Instructor’s Solutions Manual for Transport Phenomena in Biological Systems 2nd Edition

Instructor’s Solutions Manual for Transport Phenomena in Biological Systems Second Edition George A. Truskey, Fan Yuan, and David F. Katz

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ISBN-13: 978-0-13-507051-2 ISBN-10:

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0-13-507051-1

Solution to Problems in Chapter 1, Section 1.10 1.1. The relative importance of convection and diffusion is evaluated by Peclet number, vL (S1.1.1) Pe = Dij (a) Solving for L, L = PeDij/v. Assume that convection is the same as diffusion, i.e., Pe = 1, L is 0.123 cm. (b) The distance between capillaries is 10-4 m; oxygen needs to travel half of this distance, which yields a value of PE equal to 0.0455. Therefore, convection is negligible compared with diffusion. 1.2. Since HO2 = HHb, equation (1.6.4) is simplified to the following: CO2 = H O2 PO2 + 4C Hb SHct

(S1.2.1)

PO2 and S are 95 mmHg and 95% for arterial blood and 38 mmHg 70% for venous blood. CHb is

0.0203 mol L-1 x 0.45 = 0.0091 M for men, and 0.0203 mol L-1 x 0.40 = 0.0081 M for women. Based on these data, the fraction of oxygen in plasma and bound to hemoglobin is 1.5% and 98.5% in arterial blood, and 0.83% and 99.17% in venous blood for men. Corresponding values for women are 1.7% and 98.3% in arterial blood, and 0.93% and 99.07% in venous blood. Most oxygen in blood is bound to hemoglobin.

1.3. For CO2 70% is stored in plasma and 30% is in red blood cell. Therefore, the total change of CO2 is 2.27(0.70)+1.98(0.30) = 2.18 cm3 per 100 cm3. For O2, PO2 changes from 38 to 100 mmHg after blood passes through lung artery. Using data in problem (1.2), the total O2 concentration in blood is 0.0088 M in arterial blood and 0.0063 M in venous blood. At standard temperature (273.15 K) and pressure (1 atm = 101,325 Pa), 1 mole of gas occupies 22,400 cm3. Thus, the O2 concentration difference of 0.0025 M corresponds to 5.58 cm3 O2 per 100 cm3. While larger than the difference for CO2, the pressure difference driving transport is much larger for O2 than CO2. 1.4. The diffusion time is L2/Dij = (10-4 cm)2/(2x10-5 cm2 s-1) = 0.0005 s. Therefore, diffusion is much faster than reaction and does not delay the oxygenation process. 1.5. V = πR2L and the S= 2πRL where R is the vessel radius and L is the length Order volume, cm3 surface area, cm2 cumulative volume, cm3 cumulative surface area, cm2 1 0.0158 26.27 0.0158 26.27 2 0.03885 35.32 0.05 61.59 3 0.05738 31.44 0.11 92.99 4 0.09219 30.23 0.20 123.21 5 0.12788 26.64 0.33 149.86 6 0.20487 23.28 0.54 173.14 7 0.20733 15.56 0.74 188.70 8 0.24132 11.03 0.99 199.73 9 0.31010 8.17 1.30 207.89 10 0.23046 3.71 1.53 211.60 11 0.50671 3.99 2.03 215.59 3

1.6. Order

Volume (cm3)

Surface Area (cm2)

Cumulative Volume (cm3)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

30.54 11.13 4.11 1.50 3.23 3.29 3.54 4.04 4.45 5.15 6.25 7.45 9.58 11.68 16.21 22.42 30.57 42.33 60.223 90.05 138.42 213.18 326.72 553.75

67.86 36.49 19.82 10.70 28.72 37.65 50.67 70.29 95.74 133.76 192.38 273.51 403.41 569.79 876.05 1358.86 2038.28 3135.25 4817.76 7663.95 12303.82 19831.06 31874.64 54024.81

30.54 41.66 45.78 47.27 50.51 53.80 57.35 61.39 65.84 70.99 77.24 84.70 94.27 106.0 122.2 144.6 175.2 217.5 277.7 367.8 506.2 719.4 1046. 1600

Cumulative Surface Area (cm2) 67.86 104.34 124.2 134.9 163.6 201.2 251.9 322.2 418.0 551.7 744.1 1018 1421 1991 2867 4226 6264 9399 14217 21881 34185 54015 85890 139915

1.7 (a). The water content is 55% and 60% of the whole blood for men and women, respectively. Then the water flow rate through kidney is 2091 L day-1 for men and 2281 L day-1 for women (Two kidneys). Then the fraction of water filtered across the glomerulus is 8.6% for men and 7.9% for women. (b). renal vein flow rate = renal artery flow rate – excretion rate = 1.19 L min-1 Renal vein flow rate = 1.32 L min-1 – (1.5 L day -1) / (1440 min day-1) = 1.319 L min-1 (c). Na+ leaving glomerulus = 25,200 mmole day-1 / 180 L day-1 = 140 mM. Na+ in renal vein = Na+ in renal artery – Na+ excreted (1.32 L min-1 x 150mM)- (150mM day-1 /1440 min day -1) / 1.319 L min-1 = 150.035 mM There is a slight increase in sodium concentration in the renal vein due to the volume reduction. 4

1.8. (a) Bi = kmL/Dij = 5 x 10-9 cm s -1 x 0.0164cm / 1 x 10 -10 cm2 s-1 = 0.82. (b) The results indicate that the resistance to LDL transport provided by the endothelium is similar to that provided by the arterial wall. 1.9 The oxygen consumption rate is VO2 = Q ( Cv − Ca ) where Q is the pulmonary blood flow and Cv and Ca are the venous are arterial oxygen concentrations. The oxygen concentrations are obtained from Equation (1.6.4)

Co2 = Ho2 Po2 (1 − Hct ) + ( 4CHB S + H HB Po2 ) Hct The fractional saturation S is given by Equation (1.6.5). For the data given, the venous fraction saturation is 0.971. The arterial fractional saturation is 0.754 under resting conditions and 0.193 under exercise conditions. Men

Women

Rest

Ca = 0.0070 M

Ca = 0.0063 M

Exercise

Ca = 0.0019 M

Ca = 0.0017 M

CV = 0.0090 M

Cv = 0.0080 M

The oxygen consumption rates are Men

Women

Rest

0.0115 mole min-1

0.0102 mole min-1

Exercise

0.1776 mole min-1

0.1579 mole min-1

1.10. (a) To obtain the rate of oxygen removal from the lungs, we use the mass balance discussed in class that equates the oxygen removed from the inspired air with the oxygen uptake in the blood. VI ( CI − Calv ) = Q ( Cv − Ca )

(S1.10.1)

We want to assess the left hand side of Equation (S1.10.1) which represents the rate of oxygen removal from the lungs. From the data provided and the ideal gas equation: Calv =

CI =

palv (105 mm Hg ) / ( 760 mm Hg/atm ) = = 0.00543 M RT ( 0.08206 L atm/(mol K) )( 310 K )

0.21(1 atm ) palv = = 0.00826 M RT ( 0.08206 L atm/(mol K) )( 310 K )

VI = (10 breaths/min )( 0.56 − 0.19 L ) = 3.7 L/min

males

VI = (10 breaths/min )( 0.45 − 0.41 L ) = 3.1 L/min

females

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Since we have all terms on the left hand side of Equation (1), the rate of oxygen removal from the lungs is: VI ( CI − Calv ) = ( 3.7 L/min )( 0.00282 mole O 2 /L ) = 0.0104 mole O 2 /min males VI ( CI − Calv ) = ( 3.1 L/min )( 0.00282 mole O 2 /L ) = 0.00874 mole O 2 /min females

To convert to mL O2/L blood, multiply to oxygen removal rate by 22,400 L O2 per mole of O2. For males the value is 233 mL O2/min and for females the value is 196 mL O2/min. These values are a bit low but within the range of physiological values under resting conditions. (b) In this part of the problem, you are asked to find the volume inspired in each breadth or VI . Sufficient information is provided to determine the right hand side of Equation (1) which represents both the rate of oxygen delivery and oxygen consumption. First, determine the oxygen concentrations in arteries and veins. The concentration in blood is: Co2 = Ho2 Po2 (1 − Hct ) + ( 4CHb S + H Hb Po2 ) Hct Using the relation for the percent saturation to calculate the concentration in the pulmonary vein:

(P S= 1+ ( P O2

O2

P50

)

2.6

P50

)

2.6

(100 / 26 ) = 2.6 1 + (100 / 26 ) 2.6

= 0.972

Likewise for the pulmonary artery:

(P S= 1+ ( P O2

O2

P50

)

2.6

P50

)

2.6

( 20 / 26 ) = 2.6 1 + ( 20 / 26 ) 2.6

= 0.3357

This is substantially less than the value in the pulmonary artery under resting conditions, S = 0.754. The concentration in blood is: Co2 = Ho2 Po2 (1 − Hct ) + ( 4CHb S + H Hb Po2 ) Hct For men Cv = (1.33 x 10 –6 M mmHg –1 ) ( 20 mmHg ) 0.55 +

( ( 0.0203 M )( 0.3357 ) + (1.50 x 10

–6

M mmHg –1

Ca = (1.33 x 10 –6 M mmHg –1 ) (100 mmHg ) 0.55 +

( ( 0.0203 M )( 0.972) + (1.50 x 10

–6

M mmHg –1

) (100 mmHg ) ) 0.45 = 0.0090 M

For women Cv = (1.33 x 10 –6 M mmHg –1 ) ( 20 mmHg ) 0.60 +

(( 0.0203 M )( 0.3357 ) + (1.50 x 10

–6

M mmHg –1

6

) ( 20 mmHg ) ) 0.45 = 0.0031 M

) ( 20 mmHg ) ) 0.40 = 0.00275 M

Ca = (1.33 x 10 –6 M mmHg –1 ) (100 mmHg ) 0.60 +

( ( 0.0203 M )( 0.972) + (1.50 x 10

–6

M mmHg –1

Thus, the oxygen consumption rates are Q ( Cv − Ca ) 0.148 mole O2/min

) (100 mmHg ) ) 0.40 = 0.0080 M men

0.132 mole O2/min women These values are about 14 times larger than the values under resting conditions. From Equation (1) ( C − Ca ) VI = Q v 52.5 L O2/min men 46.8 L O2/min women ( CI − Calv ) For a respiration rate of 30 breaths per minutes, the net volume inspired in each breadth is: 1.75 L/min for men and 1.56 L/min for women. In terms of the total air inspired in each breadth, it is 1.94 L/min for men and 1.70 L/min for women. 1.11. CO = HR x SV where CO is the cardiac output (L min-1), SV is the stroke volume (L) and HR is the hear rate in beat min-1. Stroke Volume, L Rest Exercise Athlete 0.0833 0.238 Sedentary person 0.0694 0.2

The peripheral resistance is R = pa / CO Peripheral resistance, mm Hg/(L/min) Rest Exercise Athlete 20 5.2 Sedentary person 20 6 W = ∫ pa dV = pa ΔV since the mean arterial pressure is assumed constant. DV corresponds to

the stroke volume. Note 1 L = 1000 cm3 *(1 m/100 cm)3 = 0.001 m3 100 mm Hg = 13,333 Pa Sedentary person W = (100 mm Hg)(133.3 Pa/mm Hg)(0.069 L)(1000 cm3/L)(1 m3/1x106 cm3) = Work, J (N m) Rest Exercise Athlete 1.11 4.12 Sedentary person 0.925 4.00 Power, W (J/s) Rest Exercise Athlete 1.11 7.22 Sedentary person 0.924 8.33 7

1.12. Although the pressure drops from 760 mm Hg to 485 mm Hg, the partial pressures are unchanged. The inspired air at 3,650 m is 101.85 mm Hg. For a 30 mm Hg drop, the alveolar air is at 71.85 mm Hg.

The oxygen consumption rate is

VO2 = VI ( CI − Calv )

Assuming that the inspired air is warmed to 37 C CI =

pI (101..85 mm Hg ) / ( 760 mm Hg/atm ) = = 0.00527 M RT ( 0.08206 L atm/(mol K) )( 310 K )

Calv =

palv ( 71.85 mm Hg ) / ( 760 mm Hg/atm ) = = 0.00372 M RT ( 0.08206 L atm/(mol K) )( 310 K )

Assuming that the inspired and dead volumes are the same as at sea level VI = f (VI − Vdead ) = 20 ( 0.56 L − 0.19 L ) = 7.4 L min -1

The venous blood is at a partial pressure of 0.98(71.85) = 70.32 mm Hg The corresponding saturation is 0.930. 1.13. (1650 kcal/day)*4.184 kJ/kcal*(1day/24 h)*(1 h/3600 s) = 79.9 J/s

Athlete Sedentary person

Rest 0.014 0.014

1.14. The concentrations are found as the ratio of the solute flow rate/fluid flow rate

Sodium Potassium Glucose Urea

Urine, M 0.1042 0.0694 0.000347 0.32431

Plasma, M 0.08444 0.004 0.00444 0.005183

Urine/Plasma 1.233 17.36 0.0781 62.57

The results indicate that urine concentrates sodium to a small extent, potassium to a higher level and urea to very high levels. Glucose is at a lower concentration in urine than plasma, suggesting that its transport across the glomerulus is restricted. 1.15. Assuming that inulin is not reabsorbed by the kidneys and returned to the blood, then the mass flow rate of inulin across the glomerulus must equal the mass flow rate in urine. The mass flow rate is the product of the mass concentration (mass/volume) multiplied by the flow rate (volume/time). Thus, plasma urine Cinulin GFR = Cinulin Qurine

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Solving for the glomerular filtration rate: urine Cinulin ⎛ 0.125 ⎞ -1 -1 GFR = plasma Qurine = ⎜ ⎟ (1 mL min ) = 125 mL min Cinulin ⎝ 0.001 ⎠

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Solution to Problems in Chapter 2, Section 2.10 2.1. 3

Q = ∫ v ⋅ ndA = ∫

∫

4 3 y =0 x =0

6 ⎞ 6 ⎛ 3 ⎛ 3 2 ⎞ x+ y ⎟ dxdy = ∫ 4y =0 ⎜ x + yx ⎟ dy ⎜ 2 ⎠ 2 2 ⎠ x =0 ⎝ 2 ⎝2 2 4

18 2 ⎞ 396 ⎛ 27 18 ⎞ ⎛ 27 + Q=∫ ⎜ y ⎟ dy = ⎜ y+ y ⎟ = 2 ⎠ 2 ⎠ y =0 2 ⎝ 2 ⎝ 2 3 -1 Q = 280.01 cm s 4 y =0

n = 1 = a 2 + a 2 + a 2 = 3a

2.2.

Rearranging, a = 1/ 3 ⎛ ∂ ⎛ ∂ ∂ ∂ ⎞ ∂ ∂ ⎞ ∇ • ( ρ vv ) = ⎜ e x + e y + e z ⎟ • ( ρ vv ) = ⎜ e x + e y + e z ⎟ • ( ρ e x vx v + ρ e y v y v + ρ e z vz v ) ∂y ∂z ⎠ ∂y ∂z ⎠ ⎝ ∂x ⎝ ∂x 2.3. ∂ ∂ ∂ = ( ρ v x v ) + ( ρ v y v ) + ( ρ vz v ) ∂x ∂y ∂z Differentiating term by term, ⎛∂ ⎞ ∂ ∂ ∂ ∂ ∂ ∇ • (ρvv) = v⎜ (ρv x ) + (ρv y )+ (ρv z )⎟ + ρv x (v) + ρv y (v) + ρv z (v) ∂y ∂z ∂x ∂y ∂z ⎝ ∂x ⎠ ∇ • (ρvv) = v∇ • (ρv) + ρv • ∇v

2.4. (a) From Equation 2.2.8: ∂v ∂v ∂v ∂v ∂v + v ⋅ ∇v = + vx + vy + vz ∂t ∂t ∂x ∂y ∂z Substituting the velocity field given in the problem a=

a = U 0 ( x 2 − y 2 + x )( 2 x + 1) − U 0 ( 2 xy + y )( 2 x + 1) a = U 0 ( 2 x + 1)( x 2 + x − y 2 − y − 2 xy ) For y = 1, x = 1, a = -18 m s-1 and for y = 1, x = 3, a = 84 m s-1. (b) From equation 2.2.6, the volumetric flow rate is: Q = ∫ v ⋅ ndA A

Q=

6

∫ ∫

2

y =0 0

v x dzdy = 3U 0 ∫

6 y =0

(x

2

− y 2 + x ) dy

Therefore, Q = 288 m s-1.

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2.5. (a) From Equation 2.2.8, the x component of acceleration is: ∂v x ⎛ ∂v ⎞ ax = e x ⋅ a = e x ⋅ ⎜ v x ⎟ = vx ∂x ⎝ ∂r ⎠ From the data given and velocity field in the x-direction the derivative is: ∂v x 192 =− 3 ∂x ( x − 4) and acceleration is: −2 x ⎞ ⎛ −192 ⎞ ⎛ ax = 6 ⎜ 1 − ⎟ ⎜ ⎟ 4 ⎠ ⎜⎝ ( x − 4 )3 ⎟⎠ ⎝ When x = 2 m, acceleration in the x-direction is 144 m/s. (b). When x = 4 the velocity will not be correct.

2.6. (a) Using the definition of the volumetric flow rate, Q Q = ∫ v indA =

2π Ri

∫ ∫ v rdrdθ z

0 0

The cross-sectional area element in cylindrical coordinates is rdrdθ. Since the velocity does not vary with angular position, substitution for vz and integration in the angular direction yields: Ri 2π Ri ⎛ ⎛ r2 ⎞ r2 ⎞ Q = ∫ ∫ v max ⎜1 − 2 ⎟ rdrdθ = 2π v max ∫ ⎜1 − 2 ⎟ rdr Ri ⎠ ⎝ R ⎠ 0 0 0 ⎝ Ri is used to denote the local radius within the stenosis. Integrating in the radial direction yields: R

⎛ ⎛ r2 π Ri 2 r2 ⎞ r4 ⎞ = Q = 2π v max ∫ ⎜ 1 − 2 ⎟ rdr = 2π v max ⎜ − v max 2 ⎟ Ri ⎠ 2 ⎝ 2 4 Ri ⎠ r =0 0 ⎝ Ri

Solving for vmax:

v max =

2Q = π Ri 2

2Q 2 1/2 ⎤ ⎡ ⎛ ⎞ z ⎛ ⎞ 2 π R0 ⎢1 − 0.5 ⎜ 1 − 4 ⎜ ⎟ ⎟ ⎥ ⎜ ⎢ ⎝ L ⎠ ⎟⎠ ⎥ ⎝ ⎣ ⎦

Outside the stenosis, Ri = R0 and:

2Q π R0 2 (b) At z = 0, the velocity in the stensosis is 2Q 2Q 8Q = = v max = 2 2 π Ri π R0 2 [ 0.5] π R0 2 v max =

11

2

2 1/2 ⎤ ⎡ ⎛ ⎞ z ⎛ ⎞ Ri = R0 ⎢1 − 0.5 ⎜1 − 4 ⎜ ⎟ ⎟ ⎥ = 0.5R0 ⎜ ⎢ ⎝ L ⎠ ⎟⎠ ⎥ ⎝ ⎣ ⎦ The shear stress in the stenosis is:

τ rz

stenosis

=μ

⎛ ⎞⎤ ∂ vr r2 ∂ ⎡ ⎢ v max ⎜1 − =µ ⎟⎥ ⎜ R ( z = 0 )2 ⎟ ⎥ ∂z ∂r ⎢ i ⎝ ⎠⎦ ⎣

=− r = Ri

2 μ Ri ( z = 0 ) v max Ri ( z = 0 )

2

=−

32 μQ π R03

Outside the stenosis the shear stress is: ⎛ 2μ v max ∂ ⎡ r 2 ⎞⎤ 4μQ =− =− τ rz = µ ⎢ v max ⎜1 − 2 ⎟ ⎥ R0 ∂r ⎣ π R03 ⎝ R0 ⎠ ⎦ r = R 0

2.7. Evaluating equation 2.7.30 for y = -h/2 yields:

τ w = τ yx ( y = −h / 2 ) =

Δp h L 2

From equations 2.7.23 and 2.7.26, Δp 8μ vmax 12 μQ = = L h2 wh3 Replacing ∆p/L in equation (S2.3.1) with the expression in equation (S2.3.2) yields 6μQ τw = wh 2 6μ Q Solving for h: h = wτ w Inserting the values provided for Q, w, µ, and τw yields h = 0.036 cm.

2.8. (a). Δp = ρ gh = (1g cm −3 )( 980 cm s −1 )( 2.7 × 10−4 cm ) = 0.265dyne cm −2

(b). Rearranging equation (2.4.16) we have Δp Tc = ⎛ 1 1 ⎞ − 2⎜ ⎟⎟ ⎜R ⎝ p Rc ⎠ Tc = 3.41 x 10-5 dyne cm-1 \ 2.9. (a) From Equation (2.4.7), the pressure force corresponds to the height h of a column of fluid and is given by Δp = ρ f gh From the data given, this yields ∆p = (13.6 g cm-3)(981 cm s-1)(.03 cm Hg) = 400.25 N m-2. From Equation (2.4.16):

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⎛ 1 1 ⎞ Δp = 2Tc ⎜ − ⎜ R p Rc ⎟⎟ ⎝ ⎠ the radius of the smallest capillary can be calculated: 2Tc Rc Rp = 2Tc + ΔpRc Rp = .24 µm. (b) Yes, a cell of radius 4.0 µm could enter the capillary determined in part a. The pressure difference needed to draw the smaller cell into the micropipette is smaller than the pressure difference needed to draw in a larger cell as in part a.

2.10. A momentum balance is applied on a differential volume element, 2πrΔrΔy, as shown in the figure below.

p r 2πr Δy − p r +Δr 2π ( r + Δr ) Δy + τ yr

y +Δy

2π r Δr − τ yr 2π r Δr = 0 (S2.10.1) y

Divide each term by 2πrΔrΔy and take the limit as Δr and Δy go to zero results in the following expression: 1 d ( rp ) dτ yr = (S2.10.2) r dr dy Note that if the gap distance h is much smaller than the radial distance, then curvature is not significant. Each side is equal to a constant C1. Solving for the shear stress, τyr = C1y + C2. Substituting Newton’s law of viscosity and integrating yields: C y2 C v r = 1 + 2 y + C3 (S2.10.3) 2μ μ Applying the boundary conditions that vr = 0 at y=±h/2, C h2 C h 0 = 1 + 2 + C3 (S2.10.4a) 8μ μ 2 C h2 C h 0 = 1 − 2 + C3 (S2.10.4b) 8 μ 2 Adding Equations (S2.10.4a) and (S2.10.4b) and solving for C3, C h2 (S2.10.5) C3 = − 1 8 Inserting Equation (S2.10.5) into Equation (S2.10.4a) yields C2 = 0. Thus the velocity is: C ⎛ y 2 h2 ⎞ (S2.10.6) vr = 1 ⎜ − ⎟ μ⎝ 2 8 ⎠ 13

The volumetric flow rate is:

Q=

∫ v • ndA = ∫

h/2

2πrv r dy = y=−h / 2

2πrC1 μ

∫

⎛ y2 h2 ⎞ − ⎟dy ⎜ y=−h / 2 2 8⎠ ⎝ h/2

(S2.10.7)

h/ 2

πrC1h3 2πrC1 ⎛ y 3 h 2 y ⎞ 2πrC1h3 ⎛ 1 1 ⎞ − = − = − Q= ⎜ ⎟ ⎜ ⎟ μ ⎝ 6 μ 8 ⎠ y =− h /2 6μ ⎝ 24 8 ⎠

(S2.10.8)

Solving for C1 and inserting into equation (S2.10.6) 6Q ⎛ y 2 h 2 ⎞ vr = − ⎜ − ⎟ πrh3 ⎝ 2 8 ⎠ (S2.10.9) The shear stress can thus be written as;

τ w = τ yr

y =− h / 2

=μ

dvr dr

=− y =− h /2

6Q 3μQ = y 3 πrh y =− h /2 πrh 2

(S2.10.10)

2.11. The volumetric flow rate for each fiber can be written as: Q = v π R2 where is the average velocity for the flow through the fibers. With the data given = 19.95 cm s-1 per fiber. Reynolds number is calculated from the equation: ρ v L Re =

μ

From the data given in the problem, Re = 37356. From equation 2.7.26: Le = 0.058 DRe and the data given, Le = 65 cm. This value is much larger than the length of the hollow fiber unit.

2.12. (a) The momentum balance is the same as that used for the case of pressure-driven flow in a cylindrical tube in Section 2.7.3. dp 1 d (rτ rz ) = (S2.12.1) dz r dr (b) The velocity profile is sketched below:

Integrating the momentum balance and substituting Newton’s law of viscosity, Δp C dv (S2.12.2) τ rz = − r + 1 = μ z 2L r dr 2C1 . Assuming that C1 is Note that the shear stress and shear rate are a maximum at r = Δp / L greater than zero, then r will have a maximum in the fluid. 14

(c) Integrating Equation (S2.12.2) yields: Δp 2 C1 vz = − r + ln(r ) + C2 4μL μ Applying the boundary conditions C Δp V =− RC 2 + 1 ln( RC ) + C2 4μL μ Δp 2 C1 0=− R + ln( R) + C2 4μL μ Subtracting Δp C ⎛R ⎞ V =− RC 2 − R 2 ) + 1 ln ⎜ C ⎟ ( 4μL μ ⎝ R ⎠ Solving for C1: 2 2 μV Δp ( RC − R ) + C1 = ⎛ RC ⎞ 4 L ⎛R ⎞ ln ⎜ ln ⎜ C ⎟ ⎟ ⎝ R ⎠ ⎝ R ⎠

(S2.12.3)

(S2.12.4a) (S2.12.4b)

(S2.12.5)

(S2.12.6a)

Using this result to find C2 ⎛ ⎜ Δp 2 ⎜ V C2 = R − ⎜ ⎛ R 4μL ⎜⎜ ln ⎜ ⎝ ⎝ RC

⎞ ⎟ Δp ( RC − R ) ⎟ ln( R) + ⎞ 4μL ⎛ R ⎞ ⎟ ln ⎜ ⎟ ⎟ ⎟⎟ ⎠ ⎝ RC ⎠ ⎠ 2

2

(S2.12.6b)

The resulting expression for the velocity profile is ⎞ ΔpR 2 ⎛ Δp r2 ⎞ ⎛ − + ⎜V + vz = 1 RC 2 − R 2 ) ⎟ ( ⎜ 2 ⎟ 4μL ⎝ R ⎠ ⎝ 4μL ⎠

⎛r⎞ ln ⎜ ⎟ ⎝R⎠ ⎛ R ⎞ ln ⎜ ⎟ ⎝ RC ⎠

(S2.12.7)

(d) The shear stress is: ⎛ ⎞ ⎜ μV + Δp ( RC 2 − R 2 ) ⎟ r Δp ⎜ dv ⎟⎛ 1 ⎞ 4L + τ zr = µ z = − ⎜ ⎟ ⎜⎝ r ⎟⎠ dr 2L ⎛ R ⎞ ln ⎜ ⎜⎜ ⎟⎟ ⎟ ⎝ RC ⎠ ⎝ ⎠

(e) At r = R, the shear stress is:

15

(S2.12.8)

⎛ ⎞ ⎜ μV + Δp ( RC 2 − R 2 ) ⎟ R Δp ⎜ dv ⎟⎛ 1 ⎞ 4L + τ zr = µ z = − ⎟ ⎜⎝ R ⎟⎠ dr 2L ⎜ ⎛ R ⎞ ln ⎜⎜ ⎟⎟ ⎜ ⎟ ⎝ RC ⎠ ⎝ ⎠

(S2.12.9)

For the values provided τ zr

r=R

=−

( 0.17 cm ) (100 dyne/cm 3 ) 2

(

(

+ ( 0.03 g/cm/s ) + ( 25 dyne/cm 3 ) ( 0.15 cm ) − ( 0.17 cm )

τ zr

r=R

2

2

))

⎛ ⎞ 1 ⎜ ⎟ ⎝ ( 0.17 cm ) ⎠ ⎛ 0.17 ⎞ ln ⎜ ⎟ ⎝ 0.15 ⎠

= −16.0 dyne/cm 2

This compares with a shear stress of -8.5 dyne/cm2 in the absence of the catheter.

2.13. For a Newtonian fluid ⎛ ⎛ ΔpR 2 ⎛ r2 ⎞ r2 ⎞ r2 ⎞ − = − = − vz = 1 v 1 2 v 1 ⎜ ⎟ ⎜ max ⎜ 2 ⎟ 2 ⎟ 4μ L ⎝ R 2 ⎠ ⎝ R ⎠ ⎝ R ⎠ (S2.7.1a,b) Q U= 2πR 3

γr = −

dvz dr

= r =R

4Q πR 3

(S2.7.2)

By comparing equations (S2.7.1b) and (S2.7.2), we find that, γ Q = r = 2U 3 πR 4 As a result, γ r = 8U . For a power law fluid, the average velocity and shear rate equal (equations 2.7.52 and 2.7.55): dv Q γ =− z = ( 3 + 1/ n ) 3 = 2 ( 3 + 1/ n ) U dr r = R πR The constant relating shear rate and the reduced velocity equals 2(3n+1)/n. If n = 1, for a Newtonian fluid, then 2(3n+1)/n = 8.

2.14. A schematic of a volume element is shown below.

16

The momentum balance is:

(

ρ g cos β ΔxΔyΔz + τ xy

y +Δy

− τ xy

y

) Δ xΔ z = 0

(S2.14.1)

where the z direction is normal to the x-y plane. Dividing each term by the volume element ΔxΔyΔz and taking the limit as Δy goes to zero yields: dτ xy

= − ρ g cos β (S2.14.2) dy Note that at the air-liquid interface the shear stress is zero. This is because the viscosity of the gas is much less than the viscosity of the liquid. As a result the velocity gradient at y = 0 is zero. Integrating the momentum balance and applying the boundary condition at y =0 yields: (S2.14.3) τzx = -ρgycosβ

For a Bingham plastic, the velocity gradient is zero for a shear stress below the yield stress, τo = ρgycosβ. For the angles given, the difference in yield stress is only 6% so the manufacturer’s claim is exaggerated.

2.15. For fluid 1, since the resulting velocity is linearly related to the original velocity, the fluid is Newtonian. For fluid 2, there is no linear or power dependence between the velocities suggesting that the fluid is a Bingham plastic. Applying Equation (2.7.12b) to the base case and cases (2a) and (2b) (τ − τ ) H μV τ1 = 0 1 + τ 0 (S2.15.1a,b) or V1 = 1 0 H μ0 3V1 =

( 2τ 1 − τ 0 ) H

(S2.15.2)

μ0 ( 4τ1 − τ 0 ) H 7V1 = μ0

(S2.15.3)

Substituting Equation (S2.15.1b) for τ1 into equation (S2.15.2) ⎛ ⎛ μ0V1 ⎞ ⎞ ⎜ 2 ⎜ H + τ 0 ⎟ −τ 0 ⎟ H ⎝ ⎠ ⎠ = 2V + τ 0 H 3V1 = ⎝ 1 Solving for V1,

V1 =

τ0 H μ0

μ0

μ0

(S2.15.4) (S.2.15.5)

Substituting equations (S2.15.5) into equation (S2.15.1a), yields. 17

τ o = 0.5τ 1

(S.2.15.6) Inserting Equations (S2.15.5) and (S2.15.6) into equation (S2.9.3) verifies that equation (S2.15.6) is the correct relation. For fluid 3, the resulting velocity is proportional to the square of the original velocity. The fluid is a power law fluid. For a power law fluid, equation (2.7.7) becomes n −1

⎛ dv ⎞ dv dv x τ yx = m x = m⎜ x ⎟ dy dy ⎝ dy ⎠ Rearranging yields an expression for the velocity gradient

n

(S2.15.7)

1/ n

dv x ⎛ τ yx ⎞ =⎜ ⎟ dy ⎝ m ⎠ Integrating, and evaluating the boundary condition that vx = 0 at y = 0.

(S2.15.8)

1/ n

⎛τ ⎞ v x = ⎜ yx ⎟ y ⎝ m⎠ (S2.15.9) Evaluating the velocity at y = H for the conditions given, 1/ n

⎛ 2τ ⎞ 4V1 = ⎜ 1 ⎟ ⎝ m ⎠

(S2.15.10a)

H 1/ n

⎛ 4τ ⎞ 16V1 = ⎜ 1 ⎟ H ⎝ m ⎠ Dividing equation (S2.15.10b) by equation (S2.15.10a) yields: 1/ n 4 = ( 2) Solving, yields n = 0.5.

(S2.15.10b) (S2.15.11)

2.16. For a Bingham plastic the momentum balance is unchanged from equation (2.7.57) 1 d(r 2τ rθ ) =0 r 2 dr which after integration yields r2τrθ = C1 If the shear stress is less than τo then the shear rate is d ⎛ vθ ⎞ zero. Thus, ⎜ ⎟ = 0 or vθ = C2r where C2 is a constant. The inner cylinder is not moving. dr ⎝ r ⎠ Although the shear stress is lower on the outer cylinder, the only way the boundary condition at r = εR can be satisfied is for vθ = 0. Thus when τrθ < τo, vθ = 0.

For τrθ greater than τo, τ rθ = τ 0 + μ0γ 0 = τ 0 + μ0 Rearranging

d ⎛ vθ ⎜ dr ⎝ r

d ⎛ vθ ⎞ C1 ⎜ ⎟= dr ⎝ r ⎠ r 2

τ ⎞ C1 − o . Integrating we have: ⎟= 3 ⎠ μo r μ o r 18

vθ = −

C1 τ o r − ln r + C2 r 2 μo r μ o

Applying the boundary condition at r = R and r = εR we have τrθ > τ0, τ ⎛ r ε R ⎞ ⎡ Ωε R τ o ε R ln ε ⎤ − − vθ = − o r ln ( r / ε R ) + ⎜ ⎥ ⎟⎢ 2 μo μo 1 − ε 2 ⎦ ⎝ ε R r ⎠ ⎣1 − ε The yield stress τo can be determined from the torque required to begin rotation of the outer cylinder, T = 2πR2Lτo. Once rotation begins, the viscosity can be determined by relating the torque to the shear stress at r = R (Equation (2.7.69)). For a Bingham plastic the result is: 4πLε 2 R 2 [ μo Ω − τ o ln ε ] r=R 1− ε 2 A plot of the torque versus the rotational speed Ω will be linear with a slope proportional to µo and an intercept proportional to τo. T = 2πR 2 L τ rθ

=

2.17. For a power law fluid the shear stress is related to the shear rate as: n −1

d ⎛ v ⎞ ⎛ d ⎛ v ⎞⎞ τ rθ = m r ⎜ θ ⎟ ⎜ r ⎜ θ ⎟ ⎟ (S2.17.1) dr ⎝ r ⎠ ⎝ dr ⎝ r ⎠ ⎠ Since vq increases with r, the derivative is positive, the shear stress can be written as; ⎛ d ⎛ v ⎞⎞

n

τ rθ = m ⎜ r ⎜ θ ⎟ ⎟ = Cr ⎝ dr ⎝ r ⎠ ⎠

(S2.17.2)

1 2

Rearranging equation (S2.11.2) yields 1/ n d ⎛ vθ ⎞ 1 ⎛ C1 ⎞ = ⎜ ⎟ ⎜ ⎟ dr ⎝ r ⎠ r ⎝ mr 2 ⎠ Integrating and expressing the results in terms of vθ, 1/ n

(S2.17.3)

n−2

⎛C ⎞ vθ = ⎜ 1 ⎟ r n + rC2 ⎝m⎠ Applying the boundary conditions at r = εR and r = R, the velocity profile is: 2/ n −1 ⎤ Ωε R ⎡ r ⎛ ε R ⎞ vθ = − ⎥ ⎜ ⎟ 2/ n ⎢ 1 − ε ⎢⎣ ε R ⎝ r ⎠ ⎥⎦ For n = 1 (Newtonian fluid) this result is equal to equation (2.7.67). The shear stress is: n

(S2.17.5)

2

⎛ Ω ⎞ ⎛ 2⎞ ⎛εR⎞ τ rθ = m ⎜ ⎟ ⎜ ⎟ 2/ n ⎟ ⎜ ⎝ 1− ε ⎠ ⎝ n ⎠ ⎝ r ⎠ Finally, the torque is: ⎡ 2 ⎛ Ω ⎞n ⎛ 2 ⎞n ⎤ 2 2 T = 2πR L τ rθ r = R = 2πR L ⎢ mε ⎜ ⎟ ⎥ 2/ n ⎟ ⎜ ⎝ 1 − ε ⎠ ⎝ n ⎠ ⎦⎥ ⎣⎢ n

(S2.17.4)

(S2.17.6)

(S2.17.7)

To find m and n, take the logarithm of the left and right hand sides of equation (S2.17.7), 19

T ⎡⎛ Ωε R ⎞ ⎛ 2 ⎞ ⎤ ⎛ ⎞ ⎛2⎞ ⎛ Ωε R ⎞ log ⎜ = log ( m ) + n log ⎢⎜ ⎟ ⎥ = log ( m ) + n log ⎜ ⎟ + n log ⎜ 2 2 ⎟ 2/ n ⎟ ⎜ 2/ n ⎟ ⎝ 2πR Lε ⎠ ⎝n⎠ ⎝ 1− ε ⎠ ⎣⎝ 1 − ε ⎠ ⎝ n ⎠ ⎦ A plot of the log of the torque/2πR2Lε versus log(ΩεR/(1-ε2/n))has an intercept equal to the log(m)+nlog(2/n) and a slope equal to n. 2.18. Letting t0 = 0, the shear rate function is: ⎧γ / ε , for − ε < t < 0 γ x (t ) = ⎨ 0 ⎩0, for t < −ε or t > 0

Thus, equation (2.5.15a) becomes:

0

τ yx = ∫ G (t − t ') −ε

γ0 dt ' ε 0

Rearranging the above equation, we have:

τ yx = γ 0

∫ ε G(t − t ')dt ' −

ε

(b) According to L'Hopital's rule, the limit of this expression as ε goes to 0 is: d 0 G (t − t ')dt ' d ε ∫−ε ε =0 lim τ yx = γ 0 ε →0 1 Applying Leibnitz's rule when differentiating the integral, we have: lim τ yx = γ 0 G (t − ε ) ε =0 = γ 0G (t )

(

)

ε →0

2.19. Since the apparent viscosity depends upon the shear rate, the fluid is not Newtonian. For a power law fluid, η app = mγ n -1 . Taking the logarithm of each side yields. ln (ηapp ) = ln(m)+(n-1)ln(γ )

The data are plotted in Figure S2.19.1. From the slope n= 0.499 ≈ 0.50 so the cytoplasm is shear thinning. The value of m is 147.4 Pa s. If the results are presented in terms of the base 10 log log (η app ) = log(m)+(n-1)log(γ ) The regression line is

log (ηapp ) = 2.169+0.499log(γ )

20

Figure S2.19.1

2.20. A plot of the shear stress versus shear rate revealed that while a straight line gives a good fit, there is some curvature to the data suggesting a shear thinning fluid. A log-log plot of shear stress versus shear rate yields n = 0.855 (Figure S2.20.1). So the fluid does exhibit some shear thinning behavior. The apparent viscosity is η app = 1.42γ −0.145

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) η app = 0.01γ −0.0225 . The enzyme functions by clipping the hyaluronic acid chains decreasing their length. As a result the hyaluronic acid offers much less resistance to flow.

21

Figure S2.20.2

2.21. (a) Since the momentum balance is independent of the type of fluid, begin with Equation 2.7.36 with C2 = 0. Δp τ rz = − r 2L The shear stress is greatest at r = R. If τrz(r =R) < τ0, vz = constant for 0 ≤ r ≤ R. Since the velocity at r = R is zero, vz = 0. Thus, the following criterion must be met for the fluid to flow. Δp τ rz ( r = R ) = − R >τ0 2L (b) When τrz(r =R) > τ0, the fluid begins to move. Since τrz(r =0) = 0, the shear stress at some point in the fluid, say r0, equals the yield stress. Thus for 0 ≤ r ≤ r0, the velocity is constant. The location of r0 is determined by solving for the yield stress 2Lτ 0 r0 = Δp In the region 0 ≤ r ≤ r0, τrz < 0 and the constitutive relation is: dv Δp τ rz = − r = −τ 0 + µ0 z 2L dr Rearranging and integrating once, we have: τ Δp 2 vz = 0 r − r + C3 4µ0 L µ0 Applying the boundary condition at r = R that vz = 0, the C3 equals: τ Δp 2 C3 = − 0 R + R µ0 4µ0 L Replacing C3 yields the final expression for vz(r). vz =

ΔpR 2 ⎛ r2 ⎞ τ0R ⎛ r⎞ − − 1 ⎜ ⎜1 − ⎟ 2 ⎟ 4 µ0 L ⎝ R ⎠ µ0 ⎝ R ⎠

r0 ≤ r ≤ R

ΔpR 2 ⎛ r0 2 ⎞ τ 0 R ⎛ r0 ⎞ ⎜1 − ⎟− ⎜1 − ⎟ 4 µ0 L ⎝ R 2 ⎠ µ0 ⎝ R ⎠ The volumetric flow rate is:

0 ≤ r ≤ r0

vz =

22

R R ⎡ r0 ⎤ ⎡ ⎤ r0 2 Q = ∫ ∫ v z rdrdθ = 2π ⎢ ∫ v z rdr + ∫ v z rdr ⎥ = 2π ⎢ v z ( r = r0 ) + ∫ v z rdr ⎥ 2 r = r0 ⎢⎣ r =0 ⎥⎦ ⎢⎣ ⎥⎦ r =0 θ =0 r = r0 ⎡⎛ ΔpR 2 ⎛ r 2 ⎞ τ R ⎛ r ⎞ ⎞ r 2 R ⎛ ΔpR 2 ⎛ ⎤ r2 ⎞ τ0R ⎛ r ⎞⎞ 0 0 0 0 1 1 1 Q == 2π ⎢⎜ − − − + − − ⎟ ⎜ ⎜ ⎜ ⎜ 1 − ⎟ ⎟ rdr ⎥ ⎜ ⎟ 2 ⎟ 2 ⎟ ∫ ⎥⎦ ⎣⎢⎝ 4 µ0 L ⎝ R ⎠ µ0 ⎝ R ⎠ ⎠ 2 r = r0 ⎝ 4µ0 L ⎝ R ⎠ µ0 ⎝ R ⎠ ⎠ Replacing τ0 with r0. R ⎤ ⎛⎛ r 2 ⎞ 2r0 ⎛ r ⎞⎞ ΔpR 2 ⎡⎛ ⎛ r0 2 ⎞ 2r0 ⎛ r0 ⎞ ⎞ r0 2 Q == 2π 1 1 − + − − ⎢⎜ ⎜ 1 − 2 ⎟ − ⎟ ⎜⎜ ⎜1 − ⎟ ⎟ rdr ⎥ ⎜ ⎟ 2 ⎟ ∫ 4 µ0 L ⎣⎢⎝ ⎝ R ⎠ R ⎝ R ⎠ ⎠ 2 r = r0 ⎝ ⎝ R ⎠ R ⎝ R ⎠ ⎠ ⎦⎥ Integrating: R r 4 ⎞ 2r0 ⎛ r 2 r 3 ⎞ ⎞ ⎤⎥ ΔpR 2 ⎡⎢⎛ ⎛ r0 2 ⎞ 2r0 ⎛ r0 ⎞ ⎞ r0 2 ⎛ ⎛ r 2 Q == 2π 1− ⎜ ⎜1 − ⎟ + ⎜⎜ − ⎟− ⎟− ⎜ − ⎟⎟ 4 µ0 L ⎢⎝ ⎝ R 2 ⎠ R ⎜⎝ R ⎟⎠ ⎠ 2 ⎝ ⎝ 2 4 R 2 ⎠ R ⎝ 2 3R ⎠ ⎠ ⎥ r = r0 ⎦ ⎣ Evaluating the limits. ΔpR 2 ⎡⎛ r0 2 r0 4 ⎞ 2r0 ⎛ r0 2 r03 ⎞ ⎛ R 2 r0 2 R 2 r0 4 ⎞ 2r0 ⎛ R 2 r0 2 R 2 r03 ⎞ ⎤ − − + − − + Q = 2π ⎢⎜ − ⎟− ⎜ − ⎟+⎜ ⎟− ⎜ ⎟⎥ 4 µ0 L ⎣⎝ 2 2 R 2 ⎠ R ⎝ 2 2 R ⎠ ⎝ 2 2 4 4R2 ⎠ R ⎝ 2 2 3 2R ⎠⎦ R

2π

Collecting terms Q == 2π

ΔpR 2 ⎡⎛ r0 2 r0 4 ⎞ 2r0 ⎛ r0 2 r03 ⎞ ⎛ R 2 r0 2 r0 4 ⎞ 2r0 ⎛ R 2 r03 r0 2 ⎞ ⎤ − + 2 ⎟− + − ⎟⎥ ⎢⎜ − ⎟− ⎜ − ⎟+⎜ ⎜ 4µ0 L ⎣⎝ 2 2 R 2 ⎠ R ⎝ 2 2 R ⎠ ⎝ 4 2 4 R ⎠ R ⎝ 6 3R 2 ⎠ ⎦

⎡⎛ R 2 r0 4 ⎞ 2r0 ⎛ R 2 r03 r03 ⎞ ⎤ ΔpR 2 ⎡⎛ R 2 r0 4 ⎞ 2r0 ⎛ R 2 r03 ⎞ ⎤ − 2 ⎟− + − = − − π 2 ⎢⎜ ⎥ ⎢⎜ ⎜ ⎟ ⎟− ⎜ ⎟⎥ 4µ0 L ⎣⎝ 4 4 R 2 ⎠ R ⎝ 6 6 R ⎠ ⎦ ⎣⎝ 4 4 R ⎠ R ⎝ 6 3R 2 R ⎠ ⎦ πΔpR 4 ⎡⎛ r0 4 ⎞ 4r0 ⎛ r03 ⎞ ⎤ Q= ⎢⎜ 1 − ⎟− ⎜1 − ⎟ ⎥ 8µ0 L ⎣⎝ R 4 ⎠ 3R ⎝ R 3 ⎠ ⎦ Q == 2π

ΔpR 2 4µ0 L

Note that for r0 = 0, the result for a Newtonian fluid is obtained.

2.22. (a) Since δ = R(1-ε)<
τ yθ = μ

dv y

(S2.22.3) dr Inserting Equation (S2.22.3) into Equation (S2.22.2), integrating and evaluating the boundary conditions (y = 0 vθ = 0, y = δ vθ = ΩR) leads to the following expression for the velocity:

23

vθ =

ΩRy

τ yθ =

The shear stress is:

(S2.22.4)

δ

μΩR

δ

=

μΩ 1− ε

(S2.22.5)

The torque is 2π R 2 LμΩ (S2.22.6) δ 1− ε (b) Taking the ratio of the shear stress obtained neglecting curvature (equation (S2.22.5) to the exact result (equation (2.7.70)), yields: τ yθ 1+ ε = (S2.22.7) τ rθ r = R 2ε 2 T = 2π R 2 Lτ yθ =

2π R 3 LμΩ

=

This relation can be used to compute the error induced by neglecting curvature. The error is 0.76% for δR equal to 0.005, 1.52% for δ equal to 0.01, and 4.69% for δ equal to 0.03. (c) For a power law fluid: n −1 ⎛ dv y ⎞ dv y (S2.22.8) τ yθ = m ⎜ ⎟ ⎝ dr ⎠ dr The momentum balance indicates that the shear stress is constant and positive. n dτ yθ ⎛ dv y ⎞ = m⎜ (S2.22.9) ⎟ =0 dy ⎝ dr ⎠ Integrating once dv y =C (S2.22.10) dr Evaluating the boundary conditions, yields Equation (S2.22.4). Thus, the torque for a power law fluid is: ⎛ ΩR ⎞ T = 2π R Lτ yθ = 2π R Lm ⎜ (S2.22.11) ⎟ ⎝ δ ⎠ Taking the logarithm of both sides of Equation (S2.22.11) ⎛ ΩR ⎞ ln(T ) = ln(2π R 2 Lm) + n ln ⎜ (S2.22.12) ⎟ ⎝ δ ⎠ Thus, a plot of ln(T) versus ln(ΩR/δ) has a slope equal to n and an intercept equal to ln(2πR2Lm). n

2

2

2.23. (a) For low rotational speeds, there is only 1 velocity component, vθ. This velocity is a function of r only. There is no angular variation in velocity or pressure. Thus there is only one shear stress term, τrθ. Applying a momentum balance, Equation (2.7.57): 0=

1 d 2 ( r τ rθ ) r 2 dr

24

(b) (c) Integrating once: C1 r2 We need the velocity profile in order to determine the constant. From Equation (2.7.62b), we have. d ⎛v ⎞ C τ rθ = μ r ⎜ θ ⎟ = 21 dr ⎝ r ⎠ r d ⎛ vθ ⎞ C1 Rearranging ⎜ ⎟= dr ⎝ r ⎠ µr 3 C vθ = − 1 + C2 r Integrating 2 µr The boundary conditions are r —>∞ vθ = 0 r=R vθ = ωR

τ rθ =

From the B.C. at r —>∞, C2 = 0. At r = R, C1 = -2µ wR2. Thus, the velocity field is: vθ =

ω R2

r (d) T = (F|r=R)xRer The torque and force are determined on the surface of the cylinder. Note that the velocity is constant and the unit normal is in the - r direction. The shear stress can C be found by substituting for C1 in the relation τ rθ = 21 = −2 µω r . F|r=R = -er (ereθ) 2πRLτrq|r=R = eθ4πRLµω. The torque is T = 4πRLµωReθxer = ez4πLµwR2 The cylinder must exert an equal and opposite torque to remain in motion. (e) The torque can be measured from the electrical energy needed to keep the motion of the cylinder constant. Then, from a plot of Torque versus πLwR2, the viscosity can be found from the slope.

2.24. Using the definitions of the tube and discharge hematocrits provided in the text 25

2 HCTT = 2 RT RT −δ

HCTD =

∫ 0

RT

∫ 0

2 HCTT ( r )rdr = 2 RT

RT −δ

∫

HCTo rdr = HCTo

( RT − δ )

0

2

(S2.24.1)

RT2

⎛ r2 ⎞ ( RT − δ ) − ( RT − δ ) HCTo vz ( r )rdr HCTo ∫ ⎜1 − 2 ⎟rdr RT ⎠ 2 4 RT2 0 ⎝ = = HCT o RT RT RT2 / 4 ⎛ r2 ⎞ ∫0 vz ( r )rdr ∫0 ⎜⎝1 − RT2 ⎟⎠rdr RT −δ

2

4

⎛ 2 ( RT − δ )2 ( RT − δ )4 ⎞ HCTD = HCTo ⎜ − ⎟ 2 4 ⎜ ⎟ R R T T ⎝ ⎠ (a) Since HCTF =HCTD, the relation between HCTo and HCTF is: HCTo =

HCTF ⎛ 2 ( RT − δ )2 ( RT − δ )4 ⎞ − ⎜ ⎟ ⎜ ⎟ RT2 RT4 ⎝ ⎠

As expected, HCTo > HCTF . (b) Substituting equation (S2.24.2) into equation (S2.24.1) yields: ⎛ ⎞ RT2 HCTF ⎜ ⎟ ⎜ ( R − δ )2 ⎟ HCTF ⎝ T ⎠ HCTT = = 2 2 4 ⎛ 2 ( RT − δ ) ( RT − δ ) ⎞ RT − δ ) ( − ⎜ ⎟ 2− 2 4 ⎜ ⎟ RT2 R R T T ⎝ ⎠ δ/RT HCTT/HCTF RT, µ,m 500 0.01 0.9805 400 0.0125 0.9758 250 0.02 0.9619 100 0.05 0.9111 50 0.10 0.8403

2.25. (a) τ rz

r = RC

=µ

dv z dr

= −µ r =R

dv z dy

= −µ y =δ

VC

δ

(b) A force balance yields: Pressure X Cross-sectional area of cell = shear stress X area over which stress acts V ΔPπ RC2 = τ rz r = R 2π RC L = µ C 2π RC L

(

VC =

C

)

δ 2 ⎛ δ ⎞⎛

Δp Δp δ⎞ δ RC = R ⎜ ⎟ ⎜1 − ⎟ 2 µL 2 µL ⎝ R ⎠ ⎝ R ⎠

Alternatively, the momentum balance, Equation (2.7.34b), is: 26

(S2.24.2)

dp 1 ⎛ d ( rτ rz ) ⎞ = ⎜ ⎟ dz r ⎝ dr ⎠ Or after integration and applying the symmetry boundary condition at r = 0. Δp τ rz = − r 2L Evaluating at r = RC and using the result obtained in part (a) for the shear stress: V Δp RC −µ C = − δ 2L Δp δ RC VC = Rearranging: 2µL (c) There are two possible ways to approach this. One is to neglect the fluid in the gaps. RC

π ∫ VC rdr

RC ⎛ δ⎞ = VC ⎜ 1 − ⎟ πR R ⎝ R⎠ The more general approach is to consider the fluid in the gap. v =

0

2

= VC

R

v =

π ∫ v z rdr 0

π R2

R R ⎞ 1 ⎛ C y3 ⎛ δ⎞ = 2 ⎜ ∫ VC rdr + ∫ v z rdr ⎟ = VC ⎜ 1 − ⎟ + VC 3 ⎟ R ⎜⎝ 0 3R ⎝ R⎠ RC ⎠

R

RC

3 3 ⎛ δ ⎞ VC ⎛ RC ⎞ ⎛ δ ⎞ VC ⎛ δ ⎞ v = VC ⎜ 1 − ⎟ + ⎜1 − 3 ⎟ = VC ⎜1 − ⎟ + ⎜ 3⎟ ⎝ R⎠ 3 ⎝ R ⎠ ⎝ R⎠ 3 ⎝R ⎠

(d) From parts (b) and (c) ⎛ δ ⎞ Δp 2 ⎛ δ ⎞ ⎛ δ ⎞ v = VC ⎜ 1 − ⎟ = R ⎜ ⎟ ⎜1 − ⎟ ⎝ R ⎠ 2µL ⎝ R ⎠ ⎝ R ⎠

2

or 3 ⎛ δ ⎞ Δp 2 ⎛ δ ⎞ ⎛ δ ⎞ ⎡⎛ δ ⎞ 1 ⎛ δ ⎞ ⎤ v = VC ⎜ 1 − ⎟ = R ⎜ ⎟ ⎜ 1 − ⎟ ⎢⎜ 1 − ⎟ − ⎜ 3 ⎟ ⎥ ⎝ R ⎠ 2 µL ⎝ R ⎠ ⎝ R ⎠ ⎣⎝ R ⎠ 3 ⎝ R ⎠ ⎦

v =

Thus,

µeff µ

=

1 ⎛ δ ⎞⎛ δ ⎞ 4 ⎜ ⎟ ⎜1 − ⎟ ⎝ R ⎠⎝ R ⎠

2

R 2 Δp 8μeff L (1)

or

µeff µ

=

1 ⎛ δ ⎞⎛ δ ⎞ 4 ⎜ ⎟ ⎜1 − ⎟ ⎝ R ⎠⎝ R ⎠

2

⎡⎛ δ ⎞ 1 ⎛ δ 3 ⎞ ⎤ ⎢⎜ 1 − ⎟ − ⎜ 3 ⎟ ⎥ ⎣⎝ R ⎠ 3 ⎝ R ⎠ ⎦

(2)

(e) As shown in the table below, there is a minimum in the viscosity at δ/R between 0.2 and 0.3 and the viscosity increases as δ/R increases.

27

δ /R 0.1 0.2 0.3 0.4

µeff/µ formula 1 3.09 1.95 1.70 1.74

µeff/µ formula 2 3.43 2.45 2.46 3.00

2.26. (a) From Equation (2.7.36), the shear stress of the flow in a cylindrical tube is: Δpr τ rz = − (S2.26.1) 2L When r < rc , τ rz < τ 0 and when r > rc , τ rz > τ 0 . Therefore, 1/2

⎛ Δpr ⎞ rc < r < R, ⎜ ⎟ ⎝ 2L ⎠

= (τ 0 )

1/2

+ (η N )

1/ 2

1/ 2

⎛ dvz ⎞ ⎜ dr ⎟ ⎝ ⎠

(S2.26.2)

dv z =0 dr Rearrange Equation (S2.26.2), we have, r < rc ,

1/2 dv z 1 ⎡⎛ Δpr ⎞ 1/2 ⎤ = − τ ( ) ⎢⎜ ⎥ 0 dr η N ⎢⎣⎝ 2 L ⎟⎠ ⎦⎥

(S2.26.3)

2

(S2.26.4)

Integrating Equation (S2.26.4) from r to R: v z ( R) − v z (r ) =

1

ηN

2

∫

R

r

⎡⎛ Δpx ⎞1/2 1/2 ⎤ τ − ( ) ⎢⎜ ⎥ dx 0 ⎟ ⎣⎢⎝ 2 L ⎠ ⎦⎥ R

⎤ 1 ⎡ Δp 2 4 ⎛ Δpτ 0 ⎞ 3/2 x − ⎜ x +τ0 x⎥ = ⎢ ⎟ ηN ⎣ 4L 3 ⎝ 2L ⎠ ⎦r Applying the no slip boundary condition at r = R and evaluating the integral, we have, r 2 ⎞ 8 rc1/2 ⎛ r 3/2 ⎞ 2rc ⎛ r ⎞⎤ ΔpR 2 ⎡⎛ rc < r < R, v z = 1 1 1 − ⎟⎥ (S2.26.5) − − − + ⎢⎜ ⎜ 2 ⎟ 1/2 ⎜ 3/2 ⎟ 4η N L ⎣ ⎝ R ⎠ 3 R ⎝ R ⎠ R ⎝ R ⎠ ⎦ Integrating Equation (S2.26.3) from r to rc, vz (rc ) − vz (r ) = 0 The velocity at r = rc is obtained from Equation (S2.26.5). For r < rc , the velocity profile is: vz =

rc2 ⎤ ΔpR 2 ⎡ 8 rc1/ 2 2rc + − ⎢1 − ⎥ 4η N L ⎢⎣ 3 R1/ 2 R 3R 2 ⎥⎦

(b) Based on Equation (S2.26.1),

(S2.26.6)

τw = −

ΔpR 2L

Combining the above equation and Equation (S2.10.7), we have, τw R = τ o rc

(c) Results are plotted in Figure S.2.26.1.

28

(S2.26.7)

Figure S2.26.1 (d) The wall shear stress (r = R) can be computed from Equation (2.7.36) and is independent of the constitutive equation, τw = ΔpR/2L. The velocity profiles, shear rate and apparent viscosity are dependent on the constitutive equation. For blood, the shear rate at r = R is: 1/ 2 ⎛ τ0 ⎞ τ0 ⎤ dv z ΔpR 2 ⎡ 2 8 rc1/ 2 ⎛ 3 ⎞ 1 2rc ⎤ τ w ⎡ ⎢ = 1 2 − + − = − + ⎢ ⎜ ⎟ − ⎥ 1/2 ⎜ ⎟ 2 ⎥ dr 4η N L ⎣⎢ R 3 R ⎝ 2 ⎠ R R ⎦⎥ η N ⎢ τw ⎥ ⎝τw ⎠ ⎣ ⎦

rc τ 0 = . For τw = 15 dyne cm-2, τ0/τw = 0.0013. The shear rate is 7.2% lower than the R τw value for a Newtonian fluid. Correspondingly, the apparent fluid viscosity at r = R is 7.2% greater than the value for a Newtonian fluid. For τw = 2 dyne cm-2, τ0/τw = 0.01 and the shear rate is 19% lower than the value for a Newtonian fluid and the apparent viscosity at r = R is 19% larger than the value for a Newtonian fluid. For τw = 0.2 dyne cm-2, τ0/τw = 0.1 and the shear rate is 53.3% lower than the value for a Newtonian fluid and the apparent viscosity at r = R is 53.3% larger than the value for a Newtonian fluid.

where

2.27. The relation between flow and pressure drop is: ΔpπR 4 Q= 8μ L Assuming that flow is proportional to current and pressure drop is proportional to potential difference, this result is analogous to Ohm’s Law with a resistance equal to: 8μ L Resistance = π R4 The enzyme treatment decreased the resistance by 16%. Assuming that the enzyme decreased the inner radius of the blood vessel by removing the glycocalyx completely, the change is resistance is due solely to a change in the effective radius of the blood vessel. Thus:

29

0.84 =

R4

4 Renzyme Taking the one-fourth root yields an increase in radius of Renzyme=1.045R. Assuming that the radius after enzyme treat is 13.5 µm, then the glycocalyx thickness is 0.045*13.5 µm = 0.608 µm.

