Question

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70 kg/h of a concentrate from an evaporation system containing 55 wt.% CaCl2 at 90 °C and 101 kPa enters a continuous crystallizer. The saturated crystals of CaCl2. H20 leave at 10 °C. At 10 °C the solution contains 12.5 kg CaCl2/100 kg H20 and during the cooling 2% of the water entering the crystallizer is lost by evaporation. Additional information: 1. Atomic mass Ca = 40 g/mol, Cl = 35.5 g/mol, O = 16 g/mol, H = 1 g/mol 2. Table 2.1: Energy Data -65.9 MJ/kmol Heat of solution of CaCl2. H2O Specific heat of solution Latent heat of vaporization Mass of mild steel crystallizer Specific heat of mild steel 4.6 kJ/kg.K 2 270 kJ/kg 450 kg 0.5 kJ/kg.K 2.1 By means of a flow diagram, indicates all streams around the crystallizer. 2.2 Calculate the mass of crystals produced using the yield method and mass balance. (20) 2.3 Calculate the amount of heat required to be removed from the crystallizer. (8)

Answer 1

Shower 90°C to wgh ssh Lock Crystells, p ( cock Hw) 10°C IL (berry Callolvargeras 7=2870.13/ng. BHs - 65.9 Mill med 4.6uJ/vj-k 10°C. Ops Opste. = O-s Whyk Msreel - 450kg sool1 uglysol * 6 Scubility of call chooc= 18.5 legs lysol V = 0·08X0.46% 70 = 0·6344/h Moss Beloni F-VIL+P. 70-0°63-P=L. 69037-P-L Boone on Amount of Cock in crystal. = !!! xP. Balonie on Cccl. 111 P + Loull) . 30.5 = 0-86P + (0.11)(69-37-P). 129 0-55x70 199

3805- 00406P+ 706307. 00mp. 30.86 0075 P P 41-146 al = 88.224 ugh wifh. Heat Removed Let & be Amount of Heat Energy Bedonce on System (In) - lous) - Consupir Sen zo 70 Cpso (Tin-Tin) - gg.934 epsol Tour-Tin). + (669)x10x (0-3199) - (0-6>)(0970) -Q- X80X0-50 Tour - Ton) -0. U +88.834 x 416 + 490-2008) (10-20170970X063 = 6809x 109 x003189. O + 203860439 + 91615051, 1430 for tho. 4797-942 KJ/ W3/6 Q

using given data and required mass and energy balance we can find the required vairables

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