4 Leaching Solid Liquid Extraction

4-Leaching (Solid-Liquid Extraction) Saeed GUL, Dr.Techn, M.Sc. Engg. Associate Prof Professor essor

Department of Chemical Engineering, University of Engineering & Technology Peshawar, PAKISTAN

Introduction Leaching is concerned with the extraction of a soluble constituent from a solid by means of a solvent The desired component diffuses into the solvent from its natural solid form Typical users include: 

the metals industry for removing mineral from ores (acid solvents)



the sugar industry for removing sugar from beets (water is solvent)



the oilseeds industry for removing oil from soybeans, etc. (hexane or similar organic solvents)

General Principles Generally, the process can be considered in three parts: 1. The cha chang nge e of pha phase se of of the the solut solute e as it diss dissolv olves es in in the the solvent 2. Sol Solute ute dif diffus fusion ion thr throug ough h the sol solve vent nt in in the por pores es of the soli solid d to the outside of the particle 3. The tra transf nsfer er of the the solu solute te from from the solu solutio tion n in cont contact act with with the particles to the main bulk of the solution Any one of these three processes may be responsible for limiting the extraction rate, though the first process usually occurs so rapidly that it has a negligible effect on the overall rate

General Principles If the solute is uniformly dispersed in the solid: The material near the surface will be dissolved first, leaving a porous structure in the solid residue. The solvent will then have to penetrate this outer layer before it can reach further solute, and the process will become progressively more difficult and the extraction rate will fall. If the solute forms a very high proportion of the solid: The porous structure may break down almost immediately to give a fine deposit of insoluble residue, and access of solvent to the solute will not be impeded.

General Principles If the soluble material is distributed in small isolated pockets in a material which is impermeable to the solvent: Such as gold dispersed in rock, for example. In such cases the material is crushed so that all the soluble material is exposed to the solvent. If the solid has a cellular structure: The extraction rate will generally be comparatively low because the cell walls provide an additional resistance

Factors influencing the rate of extraction There are four important factors to be considered: 1. Particle Size I. Surface area II. Should not very small to be difficult to separate 2. Solvent I. Selective II. Viscosity 3. Temperature I. Solubility II. Diffusion Coefficient 4. Agitation of the Fluid I. Increases eddy diffusion II. Prevent Sedimentation

Mass Transfer in Leaching Operations Mass transfer rates within the porous residue are difficult to assess because it is impossible to define the shape of the channels through which transfer must take place. It is possible, however, to obtain an approximate indication of the rate of transfer from the particles to the bulk of the liquid. Using the concept of a thin film as providing the resistance to transfer, the equation for mass transfer may be written as:

Mass Transfer in Leaching Operations For a batch process in which V, the total volume of solution, is assumed to remain constant, then:

The time t taken for the concentration of the solution to rise from its initial value c0 to a value c is found by integration, on the assumption that both b and A remain constant. Rearranging:

Mass Transfer in Leaching Operations If pure solvent is used initially, c 0 = 0, and:

which shows that the solution approaches a saturated condition exponentially. In most cases the interfacial area will tend to increase during the extraction and, when the soluble material forms a very high proportion of the total solid, complete disintegration of the particles may occur. Although this results in an increase in the interfacial area, the rate of extraction will probably be reduced because the free flow of the solvent will be impeded and the effective value of b will be increased.

Example 10.1

Example 10.1

Equipment for Leaching The selection of the equipment for an extraction process is influenced by the factors which are responsible for limiting the extraction rate. Thus, if the diffusion of the solute through the porous structure of the residual solids is the controlling factor, the material should be of small size so that the distance the solute has to travel is small. On the other hand, if diffusion of the solute from the surface of the particles to the bulk of the solution is the controlling factor, a high degree of agitation of the fluid is required.

