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Hematra (Crystallization Homework)
1. A crystallizer is charged with 7500 kg of an aqueous solution at 104 degrees Celsius, 28.5% by weight which is anhydrous sodium sulphate. The solution is cooled. During cooling operation 5% of water originally present was last. As a result, the crystals of Na2SO4-10H2O crystalline out. The mother liquor is found to contain 18.2% by weight of anhydrous salt. Calculate the yield of crystals and quantity of mother liquor. Solution: Basis: 7500 kg of fed
Water present in the feed solution = 7500 x 0.715 = 5362.5 kg Amount of water lost during cooling (V) = 5362.5 x 0.05 = 268.13 kg Overall material balance: F=V+L+C 7500 = 268.13 + L + C L + C = 7500 - 278.13 L + C = 7231.87 (i) Balance for Na2SO4: F(xF) = V(xV) + L(xL) + C(xC) 7500 x 0.285 = 0 + L x 0.182 + C x 0.441 2137.5 = 0.182L + 0.441C 0.182L + 0.441C = 2137.5 (ii) Solving Eqs. (i) and (ii), we get C = 3171.43 kg L = 4059.70 kg The yield of crystals (C) = 3171.43 kg The quantity of mother liquor (L) = 4059.70 kg 2. A solution of NaNO3 in water at 40 degrees Celsius contains 48% NaNO3 by weight. Calculate the percentage saturation of this solution. Calculate the weight of NaNO3 that may be crystallised from 1000 kg of this solution by reducing the temperature to 10 degrees Celsius. Calculate the percentage yield.
Solubility of NaNO3 at 40 degrees Celsius is 51.4% by weight Solubility of NaNO3 at 10 degrees Celsius is 44.5% by weight Solution: Basis: 1000 kg of feed
Percentage saturation of NaNO3 solution = (48/51.4) x 100 = 93.38 Overall material balance: F = L + C 1000 = L + C L + C = 1000 (i) Material balance for NaNO3: 1000 x 0.48 = L x 0.445L + C 480 = 0.445L + C 480 = 0.445L + C (ii) Solving Eqs. (i) and (ii), we get C = 63.06 kg L = 936.94 kg Therefore, percentage yield of NaNO3 = (63.06/480) x 100 = 13.14 The percentage yield of NaNO3 crystals is 13.14. 3. A solution of NaCl in water is saturated at 15 degrees Celsius. Calculate the amount of NaCl that can be dissolved by 200 kg of this solution if heated to a temperature of 65 degrees Celsius. Data: Solubility of NaCl at 15 degrees Celsius = (385 kg of NaCl / 1000 kg of water) Solubility of NaCl at 65 degrees Celsius = (372.65 kg of NaCl / 1000 kg of water) Solution: Basis: 200 kg of solution
Overall material balance: F=L+C Material Balance for NaCl: F(xF) = L(xL) + C(xC)
200 x 0.278 = (F-C) x 0.2715 + C x 1 55.6 = (200-C) x 0.2715 + C 55.6 = 54.3 - 0.2715C + C 55.6 = 54.3 + 0.7285C 0.7285C = 55.6 - 54.3 0.7285C = 1.3 Therefore, C= 1.784 kg. The amount of NaCl that can be dissolved if the solution is heated to a temperature of 65 degrees Celsius is 1.784 kg. 4. A solution of CaCl2 in water contains 62 kg of CaCl2 per 100 kg of water. Calculate the weight of this solution necessary to dissolve 300 kg of CaCl2 - 16H2O at 25 degrees Celsius. Solubility of CaCl2 at 25 degrees Celsius = (7.38 kgmol CaCl2 / 1000 kg of H2O) Molecular weight of CaCl2 = 111 Molecular weight of CaCl2 - 16H2O = 219 Solution: Basis: 300 kg
Overall material balance: F = L + C Material balance for CaCl2: F(xF) = L(xL) + C(xC) 300 x 0.3827 = L x 0.45 + (FL) x 0.5068 114.81 = 0.45L + (300L) x 0.5068 114.81 = 0.45L + 152.04 0.5068L 114.81 = 152.04 0.0568L 0.0568L = 152.04 114.81 0.0568L = 37.23 L = 655.46 kg.
5.A tank holds 10,000 kg of a saturated solution of Na2CO3 at 30°C. You want to crystallize from this solution 3000 kg of Na2CO310H2O without any accompanying water. To what temperature must the solution be cooled? Solution:
Na2CO3
Na2CO3
Saturated Solution 30 degrees Celsius
H2O
Saturated Solution T=?
H2O
10,000 kg initial state
Final State Na2CO3-10H2O 3000 kg 30 Crystals Removed
Solubility data for Na2CO3 as a function of the temperature: Temperature (degrees Celsius)
Solubility (g Na2CO3 / 100g H2O)
0
7
10
12.5
20
21.5
30
38.8
Because the initial solution is saturated at 30°C, you can calculate the composition of the initial solution: ( 38.8 g Na2CO3 / (38.8 g Na2CO3 + 100 g H2O) ) = 0.280 mass fraction Na2CO3 Calculate the Composition of the crystals: Comp. Na2CO3
Mol. 1
H2O
10
Mol.Wt. 106 18
Total Basis: 10,000 kg of saturated solution at 30 degrees Celsius
Mass 106
Mass. fr 0.371
180
0.629
286
1.00
F = ? kg
10,000 kg Na2CO3 H2O 0.720
Na2CO3 H2O
0.280
mNa2CO3 mH2O
Final Initia l 3000 kg Na2CO3 H2O
0.371 0.629
Crystals Removed
Basis: I = 10,000 kg Accumulation in Tank Final Initial Transport Out Na2CO3 mFNa2CO3 10,000(0.280) = 3000(0.371) H2O mFH2O 10,000(0.720) = 3000(0.629) Total F 10,000 = 3000 Component kg mFNa2CO3 1687 mFH2O 5313 F (total) 7000 To find the temperature of the final solution, calculate the composition of the final solution in terms of grams of Na2CO3/100 grams of H2O ( 1,687 kg Na2CO3 / 5,313 kg H2O) = (31.8 g Na2CO3 / 100g H2O)
Thus, the temperature to which the solution must be cooled lies between 20°C and 30°C. By linear interpolation
