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.

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aaregbe aaregbe · Answer 1 · 2020-02-15T17:20:02+01:00

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: [tex]M_{1} = 175 + 11M_{2}[/tex]

Water: [tex]0.804M_{1} = 175 + 0.6M_{2}[/tex]

Using substitution method, we have:

[tex]M_{1}[/tex] = 220.77 kg/s

[tex]M_{2}[/tex] = 4.16 kg/s

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

Around evaporator:

[tex]0.45M_{5} = 175[/tex]

[tex]M_{5} = 175/0.45 = 389[/tex] kg/s

Around the mixing point:

[tex]M_{1} + M_{3} = M_{4} + M_{5}[/tex]

Solid crystal: [tex]0.196M_{1} + 0.4M_{3} = M_{4}[/tex]

Using the last two equations, we can obtain:

[tex]0.804M_{1} + 0.6M_{3} = M_{5}[/tex]

[tex]0.6M_{3} = 389 - 0.804*220.77 = 389 - 177.5 = 211.5[/tex]

[tex]M_{3} = 211.5/0.6 = 352.5[/tex] kg/s

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

[tex]\frac{M_{3}}{M_{1}}[/tex] = 352.5/220.77 = 1.6