🌡️ Turbulent Prandtl Number Calculator — Prt

Compute Prt = νtt using 6 correlations (Kays-Crawford, Jischa-Rieke, Kays 1994, Reynolds analogy, Cebeci, constant). Compute turbulent heat flux, effective conductivity, and Reynolds analogy factor. Ref: §7.5.

🌡️ Momentum vs Thermal Transport

📝 Configuration

⚡ Turbulence

0 = auto

0 = default 0.85
💧 Fluid Properties
🌬️ Flow / Gradients
Key Equations:

$\text{Pr}_t = \dfrac{\nu_t}{\alpha_t} = \dfrac{\varepsilon_m}{\varepsilon_H}$

$k_{\text{eff}} = k\left(1 + \text{Pr}\dfrac{\nu_t}{\text{Pr}_t\,\nu}\right)$

$\dfrac{2\text{St}}{C_f} = \dfrac{1}{\text{Pr}_t}$

📊 Results

Configure inputs and click Calculate.

ℹ️ About Prt

The turbulent Prandtl number Prt = νtt relates turbulent momentum and heat transport. For gases, Prt ≈ 0.85–0.9. For liquid metals (low Pr), Prt can reach 1–3. The Reynolds analogy assumes Prt = 1.

📘 Calculation Methodology

Theory

$$\text{Pr}_t = \frac{\nu_t}{\alpha_t} = \frac{\varepsilon_m}{\varepsilon_H}$$ $$\text{Jischa-Rieke: } \text{Pr}_t = 0.85 + \frac{0.015}{\text{Pr}}$$ $$\text{Kays (1994): } \text{Pr}_t = 0.85 + \frac{0.7}{\text{Pr}\cdot\nu_t/\nu}$$ $$k_{\text{eff}} = k\left(1+\frac{\text{Pr}\cdot\nu_t}{\text{Pr}_t\cdot\nu}\right)$$ $$\frac{2\,\text{St}}{C_f} = \frac{1}{\text{Pr}_t}\;\;\text{(Reynolds analogy)}$$

Worked Example

Problem: Air, νt=5×10⁻³, ν=1.46×10⁻⁵

1. Pr = 0.71, νt/ν = 342
2. Jischa-Rieke: Prt = 0.85 + 0.015/0.71 = 0.871
3. Kays 1994: Prt = 0.85 + 0.7/(0.71×342) = 0.853
4. αt = 5e-3/0.85 = 5.88×10⁻³
5. keff/k = 1 + 0.71×342/0.85 = 286

References

  • Kays, W. M. & Crawford, M. E. Convective Heat and Mass Transfer, McGraw-Hill.
  • Jischa, M. & Rieke, H. B. Int. J. Heat Mass Transfer, 22:1547, 1979.
  • Kays, W. M. ASME J. Heat Transfer, 116:284, 1994.
  • Gupta, S. C. Applied CFD — §7.5.