The flowchart of a steady-state process to recover crystalline chromate (K_2 CrO

Question

The flowchart of a steady-state process to recover crystalline chromate (K_2 CrO_4) from an aqueous solution of this salt is shown below: Forty-five hundred kilograms per hour of a solution that is one-third K_2 CrO_4 by mass is joined by a recycle stream containing 36.4% K_2 CrO4, and the combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains 49.4% K_2 CrO_4 this stream is fed into a crystallizer in which it is cooled (causing crystals of K_2 CrO_4 to come out of solution) and then filtered. The filter cake consists of K_2 CrO_4 crystals and a solution that contains 36.4% K_2 CrO_4 by mass: the crystal account for 95% of the total mass of the filter cake. The solution that passes through the filter, also 36.4% K_2 CrO_4, is the recycle stream. Calculate the rate of evaporation, the rate of production of crystalline K_2 CrO_4, the feed rates that the evaporator and crystallizer must be designed to handle, and the recycle ratio (mass of recycle)/(mass of fresh feed). Suppose that the filtrate were discarded instead of being recycled. Calculate the production rate of crystals. What are the benefits and costs of the recycling?

Explanation / Answer

let m= mass of filter cake

feed entering contains 4500 kg/hr containing 33.3% K2CrO4.

let m= mass of filter cake, mass of crystals= m*0.95

mass of solution = (m-m*0.95)= m*0.05

writing K2CrO4 balance for the entire plant

4500*1/3= m*0.95 +m*0.05*0.364

1500= m*0.95+m*0.0182

m=1549 mass of filter cake

mass of crystals= 1549*0.95= 1472 kg/hr mass of solution = 1549-1472= 77 kg/hr

mass of water entering the system= 4500*2/3= 3000 kg/hr, mass of water leaving with the solution of crystals= (1-0.364)*77 kg/hr=49 kg/hr

mass of water evaporated= 3000-49= 2951 kg/hr

mass of water entering from fresh feed= 3000 kg/hr

let R= recycle, mass of water in the recycle= R*(1-0.364) = R*0.636

mass of water entering = 3000+R*0.636

water leaving the tower= 3000+R*0.636-2951=49+R*0.636

let P= Prodct leaving the evaporator

hence P*(1-0.494)= water from evaporator= 49+R*0.636

P*0.506= 49+R*0.636 (1),

P= 49/0.506+R*0.636/0.506, P= 97+R*1.26 (1A)

P*0.494= 5000*1/3+R*0.364, P= 1500/0.494 +R*0.74=3036+R*0.74 (2)

from eq.1A and 2, 3036+R*0.74= 97+R*1.26

3036-97= R*(1.26-0.74)

R= 5652 kg/hr

Feed rate to evaporator= 5652+4500= 10152 kg/hr

feed rate to filter = feed enterin the evaporator-water evaporated= 10152-2951= 7201 kg/hr

Recycle ratio = 5652/4500= 1.256

When filtrate is discarded,

Crystals entering the evaporator= 4500*1/3= 1500 kg/hr

This correspond to 49.4% of K2CrO4 leaving, leaving product from evaporator= 1500/0.494= 3036 kg/hr

water evaporated= 4500-3036= 1464 kg/hr = W ( water evaporated)

let F= flow rate of filtrate

writing overall balance

4500= W+F+m

given W= 1464 kg/hr

F+m=4500-1464= 3056 kg/hr (3)

F= 3056-m (3A)

writing K2CrO4 balance

4500*1/3= m*0.95+m*0.05*0.364+F*0.364

1500= m+F*0.364

from 3A, 1500= m+(3056-m)*0.364

1500= m+3056*0.364-m*0.364

1500-1112= m*(1-0.364)

m= 610 kg/hr

The production rate is less than in absence of recycle since some of K2CrO4 is lost along with filtrate.

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