30

Solution to Problems in Chapter 3. Section 3.8 3.1. The Strouhal number is Sr =

L T vx

α=

ω R2 ν

The Womersley number is

Noting that 2R =L and T=1/ω, the definition of the Womersley number can be rearranged as follows: α=

⎛ R vx ω R2 vx = ⎜⎜ ν vx ⎝ ν

⎞ ωR ⎛ Re ⎞ Sr = ⎜ = 0.5 Re Sr ⎟⎟ ⎟ ⎝ 2 ⎠ 2 ⎠ vx

3.2. For flow in a cylindrical tube, the friction factor is defined as: f =

ΔP 2ρ v

2

D L

(S3.2.1)

Rearranging Equation (2.7.44) yields a relation between the pressure drop and average velocity: Q=

Δpπ R 4 8μ L

(S3.2.2)

Q = πR2 Δp =

8μ LQ

π R4

=

(S3.2.3) 8μ L v R2

(S3.2.4)

Substituting Equation (S3.2.4) for Δp in equation (S3.2.1) yields: f =

8μ L v 2ρ R

2

v

2

D 4μ L D 16μ 16 = = = 2 L ρ ( D/2 ) v L ρ D v Re

(S3.2.5)

3.3. The coordinate system originates at the centerline as shown in Figure 3.19. The boundary conditions are stated below. (S3.3.1a) R = εR vz = 0 (S3.3.1b) R=R vz = 0 The solution to this problem parallels the case of flow through a cylindrical tube through integration of Equation (2.7.36): τ rz = −

Δpr C2 + 2L r

(S3.3.2)

The stress is not known at any point in the fluid so the boundary condition cannot be evaluated at this point. Substituting for a Newtonian fluid: μ

dvz Δpr C2 =− + 2L dr r

(S3.3.3)

Integrating this expression yields an expression for vz(r) in terms of two constants: vz = −

Δpr 2 C2 ln r + C3 + 4μ L μ

(S3.3.4)

Evaluating the two boundary conditions results in the following two equations: 31

0=−

ΔpR 2 C2 ln R + C3 + 4μ L μ

0=−

Δpε R 2 C2 ln ε R + C3 + 4μ L μ

(S3.3.5a) (S3.3.5b)

Solving yields the following values for C2 and C3: C2 = −

(

ΔpR 2 1 − ε 2 4 L ln ( ε )

)

C3 =

The velocity vz(R) is: ΔpR 2 vz = 4μ L

⎡⎛ 2 ⎢⎜ 1 − r ⎢⎜⎝ R 2 ⎣

(

)

2 2 ΔpR 2 ΔpR 1 − ε ln R − 4μ L 4 L ln ε

(

(S3.3.6a,b)

)

2 ⎤ ⎞ 1− ε ln ( r / R ) ⎥ ⎟⎟ − ⎥ ⎠ ln ( ε ) ⎦

(S3.3.7)

Note, that as ε goes to zero, the velocity profile approaches the parabolic profile for laminar flow in a cylinder. The velocity profile is shown below for values of ε= 0.01, 0.1, 0.5. For all values of ε, there is a significant distortion of the velocity profile due to the presence of the catheter.

To find the volumetric flow rate, compute the average velocity in the fluid between r and R and compare to the value obtained in the absence of the catheter. The average velocity is: v =

2π R

1

(

π R2 1 − ε 2

) ∫ ε∫ 0

v =

v =

(

2

R2 1 − ε 2

)

ΔpR 2 4μ L

⎛⎡ 2 4 ⎜ ⎢⎜⎛ r − r ⎜ ⎢⎜⎝ 2 4 R 2 ⎝⎣

vz ( r )rdrdθ =

R

(

2

R2 1 − ε 2 r=R

(

)

ΔpR 2 4μ L

1− ε 2 ⎞⎤ − ⎟⎟ ⎥ ⎠ ⎦⎥ r =ε R ln ( ε )

32

)

(

2

R2 1 − ε 2

) ε∫

⎡⎛ r2 ∫ ⎢⎢⎜⎜⎝1 − R 2 εR ⎣ R

R

vz ( r ) rdr

(S3.3.8)

R

(

)

2 ⎤ ⎞ 1− ε ⎥ rdr − ln / r R ( ) ⎟⎟ ln ( ε ) ⎥ ⎠ ⎦

R

⎞

εR

⎠

∫ ( r ln r − r ln R ) dr ⎟⎟

(S3.3.9a,b)

The first term in the integral can be obtained by integration by parts (Appendix A1A). Let u=lnr, du=dr/r, dv =rdr and v=r2/2. Then:

∫ r ln rdr = v =

r 2 ln r 1 r 2 ln r r 2 − ∫ rdr = − 2 2 2 4

2

(

R2 1 − ε 2

)

⎛ ΔpR 2 ⎜ ⎡⎛ r 2 r4 ⎢⎜⎜ − 2 4 μ L ⎜ ⎣⎢⎝ 2 4 R ⎝

(

r=R

1− ε 2 ⎞⎤ − ⎟⎟ ⎥ ⎠ ⎦⎥ r =ε R ln ( ε )

)⎛r ⎜⎜ ⎝

2

r=R ⎞ ⎞ ln r r 2 r 2 ⎟ − − ln R ⎟⎟ 2 4 2 ⎠ r =ε R ⎟⎠

(S3.3.10)

Evaluating r between the limits yields:

v =

(

2

R2 1 − ε 2

)

(

⎛ ⎡⎛ R2 1 − ε 4 ΔpR 2 ⎜ ⎢⎜ R 2 2 2 + − − 1 ε 1 ε 4μ L ⎜⎜ ⎢⎜ 2 4 ⎝ ⎣⎝

(

)(

)

) ⎞⎟⎥⎤ + R (1 − ε ) 2

⎟⎥ ⎠⎦

2 2

4 ln ( ε )

⎞ ⎟ ⎟ ⎟ ⎠

(S3.3.11)

Collecting terms: v =

(

1

R2 1 − ε 2

v =

)

(

)

2 ⎛ 1− ε 2 ΔpR 4 ⎜ 4 1− ε + 8μ L ⎜⎜ ln ( ε ) ⎝

(

)

(

⎛ 1− ε 2 1 ΔpR 4 ⎜ 2 + + 1 ε ln ( ε ) R 2 8μ L ⎜ ⎝

(

)

The flow rate is Q = π R2 (1-ε2): Q=

(

)

(

Δpπ R 4 1 − ε 2 ⎛ 1− ε 2 ⎜ 1+ ε 2 + ⎜ 8μ L ln ( ε ) ⎝

(

)

Rearranging: 8μ LQ Δpπ R 4

(

= 1− ε

2

⎞ ⎟ ⎟ ⎟ ⎠

(S3.3.12)

) ⎟⎞

) ⎞⎟

)

(S3.3.14)

⎟ ⎠

(

⎛ 1− ε 2 ⎜ 1+ ε 2 + ⎜ ln ( ε ) ⎝

)(

(S3.3.13)

⎟ ⎠

) ⎞⎟ ⎟ ⎠

(S3.3.15)

The left hand side equals the ratio of the flow rate through the annulus to the flow rate through a cylinder. The term on the right hand side is less than or equal to one. The right hand side reduces to 1 as ε goes to zero. The presence of the annulus has a significant effect on the flow rate that is larger than the relative reduction in the cross-sectional area (1-ε2). For ε = 0.01, 0.1, 0.5 the flow rate is respectively, 78.3%, 57.4% and 12.6% of the flow rate in a cylinder.

3.4. The flow rate in the channel is w/ 2

Q=

h /2

∫ ∫y =−h / 2 v x ( y, z ) dydz

z =− w / 2

Inserting the velocity vx (equation (3.4.27)) into the expression for the flow rate. Δph 2 Q= 8μL

⎡ ⎢⎛ 4 y2 ⎢ 1 ⎜ ∫ ∫y =−h / 2 ⎢⎝⎜ h2 z =− w /2 ⎢ ⎣ w /2

h/ 2

⎞ ∞ ⎟⎟ - ∑ ⎠ n =0

n ⎛ (2n + 1)πz ⎞ ⎛ (2n + 1)πy ⎞ 32 ( −1) cosh ⎜ ⎟ cos ⎜ ⎟ h h ⎝ ⎠ ⎝ ⎠ ⎛ (2n + 1)πw ⎞ 3 3 (2n + 1) π cosh ⎜ ⎟ 2h ⎝ ⎠

33

⎤ ⎥ ⎥ dydz ⎥ ⎥ ⎦

Integrating term by term: ⎡ h/2 2 ⎢ ⎞ h n⎛ ⎛ (2n + 1)πz ⎞ ⎛ (2n + 1)πy ⎞ 32(−1) ( −1) ⎜ ⎟ sinh ⎜ ⎟ sin ⎜ ⎟ ⎢ (2n + 1)π ⎠ h h Δph 2 ⎢ 2 wh ∞ ⎝ ⎠ ⎝ ⎠ ⎝ −∑ Q= + π (2 1) n w 8μL ⎢ 3 ⎛ ⎞ 3 3 n =0 (2n + 1) π cosh ⎜ ⎟ ⎢ 2h ⎝ ⎠ ⎢ y=-h/2 ⎣

w/2

z=-w/2

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Evaluating the terms in the summation at their limits for z and y and collecting terms yields Equation (3.8.1). ⎡ n ⎛ (2n + 1)πw ⎞ 192 ( −1) tanh ⎜ ⎟ Δpwh3 ⎢ ⎛ h ⎞ ∞ 2h ⎝ ⎠ ⎢1 − ⎜ ⎟ ∑ Q= 12μL ⎢ ⎝ w ⎠ n = 0 (2n + 1)5 π5 ⎢ ⎣

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

Defining the term in brackets as g(w/h) and using the one-dimensional definition of the wall shear stress, 6µQ/wh2, the wall shear stress is estimated as: τ w = τ yx

y =− h /2

=

Δphg (h / w) 2L

The normalized shear stress τw/(Δph/2L) computed from the flow rate (=g(w/h))and the bracketed term in Equation (3.4.28), f(w/h), indicates a deviation between the two results. The error, shown in the panel on the right, decreases as the ration w/h increases. The reason for the deviation is that the one-dimensional approximation for the shear stress, 6µQ/wh2 = Δph/2L, neglects the contribution of the stresses τzx at ±w/2. Integrating Equation (3.4.7) from –h/2 to h/2 in the y direction and –w/2 to w/2 in the z direction yields: τ w = τ yx

y =− h /2

+

h τ zx 2w

z =− w /2

=

Δph 2L

The overbar represents the spatial average in the z and y directions for τyx and τzx, respectively.

34

3.5. For w/h = 50, the one-dimensional approximation (Figure 3.9) is very good. The time to reach within 1 % of the steady state value can be assessed by adjusting the dimensionless time, th2/υ until the following ration is greater than 0.99. ⎛ ( 2n + 1)2 π 2 vt ⎞ ⎛ ( 2n + 1) π y ⎞ n 32 ( −1) exp ⎜ − ⎟ cos ⎜ ⎟ 2 ⎜ ⎟ h h ⎝ ⎠ ∞ ⎝ ⎠ ∑ n=0 3 3 vx( y , z ) ( 2n + 1) π =1− 2 2 ⎛ 4y ⎞ 4y 2 ⎞ Δph ⎛ ⎜1 − 2 ⎟ ⎜1 − 2 ⎟ 8μ L ⎝ h ⎠ h ⎠ ⎝ The velocity is within 1% of its steady state value for tv/h2 = 0.5. Based on the half-thickness (4tv/h2) as shown in Figure 3.10, the time dimensionless time to reach steady state is 2.0. For v = 0.006 cm2s-1 and h = 0.04 cm, the time to reach steady state is 0.1333 s. Note that the statement on page 144, line 2 should indicate that for tv/h2 = 0.5 (or 4tv/h2 = 2.0), the solution is within 1% of its steady state value.

3.6. (a) From equation (3.4.45) the shear stress τyx(y, t) is found as: ⎡ ⎛ (2n + 1) 2 π2ν t ⎞ n ⎛ (2n + 1)πy ⎞ ⎤ 32 1 exp cos − − ( ) ⎢ ⎜ ⎟ ⎜ ⎟⎥ h h2 ⎝ ⎠⎥ ∂v 4y 2 ⎞ ∞ Δph 2 ∂ ⎢⎛ ⎝ ⎠ τ yx ( y, t ) = μ x = 1 ⎜ ∑ 2 ⎟ 3 3 ⎥ ⎟ ∂y 8L ∂ y ⎢⎜⎝ h ⎠ n =0 (2n + 1) π ⎢ ⎥ ⎢⎣ ⎥⎦ ⎡ ⎢ ∂v Δph ⎢ y τ yx ( y, t ) = μ x = − − ∂y L ⎢h ⎢ ⎢⎣

⎛ (2n + 1) 2 π2ν t ⎞ ⎛ (2n + 1)πy ⎞ ⎤ n 4 ( −1) exp ⎜ − ⎟ sin ⎜ ⎟⎥ h h2 ⎠⎥ ⎝ ⎠ ⎝ ∑ 2 2 ⎥ (2n + 1) π n =0 ⎥ ⎥⎦ ∞

The time-derivative of the shear stress gradient is: ∂τ yx ( y, t ) ∂t

=−

⎛ (2n + 1) 2 π2ν t ⎞ ⎛ (2n + 1)πy ⎞ ⎤ 4νΔp ⎡ ∞ n ⎢ ∑ ( −1) exp ⎜⎜ − ⎟⎟ sin ⎜ ⎟⎥ h hL ⎣⎢ n =0 h2 ⎠ ⎦⎥ ⎝ ⎠ ⎝

Evaluating the time derivative at the surface y=-h/2 and normalizing by hL/4νΔp yields: ⎛ (2n + 1) 2 π2ν t ⎞ ⎤ hL ∂τ yx ( y, t ) ⎡ ∞ = ⎢ ∑ exp ⎜⎜ − ⎟⎟ ⎥ 4νΔp ∂t h2 ⎝ ⎠ ⎦⎥ ⎣⎢ n =0

(b) The normalized gradient is shown below.

35

(c) The quantity, hL/4νΔp can be rewritten as h2/8ντw. Multiplying the normalized gradients by 8ντw/h2 yields the dimensional gradient. For and ν = 0.007 cm2 s-1, h = 0.015 cm, h = 0.05 cm and τw = 1 dyne cm-2, the gradients are shown below. The shear stress gradients are very different. To ensure constant shear stress for these two conditions, the flow rates must be adjusted so that 6µQ/wh2 is the same for both cases. This requires that the flow rate for h = 0.015 cm be 0.09 times the value for h = 0.05 cm.

3.7. The unsteady velocity profile for flow in a cylindrical tube of radius R is: ΔPR 2 ⎛ r2 vz = ⎜⎜1 − 2 4μ L ⎝ R

⎞ 2ΔPR 2 ⎟⎟ μL ⎠

∞

∑

(

) ( λ J (λ )

J o λn r / R exp −λn2ν t / R 2 3 n 1

n =1

)

n

In terms of the average velocity vmax/2 and the dimensionless time, τ = νt/R2. 36

(S3.7.1)

⎛ r2 ⎞ vz = 2 v ⎜⎜1 − 2 ⎟⎟ - 8 v ⎝ R ⎠

∞

∑

n =1

(

) ( λ J (λ )

J o λn r / R exp −λn2τ 3 n 1

)

(S3.7.2)

n

The following m-file was written to solve equation 2 function zz=radial(t); nn=0:0.0495:1; zz1=0; for n=1:1:10; y=fzero('bb',n*2.8); r1=-8*besselj(0,y*nn)/(y*y*y*besselj(1,y))*exp(-y*y*t); zz1=zz1+r1; end zz=1-nn.*nn+zz1; where

function z=bb(x); z=besselj(0,x);

Panel A of the figure shows the relative velocity as a function of dimensionless time. To identify when the profile is within 1% of steady state the unsteady term (second term on the right hand side of equation 2) was isolated and plotted for various values of the dimensionless time t. The solution is within 1% of steady state for dimensionless times between 0.8 and 0.9. For the rectangular channel, the corresponding time is based on the channel half-thickness, h/2 and 4tn/h2 = 2.0. The shorter time to reach steady state in the cylindrical geometry arises as a result of the deceasing cross-sectional area which improves momentum transport.

3.8. The force balance is the same that presented in equation (3.6.29) and shown in Figure 3.17. Thus, the velocity vz is obtained from equation (3.6.30) after a change in notation: vz =

2 ga 2 ( ρπ − ρ ( z ) ) 9μ

Inserting the expression for r (z) and expressing the velocity as the derivation of the z position with time. 37

dz 2 ga 2 = ( ρπ − ρo − α z ) dt 9μ

Rewriting in terms of the time constant for particle settling, tc = 2rpa2/9µ, dz = tc g (1 − ρ − α ' z ) dt

where ρ = ρ/ ρπ and α’= α/ρπ. Rearranging: −α ' dz

(1 − ρ − α ' z ) Integrating

= −α ' tc gdt

ln (1 − ρ − α ' z ) = −α ' tc gt + C1

Applying the initial condition t = 0 z = z0, C1 = ln(1-ρ-α’z0). The solution becomes: ρo + α z = ρπ − ( ρπ − ρo − α z0 ) exp ( −α ' tc gt )

The exponential is less than 0.01 and ρ0 + αz0 ≈ ρπ for t > 5/α’tcg.

3.9. The time rate of change in cell length in the pipet L’ is a function of the following variables: L’ = f(RP, Rc, µc, ΔP) There are three dimensions (m L t) and five dimensional groups. Thus, according to the Pi theorem there are two dimensionless groups. Choose the following basis group which contains all three dimensions (µc)a(ΔP)b(RP)c. The two dimensionless groups will include L’ and Rc. For L’ For mass For length For time c =-1

L’(µc)a(ΔP)b(RP)c a +b = 0 1-a-b+c = 0 -1-a-2b = 0

( l t-1)(m l-1t-1)a(m l-1 t-2)b(l)c b = -a or –1+a = 0 a = 1

and

b = -1,

L’µc(ΔP)-1RP-1 The group L’µcRP-1 represents viscous stresses in the cell which are balanced by the applied pressure. The second dimensionless group can be found, using the same approach, to be RcRP-1.

3.10. The force F acting on the leukocyte is a function of the following variables: F =f(dc,Dt, vc, vf, Hct, ρ, µ) While there are eight variables, the hematocrit Hct is dimensionless. So there are 7 dimensional variables and three characteristic dimensions, mass, length and time. Thus, in addition, there are four other dimensionless groups. If we choose the basis group as (µ)a(Vf)b(dt)c, the four other variables to make dimensionless are F, ρ, Dc and vc. While the Reynolds number is clearly one of the groups and uses the variable ρ, each group is formally derived as follows. For F F(µ)a(Vf)b(dt)c (m l t-2)(m l-1t-1)a(lt-1)b(l)c To be dimensionless, each dimension must sum to zero. For mass 38

1+a = 0 or a = -1 For length 1+1+b+c=0 or b+c = -2 For time -2+1-b = 0 or b = -1 and c = -1 -1 -1 -1 Thus the dimensionless group with force is: Fµ Vf dt . This can be rewritten as Fdt-2µ -1Vf-1 dt which represents the force per unit area normalized by the shear stress. v(µ)a(Vf)b(dt)c ( l t-1)(m l-1t-1)a(lt-1)b(l)c For vc For mass a=0 For length 1+b+c = 0 For time -1-b = 0 or b = -1 and c=0 -1 Thus the dimensionless group with vc is: vcVf which is simply the velocity ratio. For ρ ρ(µ)a(Vf)b(dt)c (m l-3)(m l-1t-1)a(lt-1)b(l)c For mass 1+a=0 or a = -1 For length -3+1+b+c = 0 or -2+b+c = 0 For time 1-b = 0 or b = 1 and c=1 Thus the dimensionless group with ρ is: ρVfdt/µ which is the Reynolds number. Dc(µ)a(Vf)b(dt)c (l)(m l-1t-1)a(lt-1)b(l)c For Dc For mass a=0 For length 1+b+c = 0 For time b=0 c = -1 Thus the dimensionless group with ρ is: Dc/dt which is the diameter ratio. Thus, the dimensionless relation is: Fµ -1Vf-1 dt-1 = f(Re, vcVf-1, Dcdt-1)

3.11. (a) For two-dimensional flow of an incompressible fluid, the conservation of mass yields: ∂ vx ∂ vy + =0 ∂x ∂y When the fluid contacts the sold surface the no slip condition results in fluid deceleration in the x-direction. Since the partial of vx with respect to x is negative for 0 ≤ y ≤δ, then ∂ vy >0 ∂y Since the derivative is positive within the boundary layer, vy is positive throughout this region. (b) Assume that vx can be represented as: vx = a1(x)y + a2(x)y2 + .… This velocity profile guarantees that vx is zero at y = 0. At y = δ we expect that vx = U. Just using the first term, application of the boundary condition at y = d yields: vx = Uy/δ The derivative of vx with respect to x is: ∂v ∂ vx Uy dδ =− 2 =− y ∂x δ dx ∂y 39

Note that δ is increasing with x so its derivative with respect to x is positive. Integrating Uy 2 dδ since vy = 0 at y=0. The resulting shape is this result yields the following: v y = 2 δ dx a parabola. (c) For the case in which the velocity profile is limited to two terms, vx = a1(x)y + a2(x)y2, the boundary conditions are satisfied when U = a1δ + a2δ2 and the first derivative at y = δ equals zero results in a1 =-2 a2δ. This latter condition ensures that the shear stress is continuous. As a result, a2 = -U/δ2 and a1 = 2U/δ. The resulting velocity profile is: vy = U(2y/δ - y2/δ2). The assumption that the velocity gradient is zero at y = δ and the parabolic shape of the profile ensure that the velocity is a maximum at y = δ.

3.12. (a) Using the dimensionless variables listed, equation (3.8.5a) becomes: ρ (ω R ) vθ* 2

r*

R

=−

ωμ ∂ P* μω R ∂ 2 v*r + α R ∂ r * (α R ) 2 ∂ z *

Simplifying: ωμ ∂ P* μω ∂ 2 v*r + α R ∂ r * α 2 R ∂ z*2 r* ραω R 2 vθ* ∂ P* ∂ 2 v*r = − + α α μ ∂ r * ∂ z*2 r* ρω 2 R

vθ*

=−

(S3.12.1)

The group ραωR2/µ represents the ratio of inertial to viscous forces. Using the same approach on equation (3.8.5b) ⎡ ρ (ω R )2 ∂ v* ρα (ω R )2 ∂ v* ⎤ μωα R ∂ 2 v*z * θ θ ⎢ ⎥ v*r v + = + z R αR ∂ r* ∂ z* ⎥⎦ ⎢⎣ (α R )2 ∂ z*2

Simplifying: 2 ρ (ω R ) ⎡

* * ⎤ μω ∂ 2 v*z * ∂ vθ * ∂ vθ v + = + ⎢ vr ⎥ z * α R ∂ z*2 ∂ z* ⎥⎦ ⎢⎣ ∂ r

R

ραω R 2 μ

* ⎤ ⎡ * ∂ vθ* ∂ 2 v*z * ∂ vθ v + = + ⎢ vr ⎥ z * ∂ z* ⎥⎦ ∂ z*2 ⎢⎣ ∂ r

(S3.12.2)

The dimensionless form of the z-component is 0=−

ωμ ∂ P* μαω R ∂ 2 v*z + α 2 R ∂ z * (α R ) 2 ∂ z * 0=−

∂ 2 v*z ∂ P* +α * ∂z ∂ z*

(S3.12.3)

(b) Assuming that ραωR2/µ << 1 and α<<1, Equations (S3.12.1), (S3.12.2) and (S3.12.3) simplify to: 0=

∂ 2 v*r ∂ z*2

(S3.12.4) 40

∂ 2 v*z ∂ z*2 ∂ P* 0=− * ∂z

(S3.12.5)

0=

(S3.12.6)

(1) If inertial forces are much smaller than viscous forces, then vr does not change in the r direction and must nut depend on r since vr = 0 at r = 0 . From equation (S3.13.4), vr* = C1z*+ C2. From the boundary condition at z = z* = 0, C2 = 0. In order to satisfy the boundary condition at z = h or z* = h/αR, C1 must equal 0. Therefore vr* = 0 throughout. (2) As a result, the conservation of mass reduces to: 0=

∂ v*z ∂ z*

(S3.12.7)

Integrating, vz* = C3. The only way in which both boundary conditions could be satisfied (vz* = 0 at z = 0 and vz* = ωr a z= h(α)) is for vz* = 0. Note that the problem statement should refer to the θ component of the conservation of (3) linear momentum, not mass. Based on the results obtained above, the only nonzero component of velocity is vθ. As a result, equation (3.8.5b) reduces to ∂ 2 vθ* =0 ∂ z*2

(S3.12.8)

Integrating twice yields: (S3.12.9) The boundary conditions are that at z = 0, vθ* = 0 and at z = h(α), vθ* = ωr. From the condition at z = 0, C2 = 0. From the condition at z = h(α),C1 = ωr/h(α). The resulting expression for vθ* is: ωrz (S3.12.10) vθ * = h (α ) which agrees with Equation (3.8.3). vθ * = C1 z + C2

(c) The shear stress at the plate surface is obtained by inserting equation (3.8.6) into equation (3.3.23e) and taking the derivative and evaluating at z = 0. ∂v μω ⎡⎣1 + 0.001ε 2 ⎤⎦ = τ zθ |z =0 = μ θ ∂z z =0 α To determine the value of w for which the shear stresses at r = 15 and 25 cm differ by 10%, the ration of τrz|z=0 for the two positions must equal 1.10: 2 ⎡ ⎛ 1( 25 )2 ω ( 0.05236 )2 ⎞ ⎤ ⎢1 + 0.001⎜ ⎟ ⎥ ⎜ ⎟ ⎥ ⎡ 2 ⎢ 0.01 ⎝ ⎠ ⎦ ⎣1 + 29.3602ω ⎦⎤ ⎣ = 1.10 = 2 ⎡ ⎡⎣1 + 3.8051ω 2 ⎤⎦ ⎛ 1(15 )2 ω ( 0.05236 )2 ⎞ ⎤ ⎢1 + 0.001⎜ ⎟ ⎥ ⎜ ⎟ ⎥ ⎢ 0.01 ⎝ ⎠ ⎦ ⎣ -1 Solving for w yields, ω = 0.0630 rad s . 41

3.13. Equation (3.6.32) is: dU o 9μ = ( vt − U o ) dt 2ρ p R2

Let u = vt –Uo. Thus, d = -du. As a result, equation (3.6.32) becomes du 9μ =u dt 2ρ p R2

Integrating yields

Or

⎛ ⎞ 9μ u = Aexp ⎜ t⎟ 2 ⎜ 2ρ p R ⎟ ⎝ ⎠ ⎛ 9μ ⎞ ⎟ t U 0 = v t -Aexp ⎜ ⎜ 2ρ p R2 ⎟ ⎝ ⎠

At t = 0, Uo = 0. As a result, A = vt. The final result is: ⎡ ⎛ 9μ ⎞⎤ ⎟⎥ U 0 = v t ⎢1-exp ⎜ t ⎜ 2ρ p R2 ⎟⎥ ⎢⎣ ⎝ ⎠⎦

3.14. For an adherent leukocyte, H/R = 1 and F*(H/R) = 1.69. Thus, the drag force is FD = 31.85μγ R 2 = 31.85τ w R 2 where τw = 6µQ/wh2 = 6µ/h where w is the channel width and h is the channel height. From the data given τw = 6(0.009 g cm-1 s-1)(2200 µm s-1)/(230 µm)= 0.517 dyne cm-2. The resulting drag force is 4.11 x 10-6 dyne = 41.1 pN. 3.15. This problem is the inverse of the example discussed in Section 3.6.2. That is, the fluid far from the sphere has a zero velocity and the sphere is moving at a constant speed U0. To find the pressure field, substitute equations (3.3.8a,b) into equations (3.6.7a,b) and compute the derivatives. These derivatives are: 2

⎛ 3R R 3 ⎞ U U ∂ vr ⎛ R ⎞ ⎛ 3r R ⎞ = − o sin θ ⎜ ⎟ ⎜ − ⎟ = − o sin θ ⎜⎜ − 3 ⎟⎟ 2 2 ∂θ r ⎠ ⎝r⎠ ⎝R r⎠ ⎝ r

∂ vr ∂ ⎛ sin θ ⎜ 2 ∂θ r sin θ ∂θ ⎝ 1

⎛ 3R R 3 ⎞ ⎞ = − U cos θ ⎜⎜ 3 − 5 ⎟⎟ o ⎟ r ⎠ ⎠ ⎝r

⎛ 3R 3R 3 ⎞ ∂ vr Uo d ⎛ 3R R 3 ⎞ U o = − 3 ⎟⎟ = cos θ ⎜⎜ cos θ ⎜⎜ − 2 + 4 ⎟⎟ 2 dr ⎝ r ∂r r ⎠ 2 r ⎠ ⎝ r ⎛ R3 ⎞ R3 ⎞ ∂ ⎛ ⎞ 3U o = − + = − cos R 3 U θ ⎜ ⎟ ⎜ 5 ⎟⎟ cos θ o⎜ ⎟ 2 ∂ r ⎜⎝ r 2 ⎟⎠ ⎠ 2r ⎝r ⎠ ⎛ R3 R ⎞ ∂ vθ 3U o = sin θ ⎜⎜ 4 + 2 ⎟⎟ 4 ∂r r ⎠ ⎝r 1 ∂ ⎛ 2 ∂ vr ⎜r ∂ r r2 ∂ r ⎝

⎞ 3U o 1 ∂ ⎛ 2 ∂ vθ ⎞ 3U o 1 ∂ ⎛ R3 = r sin θ ⎜⎜ 2 + R ⎟⎟ = − ⎜ ⎟ 2 2 4 2 ∂r ⎠ r ∂r⎝ r ∂r⎝ r ⎠

42

⎛ R3 ⎞ ⎜⎜ 5 ⎟⎟ sin θ ⎝r ⎠

⎛⎛ R ⎞ ∂ vθ U R⎞ = − o cos θ ⎜ ⎜ ⎟ + 3 ⎟ ⎜⎝ r ⎠ r ⎟⎠ ∂θ 2 ⎝ 3

∂v 1 ∂ ⎛ sin θ θ ⎜ 2 r sin θ ∂θ ⎝ ∂θ ∂v 1 ∂ ⎛ sin θ θ ⎜ 2 ∂θ r sin θ ∂θ ⎝

(

Uo ⎛ R3 R⎞ ∂ ⎞ = − ⎜ 5 + 3 3 ⎟ ( sin θ cos θ ) ⎟ r ⎠ ∂θ 4 sin θ ⎝ r ⎠

)

(

)

U o cos2 θ − sin 2 θ ⎛ R 3 U o 1 − 2 sin 2 θ ⎛ R 3 R⎞ R⎞ ⎞ = 4 + 3 = − ⎜ 5 ⎜ 5 +3 3 ⎟ ⎟ 3 ⎟ 2 sin θ 4 sin θ r ⎠ r ⎠ ⎠ ⎝r ⎝r ⎛ R R3 ⎞ = −U o sin θ ⎜⎜ 3 3 − 5 ⎟⎟ ∂θ r ⎠ ⎝ r

2 ∂ vr r2

Equation (3.6.7a) becomes 0=−

⎡ R3 ∂p + μU o cos θ ⎢ −3 5 ∂r ⎢⎣ r

⎛ 3R R 3 ⎞ ⎛ 3R R 3 ⎞ 1 ⎛ R 3 3R ⎞ 1 ⎛ R 3 R ⎞⎤ − ⎜⎜ 3 − 5 ⎟⎟ − ⎜⎜ 3 − 5 ⎟⎟ + ⎜⎜ 5 + 3 ⎟⎟ + ⎜⎜ 5 + 3 3 ⎟⎟ ⎥ r ⎠ ⎝r r ⎠ 2⎝ r r ⎠ 2⎝ r r ⎠ ⎥⎦ ⎝r

Simplifying 0=−

⎡ R3 ⎛ 3R R 3 ⎞ ⎛ R 3 3R ⎞ ⎤ ∂p ∂p ⎛ 3R ⎞ + μU o cos θ ⎢ −3 5 − 2 ⎜⎜ 3 − 5 ⎟⎟ + ⎜⎜ 5 + 3 ⎟⎟ ⎥ = − − μU o cos θ ⎜ 3 ⎟ ∂r ∂r r ⎠ ⎝r r ⎠ ⎥⎦ ⎝r ⎠ ⎢⎣ r ⎝r

(S3.15.1)

Likewise, after substituting into Equation (3.6.7b), we have 0=−

⎡ 3 ⎛ R3 ⎞ 1 ⎛ R3 1∂ p R ⎞ ⎛ R R3 ⎞ ⎤ + μU o sin θ ⎢ − ⎜⎜ 5 ⎟⎟ + ⎜⎜ 5 + 3 3 ⎟⎟ − ⎜⎜ 3 3 − 5 ⎟⎟ ⎥ r ∂θ r ⎠ ⎝ r r ⎠ ⎦⎥ ⎣⎢ 2 ⎝ r ⎠ 2 ⎝ r

This result simplifies to: 0=−

1∂ p ⎛ 3R ⎞ − μU o sin θ ⎜ 3 ⎟ r ∂θ ⎝ 2r ⎠

(S3.15.2)

Integrating Equation (S3.15.1) yields ⎛ 3R ⎞ p = C1 + μU o cos θ ⎜ 2 ⎟ ⎝ 2r ⎠

As r → ∞, p approaches p∞ which is the value of C1. p = p∞ +

μU o cos θ ⎛ 3R 2 ⎞

(S3.15.3)

⎜⎜ 2 ⎟⎟ ⎝ 2r ⎠

R

As a check, the derivative of p with respect to θ yields Equation (S3.15.2). (a) The drag force is determined by solving equation (3.6.21). First, the shear stress τrθ at r = R is determined using equation (3.3.24b). τ rθ

τ rθ

r=R

r=R

⎡ r d ⎛ R3 ⎛ ∂ ⎛ v ⎞ 1 ∂v r ⎞ R = −μU 0 sin θ ⎢ = μ⎜r ⎜ θ ⎟ + ⎜⎜ 4 + 3 2 ⎟ r ⎝ ∂r ⎝ r ⎠ r ∂θ ⎠ r = R ⎢⎣ 4 dr ⎝ r

⎞ 3R R 3 ⎤ ⎟⎟ + 2 − 4 ⎥ 2r ⎥⎦ ⎠ 2r r =R

⎡ R 3 3 R 3R R 3 ⎤ 3μU 0 sin θ + 2 − 4⎥ = = − μU 0 sin θ ⎢ − 4 − 2 2r 2R 2r 2r ⎦ r = R ⎣ r

From Equation (3.6.21), the force due to fluid shear stresses is: π

−2π ∫ τ rθ 0

π

R 2 sin 2 θ dθ = − 3πμU 0 R ∫ sin 3 θ dθ r=R 0

The integral is obtained by integration by parts 43

π

∫ sin

3

0

π

π

θ dθ = − cos θ sin 2 θ + 2∫ sin θ cos 2 θ dθ 0

0

The first term on the right hand side is zero. The second term is then integrated by parts. Let u = sin2θ and dv = sinθdθ. Thus du = 2cosθsinθdθ and v = -cosθ. π

π

π

π

π

0

0

3 2 3 ∫ sin θ dθ = − sin θ cos θ 0 + 2∫ sin θ cos θ dθ = 2∫ sin θ dθ − 2∫ sin θ dθ 0

0

π

π

π

0

0

= −2 cos θ 0 − 2 ∫ sin 3 θ dθ = 4 − 2 ∫ sin 3 θ dθ

After rearranging and solving, the integral of sin3θ = 4/3. Thus, the component of the drag due to the shear stress is: π

−2π ∫ τ rθ

r=R

R 2 sin 2 θ dθ = − 4πμU 0 R

0

The pressure at r =R is: p = p∞ +

3μU o cos θ 2R

From Equation (3.6.21), the component of the drag force due to the pressure is π π π 3μU o cos θ ⎞ 2 ⎛ R cos θ sin θ dθ = − 3πμU o R ∫ cos 2 θ sin θ dθ −2π∫ p r = R R 2 cos θ sin θ dθ = − 2π ∫ ⎜ p∞ + ⎟ 2R ⎠ 0 0⎝ 0

The integral of p∞ is zero. Since cos2θ+sin2θ = 1, the integral on the right hand side can be rewritten as: π

π

(

)

4⎞ ⎛ −2π ∫ p r = R R 2 cos θ sin θ dθ = − 3πμU o R ∫ sin θ − sin 3 θ dθ = −3πμU o R ⎜ 2 − ⎟ = −2πμU o R 3⎠ ⎝ 0 0

Summing the components of the drag force due to pressure and shear stress. FD = -6πµRU0 The magnitude of the drag force is identical to the value obtained for the case of a stationary sphere exposed to a uniform velocity U0. The minus sign indicates that the drag force is in the direction opposite the motion of the sphere.

3.16. (a). The Reynolds number for the cell at y = 0 is Re = ρDleukocyte/µ. The velocity can be obtained from the definition of the wall shear stress and equation 2.7.24. 6μ v dv 3 ⎛ 8y ⎞ τw = μ x = μ v ⎜− 2 ⎟ = dy y =− h /2 2 h ⎝ h ⎠ y =− h /2 For the data given, = (1 dyne cm-2)(0.032 cm)/(6(0.0071 g cm-1 s-1)) = 0.75 cm s-1. The Reynolds number is 0.11. Thus the assumption of low Reynolds number flow is valid. (b). Same as in unmodified solution. (c). Solve the force balances to determine the distance traveled by a cell which is initially at the centerline. To determine the distance traveled, we need to compute the change in x and y position. The cell velocity in the x direction is just the change in x position with time. Likewise, the cell velocity in the y direction is just the change in y position with time. The fluid velocity is given by equation 2.7.24. Thus, the force balances are rewritten as: dx 3 ⎛ 4 y2 ⎞ x direction = v ⎜1 − 2 ⎟ dt 2 ⎝ h ⎠ 44

dy = vt dt Where vt is the terminal velocity defined by equation 3.6.30. Notice that the displacement of the cell in the x direction depends upon the vertical position of the cell. From the force balance in the y direction, we have, y = vtt, assuming that the cell was motionless prior to t = 0. Thus, the y location of the cell is now known as a function of time. The force balance in the x direction becomes 4v 2 t 2 ⎞ dx 3 ⎛ = v ⎜1 − t2 ⎟ dt 2 ⎝ h ⎠ Integrating with respect to time and starting at x = 0, yields: 4v 2 t 3 ⎞ 3 ⎛ x = v ⎜t − t 2 ⎟ 2 ⎝ 3h ⎠ The time required for the cell to travel from y = 0 to y = -h/2 is t = -h/2vt. Note that the terminal velocity is negative so time is still a positive quantity! When y = -h/2, x = x0 and 3 ⎛ h ⎞ ⎞ 2⎛ ⎜ 4vt ⎜ − ⎟ ⎟ v 2 3 ⎜ h 3 v ⎛ 1 1⎞ 1 v t ⎠ ⎟ ⎝ − = h x0 = v ⎜ − ⎜− + ⎟ = − h 2 ⎟ 2 2vt 3h 2 vt ⎝ 2 6 ⎠ 2 vt ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ For the data given, vt = -0.0003cm s-1 and x- is 1220h =39.12 cm. (d). This result means that if you measured the number of adherent or rolling cells within 0.30 cm from the introduction of the cells you would underestimate the true frequency because not all of the cells have settled to the surface. y direction

3.17. (a) In order to determine the height, we need to know the viscosity, channel width, flow rate and either the pressure drop or shear stress. From 1, w = 20h, Q per channel is specified in 2 and 3: Q = (10 cm3/hour)(1 hour/3600 s)/(20 channels) = 1. 39 x 10-4 cm3/s From statement 5, the shear stress is specified. From equation (2.7.30) we have

τ yx = −

Δp Δph y y =− h /2 = 2L L

The pressure drop per until length is Δp 2τ yx = L h From Equations (2.7.26) and (2.7.23) Q=

2v max wh 2 Δpwh 3 Δpwh 3 = = 3 24 µL 12 µL

Replacing the pressure drop per unit length and w = 20h 45

Q=

2τ yx wh 3 12 µh

=

τ yx wh 2 6µ

=

20τ yx h 3 6µ

=

10τ yx h 3 3µ

1/3

Solving for h yields:

⎛ 3 µLQ ⎞ h=⎜ ⎜ 10τ ⎟⎟ yx ⎠ ⎝

Substituting the values provided

(

⎛ 3 ( 0.008 g/cm/s ) 6 1.39x10 −4 cm 3 /s h=⎜ ⎜ 10 1 g cm/s2 ⎝

(

)

) ⎞⎟

1/3

⎟ ⎠

= 0.0126 cm

(b) The total pressure drop per channel is Δp =

2τ yx L h

= 2 (1)

6 = 952 dyne/cm 2 0.0069

The total pressure drop is 20 times this value or 19,049 dyne = 1905 Pa. Thus, the criterion is met. The Reynolds number is

Re =

ρ v Dh µ

=

ρ Q2wh

whµ ( w + h )

=

2ρQ µ(w + h)

Re =0.131 Thus, the flow is laminar and close to the criterion for Stokes flow (Re = 0.1).

3.18. For steady, fully developed laminar flow in a cylindrical tube using rectangular coordinates, flow is in the z direction and normal to the x-y plane. For fully developed flow, vz =f(x,y) only. All other velocity components are zero (vx = vy = 0). For steady flow ∂v z (S3.18.1) =0. ∂t Assuming that the channel is horizontal, the Navier-Stokes equation reduces to: ∂p =0 ∂x

(S3.18.2a)

∂p =0 ∂y

(S3.18.2b)

⎛ ∂ 2v ∂2v ⎞ ∂p (S3.18.2c) = μ ⎜ 2z + 2z ⎟ ∂z ∂y ⎠ ⎝ ∂x From Equations (S3.18.2a,b), p(z) only. Since p(z) and vz(x,y), each side of Equation (S3.18.2c) must equal a constant. The constant is equal to –Δp/L, the pressure drop per unit length. ⎛ ∂2v ∂2v ⎞ Δp = μ ⎜ 2z + 2z ⎟ L ∂y ⎠ ⎝ ∂x (a) The boundary conditions are: −

46

(S3.18.3)

vz = finite and ∂v z ∂v z = =0 ∂x ∂y That is, the velocity is symmetric about the centerline.

x = 0, y = 0

x2 + y2 = R2 vz= 0 (b) In order to satisfy Equation (S3.18.3), vz(x,y) =A+B(x2+y2), where A and B are constants of integration. At x2 + y2 = R2, we find that A = -B. This equation already satisfies the condition at x = 0 and y = 0. (c) Substituting the expression for vz into Equation (S3.18.3): Δp B=− 4 µL The velocity profile is: Δp ΔpR 2 ⎛ x 2 + y 2 ⎞ 2 2 2 vz = R −x −y = ⎜1 − ⎟ 4 µL 4 µL ⎝ R2 ⎠

(

)

(S3.18.4)

Note: an alternate solution is: vz(x,y) =A+B(x2+y2) C x2y2. This solution does satisfy the boundary conditions. Using this in Equation (S3.18.3) yields Δp (S3.18.5) − = μ 4 B + 2C x 2 + y 2 L In order for the right hand side to equal a constant C must equal 0.

(

))

(

3.19. The solution for an elliptical tube can be obtained by a variation on the approach used in Problem 3.18. Equation (S.318.3) applies. The boundary condition at x = 0 and y = 0 is unchanged. The tube surface is given by: x 2 y2 + =1 a2 b2 Along this surface, vz = 0.

(S3.19.1)

Since this equation and boundary conditions are similar to the flow in a cylindrical tube solved using rectangular coordinates, we postulate that the solution of the form: ⎛ x 2 y2 ⎞ (S3.19.1) vz = A + B ⎜ 2 + 2 ⎟ ⎝a b ⎠ Applying the no slip condition along the ellipse surface, A = -B. ⎛ x 2 y2 ⎞ (S3.19.2) vz = B ⎜ 2 + 2 − 1⎟ b ⎝a ⎠ Using this velocity field and taking the second derivatives with respect to x and y, −

Δp ⎛ 1 1 = 2μ ⎜ 2 + 2 L ⎝a b

⎛ a2 + b2 ⎞ ⎞ = μ B 2 ⎜ 2 2 ⎟ ⎟ ⎠ ⎝ ab ⎠

Solving for B: 47

(S3.19.3)

B=−

vz =

Δp ⎛ a 2 b 2 ⎞ ⎜ ⎟ 2 µL ⎝ a 2 + b 2 ⎠

Δp ⎛ a 2 b 2 ⎞ ⎛ x 2 y2 ⎞ 1 − − ⎟ ⎜ ⎟⎜ 2 µL ⎝ a 2 + b 2 ⎠ ⎝ a 2 b 2 ⎠

(S3.19.5)

48

(S3,19,4)

Solution to Problems in Chapter 4, Section 4.10 4.1. Begin with equation (4.2.2) for the case of a control volume that changes with time: ∂ρ dV = - ∫ ∇ • ( ρ v ) dV ∂t V (t ) V(t)

∫

Applying Leibniz’s rule (Equation (A.1.29) to the term ∂⎛ ⎜ ρdV ∂t ⎜ V ∫(t ) ⎝

(S4.1.1) ∂⎛ ⎜ ρdV ∂t ⎜ V ∫(t ) ⎝

⎞ ⎟ ⎟ ⎠

⎞ ∂ (l ) ∂ (l ) ∂ρ ⎟= ∫ dV + ρ ( l2 ) dA 2 − ρ ( l1 ) dA 1 ⎟ V (t ) ∂t ∂t ∂t ⎠

(S4.1.2a)

where l2 and l1 are the limits of integration of the volume. Since integration is over the entire surface, the second and third terms on the right hand side of equation (S4.1.2a) represent the sum and (S4.1.2a) becomes ⎡ ⎛ l ∂ ( l2 ) l ∂ ( l1 ) ⎞ ⎤ ∂l − ρ ( l1 ) 1 lim l2 − l1 → 0 ⎢ dA ⎜ ρ ( l2 ) 2 ⎟ ⎥ = ∫ ∇ρ dV ∂t l2 ∂t l1 ∂t ⎠ ⎥⎦ V (t ) ⎢⎣ ⎝

(S4.1.2b)

The derivative of l with respect to time is the velocity of the surface v2. ∂⎛ ⎜ ρdV ∂t ⎜ V ∫(t ) ⎝

⎞ ∂ρ ∂l ⎟= ∫ dV + ∫ ∇ρ dV ⎟ V (t ) ∂t ∂t V ⎠

(S4.1.3)

Applying the divergence theorem to the second term on the right hand side of (S4.1.3) ∂⎛ ⎜ ρdV ∂t ⎜ V ∫(t ) ⎝

⎞ ∂ρ ⎟= ∫ dV + ∫ ρ n • v S dV ⎟ V (t ) ∂t S ⎠

(S4.1.4)

where vs is the velocity of the control volume surface (Deen, W.M., Analysis of Transport Phenomena. 1998, New York: Oxford University Press.). Rearranging and replacing ∂ρ ∫ ∂t dV in Equation (S4.1.1) V(t) ∂ρ ∂⎛ dV = ⎜ ∫ ρ dV ∂t ∂t ⎜ V (t ) V (t ) ⎝

∫

⎞ ⎟ − ∫ ρ n • v S dV = - ∫ ∇ • ( ρ v ) dV ⎟ S V(t) ⎠

(S4.1.5)

Replacing the right hand side of Equation (S4.1.5) with Equation (4.2.4) and noting that ∫ ρdV = m V(t)

dm = -∫ ρ ( v - v S ) • ndS dt S

(S4.1.6)

Equation (S4.1.4) indicates that when the control volume changes with time, the movement of mass across the surface must account for movement of the surface.

4.2. Consider a straight vessel that tapers from a radius R1 at the inlet to R2 = R1 at the outlet. Assume that the flow is laminar, the centerline is horizontal so that gravity does not influence flow and flow is steady. Equation (4.3.8) reduces to: (S4.2.1) ∫ vρ (n • v)dS = − ∫ pndS + ∫ n • τ dS S

S

S

49

Further progress can be made if the control volume is applied to a region over which fully developed flow is established as shown in the figure below.

With these simplifications, the analysis uses the same development presented in Example 4.2. The control volume can be divided into three regions, 1, 2 and 3. Mass enters through region 1 and exits through region 3. There is no flow across the walls of the vessel, region 2. The average velocity in the inlet and outlet can be related by the conservation of mass. (S4.2.2) Rearranging results in a relation for the average velocity at location 2 in terms of the average velocity at location 1 and E: v1 π R12 = v 2 π R22 = v 2 π E 2 R12

v2 =

v1

(S4.2.3)

E2

Since there is no flow across the surface 3 and flow across surfaces 1 and 2 is in the z direction the only non-zero component of equation (S4.2.1) is in the z-direction. The z component of the left hand side of equation (S4.2.1) becomes: ez • ∫ vρ (n • v )dS = − ρ v1 S

2

R1 2π

⎛ r2 − 1 r ⎜ ∫ ∫ ⎜ R2 1 r =0 θ =0 ⎝

2 ez • ∫ vρ (n • v )dS = − πR12 ρ v1 3 S

2

2

⎞ v1 ⎟⎟ drdθ + ρ 4 E ⎠ 2

2 R2 2π

⎛ r2 − 1 r ⎜ ∫ ∫ ⎜ R2 2 r =0 θ =0 ⎝

v 2 2 + πR22 ρ 14 = − πR12 ρ v1 3 3 E

2

1 ⎞ ⎛ ⎜1 − 2 ⎟ ⎝ E ⎠

2

⎞ ⎟⎟ drdθ ⎠

(S4.2.4)

The shear stress is zero over surfaces 1 and 2 and the pressure is zero over surface 3. Because the detailed velocity field through the taper is not known, the shear stress cannot be computed. 2 2 πR1 ρ v1 3

2

(

)

⎛ 1 ⎞ 2 2 ⎜ 2 − 1⎟ = p1 − p2 E π R1 + ez • ∫ n • τ dS ⎝E ⎠ S

(S4.2.5)

The right hand side of Equation (S4.2.5) represents the net force acting in the z-direction. The force depends strongly on the taper ratio E. The net pressure force acting on the fluid is obtained by rearranging equation (S4.2.5)

( p − p E )π R 1

2

2

2 1

=

2 2 πR1 ρ v1 3

2

⎛ 1 ⎞ ⎜ 2 − 1⎟ − ez • ∫ n • τ dS ⎝E ⎠ S

(S4.2.6)

The pressure drop is increased above the value for flow in a straight tube due to convective acceleration through the taper and the shear stress acting on the walls of the vessel. 50

4.3. For steady, laminar flow through the entrance region of a cylindrical tube that is horizontally so that gravity is negligible, the integral form of the conservation of linear momentum, eEquation (4.3.8), is: (S4.3.1) ∫ vρ (n • v)dS = − ∫ pndS + ∫ n • τ dS S

S

S

From the conservation of mass, U0 = , the average velocity of the fully-developed laminar flow. The integral on the left hand side is nonzero at the entrance and downstream where the flow is fully developed. The integral is zero at r = R. At the entrance, the integral is

∫ Sentrance

2

∫

vρ (n • v )dS = − ρ ez

U 0 2 dS = − ρU 0 2π R 2 ez = − ρ v π R 2 ez

Sentrance

When the flow is fully developed,

∫

∫

vρ (n • v)dS = ρ

S fully developed

∫

v z 2 dS = 4 ρ v

S fully developed

2

vρ (n • v)dS = 8πρ v

S fully developed

π R2 3

2

R 2π

⎛ r2 1 − ⎜ ∫ ∫ ⎜ R2 r =0 θ =0 ⎝

⎛ R2 R2 R2 ⎞ 4 2 ez ⎜⎜ − + ⎟⎟ = π R ρ v 2 2 6 3 ⎝ ⎠

2

⎞ ⎟⎟ rdrdθ = 8πρ v ⎠ 2

2

⎛ r3 r5 ⎞ r 2 − ⎜ ∫ ⎜ R 2 + R 4 ⎟⎟⎠ dr r =0 ⎝ R

ez

ez

2

ρ v ez = − ∫ p ndS + ∫ n • τ dS S

S

Neglecting gravity, the pressure at r = R is uniform so the integral of pressure over the control volume surface equals ez(p0 – pLe)πR2 where Le is the entrance length. The integral of the shear stress represents the net drag force acting on the fluid. Thus, the drag force equals: FD = − ( p 0 − pL ) πR 2 ez +

π R2 3

2

ρ v ez

The drag force is less than the value in fully developed flow because the flow develops in this region and the velocity gradients are not as steep.

4.4. From equations (4.3.11d) and (4.3.11e) for the z and r components of the net force on the branching vessel: 4 ρ ⎛ Q1 ez • ∫ vρ (n • v )dS = ⎜ 3π ⎜⎝ R1 S er • ∫ vρ (n • v)dS = S

4 ρ ⎛ Q1 ⎜ 3π ⎜⎝ R1

⎞ ⎟⎟ ⎠

2

{−1 + 0.3858cos φ4 + 0.2434 cos φ5 }

2

⎞ ⎟⎟ {0.3858sin φ4 + 0.2434sin φ5 } ⎠

Setting φ4 = φ5 and normalizing each component of the force by ρ(Q1/R1)2, Fz ⎛Q ρ ⎜⎜ 1 ⎝ R1 Fr

⎞ ⎟⎟ ⎠

2

= 0.4244 {−1 + 0.6292 cos φ }

⎛Q ρ ⎜⎜ 1 ⎝ R1

⎞ ⎟⎟ ⎠

2

= 0.2670sin φ

The net force is simply Fnet = (Fz2+Fr2)1/2. The results are presented in Figure S4.3.1. The net force is smallest at an angle of zero degrees. The net force at zero degrees is not zero because the 51

cross-sectional areas in sections 4 and 5 are greater than 1, resulting in a decrease in velocity through the tubes. As the branch angle increases, the net force increases, reaching a maximum at 180 degrees.

Figure S4.3.1

4.5. (a) Let δ = H/2 and x = Le, the entrance length. Rex=Le = ULe/ν. Thus, the boundary layer expression δ(x) = 5.00xRex-1/2 becomes H = 10 Le(ULe/ν)-1/2 = 10 Le1/2(U/ν)-1/2 From the conservation of mass, U0 = , the average velocity of the fully-developed laminar flow. Rearranging: Squaring both sides Using the definition of the channel Reynolds number Normalizing Le by the channel half-thickness

Le1/2 = 0.1 H(U/ν)1/2 Le = 0.01 H2U/ν Le = 0.005HRe Le/(H/2) = 0.01Re

This result provides the correct relation between the entrance length and Reynolds number, The coefficient is the correct magnitude, but is smaller than the measured values. The reasons for this discrepancy are that the centerline velocity increases as the fluid decelerates near the surfaces and the boundary layer thickness is not small. (b) From the conservation of mass: δ

H/2

Q=w

∫

v x dy = w

y=0

δ

Q=w

∫

y=0

Uy

δ

y=0 H/2

dy + w

∫

H/2

v x dy + w

∫ U ( x)dy = w

y=δ

∫

v x dy

(S4.5.1)

y=δ

Uδ ⎛H ⎞ ⎛H δ ⎞ + wU ( x) ⎜ − δ ⎟ = wU ( x) ⎜ − ⎟ 2 ⎝ 2 ⎠ ⎝ 2 2⎠

52

(S4.5.2)

Solving for U(x) U ( x) =

τw = ρ μU

δ μU

δ

∞ ⎤ ∂ ⎡ dU ⎢ ∫ vx (U − vx ) dy ⎥ +ρ ∂x ⎣⎢ 0 dx ⎦⎥

2Q w( H − δ )

(S4.5.3)

∞

∫ (U − vx ) dy

(S4.5.4)

0

=ρ

∞ ∞ Uy ⎞ ⎤ dU ⎛ Uy ⎞ dU ⎛ ∂ ⎡ Uy ⎛ ∂ ⎡ 2 ⎛ δ δ ⎞⎤ δ⎞ ⎢∫ ⎜U − ⎟ dy ⎥ +ρ ⎜U − ⎟ dy =ρ ⎢U ⎜ − ⎟ ⎥ +ρU ⎜δ − ⎟ ∫ dx 0 ⎝ dx ⎝ ∂x ⎢⎣ 0 δ ⎝ ∂x ⎣ ⎝ 2 3 ⎠ ⎦ δ ⎠ ⎥⎦ δ ⎠ 2⎠

=ρ

U 2 dδ ∂ ⎡ 2δ⎤ δ dU 5δ dU U +ρU = ρU +ρ ⎢ ⎥ ∂x ⎣ 6⎦ 2 dx 6 dx 6 dx

(S4.5.5)

Taking the derivative of U with respect to δ: dU 2Q dδ U dδ = = dx w ( H − δ )2 dx ( H − δ ) dx

(S4.5.6)

Substituting equation (S4.5.6) into equation (S4.5.5) μU

δ

=ρ

5δ U 2 d δ U 2 d δ ⎛ 4δ + H ⎞ U 2 d δ +ρ =⎜ ⎟ρ 6 ( H − δ ) dx 6 dx ⎜⎝ ( H − δ ) ⎟⎠ 6 dx

(S4.5.7)

Rewriting in terms of the derivative of δ with respect to x dδ 6ν ( H − δ ) 6μ ( H − δ ) 12 ( H − δ ) = = = dx Uδ ( 4δ + H ) Qρ wδ ( 4δ + H ) Re δ ( 4δ + H ) 2

2

(S4.5.8)

To find the distance x at which δ = H/2, make the variables x and δ dimensionless by normalizing by H/2. δ = δ/(H/2) x’ = x/(H/2) With this change of variables, equation (S4.5.8) becomes: dδ ' 12 ( 2 − δ ' ) 6 ( 2 − δ ') = = dx ' Re δ ' ( 4δ '+ 2 ) Re δ ' ( 2δ '+ 1) 2

2

The MATLAB Mfile is shown below. Integration was performed using ode23 where x(1) = δ’ and x=x’ function dx=entrance(t,x); Re=1; dx=[6*(2-x(1))*(2-x(1))/(Re*x(1)*(1+2*x(1)))]; A graph of the results is shown in Figure S4.5.1. By specifying specific elements of x(1) and the corresponding value of x’ and interpolating, Le = 0.1268(H/2)Re. This coefficient is much closer to the expected value.

