Instructor’s Solutions Manual for Transport Phenomena in Biological Systems 2nd Edition 0135070511, 9780135070512

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

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males females

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

(P S= 1+ ( P

)

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

= 0.972

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

= 0.3357

Using the relation for the percent saturation to calculate the concentration in the pulmonary vein: O2

O2

(P S= 1+ ( P

P50

P50

)

)

2.6 2.6

2.6

Likewise for the pulmonary artery: O2

O2

P50

)

2.6

P50

2.6

2.6

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

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

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

–6

–6

M mmHg –1

M mmHg –1

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

) (100 mmHg ) ) 0.45 = 0.0090 M

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 )

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

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

Calv =

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

VI = f (VI − Vdead ) = 20 ( 0.56 L − 0.19 L ) = 7.4 L min -1

Assuming that the inspired and dead volumes are the same as at sea level

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

8

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. 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 3

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

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) ∂z ∂y ∂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.

10

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 =

∫ ∫ v rdrdθ

2π Ri

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: ⎛ ⎛ 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 ⎝ R

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:

v max =

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

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

τ w = τ yx ( y = −h / 2 ) =

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

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 w 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):

12

⎛ 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, yr = C1y + C2. Substituting Newton’s law of viscosity and integrating yields: C y2 C (S2.10.3) v r = 1 + 2 y + C3 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 = ∫

2πrv r dy = y=−h / 2 h/2

2πrC1 μ

∫

⎛ y2 h2 ⎞ − ⎟dy ⎜ y=−h / 2 2 8⎠ ⎝ h/2

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

(S2.10.7)

h/ 2

(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, C dv Δp (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 ⎠ Using this result to find C2

⎛ ⎜ Δp 2 ⎜ V C2 = R − ⎜ ⎛ R 4μL ⎜⎜ ln ⎜ ⎝ ⎝ RC

(S2.12.3)

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

(S2.12.5)

(S2.12.6a)

⎞ ⎟ 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 = − ⎜ ⎟ ⎜ ⎟ dr 2L ⎛ R ⎞ ⎝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)

( 0.17 cm ) (100 dyne/cm 3 )

For the values provided τ zr

r=R

=−

(

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 βΔxΔyΔz + τ xy

y +Δy

− τ xy

y

) Δx Δ 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) zx = -ρgycosβ

For a Bingham plastic, the velocity gradient is zero for a shear stress below the yield stress, o = 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

μ0 ( 4τ1 − τ 0 ) H 7V1 = μ0

(S2.15.2) (S2.15.3)

Substituting Equation (S2.15.1b) for 1 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 ⎛ dv ⎞ dv dv x τ yx = m x = m⎜ x ⎟ dy dy ⎝ dy ⎠ Rearranging yields an expression for the velocity gradient n −1

n

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

(S2.15.7)

1/ n

(S2.15.8)

⎛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 ⎠

1/ n

(S2.15.10a)

H

⎛ 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. 1/ n

(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 r2 rθ = C1 If the shear stress is less than o 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 rθ < o, vθ = 0.

For τrθ greater than o, 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 rθ > 0, 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 o can be determined from the torque required to begin rotation of the outer cylinder, T = 2πR2L o. 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 o. T = 2πR 2 L τ rθ

=

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

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; n −1

⎛ d ⎛ v ⎞⎞

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

(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,

⎛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: 1/ n

(S2.17.3)

n−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

n

(S2.17.4)

(S2.17.5)

2

(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:

τ yx = ∫ G (t − t ') 0

−ε

Rearranging the above equation, we have:

(

)

τ yx = γ 0

γ0 dt ' ε

∫ ε G(t − t ')dt ' 0

−

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 rz(r =R) < 0, 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 rz(r =R) > 0, the fluid begins to move. Since rz(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, rz < 0 and the constitutive relation is: dv Δp τ rz = − r = −τ 0 + µ 0 z 2L dr Rearranging and integrating once, we have: τ Δp 2 vz = 0 r − r + C3 4µ 0 L µ0 Applying the boundary condition at r = R that vz = 0, the C3 equals: τ Δp 2 C3 = − 0 R + R µ0 4µ 0 L Replacing C3 yields the final expression for vz(r). vz =