Equipment for Leaching Processes involved Three distinct processes are usually involved in leaching operations: 1. Dissolving the soluble constituent. 2. Separating the solution, so formed, from the insoluble solid residue. 3. Washing the solid residue in order to free it of unwanted soluble matter or to obtain as much of the soluble material as possible as the product. The type of equipment employed depends on the nature of the solid — whether it is granular or cellular and whether it is coarse or fine. The normal distinction between coarse and fine solids is that the former have sufficiently large settling velocities for them to be readily separable from the liquid, whereas the latter can be maintained in suspension with the aid of only a small amount of agitation

Equipment for Leaching Extraction from cellular materials With seeds such as soya beans, containing only about 15 per cent of oil, solvent extraction is often used because mechanical methods are not very efficient. 

Light petroleum fractions are generally used as solvents.



Trichlorethylene has been used where fire risks are serious, and acetone or ether where the material is very wet

Extraction from cellular materials 

The upper section is filled with the charge of seeds which is sprayed with fresh solvent via a distributor



The solvent percolates through the bed of solids and drains into the lower compartment where, together with any water extracted from the seeds, it is continuously boiled off by means of a steam coil



The vapors are passed to an external condenser, and the mixed liquid is passed to a separating box from which the solvent is continuously fed back to the plant and the water is run to waste



By this means a concentrated solution of the oil is produced by the continued application of pure solvent to the seeds

Extraction from cellular materials Bollmann Extractor 

Continuous moving bed Series of perforated baskets



widely used with seeds which do not disintegrate on extraction



A typical extractor moves at about 1 revolution per hour



Each basket containing some 350 kg of seeds.



Generally, about equal masses of seeds and solvent are used.



The final solution, known as miscella, contains about 25 per cent of oil by mass

extractor

Leaching of coarse solids 

In continuous unit in which countercurrent flow is obtained: Solid is introduced near the bottom of a sloping tank and is gradually moved up by means of a rake.



The solvent enters at the top and flows in the opposite direction to the solid, and passes under a baffle before finally being discharged over a weir

Tray classifier / the Dorr classifier

Leaching of coarse solids

Leaching of fine solids Coarse solids may be leached by causing the solvent to pass through a bed of the material, fine solids offer too high a resistance to flow. Particles of less than about 200-mesh (0.075 mm) may be maintained in suspension with only a small amount of agitation, and as the total surface area is large, an adequate extraction can be effected in a reasonable time. Because of the low settling velocity of the particles and their large surface, the subsequent separation and washing operations are more difficult for fine materials than with coarse solids.

Leaching Operations

Figure 1: Single-stage leaching unit

Figure 2: Multistage cross-flow leaching unit

Figure 3: Multistage countercurrent leaching unit

Material Balance-Countercurrent Process

 

  

The stages are numbered in the direction of flow of the solid. The light phase is the liquid that overflows from stage to stage in a direction opposite to that of the flow of the solid, dissolving solute as it moves from stage N to stage 1. The heavy phase is the solid flowing from stage 1 to stage N. Exhausted solids leave stage N, while concentrated solution overflows leave from stage 1. For purposes of analysis, it is customary to assume that the solute free solid is insoluble in the solvent so that the flow rate of this solid is constant throughout the process unit

Design and Predictive Equations

The solute /solvent equilibrium and process throughput determine the cross-sectional area and the number of theoretical and /or actual stages required to achieve the desired separation. The equation for the operating line is obtained by writing a material balance. From Figure above


If the density and viscosity of the solution change considerably with solute concentration, the solids from the lower stages might retain more liquid than those in the higher stages. The slope of the operating line then varies from stage to stage.


If, however, the mass of the solution retained by the solid is independent of concentration, LN is constant, and the operating line is straight. These two mentioned conditions describe variable and constant overflow, respectively. It is usually assumed that the inerts are constant from stage to stage and insoluble in the solvent. Since no inerts are usually present in the extract (overflow) solution and the solution retained by the inerts is approximately constant, both the underflow LN and overflow VN are constant, and the equation for the operating line approaches a straight line.

