Solute Transport In Biological Systems Design of An Artificial Kidney Utilizing Urease in Polymeric Beads. Dialysate. Flow out C(z) U Q,C b UB

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1 hapte Solute Tanspot In Biological Systems.14 Design of n tificial Kidney Utilizg Uease Polymeic Beads Fo the teatment of uemia discussed section.11, fesh dialysate is used kidney dialysis to mata a concentation gadient acoss the membane that sepaates the blood fom the dialysate. s a esult, metabolic waste poducts diffuse fom the blood to the dialysis fluid that might be pocessed to elimate the waste poducts if it is gog to be eused. We will discuss a possible design of a system to educe the use of dialysate. In this design the atificial kidney acts as a eacto whee the dialysis fluid is mixed with the enzyme encapsulated solid polymeic beads. The toxic species diffuse though the membane to the dialysate becomg a pat of an agitated sluy of beads as shown Figue The enzyme/polyme bead sluy is used to keep the concentation of toxic mateials low sce any toxic species diffused to the beads will be metabolized to hamless poducts. Dialysate Hollow fibe Flow Flow out (z) U, b, b UM U Stie Figue.14-1 Schematic of solute diffusion and eaction a blood dialysis system. We will now focus ou attention on the emoval of uea that diffuses fom the blood to the dialyate. t steady state the concentation of uea the dialysate ( U ) is a constant sce the amount of uea diffuses fom the blood to the dialysis fluid will be equal to the amount diffuses to and then metabolized by the enzyme uease encapsulated solid polymeic beads. In ode to design this type of dialyze we need to specify the volume of the dialysis fluid, the bead volume faction, the amount of enzyme the eacto, and the bead size. We also need to obta expeimental data fo the ketics and mass tansfe fo this system. The polymeic beads will be spheical and will conta a unifom distibution of the enzyme uease. The beads will be contaed a well-mixed contae and suounded by the dialysis fluid with a unifom concentation of uea ( U ). The uea must dissolve the bead and -49

2 diffuse to the teio, whee eaction occus so that the concentation of uea the dialysate ( U ) can be mataed at a constant value. Makg a mole balance on a spheical bead at steady state we have Mola ate of uea eacted with the bead = Mola ate of uea ente the bead at the suface Mola ate of uea eacted with the bead = 4π ( D U du = ) Uea eacted with the bead Uea entes the bead Figue.14- onvesion of uea by uease with the bead. Hence we need to fd the concentation pofile U () of uea with the bead. If we assume that the ketic of the convesion of uea is fist ode, we can easily solve the diffeential species balance equation to obta U (). In the next section we will use a shell balance to obta a second ode diffeential equation descibg the diffusion and eaction of uea with a spheical bead Diffusion with Homogeneous Fist Ode eaction + Figue.14-3 Illustation of a spheical shell 4π We will conside the diffusion of to a spheical polymeic bead whee homogeneous chemical eaction occus. The one-dimensional mola flux of is given by the equation -50

3 " d N = D (.14-1) pplyg a mole balance on the spheical shell shown Figue.14-1 yields fo steady state 4π N " 4π " N + + 4π = 0 Dividg the equation by the contol volume (4π ) and takg the limit as 0, we obta 1 d ( " N ) + = 0 (.14-) Fo a fist ode eaction, = k and substitutg the mola flux fom equation (.14-1) to the above equation, we have 1 d D d k = 0 1 D d d k = 0 (.14-3) In this equation, D and k ae constants dependent of. We want to tansfom this equation to the fom d y α y = 0 (.14-4) whee the solution to the homogenous ODE has two foms 1) y = 1 e -α + e α ) y = B 1 sh(α) + B cosh(α) The fist exponential fom (1) is moe convenient if the doma of is fite: 0 while the second fom usg hypebolic functions () is moe convenient if the doma of is fite: 0. The constants of tegation 1,, B 1, and B ae to be detemed fom the two bounday conditions. Let α k =, we can tansfom equation (.14-3) to the fom D of equation (.14-4) by the followg algebaic manipulations 1 d d α = 0-51

4 1 d d + α = 0 d d + α = 0 d Sce becomes d ( ) d = d d + = d + d +, the above equation d d ( ) α = 0 Let y =, the equation has the same fom as equation (.14-4) with the solution y = B 1 sh(α) + B cosh(α) o = B 1 sh(α) + B cosh(α), whee α = k D The two constants of tegation B 1 and B can be obtaed fom the bounday conditions t = 0, = fite o d = 0 t =, = (a known value) pplyg the bounday at = 0 yields 0 = B pplyg the bounday at = yields = B 1 sh(α) B 1 = sh( α) Theefoe the concentation pofile fo species with the spheical bead is = sh( α) sh( α) (.14-5) t the cente of the bead, the concentation is given by -5

