🔬 Kolmogorov Scales Calculator — η, τη, uη

Compute the three Kolmogorov scales (length η, time τη, velocity uη), Taylor microscale λ, integral scale L, scale ratios, Reynolds numbers, and DNS grid estimation. Ref: §7.6.

🔬 Energy Cascade: L → λ → η

📝 Configuration

⚡ Turbulence
💧 Fluid Properties

0 = compute from ρ,μ
🌬️ Flow
Key Equations:

$\eta = \left(\dfrac{\nu^3}{\varepsilon}\right)^{1/4}$

$\tau_\eta = \left(\dfrac{\nu}{\varepsilon}\right)^{1/2}$

$u_\eta = (\nu\varepsilon)^{1/4}$

📊 Results

Configure inputs and click Calculate Scales.

ℹ️ About Kolmogorov Scales

Kolmogorov (1941) showed that the smallest scales of turbulence are uniquely determined by ν and ε. These scales are where viscous dissipation converts turbulent kinetic energy into heat.

Three scales: η (length), τη (time), uη (velocity).
DNS requirement: resolve all scales from L to η, needing ~(L/η)³ grid points.

📘 Calculation Methodology

Theory

$$\eta=\left(\frac{\nu^3}{\varepsilon}\right)^{1/4},\;\tau_\eta=\sqrt{\frac{\nu}{\varepsilon}},\;u_\eta=(\nu\varepsilon)^{1/4}$$ $$\lambda=\sqrt{\frac{10\nu k}{\varepsilon}},\;L=\frac{k^{3/2}}{\varepsilon}$$ $$\frac{L}{\eta}\sim\text{Re}_L^{3/4},\;N_{\text{DNS}}\sim\left(\frac{L}{\eta}\right)^3\sim\text{Re}_L^{9/4}$$

Worked Example

Problem: ε=100, k=2.5, air (ν=1.46×10⁻⁵)

1. η = (1.46e-5³/100)¼ = 5.5×10⁻⁵ m
2. τη = √(1.46e-5/100) = 3.8×10⁻⁴ s
3. uη = (1.46e-5×100)¼ = 0.035 m/s
4. λ = √(10×1.46e-5×2.5/100) = 6.0×10⁻⁴ m
5. L = 2.5¹·⁵/100 = 3.95×10⁻² m
6. L/η ≈ 718, NDNS ≈ 3.7×10⁸

References

  • Kolmogorov, A. N. Dokl. Akad. Nauk SSSR, 30:301, 1941.
  • Pope, S. B. Turbulent Flows, Cambridge, 2000.
  • Tennekes & Lumley A First Course in Turbulence, MIT Press.
  • Gupta, S. C. Applied CFD — §7.6.