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Items #1 #3 #4 #5 I. An evaporation-crystallization proccs is used to ohtain sol

ID: 695067 • Letter: I

Question



Items #1 #3 #4 #5

I. An evaporation-crystallization proccs is used to ohtain solid potassium sulpbate from an aqucous solution of lhis sal. Iresh-eed lo lhe process 18.6 1% K:S0.. I he wel filler cake cursisls of solid KsSC. crystals ind il 4096 K:SO4 sullion in a ratiu 10 kg uryslals kg solution lie lillrale also 40 % solution is recycled 10 juin ube fresh leed. Olhe waler leed lo lhe evaporalor. 42.66 % is evaporaled. Ihe evaoralor us a xm capacity uf 155 kg waler evaporaledmin Cakukale: a) Rarc at which fresh toed muste spplied for the maximum evanoration rat in the evaporalor b) The ralio kg recycled k^ fresh fved c) Maximm prochictio arc of solid if the et filter cake is larer dried completely W.-42.6G % of water in M uximuni evapolon- 155 kumin) T.- R 40 % K2SO1 10 ke K.So, cystal K so soution 2. A liquid mixture contains GO wtni cthanol (F), 5 % dissolvod solurc{Si and the balance water. A tream of this mixnir is tood o a continious distillton clumn opcrating at a stcady state. Proxhict stream eueral lhe lop and bollotn of lhe colun lhe colun design calls for the prducl strems to have eallo rales aud for the lop sreas to cotin9 elbanol and nu S Calculale: a> mass ruction of s in bollom slieanl h) Fracrion at the cthanol in the foed thar lcaves in the bom product scam (i., kg ofF in the bottom stream kg F in feod 90%SELL 60% DC X-mass rations Condition: TB 3. Soyhcans from which oil has been cxtracted into a hexanc sovencavc the cxtraction unit containing 0.80 kg hexanc kgsoyhcan solids. The solids arc to be dricd by being contacted ith nitrogen, which enler lhe dryer al0 M The solid leaving ube drver should conlinn 0.05 kg hexane kg solids. Ihe ga, leaving the dryer cxits at 60 C and 1.0 atal with a relative aturation of60 %, (Vapor Prc of hcxanc at (50 IC is 572.76 mm Hgl and is ted to a condenser in which it i cooled to 25 (Vapor Pressure ullexare al 25 _1s 151.28 nun lle, and compressed w 40411 , recovering parl ofthe vaporized hcxanc a a onderale. Calele etional recvery oexanekleedkg enlerinz witb the soybenus)

Explanation / Answer

1)

Water to be evaporated= 150 kg this corresponds to 42.66% water. So total water entering the evaporator= 150/0.4266=363.38 kg

Total water entering the evaporator= 363.38 kg

Let F= Feed and R= Recycle

Water entering evaporator is through recycle and through Fresh feed

F*(1-0.186)+R*0.6= 363.38 kg

F*0.814+R*0.6= 363.38 (1)

Solids entering Crystallizer F*0.186+R*0.4

If R is Recyle solution , 10R = Crystals

F*0.186+R*0.4= 10R+R*0.4

F*0.186= 10R (2)

From 1 , 10*0.814*R+R*0.6= 363.38

8.74R= 363.38 R= 41.58

F= 10*41.58/0.186

=2235.48 kg

Crystals from the system = 10*R= 10*41.58

= 415.8 kg

R/F= 41.58/2235.48

=0.0186

When t he cake is dried separately

Amount of cake in the solution = 41.58*0.4/100

=0.16 kg

Total solids = 415.8+0.16

= 415.96 kg