53

Figure S4.5.1

4.6. By comparison of equations (4.4.28) and (3.3.26), the viscous loss term is ∫ v ⋅ n ⋅ τ dS = − ∫S v ⋅ n ⋅ τ dS EV = − S A v ∫ v ⋅ ndS S

The numerator can be determined by converting to a surface integral; ∫ v ⋅ n ⋅ τ dS = − ∫V ∇ ⋅ (τ ⋅ v)dV = A = − 1 EV = − S v A v A v A v

∫

l2

∂ (τ yx v x )

l1

∂y

dx y =0

Applying the chain rule and noting that vy(y=0) = 0.

1 Ev = − v

μ

⎛ ∂v x ⎞ dx = ⎜ ⎟ ∫ l v 1 ⎝ ∂y ⎠ y =0

∂v ∫l1 τ yx ∂yx l2

2

l2

dx y =0

The velocity gradient can be estimated assuming that viscous stresses arise in a thin boundary layer near the valve surface. If the velocity is vx = U0sin(πy/2δ), then the square of velocity gradient at the valve surface (y = 0) is ⎛ ∂v x ⎞ ⎜ ⎟ ⎝ ∂y ⎠

2

y =0

⎛U π ⎛ π y ⎞⎞ = ⎜ 0 cos ⎜ ⎟⎟ ⎝ 2δ ⎠ ⎠ ⎝ 2δ

2

y =0

⎛U π ⎞ =⎜ 0 ⎟ ⎝ 2δ ⎠

2

where δ = 4.8xRex-1/2 and Rex = ρU0x/µ. In addition, = U0. Substituting equation (S4.6.3) into equation (S4.6.1c) and using the expression for results in δ.

μ

2

⎛ U 0π ⎞ ⎛ π ⎞ dx = − μU 0 ⎜ ⎜ ⎟ ⎟ ∫ U 0 l1 ⎝ 2δ ⎠ ⎝ 9.6 ⎠ Evaluating the integral, Ev = −

l2

⎛ π ⎞ Ev = − ρU 02 ⎜ ⎟ ⎝ 9.6 ⎠

2

∫

l2

l1

2

∫

l2

l1

2

dx x Rex−1 2

dx ⎛ π ⎞ ⎛ l2 ⎞ = − ρU 02 ⎜ ⎟ ln ⎜ ⎟ 2 x ⎝ 9.6 ⎠ ⎝ l1 ⎠ 54

For the data given and a density of 1 g cm-3 (=1000 kg m-3) and a valve length of 1cm and thickness 0.2 cm, l2/l1 = 1/0.8 = 1.23. EV = -0.456 N m-2. This corresponds to -0.0034 mm Hg. The viscous losses are much smaller than the pressure drops induced by conductive acceleration. Note that this analysis is quite simplified. A more thorough analysis would include the effect of viscous losses are due to the jet as discussed in [11]. 4.7. Setting v2 = vmax in equation (4.4.18), the relative error in neglecting v1 is v2 − v2 v2 error = max 2 1 = 1 − 21 ≥ 0.90 v max v max Solving for v1, v1 ≤ 0.10v max = 0.3162v max

For vmax = 3.2 m s-1, v1 must be less than 1.0118 m s-1. 4.8. From equation (4.4.18) p1 – p2 = (4 s2m-2 mmHg)(1.12-0.62)m2s-2 = 3.40 mmHg The cross-sectional area of the valve can be determined by a mass balance v1A RV = v 2 A valve

A valve = ( 0.62 m 2s −2 ) (2 × 10−4 m 2 / (1.12 m 2s −2 ) = 0.595 × m 2

4.9. This problem is similar to the previous except that the atrial area must be calculated from the diameter. (a) p1 – p2 = (4 s2m-2 mm Hg) (1.62 – 0.72)m2s-2 = 8.28 mm Hg (b) Avalve = (0.72 m2s-2)(2.27 x 10-4 m2) / (1.62 m2s-2) = 0.434 x 10-4 m2 4.10. First calculate the Reynolds number treating blood as a Newtonian fluid. Thus, from Table 2.5, ρ = 1.05 g cm-3 , µ = 0.03 g cm-1 s-1 and ν = 0.0286 cm2 s-1.

Upstream

Re = (80 cm/s)(2.50 cm)/(0.0286 cm2 s-1) = 7,000

turbulent

To compute Re downstream, first find diameter from a mass balance assuming turbulent flow. We do this assuming that the turbulent profile is uniform. If the flow is laminar, we need to recalculate assuming that the centerline is two times the average. A1 = A2 πR1 = πR2 or R2 = R1 v1 / v 2 = 1.25 80 / 500 = 0.50 cm Re = (500 cm/s)(0.50 cm)/(0.0286 cm2/s) = 8,741

turbulent

A more accurate estimate of the Reynolds number can be obtained by computing the average velocity for turbulent flow. R ⎛ ra ⎞ v max 2π ∫ ⎜ 1 − a ⎟ rdr R ⎠ 2 ⎛ R2 R2 ⎞ ⎛1 1⎞ 0⎝ vz = v = − ⎟ = 2v max ⎜ − ⎟ max 2 2 ⎜ πR R ⎝ 2 a+2⎠ ⎝2 7⎠ 55

⎛ 7 2 ⎞ 5 v z = 2v max ⎜ − ⎟ = v max ⎝ 14 14 ⎠ 7 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 dv z r a −1 = − μ v max a a τw = μ dr R

Inserting values Position 1, Position 2,

=−

μ v max a

r=R

R

τw = -0.0286(80)5/1.25 = -9.152 dyne/cm2 τw = -0.0286(500)5/0.5 = -143.0 dyne/cm2

EV = τ w2 − τ w1 = -143 +9.15 = -133.85 dyne/cm2 (1 mm Hg/1333 (dyne/cm2)) = 0.100 mm Hg.

The pressure drop from Bernoulli’s equation is: p1 – p2 = 4((0.5)2 – (0.08) 2) = 0.9744 mm Hg The viscous losses are a little more than 10% of the pressure drop due to changes in the crosssectional area. 4.11. For this problem, Bernoulli’s equation is used to relate the pressure and velocity. Choosing a streamline along the centerline, gravity can be neglected. So, Bernoulli’s equation is: p1 +

ρ 2

v12 = p2 +

ρ

2

(

v 22

)

v 22 − v12 2 From the conservation of mass, we have Q = A1=A2 Upstream, we have = Q/A1 = (2000/60 cm3/s)/(π(1.3/2)2) = 25.11 cm/s Re = ρD/µ = (1.07)(25.11)(1.3)/0.035= 998. Flow upstream is laminar. So, the centerline velocity is 2. Flow in the constriction could be laminar, transitional or turbulent. Assume that the flow is laminar. or

p1 − p2 =

ρ

From conservation of mass = A1/A2, p1 − p2 = ρ v1

Solving for the area ratio

A1 = A2

p1 − p2 +1 ρ v12

15 mm Hg = 1998 Pa ρv12= (1070 kg/m3)(0.2511 m/s)2 = 67.46 Pa 56

2

⎛ ⎛ A ⎞2 ⎞ ⎜ ⎜ 1 ⎟ − 1⎟ ⎜ ⎝ A2 ⎠ ⎟ ⎝ ⎠

A1/A2 = 5.536 Since the area drops by a factor of 5, the average velocity increases by a factor of 5. The radius decrease by 5.5361/2 = 2.35. Consequently, the Reynolds number increases by 2.35 times to 2348 which places this just above the laminar limit. Assuming laminar flow through is acceptable but there may be bursts of turbulence. If the flow were turbulent, would equal the centerline velocity. 4.12. The integral momentum balance is

∫ vρ (n ⋅ v)dS = − ∫ pn dS + ∫ τ ⋅ n dS = ∑ F S

S

S

Defining the entrance and exit as 1 and 2, respectively,

Use the conservation of mass to relate the velocities v πR = 2

2π R

2π R

∫ ∫ U rdrdθ = ∫ ∫ v rdrdθ 0

2

0 0

0 0

2π R

2π R

⎛ r3 ⎞ − v 1 ∫0 0 ∫0 ∫0 max ⎜⎝ R3 ⎟⎠ rdrdθ R ⎛ r4 ⎞ ⎛1 1⎞ 3 2 2 v π R = U 0π R = 2π v max ∫ ⎜ r − 3 ⎟ dr =2 ⎜ − ⎟ = π R 2 v max R ⎠ ⎝2 5⎠ 5 0⎝

v π R2 =

∫ U0rdrdθ =

3 v = U 0 = v max 5 The z component of the force is the only non-zero component of the integral form of the conservation of linear momentum. At 1 and 2 the result is:

∑ e iF = F

x

x

= ex i ∫ vρ (n ⋅ v)dS + ex i ∫ vρ (n ⋅ v)dS=- ∫ ρU 02 dS+ ∫ ρ v z (r) 2 d S1

S2

2π R

2π R

0 0

0 0

S1

S2

Fx = =-ρ ∫ ∫ U 0 rdrdθ +ρ ∫ ∫ v z (r)2 rdrdθ 2

R

Fx = - ρU 0 π R +2πρ ∫ v 2

0

2 max

R ⎛ r3 ⎞ ⎛ r4 r7 ⎞ 2 2 rdr ρ U π πρ 1R +2 v r-2 = ⎜ 0 3 ⎟ ∫0 max ⎜⎝ R3 + R 6 ⎟⎠ dr ⎝ R ⎠

57

5 ⎞ ⎛ 4 1⎞ ⎛ 4 Fx = - ρU 02π R 2 +ρ v 2maxπ R 2 ⎜ 1- + ⎟ = - ρU 02π R 2 +π v 2max R 2 ⎜ + ⎟ ⎝ 5 4⎠ ⎝ 20 20 ⎠ ⎛ 9 ⎞ Fx = - ρU 02π R 2 + ρ v 2maxπ R 2 ⎜ ⎟ ⎝ 20 ⎠ Replacing vmax, ⎛ 25 ⎞⎛ 9 ⎞ ⎛5⎞ ⎛1⎞ Fx = - ρU 02π R 2 +ρ U 20π R 2 ⎜ ⎟⎜ ⎟ = - ρU 02π R 2 +ρ U 20π R 2 ⎜ ⎟ = ⎜ ⎟ ρU 02π R 2 ⎝ 9 ⎠⎝ 20 ⎠ ⎝4⎠ ⎝4⎠ 4.13. Assume that the flow is laminar. The centerline is horizontal so that gravity does not influence flow and flow is steady. Equation (4.3.8) reduces to:

∫ vρ (n • v)dS

= − ∫ p ndS + ∫ n • τ dS

S

S

(S4.13.1)

S

The control volume is applied to a region over which fully developed flow as shown in the figure below.

With these simplifications, the analysis uses the same development presented in Example 4.2. The control volume can be divided into three regions, 1, 2 and 3. Mass enters through region 1 and exits through region 2. There is no flow across the walls of the vessel, region 3. The average velocity in the inlet and outlet can be related by the conservation of mass. v1 π R12 = v 2 π R22 = v 2 π E 2 R12 (S4.13.2) Rearranging results in a relation for the average velocity at location 2 in terms of the average velocity at location 1 and E: v2 =

v1

(S4.13.3)

E2

Since there is no flow across the surface 3 and flow across surfaces 1 and 2 is in the z direction the only non-zero component of Equation (S4.13.1) is in the z-direction. The z component of the left hand side of Equation (S4.13.1) becomes: ez • ∫ vρ (n • v)dS = − ρ 4 v1 S

2

2

R1 2π

v1 ⎛ r2 ⎞ 1 − r ∫r =0 θ∫=0 ⎜⎝ R12 ⎟⎠ drdθ + ρ 4 E 4

58

2 R 2π 2

2

⎛ r2 ⎞ 1 − r ∫ ∫ ⎜ R22 ⎟⎠ drdθ r =0 θ =0 ⎝

ez • ∫ vρ (n • v)dS = 2π ρ 4 v1

2

S

ez • ∫ vρ (n • v)dS = 2π ρ 4 v1 S

2

⎡ R1 ⎛ r3 r5 ⎞ 1 ⎢ − ∫ ⎜ r − 2 2 + 4 ⎟dr + 4 R1 R1 ⎠ E ⎣⎢ r =0 ⎝

⎛ r3 r5 ⎞ ⎤ ∫ ⎜1 − 2 E 2 R2 + E 4 R 4 ⎟⎠dr ⎥⎥ r =0 ⎝ ⎦ ER1

R1 ER1 ⎡ ⎛ r2 r4 r6 ⎞ 1 ⎛ r2 r4 r6 ⎞ ⎤ ⎢− ⎜ − 2 + 4 ⎟ + 4 ⎜ − 2 2 + ⎥ 4 4 ⎟ ⎢ ⎝ 2 2 R1 6 R1 ⎠ 0 E ⎝ 2 2 E R 6 E R ⎠ 0 ⎥ ⎣ ⎦

4 ez • ∫ vρ (n • v)dS = − πR12 ρ v1 3 S

2

2

v 4 4 1 ⎞ 2⎛ + πR22 ρ 12 = − πR12 ρ v1 ⎜ 1 − 2 ⎟ (4a) E 3 3 ⎝ E ⎠

Expressing in terms of Q, the volumetric flow rate: 2

4 2 1 ⎞ 4 2 ⎛ Q ⎞ ⎛ 1 ⎞ 4 ⎛ Q2 ⎞ 2⎛ ez • ∫ vρ (n • v)dS = − πR1 ρ v1 ⎜ 1 − 2 ⎟ = − πR1 ρ ⎜ 2 ⎟ ⎜ 1 − 2 ⎟ = − ρ ⎜ 2 ⎟ 3 3 3 ⎝ πR1 ⎠ ⎝ E ⎠ ⎝ πR1 ⎠ ⎝ E ⎠ S

1 ⎞ ⎛ ⎜1 − E 2 ⎟ ⎝ ⎠

The shear stress is zero over surfaces 1 and 2 and the pressure is zero over surface 3. Because the detailed velocity field through the taper is not known, the shear stress cannot be computed. 4 ⎛ Q2 ⎞ ⎛ 1 ⎞ ρ ⎜ 2 ⎟ ⎜ 2 − 1 ⎟ = p1 − p2 E 2 π R12 + ez • ∫ n • τ dS =∑ Fx 3 ⎝ πR1 ⎠ ⎝ E ⎠ S

(

)

(5)

4.14. Neglect any frictional resistance acting on the plunger. For this problem use Bernoulli’s equation along the centerline and assume that flow is steady.

The height is constant and p1 +

ρ 2

v12 = p2 +

ρ 2

v 22

From the conservation of mass: D2 d2 = v2 π Q = v1 π 4 4 Note that v1 and v2 are the maximum velocity and equal two times the local average velocity for laminar flow. D2 d2 = v 2π Q = v1π 8 8 Replacing v1 and v2 in Bernoulli’s equation. 2

ρ ⎛ 8Q ⎞ ρ ⎛ 8Q ⎞ = p2 + ⎜ 2 ⎟ p1 + ⎜ 2 ⎟ 2 ⎝πD ⎠ 2 ⎝πd ⎠

2

Note that F =p1πD2/4 and p2 = p. Making these substitutions: 59

2

4 ρ ⎛ 8Q ⎞ ρ ⎛ 8Q ⎞ + ⎜ = p+ ⎜ 2 ⎟ F 2 2 ⎟ 2 ⎝πd ⎠ πD 2 ⎝πD ⎠ Solving for F:

4 ρ F = p+ 2 πD 2 F=p

π D2 4

+8

2

⎡⎛ 8Q ⎞2 ⎛ 8Q ⎞2 ⎤ ρQ 2 ⎡ 1 1 ⎤ = p + 32 2 ⎢ 4 − 4 ⎥ ⎢⎜ 2 ⎟ − ⎜ 2 ⎟ ⎥ π ⎣d D ⎦ ⎣⎢⎝ π d ⎠ ⎝ π D ⎠ ⎦⎥

ρQ 2 D 2 π

1 ⎤ π D2 ρQ2 ⎡1 p − = + 8 ⎢ d 4 D4 ⎥ 4 π D2 ⎣ ⎦

⎡ D4 ⎤ ⎢ 4 − 1⎥ ⎣d ⎦

4.15. The expression for the pressure drop is (from Equation 4.4.28) 2 ρ dv p1 − p2 = ( v 22 − v12 ) + ρ ∫ dl 2 dt l =1 The fluid acceleration term (equation 4.4.29) is expressed in SI units:

(1050 )( 0.05) π 5cos(2π t ) = 550 cos(2π t ) ρ l dv 2 ρ l dv dl ≈ = 0.5(2π ) cos(2π t ) = dt 3 dt 3 3 l =1 3 2 The units are (kg/m )(m/s)(1/s)m = N/m = Pa. To convert to mm Hg, divide by 133.3 Pa/mm Hg. The acceleration term is then (4.12 mm Hg)cos(2πt). 2

ρ∫

The pressure drop, determined from the modified Bernoulli equation (4.4.28), neglecting viscous losses: p1-p2 = 4(25(2πt)-0.25sin2(2πt))+4.12 cos(2πt) p1-p2 = 99sin2(2πt)+4.12cos(2πt) The convective term cannot be neglected because of its magnitude and phase lag. t=0 p1-p2 =4.124 with no acceleration p1-p2 = 0 with no acceleration p1-p2 = 49.5 t= 0.125 p1-p2 =52.4 t = 0.25 p1-p2 =99 with no acceleration p1-p2 = 99

4.16. (a) Defining the control volume as:

60

From the integral form of the conservation of momentum,

∑ F =∫ ρ v ( niv )dS The adhesion force must balance the fluid forces in the z direction, so we need only focus on this direction. Fadh = ∑ Fx = − ∫ ρ v 2x dS

Since the velocity is uniform, Q = vxπR2 = vxπDp2/4 and: Fadh = − ρ v

2 x

π Dp 2 4

2

⎛ 4Q ⎞ π D p 2 4 ρQ 2 = −ρ ⎜ = − ⎟ ⎜πD 2 ⎟ 4 π Dp 2 p ⎠ ⎝

(b) Rearranging the result from part (a) Q = Dp

π Fadh 4ρ

From the data given Q = Dp

π ( 0.05 dyne )

(

4 1.05 g/cm 3

)

= 0.193D p

The pressure drop is related to the flow rate from Equation (2.7.45). 4 Δpπ R 4 Δpπ D p Q= = 8 µL 128 µL Δpπ D p 4 0.109D p = Replacing Q 128 µL ⎛ 0.193 (128 µL ) ⎞ Dp = ⎜ ⎟ Δpπ ⎝ ⎠

1/3

Solving for Dp:

61

Inserting values for the parameters and setting Δp to the limiting value of 30 mm Hg (Δp = ρHggh = (13.6 g/cm3)(980 cm/s2)(3.0 cm) = 39984 dyne/cm2 1/3

⎛ 0.193 (128 )( 0.03 g/cm/s )(10 cm ) ⎞ ⎟ Dp = ⎜ 2 ⎜ ⎟ 39984 dyne/cm π ( ) ⎝ ⎠ 4.17. (a) τ w = µ

dv z dr

=− r =R

= 0.039 cm

4µ v z R

The negative sign is ignored. The wall shear stress is reported as a positive quantity. The Reynolds number is Re =

ρ vz D

µ Using this definition and rearranging terms in the definition of τw yields 8µ v z 8µ 2 ρ v z D 8µ 2 τw = Re = = ρ D2 ρ D2 D µ (b) The cardiac output is the flow rate from the heart

Cardiac output (L min–1) Hear rate (beats per minute) Aortic root diameter (cm) Re α τw

Human 5.0 60 3.0 1061 20.6 1.10

Mouse 0.012 600 0.1 76.4 2.17 71.3

Both Re and τw are very different for the mouse and human. Thus, blood flow in the human does not represent a dynamically scaled model. 4.18. (a) From the boundary condition at y = 0, a0 = 0. From the other two conditions: 1 = a1 + a3 0 = a1/d+ 3a3/d or 0 = a1+ 3a3 a1 = - 3a3 a3 = -1/2 a1 = 3/2 vx/U0 = (3/2)(y/δ) –(1/2)(y/δ)3

(b) The momentum integral is δ ⎤ ∂v dU ∂ ⎡ τ w = µ x =ρ ⎢ ∫ v x (U − v x ) dy ⎥ +ρ dx ∂y y=0 ∂x ⎣ 0 ⎦ 62

δ

∫ (U − v ) dy x

0

∂v For flow over a flat plate, this reduces to: τ w = µ x ∂y The left had side is:

τw = µ

∂v x ∂y

y=0

y=0

δ ⎤ d ⎡ =ρ ⎢ ∫ vx (U 0 − v x ) dy ⎥ dx ⎣ 0 ⎦

⎛ 3U ⎞ = µ⎜ 0 ⎟ ⎝ 2δ ⎠

The right hand side equals

ρ

δ ⎡ δ ⎛ 3 y y3 ⎞ ⎛ 3 y y3 ⎞ ⎤ ⎤ d ⎡ 2 d ρ v v = U dy U − − + ( ) ⎢ ⎜ ⎢ x 0 ⎥ ⎟ ⎜1 − ⎟ dy ⎥ x 0 dx ⎣ ∫0 dx ⎣ ∫0 ⎝ 2δ 2δ 3 ⎠ ⎝ 2δ 2δ 3 ⎠ ⎦ ⎦

δ ⎡ δ ⎛ 3y 9 y2 ⎤ d ⎡ y3 3 y 4 y6 ⎞ ⎤ 2 d ρ ⎢ ∫ v x (U 0 − v x ) dy ⎥ = ρU 0 − 2 − 3 + 4 − 6 ⎟ dy ⎥ ⎢∫ ⎜ 2δ 2δ 4δ ⎠ ⎦ dx ⎣ 0 dx ⎣ 0 ⎝ 2δ 4δ ⎦ Integrating δ ⎤ d ⎡ d ⎛ 3δ 3δ δ 3δ δ ⎞ 39 ρU 02 dδ ρ ⎢ ∫ v x (U 0 − v x ) dy ⎥ =ρU 02 ⎜ − − + − ⎟ = dx ⎣ 0 dx ⎝ 4 4 8 10 28 ⎠ 280 dx ⎦ Equating the left and right hand sides of the von Karman momentum integral: 39 ρU 02 dδ ⎛ 3U ⎞ µ⎜ 0 ⎟ = − 280 dx ⎝ 2δ ⎠ µ Rearranging: (10.77 ) dx = δ dδ ρU 0

Using the initial condition that at x = 0, d = 0, and integrating; µ µ δ= (21.54 )x = δ 2 or ( 21.54 ) x == 4.64 x Re −x1/2 ρU 0 ρU 0 Inserting this result into the expression for the wall shear stress: ∂v τw = µ x ∂y

y=0

⎛ ⎞ ⎛ µU 0 Re1/2 ⎞ U0 ⎛ 3U 0 ⎞ x 0.323 = µ⎜ = µ1.5 ⎜ = ⎟ ⎜ ⎟ ⎟ −1/2 x ⎝ 2δ ⎠ ⎝ ⎠ ⎝ 4.64 x Re x ⎠

4.19. The flow is steady and two-dimensional. Gravity forces can be neglected. The characteristic length in the x-direction is L and the characteristic length in the y-direction is h. The characteristic velocity in the x-direction is U. The characteristic velocity in the y-direction can be determined from the dimensionless form of the conservation of mass. Define the following dimensionless variables as follows. v*x =

vx U

v*y =

vy

x* =

V

x L

y* =

y h

P =

μUL h2

As a result, the conservation of mass (Table 3.1) for an incompressible fluid in terms of the dimensionless variables is: 63

* U ∂ v*x V ∂ v y + =0 L ∂ x* h ∂ y*

(S4.19.1)

In order for the scaling to be appropriate, U/L and V/h must be of the same magnitude. Since h/L<<1, equation (S4.19.1) indicates that V~Uh/L<
ρ ⎜⎜

2

⎝ L

∂v*x

v*x

∂x

*

+

U2 L

v*y

⎛ 1 ∂ 2 v*x ∂v*x ⎞ μUL ∂P 1 ∂ 2 v*x μ = + + U ⎟ ⎜ ⎜ L2 ∂x*2 ∂y* ⎟⎠ h 2 L ∂x* h 2 ∂y*2 ⎝

⎞ ⎟⎟ ⎠

(S4.19.2)

Collecting terms and ρ

* ⎞ U 2 ⎛ h * ∂v*x μU ∂P μU ⎛ h 2 ∂ 2 v*x ∂ 2 v*x * ∂v x + + ⎜⎜ v x * + v y * ⎟⎟ = - 2 ⎜ ∂x ∂y ⎠ ∂y*2 L ⎝L h ∂x* h 2 ⎜⎝ L2 ∂x*2

⎛ ρUh ⎞ ⎛ h ⎜ ⎟⎜ ⎝ μ ⎠⎝ L

* * ⎞ ⎛ h 2 ∂ 2 v*x ∂ 2 v*x ∂P ⎞ ⎛ h * ∂v x * ∂v x v + v = + ⎜ ⎟ ⎜⎜ 2 *2 + x y ⎟ ∂x* ∂y* ⎠⎟ ∂x* ∂y*2 ⎠ ⎝⎜ L ⎝ L ∂x

⎞ ⎟⎟ ⎠

(S4.19.3)

⎞ ⎟⎟ ⎠

(S4.19.4)

The first term in parentheses on the left hand side of equation (S4.19.4) is the Reynolds number. Since we assume that the Reynolds number is small (order 1 or less), then the left hand side can be neglected because these terms are multiplied by h/L. Further the second derivative of vx with respect to x can be neglected because the terms are multiplied by (h/L)2. As a result, the dimensional form of the x-component of the Navier-Stokes equation simplifies to ∂P ∂x

=μ

∂2 vx ∂y 2

For the y-component of the Navier-Stokes equation, we have ρU 2

* ⎛ h ∂ 2 v*y 1 ∂ 2 v*y ∂v*y ⎞ μUL ∂P h ⎛ * ∂v y * ⎜ ⎟ ⎜ 3 *2 + U v + v = + μ x y ⎜ L ∂x Lh ∂y*2 h3 ∂y* L2 ⎜⎝ ∂x* ∂y* ⎟⎠ ⎝

ρU 2

* ∂v*y ⎞ μU h ⎛ * ∂v y μUL ∂P * ⎜ ⎟ =- 3 v + v + 2 y 2 ⎜ x * * ⎟ * ∂y ⎠ L ⎝ ∂x h ∂y h

3 ρUh ⎛ h ⎞ ⎛

2 * 2 2 * ⎛ h ⎞⎛ h ∂ vy ∂ vy ⎜ + ⎜ ⎟ ⎜ 2 *2 ∂y*2 ⎝ L ⎠ ⎝ L ∂x

2 * 2 2 2 * ∂v*y ∂v*y ⎞ ∂P ⎛ h ⎞ ⎛ h ∂ vy ∂ vy * * ⎜ ⎟ ⎜ v + v = + + x y ⎜ ⎟ ⎜ ⎟ μ ⎝ L ⎠ ⎝⎜ ∂x* ∂y* ⎠⎟ ∂y* ⎝ L ⎠ ⎝⎜ L2 ∂x*2 ∂y*2

64

⎞ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎠

(S4.19.5)

⎞ ⎟ ⎟ ⎠

(S4.19.6)

(S4.19.7)

The let-hand side of equation (S4.19.7) can be neglected because the terms are multiplied by (h/L)3. Likewise, viscous stresses due to vy can be neglected because the terms are multiplied by (h/L)2. As a result, the y-component of the Navier-Stokes equation reduces to: ∂P =0 ∂y

4.20. FN =

1

(1 + a ) 2

1 2

(S4.19.8)

β − 1⎞ ⎛ 6μ UW ⎞ ⎛ ⎜ ⎟ ⎜ ln β − 2 2 β + 1 ⎟⎠ ⎝ a ⎠⎝

β − 1⎞ ⎛ 2 μ UW ⎞ ⎛ FF = ⎜ a2 ⎟ ⎜ 2 ln β − 3 β + 1 ⎟⎠ ⎝ ⎠⎝ From the data given in Example 4.6, L = 5 cm, α = 0.004, β = 0.8333. For a water-like fluid and W=L ⎛ 6(0.001Pa s )0.05 m ⎞ ⎛ U 0.1667 ⎞ FN = ln(0.833) + 2 ⎜ ⎟ ⎜ ⎟ = −0.9444U N 2 2 1 (0.0004) 1.8333 ⎠ ⎝ ⎠⎝ 1 + 0.0004)

(

(

)

)

2

2(0.001Pa S )(0.05 m)U ⎛ 0.1667 ⎞ ⎜ 2 ln(0.833) + 3 ⎟ = −0.023U N 0.0004 1.8333 ⎠ ⎝ Note that the ratio given at the end of Example 4.6 is incorrect. It should be 0.0243. To support a net force of 48.93 N, the velocity U is (48.93 N) 2 = [(−0.9444) 2 + (−0.023) 2 ]U 2 U = 51.80 m s −1 For a viscosity of 1 Pa s, FN = 944.4 U and FF = 23 U. To support a net force of 48.93 N, the velocity U needs to be 0.0518 m s-1. Clearly, by increasing the fluid viscosity, the sliding velocity can be significantly reduced. FF =

4.21. Beginning with Equation (4.7.25) vw =

h3 ∂ 2 p = cons tan t 3μ ∂x 2

(S4.21.1)

Integrating twice p=

3μ v w 2 h3

(S4.21.2)

x + C1 x + C2

Applying the boundary condition that p = p0 at x = 0, yields C2 = p0. Next, at x = L, p = pL and ⎛ p − pL ⎞ 3μ v w C1 = − ⎜ 0 − L L ⎟⎠ 2h3 ⎝ p = p0 −

3μ v w 2h3

(

⎛ p − pL x 2 − xL − ⎜ 0 L ⎝

)

(S4.21.3) 2

3μ v w ⎛ L ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ p0 − pL ⎞ ⎟ x = p0 + 2h ⎜ h ⎟ ⎜ L ⎟ ⎜ 1 − L ⎟ − ⎜ L ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

65

⎞ ⎟x ⎠

(S4.21.4)

The pressure gradient is: dp 3μ v w ⎛ L ⎞ = ⎜ ⎟ 2hL ⎝ h ⎠ dx

2

⎛ 2 x ⎞ ⎛ p0 − pL ⎞ ⎜1 − ⎟ − ⎜ L ⎠ ⎝ L ⎟⎠ ⎝

(S4.21.5)

From equation (S4.21.4) we see that the presence of a flow into the walls reduces the pressure gradient below the value for a solid wall. The wall shear stress (τyx|y=h) equals 2µvxo/h= (p0pL)h/L. Te pressure gradient can be rewritten as: 2 dp 3τ w ⎛ v w ⎞ ⎛ L ⎞ ⎛ 2 x ⎞ ⎛ p0 − pL = ⎜ ⎟ ⎜ ⎟ ⎜1 − ⎟ − ⎜ dx 4 L ⎝ v xo ⎠ ⎝ h ⎠ ⎝ L ⎠ ⎝ L

⎞ ⎛ p0 − pL ⎟ = −⎜ L ⎠ ⎝

⎞ ⎛ 3 ⎛ vw ⎞ ⎛ L ⎞ ⎟ ⎜⎜ 1 − 4 ⎜ v ⎟ ⎜ h ⎟ ⎠⎝ ⎝ xo ⎠ ⎝ ⎠

⎛ 2x ⎞ ⎞ ⎜ 1 − ⎟ ⎟⎟ L ⎠⎠ ⎝

(S4.21.6)

The pressure is reduced by the second term on the right hand side of equation p4.12.6. The product (vw/vxo)(L/h) is of order unity or smaller. The pressure gradient is smaller than the value for a solid wall for 0<x
with boundary conditions

The equation for the wavy wall is

Integrating, we find

∂ 2 vx ∂y

2

=

dp dx

(S4.22.1)

∂vx = 0 at ∂y

y=0

(S4.22.2a)

vx = −c at

y = h.

(S4.22.2b)

h ( x, t ) = a + b sin

2 2 vx a 2 dp ⎡⎛ h ⎞ ⎛ y ⎞ ⎤ = 1− − ⎢⎜ ⎟ ⎜ ⎟ ⎥ 2 μ c dx ⎢⎣⎝ a ⎠ ⎝ a ⎠ ⎥⎦ c

2π

λ

( x − ct )

(S4.22.3)

(S4.22.4)

The flow rate per unit depth in the moving frame is h

q = ∫ hdy

(S4.22.5)

0

h3 dp = −ch − 3μ dx

This is constant, and independent of x and t. The flow rate per unit depth in the fixed frame is h

Q = ∫ Vx ( X , Y , t ) dy = q + ch.

(S4.22.6)

0

Combining equations (4), (5) and (6), we obtain ⎞ Vx −3 1 ⎛ y 2 = ⎜⎜ 2 − 1⎟⎟ Q c 2 h⎝h ⎠

(S4.22.7) 66

dp 3μ =− 3 Q dx h

(S4.22.8)

We now take the time averages of equations (S4.22.7) and (S4.22.8) over the period T of one peristaltic contraction. The involves the integrals T

(

1 dt 1 = 1−φ 2 T ∫0 h a

)

−

1 2

(S4.22.9a)

T 1 cos β 11⎡ 2 dt = ⎢1 − 1 − φ ∫ T0 h a φ ⎣⎢

(

T

(

)(

1 dt 1 = 3 2 + φ 2 1−φ 2 ∫ 3 T 0h 2a

)

1 cos β −3 dT = 3 φ 1 − φ 2 ∫ 3 T0 h 2a

)

T

where ϕ =

)

(

−

−

−

1 2

⎤ ⎥ ⎦⎥

(S4.22.9b)

5 2

(S4.22.9c)

5 2

(S4.22.9d)

b 2π and β = ( x − ct ) , and we have taken x = 0 for convenience since we are averaging. a λ

It follows that Vx 3 ⎧⎪ = ⎨1 − 1 − φ 2 c 2⎪ ⎩

(

)

−

1 2

(

3 + φ 2 1−φ 2 2

)

−

5 2

2 ⎡ ⎛ y⎞ 2 ⎜ ⎟ + Q ⎢ 1− φ ⎝a⎠ ⎢⎣

(

)

−

1 2

⎛ φ2 − ⎜⎜1 + 2 ⎝

⎞ 2 ⎟⎟ 1 − φ ⎠

(

)

−

5 2

⎛ y⎞ ⎜ ⎟ ⎝a⎠

2 ⎤⎫

⎪ ⎥ ⎬ (S4.22.10) ⎥⎦ ⎪⎭

where Q is the time average of the flow rate in the fixed frame. We can rewrite equation (S.4.13.10) to obtain

(

Vc 3 = 1−φ2 c 2

)

−

1 2

⎧⎪ ⎡ 2 ⎨Q − ⎢1 − 1 − φ ⎢ ⎣ ⎩⎪

(

)

1 2

⎤ ⎫⎪ 3 ⎛ φ 2 ⎥ ⎬ + ⎜⎜ 1 + 2 ⎥⎦ ⎭⎪ 2 ⎝

⎞ 2 ⎟⎟ 1 − φ ⎠

(

)

−

5 2

⎛ 3φ 2 ⎞ ⎛ y ⎞2 − Q ⎜⎜ ⎟ 2 ⎟⎜ ⎟ ⎝ 2 +φ ⎠⎝ a ⎠

(S4.22.11)

1

By inspection we see that for Q < 1 − (1 − φ 2 ) 2 we have a velocity profile shaped like

Thus there are positive and negative values of vx, and flow reflux occurs. After considerable manipulation, we find a second condition in which reflux occurs,

67

3φ 2

(

2 + φ −2 1 − φ 2 Q> 2 5 2 +φ

)

5 2

⎡ ⎢1 − 1 − φ 2 ⎢ ⎣

(

which has velocity profile

68

)

1 2

⎛ φ2 ⎜⎜1 + 2 ⎝

⎞ ⎟⎟ ⎠

−1 ⎤

⎥ ⎛⎜ 1 − 2 φ 2 ⎞⎟ ⎥⎝ 5 ⎠ ⎦

−1

(S4.22.12)

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, , a, R, L) There are seven dimensional groups and three dimensions (mass, length and time). Thus, there are, at most, four dimensionless groups. These can be identified using the procedures outlined in Section 3.5.1. Alternatively, the groups can be identified from the known dimensionless groups. The dimensionless pressure is Δp2a/(2ρ2L) as is analogous to the group developed for flow in a straight tube of diameter 2a and represents the product of two groups, Δp/(2ρ2) and a/L. The dimensionless pressure drop depends on two dimensionless groups. There are two possible groupings. One is the Reynolds number and the ratio a/R. ALernatively, one could use the Dean number and a/R.

(b) Three velocity profiles are shown in Figure S5.1.1. At the entrance to a curved region or for low values of Re and a/R (profile 1) the profile is parabolic. But as curvature and/or Reynolds number increase, the axial velocity becomes skewed towards the outer wall (profile 2). At high values of curvature, there is a steep boundary layer near the outer wall. (c) Secondary flows arise due to inertial forces which push the fluid towards the outside. (d) As curvature increases, secondary flows become more prominent as shown in Figure 5.12. Figure S5.1.1. 5.2. Using Xe = 0.113RRe and Table 2.3, the following results are obtained for the lung. Xe, cm, Vigorous Breathing Generation Xe, cm, Quiet Breathing Trachea 236.4 948.3 1 118.5 474.0 2 60.1 240.3 3 29.1 116.6 4 15.1 60.4 5 7.30 29.2 10 0.235 0.933 15 0.0071 0.0283 20 0.0020 0.0083 For quiet breathing, the length of the branch exceeds the entrance length for branch generation 10 and greater. For vigorous breathing, the length of the branch exceeds the entrance length for branch generation 15 and greater.

69

5.3. (a) This part can be solved by specifying coefficients and the number of harmonics. Alternatively, the velocity profiles can be generated from the pressure gradient analyzed in part (b) Two simplified profiles were considered

(1) a0 = 1, a1 = 0.5, a2 = -0.25 (2) a0 = 1, a1 = 0.5, a2 = -0.25, b1 = -1, b2 = -0.25 The pressure gradient and velocity profiles are shown in Figure S5.3.1 for case 1 and Figure S5.3.2 for case 2. Case (1) has positive pressures throughout each cycle. The velocity varies about the mean value and shapes are similar at all time. Pressure and velocities are symmetric about t/T=0.5. For case (2) there is a significant negative pressure during the early part of the cycle period causing fluid deceleration. Fluid accelerates during the pressure upswing and the velocity reaches a maximum by 0.8. The change in fluid velocity with time lags the pressure change due to inertia.

Figure S5.3.1. Pressure (A) and velocity (B) for a0 = 1, a1 = 0.5, a2 = -0.25.

70

Figure S5.3.2. Pressure (A) and velocity (B) for a0 = 1, a1 = 0.5, a2 = -0.25, b1 = -1, b2 = -0.25. (b) The following program was used to analyze the coronary data in reference [73]. The pressure gradient as a function of time is shown in Figure S5.3.3. The pressure pulse wave can be well represented by six to eight harmonics. The pressure is unusual in that it becomes negative during the early part of the cycle. This is due to vessel compression arising from contraction of the left ventricle. The flow tracks the pressure, although there is a slight lag. For other results see the paper by He et al. 1993. Ann. BME 21:45-49. function transform3(t); pm=[68 81 115.5 128.5 128 126 128 126 100 62 52 54 56 57 58 59 60 61.5 63 65 68]; pa=[103 102 101 103 107 112 116 121 120 119 120 119 118 116 114 112 110 108 106 104 103]; t=[0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00]; p=(pa-pm)/6; % pressure drop in mm Hg per cm; y=fft(p); pift=ifft(y); a=2*real(y)/20; b=2*imag(y)/20; tft=0:0.05:1; al=1.9263; w=2*pi; tt=0.0:0.05:1.0; tt1=1-tt; qtc=a(1)./2; qts=0; for k=2:1:8 71

qtc=0.416*ptc+(3.3325/(i*al*al*(k-1))*(1-2*besselj(1,i^1.5*al*sqrt(k-1))./(al*sqrt(k1)*i^1.5*besselj(0,i^1.5*al)))).*... a(k).*cos((k-1)*w*tt); qts=pts+(3.3325/(i*al*al*(k-1))*(1-2*besselj(1,i^1.5*al*sqrt(i-1))./(al*sqrt(k1)*i^1.5*besselj(0,i^1.5*al)))).*... b(k).*sin((k-1)*w*tt); end qt=qtc-i*qts; plot(tt,qt)

Figure S5.3.3. Pressure during the period T. Solid line are original data. Squares represent the predicted values from the inverse FFT. The dotted line represents the pressure obtained from the first 8 harmonics. (c) The FFT permits reconstruction of the pressure pulse. Six to eight harmonics are sufficient to generate reasonably accurate reconstructions of the pulse although it was not possible to duplicate the pulse even with more harmonics. Including higher harmonics in the Fourier series did not improve results. Going above the Nyquist frequency of 11 harmonics actually resulted in some degradation of the signal. 5.4. Note that rc is defined in Equation (2.10.7).

(a) To present on an appropriate scale, the velocity profile for the Newtonian fluid (equation (5.2.11) was rewritten as

(

⎛ J o i 3/ 2α r / R A* R 2 ⎜ vz = 1iα 2 μ ⎜ J o i 3/ 2α ⎝

(

)

72

) ⎞⎟ e ω ⎟ ⎠

i t

(S5.4.11)

using the definition of α. For presentation purposes, the Casson fluid velocity profile was normalized by R2A/m and the Newtonian velocity was normalized by R2A*/µ. Shown in Figure S5.4.1 are results for times of 0.0 s, and 0.50 s. The velocity profiles for the Casson fluid exhibits a slight amount of flattening. The Newtonian velocities are smaller than the corresponding values for the Casson fluid. There is a very modest effect of α.

Figure S5.4.1. (b) The pressure waveforms for the Newtonian and Casson fluid are identical. The flow rate for the Casson fluid is: R

Q = 2π ∫ vz rdr = 0

where

π R4 A ⎡

1/ 2

4 ⎛ 2τ ⎞ ⎢ 0.25exp ( iωt ) − ⎜ 0 exp ( iωt ) ⎟ 2m ⎣⎢ 7 ⎝ AR ⎠ 1/ 2

R

⎛ rc ⎞ ∫ F ( r , t ) rdr = 0.010417 + 0.142316 ⎜⎝ R ⎟⎠ 0

−

τ 0 ⎤ 4π R 4 A

⎥+ 6 AR ⎦⎥

m

R

α 2 sin ( iωt ) ∫ F ( r , t ) rdr 0

⎛r ⎞ + 0.065079 ⎜ c ⎟ ⎝R⎠

Flow rate curves are shown in Figure S5.4.2A. As expected from the velocity profiles, the flow rate for the Casson fluid is larger than the value for the Newtonian fluid. (c) The wall shear stress for a Newtonian fluid is: τw = μ

∂ vz ∂r

=r=R

(

3/ 2 ⎛ 1/ 2 A* R ⎜ i J1 i α α ⎜ J o i 3/2α ⎝

(

)

) ⎞⎟ e ω ⎟ ⎠

i t

(S5.4.11)

For a Casson fluid, the wall shear stress is, using equation (5.11.10): 1/2 1/2 1/ 2 ⎛ ⎞ ⎤ ⎛ ⎛ ∂ vz ⎞ ⎞ ⎞ ⎛ RA ⎞ ⎡⎛ r0 ⎞1/ 2 ⎛ 2τ o ∂ F (r , t ) ⎛ 2τ o ⎞ 1/ 2 2 ⎢ ⎜ ⎟ ⎟ ⎥ τ w = τ 0 + ⎜⎜ m ⎜ 2= ⎜ exp(iωt ) ⎟ + +8α sin (ωt ) ⎟ ⎟ ⎜ ⎟ + ⎜⎜ −2 exp(iωt ) - 4 ⎜ ⎟ 2 ⎜ AR ∂ r r = R ⎠⎟ ⎥ ⎝ ∂ r ⎠ r = R ⎠⎟ ⎟ ⎝ 4 ⎠ ⎢⎝ R ⎠ ⎝ AR ⎠ ⎝ ⎝ ⎝ ⎠ ⎣ ⎦

73

2

1/ 2

∂ F (r , t ) 1 ⎛ rc ⎞ =− − 16 R ⎜⎝ R ⎟⎠ ∂ r r=R

1/ 2

rc 1 3 ⎛ rc ⎞ ⎡ 2 1⎤ ⎢ 21 − 6 ⎥ − 18 R 2 = − 16 R − 14 ⎜ R ⎟ ⎣ ⎦ ⎝ ⎠

−

rc 18R 2

Shown in Figure S5.4.2B are the normalized shear stresses for the time period from 0 to 0.5 s for α = 1.0 and α = 0.5. The shear stress was calculated using the velocity profiles for the Newtonian fluid (equation (5.2.11)) and the Casson fluid (equation (5.11.10)) and normalized by AR/m. At early times, the shear stress for the Casson fluid is greater than the value for the Newtonian fluid. This difference declines as the magnitude of the velocity declines. When flow is negative, the shear stress for the Casson fluid again becomes slightly greater in magnitude than the value for the Newtonian fluid.

Figure S5.4.2. Flow (A) and shear stress (B) for Casson and Newtonian fluids. 5.5. For the right coronary artery, data in Table 5.2 are Remean = 233, De = 36, and a = 0.097 cm. The value of δ is 0.024 which may be at the limit of applicability of equations (5.4.3) and (5.4.4). The average fluid velocity is 7.059 cm s-1. Assume a Newtonian fluid of viscosity of 0.03 g cm-1 s-1. For these data, equations (5.4.3) and (5.4.4) are: τ rz | r = a = 0.546 [1 + 0,018 cosθ ]

τ rθ | r = a = 0.237 sin θ Curvature induces a shear stress component in the θ direction which is as much as 50% of the shear stress in the z direction. The effect of the curvature on the shear stress in the z direction is modest. 5.6. To solve, use equation (5.7.2) and κ = 0.120 µm-1for cells not exposed to flow and κ = 0.08 µm-1 following 24 hours of exposure to flow and a maximum height difference of 3 µm as shown in Figure 5.21. For cells not exposed to flow, τ/τw = 1.48 and τ = 23.7 dyne cm-2. For a cell exposed to flow, τ/τw = 1.24 and τ = 19.8 dyne cm-2. 5.7. The net force on the cells is: F = ∫ τ dS

(S5.7.1)

S

where S is the surface area exposed to flow. For an ellipsoid shape, this represents one-half the surface area of an ellipsoid. The general equation for an ellipsoid is 74

x2 a2

+

y2 b2

+

z2 c2

=

r 2 cos 2 θ sin 2 φ a2

+

r 2 sin 2 θ sin 2 φ b2

+

r 2 cos 2 φ c2

=1

(S5.7.2)

The surface area element is: dS = sin(φ ) b 2 c 2 sin 2 (φ ) cos 2 (θ ) + a 2 c 2 sin 2 (θ ) sin 2 (φ ) + a 2 c 2 cos 2 (φ ) dθ dφ

(S.5.7.3)

This result is derived from the general relation for the surface integral which can be found in calculus texts. In order to ensure that the shear stress on a flat surface equals τw, have=0. Integration of (S5.7.1) can be performed using the function dblquad in MATLAB. The function subroutine is function dFvec=netforce(qvec,f); a=20e-04; b=5e-04; c=2.5e-4; t0=15; k=0.12e4; have=0.0e-04; n=length(qvec); for i=1:n; dFvec(i)=t0*(1+k*(c*cos(f)have))*sin(qvec(i)).*sqrt((b*c*... sin(qvec(i)).*cos(f)).^2+(a*c*sin(qvec(i)).*sin(f)).^2+... (a*b*cos(qvec(i))).^2); end The net force on the cell is 5.78 x 10-5 dyne. The net force acting on an ellipse of minor axis lengths 5 µm and 20 µm is 4.712 x 10-5 dyne. Thus, the force is 23% higher due to the nonuniform shape of the cell. 5.8. The quasi-steady state velocity is Equation (5.2.15) 2 R2 ⎛ ⎛ r ⎞ ⎞ v z qss = A* ⎜1 − ⎟ cos(ωt ) 4 μ ⎜⎝ ⎜⎝ R ⎟⎠ ⎟⎠ The flow rate is: 2 R R ⎤ 2π A* R 2 ⎡ ⎛ ⎛ r ⎞ ⎞ 1 cos( ω ) Q qss = 2π ∫ v z (r , t )rdr = t − ⎢ ⎥ rdr . ⎜ ⎟ ∫0 ⎢ ⎜ ⎜⎝ R ⎟⎠ ⎟ 4μ ⎥⎦ 0 ⎠ ⎣⎝ ⎛ R 2 R 2 ⎞ πΔpR 4 πΔpR 2 − cos(ωt ) ⎜ cos(ωt ) ⎟= 2Lμ 4 ⎠ 8L μ ⎝ 2 πΔpR 4 The amplitude is , the value obtained from the steady flow result. 8 Lμ Q qss =

π R 2 A * ⎛ 2α i 3/2 J1 (i 3/2α ) ⎞ iωt π R 4 Δp ⎛ 2α i 3/2 J1 (i 3/2α ) ⎞ iωt Q= ⎜1 − 3 2 ⎟e = ⎜1 − 3 2 ⎟e iωρ ⎝ i α J 0 (i 3/2α ) ⎠ iµLα 2 ⎝ i α J 0 (i 3/2α ) ⎠ 75

Q=

8Qsteady ⎛ i8Qsteady ⎛ 2i1/2 J1 (i 3/2α ) ⎞ iωt 2i1/2 J1 (i 3/2α ) ⎞ iωt 1 1 e + = − + ⎜ ⎟ ⎜ ⎟e iα 2 ⎝ α J 0 (i 3/2α ) ⎠ α2 ⎝ α J 0 (i 3/2α ) ⎠

where α = R 2ω / ν The amplitude was determined as: Q = QQ This was done using the conj command in MATLAB to compute the complex conjugate of Q. function z=amp(al); % This program computes the amplitude of the complex flow rate z1=-8*i.*(1+2*sqrt(i)*besselj(1,i*sqrt(i)*al)./(al*... besselj(0,i.*sqrt(i)*al)))/al^2; z=sqrt(z1*conj(z1)); For α = 1.38, the amplitude is 0.9505. Equation (5.2.22) is used to compute the phase lag at t = 0. For the quasi-steady result, the phase lag is 0 radians. For α = 0.55, the phase lag is -0.0505 radians. 5.9. From equation (5.2.1) α =R ω/v If α ≤ 0.2 the flow is considered to be quasi-steady. With the values given in the problem α = 0.548-0.592. For this particular value of α, velocity is not in phase with the pressure waveform and is not quasi-steady. 5.10. (a) Using the definition of power provided, P = v z τ rz r = R 2π RL

(S5.10.1)

The units of the terms on the right hand side of Equation (S5.10.1) are J s (W) or dyne cm s-1, the appropriate units for power. The wall shear stress for laminar flow in a cylindrical tube is obtained from Equation (2.7.36). ΔpR τ rz r = R = (S5.10.2) 2L Combining (S5.10.2) with the definition of the average velocity, = Q/πR2, the power is restated as: ⎛ Q ⎞ ⎛ ΔpR ⎞ P = v z τ rz r = R 2π RL = ⎜ ⎟ 2π RL = QΔp (S5.10.3) 2 ⎟⎜ ⎝ π R ⎠ ⎝ 2L ⎠ The pressure drop and flow rate are related by Poiseuille’s law, Equation (2.7.45) Δpπ R 4 Q= (S5.10.4a) 8 µL Solving for Δp: 8 µLQ Δp = (S5.10.4b) π R4 Using Equation (S5.10.4b) to replace Δp in Equation (S5.10.3) yields

76

-1

2

⎛ 8 µLQ ⎞ 8 µLQ P = Q⎜ = 4 ⎟ π R4 ⎝ πR ⎠ (b) Using Equation (S5.10.5) in the metabolic cost function F yields:

F = ΔpQ + Emπ R 2 F=

(S5.10.5)

(S5.10.6a)

2

8 µLQ + E mπ R 2 L 4 πR

(S5.10.6b)

Take the first derivative of Equation (S5.10.6b) with respect to R to find when the cost function is minimized dF 32 µLQ 2 =− + 2 Emπ RL = 0 (S5.10.7a) π R5 dR Solving for Q 1/2

Q=

π R 3 ⎛ Em ⎞

⎜ ⎟ 4 ⎝ µ ⎠ The flow rate varies with the cube of the radius.

(S5.10.7b)

To establish whether Q is a minimum or maximum value, compute the second derivative of Equation (S510.7a): d 2 F 160 µLQ 2 = + 2 E mπ L (S5.10.8a) π R6 dR 2 Substituting for Q d 2 F 160 µL ⎛ π 2 R 6 ⎛ Em ⎞ ⎞ (S5.10.8b) = ⎜ ⎜ ⎟ ⎟ + 2 Emπ L = 12 Emπ L π R 6 ⎝ 16 ⎝ µ ⎠ ⎠ dR 2 Since the second derivative is positive, the cost function is minimized. Q is a monotonically increasing function of radius to the third power. Therefore, when F is minimized, Q reaches its maximum value. (c)

Use Equations (S5.10.2) and (S.5.10.4a) to relate shear stress and flow rate: ΔpR R ⎛ 8 µLQ ⎞ 4 µQ τ rz r = R = (S5.10.9) = = 2 L 2 L ⎜⎝ π R 4 ⎟⎠ π R 3

Replace Q using Equation (S5.10.7b):

τ rz r = R

1/2 1/2 4 µQ 4 µ ⎛ π R 3 ⎛ Em ⎞ ⎞ ⎜ ⎟ = ( µEm ) = = ⎜ ⎟ 3 3 πR πR ⎜ 4 ⎝ µ ⎠ ⎟ ⎝ ⎠

77

(S5.10.10)

As long as the metabolic consumption rate is constant, the shear stress is constant. 5.11. (a) The relationship between the flow rate in the parent vessel and the two daughter vessels is: Q0 = Q1 + Q2 (S5.11.1)

Using Equation (S5.10.7b) to relate Q and R and removing common terms: R03 = R13 + R23

(S5.11.2)

Solving for R0 in terms of R1 and α:

(

R0 = R1 1 + α 3

)

1/3

(S5.11.2)

Substituting for R0 is the definition of β: R12 + R22 1+α2 = β= 2/3 2/3 R12 (1 + α 3 ) (1 + α 3 )

(S5.11.3)

(b) Taking the first derivative of β with respect to α. dβ 2α = dα 1+α3

(

)

2/3

−

(

2α 2 1 + α 2

(1 + α ) 3

)

(S5.11.4)

5/3

Collecting terms and setting the derivative equal to zero 3 2 2 d β 2α 1 + α − 2α 1 + α = =0 5/3 3 dα α + 1

(

(

)

)

(

)

The derivative is equal to zero when the numerator is zero: 2α 1 + α 3 − 2α 2 1 + α 2 = 0

(

)

Simplifying

(

)

(S5.11.5)

(S5.11.6)

(1 + α ) − α (1 + α ) = 0 3

2

The solution to Equation (S5.11.7) is α = 1. The corresponding value of β is 1+α2 2 1/3 = = (2) β= 2/3 2/3 3 (2) 1+α

(

)

78

(S5.11.7)

(S.5.11.8)

Solution to Problems in Chapter 6, Section 6.12 6.1. For oxygen, the partial molar volume is 25.6 cm3 mol-1. Application of the Wilkie-Chang equation (equation 6.6.25) results in the following value for the diffusion coefficient, D = 2.0 x 10-5 cm2s-1. Comparing with the measured diffusion coefficient in Table 6.3 the use of the Wilkie-Chang equation produces a 5% error. An estimate for the radius is needed to apply the Stokes-Einstein equation. Using the radius as 1/2 of the collision diameter, the diffusion coefficient is D = 1.24 x 10-5 cm2 s-1, which is in error by 41% relative to the measured value. 6.2. If the measurements for both molecules are assumed to occur at the same temperature and both can be treated as spheres, then the diffusion coefficients are related by: D2 R 2 = D1 R1 -6 2 -1 and D2 = 2.13 x 10 cm s . 6.3. (a) For a nonconstant value of δ, Equation (6.5.5.c) is: x

(The cross term

2 1/ 2

⎡ = ⎢ xn −12 ⎣

1/ 2

1/ 2

⎤ 1 n + ∑ δi2 ⎥ n i =1 ⎦

n

∑ xn−1δ i is zero because the mean of δi is still zero. Substituting for xn-1, the root i =1

mean square displacement can be written in terms of xn-1: x2

1/ 2

⎡ = ⎢ xn − 2 2 ⎣

1/ 2

1/2

+

2 n 2⎤ ∑ δi ⎥ n i =1 ⎦

Repeating this procedure until the first term on the right hand side is x1 = 0, the root mean square displacement becomes. 1/2

1/2 ⎡n n ⎤ (6.5.6) = ⎢ ∑ δ i 2 ⎥ = n1/2 δ 2 x ⎣ n i =1 ⎦ (b) Random walks can be simulated on Excel using either of the two recursion relations: 2 1/2

or

x +1 = xj+randbetween(-1,1) x +1 = xj+2*(rand()-0.5)

The first relation, randbetween(-1,1), generates –1,- or 1 randomly and the mean square value of randbetween(-1,1) is 2/3. The second relation generates random numbers between –1 and 1 and 2*(rand()-0.5) has a root mean square value of 0.25. Shown below are four 500 step simulations using the second relation.

79

Fig S6.3.1 After 500 steps Equation (6.5.6) indicates that the root mean square displacement equals (500)1/2δ or 11.18. Forty two simulations were performed and the root mean square displacement was 11.31 ± 8.31, a 1.16% error (Figure S6.3.2). Thirty three simulations are needed to be within 10% of the theoretical value.

Figure S6.3.2 The relation between the rms displacement and the absolute value of displacement after 500 steps is linear. The scatter represents the variability arising from the random number generation. 6.4.

