Question

A two-lane urban road with one-way traffic has a maximum capacity of 1800 vehicles/hour. under the jam condition, the average length occupied by the vehicle is 5.0 m. the speed versus density relationship is linear. For a traffic volume of 1000 vehicles/hour, the density (in vehicles/km) is

1. 52
2. 58
3. 67
4. 75

Answer 1

Correct Answer - Option 3 : 67

Concept:

The relationship between speed ( u )  and density ( k ) is given by the

\({\bf{u}} = {{\bf{u}}_{\bf{f}}} - \left( {\frac{{{{\bf{v}}_{\bf{f}}}}}{{{{\bf{k}}_{\bf{j}}}}}} \right) \times {\bf{k}}\)

u - Mean speed at density k ( m/s )

k- Density of stream ( veh/km )

u_f - Free mean speed

k_j – Jam density

Relation between flow, speed, and density is given by,

q = k × u

q – Flow given in veh/hr

Space headway is defined as the distance between corresponding points of two successive vehicles at any given time. It can also be taken as the average space occupied by each vehicle.

\({\bf{k}} = \frac{{1000}}{{{\bf{Space}}\;{\bf{headway}}}}\)

At capacity or maximum flow,

\({{\rm{k}}_0} = \frac{{{{\rm{k}}_{\rm{j}}}}}{2}\)

\({u_0} = \frac{{{{\rm{u}}_{\rm{f}}}}}{2}\)

\({{\bf{q}}_{{\bf{max}}}} = {{\bf{k}}_0} \times {{\bf{u}}_0} = \frac{{{{\bf{k}}_{\bf{j}}} \times {{\bf{u}}_{\bf{f}}}}}{4}\)

q_max – Maximum flow or flow at capacity

k₀, u₀ – Density and speed at capacity

CALCULATION:

GIVEN:

Two-lane maximum capacity = 1800 veh/hr

At jam condition, space headway = 5m

\({\rm{One\;lane\;capacity}} = {\rm{\;}}{{\rm{q}}_{{\rm{max}}}} = \frac{{1800}}{2} = 900{\rm{veh}}/{\rm{hr}}\)

\({{\rm{k}}_{\rm{j}}} = \frac{{1000}}{5} = 200{\rm{\;veh}}/{\rm{km}}\)

\(900 = \frac{{200 \times {{\rm{v}}_{\rm{f}}}}}{4}\)

→ u_f = 18 km/hr

For two-lane flow = 1000 veh/hr

\({\rm{For\;one\;lane\;flow\;q}} = \frac{{1000}}{2} = 500{\rm{veh}}/{\rm{hr}}\)

\({\rm{u}} = 18 - \left( {\frac{{18}}{{200}}} \right) \times {\rm{k}}\)

\(\frac{{500}}{{\rm{k}}} = 18 - \left( {\frac{{18}}{{200}}} \right) \times {\rm{k}}\;\;\left( {\;{\bf{u}} = \frac{{\bf{q}}}{{\bf{k}}}\;} \right)\)

Solving we get,

k = 33.32 veh/km (for single lane )

For two-lane k = 2 × 33.32 = 66.64 ≈ 67