Potassium chromate (K2CrO4) is manufactured in a process in which 1.25 kg/s of a

Question

Potassium chromate (K2CrO4) is manufactured in a process in which 1.25 kg/s of a solution that is 33.3wt% K2CrO4 in the fresh feed stream is mixed with a recycle stream containing 36.36% K2CrO4, and the combined stream is fed to an evaporator. The concentrated stream that leaves the evaporator is 49.4% K2CrO4; this stream is fed to a crystallizer in which it is cooled (causing crystals of K2CrO4 to come out of solution) and then filtered. The filter cake consists of K2CrO4 crystals and a solution that contains 36.36 wt% K2CrO4. The crystals formed account for 95% of the mass of the total filter cake. The solution that passes through the filter, also 36.36 wt% K2CrO4, is the recycle stream. Calculate the weight of water removed in the evaporator, the rate of production of crystalline K2CrO4, the ratio (kg recycle)/ (kg fresh feed), and the feed rates that the evaporator and the crystallizer must be designed to handle.

Explanation / Answer

basis = 1 second of operation

take overall balance of potassium chromate

amount of potassium chromate in the system = (1.25*0.333) = 0.41625 kg

under steady state, potassium chromate in = potassium chromate out

so amount of potassium chromate in cake = 0.41625 Kg
this is 95% of cake mass, rest is water

so amount of water = (5/95)*0.41625 = 0.02191 kg

but amount of water fed in to the system = 1.25-0.41625 = 0.83375 kg
only 0.02191 kg leaves in cake
rest leaves from evaporator

mass of water leaving the evaporator = 0.83375-0.02191 = 0.81184 Kg

so,

rate of water leaving evaporator = 0.81184 kg/s

rate of production of pure potassium chromate = 0.41625 kg/s

alternatively, rate of production of cake = 0.43816 kg/s of which 95% is pure potassium chromate.

let flow rate of recycle stream be R, and the stream leaving the evaporator be E

then,

amount of potassium chromate in = (1.25*0.333)+(0.3636R)

amount of potassium chromate out = 0.494E

so, balancing mass of potassium chromate we will get

0.41625+0.3636R = 0.494E ------eqn(1)

amount of water in = (1.25*0.667)+(0.6364R)

amount of water out = 0.81184+0.506E

balance water

0.83375+0.6364R = 0.81184+0.506E ----------eqn(2)

solve eqn(1) and eqn(2)

E = 1.97

R = 1.532

so (Kg recycle/kg fresh feed) = 1.532/1.25 = 1.2256

the evaporator must be designed for (1.25+1.532) = 2.782 kg/s

cystallizer must be designed for 1.97 kg/s

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