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CE 2710 Transportation Engineering Homework 8 Solution
- A vehicle stream is interrupted and stopped by a flag-person. The traffic volume for the vehicle stream before the interruption is 1200 veh/hr and the concentration is 100 veh/mi. Assume that the jam concentration is 240 veh/mi. After five minutes the flag-person releases the traffic. The flow condition for the release is a traffic volume of 1600 veh/hr and a speed of 20 mph.
Approach Conditions: q = 1200 veh/hr u = 12 mi/hr k = 100 veh/mi
Platoon Conditions: q = 0 veh/hr u = 0 mi/hr k = kj = 240 veh/mi
Shock wave at the rear of the platoon Beginning at t = 0:
Usw1 = (0-1200)/(240 – 100) = -8.57 mi/hr
Length of Queue after 5 minutes = usw1 * t = -8.57 * (5/60) = 0.714 mi Number of vehicles in queue = L * k 2 = 0.714 * 240 = 171.
Approach Conditions: q = 1200 veh/hr u = 12 mi/hr k = 100 veh/mi
Platoon Conditions: q = 0 veh/hr u = 0 mi/hr k = kj = 240 veh/mi
Release Conditions: q = 1600 veh/hr u = 20 mi/hr k = 80 veh/mi
Shock wave at the front of platoon at t = 5 min, it was defined between the jam and release conditions:
Usw2 = (1600-0)/(80-240) = -10 mi/hr
Time for queue of 0.714 mi to dissipate Speed of dissipation = -10 – (-8.57) = - 1.43 mi/hr
Time = Length / speed = 0.714 / 1.43 = 0.5 hour
- Traffic is traveling at 40 mph and a flow rate of 1000 veh/hr when a tractor going at 15 mph turns on to the road. The platoon that develops behind the tractor has a density of 60 veh/mi. After 0.75 mile, the tractor turns off the main road. The release travel condition for the vehicles in the platoon is at a flow rate of 1200 veh/hr and a density of 40 veh/mi. Calculate the length of the platoon that develops and the time that it will take for it to dissipate after the tractor leaves.
Approach Conditions : q = 1000 veh/hr u = 40 mi/hr k = 25 veh/mi
Platoon Conditions : q = 900 veh/hr u = 15 mi/hr k = 60 veh/mi
Condition in Front of Truck: q = 0 veh/hr k = 0 veh/mi
(Free-flow condition)
Length of Platoon when truck turns off main road
The shock wave at the rear of the platoon (i.e., between the approach and platoon conditions) will have a speed of:
Usw1 = (900-1000)/(60-25) = -2.86 mi/hr
Shock wave at front of truck has same speed as truck, Usw1 = 15 mi/hr
Speed of growth of platoon, = 15 – ( - 2.86)= 17.86 mi/hr
Time the truck takes to turn off, = 0.75 / 15 = 0.05 hr
_Length of platoon after 0.05 hr, = 0.0517.86 = 0.893miles_*
Approach Conditions : q = 1000 veh/hr u = 40 mi/hr k = 25 veh/mi
Platoon Conditions : q = 900 veh/hr u = 15 mi/hr k = 60 veh/mi
Release Conditions : q = 1200 veh/hr u = 30 mi/hr k = 40 veh/mi
Shock wave at the back of platoon still has speed of -2.86 mi/hr
