Respuesta :
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
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
