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|CE361 Introduction to Transportation Engineering |Posted: Wednesday 3 December 2008 |

|Homework 12 |Due: Wednesday 10 December 2008 |

Moving Freight and USING ENERGY

Dear Consultant:

Freight transportation is often overlooked, yet it is crucial to a society’s quality of life. In this HW, you are asked to demonstrate your ability to analyze various aspects of the shipment of goods.

Please complete the exercises below completely and clearly. You may submit your assignment as a member of a group of CE361 students not to exceed three in size. Signatures of all group members must appear on the top page of the work submitted.

1. Mixed Freight Train. An 85-car container train is scheduled to move through Horseshoe Curve (pictured) near Altoona PA at 16 mph. “The 220 degree arc that forms Horseshoe Curve comprises two curves; the north side has a radius of 637 feet while the south side tightens to 609 feet.” [] Assume the tare weight of each 4-axle rail flat car is 40,000 lbs, each 40-ft container weighs 60,000 lbs (gross), and there is no gradient in the curve.

A. (7 points) If there is one container per railcar, calculate Rtt, Rcurv, and Rtotal.

Total weight each car = 40,000 + 60,000 = 100,000 lbs / 2000 lbs per ton = 50 tons; w = 50 tons/4 axles = 12.5 tons/axle

(12.5) Rtt = [pic] = [pic]

Rtt = 0.6+1.6+0.16+0.4787 = 2.8387 lb/ton

(7.8) Δ = 5729.6/R = 5729.6/609 = 9.41 degrees (or 9.15 degrees from the Horseshoe website)

(12.7) Rcurv = 0.8 Δ = 0.8 * 9.41 degrees = 7.528 lb/ton (or 7.40 lb/ton)

(12.8) Rtot = (2.8387+0+7.528) lbs/ton * 85 cars * 50 tons per car = 44,058 lbs (or 43,514 lbs)

|(7 points) If the containers are double-stacked, calculate Rtt, Rcurv, and Rtotal. |

|Total weight each car = 40,000 + (2*60,000) = 160,000 lbs / 2000 lbs per ton = 80 tons; w = 80 tons/4 axles = 20 tons/axle |

|(12.5) Rtt = [pic] = [pic] |

|Rtt = 0.6+1.0+0.16+0.4787 = 2.2387 lb/ton; note that K=0.0935 for single- or double-stack COFC (FTE 684) |

|(12.7) Rcurv = 0.8 Δ = 0.8 * 9.41 degrees = 7.528 lb/ton |

|(12.8) Rtot = (2.2387+0+7.528) lbs/ton * 85 cars * 80 tons per car = 66,414 lbs |

|(5 points) On a straight section of track before or after Horseshoe Curve, what ruling grade is equivalent to the curvature of Horseshoe|

|Curve for the single-stack container train? |

|(12.7) Rcurv = 0.8 Δ = 0.8 * 9.41 = 7.528 lbs/ton; (12.6) Rgrade = 20 G = 7.528 lbs/ton when G = 0.376 percent (or 0.37 percent). [FYI:|

|The steepest standard-gauge mainline railway grade in the United States is the Saluda Grade in Polk County, North Carolina. It reaches |

|5.1% at one point.] |

|(6 points) Each available locomotive has Rlocomotive=1536 lbs, 6660 HP, and 86 percent efficiency. How many locomotives are needed to |

|pull the single-stack container train in Part A? |

|(12.10) = (12.11): 44,058 + (1536*N) = (375*6660*N*0.86)/16; Solve for N=0.33 (or N=0.33) ( 1 engine |

2. (25 points) Horsepower for 3x5 tow. A 3 x 5 tow (each scow is 35 by 195 feet) is carrying coal and has a draft of 8.5 feet. The desired speed in still water is 5.6 knots. If the water is 16 feet deep, estimate the horsepower to be delivered to the tow if the engine/propeller system is 66 percent efficient.

Follow steps in FTE Example 12.14 so that we can use (12.14).

(12.15) f = 0.0106 L-0.031 = 0.0106 * (5*195)-0.031 = 0.008563.

(12.17) Fr = [pic] = 0.0565. Because Fr ................
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