1-71 Chapter 1 Example 1.10-1. ---------------------------------------------------------------------------------- 5Blood is flowing through a hollow fiber based bioartificial organ at a flowrate (Q) of 8 cm3/min for each fiber. Surrounding the fibers in the shell space are living cells obtained from a pig’s liver. Each fiber has a diameter of 800 microns and a length of 30 cm. At time equal to zero, equal concentrations of a permeable (relative to the fiber wall) test solute and labeled albumin (impermeable) are injected into the blood entering the device. Exiting blood samples are collected over the next few seconds and analyzed for the concentrations of the test solute (C) and the reference solute, albumin (Co). At a particular time, the value of C/Co was found to equal 0.60. Estimate the permeability (Pm) of the hollow fiber membrane in cm/s. Solution ------------------------------------------------------------------------------------------ The clearance of a given hollow fiber for the test solute is given by CL = ooCQCQC− = Q(1 − C/Co) CL = 8 cm3/min×(1 − 0.6) = 3.2 cm3/min The permeability (Pm) of the hollow fiber membrane can be determine from 1 − oCC = 1 − exp−QSPm Pm = − QSlnoCC = − 38/60 cm /0.08 cm 30 cmsπ× ×ln(0.6) = 0.009 cm/s 1.11 Solute Transport in a Vascular Bed The Krogh tissue cylinder provides a model for solute transport within a single capillary. If the solute concentration Cin the tissue space can be assumed to be zero, the solute concentration C within the blood leaving a tissue space is given by the expression derived in previous section oCC = exp−QSPm (1.11-1) In this equation Co is the solute concentration entering the tissue region, Q is the volume flow rate of blood to the region, S is the total surface area for mass transfer between the capillaries and the tissue space, and Pm is the permeability of the capillary membrane. 5 Fournier, R. L., “Basic Transport Phenomena in Biomedical Engineering”, Taylor & Francis, 2007, p. 213.1-72 {}QinQoutCCCOCCapillaryTissue space Figure 1.11-1 An idealized capillary bed. If the solute concentration Cin the tissue space is not zero, equation (1.11-1) is not valid. However, the perfect mixing model can be used to model the solute transport in this situation. We can treat the capillary bed and the tissue space as separate well-mixed regions so that the solute balance over the tissue region can be written as QinoC = PmS(C − C) + QoutC (1.11-2) The term PmS(C − C) represents the amount of solute transfer to the tissue region that must be equal to the solute reacted within the tissue volume VT. For zero order reaction with reaction rate Ro PmS(C − C) = RoVT (1.11-3) Substituting RoVT into equation (1.11-2) and solving for C we obtain C = Co − QVRTo (1.11-4) The ratio TVQ is called the tissue blood perfusion rate and is denoted by qb qb = TVQ = LrnVrnTC22ππ = 2TCrrLV (1.11-5) The tissue region is usually characterized by two parameters: the surface area capillary density (s) and the volume of capillaries per unit volume of tissue (v).1-73 s = LrnLrnTC22ππ = 22TCrr, and v = LrnLrnTC22ππ = 2TCrr If the blood resident time τ is defined as LV, the tissue blood perfusion rate can be expressed as qb = τv. The solute concentration within the tissue space Ccan be solved from equation (1.11-3) C = C − SPVRmTo = Co − QVRTo− SPVRmTo C = Co − RoT TmV VQ P S +   = Co − Ro1Tb mVq P S +   (1.11-6) Example 1.11-1. ---------------------------------------------------------------------------------- 5Consider the transport of glucose from capillary blood to exercising muscle tissue. As a basis consider 1 gram of tissue. The glucose consumption of the tissue is 0.01 µmol/g⋅s. Blood flow to the region is 0.01 cm3/g⋅s. The arterial glucose concentration is 5 µmol/cm3. The value of PmS based on capillary recruitment during exercise is 0.0033 cm3/ g⋅s. Calculate the glucose concentration in the tissue space and in the exit blood. Solution ------------------------------------------------------------------------------------------ The glucose concentration in the blood exiting the tissue region is given by C = Co − QVRTo= Co − obRq C = 5 µmol/cm3 − 30.01 mol/g sec0.01 cm /g secµ⋅⋅ = 4 µmol/cm3 The glucose concentration in the tissue space is given by C = Co − RoT TmV VQ P S +   = Co − Ro1Tb mVq P S +   C = 5 µmol/cm3 − 0.01 µmol/g⋅s3 31 1 g0.01 cm /g s 0.0033 cm /g s + ⋅ ⋅  C = 5 µmol/cm3 − 4.03 µmol/cm3 = 0.97 µµµµmol/cm3 5 Fournier, R. L., “Basic Transport Phenomena in Biomedical Engineering”, Taylor & Francis, 2007, p.