Lecture 8 Flashcards
(23 cards)
What are the different cascade configurations covered in this course?
Systems are depicted in digital notes
- co-current cascade: just what we have covered as of yet, both the solvent and feed flow in the dame directions
- crosscurrent cascade: . The MSA, S, is divided into fractions that are fed individually to each stage of the process. Each box represents an equilibrium stage with a mixer. In digital notes, the feed, F, is added to the first stage, where it is brought
to equilibrium with a fraction of the fresh solvent, S. The extracted feed from the first stage is then added to the second stage, where it is again contacted with a fraction of the fresh solvent. In this manner, the feed progresses from stage to stage. Extracted feed, R, leaving the final stage is the raffinate. The extracts, E, leaving each stage can be processed separately or combined to recover solvent and extracted solutes - countercurrent cascade: The two streams flow countercurrently to each other, with equilibrium achieved at each stage.
Important note + break moment
Cascades we talk about are single-section cascades configured and designed to achieve a desired percent recovery
of just one component, called the key component, from a feed stream that enters at one end and leaves at the other. When it is desired to achieve a percent purity or percent recovery of two feed components, called key components, a two-section cascade is necessary.
We will only focus on the case of single-section cascades and we won’t be expected to go past that. in the upcoming FC, we will derive the expression for the fraction of unextracted solute for each cascade
What is the derivation for the fraction of unextracted b for each cascade?
Look at written notes they will guide you along with digital notes under FC 1
Why do we use co-current?
Accordingly, a co current cascade
has no merit unless required residence times are so long that equilibrium is not achieved in a single stage and one or more
extra stages are needed to provide additional residence time. Long residence times may be needed to achieve sufficient contacting between feed and solvent, or to accommodate slow mass transfer of solute from feed to solvent.
Counter current cascade
In the countercurrent arrangement in digital notes FC 1iii, the feed, a
carrier-rich liquid phase, initially rich in solute, passes through the cascade countercurrently to the solvent-rich phase, which is initially solute-poor. Go back to written notes to see the material balance for each stage
Can a perfect extraction be achieved with a countercurrent cascade?
For an infinite number of equilibrium stages, the limit of the derived equations gives two results, depending on the value of the extraction factor, E:
X(∞) ∕X(F) = 0, 1 ≤ E ≤ ∞
X(∞) ∕X(F) = (1 − E), E ≤ 1
Thus, complete extraction can be achieved with a countercurrent cascade of an infinite N if the extraction factor E > 1.
Co vs cross vs counter current?
They all have different applications depending on what you want your system to do, generally, as mentioned before co-current is used when residence times are long. Crosscurrent is used when there is a temp change in the system so that the solvent feed only gets affected at the stage where the temp change happens. finally, Counter current provides the best extraction and is usually used!
A figure comparing the extraction of the cascades is shown in the notes
Break moment
We now finished the subtopic of this lecture we will now move on to the Hunter-Nash graphical equilibrium stage model, which is a method used to find the number of equilibrium stages in a liquid-liquid TERNARY system using a countercurrent system.
What is the description of the countercurrent system?
in digital notes, a schematic diagram for a countercurrent, continuous-flow, N-equilibrium-stage extraction cascade
operating isothermally at steady state. Each stage includes mixing and phase separation. Stage numbering begins at the
feed end, opposite to the end at which the solvent enters. Typically, mass, rather than molar flow rates are used. The final raffinate feed rate is RN and the final extract feed rate is E1.
Phase equilibrium is assumed at each stage, so for any intermediate stage n, the components in extract En and raffinate Rn are in equilibrium. Mass transfer of all species occurs at each stage. The feed, F, contains the carrier, C, and the solute, A. It can also contain solvent, S, up to the solubility limit. Entering solvent flow rate, S can contain C and A up to the solubility limits, but preferably contains little
of either. Because most liquid–liquid equilibrium data are listed and plotted in mass rather than mole concentrations,
let: (yi)n = mass fraction of species i in extract leaving stage n and (xi)n = mass fraction of species I in raffinate leaving
stage n.
general notes about the ternary diagram before we begin
- The bold line is the equilibrium curve (also called the binodal curve because the plait point separates the curve into an
extract to the right and a raffinate to the left). - Because the tie lines slope upward
from the C side toward the S side, at equilibrium, the solute, A, has a concentration higher in S than in C. This makes S an effective solvent for extracting A from C. If the tie lines slope downward, S is not an effective solvent. (example shown in digital notes)
What is the procedure of the Hunter–Nash
method
the procedure is just constructing three steps on the triangular diagram:
Construction 1 (Product Composition Points)
Construction 2 (Operating Point and Operating Lines)
Construction 3 (Tie Lines and Equilibrium Lines)
given feed, F, and solvent, S, flow rates and compositions we can determine the number of equilibrium stages using these three steps.
note: Hunter-Nash is more difficult than the McCabe–Thiele staircase-step method for distillation
Construction 1 (Product Composition Points)
By product composition points they mean (look at digital notes FC 9) Rn and E1.
To determine Rn and E1 we need to first determine what M (the mixing point is)
What is mixing point M and how can it be determined?
M represents the combined flow rate and composition of feed F plus entering solvent S.
To locate M:
By total and component material balances, the composition of M = F + S = 350 kg is determined as shown in digital notes.
