Respuesta :
Solution :
Given :
The number of lanes in each of the direction is N = 3
The terrain is rolling. And the traffic stream is cars and heavy vehicles. And the peak hour factor is given as 0.85
Now the observed flow speed is S = 48 mph
Flow rate of vehicles is V = 3690 veh/h
Highway operates at the maximum LOS (level of service) D condition.
The density value that is allowed at this condition is D = 35 pc/mi/lane
Therefore, [tex]$D=\frac{v_p}{S}$[/tex]
or [tex]$v_p = D \times S$[/tex]
= 35 x 48
= 1680 pcphpl
[tex]$v_s= \frac{V}{PHF \times N \times f_{HV} \times f_p}$[/tex]
[tex]$1680= \frac{3690}{0.85 \times 3 \times f_{HV} \times 1}$[/tex]
[tex]$f_{HV}=0.8613$[/tex]
[tex]$f_{HV}=\frac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$[/tex]
[tex]$0.8613=\frac{1}{1+P_T(2.5-1)}$[/tex]
[tex]$P_T=0.1073$[/tex]
So the percentage of trucks is 23.33% of the total volume and is equal to 754 vehicles per hour.
If the heavy vehicles are banned from the highway, the number of vehicles will reduce to 2936 vehicles per hour.
The value of [tex]$f_{HV} = 1$[/tex]
[tex]$v_p=\frac{2936}{0.85 \times 3 \times 1 \times 1}$[/tex]
= 1151.3 pcphpl
Density value is , [tex]$D=\frac{v_p}{S}$[/tex]
= [tex]$\frac{1151.3}{48}$[/tex]
23.98 pc/mi/lane
Therefore, the level of service improves to LOS C, when the heavy vehicles are banned in the highways while the car traffic remains the same.
