CE 353 Lecture 6: System design as a function of train performance, train resistance

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CE 353 Lecture 6 System design as a
function of train performance, train
resistance
Text (31p.) Ch. 2 (14-27), Ch. 5 (87-103) Supp.
Text Ch. 6 (69-89), Ch. 9 (140-155)

• Objectives
• Choose best route for a freight line
• Determine optimum station spacing and length

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Choose best route for a freight line
Figure 2-6 Three Ways to H With Coal (Armstrong,
p. 19) One objective might be to minimize the
total energy used (HP-hours) Another might be to
minimize travel time
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Choose best route for a freight line (cont.)

• Train Resistance
• 2 forces of resistance
• inherent or level tangent resistance (speed,
  cross-sectional area, axle load, journal type,
  winds, temperature, and track condition)
• use a Davis equation variation (from Hay, pg. 76)
• where
• Ru unit resistance in pounds per ton of train
  weight
• w weight per axle in tones - weight on rails in
  tons (W) divided by the number of axles (n)
• b an experimental coefficient based on flange
  friction, shock, sway, and concussion
• A cross-sectional area in square feet of the
  car or locomotive
• C drag coefficient based on the shape of the
  front end of the car or locomotive and the
  overall configuration, including turbulence from
  car trucks, air-brake fittings under the cars,
  space between cars, skin friction and eddy
  currents, and the turbulence and partial vacuum
  at the rear end
• incidental resistance (curvature, grades)

Ru 1.3 29 bV CAV
2
w
wn
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Choose best route for a freight line

• for a 100 car container train, 200,000 lbs/per
  car, 50 mph, 0.5 grade, 4 curve
• weight of locomotives 300 tons/each
• weight of train 100x200,000 lbs. 20,000,000
  lbs 10,000 tons
• weight of single freight car (in tons)
  200,000/2,000 100 tons/car
• inherent force 50,850 lbs. (about 5.1 lbs. per
  ton)
• w 100 tons/4 axles 25 tons/axle
• b 0.03 for locomotives and 0.045 for freight
  cars (lets use 0.045)
• A 85 - 90 sq. ft. for freight cars (lets use
  90)
• C 0.0017 for locomotives and 0.0005 for freight
  cars (lets use 0.0005)
• inherent force 1.3 (29/25) (0.0350)
  (0.000590502)/(254)
• 1.3 1.16 1.5 (112.5/100) 5.085 lbs/ton
  for the freight car
• 5.085 lbs/ton 10,000 tons 50,850 lbs
• See Figures 2-7, 2-8, and 2-10 (next slides)

Ru 1.3 29 bV CAV
2
w
wn
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Figure 2-7 How Much Energy It Takes to Move a
Car (Armstrong, p. 21)
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Figure 2-10 Energy needed to maintain speed,
accelerate and curve (Armstrong, p. 25)
Figure 2-8 Train Resistance (Armstrong, p. 22)
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Grade Resistance

• F (W X CB) / AB 20 lbs/ton for each percent
  of grade
• where
• W weight of car
• CB distance between C and B
• AB distance between A and B
• For our example
• Grade Resistance 20 0.5 10 lbs/ton 10
  lbs/ton 10,000 tons 100,000 lbs

Figure 9-1 Derivation of Grade Resistance (Hay,
p. 141)
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Curve Resistance

• Slippage of wheels along curved track contributes
  to curve resistance
• See Figure 9-2, 9-3, 9-4, and 9-5 to visualize
  curve resistance (next slide)
• Unit curve resistance values determined by test
  and experiment
• Conservative values suitable for a wide range of
  conditions 0.8 - 1.0 lbs/ton/degree of curve
• Equivalent grade resistance - divide curve
  resistance by unit grade resistance
• For our example
• Equivalent grade resistance 0.8/20 0.04 of
  the resistance offered by a 1 grade
• Curve resistance (4 0.04) 20 3.2 lbs/ton
  3.2 lbs/ton 10,000 tons 32,000 lbs
• THUS
• Total Resistance 50,850 lbs 100,000 lbs
  32,000 lbs 182,850 lbs (or 18.3 lbs/ton),
• Car Resistance 182,850 lbs/100 cars 1829
  lbs/car, and
• 182,850 lbs/80,000 lbs 2.29 gt 3 locomotives
  (assuming 80,000 lbs of pull/locomotive) would be
  needed to pull the train

