Introduction to Mass Transfer

Fick's law

Mass transfer by molecular diffusion is the movement of a species because of a concentration difference. For a binary mixture, the molar flux of A in the $x$ direction (relative to the mixture as a whole) is

$$J_A = -D_{AB}\frac{dC_A}{dx}$$

where $D_{AB}$ is the diffusion coefficient (m$^2$/s), $C_A$ the molar concentration (kmol/m$^3$). The flux is toward lower concentration.

Typical $D_{AB}$: gases about $10^{-5}$ m$^2$/s, liquids about $10^{-9}$ m$^2$/s, solids $10^{-10}$ to $10^{-14}$ m$^2$/s.

Equimolar counter diffusion

Two vessels, at the same total pressure $P$ and temperature $T$, are connected by a tube of length $L$. Gas A diffuses from vessel 1 (partial pressure $p_{A1}$) to vessel 2 ($p_{A2}$), and gas B diffuses in the opposite direction at the same molar rate: $N_A = -N_B$.

 [ vessel 1 ]=======L=======[ vessel 2 ]
   pA1 high        A -->       pA2 low
   pB1 low         <-- B       pB2 high

Since there is no net bulk flow, the flux is only diffusion:

$$N_A = -D_{AB}\frac{dC_A}{dx}$$

At steady state $N_A$ is constant. Integrate from $x = 0$ ($C_{A1}$) to $x = L$ ($C_{A2}$):

$$N_A = \frac{D_{AB}}{L}\,(C_{A1}-C_{A2})$$

For ideal gases $C_A = p_A/\bar R T$, so

$$N_A = \frac{D_{AB}}{\bar R T\,L}\,(p_{A1}-p_{A2})$$

with $\bar R = 8.314$ kJ/kmol K (or 8314 J/kmol K, with $p$ in Pa). The concentration profile of A is linear in $x$, as for conduction through a plane wall.

Model

Water vapour (A) diffuses upward through stagnant air (B). Air does not move, so $N_B = 0$ and there is a net bulk flow of the mixture (Stefan flow). For steady state, constant temperature and pressure:

$$N_A = \frac{D_{AB}P}{\bar R T\,z}\,\ln\frac{P-p_{A2}}{P-p_{A1}}$$

where $z$ is the diffusion length, $p_{A1}$ the vapour pressure at the water surface and $p_{A2}$ at the top.

Data

• $P = 101\,325$ Pa, $T = 298$ K, $z = 0.15$ m
• $p_{A1} = 3170$ Pa (saturated at the water surface), $p_{A2} = 0$ (dry air at top)
• $D_{AB} = 0.256\times10^{-4}$ m$^2$/s, $\bar R = 8.314$ J/mol K
• $A_c = \pi(0.01)^2/4 = 7.854\times10^{-5}$ m$^2$

Molar flux

$$\begin{aligned} \frac{D_{AB}P}{\bar RTz} &= \frac{0.256\times10^{-4}\times101\,325}{8.314\times298\times0.15} = 6.98\times10^{-3}\ \text{mol/m}^2\text{s}\\ \ln\frac{101\,325-0}{101\,325-3170} &= \ln(1.0323) = 0.03178\\ N_A &= 6.98\times10^{-3}\times0.03178 = 2.22\times10^{-4}\ \text{mol/m}^2\text{s} \end{aligned}$$

That is $2.22\times10^{-7}$ kmol/m$^2$s.

Evaporation rate

$$\begin{aligned} \dot n &= N_AA_c = 2.2185\times10^{-4}\times7.854\times10^{-5} = 1.742\times10^{-8}\ \text{mol/s}\\ \dot m &= \dot n\times18\ \text{g/mol} = 3.14\times10^{-7}\ \text{g/s}\\ &= 3.14\times10^{-7}\times3600 = 1.13\times10^{-3}\ \text{g/h} = 1.13\ \text{mg/h} \end{aligned}$$

Answer: $N_A \approx 2.22\times10^{-4}\ \text{mol/m}^2$s; evaporation rate about $1.13$ mg per hour.

Convective mass transfer coefficient

When a fluid flows over a surface from which species A is transferred, the flux is written in the form of Newton's law of cooling:

$$N_A = h_m\,(C_{As}-C_{A\infty})$$

$h_m$ (m/s) is the convective mass transfer coefficient, found from correlations just like $h$.

Dimensionless numbers

Analogy between momentum, heat and mass transfer

Because all three processes are carried by the same molecular and eddy motion in the boundary layer, their equations are similar. Reynolds analogy (for $Pr = Sc = 1$): $\dfrac{f}{2}=St=St_m$. For other fluids the Chilton-Colburn analogy applies ($0.6 < Pr < 60$, $0.6 < Sc < 3000$):

$$j_H = St\,Pr^{2/3}= \frac{f}{2}= j_M = St_m\,Sc^{2/3}$$

Hence, if $h$ is known, $h_m$ follows from

$$\frac{h}{h_m} = \rho c_p\,Le^{2/3}$$

Mass transfer can thus be predicted from heat transfer results by replacing $Nu \to Sh$ and $Pr \to Sc$.

Correlations

• Laminar flow over a flat plate: $Sh_L = 0.664\,Re_L^{1/2}\,Sc^{1/3}$.
• Turbulent flow in tubes: $Sh = 0.023\,Re^{0.83}\,Sc^{1/3}$ (Gilliland-Sherwood form).
• Flow past a sphere (Frossling): $Sh = 2 + 0.552\,Re^{1/2}Sc^{1/3}$.