An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an aqueous solution of this salt. The fresh feed to the process contains 19.6 wt% K2SO4. The wet filter cake consists of solid K2SO4 crystals and a 40.0 wt% K2SO4 solution, in a ratio 10 kg crystals/kg solution. The filtrate, also a 40.0% solution, is recycled to join the fresh feed. Of the water fed to the evaporator, 45.0% is evaporated. The evaporator has a maximum capacity of 175 kg water evaporated/s. Calculate the maximum production rate of solid K2SO4, the rate at which fresh feed must be supplied to achieve this production rate.

feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6

Explanation:

Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.

Overall balance:

Water:

Using substitution method, we have:

= 220.77 kg/s

 = 4.16 kg/s

The maximum production rate of solid crystal is = 10*4.16 = 416 kg/s

Around evaporator:

kg/s

Around the mixing point:

Solid crystal:

Using the last two equations, we can obtain:

kg/s

The rate of supplying fresh feed to obtain the production rate is:

= 352.5/220.77 = 1.6

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water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.