03 Train Resistance and Tractive Power (Railway Engineering Lectures هندسة السكك الحديدية & Dr. Walied A. Elsaigh)
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The document discusses railway engineering focusing on energy efficiency and resistance in train movement. It compares the energy efficiency of rail to trucks, noting that trains are over three times more efficient, requiring less energy to transport freight due to lower rolling resistance and aerodynamic advantages. The resistance affecting trains is categorized into rolling resistance, grade resistance, and curve resistance, with formulas provided for calculating each type based on various factors such as weight and speed.
![Example 2
[No. of Locomotives]
•How many 5,000 horsepower locomotives at 85%
efficiency will it take to provide the pull of the 77,440
lb. if the resistance of each engine is 1,200 lb.?
375 X HP x e
V
TE=](https://image.slidesharecdn.com/trainresistanceandtractivepower-200415163134/85/03-Train-Resistance-and-Tractive-Power-Railway-Engineering-Lectures-Dr-Walied-A-Elsaigh-32-320.jpg)
03 Train Resistance and Tractive Power (Railway Engineering Lectures هندسة السكك الحديدية & Dr. Walied A. Elsaigh)
- 1. CE 435 RAILWAY ENGINEERING Part II. Resistance and Tractive Effort Text Practical Guide to Railway Engineering, AREMA Page 52 to 59
- 2. • What are the two primary aspects of transportation energy efficiency? • Resistance How much work is required to move something. • Energy efficiency How efficiently energy is converted into useful work.
- 3. Comparison of truck vs. rail energy efficiency What distance can each mode transport a given amount of freight for a given amount of energy? i.e. how far can we transport one ton of freight with one gallon of diesel fuel? Rail is over 3 times more efficient than truck
- 4. • 1 Train = 1 ton of freight carried 435 miles on 1 gallon of diesel fuel • 1 Train = 280 truckloads • 1 Train = 3 to 4 times more fuel efficient than trucks
- 5. Trains VS. Trucks Trains require less energy to move than trucks because of smaller resistive forces: •Lower friction factor (0.18) –Smaller contact area –Steel wheels on steel rails –Rubber tires on pavement •Less wind resistance per pound of load •Size of contact area?
- 6. Energy Utilization The two primary aspects of transportation energy efficiency: • Resistance – How much work is required to move something • Energy Efficiency – How well energy is converted into useful work
- 7. Train Resistance For our purposes, Resistance is the combination of forces that work against a train’s movement. Early measurement of railcar resistance simply involved piling on weight, w, and determining how much was needed to make the car move. If the weight, w, is great enough to overcome the static friction, the car will start to move.
- 8. What is resistance and how is it measured? Resistance is typically measured in “pounds per ton” (in U.S.)
- 9. Train Resistance Two Elements: Rolling Resistance (On Level Tangent Track) Friction related wheel, track quality, track modulus Aerodynamics of equipment Grade & Curve Resistance Track Profile & Alignment related Change in potential energy head
- 10. Sources of rail vehicle resistance A varies with weight ("journal" or "bearing" resistance) B varies directly with velocity ("flange" resistance) C varies with the square of velocity (air resistance) The general expression for train resistance is thus: R = AW + BV + CV2 where: R equals total resistance W = weight V = velocity A = resistances that vary with axle load (includes bearing friction, rolling resistance and track resistance) B = resistances that vary directly with speed (primarily flange friction and effects of sway and oscillation) C = resistances that vary as the square of speed (affected by aerodynamics of the train) Cross-section of the vehicle, streamlining of the front & rear, and surface smoothness all affect air resistance
- 11. SPEED (mph) RESISTANCE(lbs./ton)Speed and Resistance by Transport Mode Rail uniquely combines High Speed and Low Resistance Airplanes Trucks Boat
- 12. Rolling Resistance •Rolling Resistance • The resistance for train movement on straight and level track can be determined by W. J. Davis formula: R=A+BV+CV² Where R= Train rolling resistance A= Rolling resistance component independent of train speed B= Train resistance dependant on speed C= Drag coefficient based on the shape of the front of the train and other features affecting air turbulance V= Train speed • The Train resistance on level tangent track is the sum of locomotive and train resistances.
- 13. Speed and resistance for conventional freight train At low speeds, journal resistance dominates, but as speed increases air resistance is increasingly the most important term C B A
- 14. Resistance versus speed for a 10,000 ton train Train resistance is calculated by multiplying the resistance per ton at each speed, by the total tonnage of the train.
