Question

The water flow rate for a particular sprinkler is 17 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?

Answer 1

we know,

1 gallon = 0.003785 m^3

so, volume flow rate = 17*0.003785 m^3/s

= 0.0643 m^3/s

let h = 12 feet

= 12*0.3048

= 3.66 m

x = 25 feet

= 25*0.3048

= 7.62 m

theta = 26 degrees

let v is the initial velocity of water from the opening.

let t is the time taken for the water to strike the ground.

t = x/vx

= 25/(v*cos(26))

y = vy*t - (1/2)*g*t^2

3.66 = v*sin(26)*25/(v*cos(26)) - (1/2)*9.8*(25/(v*cos(26)))^2

3.66 = 25*tan(26) - 4.9*25^2/(v*cos(26))^2

==> v = 21.1 m/s <<<<<<<-----------------Answer

b) we know, volume flow rate = A*v

0.0643 = (pi*d^2/4)*v

0.0643 = (pi*d^2/4)*21.1

==> d = 0.0622

= 6.22 cm

= 2.45 inch <<<<<<<-----------------Answer