PART 1 · 2010. 1. 22. · these problems because the vehicle floats above the rail without any...

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Behind the Scenes: Theme Parks

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TEAMS 2009 Sponsors

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PART 1

Winning T-Shirt Design submitted by Caroline Win, Beavercreek High School (OH)

Scenario #5

The design of public transportation systems, including the ones found in amusement parks, requires the input of civil, mechanical, and electrical engineers. Construction engineers have the opportunity to turn those ideas into reality. Monorails are sometimes used in zoological parks to bring visitors closer to the animals without disrupting their habitat. Your team is to design structural components such as a vertical support pier and the guide rail beam between two piers such that it will perform adequately in all design conditions. In a MAGLEV monorail system, the lifting force of the electro-magnetic coils is just as important as sizing the heat sink for the solid-state triac switches used to control the electrical power to the coils. Providing electric power for lighting and air-conditioning to each car offers opportunities for innovative energy solutions such as flywheel storage, LED lights, and thermal storage. Your engineering team may also be asked to identify alternative on-board power solutions. The Monorail Society http://www.monorails.org/index.html Magnetically Levitated Trains http://www.maglevpa.com/tech.html Explaining the Physics of Everyday Life http://howthingswork.virginia.edu/electric_power_generation.html

Get Active! Learn more about monorail systems and the engineering behind their power:

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TEAMS Competition 2009©

scenario # 5

Submitted by David Meredith, P.E., Pennsylvania State University, Fayette, Uniontown, PA.

Designing Transportation Systems to Bring Visitors to Zoological Parks Close to the Animals Without Disrupting Their Habitat

Assumptions and Givens

• A typical support pier with a diameter of 1.2 m stands 6-m above grade. See Figure 5-1.

Introduction

Monorails are sometimes used in zoological parks to bring visitors close to the animals with-out disrupting their habitat. Your team is to design a monorail loop to move a large number of peo-ple per hour over the exhibits. You will need to design structural components such as the vertical support pier and the guide rail beam between two piers so that it can function in all weather condi-tions. The lifting force of the electro-magnetic coils is just as important as sizing the heat sink for the solid-state triac switches used to control the electrical power. Providing electric power for lighting and air-conditioning to each car offers opportunities for innovative energy solutions such as flywheel storage, LED lights and thermal storage, respectively. Your engineering team may also be asked to consider if Maglev would be the most efficient way to solve this problem, or to identify alternative on-board power solutions.

Background

Monorails offer a safe and efficient method of moving a large number of people through con-gested downtown regions or over pristine envi-ronmental areas. Traditional monorails run on wheels with guide wheels on each side to keep the load-bearing wheels on the rail. The cars are propelled by electric motors that get their power from an energized "third rail" through brushes that must maintain continuous contact with this rail to transfer the power to the moving vehicle. Moving surfaces in contact create maintenance problems - especially when forced to operate in weather conditions that range from hot to cold and dry to rain and snow or even ice.

Magnetic Levitation or Maglev solves some of these problems because the vehicle floats above the rail without any contact. It is supported on

a thin magnetic field (“like” magnetic poles repel). The good news is that the operating efficiency of these sys-tems increases at higher velocity - the Shanghai Maglev train travels at over 250 mph (417 km/h). The ride is smooth and quiet with no environmental pollution at the point of use.

If you know how to produce an electromagnet by wrap-ping copper wire around an iron core and passing elec-tricity through the wire, you are on your way to understanding magnetic levitation. Bigger coils can support heavier vehicles, but consume a greater quan-tity of expensive copper to build and need more electric-ity to operate. So, like an airplane, we try to keep the weight of the vehicle as low as possible. The vehicle will have lots of windows to get a clear view of the animals. But that means we need a big air-conditioner to keep the visitors from getting cooked by the sun. That unit will be heavy and lots of electrical power will be required to run it. An onboard generator to run the A/C unit will also add weight to the vehicle. Then we will need fuel (more weight) and to deal with the noise and emissions. As engineers we think that there has to be a better way for a ride that only lasts a couple of minutes.

Civil engineers build the piers and monorail sections using knowledge skills perfected through decades of building bridges. The cars, suspension system, and undercarriage are designed by mechanical engineers. Electrical engineers provide both the power and the safety control system. To make this project work, all engineers involved will need to think "outside the box" with new innovations. Indeed, building a maglev monorail for a theme park takes a team effort. Let's get started.

