Transport Phenomena in Biological Systems 2nd Edition

227 !!!"$$$%+!!!"$$$%+!!!"$$$%=!!!"$$$%system within theProduction System on the Done Surfaces System AcrossTransfer system within the onAccumulati Using a rectangular control volume and the definition of the system energy per unit mass (Equation (17.2.3)) and energy flux (Equation (17.2.4)) !x!y!z"#E#t=exx$exx+!x( )!y!z+eyy$eyy+!y( )!x!z+ezz$ezz+!z( )!x!y+!Wt+!Qp*( )!x!y!z Dividing by the volume element xyz, taking

Solution Manual for Transport Phenomena in Biological Systems George A. Truskey, Fan Yuan and David F. Katz 2 Solution to Problems in Chapter 1, Section 1.10 1.1. The relative importance of convection and diffusion is evaluated by Peclet number, Pe =vL Dij (S1.1.1) (a)

21 Figure S2.20.1 After the enzyme is added the apparent viscosity decreases and is much less sensitive to shear rate, as determined by the following regression of data (Figure S2.20.2) 0225.0 01.0 = app . The enzyme functions by clipping the hyaluronic acid chains decreasing their length. As a result the

41 z z=0= v zz=0 = 10.003294 2 (S3.12.11) To determine the value of for which the shear stresses at r = 10 and 20 cm differ by 10%, the ratio of rz|z=0 is rewritten as 0.90 =10.003294 1(20)2(0.05236)20.01210.003294 1(10)2(0.05236)20.012=139.613212.47582 (S3.12.12) 0.90-2.2282 = 1 39.613 2 (S3.12.13) 37.385 2 = 0.10 Solving for yields,

55 So the Reynolds number at position 1 is 5,000 and the Reynolds number at position 2 is 6,244. The conclusion still holds. Since the flow is turbulent, the wall shear stress is w= dvz dr = vmax ara1 Ra r=R = vmax a R Inserting values Position 1, w = -0.0286(80)5/1.25 = -9.152 dyne/cm2

161 Solution to Problems in Chapter 12, Section 12.7 12.1. From the data in Table 12.2, CD = 4.255 and CM =0.77. From equations (12.3.14a,b), the drag force and torque are: FD = wApCD T= wApCMhc The exact shape is not specified. Assuming that the shape is roughly similar

181 g1 = x(k)^2; g2 = x(k)*(1+1/log(x(k))); gg1 = g2 - g1*2/dxB; gg2 = (g2 + g1*2/dxT)/s; AA(k,k+1) = 2*g1/(dxB^2)/gg1; AA(k,k-1) = -2*g1/(dxT^2)/gg2; AA(k,k) = (-2*g1/(dxB^2) - hB)/gg1 ... -(-2*g1/(dxT^2) - hT)/gg2; g1 = x(k+n)^2; AA(k+n,k+n) = -2*g1/(dxB^2) - hB; AA(k+n,k+n-1) = 2*g1/(dxB^2); C2 = inv(AA)*BB; C1 = ones((k+n),1);

201 Figure S15.4. 15.8. The governing equation and the boundary conditions are identical to those in Problem 15.7, i.e., = r C r rr 1 D t C eff (S15.8.1) )K/CC(P r C DAVpeff = , at r = a (S15.8.2) 0 r C Deff = , at r = R (S15.8.3) Here we assume that the concentration of the drug in plasma is negligible

214 and is the tissue mass density, which is approximately equal to 1 g/ml. Note that Ct is defined as the number of moles per unit tissue volume whereas Ca and CaG are defined as the number of moles per unit volume of the interstitial space. The volume fraction of

Solution to Problems in Chapter 5, Section 5.11 5.1. (a) From Figure 5.11 one can write an expression for the variables that influence the pressure drop: p =f(, , <v>, a, R, L) There are seven dimensional groups and three dimensions (mass, length and time). Thus, there are, at most,

88 Since is not a function of time, the time derivative is: ' = f =1 f (6.17.4) Equating Equations (6.17.3) and (6.17.4) 1f=12f2 (6.17.5a) f=2f2 (6.17.5b) Equating Equations (6.17.3) and (6.17.4) 1f=12f2 (6.17.5a) f=2f2 (6.17.5b) Equating Equations (6.17.3) and (6.17.4) 1 f =1 2f 2 (6.17.5a) f =2f 2 (6.17.5b) 6.18. (a) The

100 7.7. The short contact time solutions for concentration, Equation (7.6.26) and flux, Equation (7.5.28), are valid as long as zDij/vmaxR2 <0.01. For oxygen Dij = 1.10 x 10-5 cm2 s-1. From data in Table 2.4 for the canine cardiovascular system, the radii of capillaries and venules are 0.0003

111 Solution to Problems in Chapter 8, Section 8.6 8.1. The total volume of a 70-kg human body is approximately 70 liters. Therefore, the volume fraction of the vascular compartment is 6/70 or 8.6%. 8.2. The structure of the material is shown in Figure

131 dCS dt =k1CECS+k1CES (S10.4.1) dCES dt =k1CECS(k1+k2)CES +k2CECP (S10.4.2) dCP dt =k2CES k2CECP (S10.4.3) The total enzyme concentration CT is, CT = CE + CES (S10.4.4) Assuming that the enzyme-substrate complex is in a quasi-steady state, and solving for CES yields CES =CEk1CS+k2CP ( ) k1+k2 (S10.4.5) Substituting equation (S10.4.5)

146 From the data given the ratio CG*/CLR = 0.05 and CG*<< CGo. Equation (S11.5.2) reduces to: kaki=CG *CGoCLR=0.05100000=5.0 x 10-7 cell molecule1 (S11.5.3) (b) If LR dissociates without rebinding by binding to the antibody ligand, the formation of G* is prevented. As a result, G* reforms