Water flows through a horizontal pipe at a rate of 32 gpm (1 ft 3 = 7.48 gal). W

Question

Water flows through a horizontal pipe at a rate of 32 gpm (1 ft3 = 7.48 gal). What is the velocity in the larger opening of the pipe that reduces from 4.3 to 2.9 inches in diameter? Larger Pipe Velocity = [Num]ft/s

The water flow rate for a particular sprinkler is 20 gpm. The water must be projected at least 25 feet in radius. The sprinkler is mounted 12 feet above the ground and is aimed at an angle of 26 degrees above the horiztonal. With what velocity must the water leave the sprinkler? What diameter (inches) should the opening be to achieve this velocity? Water Velocity = [Num]ft/s Diameter Opening =  [Num] [Units]

Explanation / Answer

Using A1v1 = A2V2

A = pi(d/2)^2

Volume flow rate = 32gpm = 32 / 7.48 = 4.28 ft^3 per minute

area of larger pipe = pi x (0.0833x4.3 /2 ft)^2 = 0.1 ft^2

Velocity = Volum flow rate / area = 4.28 / 0.1 = 42.5 ft /min

= 42.5 x 60 ft / sec = 2548.46 ft/s

please post 2nd Question in aNOTHER QUESTION LINK

Or you can just use Bornouli's equation,

P + density x gh + density x v^2 /2 = constant

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