Answer:
a) Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]
, b) Q = 3.4 10⁻² m³ / s
, c) Q = 4.8 10⁻² m³ / s
Explanation:
We can solve this fluid problem with Bernoulli's equation.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
With the two tubes they are at the same height y₁ = y₂
P₁-P₂ = ½ ρ (v₂² - v₁²)
The flow rate is given by
A₁ v₁ = A₂ v₂
v₂ = v₁ A₁ / A₂
We replace
ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]
ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]
Let's clear the speed
v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]
The expression for the flow is
Q = A v
Q = A₁ v₁
Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]
The areas are
A₁ = π r₁
A₂ = π r₂
We replace
Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]
Let's calculate for the different pressures
r₁ = d₁ / 2 = 1.00 / 2
r₁ = 0.500 10⁻² m
r₂ = 0.250 10⁻² m
b) ΔP = 6.00 kPa = 6 10³ Pa
Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))
Q = 1.57 10⁻² √(12 10³/2550)
Q = 3.4 10⁻² m³ / s
c) ΔP = 12 10³ Pa
Q = 1.57 10⁻² √(2 12 10³ / (850 3)
Q = 4.8 10⁻² m³ / s
