๐ Mass Diffusion โ Fick\
Diffusion de masse โ Solve binary mass diffusion equations. Supports Equimolar Counter-Diffusion (Loi de Fick) and Stefan stagnant film diffusion (diffusion dans un film stagnant) with molar flux, concentration, and mole fraction profiles.
โก Fortran 90 Engine
Double Precision (IEEE 754)
โ ISO / ASME Validated
๐ Solver Telemetry
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Jun 2026
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< 1 ms
๐ Film Diffusion Schematic & Profile Simulation
๐ Configuration
Equations & Presets:
Quick Presets:
- Evaporation of water in stagnant air (25ยฐC)
- Helium and Nitrogen counter-diffusion
ECD Molar Flux:
$$N_A = \frac{D_{AB}}{R T L} (P_{A1} - P_{A2})$$
Stagnant B Molar Flux:
$$N_A = \frac{D_{AB} P}{R T L P_{B,lm}} (P_{A1} - P_{A2})$$
Quick Presets:
- Evaporation of water in stagnant air (25ยฐC)
- Helium and Nitrogen counter-diffusion
ECD Molar Flux:
$$N_A = \frac{D_{AB}}{R T L} (P_{A1} - P_{A2})$$
Stagnant B Molar Flux:
$$N_A = \frac{D_{AB} P}{R T L P_{B,lm}} (P_{A1} - P_{A2})$$
๐ Results & Visualization
Configure inputs and click Calculate to view results.
โน๏ธ Mass Diffusion Theory
Mass diffusion is the transport of species from regions of higher concentration to regions of lower concentration.
- Equimolar Counter-Diffusion (ECD): Occurs when species A and B diffuse in opposite directions at equal rates ($N_A = -N_B$). Concentration profile is linear.
- Stagnant Gas Diffusion: Occurs when species A evaporates/diffuses through B, which is stagnant or non-diffusing ($N_B = 0$). Concentration profile is logarithmic due to the bulk flow induced by A's diffusion.
Mass diffusion is the transport of species from regions of higher concentration to regions of lower concentration.
- Equimolar Counter-Diffusion (ECD): Occurs when species A and B diffuse in opposite directions at equal rates ($N_A = -N_B$). Concentration profile is linear.
- Stagnant Gas Diffusion: Occurs when species A evaporates/diffuses through B, which is stagnant or non-diffusing ($N_B = 0$). Concentration profile is logarithmic due to the bulk flow induced by A's diffusion.
๐ Calculation Methodology
Mathematical Model & Equations
For binary gas mixtures, Fick's first law defines the diffusion flux of A relative to the average velocity. For a stagnant gas ($N_B=0$), the bulk flow (advection) enhances the transport of A:
$$N_A = -D_{AB} C \frac{dy_A}{dx} + y_A(N_A + N_B)$$
$$N_A = \frac{D_{AB} C}{L} \ln\left(\frac{1 - y_{A2}}{1 - y_{A1}}\right) \quad (\text{for } N_B = 0)$$
Worked Engineering Example
Evaporation Preset:
A tube containing water at 25ยฐC ($T_f = 298.15\text{ K}$) evaporating into dry air ($P_{A2} = 0$). Water vapor pressure is $P_{A1} = 3.17\text{ kPa}$. Total pressure is $P = 101.3\text{ kPa}$. path length $L = 0.1\text{ m}$. $D_{AB} = 2.6 \times 10^{-5}\text{ m}^2/\text{s}$.
Solution:
1. Calculate boundary mole fractions:
$$y_{A1} = \frac{3.17}{101.3} = 0.0313, \quad y_{A2} = 0$$ 2. Calculate total molar concentration:
$$C = \frac{P}{RT} = \frac{101300}{8.314 \times 298.15} = 40.87\text{ mol/m}^3$$ 3. Evaluate stagnant log-mean pressure of air B:
$$P_{B1} = P - P_{A1} = 98.13\text{ kPa}, \quad P_{B2} = P - P_{A2} = 101.3\text{ kPa}$$ $$P_{B,lm} = \frac{101.3 - 98.13}{\ln(101.3/98.13)} = 99.71\text{ kPa}$$ 4. Compute molar flux $N_A$:
$$N_A = \frac{2.6 \times 10^{-5} \times 101.3}{8.314 \times 298.15 \times 0.1 \times 99.71} \times 3.17 \times 10^3 = 3.39 \times 10^{-4}\text{ mol/m}^2\text{s}$$
A tube containing water at 25ยฐC ($T_f = 298.15\text{ K}$) evaporating into dry air ($P_{A2} = 0$). Water vapor pressure is $P_{A1} = 3.17\text{ kPa}$. Total pressure is $P = 101.3\text{ kPa}$. path length $L = 0.1\text{ m}$. $D_{AB} = 2.6 \times 10^{-5}\text{ m}^2/\text{s}$.
Solution:
1. Calculate boundary mole fractions:
$$y_{A1} = \frac{3.17}{101.3} = 0.0313, \quad y_{A2} = 0$$ 2. Calculate total molar concentration:
$$C = \frac{P}{RT} = \frac{101300}{8.314 \times 298.15} = 40.87\text{ mol/m}^3$$ 3. Evaluate stagnant log-mean pressure of air B:
$$P_{B1} = P - P_{A1} = 98.13\text{ kPa}, \quad P_{B2} = P - P_{A2} = 101.3\text{ kPa}$$ $$P_{B,lm} = \frac{101.3 - 98.13}{\ln(101.3/98.13)} = 99.71\text{ kPa}$$ 4. Compute molar flux $N_A$:
$$N_A = \frac{2.6 \times 10^{-5} \times 101.3}{8.314 \times 298.15 \times 0.1 \times 99.71} \times 3.17 \times 10^3 = 3.39 \times 10^{-4}\text{ mol/m}^2\text{s}$$
Standard Assumptions & References
Assumptions: Steady state, 1D diffusion along $x$-direction, ideal gas mixture behavior, and constant pressure and temperature across the film.
References:
- Welty, J. R., Wicks, C. E., Wilson, R. E., & Rorrer, G. L. Fundamentals of Momentum, Heat, and Mass Transfer. Wiley.
- Geankoplis, C. J. Transport Processes and Separation Process Principles. Prentice Hall.