Chap 1.3 - Leaching

CHAPTER 3: SOLID – LIQUID EXTRACTION / LEACHING

CHAPTER / CONTENT Introduction to Leaching Process Rates of Leaching

Types of Equipments for Leaching Equilibrium Relations in Leaching Calculation in Leaching

Introduction to Leaching Process

Widely used in the metallurgical, natural product and food industries under batch, semi – continuous or continuous condition. The major difference between Leaching and LLE centers about the difficulty to transport the solid or the solid slurry from stage to stage. Leaching or also known as solid – liquid exraction involves dissolving soluble material from its mixture with an insoluble solid. Many biological inorganic or biological organic substances occur in a different mixture of different components in a solid. In order to separate the desired solute constituent or remove the undesirable solute component from the solid phase, the solid is contacted with a liquid phase.

Introduction to Leaching Process The two phases are in intimate contact and the solute or solutes can diffuse from the solid to the liquid phase, which causes a separation of the components originally in the solid. This process is called liquid – solid leaching or simply leaching. Leaching process for biological substances An important such process is to leach sugar from sugar beets with hot water. In production of vegetable oils, organic solvents such as hexane, acetone and ether are used to extract the oil from peanuts, soybeans, flax seeds, castor beans, sunflower seeds, etc.

In pharmaceutical industry, many different pharmaceutical products are obtained by leaching plan roots, leaves and stems. For ‘instant’ coffee, ground roasted coffee is leached with water and soluble tea is produced by water leaching of tea leaves.

Tannin is removed from tea barks by leaching with water.

Introduction to Leaching Process Leaching process for inorganic and organic materials Used in metal – processing industries In metal ores, the desired metal components usually occur with a large amount of undesirable constituents and leaching is used to obtain these metal components in the form of metal salts. E.g.: Copper salts are leached by dissolving raw copper ores by using sulfuric acid or ammoniacal solutions. E.g.: Nickel salts are leached using sulfuric acid – ammonia – oxygen mixures. Gold is leached using an aqueous sodium cyanide solution.

Rates of Leaching

PRINCIPLES OF LEACHING

RATE OF LEACHING WHEN DISSOLVING A SOLID

METHODS OF OPERATING IN LEACHING

Principles of Leaching The solvent must be transferred from the bulk solvent solution to the surface of the solids. Next, the solvent must penetrate or diffuse into the solids. The solute then diffuses through the solid solvent mixture to the surface of the particle. Finally, the solute is transferred to the bulk solution. The rate of the solvent transfer from the bulk solution to the solid surface is quite rapid.

However, the rate of transfer of the solvent into the solid can be rather slow or rapid. This solvent transfer usually occurs initially when the particle are first contacted with the solvent.

Principles of Leaching The rate of diffusion of the solute through the solid and solvent to the surface of the solid is often the controlling resistance in the overall leaching process and can depend on a number of different factors. If the solid is made of porous and solid structure with the solute and solvent in the pores in the solid, the diffusion through the porous solid can be described by an effective diffusivity. In biological or natural substances, additional complexity occurs because of the cell present, in the leaching of thin sugar beet slices, one – fifth of the cells are ruptured in the slicing of the beets. The leaching of sugar is then similar to the washing process, where in the remaining cells, sugar must diffuse out through the cell walls. With soybeans, whole bean cannot be leached effectively. The rolling and flaking of the soybeans ruptures cell walls so that the solvent can more easily penetrate by capillary action. The resistance to mass transfer to the solute from the solid surface to the bulk solvent is generally quite small compared to the resistance to the diffusion within the solid itself.

Rate of leaching when dissolving a solid When a material is being dissolved from the solid to the solvent solution, however, the rate of mass transfer from the solid surface to the liquid is the controlling factor. There is essentially no resistance in the solid phase if it is a pure materials. The equation for this can be derived as follows from a batch system. The rate of mass transfer of the solute A being dissolved to the solution of volume V in m3 is:



NA  k L c AS  c A  A

where NA

 Eq.1

 kg mol of A dissolving to the solutionper second

A

 the surface area of particles in m 2

kL

 mass tansfer coefficien t in m/s

c AS  saturationsolubilityof the solid solute A in the solutionin kg mol/m 3 c A  concentration of A in the solutionat time t sec in kg mol/m 3

Rate of leaching when dissolving a solid By material balance, the rate of accumulation of A in the solution is equal to equation (1) times the area:  dcA V  N A  AkL c AS  c A  dt

Integrating from time t = 0 and cA = cA0 to t = t and cA = cA: cA

dcA AkL  c cAS  cA V A0

t

 dt

t 0

c AS  c A  e  k L A V  t c AS  c A0 The solution approaches a saturated condition exponentially. Often the interfacial area A will increase during the extraction if the external surface becomes very irregular.