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• The top of the pier foundation is located 1 m below the ground surface.

• The foundation is a 60 cm thick square measuring 2.2 m on each side.

• A total length of 645 m of 25 mm diameter steel rebar is woven into the concrete to give it strength.

Figure 5-1. Vertical support pier

41. The total volume (m3) of concrete required to construct this pier and its foundation is closest to:

a. 7.91 d. 10.8 b. 9.37 e. 41.5 c. 10.5

Additional Assumptions and Givens

Figure 5-2. Horizontal guide rail beam

• The typical guide rail cross section between each pair of piers is 0.70 m wide by 1.3 m high. See figure 5-2.

• To make the guide rail lighter, and to provide a protected area to run power cables and controls, there is a void within the beam that is 32 cm wide and 90 cm tall, rounded at both top and bottom.

• The nominal length of each section is 35 m.

• There are 22 pieces of 25 mm steel rebar running the length of each section (four at the top, two rows of four at the bottom and five on each side).

• The density (ρ) of concrete is 2,300 kg/m3. Density is defined as the mass per unit volume (ρ = m/V).

42. The total mass (kg) of concrete required to construct this section of guide rail is closest to:

a. 51,980 d. 48,070 b. 51,060 e. 1,482 c. 48,760

More Assumptions and Givens

Lmax = ?

Figure 5-3. Guide rail beam expansion gap

• The gap between two guide rail sections is critical due to thermal expansion as the temperature changes. One end of the guide rail is rigidly attached to the pier. The other end is free to expand and contract as the ambient temperature changes.

• At maximum extension, the minimum allowable gap (to permit the insertion of an elastomeric material to keep out moisture) is 8 mm.

• The location selected for this park has temperatures that range from – 220C in the winter to +380C in the summer.

Grade

1 m

2.2m

6 m

D = 1.2m

60 cm

90 cm 32 cm

0.7m

1.3m

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• The coefficient of linear expansion for sand stone-based concrete is

1.1x10-5 m/m-0C (5-1)

• The equation for thermal expansion is

L = αLo T (5-2)

where,

α = coefficient of linear expansion (m/m-0C).

Lo = nominal length of the object being analyzed (m).

T = difference between the maximum and minimum temperatures of the object (0C).

43. The maximum gap (mm) at the other temperature extreme is closest to:

a. 15 d. 54 b. 23 e. 103 c. 31


• The equation to determine the lift of a magnetic dipole oriented vertically is given by

(5-3)

where,

FL = lifting force (N)

μo = permeability of free space (Webers/Amp-m).

m = magnetic moment of the power coil,

m = (N) (I) (A) (5-4) where,

N = number of wire turns around the coil.

I = electrical current through the wire (Amps).

A = cross-sectional area of the electric coil (m2).

h = levitation height of monorail car above the coil (m).

w = penetration into the receiving plate.

(5-5) where,

σ = conductivity of the receiving plate (1/Ohms-m).

d = thickness of the receiving plate (m).

v = nominal velocity of vehicle (m/s).

• The nominal traveling speed of each unit throughout the ride is 6 m/s.

• At nominal speed, the cars are to levitate 8 mm above the guide rail.

• The power coil has 120 turns of wire around a 20 cm diameter iron core with 8 amps of electric current flowing through the wire.

• Each receiving plate (mounted on the car) is a 30 cm plate of 1 cm thick aluminum with a conductivity of 1.7x107 (1/Ohms-m).

• The permeability of free space is 1.256x10-6 (Webers/Amp-m).

• Permeability measures the magnetization of a material that responds to an applied magnetic field.

44. The lifting force (N) for each power coil is closest to:

a. 43.6 d. 1314 b. 118 e. 7712 c. 521


• The copper wire selected to build each coil is AWG 16-gauge which has a diameter of 1.29 mm and a resistance of 13.1725 Ohms per km.

• The insulating varnish on the wire is thick enough that the wire is wrapped around the coil with a 1.35 mm centerline.

