,100kg
Frontal area	2.596m x 0.35m = 0.7476`m^2`
Towing speed	1.05m/s
Assumptions
Coefficient of rolling resistance	0.02
Air density	1.225 `kgm^-3`
Coefficient of drag	1.20
Towing time	20 seconds

a. Rolling resistance force of aircraft:

`F_(rr)=mu_(rr)mg`

= 0.02*68000*9.81

= 13,341.6N

 

b. Rolling resistance force of ATS

`F_(rr)=mu_(rr)mg`

= 0.02*2100*9.81

= 412.02N

 

c. Inertial force of aircraft

`F=m(v/t)`

68000*(1.05/20)

= 3570N

 

d. Inertial force of ATS

`F=m(v/t)`

2100*(1.05/20)

= 110.25N

 

e. Drag force aircraft experiences

`=1/2rhoAC_dv^2`

`=1/2*1.225*105.4*0.019*1.05^2`

= 1.35N

 

f. Drag force ATS experiences

`=1/2rhoAC_dv^2`

`=1/2*1.225*0.7476*1.2*1.05^2`

= 0.61N

 

Total traction force the ATS needs to produce = 13,341.6 + 412.02 + 3570 + 110.25 + 1.35 + 0.61 = 17,435.83N or approximately 17.5kN

Total traction power = Fv = 17,435.96*1.05 = 18,307.76W or approximately 18.3kW

 

Note: The force and power values are for both pushing and pulling the aircraft, which is not always the same. Push and pull forces are often applied at different angles (consider a box for instance), making the pull force smaller than push force. However, with an ATS towing an aircraft, both pushing and pulling are applied at the same angle, which leads to no difference in the amount of force needed.

 

2) Design an electric powertrain with type of motor, its power rating, and energy requirement to fulfill the aircraft towing application. Estimate the duty cycle range to control the aircraft speed from zero to highest. Make all necessary assumptions and illustrate the powertrain with a block diagram.

 

Motor:

 

Continuing on from the previous question, 18.3kW is the minimum amount of power rating required in the motor.

How much torque the motor can produce is also a very important characteristic, hence before choosing a motor it is also necessary to know the rated torque needed by the ATS to tow the aircraft:

`T=(Fr)/(Geta_g)`

where;

G = gear ratio of the transmission system -------------- Mototok's ATS' are known to have extremely low gear ratios (around G = 27) to achieve the kind of massive torque needed to tow an aircraft of the size of Boeing 737-400

r =  radius of the wheels; given as 0.15m

 

Assuming the efficiency of the gear transmission is 95%:

`T=(17435.83*0.15)/(27*0.95) = 101.96Nm`

`omega=P_(out)/T_(out)`

`omega=18307.76/101.96 = 179.55(rad)/(sec)=1714.65 RPM`

 

The motor selected keeping the above criterias in mind is 3-phase AC induction motor H-MA-MTS series:

• Power rating of 22kW
• Rated torque of 120Nm
• Max rotational speed of 2940 RPM
• Voltage rating of 480V

 

The motor needs to generate an output power of 18.3kW. Assuming the motor is 90% efficient:

`P_(i) = P_(o)/(eta_m)`

`P_(i)=18307.76/0.9=20,341.96W` -------------- this is the power input it needs to be given from the battery

 

 

 

Above is a basic diagram of an electric powertrain. Motor properties have been determined, and the motor must receive power from the battery through a motor controller which is a power converter. The power converter has been assumed to be 100% efficient.

 

Battery:

Everything calculated so far has been done with the idea of towing the aircraft at 1.05m/s, which is the highest speed the towing machine can travel at. To move at this speed, full load current must be supplied by the battery to the motor.