If the soluble materials forms a high proportion of the total solid, disintegration of the particles may occur.

Rate of leaching when dissolving a solid Example 12.8 – 1. Prediction of Time for Batch Leaching Particles having an average diameter of approximately 2.0 – mm are leached in a batch type apparatus with a large volume of solvent. The concentration of the solute A in the solvent is kept approximately constant. A time of 3.11 hour is needed to leach 80% of the available solute from the solid. Assuming that diffusion in the solid is controlling and the effective diffusivity is constant, calculate the time of leaching if the particle size is reduced to 1.5 mm.

Rate of leaching when dissolving a solid Solution 12.8 – 1. Prediction of Time for Batch Leaching For 80% extraction, the fraction unextracted ES is 0.20. Using Figure 5.3 – 13 for a sphere, for ES = 0.20, a value of DAeff t/a2 = 0.112 is obtained, where DAeff is the effective diffusivity in mm2/s, t is time in s, and a is radius in mm. For the same fraction ES, the value of DAeff t/a2 is constant for a different size. Hence, t1a22 t2  2  Eq.3 a1 where t2 is time for leaching with a particle size a2. Substituting in equation above: 2  1.5 2 t2  3.11 2.0 22

 1.75 h

Methods of operating in leaching Can be carried out in batch or unsteady state conditions, continuous and steady state conditions. Both continuous and stage wise types of equipment are used in steady or unsteady state operations. In steady state leaching a common method used in the mineral industries is in – place leaching, where the solvent is allowed to percolate through the actual ore body. For example, copper is leached by sulfuric acid from sulfides ore by leach liquor is pumped over a pile of crushed ore and collected at the ground level as it drains from the heap.

Types of Equipments for Leaching

FIXED – BED LEACHING

MOVING BED LEACHING

AGITATED SOLID LEACHING

Fixed – Bed Leaching Used in beet sugar industry and is also used for extraction of tanning extracts from the tanbark, extraction of pharmaceuticals from barks and seeds and other processes. Figure 12.8-1 shows a typical sugar beet diffuser or extractor. The cover is removable so that sugar beet slices called cossettes can be dumped into the bed.

Heated water at 344 K to 350 K flows into the bed to leach out the sugar. The leached sugar solution flows out the bottom onto the next tank in series. About 95% of the sugar in beets is leached to yield an outlet solution from the system of about 12 wt%.

Moving – Bed Leaching There are number of devices for stagewise countercurrent leaching where the bed or stages moves. Used widely in extracting oil from vegetable seeds such as cottonseeds, peanuts and soybeans. The seeds are usually dehulled first, sometimes precooked, often partially dried and rolled or flaked. The solvents used are particularly hydrocarbons such as hexane and the final solvent – vegetable solution called miscella may contain some finely divided solids.

Agitated Solid Leaching When the solid can be ground fine abou 200 mesh (0.074 mm), it can kept in suspension by small amounts of agitation. Continuous countercurrent leaching can be accomplished by placing the number of agitator in series, with setttling tanks or thickeners between each agitator. Sometimes thickeners are used as combination contactor – agitators and settlers – shown in Figure 12.8-3.

Equilibrium Relations in Leaching To analyze single – stage and countercurrent – stage leaching, an operating line equation, or material balance relation and the equilibrium relations between the two streams are needed as in LLE. Assumptions made by achieving the equilibrium relations: Sufficient solvent is present so that all the solute in the entering solid dissolved in the solvent. The solute in the entering solid dissolved completely in the first stage. No adsorption of the solute by the solid. * This means the solution in the liquid phase leaving a stage is the same as the solution that remains with the solid matrix in the settled slurry leaving the stage. The settled solid leaving a stage always contains some liquid in which dissolved solids is present. The solid – liquid stream is called underflow or slurry stream.

Equilibrium Relations in Leaching Consequently, the concentration of oil or solute in the liquid or overflow stream is equal to the concentration of solute in the liquid solution accompanying the slurry or underflow stream. The amount of solution retained with the solids in the settling portion of each stage may depend the density and viscosity of liquid in which the solid is suspended. Equilibrium diagrams for leaching: The concentration of inert or insoluble solid B in the solution mixture or the slurry mixture can be expressed in kg (lbm) units: N

kg B kg solid lb solid   kg A  kg C kg solution lb solution

For overflow, N = 0 For underflow, N value depending on the solute concentration in the liquid.

