Mass and heat transfer fronts

In any adsorption system, the mass transfer and heat transfer fronts move at different speeds depending on equilibrium parameters (e.g. heat capacities and adsorbed concentration), as well as on kinetic parameters (e.g. mass and heat transfer coefficients). Depending on the adsorbent and feed conditions, the mass transfer front can move more slowly than the heat transfer front, which has practical implications worth considering.

If the mass transfer front moves behind the temperature front, the temperature of adsorption is simply occurring at the feed temperature. This is because as the adsorbate is loaded onto the solid, it generates heat and is transported ahead of the adsorption location. In contrast, if the mass transfer front moves ahead of the temperature front, the adsorption location will be at a temperature higher than the feed temperature due to the heat of adsorption.

To evaluate this, let's assume instantaneous mass transfer. If the adsorbent initially has zero loading, then the time (per unit mass of adsorbent) for component $i$ to reach the outlet of the vessel is:

$$t_{i} = \frac{q_i^{sat}}{Fy_i}$$

Assuming adiabatic operation with no heat generation, the energy balance simply states the transfer of energy from the gas phase to the solid phase:

$$FC_{p,g}t = m_{ads}C_{p,ads}$$

Therefore, the time (per unit mass of adsorbent) for the temperature front to reach the end of the bed is:

$$t_T = \frac{C_{p,ads}}{FC_{p,g}}$$

Defining a ratio:

$$\frac{t_T}{t_i} = \frac{C_{p,ads} y_i}{C_{p,g} q_i^{sat}}$$

Notice that the above equation is independent of the flow rate, which cancels out.

If the ratio $\frac{t_T}{t_i} < 1$, then the mass transfer front lags behind the heat transfer front which occurs in dilute/trace systems where $y_i$ is small.