Dij =

k BT f

For a prolate ellipsoid (p=a/b>1) f =

Dij = 2.511 x 10-7 cm2 s-1

6πμb(p 2 -1)1/2 ln[p+(p 2 -1)1/2 ]

Comparing with the value in Table 6.5, there is a 25.5% error

80

For a cylinder of radius a and length L f =

8πμL 3ln(L/a) − 0.94

Dij = 2.488 x 10-7 cm2 s-1 There is a 24% error between the calculated and measured values. The cylinder is a slightly more accurate model than the prolate ellipsoid but both shapes do not accurately predict the diffusion coefficient. 6.5. (a) Assume steady state, no reactions, no convection and a dilute solution. A mass balance

on the volume element shown in Figure 6.28 is: Nix|x A(x) –Nix|x+ΔxA(x+Δx) = 0 Dividing by the volume element which is approximately equal to AΔx and taking the limit as Δx approaches zero, we have: 1 d ( Nix A( x ) ) =0 A( x) dx

or

d ( N ix A( x) ) dx

=0

This result indicates that the total number of moles passing through any cross section is constant. Because the area changes with x, the flux decreases with increasing cross-sectional area. (b) Integrating the material balance equation: Nix = − Dij

dCi K = 1 dx A( x)

The cross-sectional area is equal to πr(x)2. Using this relation and Equation (6.12.2) for r(x) the integrated material balance equation is: dCi =− dx

K1 x⎞ ⎛ Dij πro 2 ⎜ 1 + ⎟ ⎝ L⎠

2

Integrating this expression yields: Ci =

K1 + K2 x⎞ ⎛ LDij πro 2 ⎜ 1 + ⎟ ⎝ L⎠

The flux equals the following: N ix = − Dij

dCi 2 Dij ( Co − CL ) = 2 dx x⎞ ⎛ L ⎜1 + ⎟ ⎝ L⎠

6.6. Since the mole fraction of NO is 0.0001, the error is approximately of this level. The term 1xi in the expression for the flux is initially 0.9999 and grows. Thus, the flux can be approximated as: CDij dxi dx ≈ CDij i Ni = − 1 − xi dr dr

81

In spherical coordinates, the time to reach steady state is R2/Dij. For the values given the time is 1.51 x 10-4 s. For the problem given this is the time that it would take for the gas to leave the alveolus. This is much shorter than the time between breathes, 5 sec. 6.7. K = Cm/C. Since the same solvent is on both sides of the membrane, K should be the same on both sides. (a) This is feasible and K < 1. (b) This is not feasible since K>1 on left side and K<1 on the right side. Further, the fluxes on either side of the membrane are in different directions. (c) This case is not feasible because K < 1 on the left side and K>1 on the right side. Further the flux through the membrane is in the opposite direction as the flux in the fluid phase. 6. 8. For this problem there is a steady state, and one-dimensional diffusion in a dilute solution. The two regions have cross-sectional areas A1 and A2, respectively. The conservation relations for each phase are: d 2 C1 dx

2

=0

d 2 C2 dx 2

=0

The solutions are: C1=A1x +B1 C2=A2x+B2 Applying the boundary conditions at x = 0 and x = L, yields: C1 = K1Co + K1 ( CL − Co )

x L

C2 = K 2Co + ( CL − Co )

x L

The resulting flux in each phase is given by: N i1 = − Di1

D dC1 = − i1 K1 ( CL − Co ) dx L

N i 2 = − Di 2

D dC2 = − i 2 K 2 ( CL − Co ) dx L

A mass balance yields: Moles of i entering surfaces = moles of i entering phase 1 + moles of i entering phase 2. This can be written as: Nitot(A1+A2) = A1Ni1+A2Ni2. The total flux can be written as: ⎛ A K D + A 2 K 2 Di 2 ⎞ ⎛ Co − CL ⎞ N iTOT = ⎜ 1 1 i1 ⎟⎜ A1 +A 2 L ⎟⎠ ⎝ ⎠⎝

The terms A1K1Di1/L(A1+A2) and A2K2Di2/L(A1+A2) represent conductances. So their reciprocals are in diffusive resistances. Thus, diffusion through two media in parallel is analogous to two electrical resistances in parallel. 6.9. (a). Assume K1=K2 = 1. From equation 6.7.16b, Deff = 8.97 x 10-7 cm2 s-1. (b). If K1=K2 and D1=D2 then, from the standpoint of diffusion, the medium would be effectively a single phase. 6.10. The steady state mass balances for one dimensional radial diffusion in cylindrical coordinates are: 82

1 d ⎛ dCij ⎞ ⎜r ⎟=0 r dr ⎝ dr ⎠

where j=1 or 2 for the two layers.

The boundary conditions are r = Ri = Ci1 = Ci, r = Ro = Ci2 = Co, r = R1 C1=C2 (assume partition coefficients are equal) and Ni1 = Ni2. The solution of the mass balance is: Cij = Ajlnr + Bj Applying the condition at r = R1 that the fluxes are the same yields: Di1A1 = Di2A2. The conditions that the concentrations are equal at r = R1 yields, ⎛ ⎞ D B2 = A1 ⎜ ln R1 − 1 ln R1 ⎟ + B1 D ⎝ ⎠ 2

Applying the other two boundary conditions results in the following: A1 =

Di 2 ( Co − Ci )

⎛R Di1 ln ⎜ o ⎝ R1

B1 = Ci −

A2 =

⎞ ⎛ Ri ⎞ ⎟ − Di 2 ln ⎜ ⎟ ⎠ ⎝ R1 ⎠

Di1 ( Co − Ci )

⎛R Di1 ln ⎜ o ⎝ R1

⎞ ⎛ Ri ⎞ ⎟ − Di 2 ln ⎜ ⎟ ⎠ ⎝ R1 ⎠

Di1 ( Co − Ci ) ln Ri ⎛R Di1 ln ⎜ o ⎝ R1

⎞ ⎛ Ri ⎞ ⎟ − Di 2 ln ⎜ ⎟ ⎠ ⎝ R1 ⎠

The concentrations are: Ci1 =

Ci 2 =

Di 2 ( Co − Ci )

ln ⎛⎜ r ⎞⎟ + Ci ⎛R ⎞ ⎛ R ⎞ ⎝ Ri ⎠ Di1 ln ⎜ o ⎟ − Di 2 ln ⎜ i ⎟ ⎝ R1 ⎠ ⎝ R1 ⎠ Di1 ( Co − Ci )

Di 2 ( Co − Ci ) R ln ⎛⎜ r ⎞⎟ + ln ⎛⎜ 1 ⎞⎟ + Ci R i⎠ ⎛ Ro ⎞ ⎛ Ri ⎞ ⎝ ⎛ Ro ⎞ ⎛ Ri ⎞ ⎝ Ri ⎠ Di1 ln ⎜ ⎟ − Di 2 ⎜ ⎟ Di1 ln ⎜ ⎟ − Di 2 ⎜ ⎟ ⎝ R1 ⎠ ⎝ R1 ⎠ ⎝ R1 ⎠ ⎝ R1 ⎠

Note that if Di1=Di2, then the concentration distribution in each phase is continuous and equal to Equation (6.6.36). The flux in each phase is: Ni1 = − Di1

dCi1 = dr

Ni 2 = − Di 2

dCi1 = dr

− Di1 Di 2 ( Co − Ci )

⎛1⎞ ⎜ ⎟ ⎛R ⎞ ⎛ R ⎞⎝r ⎠ Di1 ln ⎜ o ⎟ − Di 2 ln ⎜ i ⎟ ⎝ R1 ⎠ ⎝ R1 ⎠ − Di1 Di 2 ( Co − Ci )

⎛1⎞ ⎜ ⎟ ⎛ Ro ⎞ ⎛ Ri ⎞ ⎝ r ⎠ Di1 ln ⎜ ⎟ − Di 2 ln ⎜ ⎟ ⎝ R1 ⎠ ⎝ R1 ⎠

By comparing each flux expression, with that for a single component medium, the following relation can be made for the effective diffusion coefficient: 83

Deff ln ⎛⎜ Ro ⎞⎟ ⎝ Ri ⎠

=

Di1 Di 2 ⎛ Ro ⎞ ⎛R ⎞ Di1 ln ⎜ ⎟ − Di 2 ln ⎜ i ⎟ ⎝ R1 ⎠ ⎝ R1 ⎠

⎛ Ro ⎞ ⎛ Ro ⎞ ⎛ Ri ⎞ ln ⎜ ⎟ ln ⎜ ⎟ ln ⎜ ⎟ R R ⎝ i ⎠ = ⎝ 1 ⎠ − ⎝ R1 ⎠ Deff Di 2 Di1

or

If the thickness of each layer is much less than the inner radius (i.e. R1-Ri <
dC1 (C - C ) = A m Dm K 1 2 dt L

(S6.11.1)

In order to integrate this expression, we need to relate the concentration C1 and C2. This can be done by noting that once solute leaves side 1 it is either in the membrane or on side 2. Thus, the loss of solute from side is balanced by the gain of solute in the membrane or side 2. V1

dC1 dC 2 ⎞ ⎛ V dC m = −⎜ m +V2 dt dt ⎟⎠ ⎝ K dt

(S6.11.2)

If Vm << V2 and Vm << V1 then the first term on the right hand side is much less than the other two terms. This means that the amount of solute in the membrane is small relative to the amount of solute in either reservoir. As a result, equation (S6.11.2) can be simplified to dC2 V dC1 = − 1 dt V2 dt

(S6.11.3)

Using the initial conditions C1 = Co and C2 = 0, equation (S6.11.3) can be integrated to yield C2 = −

V2 ( C1 − Co ) V1

(S6.11.4a)

Substituting Equation (S6.11.4a) into Equation (S6.11.1) leads to a differential equation in one variable. −V1

A D K dC1 = m m dt L

⎡⎛ V1 ⎢⎜ 1 + ⎢⎣⎝ V2

⎤ ⎞ V1 Co ⎥ ⎟ C1 V2 ⎥⎦ ⎠

(S6.11.5)

Equation (S6.11.5) is a first order ordinary differentiation equation that can be solved with the use of an integrating factor. Solving Equation (6.11.5) subject to the initial condition that C1 = Co and C2 = 0, yields: ⎛ A D K⎛ V C1 =Co exp ⎜⎜ − m m ⎜ 1 + 1 V1 L ⎝ V2 ⎝

⎛V ⎞ Co ⎜ 1 ⎟ ⎞ ⎞ ⎝ V2 ⎠ ⎡1 − exp ⎛ − A m Dm K ⎛ 1 + V1 ⎢ ⎜⎜ ⎟ t ⎟⎟ + ⎜ V1 L ⎝ V2 ⎠ ⎠ ⎛1 + V1 ⎞ ⎣⎢ ⎝ ⎜ ⎟ ⎝ V2 ⎠

84

⎞ ⎞⎤ ⎟ t ⎟⎟ ⎥ ⎠ ⎠ ⎦⎥

(S6.11.6a)

This equation needs to be rearranged in order to be used practically for the estimation of the permeability (DmK/L). After such rearrangement, the analog of Equation (6.8.107) is: ⎡C ⎛ V ⎞ V ⎤ A D K⎛ V ⎞ ln ⎢ 1 ⎜ 1 + 1 ⎟ − 1 ⎥ = − m m ⎜ 1 + 1 ⎟ t V1 L ⎝ V2 ⎠ ⎢⎣ Co ⎝ V2 ⎠ V2 ⎥⎦

(S6.11.7)

This expression equals Equation (6.8.107) when V1 = V2. Thus plotting the left hand side of Equation (S6.11.7) versus time yields a slope equal to –AmP(1+V1/V2). 6.12. For diffusion-limited reactions on the cell surface, equation (6.9.23) applies, for which s is the coated pit radius and b is the distance between coated pits. This distance is 1.755 µm and k = 0.14 s-1. What this means is that if measured values of the dissociation constant are less than 0.14 s-1, then the dissociation step is not diffusion-limited. 6.13. a. Assume that adsorption is diffusion-limited. Equation 6.7.29 can be used to predict the time to cover the surface. Rearranging this equation gives: 2

⎛C ⎞ π t = ⎜ ads ⎟ ⎝ 2C0 ⎠ Dij Substituting the values provided for albumin: 2

⎛ 0.376 × 10−3 mg/cm 2 ⎞ π ⎟ t=⎜ = 0.072 ms ⎜ 2 ( 50 mg/cm3 ) ⎟ 6.2 × 10−7 cm 2 / s ⎝ ⎠ For fibronectin, the time is 2

⎛ 0.352 × 10−3 mg/cm 2 ⎞ π ⎟ t=⎜ = 14.2 s −7 3 ⎜ 2 ( 0.2mg/cm ) ⎟ 1.7 × 10 cm 2 / s ⎝ ⎠ b. If adsorption is irreversible then albumin, since it is the most abundant protein, will rapidly cover the entire surface. During the time that albumin absorbs, the fibronectin surface concentration only reaches 0.79 ng/cm2, 0.2 % of the albumin surface. c. Clearly, using a purified solution of fibronectin will permit the complete coverage of the surface with this protein.

⎛ Rate of dissolution ⎞ ⎛ Rate of transport away ⎞ 6.14. a. The mass balance is – ⎜ ⎟=⎜ ⎟ of polymer ⎝ ⎠ ⎝ from polymer surface ⎠ d ( VCp ) − = AN r |r = R dt d ( 4 / 3π R 3 ) −C p = 4π R 2 N r |r = R dt dR 3 −C p = 3R 2 N r |r = R dt dR −C p = N r |r = R dt DKCp b. Evaluating equation 6.6.49 at r=R, the flux is given by Nr|r=R = R 85

c. Inserting the expression for the flux into the result from part a, solving, and applying the boundary conditions yields R ( t ) = R 0 − 2DKt d. For the data given, the time for the polymer to dissolve is 184 days. Under this condition the quasi-steady state assumption is valid. 6.15. In this problem there is unsteady, diffusion in the radial direction with no convection or chemical reactions. Spherical coordinates should be used and the coordinate origin corresponds to the center of the sphere. The conservation relation is the same as Equation (6.7.48) with Ri = 0. The initial condition is 0 < r < R t ≤ 0 C = Co and the boundary condition at r = 0 is ∂C =0 ∂r

t ≥0 The boundary condition at r = R is: − Di

∂C = P ( C1 − C ) ∂r

The transformations for θ’, η and τ yield the following dimensionless conservation relation and boundary conditions: ∂θ ' 1 ∂ ⎛ 2 ∂θ ' ⎞ = 2 ⎜η ⎟ ∂τ η ∂η ⎝ ∂η ⎠

0<η<1

τ ≤ 0 θ’ = 

η=0

τ ≥0

η=1

τ ≥0

∂θ ' =0 ∂r ∂θ ' − = Biθ ' ∂r

The problem is solved using the approach outlined in Equations (6.8.55)-(6.8.57). The solution to the conservation relation is: ⎛A

θ' = ⎜

⎝η

sin(λη ) +

⎞ cos(λη ) ⎟ exp(−λ 2τ ) η ⎠ B

Applying the boundary condition at η = 0 yields B = 0 (cf. Equation 6.8.58 and following). θ’ simplifies to: θ' =

A

η

sin(λη ) exp(−λ 2τ )

Applying the boundary condition at η = 1 yields: Asin(λ ) exp(−λ 2τ ) − Aλ cos (λ ) exp(−λ 2τ ) = BiAsin(λη ) exp(−λ 2τ )

Rearranging this expression results in the following relation for λ: λn

1 − Bi

= tan λn

The solution for θ' is a series with coefficients An. θ' =

∞

A

∑ ηn sin(λnη ) exp(−λn2τ ) n=1

Applying the orthogonality relation to the initial condition: 86

n=1,2,3…

1

1

∞

An ∫ sin[λnη ]sin[λmη ]dη ∫η sin[λnη ]dη =∑ n =1 0

0

Using integration by parts, the left hand side equals(sinλn – λncosλn)/λn2. By orthogonality, the integral on the right hand side is nonzero for m=n. This integral equals 1/2(1-cosλnsinλn /λn). The coefficients An are thus equal to the following: An =

2 ( sin λn − λn cos λn )

λn ( λn − cos λn sin λn )

After substituting for An, the concentration distribution is: θ' =

∞

2 ( sin λn − λn cos λn )

∑ λ η (λ n=1

n

n

− cos λn sin λn )

sin(λnη ) exp(−λn 2τ )

In order to determine the time to reach steady state, the average concentration is calculated and the time for its value to reach 0.01 is assumed to be the time to reach steady state. θ ' = 3∫ θ 'η 2 dη =

∞

6 ( sin λn − λn cos λn )

∑ λ 3 (λ n=1

n

n

2

− cos λn sin λn )

exp(−λn 2τ )

The result is plotted in Figure S6.15.1 for Bi = 0.1, 10 and 1000 which correspond to the range of parameters given. The dimensionless times to reach steady state are 15 for Bi = 0.1, 0.5 for Bi=10 and 0.4 for Bi=1000. The time increases as the membrane resistance increases relative to diffusion within the cell (smaller Bi).

Figure S6.15.1 6.16. Using equation 6.7.22, the product of the flux times the surface area for C1 = 0 is: D N ix = −Co π R2 πt For a five hour flight, the release rate is 7.62 x 10-9 mole s-1. For a 14 hour flight, the release rate is 4.55 x 10-9 mole s-1. Thus the criterion is met.

87

6.17. For a cell of radius 17 µm, there are 3.03 x 109 receptors cm-2. The distance between receptors is 1/(3.03 x 109 receptors cm-2)1/2 = 182 nm. From equation 6.9.15, the fractional reduction in the rate constant for a surface containing this density of receptors is: Ns = 0.22 Ns + πR 6.18. (a) The conservation relation for unsteady, one-dimensional diffusion without chemical reaction in a rectangular coordinate system is given by Equation (6.8.30). The boundary conditions are: x=L Cm = 0 x=0 Cm = Co The initial condition is: t≤0 Cm = 0

Using the dimensionless variables listed in the problem statement, the conservation relation and boundary conditions are: ∂θ ∂ 2θ = ∂τ ∂η 2

τ ≤0 θ=0 0 ≤ η≤ 1 η=0 τ ≥0 θ = 1 η=1 τ ≥0 θ=0 In order to apply separation of variables, both boundary condition need to be homogeneous. The boundary condition at η = 0 is not homogeneous. A homogeneous problem can be developed by noting that at long times, a steady state concentration gradient exists across the membrane, i.e., θ(η,τ→∞)=ψ(η). In order for this to be the case, the time dependent solution must consist of two parts, one ψ(η) which represents the steady state solution and a second term φ(η,τ) which represents the transient change in concentration. Thus, θ(η, τ) = ψ(η) + φ(η, τ) and φ(η, τ →∞) = 0. Further, the steady state solution satisfies the same boundary conditions: η=0 ψ =1 η=1 ψ=0 In order to satisfy the boundary conditions for ψ and θ, the boundary conditions for φ are: η=0 τ ≥0 φ = 0 η=1 τ ≥0 φ=0 The initial condition is that τ ≤ 0 φ + ψ = 0 or φ = − ψ 0 ≤ η≤ 1 (b) The solution for ψ is obtained by integrating the steady state conservation relation: d 2ψ dη 2

=0

θ = Aη + B To yield: Evaluating the boundary condition at η = 0 yields B = 1. The boundary condition at η = 0 indicates that A = -1. Thus, θ = 1-η Substituting θ(η, τ) = ψ(η) + φ(η, τ) and using the conservation relation for ψ results in the following conservation relation for φ: 88

∂φ ∂ 2φ = ∂τ ∂η 2

The boundary conditions are homogeneous and the initial condition is nonhomogeneous, so the conservation relation can be solved by separation of variables. The process is identical to that followed to solve Equations (6.8.37) and (6.8.38). The result is: φ = ( Asin(λη ) + Bcos(λη ) ) exp(−λ 2τ )

The boundary condition at η = 0 indicates that B = 0. The boundary condition at η = 1 results in the following relation: Asin(λη) = 0 Setting A = 0 leads to a trivial solution. Alternatively, the boundary condition is satisfied if λ is a multiple of π. λn = nπ n = 1,2,3,… The solution for φ becomes: ∞

φ = ∑ A n sin(nπη ) exp(− n 2 π2τ ) n=1

The coefficients An are found by applying the initial condition. ∞

1-η = -∑ A n sin(nπη ) n=1

Applying the orthogonality relation to the initial condition: 1

∞

1

0

n =1

0

∫ (1 − η ) sin[nπη ]dη = − ∑ An ∫ sin[nπη ]sin[mπη ]dη Using integration by parts, the left hand side equals 1

∫ (1 − η ) sin(nπη )dη = 0

η =1

η =1

1

η =1

cos(nπη ) cos(nπη ) sin(nπη ) 1 − ∫ η sin(nπη )dη = (1 − η ) + = 2 2 nπ n π n π n π η =0 0 η =0 η =0

The integral right hand side is nonzero for m= n and equals1/2. Thus, An = −

1 nπ

The concentration θ now becomes: φ = 1-η -

2 ∞ sin(nπη ) ∑ n exp(−n2 π2τ ) π n=1

(c) The following MATLAB program was used to generate the concentration profile. function zz=memb(t); n=1:1:10; y=n*pi; for x=0.0:0.05:1.0 j=20*x+1; zz(j)=1-x-2*sum(sin(y.*x).*exp(-y.*y.*t)./y); end Steady state is reached for dimensionless times greater than 0.5 or a dimensionless time of 0.5L2/Dm. This is similar to the estimate used in deriving the quasi-steady state assumption across the membrane. 89

(d) To derive the mass balance, equate the rate of loss of solute from side with the rate of transport into the membrane. The rate of loss of solute is the change in mass of the solute in side 1. The rate of transport into the membrane is the product of the flux times the surface area. −V

dC1 =N dt

−V

∂C dC1 = − ADm m dt ∂x

x =o

A

Substituting for the flux: x =0

The concentration gradient at x = 0 is obtained by differentiating the solution for C(x,t) = θ(x/L, τ)Co.

∂Cm ∂x

= x =0

∞ ∞ ⎞ ⎞ Co ∂θ C ⎛ C ⎛ = o ⎜ -1-2∑ cos(nπη ) exp(− n 2 π2τ ) ⎟ = − o ⎜ 1+2∑ exp(− n 2 π 2τ ) ⎟ ⎜ ⎟ L ∂η η =0 L ⎝ L ⎝ n=1 n=1 ⎠ ⎠η =0

Substituting this expression into the mass balance and rearranging terms gives the following expression for the derivative of C1 with respect to time. ∞ ⎛ C AD ⎡ dC1 tL2 ⎞ ⎤ = − o m ⎢1 + 2∑ exp ⎜⎜ − n 2 π2 ⎟⎥ dt LV ⎣⎢ Dm ⎟⎠ ⎥⎦ n=1 ⎝

The exponential term decays with time as follows ∞ ⎛ tL2 ⎞ 2∑ exp ⎜⎜ −n 2 π2 ⎟ Dm ⎟⎠ n=1 ⎝

0.1 0.7843 0.3 0.1036 0.4 0.0386 0.5 0.0144 0.6 0.0054 Likewise, a similar mass balance at x = L yields that the rate of transport from the membrane at x =L equals the rate of increase of moles on side 2. V

∂C dC2 = − ADm m dt ∂x

x=L

The concentration gradient at the surface x = L is: ∂Cm ∂x

= x =0

Co ∂θ C = o L ∂η η = 0 L

∞ ∞ ⎛ ⎞ ⎞ C ⎛ 2 2 = − o ⎜1+2∑ (−1) n exp(− n 2 π 2τ ) ⎟ ⎜⎜ -1-2∑ cos(nπη ) exp(− n π τ ) ⎟⎟ L ⎝ n=1 n=1 ⎠ ⎝ ⎠η =0

The change in C2 with time is simply: ∞ ⎛ 2 2 tL2 ⎞ ⎤ dC2 Co ADm ⎡ n = ⎢1 + 2∑ (−1) exp ⎜⎜ − n π ⎟⎥ dt LV ⎣⎢ Dm ⎟⎠ ⎥⎦ n=1 ⎝

Thus, after a dimensionless time interval of 0.5, the exponential terms decline to 1% or less of 1 and the rate of loss from compartment 1 equals the rate of gain in compartment 2. Further, the change in concentration in either reservoir is proportional to LV/ADm. Thus, the quasi-steady state approximation is valid for times longer than 0.5L2/Dm. Note that a more accurate answer can be determined by solving for the time-dependent boundary conditions. 6.19. For a cell of radius 15 µm, there are 3.54 x 109 receptors cm-2. The distance between receptors is 1/(3.54 x 109 receptors cm-2)1/2 = 168 nm. From Equation (6.9.15), the fractional reduction in the rate constant for a surface containing this density of receptors is:

90

Ns = 0.24 Ns + πR

6.20. For diffusion-limited dissociation of a ligand from a receptor, Equation (6.9.17) applies subject to the boundary conditions (6.9.22a,b). Integrating Equation (6.9.17) yields: C = A ln r + B

Applying the boundary conditions yields the following two equations; Co = A ln s + B

Cb = A ln b + B

solving for A and B yields A=

Co − Cb ln ( s / b )

(Co − Cb ) ln b ln ( s / b ) (C − Cb ) ln ( r / b ) C = Cb + o ln ( s / b ) B = Cb −

The concentration profile is;

The diffusion-limited dissociation rate constant is found by equation the rate of dissociation from the perimeter with the product of the flux times the perimeter: Nr

r =s

2 πs = k− ( Co − Cb ) πs 2

Nr

r =s

= − DL

The flux at r = s is dC dr

=

DL ( Co − Cb )

r =s

s ln ( b / s )

The diffusion-limited rate constant k- is: k− =

2 DL s ln ( b / s ) 2

6.21. Begin with the definition of the error function: η 2 − x2 erfc (η ) = 1 − erf (η ) = 1 − ∫ e dx

π

2

(S6.21.1)

0

Note that the integral of e-x from 0 to ∞ is ∞

∫e

− x2

dx =

π

2 Further, this integral can be divided into two portions: 0

91

(S6.21.2)

∞

η

∞

−x −x −x ∫ e dx = ∫ e dx + ∫ e dx = 2

0

-x2

Thus, the integral of e

η

2

2

η

0

π

(S6.21.3)

2

from 0 to η can be rewritten as:

−x ∫ e dx =

π

2

0

2

∞

− ∫ e − x dx 2

(S6.21.4)

η

Use Equation (S6.21.4) to replace the in Equation (S6.21.1). ∞ ⎞ 2 2 ⎛ π 2 erfc (η ) = 1 − − ∫ e − x dx ⎟ = ⎜ ⎟ π ⎜⎝ 2 η π ⎠

∞

∫e η

− x2

dx

(S6.21.5)

6.22. For a three-dimensional random walk, the mean square displacement is related to time and the diffusion coefficient by Equation (6.5.10). 〈 r 2 〉 = 6 Dij t.

Dij = /6t , cm2 4.32 x 10-8 8.64 x 10-8

or t, s 720 1440

Dij, cm2/s 1 x 10-11 1 x 10-11

6.23. (a) Beginning with Equation (6.8.22) for the flux of solute at the surface with concentration C1 diffusing into a semi-infinite medium initially at a concentration C0. N ix ( x = 0) =

Dij

πt

(C0 − C1 ).

Since, there is no drug initially in the tissue, C1 = 0. The total amount of drug entering the tissue by time t, M, is the integral of the flux times surface area over time. t

t

M = ∫ N ix ( x = 0)Adt = ∫ 0

0

Dij

πt

C0 Adt = 2

Dij t

π

C0 A

Solving for Dij yields: Dij =

π ⎛ M ⎞

2

⎜ ⎟ 4t ⎝ C 0 A ⎠ Converting t = (25 min)(60 s/min) = 1500 s C1 = 1 x 10-9 mole/cm3

M/C1A = (5 x 10-13 moles)/[(1 x 10-9 mole/cm3)( 0.75 cm2)] = 0.000667 cm Dij =

3.14159 2 ( 0.000667 ) = 2.33 x 10 −10 cm 2 /s 6000

t = L2/36D = 1907 s > 1500 s 92

6.24. This is a case of unsteady diffusion in finite dimensions. C1 = 0, C0 = 0.1 mg/ml. We want to find the time at which C/C0 = 0.05. For the results in Figure 6.20, this corresponds to (CC0)/(C1 - C0) =(C- C0)/( - C0) = 1- C/C0 = 0.95. From the graph this corresponds to about 0.4 = Dijt/R2.

As a result, t = 0.4R2/Dij = 0.4(0.04 cm2)/(5 x 10-10 cm2 s-1) = 3.2 x 107 s = 8,889 hours = 370 days. The result depends upon the interpolated value. To get an exact result, evaluate equation (6.8.63) at η = 0. This is done by applying L’Hopital’s rule. The result is ∞

θ (η = 0 ) = 1 + ∑ 2(−1) n exp(− n2π 2τ ). n =1

Programming this equation in MATLAB (or a spreadsheet) yields q = 0.9503 at t = 0.375. Revising the calculation yields 347.2 days. 6.25. Taking the logarithm of both sides of Equation (6.8.69): x2 4 Dij t The slope is equal to -1/4Dij. From the data at 3,000 s, the slope equals - 2.51 x 107 s cm-2. From this Dij = 9.96 x 10-9 cm-2 s. At 63,000 s the slope equals - 2.47 x 107 s cm-2 and Dij = 10.8 x 10-9 cm-2 s. The two values differ slightly, but the mean value is 10.38 x 10-9 cm-2 s.

ln ( Ci / M 0 ) = − ln 2 (π Dij t )

1/2

−

6.26. (a) At steady state, the exponential term in the equation given in the problem statement is zero and this expression reduces to: C = ΦC0(1-y/l) DΦC0 dC The flux at y = L is N y =l = − D = dy y =l L R = AN y =l =

The release rate R is:

ADΦC0 L

(b) The time to reach steady state is t = 0.5 = tD/l2. Solving for l: l = ( 2 Dt )

1/2

(

= 2 ( 5 x 10 −9 cm 2 /s ) ( 600 s )

)

1/2

= 0.0024 cm

Rearranging the expression for the release rate and solving for C0: C0 =

( 0.0024 cm )( 0.15 mg/h )(1 h / 3600 s ) = 100 mg/cm 3 LR = ADΦ 2 cm 2 5 x 10 −9 cm 2 /s 0.10

(

)(

)

(c) After 10 hours, 1.5 mg is released, assume a volume of 1 cm3. This is only 1.5% of the initial amount and the criterion is met.

93

6.27. Assuming steady state, then the problem is identical to the two membranes in series discussed in Section 6.7. Using Equation (6.7.17b) for the effective permeability: L1 L2 1 L . = = + Peff Φ eff Deff Φ1 Di , 1 Φ 2 Di , 2

For the membrane, For the cytoplasm

Pm = (3.50)(3.9x 10-6 cm2/s)/(9 x 10-7 cm) = 15.167 cm/s. Pc = (1)(3.3 x 10-5 cm2/s)/(5 x 10-4 cm) = 0.066 cm/s

Peff = 0.0657 cm/s The membrane accounts for 0.4% of the resistance (1/Pm)/(1/Peff). 6.28. (a)

(b)

D d ⎛ 2 dC ⎞ =0 r r 2 dr ⎜⎝ dr ⎟⎠

r = Ri r = R0

C = C0 C=0

dC A = dr r 2 A C =− +B Integrating again r A Applying the boundary conditions C0 = − + B Ri A Solving for B B = R0

(c) Integrating once yields:

Substituting Solving for A: A =

C0 =

0=−

A +B R0

A A A ( Ri − R0 ) − = R0 Ri R0 Ri

C0 R0 Ri ( Ri − R0 )

⎛ 1 1⎞ C R R Thus, C = C0 ⎜ − ⎟ 0 0 i ⎝ R0 r ⎠ ( Ri − R0 )

(d) The flux at r = R0. N r

r = R0

= −D

dC dr

= r = R0

DC0 R0 Ri DC0 Ri = r ( Ri − R0 ) r = R R0 ( Ri − R0 ) 2

0

6.29. (a) This problem is identical to the one discussed in Section 6.8.1, except that C0 = 0. Thus, the flux is given by Equation (6.8.22) Nx ( x = 0) =

D C πt 1

94

(b) Inserting the expression for the flux into the integral and integrating t

t

D D −1/2 Dt C1dt = C1 A t dt = 2C1 A ∫ πt π 0 π

M (t ) = A∫ 0

where A is the surface area. (c) Rearranging the uptake expression D=

π M (t ) 2

t 4 A C1

π ( 8.74 x 10 −13 mole )

2 2

=

(1 cm ) ( 600 s ) 4 (1 x 10 2

2

2

2

-9

−9

2

mole/cm

3

)

2

= 1 x 10 −9 cm 2 / s

2

(d) t< L /36D = (0.015 cm) /(36 x 10 cm /s) = 6,250 s. The 600 s are much smaller than this time, so the assumption of a semi-infinite medium is valid. 6.30. (a) At steady state the concentration in each reservoir is the same. From equation (6.8.107) we obtain C1 ⎛ 2A D Φt ⎞ = 0.5+0.5*exp ⎜ - m m ⎟ Co VL ⎝ ⎠

At long times, C1 = 0.5C0 = 0.00075 M. Since Cm = FC1, F = Cm/C1 = (2 x 10-5)/ (75 x 10-5) = 0.02667. (b) From Equation (6.8.107), ⎛ 2C − Co ⎞ 2A m Dm Φt ln ⎜ 1 ⎟ =C VL o ⎝ ⎠

Rearranging and substituting Dm = −

⎛ 2C − Co ⎞ VL (100 cm3 )(40 x 10-4 cm) ⎛ 0.0021 − 0.0015 ⎞ ln ⎜ 1 = ln ⎜ ⎟ ⎟ 2 2A m Φt ⎝ Co 0.0015 2(50 cm ) 0.026667 48 h*3600 s/h ⎝ ⎠ Dm = 7.95 x 10-7 cm2 /s ( )( ) ⎠

95

Solution to Problems in Chapter 7, Section 7.11 7.1. The effective diffusion coefficient is given by Equation (7.5.23) and the diffusion potential is given by Equation (7.5.27). Deff =

( z + − z− ) D+ D−

(7.5.23)

z + D+ − z - D− ⎛

D+ − D− ⎞ RT ⎛ Co ⎞ ln ⎜ ⎟ ⎟ ⎝ z+ D+ − z− D− ⎠ F ⎝ CL ⎠

(7.5.27)

ψ ( L) −ψ (0)= ⎜

Diffusion coefficients for the ions are provided in Table 7.4. (a) CuSO4

Cu++ D+ = 0.72 x 105 cm2 s-1 SO4 Deff = 0.8592 x 105 cm2 s-1

D- = 1.065 x 105 cm2 s-1

⎛

⎞ 0.72 − 1.065 ⎛ 1 mM ⎞ ⎟ ( 25.69 mV ) ln ⎜ ⎟ =-5.72 mV ⎝ 0.1 mM ⎠ ⎝ 2(0.72) − (−2)(1.065) ⎠

ψ ( L) −ψ (0)= ⎜

(b) MgCl2

Mg++ D+ = 0.7063 x 105 cm2 s-1 Deff = 1.25 x 105 cm2 s-1

Cl-

⎛

D- = 2.032 x 105 cm2 s-1

⎞ 0.7063 − 2.032 ⎛ 1 mM ⎟ ( 25.69 mV ) ln ⎜ ⎝ 0.1 mM ⎝ 2(0.7063) − (−1)(2.032) ⎠

ψ ( L) −ψ (0)= ⎜

⎞ ⎟ =-22.77 mV ⎠

7.2. a. Using the definition of the permeability and equation ⎡ ⎤ D N iz CO 1 Pi = = ij PeL ⎢ − ⎥ CO − CL L ⎣ ( CO − CL ) 1 − exp ( PeL ) ⎦

b. For small values of PeL, exp(PeL)≈1+ PeL. Pi =

N iz Co − CL

=

⎡ ⎤ Dij ⎡ Co PeL ⎤ Co 1 PeL ⎢ − + 1⎥ ⎥= ⎢ L L ⎣ ( Co − CL ) ⎣ ( Co − CL ) 1 + (1 + PeL ) ⎦ ⎦

Di j

Thus, in the limit of PeL going to zero, the terms PeLCo/(Co-CL) reduce to Dij/L. c. For the data given, PeL = 10. The permeability from part a for CL = 0 is: Pi=DijPeL/L=1.1 x 10-6 cm s-1. The diffusive permeability is Pi = DijPeL= 1 x 10-7 cm s-1. Thus, convection leads to a permeability ten times larger than the permeability estimated by diffusion alone. 7.3. Assuming steady state, no convection, no chemical reactions, and one-dimensional transport, the material balance is: dNiz dz

where

Ni = - Dij

=0

dCi Dij Ci zi F dψ dz RT dz

96

(S7.3.1) (S7.3.2)

The subscript i represents Na+, K+ or Cl- for zi is +1,+1, and –1 respectively. The net current is zero,

(

3

i = 0 = F ∑ zi Niz = F N Na + + N K + − NCl − i =1

z

z

z

)

(S7.3.4)

Electroneutrality requires that: 3

0 = ∑ zi Ci = CNa + + CK + − CCl −

(S7.3.5)

i =1

The boundary condition for each ion is that: z=L C = ΦiCL (S7.3.6a,b) z = 0 Ci = ΦiCo Assuming a linear potential variation across the membrane, the potential gradient simplifies to: dψ Δψ (ψ o −ψ L ) Vm = = = dz L L L

(S7.3.7)

The flux is: Ni = - Dij

dCi Dij Ci zi F Vm dz RT L

(S7.3.7)

Inserting this expression for the flux into the material balance leads to the following expression: d 2 Ci dz

2

=

−

zi F Vm dCi RT L dz

(S7.3.8)

Integrating twice yields: Ci = A

RTL ⎛ z FV exp ⎜ − i m zi F Δψ ⎝ RTL

⎞ z⎟+ B ⎠

(S7.3.9)

Applying the boundary conditions results in the following two equations: Φ i Co = A

RTL +B zi FVm

Φ i CL = A

RTL ⎛ z FV exp ⎜ − i m zi F Δψ RT ⎝

(S7.3.10a) ⎞ ⎟+ B ⎠

(S7.3.10b)

Using equation (S7.3.10a) to solve for B: B = Φ i Co − A

RTL zi FVm

(S7.3.10c)

This result is then substituted into the expression arising from the boundary condition at z = L. Φ i CL = A

RTL zi FVm

Solving for A yields: A=

⎛ ⎛ zi FVm ⎜ − exp ⎜ ⎝ RT ⎝

Φ i ( CL − Co ) RTL zi FVm

⎛ ⎛ zi FVm ⎜ exp ⎜ − RT ⎝ ⎝

97

⎞ ⎞ ⎟ − 1⎟ ⎠ ⎠

⎞ ⎞ ⎟ − 1⎟ + Φ i Co ⎠ ⎠

(S7.3.11a)

(S7.3.11b)

B = Φ i Co −

Φi ( CL − Co ) RTL zi FVm

⎛ ⎛ zi FVm ⎜ exp ⎜ − RT ⎝ ⎝

(S7.3.11c)

⎞ ⎞ ⎟ − 1⎟ ⎠ ⎠

The concentration distribution now becomes ⎛ ⎛ z FV ⎞ ⎞ Φ i ( CL − Co ) ⎜ exp ⎜ − i m z ⎟ − 1⎟ ⎝ RTL ⎠ ⎠ ⎝ + Φ i Co Ci = − ⎛ ⎛ zi FVm ⎞ ⎞ ⎜ 1 − exp ⎜ − ⎟ RT ⎟⎠ ⎠ ⎝ ⎝

(S7.3.12a)

This result agrees with equation (7.4.30a). Taking the limit as ziFVm/RT approaches zero. ⎛ z FV z ⎞ ⎛ z FV Φi ( CL − Co ) ⎜ i m ⎟ exp ⎜ − i m ⎝ RTL ⎠ ⎝ RTL Ci = − ⎛ zi FVm ⎞ ⎛ zi FVm ⎞ ⎜ − RT ⎟ exp ⎜ − RT ⎟ ⎝ ⎠ ⎝ ⎠

⎞ z⎟ ⎠ +Φ C = Φ C +Φ C −C z i o i o i( L o) L

(S7.3.12b)

which agrees with Equation (6.7.9). To obtain the flux, first calculate the derivative of Ci with respect to z: Φ i ( CL − Co ) dCi zi FVm = dz RTL ⎛ ⎛ zi FVm ⎜ 1 − exp ⎜ − RT ⎝ ⎝

⎛ z FV exp ⎜ − i m ⎞⎞ ⎝ RTL ⎟⎟ ⎠⎠

⎞ z⎟ ⎠

(S7.3.13)

The flux of each ion is obtained using the Nernst-Planck equation, Equation (7.5.4):

N iz = −

zi Dij FVm RTL

⎛ z FV Φ i ( CL − Co ) exp ⎜ − i m ⎝ RTL z ⎛ ⎛ i FVm ⎞ ⎞ ⎜ 1 − exp ⎜ − ⎟ RT ⎟⎠ ⎠ ⎝ ⎝

⎞ z⎟ ⎠ + zi Dij FVm RTL

⎡ ⎤ ⎛ ⎛ zi FVm ⎞ ⎞ z ⎟ − 1⎟ ⎢ Φ i ( CL − Co ) ⎜ exp ⎜ − ⎥ ⎝ RTL ⎠ ⎠ ⎝ ⎢ − Φ i Co ⎥ ⎢ ⎥ ⎛ ⎛ zi FVm ⎞ ⎞ ⎢ ⎥ ⎜ 1 − exp ⎜ − ⎟ ⎟ RT ⎠ ⎠ ⎢⎣ ⎥⎦ ⎝ ⎝

⎡ ⎤ ⎢ ⎥ Φ i zi Dij FVm zi Dij FVm Φ i ( CL − Co ) ⎢ + Φ i Co ⎥ = − Niz = − ⎥ RTL ⎢ ⎛ RTL ⎛ zi FVm ⎞ ⎞ ⎢ ⎜ 1 − exp ⎜ − ⎥ ⎟ ⎟ RT ⎠ ⎠ ⎢⎣ ⎝ ⎥⎦ ⎝

⎡ ⎛ zi FVm ⎞ ⎤ ⎢ CL − Co exp ⎜ − ⎥ RT ⎟⎠ ⎥ ⎝ ⎢ ⎢ ⎛ ⎛ zi FVm ⎞ ⎞ ⎥ ⎢ ⎜ 1 − exp ⎜ − ⎟ ⎥ RT ⎟⎠ ⎠ ⎥⎦ ⎢⎣ ⎝ ⎝

⎡ ⎤ ⎛ z FV ⎞ C exp ⎜ − i m ⎟ − CL ⎥ Φ i zi Dij FVm ⎢ o RT ⎠ ⎝ ⎢ ⎥ Niz = − RTL ⎢ ⎥ ⎛ zi FVm ⎞ ⎢ exp ⎜ − RT ⎟ − 1 ⎥ ⎝ ⎠ ⎣ ⎦

In the limit as ziFVm/RT approaches zero, the flux becomes ⎧ ⎡ ⎤⎫ ⎛ zi FVm ⎞ − CL ⎥ ⎪ lim ⎪ Φ z D FV ⎢ Co exp ⎜ − ⎟ ΦD ⎪ i i ij m ⎝ RT ⎠ ⎢ ⎥ ⎪⎬ ≈ i ij ( Co − CL ) N iz = zi FVm ⎨− z FV RTL L →0 ⎪ ⎢ ⎥⎪ 1 − i m −1 RT ⎢ ⎥⎪ RT ⎪⎩ ⎣ ⎦⎭ 98

Typically, the derivation performed in the electrophysiology literature takes the origin of the coordinate system at the intracellular surface and the flux and ion currents are have a sign that is opposite to that noted above. 7.4. Based on concentration differences, potassium is transported from the cell and sodium and chlorine are transport into the cell. If potassium transport is associated with the outward current, then the outward current is FNK+z. In the limit as Vm goes to zero, the outward current becomes: iK + z = FN K + = FPK + ( CK + o − CK + L ) z

Rearranging yields the following expression for the permeability: PK + =

iK + z

F ( CK + o − CK + L )

7.5. To use the data provided, equation (7.4.32) is rearranged by dividing each permeability by PK+. RT ⎡ CK + L + PNa + / PK + C Na + L + PCl− / PK + CCl−o ⎤ ⎥ Vm = ln ⎢ F ⎢ CK + o + PNa + / PK + C Na + + PCl− / PK + CCl− ⎥ o L ⎦ ⎣

( (

) )

( (

) )

Substituting data provided: ⎡145 + ( 0.1)14 + ( 0.1)124 ⎤ Vm = ( 25.69 mV ) ln ⎢ ⎥ = 54.41 mV ⎣ 5 + ( 0.1)135 + ( 0.1) 6 ⎦

7.6. If we assume that sodium, potassium and chlorine are the major small molecule ions within the cell, then proteins must balance the excess positive charge from cations. To satisfy electroneutrality, equation (7.4.5), the following relation must hold CK + + C Na + − CCl− − z p Cp = 0

(

)

Cp = CK + + C Na + − CCl− / z p For concentrations inside the cell from the problem 7.5, Cp = (153 mM ) / z p Likewise outside the cell Cp = (16 mM ) / z p The concentration CP has to be viewed as the concentration of all charged molecules within the cell and not simply proteins. Other contributors to the net charge within the cell are many small molecules such as ATP, amino acids, and nucleotides. Since these molecules are present at higher concentrations than proteins, they contribute significantly to the balance needed to maintain electroneutrality. 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 99

and 0.002 cm, respectively. The corresponding values for lengths are 0.06 and 0.15 cm, respectively. The peak velocities are 7 and 35 cm s-1, respectively. For capillaries, z must be less than 0.00057 cm for the solution to be applicable. This is much less than the length of the capillary. The correction for the flux, equation (7.5.29), is still only valid for 0.0057 cm, so the short contact time solution does not apply. In contrast, z must be less than 0.127 cm for short contact time to be applicable in venules. This is slightly less than the venule length, the short contact time solution applies for most of the venule length. The corrected solution would apply in the venules. 7.8. By substituting various values for y/R, the approximation of the velocity profile by vmaxy/R is within 10% of the true value for y/R less than 0.19. vz

y/R

v max

0.05 0.10 0.15 0.19 0.20

=

y⎛ y⎞ ⎜2 − ⎟ R⎝ R⎠

vz v max

0.0975 0.19 0.2775 0.3439 0.3600

=2

y R

0.1 0.20 0.30 0.38 0.40

Error 2.56% 5.26% 8.11% 10.5% 11.11%

A value of y/R = 0.19 and zDi/vmaxR2 = 0.01 yields η = 0.19(200/9)1/3 = 0.5342. concentration in equation (7.5.26) can be determined by rewriting as:

The

η

∞

C = 1.1198 ∫ exp( − z 3 ) dz = 1 − 1.1198∫ exp(− z 3 ) dz C0 η 0

The integral can be determined using the MATLAB function quad and equals 0.5341. This corresponds to a relative concentration C/Co equal to 0.4853. Accurate determination of the average concentration at the outlet requires a numerical solution valid over the entire channel radius. 7.9. Short contact time release in a rectangular channel. The geometry is shown in the schematic below. y H

vz

z

Assuming that the width of the channel is much greater than the height, transport in the xdirection is negligible. vz

⎛ ∂ 2C ∂Ci ∂ 2 Ci = Dij ⎜ 2 i + ⎜ ∂z ∂y 2 ⎝ ∂z

where v z =

⎞ ⎟⎟ ⎠

⎛ 4 y2 ⎞ 3 v ⎜⎜ 1 − 2 ⎟⎟ 2 H ⎠ ⎝

100

An order of magnitude analysis identical to that used in Equation (7.6.2) indicates that diffusion in the x-direction is negligible. Since the concentration change is confined to the region near y = ±H/2, the y coordinate can be shifted to the walls. Using the wall at y=H/2 as a reference., y’ = H/2-y. ⎛ 3 2 ⎛H 2 ⎛H ⎛ 2 y ⎞⎛ 2 y ⎞ 3 ⎞⎞⎛ ⎞⎞ v z = v ⎜ 1 + ⎟⎜ 1 − ⎟ = v ⎜ 1 + ⎜ − y ' ⎟ ⎟ ⎜ 1 − ⎜ − y ' ⎟ ⎟ H ⎠⎝ H⎠ 2 2 ⎝ ⎠⎠⎝ H ⎝ 2 ⎠⎠ ⎝ H⎝ 2 3 2y ' ⎞ ⎛ 2y ' ⎞ y ' ⎞⎛ 2y ' ⎞ ⎛ ⎛ ⎛ y' ⎞ = 3 v ⎜1 − ⎟ ⎜ ≈6 v ⎜ ⎟ v ⎜2 − ⎟ ⎜ ⎟ ⎟ H ⎠⎝ H ⎠ 2 ⎝ ⎝ H ⎠⎝ H ⎠ ⎝H⎠ since y’ <
As a result, the conservation relation simplifies to the following: 6 vz

∂ 2 Ci y ' ∂Ci = Dij H ∂z ∂y 2

7.10. The local mass transfer coefficient for laminar flow along a flat plate is: kloc z 1/2 = 0.323Reloc Sc1/3 Dij 1/ 2

v kloc = 0.323Dij Sc1/3 vz The length average mass transfer coefficient is defined as (equation (7.7.6b)) 1 k f = ∫ kloc dS SS where S is the area over which mass-transfer occurs. For this problem dS = wdz where w is the width of the plate and L is the plate length. L 1 k f = ∫ kloc dz L0 Substituting equation (S7.10.1) into equation (S7.10.4) and integrating: 1/2

0.323Dij ⎛ v ⎞ kf = ⎜ ⎟ L ⎝ v ⎠

L

1/2

Sc

∫z

1/2

−1/2

0

0.646 Dij ⎛ v z ⎞ dz = ⎜ ⎟ L ⎝ v ⎠

Sc1/3

z=L z =0

1/2

0.646 Dij ⎛ v z ⎞ 0.646 Dij 1/2 1/3 1/3 Re Sc ⎜ ⎟ Sc = L L ⎝ v ⎠ For the data given, Re = 2400, Sc = 3906, kf = 2.13 x 10-4 cm s-1 kf =

7.11. From the data in Example 7.6, the diffusion coefficients of urea and albumin in aqueous solution are 1.81 x 10-5 cm2 s-1 and 6.34 x 10-7 cm2 s-1, respectively. Urea Albumin Shear rate, s-1 101

0.962 x 10-5 cm2 s-1 0.967 x 10-5 cm2 s-1 1.75 x 10-5 cm2 s-1

5 15 1500

0.0362 x 10-5 cm2 s-1 0.0415 x 10-5 cm2 s-1 0.827 x 10-5 cm2 s-1

7.12. The flux across the membrane equals the flux through the fluid to the membrane surfaces Nix = kf1(Cb1-C1) = Pi(C1-C2) = kf2(C2 - Cb2) where Pi = ΦDij/L. Equating the first two expressions and solving for C1: C1 =

k f 1Cb1 + PC i 2 k f 1 + Pi

Substituting into the second expression for transport across the membrane yields: ⎛ k f 1Cb1 + PC ⎞ Pk f 1 i 2 − C2 ⎟ = N ix = Pi ⎜ C − C2 ) = k f 2 ( C2 − Cb 2 ) ⎜ k f 1 + Pi ⎟ k f 1 + Pi ( b1 ⎝ ⎠

Solving for C2 yields: Pk f 1Cb1 k f 1 + Pi C2 = Pk f 1

+ k f 1Cb 2

k f 1 + Pi

+ kf1

Pk f 1 ⎛ Pk f 1Cb1 ⎞ + k f 1Cb1 ( Cb1 − Cb 2 ) ⎜ ⎟ kf 2 k f 1 + Pi k f 1 + Pi ( Cb1 − Cb 2 ) ⎜ ⎟ Nix = k f 2 − Cb 2 = = ⎜ ⎟ 1 1 1 Pk f 1 Pk f 1 + + + kf1 + kf1 ⎜ ⎟ ⎜ k f 1 + Pi ⎟ k f 2 Pi k f 2 k f 1 + Pi ⎝ ⎠

Let,

1 1 1 1 = + + k0 k f 2 Pi k f 2

And

Nix = k0(Cb1 - Cb2)

7.13. For countercurrent exchange, the dialysate enters at z = L and exits at z = 0. The clearance and the extraction ratio are: K=

Mi CiB (0) − CiD ( L)

102

E=

(

Ci (0) − CiB ( L) Q CiD (0) − CiD ( L) K = B = D QB CiB (0) − CiD ( L) Q B CiB (0) − CiD ( L)

)

The quantities Z and FR are as defined by Equations (7.9.19a) and (7.9.19b), respectively. Equation (7.8.14) is transformed to:

⎛ Ci (0) − CiD (0) ⎞ ⎛ 1 1 ⎞ − ln ⎜ B ⎟ = ko Am ⎜ ⎟ = FR(1 − Z ) ⎜ Ci ( L ) − Ci ( L ) ⎟ Q Q B D ⎝ ⎠ D ⎝ B ⎠

(S7.13.1)

The term in the logarithm on the left-hand side of equation (S7.13.1) is written as: CiB (0) − CiD (0) CiB ( L) − CiD ( L) =E

( CiB (0) − Ci

D

=

( L)

CiB ( L) − CiD ( L)

(

CiB (0) − CiB ( L) − CiD (0) − CiB ( L) CiB ( L) − CiD ( L)

) + ( Ci ( L) − Ci B

D

) = E (CiB (0) − Ci

D

( L)

CiB ( L) − CiD ( L)

(

( L) − CiD (0) − CiD ( L)

CiB ( L) − CiD ( L)

) − (Ci

) ) = 1 + E (1 − Z ) ⎛ CiB (0) − Ci

D

(0) − CiB ( L)

)

CiB ( L) − CiD ( L)

( L) ⎞ D ⎜ ⎟ ⎜ Ci ( L) − Ci ( L) ⎟ ⎝ B ⎠ D

(S7.13.2)

The reciprocal of the concentration ratio on the right hand side of Equation (S7.13.2) is expanded as follows: CiB ( L) − CiD ( L) CiB (0) − CiD ( L)

=

CiB ( L) − CiB (0) + CiB (0) − CiD ( L) CiB (0) − CiD ( L)

= 1− E

Thus, the term in the logarithm of Equation (S7.13.1) becomes: ⎛ CiB (0) − CiD ( L) ⎞ E (1 − Z ) = 1 + E (1 − Z ) ⎜ = exp ⎡⎣ E (1 − Z ) ) ⎤⎦ ⎟ = 1+ ⎜ Ci ( L) − Ci ( L) ⎟ CiB ( L) − CiD ( L) 1− E D ⎝ B ⎠ CiB (0) − CiD (0)

Rearranging this expression one obtains:

(

)

E (1 − Z ) = (1 − E ) exp ⎡⎣ E (1 − Z ) ) ⎤⎦ − 1

Collecting terms multiplied by E and rearranging yields equation (7.8.18b): E=

exp ⎡⎣ E (1 − Z ) ) ⎤⎦ − 1

exp ⎡⎣ E (1 − Z ) ) ⎤⎦ − Z

7.14. (a) The extraction fraction E = K/QB. From the data given: Urea E = 200/208 = 0.96 E = 136/208 = 0.65 without protein E = 24/208 = 0.115 with protein 103

(b) Since CiBbound = KACiBf, the total concentration of the protein is CiBT = CiBbound + CiBf= CiBf(1 + KA)

dMi = −QB dCiBT = −QB dCiBf (1 + K A )

Thus, Alternatively,

dMi QB (1 + K A ) The expression for the dialysate side is unchanged, equation (7.8.11). Subtracting the change in concentrations on the dialysate and blood sides ⎛ 1 1 ⎞ dCiBf − dCiD = − dMi ⎜ − ⎜ Q (1 + K ) Q ⎟⎟ A D ⎠ ⎝ B The dMi quantity can be eliminated using the expression for exchange across the membrane dCiBf = −

(

)

dMi = k0 CiB f − CiD dAm

Thus:

⎛ 1 1 ⎞ dCiBf − dCiD = −k0 CiB f − CiD dAm ⎜ . − ⎜ Q (1 + K ) Q ⎟⎟ A D ⎠ ⎝ B (c) From the above result, when a solute in blood binds to a protein, QB can be replaced with QB(1+KA). Thus, the extraction fraction is given by either equation (7.8.17) or (7.8.18b) with QB(1+KA) replacing QB. From equation (7.8.17), we have E = K/QB(1+KA) where E is the result determined in part (a) and K is now the clearance in the absence of protein. Solving for KA, K 136 −1 = − 1 = 4.686 KA = 208* 0.115 QB E To check, calculate E from Equation (7.8.18b). We do need to know k0Am. Assume that the values are the same with and without protein. Thus, for the case of the intermediate molecular weight molecule without protein, Z = 0.28 and we have:

(

E=

)

1 − exp[ FR(1 − Z )] (counter-current exchange). Z − exp[ FR (1 − Z )]

0.65 ( 0.28 − exp[ FR(1 − Z )]) = 1 − exp[ FR(1 − Z )]

0.182 =1− 0.35exp[FR(1− Z)] exp[ FR 0.72] = 2.337 FR = k0Am/QB =1.18 k0Am = 245 ml/min Now for the case of protein binding, Z = 0.28*8.304 = 2.33 FR = k0Am/QB(1+KA) = 245/208*8.304= 0.142 1 − exp[ FR(1 − Z )] 1 − exp[ −0.189] = = 0.115 Z − exp[ FR(1 − Z )] 2.33 − exp[ −0.189] Note: An alternative way to obtain KA is to compare E obtained with and without proteins. For this case: 104

E=

K K = with protein QB (1 + K A ) QB Rearranging:

KA =

K K with protein

− 1 = 4.667

7.15. (a) First determine the average velocity and Reynolds number per channel as well as the diffusion coefficient in blood and the Schmidt number. 2 (10 )( 0.01) w2 H 4 Area = = = 0.01998 . Dh = (10.01) Perimeter (W + 2 H ) Q 500 / 60 = = 0.926 cm/s NW 2 H 90 ( 20 ) 0.01 For diffusion in blood, equation (7.9.20) applies: v blood =

Reblood = 0.926(0.01998)/0.03 = 0.6167

Deff = 0.53Dij + 5.292 × 10−9 γ w .

For laminar flow in a channel of height 2 H ⎛ 3 y2 ⎞ v x = v ⎜1 − 2 ⎟ 2 H ⎠ ⎝ where the y coordinate is from the centerline. The shear rate at the surface y = - H is:

γ=

dv x dy

= −3 v y =− H

y H2

= y =− H

3 v

H

dv x 3 ( 0.926 ) = = 555 s-1. The effective diffusion coefficient is 4.53 x 10-6 dy 0.005 2 cm /s and Sc = 0.03/4.53 x 10-6 = 6623. −6 4 Deff L 4 ( 4.5 x10 ) 25 = = 4.89 >> 0.01 2 v H2 0.926 ( 0.01) For the blood γ =

For the dialysate, Q 750 / 60 = = 1.39 cm/s v dialysate = NW 2 H 90 ( 20 ) 0.01 Sc = 0.009/3 x 10-6 = 3000. −6 4 Deff L 4 ( 3 x10 ) 25 = = 2.16 >> 0.01 v H 2 1.39 ( 0.01)2

Redialysate = 1.39(0.01998)/0.009 = 3.09 and

For the dialysate and blood, the following result should be used for Sh For blood, kB = 3.77(4.53 x 10-6 cm2/s)/(0.01 cm) = 1.71 x 10-3 cm/s 105

k 2H = 3.770 Dij

For dialysate, kD = 3.77(3.0 x 10-6 cm2/s)/(0.01 cm) = 1.13 x 10-3 cm/s The overall mass transfer coefficient k0 is: 6.0 x 10-4 cm/s FR = k0Am/QB = (6.0 x 10-4 cm/s)(25*10*2*90)/(500/60)= 3.234 Where Am = LW*2*90, since each channel has an upper and lower surface for exchange. Z = QB/QD = 2/3. 1 − exp[3.234(0.333)] = 0.853 0.667 − exp[3.234(0.333)] K = EQB = 0.853(500) = 426.5 cm3 min-1

E=

To determine the concentration in blood, use the single compartment model (equation (7.9.24)):

⎛ Kt ⎞ CiB (t ) = CiB exp ⎜ − ⎟ . 0 ⎝ V ⎠ Solving for time

t=−

V ⎡ 10000 ln CiB (t ) / CiB ⎤ = − ln 10 −12 / 3 x 10 −9 = 187.72 min . 0 ⎦ ⎣ 426.5 K

(

)

(b.1) If the dialysate were recycled, then the concentrations at the end of dialysis can be determined from a mass balance. Initial mass = mass in blood + mass in dialysate The mass is the product of the concentration times the volume times the MW. C0VB = CBVB+CDVD At equilibrium, CB= CD = C. So, C = C0VB/(VB+VD) = (3 x 10-9 mole/cm3)10L/(10L + 50L) = 0.5 x 10-9 mole/cm3. Thus, even if the process were fast enough recycling would not permit sufficient volume to reduce the concentration in the blood to the desired level. In fact, the recycle volume needed to reduce the concentration to 1 x 10-12 mole/cm3 is 29,990 L! ⎛ Kt ⎞ (b.2) Setting t = 360 minutes, then one can solve CiB (t ) = CiB exp ⎜ − ⎟ for K which equals 0 ⎝ V ⎠ V 10000 ln 10 −12 / 3 x10 −9 = 222.31 cm 3 min -1 . K = − ln ⎡CiB (t ) / CiB ⎤ = − 0 ⎦ ⎣ 360 t This corresponds to E = 222.31/500 = 0.4446. To find the value of Z that results in a solution, assume that FR is unchanged. This is likely because, QD will be lower that 750 ml min-1 which 4 Deff L means that is smaller and > 2.16 >> 0.01 v H2

(

E=

)

1 − exp[ FR(1 − Z )] 1 − exp[145.53(1 − Z )] = = 0.4446 . Z − exp[ FR(1 − Z )] Z − exp[145.53(1 − Z )]

Examining the table below for E as a function of Z, we see that E = 0.4446 for Z =1.875. This corresponds to QD = 266.7 ml/min. This corresponds to 80 L in 3 hours. This is the smallest volume that can be exchanged in the dialysate for the conditions provided. 106

Z 0.1 0.3 0.5 0.7 0.8 0.9 1.1 1.3 1.5 1.7 1.9 2 2.1 2.2 2.243

E 1 1 1.000 1.000 1.000 1.000 0.909 0.769 0.667 0.588 0.526 0.500 0.476 0.455 0.446

7.16. (a) At steady state the solute conservation relation near the membrane surface is: dN y

=0 (S7.16.1) dy This is a simplification in that we assume that the balance of convection and diffusion can be characterized in terms of a mass transfer coefficient, kf = Dij/d. This result indicates that the flux is constant. Thus the flux through the fluid to the membrane surface equals the flux leaving the membrane, -vfCf. The negative sign arises because the velocity is in the negative y direction. The flux through the fluid is.