From these two (xi)M values, point M is located. Based on the properties of the triangular diagram, point M must be located on the straight line connecting F and S. Therefore, M can be located knowing just one value of (xi)M, say, (xS)M and the ratio S∕F is given by the inverse-lever-arm rule as
S∕F = FM∕MS =100∕250 = 0.400
Thus, point M can be located by two composition points or by measurement, employing either of these ratios as shown in digital notes
How can we determine Rn and E1
First, specify (xA)RN = 0.025. Because it must lie on the equilibrium curve, RN can be located and the values of (xC)RN and
(xS)RN can be read from the triangular diagram. A straight line drawn from RN through M locates E1 at the equilibrium-curve intersection, from which the composition of E1 can be read as shown in digital notes.
Values of the flow rates RN and E1 can then be determined from the overall material balances in digital notes, or from Figure in digital notes (FC14I) by the inverse-lever-arm rule:
E1∕Rn = RbM∕ME1
Again, the composition mass fractions for the two products are seen to sum to 1.000.
note
The combined flow rates and compositions of entering streams, F and S, must also equal the combined flow rates and
compositions of the exiting streams, RN and E1. This is shown in digital notes
What is Mmax
Mmax, lies on the equilibrium curve along the straight line connecting F to S (Shown in digital notes). It corresponds to the maximum possible solvent addition if two liquid phases are to exist. By the inverse-lever-arm rule, it is seen that a very large ratio of S to F exists
S∕F = FMmax∕MmaxS
Construction 2 (Operating Point and Operating Lines)
To obtain more information on the passing streams in the multistage cascade, material balances around groups of stages from the feed end can be written as differences:
𝐹 − 𝐸1 = 𝑅1 − 𝐸2 = ⋯ = 𝑅(𝑛−1) − 𝐸𝑛 = ⋯ = 𝑅𝑁 − 𝑆 = 𝑃
Because the passing streams are differenced, P defines a
difference point, not a mixing point, M. From the same geometric considerations that apply to a mixing point, a difference
point also lies on a line through the points involved. However, whereas M lies inside the diagram and between the two end
points, P usually lies outside the triangular diagram along an extrapolation of the line through a pair of passing-stream
points forming differences, such as F AND E1, RN AND S, and so on.
How can you locate the difference point?
To locate the difference point, P, two straight lines are drawn through the passing-stream point pairs (E1, F) and
(S, RN) established by Construction 2, as shown in digital notes. These lines are extrapolated until they intersect at difference (operating) point P. The figure shows the intersection at point P. From the difference equation in the Last FC (also as we explained before), straight lines drawn through points for any other pair of passing streams, such as (En, Rn-1), must also pass through point P. Thus, the difference point becomes an operating point, and lines drawn through pairs of points for passing streams and extrapolated to point P are operating lines.
The difference point has properties similar to a mixing point. If F − E1 = P is rewritten as F = E1 + P, F can be interpreted as the mixing point for P and E1. Therefore, by
the inverse-lever-arm rule, the length of line E1P relative to the length of line FP is
E1F/FP = (E1 + P)/E1 = F/E1
Thus, point P can be located with a ruler using either pair of feed-product passing streams.
Construction 3 (Tie Lines and Equilibrium Lines)
Involves stepping-off stages such that equilibrium stages are stepped off by alternate use of tie lines and operating lines.
Constructions 1 and 2 have been employed to locate points F, E, S, R1, and P. Starting at
the feed end of the cascade, Construction 3 is used to draw a tie line from point E1 to equilibrium phase R1. Because R1 and E2 are passing streams (Look at FC 9) Construction 2 requires that E2 lie at the intersection of a straight operating line drawn through points R1 and P, and back to the extract side of the equilibrium curve. R2 is located with a tie line from E2 by Construction 3. From R2, E3 is located with an operating line through P by Construction 2.
Continuing in this fashion by alternating between equilibrium tie lines and operating lines, the specified point RN is reached
or passed. If the latter, a fraction of the last stage is taken. All of this is depicted in digital notes.
Break moment
Okies since he didn’t have time, He split this lecture between Monday and Thursday, so now I’m going to continue and finish it off rn
What does it mean to half a maximum and minimum solvent-to-feed ratio
A maximum solvent-to-feed ratio means that you will only need 1 equilibrium stage.
A minimum solvent-to-feed ratio means that you need an infinite amount of equilibrium stages. The infinity of stages occurs at an equilibrium-curve
and operating-line pinch point
How can you determine the maximum solvent-to-feed ratio?
As we talked about the mixing point lies in a two-phase region as seen in Flashcard 16.
As this point moves along the line SF toward S, the ratio S∕F increases according to the inverse-lever-arm rule. In the limit, a
maximum S∕F ratio is reached when M = Mmax arrives at the equilibrium curve on the extract side. Now all of the feed is
dissolved in the solvent, no raffinate is obtained, and only one stage is required.
Now simply we can use the inverse-lever-arm-rule to determine the maximum ratio as shown in digital notes.
How can we determine the minimum solvent-to-feed ratio
Remember how we said that in ternary systems the pinch point becomes the pinch line…the pinch line is basically all the tie line go look at digital notes (The first step determine E1 then find S/F min)
