Deacidification of biological oils can be accomplished using solvent extraction. 150 kg/h of a mixture of...

Deacidification of biological oils can be accomplished using solvent extraction. 150 kg/h of a mixture of 15 wt% oleic acid in cottonseed oil mixture is fed to an extractor, where the acid is extracted using propane. At the extraction temperature of 75 °C, propane is immiscible with cottonseed oil. The equilibrium distribution coefficient is 0.20 (mass fraction oleic acid in oil / mass fraction oleic acid in propane). 75 % of the oleic acid in the feed is removed. Determine the flow rate and composition of the extract (propane-rich stream).

Solutions

Expert Solution

Feed to the process contains = 150 kg/hr, Oleic acid= 150*15/100= 22.5 kg/hr

Cotton seed oil in the feed= 150-22.5=122.5 kg/hr

Oleic acid removed is 75%, oleic acid removed= 22.5*0.75= 16.875 kg/hr

Mass of oleic acid in oil= 22.5-16.875= 5.625

Since propane is immiscible with oil

Mass fraction of oleic acid in oil= 5.625/122.5 =0.046

Let P= flow rate of propane, mass fraction of oleic acid in propane= 16.875/(16.875+P)

Given distribution coefficient mass fraction of oil in oleic acid/ mass fraction of oleic acid in proane=0.2

0.046*(16.875+P)/16.875=0.2

0.046*(16.875+P)= 16.875*0.2

.046*P= 16.875*(0.2-0.046)= 16.875*0.154

P= 16.875*0.154/0.046 =56.5 Kg/hr

Flow rate of propane rich stream= 56.5+16.875= 73.375 kg/hr

Extract composition: Mass fraction of propane= 56/73.375= 0.76 and oleic acid =1-0.76=0.24

queen_honey_blossom answered 1 year ago

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