5 α ( = 0) = sh( α).14- Paametes Specification fo an tificial Kidney We want to design a hemodialyze that will maximize the utilization of the enzyme and pemit the convesion of uea to poceed at a ate that is not limited by mass tansfe. We encapsulate the enzyme spheical beads fo which the diamete must be specified. If the total amount of enzyme is fixed and if the concentation of enzyme each bead dependent of the bead adius then the total bead volume is a constant M = N 3 4 π 3 = constant (.14-6) In this equation N is the numbe of beads and is the bead adius. When vaies, N will change. Howeve the total bead volume M emas a constant. The total mola ate of uea metabolized with the beads is given by Total ate of uea eacted with the beads = 4π N ( D U du = ) The uea concentation with the bead is given by equation (.14-5) ewitten with the subscipt U denotg uea U = sh( α) sh( α) (.14-5) d U = sh( α) 1 α sh( α ) + cosh( α) du = = sh( α) α 1 cosh( α) sh( α) du = = [ ( α ) coth( α) 1) ] 1/ k Let φ = α = = Thiele modulus fo a fist ode eaction. Ignog the mus sign, DU the total mola ate of convesion of uea is then Total ate = 4π N D U (φ cothφ - 1) -53

6 Fom equation (.14-6), 4π N = 3M/, theefoe Total ate = 3M D U (φ cothφ - 1) We now defe an effectiveness facto to compae the effect of diffusion with that of eaction. η = actual ideal convestion convesion ate ate = 3M DU ( / )( φ cothφ 1) Mk η = 3( φ cothφ 1) k D U 3 = (φ cothφ - 1) (.14-7) φ The ideal convesion ate is the maximum convesion ate that can be achieved when the mass tansfe ate is much lage than the eaction ate. In this condition, the concentation of uea eveywhee with the bead is the same as the concentation at the suface ( ). Theefoe the total convesion ate is simply the total volume (M) time the convesion ate pe unit volume (k ). The effectiveness facto η is toduced sce it is moe convenient to cast mathematical models dimensionless fomat to dicate the mimum numbe of dimensionless goups that affect the physics. This knowledge is useful guidg the design of expeiments and the coelation of data. The Thiele modulus is essentially the atio of eaction to diffusion ates. We should choose a value of so that the effectiveness is close to one. This means that the eaction ate is not limited by the diffusion of uea to the beads. The followg Matlab statements ae used to plot equation (.14-7). 3 Matlab pogam to plot η = φ % Effectiveness as a function of Thiele modulus % phi1=0.1:.1:1; phi=:10; phi3=0:10:100; phi=[phi1 phi phi3]; ena=3*(phi.*coth(phi)-1)./phi.^; loglog(phi,ena) gid on xlabel('thiele Modulus');ylabel('Effectiveness Facto') (φ cothφ - 1) Figue.14-4 plots the effectiveness facto η vesus the Thiele modulus φ. We see that η has almost eaches its maximum value as soon as φ is less than unity. Thee is no need to educe any futhe once φ = 1. The bead adius is then detemed -54

7 10 0 Effectiveness Facto Thiele Modulus Figue.14-4 The effectiveness facto fo fist ode eaction beads. 1/ 1/ k D φ = 1 = design = U DU k This is the design design we should use. Fo any > design, the convesion of uea will be limited by diffusion, and fo < design, the cost will be highe without an cease oveall eaction ate. We still need to know the values of the ketic eaction coefficient k and the diffusivity D U fo the detemation of design and the calculation of the uea concentation leavg the atificial kidney. If we cay out expeiments with beads smalle than design we will fd a convesion ate that is dependent of and dependent of D U. Fo < design, η 1, Total convesion ate = ηmk = Mk If we change the concentation of enzyme but keep M constant though expeiments usg beads smalle than design, the total convesion ate should be a lea function of. Theefoe k can be obtaed fom the slope of the le of convesion ate vesus. Once the ate constant k has been detemed, we may fd D U by pefomg expeiments with lage beads such that φ >