Equilibrium Relations in Leaching The composition of solute A in liquid will be expressed as wt fractions: xA 

kg A kg solute overflow liquid  kg A  kg C kg solution

yA 

kg A kg solute  kg A  kg C kg solution

 liquid in slurryor    underflow liquid  

Calculation in Leaching

SINGLE – STAGE LEACHING

COUNTER – CURRENT MULTISTAGE LEACHING

Single – stage Leaching Process flow Overflow solution

Solvent Feed V2, x2

V1, x1

Underflow solution

Feed Slurry

L1, N1, y1, B

L0, N0, y0, B

V L B

Mass of overflow solution Mass of liquid in slurry solution Mass of dry, solute – free solid.

xA yA

Composition of A at overflow solution Composition of A at slurry solution

Material balance is divided into 3 parts:

Totalsolution balance L0 y A0  V2 x A2  L1 y A1  V1 x A1  MxAM Comp. A balance Solid balance B  N 0 L0  0  N1 L1  0  N M M L0  V2  L1  V1  M

Single – stage calculations Example 1 In a single – stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of soybean containing 20 wt% oil is leached with 100 kg of fresh hexane solvent. The value of N for the slurry underflow is essentially constant at 1.5 kg insoluble solid/kg solution retained. Calculate the amounts and compositions of the overflow V1 and the underflow slurry L1 leaving the stage.

Single – stage calculations Solution 1 Overflow solution V1, x1 Feed Slurry L0, N0, y0, B

Solvent Feed V2, x2 Underflow solution L1, N1, y1, B

Information given:

Feed slurry = 100 kg containing 20 wt% oil Entering solvent, V2 = 100 kg

N = 1.5 kg B/kg (A+C)

Single – stage calculations Solution 1 Find coordinate at L0. Mass of A = 0.20 x 100

A = 20 kg

Mass of B = 0.80 x 100

B = 80 kg

Mass of C = 0 kg

C = 0 kg

y A0 

A A 20    1.0 L0 A  C 20  0

N0 

B B 80    4.0 L0 A  C 20  0

Coordinate for L0

(yA0 , N0) = (1.0 , 4.0)

Single – stage calculations Solution 1 Find coordinate at V2. Mass of A = 0

A = 0 kg

Mass of B = 0

B = 0 kg

Mass of C = 100 kg

C = 100 kg

x2 

A A 0   0 V2 A  C 0  100

N2 

B B 0   0 V2 A  C 0  100

Coordinate for V2 (x2 , N2) = (0 , 0)

Single – stage calculations Solution 1 From material balance calculations: Total solution balance:

L0  V2  L1  V1  M 20  100  M

M  120 kg

Component A balance:

L0 y A0  V2 x A2  MxAM

201.0  1000  120x AM x AM  0.167

Single – stage calculations Solution 1 Solid balance:

B  N 0 L0  N1 L1  N M M N 0 L0  N M M

420  N M 120

N M  0.667

Coordinate for M (xM , NM) = (0.167 , 0.667) Plot coordinate M in the graph. Construct straight vertical line through point M in order to find value V1 and L1

Single – stage calculations Solution 1 L0

4 3.5 3

N

2.5 2

N versus y A

L1

1.5 1

M

0.5

N versus x A

V1

0 0

V2

0.2

0.4

0.6 x A, y A

From figure, Coordinate for V1 (x1 , N1) = (0.167 , 0)

Coordinate for L1

(y1 , N1) = (0.167 , 1.5)

0.8

1

Single – stage calculations Solution 1 From material balance calculations: Total solution balance:

L1  V1  M L1  V1  120 V1  120  L1



Eq.1

Solid balance:

B  N 0 L0  N1 L1  N M M N1 L1  N M M

1.5L1   0.667120

L1  53.36 kg

Single – stage calculations Solution 1 From material balance calculations: From Eq. (1)

V1  120  L1 V1  120  53.36



Eq.1 V1  66.64 kg

Multi – stage counter current Leaching Process flow Overflow solution

Solvent Feed VN+1, xN+1

V1, x1

Underflow solution

Feed Slurry

LN, NN, yN, B

L0, N0, y0, B V L B

Mass of overflow solution Mass of liquid in slurry solution Mass of dry, solute – free solid.

xA yA

Composition of A at overflow solution Composition of A at slurry solution

Multi – stage counter current Leaching The ideal stages are numbered in the direction of the solids or underflow stream. The ideal stages are numbered in the direction of the solids or underflow stream. The solvent (C) – solute (A) phase or V phase is the liquid phase that overflows continuously from stage to stage countercurrently to the solid phase, and it dissolves solute as it moves along. The slurry phase L composed of inert solid (B) and liquid phase of A and C is the continuous underflow from each stage. Composition of V – denoted by x Composition of L – denoted by y Assumption: The solid B is insoluble and is not lost in the liquid V phase. The flow rate of solid is constant throughout the process

Multi – stage counter current Leaching L0  VN 1  LN  V1  M

Totalsolution balance

L0 y A0  VN 1 x AN 1  LN y AN  V1 x A1  MxAM B  N 0 L0  N N LN  N M M

Comp. A balance

Solid balance

Multi – stage counter current Leaching Example 2 A continuous countercurrent multistage system is to be used to leach oil from meal by benzene solvent (B3).

