A pre-timed three phase signal has critical lane flow rate for first two phases as $300, 278 v e h / h r$ with saturation flow rate as $1800 v e h / h r / l a n d$ for all phases. The lost time is $5$ seconds for each phase. If cycle length is $60$ seconds, then the effective green time of the third phase is seconds.
18.35

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Solution

The correct option is A 18.35
Lost time $= 3 \times 5 = 15$ seconds
Length of cycle, $C_{0} = 60$ seconds
$y_{1} = \frac{q_{1}}{S_{1}} = \frac{300}{1800}$
$y_{2} = \frac{q_{2}}{S_{2}} = \frac{278}{1800}$
$∴$ $C_{0} = \frac{1.5 \times 15 + 5}{1 - y}$
$\Rightarrow$ $60 = \frac{1.5 \times 15 + 5}{1 - y}$
$y = 0.542$ and $y = y_{1} + y_{2} + y_{3}$
$\Rightarrow 0.542 = \frac{578}{1800} + y_{3}$
$\Rightarrow y_{3} = 0.221$
$∴$ $G = \frac{y_{3}}{y} (C_{0} - L)$
$= \frac{0.221}{0.542} (60 - 15) = 18.35$ sec

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A pre-timed three phase signal has critical lane flow rate for first two phases as 300,278 veh/hr with saturation flow rate as 1800veh/hr/land for all phases. The lost time is 5 seconds for each phase. If cycle length is 60 seconds, then the effective green time of the third phase is seconds.18.35

The correct option is A 18.35Lost time =3×5=15 seconds Length of cycle, C0=60 seconds y1=q1S1=3001800 y2=q2S2=2781800 ∴ C0=1.5×15+51−y ⇒ 60=1.5×15+51−y y=0.542 and y=y1+y2+y3 ⇒ 0.542=5781800+y3 ⇒ y3=0.221 ∴ G=y3y(C0−L) =0.2210.542(60−15)=18.35 sec