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)
3.3779e-4 mol/mยฒยทs
Mass Flux (n_A): 6.0853e-3 g/mยฒยทs
Mass Transfer Coefficient (hm)
2.6400e-4 m/s
Total Concentration: 40.87 mol/mยณ
Diffusion Path (L)
0.1000 m
Log-Mean P_Blm: 99.732 kPa
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MASS DIFFUSION CALCULATION ENGINE
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Calculation Mode = Diffusion through a Stagnant Gas B
Mode Code = 2
Temperature = 25.00 deg-C
Total Pressure = 101.33 kPa
Binary Diffusivity D_AB = 2.6000E-05 m2/s
Film Thickness (L) = 0.1000 m
Molecular Weight M_A = 18.02 g/mol
--- BOUNDARY CONDITIONS -------------------------------------
Boundary 1 Partial P_A1 = 3.1700 kPa
Boundary 2 Partial P_A2 = 0.0000 kPa
Boundary 1 Mole Frac yA1 = 0.031285
Boundary 2 Mole Frac yA2 = 0.000000
--- ENGINE RESULTS ------------------------------------------
Total Concentration C = 40.8740 mol/m3
Log-Mean Pressure P_Blm = 99.7316 kPa
Mass Transfer Coeff (hm) = 0.000264 m/s
Molar Flux of A (N_A) = 3.3779E-04 mol/m2.s
Mass Flux of A (n_A) = 6.0853E-03 g/m2.s
--- LOCAL PROFILE ALONG FILM --------------------------------
x [m] y_A [-] C_A [mol/m3] P_A [kPa]
-----------------------------------------------------------
0.0000 0.031285 1.2788 3.1700
0.0025 0.030515 1.2473 3.0920
0.0050 0.029745 1.2158 3.0139
0.0075 0.028973 1.1843 2.9357
0.0100 0.028201 1.1527 2.8575
0.0125 0.027429 1.1211 2.7792
0.0150 0.026656 1.0895 2.7009
0.0175 0.025882 1.0579 2.6225
0.0200 0.025108 1.0263 2.5440
0.0225 0.024333 0.9946 2.4655
0.0250 0.023557 0.9629 2.3869
0.0275 0.022781 0.9311 2.3083
0.0300 0.022004 0.8994 2.2296
0.0325 0.021227 0.8676 2.1508
0.0350 0.020448 0.8358 2.0719
0.0375 0.019670 0.8040 1.9930
0.0400 0.018890 0.7721 1.9141
0.0425 0.018111 0.7403 1.8351
0.0450 0.017330 0.7083 1.7560
0.0475 0.016549 0.6764 1.6768
0.0500 0.015767 0.6445 1.5976
0.0525 0.014985 0.6125 1.5183
0.0550 0.014202 0.5805 1.4390
0.0575 0.013418 0.5484 1.3596
0.0600 0.012634 0.5164 1.2801
0.0625 0.011849 0.4843 1.2006
0.0650 0.011063 0.4522 1.1210
0.0675 0.010277 0.4201 1.0413
0.0700 0.009490 0.3879 0.9616
0.0725 0.008703 0.3557 0.8818
0.0750 0.007915 0.3235 0.8020
0.0775 0.007126 0.2913 0.7221
0.0800 0.006337 0.2590 0.6421
0.0825 0.005547 0.2267 0.5620
0.0850 0.004756 0.1944 0.4819
0.0875 0.003965 0.1621 0.4018
0.0900 0.003173 0.1297 0.3216
0.0925 0.002381 0.0973 0.2413
0.0950 0.001588 0.0649 0.1609
0.0975 0.000794 0.0325 0.0805
0.1000 0.000000 0.0000 0.0000
--- CORRELATIONS USED ---------------------------------------
Stefan's Diffusion Law (diffusion through stagnant gas B, logarithmic profile).
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.