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katrin2010 [14]
3 years ago
5

A multilane highway (two lanes in each direction) is on level terrain. The free-flow speed has been measured at 45 mi/h. The pea

k-hour directional traffic flow is 1300 vehicles with 8% heavy vehicles. If the peak-hour factor is 0.85, determine the highway's level of service.
Engineering
1 answer:
Pani-rosa [81]3 years ago
7 0

Answer:

17.71 veh/mi/ln

Explanation:

Given,

Free Flow Speed, FFS = 45 mph

Number of lanes in each direction = 2

Peak Flow, V = 1300 veh/hr

Peak-hour factor = 0.85

HV = 8 %

Level Terrain

fHV = 1/ (1 + 0.08 (1.5-1)) = 1/1.04 = 0.96

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 1300/(0.96*2*0.85*1.0) = 796.57 ~ 797 veh/hr/ln

Vp < 1400

S = FFS = 45 mph

Density = Vp/S = (797) / (45) = 17.71 veh/mi/ln

Density of LOS B should lie between 11 – 18 veh/mi/ln

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