Question

1. An evaporation-crystallization process is used to obtain solid potassium sulphate from an aqueous solution of this salt. Fresh feed to the process 18.6 wt% K₂SO₄. The wet filter cake consists of solid K₂SO₄ crystals and a 40 wt% K₂SO₄ solution, in a ratio 10 kg crystals/kg solution. The filtrate also a 40 % solution, is recycled to join the fresh feed. Of the water feed to the evaporator, 42.66 % is evaporated. The evaporator has a maximum capacity of 155 kg water evaporated/min. Calculate:

Rate at which fresh feed must be supplied for the maximum evaporation rate in the evaporator

The ratio kg recycled / kg fresh feed

Maximum production rate of solid if the wet filter cake is later dried completely.

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