Archives

978-0131569881 Chapter 17

978-0131569881 Chapter 17

Solution to Problems in Chapter 17, Section 17.10 17.1. In words, the conservation relation is: 227 ! ! ! “ $ $ $ % + ! ! ! “ $ $ $ % + ! ! ! “ $ $ […]

9 Pages | August 23, 2020
978-0131569881 Solution Chapters 1-4 Part 1

978-0131569881 Solution Chapters 1-4 Part 1

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 […]

14 Pages | August 23, 2020
978-0131569881 Solution Chapters 1-4 Part 2

978-0131569881 Solution Chapters 1-4 Part 2

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 − = γη  app . The enzyme […]

Pages | August 23, 2020
978-0131569881 Solution Chapters 1-4 Part 3

978-0131569881 Solution Chapters 1-4 Part 3

41 τ z θ z=0= µ∂ v θ ∂ zz=0 = µω α 1−0.003294 ε 2 ⎡ ⎣⎤ ⎦ (S3.12.11) To determine the value of ω for which the shear stresses at r = 10 and 20 cm differ by […]

9 Pages | August 23, 2020
978-0131569881 Solution Chapters 1-4 Part 4

978-0131569881 Solution Chapters 1-4 Part 4

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 τ w= µ dvz dr =− µ […]

Pages | August 23, 2020
978-0131569881 Solution Chapters 12-16 Part 1

978-0131569881 Solution Chapters 12-16 Part 1

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 = τwApCD T= τwApCMhc 12.2. For a spherical […]

14 Pages | August 23, 2020
978-0131569881 Solution Chapters 12-16 Part 2

978-0131569881 Solution Chapters 12-16 Part 2

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; […]

14 Pages | August 23, 2020
978-0131569881 Solution Chapters 12-16 Part 3

978-0131569881 Solution Chapters 12-16 Part 3

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 […]

Pages | August 23, 2020
978-0131569881 Solution Chapters 12-16 Part 4

978-0131569881 Solution Chapters 12-16 Part 4

214 and ρ 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 […]

Pages | August 23, 2020
978-0131569881 Solution Chapters 5-7 Part 1

978-0131569881 Solution Chapters 5-7 Part 1

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: Δp =f(ρ, µ, <v>, a, R, L) There are seven dimensional groups and three […]

14 Pages | August 23, 2020
978-0131569881 Solution Chapters 5-7 Part 2

978-0131569881 Solution Chapters 5-7 Part 2

88 Since η is not a function of time, the time derivative is: ∂ θ ‘ ∂ τ =∂ ∂ τ f η ⎛ ⎝ ⎜⎞ ⎠ ⎟=1 η ∂f ∂ τ (6.17.4) Equating Equations (6.17.3) and (6.17.4) 1 η […]

9 Pages | August 23, 2020
978-0131569881 Solution Chapters 5-7 Part 3

978-0131569881 Solution Chapters 5-7 Part 3

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 […]

9 Pages | August 23, 2020
978-0131569881 Solution Chapters 8-11 Part 1

978-0131569881 Solution Chapters 8-11 Part 1

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 […]

14 Pages | August 23, 2020
978-0131569881 Solution Chapters 8-11 Part 2

978-0131569881 Solution Chapters 8-11 Part 2

131 dCS dt =−k1CECS+k−1CES (S10.4.1) dCES dt =k1CECS−(k−1+k2)CES +k−2CECP (S10.4.2) dCP dt =k2CES −k−2CECP (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 […]

Pages | August 23, 2020
978-0131569881 Solution Chapters 8-11 Part 3

978-0131569881 Solution Chapters 8-11 Part 3

146 From the data given the ratio CG*/CLR = 0.05 and CG*<< CGo. Equation (S11.5.2) reduces to: ka ki =CG * CGo CLR =0.05 100000 =5.0 x 10-7 cell molecule−1 (S11.5.3) (b) If LR dissociates without rebinding by binding to […]

9 Pages | August 23, 2020