Solution Manual for Transport Phenomena in Biological Systems George A. Truskey, Fan Yuan and David F. Katz 2 Solution to Problems in Chapter 1, Section 1.10 1.1. The relative importance of convection and diffusion is evaluated by Peclet number, Pe = v L D ij (S1.1.1) (a) Solving for L, L = PeD ij /v. When convection is the same as diffusion, Pe =1, L is 0.11cm. (b) The distance between capillaries is 10 -4 m, O 2 needs to travel half of this distance, and Pe = 0.0455. Therefore, convection is negligible compared with diffusion. 1.2. Since H O 2 = H Hb , equation (1.6.4) is simplified to the following: C O 2 = H O 2 P O 2 + 4C Hb S Hct (S1.2.1) 2 O P and S are 95 mmHg and 95% for arterial blood and 38 mmHg 70% for venous blood. C Hb 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 CO 2 70% is stored in plasma and 30% is in red blood cell. Therefore, the total change of CO 2 is 2.27(0.70)+1.98(0.30) = 2.18 cm 3 per 100 cm 3 . For O 2 , 2 O P changes from 38 to 100 mmHg after blood passes through lung artery. Using data in problem (1.2), the total O 2 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 cm 3 . Thus, the O 2 concentration difference of 0.0025 M corresponds to 5.58 cm 3 O 2 per 100 cm 3 . While larger than the difference for CO 2 , the pressure difference driving transport is much larger for O 2 than CO 2 . 1.4. The diffusion time is L 2 /D ij = (10 -4 cm) 2 /(2x10 -5 cm 2 s -1 ) = 0.0005 s. Therefore, diffusion is much faster than reaction and does not delay the oxygenation process. 1.5. V = π R 2 L and the S= 2 π RL where R is the vessel radius and L is the length Order volume, cm 3 surface area, cm 2 cumulative volume, cm 3 cumulative surface area, cm 2 1 0.01 58 26.27 0.0 158 26.27 2 0.03885 35.32 0.05 61.59 3 0.05738 31.44 0.11 9 2 . 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 1.53 211.6 0 11 0.50671 3.99 2.03 215. 59 3 1.6. Order Volume ( c m 3 ) Surface Area ( c m 2 ) Cumulative Volume ( c m 3 ) Cumulative Surface Area ( c m 2 ) 0 30 . 5 4 67 . 86 30 . 54 67 . 86 1 11 . 1 3 36 . 4 9 41 . 6 6 104 . 34 2 4 . 1 1 19.82 45.78 124.2 3 1 . 50 10 . 70 47.27 134.9 4 3 . 2 3 28.72 50.51 163.6 5 3 . 29 37.65 53.80 201.2 6 3 . 54 50.67 57.35 251.9 7 4 . 0 4 70.29 61.39 322.2 8 4 . 45 95.74 65.84 418.0 9 5.1 5 133.76 70.99 551.7 10 6. 25 192.38 77.24 744.1 11 7.45 273.51 84.70 1018 12 9.58 403.41 94.27 1421 13 11.68 569.79 106.0 1991 14 16.2 1 876.05 122.2 2867 15 22.42 1358.86 144.6 4226 16 30.57 2038.28 175.2 6264 17 42.33 3135.25 217.5 9399 18 60.223 4817.76 277.7 14217 19 90.05 7663.95 367.8 21881 20 138.42 12303.82 506.2 34185 21 213.18 19831.06 719.4 54015 22 326.72 31874.64 1046. 85890 23 553.75 54024.81 1600 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 990 L day -1 for men and 1,080 L day -1 for women. Then the fraction of water filtered across the glomerulus is 18.2% for men and 16.67% for women. (b) renal vein flow rate = renal artery flow rate – excretion rate = 1.19 L min -1 renal vein flow rate = 1.25 L min -1 – (1.5 L day -1 )/(1440 min day -1 ) = 1.249 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 (1.25 L min -1 x 150mM – 150 mM day -1 /(1440 min day -1 ))/1.249 L min -1 = 150.037 mM There is a slight increase in sodium concentration in the renal vein due to the volume reduction. 