Chemical Engineering Mass Transfer Notes

CHE3053S – Separation Processes – Course Notes

Section 6:

Stage-wise Extraction and Absorption Processes

Material balances

Assume the following

Perfect contact between two miscible phases Steady state, isothermal Concentration changes with position in the apparatus (or in the

case of a batch operations with time For a big enough apparatus or a long enough time, the exiting

concentrations will reach equilibrium.

Steady state co-current processes (Treybal)

Fig. 6-1 : Co-current process nomenclature

Consider a unit operation in which two immiscible streams E and R are contacted with each other in a co-current operation. (gas-liq, liq-liq, solid-gas, solid-liq). A single component A is transferred across from phase R to phase E.

E1, E2, R1, R2 : total flow rate : mole/time

y1, y2 : mole fractions of species A in phase E,

x1, x2 : mole fractions of species A in phase R.

Y1, Y2 : mole ratios of species in phase E;

Section 6: Stage-wise Extraction 90


X1, X2 : mole ratios of species in phase R;

Es, Rs : flow rate of the non-transferred species (inerts or the tie component)

The definition for Yi and Xi is

Overall mass balance between the inlet (1) and the outlet (2) :

mass balance for A between the inlet (1) and the outlet (2) :

Now use can be made of Rs and Es to eliminate one equation (tie substance):

thus

(1)

An mass balance from the inlet (1) to any point in the apparatus (the dashed line in Fig 1) yields :

(2)

Equations (1) and (2) represent the operating lines for this operation. This is nothing other than the mass balance across the unit. They can be used to calculate the variation of the concentration through the apparatus.



Fig 6-2 : Operating line for the transfer of A from R to E

Notes :

The Y vs X diagram shows the equilibrium line. This represents the phase equilibrium for component A between phase E and Phase R. It is obtained from thermodynamic calculations (i.e. VLE, Henrys Law, solubility data, etc.)

The operating line is straight on Y vs. X coordinates ONLY. It represents the graphical solution to equation (2) with the end points given by equation (1).

X, Y is the composition anywhere in the apparatus Xi, Yi : is equilibrium composition that would be achieved in the

system if it was left in contact for an infinite amount of time, or the apparatus was sufficiently large to provide a long enough contact time.



For transfer of species A from E to R, the same derivation can be made. This is left as a exercise. The graphical solution is schematically shown on Y vs. X co-ordinates, only the co-ordinates will change to yield:

Fig 6-3 : Transfer from E to R

Representing the X-Y diagram in x-y coordinates:

Fig 6-4: Operating lines on y vs. x diagram



In this case the operating line is no longer straight. This is because the flow rates R and E change through the unit due to the changing total number of moles in each stream. To get the operating line it is necessary to solve

simultaneously to account for the changing flows through the apparatus:

Note that x and X (or y and Y) become equivalent at very low concentrations – usually for x < 0.05 (then X < 0.052) this is acceptable.

Single section liquid-liquid extraction cascades (S&H chapter 4&5)

In the special case of LINEAR EQUILIBRIUM the single stage mass balance becomes:

Figure 6-5: A single Liquid-Liquid extraction stage



Assumptions:

Components C and A are mutually insoluble

The solvent stream (S) entering contains no B, i.e. YBS = 0 The equilibrium between the two streams is described by a linear

equilibrium, i.e. YB=KB·XB

FA: Feed rate of A, FS: Feed rate of S

XB and YB represent mole ratios of component B in raffinate and extractstreams, respectively, i.e.

XB = (moles B)/(moles A), andYB = (moles B)/(moles C)

mass balance on solute (B):

equilibrium:

Elimination of YBE yields:

Now define the extraction factor E:

then

This represents the degree of extraction that can be achieved by a single stage. The larger E, the greater the extent of extraction. The fraction of B not extracted is:



Continuous cross flow cascades (or staged systems)

Purpose: connected in such a way to enhance the mass transfer over that achieved in a single stage. Often applied in testing a solvent extraction process by repeatedly contacting the raffinate with fresh solvent – can generate the X-Y curve in this way.

Fig. 6-6 : Cross flow stages

Mass balance for each stage (each stage is in co-current flow, thus use the previous equations):

The composition of the E phase for each stage is assumed to be the same. Graphically :



Fig. 6-7: Cross flow operating lines

NOTES :

If Es1 = Es2 = Es3 = Es then the slopes of all the operating lines are equal

Point A, is the lowest concentration that can be achieved in X using the given feed with a composition Y0.

General equations for a stage:



Special case cross-current with linear equilibrium and fresh solvent feed

Fig 6-8: Cross flow liquid-liquid extraction stages

Assume that the solvent is sent to each stage in equal amounts, thus for N stages the solvent flow to each stage will Fs/N. Thus the extraction factor for each stage n will be

Doing the mass balance as for connected co-current stages:

Eliminate of the inter-stage compositions by multiplying:

For all stages N:



For any intermediate stage n:

If there was an infinite number of stages then N → ∞

This says that for a fixed E, which means that the flow rates of FA and FS

are fixed, it is not possible to complete extraction using cross-current stages. If the extraction factor is large (for example E=10 gives XB∞/XBR = 0.0005), so extraction is almost complete.

