Question

Four lawn sprinkler heads are fed by a 1.9-cm-diameter pipe. The water comes out of the heads at an angle of 35° above the horizontal and covers a radius of 6.0 m. (a) What is the velocity of the water coming out of each sprinkler head? (Assume zero air resistance.) (b) If the output diameter of each head is 3.0 mm, how many liters of water do the four heads deliver per second? (c) How fast is the water flowing inside the 1.9-cm-diameter pipe?

Answer 1

Answer:

A) Vo = 9.36 m/s

B) Volumetric flow rate = 2.65 l/s

C) Water velocity = 0.933 m/s

Step-by-step explanation:

We are given;

Vertical distance Δx= 6m

Angle = 35°

Output diameter of each head = 3mm = 0.003m

A) Let Vo be the velocity of water from each sprinkler

Voy = initial vertical component of water velocity from sprinkler head = Vo sin 35

Vox = initial horizontal component of water velocity from sprinkler head = Vo cos 35

Time for water to reach 8.4 m from sprinkler head is given as t = 2Voy/g. Thus, t = 2Vo sin 35/g

Distance water reaches from sprinkler head = (t)(Vox) = 2Vo sin 35/g (Vo cos 35)

Thus;

[2Vo²(sin 35)(cos 35)]/g = 8.4

Vo² = (8.4 x 9.81)/(2)(sin 35)(cos 35) =√87.6927

Vo = 9.36 m/s

B) Area of sprinkler head (A) = πD²/4 = π(0.003)²/4 = 7.0686 x 10^(-6) m²

volume rate of flow thru one sprinkler head is given as;

VoA

Thus;

= (9.36)(7.0686 x 10^(-6)) = 66.16 x 10^(-6) m³/s

Question asks for unit in l/s. Thus, let's convert 66.16 x 10^(-6) m³/s to l/s. So, 66.16 x 10^(-6) m³/s = 66.16 x 10^(-6) x 1000 = 0.0662 liter/s

This is for just one sprinkler

Therefore, for total amount of liters/s flowing in all 4 sprinklers, we have; flow rate = 4 x 0.662 = 2.65 l/s

C) Area of 1.9 cm or 0.019m diameter pipe is;

πD²/4 = π(0.019)²/4 = 2.835 x 10^(-4) m²

From last answer,we saw that flow rate across one sprinkler = 66.16 x 10^(-6) m³/s and for 4 in m³/s will be; 4 x 66.16 x 10^(-6) m³/s = 264.6 x 10^(-6) m³/s

water velocity in pipe is given by; velocity = volumetric flow rate/Area.

Thus,

Velocity = 264.6 x 10^(-6)/2.835 x 10^(-4) = 0.933 m/s

Answer 2

Answer:

Step-by-step explanation:

Given:

Angle, θ = 35°

Vertical distance, Δx = 6 m

Diameter, d = 1.9 cm

= 0.019 m

A.

When the water leaves the sprinkler, it does so at a projectile motion.

Therefore,

Using equation of motion,

(t × Vox) = 2Vo²(sin θ × cos θ)/g

= Δx = 2Vo²(sin 35 × cos 35)/g

Vo² = (6 × 9.8)/(2 × sin 35 × cos 35)

= 62.57

Vo = 7.91 m/s

B.

Area of sprinkler, As = πD²/4

Diameter, D = 3 × 10^-3 m

As = π × (3 × 10^-3)²/4

= 7.069 × 10^-6 m²

V_ = volume rate of the sprinkler

= area, As × velocity, Vo

= (7.069 × 10^-6) × 7.91

= 5.59 × 10^-5 m³/s

Remember,

1 m³ = 1000 liters

= 5.59 × 10^-5 m³/s × 1000 liters/1 m³

= 5.59 × 10^-2 liters/s

= 0.0559 liters/s.

For the 4 sprinklers,

The rate at which volume is flowing in the 4 sprinklers = 4 × 0.0559

= 0.224 liters/s

C.

Area of 1.9 cm pipe, Ap = πD²/4

= π × (0.019)²/4

= 2.84 × 10^-4 m²

Volumetric flowrate of the four sprinklers = 4 × 5.59 × 10^-5 m³/s

= 2.24 × 10^-4 m³/s

Velocity of the water, Vw = volumetric flowrate/area

= 2.24 × 10^-4/2.84 × 10^-4

= 0.787 m/s