Question

3-lane freeway reduces to 2 lanes

total flow in section = 5000 veh/hr

lane saturation flow rate= 2400 veh/hr

traffic density at A: 10veh/km/ln

shockwabe speed Wab between A nad B = -10km/hr

Find traffic density at B (immediately upstream of the lane reduction) and the stream flows at locations A (far upstream from lane reduction), B, and C (within the lane reduction)

Answer 1

Pt A is far upstream of lane reduction where

we have 3 lanes, upstream arrival rate = qapproach = 5000 vph and this is over three lanes

Pt B iis the point where the effects of the lane reduction are felt. It is the point far upstream from thelane reduction, basically it is the back end of the queue.

So at A,

densityA = kA = 10 veh/km/n

flow at A = qA = 5000 veh/hr andn this is over 3 lanes, so flow/lane = 1666.667veh/.hr/lane

so speed at A = flow(over one lane)/density = 1666.667/10 = 16.667 km/hr

So at B,

densityB = kB = we need to find this.

flow at B = qB = 2400 veh/hr/lane, so total flow at B = 4800 veh/hr

Speed of the backward shockwave (i.e at which queue is forming and travelling backward ie from A to B towards B) =

uwr = (qapproach arrivals at A -qarrivals at B i.e after one lane is closed)/(karrivals at A  - karrivals at B i.e after one lane is closed)

where

uwr is given to be -10km/hr

qapproach arrivals at A = 5000 veh/hr

kapproach arrivals at A  = 10 veh/km/lane

qafter one lane is closed  = 2400 veh/hr - this is given to be over one lane so total = 2400*2 = 4800

kafter one lane is closed = ????

So put these into the equation

uwr = (qarrivals at A -qarrivals at B i.e after one lane is closed)/(karrivals at A  - karrivals at Bi.e after one lane is closed)

-10 = (5000- (2400*2)/(10 - karrivals at B) i.e -10 = (5000- 4800)/(10 - karrivals at B)

(so solving we get karrivals at B = 30 veh/km/lane)

traffic density at B = 30 veh/km/ln

Flow at A = 5000 veh/hour

Flow at B = 2400 * 2 = 4800 veh/hour

so speed at B = flow at B(over one lane)/density = 2400/30 = 80 km/hr

speed at A * density at A = speed at B (of forward moving cars) * density at B

so 16.667 * 10 = speed at B * 30 - so speed at B = 5.556 km/hour ( this will be the speed within the lane reduction)

Flow at C = speed of vehicles arriving at B * density at B = 16.667 * 30 = 500 veh/hr/lane, so over two lanes it will be 1000 veh/hr

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