HW 7 solutions

# 23 20 010 8 0476 destination 4 017 50 023

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Unformatted text preview: .10 (8) = 0.476 Destination 4 = 0.17 (5.0) -0.23 (7.0) + 0.10 (15) = 0.74 Total = ∑ = + + + = + = = 0.4020 therefore trips to destination one: 0.4020 * 725 = 291 trips = = 0.2343 therefore trips to destination two: 0.2343 * 725 = 170 trips = = 0.1580 therefore trips to destination three: 0.1580 * 725 = 115 trips = = 0.2057 therefore trips to destination four: 0.2057*725 = 149 trips 3 CEE 362 – Fall 2013 - Homework # 7, Solution Manual Problem # 4 = 0.17 (6.0) - 0.23*(5) +0.10*(10+x) = 0.87 +0.1x = (0.87 + 0.1x) = 220 Solving for x x = 2191.4 sf should be added 4 CEE 362 – Fall 2013 - Homework # 7, Solution Manual Problem # 5 a) = 4 + 3 (9.780) = 33.34 min = 33.34 = b + 6 q= 15 = 9.78 + Therefore: B = 2.02 min b) q = 7000 veh/h q= + therefore 7 = = 74+3 + or (equation 1) = 2.02 + 6 using equation 1: 4 + 3 = 2.02 + 6 (7- Therefore: = 7- Therefore: =4+3 = 4 + 3 (4.447) = 17.3 min = 2.02 + 6 (2.553) = 17.3 min 5 ) CEE 362 – Fall 2013 - Homework # 7, Solution Manual Problem # 6 + 2+ =1+ P (h<6) = 0.30 = 1 - Solving for = 0.70 = -3600 / 6 *ln 0.70 = 1214 veh/hr 6 CEE 362 – Fall 2013 - Homework # 7, Solution Manual Problem # 7 Route 2 will not be used because it takes more time to pass it even if the flow is 0 in it, To continue routes 1 and 3 will be considered 5+ 1.5 = 2 + 0.2 = 4.3 (equation 1) (equation 2) Using equations 1 and 2: t= 5.33 min 7 CEE 362 – Fall 2013 - Homework # 7, Solution Manual Problem # 8 ∫ ∫ = q- = 5q – 5 = 1.5 Solving for q: q= 7 -0.5 + 3.5 1.5 = 19843 Δt = 22314 – 19483 Δt = 2831 8...
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## This document was uploaded on 02/11/2014.

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