∆

∆ ∆

2(μo)(σ)(d)

32 • π • h4FL = (3μo • m2 ) w1 - (v2 + w2)1/2][

w =

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1.35 mm

0.25 m

Figure 5-4. Assembly detail and spacing of electromagnetic coils

45. The mass of copper (kg) required by the coil design per km of monorail is closest to:

a. 395 d. 12,750 b. 1,027 e. 31,500 c. 3,230


• The coils are spaced such that one power switch can control a string twelve (12) coils wired in series.

• The wire connecting each pair of coils is large enough to have negligible electric resistance.

• The electrical voltage required is given by

V = IR (Ohm’s Law) (5-6)

where,

V = voltage (Volts).

I = electrical current in the wire (Amps).

R = total electrical resistance for the total length of the 16-gauge wire.

1V = (1A) (1 Ohm)

Figure 5-5. Coiled string wired in series

46. The voltage (Volts) required to control one string of coils is closest to:

a. 10,000 d. 12 b. 95 e. 8 c. 31


• The power to each string of twelve coils is switched on and off as required by a solid state electronic component called a triac, or bidirectional triode thyrister. One design constraint is that each triac must be mounted on a metal heat sink to dissipate thermal losses equaling approximately one percent of the connected electrical power. Since each string is only energized intermittently (with time to recover before the next train arrives), effects of power factor and rms voltage can be ignored.

• The connected power is given by the equation:

P = I2R (5-7)

where,

P = connected power (Watts).

I = electrical current in the wire (Amps).

R = total electrical resistance for the full length of the wire (Ohms).

• 1 watt (W) = (1V) (1A)

47. The rate that thermal energy (W) must be dissipated from each triac is closest to:

a. 7.6 d. 760 b. 9.8 e. 3,140 c. 11.9


• Each car in each two-car unit normally seats 36 people (plus room for two wheelchairs in each car).

20 cm

• Coils are imbedded every 0.25 m for the length of the monorail loop.

• The density of copper is 8,200 kg/m3.

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• It takes 90 seconds to unload and load each unit.

• Units arrive and leave the station every two minutes.

• The average speed around the 3.5 km loop is 6 m/s.

48. Not including wheelchair passengers, the number of visitors per hour that this ride can hold is closest to:

a. 241 d. 1,080 b. 386 e. 2,160 c. 482


2 m 3 m x 1.2 m 3 m x 1.2 m

3 m x 1.2 m 3 m x 1.2 m

15 m

2 m

3 m x 0.5 m 3 m x 0.5 m 3 m x 0.5 m 3 m x 0.5 m

0.8m x 1.2m

0.8m x 0.5m

Figure 5-6. End view and side view of car showing windows and glass doors

• The air-conditioning load for each car is caused by the sunlight steaming through the windows plus the body heat produced by the onboard visitors plus the lights onboard plus the energy gained because the outside ambient air temperature is higher than the control temperature inside the car.

• The solar gain to the vertical windows mounted in each car is a complex Trigonometric problem. As the cars weave through the exhibits, the sun shines at varying angles. The sun’s intensity also depends on the cloud cover and how high it is in the sky for that date and time of day.

• To simplify the problem we are assuming that the train is running straight in an east- west direction (the windows are facing due north and south). We also consider the solar gain at noon on a sunny day during the solar equinox. Under these conditions, the sun is directly south and the angle between a vertical line and

the sun’s angular position is equal to the local latitude.

• The solar gain under these conditions is given by: Q = I • A • SC • cos β (5-8)

where,

Q = thermal gain from solar (W).

I = incident solar insolation (intensity) (W/m2).

A = window area exposed to sunlight (m2).

SC = Shading coefficient or the ratio of solar energy passing through the window to the solar energy striking the outside of the window (-).

β = solar angle above the horizontal (0).

• Horizontal and vertical refer to the relative position of the train.

• At solar equinox the sun can be observed to be directly above the earth’s equator.

• Latitude measures the location (angle) of a place from the equator. Zero degrees at the equator, 900 N for North Pole, and 900 S for South Pole.

• The local latitude is 300N.

• The incident solar isolation is 980 w/m2.

• The shading coefficient for the selected window is 0.45.

49. The solar gain (W) to each car is closest to:

a. 4,410 d. 8,830 b. 5,095 e. 10,520 c. 6,620


• Lighting within the cars at night is provided by Light Emitting Diodes (LEDs), which are almost ten times as efficient as conventional incandescent light bulbs and last over 20 times as many hours.