N y = −v f C − Dij (b) To solve, let u = C-Cf. Then − Dij

dC = −v f C f dy

du = v f u . Rearranging and integrating from y = 0 to y = δ. dy

v du = − f dy u Dij C = Cb

d (C − C f )

∫ (C − C )

(S7.16.2)

(S7.16.3)

δ

=−

∫

C =CW

f

y =0

⎛ C −Cf ln ⎜ b ⎜ C −C f ⎝ w

⎞ v fδ ⎟⎟ = − Dij ⎠

vf

Dij

107

dy

(S7.16.4)

(S7.16.5)

⎛v δ = exp ⎜ f ⎜ D Cb − C f ⎝ ij

Cw − C f

or

⎞ ⎟⎟ ⎠

(S7.15.6)

Using the definition of the mass transfer coefficient Cw − C f ⎛v ⎞ = exp ⎜ f ⎟ Cb − C f ⎝ k ⎠

(S7.16.7))

(c) Sh = kfD/Dij =3.657 or kf =3.657 Dij /D. ⎛ vfD ⎞ Cw − C f ⎛ Pe ⎞ = exp ⎜ = exp ⎜ ⎟ ⎟ ⎜ 3.657 D ⎟ Cb − C f ⎝ 3.657 ⎠ ij ⎠ ⎝ Cw − C f ⎛ 10 ⎞ For Pe = 10 = e exp ⎜ ⎟ = 15.407 Cb − C f ⎝ 3.657 ⎠

(S7.16.8) (S7.16.9)

(

)

(d) Including the osmotic pressure in the velocity terms v f = L p Δpc − σ (π − π f ) . Substituting,

(

2 v f = L p Δpc − σ ⎡ a1 ( Cw − C f ) + a2 ( Cw − C f ) ⎤ ⎣⎢ ⎦⎥

)

(S7.16.10)

The effect of concentration polarization is to raise the concentration at the wall and thereby increase osmotic pressure. Consequently, net pressure difference will be reduced resulting in a decrease in the filtration velocity.

7.17. (a) Accumulation of solute in = peritoneal cavity

rate of transport into peritoneum from body

-

rate of transport from peritoneum to body

From this word statement, the following mass balance is derived: dCD VD = k0 Am ( CB − CD ) dt (b) Let u = CB- CD. Thus, dCD = -du. k A du =− 0 mu dt VD

CD t d (CB − CD ) k A du Integrating ∫ = ∫ = − ∫ 0 m dt u CD = 0 ( C B − C D ) VD 0

⎡ C ⎤ k A ln ⎢1 − D ⎥ = − 0 m t VD ⎣ CB ⎦

108

or

⎛ k A CD = 1 − exp ⎜ − 0 m CB ⎝ VD

⎞ t⎟ ⎠

(c) For urea k0Am/ VD = (21 cm3/min)/(2000 cm3) = 0.0105. ⎛ k A CD = 1 − exp ⎜ − 0 m CB ⎝ VD

⎞ t ⎟ = 1 − exp ( −0.0105* 300 ) = 0.957 ⎠

For vitamin B-12 k0Am/ VD = (5 cm3/min)/(2000 cm3) = 0.0025. ⎛ k A CD = 1 − exp ⎜ − 0 m CB ⎝ VD

⎞ t ⎟ = 1 − exp ( −0.0025* 300 ) = 0.5276 ⎠

7.18. a. The mass balance on the solute is, in words, Accumulation of Solute

= Rate of production –

rate of removal – by kidneys

rate of removal by dialyzer

dCi = G − K RCi − KCi = G − ( K R + K ) Ci dt dCi Ri ( K R + K ) Ci = − dt V V

V

(b) The general solution, from Equation (A.1.12) ⎛ (K + K ) ⎞ ⎛ (K + K ) ⎞ ⎛ (K + K ) ⎞ G Ci = exp ⎜ − R t ⎟ ∫ exp ⎜ R t ⎟ dt + A exp ⎜ − R t⎟ V V V V ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ where A is a constant of integration. After integration, ⎛ (K + K ) ⎞ G Ci = t⎟ + A exp ⎜ − R KR + K V ⎝ ⎠ G t = 0, Ci = C0. A = C0 − KR + K

⎛ (K + K ) ⎞ ⎛ (KR + K ) ⎞ ⎞ G ⎛ Curea = C0 exp ⎜ − R t⎟+ t ⎟⎟ ⎜⎜ 1 − exp ⎜ − ⎟ V K K V + R ⎝ ⎠ ⎝ ⎠⎠ ⎝ Note that for G = 0 and KR = 0, this result reduces to Equation (7.9.24). For the values given: ⎛ 0.173 ⎞ 0.00022 ⎛ ⎛ 0.173 ⎞⎞ Ci = 37 x 10 -3 exp ⎜ − 240 ⎟ + 240 ⎟ ⎟ ⎜ 1 − exp ⎜ − ⎝ 40 ⎠ 0.173 ⎝ ⎝ 40 ⎠⎠ 109

Ci(4 hours) = 0.0139 M (c) For the period between dialysis, K = 0. The solution is: ⎛ K ⎞ G ⎛ ⎛ KR ⎞ ⎞ Ci = Cdialysis ( t = 4 hr ) exp ⎜ − KR t ⎟ + t ⎟⎟ ⎜ 1 − exp ⎜ − ⎝ V ⎠ KR ⎝ ⎝ V ⎠⎠ Cdialysis(4 hr) = 0.0139 M ⎛ 0.006 ⎞ 0.00022 ⎛ ⎛ 0.006 ⎞⎞ Ci = 0.0139 exp ⎜ − 4320 ⎟ + 4320 ⎟ ⎟ ⎜ 1 − exp ⎜ − 40 40 ⎝ ⎠ 0.006 ⎝ ⎝ ⎠⎠ Ci(3 days) = 0.02476 M After the second dialysis Cdialysis(4 hr) = 0.00959 M

Ci(3 days) = 0.02250 M

After the third dialysis Cdialysis(4 hr) = 0.00879 M

Ci(3 days) = 0.02209 M

After the fourth dialysis Cdialysis(4 hr) = 0.00864 M

Ci(3 days) = 0.02201 M

After the fifth dialysis Cdialysis(4 hr) = 0.00862 M

Ci(3 days) = 0.02199 M

After the sixth dialysis Cdialysis(4 hr) = 0.00861 M

Ci(3 days) = 0.02199 M

110

Solution to Problems in Chapter 8, Section 8.6 8.1. The total volume of a 85-kg human body is approximately 85 liters. Therefore, the volume fraction of the vascular compartment is 7/85 is 7.3%. 8.2. The structure of the material is shown in Figure S8.2.1. The porosity of the material is 2h nL. The available volume fraction of the solute in the material is (2h - b) nL. The partition coefficient is (2h - b) / (2h).

Figure S8.2.1

8.3. The void volume and the available volume are d3 N 6

Void volume = V- π

Available volume = V- π

(S8.3.1) ( d + b) 3 N 6

(S8.3.2)

Here N is the total number of spheres in the material and V is the total volume of the porous material. Thus, KAV = 1 − π nv ⎛

Φ = ⎜1 − π nv ⎝

( d + b)3 6

( d + b )3 ⎞ ⎟ 6 ⎠

(S8.3.3) ⎛ d3 ⎞ π − 1 n ⎜ ⎟ v 6 ⎠ ⎝

(S8.3.4)

8.4. Assume the initial volume of the tissue is V0. Then the total volume of cells and extracellular matrix is V0(1-ε0). The volume dilatation is defined as e = (V-V0) / V0

(S8.4.1)

Thus, the porosity after tissue deformation is ε=

V − V0 (1 − ε 0 ) e + ε 0 = V 1+ e

8.5. (a) The porosity of the tissue is

111

(S8.4.2)

ε = 1 − π nv

d 2f d c3 −π 6 4

(S8.5.1)

(b) Using the Kozeny-Carman equation, we have K=

ε3

(S8.5.2)

G μ s02 (1 − ε ) 2

where ε is determined by Equation (S8.5.1). By definition, s0 =

π (nv d c2 + d f ) 1− ε

(S8.5.3)

Thus, ⎛ d 2f ⎞ d3 ⎜ 1 − π nv c − π ⎟ ⎜ ⎟ 6 4 ⎠ K=⎝ G μπ 2 (nv d c2 + d f ) 2

(c)

3

(S8.5.4) (S8.5.5)

v = − K ∇p

8.6. The velocity profiles for h k equal to 1, 4, 20 and 100 are plotted in Figure S8.6.1. The velocity profile approaches a parabolic curve when h k is small and approaches a uniform profile when h k is larger than 20. In the latter case, the figure shows that the no-slip boundary condition only affects the flow close to the boundary. The thickness of the boundary layer is on order of magnitude of k .

Figure S8.6.1 The ratio of the flow rate q and -Bh3/(12μ) is plotted in Figure S8.6.2. -Bh3/(12μ) is the rate of the fluid flow between the two parallel plates without porous media.

112

Figure S8.6.2 If h k < 1, the effect of porous material on the fluid flow in the channel is negligible. From Example 8.7, the effect of the channel wall on the fluid flow is negligible if h k > 20. When 1 < h k < 20, both effects should be considered in the analysis of fluid flow.

8.7. (a) The geometry of the pipe is shown in Figure S8.7.1. R

r x

Figure S8.7.1 The flow is unidirectional in the pipe. The mass balance equation is, ∂ vx =0 ∂x

(S8.7.1)

In the cylindrical system, the Brinkman equations are −

∂p =0 ∂r

(S8.7.2) μ

∂p 1 ∂ ⎛ ∂ vx ⎞ μ =0 ⎜r ⎟ − vx − r ∂r ⎝ ∂r ⎠ k ∂x

(S8.7.3)

As shown in Example 8.7 in the textbook, Equation (S8.7.3) becomes 1 ∂ ⎛ ∂ vx ⎞ 1 B ⎜r ⎟ − vx = − μ r ∂r ⎝ ∂r ⎠ k

(S8.7.4a)

Here -B is the pressure gradient in x direction, (p2-p1) / L. The boundary conditions of Equation (S8.7.4) are 113

∂ vx = 0 at ∂r

vx = 0

r=0

(S8.7.4b)

at r = R

(S8.7.4c)

To solve Equation (S8.7.4), we substitute vx with v’+kB/μ. Therefore, Equation (S8.7.4) and the boundary conditions become, 1 ∂ ⎛ ∂v ' ⎞ 1 ⎜r ⎟− v' = 0 r ∂r ⎝ ∂r ⎠ k ∂v ' =0 ∂r

v' = −

kB

μ

(S8.7.5)

at r = 0

(S8.7.5a)

at r = R

(S8.7.5b)

The solution of Equation (S8.7.5) is v' = −

kB

μ

I0 ( r / k ) I0 ( R / k )

(S8.7.6)

Here I 0 ( r / k ) is the zeroth order modified Bessel function of r / k . Thus, the solution of vx is vx =

kB

μ

(1 − I0 ( r / k ) I 0 ( R / k ))

(S8.7.7)

(b) The velocity profile is plotted in Figure S8.7.2.

Figure S8.7.2 (c) The flow rate through the cross section of the pipe can be obtained by integrating Equation (S8.7.5). Note that

d ( xI1 ( x) ) = xI 0 ( x) . The flow rate predicted by Brinkman equation is dx

114

q=∫

R

0

=

kB

=

kB

=

μ μ kB

μ

vx ⋅ 2π rdr

∫0 (1 − I 0 ( r / R

(

)

k ) I 0 ( R / k ) ⋅ 2π rdr

π r − 2k ⋅ ( r / k ) I1 ( r / k ) I 0 ( R / k ) 2

)

R

(S8.7.8)

0

π R 2 (1 − 2 k / R ⋅ I1 ( R / k ) I 0 ( R / k ) )

For the same problem, the flow rate predicted by Darcy’s law is kBπ R 2 μ . The difference between the two predictions is the second term in the parenthesis in Equation (S8.7.8). The difference decreases as R k increases. When R k equals to 20, the result given by Brinkman equation is only 9.7% less than that by Darcy’s law.

8.8. (a) R

r x

Figure S8.8.1 The geometry of the tube is shown in Figure S8.8.1. Assume the flow is unidirectional. The governing equation and boundary conditions are, μ

1 ∂ ⎛ ∂u ⎞ ∂ p =0 ⎜r ⎟− r ∂r ⎝ ∂r ⎠ ∂ x

∂u =0 ∂r

k

∂u = −α u, ∂r

(S8.8.1)

at r = 0

(S8.8.2a)

at r = R

(S8.8.2b)

The pressure gradient is a constant. The solution of Equation (S8.8.1) is, u=

∂ p r2 + C1 ln r + C2 ∂ x 4μ

(S8.8.3)

C1 and C2 are constants, which can be determined by applying the boundary conditions. The final solution is 115

u=

1 ∂p 2 k R ∂p r − R2 − α 2μ ∂ x 4μ ∂ x

(

)

(S8.8.4)

(b) The flow rate q is R

q = ∫ u ⋅ 2π rdr 0

=−

(S8.8.5)

∂ p ⎛ π R4 k π R3 ⎞ + ⎜⎜ ⎟ α 2μ ⎟⎠ ∂ x ⎝ 8μ

According to Equation (S8.8.5), the slip effect reduces the pressure gradient by a factor of 1+

4 k αR

, if the flow rate is fixed.

8.9.

r

Figure S8.9.1 The fluid flow is spherically symmetric. Therefore, the velocity is unidirectional along the radial direction. In this case, the spherical coordinate system is used, in which Equations (8.3.2) and (8.3.3) become, vr = − K

dp dr

(S8.9.1)

1 d ⎛ 2 dp ⎞ ⎜r ⎟=0 r 2 dr ⎝ dr ⎠

(S8.9.2)

The boundary conditions for Equation (S8.9.2) are, p = p0,

at r = δ.

(S8.9.2a)

p = 0,

at r = a.

(S8.9.2b)

By solving above equations, we have, p=

p0δ a −δ

vr =

p0 aδ K a − δ r2

⎛a ⎞ ⎜ − 1⎟ ⎝r ⎠

(S8.9.3) (S8.9.4)

(b). The infusion rate can be obtained by integrating the velocity at the surface of the fluid cavity. 116

2π π

q=

∫ ∫ vrδ

2

sin θ dθ dϕ = 4πδ 2 vr =

0 0

4π p0 aδ K a −δ

(S8.9.5)

i.e., q = 4πr2vr at any location.

8.10. (a) The half distance h is a function of time,

(

)

h = h0 − ∫ Vh dt = h0 ⎡1 − ε 0 1 − e −α t ⎤ ⎣ ⎦ 0 t

The tissue dilatation is

e=

(S8.10.1)

h − h0 h0

(S8.10.2)

Substitute the above equation into Equation (S8.4.2), we have, ε=

e + ε0 h = 1 − (1 − ε 0 ) 0 1+ e h

(S8.10.3)

The specific hydraulic conductivity is, n

n ⎛ ⎞ ⎛ 1 − ε 0 (1 − e−α t ) ⎞ h ⎛ ε ⎞ k =β⎜ − 1⎟ = β ⎜ − 1⎟ ⎟ = β ⎜⎜ ⎟ ⎜ ⎟ 1− ε0 ⎝ 1− ε ⎠ ⎝ (1 − ε 0 ) h0 ⎠ ⎝ ⎠

n

(S8.10.4)

Define the dimensionless pressure p*, velocity v*, and time t* as p* =

1 k0 p − p0 Vh 0 μ R

, v* = vr/Vh0,

and t* = αt, respectively. Substituting Equations (S8.10.1) and (S8.10.4) as well as the expression of Vh into Equation (8.3.51), we can get the expression of p* as a function of t* and r. ln(p*) at r = 0 is plotted as a function of t* in Figure S8.10.1. 20

ln(p*)

15

10

5

0

0

1

2

3

4

5

t*

Figure S8.10.1 Similarly, Substituting Equations (S8.10.1) and (S8.10.4) as well as the expression of Vh into Equation (8.3.52), we can get the expression of v* as a function of t*, r, and z. v* at r = R is plotted as a function of z/h for t* = 0, 0.1, 1, and 3, respectively.

117

40

t* = 0 30

v*

t* = 1 20

t* = 3 10

0 0

0.2

0.4

0.6

0.8

1

z/h Figure S8.10.2 There is no significant difference in the velocity profile between t* = 0 and 0.1. (b) Assume that the specific hydraulic permeability k is equal to k0. The dependence of h on t is still the same, i.e.,

(

)

h = h0 − ∫ Vh dt = h0 ⎡1 − ε 0 1 − e −α t ⎤ ⎣ ⎦ 0 t

(S8.10.5)

Then, 20

ln(p*)

15

10

5

0

0

1

2

3

t*

Figure S8.10.3

118

4

5

40 t* = 1 30

v*

t* = 0 t* = 3

20 10 0 0

0.2

0.4

0.6

0.8

1

z/h Figure S8.10.4 Again, there is no significant difference in the velocity profile between t* = 0 and 0.1. However, both the profiles of ln(p*) and v* are significantly different from those in (a) where k depends on porosity of the medium.

8.11. Using Equations (8.3.51) and (8.3.52), the effective hydraulic conductivity in the channel (Kchannel) can be derived as, K channel =

k⎡ k ⎛ h ⎞⎤ 1− tanh ⎜ ⎟⎥ ⎢ μ⎣ h ⎝ k ⎠⎦

(S8.11.1)

which is the same as that derived for one-dimensional flow (see Equation (8.3.41)). From Problem 8.10, the specific hydraulic conductivity is, ⎛ ⎞ h k =β⎜ − 1⎟ ⎜ (1 − ε ) h ⎟ 0 0 ⎝ ⎠

n

(S8.11.2)

where (1 - ε0) ≤ h/ h0 ≤.1. Substituting Equation (S8.11.2) into Equation (S8.11.1) and assuming the values of constants to be the same as those in Problem 8.10, we can plot Kchannel/(k0/μ) as a function of h/h0.

119

Figure S8.11.1 The results shown in Figure S8.11.1 indicate that when h decreases, Kchannel decreases faster if k depends on h. In this case, the decrease in Kchannel is due to both the reduction in h and the consolidation of the extracellular matrix. If k is fixed at k0, the decrease in Kchannel is caused only by the reduction in h.

8.12.

Figure S8.12.1 Drug transport in the tissues can be considered by a one-dimensional diffusion, since the diameter of the polymer membrane is much larger than its thickness. Therefore, the mass balance equation of the drug is, ∂C ∂ 2C = Deff ∂t ∂ x2

(S8.12.1)

Before the membrane is implanted, there is no drug in the tissues. Thus, the initial condition is, C = 0,

when t = 0

(S8.12.2)

The concentration of the drug at the surface of the membrane is assumed to be C0. We also assume that the tissue is infinitely large in the x-direction since its dimension is much larger than the thickness of the membrane. In this case, the boundary conditions are, C = C0,

at x = 0

(S8.12.3)

C = 0,

at x → ∞ 120

(S8.12.4)

Based on Equation (6.8.18), the solution of C can be obtained as, ⎛ x C = C0 erfc ⎜ ⎜ 4 Deff t ⎝

⎞ ⎟ ⎟ ⎠

(S8.12.5)

8.13. For the spherical implant, it is more convenient to use spherical coordinates. The diffusion is only in the radial direction. Thus, the mass conservation equation is 1 ∂ ⎛ 2 ∂C ⎞ ∂C = Deff 2 ⎜r ⎟−kf C ∂t r ∂ r ⎝ ∂r ⎠

(S8.13.1)

which considers both diffusion and metabolism of the drug in the tissue. Similar to Problem 8.12, the initial and boundary conditions are, C = 0,

when t = 0

(S8.13.2)

C = C0,

at r = a

(S8.13.3)

C = 0,

as r → ∞

(S8.13.4)

Let C = y/r, we have, 1 ∂ ⎛ 2 ∂C ⎞ 1 ∂ ⎛ 2 ∂ ⎛ y ⎞ ⎞ ⎜r ⎟= ⎜r ⎜ ⎟⎟ r2 ∂r ⎝ ∂r ⎠ r 2 ∂r ⎝ ∂r ⎝ r ⎠ ⎠ ⎞ 1 ∂ ⎛ ∂y = 2 − y⎟ ⎜r r ∂r ⎝ ∂r ⎠ =

(S8.13.5)

1 ∂2 y r ∂r2

Substituting C = y/r and Equation (S8.13.5) into Equation (S8.13.1), we have, ∂ ⎛ y⎞ 1 ∂2 y y −kf ⎜ ⎟ = Deff 2 ∂t ⎝ r ⎠ r ∂r r

(S8.13.6)

∂y ∂2 y = Deff −kf y ∂t ∂ r2

(S8.13.7)

The corresponding initial and boundary conditions are, y = 0,

when t = 0

(S8.13.8)

y = aC0,

at r = a

(S8.13.9)

y = 0,

as r → ∞

(S8.13.10)

The equations above can be solved by the Laplace transform method if r is replaced by x + a. The resulting governing equations are ∂y ∂2 y = Deff −kf y ∂t ∂ x2

(S8.13.11)

y = 0,

when t = 0

(S8.13.12)

y = aC0,

at x = 0

(S8.13.13) 121

as x → ∞

y = 0,

(S8.13.14)

Define the Laplace transform as ∞

Y (p, x ) = L(y(t, x )) = ∫ y(t, x )e − pt dt

(S8.13.15)

0

The Laplace transform of Equation (S8.13.11) yields, pY = Deff

∂ 2Y ∂ x2

(S8.13.16)

−kfY

the boundary conditions become, Y = aC0/p,

at x = 0

(S8.13.17)

Y = 0,

as x → ∞

(S8.13.18)

The solution of Equation (8.13.16) is, ⎛ p + kf Y = A1 exp ⎜ x ⎜ Deff ⎝

⎞ ⎛ p +kf ⎟ + A2 exp ⎜ − x ⎟ ⎜ Deff ⎠ ⎝

⎞ ⎟ ⎟ ⎠

(S8.13.19)

where A1 and A2 are constants which are determined from the boundary conditions. The final solution of Y is, Y ( p, x ) =

⎛ p + kf aC0 exp ⎜ − x ⎜ p Deff ⎝

⎞ ⎟ ⎟ ⎠

(S8.13.20)

y(t, x) can be obtained by the inverse Laplace transform of Y(p,x), y ( t , x ) = L−1 (Y ( p, x ) )

i.e.,

(S8.13.21)

⎛ aC ⎛ p + kf y ( t , x ) = L−1 ⎜ 0 exp ⎜ − x ⎜ ⎜ p Deff ⎝ ⎝

⎞⎞ ⎟⎟ ⎟⎟ ⎠⎠

(S8.13.22)

⎛ aC ⎛ p + k f ⎞⎞ 0 ⎟⎟ y ( t , x ) = L−1 ⎜ exp ⎜ − x ⎜ ⎜ p +kf Deff ⎟⎠ ⎟ ⎝ ⎝ ⎠ ⎛ k f aC ⎛ p + kf 0 + L−1 ⎜ exp ⎜ − x ⎜ ⎜ p p + kf Deff ⎝ ⎝

(S8.13.23)

⎞⎞ ⎟⎟ ⎟⎟ ⎠⎠

The result is ⎛ x y ( t , x ) = aC0 erfc ⎜ ⎜ 4 Deff t ⎝ ⎛ t + aC0 k f ∫ erfc ⎜ 0 ⎜ ⎝

Therefore, the concentration C is,

122

⎞ ⎟ exp − k f t ⎟ ⎠ ⎞ x ⎟ exp −k f τ dτ 4 Deff τ ⎟⎠

(

)

(

)

(S8.13.23)

⎛ r−a aC0 erfc ⎜ ⎜ 4 Deff t r ⎝ ⎛ aC0 k f t + erfc ⎜ ∫ 0 ⎜ r ⎝

C (t, r ) =

⎞ ⎟ exp − k f t ⎟ ⎠ r −a ⎞ ⎟ exp − k f τ dτ 4 Deff τ ⎟⎠

(

)

(

(S8.13.24)

)

If there is no chemical reaction, kf = 0, Equation (S8.13.24) reduces to C (t, r ) =

⎛ r−a aC0 erfc ⎜ ⎜ 4 Deff t r ⎝

⎞ ⎟ ⎟ ⎠

(S8.13.25)

8.14. To simplify, assume that the extracellular matrix is highly compressible compared with the intracellular volume. Therefore, the normal forces generated by elastic compression of the fibers are negligible compared to the pressure generated within the porous layer. Without this assumption, the problem can still be solved, but it is too difficult for most undergraduate students. The total force on the cell membrane F is in the z direction. It can be calculated as R

F = − ∫ σ zz 2π rdr

(S8.14.1)

0

where σzz is the effective solid stress on the cell membrane in the z direction and is equal to σ zz = 2 μG Ezz + μλ e − p (S8.14.2) If the normal forces generated by elastic compression of the fibers are negligible, then σ zz ≈ − p (S8.14.3) and the pressure distribution on the cell membrane can be determined by solving the Brinkman equation as shown in Section 8.3.3. The result is Equation (8.3.51), p = p0 +

Vh 4k k ⎡ ⎛ h ⎞ h ⎤ tanh ⎜ ⎟− k ⎥⎦ μ ⎢⎣ ⎝ k⎠

(r 2 − R 2 )

(S8.14.4)

It is independent of z. Substituting Equation (S8.14.4) into Equation (S8.14.1) yields, Vh

R

F = 2π ∫ { p0 + 0

4k k ⎡ ⎛ h ⎞ h ⎤ tanh ⎜ ⎟− μ ⎢⎣ k ⎥⎦ ⎝ k⎠

⎧ ⎪ Vh πR ⎪ =− ⎨ p0 + 2 ⎪ 4k k ⎡ ⎛ h tanh ⎜ ⎢ ⎪⎩ μ ⎣ ⎝ k 4

123

(r 2 − R 2 )r}dr

⎫ ⎪ ⎪ ⎬ ⎤ h ⎞ ⎪ − ⎟ k ⎥⎦ ⎪⎭ ⎠

(S8.14.5)

Solution to Problems in Chapter 9, Section 9.6 9.1. From Equation (9.3.36), J s = J v (1 − σ f )

Ci − CL exp ( Pe )

(9.3.36)

1 − exp ( Pe )

Substitute Ci with -ΔC+CL, we have J s = J v (1 − σ f )

−ΔC + CL − CL exp ( Pe ) 1 − exp ( Pe )

(9.3.37)

⎛ ⎞ ΔC = J v (1 − σ f ) ⎜ CL − ⎟ ⎜ 1 − exp ( Pe ) ⎟⎠ ⎝

Expand Equation (9.3.37) and substitute Jv(1-σf) with Pe PS in the second term, J s = J v (1 − σ f )CL − J v (1 − σ f )

ΔC 1 − exp ( Pe )

P PS ΔC = J v (1 − σ f )CL + e exp ( Pe ) − 1

(9.3.38)

Substitute CL with (CL+Ci+ΔC)/2 in the first term in Equation (9.3.38), we have J s = J v (1 − σ f ) = J v (1 − σ f ) = J v (1 − σ f )

CL + Ci ΔC ΔC + J v (1 − σ f ) − J v (1 − σ f ) 2 2 1 − exp ( Pe ) ⎛ ΔC ⎞ CL + Ci ΔC + J v (1 − σ f ) ⎜ − ⎟⎟ ⎜ 2 ⎝ 2 1 − exp ( Pe ) ⎠

(9.3.39)

CL + Ci 1 ⎪⎧ J v (1 − σ f )ΔC ( exp ( Pe ) + 1) ⎪⎫ + ⎨ ⎬ 2 2 ⎪⎩ exp ( Pe ) − 1 ⎭⎪

9.2. The thickness of a membrane is very small. As a result, the transport of the solute can reach steady state rapidly. Therefore, the flux of the solute in the membrane can be calculated based on the steady state mass balance equation, 0 = DA

∂ 2C

(S9.2.1)

∂ x2

where C is the concentration of solute in the membrane. At the surface of the porous membrane, the continuity of the concentration is, C = C1 KA, at x = 0

(S9.2.2a)

C = C2 KA, at x = L

(S9.2.2b)

By solving Equation (S9.2.1) with its boundary conditions, we obtain, C = K AC1 + K A ( C2 − C1 )

x x = K AC1 − K A ΔC L L

The flux of solute in the membrane is 124

(S9.2.3)

J s = − DA

∂C ΔC = DA K A ∂x L

(S9.2.4)

Therefore, the permeability of the membrane is (S9.2.5)

P = DA K A / L

9.3. (a) For an electrolyte, the mass balance equation is (see Sec. 7.4), 0=

Here

∂Ψ Ψ 2 − Ψ1 = ∂x L

∂ ⎛ ∂C ⎛ F ⎞ ∂Ψ ⎞ + zBC ⎜ ⎜ DB ⎟ ⎟ ∂x ⎝ ∂x ⎝ RT ⎠ ∂ x ⎠

(S9.3.1)

. The boundary conditions for Equation S9.3.1 are,

C = C1 KB, at x = 0

(S9.3.2a)

C = C2 KB, at x = L

(S9.3.2b)

The solution of Equation (S9.3.1) is ⎧⎪ ⎡ ⎛ E C = K B ⎨C1 + ( C2 − C1 ) ⎢ exp ⎜ ⎝ DB ⎣ ⎩⎪

where E =

z B F ( Ψ1 − Ψ 2 ) RTL

⎞ ⎤ x ⎟ − 1⎥ ⎠ ⎦

⎡ ⎛ E ⎞ ⎤ ⎫⎪ L ⎟ − 1⎥ ⎬ ⎢exp ⎜ ⎝ DB ⎠ ⎦ ⎭⎪ ⎣

(S9.3.3)

.

(b) The flux of the solute B in the membrane is, ⎡ ⎛ E J s = K B E ⎢C1 exp ⎜ ⎝ DB ⎣

⎤ ⎞ ⎟ − C2 ⎥ ⎠ ⎦

⎡ ⎛ E ⎢ exp ⎜ ⎝ DB ⎣

⎞ ⎤ ⎟ − 1⎥ ⎠ ⎦

(S9.3.4)

9.4. In each pore, the diffusion flux of the solute is (S9.4.1)

J s = DΔC / h

Assume the pores are identical, the total flux of the solute through the vessel wall equals to the diffusion flux in one pore times the surface area ratio of the pore, J = J s ⋅ nA

π d2 4

=

n Aπ d 2 DΔC 4h

(S9.4.2)

Therefore, the permeability of the vessel wall is, P = J / ΔC =

n Aπ d 2 D 4h

(S9.4.3)

9.5. (a) The porosity of the porous clefts is (S9.5.1)

ε = 1 − π rf2 l

(b) Using the Kozeny-Carman theory (Equation 8.3.27), the hydraulic conductivity of a fiber matrix material is 125

K=

r f2ε 3 4G μ (1 − ε ) 2

=

(1 − π r l ) 2 f

3

4G μ (π r 2 l ) 2

(S9.5.2)

Here G is Kozeny constant. For fiber matrix materials, G is determined by Equation 8.3.28. (c) First, we consider the flux in one cleft. According to Darcy’s law, (S9.5.3)

V = − K ∇P

The width of the cleft is much less than its length. In this case, the flow in the cleft is semiunidirectional, i.e. the flow is always parallel to the wall of the cleft. The press gradient should also be calculated along the route of the cleft. Therefore, (S9.5.4)

V = K ΔP / h

Taken together, the flux across the vessel wall is, J=

(1 − π r l ) 2 f

3

ΔP Ap 4G μ (π r l ) h 2

2

(S9.5.5)

9.6. In a spherical system, Equation (8.3.4) becomes 1 ∂ ⎛ 2 ∂p⎞ ⎜r ⎟=0 r2 ∂r ⎝ ∂r ⎠

(S9.6.1)

The boundary conditions for Equation (S9.6.1) are,

p = p 0 , at r = b

(S9.6.1a)

p = 0 , at r → ∞

(S9.6.1a)

Integrate Equation S9.6.1 and apply the boundary conditions, we have, (S9.6.2)

p = p0 ⋅ b / r

The velocity can be calculated by Darcy’s law, (S9.6.3)

V = − K ∇p = Kp0 b / r 2

In a spherical system, the mass balance equation (Equation 8.4.5) becomes, 1 ∂ 2 1 ∂ ⎛ 2 ∂C ⎞ r VC = D 2 ⎜r ⎟ 2 r ∂r r ∂r ⎝ ∂r ⎠

(

)

(S9.6.4)

The boundary conditions for above equation are,

C = C p ε , at r = b

(S9.6.4a)

C = 0 , at r → ∞

(S9.6.4a)

Here ε is the porosity of the subendothelial tissue. Solve Equation S9.6.4, we have, ⎡ ⎛ kp b ⎞ ⎤ C = C p ε ⎢ exp ⎜ − 0 ⎟ − 1⎥ ⎝ Dr ⎠ ⎦ ⎣

126

⎡ ⎛ kp0 ⎞ ⎤ ⎢exp ⎜ − ⎟ − 1⎥ ⎝ D ⎠ ⎦ ⎣

(S9.6.5)

9.7. The fluorescence intensity is a linear function of the amount of the solute. Thus, A = α ( I − Ib )

(S9.7.1)

Here A is the total amount of the solute, α is a rate constant. The concentration of the solute in the vessel can be estimated by,

(

C p = α ( I 0 − Ib ) / π r 2l

)

(S9.7.2)

The concentration of the solute in the extravascular space is much smaller than that in the vessel. Thus, the flow rate of the solute across the vessel wall is approximately, (S9.7.3)

Q = Papp C p ⋅ 2π rl

The mass balance of the extravascular space is d 2 α ( I − I 0 ) ) = Q = Pappα ( I 0 − I b ) ⋅ ( dt r

(S9.7.4)

Rearrange the equation, we have Papp =

r 1 dI 2 ΔI 0 dt

(S9.7.5)

127

Solution to Problems in Chapter 10 Problem, Section 10.8 10.1 The generalized rate expression is: R = kC Aa CBb a. For a fixed concentration of B, tripling the concentration of A causes a nine-fold increase in rate. Thus, a = 2 b. For a fixed concentration of A, doubling the concentration of B causes a four-fold increase in rate. Thus, b = 2. c. For the data given, R 5 × 10−7 Ms −1 = 5 × 10−7 M −2 s −1 k = a b = C ACB (1.0 M )2 (1.0 M )2 Using another concentration yields the same result for k.

10.2. (a) For fixed O2, tripling the concentration of NO leads to a nine-fold increase in the rate. Therefore, the rate is proportional to the square of the NO concentration. When the oxygen concentration is reduced by a factor of 4.2 at a fixed NO concentration, the rate declines by a factor of 4.2. Thus the rate is first order in O2. The rate law for the appearance of NO2- is: 2 RNO − = kC NO CO2 2

(b) Based on the mechanism provided and the overall stoichiometry, the following expressions are written for the time rate of change of NO2, N2O3 and NO2-. 1 dC NO2 = k1C NO 2 CO2 - k 2 C NO C NO2 + k3C N 2O3 2 dt

(S10.2.1)

1 dC N2O3 = k 2 C NO C NO2 − k3C N2O3 − k4 C N 2O3 2 dt

(S10.2.2)

1 dCNO2− = k4 CN 2O3 4 dt

(S10.2.3)

Assuming a quasi-steady state for NO2 and N2O3, dC NO2

dC N 2O3 dt

(S10.2.4)

≈0

dt

(S10.2.5)

≈0

By applying Equation (S10.2.5) to Equation (S10.2.2) and solving for the concentration of N2O3 yields: C N2O3 =

k 2 C NO C NO2

(S10.2.6)

k3 + k 4

By applying Equation (S10.2.4) to Equation (S10.2.1) and substituting for N2O3 yields: 0 = k1C NO 2 CO2 - k 2 C NO C NO2 + k3C N 2O3 = k1C NO 2 CO2 - k 2 C NO C NO2 +

128

k 2 k3C NO C NO2 k3 + k 4

(S10.2.7)

Rearranging and simplifying: 0 = k1C NO 2 CO2 +

- ( k3 + k4 ) k 2 C NO C NO2 +k 2 k3C NO C NO2 k3 + k 4

= k1C NO 2 CO2 −

k4 k 2 CNO CNO2 k3 + k 4

(S10.2.8)

Solving for NO2 yields the following relation: C NO2 =

k1 ( k3 + k4 ) C NO CO2

(S10.2.9)

k4 k 2

Using equation (S10.2.9), the N2O3 concentration, equation (S10.2.6), can be expressed in terms of the NO and O2 concentrations: C N2O3 =

k1 C NO 2 C NO2 k4

(S10.2.10)

Thus, the rate of appearance of NO2- is: dC NO − 2

dt

(S10.2.11)

= 4k1C NO 2 C NO2

10.3. Equation (10.2.32) can be written as:

(C

dCC

A0

− CC

)( C

B0

(S10.3.1)

= k2 dt

)

− CC − CC / K

Rewriting the denominator on the left hand side of equation (S10.3.1) as a quadratic dCC

(

)

C A0 CB0 − C A0 + CB0 + 1 / K CC + CC

2

=

dCC

( CC − r1 )( CC − r2 )

(S10.3.2)

= k2 dt

where the roots r1 and r2 are given by: r1,2 =

(C

A0

) (C

+ CB0 + 1 / K ±

A0

+ CB0 + 1 / K

)

2

− 4C A0 CB0

(S10.3.3)

2

Equation (S10.3.2) can be expressed in the following form: dCC AdCC BdCC = + = k dt ( CC − r1 )( CC − r2 ) ( CC − r1 ) ( CC − r2 ) 2

Solving for A and B, we have A = − B =

1 = r1 − r2

(S10.3.4) 1

(C

A0

+ CB0 + 1 / K

)

2

. − 4C A0 CB0

As a result, equation (S10.3.4) becomes dCC dCC − =k ( CC − r1 ) ( CC − r2 ) 2

(C

A0

+ CB0 + 1 / K

)

2

− 4C A0 CB0 dt

(S10.3.5)

Integrating and using the condition that CC = 0 at t = 0, ⎡ C / r −1 ⎤ ln ⎢ C 1 ⎥ = k2 ⎣ CC / r2 − 1 ⎦

(C

A0

+ CB0 + 1 / K

)

2

− 4C A0 CB0 t

(S10.3.6)

Substituting for r1 and r2 yields equation (10.2.33)

10.4. Rate expressions for the substrate, enzyme-substrate complex and product are: 129

dCS = − k1CE CS + k−1CES dt dCES = k1CE CS − (k−1 + k2 )CES + k−2 CE CP dt dCP = k2 CES − k−2 CE CP dt

(S10.4.1) (S10.4.2) (S10.4.3)

The total enzyme concentration CT is, (S10.4.4) CT = CE + CES Assuming that the enzyme-substrate complex is in a quasi-steady state, and solving for CES yields CE ( k1CS + k−2 CP )

CES =

k−1 + k2

(S10.4.5)

Substituting equation (S10.4.5) into equation (S10.4.4) and solving for CE yields CE =

CET ( k−1 + k2 )

k−1 + k2 + k1CS − k−2 CP

(S10.4.6)

The concentration of enzyme-substrate complex can then be written as CET ( k1CS + k−2 CP )

CES =

k−1 + k2 + k1CS − k−2 CP

(S10.4.7)

Replacing CE and CES using equations (S10.4.6) and (S10.4.7) results in the following expression for the rate of product formation: k C ( k C + k −2 C P ) k C C (k + k ) dCP = 2 ET 1 S − −2 ET P −1 2 dt k−1 + k2 + k1CS − k−2 CP k−1 + k2 + k1CS − k−2 CP

dCP CET [ k1k2 CS − k−1k−2 CP ] = dt k−1 + k2 + k1CS − k−2 CP

(S10.4.8) (S10.4.9)

This result applies to the initial rate of product formation with and without any product present initially.

10.5. (a) The rate of formation of ECS complex is: dCECS = k1CEC CS − (k−1 + k2 )CECS dt

(S10.5.1)

The total enzyme concentration is CEo = CE +CEC +CECS. From the equilibrium relation for the enzyme-cofactor complex: KC =

CE CC CEC

(S10.5.2)

The total cofactor concentration is: CCo = CC +CEC . Substituting this expression into Equation (S10.5.2) and rearranging yields the following expression for the enzyme-cofactor complex CEC =

CE CCo K C + CE

(S10.5.3)

If CE << KC then equation (S10.5.3) simplifies to: CEC ≈

CE CCo

(S10.5.4)

KC

130

Using Equation (S10.5.4) to replace CE in the expression for total enzyme concentration yields; ⎛ K CEo = CE + CEC + CECS = CEC ⎜1 + C ⎜ CC o ⎝

⎞ ⎟ + CECS ⎟ ⎠

(S10.5.5)

Solving for CEC yields: CEC =

CEo − CECS

1+

(S10.5.6)

KC CCo

Substituting Equation (S10.5.6) into Equation (S10.5.1) yields: ⎛ ⎜C −C dCECS E ECS = k1 ⎜ o ⎜ KC dt ⎜⎜ 1 + CCo ⎝

dCECS dt

⎛ ⎜ C Eo = k1 ⎜ ⎜ KC ⎜⎜ 1 + C Co ⎝

⎞ ⎟ ⎟ C − (k + k )C −1 2 ECS ⎟ S ⎟⎟ ⎠

⎞ ⎛ ⎟ ⎜ ⎟ C − ⎜ k + k + k CS ⎟ S ⎜ −1 2 1 K 1+ C ⎟⎟ ⎜⎜ C Co ⎠ ⎝

(S10.5.7a)

⎞ ⎟ ⎟C ⎟ ECS ⎟⎟ ⎠

(S10.5.7b)

Applying the quasi-steady state assumption to Equation (S10.5.7b) and solving for CECS

CECS =

CEo CS ⎛ K K M ⎜1 + C ⎜ CC o ⎝

(S10.5.8)

⎞ ⎟ + CS ⎟ ⎠

The rate of product formation is: dC p dt

= k2 CECS =

k2 CEo CS ⎛ K K M ⎜1 + C ⎜ CC o ⎝

(S10.5.10)

⎞ ⎟ + CS ⎟ ⎠

The cofactor affects KM by increasing the apparent affinity. Increasing the cofactor concentration causes the apparent KM to decline.

10.6. The definition of the effectiveness factor is, from Equation (10.6.9c), for a first order reaction in which the reactant is consumed η=

R R

'''

'''

( K AV C0 )

=

1 R ''' dV V V∫

(S10.6.1)

−k ''' K AV C0

Using equation (10.6.9b) for the integral of the rate over the tissue volume 131

η=

1 R ''' dV V V∫ − k ''' K AV C0

=

1 Deff n • ∇CdS V ∫S

(S10.6.2)

k ''' K AV C0

The unit normal points in the same direction as the gradient vector. For a rectangular geometry, the gradient is just the derivative in the x-direction and this derivative does not change on the surface S. η=

1 dC Deff dS ∫ VS dx '''

k K AV C0

=

2 SDeff

dC Vk K AV C0 dx '''

= x=L

Deff

dC Lk K AV C0 dx '''

(S10.6.3) x=L

The integral is evaluated on the surface of area S at x = L (or x = -L) and S/V = 2L.

10.7. Since the lethal dose is much smaller than KM, it is likely that toxin levels in the blood are less than the lethal dose and the reaction can be assumed to be first order. Thus, the Thiele modulus is:

φ=L

Rmax ⎛ .006 cm ⎞ =⎜ ⎟ 3 ⎠ K M Di ⎝

1.1 × 10−9 molecm −3 s −1 = 2.77 (1.1 × 10−9 molecm−3 )( 5.2 × 10−7 cm2 s −1 )

For a first-order reaction, η = tanh( φ ) φ =0.358. For beads of radius 6 µm, φ = 0.277 and η = 0.975. Thus, the 6 µm beads should be used. 10.8. The results for rectangular geometry can be used, assuming that L = (Λ – R)/2. The Thiele modulus is 6.80 for the anti-cancer drug and 29.53 for the nerve growth factor. The corresponding values for the effectiveness factors are 0.147 for the anticancer drug and 0.034 for the nerve growth factor.

10.9. From equation (10.6.53), the observable modulus is: 2 5.1 × 10−6 moles cm −3s −1 ) ( Robs L2 ⎛ .0065 cm ⎞ φ= = ⎜ ⎟ = 24 Deff K AV Co (1 × 10−6 cm −2s −1 )(1 × 10−6 moles cm −3s −1 ) ⎝ 3 ⎠ Since C0 << KM, the reaction is first order. As a result, Robs=ηRmaxC0/KM. For diffusion-limited reaction with φ > 3, Φ ≈ φ. As a result η = 1/φ=1/Φ=0.05. As a result, Rmax = RobsKM/ηC0= (5.1 x 10-6 mole cm-3 s-1)(17 x 10-6 mole cm-3)/(0.05)(1 x 10-6 mole cm-3) = 0.0017 mole cm-3.

10.10. For this problem there is steady diffusion and reaction in spherical coordinates. We do not know the magnitude of the rate coefficient. First, we will determine whether or not the reaction is diffusion-limited. 2 −7 −3 −1 R′′′L2 ( 4.2 x10 molecm s ) ( 0.0125 cm ) Φ= = = 28.4 Deff Co ( 2.1 x10−5 cm 2 s −1 )(1.1 x10−7 cm −3 s −1 ) a. Thus, the reaction is diffusion-limited. b. For diffusion-limited reactions, Φ = φ. A worst-case scenario is to assume that the reaction is first order. (This is a worst-case scenario for the effectiveness factor is the smallest.) For this case, η=1/φ. The intrinsic rate Qo2 is related to the observed rate as R obs = η Qo2 . Thus, R o2 =28.4, Robs = 12 x 10-6 moles cm-3s-1. For a zero order reaction, equation 11.6.56 yields: 132

φ=

R ′′ L2 2 =φ β Deff Co

and η =

2β

φ2

Rearranging equation (S10.10.2) yields, η = 0.07, and R o2 =14.2, Robs=5.97 x 10-6 moles

cm-3 s-1. c. The cell density is X = Ro2 /Qo2 = 4.3 × 109 cells/cm3.

10.11. For a surface reaction, the first-order reaction coefficient has units of cm s-1. The rate coefficient is: X ( 24 x 10 =

−9

R k ′′ = max KM

mole min −1 (106 cells )

−1

) ( 0.11 x10 cells cm 6

−2

)

2.46 x 10−7 mole cm −3

10.12. The rate coefficient is from problem 10.10. For the data given, Sc = 6000, and Pe = 4.5 x 106. The mass transfer coefficient is 1/3 k f = ( 5 × 10-6 cm 2s -1 ) (1.615 )( 96923) / (1.4cm ) = 2.65 × 10-4 cm s -1 Da = 0.017/2.65 x 10-4 = 40.39. The reaction rate is reduced to 2.42% of its maximum value.

10.13. (a) Begin by using the definition of s and factoring k1 on the right hand side of Equation (10.4.5): C S0

⎡ ⎤ ⎛ k ⎞ ds = k1 ⎢ −CS0 sCE0 + ⎜ CS0 s + −1 ⎟ CES ⎥ dt k1 ⎠ ⎝ ⎣⎢ ⎦⎥

(S10.13.1)

From the definition of KM = (k-1+k2)/k1 we have for k1, k1 =

k−1 + k2 KM

(S10.13.2a)

k−1 k−1 K M KM KM κ KM = = = = 1+ κ k1 k−1 + k2 1 + k2 1+ 1 κ k−1

and

(S10.13.2b)

Substituting for k-1/k1 and C S0

⎡ κ KM ds ⎛ = k1 ⎢ −CS0 sCE0 + ⎜ CS0 s + 1+ κ dt ⎝ ⎢⎣

⎞ ⎛ CE0 CS0 ⎟ c ⎜⎜ K + C ⎠ ⎝ M S0

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(S10.13.3)

Eliminating CS0. ⎡ ⎛ ⎞ ⎛ CE0 CS0 κ ds = k1 ⎢ − sCE0 + ⎜⎜ s + ⎟⎟ c ⎜⎜ dt ⎢⎣ ⎝ σ (1 + κ ) ⎠ ⎝ K M + CS0

Using the definition of τ,

133

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(S10.13.4)

⎛ CE0 k2 ⎜ ⎜ K M + CS 0 ⎝

⎡ ⎞ ds ⎛ ⎞ ⎛ CE0 CS0 κ c⎜ = k1 ⎢ − sCE0 + ⎜⎜ s + ⎟ ⎟ ⎟ ⎟ dτ ⎜ ⎢⎣ ⎝ σ (1 + κ ) ⎠ ⎝ K M + CS0 ⎠

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(S10.13.5) Eliminating CE0 and using the definitions of σ, κ, and k-1/k1. ⎛ ⎞ ds ⎛ 1+ κ 1 =κ⎜ ⎜ ⎟ ⎜ K M + CS ⎟ dτ ⎝ κ KM 0 ⎠ ⎝

⎛ ⎞ ⎛ CS0 ⎞⎡ κ ⎟⎟ c ⎜⎜ ⎟ ⎢ − s + ⎜⎜ s + ⎠ ⎢⎣ ⎝ σ (1 + κ ) ⎠ ⎝ K M + CS0

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(S10.13.6)

Making the remaining groups dimensionless: ⎡ ⎛ ⎞ ⎛ σ κ ⎛ 1 ⎞ ds = (1 + κ ) ⎢ − s + ⎜⎜ s + ⎟⎟ c ⎜ ⎜ ⎟ ⎝ 1 + σ ⎠ dτ ⎢⎣ ⎝ σ (1 + κ ) ⎠ ⎝ 1 + σ

⎞⎤ ⎟⎥ ⎠ ⎥⎦

(S10.13.7)

Multiplying both sides of Equation (S10.13.7) by 1+σ and rearranging yields: ⎡ ⎤ ds σ sc κc = (1 + σ )(1 + κ ) ⎢ − s + + ⎥ dτ 1 + σ (1 + σ )(1 + κ ) ⎦⎥ ⎣⎢

(S10.13.8)

This result is identical to Equation (10.8.10). For the enzyme-substrate complex, begin with equation (10.4.6) and use definition of s, factor k1 on the right hand side, and apply the definition of KM. dCES = k1 ⎡⎣CE CS0 s − K M CES ⎤⎦ dt

(S10.13.9)

Applying the definitions of t and c to equation (S10.13.9) ⎛ CE0 k2 ⎜ ⎜ K M + CS 0 ⎝

⎞ ⎛ CE0 CS0 ⎟⎜ ⎟ ⎜ K M + CS 0 ⎠⎝

⎡ ⎞ dc ⎛ CE0 CS0 = k1 ⎢CE0 CS0 s − CS0 s + K M c ⎜ ⎟ ⎟ dτ ⎜ K M + CS ⎢⎣ 0 ⎠ ⎝

(

)

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(S10.13.10)

Eliminating CE0 and using the definition of ε. ⎛ CS0 ⎜ K M + CS 0 ⎝

ε⎜

⎞ dc k1 = C ⎟ ⎟ dτ k2 S0 ⎠

⎡ ⎛ 1 ⎞ ⎛ CS0 ⎢s − ⎜ s + ⎟ c ⎜ ⎢⎣ ⎝ σ ⎠ ⎜⎝ K M + CS0

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(S10.13.11)

Using the definitions of κ ε, and k-1/k1. ⎛ ⎞ dc ⎛ 1+ κ 1 =κ⎜ ⎟ ⎜ K M + CS ⎟ dτ ⎝ κ KM 0 ⎠ ⎝

ε⎜

⎞⎡ ⎛ 1 ⎞ ⎛ CS0 ⎟ ⎢ s − ⎜ s + ⎟ c ⎜⎜ ⎠ ⎢⎣ ⎝ σ ⎠ ⎝ K M + CS0

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

(S10.13.12)

Using the definition of σ ⎛ 1 ⎝ 1+ σ

ε⎜

⎡ ⎛ 1 ⎞ ⎛ σ ⎞⎤ ⎞ dc = (1 + κ ) ⎢ s − ⎜ s + ⎟ c ⎜ ⎟ ⎟⎥ d τ σ ⎠ ⎠ ⎝ 1 + σ ⎠⎦ ⎣ ⎝

(S10.13.13)

Rearranging:

dc (1 + σ )(1 + κ ) ⎡ c ⎤ σ cs s− = − ⎢ ⎥ dτ ε ⎣ 1+ σ 1+ σ ⎦

(S10.13.14)

Upon application of the quasi-steady state assumption, the rate of substrate disappearance is. 134

−

dCS k2 CE0 CS = dt K M + CS

(S10.13.15)

Applying the definition of s’, σ and τ. ⎛ CE0 CS0 −k2 ⎜ ⎜ K M + CS 0 ⎝

⎞ ds ' k2 CE0 s ' = ⎟ ⎟ dτ 1/ σ + s ' ⎠

(S10.13.16)

Eliminating common terms from the left and right sides of Equation (S10.13.16) σs' ⎛ σ ⎞ ds ' −⎜ = ⎟ ⎝ 1 + σ ⎠ dτ 1 + σ s '

(S10.13.17)

Rearranging yields the final result: −

ds ' (1 + σ ) s ' = dτ 1+σ s '

(S10.13.18)

From equation (10.4.12), the enzyme-substrate complex concentration is CES =

CE0 CS

(S10.13.19)

K M + CS

Using the definition of c’ (Equation (10.8.8)) CE0 CS c' =

K M + CS CS = CE0 CS0 K M + CS K M + CS 0 c' =

⎛ K M + CS 0 ⎜ ⎜ CS 0 ⎝

⎞ (1 + 1/ σ ) s ' ⎟= ⎟ s '+ 1/ σ ⎠

(σ + ) s '

(S10.13.20)

(S10.13.21)

σ s '+ 1

(b) Figures S10.13.1 and S10.13.2 provide comparisons between the exact and the quasi-steady solutions for σ = 0.1, κ =10 and ε = 0.1 or ε = 1 for the substrate (Figure S10.13.1) and enzymesubstrate complex (Figure S10.13.2). For ε = 0.1, agreement between the exact and the quasisteady solutions is quite good except at short times. For ε = 1, agreement is poor throughout most of the time during which substrate is consumed. Figure S10.13.3 shows phase plots for σ = 0.1, 1.0 and 10 and ε = 0.1 or ε = 1. The deviations between exact and quasi-steady state are very pronounced for all values of σ at ε = 1.

135

Figure S10.13.1

Figure S10.13.2

136

Figure S10.13.3 (c) Increasing κ had a minor effect on the substrate concentration (Figure S10.13.4). The major effect was a more rapid initial decline of substrate concentration at early times. The effect of κ on the enzyme-substrate complex concentration was more pronounced but confined to early times.

Figure S10.13.4. The effect of κ for ε = 0.1 and σ = 0.1. 137

10.14. Starting with Equation (10.6.43) Φ = zero order reaction, η =

Φ=

2β

φ2

R′′′L2 ηφ 2 and the effectiveness factor for a = Deff K AVC0 1 + β

, the observable modulus is:

2β ⎛ φ 2 ⎞ ⎛ φ 2β ηφ 2 = ⎜ ⎟=⎜ φ 2 ⎝ 1 + β ⎠ ⎜⎝ 1 + β 1+ β

⎞ ⎟⎟ . ⎠

When β >> 1, this result reduces to Φ =

lim ⎛ φ 2 β ⎜ β >> 1 ⎜⎝ 1 + β

⎞ 2 . ⎟⎟ = φ β ⎠

10.15. From equation 10.6.29:

φ= From the data given, 10.6.39:

=1.28. Because

R ′′′L2 Deff k AV C0 >>1, the reaction is diffusion limited. From equation

R max k M Deff Rearranging the equation and using the data given, Rmax = 1.50 x 10-12 M s-1.

φ =φ = L

10.16. (a) In this problem, there is radial diffusion in cylindrical coordinates and zero order reaction. The conservation relation is: Deff d ⎛ dC ⎞ = RO2 r (10.16.1) r dr ⎜⎝ dr ⎟⎠ The boundary conditions are: dC r = RT C = C0 (10.16.2) r=0 =0 dr Integrating equation (1.1) once yields; dC RO2 r A = + (10.16.3) dr 2 Deff r To satisfy the boundary condition at r = 0, A must equal 0. Integrating again yields: RO2 r 2 +B (10.16.4) C= 4 Deff Applying the boundary condition at r = RT yields: RO2 RT2 (10.16.5) B = C0 − 4 Deff The resulting concentration is;

138

RO2 RT2 ⎛ r2 ⎞ C = C0 − ⎜1 − ⎟ 4 Deff ⎝ RT2 ⎠

(10.16.6)

To find the maximum size, set C = 0 at r = 0.

RT =

4 Deff C0

(10.16.7)

RO2

(b) The solution consists of dissolved oxygen and fluorocarbon: C0 = (1-f)Caqueous + fCPFC Since K =CPFC/Caqueous, C0 = [(1-f) + fK] Caqueous C0 =[0.70+20*0.3](2.01 x 10-7 mol/cm3)

(

)(

72 x 10

−9

(10.16.9)

=1.35 x 10-6 mol/cm3 (10.16.10)

4 2.0 x 10 −5 cm 2 /s 1.35 x 10 −6 mole/cm 3

R=

(10.16.8)

3

mole/cm / s

) = 0.039 cm

(10.16.11)

10.17. The reaction rate is

k ′′C0 1 + Da where Da = k ′′ / k f . . For the data given, the mass transfer coefficient is

− R′′ =

1/3

⎛ ⎞ 4.24 cm s −1 ⎟ k f = 1.4674 (1 × 10 cm s ) ⎜ ⎜ (1 × 10−6 cm 2s −1 ) ( 0.27 cm )( 0.007 cm ) ⎟ ⎝ ⎠ -6 -1 and k’’ = 1.59 x 10 cm s . Because Da = 0.00083 << 1, transport is reaction limited. The rate is reduced by 1/(1 + Da) or 99.9% −6

2 −1

10.18. Note that the rates for C0 = 5 x 10-8 M are ½ the rates for C0 = 1 x 10-7 M . This concentration is in the first order region. Further, the rates for 5 µm particles are just slightly greater than those for 10 µm particles suggesting that η ≈ 1 for 5 µm particles. Finally, C0 = 1 x 10-4 M is probably close to zero order since the rate is not too sensitive to size and rates are much larger than those for the lower concentrations. Thus for the 5 µm particles, taking the ratio of rates at C0 = 1 x 10-4 M and C0 = 5 x 10-8 M yields.

