Answer
Rotating Disc Torque and Power Calculation (Fluid Mechanics – Question 8)
🧠 Detailed Solution
✅ Given Data:
- Disc diameter in water: D₁ = 0.225 m
- Rotational speed in water: N₁ = 23 rev/s
- Torque in water: T₁ = 1.1 Nm
- Density and viscosity:
- Water: ρ = 1000 kg/m³, μ = 1.01 × 10⁻³ Ns/m²
- Air: ρ = 1.2 kg/m³, μ = 1.86 × 10⁻⁵ Ns/m²
🔍 Concept Used:
We use dynamic similarity through Reynolds number to relate conditions between water and air:
Re = (ρ × N × D²) / μ
We require: Rewater = Reair
⚙️ Step-by-Step Calculations:
1️⃣ Power in Water:
P = 2π × N₁ × T₁ = 2π × 23 × 1.1 ≈ 158.93 W
2️⃣ Match Reynolds Number:
Air disc diameter: D₂ = 0.675 m
(1000 × 23 × 0.225²) / (1.01 × 10⁻³) = (1.2 × N₂ × 0.675²) / (1.86 × 10⁻⁵)
Solve for N₂:
N₂ ≈ 39.15 rev/s
3️⃣ Power in Air:
Pair = Pwater × (ρair / ρwater) × (N₂ / N₁)³ × (D₂ / D₁)⁵
Pair = 158.93 × 0.0012 × (1.702)³ × (3)⁵ ≈ 227.3 W
4️⃣ Torque in Air:
T = P / (2πN) = 227.3 / (2π × 39.15) ≈ 0.933 Nm
✅ Final Answers:
• Required speed in air: 39.22 rev/s
• Torque in air: 0.933 Nm
• Required speed in air: 39.22 rev/s
• Torque in air: 0.933 Nm
💡 Explanation:
This fluid mechanics problem evaluates the power and torque needed to rotate a disc in air based on a water model test. By applying dynamic similarity and maintaining equal Reynolds numbers, the conditions in two different fluids are made equivalent. This helps engineers accurately predict full-scale behavior without full-scale testing.