Respuesta :
Answer:
Q = 165.95 cm³ / s, 1) v = [tex]\sqrt{0.55^2 + 19.6 y}[/tex], 2) v = 2.05 m / s,
3) d₂ = 1.014 cm
Explanation:
This is a fluid mechanics exercise
1) the continuity equation is
Q = v A
where Q is the flow rate, A is area and v is the velocity
the area of a circle is
A = π r²
radius and diameter are related
r = d / 2
substituting
A = π d²/4
Q = π/4 v d²
let's reduce the magnitudes
v = 0.55 m / s = 55 cm / s
let's calculate
Q = π/4 55 1.96²
Q = 165.95 cm³ / s
If we focus on a water particle and apply the zimematics equations
v² = v₀² + 2 g y
where the initial velocity is v₀ = 0.55 m / s
v = [tex]\sqrt{0.55^2 + 2 \ 9.8\ y}[/tex]
v = [tex]\sqrt{0.55^2 + 19.6 y}[/tex]
2) ask to calculate the velocity for y = 0.2 m
v = [tex]\sqrt{0.55^2 + 19.6 \ 0.2}[/tex]
v = 2.05 m / s
3) We write the continuous equation for this point 2
Q = v₂ A₂
A₂ = Q / v₂
let us reduce to the same units of the SI system
Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s
A₂ = 165.95 10⁻⁶ / 2.05
A₂ = 80,759 10⁻⁶ m²
area is
A₂ = π/4 d₂²
d₂ = [tex]\sqrt{4 A_2 / \pi }[/tex]
d₂ = [tex]\sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }[/tex]
d₂ = 10.14 10⁻³ m
d₂ = 1.014 cm