(

)

Rmax 1 x 10 −4 ⎛ K M + 5 x 10 −8 9.20 x 10 −5 ⎜ = 153.85 = K M + 1 x 10 −4 ⎜ Rmax 5 x 10 −8 5.98 x 10 −7 ⎝ Note that Rmax cancels and we can solve for KM. ⎛ 5 x 10 −8 ⎞ K M + 5 x 10 −8 = = 153.85 ⎜ 0.0769 −4 ⎟ K M + 1 x 10 −4 ⎝ 1 x 10 ⎠

(

139

⎞ ⎟ ) ⎟⎠

(

)

0.0769 K M + 1 x 10 −4 = K M + 5 x 10 −8 Solving, KM = 7.68 x 10-6 M Rmax = 9.22 x 10-5 M/s Checking at C0 = 1 x 10-7 M and 5 µm. Robs = (9.22 x 10-5 M/s)( 1 x 10-7 M)/(7.68 x 10-6 M +1 x 10-7 M)= 1.185 x 10-6 M s-1 Comparing rates at 5 and 50 µm, C0 = 5 x 10-8 M Robs(50 µm) = ηRobs(5 µm), η = 1.82/5.98 = 0.3043 For the first order reaction, C0 = 5 x 10-8 M, η = 1/φ and φ = 3.286. From the definition of the Thiele modulus, 2

⎞ ⎛R⎞ ⎛ R Deff = ⎜ ⎟ ⎜ max2 ⎟ ⎝ 3 ⎠ ⎝ K Mφ ⎠ For the 50 µm particles,

Deff

2⎛ 9.22 x 10 −5 ⎛ 0.005 ⎞ ⎜ =⎜ ⎟ 2 ⎝ 3 ⎠ ⎜ 7.68 x 10 −6 ( 3.286 ) ⎝

(

)

⎞ ⎟ = 3.08 x 10 −6 cm 2 s -1 ⎟ ⎠

Check using the data for C0 = 1 x 10-7 M. Rate = ηRobs(5 µm) = 0.3043(1.19 x 10-6 M/s) = 0.3672 x 10-6 M/s OK!

140

Solution to Problems in Chapter 11, Section 11.10 11.1. Beginning with Equation (11.3.3) NC =

CL0 N RT

(11.3.3)

K D + CL0

Multiply each side by KD + CL0 and divide each side by CL0 ⎛K ⎞ NC ⎜ D + 1⎟ = N RT ⎜ CL ⎟ ⎝ 0 ⎠

(S11.1.1)

Divide each side of Equation (S11.1.1) by KD NC N C N RT + = CL0 K D K D

(S11.1.2)

Subtract NC/KD from each side of Equation (S11.1.2) to obtain the final result NC N RT NC = − CL0 KD KD

(S11.1.3)

11.2. The total number of type 1 receptors on the cell surface is: (S11.2.1)

N RT = N R1 + NC1 + N *C1 1

Solving the steady state form of Equation (11.4.13c) for NC1 yields:

(

)

k11 N RT − N *C1 CLo

NC1 =

1

1

k −1

(S11.2.2)

+ k11CLo

Inserting this result into the steady state form of Equation (11.4.13a) and using the definition of

K 1D =

(

)

⎛ N RT − N *C1 CLo ⎞ 1 ⎟ C* − K1 + C* N * 0 = ⎜ N RT − D Lo C1 1 ⎜⎜ 1 ⎟⎟ Lo K D + CLo ⎝ ⎠

k−11 k11

(

)

(S11.2.3)

Collecting terms multiplied by N*C1 0=

(

)

N RT C *L K 1D + CLo − N RT CLo C *L 1

o

1

o

K 1D + CLo

⎡ CL C * o Lo +⎢ 1 − K 1D + C *L o ⎢ K D + CLo ⎣

(

⎤

)⎥⎥ N ⎦

(S11.2.4)

* C1

Solving for N*C1

(

)

N RT C *L K 1D + CLo − N RT CLo C *L 1

N *C1

=−

o

*

CLo C L − o

(

K 1D K 1D

1

o

+ CLo *

+ CL

o

)(

K 1D

+ CLo

)

=−

(

)

N RT C *L K 1D + CLo − N RT CLo C *L 1

o

*

CLo C L − o

K 1D + CLo

Simplifying: 141

(

K 1D

*

+ CL

o

)(

1

K 1D

o

+ CLo

)

(S11.2.5)

N *C1

=−

(

)

N RT C *L K 1D + CLo − N RT CLo C *L 1

o

(

CLo C *L − K 1D + C *L o

o

)(

1

o

K 1D + CLo

)

=

(

N RT C *L K 1D 1

o

K 1D K 1D + C *L + CLo o

)

=

N RT C *L 1

o

K 1D + C *L + CLo

(S11.2.6)

o

For a competitive inhibitor with inhibitor constant KI we have from equation (10.5.6). N *C1

N RT C *L

=

1

K 1D

(S11.2.7)

o

⎛ CLo ⎞ * ⎜1 + ⎟ + C Lo KI ⎠ ⎝

If K I = K 1D then Equation (S11.2.7) reduces to equation (S11.2.6). A labeled ligand is expected to bind to the receptor with the same affinity as the unlabeled form. A similar analysis can be performed for the second class of receptors.

11.3. Assume that all reactions shown in Figure 11.14 are at steady state. First consider the interconversion between C1 and C2 as show in Figure 11.14. At steady state: (S11.3.1)

0 = k12 N C 1 − k−12 N C 2

Using the definition of KD12, the equilibrium relation between C1 and C2 is K D12 =

k−12 NC 1 = k12 NC 2

(S11.3.2)

As a result, the total concentration of complex is NC = NC1 + NC2 = NC2 (1 + K D12 )

(S11.3.3)

Likewise, one can show that

N R1 + N R2 = N R2 (1 + K D 21 )

(S11.3.4)

The total receptor concentration is

N RT = N R1 + N R2 + NC1 + NC2 = N R2 (1 + K D 21 ) + NC2 (1 + K D12 )

(S11.3.5)

NC2 and NR2 can be related from the reaction for the conversion of NC2 to NR2. NC2 =

N R2 CL0 k22 N R2 CL0 = k−22 K D 22

(S11.3.6)

The total number of receptors can be written as N RT = N R2 (1 + K D 21 ) +

N R2 CL0 K D 22

(1 + K D12 ) = N R

2

CL0 ⎛ ⎞ (1 + K D12 ) ⎟⎟ ⎜⎜1 + K D 21 + K D 22 ⎝ ⎠

(S11.3.7)

Solving for NR2 N R2 =

N RT K D 22

K D 22 (1 + K D 21 ) + CL0 (1 + K D12 )

(S11.3.8)

Substituting Equation (S11.3.8) into Equation (S11.3.6) and replacing with NC2 in Equation (S11.3.3) yields

142

NC2 =

N RT CL0

K D 22 (1 + K D 21 ) + CL0 (1 + K D12 )

=

NC (1 + K D12 )

(S11.3.9)

Rearranging yields: NC =

N RT CL0

(S11.3.10)

(1 + K D 21 ) + C K D 22 (1 + K D12 ) L

0

11.4. (a) Assuming that loss of EGF from the surface is first order, dissociation from the receptor is negligible and degradation is not important on the time scale of the problem. Thus, for a first order rate process: ⎞ 1 ⎛N ki = − ln ⎜ EGF − R ⎟ t ⎜⎝ N EGF − R0 ⎟⎠ The data at 15 and 25 minutes both yield k2 = 0.10 min-1. (b) If the transfer from the cytoplasm can be assumed to be much slower than the release of the polyplex from the endosome, then the process can be treated as a sequential reaction with two steps, internalization with rate constant ki (equal to λξ) and transfer to the cytoplasm with rate coefficient ktr. krec ki surface ⎯⎯ → endosome ⎯⎯→ cytoplasm From section 10.3, which deals with sequential reaction A → B → C, the endosomal EGFpolyplex (EGF-Pe) complex concentration corresponds to the intermediate and its concentration is given from equation (10.3.6), after replacement of the appropriate rate coefficients: ki N EGF − R0 N endosome = ⎡ exp ( −ki t ) − exp ( − ktr t ) ⎤⎦ k −k ⎣ ec

i

From equation (10.3.8), the time for the endosomal concentration to reach a maximum is: ln ( ki ktr ) tmax = − ktr − ki Using tmax = 18.2 min and ki = 0.10 min-1 and solving iteratively yields ktr = 0.026 min-1. (c) For a sequential reaction, the cytoplasmic concentration is given by equation (10.3.7) ki ktr N EGF − R0 ⎡1 − exp ( −ki t ) 1 − exp ( −ktr t ) ⎤ N cytoplasm = − ⎢ ⎥ ktr − ki ⎣ ki ktr ⎦ Using N cytoplasm / N EGF-R 0 = 0.96 and values of ki and ktr determined above, t is found iteratively to be 135.38 min.

11.5. (a) A material balance on the intermediate G* is: dCG* = k a CG CLR − k i CG* dt A steady state can arise if the level of CLR is maintained fixed by incubating the cells with a constant concentration of ligand L. At steady state the ratio ka/ki is: 143

ka CG* CG* = = ki CG CLR CG0 − CG* CLR

(

)

From the data given, the ratio CG*/CLR = 0.05 and CG*<< CG0 . Equation (S11.5.2) reduces to: ka CG* 0.05 = = = 2.5 × 10−7 cell molecule −1 ki CG CLR 200000 (b) If LR dissociates without rebinding by binding to the antibody ligand, the formation of G* is prevented. As a result, G* reforms G by a first order process with rate coefficient ki. A semilog plot of ln(CG*) versus t should fit the following equation: ln ( CG* ) = ln CG0* − k i t

(

)

A graph of the data (Figure S11.5.1) provided shows that the process is first order. The rate coefficient ki is 0.20 min-1.

(c) From parts a and b, ka = ki(2.5 x 10-7 cell molecule-1) = 5 x 10-8 cell molecule-1min-1

11.6. (a) If the total amount of arrestin exceeds the total amount of bound complex LR then the free arrestin concentration can be assumed to be constant and equal to the total arrestin concentration. The equilibrium expression for LRA formation is; CLR T − CLRA CA C C K A = LR A = CLRA CLRA After rearranging this becomes: CLR T CA CLRA = K A + CA As a result, the concentration of LR is: ⎛ ⎞ CA KA CLR = CLR T − CLRA = CLR T ⎜1 − ⎟ = CLR T K A + CA ⎠ K A + CA ⎝ (b) Inserting equation (S11.6.3) into equation (S11.6.1) yields the following relation between G* and arrestin:

(

)

144

ka CG* CG* ⎛ K A + C A ⎞ = = ⎜ ⎟ ki CG0 CLR CG0 CLR T ⎝ K A ⎠ or k a CG0 C LRT K A k i K A + CA 5 When 5.3 x 10 arrestin molecules are added, G* levels drop to 45% of the value in the absence of arrestin. From equation (S11.6.4b), we have: k a CG0 CLR T K A CG* k K A + CA KA = i = ka CG* ( CA = 0 ) K A +CA CG0 CLR T ki The left hand side of equation (S11.6.5) equals 0.45 and KA equals 4.34 x 105 molecules ccell-1. CG* =

11.7. There are two different receptors on the cell surface. The unlabeled ligand binds to all of the high affinity sites so that dissociation is associated with the low affinity receptor. In the absence of unlabeled ligand during binding dissociation follows two exponentials. To assess whether this is correct, another binding and dissociation experiment can be performed at an unlabeled ligand concentration between 10-5 M and 10-12 M. 11.8. The rate expression for a noncompetitive inhibitor is: − RS = RP =

KM

Rmax CS + CS (1 + CS / K D11 )

(S11.8.1)

To find the sensitivity, first compute d(-RS)/dCS. dRS Rmax Rmax CS = − dCS K M + CS (1 + CS / K D11 ) ⎡ K + C (1 + C / K ) ⎤ 2 S S D11 ⎦ ⎣ M

⎛ 2CS ⎞ ⎜1 + ⎟ K D11 ⎠ ⎝

(S11.8.2)

Rearrange and simplify ⎛ C S2 ⎞ ⎛ ⎛ 2CS ⎞ ⎞ ⎜ ⎟ R K − Rmax ⎜⎜ K M + CS (1 + CS / K D11 ) − CS ⎜1 + max M ⎟ ⎟⎟ ⎜ K D11 ⎟ dRS ⎝ K D11 ⎠ ⎠ ⎝ ⎝ ⎠ = = 2 2 dCS ⎡⎣ K M + CS (1 + CS / K D11 ) ⎤⎦ ⎡⎣ K M + CS (1 + CS / K D11 ) ⎤⎦

(S11.8.3)

The sensitivity coefficient is S=

⎛ C2 ⎞ Rmax ⎜ K M − S ⎟ ⎜ K D11 ⎟ ⎝ ⎠

CS dRS CS = = RS dCS ⎛ ⎞ ⎡ K + C (1 + C / K ) ⎤ 2 K M Rmax CS S S D11 ⎦ ⎜⎜ ⎟⎟ ⎣ M ⎝ K M + CS (1 + CS / K D11 ) ⎠

Expressing in terms of the ratio of CS/KM.

145

⎛ C2 ⎞ ⎜ KM − S ⎟ ⎜ K D11 ⎟ ⎝ ⎠ (S11.8.4) + CS (1 + CS / K D11 )

⎛ ⎛C ⎞ ⎞ ⎜1 − ⎜ S ⎟ K M ⎟ ⎜⎜ ⎜ K M ⎟ K D11 ⎟⎟ ⎠ ⎝ ⎝ ⎠ S= CS ⎛ ⎛ CS ⎞ K M ⎞ 1+ ⎜1 + ⎜ ⎟ ⎟ K M ⎜⎝ ⎝ K M ⎠ K D11 ⎟⎠

(S11.8.5)

Results are plotted in Figure S11.8.1 for different values of R = KM/KD11. In all cases, the limit of the sensitivity approaches 1 as CS/KM approaches zero, and zero as CS/KM gets larger. Small values of R indicate that inhibition is not significant and the curve does not deviate significantly from to case without inhibition. As R gets larger, inhibition is more significant and the sensitivity becomes larger when CS/KM is less than one.

Figure S11.8.1. Logarithmic sensitivity coefficient for noncompetitive inhibition of an enzyme reaction by the substrate. Results are plotted as a function of R = KM/KD11. 11.9. Assume a quasi-steady state for CE1P and CE2P*. CE1 P =

k11CP CE1

CE2 P* =

k−11 + k21

k12 CP*CE2

(S11.9.1)

k−21 + k22

The total concentration of enzyme 1 of is ⎛ k1C ⎞ CE1T = CE1 + CE1 P = CE1 ⎜⎜1 + 1 1 P 1 ⎟⎟ ⎝ k−1 + k2 ⎠

(S11.9.2)

As a result, CE1P can be expressed in terms of CE1T and K M =

k−11 + k21

1

CE1 P =

k11CP CE1T k−11

+ k21

1 ⎛ ⎜⎜1 + ⎝

k11CP k−11 + k21

146

⎞ ⎟⎟ ⎠

=

k11CP CE1T k−11

+ k21

+ k11CP

k11 =

CP CE1T K M1 + CP

(S11.9.3)

Likewise for CE2 and CE2P* CE2 =

CE2T K M 2

CE2 P* =

K M 2 + C P*

CP*CE2T K M 2 + C P*

Substituting equations (S11.9.3) and (S11.9.4) into equation (11.6.8) k12CP CE1T k12CP* CE2 T K M2 k1-1CP*CE2 T dCP* = + dt K M1 + CP K M2 + C P* K M2 + C P*

=

k12CP CE1T K M1 + CP

-

(

)

k -12 + k 22 CP* CE2 T K M2 + C P*

+

k1-1CP*CE2 T

(S11.9.4)

(S11.9.5)

K M2 + C P*

Simplifying yields the desired result: dCP* dt

=

k12 CP CE1T

-

k 22 CP* CE2T

K M 1 + C P K M 2 + C P*

(S11.9.6)

11.10. (a) At steady state with no ligand present (CLo = NC = 0), equation (11.10.7) for NRi is solved to yield: N Ri =

keR NR kdeg f R + krec (1 − f R ) S

(S11.10.1)

Substituting this relation into equation (11.6.5) yields the following expression for the rate of synthesis VS: Vs = keR N RS − krec (1 − f R )

keR kdeg f R keR N RS = NR kdeg f R + krec (1 − f R ) kdeg f R + krec (1 − f R ) S

(S11.10.2)

Substituting values provided yields the following: NRi = 0.4656NR = 83,799 receptors per cell VS = 36.97 receptors per cell per min (b) Using values for NRi and VS obtained in part a, NRS and NRi do not change with time when ligand is not present, CLo = NC = 0 (Figure S11.10.1).

147

Figure S11.10.1 (c) Representative plots are shown in Figure S11.10.2. For all cases, the concentration of intracellular EGF exceeds the concentration of surface bound EGF. Regulation of surface bound EGF due to more rapid internalization of complex is evident at ligand concentrations of 1 x 10-8 M and 1 x 10-7 M. After initial rise in surface bound EGF, there is a rapid decline to a new steady state level.

Figure S11.10.2. (d) For negligible ligand depletion, nNRo/NACLo < 0.1. For the conditions given, the ratio ranges from 29.9 for CLo = 1 x 10-9 M to 0.299 for CLo = 1 x 10-9 M. However, due to down-regulation of the receptor, ligand depletion is not a significant problem at higher concentrations.

148

Figure S11.10.3 (e) Steady state levels were calculated for binding at 240 minutes and are plotted in Figure S11.10.4. The curves resemble a binding isotherm but the amount bound to the surface receptors is far less than the total number of receptors. This is because receptors are internalized. However, the level at 240 minutes can be predicted assuming steady state and using the values of CL and NRS at 240 minutes. Solving the steady state form of Equation (11.10.6) yields: N CS =

k1CL N RS k−1 + keC

(S11.10.3)

Agreement between this result and simulated values at 240 minutes is very good. Using this result and the steady state form of Equation (11.10.8) yields; keC k1CL N RS N Li =

k−1 + keC kdeg f R + krec (1 − f R )

Agreement with simulations should also be very good.

149

(S11.10.4)

Figure S11.10.4. The major difference between class 1 and 2 receptors is that the binding curve for class 2 receptors can be used to determine the total number of receptors on the cell. This can only be done with class 1 receptors by doing the experiment at 4 C when internalization is blocked. (Other answers that contrast binding and internalization for the two receptor types are acceptable.) (f) There are multiple ways to do this. Essentially one must isolate individual steps so that rate constants can be uniquely determined. For example, k-1 and keC can be determined by performing binding at 4 C, rinsing cells to 37 and then measuring loss of bound receptor and appearance of ligand in the medium at 37 C. The initial rate of disappearance from the surface equals (k-1 + keC)NCS and the initial rate of appearance in the medium would equal k-1NCS. Thus k-1 and keC can be separated. keR is more difficult to determine and requires separate measurements of loss of receptors from the surface and blocking recycling. If the receptor can be labeled, then the disappearance from the surface can be studied. The total receptor number on the surface in the absence of ligand will be constant, but the amount of labeled receptor will decline and the rate of initial disappearance will equal keRNCS. 11.11. The mechanism is shown schematically as: LR + A

k+ ⎯⎯ →

←⎯ ⎯ k

λ RA ⎯⎯ → Li

−

The kinetics of binding of receptor to adaptor protein and internalization of the receptor-ligandadaptor protein complex are: dN LR = −k+ N A N LR + k− N RA dt dN RA = k+ N A N LR − ( λ + k− ) N RA dt dN Li = λ N RA dt

Let NAT = NA + NRA and NRT = NR + NRA+ NRL. 150

(S11.11.1) (S11.11.2) (S11.11.3)

Assuming that the first reaction is rate-limiting, the second reaction is at steady state. As a result, the number of receptors bound to adaptor protein is: k+ N A T N LR

N RA =

(S11.11.4)

λ + k− + k + N LR

Substituting this result into equations (S11.11.1) and (S11.11.3) yields λ k+ N AT N LR ( k− + k+ N LR ) k+ N AT N LR dN LR \ = −k+ N AT N LR + =− λ + k− + k+ N LR λ + k− + k+ N LR dt dN Li dt

=−

(S11.11.5)

λ k+ N AT N LR dN LR = dt λ + k− + k+ N LR

(S11.11.6)

By comparison of this result with the expression for internalization of class 1 receptors in terms of keC, λ k+ N A

keC =

(S11.11.16)

T

λ + k− + k+ N LR

This analysis predicts that the rate coefficient keC should change with time as receptor-ligand complex binds to adaptor protein. 11.12. The reaction rate of each step can be written by adapting Equation (11.7.10). For the first reaction in the cascade. RP1 * =

dCP1 * dt

( ) − Rmax CP * + ( CT − CP * ) K M + CP *

Rmax1 CT1 − CP1 *

=

K M1

1

2

2

1

(S11.12.1)

1

1

The derivative of the rate with respect to CP1* is: dRP1 * dCP1 *

=−

Rmax1

(

K M1 + CT1 − CP1 *

( ) − Rmax + Rmax CP * 2 2 K M + CP * + ( CT − CP * ) ) ( K M + CP * )

Rmax1 CT1 − CP1 *

+

) ( KM

2

2

1

1

2

1

2

1

(S11.12.2a)

1

1

Collecting terms simplifies Equation (S11.12.2a) dRP1 *

=−

dCP1 *

dRP1 * dCP1 *

=−

(K

(

Rmax1 K M1 M1

Rmax1 K M1 K M 2 + CP1 *

(K

M1

)

2

(

+ CT1 − CP1 *

Rmax 2 K M 2

−

)) ( KM 2

(

2

+ CP1 *

( 2 + CP * )

)

+ Rmax 2 K M 2 K M1 + CT1 − CP1 *

(

+ CT1 − CP1 *

)) ( KM 2

2

(S11.12.2b)

2

))

2

(S11.12.2c)

1

The logarithmic sensitivity coefficient for these reactions is: S P1 * =

CP1 * dRP1 * RP1 * dCP1 *

=−

(

( R (C max1

T1

⎡ CP1 * ⎢ Rmax1 K M1 K M 2 + CP1 * ⎣ − CP1 *

)( K M

2

)

(

)

2

(

(

+ CP1 * − Rmax 2 CP1 * K M1 + CT1 − CP1 *

(S11.12.3) 151

(

+ Rmax 2 K M 2 K M1 + CT1 − CP1 *

))

2⎤

⎥⎦

) ) ) ( K M + ( CT − CP * ) ) ( K M 1

1

1

2

+ CP1 *

)

Using the values from the legend to Figure 11.32a, KM1 =KM2 = 0.1CT1, R max1 = Rmax2, and C= CP1*/CT1, the sensitivity is: S P1 *

2 2 C ⎡ 0.1( 0.1 + C ) + 0.1(1.1 − C ) ⎤ ⎣ ⎦ =− ( (1 − C )( 0.1 + C ) − C (1.1 − C ) ) (1.1 − C )( 0.1 + C )

(S11.12.4)

For reactions 2 and 3 we have

( ) − Rmax CP * dt K M + ( CT − CP * ) K M + CP * 5 dCP * k CP * ( CT − CP * ) Rmax CP * RP * = = − dt K M + ( CT − CP * ) K M + CP *

RP2 * =

dCP1 *

=

k 23CP1 * CT2 − CP2 * 3

2

1

2

2

2

3

3

3

3

4

(S11.12.5a)

2

4

2

6

(S11.12.5b)

3

3

5

6

3

The corresponding sensitivities are

S P2 * = −

(

( k C (C 3

2

S P3 * = −

⎡ CP2 * ⎢ k 23CP1 * K M 3 K M 3 + CP2 * ⎣

P1 *

T2

− CP2 *

)( K M

4

5

2

P2 *

T3

(

)

⎡ CP3 * ⎢ k 25CP2 * K M 5 K M 5 + CP3 * ⎣ − CP3 *

)( K

M6

2

(

(

)

2

(

)

(

+ Rmax 4 K M 4 K M 3 + CT2 − CP2 *

+ CP2 * − Rmax 4 CP2 * K M 3 + CT2 − CP2 *

(

( k C (C

)

(

) )) ( K M + ( CT 3

2

(

+ Rmax 6 K M 6 K M 5 + CT3 − CP3 *

(

+ CP3 * − Rmax 6 CP3 * K M 5 + CT3 − CP3 *

))) ( K

M5

(

))

2⎤

⎥⎦

− CP2 *

))

)) ( KM

4

+ CP2 *

)

2⎤

⎥⎦

+ CT3 − CP3 *

)) ( K

M6

+ CP3 *

)

Using the parameter values from the legend to Figure 11.32 S P2 *

S P3 *

2 2 C2 ⎡ 0.1C ( 0.1 + C2 ) + 0.1(1.1 − C2 ) ⎤ ⎣ ⎦ =− ( C (1 − C2 )( 0.1 + C2 ) − C2 (1.1 − C2 ) ) (1.1 − C2 )( 0.1 + C2 ) 2 2 C3 ⎡ 0.1C2 ( 0.1 + C3 ) + 0.1(1.1 − C3 ) ⎤ ⎣ ⎦ =− C2 CT3 − CP3 * ( 0.1 + C3 ) − C3 (1.1 − C3 ) (1.1 − C3 )( 0.1 + C3 )

( (

)

)

(S11.12.6a) (S11.12.6b)

The sensitivities of reactions 1 and 2 are shown in Figure S11.12.1 for values from Figure 11.32a. Reaction 1 has high sensitivity in two regions where the denominator in equation (S11.12.4) approaches zero. The sensitivity of reaction cascade 2 depends on CP1* and CP2*. The When CP2* is much smaller or larger than 1, the sensitivity is small except for an isolated region near CP1* = 1. However, when CP2* is close to 1, the sensitivity is larger over an extended region of CP1*. The sensitivity of reaction cascade 3 shows a similar trend with CP2*.

152

Figure S11.12.1. 11.13. There are three reactions in the cascade that can be assembled from Figure 11.38 and equation (11.6.10). For the first set of kinase and phosphatase reactions: dCP1 * dt

=

Rmax1 CP1 K M1 + CP1

−

Rmax 2 CP1 *

(S11.13.1)

K M 2 + CP1 *

Since CT1 = CP1 + CP1*, equation P1.1 is rewritten in terms of P1*. dCP1 * dt

=

( ) − Rmax CP * + ( CT − CP * ) K M + CP *

Rmax1 CT1 − CP1 * K M1

1

2

1

2

1

(S11.13.2)

1

For the second cascade, CP1* is the kinase. Thus, Rmax3 = k23CP1*, and the second reaction pair is represented as: dCP2 * dt

=

( + ( CT

) − Rmax CP * − CP * ) K M + CP *

k23CP1 * CT2 − CP2 * KM3

2

4

4

2

2

(S11.13.3)

2

Likewise for the third cascade, CP2* is the kinase. Thus, Rmax5 = k25CP2* and the reaction pair is represented as: dCP3 * dt

=

( + (C

)− R ) K

k25CP3 * CT3 − CP3 * KM5

T3

− CP3 *

max 6 C P3 *

M6

+ CP3 *

(S11.13.4)

(a.1) Equations (S11.13.1), (S11.13.3) and (S11.13.4) can be solved numerically using the MATLAB solver for ordinary differential equations, ode45. The kinetics are shown in Figure S11.13.1 for the baseline set of parameters. Steady state is achieved in less than 10 seconds and the steady state levels of CP2* and CP3* are less than the value of CP1*. 153

Figure S11.13.1. Kinetics of appearance of P1*, P2*, and P3*. (a.2) Steady state values of the enzymes are CP1* = 0.99002 CP2*= 0.86714 and CP3* = 0.84758. These levels were obtained from equations S11.13.1 to S11.13.4 by setting the derivatives equal to zero and solving the resulting quadratic equations. These values are, to four decimal places identical to those obtained from the numerical simulation. (a.3) There are two major approaches in the analysis, one is to lower P1* and maintain high levels of P2* and P3* or, elevate P2* and P3* levels without changing the levels of P1*. If the kinetic parameters for production of P1* are not altered from those given, then amplification of CP2* and CP3* can only occur if CT2 and CT3 are greater than CT1. Since the behavior of CP3 is similar to that for CP2, results for CP2 are presented. Rearranging the steady state form of equation (S11.13.3) yields: ⎡ ⎛ RMax ⎞ CT2 K M 4 + ⎢CT2 − K M 4 − ⎜ 3 4 ⎟ CT2 + K M 3 ⎜ k2 CP * ⎟ ⎢ 1 ⎝ ⎠ ⎣

(

⎤

)⎥⎥ C ⎦

P2 *

⎛ RMax ⎞ + ⎜ 3 4 − 1⎟ CP2 * ⎜ k2 CP * ⎟ 1 ⎝ ⎠

(

)

2

(S11.13.5)

This equation can be cast into dimensionless for by letting α2 =

RMax4 k23CP1 *

C2 * =

CP2 * CT2

K3 =

CP3 * CT2

K4 =

CP2 * CT2

(S11.13.6a,b, c, d)

As a result Equation (S11.13.5) has the following solution, C2 * =

− β 2 − β 2 2 − 4 (α 2 − 1) K 4 2 (α 2 − 1)

(S11.13.7)

where β2 = 1- K4– α2(K3+1). Results for various parameter manipulations are shown at steady state as a function of α2. In general, the levels of CP2* are greatest for small values of α2, when Rmax4 is much smaller than k23CP1*. Under these conditions formation of CP2* is greater than its removal leading to values of CP2* that approach CT2. Raising CT2 above 1 µM leads to amplification (Figure S11.13.2). 154

Figure S11.13.2. Effect of CT2 on the levels of C2*. Alternatively, amplification can arise by variation of parameters affecting P1*. At steady state, the concentration of P1* is: C1* =

where

α1 =

CP1 * CT1

=

− β1 − β12 − 4 (α1 − 1) K 2

β1 = 1 − K − α1 (1 + K

RMax2

1

RMax1

(S11.13.8)

2 (α1 − 1)

1

)

(S11.13.9a,b)

K1 and K2 are defined analogously to K3 and K4 in equations (S11.13.6c, d). dimensionless concentration C1* and letting α 2 =

RMax4 k23CT1

Using the

and β 2 = 1 − K − (α 2 / C1 *) (1 + K 2

3

solutions for P2* and P3* are: C2 * =

CP2 * CT2

=

− β 2 − β 2 2 − 4 (α1 / C1 * −1) K 4 2 (α1 C1 * − 1)

C3 * =

where

α3 =

CP3 * CT3

RMax6 k25CT2

=

− β3 − β32 − 4 (α 3 / C2 * −1) K 6 2 (α 3 / C2 * −1)

β3 = 1 − K 6 − (α 3 / C2 *)(1 + K5 )

155

(S11.13.10) (S11.13.11)

(S11.13.12a,b)

),

the

Figure S11.13.3. Effect of KM1 upon P1*, P2* and P3*. All other parameters are listed in Table 11.7.

Amplification arises when the ratio KM1/KM2 rises above 0.5 (Figure S11.13.3). As this ratio rises, dephosphorylation is favored over phosphorylation and the concentration of P1* declines. The concentrations of P2* and P3* also decline, but more slowly. As a result, the ratio of these concentrations relative to P1* increases above 1. Amplification for P2* is greater than amplification for P3*. Amplification also arises for different values of α1 = Rmax2/R max1 (Figure S11.13.4). P1*, P2* and P3* decline with increasing values of the ratio Rmax2/R max1 due to increasing phosphorylation but P2* and P3* decline more slowly with increasing Rmax2/Rmax1. Note that the level of amplifcation is similar to that observed with large values of KM1. BASed on these results, we can conclude that the level of amplification is greater when the phosphorylation reaction is favored.

156

Figure S11.13.4. Effect of the ratio Rmax2/Rmax1 on amplification. All other parameters are listed in Table 11.7.

(b) Product Inhibition. If CP3* noncompetitively inhibits enzyme E1, Equation (S11.13.1) becomes: dCP1 * dt

=

(

Rmax1 CT1 − CP1 *

(

K M1 + CT1 − CP1 *

)

)

⎛ CP3 * ⎞ ⎜1 + ⎟ KI ⎠ ⎝

−

Rmax 2 CP1 * K M 2 + CP1 *

(S11.13.13)

Noncompetitive inhibition produces two significant changes to the results (compare Figures S11.13.5 and S11.13.1), First the concentrations peak between 1 and 10 seconds and decline to a new steady state level below the levels obtained in the absence of inhibition. At early times, little P3* is produced and inhibition is negligible. As more P3* appears, the further production of P1* is reduced. The peak level of P1* is above the steady state value of P1*, so the enzyme concentration decreases. Declines in P1* affect P2* and P3*. The second effect of noncompetitive inhibition is that the steady state levels of P2* and P3* exceed the value of P1* for CT1 equal to 1 µM. This effect arises because inhibition reduces Rmax1, causing less P1* to be produced.

157

Figure S11.13.5. Effect of noncompetitive inhibition of enzyme E1 by CP3*.

(a.3) Many signals generated by second messengers are transient, in part, because the second messengers are short-lived. The major benefit of a transient response is to limit the action of the response in space and time. For example, acetylcholine will increase heart rate and redirect blood flow to the muscles. This stimulus is only needed when the animal needs fast motion in response to a threat. Once the danger has passed, blood flow should be redirected and heart rate decline. This will only happen if the second messengers return to their level in the absence of acetylcholine. This can be done by removal of one or more intermediates or by binding, and inactivation of one of the kinase intermediates. (b) To simulate attenuation, removal of P1* was assumed to be first order with a rate constant of 0.05 s-1. Because P1* is removed, the relation CTi = CPi + CPi* no longer applies. Thus, equation P1.2 is replaced with the following: dCP1 dt dCP1 * dt

=− =

Rmax1 CP1 K M1 + CP1 Rmax1 CP1

K M1 + CP1

+ −

Rmax 2 CP1 *

(S11.13.13)

K M 2 + CP1 * Rmax 2 CP1 *

K M 2 + CP1 *

− 0.05CP1 *

(S11.13.13)

For the base case of parameters, P1 declines to zero over a 20 s period as P1* reacts. Curves for P2* and P3* resemble those for P1* although there is amplification of the signal (Figure S11.13.6). The dynamics of P1* removal can be affected by reducing the value for the maximum rate of dephosphorylation, Rmax2 and decreasing KM3 and K M5 to 0.01 µM. Although P1 declines faster, higher levels of P2* and P3* result and these molecules last longer than P1* which catalyzes their formation. 158

Figure S11.13.6. Effect of removal of P1* on concentrations.

159

Solution to Problems in Chapter 12, Section 12.7 12.1. From the data in Table 12.2, CD = 3.284 and CM = 0.31. From equations (12.3.14a,b), the drag force and torque are: FD = τ w A p CD T = τ w A p CM H c

The exact shape is not specified. Assuming that the shape is roughly similar to the 4.5 µm high cell in Table 12.2, then the projected area is 511.94 µm2 and the force and torque are 23.54 x 10-5 dyne and 1.000 x 10-8 dyne cm, respectively. From equation (12.3.17) the net force is 24.46 x 105 dyne. If the cell is circular, the projected area is 530.93 µm2. This raises the force and torque by 3.58%. 12.2. For a spherical cell adherent via a single microvillus, the force balance equations are given by equations (12.4.10) to (12.4.12). The orientations area is shown in Figure S12.2.1. Since the angle θ is given, the force can be determined from equation (12.3.10) Fb = 32.054 R 2τ W / cosθ For a angle of 0°, Fb = 32.054(7 x 10-4 cm)2 (3 dyne cm-2) = 4.71 x 10-5 dyne = 471 pN. For an angle of 18.2°, Fb=496 pN 12.3. To simplify the analysis, let κ =

k−01 k1 N R0

,α=

xa F k BTN R0

, and θ = NC/NR0. For a single bond, the

steady state form of equation (12.3.18) is: (S12.3.1) (1-θ)2 = Nκexp(αθ). Equation (S12.3.1) can be solved using the MATLAB function fzero. Results are shown in Figure S12.3.1 as a function of the dimensionless equilibrium constant κ for values of the dimensionless forces α As κ increases in magnitude, the receptor-ligand affinity decreases and dissociation increases. As a result cell attachment decreases. Increasing the force, shifts the equilibrium curve to lower values of attached cells. While the curve for α = 0 gradually goes to zero at large values of κ, the presence of an applied force leads to a rapid decline in attachment above a critical value of κ. As a result the curves for α > 0 end indicating that adhesion is not supported above that value of κ.

160

Figure S12.3.1. Fraction of adherent cells as a function of the dimensionless equilibrium constant κ and force α. 12.4. The rate coefficient k-1 equals ln(2).t1/2. Taking the logarithms of equation (12.2.4), xF ln ( k−1 ) = ln ( k−01 ) + a k BT Shown in Figure S12.4.1 is a plot of ln(k-1) versus F/kBT. The data are well fit by a straight line. From the intercept, the unstressed rate coefficient k−01 equals 1.9 and, from the slope, xa equals 0.08 A.

12.5. The force balance is described by Equations (12.4.10) and (12.4.11). Equation (12.4.12) provides a geometric relation for the angle θ between the microvillus and the surface (Figure 12.15). The bond force is given by Equations (12.4.13) and (12.4.14). Equation (12.4.13) 161

applies if the bond force is less than the critical force of 45 pN. Otherwise equation (12.4.14) applies. The angle θ can be eliminated from Equations (12.4.10) to (12.4.12) by dividing equation (12.4.11) by Equation (12.4.10) and substituting into Equation (12.4.12). The result is: l 2 + R2 ⎛ 1.3701R ⎞ −1 ⎛ R ⎞ −1 ⎛ tan −1 ⎜ ⎟ = tan ⎜ ⎟ + cos ⎜⎜ 2 2 l ⎝ ⎠ ⎝l ⎠ ⎝ 2L L + R

⎞ ⎟⎟ ⎠

(S12.5.1)

R and L are specified, so the only unknown is l. This equation can be solved using the MATLAB subroutine fzero. For ease of solution it is best to rearrange in the following form. ⎡ l 2 + R2 ⎛ 1.3701R ⎞ −1 ⎛ R ⎞ ⎤ tan f ( l ) = cos ⎢ tan −1 ⎜ − − ⎟ ⎜ ⎟⎥ l ⎝ ⎠ ⎝ l ⎠ ⎦ 2 L L2 + R 2 ⎣

(S12.5.2)

The MATLAB codes are listed below. The tether length is shown as a function of time in Figure S12.5.1. The smallest force is 115 pN above the critical value (Figure S12.5.2). function tether(z,T); global R L T t=0:0.0005:1.2; L=0.35; T=0; x=2.075; for i=1:1:2401 Y=fzero('angle',x); Fb=32.054*R*R*1e-13./cos(atan(1.3701*R./Y)); z=L+0.0005*(Fb-45e-12)/11e-12; L=z; LL(i)=z; LY(i)=Y; x=Y-0.001; T=T+0.0005; end plot(t,LL,t,LY) function y=angle(l) global R L y=cos(atan(1.3701*R/l)-atan(R/l))-((l*l+L*L)/(2*L*sqrt(l*l+R*R)));

162

Figure S12.5.1. Change in tether length with time.

Figure S12.5.2. Force acting on the microvillus as the tether elongates. The bond lifetime is given by

⎛x F dNb = − kr0 exp ⎜ a b dt ⎝ k BT

163

⎞ ⎟ Nb ⎠

(S12.5.3)

Since the force per bond is declining due to tether formation, this expression cannot be integrated analytically. Integration was done using a simple Euler method with the same time step used for the integration of the force and tether length. Results are shown in Figure S12.5.3 for kr0 = 1 s-1 and xa = 0.005 nm and 0.001 nm. Since tether formation reduces the force on the bonds (Figure S12.5.2), the bond lifetimes are longer when tethers form.

Figure S12.5.3. Cell lifetimes for adhesion with and without tether formation. 12.6. (a) On a basis of 2 moles of A being produced, 0.5 moles of A2 react with another 0.5 moles of A2. In a mixture of x0 initial numbers of molecules of A2, the probability that there are x molecules of A2 at time t + Δt, Px(t + Δt), is equal to the probability that there are x molecules of A2 at time t, Px(t), plus the rate that x+2 molecules have collided with x+1 molecules to leave x molecules in time interval Δt, minus the probability that x molecules have reacted in time interval Δt. Mathematically, we have;

Px(t + Δt) = Px(t) +0.5k(x+2)(x+1)Px+2(t)Δt - 0.5kx(x-1)Px(t) Δt

(S12.6.1)

Subtracting Px(t) from both sides of equation (S12.6.1), dividing by Δt and taking the limit as Δt goes to zero yields; dPx = 0.5k ( x + 2 )( x + 1) Px + 2 − 0.5kx ( x − 1) Px dt

(S12.6.2)

dP0 dt dP2 dt dP4 dt dP6 dt

(S12.6.3a)

For x0 = 6, = kP2

(S12.6.3b)

= 6kP4 − kP2 = 15kP6 − 6kP4

(S12.6.3c)

= −15kP6

(S12.6.3d) 164

The solution to equation (S12.6.3.d) for the initial condition that P6(t=0) = 0 is: P6 ( t ) = exp ( −15kt )

(S12.6.7a) Inserting this result into equation (S12.6.3c) yields the following result for P4(t), which is obtained using the approach outlined in Section 12.5. P4 ( t ) =

5 ( exp ( −6kt ) − exp ( −15kt ) ) 3

(S12.6.7b)

Using the same approach, solutions for P2(t) and P0(t) are; 5 ( exp ( −kt ) − exp ( −15kt ) ) 7 ⎛ 1 − exp ( −6kt ) ⎞ 5 ⎛ 1 − exp ( −15kt ) ⎞ P0 ( t ) = 2 ⎜ 1 − exp ( − kt ) − ⎟ − ⎜ 1 − exp ( − kt ) − ⎟ 6 15 ⎝ ⎠ 7⎝ ⎠

P2 ( t ) = 2 ( exp ( −kt ) − exp ( −6kt ) ) −

(S12.6.7c) (S12.6.7d)

Solutions are presented in Figure S12.6.1 in terms of t’ = kx0t = 6kt. The shapes are qualitatively similar to those shown in Figure 12.17, except that the overall kinetics are slower. The slower kinetics are expected when plotted versus dimensionless time given results presented in Section 10.2.4.

Figure S12.6.1. (b) To find <x>, multiply each term in equations (S12.6.7a) to (S12.5.7d) by x and sum. The result is: x = ∑ x =0 xPx = 2 P2 + 4 P4 + 6 P6 = 6

16 8 18 exp ( −15kt ) + exp ( −6kt ) + exp ( − kt ) 21 3 7

(S12.6.8)

Results are shown in Figure (12.6.2) and compared with the result for a deterministic second order reaction, Equation (10.2.16.b). The results are very similar, although the stochastic model 165

predicts lower concentrations of dimmer at longer times. The variance for the stochastic model is obtained in a similar fashion to the mean and declines very rapidly from its initial value.

Figure S12.6.2.

166

Solution to Problems in Chapter 13, Section 13.8 13.1. The following MATLAB code was developed to solve Equation (13.3.2) with constant k’1 and with k’1 given by equation given by equation (13.3.5). k1=3.5e6; k11=44; CHBT=0.45*0.020; CHBT1=0.45*0.020; n=2.7; C50=1.005e-08*133.3*26; CO2=1.005e-08*133.3*100; x=CO2/C50; S=x.^n./(1+x.^n); k1p=k11/C50*((S./(1-S)).^(1-1./n)); dc=[k1*CO2*(CHBT-y(1))-k11*y(1);k1p*CO2*(CHBT1-y(2))-k11*y(2)];

The code for equation (13.3.4) is: function dc=HBO2S(t,y); k1=17.7e4; k11=12.0; k2=33.2e4; k22=15.8; k3=4.89e4; k33=53.9; k4=33.0e4; k44=5.0; CHBT=0.45*0.020; C50=1.005e-08*133.3*26; CO2=1.005e-08*133.3*100; dc=[k1*CO2*(CHBT-y(1)-y(2)-y(3)-y(4))-k11*y(1);k2*CO2*y(1)-... k22*y(2) ;k3*CO2*y(2)-k33*y(3);k4*CO2*y(3)-k44*y(4)]; The solutions obtained for Equation (13.3.2) worked well for the parameters given (Figure S13.1.1). For Equation (13.3.4), the best results are for k-1 = 12 s-1. At high oxygen concentrations, all three models give similar results. However, at lower concentrations, equation (13.3.2) overestimates the steady state value and predicts a more rapid rise that do the other two models.

167

Figure S13.1.1 13.2. The molecular weight of hemoglobin is 64,000. A mass concentration of hemoglobin of 0.14 g ml-1 corresponds to a molar concentration of 2.188 mM. Assuming 97% saturation, which occurs at 100 mm Hg and normal oxygen concentration, is blood (1.34 x 10-4 M) and a Hct = 0.45, equation (13.2.3) becomes CBT = Cpl + Hct4CHbS = 0.0395 M 13.3. For a thickness of 23 µm, the permeability through the layer of edema is 1.04 x 10-2 cm s-1. To incorporate this permeability into the model, add an additional mass transfer resistance. The overall permeability of the edema layer plus the resistance of the cell and plasma layers is (Figure S13.3.1) 1 1 1 1 = + + Ped + ce + pl Ped Pce Ppl Using the data in Section 13.4, Ped+ce+pl = 0.9931 x 10-2 cm s-1. The corresponding Biot number is Bi = Ped+ce+pl L RBC /D02 ,RBC = 0.223. The Thiele modulus is

0.6356. The effectiveness factor is 0.3393. The ratio vz/kη is 50.96 µm versus 24.4 for a normal 168

alveolus. The overall effect of the edema layer is doubling of the distance required for oxygenation of the red cells. 13.4. Using equation (13.5.7)

⎛ R ⎞ 4CRC DO2 RC2 − R02 + 2 R02 ln ⎜ 0 ⎟ = RO2 ⎝ RC ⎠ Substituting

( 4 × 10 )

−4 2

−6 −5 ⎡ ⎛ R0 ⎞ ⎤ 4 ( 0.16 × 10 )(1.92 × 10 ) + R ⎢ 2 ln ⎜ − 1⎥ = −4 ⎟ ⎝ 4 × 10 ⎠ ⎦ ( 2 × 10−6 ) ⎣

2 0

Rearranging yields: ⎡ ⎛ R0 ⎞ ⎤ − 1⎥ 5.984 × 10−6 = R02 ⎢ 2 ln ⎜ −4 ⎟ ⎝ 4 × 10 ⎠ ⎦ ⎣ Solving iteratively yields R0 = 17.5 µm. 13.5. For a venular concentration of 9 mM, solving equation (13.5.12) for L yields.

(C − C ) R v L= (R − R ) R Ba

2 0

Bv

2 C z

2 C

O2

(13 × 10 )( 4 × 10 ) ( 0.2 ) = 2.08 × 10 = ( R − ( 4 × 10 ) ) ( 2 × 10 ) ( R − ( 4 × 10 ) ) −4 2

−6

2 0

−4 2

−7

−6

2 0

−4 2

The venular concentration in blood is related to the concentration in plasma by equation (13.2.3) 2.7 ⎛ ⎞ CRC ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 3.617 × 10−8 ⎠ ⎜ ⎟ ⎝ −6 −6 9 × 10 = 0.55CRC + 0.45 ⎜1.1194CRC + 4 ( 5.1 × 10 ) 2.7 ⎟ CRC ⎛ ⎞ ⎟ ⎜ + 1 ⎜ ⎟ ⎟ − 8 ⎜ ⎝ 3.617 × 10 ⎠ ⎠ ⎝ Solving iteratively yields C R C = 12.5 x 10-8 mole cm-3. Using this oxygen concentration, equation (13.5.7) becomes. ⎡ ⎛ R0 ⎞ ⎤ 4.64 × 10−6 = R02 ⎢ 2 ln ⎜ − 1⎥ −4 ⎟ ⎝ 4 × 10 ⎠ ⎦ ⎣ Solving iteratively yields R0 = 16.11 µm. The capillary spacing is not this close. As a result, during heavy exercise, anoxic regions develop. Energy is provided by anaerobic metabolism resulting in lactic acid production.

13.6. Integrating equation (13.5.14) using the following MATLAB program, the saturation drops to 0.3 after 0.0125 s (FigureS13.6.1). function dc=unload2(t,y); tu-0.053; n=2.7; dc=[-sqrt(2*n*y(1).^(1+1./n)))./(tu*(n+1).*(1-y(1)).^1./n)]; [T,Y]=ode23s(‘unload2’,[0 0.05],[0.95]);

169

13.7. The transport processes are the same in each tissue, radial diffusion and zero order reaction. The only difference between the two regions is the oxygen consumption rate. Rc < r < R1

DO2 d ⎛ dCO2 ⎜r r dr ⎜⎝ dr

R1 < r < R0

DO2 d ⎛ dC 'O2 ⎜r r dr ⎜⎝ dr

The boundary conditions are R = RC

⎞ ⎟⎟ = RO2 ⎠

(S13.7.1a)

⎞ ⎟⎟ = R 'O2 ⎠

(S13.7.1b)

CO2 = CRC dC 'O2

R = RO

dr

R = RC

(S13.7.2a) (S13.7.2b)

=0

CO2 = C’O2 dCO2 dr

=

(S13.7.2c)

dC 'O2

(S13.7.2d)

dr

Integrating Equations (S13.7.1a,b) dCO2 dr

=

dC 'O2 dr

RO2 r

+

2DO2

=

R 'O2 r

2DO2

A r

+

(S13.7.3a)

A' r

(S13.7.3b)

From the boundary condition at r = R0, A' = −

R 'O2 R02

(S13.7.3b)

2DO2

Applying Equation (S13.7.2d)

( RO A = A '+

2

)

− R 'O2 R12 2DO2

=−

R 'O2 R02 2DO2

( RO −

2

)

− R 'O2 R12

(S13.7.4)

2DO2

As a result, Equation (S13.7.3a, b) becomes dCO2 dr

=

RO2 r 2DO2

(

)

2 RO2 − R 'O2 R12 ⎤ R2 1 ⎡ R 'O2 R0 ⎥= 0 − ⎢ + r ⎢ 2DO2 2DO2 ⎥ 2DO2 ⎣ ⎦

dC 'O2 dr

=

R 'O2 r 2DO2

(

)

⎡ ⎛ R − R 'O2 R12 ⎞ ⎤ ⎢ RO r − 1 ⎜ R 'O + O2 ⎟⎥ 2 ⎢ 2 R02 r ⎜ ⎟⎥ R02 ⎝ ⎠⎦ ⎣

2 2 1 ⎡ R 'O2 R0 ⎤ R 'O2 R0 ⎥= − ⎢ r ⎢ 2DO2 ⎥ 2DO2 ⎣ ⎦

Integrating Equations (S13.7.3a) and (S13.7.5)

170

⎡ r 1⎤ ⎢ 2− ⎥ ⎣⎢ R0 r ⎥⎦

(S13.7.5a) (S13.7.5b)

CO2 =

(

)

⎡ R ' R2 RO2 − R 'O2 R12 ⎤ O2 0 ⎢ ⎥+B − ln ( r ) + 2DO2 ⎢ 2DO2 ⎥ ⎣ ⎦

RO2 r 2 4DO2

(S13.7.6a) C 'O2 =

R 'O2 R02 ⎡ r 2 ⎤ ⎢ 2 − ln ( r ) ⎥ + B ' 2DO2 ⎢⎣ 2 R0 ⎥⎦

(S13.7.6b)

The constant B is determined from the boundary condition at r = Rc. B = CRC −

RO2 RC2 4DO2

(

)

⎡ R ' R2 RO2 − R 'O2 R12 ⎤ O2 0 ⎥ + ln ( RC ) ⎢ + 2DO2 ⎢ 2DO2 ⎥ ⎣ ⎦

The concentration CO2 is CO2 = CRC

RO2 R02 ⎛ r 2 RC2 + − ⎜ 4DO2 ⎜⎝ R02 R02

⎞ ⎛ r ⎟⎟ − ln ⎜ ⎝ RC ⎠

(S13.7.7)

(

)

2 2 ⎞ ⎡ R 'O2 R0 + RO2 − R 'O2 R1 ⎤ ⎢ ⎥ ⎟ 2DO2 ⎥ ⎠ ⎢⎣ ⎦

(S13.7.8)

The constant B’ is determined from the boundary condition represented by equation (S13.7.2c). CRC +

RO2 4DO2

B ' = CRC

(

⎛R R12 − RC2 − ln ⎜ 1 ⎜R ⎝ C

)

RO2 R02 ⎡ R12 ⎛ R 'O2 ⎢ ⎜1 − − 4DO2 ⎢ R02 ⎜⎝ RO2 ⎣

(

)

2 2 ⎞ ⎡ R 'O2 R0 + RO2 − R 'O2 R1 ⎤ R 'O2 R02 ⎢ ⎥= ⎟⎟ 2DO2 2DO2 ⎥ ⎠ ⎢⎣ ⎦

⎞ RC2 ⎤ ⎛R ⎟ − 2 ⎥ − ln ⎜⎜ 1 ⎟ R ⎥ ⎝ RC 0 ⎦ ⎠

(

⎡ R12 ⎤ ⎢ 2 − ln R1 ⎥ + B ' ⎥⎦ ⎣⎢ 2 R0

( )

)

2 2 2 ⎞ ⎡ R 'O2 R0 + RO2 − R 'O2 R1 ⎤ R 'O2 R0 ⎢ ⎥ + ln R1 ⎟⎟ 2DO2 ⎥ 2DO2 ⎠ ⎣⎢ ⎦

( )

Finally, the concentration C’O2 is C 'O2 = CRC +

R 'O2 R02 ⎡ r 2 R12 ⎛ r ⎢ 2 − 2 − 2 ln ⎜⎜ 4DO2 ⎣⎢ R0 R0 ⎝ R1

⎞ ⎤ RO2 R02 ⎟⎟ ⎥ + ⎠ ⎦⎥ 4DO2

⎡ R12 RC2 ⎤ ⎛ R1 ⎢ 2 − 2 ⎥ − ln ⎜⎜ ⎢⎣ R0 R0 ⎥⎦ ⎝ RC

(

(S13.7.9) (S13.7.10)

)

2 2 ⎞ ⎡ R 'O2 R0 + RO2 − R 'O2 R1 ⎤ ⎥ ⎟⎟ ⎢ 2DO2 ⎥ ⎠ ⎢⎣ ⎦

(S13.7.11)

Note that when RO2= R’O2, Equations (S13.7.8) and (S13.7.11) reduce to: CO2 = CRC +

RO2 R02 ⎛ r 2 RC2 ⎛ r ⎜⎜ 2 − 2 − 2 ln ⎜ 4DO2 ⎝ R0 R0 ⎝ RC

C 'O2 = CRC +

RO2 R02 ⎡ r 2 RC2 ⎛ r ⎢ 2 − 2 − 2 ln ⎜⎜ 4DO2 ⎣⎢ R0 R0 ⎝ RC

⎞⎞ ⎟ ⎟⎟ ⎠⎠

(S13.7.12)

⎞⎤ ⎟⎟ ⎥ ⎠ ⎦⎥

(S13.7.13)

As a result, these equations reduce to Equation (13.5.4).

13.8. Shown in Figure S13.8.1 are the concentration profiles for the two-layer model (Equations (S13.7.8) and (S13.7.11)) using D O2 = 2.0 x 10-5 cm2 s-1 and CRC = 0.14 mM. For the two layer model, the decline in concentration occurs primarily near the capillary surface due to the higher oxygen consumption rate. Nevertheless, the decrease in concentration is less than the change in concentration for a single layer model with an oxygen consumption rate equal to (RO2+ R’O2)/2. For the composite model, the flux at r =RC is: N

r = RC

= − DO2

dCO2 dr

=− r = RC

RO2 R02 ⎛ RC ⎜ 2 ⎜⎝ R02

⎞ 1 ⎡ R ' R 2 + RO2 − R 'O2 R12 ⎤ ⎟⎟ + ⎣ O2 0 ⎦ R 2 C ⎠

(

171

)

(S13.8.1)

When RO2= R’O2, The flux at r =RC is: N

r = RC

= − DO2

dCO2 dr

=− r = RC

RO2 R02 ⎛ RC 1 ⎞ ⎜⎜ 2 − ⎟ 2 ⎝ R0 RC ⎟⎠

(S13.8.2)

The fluxes at the blood vessel surface are different. This difference declines as the oxygen consumption rates approach one another or as RI approaches R0.

Figure S13.8.1

13.9. For steady one-dimensional radial diffusion with a constant oxygen uptake rate, the conservation relation for oxygen transport is: DO2 d ⎛ dCO2 ⎞ ⎜r ⎟ = RO2 r dr ⎝ dr ⎠ The boundary condition at r = Rc with NO consumption at the endothelial cell surface: dCO′ 2 − DO2 = − k NO C pi + Pec ( C pl − C ( r = Rc ) ) dr where Cpl is the plasma concentration at the surface and C(r=Rc) is the concentration of oxygen in the tissue at the capillary surface. In the absence of reaction and infinite permeability, C(r=Rc) = Cpl = CRc. The other boundary condition is: dCO′ 2 R = RO =0 dr Integrating equations (S13.7.1a,b) dCO2 RO2 r A = + dr r 2 DO2 From the boundary condition at r = R0, 172

A=−

RO2 R02 2 DO2

Equation (S13.9.3) is: dCO2 dr

=

RO2 r 2 DO2

−

2 2 1 ⎡ RO2 R0 ⎤ RO2 R0 ⎡ r 1⎤ = ⎢ ⎥ ⎢ 2 − ⎥ 2 DO2 ⎣ R0 r ⎦ r ⎢⎣ 2 DO2 ⎥⎦

Integrating equation (S13.9.5) RO2 R02 CO2 = 4 DO2

⎡ r2 ⎤ ⎢ 2 − 2 ln ( r ) ⎥ + B ⎣ R0 ⎦ Applying the boundary condition at r = Rc, ⎛ ⎞ RO2 R02 ⎡ RC RO2 R02 ⎡ RC2 ⎤ 1 ⎤ − ⎢ 2 − ⎥ = − k NO C pl + Pec ⎜⎜ C pl − ⎢ 2 − 2ln ( RC ) ⎥ − B ⎟⎟ 2 ⎣ R0 RC ⎦ 4 DO2 ⎣ R0 ⎦ ⎝ ⎠ Solving for B 2 ⎞ RO R02 ⎡ R 2 ⎤ 1 ⎛ RO2 R0 ⎡ RC 1 ⎤ B = C pl − 2 ⎢ C2 − 2ln ( RC ) ⎥ + ⎜⎜ ⎢ 2 − ⎥ − k NO C pl ⎟⎟ 4 DO2 ⎣ R0 ⎦ Pec ⎝ 2 ⎣ R0 RC ⎦ ⎠ 2 ⎞ RO2 R02 ⎡ r ⎛ r ⎞ ⎤ 1 ⎛ RO2 R0 ⎡ RC RC2 1 ⎤ 2ln − − + ⎜⎜ ⎢ 2 ⎜ ⎟⎥ ⎢ 2 − ⎥ − k NO C pl ⎟⎟ 2 4 DO2 ⎣ R0 R0 ⎝ RC ⎠ ⎦ Pec ⎝ 2 ⎣ R0 RC ⎦ ⎠ Using typical values for small vessels, RC = 1.5 µm, R0 = 25 µm, RO2 = 5 x 10-8 mole cm-3 s-1, Cpl = 0.14 mM and the values of kNO and Pec, provided the reaction and finite permeability reduce the surface concentration by 7.65%. Since kNO << Pec, the overall effect of reaction is rather modest relative to permeability changes. The ratio kNO/Pec will need to be above 0.1 to significantly reduce the concentration at the surface.

