# Derivation of Overall Mass Transfer Coefficient

The flux in the gas can be written as:

$$N_{1}=k_{p}(p_{1}-p_{1i})$$

And the flux in the liquid can be written as: $$N_{1}=k_{x}(x_{1i}-x_{1})$$

And we know that $$x^*_{1}=\frac{p_{1}}{H}$$

How can I use this to find the overall mass transfer coefficient? I know that the driving force will just be $(p_{1}-Hx_{1})$ but what's $K_{g}$ and more importantly, how is it derived?

EDIT: I'll give the best answer to anyone who can even give a hint

Regarding $k_p$:

So, you've got $N_1$ a flux in $mol/ s \cdot m^2$ and a partial pressures $p_i$ in $Pa$. $k_p$ has got to connect those, and make the units work out.

Exactly what $k_p$ is depends on the flow of the gas. If the gas is stationary, mass transfer will by diffusion alone. Unfortunately, that process doesn't have a steady-state solution in a semi-infinite domain. If you had a thickness for the gas layer ($L_x$), $k_p$ would be:

$$\frac{D_{AB}}{L_x R T}$$

where $D_{AB}$, $R$, and $T$, are the diffusion coefficient, universal gas constant and temperature. $D_{AB}$ is a property of the species that you're tracking ($A$) and the remaining species that make up the gas ($B$). You can look those up. In a pinch, I assume $2 \cdot 10^{-5}$ for air-like gases.

Now, go with the idea that you've got some kind of flow.

With flow over a surface you'll end up with boundary layer, a thin region where velocity and species concentrations vary between their values at the surface and in the free stream. That flow is generally going to enhance the mass transfer, by actually moving things around a lot faster than diffusion (diffusion is wicked slow).

It is possible to cook up some solutions to the boundary layer equations and actually derive $k_p$. But that's tedious, only works in a few situations and the results are still approximate. What you actually want to do is look up a correlation. Like these:

Those'll give you the Sherwood number based on the Reynolds number and Schmidt number.

$$Sh = \frac{k_g L}{D_{AB}}, Re = \frac{u L}{\nu}, Sc = \frac{\nu}{D_{AB}}$$

If you aren't familiar, the only thing that might be tricky here is choosing $L$. It's the characteristic length of the flow configurations. Usually, it's just the length of the object in the direction of flow. Except inside pipes, use the diameter.

Convert both the pressure and concentration terms as mole fractions. Now you have y for gas phase mole fraction and x for liquid phase mole fraction. Henry's law is taken as am equilibrium relation in your question, so the equilibrium diagram(y vs x) is just a straight line through the origin with slope H'. Taking the gas side mass transfer coefficient on mole fraction basis as ky. So, on the equilibrium diagram, the bulk point is given as M(x0,y0) and interfacial concentration as N(xi,yi).[Note: N must be on the equilibrium curve as the interface concentrations are always in equilibrium.] Now, we take a quantity (yb*) that is defined as the mole fraction of solute in gas phase that can remain in equilibrium with (xb), the mole fraction in bulk liquid phase.

N = ky(yb-yi) = kx(xi-xb) = Kg(yb-yb*)

(yb-yb*) = (yb-yi) + ((yi-yb*)/(xi-xb))*(xi-xb)

N/Kg = N/ky = H'*N/kx (since yb is in equilibrium with xb, by definition)*

This give the overall gas phase coefficient Kg = (1/ky) + (H'/kx). Overall liquid phase coefficient can be determined similarly.

NOTE : The equilibrium diagram described above is used when the equilibrium relationship isn't linear.