TBP01x 5.4 Separation principles
Welcome to this unit, where we will look a bit deeper in the design equations for
downstream processing unit operations. This unit will require a bit more mathematics, but
the approach is reasonably simple and can be applied to calculate flows and approximate
equipment size. Note that all variables must be consistent.
We will explain separation principles for molecular separations in the concentration and
purification sections of a DSP first. Then we show how the same principles can be applied to
mechanical separations such as for cell separation and  further downstream also to recover
and wash solid products.
The starting point for this (simplified) approach is the equilibrium assumption for the
OUTGOING flows of the (equilibrium) stage. An important parameter that appears in
multiple operations is the so called SEPARATION FACTOR, defined as the ratio of flows of the
auxiliary phase (V) and processed feedstock (L), multiplied with the concentrations in each
flow: y and x. If you look more closely, the separation factor represents the ratio of transport
capacities of both flows. It indicates how much auxiliary phase, like hot water, is required to
extract product from the feed, like in the case of the coffee beans. We come back to this
aspect later. It is good to realise that the ratio of equilibrium concentrations  y divided by x,
in this case y1 and x1 equals the partition coefficient K. The auxiliary phase goes on top of
the equation.
As we have seen in the coffee example, the separation factor follows from solving the mass
balance over a single equilibrium stage. What goes in via the feed and solvent, must come
out via raffinate or extract. The result of solving the mass balance can then be used to
calculate (for instance) the extraction yield. When we plot the yield as a function of S, we
see that the extraction yield increases with increasing S. This is positive since there is more
productivity, so more revenue. But it also indicates that there are more operational costs
(OPEX) associated to (handling) the auxiliary phase, which can be solvent or adsorbent
material. I will also come back to this economic point.
When we made coffee earlier this week, we saw that multistage, counter current separation
processes can be more efficient in recovering product with the same or smaller amount of
auxiliary phase. To calculate how much more efficient we have to set up the mass balance
over each of the N stages. We can solve these mass balances for either the concentrations x
or y by repetitive substitution of the equilibrium relation. This gives a new extraction yield,
as a function of separation factor S and the number of stages N. This is called the Kremser
equation. When this equation is plotted, you indeed see that the yield at the same S
improves a lot with increasing the number of equilibrium stages.
This gives us a possibility to link the calculations to process economics already in this stage.
Before, we saw that increasing the separation factor S implies increasing (handling of) the
auxiliary phase, and thereby is roughly proportional to operational costs or OPEX. This is
plotted in the figure.
The number of stages N is roughly proportional to equipment size and therefore roughly
proportional to equipment investment and thereby to capital costs or CAPEX. At a present
extraction yield (and feed), we can now use the equation of the previous slide to calculate
the required number of stages at a certain separation factor. That gives us the declining
CAPEX curve that is also plotted in this diagram. Obviously you have to pay both operational
and capital expenditures. Adding both gives typically the dotted curve with a minimum cost
at a certain S. The optimal S corresponding to the total cost minimum is often somewhere
between 1.3 and 3, but the exact value of course depends on exact conditions.
Depending on the required level of accuracy, we can obtain information for estimating exact
costs from similarity to existing plants somewhere else, from experience or by detailing the
process design more and more. Clearly more accuracy requires more detailed information
and more time of engineers. We will discuss this aspect in one of the upcoming modules.
It is useful to have some guidelines for a reasonable first estimate of the working range of a
process, which is called  window of operation . One such guideline can be obtained from
minimizing use of the auxiliary phase (and thus S) in a multistage unit operation. For
instance if we assume complete recovery, all product entering with feed L, leaves with the
auxiliary flow V as extract at stage 1. This is the first bullet. Secondly, minimum use of
auxiliary phase implies working at the highest concentrations possible. For the feed stage 1,
this is equal to the feed concentration or x1 = x0. And thirdly, we have the equilibrium
assumption that we can substitute in the 1st relation. Solving this gives a minimum S of 1.
But then the number of stages has to be infinite. Use the relation for extraction yield to
verify this.
To understand the implications, let s go back once more to the result of the extraction yield
as a function of S and N. When S>1, the product is primarily moving with the auxiliary phase,
whereas if S<1, the product stays in the original feed phase. We can use this understanding
when designing more complicated multistage processes such as distillation and continuous
chromatography. These are basically combinations of different multistage units coupled
together with intermediate feed streams or other flows, to separate different components.
Such an example is shown here, which is a (very) simplified model of a distillation column of
continuous chromatography system. The feed containing products A and B is between the
 top and  bottom sections. Each have different ratios of auxiliary and feed flows. If A and B
have different affinities for the auxiliary phase (relative to the feed) their partition
coefficients will be different. For instance KA > KB.
If we want to separate A and B, the one (say A) has to move to the top with the auxiliary
flow, and the other (B) to the bottom. This means that the separation factor S for A in both
top and bottom section have to be larger than unity. And of course the opposite for B (both
SB s <1). We don t elaborate this here further, but the flanking written material gives more
examples.
So far, we did not indicate whether we were dealing with liquid liquid system (extraction),
solid liquid (adsorption) systems or vapour/liquid systems (distillation, absorption, and
stripping), or whether the phases were separated by means of a membrane. So the
approach is very, very general and can always be applied. Of course, the actual equipment
will look very different for all these systems. Discussing these aspects is more then we can
do in this MOOC  come to Delft for more advanced courses on this topic.
Now, let s extend this generalised approach to include mechanical separations such as
filtration and centrifugation. The feed flow F contains an amount of particles C such as cell or
crystals. When these are separated from the feed, there is a certain amount of solution
adhering to these particles. This is proportional to the amount of particles (alpha C). hence
the wet particle flow is equal to C + alpha C. Often alpha is approximately unity, but it
depends on how the separation is done. The resulting particle free liquid flow is supernatant
of filtrate. It is more practical (and more common) to describe mechanical separations in
volume balances, when the various components are not compressible.
Then we can derive completely analogous equations for distribution coefficients, separation
factors and extraction yields. This can be done for single stages as is shown here. But often
the adherent liquid also contains dissolved product that need to be recovered as well. Then
several units are connected in a countercurrent manner.
We can describe this with the same Kremser equations as used before. Now, L is the  wet
cells flow and V is the supernatant or filtrate flow.  extraction yield describes the yield of
dissolved product (solutes) in the filtrate due to wash stream W. And it has the same
dependency on S and N as plotted before, and all other conclusions such as about those
about balancing OPEX and CAPEX hold here as well.
Summarizing: the design recipe always works. Step 1 and 2 (balance equations and
equilibrium relations) result in first estimates of feed flow. This helps to do the first
economic evaluations. The last two steps will help to calculate more precise equipment
dimensions. This is beyond the scope of this course, but you are welcome at TU Delft to
study these topics. In the next unit, we will apply the above to the PDO case.
So, see you in the next unit!