CO2 = C pl +

13.10. (a) The conservation relations differ slightly for each region due to the presence of capillaries in part of the tumor. 0 < r < R1

R1 < r < RT

DO2 d ⎛ 2 dC1 ⎞ ⎜r ⎟ = RO2 r 2 dr ⎝ dr ⎠ DO2 d ⎛ 2 dC2 ⎞ S r = RO2 − Pec C pl ⎜ ⎟ 2 dr ⎠ V r dr ⎝

(S13.10.1a)

(S13.10.1b)

Since Cpl is treated as constant, the presence of capillaries is to produce an effective reduction in the oxygen consumption rate. Let Reff = RO2 – PecCplS/V. Boundary conditions are r=0 r = R1 r = RT

dC1 =0 dr dC1 dC2 = dr dr

C1 = C2 C2 = CT 173

(S13.10.2a) (S13.10.2b) (S13.10.2c) (S13.10.2d)

CT is the oxygen concentration in the surrounding tissue. Integrating Equations (S13.101a) and (S13.101b) once dC1 RO2 r A1 = + dr 3DO2 r 2

(S13.10.3a)

dC2 Reff r A2 = + dr 3DO2 r 2

(S13.10.3b)

From the boundary condition at r = 0, A1 = 0. Since the derivatives at r = R1 are equal, A2 equals A2 =

( RO

)

− Reff R13

2

(S13.10.4)

3DO2

After substituting for A2 in equation (S13.103b) and integrating Equations (S13.103a) and (S13.103b), we have C1 = C2 =

RO2 r 2

(S13.10.5a)

+ B1

6DO2 Reff r 2 6DO2

−

(

)

3 1 RO2 − Reff R1 + B2 3DO2 r

(S13.10.5b)

The constant B2 is found by evaluating the boundary condition at r = RT. B2 = CT −

Reff RT2 6DO2

( RO +

2

Reff RT2 ⎛ r2 ⎜⎜1 − 2 6DO2 ⎝ RT

(S13.10.6)

3DO2 RT

As a result C2 is C2 = CT −

)

− Reff R13

(

)

2 ⎞ RO2 − Reff R1 ⎛ R1 R1 ⎞ + − ⎟⎟ ⎜⎜ ⎟ 3DO2 r ⎟⎠ ⎠ ⎝ RT

(S13.10.7)

B1 is found using the second boundary condition at r = R1, Equation (S13.10.2c). B1 = CT −

Reff RT2 6DO2

+

( RO

2

)

− Reff R12 ⎛ R ⎞ 1 − 1⎟ ⎜⎜ ⎟ 3DO2 ⎝ RT ⎠

As a result, C1 is C1 = CT −

Reff RT2 ⎛ r2 ⎜⎜ 1 − 2 6DO2 ⎝ RT

(

)

2 ⎞ RO2 − Reff R1 ⎛ R1 ⎞ − 1⎟ ⎟⎟ + ⎜⎜ ⎟ 3DO2 ⎠ ⎝ RT ⎠

(S13.10.8)

(S13.10.10)

(b) Necrosis results if the concentration drops to zero at R0 >0. Using this relation, equation (S13.10.10) can be solved to yield R0 RT

= 1−

6CT DO2 Reff RT2

+

(

)

2 RO2 − Reff R12 ⎛ R ⎞ 1 − 1⎟ ⎜⎜ 2 ⎟ Reff RT ⎝ RT ⎠

(S13.10.11)

13.11. Since the initial oxygen concentration is much higher than the concentration of NO, oxygen can be assumed to be constant at its initial value. For a second order reaction in NO, the solution is given by equation (10.2.16a) 174

1 1 − = 4kCO2 t CNO CNO0 At the half-time C NO = 0.5C NO0 . t1/2 =

1 4kCO2 CNO0

For k = 2 x 106 M-2s-1, C NO0 =0.09 x 10-6 M, and CO2 =120 x 10-6 M, t1/2 = 11574 s = 3.22 h.

13.12. Assume that the reaction mechanism is k NO + S − OHb ⎯⎯ → SNO − Hb where OHb represents oxygenation hemoglobin. The rate expression is: dC − OHb − s = kCS −OHb CNO dt The solution to equation (S13.12.1) is given by equation (10.2.28) ⎛ CS −OHb0 CNO0 ⎞ − ln ⎜ ⎟ = k2 CS −OHb0 − CNO0 t CNO ⎠ ⎝ CS −OHb From the data given k2 is ⎛ CS −OHb0 ⎞ ⎛ CNO0 ⎞ ⎛ 54 ⎛ 6 ⎞⎞ ln ⎜ − ln ⎜ ln ⎜ ⎟ − ln ⎜ ⎟ ⎟⎟ Cs −OHb ⎠ CNO ⎠ 53.94 ⎝ 5.94 ⎠ ⎠ ⎝ ⎝ ⎝ k2 = = = 0.558 M −1s −1 −6 −6 CNO0 − CS −OHb0 t ( 0.6 × 10 − 54 × 10 ) 300

(

(

)

)

13.13. Equation (13.6.5) governs NO transport in the lumen and the surrounding tissue of a capillary. It can be re-written as DB d ⎛ dCB ⎞ − kB C = 0 r r dr ⎜⎝ dr ⎟⎠ DT d ⎛ dCT ⎞ r − kT CT2 = 0 ⎜ ⎟ r dr ⎝ dr ⎠

(0 < r < RC)

(r > RC)

(S13.13.1)

(S13.13.2)

Here we have dropped the subscript “NO” in the. In addition, we assume that the reaction of NO follows first-order kinetics in the lumen and second-order kinetics in the surrounding tissue. The boundary conditions are the same as those in the textbook. They are dCB =0 dr •

q NO = DB CB = CT dCT =0 dr

dCB dCT − DT dr dr

at r = 0

(S13.13.3)

at r = RC

(S13.13.4)

at r = RC

(S13.13.5)

as r → ∞

(S13.13.6)

Substituting Equation (S13.13.6) into Equation (S13.13.2) yields CT = 0 as r → ∞

175

(S13.13.7)

since CT and its derivatives are finite in tissues as r → ∞. Thus, the boundary conditions in Equations (S13.13.6) and (S13.13.7) are equivalent and we will use the latter. In order to solve this problem using the finite difference method, replace r with x, r=−

RC

α

(S13.13.8)

ln( x )

where α is an arbitrary constant which controls the distribution of vertices as r → ∞ if the vertices on the x-axis are uniformly distributed. The new derivatives are, dCi dCi dx α x dCi = =− dr dx dr RC dx d 2 Ci dr 2

(S13.13.9)

2

⎛ α ⎞ ⎡ d 2 Ci dCi ⎤ =⎜ ⎥ ⎟ x ⎢x 2 + dx ⎥⎦ ⎝ RC ⎠ ⎢⎣ dx

(S13.13.10)

With the new axis, the governing equations and boundary conditions are transformed as 2

⎛ α ⎞ ⎡ d 2 CB ⎛ 1 ⎞ dCB ⎤ + ⎜1 + DB ⎜ ⎥ − kBC = 0 ⎟ x ⎢x ⎟ 2 ⎝ ln( x) ⎠ dx ⎥⎦ ⎝ RC ⎠ ⎢⎣ dx 2

⎛ α ⎞ ⎡ d 2 CT ⎛ 1 ⎞ dCT + ⎜1 + DT ⎜ ⎟ x ⎢x ⎟ 2 R x) ⎠ dx ln( dx ⎝ ⎝ C ⎠ ⎢⎣ dCB =0 dx

⎤ 2 ⎥ − kT CT = 0 ⎦⎥

[exp(-α) < x < 1]

(S13.13.11)

[0 < x < exp(-α)]

(S13.13.12)

at x = 1

(S13.13.13)

at x = exp(-α)

(S13.13.14)

•

q NO RC dCB dCT = − DB + DT α exp(−α ) dx dx

at x = exp(-α) (S13.13.15) CT = 0 at x = 0 (S13.13.16) The equations listed above are solved by a finite difference method based on the MATLAB code attached at the end of this problem. The procedure is as follows. Let 2 2 ⎛ 1 ⎞ ϕ ϕ ⎟⎟ ; hB = ⎛⎜ B ⎞⎟ ; hT = ⎛⎜ T ⎞⎟ g1 ( x) = x 2 ; g 2 ( x ) = x ⎜⎜1 + ⎝α ⎠ ⎝α ⎠ ⎝ ln(x ) ⎠ CB = CT

•

a = exp(−α) ;

q R q = NO C ; α aDB

s=

DT DB

Then, g1 g1

d 2 CB 2

+ g2

dCB − hB C = 0 dx

2

+ g2

dCT − hT CT2 = 0 dx

dx d 2CT dx

dCB =0 dx dC dC q=− B +s T dx dx CB = CT

(a < x < 1)

(S13.13.17)

(0 < x < a)

(S13.13.18)

at x = 1

(S13.13.19)

at x = a

(S13.13.20)

at x = a (S13.13.21) CT = 0 at x = 0 (S13.13.22) Divide the domain [0 1] into k + n sections, with i = k when x = α and i = k+n when x = 1. Thus, 176

ΔxT =

a k

;

ΔxB =

1− a ; n

xT = iΔxT; (i = 1, 2, …, k) g1 ( xi )

CTi +1 − 2CTi + CTi −1 ΔxT2

xB = a + (i - k)ΔxB.

+ g 2 ( xi )

CTi +1 − CTi

i = k+1, k+2, …, k+n

− hT (CTi ) 2 = 0

ΔxT

(S13.13.23)

i = 2, …, k-1 g1 ( xi )

CBi +1 − 2CBi + CBi −1 ΔxB2

+ g 2 ( xi )

CBi +1 − CBi Δx B

(S13.13.24)

− hB CBi = 0

i = k+1, k+2, …, k+n-1 g1 ( x1 )

CT2

2 g1 ( xk + n )

− 2CT1 ΔxT2

+0

+ g 2 ( x1 )

CBk + n −1 − CBk + n ΔxB2

CT2

− CT1

ΔxT

− hT (CT1 ) 2 = 0

aa

− hB CBk + n = 0

(S13.13.25) (S13.13.26)

CBk +1 − C k C k −1 − C k 2 g1 ( xk ) T 2 − hB C k − hT (C k ) 2 2 ΔxB ΔxT −s =q 2 2 g 2 ( xk ) − g1 ( xk ) g 2 ( xk ) + g1 ( xk ) Δx B ΔxT

2 g1 ( xk )

(S13.13.27)

where C k = CTk = CBk . The resulting nonlinear algebraic equations can be solved by the iteration method. For the parameter values given in Table 13.1 and kT = 0.05 μM-1 s-1, the simulation results are shown in the following figure. They are similar to those shown in Figure S13.13.1, based on the first-order kinetics of NO reaction in the surrounding tissue. The concentration gradients are slightly smaller than those predicted by the first-order kinetics of NO reaction.

177

Figure S13.13.1

MATLAB code clear all; eps = 0.001; % Model Parameters qNO = 5.5e-9; % M*cm/sec or 5.5e-12 mol/cm2/sec DB = 4.5e-5; % cm2/sec DT = 3.3e-5; % cm2/sec RC = 25e-4; % cm kB = 454; kT = 0.05;

% 1/sec 454 to 4.54e5 % 1/uM/sec

alpha = 0.05; a = exp(-alpha)

% dimensionless % dimensionless

fB2 = RC*RC*kB/DB; fT2 = RC*RC*kT/DT;

% dimensionless % dimensionless

hB = fB2/(alpha*alpha); % dimensionless hT = fT2/(alpha*alpha); % dimensionless q = 1e6*qNO*RC/(alpha*a*DB); % uM s = DT/DB; % dimensionless % Mesh definition n = 1000; k = 200; dxT = a/k; dxB = (1-a)/n; x = zeros((k+n),1); for i = 1:k x(i) = i*dxT; end for i = (k+1):(k+n) x(i) = a + (i-k)*dxB; end AA = zeros((k+n),(k+n),1); BB = zeros((k+n),1); BB(k) = q; for i = 2:(k-1) g1 = x(i)^2; 178

g2 = x(i)*(1+1/log(x(i))); AA(i,i+1) = g1/(dxT^2)+g2/dxT; AA(i,i) = - 2*g1/(dxT^2) - g2/dxT - hT; AA(i,i-1) = g1/(dxT^2); end for i = (k+1):(k+n-1) g1 = x(i)^2; g2 = x(i)*(1+1/log(x(i))); AA(i,i+1) = g1/(dxB^2)+g2/dxB; AA(i,i) = - 2*g1/(dxB^2) - g2/dxB - hB; AA(i,i-1) = g1/(dxB^2); end g1 = x(1)^2; g2 = x(1)*(1+1/log(x(1))); AA(1,2) = g1/(dxT^2) + g2/dxT; AA(1,1) = - 2*g1/(dxT^2) - g2/dxT - hT; 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); delta = 1; g1 = x(k)^2; g2 = x(k)*(1+1/log(x(k))); gg1 = g2 - g1*2/dxB; gg2 = (g2 + g1*2/dxT)/s; while delta > eps for i = 2:(k-1) AA(i,i) = AA(i,i) + hT*(C1(i) - C2(i)); end AA(1,1) = AA(1,1) + hT*(C1(1) - C2(1)); 179

AA(k,k) = AA(k,k) - hT*(C1(k) - C2(k))/gg2; C3 = inv(AA)*BB; % calculate delta delta = 0; mm = 0; for i = 1:(n+k) delta = delta + (C3(i) - C2(i))^2; mm = mm + C3(i); end delta = sqrt(delta)*(n+k)/mm C1 = C2; C2 = C3; end r = zeros((k+n),1); for i = 1:(k+n) r(i) = -1e4*RC*log(x(i))/alpha; end plot(r,C3),axis([0 100 0 5]); nn=0; for i = 1:(k+n) if r(i) > 100 nn = i; end end fid = fopen('data.txt', 'w'); fprintf(fid,'r, C\n'); for i=nn:(n+k) fprintf(fid,'%8.4f,%8.4f\n',r(i),C3(i)); end fclose(fid);

13.14. This is a case of diffusion in spherical coordinates and zero order reaction. To find the size at which necrosis begins, determine the concentration as a function of radial position. Then set C to 0 at the center, r = 0 and solve for the radius. At this point oxygen is limiting. Any tumors bigger than this will have necrosis unless the capillaries penetrate into the tumor. (This often happens but makes the analysis much more complicated.) The conservation relation for steady state diffusion and reaction is: Deff d ⎛ 2 dC ⎞ r = RO2 (S13.14.1) r 2 dr ⎜⎝ dr ⎟⎠ 180

The boundary conditions are: dC r=R C = CB (S13.14.2) =0 dr Integrating equation (3.1) once yields: dC RO2 r A (S13.14.3) = + dr 3Deff r 2 To satisfy the boundary condition at r = 0, A must equal 0. Integrating again yields: RO2 r 2 +B (S13.14.4) C= 6 Deff Applying the boundary condition at r = R results in the following value for B: RO2 R 2 (S13.14.5) B = CB − 6 Deff The resulting expression for the concentration is: RO R 2 ⎛ r2 ⎞ (S13.14.6) C = CB − 2 ⎜ 1 − 2 ⎟ 6 Deff ⎝ R ⎠ To find the maximum radius before necrosis begins, C = 0 at r = 0. Solving for R yields: 6 Deff CB (S13.14.7) R= RO2 r=0

Inserting the values provided: R=

(

)(

6 1.5 x 10 −5 cm 2 /s 13 x 10 −6 mole/cm 3 120 x 10

−9

3

mole/cm / s

) = 0.0987 cm

(S13.14.8)

13.15. (a) Since the reaction only occurs on the surface of the mitochondrion, the reaction term does not appear in the conservation relation and is included in the boundary condition. The steady state conservation relation in spherical coordinates with no reaction in solution and no convection is: Deff d ⎛ 2 dC ⎞ r =0 r 2 dr ⎜⎝ dr ⎟⎠

(S13.15.1)

The boundary conditions are dC r = RM − Deff =Γ dr r =RC

(S13.15.2)

C = C0

(S13.15.3)

dC A = dr r 2 A can be found from the boundary condition at r = RM. dC A Or − Deff = − Deff 2 = Γ dr RM

Integrating once:

181

ΓRM2 A=− Deff

(S13.15.4)

ΓRM2 dC =− dr Deff r 2

(S13.15.5)

ΓRM2 +B Deff r

(S13.15.6)

C=

Integrating again

Applying the boundary condition at r = RC, ΓRM2 B=C− Deff RC

(S13.15.7)

ΓRM2 ΓRM2 ΓRM2 ⎛ RC ⎞ 1− C = C0 + − = C0 − Deff r Deff RC Deff RC ⎜⎝ r ⎟⎠ (b) Since the V = fVc =4/3πRc3, the mitochondrial radius is 1/3

(S13.15.8)

1/3 ⎛ 3V ⎞ RM = ⎜ = fRC3 = f 1/3 RC ⎟ 4 π ⎝ ⎠ RM = 0.0851/3(0.0005 cm) = 0.0002198 cm

(

)

(1.02 x 10 mole/cm / s ) ( 0.0002198 cm ) + (1.92 x 10 cm / s ) ( 0.0005 cm ) −10

C = 2.68 x 10

−8

mole/cm

3

2

−5

2

(S13.15.9)

2

0.0005 ⎞ ⎛ ⎜1 − ⎟ ⎝ 0.0002198 ⎠

C = 2.6145 x 10-8 mole/cm3 which is only 2.5% below the surface concentration. So the effect of diffusion is modest. Since 2.6145 x 10-8 mole/cm3 >> KM = 9 x 10-11 mole/cm3, the assumption of zero order kinetics is OK.

13.16. Using the Krogh model with axial variation in concentration, equation (13.5.12) applies from z = 0 to L. The only difference is that the oxygen concentration represents the concentration in solution. Ro 2 − Rc 2 C ( L ) = C (0) RO2 L Rc 2 v z The oxygen consumption rate is the rate per million cells times the cell number

(

)

RO2 = RO2/NNc. Solving for Nc

(R C ( L ) = C (0) -

2

o

NC =

− Rc 2 2

Rc v z

)R

O2 / N

C ( 0 ) -C ( L ) ⎛ Rc 2 v z ⎜ RO2 / N L ⎜ Ro 2 − Rc 2 ⎝

(

)

NC L ⎞ ⎟ ⎟ ⎠

Substituting values

182

NC =

2 ⎛ ⎞ (140 mm Hg ) (1.34 x 10 −9 mole/cm 3 ) ⎜ ( 25 ) ( 0.3 cm/s ) ⎟ = 89333 cells (1 x 10 −9 mole/cm 3 / s / (1 x 10 6 cells) ) (10 cm ) ⎜⎜ ( ( 200 )2 − ( 25)2 ) ⎟⎟

⎝

⎠

Volume 2πL(R02 – Rc2) = 2π(10 cm) (2002-252) x 10-8 cm2 = 0.0235 cm3. The overall cell density is 89333 cells/0.0235 cm3 = 3.80 x 106 cells/cm3.

183

Solution to Problems in Chapter 14, Section 14.7 14.1. At steady state, the one-dimensional diffusion equation in the membrane is 0 = Dm

∂ 2 Cm

(S14.1.1)

∂ x2

where Dm and Cm are the diffusion coefficient and concentration of the solute in the membrane, respectively. At the surface of the membrane, the boundary conditions are, Cm = C1 φ, at x = 0

(S14.1.1a)

Cm = C2 φ, at x = l

(S14.1.1b)

Solve Equation S14.1.1, we have, Cm = ϕ C1 + ϕ ( C2 − C1 )

x x = ϕ C1 − ϕΔC l l

(S14.1.2)

Therefore, the flux of the solute across the membrane is J s = − Dm

∂Cm ΔC = Dmϕ l ∂x

(S14.1.3)

Finally, the permeability of the membrane is, P=

Js D ϕ = m ΔC l

(S14.1.4)

14.2. The charge of H3O+ is q = 1 x (1.602x10-19 C) = 1.602x10-19 C Using Equation 14.2.5, the partition coefficient of H3O+ in the membrane is ⎡ 1 q2 ⎛ 1 1 ⎞⎤ φ = exp ⎢ − − ⎟⎥ ⎜ ⎣ k BTr 8π rε 0 ⎝ κ m κ s ⎠ ⎦ 2 ⎡ ⎤ 1.602 × 10−19 ) × (1 / 2 − 1 / 78 ) ( ⎢ ⎥ = exp − ⎢ 1.38 × 10−23 × 310 × 8 × 3.14 × 0.12 × 10−9 × 8.9 × 10−12 ⎥ ⎣ ⎦ −48 = 5.16 × 10 Therefore, the permeability of H3O+ in the membrane is D φ 1.4 × 10−4 × 5.16 × 10−48 = 1.72 × 10−45 cm / sec P= m = 4.2 × 10−7 l 14.3. The net pressure difference between the blood and the Bowman’s space is (Δp – σsΔπ). Δp equals 48 mmHg – 12 mmHg. Δπ equals to 25 mmHg at the afferent arteriole end and 35 mmHg at the efferent arteriole end. σs equals to unity. Therefore, the net pressure difference is 9 mmHg at the afferent arteriole end and one at the efferent arteriole end. The average net pressure difference is 4 mmHg since the distribution of net pressure difference is a linear function of the distance along the glomerular capillary. 14.4. From Equation (14.4.6), we have 184

N s = − HD∞

dC + WCvm dz

(S14.4.1)

The boundary conditions for Equation (S14.4.1) are, C = C0, at z = 0

(S14.4.1a)

C = Cl, at z = l

(S14.4.1b)

Integrate Equation (S14.4.1), ⎛ z⎞ N C = A exp ⎜ Pe ⎟ + s ⎝ l ⎠ Wvm

Here A is a constant and Pe =

Wvm l HD∞

(S14.4.2)

. Substitute Equation (S14.4.2) into Equations (S14.4.1a) and

(S14.4.1b), we have, C0 = A +

Ns Wvm

Cl = A exp ( Pe ) +

(S14.4.3a) Ns Wvm

(S14.4.3b)

Subtract (S14.4.3a) and (S14.4.3b)×exp(-Pe), we have, N s = Wvm C0

1 − ( Cl / C0 ) exp ( − Pe )

(S14.4.4)

1 − exp ( − Pe )

(b) From Equation (14.4.11), Js =

W φ J v (C0 − C e − Pe )

(S14.4.5)

1 − e − Pe

Substitute C in Equation (S14.4.5) with Js/Jv, ⎡W φ J v C0 − W φ J s e − Pe ⎤ ⎦ Js = ⎣ 1 − e − Pe

(

)

J s 1 − e − Pe + W φ e− Pe = W φ J v C0 Js =

W ϕ J v C0

(S14.4.6) (S14.4.7) (S14.4.8)

1 − e − Pe (1 − W ϕ )

14.5. The continuity equation and the momentum equations in the x and r directions at steady state are, ∂ vx 1 ∂ + ( rvr ) = 0 ∂x r ∂r ⎛

ρ ⎜ vr ⎝

(S14.5.1)

⎡ 1 ∂ ⎛ ∂ vx ⎞ ∂ 2 vx ⎤ ∂ vx ∂v ⎞ ∂p + vx x ⎟ = − +μ⎢ ⎜r ⎟+ 2 ⎥ ∂r ∂x ⎠ ∂x ⎢⎣ r ∂ r ⎝ ∂ r ⎠ ∂ x ⎥⎦

185

(S14.5.2)

⎛

ρ ⎜ vr ⎝

⎡ ∂ ⎛1 ∂ ⎞ ∂2 v ⎤ ∂ vr ∂v ⎞ ∂p + vx r ⎟ = − +μ⎢ ⎜ ( rvr ) ⎟ + 2r ⎥ ∂r ∂x ⎠ ∂r ⎢⎣ ∂ r ⎝ r ∂ r ⎠ ∂ x ⎥⎦

(S14.5.3)

The boundary conditions are, ∂ vx =0, ∂r

at r = 0

(S14.5.4)

vx = 0, at r = R vr = vw =

(

K p l

(S14.5.5) fluid at wall

− pi

) , at r = R

(S14.5.6)

Here l is the thickness of the wall. Similar to Example 4.7, we can simplify Equations (S14.5.2) and (S14.5.3) by examining them in dimensionless form. We assume that the velocity and length scales are Vx0 and L in x direction, and Vr0 and R in r direction. From Equation (S14.5.1), we have Vr0<
(S14.5.7)

∂p =0 ∂r

(S14.5.8)

Equation (S14.5.8) indicated that p is independent of r. Therefore, we can directly integrate Equation (S14.5.7) and obtain the following expression, vx =

1 ∂p 2 r − R2 4μ ∂ x

(

)

(S14.5.9)

Therefore, the flux in the x direction is, J vL =

1

R2 ∂ p

R

vx ⋅ 2π rdr = − 8μ ∂ x π R 2 ∫0

(S14.5.10)

14.6. Based on Equation (14.5.8b), JiL ≈ CiLJvL. Therefore,

∑ zi J iL = ∑ zi CiL J vL = J vL ∑ zi CiL = 0 i

i

(S14.6.1)

i

If the electric charge is conserved after chemical reactions, then

∑ zi QiL = ∑ zi Qic = 0 i

(S14.6.2)

i

Multiplying both sides of Equations (14.5.1) and (14.5.3) and summing them for all solutes yield, d 2 1 ∑ zi J iL = − R ∑ zi J ic + π R 2 ∑ zi QiL dx i i i 1 ∑ zi J ic = ∑ zi J ip − 2π R ∑ zi QiC i i i

(S14.6.3) (S14.6.4)

Substituting Equations (S14.6.1) and (S14.6.2) into Substituting Equations (S14.6.3) and (S14.6.4), we have, 186

∑ zi J ic = 0

(S14.6.5)

i

∑ zi J ip = 0

(S14.6.6)

i

14.7. The initial concentration after intravenous injection is 120mg/4L. According to equation 14.4.2, the filtration rate of inulin is, J s = GFR ⋅ C p where Cp is the plasma concentration of inulin. Since inulin is not reabsorbed or secreted, the mass balance of inulin in blood is, dCLp Vp = − J s = −GFR ⋅ C p dt where Vp is the volume of plasma. The initial condition for Equation S14.7.2 is C p = 120mg/4L, when t=0. Solving Equation (S14.7.2) yields, ⎛ GFR ⎞ 120mg Cp = exp ⎜ − ⋅ t ⎟ = 30exp ( − t / 29.4 min )( mg/L ) ⎜ Vp ⎟ 4L ⎝ ⎠ The half life of inulin in the blood is 29.4ln2 = 20.4 min.

14.8. The total rate of PAH excretion from the blood equals to the sum of the rates of filtration and secretion. Therefore, the mass balance equation is, dC p TmC p Vp = −GFR ⋅ C p − dt Km + C p where the values of GFR, Tm and Km are 120 ml/min, 80 mg/min and 0.07 mg/ml, respectively. The initial condition of Equation S14.8.1 is, Cp = 110 mg/3000 ml = 0.037 mg/ml at t = 0 Substituting the values of GFR, Tm, Km and Vp into Equation S14.8.1, we have, dC p 0.67C p ⎞ 1 ⎛ = − ⎜ Cp + ⎟ 25 ⎜⎝ 0.07 + C p ⎟⎠ dt 0.07 + C p dt dC p = − 25 C p ( C p + 0.74 ) ⎛ 22.6 2.36 ⎞ + ⎜⎜ ⎟⎟ dC p = −dt 0.74 C C + p ⎠ ⎝ p 22.6 ln ⎡⎣( C p + 0.74 ) / 0.777 ⎤⎦ + 2.36 ln ( C p / 0.037 ) = −t Letting Cp=Cp0/2=0.0184 mg/ml in Equation S14.8.5 yields the half life of PAH being 2.18 min.

14.9. Assume that the water reabsorption in renal tubules depends on the osmotic pressure difference of Cl- and HCO3- across the epithelial layer. The rate of water flow across the epithelial layer is, 187

(

J v = − LP S σ sCl − Δπ Cl − + σ sHCO − Δπ HCO − 3

3

)

(S14.9.1)

where Δπ Cl and Δπ HCO are osmotic pressure difference determined by Cl- and HCO3-, −

− 3

respectively, σ sCl and σ sHCO are osmotic reflection coefficient of epithelial layer to Cl- and −

− 3

-

HCO3 , respectively, and LP is the hydraulic conductivity. Based on Equation (9.3.15), (S14.9.2)

Δπ = (ΔC ) RT

Thus,

(

J v = − RTLP S σ sCl − ΔCCl − + σ sHCO− ΔCHCO − 3

3

)

(S14.9.3)

Using the concentration data provided in this problem, we obtain,

(

)

(S14.9.4)

J v = − RTLP S σ sCl − − σ sHCO − ⋅ 20mM 3

Since the epithelial layer is more permeable to Cl- than HCO3-, σ sCl − is less than σ sHCO − . 3

Therefore, Jv > 0, indicating that the direction of the driving force is from the lumen of renal tubule to the interstitial space and water is reabsorbed.

14.10. r

Q z Js

Figure S14.10. 1 In the steady state, the mass balance equation of urea in the renal tubule is 0 = −Q

dC − P ⋅ C ⋅ 2π R dz

(S14.10.1)

The boundary condition of Equation S14.10.1 is, C = C0,

at z = 0.

(S14.10.2)

Solving Equation (S14.10.1) yields ⎛ 2π RP ⎞ C = C0 exp ⎜ − z⎟ Q ⎝ ⎠

(S14.10.3)

At the end of the renal tubule, the concentration of urea is ⎛ 2π RPL ⎞ CL = C0 exp ⎜ − ⎟ Q ⎠ ⎝

(S14.10.4)

Therefore, the total rate of reabsorption is, ⎛ ⎛ 2π RPL ⎞ ⎞ R = QC0 − QCL = QC0 ⎜1 − exp ⎜ − ⎟⎟ Q ⎠⎠ ⎝ ⎝

188

(S14.10.5)

14.11. If the translocation of uniporter is not the rate-limiting step, we have to consider the effects of all reactions simultaneous. In this case, two general equilibrium for [SE0] and [SEi] are, ko [ So ][ Eo ] − k− o [ SEo ] − k s [ SEo ] + k− s [ SEi ] = 0

(S14.11.1)

ki [ Si ][ Ei ] − k−i [ SEi ] + k s [ SEo ] − k− s [ SEi ] = 0

(S14.11.2)

Similar to that in the textbook, we assume that the total number of uniporters in the plasma membrane is a constant, ET. In addition, we assume that the total number of uniporters on both sides of the membrane are time-independent. Therefore, we have two more equations, [ Eo ] + [ Ei ] + [ SEo ] + [ SEi ] = ET

(S14.11.3)

k− s [ SEi ] − k s [ SEo ] + k− E [ Ei ] − k E [ Eo ] = 0

(S14.11.4)

We also assume that [Si] equals to zero. Then from Equation S14.11.2, we can derive, [ SEi ] =

k s [ SEo ] k−i + k− s

(S14.11.5)

Subtract (S14.11.3)k-E by (S14.11.4), we have, [ Eo ] =

1 ⎡ k− E ET − ( k− E + k s ) [ SEo ] − ( k− E − k− s ) [ SEi ]⎤⎦ k E + k− E ⎣

1 = k E + k− E

⎡ ⎤ ⎛ k s ( k− E − k− s ) ⎞ ⎢ k− E ET − ⎜⎜ k− E + k s + ⎟⎟ [ SEo ]⎥ k− i + k− s ⎠ ⎢⎣ ⎥⎦ ⎝

(S14.11.6)

By substituting (S14.11.5) and (S14.11.6) into (S14.11.1), we have [ SEo ] =

A1 [ S 0 ]

(S14.11.7)

A2 + [ S 0 ]

Here A1 and A2 are two constants, which are A1 =

( k− E

k− E ET k ( k − k− s ) + ks ) + s − E k −i + k − s

⎛ ⎞ kk A2 = ⎜ k−o + k s + s − s ⎟ ( k E + k− E ) k−i + k− s ⎠ ⎝

(S14.11.8)

⎡ ko k s ( k − E − k − s ) ⎤ ⎢ ko ( k − E + k s ) + ⎥ k− i + k− s ⎣ ⎦

(S14.11.9)

The net flux of S into the cell is V [S ] ⎛ ⎞ kk J s = k s [ SEo ] - k-s [ SEi ] = ⎜ ks − s − s ⎟ [ SEo ] = max o k−i + k− s ⎠ K m + [ So ] ⎝ ⎛ ⎞ kk Vmax = A1 ⎜ k s − s − s ⎟ + k k −i −s ⎠ ⎝

(S14.11.10) (S14.11.11) (S14.11.12)

K m = A2

189

14.12. Similar to the analysis in the textbook, we assume that the rate-limiting process in symport is the translocation of substrates between intracellular and extracellular spaces. Therefore, the binding between substrates and the symporter is at the equilibrium state. Thus, we have, [ ATo ] = [ Ao ][To ] / K oA

(S14.12.1)

[ AST0 ] = [ AT0 ][ S0 ] / K oS

(S14.12.2)

Figure S14.12.1. Substitute Equation (S14.12.1) into Equation (S14.12.2), [ AST ]0 = [ Ao ][ So ][To ] / ( K oA K oS )

Similarly,

(S14.12.3) (S14.12.4)

[ ATi ] = [ Ai ][Ti ] / K iA [ ASTi ] = [ Ai ][ Si ][Ti ] / ( K iA K iS )

(S14.12.5)

The total number of symporters in the plasma membrane is a constant, Tt. [To] + [ATo] + [STo] +[ASTo]+ [Ti] + [ATi] + [STi]+ [ASTi] = Tt

(S14.12.6)

In addition, we assume that the total number of symporters on both sides of the membrane are time-independent. Thus, kT[To] - k-T [Ti] + kC [ASTo] - k-C[ASTi] = 0

(S14.12.7)

By solving Equations (S14.12.3), (S14.12.5), (S14.12.6) and (S14.12.7), we can obtain the expression of the net fluxes of A and S. The above equations are equivalent to the random reactions with αo and αi equal to unity. Therefore, the net fluxes of A and S are, J A = J S = kC [ ASTo ] − k−C [ ASTi ] =

Tt X

⎛ [Ao ][So ] [A ][S ] ⎞ − k−C kT i A Si ⎟ ⎜⎜ kC k−T A S Ko Ko K i K i ⎟⎠ ⎝

where X is a constant,

190

(S14.12.8)

⎛ [A ] [S ] [A ][S ] ⎞ ⎛ [A ][S ] ⎞ X = ⎜1 + oA + oS + o A So ⎟ ⎜ k−T + k−C i A Si ⎟ + ⎜ ⎟ ⎜ Ko Ko Ko Ko ⎠ ⎝ K i K i ⎟⎠ ⎝ ⎛ [Ai ] [Si ] [Ai ][Si ] ⎞ ⎛ [A ][S ] ⎞ kT + kC o A So ⎟ ⎜⎜1 + A + S + ⎜ A S ⎟ Ki Ki K i K i ⎟⎠ ⎜⎝ K o K o ⎟⎠ ⎝

191

(S14.12.9)

Solution to Problems in Chapter 15, Section 15.6 15.1. The significance of convection vs. diffusion can be evaluated by the Peclet number, Pe = vL/D (see section 1.3). The Peclet numbers are 0.30, 0.00030, 30, 0.030 in cases (a), (b), (c) and (d), respectively. Therefore, diffusion is the dominant mode of transport in cases (a), (b) and (d), whereas convection is the dominant mode of transport in case (c). 15.2. (i) microvascular wall; (ii) extracellular matrix; (iii) plasma membrane of cells; (iv) cytoskeleton; and (v) nuclear envelope. 15.3. In a spherical coordinate system, the mass and momentum balance equations are ε

1 ∂ 2 1 ∂2 + − r v ε r 2u = ϕ B (1 ) r 2 ∂r r 2 ∂r ∂t

( )

⎛ ⎝

ε ⎜v −

( )

∂pi ∂u⎞ = −K ∂ t ⎟⎠ ∂r

(S15.3.2)

∂e 1 ∂ ⎛ ∂e ⎞ − K (2μG + μλ ) 2 ⎜ r 2 ⎟ = ϕ B ∂t r ∂r ⎝ ∂r ⎠

ϕB =

Lp S V

(S15.3.1)

(S15.3.3) (S15.3.4)

( pe1 − pi )

The boundary conditions are, ∂pi = 0 , at r = 0 ∂r pi = 0 ,

(S15.3.5a)

at r = R

(S15.3.5b)

∂e = 0 , at r = 0 ∂r e = 0,

(S15.3.5c)

at r = R

(S15.3.5d)

Similar to the derivations in Section 15.3.2, we assume ε is a constant. In the spherical system, it can be written as e = ∇•u =

1 ∂ (r 2 u ) r 2 ∂r

(S15.3.6)

Substituting Equation (S15.3.6) into Equation (S15.3.1) and taking the divergence of Equation (S15.3.2), we have the following two equations, 1 ∂ 2 ∂e r v + (1 − ε ) = ϕB 2 ∂ r ∂t r

(S15.3.7)

1 ∂ ∂e 1 ∂ ⎛ 2 ∂ pi ⎞ = −K 2 r 2v − ε ⎜ r ∂r ⎟ ∂t r2 ∂r r ∂r ⎝ ⎠

(S15.3.8)

ε ε

( )

( )

Subtracting Equation (S15.3.8) from Equation (S15.3.7), we have

192

∂e 1 ∂ ⎛ 2 ∂ pi ⎞ = ϕB + K 2 ⎜ r ∂r ⎟ ∂t r ∂r ⎝ ⎠

(S15.3.9)

Substituting Equation (S15.3.9) into Equation (S15.3.3), we have K

1 ∂ ⎛ 2 ∂ pi ⎞ 1 ∂ ⎛ 2 ∂e ⎞ ⎜r ⎟ − K (2μG + μλ ) 2 ⎜ r ⎟=0 2 r ∂r ⎝ ∂r ⎠ r ∂r ⎝ ∂r ⎠

(S15.3.10)

Rearranging Equation (S15.3.10), we have, ⎞ 1 ∂ ⎛ 2 ∂ ( pi − (2μG + μλ )e ) ⎟ = 0 ⎜r 2 r ∂r ⎝ ∂r ⎠

(S15.3.11)

Integrating Equation (S15.3.11), we have, pi − (2μG + μλ )e = −

C1 + C2 r

(S15.3.12)

where C1 and C2 are constants. Using the boundary conditions described by Equations (S15.3.5a through d), we have, (S15.3.13)

pi = (2μG + μλ )e

Substituting Equations (S15.3.13) and (S15.3.4) into Equation (S15.3.3), we have, ∂ pi 1 1 ∂ ⎛ ∂p − K 2 ⎜ r2 i 2μG + μλ ∂ t ∂r r ∂r ⎝

⎞ Lp S ⎟ = V ( pe1 − pi ) ⎠

(S15.3.14)

Equation (S15.3.14) can be transformed into a homogeneous equation by substituting pi – pe1 with p, ∂p 1 ∂ ⎛ ∂ p ⎞ Lp S = K ( 2μG + μλ ) 2 ⎜ r 2 ( 2μG + μλ ) p ⎟− ∂t r ∂r ⎝ ∂r ⎠ V

(S15.3.15a)

Accordingly, the boundary conditions become, ∂p =0, ∂r

at r = 0

(S15.3.15b)

p = − pe1 ,

at r = R

(S15.3.15c)

To solve Equations (S15.3.15a) by using the separation of variables method, we need to make both the equation and boundary conditions homogeneous. Let p = X(r,t) + f(r), and f(r) satisfies the following equation and boundary conditions, ⎞ Lp S f ⎟− ⎠ VK

(S15.3.16a)

df = 0, dr

at r = 0

(S15.3.16b)

f = − pe1 ,

at r = R

(S15.3.16c)

0=

1 d ⎛ 2df ⎜r r 2 dr ⎝ dr

Then, Equations (S15.3.15a,b,c) become, ∂X 1 ∂ ⎛ ∂ X ⎞ Lp S = K ( 2μG + μλ ) 2 ⎜ r 2 ( 2 μG + μ λ ) X ⎟− ∂t ∂r ⎠ V r ∂r ⎝

193

(S15.3.17a)

∂X =0, ∂r X =0,

at r = 0

(S15.3.17b)

at r = R

(S15.3.17c)

The solution of Equation (15.3.16a) is ⎛ r⎞ sinh ⎜ α ⎟ pe1 R ⎝ R⎠ f =− r sinh(α )

(S15.3.18)

where α is α=R

LP S KV

(S15.3.19)

The initial condition of pi is determined by the steady state solution of the equation when pe = pe0. 0=

1 ∂ ⎛ 2 ∂ pi ⎞ Lp S ⎜ r ∂r ⎟ − KV ( pi − pe0 ) r 2 ∂r ⎝ ⎠

(S15.3.20)

According to Equation (15.3.15), the solution of pi at steady state is α ⎞ ⎛ ⎜ R sinh( R r ) ⎟ pi = pe 0 ⎜ 1 − ⎟ r sinh(α ) ⎟ ⎜⎜ ⎟ ⎝ ⎠

(S15.3.21)

Therefore, the initial condition of X is, α ⎞ ⎛ ⎜ R sinh( R r ) ⎟ X = ( pe 0 − pe1 ) ⎜ 1 − ⎟, r sinh(α ) ⎟ ⎜⎜ ⎟ ⎝ ⎠

(S15.3.22)

Equation (S15.3.17a) with the initial condition of Equation (S15.3.22) can be solved by using the separation of variables method. The result is, r⎞ ⎛ (−1) n sin ⎜ nπ ⎟ 2α 2 ( pe1 − pe0 ) R ∞ R ⎠ − fnt ⎝ X = e ∑ 2 2 r n =1 nπ (( nπ ) + α )

(S15.3.23)

pi = pe1 + X + f ⎡ r⎞ ⎛α r ⎞⎤ ⎛ n ⎢ R sinh ⎜ R ⎟ ⎥ 2α 2 ( p − p ) R ∞ (−1) sin ⎜ nπ R ⎟ ⎝ ⎠⎥ + ⎝ ⎠e − f n t e1 e0 = pe1 ⎢1 − ∑ 2 2 r ⎢ r sinh(α ) ⎥ n =1 nπ (( nπ ) + α ) ⎢ ⎥ ⎣ ⎦

(S15.3.24)

Equation (S15.3.24) is a different expression of Equation (15.3.25) since r⎞ ⎛αr ⎞ ⎛ n sinh ⎜ ⎟ r ⎟ ∞ ( −1) sin ⎜ nπ R⎠ ⎝ R ⎠ − ≡ 2α 2 ⎝ ∑ 2 2 sinh(α ) R n =1 nπ (( nπ ) + α )

194

(S15.3.25)

Tissue dilatation can be calculated by substituting Equation (S15.3.13) into Equation (15.3.25). The result is, ⎡ r⎞ ⎛αr ⎞⎤ ⎛ n sinh ⎜ 2 ⎟ ⎟ ∞ ( −1) sin ⎜ nπ ⎢ ⎥ pe 0 R R⎠ ⎝ R ⎠ ⎥ + 2α ( pe 0 − pe1 ) R ⎝ ⎢1 − (1 − e − fn t ) e= ∑ 2μG + μλ ⎢ r sinh(α ) ⎥ π r ( 2 μG + μλ ) n =1 n(( nπ )2 + α 2 ) ⎢ ⎥ ⎣ ⎦

(S15.3.26) where fn are the coefficients defined in Equation (15.3.26). Tissue displacement is determined by integrating Equation (S15.3.6). The result is u=A

+

1 r

2

+

pe 0 2 μG + μ λ

2 ⎡r ⎛ Rr R ⎛αr ⎞ R ⎛ α r ⎞ ⎞⎤ cosh ⎜ ⎢ − 2 ⎜ ⎟ − 2 sinh ⎜ ⎟ ⎟⎥ ⎜ ⎝ R ⎠ α ⎝ R ⎠ ⎟⎠ ⎦⎥ ⎣⎢ 3 r sinh(α ) ⎝ α

2α 2 ( pe 0 − pe1 ) R r 2π ( 2 μ G + μ λ )

r⎞ r r ⎞⎤ ⎡ ⎛ ⎛ sin nπ ⎟ cos ⎜ nπ ⎟ ⎥ (−1) n (1 − e − fn t ) ⎢ ⎜⎝ R⎠ R R ⎝ ⎠⎥ ⎢ − ∑ 2 2 2 ⎢ n π ⎥ n n n ( ) + π α π n =1 ( ) ( ) ⎢ ⎥ ⎣ ⎦ ∞

(S15.3.27)

where A is a constant. Since u is zero at the center of the tumor, A is equal to zero. Substituting Equation (15.3.24) and Equation (S15.3.27) into (S15.3.2), we obtain the velocity profile, v=

∂ u K ∂pi − ε ∂r ∂t

(S15.3.25)

15.4. If the initial interstitial fluid pressure at r = 0 is p0, then the microvascular pressure can be determined by Equation (15.3.15), pe =

p0

(S15.4.1)

⎛ α ⎞ ⎜1 − sinh( α ) ⎟⎠ ⎝

After animal is sacrificed, pe jumps from p0/(1-α/sinh(α)) to zero immediately. In this case, pi is determined by Equation (15.3.25) with pe0 = p0/(1-α/sinh(α)) and pe1 = 0. Thus, ⎡ r⎞ ⎛αr ⎞⎤ ⎛ n ⎢ R sinh ⎜ R ⎟ ⎥ 2α 2 p R ∞ (−1) sin ⎜ nπ R ⎟ ⎝ ⎠⎥ + ⎝ ⎠ (1 − e− fn t ) e0 pi = pe 0 ⎢1 − ∑ 2 2 r ⎢ r sinh(α ) ⎥ π π α ( ) + n n n =1 ( ) ⎢ ⎥ ⎣ ⎦

(S15.4.2)

When r → 0, pi = p0 + 2α 2

p0

∞

∑

(−1) n

⎛ α ⎞ n =1 ( nπ )2 + α 2 ⎜1 − ⎟ ⎝ sinh(α ) ⎠

(1 − e − fn t )

(S15.4.2)

If μG and μλ are 684 mmHg and 15.2 mmHg, respectively, LP = 3.6x10-7 cm mmHg-1 sec-1, S/V = 200 cm-1, then p/p0 vs time is plotted in Figure S15.1 for α = 1, 3, and 30. When α > 10, p/p0 is nearly independent of the value of α.

195

Figure S15.4.1.

15.5. The mass balance equation is, ∂C 1 ∂ (r 2 vr fC ) 1 ∂ ⎛ ∂C ⎞ PS (C p − C / K AV ) + Rrxn + 2 = D 2 ⎜ r2 ⎟+ ∂t r ∂r r ∂r ⎝ ∂r ⎠ V

(S15.5.1)

The boundary condition at r = 0 is −D

∂C + vr fC = 0 ∂r

(S15.5.2)

Assuming that there is no concentration gradient at r = R, i.e., ∂C =0, ∂r

at r = R

(S15.5.3)

Before the infusion starts, there is no drug in the tumor tissue, i.e., C = 0, at t = 0

(S15.5.4)

The velocity profile in the tumor tissue is given by Equation 15.3.16, vr =

⎡ rα ⎛ rα ⎞ ⎛ rα ⎞ ⎤ ⎢ R cosh ⎜ R ⎟ − sinh ⎜ R ⎟ ⎥ ε r sinh(α ) ⎣ ⎝ ⎠ ⎝ ⎠⎦ Kpe R

2

(S15.5.5)

The equations described above for determining the concentration distribution can be solved numerically using the MATLAB. (The MATLAB code is attached at the end of this chapter.) The numerical solutions are plotted in Figure S15.2 for t = 10, 20, 50 and 100 hours.

196

Figure S15.5.1

15.6. Statements (a), (c) and (d) are true. Statement (d) is false.

197

15.7. (a) Assuming that convection is negligible, and that there are no chemical reactions of drugs in the tumor, the mass conservation equation in a Krogh cylinder is, ∂C 1 ∂ ⎛ ∂C ⎞ = Deff ⎜r ⎟ ∂t r ∂r ⎝ ∂r ⎠

(S15.7.1)

The boundary condition on the vessel wall is, − Deff

∂C = P(C p − C / K AV ) , ∂r

at r = a

(S15.7.2)

where the concentration of the drug in the plasma, Cp, is a bi-exponential function of time, C p = C p 0 [α exp(−λ1t ) + (1 − α ) exp(−λ2 t ) ]

(S15.7.3)

At the outer surface of the Krogh cylinder, − Deff

∂C = 0, ∂r

at r = R

(S15.7.4)

The initial condition is, C = 0,

(S15.7.5)

Equations (S15.7.1) through (S15.7.5) are solved numerically using the MATLAB. (The MATLAB code is attached at the end of this chapter.) The results are shown in Figure S15.7.1. 0.4 10 min 60 min 12 hr 48 hr 100 hr

0.3

0.2

0.1

0 0

50

100

150

r (μm)

Figure S15.7.1. (b) The dependence of drug distribution on Deff and P can be evaluated, based the MATLAB program developed in (a). (c) The index of spatial nonuniformity (ISN) is defined as N

∑ (C − C )2 ISN =

j =1

(S15.7.6)

C N ( N − 1)

198

where C is the mean of C in tumors. The mean concentration can be obtained by numerical integration of 2πrC from r = a to r = R. The result is then divided by the cross-sectional area π(R2- a2). At 12 hr, the concentration distribution is nonuniform for low Deff and the nonuniformity decreases when Deff is increased. At 48 hr, the concentration distribution in tumors becomes uniform.

Figure S15.4.

15.8. The governing equation and the boundary conditions are identical to those in Problem 15.7, i.e., ∂C 1 ∂ ⎛ ∂C ⎞ = Deff ⎜r ⎟ ∂t r ∂r ⎝ ∂r ⎠

(S15.8.1)

− Deff

∂C = P(C p − C / K AV ) , ∂r

− Deff

∂C = 0 , at r = R ∂r

at r = a

(S15.8.2) (S15.8.3)

Here we assume that the concentration of the drug in plasma is negligible compared with that in the tissue, i.e., Cp = 0. The initial condition is, C = 10 μM, at t = 0

(S15.8.4)

Equations (S15.8.1) through (S15.8.4) are solved numerically using the MATLAB. (The MATLAB code is attached at the end of this chapter.) The results are shown in Figure S15.8.1.

199

Figure S15.8.1.

15.9. The governing equation and the boundary conditions are, ∂C 1 ∂ ⎛ ∂C ⎞ = Deff ⎜r ⎟ − kC ∂t r ∂r ⎝ ∂r ⎠

(S15.9.1)

− Deff

∂C = P(C p − C / K AV ) , ∂r

− Deff

∂C = 0, ∂r

at r = a

at r = R

(S15.9.2) (S15.9.3)

The concentration of the drug in plasma is negligible compared with that in the tissue, i.e., Cp = 0. The initial condition is, C = 10 μM,

at t = 0

(S15.9.4)

Equations (S15.9.1) through (S15.9.4) are solved numerically using the MATLAB. (The MATLAB code is attached at the end of this chapter.) The results are shown in Figure S15.9.1.

200

Figure S15.9.1. Comparing with the results shown in Figure S15.8.1, drug degradation reduces the concentration but has minimal effects of the shape of concentration profiles in tumors.

15.10. (a) At steady state, the mass balance equation is, 0 = Deff

∂ 2C ∂x

2

+

(

PS C p − C / K AV V

)

(S15.10.1)

The boundary conditions are, C = C0,

at x = 0

(S15.10.2)

C = 0,

as x → ∞

(S15.10.3)

Assuming that the plasma concentration is negligible compared with the drug concentration in the tumor tissue, we have, 0 = Deff

∂ 2C ∂x

2

−

PS C VK AV

(S15.10.4)

The solution of Equation (S15.10.4) is, C = A 1 exp(β x ) + A 2 exp(− βx )

(S15.10.5)

where A1 and A2 are constants and β=

PS VDeff K AV

(S15.10.6)

Applying the boundary conditions, we have, C = C0 exp ( − β x )

(S15.10.6)

(b) The penetration depth L is defined as, 0.01C0 = C0 exp ( − β L )

(S15.10.7) 201

L=

4.61

β

= 4.61

Deff K AV

(S15.10.8)

PS / V

Although L increases with Deff and Deff can be increased by reducing the size of drugs, the reduction in drug size also increases P. Thus, a better way to increase the penetration depth is to increase the size of drugs or using large carriers of drugs.

MATLAB CODES Problem 15.5 function main global N s vol dr vr Deff PSV Cp f Kav Q % Model Parameters K = 4.13e-8*60; %cm2/mmHg/min Deff = 2e-7*60; %cm2/min PSV = 1.26e-4*60; %1/min pe = 8; %mmHg Cp = 1; %um R = 1; %cm Kav = 0.3; f = 1; e = 0.3; alfa = 100; Q = 0; % Mesh definition N = 50; dr = R/N; drh = dr/2; rm = 0:dr:R; r = drh:dr:(R-drh); s = zeros(N+1,1); for i = 1:N+1 s(i) = 4*pi*rm(i)*rm(i); end vol = zeros(N,1); for i = 1:N vol(i) = 4/3*pi*(rm(i+1)*rm(i+1)*rm(i+1)-rm(i)*rm(i)*rm(i)); end 202

%Velocity vr = zeros(N+1,1); for i=2:N+1 x = alfa*rm(i)/R vr(i) = K*pe*R/(e*rm(i)*rm(i)*sinh(alfa))*(x*cosh(x)-sinh(x)) end % c0 = zeros(N,1); tspan = [0 10 20 50 100]; [t,c] = ode15s(@func,tspan,c0); figure; plot(rm,vr); figure; plot(r,c(2,:),r,c(3,:),r,c(4,:),r,c(5,:)); fid = fopen('data.txt', 'w'); fprintf(fid,'r, 10, 20, 50, 100\n'); for i=1:N fprintf(fid,'%8.4f,%8.4f,%8.4f,%8.4f,%8.4f\n',r(i),c(2,i),c(3,i),c(4,i),c(5,i)); end fclose(fid);

function dcdt = func(t,c) global N s vol dr vr Deff PSV Cp f Kav Q dcdt = zeros(N,1); dcdt(1) = 1/vol(1)*(-(Deff*(c(1)-c(2))/dr + f*vr(2)*0.5*(c(1)+c(2)))*s(2))… + PSV*(Cp-c(1)/Kav)+Q; for i=2:N-1 iu=i-1; id=i+1; dcdt(i)=1/vol(i)*((Deff*(c(iu)-c(i))/dr + f*vr(i)*0.5*(c(iu)+c(i)))*s(i)... -(Deff*(c(i)-c(id))/dr + f*vr(id)*0.5*(c(i)+c(id)))*s(id))... +PSV*(Cp-c(i)/Kav)+Q; end dcdt(N) = 1/vol(N)*((Deff*(c(N-1)-c(N))/dr + f*vr(N)*0.5*(c(N-1)+c(N)))*s(N)... -(f*vr(N+1)*c(N))*s(N+1)) + PSV*(Cp-c(i)/Kav)+Q;

203

Problem 15.7 function main global N dr s v Deff P Kav Cp0 alpha L1 L2 % Model Parameters Deff = 1.0e-7*60e8; P = 1.5e-7*60e8; Kav = 0.3; Cp0 = 1; %uM alpha = 0.7; L1 = 4e-4; %1/min L2 = 7e-6; %1/min a = 5; %um R = 150; %um

%cm2/s->um2/min %cm/s->um2/min

% Mesh definition N = 50; dr = (R-a)/N; drh = dr/2; rm = a:dr:R; r = (a+drh):dr:(R-drh); s = zeros(N+1,1); for i = 1:N+1 s(i) = 2*pi*rm(i); end

%*L

v = zeros(N,1); for i = 1:N v(i) = pi*(rm(i+1)*rm(i+1)-rm(i)*rm(i)); end % c0 = zeros(N,1); tspan = [0 10 60 720 2880 6000]; [t,c] = ode15s(@func,tspan,c0); figure; plot(r,c(2,:),r,c(3,:),r,c(4,:),r,c(5,:),r,c(6,:)); aa1 = (c(4,1)*r(1)+c(4,N)*r(N))/2; aa2 = (c(5,1)*r(1)+c(5,N)*r(N))/2; for i = 2:(N-1) aa1 = aa1 + c(4,i)*r(i); aa2 = aa2 + c(5,i)*r(i); 204

%*L

end mean12 = aa1*dr*2/(R*R-a*a); mean48 = aa2*dr*2/(R*R-a*a); aa1 = 0; aa2 = 0; for i = 1:N aa1 = aa1 + (c(4,i) - mean12)^2; aa2 = aa2 + (c(5,i) - mean48)^2; end ISN1 = sqrt(aa1)/(mean12*sqrt(N*N-1)) ISN2 = sqrt(aa2)/(mean48*sqrt(N*N-1)) fid = fopen('data.txt', 'w'); fprintf(fid,'r, 10, 60, 720, 2880, 6000\n'); for i=1:N fprintf(fid,'%8.4f,%8.4f,%8.4f,%8.4f,%8.4f,%8.4f\n',r(i),c(2,i),c(3,i),c(4,i),c(5,i),c(6,i)); end fclose(fid);

function dcdt = func(t,c) global N dr s v Deff P Kav Cp0 alpha L1 L2 Cp = Cp0*(alpha*exp(-L1*t)+(1-alpha)*exp(-L2*t)); % dcdt = zeros(N,1); dcdt(1) = 1/v(1)*(P*(Cp-c(1)/Kav)*s(1)+Deff*(c(2)-c(1))/dr*s(2)); for i=2:N-1 iu=i-1; id=i+1; dcdt(i)= 1/v(i)*(Deff*(c(iu)-c(i))/dr*s(i)+Deff*(c(id)-c(i))/dr*s(id)); end dcdt(N) = 1/v(N)*(Deff*(c(N-1)-c(N))/dr*s(N));

Problem 15.8 function main global N dr s v Deff P Kav % Model Parameters Deff = 2.0e-7*60e8;

%cm2/s->um2/min 205

P = 1.5e-7*60e8; Kav = 0.3; c_init = 10; a = 5; R = 150;

%cm/s->um2/min % uM %um %um

% Mesh definition N = 50; dr = (R-a)/N; drh = dr/2; rm = a:dr:R; r = (a+drh):dr:(R-drh); s = zeros(N+1,1); for i = 1:N+1 s(i) = 2*pi*rm(i); end

%*L

v = zeros(N,1); for i = 1:N v(i) = pi*(rm(i+1)*rm(i+1)-rm(i)*rm(i)); end

%*L

% c0 = zeros(N,1); for i=1:N c0(i) = c_init; end tspan = [0 10 20 60 120 600]; [t,c] = ode15s(@func,tspan,c0); figure; plot(r,c(2,:),r,c(3,:),r,c(4,:),r,c(5,:),r,c(6,:)); fid = fopen('data.txt', 'w'); fprintf(fid,'r, 10, 20, 60, 120, 600\n'); for i=1:N fprintf(fid,'%8.4f,%8.4f,%8.4f,%8.4f,%8.4f,%8.4f\n',r(i),c(2,i),c(3,i),c(4,i),c(5,i),c(6,i)); end fclose(fid);

function dcdt = func(t,c) global N dr s v Deff P Kav % 206

dcdt = zeros(N,1); dcdt(1) = 1/v(1)*(-P*c(1)/Kav*s(1)+Deff*(c(2)-c(1))/dr*s(2)); for i=2:N-1 iu=i-1; id=i+1; dcdt(i)= 1/v(i)*(Deff*(c(iu)-c(i))/dr*s(i)+Deff*(c(id)-c(i))/dr*s(id)); end dcdt(N) = 1/v(N)*(Deff*(c(N-1)-c(N))/dr*s(N));

Problem 15.9 function main global N dr s v Deff P Kav k % Model Parameters Deff = 2.0e-7*60e8; P = 1.5e-7*60e8; Kav = 0.3; a = 5; R = 150; k = 0.5/60; c_init = 10;

%cm2/s->um2/min %cm/s->um2/min %um %um %1/hour->1/min %uM

% Mesh definition N = 50; dr = (R-a)/N; drh = dr/2; rm = a:dr:R; r = (a+drh):dr:(R-drh); s = zeros(N+1,1); for i = 1:N+1 s(i) = 2*pi*rm(i); end

%*L

v = zeros(N,1); for i = 1:N v(i) = pi*(rm(i+1)*rm(i+1)-rm(i)*rm(i)); end % c0 = zeros(N,1); for i=1:N 207

%*L

c0(i)= c_init; end tspan = [0 10 20 60 120 600]; [t,c] = ode15s(@func,tspan,c0); figure; plot(r,c(2,:),r,c(3,:),r,c(4,:),r,c(5,:),r,c(6,:)); fid = fopen('data.txt', 'w'); fprintf(fid,'r, 10, 20, 60, 120, 600\n'); for i=1:N fprintf(fid,'%8.4f,%8.4f,%8.4f,%8.4f,%8.4f,%8.4f\n',r(i),c(2,i),c(3,i),c(4,i),c(5,i),c(6,i)); end fclose(fid);

function dcdt = func(t,c) global N dr s v Deff P Kav k % dcdt = zeros(N,1); dcdt(1) = 1/v(1)*(-P*c(1)/Kav*s(1)+Deff*(c(2)-c(1))/dr*s(2))-k*c(1); for i=2:N-1 iu=i-1; id=i+1; dcdt(i)= 1/v(i)*(Deff*(c(iu)-c(i))/dr*s(i)+Deff*(c(id)-c(i))/dr*s(id))-k*c(i); end dcdt(N) = 1/v(N)*(Deff*(c(N-1)-c(N))/dr*s(N))-k*c(N);

208

Solution to Problems in Chapter 16, Section 16.6 16.1. The mass balance equation for the drug in the compartment is, V dC = QC0 − keVC dt

(S16.1.1)

The initial condition is, C = 0, at t = 0

(S16.1.2)

(1) The solution of Equation (S16.1.1) is, C=

QC0 ⎡1 − exp ( − ke t ) ⎤⎦ keV ⎣

(S16.1.3)

(2) The clearance of the drug is, ClB = Vke C / C = Vke

(3) The area under the curve (AUC) is infinite when the drug is infused continuously. (4) The concentration at the steady state Css is QC0/(keV). When ket > 5, C/Css > 0.99. Thus, the condition required for concentration to reach the steady state is t > 5/ke.