- 15. Measurement of train resistance Substantial research early in the 20th century led to the development of a general formula for train resistance. Developed by W.J. Davis, it is still sometimes referred to as the “Davis” equation. Ru = 1.3 + 29/w + bV + CAV2 /wn where: Ru = Unit resistance in lbs. per ton w = weight per axle (Weight, tons/ Number of axle) n = Number of axles V= velocity mi/h b = an experimental friction coefficient for flanges, shock, etc. A = cross-sectional area of vehicle C = drag coefficient based on the shape of the front of the train and other features affecting air turbulence etc. The Davis Equation has been substantially updated to reflect modern developments, but its basic form is the same.
- 16. “Davis” equation (Total Resistance) R =1.3 W + 29N + bWV + CAV2 R = Resistance in lbs W = Weight, tons N = Number of axles V= velocity mi/h b = an experimental friction coefficient. A = cross-sectional area of vehicle C = drag coefficient based on the shape of the front of the train and other features affecting air turbulence etc. See AREMA Manual for Railway Engineering Chapter 16 - 2.1.3 for latest version of formula.
- 17. Recommended values for “Davis” equation CbA sq.ftEquipment Type 0.0024 0.0024 0.0024 0.03 0.03 0.03 105 110 120 Locomotives -50 tons 70 tons 100 tons and over 0.00050.04585-90Freight cars 0.000340.03120Passenger car
- 18. Grade Resistance 1% Grade = Approximately 20 lbf / ton
- 19. Grade Resistance is the force required to overcome gradient and is equal to 20 lb. per ton per percent grade. This force is derived from the resolution of force vectors and is independent of train speed. An up grade produces a resistive force while a down grade produces an accelerating (negative resistive) force. Grade Resistance
- 20. 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 example: Grade Resistance = 20 * 0.5 = 10 lbs/ton = 10 lbs/ton * 10,000 tons = 100,000 lbs
- 21. Curve Resistance 10 = 0.8 lbf / ton
- 22. Curve Resistance Curve Resistance is an estimate of the added resistance a locomotive or car must overcome when operating through a horizontal curve. The exact details of the mechanics contributing to curve resistance are not easy to define. It is generally accepted in the railway industry that curve resistance is approximately the same as a 0.04% up grade per degree of curvature (which equals 0.8 lb. per ton per degree of curvature) for standard gauge tracks.
- 23. Curve Resistance 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
- 24. Tractive Effort or Tractive Force, is the amount of force at the wheels available for moving a train. Tractive effort is the amount of force in foot-pounds that the motive power must produce to move a train. Tractive Effort Tractive effort is simply the force available, at the roadway surface, to perform work.
- 25. POWER! Q: How do railroads overcome such large resistances?
- 26. Physics of power, force and speed Work = force x distance W = F x D Power is the rate at which work is done Power = work/time P = W/T = FD/T P = FV The relationship between Force and Distance/Time (i.e. speed) is thus inverse Force Speed The higher the speed, the less force available Curves like this are used to describe the performance capabilities of locomotives
- 27. P = F V P= (1/375) F V P = Power, HP F = Tractive effort , lb V= velocity m/h
- 28. Speed/tractive effort curve for a modern locomotive Tractive effort is measured in pounds of force available as a function of speed At low speed, tractive effort is limited by adhesion, not power
- 29. Train resistance and tractive effort are both measurements of force (typically in pounds in North America) so we can simply overlay the curves Tractive effort & train resistance curves Q: What is the maximum speed possible for this train with this locomotive? A: About 58 mph. This is referred to as the “balancing speed”. Tractive force = resistance 35,000 lbs.
- 30. Force available for acceleration declines with speed until the balancing speed is reached, where it is zero. Consequently, the rate of acceleration declines with speed. The difference between tractive effort and resistance is the force available for acceleration Q: How much force is available for acceleration at 15 mph? A: 135,000 - 15,000 = 120,000 lbs. Q: How much force is available for acceleration at 35 mph? A: 59,000 - 21,000 = 38,000 lbs.
- 31. Example 1 A train composed of 2-diesl locomotives, each weighing 136 tons, having 4-axles. 20-empty freight cars, each weighing 30 tons, with 4 –axles. At a speed of 50 mi/h on a level straight route. calculate train resistance.
- 32. Example 2 [No. of Locomotives] •How many 5,000 horsepower locomotives at 85% efficiency will it take to provide the pull of the 77,440 lb. if the resistance of each engine is 1,200 lb.? 375 X HP x e V TE=
- 33. 375 X HP x e V TE= 375 X (5,000 X N) x 0.85 7077,440 + 1,200 X N= N = 3.59 Take 4 Locomotives
- 35. Thank You