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- Search.. jerning.comitiscrm/mod/ibis/viewphpdssoum e Univer.. X Jump to.. ng Learning mpts Score Question 6 of Map Sapling Learning A round pipe of varying diameter diameter is 52.9 am (0529 m) and the flow speed of the petroleum is 12.1 mis petroleum flows at 6.97 mis. What pipe's diameter at the refinery? carries petroleum from a wellhead to a refinery. At the wellhead the pipe's At the the is the volume flow rate of the petroleum along the pipe and what is the Volume flow rate: 2.7 m'/s Diameter 50 .118 can There is a hint available View the hint divider bar again to hide the hint OPrevios Check Answer Next Exit 0

Explanation / Answer

here,

diameter 1 , d1 = 0.529 m

speed 1 , v1 = 12.1 m/s

the volume flow rate , V = pi * (d1/2)^2 * v1

V = pi * ( 0.529/2)^2 * 12.1 m^3/s

V = 2.66 m^3 / s

v2 = 6.97 m/s

let the diameter of refinery be d2

Volume flow rate , V = pi * (d2 /2)^2 * v2

2.66 = pi * ( d2 /2)^2 * 6.97

solving for d2

d2 = 0.7 m

the diameter of refinery is 0.7 m

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