SAMPLE PROBLEM #8:SAMPLE PROBLEM #8:

Oil is to be extracted from meal by means of Oil is to be extracted from meal by means of benzene using a continuous countercurrent benzene using a continuous countercurrent extractor. The unit is to treat 1,000 kg of meal extractor. The unit is to treat 1,000 kg of meal (based on completely exhausted solid) per hour. (based on completely exhausted solid) per hour. The untreated meal contains 400 kg of oil and is The untreated meal contains 400 kg of oil and is contaminated with 25 kg of Benzene. The fresh contaminated with 25 kg of Benzene. The fresh solvent mixture contains 10 kg of oil and 655 kg of solvent mixture contains 10 kg of oil and 655 kg of benzene. The exhausted solids are to contain 60 kg benzene. The exhausted solids are to contain 60 kg of un-extracted oil. Experiments carried out under of un-extracted oil. Experiments carried out under conditions identical with those of the projected conditions identical with those of the projected battery show in Table 1.battery show in Table 1.

REQUIRED

a.) the concentration of the strong solution, or extractb.) the concentration of the solution adhering to the extracted solids

c.) the mass of solution leaving with the extracted meal

d.) the mass of extract

e.) the number of stages required

ConcentrationConcentration

Kg oil / kg solutionKg oil / kg solution

Solution retainedSolution retained

kg / kg solidkg / kg solid

0.00.0 0.5000.500

0.10.1 0.5050.505

0.20.2 0.5150.515

0.30.3 0.5300.530

0.40.4 0.5500.550

0.50.5 0.5710.571

0.60.6 0.5950.595

0.70.7 0.6200.620

Table 1

DIAGRAM:

10 kg Oil655 kg Benzene

1000 kg meal/h400 kg oil25 kg Benzene

60 kgUn-extractedoil

V2

Lb

Va

L2

V1

L1

Stage 1 Stage n

SOLUTION:

At the solvent inlet: Vb= 10 + 655 = 665 kg solution/hour

yybb ==1010

== 0.0150.015665665

By trial and error:

If xb = 0.1, the solution retained (from table 1) is 0.505. Then

Lb= 0.505 (1,000) = 505 kg/h

xxbb ==6060

== 0.1190.119505505

From table 1, the solution retained is 0.507 kg/kg:

Lb = 0.507 (1,000) = 507 kg/h

xxbb ==6060

== 0.1180.118507507

Close enough0.119 ≈ 0.118

Benzene in the underflow at Lb is 507 – 60 = 447 kg/h

At the solid inlet,La = 400 + 25 = 425 kg solution/h

xxbb ==400400

== 0.9410.941425425

Oil in extract = oil in – 60 = 10 + 400 – 60 = 350kg/h

Benzene in extract = 655 + 28 -447 = 233 kg/h

Va = 350 + 233 = 583 kg / h

yyaa ==350350

== 0.6000.600585585

(a) ya = 0.60(b) xb = 0.118(c) Lb = 507 kg/h(d) Va = 583 kg/h

For (e)

x1 = ya = 0.60, solution retained is 0.595 kg/kg solid.

L1 = 0.505 (1,000) = 595

Overall material balance:

V2 = L1 +Va – La = 595 + 583 – 425 = 753 kg/h

Oil balance:

Laxa + V2y2 = L1x1 + Vaya

V2y2 = 595(0.60) + 583(0.60) – 425(0.941) = 307y2 = 307/753 = 0.408

Ln = solution retained = 0.53 (1,000) = 530 kg/h

By an overall balance,

Vn+1 = 530 + 583 – 425 = 688 kg/h

An oil balance gives

Vn+1 yn+1 = Lnxn + Vaya – Laxa

= 530(0.30) + 583(0.60) – 400 = 108.8 yn+1 = 108.8/688 = 0.158

The points xn, yn+1, xa, ya and xb, yb define a slightly curved operating line (see figure 1)

1.0

0.8

0.6

0.4

0.2

0 0.2 0.4 0.6 0.8 1.0

y

x

Equilibriu

m line

Operating lin

e

Stage 1

Hence, Four ideal stages are required.