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Introduction To Material and Energy Balances by  G.V. Reklaitis Problem 2.17 A slurry consisting of CaCO3 precipitate in a solution of NaOH and H2O is washed with an equal mass of a dilute solution of 5% (wt) NaOH in H2O. The washed and settled slurry which is withdrawn from the unit contains 2 lbm of solution per 1 lbm of solid (CaCO3). The clear solution withdrawn from the unit can be assumed to have the same concentration as the solution withdrawn with the solids (see the Figure P2.17). If the feed slurry contains equal mass fractions of all components, calculate the concentration of the clear solution. Answer (Figure P2.17) Let's make DOF (Degree Of Freedom) Subsidiary Relation 1. A slurry consisting of CaCO3 precipitate in a solution of NaOH and H2O is washed with an equal mass of a dilute solution of 5% (wt) NaOH in H2O. It’s made F1= F2 2. The washed and settled slurry which is withdrawn from the unit contains 2 lbm of solution per 1 lbm of solid (CaCO3...

Introduction To Material and Energy Balances by  G.V. Reklaitis Problem 2.12 The feed to a distillation column contains 36% benzene by weight, the remainder being toluene. The overhead distillate is to contain 52% benzene by weight, while the bottoms are contain 5% benzene by weight. Calculate: (a) The percentage of the benzene feed which is contained in the distillate. (b) The percentage of the total feed which leaves as distillate. Answer (Ilustration) We can answer this question without degree of Freedom. Let we assume W1 = W Total Material Balance W1 = W2 + W3 W = W2 +W3 … (1) Material Balance of Benzene In = Out 0.36xW = 0.52xW2 + 0.05 W3 (both side we x100) Then we got 36W = 52W2 + 5W3 … (2) Material Balance of Toluene In = Out 0.64xW = 0.48xW2 + 0.95xW3 (both side we x100) Then we got 64W = 48W2 + 95W3 … (3) Eliminate … (2) and (3) 36W = 52W2 + 5W3   |x95 64W = 48W2 + 95W3 |x5 Then we got, 3420W = 4940W2 + 475W3   320W...