The process is to treat 2000 kg/h of inert solid meal (B) containing 800 kg oil (A) and also 50 kg benzene (C). The inlet flow per hour of fresh solvent mixture contains 1310 kg benzene and 20 kg oil. The leached solids are to contain 120 kg oil. Data (B3) are tabulated below as N kg inert solid B/kg solution and yA kg oil A/kg solution Calculate the amounts and concentrations of the stream leaving the process and the number of stages required.

Multi – stage counter current Leaching Solution 2 Overflow solution

Solvent Feed VN+1, xN+1

V1, x1

Underflow solution

Feed Slurry

LN, NN, yN, B

L0, N0, y0, B

Information given:

Feed slurry (L0): A = 800 kg/h

B = 2000 kg/h

C = 50 kg/h

Entering solvent (VN+1 ) A = 20 kg/h

B = 0 kg/h

C = 1310 kg/h

Multi – stage counter current Leaching Solution 2 Overflow solution

Solvent Feed VN+1, xN+1

V1, x1

Underflow solution

Feed Slurry

LN, NN, yN, B

L0, N0, y0, B

Information given:

Underflow solution (LN): A =120 kg/h

B = 2000 kg/h

C = ?? kg/h

Multi – stage counter current Leaching Solution 2 Find coordinate at L0. Mass of A = 800 kg/h Mass of B = 2000 kg/h Mass of C = 50 kg/h

y A0 

A A 800 800     0.94 L0 A  C 800  50 850

N0 

B B 2000 2000     2.35 L0 A  C 800  50 850

Coordinate for L0

(yA0 , N0) = (0.94 , 2.35)

Multi – stage counter current Leaching Solution 2 Find coordinate at VN+1. Mass of A = 20 kg/h Mass of B = 0 kg/h Mass of C = 1310 kg/h xN 1 

A VN 1

N N 1 

B VN 1



A 20 20    0.015 A  C 20  1310 1330



B 0  0 A  C 20  1310

Coordinate for VN+1 (xN+1 , NN+1) = (0.015 , 0)

Multi – stage counter current Leaching Solution 2 Find coordinate at LN. Mass of A = 120 kg/h Mass of B = 2000 kg/h Mass of C = ?? kg/h Slope of graph,

NN yN

B N N LN B 2000     16.67 A yN A 120 LN



N  16.67 y

If x = 0.1, N = 16.67 x 0.1 = 1.67 Plot New Coordinate (x , N) = (0.1 , 1.67)

Multi – stage counter current Leaching Solution 2 3 2

LN L0

1 0

N

-0.4

-0.2

-1 0

VN 1

0.2

-2 -3 -4 -5 -6 -7 x A, y A

0.4

0.6

0.8

1

Multi – stage counter current Leaching Solution 2 From material balance calculations: Total solution balance:

L0  VN 1  LN  V1  M L0  VN 1  M 850  1330  M

M  2180 kg

Component A balance:

L0 y A0  VN 1 xN 1  MxAM

8500.94  13300.015  2180x AM x AM  0.376

Multi – stage counter current Leaching Solution 2 From material balance calculations: Solid balance:

B  N 0 L0  N N LN  N M M N 0 L0  N M M

2.35850  N M 2180

N M  0.916

Coordinate for M (xM , NM) = (0.376 , 0.916) Plot coordinate M in the graph. Construct line from point LN to point M until it cross at x – axis. Point at x – axis = V1

Multi – stage counter current Leaching 3

LN

2

M

1

V1

0 -0.2

N

-0.4

-1

0

VN 1

0.2

0.4

-2 -3 -4 -5 -6 -7

From figure,

L0

x A, y A

Coordinate for V1 (x1 , N1) = (0.592 , 0) Coordinate for LN (y1 , N1) = (0.12 , 2.0)

0.6

0.8

1

Multi – stage counter current Leaching Solution 2 From material balance calculations: Total solution balance:

V1  2180  LN



Eq. 1

Component A balance:

LN y N  V1 x1  MxAM

LN 0.12  V1 0.592  21800.376 Insert Eq. 1 into equation above

LN 0.12  2180  LN 0.592  21800.376

0.12LN  1290.56  0.592LN  819.68 0.472LN  470.88

 LN  997.62 kg

Multi – stage counter current Leaching Solution 2 From material balance calculations: Total solution balance:

V1  2180  LN



Eq.1

V1  2180  997.62 V1  1182.38 kg Construct operating point: Connect L0 with V1 & LN with VN+1. The cross line – operating point.

Total stages: 4 stages

Multi – stage counter current Leaching Solution 2 Construct the stages: 3

LN

2

L3

N

-1

0

VN 1

0.2

-2 -3 -4 -5

P

L0

V1

0 -0.2

L1

M

1

-0.4

L2

-6 -7 x A, y A

0.4

0.6

0.8

1