8 η = total ate Mk 3 = φ (φ cothφ - 1) 3 φ (φ - 1) 3 3 = φ D U k 1/ ll paametes this equation ae known, once k has been established, so a measuement of the total ate will yield a value fo the diffusivity D U. In the design of the atificial kidney shown Figue.14-1, we assume that thee is no esistance to mass tansfe of uea to the beads, i.e., = α b U, whee α b is the solubility of uea the bead. The stig ate of the eacto must be high enough so that the uea concentation with the dialysis fluid can be assumed to be unifom. Dialysate Hollow fibe Flow Flow out (z) U, b, b UM U Stie Figue.14-1 Schematic of solute diffusion and eaction a blood dialysis system. The next step that we need to esolve is to fd a elationship between the uea concentation UM exitg the hemodialyze to the uea concentation enteg the device and the uea concentation the dialysate U Mass Tansfe fom Inside a ylical Tube to the Shell Side U out U UM f f z dz Figue.14-5 Schematic of a sgle hollow fibe a blood dialysis system -56

9 We model the membane sepaation system to consist of N f numbe of paallel hollow fibes each of axial length L, side adius, and outside adius out. The blood flow though each of the hollow fibe is f that is elated to the blood flow though the atificial kidney b by the expession b = N f f Makg an uea balance ove the contol volume π dz on a sgle hollow fibe yields f U f z U z dz + N " π dz = 0 (.14-8) U In this equation N " is the mola flux of uea enteg the hollow fibe wall (membane). U The mola flux of uea at any position with the membane is given by " d N U = D Um Um Theefoe, the mola tansfe ate fo steady state is N U = N " U πdz = D Um πdz d Um = N " π dz (.14-9) U Blood U Membane U, Dialysate Blood Membane Dialysate U U,out out Figue.14-6 Depiction of the adial concentation pofile. Sepaatg the vaiables and pefomg the tegation yields out N U = DUm πdz U, out d Um U, -57

10 N U ln out = D Um πdz( U, U,out ) In this equation, U, and U,out ae the concentations of uea with the membane at and out espectively. Let κ = and solve fo N U out N U = D Um πdz( U, U,out )/lnκ (.14-10) The uea concentations the fluid and the membane at the teface ae elated by a distibution coefficient α m whee α m = U, = U, out U U Equation (.14-10) becomes N U = α m D Um πdz( U U )/lnκ We assume that the uea concentation the dialysate U is much less than that the blood ( U << U ) N U = N " U π dz α m D Um πdz U /lnκ Substitutg the mola ate of uea leavg the contol volume π dz shown Figue.14-5 to equation (.14-8) we have f U f z U z dz + + α m D Um πdz U /lnκ = 0 Dividg the equation by f dz and the limit as dz 0, we obta d U πα = mdu, m U = β U (.14-11) dz lnκ f whee β is a positive quantity defed by β = tegated fom 0 to the axial position z U = exp( βz) πα D m U, m f lnκ. Equation (.14-11) can be t z = L, U (L) = UM = exp( βl) (.14-1).14-4 aiation of Uea oncentation with Blood dug Dialysis -58

11 Body B b U Membane UM Figue.14-7 Schematic of a blood dialysis system. Figue.14-7 shows the schematic of a blood dialysis system, whee, U, and UM ae, espectively, the concentations of uea the body, the esevoi, and the blood leavg the membane unit. UM is elated to by the equation (.14-1) deived ealie UM = exp( βl) (.14-1) We assume that the uea concentation the blood with the body is unifom so that d ( B ) = b ( UM ) dt The cadiac output is about 5.6 lite/m theefoe it takes about 1 mute fo the blood to ciculate though the body and the assumption of unifom concentation of uea with the body blood is easonable. B b d = UM = exp( βl) dt Let τ B = B b = a time constant fo the system, the above equation becomes τ B d = [exp( βl) 1] dt This equation can be tegated fom the itial uea concentation with the blood o τ B o d = [exp( βl) 1] t dt 0 τ B ln o = λ t, whee λ = 1 exp( βl) λ' Theefoe = o exp t τ B (.14-13) -59

12 We now need to veify the assumption that U << UM. If we choose the system to be the dialysate with the eacto, then fo steady state the amount of uea diffuses fom the blood to the dialysis fluid though the hollow fibe wall is equal to that exits the fluid and entes the beads. The mass tansfe of uea is illustated Figue.14-1 Dialysate Hollow fibe Flow Flow out (z) U, b, b UM U Stie Figue.14-1 Schematic of solute diffusion and eaction a blood dialysis system. With the beads, the convesion of uea is given by ηmk = Mk sce η 1 fo a choice of < design. The convesion of uea with the bead must be equal to the amount of uea leavg the blood steam 4 Mk = b ( UM ) (.14-14) The solubility of uea α b with the bead is defed as α b = U U U = α b U Let f be the volume faction of beads the esevoi that contas the dialysate so that M = f. f should not be geate than 0. to pomote good mixg of the bead sluy that povides negligible mass tansfe esistance to the bead. In tems of U and, equation (.14-14) becomes kα b U f = b ( UM ) Fom equation (.14-1) UM = exp( βl) (.14-1) 4 Middleman S., n Intoduction to Mass and Heat Tansfe, Wiley, 1998, p