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• Each car has ten ceiling-mounted LEDs to provide ambient light within the car.

• Each LED draws 70-milliamps at a resistance of 1,714-Ohms. In addition, each car has two (2) computer-controlled movable 8-Watt LED spot lights on the outside of each car. These spotlights can be pointed toward the animals based on electronic tracking signals from their radio-transmitting collars.

• There are 200 decorative LEDs mounted on the exterior of each car to outline the location of each car to the rest of the park’s visitors.

• Each exterior decorative bulb draws a one milliamp current at a resistance of 120,000 Ohms.

50. The total electrical power (W) dedicated to the lighting system in each car is closest to: a. 16 d. 84 b. 24 e. 124 c. 40

41

2009 Competition

Part 1 Solution

Designing Transportation Systems for Park Visitors

Scenario #5 Designing Transportation Systems t rs to Zoological Parks Close to the

1. Answer c

The length of the pier to the top of the foundation is: 7m

The total volume is:

V = (L)(W)(H)

= (2.2m)(2.2m)(0.60m)

= 2.9 m

The volume of steel is:

V = (π/4)(D (H)

= (3.14/4)(0.025m (645m)

= 0.32m

Subtracting the volume of steel from the total volume yields:

.92m +2.9m -0.32m =10.5m

2. Answer b

The total volume of the guide rail is:

V = (L)(W)(H)

= (35m)(0.7m)(1.3m)

= 31.85m

The void cross section can be considered as the sum of a 0.32m diameter circle,

90cm – 16cm -16cm = 0.58m).

he volume of the void is:

o Bring VisitoAnimals Without Disrupting their Habitat

4 3 s

)2

) 2

3

3 ]3 3 3 [7

4 3

plus a rectangle with 0.32m by 0.58m ( T

21

V = (π/4),(D )(L)+(W)(H)(L)

(π/4)(0.32m (35m)+(0.58m)(0.32m)(35m)

9.31m

he volume of steel is

= (# of pieces)(π/4)(D )(L)

(22 pcs)(3.14/4)(0.025m (35m)

0.377m

he net volume is calculated as the total volume minus the volume of the void

= 31.85m -9.31m -0.377m

22.16m

ass is the density multiplied by volume:

= (ρ)(V)

(2,300kg/m )(22.16m )

50,968kg

3. Answer c

The temperature variation is:

T = 38 C - (-22 C) = 60C

The guide rail length is:

35m

The coefficient of linear expansion is given as:

1.1x1 m/m- C.

Inserting into the equation yields:

2

= ) 2

= 3 T V 2

= ) 2

= 3 Tand minus the volume of the steel. V 3 3 3 = 3 M M = 3 3

=

4 Δ o o o

0 5− o

22

Δ L =

= (1.1x10

= 0.023m = 23mm

The maximum gap must also include the minimum gap of 8mm.

Therefore, Maximum gap is:

23mm+8mm=31.mm

4. Answer d

To determine the magnetic moment:

m = (N)(I)(A)

= (120 turns)(8 Amps)[(π/4)(0.2m ]

= 30.14

To determine the penetration:

W = 2/[(

ΔTαLo )CC)(35m)(60m/m oo5 −− 4 ) 2

oμ )( )(d)]σ = 2/[(1.256x10 webers/amp-m)

(1.7x10 )(1/Ohms-m)(0.01m)

= 9.337

Inserting these values into equation (5-3) yields:

F = (A)(B)

where A = [(3 ]

and B = [1-w/(v

A = (3)(1.256x10 webers/amp-m)(30.14) ]/[(32)(3.14)(0.008) ] = 8,316

B = 0.1587

6−

7

L

2 4

0μ )(m )]/[(32π)(h ) ])w 1/222+ 6− 2 4

23

FL = (8,316)(0.1587) = 1,319 N

5. Answer c

The wire length of each coil is given by:

L = (N)(π)(D)

= (120 turns)(3.14)(0.2m)

= 75.35 m/coil.

There is one coil embedded every 25cm or 4 coils/m or 4,000 coils per km.

