Fundamentals of Transportation/Shockwaves/Solution

TProblem

Problem:

Flow on a road is $q_{1} = 1800 v e h / h r / l a n e$ , and the density of $k_{1} = 14.4 v e h / k m / l a n e$ . To reduce speeding on a section of highway, a police cruiser decides to implement a rolling roadblock, and to travel in the left lane at the speed limit ( $v_{2} = 88 k m / h r$ ) for 10 km. No one dares pass. After the police cruiser joins, the platoon density increases to 20 veh/km/lane and flow drops. How many vehicles (per lane) will be in the platoon when the police car leaves the highway?

How long will it take for the queue to dissipate?

Example

Solution:

Step 0

Solve for Unknowns:

Original speed

$v_{1} = \frac{q_{1}}{k_{1}} = \frac{1800}{14.4} = 125 k m / h$

Flow after police cruiser joins

$q_{2} = k_{2} v_{2} = 88 * 20 = 1760 v e h / h$

Step 1

Calculate the wave velocity:

$v_{w} = \frac{q_{2} - q_{1}}{k_{2} - k_{1}} = \frac{1760 - 1800}{20 - 14.4} = - 7.14 k m / h r$

Step 2

Determine the growth rate of the platoon (relative speed)

$v_{r 2} = v_{2} - v_{w} = 88 - (- 7.14) = 95.1 k m / h r$

Step 3

Determine the time spent by the police cruiser on the highway

$t = d / v = 10 k m / 88 k m / h r = 0.11 h r = 6.8 m i n u t e s$

Step 4

Calculate the Length of platoon (not a standing queue)

$L = v t = 95.1 k m / h r * 0.11 h r = 10.46 k m$

Step 5

What is the rate at which the queue grows, in units of vehicles per hour?

$Δ q = q_{1} - k_{1} v_{w} = q_{2} - k_{2} v_{w} = 1800 - (14.4 * - 7.14) = 1760 - (20 * - 7.14) = 1902 v e h / h r$

Step 6

The number of vehicles in platoon

$L_{p} k_{2} = 10.46 k m * 20 v e h / k m = 209.2 v e h i c l e s$

OR

$Δ q t = 1902 v e h / h r * 0.11 h r = 209.2 v e h$

How long will it take for the queue to dissipate?

(a) Where does the shockwave go after the police cruiser leaves? Just reverse everything?

$v_{w} = \frac{q_{2} - q_{1}}{k_{2} - k_{1}} = \frac{1800 - 1760}{14.4 - 20} = - 7.14 k m / h r$

Second wave never catches first.

(b) Return speed to 125, keep density @ 20? $q_{2} = 20 * 125 = 2500$

$v_{w} = \frac{q_{2} - q_{1}}{k_{2} - k_{1}} = \frac{2500 - 1760}{20 - 20} = \infty k m / h r$

Second wave instantaneously catches first

(c) Return speed to 125, keep density @ halfway between 14.4 and 20? $q_{2} = 20 * 125 = 2500$

$v_{w} = \frac{q_{2} - q_{1}}{k_{2} - k_{1}} = \frac{2150 - 1760}{17.2 - 20} = 139 k m / h r$

Second wave quickly catches first

$\begin{matrix} 139 * t = 7.14 * (.11 + t) \\ (139 - 7.14) t = 7.14 * 11 \\ t = 0.006 h = .35 min = 21 \sec \end{matrix}$

(d) Drop upstream q,k to slow down formation curve.

$v_{w} = \frac{q_{2} - q_{1}}{k_{2} - k_{1}} = \frac{1780 - 1760}{14.24 - 20} = - 3.47 k m / h r$

Second wave eventually catches first

(e) If $q_{u p s t r e a m}$ falls below downstream capacity, wave dissipates

$v_{w} = \frac{q_{2} - q_{1}}{k_{2} - k_{1}} = \frac{1250 - 1760}{10 - 20} = + 51 k m / h r$

Forward moving wave.