Theories For Mass Transfer Coefficients
2026-02-06
After today’s lecture, you will be able to:
Question: Consider steady-state, one-dimensional mass transfer of a binary gas mixture consisting of species \(A\) and \(B\) between two parallel planes, labeled (1) and (2). The total pressure of the system is \(p_T = 700\ \text{kPa}\) and the temperature is maintained at \(T = 100 ^\circ\text{C}\). The gas-phase mass transfer coefficient from molar fraction is \(k_y’ = 1.5 \times 10^{-4}\ \text{kg mol}\cdot \text{m}^{-2} \cdot \text{s}^{-1}\). Flow in pipe 1 contains 65 mol% A, while flow in pipe 2 contains 45 mol% B.
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Geankoplis Ex 7.2-1 A large volume of pure gas \(B\) at 2 atm pressure flows over a flat surface. Species \(A\) is vaporizing from the surface into the gas stream. The liquid \(A\) completely wets the surface of a blotting paper, so that the interface is always covered with liquid \(A\). The partial pressure of species \(A\) at \(298\;\text{K}\) is \(0.20\;\text{atm}\). From textbook table, \[ k_y' = 6.78 \times 10^{-5} \;\frac{\text{kg mol}}{\text{m}^2 \cdot \text{s}}\].
Calculate the evaporation rate \(N_A\) and determine the values of the gas-phase mass transfer coefficients \(k_y\) and \(k_G\) for this system.
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Since we have \(\text{[Flux]} = k \times \text{[Driving Force]}\), we can measure \(k\) using given geometries.
Question: solid benzoic acid (A) in a sphere with radius \(r_0\) was place in a tube with water flowing in the system. The fluid was turned on for a certain time interval \(\Delta t\) to remove dissolved A from the liquid. The masses of a dried sphere before and after the experiment were measured to be \(m_1\) and \(m_2\) (\(m_2 - m_1 \ll m_1\)). The solubility of A in water is \(c_{As}\) and water contain no benzoid acid. Calculate the mass transfer coefficient of this system. What form of \(k\) did you use?
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Inside the thin film region, we have EMCD-like mass transfer
\[\begin{align} N_A &= J_{Az}^*\\ &= -D_{AB} \frac{d c_A}{dz}\\ &= \boxed{\frac{D_{AB}}{\delta_f}} (c_{A, i} - c_{A, b}) \\ &= k_c' (c_{A, i} - c_{A, b}) \end{align}\]Governing equation for mass transfer (no convection in the film)
\[\begin{align} \frac{\partial c_A}{\partial t} = D_{AB}\frac{\partial^2 c_A}{\partial z^2} \end{align}\]B.C. \(c_A(z=0)=c_{A,s},\ c_A(z=\infty)=c_{A,b}\)
The concentration inside the liquid film has the following solution:
\[\begin{align} \frac{c_A(z, t) - c_{A, b}}{c_{A,s} - c_{A,b}} = 1 - \text{erf}\left( \frac{z}{\sqrt{4 D_{AB} t}} \right) \end{align}\]We can directly get \(N_A\) from \(\frac{\partial c_A}{\partial z}\vert_{\text{surf}}\):
\[\begin{align} N_A(z=0, t) &= -D_{AB} \frac{d c_A}{d t} \\ &= \sqrt{\frac{D_{AB}}{\pi t}} (c_{A,s} - c_{A,b}) \end{align}\]We can calculate the average \(N_A\) in the penetration period (\(0 < t < t_L\)):
\[\begin{align} N_A &= \frac{1}{t_L} \int_0^{t_L} -D_{AB} \frac{d c_A}{d t'} \\ &= \boxed{\sqrt{\frac{4D_{AB}}{\pi t_L}}} (c_{A,s} - c_{A,b}) \\ &= k_c' (c_{A,s} - c_{A,b}) \end{align}\]The mass balance equation for a steady-state 2D problem without generation
\[\begin{align} -\nabla \cdot \vec{N_A} &= 0 \\ -\nabla \cdot (\vec{J}_A^* + c_A \vec{v}) &= 0 \\ \vec{v} \cdot \nabla c_A &= D_{AB} \nabla^2 c_A \\ v_x \frac{\partial c_A}{\partial x} + v_y \frac{\partial c_A}{\partial y} &= D_{AB} \frac{\partial^2 c_A}{\partial y^2} \end{align}\]If the boundary thickness for fluid \(\delta\) and concentration \(\delta_c\) are same, can use the solution to velocity field:
\[\begin{align} \left( \frac{\partial c_A}{\partial y} \right)_{y=0} = (c_{A, \infty} - c_{A, s}) \left( \frac{0.332}{x} N_{Re, x}^{0.5} \right) \end{align}\]Boundary layer theory links diffusion in \(y\)-direction with:
\[\begin{align} N_{Ay} &= k_c' (c_{A,s} - c_{A,b}) \\ &= \boxed{-D_{AB} \left( \frac{0.332}{x} N_{Re, x}^{0.5} \right)} (c_{A, \infty} - c_{A, s}) \end{align}\]For \(\delta=\delta_c\), we have \(k_c'\) expression:
\[\begin{align} k_c' = \frac{0.332 D_{AB}}{x} N_{Re, x}^{0.5} \end{align}\]Generally, we have Schmidt number \(N_{Sc} = (\delta/\delta_c)^3\), so
\[\begin{align} k_c' = \frac{0.332 D_{AB}}{x} N_{Re, x}^{0.5} N_{Sc}^{1/3} \end{align}\]The global mass transfer coefficient should integrate over \(0 \sim L\):
\[\begin{align} k_c' = \frac{0.664 D_{AB}}{L} N_{Re}^{0.5} N_{Sc}^{1/3} \end{align}\]| Approach | Governing expression for \(N_A\) | Scaling of \(k_c'\) with \(D_{AB}\) |
|---|---|---|
| Film theory | \(N_A = \frac{D_{AB}}{\delta}\,(c_{A,s} - c_{A,b})\) | \(k_c' \propto D_{AB}\) |
| Penetration theory | \(N_A = \sqrt{\frac{4D_{AB}}{\pi t_L}}\,(c_{A,s} - c_{A,b})\) | \(k_c' \propto D_{AB}^{0.5}\) |
| Boundary layer theory | \(N_A = 0.664\,D_{AB}/L\,N_{Sc}^{1/3}\,N_{Re}^{1/2} (c_{As} - c_{A,b})\) | \(k_c' \propto D_{AB}^{2/3}\) |
| General formula | – | \(k_c' \propto D_{AB}^n\) \(n=0.5\sim 0.9\) |
In this lecture, we showed a few theories that can elucidate the physics behind \(k_c'\), mostly relating to \(D_{AB}\) by different boundary conditions
Why do we need dimensionless numbers? We will see in the next lecture.