Answer:
6.0 m/s
3.3×10⁵ Pa
4.7×10⁻⁴ m³/s
Explanation:
Mass is conserved, so the mass flow at the inlet equals the mass flow in the bathroom. Since the density is the same, that means the volume flow rate is the same.
Q₁ = Q₂
v₁ A₁ = v₂ A₂
v₁ (π D₁² / 4) = v₂ (π D₂² / 4)
v₁ D₁² = v₂ D₂²
Given v₁ = 1.5 m/s, D₁ = 2.0 cm, and D₂ = 1.0 cm:
(1.5 m/s) (2.0 cm)² = v (1.0 cm)²
v = 6.0 m/s
To find the pressure, we use the Bernoulli principle:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
Given:
P₁ = 4.0×10⁵ Pa
ρ = 1000 kg/m³
v₁ = 1.5 m/s
h₁ = 0 m
v₂ = 6.0 m/s
h₂ = 5.0 m
4.0×10⁵ + ½ (1000) (1.5)² + 0 = P + ½ (1000) (6.0)² + (1000) (9.8) (5.0)
P = 3.3×10⁵ Pa
Finally, the volume flow rate is:
Q = vA
Q = v (π D² / 4)
Q = π/4 (6.0 m/s) (0.01 m)²
Q = 4.7×10⁻⁴ m³/s