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: Re_water = Re_air

⚙️ 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:

P_air = P_water × (ρ_air / ρ_water) × (N₂ / N₁)³ × (D₂ / D₁)⁵

P_air = 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

💡 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.

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