Important equations in an absorption tower

 Well, these equations correspond to a general overview. These relevant equations are given for each stream. In this case, only one component is being separated from the mixture.

Simple sketch of an absorption equipment showing its streams.

Notice that in the figure above, $x$ is used for all mol fractions in the streams for the sake of clarity. However, $x$ for the liquid and $y$ for the vapor phase are to be used. Also, $L_S$ and $G_S$ represent the solvent streams.

For the gas mixture inlet at the bottom

For this stream, the mol ratio is defined as,

$Y_1=\dfrac{y_1}{1-y_1}=\dfrac{\bar{p_1}}{p_t-\bar{p_1}}$        Eq. (01)

The stream of gas mixture entering the tower is composed of,

$G_1=\text{component of interest} + \text{solvent}$        Eq. (02)

The solvent part of $G_1$ is represented as $G_S$. The flow rate of solvent $G_S$ in $G_1$ or $G_2$ is defined as,

$G_S=G_1y_{1S}=G_1\left( 1-y_1 \right)=\dfrac{G_1}{1+Y_1}$        Eq. (03)

where $y_{1S}$ is the mol fraction corresponding to the solvent in the gas mixture at the inlet. Since $G_1$ is a gas stream, it is natural to think that $Y_1$ can be expressed in terms of the total $p_t$ and partial $\bar{p}$ pressures too.

For the gas mixture outlet at the top

The gas stream at the top is $G_2$. Since no solvent in $G_2$ is passed to the liquid phase, it is straightforward that $G_S$ is the same at the bottom and at the top of the tower. Also, $y_{1S}=y_{2S}$, too.

For this stream, the mol ratio is defined as,

$Y_2=\dfrac{y_2}{1-y_2}=\dfrac{\bar{p_2}}{p_t-\bar{p_2}}$        Eq. (04)

The solvent stream can also be estimated using the top data, with,

$G_S=G_2y_{2S}=G_2\left( 1-y_2 \right)=\dfrac{G_2}{1+Y_2}$        Eq. (05)

For the liquid mixture, or not, inlet at the top

For the liquid phase, the story is very similar to the phase stream. The mol ratio is defined as,

$X_2=\dfrac{x_2}{1-x_2}$        Eq. (06)

while the liquid solvent stream is defined as,

$L_S=L_2x_{2S}=L\left( 1-x_2 \right)=\dfrac{L_2}{1+X_2}$        Eq. (07)

where again $x_{2S}$ is the mol fraction corresponding to the solvent in the liquid mixture at the inlet $L_2$. Since no part of the liquid solvent is passed to the gas stream,

$x_{2S}=x_{1S}$        Eq. (08)

too.

For the liquid mixture outlet at the bottom

For the bottom of the tower, the equations are barely the same as Eqs. (06-07).

Ildebrando.