Question #169570

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.


1
Expert's answer
2021-03-08T05:54:09-0500

Let mm 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×m0.04×10,000=0.25×m'


m=1600kg/hm'=1600kg/h


Overall material balance ;

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


10,000=kg/hofwaterevaporated+1,600kg/h10,000=kg/h of water evaporated +1,600kg/h

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


Capacity of Evaporator =8,400Kg/h.

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