A protein solution at a concentration of 0.5 g/liter and volume of 1000 liters m

Question

A protein solution at a concentration of 0.5 g/liter and volume of 1000 liters must be concentrated on a crossflow ultrafilter to a concentration of 10 g/liter. The ultrafilter has an area of 100 m^2 and operates at 5 degree C with an inlet feed pressure of 16.0 bar, outlet feed pressure of 14.0 bar, and back pressure on the permeate of 1.4 bar. The protein has a molecular weight of 20,000 Da and will not pass through the ultrafiltration membranes. The feed has a viscosity of 1.2 cp The membrane resistance (R_m) of the ultrafilter has been determined to be 2 times 10^13 cm^-1 The Reynolds number for flow within the ultrafilter is o large enough to render concentration polarization negligible. Determine the time required to perform the ultrafiltration.

Explanation / Answer

Time required can be found using the formula,

J=TMP / Rt

Where J=Flux (flow rate per membrane area) = Feed rate / Membrane area

Membrane area = 100 m2

TMP is the transmembrane pressure, TMP=(Inlet Presssure + Outlet Pressure) / (2 - Permeate Pressure)

TMP = (16-14) / (2-1.4) = 3.33 bar

= Viscosity = 1.2 cp

Rt = Resistance = 2 X 1013 cm-1

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