Question

An evaporator operating continuously is fed a caustic brine solution at 8000 kg/hr flow rate. The solution consists of 6% NaCl, 90% water and 4% NaOH by mass. In the evaporator, water is boiled away, some of the NaCl precipitates and is removed by filtration, and NaOH remains in the concentrated liquor. The liquid exiting the evaporator has the following analysis: 1% NaCl, 42% NaOH and 57% water. Draw the schematic of the process and then determine the water vaporized, NaCl precipitated, and concentrated liquid produced in kg/h.

Answer 1

1.0 Water 0.01 NaCI 0.42 NaOH 0.57 water 8000 1.0 NACI 0.06 NaCl 0.90 Water 0.04 NaOH

Apply balances:

NaOh balance:

F*0.04 = X*0 + W*0 + 0.42*P

8000*0.04 = 0.42*P

P = 8000*0.04/0.42 = 761.9047 kg/h of product...

balance of water:

8000*0.90 = W + 0.57*761.9047

W =8000*0.90 -0.57*761.9047 = 6765.714321 kg/h of vpor water

finally

X:

F = X + W + P

8000 = X + 6765.714321+ 761.9047

X = 8000 - (6765.714321+ 761.9047 = 472.380 kg/h