Optimum albumin concentration of supplementation fluid for double filtration plasmapheresis.

Until recently, the albumin concentration of supplementation fluid for double filtration plasmapheresis (DFPP) has been empirically determined. Inadequate albumin infusion often leads to hypoproteinemic symptoms such as edema. In the current study, an aimed condensation coefficient (CC(aimed)) was introduced in an attempt to estimate the appropriate plasma albumin level for each patient. This coefficient is theoretically derived from a one-compartment model for the patient's plasma albumin: CC(aimed) = C(S)/C(D) = 1 - (1 - CR)/[ 1 - exp(- CC - VR)] where C(D) and C(S) are albumin concentrations in discarded plasma and supplementation fluid, CR is the change ratio of albumin concentration in the patient's plasma during a DFPP treatment, and VR( = V(S)/V(P)) is the ratio of supplementation fluid volume (V(S)) to the patient's total plasma volume (V(P)). And CC denotes the albumin condensation coefficient in a DFPP line, which depends on the filtration fraction of the plasma fractionator (FF(PF)) and the sieving coefficients of both the plasma separator (SC(PS)) and the plasma fractionator (SC(PF)): CC = C(D)/C(P) = SC(PS) . (1 - FF(PF) . SC(PF))/(1 - FF(PF)) where C(P) is the albumin concentration of the patient's plasma. From the above relations, C(S) can be determined as follows: C(S) = CC . CC(aimed) . C(P) Because many kinds of proteins are removed during a single DFPP treatment, a slightly higher albumin concentration in the supplementation fluid is needed to maintain an appropriate plasma level. Therefore, the CR value should be more than unity. For a patient with hematocrit (HCT) of 30%, body weight (BW) of 50 kg, and C(P) of 3.0 g/dl, who is receiving a DFPP treatment using AP-05H (SC(PS) of 0.970) and Evaflux 2A (SC(PF) of 0.526) under FF(PF) of 0.8 with V(S) of 500 ml, V(P) = BW(1 -HCT/100)/13 = 50 x (1 - 30/100)/13 = 2.69 L, VR = 500/(2.69 x 1,000) = 0.186, CC = 2.81, and CC(aimed) = 1.25 assuming 1.1 for CR. Therefore, C(S) = 2.81 x 1.25 x 3.0 = 10.5 g/dl using the above equations.