Design and Predictive Equations Since the equilibrium line is also straight, the number of stages can be shown to be (with reference to Fig. 12.12)

The above equation should not be used for the entire extraction cascade if Lo differs from L1, L2, . . . , LN (i.e., the underflows vary within the system). For this case, the compositions of all the streams entering and leaving the first stage should first be calculated before applying this equation to the remaining cascade.

Single stage problems EXAMPLE Calculate the grams of water that need to be added to 40 g of sand containing 9.1 g salt to obtain a 17 % salt solution. If the salt solution is to be reduced to 0.015, calculate the amount of salt that must be removed (“leached”) from the solution. SOLUTION: Set V to be the grams of water required. The describing equation is:

Let Z be equal to the final amount of salt in the sand–water–salt solution. The describing equation is

The amount of salt removed is therefore 9.1 2 0.61 ¼ 8.49 g.

Single stage problems A sand–salt mixture containing 20.4% salt enters a solid–liquid extraction at a rate of 2500 lb /h. Calculate the hourly rate of fresh water that must be added for 99% of the salt to be “leached” from the sand–salt mixture if the discharge salt–water solution contains 0.153 (mass) fraction salt.

SOLUTION: Let W equal to the hourly rate of water. The describing equation from a mass balance is

Also note that the feed consists of 510 lb salt and 1990 lb sand. On discharge, the sand contains only 5.1 lb salt. The discharge water solution consists of the 2820 lb water plus 504.9 lb salt.

Number of Stages for Countercurrent Washing by Graphical Methods The separation of Soluble constituents from a solid by extraction with a solvent may be considered to consist of two steps: 1. Contacting the solid with liquid phase and 2. Separation of the liquid phase from the solid In actual operation it is impossible to completely separate the liquid phase from the solid. Consequently, the streams resulting from the 2nd step will consist of: 

A liquid phase stream (solution) which during normal operation does not contain any solid. This stream will be called overflow



A slurry consisting of the solid plus adhering solution

Number of Stages for Countercurrent Washing by Graphical Methods It is convenient to use the concept of a stage in performing the calculations. A stage consist of two steps: Contacting solid with liquid and Separation of the

overflow from the underflow

In Solid-Liquid extraction , an ideal stage is defend as : A stage in which the solution leaving in the overflow is of the same composition as the solution retained by the solid in the underflow For the purpose of calculations, the solid-liquid extraction system may be considered to consist of three components: 1. The solute component (A) 2. The inert solid component (B) 3. The solvent component (S)

Number of Stages for Countercurrent Washing by Graphical Methods 

The calculations for leaching system may be based on the material balance and the ideal stage concept.



Both mathematical and graphical methods of solution may be used.



The graphical solution possesses advantage in presenting a generalized treatment of the more complex cases and in permitting a better visualization of what is occurring in the process.



Although it may be inconvenient to use if a large number of stages are involved but typically in most of the cases the number of stages used is not large

Rectangular Diagram S

s

x , S t n e n o p m o C f o n o i t c a r f e l o M

0.8

100 % S

1

0.6 0.4

0.2 0.4

0.2 0.8

B 100 % B

0.6

0.2

0.4

0.6

0.8

Mole fraction of Component A, x A

A 100 % A

Single Stage Extraction system Overflow Product V1, y1 Feed

Solvent Feed

Single Stage Extraction System

V2, y2 Underflow Product

L0, x0

L1, x1

Overall Material Balance L0 + V2 = L1 + V1 The Material Balance for components A, B and S: L0(xA)0 + V2(yA)2 = L1(xA)1 + V1(yA)1 L0(xB)0 + V2(yB)2 = L1(xB)1 + V1(yB)1 L0(xS)0 + V2(yS)2 = L1(xS)1 + V1(yS)1

