= + (3) xFnF xRnR yPnP where and are the molar flow rates of the feed, retentate, and permeate nF, nR, nP streams, respectively, and and are the feed, retentate, and permeate O2 mole xF, xR, xP fraction, respectively. The molar flux of oxygen through a differential area in the dA membrane is given by equation (1) or by = = (xP = NO2 yp) Q’O2(xP yp) Therefore = (4) d(xn) yp)dA = d(yn) Q’O2(xP The above equation is just the O2 species balance around the differential volume element in the membrane. The reduction in the O2 molar flow rate of the retentate stream d(xn) provides the same O2 molar flow rate through the membrane. and are the d(yn) P p average retentate and permeate side pressures, respectively. Similar species balance for nitrogen around the differential volume element in the membrane yields = [(1x)P (1y)p)]dA (5) d[(1x)n] Q’N2 Dividing equation (4) by equation (5), we obtain = (6) The ratio is just the molar flow rate of oxygen over that of nitrogen in the permeate stream, therefore it is equal to the ratio of the mole fraction of oxygen over that of nitrogen as shown schematically in Figure 4. Let * = , equation (6) becomes = * (7) Figure 4. Molar flow rate ratio is equal to mole fraction ratio. The separation factor * is assumed to be constant. The permeate composition at the capped end of the hollow fibers is obtained from equation (7) by replacing with and y yi x with xR. = * (7) When the change in feed mole fraction of oxygen is less than 50%, the driving force for diffusion across the membrane, = is assumed to be a linear function of the change xP yp, in the molar flow on the feed side of the membrane = (8) d(xn) d
xFnF = xRnR + yPnP (3) where nF, nR, and nP are the molar flow rates of the feed, retentate, and permeate streams, respectively, and xF, xR, and xP are the feed, retentate, and permeate O2 mole fraction, respectively. The molar flux of oxygen through a differential area dA in the membrane is given by equation (1) or by NO2 = = (xP yp) = Q’O2(xP yp) Therefore d(xn) = Q’O2(xP yp)dA = d(yn) (4) The above equation is just the O2 species balance around the differential volume element in the membrane. The reduction in the O2 molar flow rate d(xn) of the retentate stream provides the same O2 molar flow rate d(yn) through the membrane. P and p are the average retentate and permeate side pressures, respectively. Similar species balance for nitrogen around the differential volume element in the membrane yields d[(1x)n] = Q’N2 [(1x)P (1y)p)]dA (5) Dividing equation (4) by equation (5), we obtain = (6) The ratio is just the molar flow rate of oxygen over that of nitrogen in the permeate stream, therefore it is equal to the ratio of the mole fraction of oxygen over that of nitrogen as shown schematically in Figure 4. Let * = , equation (6) becomes = * (7) Figure 4. Molar flow rate ratio is equal to mole fraction ratio. The separation factor * is assumed to be constant. The permeate composition at the capped end of the hollow fibers is obtained from equation (7) by replacing y with yi and x with xR. = * (7) When the change in feed mole fraction of oxygen is less than 50%, the driving force for diffusion across the membrane, = xP yp, is assumed to be a linear function of the change in the molar flow on the feed side of the membrane d(xn) = d (8)
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