 

Full load current (`I_(i)`) = `P_(i)/(sqrt3*V_(m_(rated))*PF)`

where;

PF = power factor; generally ranging between 0.85 - 0.9 for full load

`V_(m_(rated))` = 480V (as given in the motor specification)

 

`I_(i)=20341.96/(sqrt3*480*0.87)=28.12A` ----------------- current drawn by the motor when operating at full load

 

Now that `I_(i)` is known, voltage of the battery `V_(bat)` can be calculated:

`V_(bat)=P_(i)/I_(i)`

`V_(bat)=20341.96/28.12=723.40V`

 

`V_o=V_(bat)*d` ----------- `V_o` refers to the voltage coming out of the power converter and given to the motor

where;

d = duty cycle

Since the motor rating is 480V, `V_o` cannot exceed 480V, especially since the power converter is assumed to be 100% efficient. Thus:

`d=V_o/V_(bat) =480/723.4=0.6635` --------------- duty cycle range from 0 - 0.6635 to control speed of ATS from zero to highest

 

Energy requirement of the battery:

`E=P_i*t`

Assuming the total time the battery needs to discharge its energy to the power converter and thus to the motor is 60 seconds (including the 20 seconds of towing at top speed of 1.05m/s):

`E=20341.96*60=1,220,517.6J`

= `(1,220,517.6)/(1000*3600)=0.339kWh` ------------------ this is the minimal rating for the battery

 

 

SIMULATED MODEL: A full model has been simulated to present the electric powertrain of the chosen Mototok 7500NG towing a Boeing 737-400. All the above calculated parameters (sourced from actual data) and stated assumptions have been inputted to evaluate the performance of the system.

 

 

MOTOTOK 7500NG SUBSYSTEM:

 

 

BOEING 737-400 VEHICLE BODY SUBSYSTEM:

 

OUTPUT:

 

 

• In 120 seconds, the ATS is able to tow the aircraft for 78.22m

 

DISTANCE TRAVELLED:

 

• Since the calculations done in this project are for 20 seconds of towing at top speed, the above graph showcases the distance travelled in the final 20 seconds of the simulation, which happens at basically the top speed the ATS can travel at. In the final 20 seconds, the ATS is able to move the aircraft by 16.22m, giving a speed of about 0.81m/s. According to ATS data, it can travel at a top speed of 1.05m/s, which is not too far from our results here and is fairly accurate.

 

STATE OF CHARGE:

 

 

• In the 120 seconds of towing, the battery has depleted from 100 to 90%

 

TOTAL TOWING DISTANCE:

 

 

• The current powertrain is capable of towing the aircraft for about 1190 seconds, which is about 20 minutes before the battery is completely depleted to 0% and needs to be recharged.
• The total distance that this powertrain can tow the aircraft is about 950m in those 1190 seconds.

 

VELOCITY:

 

 

• The ATS reaches its top speed of about 0.82m/s at around 100 seconds, and travels constantly at that speed until the battery runs out.

 

 

Conclusion:

An electric powertrain is designed and simulated to tow a Boeing 737-400. Actual industrial data (barring the difference in motor used in the ATS) is used from verified sources and results obtained are quite close to actual data, making this quite an accurate model. The designed electric powertrain is capable of towing the craft for about 20 minutes. It reaches its top towing speed of 0.82m/s at about 100 seconds and is able to maintain that until the battery runs out. With a more powerful motor and battery system, this simulation can be further built upon to achieve even more detailed results with higher precision.

 

 

 

------------------------------------------------------ Questions and Answers------------------------------------------------------

 

3) Search and list out the total weight of various types of aircraft

 

Type	MTOW [kg]	MLW [tonnes]
Antonov An-225	640,000	591.7
Airbus A380-800	575,000	394
Boeing 747-8F	447,700	346.091
Boeing 747-8	443,613	306.175
Boeing 747-400ER	412,770	295.742
Antonov An-124-100M	405,060	330
Boeing 747-400	396,900	295.742
Lockheed C-5 Galaxy	381,000	288.417
Boeing 747-200	377,840	285.700
Boeing 747-300	377,840	260.320
Airbus A340-500	371,950	240
Airbus A340-600	367,400	256
Boeing 777F	347,800	260.816
Boeing 777-300ER	351,800	251.29
Boeing 777-200LR	347,450	223.168
Boeing 747-100	340,200	265.300
Airbus A350-1000	308,000	233.5
Boeing 777-300	299,370	237.683
Boeing 777-200ER	297,550	213.00