Counter Current Multistage Extraction System V1

L0

V2 Stage 1

L1

V3 Stage 2 L2

V j+1

V j

L j -1

Stage j

L j

Vn

Ln-1

Overall Material Balance L0 + Vn+1 = Ln + V1 The Material Balance for components A, B and S: L0(xA)0 + Vn+1(yA)n+1 = Ln(xA)n + V1(yA)1 L0(xB)0 + Vn+1(yB)n+1 = Ln(xB)n + V1(yB)1 L0(xS)0 + Vn+1(yS)n+1 = Ln(xS)n + V1(yS)1

Vn+1 Stage n Ln

PROBLEM 10.5 Seeds, containing 20 per cent by mass of oil, are extracted in a counter current plant and 90 per cent of the oil is recovered in a solution containing 50 per cent by mass of oil. If the seeds are extracted with fresh solvent and 1 kg of solution is removed in the underflow in association with every 2 kg of insoluble matter, how many ideal stages are required? GIVEN: Underflow feed: Seeds contain 20% by mass of oil

Underflow product: 1 kg solution is associated with 2 kg of insoluble matter.

Overflow feed: Fresh Solvent

Overflow product: 90% of oil is recovered in solution containing 50% by mass of oil

REQUIRED: Number of ideal stages=?

Underflow Line: The initial step is to obtain the underflow lone for which we proceeds as follows E-Intercept: E-intercept=intercept of absissa and ordinate E=



+1 K term is used where underflow is constant and K is given by 

=

        1



=

2

= 0.5

0.5



=

0.5 + 1

= 0.33

E= 0.333

Note: The composition of the underflow is therefore represented on the graph by a straight line parallel to the hypotenuse of the triangle with an intercept of 0.333 on the main two axis.

Underflow feed composition: Underflow feed contain solute and inert, so: XAO = 0.2

XB0=0.8

XSO=0

This point is marked as X 1 on graph i.e. X 1 = 0.2

Overflow feed composition: Since overflow feed is pure solvent, so: YAn+1 = 0

YBn+1= 0

YSn+1= 1

This point is marked as Yn+1 on graph i.e. Y n+1=1

Overflow Product composition: Overflow product contains solute and solvent and no inert, so: YA1= 0.5

YB1=0

YS1=0.5

This point is marked as Y1 on the graph , i.e. Y 1= (0.5,0.5)

Underflow Product composition: 2kg of inert relates to 1kg of solution so 1 kg of inert relates to 0.5 kg of solution. Oil recovery is 90% in overflow showing that 10% is left which is being taken up with the under flow product, XAn=10%*XA0 XAn= 0.1*0.2 = 0.02 The mass fraction of inert is same in both underflow feed and product i.e. X Bn = XBo XBn=0.8

To find out point a: Calculation on solvent free basis : a(A) =

 +

=

0.02 0.02+0.8

a(A) = 0.024

Difference Point: The difference point is now find by drawing in the two lines connecting X 1 with Y1 and point a with Y n+1

Yn+1 Y5 Y4

1. Locate underflow line (0.333,0.333) and join it through line.

Y3 Y2 Y1

3. Join Y1 to B and locate X 2, join X2 and Z and extend the line to fine Y2 and repeat the procedure.

0.333

X5

Xn

X4

2. Locate points, X1, Y1, Yn+1 and a and joint Y1 to X1 and point Yn+1 to ‘a’

X3 X2

a

X1

Z

0.333

4. From graph, X n lies in between X 5 and X 6. so number of stages are 5.

Problem 10.11 Seeds containing 25 percent by mass of oil are extracted in a countercurrent plant and 90 percent of the oil is to be recovered in a solution containing 50 per cent of oil. It has been found that the amount of solution removed in the underflow in association with every kilogram of insoluble matter is given by: k = 0.7 + 0.5ys + 3 2

where ys is the concentration of the overflow solution in terms of mass fraction of solute. If the seeds are extracted with fresh solvent, how many ideal stages are required?

GIVEN: Underflow feed: Seeds contain 25% by mass of oil

Overflow product: 90% of oil is recovered in solution containing 50% by mass of oil

Overflow feed: Fresh Solvent

Underflow product: solution removed in the underflow in association with every kilogram of insoluble matter is given by: k = 0.7 + 0.5ys + 3 2

REQUIRED: Number of ideal stages=?

