Question

A hollow fiber membrane separator with a nominal molecular weight cutoff of 100,000 is fed a solution of proteins at the rate of 250 mL min-1 . The composition of the protein solution is protein A (4 g/L, MW=20,000), protein B (7 g/L, MW=150,000), and protein C (6 g/L, MW=300,000). The filtrate flow rate is found to be 116.4 mL min-1 and the flow rate of the retentate is 133.5 mL min-1 . The concentration of protein A in the retentate is found to be 5.84 g/L. What is the concentration of protein A in the filtrate? Also, what are the concentrations of proteins B and C in the filtrate and in the retentate?

Answer 1

protein fed rate - 250ml/min

filterate flow rate= 116.4ml/min

retentate flow rate= 133.5ml/min

compositon of protein solution is

protein A (MW 20000) = 4g/l

protein B (MW 150000)=7g/l

Protein C(MW 300000) = 6g/l

protein A in retentate= 5.84g/l

using the material balance for protein A

total protein in feed= sum of total protein in filtrate and retentate

i.e. concentration of Protein A in feed x feed flow rate = ( concentration of protein A in filterate x filtrate flow rate)+(concentration of protein A in retentate x retentate flow rate) -------(i)

therefore putting all the values regarding protein A in equation (i)

4x 250 = (concentration of protein A in filterate x116.4 )+(5.84 x 133.5)

concentation of protein A in filterate = 1.89g/L

However, protein B and C concentration in filtrate will be zero as the nominal molecular cutoff of the membrane is 100000 which means membrane will allow only those protein to pass through the membrane which has the maximum molecular weight of 100KDa. However, here Protein B and C has a molecular weight higher than 100KDa so it will not pass through the membrane and entire protein will be present in retentate

so using material balance

concentration of Protein B in filtrate= 0 g/l

concentration of Protein B in retentate= 13.109g/l ( use value of protein B in equation (i) )

similarly

concentration of Protein C in filtrate= 0g/l

concentration of Protein B in retentate= 11.236g/l