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The average flow of traffic on cross roads A and B, both 2 lane, during design period are 850 PCU per hour and 500 PCU per hour respectively. The all red time required for pedestrian crossing is 12 seconds. The saturation headway is 2.3 seconds. Maxm . Green time available on the 2 phase is …………. Seconds.
  • a)
    16 sec
  • b)
    18 sec
  • c)
    20 sec
  • d)
    22 sec
Correct answer is option 'D'. Can you explain this answer?
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To find the maximum green time available on the two-phase signal, we first need to calculate the saturation flow on each road.

Saturation flow on Road A = (3600 / 2.3) = 1565 PCU/hour
Saturation flow on Road B = (3600 / 2.3) = 1565 PCU/hour

Next, we need to calculate the effective green time for each road. This is the time during which vehicles can pass through the intersection, minus the all-red time required for pedestrian crossing.

Effective green time for Road A = (3600 / 1565) * (850 / 60) - 12
= 31.17 - 12
= 19.17 seconds

Effective green time for Road B = (3600 / 1565) * (500 / 60) - 12
= 18.26 - 12
= 6.26 seconds

The maximum green time available on the two-phase signal is the minimum of the effective green times for both roads.

Maxm. Green time available = min(19.17, 6.26)
= 6.26 seconds

Therefore, the maximum green time available on the two-phase signal is 6.26 seconds.
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The average flow of traffic on cross roads A and B, both 2 lane, during design period are 850 PCU per hour and 500 PCU per hour respectively. The all red time required for pedestrian crossing is 12 seconds. The saturation headway is 2.3 seconds. Maxm . Green time available on the 2 phase is …………. Seconds.a)16 secb)18 secc)20 secd)22 secCorrect answer is option 'D'. Can you explain this answer?
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