SOLUTION: Basis = 100 kg underflow feed to 1 st stage

Underflow Line The first step is to obtain the underflow line, that is a plot of xs against x A. The calculations are made as follows: Ratio (kg/kg inert) Ys

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

K

0.7 0.78 0.92 1.12 1.38 1.70 2.08 2.52 3.02 3.58 4.2

oil

solvent

underflow line

K*ys

K(1-ys)

K+1

0 0.078 0.184 0.336 0.552 0.850 1.248 1.764 2.416 3.222 4.2

0.700 0.702 0.736 0.784 0.828 0.850 0.832 0.756 0.604 0.358 0

1.7 1.78 1.92 2.12 2.38 2.70 3.08 3.52 4.02 4.58 5.2

Mass fraction oil

solvent

XA = K*ys/(K+1) Xs = K*(1-ys)/(K+1) 0 0.044 0.096 0.158 0.232 0.315 0.405 0.501 0.601 0.703 0.808

0.412 0.394 0.383 0.370 0.348 0.315 0.270 0.215 0.150 0.078 0


XB0=0.75

XSO=0



YBn+1= 0

YSn+1= 1



YB1=0

YS1=0.5


Underflow Product composition: Oil recovery is 90% in overflow showing that 10% is left which is being taken up with the under flow product, XAn=10%*XA0 XAn= 0.1*0.25 = 0.025 The mass fraction of inert is same in both underflow feed and product i.e. X Bn = XBo XBn=0.75


 +

=

0.025 0.025+0.75

a(A) = 0.0322 Since there is no solvent so; a(s) = 0 Point “a” is therefore;

a = ( 0.0322 , 0 )

Difference Point: The difference point is now find by drawing in the two lines connecting X 1 with Y1 and point “a” with Yn+1

1. Locate underflow points and join it through line.

Yn+1 Y6 Y5


Y4 Y3 Y2

Y1 X7 X6

a

Z

X5

X4

X3

X2

X1

2. Locate points, X1, Y1, Yn+1 and a and joint Y1 to X1 and point Yn+1 to ‘a’ and locate point Z. 3. Join Y1 to B and locate X2, join X 2 and Z and extend the line to fine Y2 and repeat the procedure. 4. From graph, X n and X 7 are almost at same point so number of stages are 7.

Problem 10.12 Halibut oil is extracted from granulated halibut livers in a countercurrent multibatch arrangement using ether as the solvent. The solids charge contains 0.35 kg oil/kg of exhausted livers and it is desired to obtain a 90 per cent oil recovery. How many theoretical stages are required if 50 kg of ether are used/100 kg of untreated solids. The entrainment data are:

GIVEN: Underflow feed: The solid charge contains 0.35 kg oil per kg of exhausted livers Overflow product: 90% of oil is recovered Overflow feed: Fresh Solvent

REQUIRED: Number of theoretical stages=?

SOLUTION: Basis = 100 kg unreacted solids

Underflow Line The first step is to obtain the underflow line, that is a plot of xs against x A. The calculations are made as follows: Entrainment overflow conc. kg soln/kg Kg oil/kg soln livers

Ratio (kg/kg inerts) oil

solvent underflow line

Mass fraction oil

solvent

ys

K

K*ys

K(1-ys)

K+1

X A= K*ys/(K+1)

Xs=K*(1-ys)/(K+1)

0

0.28

0

0.28

1.28

0

0.219

0.1

0.34

0.034

0.306

1.34

0.025

0.228

0.2

0.4

0.080

0.32

1.4

0.057

0.229

0.3

0.47

0.141

0.329

1.47

0.096

0.224

0.4

0.55

0.220

0.33

1.55

0.142

0.213

0.5

0.66

0.330

0.33

1.66

0.199

0.199

0.6

0.8

0.480

0.32

1.8

0.267

0.178

0.67

0.96

0.643

0.3168

1.96

0.328

0.162

Plot the XA against XS so that we get the underflow line

Underflow feed composition: Solid charge= 0.35 kg / kg exhausted livers Total feed = 0.35+1 = 1.35 kg XAo = 0.35/1.35 = 0.26 XAO = 0.25

