Transcript CE 353 Lecture 6: System design as a function of train

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
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
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)
CAV 2
Ru = 1.3 + 29
+
bV
+
w
wn
– 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)
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)
CAV
Ru = 1.3 + 29
+
bV
+
w
wn
–
–
–
–
2
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*50 2)/(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)
Figure 2-7: How Much Energy It Takes to Move a Car (Armstrong, p. 21)
Figure 2-10: Energy needed to
maintain speed, accelerate and
curve (Armstrong, p. 25)
Figure 2-8: Train Resistance (Armstrong, p. 22)
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)
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
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)
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)
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)
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
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.
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
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
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 highdensity central areas. Using constant acceleration rates throughout a system,
overall travel speeds will vary significantly from central business districts to
suburban areas.