ΔpR 2 ⎛ r2 ⎞ τ0R ⎛ r⎞ − − 1 ⎜ ⎜1 − ⎟ 2 ⎟ 4µ 0 L ⎝ R ⎠ µ 0 ⎝ R ⎠

r0 ≤ r ≤ R

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

0 ≤ r ≤ r0 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 µ 0 L ⎝ R ⎠ µ 0 ⎝ R ⎠ ⎠ 2 r = r0 ⎝ 4µ 0 L ⎝ R ⎠ µ 0 ⎝ R ⎠ ⎠ ⎦⎥ Replacing 0 with r0. R ⎤ ⎛⎛ r 2 ⎞ 2r0 ⎛ r ⎞⎞ ΔpR 2 ⎡⎛ ⎛ r0 2 ⎞ 2r0 ⎛ r0 ⎞ ⎞ r0 2 1 1 Q == 2π − + − − ⎢⎜ ⎜ 1 − 2 ⎟ − ⎟ ⎜⎜ ⎜1 − ⎟ ⎟ rdr ⎥ ⎜ ⎟ 2 ⎟ ∫ 4 µ 0 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 µ 0 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 µ 0 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 ⎟− + − ⎟⎥ ⎢⎜ − ⎟− ⎜ − ⎟+⎜ ⎜ 4 µ 0 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 ⎢⎜ ⎥ ⎢⎜ ⎜ ⎟ ⎟− ⎜ ⎟⎥ 4 µ 0 L ⎣⎝ 4 4 R 2 ⎠ R ⎝ 6 6 R ⎠ ⎦ ⎣⎝ 4 4 R ⎠ R ⎝ 6 3 R 2 R ⎠ ⎦ πΔpR 4 ⎡⎛ r0 4 ⎞ 4r0 ⎛ r03 ⎞ ⎤ Q= ⎢⎜ 1 − ⎟− ⎜1 − ⎟ ⎥ 8µ 0 L ⎣⎝ R 4 ⎠ 3R ⎝ R 3 ⎠ ⎦ Q == 2π

ΔpR 2 4µ 0 L

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

2.22. (a) Since δ = R(1-ε)∞, 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) 2πRL rq|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

r =R

)

= −µ

dv z dy

y =δ

= −µ

δ

VC

(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 =

δ 2 ⎛ δ ⎞⎛

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

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 Rearranging: VC = 2µL (c) There are two possible ways to approach this. One is to neglect the fluid in the gaps.

π ∫ VC rdr RC

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

0

2

π ∫ v z rdr R

v =

π R2

0

= VC

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

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,

2

µ eff µ

=

1

⎛ δ ⎞⎛ δ ⎞ 4 ⎜ ⎟ ⎜1 − ⎟ ⎝ R ⎠⎝ R ⎠

2

R 2 Δp 8μeff L (1)

or

µ eff µ

=

⎛ δ ⎞⎛ δ ⎞ 4 ⎜ ⎟⎜ 1 − ⎟ ⎝ R ⎠⎝ R ⎠

1

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, ⎛ Δpr ⎞ rc < r < R, ⎜ ⎟ ⎝ 2L ⎠

1/2

= (τ 0 )

1/2

+ (η N )

1/ 2

r < rc ,

dv z =0 dr Rearrange Equation (S2.26.2), we have, 1/2 dv z 1 ⎡⎛ Δpr ⎞ 1/2 ⎤ = − τ ( ) ⎢⎜ ⎥ 0 dr η N ⎢⎣⎝ 2 L ⎟⎠ ⎦⎥

⎛ dvz ⎞ ⎜ dr ⎟ ⎝ ⎠

1/ 2

(S2.26.2) (S2.26.3)

2

(S2.26.4)

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

ηN 1

∫

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

R

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 = (S2.26.5) 1 1 1 − ⎟⎥ − − − + ⎢⎜ ⎜ 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 R

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

τw R = τ o rc

Combining the above equation and Equation (S2.10.7), we have,

(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,

w

= 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 w = 15 dyne cm-2, 0/ w = 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 w = 2 dyne cm-2, 0/ w = 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 w = 0.2 dyne cm-2, 0/ w = 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 =

The Womersley number is

α=

L T vx

ω R2 ν

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

=

8μ L v R2

(S3.2.3) (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=−

0=−

(

ΔpR 2 C2 ln R + C3 + 4μ L μ

(S3.3.5a)

Δpε R 2 C2 ln ε R + C3 + 4μ L μ

)

(S3.3.5b)

(

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

ΔpR 2 1 − ε 2 4 L ln ( ε )

The velocity vz(R) is:

ΔpR 2 vz = 4μ L

C3 = ⎡⎛ 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 =

(

1

π R2 1 − ε 2

) ∫ ε∫

2π R 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

(

)

ΔpR 2 4μ L

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

32

)

(

2

R2 1 − ε 2

) ε∫

⎡⎛ r2 ∫ ⎢⎢⎝⎜⎜1 − R 2 εR ⎣ R

R

R

vz ( r ) rdr

(

)

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

⎞

∫ ( r ln r − r ln R ) dr ⎟⎟ R

εR

(S3.3.8)