XB0 = 1-0.26 XB0=0.75

= 0.74

XSO=0



YBn+1= 0

YSn+1= 1


Underflow Product composition: Oil recovery is 90% in overflow showing that 10% is left which is being taken up with the under flow product, XAn=10%*XA0 XAn= 0.1*0.26 = 0.026

The mass fraction of inert is same in both underflow feed and product i.e. X Bn = XBo XBn=0.74


 +

=

0.026 0.026+0.74


a = ( 0, 0.034 )

To find Y1: Since the recovery of oil is 90% so overall mass balance becomes:

To find e: To find “e” i.e. ether in underflow product In the underflow product: the ratio i.e.

2 .6 oil = = 0.035 kg/kg exhausted livers 74

which, from the entrainment data, is equivalent to The ratio Or

ether

exhausted livers e

= 0.306 kg/kg

x A = 0.025, xs = 0.228

(through interpolation) interpolation)

= 0.306 × 74 = 22.6 kg

Overflow product composition: The mass of ether = (50 − 22.6) = 27 .4 kg

Y A =

23.4 23.4+27.4

= 0.46,

from which Ys = 0.54 which is marked in as Y 1. This point is marked as Y1 on the graph , i.e. Y 1= ( 0.46 , 0.54 )

1. Loca Locate te und under erfl flow ow poi point ntss and join it through through line.

Yn+1 Y3

1. Locate underflow points and join it through line. Y2

Y1

X4 X3

a

2. Locate points, X1, Y1, Yn+1 and a and joint Y1 to X1 and point Yn+1 to ‘a’ and locate point Z. 3. Join Y1 to B and locate X2, join X2 and Z and extend the line to fine Y2 and repeat the procedure.

X2

X1

4. From graph, X n lies between X 3 and X 4 so number of stages are 3.

Assignment / Do yourself Oil is extracted from meal by means of benzene in a continuous

counter current extractor. The unit is to treat

1000 lb of meal based on completely exhausted solids per hour. The untreated meal contains 400 lb of oil and no benzene. The final product obtained from the operation is to contain 60% oil and 90% of oil from the underflow is recovered. Assume no carry over of inert in overflow. Test data results that constant under flow solution is not possible. Ys

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

K

0.5

0.505

0.515

0.53

0.55

0.571

0.595

0.62

Problem It is desired to extract the oil from halibut liver by continuous counter current multiple contact extraction with ethyl ether. The quality of solution retained by granulated liver has been determined experimentally as a function of composition of: lb of oil/lb of soln 0 Lb of soln/lb of extracted liver 0.205

0.1

0.2

0.3

0.4

0.5

0.6

0.65

0.70

0.72

0.242

0.286

0.339

0.405

0.489

0.600

0.676

0.765

0.810

The fresh halibut liver contains 25.7 mass % oil. If 95% of oil is to be extracted and the final solution obtained from the operation is to contain 70 mass % oil. Compute; i.

The lb of oil free ether per 1000 lb charge of fresh liver

ii.

The no. of ideal stages required

iii.

The no. actual stages required if overall efficiency is 70%

iv.

The quantity of discharge solid.

GIVEN: Underflow feed: Seeds contain 25.7% by mass of oil Overflow feed: Fresh Solvent

Overflow product: The final product contains 70 mass % oil

Underflow product: 95% oil is recovered. So 5% is left in the underflow product

REQUIRED: i. ii. iii. iv.

The lb of oil free ether per 1000 lb charge of fresh livers = ? Number of ideal stages=? If Ƞ = 70%, number of actual stages = ? The quantity of discharged solids = ?

