Basis: In the slurry feed, solvent: C = 1 kg H2O, solid: B = 1.667 kg
Transcription
Basis: In the slurry feed, solvent: C = 1 kg H2O, solid: B = 1.667 kg
CENG 3210 Separation Processes Tutorial 9 Solution Multistage countercurrent leaching A multistage countercurrent leaching battery is used to extract the sludge from the reaction Na2CO3 + CaO + H2O CaCO3 + 2NaOH The products from the reaction enter the first unit with no excess reactants but with 1.667 kg CaCO3/kg H2O. Pure water is used to wash NaOH from the sludge. CaCO3 is assumed to be completely insoluble in water. The sludge retains solution varying with the concentration as shown in the following table. NaOH, wt % 0 5 10 15 20 Kg CaCO3/kg solution 0.667 0.571 0.455 0.370 0.278 If a 20% solution of the NaOH is to be produced, how many stages must be used to recover 97% of the NaOH? Basis: In the slurry feed, solvent: C = 1 kg H2O, solid: B = 1.667 kg CaCO3. The solute A (NaOH) is found from the reaction formula. For 1 mol CaCO3 produced, 2 mol NaOH is generated. So A = (2B)(MWNaOH/MWCaCO3) = 2x1.667(40/100) = 1.333 kg L0 = A + C = 1.333 + 1 = 2.333 kg y0 = A/L0 = 1.333/2.333 = 0.571 N0 = B/L0 = 1.667/2.333 = 0.714 xN+1 = 0 x1 = 0.2 (20% solution of NaOH) 1 Since 97% of the NaOH is to be recovered, NaOH in strong liquor (V1) = S1 = 1.333(0.97) = 1.293 kg NaOH in washed solid (LN) = AN = 1.333(1-0.97) = 0.04 kg Water in strong liquor = m1 = (S1/x1)(1-x1) = (1.293/0.2)(1-0.2) = 5.172 kg V1 = 1.293 + 5.172 = 6.465 kg To find yN, do a water balance: mN + 5.172 = mf + 1 mN = mf – 4.172 yN = AN/( mN + AN) = 0.04/( mN + 0.04) NN = B/( mN + AN) = 1.667/( mN + 0.04) NN, yN and mN are obtained by trial from the table. yN = 0.0153 NN = 0.638 mN = 2.574 LN = mN + AN = 2.574 + 0.04 = 2.614 kg To determine an intermediate point on the operating curve, choose yn = 0.1 (assumed), N = 0.455 = B/Ln = 1.667/Ln Ln = 3.664 kg Vn1 Ln V1 L0 3.664 6.465 2.333 7.796 kg L V x L0 y0 x A,n 1 n yn 1 1 Vn 1 Vn 1 3.664 6.465 0.20 2.333 0.571 (0.1) 0.042 7.796 7.796 Similarly, with yn = 0.15, xn+1 = 0.0735, with yn = 0.2, xn+1 = 0.114. 2 Using the points of various (yn, xn+1), and (yN, xN+1) to obtain a slightly curved operating line. By the McCabeThiele method, the number of ideal stages is determined to be 4. 3