⎠

(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 ⎝

(

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

)⎛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

)

)

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

Collecting terms: v =

(

1

R2 1 − ε 2 v =

(

(

)(

)

(

)

)

(

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

(

)

(

)

(

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

Rearranging:

)

(

The flow rate is Q = π R2 (1-ε2): Q=

(

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

8μ LQ

Δpπ R 4

(

= 1− ε

2

)(

)

) ⎞⎟⎤⎥ + R (1 − ε ) ⎟⎥ ⎠⎦

2

2 2

4 ln ( ε )

⎞ ⎟ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎟ ⎠

(S3.3.11)

(S3.3.12)

) ⎟⎞ ⎟ ⎠

(S3.3.13)

) ⎞⎟ ⎟ ⎠

(

⎛ 1− ε 2 ⎜ 1+ ε 2 + ⎜ ln ( ε ) ⎝

(S3.3.14)

) ⎞⎟ ⎟ ⎠

(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 Q=

∫ ∫y =−h / 2 v x ( y, z ) dydz

w/ 2

h /2

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 ⎞ (2n + 1)3 π3 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 w/(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 zx 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

34

yx

and

zx,

respectively.

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

yx(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 ∂ ⎢⎛ ⎝ ⎠ 1 τ yx ( y, t ) = μ x = ⎜ ∑ 2 ⎟ 3 3 ⎥ ⎟ 8L ∂ y ⎢⎜⎝ ∂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 ⎜⎜ − ⎟⎟ ⎥ ∂t 4νΔp h2 ⎝ ⎠ ⎦⎥ ⎣⎢ n =0

(b) The normalized gradient is shown below.

35

(c) The quantity, hL/4νΔp can be rewritten as h2/8ν w. Multiplying the normalized gradients by 8ν w/h2 yields the dimensional gradient. For and ν = 0.007 cm2 s-1, h = 0.015 cm, h = 0.05 cm and -2 w = 1 dyne cm , 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.

(

) ( λ J (λ )

)

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

⎞ 2ΔPR 2 ⎟⎟ μL ⎠

∑ ∞

n =1

J o λn r / R exp −λn2ν t / R 2 3 n 1

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

n

(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. ρ (ω R ) vθ*

3.12. (a) Using the dimensionless variables listed, equation (3.8.5a) becomes: 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. ⎡ ρ (ω R )2 ∂ v* ρα (ω R )2 ∂ v* ⎤ μωα R ∂ 2 v*z * θ θ ⎢ ⎥ v*r v + = + z αR R ∂ r* ∂ z* ⎥⎦ ⎢⎣ (α R )2 ∂ z*2

Using the same approach on equation (3.8.5b)

2 ρ (ω R ) ⎡

Simplifying: R

ραω R 2 μ

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

* ⎤ ⎡ * ∂ 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 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

C=

Integrating again

(S13.15.5)

ΓRM2 +B Deff r

(S13.15.6)

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

(

(S13.15.8)

)

1/3 ⎛ 3V ⎞ RM = ⎜ = fRC3 = f 1/3 RC ⎟ 4 π ⎝ ⎠ RM = 0.0851/3(0.0005 cm) = 0.0002198 cm 1/3

C = 2.68 x 10

(1.02 x 10 mole/cm / s ) ( 0.0002198 cm ) + (1.92 x 10 cm / s ) ( 0.0005 cm ) −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

(

)

(R C ( L ) = C (0) -

)R

RO2 = RO2/NNc. Solving for Nc 2

o

NC =

− Rc 2 2

O2 / N

C ( 0 ) -C ( L ) ⎛ Rc 2 v z ⎜ RO2 / N L ⎜ Ro 2 − Rc 2 ⎝ Rc v z

(

)

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 ∂ x2

(S14.1.1)

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 1 ⎞⎤ q2 ⎛ 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 – sΔπ). Δp equals 48 mmHg – 12 mmHg. Δπ equals to 25 mmHg at the afferent arteriole end and 35 mmHg at the efferent arteriole end. s 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 ) 1 − exp ( − Pe )

(S14.4.4)

(b) From Equation (14.4.11), Js =

W φ J v (C0 − C e − Pe ) 1 − e − Pe

(S14.4.5)

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

(

)

(S14.4.6)

J s 1 − e − Pe + W φ e− Pe = W φ J v C0

(S14.4.7)

Js =

(S14.4.8)

W ϕ J v C0

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

(S14.5.1)

ρ ⎜ vr

(S14.5.2)