So the total length of wire per km of monorail is:

(4,000 coils/km)(75.35m/coil) = 301,440 m/km

The cross-sectional area of the wire is:

A = (π/4)(D

= (3.14/4)(0.00129m

= 1.306x10

4 )2 ) 2 26 m− The volume is given by:

V = (A)(L)

x10-6m2)(301,440m/km)

= 0.3937m km

Mass is given by density multiplied by volume:

m = (ρ)(v) = (8,200 kg/m )(0.3937m /km)

= 3,229 kg/km of monorail

6. Answer b

The length of wire per coil was determined to be: 75.35m.

= (1.306 /3

3 3

4

24

In twelve coils, the total wire length is:

(75.35m)(12) = 904.2m = 0.905 km

In previous assumptions we find:

The resistance per km is given as:

13.1725 Ohms/km

So the total resistance is:

R = (13.1725 Ohms/km)(0.905km)

= 11.91 Ohms

The voltage is calculated by multiplying the resistance times the current of 8

= (I)(R)

(8 Amps)(11.91Ohms)

95.2 Volts.

7. Answer a

With the resistance calculated at:

R = 11.91 Ohms

and the current at:

I = 8 Amps.

The total switched power is:

P = (I) (R)

= (8 Amps) (11.91 Ohms)

= 762 Watts

One percent of this value is:

7.6 Watts

Amps. V = =

4 2

2

25

48. Answer e

A two-car unit with each car holding 36 people will handle: 72 people per trip.

Leaving every 2 minutes:

Indicates 30 minutes/hr.

Therefore,

(72 people/unit)(30 units/hr) = 2,160 people/hr

9. Answer b

The window area on the south side of the car is:

A = 4[(3m)(1.2m)]+4[(3m)(0.5m)] + 2[(0.8m)(1.2m)]+2[(10.8m)(0.5m)]

= 23.12m

The solar angle above the horizontal is the complement of the vertical angle at the

The vertical angle is: 30

Using equation (5-8) the solar gain is:

Q = (I)(A)(SC)(cos

= (980W/m

= 5,098 W

0. Answer e

The power of each interior LED is:

P = (I) (R)

Amps (1,714 Ohms)

= 8.4 Watts

4 2 equinox. o ooo 603090β =−= β) )s60)(0.45)(co)(23.12m o22 5 2

= (0.070 ) 2

26

So ten lights require:

84 Watts of electric power.

The two spotlights total to:

(2)(8Watts) = 16 Watts.

Each exterior LED requires:

P = (I) (R)

= (0.001 Amps) (120,000 Ohms)

= 0.12 Watts

The 200-LED string requires:

(200)(0.12W) = 24W

Adding all LEDs yields:

84W+16W+24W = 124W

2

2

27

Shell is the exclusive sponsor of theTEAMS 2009 T-Shirt Design Challenge

Behind the Scenes: Theme Parks

ww

w.j

ets

.org

/te

am

s

TEAMS 2009 Sponsors

S.D. Bechtel, Jr. FoundationShellTyco Electronics

JETS Sponsors

AbilityOneBechtel Group FoundationBlack & VeatchCH2M HillDisney Y.E.S.

MicrosoftMotorola FoundationRockwell CollinsSolidWorksUnited Engineering Foundation

PART 2

Winning T-Shirt Design submitted by Caroline Win, Beavercreek High School (OH)

JETS, Inc.


extension to scenario # 5

Submitted by David Meredith, P.E., Pennsylvania State University, Fayette, Uniontown, PA.

Designing Transportation Systems to Bring Visitors to Zoological Parks Close to the Animals Without Disrupting their Habitat

Background

To reduce the electric power requirements and weight of each car, there is no air-conditioning unit aboard. It has been replaced by an insulated tank that holds ice water at 0OC. A small pump and fan circulate cold water to the cooling coil and cool air to the passengers. Each time the car returns to the unload/load station, fresh ice is added to replace the quantity that melted during the previous trip. Excess water is simply dumped from the car through an overflow pipe.

Assumptions and Givens

Overflow

Fan

Pump

Ice + water tank

Figure 5-1x. The cooling systemof the train cars

The quantity of thermal energy absorbed by melt-ing ice is given by:

Q = (m) (Δhf) (5-1x)

where,

Q = quantity of thermal energy (kJ)

m = mass of melted ice (kg)

Δhf = latent heat of fusion for ice (kJ/kg)

• The latent heat of fusion for ice is 333.6 kJ/kg. • The total thermal load on each car is 21,500 kJ/hr.

Task 1

Calculate the total mass (kg) of ice that must be added after each trip.

