Answer:
a. 24 m³, 7.0×10⁴ Pa
Explanation:
Use Bernoulli's equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₃ + ½ ρ v₃² + ρgh₃
Given or assumed:
P₁ = 0 Pa
v₁ ≈ 0 m/s
h₁ = 10 m
P₃ = 0 Pa
h₃ = 2 m
ρ = 1000 kg/m³
g = 9.8 m/s²
Substituting and simplifying:
ρg (10) = ½ ρ v₃² + ρg (2)
10g = ½ v₃² + 2g
(Notice this is simply an energy balance.)
Solving for v₃:
v₃ = √(16g)
v₃ = 12.5 m/s
Volumetric flow rate is velocity times area:
Q = v₃ A₃
Q = (12.5 m/s) (0.0160 m²)
Q = 0.200 m³/s
After 2 minutes, the volume discharged is:
V = (0.200 m³/s) (120 s)
V = 24.0 m³
Using Bernoulli's principle again:
P₂ + ½ ρ v₂² + ρgh₂ = P₃ + ½ ρ v₃² + ρgh₃
Since h₂ = h₃ and P₃ = 0:
P₂ + ½ ρ v₂² = ½ ρ v₃²
P₂ = ½ ρ (v₃² − v₂²)
Since volumetric flow is constant (mass is conserved):
Q = Q
v₂ A₂ = v₃ A₃
v₂ = v₃ A₃ / A₂
Substituting:
P₂ = ½ ρ (v₃² − (v₃ A₃ / A₂)²)
P₂ = ½ ρ v₃² (1 − (A₃ / A₂)²)
P₂ = ½ (1000 kg/m³) (12.5 m/s)² (1 − (0.0160 / 0.0480)²)
P₂ = 69,700 Pa
P₂ ≈ 7.0×10⁴ Pa
