1 - Purdue University College of Engineering



CE361 Introduction to Transportation Engineering |Posted: Saturday 26 November 2005 | |

|Homework 10 |Due: Wednesday 7 December 2005 |

Moving Freight and People

You may submit your assignment as a member of a group of CE361 students not to exceed four in size. Signatures of all group members must appear on the top page of the work submitted.

1. (20 points) Trade Between Countries – Cheese and Cell Phones. Repeat FTE Example 12.1, but with the data contained in the table below.

| |Country E |Country A |

|Population |322,000 |300,000 |

|Cheese Consumption |48.5 lbs/person per year |38.2 lbs per person per year |

|Cost to produce cheese |$15.30 per 40-lb block |$14.20 per 40-lb block |

|Cell phone purchases |0.124 per person per year |0.120 per person per year |

|Cost to produce cell phones |$133 per unit |$166 per unit |

It costs $9.90/cwt to ship cheese and $119/cwt to ship cell phones. Each cell phone is shipped in a 5-lb box. How much will the consumers of each country save by importing the good with the lower production cost? What is the maximum logical tariff that each country could apply to the imported product?

Calculations are summarized in the table below. One key calculation is for cost to ship cell phones from Region E to Region A: 300,000 persons * 0.120 cell phones/person * 5 lbs/cell phone * $119.00 per cwt = $214,200.

| |Goods |Units |

|X = |cheese |lbs |

|Y = |cell phones |each |

|Shipping costs | | |

| $ 9.90 |per cwt. X | |

| $ 119.00 |per cwt. Y |5 lbs/box |

| | | |

| |Region E |Region A |

|Population |322,000 |300,000 |

| | | |

|Units X consumed per person |48.5 |31.2 |

|$/unit to produce X |$0.3825 |$0.3550 |

|Total cost to buy X |5.97E+06 |3.32E+06 |

| | | |

|Units Y consumed per person |0.124 |0.120 |

|$/unit to produce Y |133.00 |166.00 |

|Total cost to buy Y |5.31E+06 |5.98E+06 |

| | | |

|Total cost w/o trade |1.13E+07 |9.30E+06 |

| | | |

|Cost of buying other region's | | |

| cheaper good (w/o tp.) |5.54E+06 |4.79E+06 |

|Tpt cost to import cheaper good |1.55E+06 |2.14E+05 |

|Tot import costs |7.09E+06 |5.00E+06 |

| | | |

|Money saved by importing |-1.12E+06 |9.74E+05 |

Persons in Region A will save $974,000 by importing cheese from Region E, but persons in Region E will have to pay $1,120,000 more to import cell phones from Region A. The transport costs are too high. A tariff should never exceed the potential saving, or else it will prevent the imports and no tariff revenues will be realized.

2. Rail locomotive power. Uwanna Railroad using a standard consist of two 5800-HP locomotives pulling 100 container cars at 90 percent efficiency. Each locomotive offers 1250 lbs resistance. Each car has four axles, weighs 64,000 lbs unloaded, and carries two containers that each weigh 53,000 lbs loaded.

A. (10 points) In the absence of any ruling grade or horizontal curve, at what speed can the standard consist operate?

B. (5 points) At V = 35 mph, what is the maximum ruling grade that the standard consist can climb?

C. (5 points) At V = 35 mph with G = 0, what is the most severe horizontal curve that the standard consist can negotiate?

A. Each railcar weighs 64,000 + (2*53,000) = 170,000 lbs = 85 tons. Therefore, w = 21.25 tons in (12.5): [pic]

[pic]. In (12.10), Rloco = 2 * 1250 = 2500 lbf.

TE = [Rtt * (C * w * n)] + Rloco. = [pic] = [pic]= 15,590 +85V + 9.35V2. On the “propulsion side” (12.11) leads to TE = [pic]. Setting the two TE equations equal and rearranging terms: 9.35V3 + 85V2 + 15,090V = 3,915,000. Solve for V = 64.82 mph.