Additional Background

Electrical generating equipment is usually very large and heavy. To minimize both space and weight, the design team has decided to install a flywheel generator. By spinning a disk at very high speeds (40,000 rpm in this case), kinetic energy is stored for later use. In this case it is converted back into electricity during each trip. The disk is re-spun to its initial speed each time the car returns to the unload/load station.


The kinetic energy of a spinning object is given by:

E = ½ m(R2 – r2) ω2 (5-2x)where,

E = energy (J)

m = mass of rotating disk (kg)

R = outside radius to center of mass for a rotating disk (m)

r = inside radius to center of mass for a rotating disk (m)

ω = angular velocity (radians/s, where there are 2π radians in a full rotation)

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30 cm

24 cm

32 cm

Figure 5-2x. The spinning disk ofa flywheel generator

• The outside and inside diameters of the disk are 30 cm and 24 cm respectively.

• The disk is 32 cm high.

• The mass of the ribs connecting the disk to the center shaft is negligible. • The carbon fiber material it is constructed from has a density of 1,100 kg/m3.

Task 2

Calculate the kinetic energy (J) stored in this disk at design speed.

Additional Background

Providing both mobile electric power as well as comfort conditioning (heating or cooling) to ve-hicular passengers has a unique set of design constraints. The size and weight of the required equipment will create greater wind resistance and lower fuel efficiency. The chosen system should have low operating cost and minimal environmental emissions - especially near the vehicle. Because of the high visibility of this project to the public, the management team has said not to worry about the initial cost of the system. As a demonstration model project, they are willing to fund a design that meets the above constraints.

Task 3

Name several alternative on-board electric power sources to the flywheel system. Discuss the pros

and cons of each and make a final recommendation on which system should be selected.

8

2009 Competition

Part 2 Solution

Designing Transportation Systems for Park Visitors

Extension to Scenario #5 Designing Transportation Systems to Bring Visitors to Zoological Parks Close to the

Animals Without Disrupting their Habitat

Task 1 The ride is 3.5km long and is covered at an average speed of 6m/s. The time duration for each ride is:

t = 6m/s

3,500m = 583s = 9.7 minutes

In addition it takes 1.5 minutes to unload and load. Therefore, the total cycle time is: 9.7 minutes + 1.5 minutes = 11.2 minutes = 0.187 hr The amount of energy required for 0.187 hr is: (21,500 kJ/hr)(0.187 hr) = 4,021 kJ Solving equation (5-1x) for mass gives:

m = f

Q 4,201kJ 12.1kgΔh 333.6kJ/kg

= =

Task 2 The volume of the disk is: V = ( )(H)dπ/4)(D 22 − = (3.14/4)(0.3 )(0.32m)0.24m 22 − = 0.00814m 2

Using equation: m = (ρ)(V) yields: m = (1,100kg/m = 8.95 kg ))(0.00814m 23

To calculate the angular velocity (ω in radians/s we need to convert the spinning disk of 40,000 rpm to revolutions per second:

)

50

(40,000rpm (1min/60s) = 666.7 revs/s. (666.7 revs/s)(2 radians/rev) = 4,187 radians/s π Replacing all these values into equation (5-2x) yields: E = (0.5)(m)(R 222 )ωr− = (0.5)(8.95kg)(0.15 (4,197 radians/s) )m0.12m 2222 − 2

= 635,453 J Task 3 1. Large car batteries – well understood technology, but very heavy and (since they usually involve acid) somewhat of an environmental hazard. 2. Capacitors – some progress has been made recently in storing large quantities of electric power in capacitors using nano technology. But it is new and relatively unproven. Therefore the cost will probably be fairly expensive. 3. Solar panels on the roof – energy would be free, but panels are very expensive and probably not enough roof area is available. Nor would they work at night or on cloudy days. 4. Fuel cells – Very efficient and compact, but the technology today is still very expensive. And the storage of Hydrogen fuel is not well understood yet. 5. Combustion engine (could be vapor-compression, gas turbine, diesel, etc) – well known technology and fairly compact, but noise and vibration and on-board fuel storage could be problems. 6. Compressed air – good for short events, but not for continuous service like this train. 7. You could also install a 3rd rail (like a subway or electric bus) but that would no longer be defined as an on-board electric power source.

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