The peak hour volume is obtained as 865 PCU/hr and the design traffic flow rate obtained as 995PCU/hr. The peak 10 minute volume is

82.91PCU/hr

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144.17PCU/hr

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165.833 PCU/hr

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72.08 PCU/hr

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Solution

The correct option is C 165.833 PCU/hr
Design traffic flow rate = $\frac{p e a k 10 m i n u t e v o l u m e}{\frac{60}{10}}$
$= \frac{995}{6} = 165.833 \frac{P C U}{h r}$

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The peak hour volume is obtained as 865 PCU/hr and the design traffic flow rate obtained as 995PCU/hr. The peak 10 minute volume is

The correct option is C 165.833 PCU/hrDesign traffic flow rate = peak10minutevolume6010 =9956=165.833PCUhr

The correct option is C 165.833 PCU/hr
Design traffic flow rate = $\frac{p e a k 10 m i n u t e v o l u m e}{\frac{60}{10}}$
$= \frac{995}{6} = 165.833 \frac{P C U}{h r}$