13 Theefoe kα b U f = b [ exp( βl)] o U = b[ 1 exp( β L)] kα f b (.14-15) Sce this is a design poblem we might have the feedom to select some design and opeatg vaiables to satisfy the assumption that U << UM o U <<. Let pick an U abitay value = 0.05, then the constat fo the dialysis unit is b[ 1 exp( β L)] kα f b = 0.05 (.14-16) Let op the subscipt b b so that equation (.14-16) becomes [ 1 exp( β L)] kα f b = 0.05 (.14-16) We now need to deteme the design paamete and the opeatg paamete. Equation (.14-16) povides a constat on these paametes to satisfy the assumption that U << UM. Suppose that dialysis unit should educe the uea concentation the blood by a half afte two hous of opeation given the followg paametes α m D U,m = 10-5 cm /s, out = 50 µm, = 50 µm = πn f L = 500 cm, k = 10-3 s -1, α b = 1, B = 5 L, κ = / out = 0.5 Sce o = 0.5 Fom equation (.14-13) We have = o exp λ' τ B t, whee λ = 1 exp( βl) (.14-13) o βl t = exp ( ) 1 e = 0.5 (.14-17) τ B whee τ B = B / = 5000cm 3, = N f f -61

14 βl = πlα D m U, m f lnκ = πln f α mdu, lnκ m α mdu, m = lnκ βl = 5 (500)(10 ) 4 (50 10 ) ln(0.5) = Fom equation (.14-17) Substitutg βl = βl ln() = = ( 1 e ) t τ and τ B = B / = B 5000cm 3 at t = 700 s, we have = exp (.14-18) Equation (.14-18) can be solved fo usg Newton method whee a function of is defed as f() = exp 7 ln() (.14-19) 50 Takg deivative of the function with espect to yields f () = exp exp is then given by the followg fomula fom the pevious guess value i = i f( i )/f ( i ) The followg table lists the Matlab pogam to solve fo usg an itial guessed value of 1 cm 3 /s. Table % dialysis.m % Blood Dialysis % =1; fo i=1:0 f=(1-exp(-1.443/))*7*/50-log(); fp=(-1.443*exp(-1.443/)/+(1-exp(-1.443/)))*7/50; e=f/fp; -6

15 =-e; if abs(e)<.0001, beak, end end h=*3.6; fptf('(lites/h) = %8.f, (cm3/s) = %8.f\n',h,) >> dialysis (lites/h) = 1.84, (cm3/s) = 0.51 Sce blood can be withawn fom the body at a maximum ate of 400 cm 3 /m o 6.7 cm 3 /s, the blood flow ate of 0.51 cm 3 /s is acceptable. Equation (.14-16) is then used to solve fo. βl ( 1 e ) f kα b = 0.05 βl = = =.8 1 e βl ( ) f = 0.05 kαb = ( 1 exp(.8)) (10 )(1) f = cm = 161 lites If the volume faction of the bead side the eacto is 0., the eacto volume equied is = 800 lites This volume is quite lage so we should exploe othe design and opeatg paametes that will esult a much smalle eacto. We could cease the aea fo mass tansfe, the eaction ate constant k, and/o the blood flow ate. These actions will cease the ate of mass tansfe fom the blood to the dialysis fluid so that the eacto volume equied should be smalle. -63

16 Example Figue.13-1 shows the schematic of a blood dialysis system, whee, U, and UM ae, espectively, the concentations of uea the body, the esevoi, and the blood leavg the membane unit. The opeatg paamete and the design paamete ae fixed by the followg two conditions [ 1 exp( β L)] kα f b = 0.05 and o βl t = exp ( ) 1 e = 0.5 τ B Othe paametes ae given as α m D U,m = 10-5 cm /s, out = 50 µm, = 50 µm, f = 0. = πn f L = 500 cm, k = 10-3 s -1, α b = 1, B = 5 L, κ = / out = 0.5 1) Pepae plots of and as a function of total aea fo the ange [100, 1000] cm. ) Pepae plots of and as a function of eaction constant k the ange [0.1, 0.5] s -1, fo = 500 cm. Body B U Membane UM Figue.13-1 Schematic of a blood dialysis system. Solution We can solve he poblem by simply pluggg numbes to the appopiate equations and calculatg answes. Howeve the goal of mathematical modelg is to study paametic sensitivity, the manne which impotant featues of a system depend on the design and opeatg paametes of the system. 1) Pepae plots of and as a function of total aea fo the ange [100, 1000] cm. Fom the fist constat of the system [ 1 exp( β L)] = 0.05 kα f b -64

