Question

An evaporator-crystallizer process is used to obtain solid potassium sulfate. The fresh feed to the evaporator contains 19.60% K₂SO₄ and is mixed with the recycled mother liquor stream before entering the evaporator. The crystal product of the crystallizer is still wet as it contains 10.00 kg solid K₂SO₄ per kg adhering mother liquor solution. The filtered mother liquor contains 40.00% K₂SO₄ and is recycled to join the fresh feed. Of all the water entering the evaporator, 45.00% is evaporated. The evaporator has a maximum capacity of 175.0 kg/s water evaporated. If the system is kept running at maximum capacity, determine the following:

1. Flow rates of feed, crystal product, recycle, and mixed feed
2. Composition of mixed feed in % by mass
3. If we want to have a higher quality crystal product (meaning, K₂SO₄ crystals with a lesser water content), what unit do you suggest we add to the process? Where do we add the unit?
4. How much water needs to be removed from the crystal product to obtain pure K₂SO₄ crystals?

Answer 1

The fresh feed contains 19.60% K₂SO₄

F- fresh feed flowrate

xf = wt fraction of K₂SO₄ in fresh feed

V- water evaporated

R- recycle or mother liquor

x_R - wt fraction of potassium sulphate in recycle

C- crystal product

x_c - wt fraction of potassium sulphate in C

M- mixed stream entering evaporator

xm - wt fraction of potassium sulphate in mixed stream

The schematic diagram us given below for better understanding

The crystal product contain

10 Kg solid K₂SO₄ /Kg mother liquor

x_R = 0.40

In one Kg mother liquor, 0.4 Kg potassium sulphate is present

So mole ratio of potassium sulphate/water in product(C) = 10.4/0.6 = 17.333

Mole fraction (x_c) = 17.333/(1+17.333) = 0.94545

x_f = 0.1960

V = 175 Kg/s

Doing overall material balance

F = 175 + C (1)

Doing overall water balance we get

F(1-0.1960) = 175(1) + C(1-0.94545)

F(0.804) = 175 + C(0.05455) (2)

Solving both equations simultaneously we get

F = 220.766 Kg/s

C = 45.768 Kg/s

In evaporator 45% of total water is evaporated

Water evaporated = 175 Kg/s

Total water present at evapoator entry(mixed feed) = 175/(0.45) = 388.889 Kg/s

Doing water balance around mixed stream

We get

F(1-0.1960) + R(1-0.40) = 388.889

220.766(1-0.1960) + R(1-0.40) = 388.889

R = 352.321 Kg/s

F + R = M

M = 220.766 + 352.321 = 573.087 Kg/s

Doing potassium sulphate balance around mixed stream

220.766 (0.1960) +352.321(0.40) = 573.087(x_m)

x_m = 0.32141

Composition of mixed stream

32.141% potassium sulphate

67.858% water

B)

To gave higher quality of potassium sulphate crystal

We can add another evaporator of smaller capacity to further concentrate the crystal.

The evaporator has to be added after crystallizer to further evaporate water in adhering mother liquor

In this case

C = 45.768 Kg/s

Wt fraction of pure potassium sulphate present = 0.94545

In order to obtain pure crystal, amount of water to be evaporated = 45.768(1-0.94545)

= 2.49664 Kg/s

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