Counter-current staged cascades

Fig 6-9: Counter-current staged operation

Mass balances for each stage:

Each of these equations represents the co-current mass balance for each stage, i.e. the operating line for each stage. For the entire cascade :



This gives the operating line for the overall process. For a balance from the last stage N to stage n :

This gives the O/L from stage N to any other stage in the system. Graphically:

Fig 6-10 : Staged counter-current operation

Notes:

The operation line for each stage n follows the dotted line The number of intersections with the equilibrium line (or imaginary

efficiency line) represents the number of units that will be required to achieve the separation.

The overall operating line is not continuous and only represents the behaviour of the overall "black box" mass balance.

The stage n operating lines need not reach the equilibrium line i.e. there can be a Murphree-type efficiency factor. The overall operating line still applies but more stages will be needed (smaller steps, see Fig. 6-11)



Fig 6-11: Staged-counter-current for efficiency < 100%

The feed to each stage n is given by (Yn+1, Xn-1) and the leaving stream is given by (Yn, Xn). A simplified diagram (Fig 6-12) can be used to develop the same equations:

Fig 6-12 : Simplified counter current stages.

Solution of the system of equations on a computer

The mass balance equations below will give the composition for each stage if a relationship for the exiting streams (Xn, Yn) is available:

As an example assuming that each stage goes to equilibrium and the equilibrium line is represented by the following local relationship:



For 3 stages the following balances must be solved:

There are 6 equations to solve for the 6 unknown compositions (X1, Y1, X2, Y2, X3, Y3) leaving each of the stages given the composition of the input streams X0, Y4 and the flow rates RS and ES, in fact only the ratio Rs/Es is needed. Any one of the unspecified variables can be exchanged with the specified variables, e.g. If the recovery of Y1 is given, then the flow rate of ES can be found instead of Y1. If for example Y1 is assumed to be in equilibrium with X0, this is the maximum separation that can be achieved, and then the minimum flow rate of RS can be obtained when RS is specified, this leads to the concept of minimum flow rates. If the equilibrium is linear (as above), a system of 6 linear equations forming a 6 x 6 matrix needs to be solved.

Analysis of counter-current extraction with linear equilibrium and fresh solvent feed

Fig 6-13: Counter current liquid-liquid extraction cascades



Mass balances:

Solving these proceeds as follows, or simply arrange as a matrix and use a computer package.

The general formula for N stages is



NOTES:

The decrease in concentration per state is greater than in cross flow.

For an infinite number of stages and E > 1, = 0, thus complete

extraction can be achieved.

When E < 1, extraction will not be complete.

Kremser Equations: Special case of linear equilibrium (Y i = Ki·Xi), but any feed concentration



Minimum flow rates of E and R

As the operating line cannot cross the equilibrium line, it might be expected that there will be some situations at which the slope of the equilibrium line might need to cross the equilibrium curve to achieve the desired recovery. This situation puts constraints (upper and lower) on the flow rates of ES and RS. Consider the following:

Fig 6-13: Minimum flow rates shown graphically



This analysis applies to both continuous and stages operations. Before starting, it is necessary to know at least three of the compositions of the streams entering and exiting the unit. The 4th composition and the ratio RS/ES can then be found for the following situations:

Point A : X1, Y1, X2 are specified and is chosen such that X2 is still obtained. The maximum slope that the operating line can have is when X2 is in equilibrium with YA. This happens when the operating line touches the equilibrium curve at point A. This requires that the unit has an infinite number of stages or that the unit is infinitely large. Any

slope greater than will mean that the operating will have to cross the equilibrium line to reach X2. Because the operating line cannot cross the equilibrium line, X2 will never be reached and the system will pinch before reaching X2. For feasible operation of the

unit, will have to be decreased, either by decreasing RS or by increasing ES. Generally, given the flow rate of RS, this analysis yields the minimum flow rate of ES.

Point B: X1, Y1, Y2 are specified and is chosen such that Y2 is still obtained. The maximum slope that the operating line can have is when Y2 is in equilibrium with XB.

Point C: X2, Y2, and X1 are specified and is chosen such that X1 is still obtained. The minimum slope that the operating line can have is when X1 is in equilibrium with YC. By the same argument, any

slope less than will mean that X1 will not be achieved because the operating line cannot cross the equilibrium curve, and the system will pinch before X$_{1}$ is reached. For feasible operation the

slope of the operating line must be increased to larger than by increasing RS or decreasing ES. Generally, ES will be specified and this representsthe minimum flow rate of ES.



Point D : X2, Y2, Y1 are specified and is chosen such that Y1 is still obtained. The minimum slope that the operating line can have is when Y1 is in equilibrium with XD.

Solution on a computer

The overall mass balance and the equilibrium line equation have to be solved simultaneously. Consider Point A, solving then requires:

When YA=K·X2 the system is linear and solution is easy. When the equilibrium is non-linear, then its best to use a computer, where the unknowns are RS/ES and YA.

Special case: When the regular method does not work

Fig 6-14 : Minimum flow rates from the tangent case



Consider the situation when Y1, X1, Y2 are specified. Them in order to

find the maximum , the regular construction yields point B. This is not possible as the operating line cannot cross the equilibrium line, thus point B will never be reached. It is possible to choose point A such that the operating line is tangent to the equilibrium line at (XP,YP). Although in this case point A will not be reached as the unit will pinch at (XP,YP). This represents the correct limiting behaviour in finding

.



Section 7:

Staged Absorption and Stripping of Gases into Liquids

Section 7: Gas-liquid Absorption 109

































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Chemical Engineering Mass Transfer Notes

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