Multi-Stage Absorption: Part 1

  • So far we've considered a single theoretical / ideal absorber stage…
  • Two streams enter the system, $V_2$ and $L_0$, and equilibrate before leaving the stage (this is the definition of an ideal stage).
  • In real life, we would not use a single stage of absorption, but many stages.
  • Another consideration is that no real stage is ideal.
  • The main limitation of real stages is the residence time of the liquid and vapour is not long enough for the two to come into full equilibrium (poor mixing).
  • In a multi-stage absorber, the liquid and vapour from one stage is passed to another stage.
  • The diagram above of a counter current absorber reveals the logic behind the unusual numbering of the streams in the previous examples.
  • Vapour flows from the next stage to the current stage, and liquid flows in the opposite direction.
  • We will now discuss why counter-current flow is the natural choice for absorption/distillation
  • Assuming each stage does not quite reach equilibrium, we could use as many real stages as needed until the fluids do reach equilibrium.
  • This is co-current flow, and is illustrated above with vapour and liquid streams.
  • The separation achieved by this configuration is limited by the equilibrium state of the fluids (little/no mass transfer as equilibrium is reached).
  • Just like in heat transfer systems, a far more efficient operation would be to operate many absorber stages in counter-current flow.
  • Counter-current absorbers can achieve higher loadings of the liquid stream, as the equilibrium point varies along the absorber.
  • We can maintain a significant equilibrium (and dynamic) driving force for mass transfer along the whole length of the absorber.
  • All real absorbers are counter current, but not only for the better mass transfer characteristics.
  • Counter-current flow within a vertical gas absorber column is a natural consequence of the bouyancy forces.
  • The gas stream will naturally bubble up through the liquid stream.
  • On the right is a diagram of a packed-bed column, where liquid flows down over the packing and gas rises, and the counter current flow is obvious.
  • Packing consists of a large number of small objects with high surface-to-volume ratios.
  • They range from the simple Raschig ring “tubes”, to the more complex Berl saddle.
  • They are designed to form a self-supporting, lightweight porous structure which maximises the contact between the liquid and the gas passing over the bed.
  • For an initial design, we could assume that a certain height of packing achieves the same effect as a single ideal stage ( height of an equivalent theoretical plate (HETP)).
  • The other common form of absorber/distillation tower is that of a plate or tray column.
  • In these columns, the liquid is held up on trays which have perforations to allow the gas/vapour phase to bubble through.
  • Although this appears to have an easily identifiable “stage”, typically the plate efficiency is less than one.
  • A single plate might only achieve 70-85% of the separation of an ideal stage (plate efficiency=70-85%).
  • Both packing and trays have their advantages and disadvantages, and their selection depends on the viscosity, flow-rates and fouling of the two phases.
  • A major consideration is that we have more accurate design methods for plate columns.
  • However, packing can be significantly cheaper to construct.
  • But we will ignore the internal structure of the column for now, and just look at the concept of ideal stages.
  • To translate these ideal designs into real designs, we will need to consider tray efficiencies and the packing height of an equivalent theoretical plate.
  • When designing an absorber column, we often know the conditions of the available absorbing liquid $L_0$, and the target vapour outlet conditions $V_1$.
  • We then need to solve from the left to the right, stage by stage, to determine the outlet liquid and inlet vapour conditions.
  • Sometimes the problem is specified in reverse, we have a target loading of the outlet liquid $L_N$, and an inlet specification of the gas/vapour $V_{N+1}$.
  • Here we would need to solve from right to left, but in both cases we need the governing equations of the system.
  • Just as before, around a single stage we can create a total mass/molar balance… \begin{align*} L_0+V_2 = L_1+V_1 \end{align*} … and a component mass/molar balance. \begin{align*} L_0 x_{A,0}+V_2 y_{A,2}&=L_1 x_{A,1} + V_1 y_{A,1} \end{align*}
  • We can create $N$ of these balances, where $N$ is the total number of stages.
  • These $N$ balances will then need to be combined with the equilibrium data to close the system of equations.
  • However, there are several other balances that we can create.
  • A similar mass balances could be created around all $N$ stages… \begin{align*} L_0+V_{N+1} &= L_N+V_{1}\\ L_0 x_{A,0}+V_{N+1} y_{A,N+1}&=L_N x_{A,N} + V_1 y_{A,1} \end{align*}
  • But this balance alone cannot specify the outlet stream conditions, as we need to examine the equilibrium state of each individual stage.
  • We can of course create a balance around some smaller number, $n$, of the stages. \begin{align*} L_0+V_{n+1} &= L_n+V_{1}\\ L_0 x_{A,0}+V_{n+1} y_{A,n+1}&=L_n x_{A,n} + V_1 y_{A,1} \end{align*}
  • The key thing to note is that common to the balances we just created is that the input liquid $L_0$, and output vapour $V_1$ are always included.
  • If we take the last balance equation, and assume the vapour and liquid streams are binary and contain inert components (as for the single stage absorber example), we can transform the balance equation to \begin{align*} L'\frac{x_{A,0}}{1-x_{A,0}}+V'\frac{y_{A,n+1}}{1-y_{A,n+1}}&=L'\frac{x_{A,n}}{1-x_{A,n}} + V'\frac{y_{A,1}}{1-y_{A,1}} \end{align*}
  • This equation is powerful, as it relates the gas composition entering any stage $n$ ( $y_{A,n+1}$), to the liquid composition in that stage ( $x_{A,n}$) (assuming $x_{A,0}$ and $y_{A,1}$ are known).
\begin{align*} V'\frac{y_{A,n+1}}{1-y_{A,n+1}}&=L'\frac{x_{A,n}}{1-x_{A,n}} + V'\frac{y_{A,1}}{1-y_{A,1}} - L'\frac{x_{A,0}}{1-x_{A,0}} \end{align*}
  • We only need to add the VLE data ($y_{A,n} = f(x_{A,n})$), and we can solve our problem stage by stage.
  • This can be done by calculator or, as will be shown in the next lecture, it can be done graphically!