Question

The water flow rate for a particular sprinkler is 18 gpm. The water must be projected at least 22 feet in radius. The sprinkler is mounted 10 feet above the ground and is aimed at an angle of 29 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 = ft/s

Diameter Opening =  inches

Answer 1

As there is no acceleration in the horizontal direction

S_x = u_xt + (1/2)a_xt²

R = vcos×t + 0 ( as S_x = R and a_{​​​​​​x} = 0 )

t = R/vcos    .............. (1)

For vertical motion

S_y = u_y​​​​t + (1/2)a_yt²

h = -vsint + (1/2)gt​​​​​​² ..................  (2)

( as vsin is upwards so negative)

From equation (1) and (2)

h = -vsin(R/vcos) + (1/2)g×(R/vcos)²

h = -Rtan + gR² / 2v²​​​​​​cos²

On putting the numerical values

10 = -22*tan29° + 32*(22)²/ 2*v²*cos²29°

10 = -12.194 + 15488 / 2*v^2*0.765

22.194 = 15488 / 2*v²*0.765

v² = 456.116

v = 21.35 ft/s

From equation of continuity

Area of cross section × velocity = rate of flow of liquid

A * v = 18 gpm (gpm is gallon per minute)

A * 21.35 = (18*0.1336 /60 ) ft³/s

d²/4 = 0.04008 / 21.35 ( d = diameter of the pipe)

d^{​​​​​​2}/4 = 0.001876

d² = 0.002389

d = 0.0488 ft

d = 0.0488 * 12 inches

d = 0.586 inches