⎛ ⎝

⎡ 1 ∂ ⎛ ∂ vx ⎞ ∂ 2 vx ⎤ ∂ vx ∂v ⎞ ∂p + vx x ⎟ = − +μ⎢ ⎜r ⎟+ 2 ⎥ ∂r ∂x ⎠ ∂x ⎢⎣ r ∂ r ⎝ ∂ r ⎠ ∂ x ⎥⎦

185

⎛

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

fluid at wall

− pi

) , at r = R

(S14.5.5) (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 Vr0um2/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 =

λ2 = α=

(S16.2.7)

2

(k1 + k2 + ke1 + ke 2 ) − (k1 + k2 + ke1 + ke 2 ) 2 − 4 ( k2 ke1 + k1ke 2 + ke1ke 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 ⎝ ⎠

VC = 1 0 ⎣⎡( k1 + ke1 − λ1 ) α exp ( −λ1t ) + ( k1 + ke1 − λ2 ) (1 − α ) exp ( −λ2 t ) ⎦⎤ k2V2

(S16.2.10)

(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, AUC = ∫ C1dt = ∫ C0 ⎣⎡α exp ( −λ1t ) + (1 − α ) exp ( −λ2 t ) ⎦⎤ dt ∞

∞

⎛ α 1−α ⎞ = C0 ⎜ + ⎟ λ2 ⎠ ⎝ λ1 0

0

(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 ) ]

where λ1 = λ2 =

α=

(S16.3.6)

( 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

λ1 − λ2

(S16.3.9)

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

aG

C 100

a

C

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

where

3

+ A1

d 2 C1 dt

2

+ A2

d 2 C1 dt2

+ A3C1 = 0

(S16.5.7)

A 1 = k 12 + k 13 + k e + k 21 + k 31

A 2 = k 21 k 13 + k 21 k e + k 31 k 12 + k 31 k e + k 21 k 31

A 3 = k 21 k 31 k e

(S16.5.8) (S16.5.9) (S16.5.10)

The initial conditions of Equation (S16.5.7) are, C1 = C0, at t = 0

d C1 = −(k 12 + k 13 + k e )C 0 , at t = 0 dt

[

(S16.5.11)

]

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

(S16.5.12)

(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

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 ) ⎤⎦

where

(S16.5.14)

(S16.5.15)

( 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 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 ∞, 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 T = ( T0 − T∞ )

The temperature profile is:

R + T∞ r

(S17.4.3)

(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 ⎞ (S17.5.1) β= ⎜ 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

Thus,

(S17.5.2)

β=

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.

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)

⎛ 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

The solution is:

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 =

−5

) 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. -3

N y=h

m 2 s-1

⎟ ⎠

-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 ≈ ρ0 − ρ0 βΔT 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:

ρ0 2 gβΔTL3 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

y=h

= ΔH vap N y = h

Ta = Tair

(17.5.21) (17.5.24)

(S17.15.1a) (S17.15.1b) (S17.15.1c) (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 ⎜ ⎜ ⎟ ⎜ ⎟ N y = h ρ Cˆ p ⎣⎢ ⎝ Cvap ka ⎠ ⎝ Cvap ka ⎠ ⎦⎥ Ta = Tair +

a3 = −

⎛ ρ Cˆ p ⎞⎡ ⎛ ρCˆ p ⎞⎤ aC k ΔT N y = h h ⎟ ⎢1 − exp ⎜ N y = hδ ⎟ ⎥ + 1 vap a exp ⎜ ⎜ ⎟ ⎜ ⎟ h hN y = h ρ Cˆ p ⎝ Cvap ka ⎠ ⎢⎣ ⎝ Cvap ka ⎠ ⎥⎦

where ΔT=Tb -Tair .

233

(S17.15.4b)

⎛ 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 ⎝ ⎠⎣ ⎝ ⎠⎦

The liquid temperature is

(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 ⎜ ⎜ ⎟ ⎜ ⎟ ⎛ ΔH vap N y = h h ⎞⎢ ⎝ Cvap ka ⎠ ⎝ Cvap ka ⎠ ⎦⎥ (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 ⎠ ⎦⎥ ⎣⎢ − ΔT ⎟ Ta = Tair + ⎜ (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 ⎝ ⎠⎦ ⎣

234

The group hN y = h ρCˆ p

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

=

hv y ρ Cˆ p

= Peair

⎛ ΔH vap N y = h h ⎞ ⎣⎡1 − exp ( Peairδ / h ) ⎦⎤ Ta = Tair + ⎜ − ΔT ⎟ ⎜ ⎟ ka kl ⎝ ⎠ ( Peair ) − ⎡⎣1 − exp ( Peairδ / h ) ⎤⎦ kl

Cvap k a

Cvap k a

ka

(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 ⎝

235

(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