Reynolds and Prandtl Numbers
Reynolds and Prandtl Numbers
Dimensionless numbers let us collapse complex fluid dynamics into universal relationships. The Reynolds number tells us the flow regime; the Prandtl number describes the fluid’s thermal behavior.
$$Re = \frac{\rho V L_c}{\mu} = \frac{V L_c}{\nu}$$
Ratio of inertial to viscous forces. $V$ = velocity, $L_c$ = characteristic length, $\mu$ = dynamic viscosity, $\nu = \mu/\rho$ = kinematic viscosity.
Internal flow (pipe): $L_c = D$ (diameter). Transition at $Re_D \approx 2300$.
External flow (flat plate): $L_c = x$ (distance from leading edge). Transition at $Re_x \approx 5 \times 10^5$.
$$Pr = \frac{\nu}{\alpha} = \frac{\mu c_p}{k}$$
Ratio of momentum diffusivity to thermal diffusivity. $\alpha = k/(\rho c_p)$.
Liquid metals: $Pr \ll 1$ • Gases: $Pr \approx 0.7$ • Water: $Pr \approx 7$ • Oils: $Pr \gg 1$ (50–100,000)
Water ($\nu = 1.0 \times 10^{-6}$ m²/s) flows at 2 m/s in a 5 cm pipe. Laminar or turbulent?
$$Re = \frac{VD}{\nu} = \frac{2 \times 0.05}{1.0 \times 10^{-6}} = 100{,}000$$
$Re = 100{,}000 \gg 2300$ → turbulent.
Air ($\nu = 1.5 \times 10^{-5}$ m²/s) flows at 10 m/s over a flat plate. At what distance $x$ does transition occur?
$$x_{\text{cr}} = \frac{Re_{\text{cr}} \cdot \nu}{V} = \frac{5 \times 10^5 \times 1.5 \times 10^{-5}}{10} = 0.75 \text{ m}$$
Water at 20°C: $\mu = 1.002 \times 10^{-3}$ Pa·s, $c_p = 4182$ J/(kg·K), $k = 0.598$ W/(m·K). Find $Pr$.
$$Pr = \frac{\mu c_p}{k} = \frac{1.002 \times 10^{-3} \times 4182}{0.598} = 7.01$$
This matches the known value. Water’s velocity boundary layer is ~7× thinner than its thermal boundary layer.
Practice Problems
Show Answer Key
1. $Re = 0.5 \times 0.1 / (5 \times 10^{-4}) = 100$. Laminar.
2. $Re = 30 \times 1 / (1.6 \times 10^{-5}) = 1{,}875{,}000$. Turbulent ($> 5 \times 10^5$).
3. $Pr = 1.85 \times 10^{-5} \times 1005 / 0.026 = 0.715$
4. Thicker — heat diffuses faster than momentum when $Pr < 1$.
5. $V = Re \cdot \nu / D = 2300 \times 10^{-6} / 0.02 = 0.115$ m/s
6. $V = 2300 \times 1.5 \times 10^{-5} / 0.15 = 0.23$ m/s