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CRYSTALLIZATION

•  important industrially because:

•  solid-liquid separation process

i) a crystal formed from an impure solution and itself pure

ii) Practical method of obtaining pure chemical substances in a satisfactory condition for packaging and storing

•  yield , purity, sizes & shapes of crystals important

•  crystals - uniform in size

•  saturated solution cooled/evaporated forms crystal

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SOLUBILITY CURVE

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SOLUBILITY CURVE

•  solubilities of most salts increase slightly or markedly with temperature

Solubility curve for some typical salts in water Solubility curve for sodium thiosulfate

•  equilibrium is attained when the solution or mother liquor is saturated (represented by solubility curve)

•  solubility dependent mainly on temperature (pressure negligible effect)

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SOLID-LIQUID PHASE DIAGRAM

Solid-liquid phase diagram for the MgSO4-nH2O system at 1 atm

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YIELD & HEAT & MATERIAL BALANCES

initial concentration of solute

final temperature solubility at this temperature

•  material balances

crystals are hydrated - some water in the solution is removed with crystals

•  yield of crystals can be calculated by knowing:

solute crystals are anhydrous - simple water & solute material balances

F kg Hot solution X1 kg solute/ 100 kg H2O

C kg Crystal

X2 kg solute/ 100 kg H2O

S kg solution

kg H2O evaporate Mass balance :

Water balance : Input = Output

Solute balance : Input = Output

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EXAMPLE 12.11-1

A salt solution weighing 10 000 kg with 30 wt % Na2CO3 is cooled to 293K (20oC). The salt crystallized as the decahydrate. What will be the yield of Na2CO3.10H2O crystals if the solubility is 21.5 kg anhydrous Na2CO3/100 kg of total water?

a)  Assume that no water evaporated.

b)  Assume that 3% of the total weight of the solution is lost by evaporation of water in cooling.

Ans: 6370 kg, 6630 kg

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EXAMPLE 12.11-1 Solution:    Molecular weights: Na2CO3 = 106.0, 10H2O = 180.2, Na2CO3 •10H2O = 286.2

(a)  W = 0, water balance:

Na2CO3 balance:

C = 6370 kg of Na2CO3 •10H2O crystals and S = 3630 kg solution (b) W = 0.03(10000) = 300 kg, water balance: C = 6630 kg of Na2CO3 •10H2O crystals and S = 3070 kg solution.

         

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EXAMPLE 27.1(McCabe, Smith & Harriott) A solution consisting of 30 wt% MgSO4 is cooled to 60oF. During cooling, 5 % of the total water in the system evaporates. How kg of crystals are obtained per 1000 kg of original mixture?

Data: Molecular weight of MgSO4 = 120.4

Ans: 261 kg

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EXAMPLE (P12.11-3)

A hot solution containing 1000 kg of MgSO4 and water having a concentration of 30 wt% MgSO4 is cooled to 288.8 K (60oF) where crystals of MgSO4.7H2O are precipitated. The solubility at 288.8 K is 24.5 wt % anhydrous MgSO4 in the solution. Calculate the yield of crystals obtained if 5 % of the original water in the system evaporates on cooling.

Data: Molecular weight of MgSO4 = 120.4

Ans: 261 kg

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HEAT BALANCES IN CRYSTALLIZATION normally, crystallization is exothermic

heat of crystallization, ΔHcrys = - heat of solution at infinite dilution,

!

"Hsoln

#

When Tdatum = Tequil./sat. , FhF + q = Wλ + ChC = Wλ + CΔH∞crys

FhF + q = (S + C)hM + WHV

Total heat absorbed, q (kJ):

F kg Hot solution hF kJ/kg

C kg Crystal hC kJ/kg

hS kJ/kg

S kg solution

W kg H2O HV kJ/kg

Heat absorbed, q = +’ve ,Heat given off , q = -’ve

When Tdatum = 32oF ,

FhF + q = ShS + ChC + WHV

or

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ENTHALPY-CONCENTRATION DIAGRAM

Enthalpy-concentration diagram for the MgSO4-nH2O system at 1 atm

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ENTHALPY-CONCENTRATION DIAGRAM

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EXAMPLE A 32.5% solution of MgSO4 at 120oF (48.9oC) is cooled, without appreciable evaporation to 70oF (21.1oC) in a batch-cooled crystallizer. How much heat must be removed from the solution per 100 Ib of the feed solution? The average heat capacity of the feed solution is 0.72 Btu/Ib oF and the heat of solution at 18oC is 23.2 Btu/Ib of MgSO4.7H20.

Ans: -4260 Btu

Enthalpy-concentration diagram for the MgSO4-nH2O system at 1 atm

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EXAMPLE 12.11-2

A feed solution of 2268 kg at 327.6 K (54.4oC) containing 48.2 kg MgSO4/100 kg total water is cooled to 293.2 K (20oC) where MgSO4.7H2O crystals are removed. The solubility of the salt is 35.5 kg MgSO4/100 kg total water. The average heat capacity of the feed solution can be assumed as 2.93 kJ/kg.K. The heat of solution at 291.2 K (18oC) is 13.31 x 103 kJ/kg mol MgSO4.7H2O. Calculate the yield of crystals and make a heat balance to determine the total heat evolved and removed from the crystalliser, q, assuming that no water is vaporized. Data: Molecular weight of MgSO4•7H20 is 246.49 Ans: 616.9 kg crystal, -261 912 kJ

Enthalpy-concentration diagram for the MgSO4-nH2O system at 1 atm

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EXAMPLE 12.11-2 Solution:    Molecular weights: MgSO4 = 120.4, 7H2O = 126, MgSO4 •7H2O = 246.4 W = 0, water balance:

MgSO4 balance: Total balance: 2268 = C + S C = 631.9 kg MgSO4•7H2O crystals, S = 1636.1 kg solution.

         

2268 kg 48.2 kg MgSO4/100 kg H2O S kg solution

35.5 kg MgSO4/100 kg H2O

C kg MgSO4 •7H2O

2268 10048.2+100!

"#

$

%&= S

10035.5+100!

"#

$

%&+C

126246.4!

"#

$

%&

2268 48.248.2+100!

"#

$

%&= S

35.535.5+100!

"#

$

%&+C

120.4246.4!

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$

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EXAMPLE 12.11

Solu&on   Tref. = 293.2K (Tdatum)

Heat balance : Enthalpy of the feed , hF = CpΔT = (2.93)(327.6-293) Heat of solution = 13.31x103/246.49 = 54.0 kJ/kg crystals. Heat of crystallization = -(54.0) = -54.0 kJ/kg crystals Total heat absorbed, q

FhF + q = Wλ + ChC = Wλ + CΔH∞crys

2268(2.93)(327.6-293) + q = 631.9(54.0)

q = 631.9(54.0) - 2268(2.93)(327.6-293) = -262 689 kJ

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EXAMPLE

A solution of 500 kg of Na2SO4 in 2500 kg water is cooled from 333K to 283K in an agitated mild steel vessel. At 283K, the solubility of the anhydrous salt is 8.9 kg/100kg water and the stable crystalline phase is Na2SO4.10H2O. At 291K, the heat of solution is -78.5 MJ/kmol and the specific heat capacity of the solution is 3.6 kJ/kg.K. If, during cooling, 2% of the water initially present is lost by evaporation, estimate the heat which must be removed.

Ans: 240 407 kJ

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CRYSTALLIZER