⚖️ Heat & Mass Transfer Analogy Calculator
Verify boundary layer analogies and predict mass transfer coefficients from thermal coefficients. Computes Sherwood ($Sh$), Schmidt ($Sc$), Lewis ($Le$) numbers, and Chilton-Colburn $j$-factors.
🌊 Boundary Layers & Relative Thickness Visualizer
📝 Configuration
Quick Presets:
- Wet plate drying in air flow (20°C)
- Oxygen transport in water pipe flow
Prandtl Number: $$Pr = \frac{\nu}{\alpha}$$
Schmidt Number: $$Sc = \frac{\nu}{D_{AB}}$$
Lewis Number: $$Le = \frac{\alpha}{D_{AB}} = \frac{Sc}{Pr}$$
Chilton-Colburn Analogy:
$$\frac{h}{h_m} = \rho C_p Le^{2/3}$$
📊 Results & Analogy Verification
Configure inputs and click Calculate to view results.
The transport equations governing convective heat and mass transfer are mathematically identical under appropriate boundary conditions. This allows engineers to predict mass transfer coefficients from thermal measurements:
- Lewis Number ($Le$): Represents the ratio of thermal diffusivity to mass diffusivity. - $Le \approx 1$: Thermal and concentration boundary layers grow at the same rate. - $Le > 1$: Mass diffusivity is slower, concentration boundary layer is thinner than thermal. - $Le < 1$: Mass diffusivity is faster, concentration boundary layer is thicker.
📘 Calculation Methodology
Mathematical Model & Theory
Under boundary layer assumptions, the Nusselt ($Nu$) and Sherwood ($Sh$) numbers represent the dimensionless temperature and concentration gradients at the wall. The Chilton-Colburn analogy correlates these parameters directly using the Lewis number:
Worked Engineering Example
Air at 20°C ($\rho = 1.2\text{ kg/m}^3$, $C_p = 1007\text{ J/kg·K}$, $Pr = 0.73$, $\nu = 1.5 \times 10^{-5}\text{ m}^2/\text{s}$) flows over a flat plate ($L = 0.5\text{ m}$) at $U = 5\text{ m/s}$. Water binary diffusivity $D_{AB} = 2.6 \times 10^{-5}\text{ m}^2/\text{s}$. Calculate $h$ and $h_m$.
Solution:
1. Calculate Reynolds number:
$$Re = \frac{U L}{\nu} = \frac{5 \times 0.5}{1.5 \times 10^{-5}} = 1.67 \times 10^5 \quad (\text{Laminar})$$ 2. Calculate Nusselt number ($Nu_L$):
$$Nu_L = 0.664 Re^{0.5} Pr^{1/3} = 0.664 \times (1.67 \times 10^5)^{0.5} \times 0.73^{1/3} = 244.1$$ 3. Calculate Schmidt & Lewis numbers:
$$Sc = \frac{1.5 \times 10^{-5}}{2.6 \times 10^{-5}} = 0.577, \quad Le = \frac{0.577}{0.73} = 0.79$$ 4. Compute Sherwood number ($Sh_L$):
$$Sh_L = 0.664 Re^{0.5} Sc^{1/3} = 225.8$$ 5. Compute average transfer coefficients:
$$h = \frac{Nu_L k_f}{L} = 12.3\text{ W/m}^2\text{K}$$ $$h_m = \frac{Sh_L D_{AB}}{L} = 0.0117\text{ m/s}$$
Standard Assumptions & References
Assumptions: Constant fluid and species properties, negligible viscous dissipation, negligible radiative heat transfer, and low mass transfer rate (so bulk diffusion velocity does not disturb velocity boundary layer profile).
References:
- Chilton, T. H., & Colburn, A. P. Mass transfer coefficients; correlation of data. Industrial & Engineering Chemistry, 1934.
- Incropera, F. P., et al. Fundamentals of Heat and Mass Transfer. Wiley.