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A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream and downstream of the segment midpoint. It is on rolling terrain with 10% heavy vehicles and is operating at capacity with a peak-hour factor of 0.9. If the road is expanded to four 11-foot lanes with a 2-foot right shoulder, and traffic after the expansion is projected to increase by 10% with the same heavy vehicle percentage and peak-hour factor, what is the new LOS and estimated density?

Respuesta :

Chrisnando

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

[tex] F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)} [/tex]

Substituting figures in the equation we have:

[tex] = \frac{1}{1+0.1(2.5-1)+0(2-1)}[/tex]

= 0.75

Let's now calculate equivalent flow rate of the car using:

[tex]Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)} [/tex]

[tex] = \frac{7500}{0.9*4*0.75*1} [/tex]

= 2777.7 pc/h/en

Calculating for traffic density, we have:

[tex] D = \frac{Vp}{FFS} [/tex]

[tex] D = \frac{2777.7}{70} [/tex]

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

Lanuel

The estimated density for this westbound section of freeway is equal to 39.685 Pc/mi/en.

Given the following data:

  • Lane width = 12ft
  • Right shoulder width = 6ft
  • Percentage of heavy vehicles = 10℅
  • Peak hour factor = 0.9.

How to determine the density.

First of all, we would calculate the heavy adjustment factor and equivalent flow rate of the car. Also, we would use the following values:

  • Passenger-car equivalent for recreational purpose, [tex]E_R[/tex] = 2
  • Passenger-car for trucks and buses, [tex]E_{T}[/tex] = 2.5
  • Number of lanes, N = 4.
  • Percentage of RV = 0.18.
  • Traffic flow, V = 7500 veh/h.
  • Freeflow speed, FFS = 70mph.

For heavy adjustment factor:

[tex]F_{hv}=\frac{1}{1+P_{T}(E_{T}-1)+P_R(E_R-1)} \\\\F_{hv}=\frac{1}{1+0.1(2.5-1)+0.18(2-1)}\\\\F_{hv}=\frac{1}{1+0.1(1.5)+0.18}\\\\F_{hv}=\frac{1}{1.33}\\\\F_{hv}=0.75[/tex]

For the equivalent flow rate:

[tex]V_P=\frac{V}{P_{HF}NF_{HV}F_P} \\\\V_P=\frac{7500}{0.9 \times 4 \times 0.75 \times 1}\\\\V_P=\frac{7500}{2.7}\\\\V_P= 2777.78\;pc/h/en[/tex]

Now, we can determine the density:

[tex]D=\frac{V_P}{FFS} \\\\D=\frac{2777.78}{70}[/tex]

Density, D = 39.685 Pc/mi/en.

Based on the table for level of service (LOS) criteria, the new LOS is E.

Read more on traffic flow here: https://brainly.com/question/15236911

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