Module 5 : Uninterrupted Flow

Lecture 24 : Freeway Operations

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	Numerical example 1 Consider an existing four lane freeway in rural area, having very restricted geometry with rolling terrain. Peak hour volume is 2000 veh/h with 5% trucks. The traffic is commuter type with peak hour factor 0.92 and interchange density as 0.6 interchanges per kilometer. Freeway consists of two lanes in each direction of 3.3 m width with lateral clearance of 0.6 m. Find the LOS of freeway during peak hour. Solution Assumptions: Assume 0 percent buses and RVs since none are indicated. Assume BFFS of 120 km/h for rural areas. Since the freeway is in a rural area assume that the number of lanes does not affect free-flow speed. Assume $f_p$ = 1.00 for commuter traffic. We can get the corresponding values of adjustment factors from the tables as - $f_{LW}$=3.1, $f_{LC}$=3.9, $f_{ID}$=3.9 and $f_N$=0. Step 1 Find $f_{HV}$ using equation  as given below - $$f_{HV} = \frac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$$ $$= \frac{1}{1+0.05(2.5-1)+0}$$ $$= 0.930$$ Step 2 Convert volume (veh/h) to flow rate (pc/h/ln) using equation as given below $$V_p = \frac{V}{PHF \times N \times f_{HV} \times f_P}$$ $$= \frac{2000}{0.92\times2\times0.930\times1.00}$$ $$= 1,169~\mathrm{pc/h/ln}$$ Step 3 Compute free-flow speed from equation  as given below and putting the respective values of adjustment factors we get $FFS$ as $$FFS = BFFS - f_{LW} -f_{LC} -f_N - f_{ID}$$ $$= 120 - 3.1 - 3.9 - 0.0 - 3.9$$ $$= 109.1~\mathrm{kmph}.$$ Step 4 Determine the density using the equation  as - $$D = \frac{V_p}{S}$$ Since, $90 \leq FFS \leq 120$ and $V_p \leq (3100-15FFS)$ we can take $S=FFS$ (from equation ). Keeping values of $V_p$ and $S$ we can get the value of density as - $$D = \frac{1169}{109.1} = 10.7~\mathrm{pc/km/ln}$$ Step 5 Find Level of service, for the calculated value of density we can get the level of service from the LOS table. i.e for $D$ = 10.7 pc/km/ln we get LOS = B