๐ Mass Diffusion โ Fick's Law & Stagnant Film
Calculate binary molar and mass diffusion rates. Compare Equimolar Counter-Diffusion (Fick's law) with evaporation through a Stagnant Gas (Stefan's diffusion).
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
๐ Calculation Output
๐ Molar Concentration Profile C_A(x)
๐ Mole Fraction Profile y_A(x)
Molar Diffusion Flux (N_A)
5.9080e-3 mol/mยฒยทs
Mass Flux (n_A): 2.3650e-2 g/mยฒยทs
Mass Transfer Coefficient (hm)
3.6000e-4 m/s
Total Concentration: 41.57 mol/mยณ
Diffusion Path (L)
0.2000 m
ECD Model
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MASS DIFFUSION CALCULATION ENGINE
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Calculation Mode = Equimolar Counter-Diffusion
Mode Code = 1
Temperature = 20.00 deg-C
Total Pressure = 101.33 kPa
Binary Diffusivity D_AB = 7.2000E-05 m2/s
Film Thickness (L) = 0.2000 m
Molecular Weight M_A = 4.00 g/mol
--- BOUNDARY CONDITIONS -------------------------------------
Boundary 1 Partial P_A1 = 60.0000 kPa
Boundary 2 Partial P_A2 = 20.0000 kPa
Boundary 1 Mole Frac yA1 = 0.592154
Boundary 2 Mole Frac yA2 = 0.197385
--- ENGINE RESULTS ------------------------------------------
Total Concentration C = 41.5712 mol/m3
Mass Transfer Coeff (hm) = 0.000360 m/s
Molar Flux of A (N_A) = 5.9080E-03 mol/m2.s
Mass Flux of A (n_A) = 2.3650E-02 g/m2.s
--- LOCAL PROFILE ALONG FILM --------------------------------
x [m] y_A [-] C_A [mol/m3] P_A [kPa]
-----------------------------------------------------------
0.0000 0.592154 24.6165 60.0000
0.0050 0.582285 24.2063 59.0000
0.0100 0.572415 23.7960 58.0000
0.0150 0.562546 23.3857 57.0000
0.0200 0.552677 22.9754 56.0000
0.0250 0.542808 22.5652 55.0000
0.0300 0.532939 22.1549 54.0000
0.0350 0.523069 21.7446 53.0000
0.0400 0.513200 21.3343 52.0000
0.0450 0.503331 20.9241 51.0000
0.0500 0.493462 20.5138 50.0000
0.0550 0.483592 20.1035 49.0000
0.0600 0.473723 19.6932 48.0000
0.0650 0.463854 19.2830 47.0000
0.0700 0.453985 18.8727 46.0000
0.0750 0.444115 18.4624 45.0000
0.0800 0.434246 18.0521 44.0000
0.0850 0.424377 17.6419 43.0000
0.0900 0.414508 17.2316 42.0000
0.0950 0.404639 16.8213 41.0000
0.1000 0.394769 16.4110 40.0000
0.1050 0.384900 16.0008 39.0000
0.1100 0.375031 15.5905 38.0000
0.1150 0.365162 15.1802 37.0000
0.1200 0.355292 14.7699 36.0000
0.1250 0.345423 14.3597 35.0000
0.1300 0.335554 13.9494 34.0000
0.1350 0.325685 13.5391 33.0000
0.1400 0.315815 13.1288 32.0000
0.1450 0.305946 12.7186 31.0000
0.1500 0.296077 12.3083 30.0000
0.1550 0.286208 11.8980 29.0000
0.1600 0.276339 11.4877 28.0000
0.1650 0.266469 11.0774 27.0000
0.1700 0.256600 10.6672 26.0000
0.1750 0.246731 10.2569 25.0000
0.1800 0.236862 9.8466 24.0000
0.1850 0.226992 9.4363 23.0000
0.1900 0.217123 9.0261 22.0000
0.1950 0.207254 8.6158 21.0000
0.2000 0.197385 8.2055 20.0000
--- CORRELATIONS USED ---------------------------------------
Fick's First Law for Equimolar Counter-Diffusion (linear gradient).
Total concentration computed from ideal gas equation of state.
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.