1-51 Chapter 1 Example 1.6-3. ---------------------------------------------------------------------------------- 5A single hollow fiber is placed within a very large glass tube. The hollow fiber is 20 cm in length and has a diameter of 400 microns. The flow rate of a liquid containing a permeable solute through the hollow fiber is 1 ml/min. It is found that the concentration of the permeable solute exiting the hollow fiber is 10% of the concentration of this solute when entering the hollow fiber. Estimate the permeability of the hollow fiber membrane for this solute. Solution ------------------------------------------------------------------------------------------ ∆zzrfC=0CHollow fiber We will consider a simplified model for solute transport in the hollow fiber. We assume that the flow rate in the shell space is much higher than that of the fiber so that the solute concentration in the shell space is zero and the resistance to the solute within the fiber is negligible so that Ko ≈ Pm. The steady state solute balance on the control volume πrC2∆z gives Vπrf2zC − Vπrf2zzC∆+ = 2πrC∆zKo (C − 0) (E-1) where Ko is the overall mass transfer coefficient between the solute in the hollow fiber and the solute in the shell space next to the hollow fiber wall, and C is the solute concentration in the fiber. The solute concentration in the surrounding shell space is zero. The overall mass transfer coefficient is related to the film mass transfer coefficient and the permeability by the expression oK1 = mk1 + mP1 5 Fournier, R. L., “Basic Transport Phenomena in Biomedical Engineering”, Taylor & Francis, 2007, p. 220.1-52 where oK1, mk1, and mP1are the total mass transfer resistance of solute from the fiber to the outside surface of the fiber, the mass transfer resistance in the fiber side, and the mass transfer resistance through the fiber wall, respectively. Since mk1 << mP1 oK1 ≈ mP1 ⇒ Ko ≈ Pm Dividing equation (E-1) by πrf2∆z and taking the limit as ∆z → 0, we obtain − VdzdC = 2fr PmC (E-2) dzdC = − 22ffrV rππPmC = − 2frQπPmC (E-3) We have replaced the liquid velocity V with the volumetric flow rate Q. Equation (E-3) can be integrated ∫)(zCCOCdC = − 2frQπPm∫zdz0 ⇒ lnoCzC )(= − 2frQπPmz At the exit of the hollow fiber, z = 20 cm and oCzC )( = 0.1. Therefore Pm = − ln(0.1)2 (20)fQrπ = 42.303 (1/ 60)2 (200 10 )(20)π−×× = 0.0153 cm/s 1.7 Solute Permeability MembraneBulk solutionBulk solutionChighClowNSx=0x=tm Figure 4.5b-1 Solute transport across a membrane.1-53 Fick’s law for the diffusion of solute across a membrane is represented by JS = − DmAPτ1dxdC = − DAPτωKdxdC (1.7-1) This equation is integrated across the membrane to obtain JS = DAPτωKmlowhightCC− (1.7-2) The solute transport can also be expressed in terms of the permeability Pm of the membrane JS = PmS(Chigh − Clow) (4.5b-2) Comparing equation (4.5b-1) with equation (4.5b-2) we have a relation between permeability Pm and diffusivity D Pm = mtDSApτωK In some cases, it is difficult to estimate the surface areas AP of the pores or the capillaries involved. Equation (4.5b-2) is then more convenient to use since you can work with the product PmS, rather than the specific values of Pm or S. The following correlation may be used to estimate capillary PmS values for a given solute of radius a PmS = 0.0184a-1.223 a < 1 nm PmS = 0.0287a-2.92 a > 1 nm In these expressions, the unit of a is nm and the units of PmS are cm3/sec/100g of tissue. 1.8 Solute Transport between Capillary and Tissue Space The molecules required for tissues are carried in the blood vessels to the capillaries where they diffuse through the capillary wall to the tissue space. To develop the model for solute transport from the capillary to the surrounding tissues, we will use a shell balance and the Krogh tissue cylinder model. The Krogh tissue cylinder model is a simplified model of the tissue surrounding the capillary. It assumes a cylindrical layer of tissue surrounding each capillary with the solute transferred only from that capillary. The capillary is assumed to be cylindrical and of constant radius. The Krogh tissue cylinder is shown graphically in Figure 1.8-1. Tissue cylinderCapillary Figure 1.8-1 The Krogh tissue cylinder1-54 As the solute move moves along the capillary, its concentration C decreases because of solute transport through the capillary wall. We can make a solute balance on the control volume πrC2∆z shown in Figure 1.8-2 assuming that the blood flows through the capillary with an average velocity V. ∆z∆rr∆zzrcr +tc mCapillary wallControl volumein the tissue space = 2 r r zπ ∆ ∆ Figure 1.8-2 Control volumes for the capillary and the tissue space. The steady state solute balance on the control volume πrC2∆z gives VπrC2zC − VπrC2zzC∆+ = 2πrC∆zKo (C − mctrC+) (1.8-1) where Ko is the overall mass transfer coefficient between the solute in the capillary and the solute in the tissue space next to the capillary wall, C is the solute concentration in the capillary, and Cis the solute concentration in the surrounding tissue space. Dividing equation (1.8-1) by πrC2∆z and taking the limit as ∆z → 0, we obtain − VdzdC = Cr2Ko (C − mctrC+) (1.8-2) There are two dependent variables C and mctrC+in this equation. Therefore, we need more information before we can solve for C(z). Making a steady state shell balance around control volume 2πr∆z∆r shown in Figure 1.8-2 for solute in the tissue space, we have − DT(2πr∆z)rdrCd + DT(2πr∆z)rrdrCd∆+= R(C)2πr∆r∆z In this equation, R(C) is the consumption rate per unit volume of solute in the tissue space. We will assume that the reaction rate is zero-order in the solute concentration, R(C) = Ro =1-55 constant. Dividing the equation by the control volume 2πr∆z∆r and taking the limit as the control volume approaches zero, we obtain DTr1drddrCdr = Ro (1.8-3) With the assumption of zero-order reaction, the solute concentration C(z) along the axial position can be obtained from the balance over part of the capillary from the entrance to z rTtissue spacecapillaryzCoC Figure 1.8-3 Solute leaving in the first part of the capillary. VπrC2Co − VπrC2 C(z) =