Question

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.

Answer 1

Answer:

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: $M_{1} = 175 + 11M_{2}$

Water: $0.804M_{1} = 175 + 0.6M_{2}$

Using substitution method, we have:

$M_{1}$ = 220.77 kg/s

$M_{2}$  = 4.16 kg/s

The maximum production rate of solid crystal is $10M_{2}$ = 10*4.16 = 416 kg/s

Around evaporator:

$0.45M_{5} = 175$

$M_{5} = 175/0.45 = 389$ kg/s

Around the mixing point:

$M_{1} + M_{3} = M_{4} + M_{5}$

Solid crystal: $0.196M_{1} + 0.4M_{3} = M_{4}$

Using the last two equations, we can obtain:

$0.804M_{1} + 0.6M_{3} = M_{5}$

$0.6M_{3} = 389 - 0.804*220.77 = 389 - 177.5 = 211.5$

$M_{3} = 211.5/0.6 = 352.5$ kg/s

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

$\frac{M_{3}}{M_{1}}$ = 352.5/220.77 = 1.6