Answer to Question #169570 in Organic Chemistry for Tony

Question

An evaporator operating at atmospheric pressure (101.325 kPa) is fed at the rate of 10000 kg/h of weak liquor containing 4% caustic soda. Thick liquor leaving the evaporator contains 25% caustic soda. Find the capacity of the evaporator.

Answer 1

Let $m$ be the Kg/h of thick liquor leaving the evaporator.

Material balance of Caustic Soda ;

Caustic soda in feed =Caustic soda in thick liquor

$0.04×10,000=0.25×m'$

$m'=1600kg/h$

Overall material balance ;

kg/h of feed=kg/h of water evaporated +kg/h of thick liquor.

$10,000=kg/h of water evaporated +1,600kg/h$

Water evaporated =10,000-1,600=8,400kg/h

Capacity of Evaporator =8,400Kg/h.