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Figure 9-2 Rolling cylinder concept (Hay, p. 143)
Figure 9-3 Position of new wheel on new rail
(Hay, p. 143)
Figure 9-4 Possible car/truck attitudes to rail
and lateral forces (Hay, p. 145)
Figure 9-3 Lateral slippage across rail head
(Hay, p. 145)
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Choosing the preferred route

• From Figure 2-6, resistance calculations would be
  done for each route and compared, as in Figure
  2-9

Figure 2-9 Comparing the Energy It Takes to
Deliver the Goods (Armstrong, p. 24)
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Determine optimum station spacing and length

• Urban Rail Transit
• use a max of 8 fps acceleration (5 for standees)
• station spacing usually impacts average speed the
  most
• An example (following slides)

(example taken from Wright and Ashford, p.
106-108)
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Determine Optimum Station Spacing and Length

• A rapid-transit vehicle has a top-speed
  capability of 60 mph, a capacity of 63 persons
  seated, anda length of 60 3. The maximum
  passenger volume to be carries is 6000 persons in
  one direction. Manual control will be used, and
  the minimum headway will be 5 min. Station stops
  of 15 sec are to be used. What is the minimum
  station spacing if the full top-speed capability
  of the vehicle is to be used and what station
  lengths must be designed?
• Number of trains per hour 12 at 5-min headway
• Required train capacity hourly passenger volume
  6000
• number of
  trains per hour 12
• 500 passengers
• Required number of cars per train train
  capacity 500 8 cars
• car
  capacity 63

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• Platform length required car length x number of
  cars per train
• 60.25 x 8 482 ft
• Assuming a maximum acceleration and that a
  deceleration rate of 5 ft/sec2 was used
  throughout the run, the distance-speed diagram
  would be as shown, attaining a maximum speed of
  60 mph, or 88 ft/sec, at the halfway point.

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• Time to reach midpoint velocity 88
  17.6 sec
• acceleration 5
• Total running time 17.6 x 2 35.3 sec
• Distance to midpoint .5 x acceleration x time
  squared
• .5 x5x17.62 774 ft
• Minimum station spacing 774 x 2 1548 ft 0.3
  mile
• Average running speed 30 mph
• Time to travel between stations acceleration
  time deceleration time
• 2 x 17.6 35.3 sec
• Total time running time station stop time
  35.2 15.0 50.2 sec
• 1548 ft of line
• Average overall spacing station spacing 1548
  30.8 ft/sec 20.9 mph

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• It is worth noting that with stations set at
  1548 ft spacing there is no reason to use more
  powerful equipment, since overall travel speeds
  are limited by accelerations rates and station
  stop time. If high-speed equipment with limiting
  speeds of 90 mph were used for this example, the
  overall travel speed would be identical. It is
  interesting to note the effect of increasing the
  station spacing to 1/2 mile. The additional
  distance is covered at top speed. The
  speed-distance diagram is now as follows
• Additional travel time 1092/88 12.4 sec
• Total travel time 50.2 12.4 62/6 sec
• Overall speed 42.4 ft/sec or 29 mph

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• An increase of approximately .2 mile to the
  station spacing results in a 37 increase in
  overall line-haul speeds. In practice, the
  calculations may become slightly more complicated
  by the non uniform acceleration characteristics
  of the transit vehicles. Acceleration and
  tractive effort decrease with speed, as has been
  indicated in the discussion of electric
  locomotives. Speed -distance relationships must
  be computed from the characteristic
  speed-velocity relationships of the individual
  pieces of equipment. The overall approach is
  similar to the somewhat simplified example. It
  should be noted that the station spacings are
  normally greater in outlying areas than in the
  congested high-density central areas. Using
  constant acceleration rates throughout a system,
  overall travel speeds will vary significantly
  from central business districts to suburban
  areas.