 

Maximum take off weight (MTOW) - max weight the pilot is permitted to take off with

Maximum landing weight (MLW) - max weight the pilot is permitted to land with. This would be lower than MTOW because of the reduction in mass of fuel that would be used up during the journey

Manufacturers empty weight (MEW) - includes weight of the structure, and everything that was "built in" for the operation of the aircraft. Excludes passenger or payload weight and fuel.

Maximum ramp weight (MRW) - max weight permitted for manuevering on the ground (taxiing or towing)

Aircraft gross weight - total weight at any given time during in-flight or ground operations. Decreases during flight due to fuel usage.

 

4) Is there any difference between ground speed and airspeed?

 

Airspeed - speed at which the aircraft is moving relative to the air it is flying in

Ground speed - speed at which the aircraft is moving relative to the ground. This term can only be used when the aircraft is moving in a horizontal direction. It will not be applicable for any vertical movement such as during lift or takeoff.

Ground speed = Airspeed + Windspeed

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.

For instance, 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 it would thus take the aircraft five hours to reach its destination.

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.

 

5) Why is it not recommended to use aircraft engine power to move it on the ground at the airport?

 

Aircrafts use "pushback" to steer the vehicle away from the terminal/parking area. Pushback is the procedure where the aircraft is attached to a towing vehicle and is 'pushed back' i.e. reversed from the terminal/parked area towards the taxiway (refer to figure below). Although these towing vehicles are much smaller than the aircraft itself, they produce high torque and carry high mass.

 

 

Aircrafts are perfectly capable of moving themselves backwards using 'reverse thrust', and this apparently could save a lot of time that is required in connecting the towing truck, attaching the tug, etc. But why don't they?

Planes are different from ground based vehicles in the sense that they do not have a direct connection between their engines and wheels. The engines produce a thrust by sucking in air from the front and propelling it through the back, which provides the force to move forward. In reverse thrust, however, air is sucked into the engines but instead of moving to the rear, it is ejected through new openings that reverse the thrust (refer to the media below).

 

 

Reverse thrust is very commonly used to decelerate the plane once it has landed, however there are several reasons as to why this method isn't optimal for pushing back:

• The air blast around the aircraft can stir debris that may cause damage to other ground vehicles, groundsmen, equipment, etc.
• Since the sucking in of air as the engine spins with increasing power creates a vortex, anything else in fairly close proximity could also be sucked in. This can be very dangerous and expensive, especially if someone gets injured or a sharp tool sucked into the engine damages it.
• This operation also uses a lot of fuel and is very noisy.
• Aircrafts don't have rearview mirrors like other ground vehicles, so a spotter would anyway be needed to guide the pilot, which negates the whole point of doing this without any manpower.
• To use this procedure, the groundsmen would need to clear everything near the terminal gate for safety reasons, and this would not save any time compared to using a towing vehicle.

 

Using a towing vehicle for pushback until the plane is a safe distance away from the terminal gates, passengers, tools, and groundsmen, is therefore recommended over using aircraft engine power to manuever it on the ground.

 

6) How does an aircraft get pushed to the runway when it's ready to take off? 

 

Once all passengers have boarded and pilots have been given authorisation to steer the aircraft away from the parking area, the first procedure the aircraft undertakes is pushback. Once it is a fair distance away from the terminal gate, groundstaff and any equipment, it begins taxiing. Taxiing is the procedure before the aircraft takes flight, in which the aircraft moves on the taxiway using its own engine power without the assistance of any towing vehicles. Taxiways are well designed paths for the aircraft to navigate through to get to the runway (refer to figure below).

