TKP 4105. Exercise on extraction

Problem 1. Equilibrium-stage extraction

We have a feed of 1000 kg/h acetic-water solution with 37.7 wt% acetic acid.We want to remove (extract) the acetic acid using 3000 kg/h of pure isopropylether as the solvent. A triangular diagram (or more precisely, part of it) is givenbelow (you can make a better diagram with data from Appendix A.3-24).

(a) What is the amount of aqueous product (L1) and its concentration ofacetic acid with single-stage extraction (mixer-settler)?

(b) To extract more acetic acid we want to use countercurrent multi-stageextraction. What is the amount of aqueous product (LN ) and how many stagesN are required if the aqueous product contains 5 wt% acetic acid?

(c) With the given feed rates (1000 kg/h and 3000 kg/h), how small can theacetic acid concentration in the aqueous product get with multi-stage extraction(with an infinite number of stages)?

(d) With the given feed rate of water/acetic acid (1000 kg/h) and a givenconcentration of 5% acetic acid in the aqueous product, what is the minimumamount of solvent that is needed with an infinite number of stages?

(e) You have a choice between the following: (1) 3-stage countercurrentextraction with 3000 kg/h solvent (similar to what you studied in (b)). (2)3-stage extraction with 1000 kg/h pure solvent in each stage. Make a flow sheetof the two alternatives. Which is the best?

0 0.1 0.2 0.3 0.4 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1A = acetic acid, C = isopropyl ether

xA

, yA

xC

, yC

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Solution

Problem 1. Extraction

Figure 1: Single-stage extraction

(a) Single-stage extraction.Fist we locate the two feeds (L0 and V0). The overall feed M is located on

a line between these two points and from the lever arm rule

L0

V0=

V0M

L0M=

1000kg/h

3000kg/h=

1

3

From this we find that M is located 1/4 of the distance from V0 to L0.The two products L1 and V1 are in equilibrium and thus lie on a tie line. We

find the tie line through M and identify the products V1 and L1. The amountof L1 can be found from the lever arm rule.

We find: The aqueous product L1 = 718 kg/h and it contains 17.5% aceticacid.(b) Multi-stage countercurrent extraction.

2

L0

VN+1=V

0

Δ

LN

=L3

V1

L2

L1

V2

V3

M

Figure 2: Multi-stage countercurrent extraction

Refer to Figure 12.7-1 in the book for the flowsheet.1. Locate the aqueous feed L0

2. Locate the solvent feed VN+1 = V0.3. Locate the overall feed M (same as we did for the single-stage case)4. Locate the aqueous product (raffinate) LN . The products V1 and LN

are not on a tie line as in the single-stage case. However, it is given that theaqueous product LN contains 5% acetic acid and it can then be located on thephase boundary.

5. Locate the extract product V1. From the total mass balance, V1 must beon a straight line through LN and M and it must be on the phase boundary.

6. Locate the operating point (difference point) ∆ as the intersection be-tween the lines L0V1 and LNVN+1

The remaining compositions are obtained by alternating between the oper-ating line (between stages) and equilibrium tie line (on stages).

7. Locate L1 on a tie line through V1.8. Locate V2 on the operating line through L1 and ∆.9. Locate L2 on a tie line through V2.

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10. Locate V3 on the operating line through L2 and ∆.11. Locate L3 on a tie line through V3.We find that L3 is approximately the same point as the desired product LN

so we conclude that we need N = 3 equilibrium stages.We should also find the amount of product LN = L3. From the lever arm

rule (mass balance) we see that V1M is about 1/7 of the total distance V1LN ,so we have that LN ≈ 4000kg/h(M)/7 = 571 kg/h.(c) Countercurrent extraction with infinite stages.

In this case the feed rates, including the amount of solvent, is fixed.In general, we get infinite stages when the operating line and tie line are

parallel (or actually coincide) so that we get no improvement from stage tostage. In our case, this corresponds to moving LN to the left until we have thatthe operating line LNV0 coincides with a tie line. In our case we find that wecan move LN all the way to 0 % acetic acid; the tie line and the operating lineare then both vertical.

Conclusion: With infinite stages we can get an aqueous product with 0%acetic acid. The reason is that the solvent is pure.(d) Minimum amount of solvent.

The feed L0 and the product LN is fixed. In this case, the operating line andthe tie line will coincide when the extension of tie line through V1 goes throughL0.

We extend the tie line through L0 and find the point V1. We then make aline from V1 to LN and identify the point M We can now use the lever arm ruleto find the amount of solvent feed V0. We find

L0

V0=

V0M

L0M=

5.4cm

6.4cm= 0.843

That is, V0 = L0/0.843 = 1000 kg/h / 0.843 = 1185 kg/h (which is much lessthan our actual feed of 3000 kg/h).(e) Extraction with pure solvent in each stage.

Countercurrent operation is always best so alternative (1) is the best!This is confirmed by more detailed calculations.(1) With 3-stage countercurrent extraction we already showed that we can

get down to 5% in the aqueous product.(2) With 3-stage extraction with pure solvent in each stage we see from the

Figure that we are only able to get down to 12% acetic acid (which is betterthan single-stage at 17.5% but far from countercurrent operation!)

Comment: Problem (e) is similar to a problem that was given at the examin 2005. In the solution to that exam is is claimed that it is better with puresolvent “because the driving force is then better”. This would be true if weadded 3000 kg/h of pure solvent in each stage but this is not really a faircomparison because the total use of solvent (9000 kg/h) is then 3 times larger.In summary, if we use the same total amount of pure solvent, countercurrentoperation is better than using pure solvent in each stage.

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L0

V1

LN

M

V0

tieline

Figure 3: Determination of minimum amount of solvent (V0)

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Figure 4: (e2) 3-stage extraction with pure solvent at each stage

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