Question

A hazardous material spill on Interstate I-76 caused the eastbound traffic 20-minute delay. Traffic flow is completely shut down on a three-lane section creating a jam density of 160vpm/lane. If the arriving flow on this roadway section is 3300vph at a density of 35 vpm/lane, determine the following:

a) the total number of vehicles in the queue and the queue length (miles) after 20 minutes.

b) spacing (in ft/Veh) between vehicles in the jammed traffic.

c) if the roadway is re-opened after 20 minutes and the releasing flow at the bottleneck is 1800vph/lane at a density of 60vpm/lane, how long will it take for the releasing wave (moving upstream from the bottleneck) to catch up with the jam density shockwave?

Answer 1

a) Arrival rate = 3300 vph

so in 20 minutes # of vehicles which arrived = 3300*20/60 = 1100 vehicles

Queue length after 20 minutes can be calculated by calculating the shockwave speed and then doing some calculations

a) q1 = 3300 veh/hr, k1 = 35vpm

kjam = 160 veh/mi/lane

Speed of the backward shockwave (i.e at which queue is forming and travelling backward) =

where

qapproach = 3300 veh/hr this is over three lanes so its basically 1100 veh/hr

kapproach = 35 veh/mi/lane

kjam = 160 veh/mi/lane

So speed of backward shockwave = uwr = (1100-0)/35-160 = -3300/125 = -8.8 miles/hour

so after 20 minutes the length of the queue = Speed of queue shockwave * time

= 8.8 * 20/60 = 2.93 miles

Jam density = 160 veh/mil/lane

length of queue = 2.93 miles

So # of vehicles queued up = 160 * 2.93 * number of lanes = 160 *2.93 * 3 = 1406 vehicles.

--------------------------------------------------------------------------------------------------------------------------------------------------------

b) Jammed traffic has a density of 160vpm/lane i.e 160 vehicles/mile = 160/5280 veh/ft

So spacifing in ft/veh = 5280/160 = 33 ft/vehicle

-----------------------------------------------------------------------------------------------------------------------------------------------

c) Now after the roadway is opened up, you will have a shockwave starting from the head of the queue and coming back upto the end of the queue. The speed of this shock wave is given by

Uwe = (0 - Qveh being released)/ (kjam - kleaving)

Speed of forward shock wave = (0 - Qveh being released)/ (kjam - kleaving) = (0 - 1800)/(160 - 60 ) = -18 mph

So now we have two shockwaves

Speed at which queue is forming from front of queue = -8.8 miles/hour

Speed at which front of queue is shortening = -18mph

Now do a space time diagram of both the shockwaves as below

So Uwr = -8.8 mph, Uwe = -18 mph

so time to maximum queue = (20/60 * uwe)/Uwe - Uwr  (use absolute values without sign)

= 6/9.2 = 0.6521 hours = 39.13 minutes

So it takes 39.13 minutes for the releasing wave moving upstream from bottleneck to catch up with jam density sockwae