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Caustic concentration evaporator problem
Question - 22 25 tons of caustic solution having 10 % concentration was fed to an evaporator . the concentrated liquor leaving the evaporator contains 50% NaOH , 2 % Nacl and 48 % H2O .
Calculate a - the pounds of water evaporated per hour
b- the pounds of salt precipitated per hour
c- pounds of concentrated liquor produced per hour solution -
PART A- as the NaOH is the only constituent passing through the evaporator unchanged, it is convenient to base the concentrations of the H2O and Nacl on the NaOH. INPUT OF NaOH = 25 *2000 * 10/100 = 5000 LB/HR WATER REMOVED = 5000 *( 80/10 - 48/50) = 5000*( 8-0.96 ) = 35200 LB/HR
PART B- AMOUNT OF SALT = 5000 (10/10- 2/50 ) = 5000 ( 1-0.098) = 4800 LB/HR PART -C THE RATIO OF CONCETRATED LIQOUR = 100/50 AMOUNT OF Na OH = 5000 *100/50 =10000 LB/HR OR FEED BALANCE = 2000*25 = 50000 LB/HR SO , CONCENTRATED LIQUO R = 50000 - (35200 +4800 ) = 10000 LB/HR
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