Module 5 : Uninterrupted Flow
Lecture 24 : Freeway Operations
1 2 3 4 5 6 7 8 9 10
 

Numerical example 2

A new suburban freeway is designed in the level terrain. Peak hour volume is 4,000 veh/h and the flow consists of 15% trucks and 3% recreational vehicles (RV's). The traffic is commuter type with peak hour factor 0.85 and interchange density as 0.9 interchanges per kilometer. Lane width is proposed to be 3.6 m with lateral clearance of 1.8 m. How many lanes are needed to provide LOS C during the peak hour?

Solution

Assumptions: Assume $ BFFS$ of 120 km/h. Since the freeway is being designed in a suburban area assume that the number of lanes affects free-flow speed. For commuter traffic we can take $ f_p$ = 1.00. We can get the corresponding values of adjustment factors from the tables as - $ f_{LW}=0, f_{LC}=0, f_{ID}=8.1$ and $ f_N=4.8$.

Step 1

Find $ f_{HV}$ using equation [*] as given below:
$\displaystyle f_{HV}$ $\displaystyle =$ $\displaystyle \frac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$  
  $\displaystyle =$ $\displaystyle \frac{1}{1+0.15(1.5-1)+0.03(1.2-1)}$  
  $\displaystyle =$ $\displaystyle 0.925$  

Step 2

Convert volume (veh/h) to flow rate (pc/h/ln) using equation [*]. Consider a four lane option, for four lane $ N=2$, keeping value of $ f_{HV}$ and $ N$ in equation [*] we get $ V_p$ as:
$\displaystyle V_p$ $\displaystyle =$ $\displaystyle \frac{V}{PHF \times N \times f_{HV} \times f_P}$  
  $\displaystyle =$ $\displaystyle \frac{4000}{0.85\times2\times0.925\times1.00}$  
  $\displaystyle =$ $\displaystyle 2,544~\mathrm{pc/h/ln}.$  

Four lane option is not acceptable as 2544 pc/h/ln exceeds capacity of 2400 pc/h/ln. Here 2400 pc/h/ln is the capacity of a single lane under standard conditions.

Step 4

Consider a six lane option
$\displaystyle V_p$ $\displaystyle =$ $\displaystyle \frac{4000}{(.85\times3\times0.925\times1.00}$  
  $\displaystyle =$ $\displaystyle 1,696~\mathrm{pc/h/ln}.$  

Step 5

Compute FFS for a six-lane freeway from equation [*] and putting the respective values of adjustment factors we get $ FFS$ as:
$\displaystyle FFS$ $\displaystyle =$ $\displaystyle BFFS - f_{LW} -f_{LC} -f_N - f_{ID}$  
  $\displaystyle =$ $\displaystyle 120-0-0-4.8-8.1$  
  $\displaystyle =$ $\displaystyle 107.1.~\mathrm{km/h}.$  

Step 6

Determine density from equation [*]
$\displaystyle D$ $\displaystyle =$ $\displaystyle \frac{V_p}{S}$  

Since, $ 90 \leq FFS \leq 120$ and $ v-p \leq (3100-15FFS)$ we can take $ S=FFS$ (from equation [*]) Putting values of $ V_p$ and $ S$ we get density as
$\displaystyle D$ $\displaystyle =$ $\displaystyle \frac{1696}{107.1}~=~15.8~\mathrm{pc/km/ln}$  

Step 7

Check the LOS, for the calculated value of density we can get the level of service from the LOS table; i.e for $ D$ = 15.8 pc/km/ln we get LOS = C. Hence number of lanes to be provided to satisfy LOS C during peak hour = 6.