Question

The water flow rate for a particular sprinkler is 16 gpm. The water must be projected at least 25 feet in radius. The sprinkler is mounted 11 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?

Answer 1

Let ,

R = maximum distance reached by the water

v = velocity of water leaving the sprinkler

h = height from which the water is projected

= angle of projection from the horizontal

Horizontal component of the velocity = vcos

Vertical component of the velocity = vsin

For horizontal motion

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 ..............equation (1)

For vertical motion

S_y = u_y​​​​t + (1/2)a_yt²

h = -vsint + (1/2)gt​​​​​​² ....  eq(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

11 = -25×tan29° + 32×(25)²/ 2×v²×cos²29°

11 = -13.86 + 20000 / 2×v²×0.765

24.86 = 20000 / 2×v²×0.765

v² = 525.82

v = 22.93 ft/s

From equation of continuity

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

A × v = 16 gpm (gpm is gallon per minute)

A × 22.93 = (16×0.1336 /60 ) ft³/s

d²/4 = 0.0356 / 22.93 ( d = diameter of the pipe)

d^{​​​​​​2}/4 = 0.00155

d² = 0.001979

d = 0.0445 ft

d = 0.0445 × 12 inches

d = 0.534 inches