Homework 12 Solutions / Due: Wednesday 8 December 2010
Moving Freight and USING ENERGY
CE361 HW12-1-
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.
- Freight modes in the US and Europe. A friend of the City Engineer returns from Europe with several questions. They concern the relative use of truck and rail as freight modes, especially in relation to the available supply roadways and railways.
- (10 points) What is the freight mode share between trucks and rail in the US? In Europe? Do not include other freight modes in your calculations.
The values in the first two rows of the table below were found in Tables 1 and 2 of the FHWA report “pointed to” in Part C of this problem. US truck share = = 0.427; EU-15 truck share = = 0.889.
- (10 points) Using some reasonable measure, how much of the available road or rail “supply” do the truck and rail modes use in the US and Europe? What name would you give your measure, and why?
(8 pts) Try for each mode. For US trucks, = 0.232*106; for EU-15 trucks, =0.358*106. For US rail, = 8.375*106; for EU-15 rail, =1.538*106.
(2 pts) I would call it truck mode densityor rail mode density, because it seems analogous to vehicles per mile (FTE 60) or population per unit area.
- (10 points) To answer Questions 1A and 1B, first fill in the entries in the table below with values found in an FHWAreport that compares US and “EU-15”. What does “EU-15” stand for?
(8 pts) The values in the first two rows of the table below were found in Tables 1 and 2 of the FHWA report “pointed to” by the link provided.
(2 pts) Unfortunately, the report did not define “EU-15”, so I had to use bing.com to find the definition. (That would cost the authors points in CE361.) Among several sources, I used a European Commission website at , which said that EU-15 refers to“the 15 countries that were members of the EU [European Union] before the enlargement on 1st May 2004.”
Measure / US truck / US rail / Europe truck / Europe railTon-kilometers / 1499*109 / 2010*109 / 1254*109 / 240*109
Highway road-km or Rail track-km / 6460*103 / 240*103 / 3500*103 / 156*103
Mode share / 0.232 / 0.768 / 0.358 / 0.642
Measure of …
freight mode density / 0.232*106 / 8.375*106 / 0.358*106 / 1.538*106
- TOFC Freight Train. The City Engineer is talking with a railroad buff as they leave City Hall, when they see a freight train passing by on the tracksnearby. The buff observes, authoritatively, “One engine, probably pulling 150 cars.” The City Engineer can’t see the front end of the train, but he notes that every car he can see has a semi-trailer on it, as shown in the photo below. After he went back to his office, the City Engineer thought of several questions. They (and the interim questions he had to answer) are listed below.
Source:
- (15 points) The City Engineer looked up the empty weight of the 4-axle spine car shown above. It is 60,000 lbs. He recalls (from HW9) that a fully-loaded semi-trailer cannot exceed 80,000 – 12,000 – 2,500 = 65,500 lbs to be legal on most highways. He also recalls that the train was moving at about 30 mph. If there are no curves or grades, calculate Rtotal. If the locomotivesused by the railroad haveRlocomotive=1555 lbs, 5995 HP, and 90 percent efficiency, will one locomotive be enough to pull 150 TOFC cars?
Resistance calcs. (12.8) Rtotal = (Rtt + Rgrade + Rcurv) lb/ton * (C*w*n) tons, with C=150 cars, w== 15.69 tons/axle and n=4 axles per car.
(12.5) Rtt = with V=30 mph and K=0.16 for TOFC (FTE p. 684). Rtt = = 0.6+1.275+0.30+(0.16*14.34) = 4.47 lb/ton.
Rtotal = (4.47 + 0 +0) lb/ton * (150*15.69*4) tons = 4.47*9414= 42,081 lbs = TEdrawbar
Propulsion calcs. As in FTE Example 12.9, TEdrawbar = TEtractive:
42,081 +1555N = =67,444N;
N==0.639 locomotives needed. One engine is enough!
- Although Part A assumed no grades or curves, …
- (15 points) What is the maximum ruling grade (to 0.01 percent) that one locomotive can overcome? Using a worksheet Solver is permitted.
(12.6) Rgrade=20*G. Add this term to (12.5) in Part A.
Rtotal = (4.47 + 20G +0) lb/ton * (150*15.69*4) tons = (4.47+20G)*9414= (42,081+188,280G) lbs = TEdrawbar
As in FTE Example 12.9, TEdrawbar = TEtractive with N=1:
42,081+188,280G+(1555*1) = ; 43,636+188,280G=67,444
G==0.126 percent 0.13 percent ruling grade
- (10 points) What is the maximum curve (to 0.1 degree) that one locomotive can overcome? Using a worksheet Solver is permitted.
(12.7) Rcurv=0.8*. Add this term to (12.5) in Part A.
Rtotal = (4.47 + 0 +0.8) lb/ton * (150*15.69*4) tons = (4.47+0.8)*9414= (42,081+7531.2) lbs = TEdrawbar
As in FTE Example 12.9, TEdrawbar = TEtractive with N=1:
42,081+7531.2+(1555*1) = ; 43,636+7531.2=67,444
G==3.16 degrees3.1 degree maximum horizontal curve
3.(30 points) Horsepower for 2x3 tow. A 2 x 3 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 knots. If the water is at the depth of 13 feet, estimate the horsepower to be delivered to the tow.
Follow Example 12.14. (12.17) Fr = 0.298 = 0.298= 0.0616.
