Question

In your garden, you have a hose with a radius of 3 cm connected to a sprinkler with twelve openings, each of which has a radius of 0.2 cm. The sprinkler openings are aimed 25• above the horizon, and the water travels a distance of 4 m before hitting your flowers.

a) What is the exit velocity at the sprinkler openings?

b) What is the flow rate of the sprinkler?

c) What must the water pressure in the hose be?

Answer 1

The water leaves sprinkler at an angle of 25 deg above horizontal

Horizontal distance travelled by water= 4m

Vertical distance travelled = 0 m ( assuming that hose and sprinkler are at ground level )

(a)

Let the exit velocity of water be V

Horizontal velocity = V cos25 and vertical velocity = V sin25

vertical distance = 0m = V sin25 X t - 0.5 g t^2 , where t = time of flight of water ( assuming hose + sprinkler is almost at ground level)

V sin25 X t = 0.5 g t^2 Or t = Vsin25 / 0.5 g

Horizontal distance = 4m = V cos25 X t = Vcos 25 X ( V sin25 /0.5g) = V^2 X cos 25 X sin 25 /0.5 g

V^2 = (4m X 0.5 X g)/ ( cos25 X sin 25 ) = 51.17

V = 7.15 m/s

(b) Sprinkler hole radius = 0.2 cm = 0.002 m Sprinler hole area = 3.14 X (0.002m)^2 = 1.26X 10^(-5) m^2

flow rate per hole = Density of water X Area of the sprinkler hole X velocity of water at the hole

= 1000 Kg/m^3 X 1.26 X 10^(-5)m^2 X 7.15 m/s = 0.09 Kg/sec

so flow rate for 12 openings = 12 X 0.09 = 1.08Kg/s

(c) Force of emerging water = 1.08Kg/s X 7.15 m/s = 7.72N

So this force is coming from water pressure in hose p

p X area of hose = 8.58 N

area of hose = Pi X (0.03 m)^2

p = pressure = 7.72N/ (pi X 0.03m^2) = 2730 N/m^2