17 [1 exp( βl)] = 0.05kα b f [1 exp( βl)] = 0.05(10-3 )(1)(0.) = 10-5 Fom the second constat of the system o βl t = exp ( ) 1 e = 0.5, sce τ B = B τ B [1 exp( βl)] We can solve fo diectly 700 = = = 48,15 cm Hence thee is only one eacto volume that will satisfy the two constats egadless of the choice of. On the othe hand, will vay with sce βl = πlα D m U, m f lnκ = πln f α mdu, lnκ m α mdu, m = lnκ and [1 exp( βl)] = 10-5 = βl = ln 0.5 = Theefoe [1 exp( βl)] = [1 exp( )] = (.14-0) t a given value of we might solve the above equation fo. Sce the equation is nonlea, it might not possess a solution until the values of the paametes and ae with a ceta ange. Fo βl << 1, 1 exp( βl) βl [1 exp( βl)] = βl = = = 168 cm Theefoe must be geate than 168 cm befoe equation (.14-0) will have a solution o the constats can be satisfied. Fo βl >> 1, 1 exp( βl) 1 = cm 3 /s -65

18 Hence we cannot use a below this value and satisfy the constats on the design. Equation (.14-0) can be solved fo at a given value usg Newton method whee a function of is defed as f() = 1 exp (.14-1) Takg deivative of the function with espect to yields f () = exp exp is then given by the followg fomula fom the pevious guessed value i = i f( i )/f ( i ) Figue.14-8 shows the plot of vesus. Fom the esults, it can be seen that a membane aea highe than 500 cm will have little effect on the opeatg paamete. Table.14- lists the Matlab pogam to solve fo with the ange [180, 1000] (cm3/s) (cm) Figue.14-8 Blood flow ate at vaious membane aeas. -66

19 Table % Blood Dialysis % =1; =300:100:1000;=[ ]; na=length(); p=; fo k=1:na =(k);=0.009*; fo i=1:0 f=*(1-exp(-/)) ; fp=(1-exp(-/))-*exp(-/)/; e=f/fp; =-e; if abs(e)<.0001, beak, end end fptf('# of iteation = %g, (cm3/s) = %8.f\n',i,) p(k)=; =/; end plot(,p,'-d') gid on xlabel('_i_n(cm)');ylabel('(cm3/s)') >> dialysis # of iteations = 6,(cm) = , (cm3/s) = 3.17 # of iteations = 4,(cm) = , (cm3/s) = 1.99 # of iteations = 5,(cm) = 00.00, (cm3/s) = 1.50 # of iteations = 4,(cm) = 30.00, (cm3/s) = 0.96 # of iteations = 4,(cm) = 60.00, (cm3/s) = 0.77 # of iteations = 4,(cm) = , (cm3/s) = 0.65 # of iteations = 4,(cm) = , (cm3/s) = 0.55 # of iteations = 4,(cm) = , (cm3/s) = 0.51 # of iteations = 3,(cm) = , (cm3/s) = 0.50 # of iteations = 3,(cm) = , (cm3/s) = 0.49 # of iteations = 3,(cm) = , (cm3/s) = 0.49 # of iteations = 3,(cm) = , (cm3/s) = 0.48 # of iteations = 3,(cm) = , (cm3/s) = 0.48 ) Pepae plots of and as a function of eaction constant k the ange [0.1, 0.5] s -1, fo = 500 cm. We assume that thee ae some ways to poduce enzyme/beads of high eactivity and to select a bead adius small enough so that diffusion esistance is negligible (φ < 1) the beads. -67

20 Fom the fist constat of the system [ 1 exp( β L)] kα f b = [1 exp( βl)] = 0.05kα b f [1 exp( βl)] = 0.05(k)(1)(0.) = 10 - k Fom the second constat of the system o βl t = exp ( ) 1 e = 0.5, sce τ B = B τ B [1 exp( βl)] We can solve fo k diectly 700 = k 700 = k = cm 3 /s 5000 Fo k = 0.1 s -1, = 48.15/0.1 = cm 3 /s The lage value of k pemits the use of a vey small eacto, as expected. can then be solved fom the fist constat [1 exp( βl)] = 10 - k = βl = πlα D m U, m f lnκ = πln f α mdu, lnκ m α mdu, m = lnκ βl = ln 0.5 = Theefoe exp = This is the case whee = 500 cm fo equation (.14-1) f() = 1 exp (.14-1) The solution is given Table.14- whee = 0.51 cm 3 /s dependent of k. -68