SOLUTION: Basis = 1000 lb of fresh halibut livers as underflow feed to 1 st stage Part 1: Lbs of free oil ether per 1000 lb charge of fresh livers = ? First of all we find the underflow line and compositions, for which we proceed as follows

Underflow line 1st step is to draw underflow line i.e. X A vs Xs From the given data of K and Ys, we find X A and Xs as follows. Ratio (kg/kg inerts)

Mass fraction

oil/kg soln

Lb soln/lb inert

oil

solvent

underflow line

Oil

solvent

ys

K

K*ys

K(1-ys)

K+1

XA= K*ys/(K+1)

Xs=K*(1-ys)/(K+1)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.65 0.7 0.72

0.205 0.242 0.286 0.339 0.405 0.489 0.600 0.676 0.765 0.810

0 0.0242 0.0572 0.1017 0.1620 0.2445 0.3600 0.4394 0.5355 0.5832

0.2050 0.2178 0.2288 0.2373 0.2430 0.2445 0.2400 0.2366 0.2295 0.2268

1.205 1.242 1.286 1.339 1.405 1.489 1.600 1.676 1.765 1.810

0 0.0195 0.0445 0.0759 0.1153 0.1642 0.2250 0.2621 0.3034 0.3221

0.1701 0.1754 0.1779 0.1772 0.1729 0.1642 0.1500 0.1412 0.1300 0.1253


XB0=0.743

XSO=0



YBn+1= 0

YSn+1= 1



YB1=0

YS1=0.3


Underflow Product composition: Oil recovery is 95% in overflow showing that 5% is left which is being taken up with the under flow product, XAn=5%*XA0 XAn= 0.05*0.257 = 0.01285 The mass fraction of inert is same in both underflow feed and product i.e. X Bn = XBo XBn=0.743


 +

=

0.01285 0.01285+0.743


a = ( 0.02 , 0 )

Difference Point: The difference point is now find by drawing in the two lines connecting X 1 with Y1 and point “a” with Yn+1


Yn+1

2. Locate points, X1, Y1, Yn+1 and a and joint Y1 to X1 and point Yn+1 to ‘a’ and locate point Z.

Y1

a

X1

Z


Yn+1 Y5

2. Locate points, X1, Y1, Yn+1 and a and joint Y1 to X1 and point Yn+1 to ‘a’ and locate point Z.

Y4 Y3 Y2 Y1

Xm X5 X4

Xn

a

X3

X2 X1

Z

3. Join Y1 to B and locate X2, join X2 and Z and extend the line to fine Y2 and repeat the procedure. 4. Join Yn+1 with X1 and Xn with Y1 so that we get point Xm. X n lies 5. From graph, between X 5 and X 6 so number of stages are 5.

Now by lever rule Yn+1*(Yn+1*Xm) = X1* (X1*Xm) Yn+1 X1∗Xm = Yn+1∗Xm X1 Where, Yn+1/X1 = lbs of oil free ether removed per 1000 lb of fresh liver From graph;

Yn+1.Xm = 10.4

X1.Xm = 2.7

Putting these values in equation Yn+1 X1∗Xm = X1 Yn+1∗Xm Yn+1 2.7 = = 0.259 lb 10.4 X1 Here; X1 = 1000 lb So; Yn+1 = 0.259 * X 1 Yn+1 = 0.259*1000 Yn+1 = 259 lb Thus the charge of ether per 1000 lb of fresh liver is 259 lbs

Part 2: From graph, it is seen that X n lies between X 5 and X 6 so number of stages are 5. Part 3: No. of actual stages = ? Ƞ = 0.7

Actual number of stages = 5/0.7 = 7.2 i.e. 8 number of stages are required Part 4: The quantity of discharge solids = ? Quantity of discharge solids = Quantity of discharge solids =

        0.743 0.82

∗ 1000

Quantity of discharge solids =906.9 lb

∗   

4 Leaching Solid Liquid Extraction

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