Mass balance involving crystallized substances in manufacturing industries






A solution of sucrose in water contains 14% of sucrose by weight. 150kg of the solution is evaporated to remove some amount of water; the remaining solution is cooled to 18˚C. If the yield of sucrose is 90%, calculate the amount of water evaporated. Given that; molecular formula of sucrose (C12H22O11), C=12, H=1, O=16 and solubility of sucrose is 0.40kg/mole. 

This is a mass balance of crystallized substance question in most industries, for instance sugar processing industry. To this question you have to come up with two equations: one for general material balance and the other is for crystallized substance (in decimal points or in solid fraction). If you have been in sugar processing industry you have realized that cane juice coming to the evaporators, to pans then massecuites to the crystallizers and final to the centrifugal machines where the molasses is separated from sugar (sucrose). This question is derived from that process. Let material charged in the process (150kg of the solution) to be F, crystallized substance(sucrose) to be C, amount water evaporated/evaporation(steam) to be E, and finally mother liquor(molasses) to be M. 

Overall material balance will be;
F= C + E + M (according to the Law of mass balance)
Material balance for crystallized substance (sucrose), in solid fraction
Ffr = Ccr + Eer + Mmr, where fr, cr, er, mr are the fractions of crystals formed in the process.
Molecular weight of sucrose will be;
(12 X 12) + (22 X 1) + (11 X 16) = 342 kilograms/mole, in the question the material charged in the process is 150kg (solution) but how much amount of sucrose in the process? This 14% of the solution (material charged); 

(14/100 X 150) = 21 kg (amount of sucrose charged in the process), the amount of sucrose obtained at the end of process will be;
(90/100 X 21 grams) = 18.9 kg (crystals of sucrose formed in the process). Now inserting these values in the overall material balance, this gives; 

F= M + C + E, M=? C = 18.9 kilograms, E=?
150kg = M + 18.9 + E, making M the subject of the formula, this result in;
M= (150-18.9) - E
M= 131.1 kg – E…………………………………… equation (i)    

Equation of crystals formed in the process (in solid fraction)
Ffr = Ccr + Eer + Mmr, before substituting appropriate values in these should be identified;
Ffr = (14/100 X 150) kg = 21 kg
Evaporation (Eer) = 0
Crystals (Ccr) {Sucrose} = 100/100 = 1

Mother liquor (M) = solubility in grams (or kilograms)/ (solubility in grams + solvent soluble);
(0.4 X 342)/ {(0.4 X 342) + 150) = 0.47699 = 0.5
Inserting these values in the equation gives; 

21kg = 0.5M + 18.9 + E X (0)
21 = 0.5M + 18.9 + 0; making M the subject of the formula gives;
2.1 = 0.5M

M = (2.1/0.5)……………………………………………………equation(ii),then substituting equation (ii) in equation (i) gives; 

4.2 = 131-E


-126.9kg = -E



E = 126.9 kg and this is the amount of water evaporated.






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