16.2. The mass balance equations of the drug in the central compartment and the peripheral compartments are, respectively, V1

dC1 = − k1V1C1 + k2V2 C2 − ke1V1C1 dt

(S16.2.1)

V2

dC2 = k1V1C1 − k2V2 C2 − ke 2V2 C2 dt

(S16.2.2)

The initial conditions are, C1 = C0, at t = 0

(S16.2.3)

C2 = 0, at t = 0

(S16.2.4)

where C0 is the drug concentration in the central compartment immediately after a bolus injection. It is equal to Dose/V1. Rearranging Equations (S16.2.1) and (S16.2.2) to eliminate C2, we have, d 2 C1 dt

2

+ (k1 + k2 + ke1 + ke 2 )

dC1 + ( k2 ke1 + k1ke 2 + ke1ke 2 ) C1 = 0 dt

(S16.2.5)

The initial conditions of Equation (S16.2.5) can be derived from Equations (S16.2.1) (S16.2.3), and (S16.2.4). They are, C1 = C0, dC1 = −(k1 + ke )C0 , dt

at t = 0

(S16.2.5a)

at t = 0

(S16.2.5b)

(1) Solving Equation (S16.2.5) gives, C1 = C0 [α exp(−λ1t ) + (1 − α ) exp(−λ2 t ) ]

where 209

(S16.2.6)

(k1 + k2 + ke1 + ke 2 ) + (k1 + k2 + ke1 + ke 2 ) 2 − 4 ( k2 ke1 + k1ke 2 + ke1ke 2 )

λ1 =

(k1 + k2 + ke1 + ke 2 ) − (k1 + k2 + ke1 + ke 2 ) 2 − 4 ( k2 ke1 + k1ke 2 + ke1ke 2 )

λ2 = α=

(S16.2.7)

2

(S16.2.8)

2 k1 + ke − λ2 λ1 − λ2

(S16.2.9)

Substituting Equation (S16.2.6) into Equation (S16.2.1) yields, C2 =

V1 ⎛ dC1 ⎞ + k1C1 + ke1C1 ⎟ ⎜ dt k2V2 ⎝ ⎠

(S16.2.10)

VC = 1 0 ⎡⎣( k1 + ke1 − λ1 ) α exp ( −λ1t ) + ( k1 + ke1 − λ2 ) (1 − α ) exp ( −λ2 t ) ⎤⎦ k2V2

(2) The half lives of the two phases in the bi-exponential concentration profile are t1/2λ1 = 0.693 / λ1

(S16.2.11)

t1/2λ2 = 0.693 / λ2

(S16.2.12)

(3) The drug is cleared from both compartments. Thus the clearance is, ClB = (V1ke1C1 + V2 ke 2 C2 ) / C1

(S16.2.13)

which is a function of time. (4) The AUC of the drug in the central compartment is, ∞

∞

0

0

AUC = ∫ C1dt = ∫ C0 ⎣⎡α exp ( −λ1t ) + (1 − α ) exp ( −λ2 t ) ⎦⎤ dt ⎛ α 1−α ⎞ = C0 ⎜ + ⎟ λ2 ⎠ ⎝ λ1

(S16.2.14)

16.3. The change in CD is due to the binding of the free drug to tissues, the release of the bound drug, and the clearance of the free drug from the compartment, whereas the change in CB is solely due to the reversible binding. Therefore, the mass balance equations of CD and CB are, dCD = − k f CD + k r C B − ke C D dt

(S16.3.1)

dCB = k f CD − k r C B dt

(S16.3.2)

The initial conditions are, CD = C0,

at t = 0

(S16.3.3)

CB = 0,

at t = 0

(S16.3.4)

Here C0 is the initial concentration immediately after the bolus injection. Rearranging Equations (S16.3.1) and (S16.3.2) to eliminate CB, we have, 210

d 2 CD dt

2

+ ( k f + k r + ke )

dCD + k r ke C D = 0 dt

(S16.3.5a)

The initial conditions of Equation (S16.3.5) are, CD = C0, dCD = −(k f + ke )C0 , dt

at t = 0

(S16.3.5b)

at t = 0

(S16.3.5c)

Solving Equation (S16.3.5) gives, CD = C0 [α exp(−λ1t ) + (1 − α ) exp(−λ2 t ) ]

(S16.3.6)

where λ1 = λ2 =

α=

( k f + k r + k e ) + ( k f + k r + ke ) 2 − 4 k r ke

(S16.3.7)

2 ( k f + k r + ke ) − ( k f + k r + ke ) 2 − 4 k r ke

(S16.3.8)

2 k f + ke − λ2

(S16.3.9)

λ1 − λ2

Substituting Equation (S16.3.6) into Equation (S16.3.1), we have, CB = =

1 kr

⎛ dCD ⎞ ⎜ dt + k f CD + ke CD ⎟ ⎝ ⎠

(

)

(

)

C0 ⎡ k f + ke − λ1 α exp ( −λ1t ) + k f + ke − λ2 (1 − α ) exp ( −λ1t ) ⎤ ⎦ kr ⎣

(S16.3.10)

16.4. The mass balance of the antibody in the tumor tissue is, ∂C t = ρ ( k Cp − k e Ca φ ) ∂t where Ct is the total concentration of the antibody, C t = φ ( Ca + CaG ) 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 per unit volume of the interstitial space. The volume fraction of the interstitial space is φ. To solve Equation (S16.4.1), we need to determine how Ct depends on Ca. Since the binding between the antibody and the tumor-associated antigen is at equilibrium, CaG = K a Ca CG In addition, the total concentration of the antigen (free plus bound) does not change during the reaction. Thus, CG + CaG = G 0 Substituting Equations (S16.4.3) and (S16.4.4) into Equation (S16.4.2) yields, 211

⎛ K Ca G 0 ⎞ C t = φ ⎜ Ca + a ⎟ 1 + K a Ca ⎠ ⎝ Substituting Equation (S16.4.5) into Equation (S16.4.1) gives, ⎡ ⎤ a G0Ka ⎢1 + ⎥ ∂C = ρ ⎛⎜ k C p − k e Ca ⎞⎟ 2 ⎢ (1 + K Ca ) ⎥ ∂t ⎝φ ⎠ a ⎣ ⎦ The initial concentration is, at t = 0 Ca = 0, The plasma concentration of the antibody is, Cp = Cp0 ⎡⎣α exp ( −λ1t ) + (1 − α ) exp ( −λ 2 t ) ⎤⎦

Equations (S16.4.6) through (S16.4.8) are solved numerically using the MATLAB. (The MATLAB code is attached at the end of this chapter.) The concentration profiles of the drug are shown in Fig. S16.4.1. Figure S16.4.1 120

CaG Ca

100

Concentrations (nM)

80

60

40

Cp 20

0

0

2000

4,000 Time (min)

MATLAB CODES Problem 16.4 function main global cp0 alpha lamda1 lamda2 k ke ka fi g0 ro cp0=110; % nM alpha=0.459; lamda1=1.11e-2; % l/min lamda2=1.38e-4; % l/min 212

6,000

8,000

k=0.6; ke=0.9; ka=1; fi=0.243; g0=110; ro=1e-3;

% ul/min/g % ul/min/g % l/nm % nM % g/ul

c0=0; tspan=[0 7200]; [t,ca]=ode23(@func,tspan,c0); [m,n]=size(t); cag = zeros(m,1); cp=zeros(m,1); for i=1:m cag(i)=ka*ca(i)*g0/(1+ka*ca(i)); cp(i)=cp0*(alpha*exp(-lamda1*t(i))+(1-alpha)*exp(-lamda2*t(i))); end plot(t,ca,t,cag,t,cp) fid=fopen('data.txt','w'); for i=1:m fprintf(fid,'%7.3f,%7.3f,%7.3f,%7.3f\n',t(i),ca(i),cag(i),cp(i)); end fclose(fid); function dcdt=func(t,c) global cp0 alpha lamda1 lamda2 k ke ka fi g0 ro cp = cp0*(alpha*exp(-lamda1*t)+(1-alpha)*exp(-lamda2*t)); dcdt=ro*(k*cp/fi-ke*c)/(1+g0*ka/((1+ka*c)^2)); 16.5. The mass balance equations in the three compartments are, V1

dC1 = − ( k12 + k13 + ke ) V1C1 + k21V2 C2 + k31V3C3 dt

(S16.5.1)

V2

dC2 = k12V1C1 − k21V2 C2 dt

(S16.5.2)

V3

dC3 = k13V1C1 − k31V3C3 dt

(S16.5.3)

The initial conditions in the three compartments are, C1 = C0, at t = 0

(S16.5.4)

C2 = 0, at t = 0

(S16.5.5)

C3 = 0, at t = 0

(S16.5.6) 213

where C0 is the concentration of the drug in the first compartment immediately after the bolus injection. Similar to the two-compartment model, we can eliminate C2 and C3 in Equations (S16.5.1) through (S16.5.3). The resulting equation is, d 3C1 dt

3

+ A1

d 2 C1 dt

2

+ A2

d 2 C1 dt2

(S16.5.7)

+ A3C1 = 0

where A 1 = k 12 + k 13 + k e + k 21 + k 31

(S16.5.8)

A 2 = k 21 k 13 + k 21 k e + k 31 k 12 + k 31 k e + k 21 k 31

(S16.5.9)

A 3 = k 21 k 31 k e

(S16.5.10)

The initial conditions of Equation (S16.5.7) are, C1 = C0, at t = 0

(S16.5.11)

d C1 = −(k 12 + k 13 + k e )C 0 , at t = 0 dt

(S16.5.12)

[

]

d 2 C1 2 = (k 12 + k 13 + k e ) + k 12 k 21 + k 13 k 31 C 0 , at t = 0 2 dt

(S16.5.13)

The three characteristic values (λ1, λ2 and λ3) of Equation (S16.5.7) are obtained by solving the following equation, X 3 + A1X 2 + A 2 X + A 3 = 0

(S16.5.14)

The solution of Equation (S16.5.7) in terms of λ1, λ2, λ3, and the kinetic constants is, C1 = C0 ⎡⎣α exp ( λ1t ) + β exp ( λ2 t ) + (1 − α − β ) exp ( λ3t ) ⎤⎦

(S16.5.15)

where

( k12 + k13 + ke + λ3 )( λ2 + λ3 ) + ⎡⎣( k12 + k13 + ke )2 + k12 k21 + k13 k31 − λ32 ⎤⎦ α= (S16.5.16) ( λ1 − λ3 ) ( λ1 − λ2 ) ( k12 + k13 + ke + λ3 )( λ1 + λ3 ) + ⎡⎣( k12 + k13 + ke )2 + k12 k21 + k13 k31 − λ32 ⎤⎦ β= (S16.5.17) ( λ2 − λ3 ) ( λ2 − λ1 ) 16.6. With the alternative approach to modeling the process of methotrexate transport in the hepatic duct, the only change in the PBPK analysis is to replace Equations (16.4.10a, b, and c) by the following one. (S16.6.1) r3 = r(t – L/v) 2 where v = 4qB/(πd ), which is the flux of bile fluid, qB is the total bile flow rate, and d and L are the diameter and length of the duct, respectively. Equation (S16.5.17) means that r3 is equal to the rate of liver excretion with a time delay of L/v which is the time required for bile fluid to flow through the hepatic duct. Other equations for concentrations of methotrexate in different 214

organs described in Section 16.4.2 are not changed. These equations are solved numerically using the MATLAB. (The MATLAB code is attached at the end of this chapter.) The results are shown in Figure S16.5.1b. For comparisons, Figure 16.6 is re-plotted here as Figure S16.6.1a. The difference between Figure S16.5.1a and Figure S16.5.1b is mainly caused by the bile flow rate. In this homework problem, qB = 20 ml/hr based on the data in the literature. However, Figure 16.6 is obtained with qB = kL/KL = 200 ml/min (see Table 16.1). If qB is also equal to 200 ml/min, the concentration profiles are shown in Figure S16.6.1c. These concentration profiles are very similar to those shown in Figure S16.2a. The only noticeable difference is the concentration in the gut lumen during the first 100 min. The concentration increases faster in the alternative model, compared with the old one, presumably due to a reduction in the holding time of methotrexate in the hepatic duct.

Figure S16.6.1. (a) A re-plot of Figure 16.6. (b) PBPK analysis based on the alternative model of methotrexate transport in the bile duct with qB = 20 ml/hr. (b) The same PBPK analysis with qB = 200 ml/min. The dose of methotrexate in all simulations is 70 mg.

16.7. The concentration distributions of methotrexate in a mouse, a rat, a dog, and a monkey are determined by solving the equations for its concentrations in different organs described in Section 16.4.2. The solutions are obtained numerically using the MATLAB. (The MATLAB code is attached at the end of this chapter.) The results are shown in Fig. S16.5.1. The concentration profiles in the plasma, the muscle and the liver are similar in all species. The accumulation of methotrexate in the gut lumen in the mouse and the rat is more significant than that in the monkey and the dog. On the contrary, the concentrations of methotrexate in the kidney in the monkey and the dog are higher than those in the mouse and the rat.

215

Figure S16.7.1. The concentration distributions of methotrexate in (a) Mouse, (b) Rat, (c) Monkey and (d) Dog. The doses of methotrexate in all simulations are fixed at 1mg/kg body weight. P, M, K, L and GL denote plasma, muscle, kidney, liver and gut lumen, respectively.

16.8. The concentration distribution of methotrexate in a mouse is determined by the same MATLAB code as that for Problem 16.7. The only difference is the values of model constants. 216

The simulated concentration profiles of methotrexate are shown in Figure S16.4. They are based on either the predicted parameters or experimentally measured parameters.

Figure S16.8.1. The numerical solutions of the concentration distributions of methotrexate in a mouse based on (a) experimentally measured parameters in Table 16.3 plus data in Table 16.4 or (b) predicted parameters in Table 16.3 plus data in Table 16.4. The doses of methotrexate in both cases are 1 mg/kg body weight. P, M, K, L and GL denote plasma, muscle, kidney, liver and gut lumen, respectively. The shapes of concentration profiles in Panel (a) are similar to those in Panel (b); and the concentrations during the first few minutes are similar between the two panels in corresponding organs. However, the concentrations in all compartments in Panel (b) are significantly lower than those in Panel (a) in later time points, due to the higher value of predicted clearance rate in the kidney. The AUC during the first 6 hours is listed in the table. Model Plasma Based on predicted 43.3* parameters Based on measured 73.8 parameters * The unit of AUC is μg/ml×min.

Muscle

Kidney

Liver

Gut Lumen

7.14

107.3

458.8

545.2

12.0

190.2

726.9

962.3

MATLAB CODES Problem 16.4 217

function main global cp0 alpha lamda1 lamda2 k ke ka fi g0 ro cp0 = 100; % nM alpha = 0.459; lamda1 = 1.11e-2; % 1/min lamda2 = 1.38e-4; % 1/min k = 0.5; % ul/min/g ke = 0.8; % ul/min/g ka = 1; %1/nM fi = 0.243; g0 = 100; % nM ro = 1e-3; %g/ul c0 = 0; tspan = [0 7200]; [t,ca] = ode23(@func,tspan,c0); [m, n] = size(t); cag = zeros(m,1); cp = zeros(m,1); for i =1:m cag(i) = ka*ca(i)*g0/(1+ka*ca(i)); cp(i) = cp0*(alpha*exp(-lamda1*t(i))+(1-alpha)*exp(-lamda2*t(i))); end plot(t,ca, t, cag, t, cp) fid = fopen('data.txt', 'w'); for i=1:m fprintf(fid,'%7.3f, %7.3f, %7.3f, %7.3f\n', t(i), ca(i), cag(i), cp(i)); end fclose(fid);

function dcdt = func(t,c) global cp0 alpha lamda1 lamda2 k ke ka fi g0 ro cp = cp0*(alpha*exp(-lamda1*t)+(1-alpha)*exp(-lamda2*t)); dcdt = ro*(k*cp/fi-ke*c)/(1+g0*ka/((1+ka*c)^2));

Problem 16.6 % Physiologically Based Pharmacokinetic Analysis of Methotrexate clear all; global para n ct m td 218

h = 0.1; tend = 400; N = fix(tend/h)+1;

% delta t (min) % last time point (min) % total steps in time

%Initialization para = zeros(22,1); %Volume of different compartment (ml) para(1) = 3000; %plasma para(2) = 35000; %muscle para(3) = 280; %kidney para(4) = 1350; %liver para(5) = 2100; %G.I. tract para(6) = 2100; %gut lumen %Plasma flow rate (ml/min) para(7) = 420; %muscle para(8) = 700; %kidney para(9) = 800; %liver para(10) = 700; %G.I. tract %Tissue/plasma equilibrium distribution rationfor linear binding para(11) = 0.15; %muscle para(12) = 3.0; %kidney para(13) = 3.0; %liver para(14) = 1.0; %G.I. tract %Kidney clearance (ml/min) para(15) = 190; %Bile secretion parameters para(16) = 20/60; % clearance rate ml/min % para(17) = 10; % holding time %Gut-lumen parameters para(18) = 0.001; para(19) = 1900; para(20) = 200; %Dose para(21) = 70; para(22) = 70;

%kf %kg %KG % mg %Bodyweight (kg)

% modeling convection in the bile duct L = 20; d = 1; m = round(L*3.1415926*d*d/(4*para(16)*h)) 219

% bile duct length, cm % bile duct diameter, cm %time delay in r3

td = m*h;

%time delay in r3

c0 = zeros(9,1); c0(1) = para(21)*1000/para(1);

% microgram/ml

% % % % % % % % %

1. Plasma 2. Muscle 3. Kidney 4. Liver 5. Gut tissue 6. C1 7. C2 8. C3 9. C4

%initialize ct = zeros(9,N+1); t = zeros(1,N+1); k1 = zeros(9,1); k2 = zeros(9,1); k3 = zeros(9,1); k4 = zeros(9,1); n = 0; while t <= tend n = n + 1; k1 = h*f16_6(c0, t(n)); k2 = h*f16_6(c0+k1/2,t(n)+h/2); k3 = h*f16_6(c0+k2/2,t(n)+h/2); k4 = h*f16_6(c0+k3,t(n)+h); ct(:,n+1) = c0 + (k1 + 2*k2 + 2*k3 + k4)/6; c0 = ct(:,n+1); t(n+1) = n*h; end cGL = ct(6,:) + ct(7,:) + ct(8,:) + ct(9,:); semilogy(t, ct(1,:), t, ct(2,:), t, ct(3,:), t, ct(4,:), t, cGL) ylim([0.001, 100]) fid = fopen('datan.txt', 'w'); for i=1:2:N+1 fprintf(fid,'%7.4f, %7.4f, %7.4f, %7.4f, %7.4f, %7.4f\n', t(i), ct(1,i), … ct(2,i), ct(3,i), ct(4,i), cGL(i)); end fclose(fid); 220

function y = f16_6(c,t) global para n ct m td y = zeros(9,1); y(1) = (para(7)*c(2)/para(11) + para(8)*c(3)/para(12) + para(9)*c(4)/para(13)... - (para(9)+para(8)+para(7))*c(1))/para(1); y(2) = para(7)*(c(1) - c(2)/para(11))/para(2); y(3) = (para(8)*(c(1) - c(3)/para(12)) - para(15)*c(3)/para(12)) / para(3); r = para(16) * c(4) / para(13); y(4) = ((para(9) - para(10))*(c(1) - c(4)/para(13)) + para(10) * (c(5)/para(14)… - c(4)/para(13)) - r) / para(4); y(5) = (para(10)*(c(1)-c(5)/para(14)) + 0.25*para(19)*(c(6)/(para(20)+c(6))... + c(7)/(para(20)+c(7)) + c(8)/(para(20)+c(8)) + c(9)/(para(20)+c(9))))/para(5); if t > td r3 = para(16) * ct(4,(n - m)) / para(13); else r3 = 0; end y(6) = (r3 - para(18)*para(6)*c(6) - 0.25*para(19)*c(6)/(para(20)+c(6)))*4/para(6); y(7) = (para(18)*para(6)*(c(6)-c(7)) - 0.25*para(19)*c(7)/(para(20)+c(7)))*4/para(6); y(8) = (para(18)*para(6)*(c(7)-c(8)) - 0.25*para(19)*c(8)/(para(20)+c(8)))*4/para(6); y(9) = (para(18)*para(6)*(c(8)-c(9)) - 0.25*para(19)*c(9)/(para(20)+c(9)))*4/para(6);

Problem 16.7 % Physiologically Based Pharmacokinetic Analysis of Methotrexate clear all; global para n ct h = 0.1; tend = 400; N = fix(tend/h)+1;

% delta t (min) % last time point (min) % total steps in time

%Initialization para = zeros(22,1); %Volume of different compartment (ml) para(1) = 3000; %plasma 221

para(2) = 35000; para(3) = 280; para(4) = 1350; para(5) = 2100; para(6) = 2100;

%muscle %kidney %liver %G.I. tract %gut lumen

%Plasma flow rate (ml/min) para(7) = 420; %muscle para(8) = 700; %kidney para(9) = 800; %liver para(10) = 700; %G.I. tract %Tissue/plasma equilibrium distribution rationfor linear binding para(11) = 0.15; %muscle para(12) = 3.0; %kidney para(13) = 3.0; %liver para(14) = 1.0; %G.I. tract %Kidney clearance (ml/min) para(15) = 190; %Bile secretion parameters para(16) = 200; para(17) = 10; %Gut-lumen parameters para(18) = 0.001; para(19) = 1900; para(20) = 200; %Dose para(21) = 70; para(22) = 70;

% clearance rate ml/min % holding time %kf %kg %KG % mg % Bodyweight (kg)

c0 = zeros(12,1); c0(1) = para(21)*1000/para(1);

% microgram/ml

% 1. Plasma % 2. Muscle % 3. Kidney % 4. Liver % 5. r1 % 6. r2 % 7. r3 % 8. Gut tissue % 9. C1 % 10. C2 222

% 11. C3 % 12. C4 %initialize ct = zeros(12,N+1); t = zeros(1,N+1); k1 = zeros(12,1); k2 = zeros(12,1); k3 = zeros(12,1); k4 = zeros(12,1); n = 0; while t <= tend n = n + 1; k1 = h*f16_7(c0, t(n)); k2 = h*f16_7(c0+k1/2,t(n)+h/2); k3 = h*f16_7(c0+k2/2,t(n)+h/2); k4 = h*f16_7(c0+k3,t(n)+h); ct(:,n+1) = c0 + (k1 + 2*k2 + 2*k3 + k4)/6; c0 = ct(:,n+1); t(n+1) = n*h; end cGL = ct(9,:) + ct(10,:) + ct(11,:) + ct(12,:); semilogy(t, ct(1,:), t, ct(2,:), t, ct(3,:), t, ct(4,:), t, cGL) ylim([0.001, 100]) fid = fopen('datan.txt', 'w'); for i=1:2:N+1 fprintf(fid,'%7.4f, %7.4f, %7.4f, %7.4f, %7.4f, %7.4f\n', t(i), ct(1,i), … ct(2,i), ct(3,i), ct(4,i), cGL(i)); end fclose(fid);

function y = f16_7(c,t) global para n ct y = zeros(12,1); y(1) = (para(7)*c(2)/para(11) + para(8)*c(3)/para(12) + para(9)*c(4)/para(13)... - (para(9)+para(8)+para(7))*c(1))/para(1); y(2) = para(7)*(c(1) - c(2)/para(11))/para(2); y(3) = (para(8)*(c(1) - c(3)/para(12)) - para(15)*c(3)/para(12)) / para(3); r = para(16) * c(4) / para(13); y(4) = ((para(9) - para(10))*(c(1) - c(4)/para(13)) + para(10) * (c(8)/para(14) … 223

- c(4)/para(13)) - r) / para(4); y(5) = (r - c(5))/para(17); y(6) = (c(5) - c(6))/para(17); y(7) = (c(6) - c(7))/para(17); y(8) = (para(10)*(c(1)-c(8)/para(14)) + 0.25*para(19)*(c(9)/(para(20)+c(9))... + c(10)/(para(20)+c(10)) + c(11)/(para(20)+c(11)) … + c(12)/(para(20)+c(12))))/para(5); y(9) = (c(7) - para(18)*para(6)*c(9) - 0.25*para(19)*c(9)/(para(20)+c(9)))*4/para(6); y(10) = (para(18)*para(6)*(c(9)-c(10)) - 0.25*para(19)*c(10)/(para(20)+c(10)))*4/para(6); y(11) = (para(18)*para(6)*(c(10)-c(11)) … - 0.25*para(19)*c(11)/(para(20)+c(11)))*4/para(6); y(12) = (para(18)*para(6)*(c(11)-c(12)) … - 0.25*para(19)*c(12)/(para(20)+c(12)))*4/para(6);

224

Solution to Problems in Chapter 17, Section 17.10 17.1. In words, the conservation relation is: ⎡ Rate of Energy ⎤ ⎡ Net Rate of Energy ⎤ ⎡ Rate of Work ⎤ ⎡ Rate of Energy ⎤ ⎢ Accumulation ⎥ = ⎢ Transfer Across ⎥ + ⎢ Done on the ⎥ + ⎢ Production ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ within the system ⎥⎦ ⎢⎣System Surfaces ⎥⎦ ⎢⎣System ⎥⎦ ⎢⎣ within the system ⎥⎦

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))

)

(

∂Eˆ = ex x − ex x +Δx ΔyΔz + ey − ey Δx Δz + ez z − ez z +Δz ΔxΔy + Wt + Q*p ΔxΔyΔz +Δ y y y ∂t Dividing by the volume element ΔxΔyΔz, taking the limit as the volume goes to zero and using the definition of the derivative yields: Δx ΔyΔz ρ

(

)

ρ

(

∂e ∂e ∂e ∂Eˆ = − x − y − z + Wt + Q*p ∂t ∂x ∂y ∂x

)

(

)

(S17.1.1)

Using the definition of the divergence of a vector (Equation (A.3.10), Equation (S17.1.1) becomes ρ

∂Eˆ = −∇i e + Wt + Q*p ∂t

Using the definition of e, Equation (17.2.4), the divergence of e is: ∇i e = ∇i( ρ Eˆ v + q ) = ρ Eˆ ∇iv + v i∇ρ Eˆ + ∇i q

(S17.1.2) (S17.1.3)

For an incompressible fluid, ∇i v = 0 and ρ is a constant. As a result, Equation (S17.1.3) reduces to: (S17.1.4) ∇i e = ∇i( ρ Eˆ v + q ) = ρ vi∇Eˆ + ∇i q Inserting Equation (S17.1.4) into Equation (S17.1.2) ρ

∂Eˆ = − ρ vi∇Eˆ − ∇i q + Wt + Q*p ∂t

(S17.1.5)

Moving both terms with the system energy to the left hand side of Equation (S17.1.5) yields: ⎛ ∂Eˆ

⎞ + v i∇Eˆ ⎟ = −∇i q + Wt + Q*p ⎝ ∂t ⎠

ρ⎜

(S17.1.5)

Lastly, the total rate of work represents work done by fluid stresses ( ∇i(σ iv ) = −∇i( pv ) + ∇i(τ iv ) ), body forces ( ∇i v ) and other types of mechanical work by the body ( W ). Inserting these terms into Equation (S17.1.5) yields Equation (17.2.6) ⎛ ∂Eˆ

⎞ + v i∇Eˆ ⎟ = −∇i q − ∇i( pv ) + ∇ i(τ iv ) + F iv + W + Q*p ⎝ ∂t ⎠

ρ⎜

17.2. The work is:

(S17.1.6)

W = ∫ F i ndx = ∫ Fdx since the force and unit outward normal are both

positive. Normally, a protein is present in a specific conformation which is much less than the maximum length, know as the contour length, L. The contour length is the length of the polymer if each chain element were aligned along a line. 225

Substituting for the wormlike chain model: −2 −1 ⎛ k BT ⎞ x ⎛ ⎛ k BT ⎞ ⎛ x⎞ x⎞ x⎞ x2 ⎞ ⎛ ⎛ 0.25 ⎜ 1 − ⎟ − 0.25 + ⎟dx = ⎜ 0.25L ⎜ 1 − ⎟ − 0.25 x + W =⎜ ⎟ ⎜ L ⎟⎟ ∫ ⎜⎜ ⎜ L ⎟⎟ ⎜⎜ 2 L ⎟⎠ L ⎟⎠ ⎝ L⎠ ⎝ L⎠ ⎝ p ⎠0⎝ ⎝ p ⎠⎝ −1 ⎛ k BTL ⎞ ⎛ x⎞ x x2 ⎞ ⎛ 0.25 ⎜ 1 − ⎟ − 0.25 + 2 ⎟ W =⎜ ⎜ L ⎟⎟ ⎜⎜ L 2 L ⎟⎠ ⎝ L⎠ ⎝ p ⎠⎝ This result is shown in the graph below. At low extensions, the relation is linear. However, as x approaches L, the work increases dramatically. The flexibility of the polymer arises from the arrangement of the chains. As the polymer elongates, more work must be done to extend the polymer to overcome the tendency for the chains to move freely and to extend each element.

6 5 4 3 2 1 0 0

0.2

0.4

0.6

0.8

x/L

17.3. Note: The equation listed in the problem statement should be: 2

⎛ dv ⎞ Φ v = τ : ∇v = μ ⎜ z ⎟ ⎝ dr ⎠ The shear stress tensor for a Newtonian fluid is:

226

1

(

τ = μ ∇v + ( ∇v )

(

T

)

(S17.3.1)

Φ v = τ : ∇ v = μ ∇v + ( ∇v )

T

) : ∇v

(S17.3.2)

Using the summation convention for vectors and tensors ⎛ 3 3 ⎛ ∂v ∂v ⎞ ⎞ ⎛ 3 3 ∂v ⎞ T μ ∇v + ( ∇v ) : ∇v = μ ⎜ ∑∑ ⎜ i + j ⎟ ei e j ⎟ : ⎜ ∑∑ k ek el ⎟ ⎜ i =1 j =1 ⎜ ∂x j ∂xi ⎟ ⎟ ⎝ k =1 l =1 ∂xl ⎠ ⎝ ⎠ ⎝ ⎠ Since e i e j : e k e l = ( e j i e k ) ( e i i e l ) = δ jk δ il , Equation (S17.3.3) becomes:

(

)

⎛ ⎛ ∂v ⎞ ⎛ ∂v ⎞ ⎛ ∂v ⎞2 ⎞ μ ∇v + ( ∇v ) : ∇v = μ ∑∑ ⎜ ⎜ i ⎟ ⎜ j ⎟ + ⎜ j ⎟ ⎟ ⎜ ⎟ i =1 j =1 ⎜ ⎝ ∂x j ⎠ ⎝ ∂xi ⎠ ⎝ ∂xi ⎠ ⎟⎠ ⎝ For fully developed steady, laminar flow in a cylindrical tube of radius R, ⎛ ∂v ∂v ⎞ τ = μ ⎜ z + r ⎟ er ez ∂z ⎠ ⎝ ∂r Using the symmetry property of the shear stress, τij = τji: ∂v ⎛ ∂v ⎞ ∂v Φ v = μ ⎜ z er ez + r ez er ⎟ : z er ez ∂z ⎝ ∂r ⎠ ∂r

(

T

)

3

3

∂v ⎛ ∂v ⎞ ∂v ⎛ ∂v ∂v ⎞ ⎛ ∂v ⎞ = μ ⎜ z er ez + z ez er ⎟ : z er ez = μ ⎜ z z ⎟ = μ ⎜ z ⎟ ∂r ⎝ ∂r ⎠ ∂r ⎝ ∂r ∂r ⎠ ⎝ ∂r ⎠

2

(S17.3.3)

(S17.3.4)

(S17.3.5)

For laminar flow in a tube ⎛ r2 ⎞ (S17.3.6a) v z = 2 vz ⎜1 − 2 ⎟ ⎝ R ⎠ ∂v z 2 v z r = (S17.3.6b) ∂r R2 The velocity gradient is maximum at r = R. Thus, the maximum value of viscous dissipation is: 2

⎛ 4 v z ⎞ 16 µ v z Φv = µ⎜ = ⎜ R ⎟⎟ R2 ⎝ ⎠

2

(S17.3.7)

In terms of flow rate Φv

max

=

For the data given: Φv Φv

max

max

=

16 µQ 2 π 2 R6

(S17.3.8)

(

)(

16 0.01 g cm -1 s-1 83.3 cm 3 s-1

π (1.5 cm ) 2

6

= 2.22 Pa s-1 = 2.22 J m -3 s-1 = 2.22 W m -3

227

)

2

= 22.22 g cm -1 s-3 = 2.22 kg m -1 s-3

To compute the maximum heating in blood arising from viscous dissipation, assume steady radial conduction with viscous dissipation. From Equations (17.2.8), (17.2.9), (17.2.12) and (S17.3.6b), the following result is obtained. k d ⎛ dT ⎞ 16 µQ 2r 2 (S17.3.9) r = −Φ = − v π 2 R8 r dr ⎜⎝ dr ⎟⎠ The boundary conditions are that for r = 0, the flux is zero and at r = R, T = T0. Integrating Equation (S17.3.9) once yields:

dT 4 µQ 2r 3 C =− + dr kπ 2 R8 r

(S17.3.10)

From the boundary condition at r = 0, C = 0. Integrating Equation (S17.3.10) yields: Φ v max r 4 µQ 2 r 4 T = − 2 8 + C2 = − + C2 (S17.3.11) kπ R kR 2 From the boundary condition at r = R, C2 is C2 = T0 +

Φv

max

R2

(S17.3.12)

k

The temperature profile is: Φv

R2 ⎛ r4 ⎞ 1 T = T0 + − ⎜ 4 ⎟ k ⎝ R ⎠ The maximum temperature difference occurs between r = 0 and r = R: Φ v max R 2 ΔTmax = T ( r = 0 ) − T0 = k max

(S17.3.13)

(S17.3.14)

For the value of the viscous dissipation obtained above and the thermal conductivity of blood (Table 17.2):

( 2.22 W m ) ( 0.015 m ) = -3

ΔTmax

0.642 W m -1K -1

2

= 0.00078 K

(S17.3.15)

Thus, viscous dissipation has a very minor effect on the temperature of blood and can be neglected.

17.4. For steady conduction for a spherical surface of radius R, Equation (17.2.14c) simplifies to: k d ⎛ 2 dT ⎞ (S17.4.1) ⎜r ⎟=0 r 2 dr ⎝ dr ⎠ The boundary conditions are at r = R, T = T0 and as r —> ∞, T = T∞. Integrating equation (S17.4.1) twice yields: T=

C1 + C2 r

(S17.4.2) 228

From the boundary condition as r —> ∞, C2 = T∞. At r = R C1 = ( T0 − T∞ ) R

(S17.4.3)

The temperature profile is: T = ( T0 − T∞ )

R + T∞ r

(S17.4.4)

To obtain the Nusselt number, compute the flux at r = R and apply the definition of the heat transfer coefficient:

(T − T ) dT = k 0 ∞ = hm ( T0 − T∞ ) dr r = R R The heat transfer coefficient for conduction is: k k hm = = 2 R D Using this result in the definition of the Nusselt number: −k

Num =

hm D kD =2 =2 k kD

(S17.4.5)

(S17.4.6)

(S17.4.7)

17.5. The definition of β is given by Equation (17.4.7) 1 ⎛ ∂V ⎞ β= ⎜ (S17.5.1) V ⎝ ∂T ⎟⎠ P From the ideal gas relationship, PV = nRT. For a fixed number of moles, V=nRT/P and the derivative in Equation (S17.5.1) is: nR ⎛ ∂V ⎞ ⎜ ∂T ⎟ = P ⎝ ⎠P

(S17.5.2)

Thus, nR 1 = (S17.5.3) VP T since T = PV/nR for an ideal gas. 17.6. For this problem, assume unsteady conduction in a tissue of thickness 2L. Based upon analogy with unsteady diffusion in a region of half thickness of L, the time to reach steady state is 2L2/α. While specific thermal diffusivities for tissue are not provided in Table 17.2, a reasonable value, between water and fat, is 1.1 x 10-7 m2 s-1. For the half-thickness of 125 µm = 1.25 x 10-4 m, the time to reach steady state is 0.284 s. So, one would expect uniform temperatures in well perfused tissues.

β=

17.7. Note: The phase change during freezing is discussed in Section 17.3.4, not Section 17.3.3. The rate of growth of the ice front is

dX . X is given by Equation (17.3.26b). Thus, dt

229

α dX =C S (S17.7.1) dt t C is dimensionless and is provided by solving Equation (17.3.31) or Equation (17.3.33). Values of C are tabulated in Table 17.3 for several different values of Tm-T0 and αS is given in Table 17.2 as 1.06 x 10-6 m2 s-1. For a value of Tm-T0 =10 C, C = 0.183 and the derivative in Equation (S17.7.1) is (1.8448 x 104 -1/2 )t m s-1. 17.8. This problem is a modification of the problem presented in Example 6.6. Thus, Equation (6.7.25) applies for the distribution of vapor concentration in a column of height δ. d ⎛ 1 dx ⎞ (S17.8.1) ⎜ ⎟=0 dy ⎝ 1 − x dy ⎠ The boundary conditions are that, at y = h, x = xa which is the vapor pressure at the given temperature and pressure. At y = h + δ, x = xs, the relative humidity in the air. Integrating Equation (S17.8.1) once yields: dx = C1 (1 − x ) (S17.8.2) dy Integrating again, (S17.8.3) ln (1 − x ) = −C1 y + C2 Applying the boundary conditions: ln (1 − x s ) = −C1h + C2

ln (1 − x a ) = −C1 ( h + δ ) + C2

Subtracting (S17.8.4b) from Equation (S17.8.4a) 1 ⎛ 1 − xa ⎞ C1 = ln ⎜ ⎟ δ ⎝ 1 − xa ⎠ Inserting Equation (S17.8.4c) in Equation (S17.8.4b) and solving for C2 yields; ⎛ 1 − xs ⎞ ⎛ h + δ ⎞ ln (1 − x a ) = − ln ⎜ ⎟⎜ ⎟ + C2 ⎝ 1 − xs ⎠ ⎝ δ ⎠ ⎛ 1 − xs ⎞ ⎛ h + δ ⎞ C2 = ln (1 − x a ) + ln ⎜ ⎟⎜ ⎟ ⎝ 1 − xa ⎠ ⎝ δ ⎠

(S17.8.4a) (S17.8.4b) (S17.8.4c)

(S17.8.4d)

(S17.8.4d)

The solution is: ⎛ 1− x ⎞ y ⎛ 1 − xs ⎞ ⎛ h + δ ⎞ ⎛ 1 − xs ⎞ ln ⎜ ln ⎜ ⎟ = − ln ⎜ ⎟+ ⎟ (S17.8.4d) δ ⎝ 1 − xa ⎠ ⎜⎝ δ ⎟⎠ ⎝ 1 − xa ⎠ ⎝ 1 − xa ⎠ Add the term ln((1-xa)/(1-xs)) to each side: ⎛ 1− x ⎞ ⎛ 1 − xa ⎞ ⎛ 1 − xa ⎞ y ⎛ 1 − xs ⎞ ⎛ h ⎞ ⎛ 1 − xs ⎞ ln ⎜ ⎟ + ln ⎜ ⎟ = − ln ⎜ ⎟ + ⎜ + 1 ⎟ ln ⎜ ⎟ + ln ⎜ ⎟ (S17.8.5) δ ⎝ 1 − xa ⎠ ⎝ δ ⎠ ⎝ 1 − xa ⎠ ⎝ 1 − xa ⎠ ⎝ 1 − xs ⎠ ⎝ 1 − xs ⎠ Collect terms

230

⎛ 1 − x ⎞ y ⎛ 1 − xa ⎞ h ⎛ 1 − xa ⎞ y − h ⎛ 1 − xa ⎞ ln ⎜ ln ⎜ ⎟ = ln ⎜ ⎟ − ln ⎜ ⎟= ⎟ δ ⎝ 1 − xs ⎠ δ ⎝ 1 − xs ⎠ δ ⎝ 1 − xs ⎠ ⎝ 1 − xs ⎠ Raising each side to the power e:

⎛ 1− x ⎜ ⎝ 1 − xs

⎞ ⎛ 1 − xa ⎞ ⎟=⎜ ⎟ ⎠ ⎝ 1 − xs ⎠

(S17.8.6)

y −h

δ

(S17.8.7)

17.9. The vapor flux is given by Equation (17.5.11) cD ⎛ 1 − xa ⎞ N y = h = w,air ln ⎜ ⎟ δ ⎝ 1 − xs ⎠ where xs is the partial pressure of water in air at saturation (vapor pressure/total air pressure) and xa is the partial pressure of water/total air pressure. The quantity xa can be expressed as xHxs, where xH is the relative humidity. Using the data for Problem 17.10 and a total air pressure of 101,325 Pa. The quantity c = ptot/RT = 101,325 Pa/(8.314 N m K-1 mol-1)(298 K) = 40.90 mole m-3. The diffusivity of water in air is provided in the text, p. 797, as 2.6 x 10-5 m2 s-1. Thus, xs = 0.0310 at 25 C and 0.0728 at 40 C. For 20% relative humidity at 25 C.

( 40.90 mol m )( 2.6 x 10 = -3

−5

m 2 s-1

) ln ⎛ 1 − 0.20 ( 0.031) ) ⎞ = 0.0020 mol m

⎜ 0.0136 m 1 − 0.031 ⎝ For 80% relative humidity, the flux is 0.00050 mol m-2 s-1. N y=h

⎟ ⎠

-2 -1

s

17.10. The error can be computed from the ratio of Equations (17.5.12) to Equation (17.5.13): ⎛ 1 − xs ⎞ ⎛ 1 − xs ⎞ error = x s − x a ln ⎜ ⎟ = x s (1 − x H ) ln ⎜ ⎟ ⎝ 1 − xa ⎠ ⎝ 1 − xs x H ⎠ At 25 C the error is -0.000226 and at 40 C the error rises to -0.00128. Thus, Equation (17.5.13) is a good approximation.

17.11. Since the enthalpy of vaporization is a function of temperature, application of Equation (17.5.25) or Equation (17.5.26) is done iteratively. That is, the enthalpy of vaporization is updated, once the temperature at the air-sweat interface is calculated. The flux for the evaporating liquid is temperature independent and was found to be 0.001 mol m-2 s-1 for 60% relative humidity. For the calculation reported in the text, Equation (17.5.5a) was used and ΔH vap was determined for a temperature of 25 C. Using T equal to 37 C, ΔH vap = 54047.6 J mol-1. The temperature drop is 0.444 C and the energy flux is 54.05 J m-2 s-1. Updating the values at T = 309.7 K, the temperature drop is 0.444 C and the energy flux is 54.15 J m-2 s-1. These values are within 1% of the values obtained for T = 310.00 K.

17.12. Use Equation (17.4.3) to calculate the Nusselt number. The Prandtl number does not vary significantly with temperature and a value of 0.72 is commonly used for air. The kinemtic viscosity of air 0.1327 cm2 s-1 = 1.327 x 10-5 m2 s-1. As noted on page 797, a characteristic diameter for a typical female is 0.304 m. The following table lists values of Re, Nu, h and q for various wind speeds. The energy flux can be quite substantial and is reduced significantly by clothing. 231

v, miles/h 1 2 5 10 25

v, m/s 0.447 0.894 2.235 4.470 11.176

Re 10241 20482 51206 102412 256029

Nu 54.64 80.86 140.33 220.05 420.46

h, W m-2 K-1 4.49 6.65 11.54 18.10 34.58

q, W m-2 143.80 212.79 369.30 579.08 1106.48

17.13. Start with the definition of the Grashof number, Equation (17.4.22) ρ 2 gβΔTL3 Gr = 2

μ

The definition of β in terms of the density is given by Equation (17.4.6) ρ ≈ ρ 0 − ρ 0 β ΔT Let Δρ = ρ 0 − ρ . Thus, Δρ ≈ ρ0 βΔT . Assuming that density in the definition of the Grashof number is the value at the reference temperature, ρ0, the Grashof number becomes:

ρ 0 2 gβ ΔTL3 ρ0 2 gΔρ L3 ρ 0 gΔρ L3 Gr = == = μ2 μ 2 ρ0 μ2 17.14. For free convection, Equation (17.4.5) is used for flow over a sphere. The viscosity ratio is 0.900 and Pr = 0.72. v, miles/h v, m/s Diameter, m Re Nu adult 10 4.47 0.178 60050 164.29 child 10 4.47 0.124 41820 133.60 For free convection, the Grashof number is calculated using Equation (17.4.22) with L equal to the diameter and β = 1/T where T is the air temperature (273.15 K). Equation (17.4.24) is used to determine the Nusselt number for a flat plate. The correlation for spheres is found in reference [18], page 301. Nu = 2.0 + 0.43(Pr Gr )1/4

adult child

Diameter, m Gr Nu, flat plate Nu, sphere 0.178 42697290 38.57 34.02 0.124 14422151 29.40 26.41

For radiation, the energy flux is given by Equation (17.2.19c). Treating the absorptivity and emissivity as the same, the flux equals q = σe(Tb4-Tair4). A heat transfer coefficient can be defined as h=q/ΔT and a Nussel number determined. Results are: h

qrad 193.44

Nu adult 5.23

Nu child 37.22

25.93

Comparing results, the free convection and radiation terms are comparable and are about 20% of the value for forced convection. 232

17.15. Note, there is a typographical error in the text and Equation (17.5.25) should be: ⎡ ⎛ ρ Cˆ p ⎞ ⎛ ρCˆ p ⎞⎤ N y = h ( y-h ) ⎟ − exp ⎜ N y = hδ ⎟ ⎥ ⎢ exp ⎜ ⎜ Cvap ka ⎟ ⎜ Cvap ka ⎟⎥ ⎛ ΔH vap N y = h h ⎞⎢ ⎝ ⎠ ⎝ ⎠⎦ Ta = Tair + ⎜ − ΔT ⎟ ⎣ (17.5.25) ⎜ ⎟ ˆ ˆ k ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ h C C N ρ ρ l ⎝ ⎠ ka y =h p p N y = hδ ⎟ ⎥ ⎜ ⎟ − ⎢1 − exp ⎜ ⎜ ⎟ ⎜ ⎟⎥ kl ⎝ Cvap k a ⎠ ⎢ C k vap a ⎝ ⎠⎦ ⎣ Begin with Equation (17.5.21) for air and Equation (17.5.24) for the liquid. ⎛ ρ Cˆ p ⎞ aC k Ta = 1 vap a exp ⎜ N y=h y ⎟ + a 2 ⎜ Cvap ka ⎟ N y = h ρCˆ p ⎝ ⎠ Tl = a3 y + a4 The boundary conditions are: y=0 Tl = Tb y=h Tl = Ta dT dT − kl l ka a dy y = h dy y = h+δ

(17.5.21) (17.5.24)

(S17.15.1a) (S17.15.1b) = ΔH vap N y = h

(S17.15.1c)

y=h

Ta = Tair

(S17.15.1a)

From the boundary condition at y = 0 a 4 =Tb Tl = a3 y + Tb and

(S17.15.2a) (S17.15.2b)

From the boundary condition at y = h+δ, ⎛ ρCˆ p ⎞ aC k a 2 =Tair − 1 vap a exp ⎜ N y=h ( h + δ ) ⎟ ⎜ Cvap ka ⎟ N y = h ρ Cˆ p ⎝ ⎠

(S17.15.3a)

⎛ ρCˆ p ⎞ ⎛ ρ Cˆ p ⎞⎤ a1Cvap k a ⎡ ⎢exp ⎜ (S17.15.3b) N y = h y ⎟ − exp ⎜ N y=h ( h + δ ) ⎟ ⎥ ⎜ Cvap ka ⎟ ⎜ Cvap ka ⎟⎥ N y = h ρ Cˆ p ⎢⎣ ⎝ ⎠ ⎝ ⎠⎦ Equating Equations (S17.15.2b) and (S17.15.3b) at y = h, ⎛ ρ Cˆ p ⎞ ⎛ ρCˆ p ⎞⎤ aC k ⎡ N y = h h ⎟ − exp ⎜ N y = h ( h + δ ) ⎟ ⎥ (S17.15.4a) a3 h + Tb = Tair + 1 vap a ⎢exp ⎜ ⎜ Cvap ka ⎟ ⎜ Cvap ka ⎟⎥ N y = h ρ Cˆ p ⎢⎣ ⎝ ⎠ ⎝ ⎠⎦ Ta = Tair +

a3 = −

⎛ ρ Cˆ p ⎞⎡ ⎛ ρCˆ p ⎞⎤ aC k ΔT N y = h h ⎟ ⎢1 − exp ⎜ N y = hδ ⎟ ⎥ + 1 vap a exp ⎜ ⎜ Cvap ka ⎟⎢ ⎜ Cvap ka ⎟⎥ h hN y = h ρ Cˆ p ⎝ ⎠⎣ ⎝ ⎠⎦

where ΔT=Tb -Tair . 233

(S17.15.4b)

The liquid temperature is ⎛ ρ Cˆ p ⎞⎡ ⎛ ρ Cˆ p ⎞⎤ y ya1Cvap k a ⎢ exp ⎜ N y = h h ⎟ 1 − exp ⎜ N y = hδ ⎟ ⎥ + Tb Tl = −ΔT + ⎜ Cvap ka ⎟⎢ ⎜ Cvap ka ⎟⎥ h hN y = h ρ Cˆ p ⎝ ⎠⎣ ⎝ ⎠⎦

(S17.15.5)

Use Equations (S17.15.3b) and (S17.15.5) to compute the derivatives of the temperature. The boundary condition, Equation (S17.15.1c), becomes: ⎛ ρ Cˆ p ⎞ k ΔT a1 k a exp ⎜ N y=h h ⎟ + l ⎜ Cvap ka ⎟ h ⎝ ⎠

(S17.15.6) ⎛ ρCˆ p ⎞⎡ ⎛ ρ Cˆ p ⎞⎤ kl a1Cvap k a exp ⎜ N y = h h ⎟ ⎢1 − exp ⎜ N y = hδ ⎟ ⎥ = ΔH vap N y = h − ˆ ⎜ ⎟ ⎜ ⎟⎥ hN y = h ρ C p ⎝ Cvap ka ⎠ ⎢⎣ ⎝ Cvap ka ⎠⎦ Solving for a1: ⎛ ρ Cˆ p ⎞⎛ k ΔT ⎞ exp ⎜ − N y = h h ⎟ ⎜ ΔH vap N y = h − l ⎜ Cvap ka ⎟⎝ h ⎟⎠ ⎝ ⎠ a1 = (S17.15.7) ⎛ ρ Cˆ p ⎞⎤ klCvap k a ⎡ ka − N δ ⎟⎥ ⎢1 − exp ⎜ ⎜ Cvap ka y = h ⎟ ⎥ hN y = h ρ Cˆ p ⎢⎣ ⎝ ⎠⎦ Inserting this expression for a1 into Equation (S17.15.3b) yields the final result for the air temperature. ⎛ ρ Cˆ p ⎞ ⎛ ρ Cˆ p ⎞⎤ Cvap k a ⎡ N y = h ( y-h ) ⎟ − exp ⎜ N y = hδ ⎟ ⎥ ⎢exp ⎜ ˆ ⎜ ⎟ ⎜ ⎟ kl ΔT ⎞ N y = h ρ C p ⎢⎣ ⎛ ⎝ Cvap ka ⎠ ⎝ Cvap ka ⎠ ⎥⎦ vap Ta = Tair + ⎜ ΔH N y = h − h ⎟⎠ ⎛ ρCˆ p ⎞⎤ ⎝ kC k ⎡ k a − l vap a ⎢1 − exp ⎜ N y = hδ ⎟ ⎥ ⎜ Cvap ka ⎟⎥ hN y = h ρCˆ p ⎢⎣ ⎝ ⎠⎦ (S17.15.8a)

Rearrange to yield the correct form of Equation (17.5.25) ⎡ ⎛ ρ Cˆ p ⎞ ⎛ ρCˆ p ⎞⎤ N y = h ( y-h ) ⎟ − exp ⎜ N y = hδ ⎟ ⎥ ⎢ exp ⎜ ⎜ Cvap ka ⎟ ⎜ Cvap ka ⎟⎥ ⎛ ΔH vap N y = h h ⎞⎢ ⎝ ⎠ ⎝ ⎠⎦ (17.5.25) Ta = Tair + ⎜ − ΔT ⎟ ⎣ ⎜ ⎟ ˆ ˆ k ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ N h C C ρ ρ l k ⎝ ⎠ a y=h p p N y = hδ ⎟ ⎥ ⎜ ⎟ − ⎢1 − exp ⎜ ⎜ ⎟ ⎜ ⎟⎥ kl ⎝ Cvap k a ⎠ ⎢ C k ⎝ vap a ⎠⎦ ⎣ At y = h, Equation (17.5.25) is: ⎡ ⎛ ρ Cˆ p ⎞⎤ N y = hδ ⎟ ⎥ ⎢1 − exp ⎜ ⎜ ⎟ ⎛ ΔH vap N y = h h ⎞ ⎝ Cvap ka ⎠ ⎦⎥ ⎣⎢ Ta = Tair + ⎜ − ΔT ⎟ (S17.15.8b) ⎜ ⎟ ⎛ hN ρCˆ ⎞ ⎡ ˆ k ⎤ ⎛ ⎞ C ρ l k ⎝ ⎠ a y=h p p N y = hδ ⎟ ⎥ ⎜ ⎟ − ⎢1 − exp ⎜ ⎜ ⎟ ⎜ ⎟⎥ kl ⎝ Cvap k a ⎠ ⎢ C k a vap ⎝ ⎠⎦ ⎣

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The group hN y = h ρCˆ p Cvap k a

hN y = h ρCˆ p Cvap k a =

can be rewritten as the following by using Equation (17.5.17):

hCvap v y ρ Cˆ p Cvap k a

=

hv y ρ Cˆ p ka

= Peair

⎛ ΔH vap N y = h h ⎞ ⎣⎡1 − exp ( Peairδ / h ) ⎦⎤ − ΔT ⎟ Ta = Tair + ⎜ ⎜ ⎟ ka kl ⎝ ⎠ ( Peair ) − ⎡⎣1 − exp ( Peairδ / h ) ⎤⎦ kl

(S17.15.8c)

The thermal Peclet number for air is 0.20, which is larger than the value for sweat, but still much less than 1. For the case of conduction only, energy transport through the liquid is unchanged. Equation (17.5.17) for the air simplifies to: d 2Ta =0 dy 2

(S17.15.9)

After integration we obtain: Ta = a1 y + a2 At y = h+δ, a2 = Tair − a1 ( h + δ )

Ta = a1 ( y − ( h + δ ) ) + Tair

Tl = a3 y + Tb Equating the air and sweat temperatures at y = h: a3 h + Tb = −a1δ + Tair Tair − Tb δ ΔT = − a1 − h h h h δ ΔT ⎤ ⎡ Tl = ⎢ − a1 − y + Tb h h ⎥⎦ ⎣ Use these results for Ta and Tl to compute the derivatives in Equation (S17.15.1c) ⎛ δ ΔT ⎞ vap ka a1 + kl ⎜ a1 + (S17.15.10a) ⎟ = ΔH N y = h ⎝ h h ⎠ Solving for a1: a3 = −a1

δ

+

δ⎞ ΔT ⎛ a1 ⎜ ka + kl ⎟ = ΔH vap N y = h − kl h⎠ h ⎝

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(S17.15.10b)

a1 =

ΔH vap N y = h − kl

ΔT h

δ⎞ ⎛ ⎜ k a + kl ⎟ h⎠ ⎝ The resulting expression for the air temperature is: ⎛ ⎜ kl Ta = Tair + ⎜ ⎜ ⎜ ⎝

ΔT − ΔH vap N y = h h δ⎞ ⎛ ⎜ k a + kl h ⎟ ⎝ ⎠

⎛ hΔH vap N y = h ⎞ ⎜ ΔT − ⎟ kl ⎟ ( ( h + δ ) − y ) = Tair + ⎜ ⎜ ⎛ ka ⎞ ⎟ ⎜⎜ ⎟ ⎜ ⎟h +δ ⎠ ⎝ kl ⎠ ⎝

(S17.15.10c)

⎞ ⎟ ⎟ (h + δ ) − y ) ⎟( ⎟⎟ ⎠

for y = h ⎛ hΔH vap N y = h ⎜ ΔT − kl Ta ( y = h ) = Tair + ⎜ ⎜ ⎛ ka ⎞ ⎜⎜ ⎜ ⎟h +δ ⎝ kl ⎠ ⎝

⎞ ⎟ ⎟δ (S17.15.11) ⎟ ⎟⎟ ⎠ For values of h (0.005 m) and δ (0.0136 m) provided in Section 17.5 and thermal conductivities of air

δ

= 0.985 . Thus, the approximation presented in Equation (17.5.26) ⎛ ka ⎞ ⎜ ⎟h +δ ⎝ kl ⎠ is reasonable. Further, Equation (17.5.26) arises as a limiting value of Equation (17.5.25) when ka/klPe << 1.

and water in Table 17.2,

If vaporization does not occur, then ΔH vap = 0 and Equation (17.5.27) results.

17.16. From Table 2.4, the blood vessel diameters range from 6 x 10-6 m to 5 x 10-5 m. Corresponding mean velocities range from 2 x 10-4 to 0.001 m s-1. The Pe ranges from 0.0068 to 0.284. Blood vessel densities range from 2.0 x 108 vessels m-2 to 2.22 x 109 vessels m-2. The ratio of thermal conductivities between blood and tissue range from 1.5 to 3. From Equation (17.7.4), keff/ktissue ranges from 1.00 to 5.21.

236