IdealLeaching

7/30/2019 IdealLeaching

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Lecture 8: Ideal Leaching1

Leaching and Crystallization

This lecture will cover:

Ideal leaching

Non-ideal leaching

Crystallization

Last lecture covered: Ternary Liquid-Liquid extractions.

Ternary phase diagrams.

A procedure to determine the product compositions and

flow rates of a liquid-liquid extraction separation.


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Ideal Leaching

A,B

BC

Leaching is a solid-liquid separation method which involvespreferential solvation of

one component, the solute, of a solid mixture into a liquid solvent.

Diffusion of the solute through the solid is slow and complicates the separation process

in practice because equilibrium is difficult to attain .

It is also difficult to completely disengage the solid and liquid phases.

B

A

A

slow

Solid exits wet

Diffusion through solids

is slow

The ideal case is one where all of B is leached out of solid A.


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Ideal Leaching

Solvent

C

Solid Feed

A (carrier)

B (solute)

Overflow

B,C

UnderflowB,C (liquid)

A (solid)

Solid-free

liquid

Solid-liquid

slurry

(B only in liquid)


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Solvent

yS0, xS0

Solids

yL0, xL0 Underflow

S1, yS1, xS1

Overflow

L1, y

L1, x

L1

Liquid B and C

Solid A (insoluble)

Liquid B and C

No Solids

Liquid Solvent C

Solid Carrier Aand Solute B

Ideal Leaching

Solute completely

dissolves

XS = solute/(solvent + solute) in overflow

YS = solute/(solvent + solute) in underflow

YI = carrier/(solvent + solute + carrier) in overflow or

underflow

y= fraction of solvent in overflow or underflow

x= fraction of solute in overflow or underflow

Ratio of solid to liquid

in the underflow is a constant


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Ideal Leaching

Nomenclature:

XS = solute/(solvent + solute) in overflow

YS = solute/(solvent + solute) in underflowYI = carrier/(solvent + solute+carrier) in overflow

or underflow



YI = carrier/(solvent + solute + carrier) is constant so if we increase the solvent flow rate, theflow rate of the overflow increases while the underflow remains constant.

If solvent flow rate increases, the liquid becomes less concentrated in solute.

The composition of the overflow and the liquid in the underfloware the same.

Since all the solute is dissolved, adding solvent only dilutes the liquid in the solute. The dilution

is the same for the overflow and underflow liquids since the liquids have the same concentration.

Since the underflow also has insoluble solid A the concentration of B in the underflow islower.

Givens:

The fraction of solids in the underflow YI.

All of the solute is dissolved (ideal leaching).The composition of the solid feed.

The fraction of solids in the overflow YI = 0.


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Ideal Leaching



Overflow: y = solvent/(solvent+solute)x= solute/(solvent+solute)

y/x = solvent/solute

Underflow: y = solvent/(solvent+solute+carrier)

x = solute/(solvent+solute+carrier)

y/x = solvent/solute

Underflow: y = solvent/(solvent+solute+carrier)

x = solute/(solvent+solute+carrier)

YI = carrier/ (solvent+solute+carrier)x+y+YI =1 YI is a constant.

y = 1xYI So underflow has slope of1

So tie lines joining the overflow and underflow lines pass through the origin.

The overflow and underflow lines both have slope of1

y

x

Overflow: y = solvent/(solvent+solute)x = solute/(solvent+solute)

x+y=1

y = 1x So overflow has slope1

overflow

underflow


7/15Lecture 8: Ideal Leaching7

Ideal Leaching

Solvent

yS0, xS0

Solids

yL0, xL0

Underflow

S1, yS1, xS1

Overflow

L1, yL1, xL1

Liquid B and C

Solid A (insoluble)

Liquid B and C

No Solids

1) Locate the feed point L0. Locate the underflow line using the fraction of solids given.

Draw overflow and underflow lines

2) Draw the mixing line.

3) Locate the mixing point M using the inverse Lever Rule

4) Draw the tie line through M.

5) Determine the compositions of the flow streams

6) Locate the overflow and underflow rates L1 and S1 using the inverse lever rule mass balances

OverflowUnderflow

S0

L0 (Feed)

M

L1

S1

(1fraction solids in underflow)

yC

xB1

1



Example

Caffeine is leached out of coffee grounds using water. If 1000kg/hr of pure water is

mixed with 1000 kg/hr of ground coffee beans which are 80% by weight insoluble

and 20 % caffeine, calculate the amounts and compositions of underflow and overflow

from an ideal separator. A test of the underflow shows that it is 70% solids.

Step 1. Draw the y-x diagram with the overflow line. Draw the underflow line

parallel to the overflow line and intersecting 0.3. Locate feed L0 at solid

mixture composition and S0at the solvent concentration.

1

S0

0

y

xL0 1

overflow line

underflow line

Feed point

Solvent point

YI



Example

Step 2 Draw the mixing line through the S0, and L0. The mixing point will lie on

this line and depend on the ratio of solvent to solid feed.

Step 3 Apply the lever rule to a solute (caffeine) balance or solvent (water) balance.L0S0

xS0 xM

xM xL0

1000kg

1000kg 1

0 xMxM 0.2

xM 0.1

L0S0

yS0 yM

yMyL0

1000kg

1000kg 1

1yMyM0

yM 0.5Solvent

balance

Solute

balance

1S0

0

y

xL0 1

M

1S0

0

y

xL0 1

M

xM

yM

overflow

underflow



Example

Step 4 Draw the tie line by extending a line from the origin through the mixing

point to the overflow line.

Step 5 From point M we know y/x = 5. Using the equations of the overflow andunderflow lines allows us to determine the stream composition.

1

S0

0

y

xL

0 1

M

L1

S1

y

x 5

y 1 x 5x

y 0.8333

x 0.1667

Overflow

yx

5

y 1 x 0.7 5x

y 0.25

Underflow

x 0.05

overflow

underflow



Example

Step 6 Determine the overflow and underflow flow rates

1

S0

0

y

xL

0

1

M

L1

S1

L1 fL1M

fL1 xMxS1xL1 xS1

0.1 0.05

0.1667 0.05 0.429 L1 0.429 2000kg/ hr 858kg/ hr

S1 1142kg/ hr

overflow

underflow



Non-Ideal Leaching

Often, the conditions in a leaching system are such that equilibrium is not

obtained. The overflow and underflow curves become non-ideal.

1

S0

0

y

xL0

1

M

L1

S1

underflow

overflow



Ideal Leaching

For systems with low solute to carrier ratios, a finite amount of solvent

may be needed to create a two phase equilibrium condition.

1

S0

0

y

xL0

1

M

L1

S1

M



Crystallization

Separation using crystallization involves cooling a liquid solution into

a two-phase solid-liquid equilibrium condition. The solid precipitate is

very concentrated in the higher melting point material and can be separated

from the remaining liquid solution.

T

xnaphthalene

M

benzene

L

naphthalene

L + Solid Naph

L + Solid Benz

Solid Naph + Solid Benz

MS xnM xn

L

xnSxn

L

MF

The mass of the solid naphthalene

produced is determined using the

lever rule. MF is the mass of the

original feed. The solid concentration

is usually assumed to be pure.



Summary

This lecture covered:

Ternary Ideal leaching.

Graphical interpretation of mass balances and equilibrium. A procedure to determine the product compositions and

flow rates of a leaching separation.

Simple description of crystallization separations

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