ERT 313BIOSEPARATION ENGINEERING

LEACHING

Prepared by:Pn. Hairul Nazirah Abdul Halim

Introduction

• Leaching / solid extraction – methods of removing one constituent from a solid by means of a liquid solvent.

• Examples:

1) Extraction of oil from soybean flakes

2) Extraction of coconut milk from coconut• The amount of soluble material removed is often

greater than in ordinary filtration washing• The granular feed solids may disintegrate in pulp or

mush wheb their content of soluble material is removed.

Leaching Equipment: Bollman extractor• Contains a bucket elevator in a closed casing• The buckets are loaded with flaky solids such as

soybeans• The solids are sprayed with appropriate amount of

half miscella as they travel downward.• Half miscella is the intermediate solvent containing

some extracted oil and some small solid particles.• As solids and solvent flow concurrently down the

right-hand side of the machine, the solvent extracts more oil from the beans.

Fig. 23.1 (a) Bollman extractor

Rotocel extractor

• A horizontal basket is divided into walled compartments with a floor that is permeable to the liquid

• The basket rotates slowly about a vertical axis• Solid are admitted to each compartment at feed

point• The compartments then pass a number of solvent

sprays, a drainage section and a discharge point.• To give countercurrent extraction, the fresh solvent

is fed only to the last compartment before the discharge point.

Principle of Continuous Countercurrent Leaching

• The most important method of leaching is the continuous countercurrent method suing stages.

• The solid is not moved physically from stage to stage• The liquids is being moved from stage to stage

• V phase = the liquid phase (from stage N to stage 1)

• L phase = the liquid carried with the solid (from stage 1 to stage N)

• Exhausted solids leave Stage N

• Concentrated solution overflow from Stage 1

• Equilibrium is attained when the solute is completely dissolved and the concentration of the solution so formed is uniform.

• The concentration of the liquid retained by the solid leaving any stage is the same as that of the liquid overflow from the same stage.

• The equilibrium relationship is simply xe = y.

Equilibrium

Operating Line

• Total solution:

• Solute:

• Operating line equation:

Number of Ideal Stages for Constant Underflow

• Use McCabe Thiele Method if the operating line is straight

• In leaching, the operating line is always straight

• The equilibrium line is on 450 line.

ERT 313BIOSEPARATION ENGINEERING

TUTORIAL 5 - LEACHING

Prepared by:Miss Hairul Nazirah Abdul Halim

QUESTION 1EXAMPLE 23.1.

Oil is to be extracted from meal by means of benzene using a continuous countercurrent extractor. The unit is to treat 1,000 kg of meal (based on completely exhausted solid) per hour. The untreated meal contains 400 kg of oil and is contaminated with 25kg of benzene. The fresh solvent mixture contains 10kg of oil and 655 kg of benzene. The exhausted solids are to contain 60kg of unextracted oil. Experiments carried out under conditions identical with those of the projected battery show that the solution retained depends on the concentration of the solution, as shown in Table 23.1.

Find (a) the concentration of the strong solution, or extract;

(b) 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.

All quantities are given on an hourly basis.

Solution

Let x and y be the mass fractions of oil in the underflow and overflow solutions. At the solvent inlet,

Determine the amount and composition of the solution in the spent solids by trial. If Xb = 0.1, the solution retained, from Table 23.1, is 0.505 kg/kg. Then

From Table 23.1, the solution retained is 0.507 kg/kg:

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

At the solid inlet,

e) Determine the inlet and exit concentrations for the first stage and locate the operating line for the remaining stages.

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

Overall material balance:

Oil balance:

• The point x1 = 0.60, y2 = 0.408 is at one end of the operating line for the remaining stages.

• To determine an intermediate point on the operating line, choose xn = 0.30.

• By an overall balance,

• An oil balance gives

• The points xn , xn+1 , xa , ya and xb , yb define a slightly curved operating line, as shown in Fig. 23.3.

• Four ideal stages are required.