 

 

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

 

Take-off power:

Where this power comes from is the interesting part. Once the aircraft is on the runway, it starts to accelerate using jet fuel to power the engine that now acts as a vortex sucking air in. As the air is sucked in and ejected out the rear of the engine, this creates the thrust to accelerate forward. As the airplane gains speed, lift is induced, and when the lift force exceeds the weight force of the aircraft, the aircraft is able to produce vertical motion. 

`F_(l)=1/2rhoAC_Lv^2`

To increase the lift force, properties such as shape of airfoil, angle of attack, wingspan area, and the speed of the plane can be altered. The aircraft is already accelerating on the ground by the thrust produced, but to increase the lift further, slats and flaps are activated. This changes the shape of the airfoil in such a way that the wingspan area is increased. In addition, the angle of attack (the nose of the plane) is increased, tilting the craft to about 15 degrees, which provides the plane with enough lift force to make it takeoff.

 

Tyre design:

The number of tires increase with the aircraft weight, as the weight is required to be distributed as evenly as possible for stability of the entire structure. There are mainly two types of tire designs:

• Bias ply tire

- Casing plies running diagonally provide the strength

- Inter-tread reinforcing fabric: provides additional high-speed stability, reduces tread distortion under load, protects casing plies, etc.

 

• Radial tire

- Offers more landings per tread

- Lighter weight than bias tires, sometimes at the expense of retreading

- Plies run radially 

 

Rolling resistance:

The force that resists the motion of a body that rolls on a surface. With reference to this topic, the rolling resistance force exists between the aircraft wheels and the tarmac it is moving on.

`F_(rr)=mu_(rr)N`

This force is proportional to the normal force on the tires and also to the coefficient of the rolling resistance, which depends on:

• Tire material
• Tire dimensions
• Tread thickness and shape
• Tire pressure
• Surface roughness
• Tire temperature
• Dryness/wetness of the surface

 

Tyre pressure:

Aircraft tires are inflated with an inert gas, nitrogen being the most common one. This is because they are non-reactive and not flammable. Tires experience high magnitudes of forces, especially during takeoff and landing, and to minimize expansion and contraction due to the pressure and temperature changes, it is important to use an inert gas. If oxygen, a flammable gas, was used, then there could be a serious risk of tire explosion. 

Making sure that the aircraft tires are properly inflated is a crucial part of preflight and routine service checks which can go a long way to maximizing tire utilization and minimizing risks to safe operations. 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. If the tire operates underinflated or overdeflected/overloaded, the nylon cords which form the structure of the tire go in and out of compression as the tire rotates. This weakens the cords – much like a paperclip which is bent back and forth – until eventually the cords break. If enough cords break, the entire structure will eventually fail.

 

Brake forces when landing:

During braking, the kinetic energy of the moving plane is converted to heat as the tires make contact with the tarmac. Braking is done through disc brakes, the same type of mechanism used in cars. As soon as the pilot presses the brake pedal, a signal is sent which induce the actuators to press the brakes against the moving wheels, thus creating friction and slowing down the angular velocity of the wheels. This whole process produces a lot of heat, because of the amount of energy required to stop a plane, therefore these disc brakes are designed as heat sinks in order to thermally withstand the changes in temperature.

To be able to generate this magnitude of friction, several braking mechanisms are needed. The most obvious ones are the friction created by the wheels on tarmac, and the disc brakes applied on the wheels. However, to safely execute this process and reduce the braking distance once the plane has landed, it becomes vital to have other mechanisms.

• Reverse thrust - this allows the slowing of the plane by ejecting wind from the engine in the direction the plane is moving, as opposed to ejecting it behind the engine (which is used for thrust).
• Flaps/spoilers - located at the end of the wings, they reduce aerodynamic lift (just like how they do in cars) and increase drag