B. (12.10) with V = 35 mph: TE = [(Rtt + Rgrade) * (C*w*n)] + Rloco. = [pic] =

((3.2375 + 20G) * 8500) + 2500 = 27,519 + 170,000G + 2500. (12.11) with V = 35 mph: TE =[pic] = 111,857 lbf. After setting (12.10) = (12.11), [pic]

C. (12.10) with V = 35 mph: TE = [(Rtt + Rcurv) * (C*w*n)] + Rloco. = ((3.2375 + 0.8Δ) * 8500) + 2500 = 27,519 + 6800Δ + 2500. (12.11) with V = 35 mph: TE =[pic] = 111,857 lbf. After setting (12.10) = (12.11), [pic]

3. (20 points) Horsepower for 3x2 tow. FTE Exercise 12.16 with e = 0.65.

L = 2*195 = 390 ft, W = 3 * 35 = 105 ft, V = 5 kts.

HP is in (12.21), which needs Rtotal. By (12.20), Rtotal = Rsf + RWM + Raero. By (12.14), Rsf = f * Swet * V1.825. By (12.15), f = 0.0106 L-0.031 = 0.0106*390-0.031 = 0.0088. Because scows have sides that are essentially vertical and are lashed together with a draft of 8.5 feet, the perimeter of the tow is (2*3*35) + (2*2*195) = 990 ft., the submerged sides have area 990 * 8.5 = 8415 sq ft. The tow’s bottom surface is 35*195*6 = 40,950 sq ft. Its Swet = 8415 + 40,950 = 49,365 sq ft. So Rsf = 0.0088 * 49,365 * 51.825 = 8194 lbf. Following FTE Example 12.14, (12.17) Fr = [pic] = [pic]= 0.075. In FTE Figure 12.35, values of Fr < 0.15 are not shown, because the corresponding Cw values are so small. RWM can be ignored. Just above (12.19), it is stated that Raero can be ignored for slow vessels such as tows. So Rtotal = Rsf = 8194lbf. By (12.21) [pic][pic]. Rearrange to find [pic]= 194 horsepower, but this is for deep water. Follow FTE Example 12.17 as corrected in the Errata webpage. 5 kts = 5.75 mph. In FTE Figure 12.36, 5.75 mph in deep water as a value of about 300 HP. At channel depth 13 ft, the value for 5.75 mph is about 1200 HP. Use the ratio of 1200/300 = 4 to convert 194 HP in deep water to 4*194 = 776 HP in a channel 13 ft deep.

4. (20 points) Energy Intensity of freight movement. Coal for power plant. “Airline distance” 144 miles. The actual distance is 183 miles by rail, 177 miles by truck, and 160 miles by barge. How much energy is expended for each 1000 tons of coal delivered to the power plant by each mode. Use the Year 2000 EI values in FTE Table 12.16 and assume the backhaul is empty for each mode. The backhaul energy values with respect to their loaded energy values are 70 percent for rail, 60 percent for truck, and 50 percent for barge.

BTU/TM values from table on FTE page 702: Truck 3200, Rail 352, Water 508.

Truck E1 = 3200 BTU/TM * 1000 tons * 177 mi = 566.4 MBTU; with backhaul: 1.60 * 566.4 = 906.24 MBTU

Rail E1 = 352 * 1000 * 183 = 64.416 MBTU; with backhaul: 1.70 * 64.416 = 109.507 MBTU

Barge E1 = 508 * 1000 * 160 = 81.28 MBTU; with backhaul: 1.50 * 81.28 = 121.92 MBTU

Unlike a river, Murdock Bay offers no current for the barge to fight against.

5. (20 points) Energy Intensity of passenger travel. FTE Exercise 13.3. Parts (a) and (b) = 5 points each. Part (c) = 10 points.

a) EI if auto achieves 30 mpg and carries 2 persons

[pic]

b) At 21 or 17 mpg, occupancy of SUV to achieve the same EI as in Part A.

Occupancy [pic]

c) HEV at 50 mpg with driver alone, with 2 people, with 3 people, with 4 people:

[pic]

With 2 persons in vehicle, 2500/2 = 1250 BTU/Pax-Mi;

with 3 occupants, 2500/3 = 833 BTU/Pax-Mi;

with 4 occupants, 2500/4 = 625 BTU/Pax-Mi

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