Because Fr < 0.15, we can ignore RWM and Raero. As a unit, the tow has underwater dimensions 585 ft long, 70 ft wide, and 8.5 ft deep.
Bottom area of tow = 6 barges * 195 ft * 35 ft = 40,950 ft2. Perimeter of tow = (6*195)+(4*35) = 1310 ft. Underwater outside area = 1310*8.5 = 11,135 ft2. Wetted surface = Swet = 40,950 + 11,135 = 52,085 ft2.
(12.14) with f = 0.0106L-0.031 = 0.0106(3*195)-0.031 = 0.0087.
= 8548 lbf. (12.21)’ 131.5 HP with e=1.00 (the default value).
Adjust for channel depth. In Figure 12.36 (see below), HP at 13 feet and 5 knots (5.75 mph) is about 1200 HP. In deep water at 5.75 mph, HP = 300. Apply the ratio 1200/300 to 131.5 HP: 4*131.5 = 526 HP needed to move the tow at 5 knots.
But Figure 12.36 is for a 3x2 “barge” or tow. Our problem is for a 3x2 tow. Is there any difference in the Swet of each? Check to make sure. Perimeter for 2x3 = (4*35)+(6*195) = 1310 ft. Perimeter for 3x2 = (6*35)+(4*195) = 990 ft. Side area under water for 2x3 tow = 8.5 ft draft * 1310 ft perimeter = 11,135 ft2. Side area under water for 3x2 tow = 8.5 ft draft * 990 ft perimeter = 8,415 ft2. Bottom area for 2x3 tow = (2*35)*(3*195) = 40,950 ft2. Bottom area for 3x2 tow = (3*35)*(2*195) = 40,950 ft2. So Swet of 2x3 tow = 11,135+40,950 = 52,085 ft2 and Swet of 3x2 tow = 8,415+40,950 = 49,365 ft2 (as calculated in FTE Example 12.17). Because the 3x2 tow has more wetted surface than the 2x3 tow in the chart, adjust the 526 hp needed as follows: 526 hp*= 526*1.055 = 555 hp needed.
If you interpreted “3x2 Barge Configuration, 195’x35’ barge” above FTE Figure 12.36 as meaning to indicate a 2x3 tow, the last conversion above is not necessary.
Unassigned problem:
- Transportation and Comparative Advantage. Region C made 767,000 cell phones last year for local consumption at a price of $171.42 each and 93,000 tons of dry wall last year for local consumption at a price of $79.35 per ton. Region D made 1,400,000 cell phones last year for local consumption at a price of $192.95 each and 80,000 tons of dry wall last year for local consumption at a price of $54.72 per ton.
- (10 points) How much were customers paying for the two products in their respective regions last year? Fill in the table below and show your calculations clearly.
Region C cell phones: 767,000*$171.42=$131,479,140
Region C dry wall: 93,000 tons*$79.35 per ton=$7,379,550
Total costs to Region C consumers=$131,479,140+$7,379,550=$138,856,690
Region D cell phones: 1,400,000*$192.95=$270,130,000
Region D dry wall: 80,000 tons*$54.72 per ton=$4,377,600
Total costs to Region D consumers=$270,130,000+$4,377,600=$274,507,600
- (10 points) Each of the two regions agrees to reallocate all of its resources now devoted to cell phones and dry wall to makingthe good it can produce more efficiently. Next year, they will export their additional production to the other region. How much will customers pay for the two products in their respective regions? Fill in the table below and show your calculations clearly. Include transport costs of $5.72 per cell phone and $0.90 per cwt dry wall in your calculations.
Region C cell phones: 767,000*$171.42=$131,479,140 (no change)
Region C dry wall: 93,000 tons*$54.72 per ton=$5,088,960 w/o transport
Add dry wall transport cost: 93,000 tons*$0.90/100lbs*(2000lbs/ton)= $1,674,000
Total costs to Region C consumers w/ transport costs= $131,479,140+$5,088,960+$1,674,000=$138,242,100
Region D cell phones: 1,400,000*$171.42=$239,988,000 w/o transport
Region D dry wall: 80,000 tons*$54.72 per ton=$4,377,600 (no change)
Add cell phone transport cost: 1,400,000*$5.72= $8,008,000
Total costs to Region D consumersw/ transport costs= $239,988,000+$4,377,600+$8,008,000=$252,373,600.
- (10 points) To what extent do the consumers in both regions benefit from the reallocation in resources? Fill in the table below and show your calculations clearly.
Region C benefits: $138,856,690-$138,242,100=$614,590
Region D benefits: $274,507,600-$252,373,600=$22,134,000
Consumers in both regions benefited, but Region D consumers benefited much more.
Region C consumers / Region D consumersCost last year for cell phones / $131,479,140 / $270,130,000
Cost last year for dry wall / $7,379,550 / $4,377,600
Total cost to region’s consumers last year / $138,856,690 / $274,507,600
Cost of cell phones bought next year w/ any transport costs / $131,479,140 / $239,988,000+$8,008,000=
$247,996,000
Cost of dry wall bought next year w/ any transport costs / $5,088,960+$1,674,000=
$6,762,960 / $4,377,600
Total cost of both products to region’s consumers next year / $138,242,100 / $252,373,600
Savings to region’s consumers next year / $614,590 / $22,134,000