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

1. 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

2. 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

3. Choose best route for a freight line (cont.) • Train Resistance • 2 forces of resistance: • inherent or level tangent resistance = ƒ(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 = ƒ(curvature, grades) Ru = 1.3 + 29 + bV + CAV 2 w wn

4. 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 (let’s use 0.045) • A = 85 - 90 sq. ft. for freight cars (let’s use 90) • C = 0.0017 for locomotives and 0.0005 for freight cars (let’s use 0.0005) • inherent force = 1.3 + (29/25) + (0.03*50) + [(0.0005*90*502)/(25*4)] • = 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

5. Figure 2-7: How Much Energy It Takes to Move a Car (Armstrong, p. 21)

6. Figure 2-10: Energy needed to maintain speed, accelerate and curve (Armstrong, p. 25) Figure 2-8: Train Resistance (Armstrong, p. 22)

7. 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)

8. 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 => 3 locomotives (assuming 80,000 lbs of pull/locomotive) would be needed to pull the train

9. 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)

10. 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)

11. 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)

12. 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 Determine Optimum Station Spacing and Length

13. 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.

14. 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 travel time 50.2

15. 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

16. 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.