Answer:
The inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s
Explanation:
Using Bernoulli's equation
P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²
P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²
ΔP + ρgΔh = 1/2ρ(v₂² - v₁²) (1)
where ΔP = pressure difference = 12.35 kPa = 12350 Pa
Δh = height difference = 1.35 m
From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.
Substituting v₁ into (1), we have
ΔP + ρgΔh = 1/2ρ(v₂² - (d₂²v₂/d₁² )²)
ΔP + ρgΔh = 1/2ρ(v₂² - (d₂/d₁)⁴v₂²)
v₂ = √[2(ΔP + ρgΔh)/ρ(1 - (d₂/d₁)⁴)}
substituting the values of the variables, we have
v₂ = √[2(12350 Pa + 1000 kg/m³ × 9.8 m/s² × 1.35 m)/(1000 kg/m³ (1 - (0.44 m/0.95 m)⁴))}
= √[2(12350 Pa + 13230 Pa)/(1000 kg/m³ × 0.954)]
= √[2(25580 Pa)/954 kg/m³]
= √[51160 Pa/954 kg/m³]
= √53.627
= 7.32 m/s
v₁ = d₂²v₂/d₁²
= (0.44 m/0.95 m)² × 7.32 m/s
= (0.954)² × 7.32 m/s
= 6.66 m/s
The volume flow rate Q = A₁v₁
= πd₁²v₁/4
= π(0.95 m)² × 6.66 m/s ÷ 4
= 18.883 m³/s ÷ 4
= 4.72 m³/s
So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s