1.8. (a) Bi = k m L/D ij = 5 x 10 -9 cm s -1 x 0.0150cm/(1 x 10 -10 cm 2 s -1 ) = 0.75. (b) The results indicate that the resistance to LDL transport provided by the endothelium is similar to that provided by the arterial wall. 4 1.9 The oxygen consumption rate is  V O 2 = Q C v − C a ( ) where Q is the pulmonary blood flow and C v and C a are the venous are arterial oxygen concentrations. The oxygen concentrations are obtained from Equation (1.6.4) 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 C a = 0.0070 M C a = 0.0063 M Exercise C a = 0.0019 M C a = 0.0017 M C V = 0.0090 M C v = 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.  V I C I − C alv ( ) = Q C v − C a ( ) (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: C alv = p alv RT = 105 mm Hg ( ) / 760 mm Hg/atm ( ) 0.08206 L atm/(mol K) ( ) 310 K ( ) = 0.00543 M C I = p alv RT = 0.21 1 atm ( ) 0.08206 L atm/(mol K) ( ) 310 K ( ) = 0.00826 M  V I = 10 breaths/min ( ) 0.56 − 0.19 L ( ) = 3.7 L/min males  V I = 10 breaths/min ( ) 0.45 − 0.41 L ( ) = 3.1 L/min females Since we have all terms on the left hand side of Equation (1), the rate of oxygen removal from the lungs is:  V I C I − C alv ( ) = 3.7 L/min ( ) 0.00282 mole O 2 /L ( ) = 0.0104 mole O 2 /min males  V I C I − C alv ( ) = 3.1 L/min ( ) 0.00282 mole O 2 /L ( ) = 0.00874 mole O 2 /min females To convert to mL O 2 /L blood, multiply to oxygen removal rate by 22,400 L O 2 per mole of O 2 . ( ) ( ) 2 2 2 2 O O O O 1 Hct 4 Hct Hb Hb C H P C S H P = ! + + 5 For males the value is 233 mL O 2 /min and for females the value is 196 mL O 2 /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  V I . 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: Using the relation for the percent saturation to calculate the concentration in the pulmonary vein: S = P O 2 P 50 ( ) 2.6 1 + P O 2 P 50 ( ) 2.6 = 100 / 26 ( ) 2.6 1 + 100 / 26 ( ) 2.6 = 0.972 Likewise for the pulmonary artery: S = P O 2 P 50 ( ) 2.6 1 + P O 2 P 50 ( ) 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: For men C v = 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 ( ) 20 mmHg ( ) ( ) 0.45 = 0.0031 M C a = 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 C v = 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 ( ) 20 mmHg ( ) ( ) 0.40 = 0.00275 M C a = 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 ( ) 100 mmHg ( ) ( ) 0.40 = 0.0080 M Thus, the oxygen consumption rates are ( ) ( ) 2 2 2 2 O O O O 1 Hct 4 Hct Hb Hb C H P C S H P = ! + + ( ) ( ) 2 2 2 2 O O O O 1 Hct 4 Hct Hb Hb C H P C S H P = ! + + 6 Q C v − C a ( ) 0.148 mole O 2 /min men 0.132 mole O 2 /min women These values are about 14 times larger than the values under resting conditions. From Equation (1)  V I = Q C v − C a ( ) C I − C alv ( ) 52.5 L O 2 /min men 46.8 L O 2 /min women 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 = p a / CO Peripheral resistance, mm Hg/(L/min) Rest Exercise Athlete 20 5.2 Sedentary person 20 6 W = p a dV ∫ = p a Δ V since the mean arterial pressure is assumed constant. DV corresponds to the stroke volume. Note 1 L = 1000 cm 3 *(1 m/100 cm) 3 = 0.001 m 3 100 mm Hg = 13,333 Pa Sedentary person W = (100 mm Hg)(133.3 Pa/mm Hg)(0.069 L)(1000 cm 3 /L)(1 m 3 /1x10 6 cm 3 ) = 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  V O 2 =  V I C I − C alv ( ) Assuming that the inspired air is warmed to 37 C C I = p I RT = 101..85 mm Hg ( ) / 760 mm Hg/atm ( ) 0.08206 L atm/(mol K) ( ) 310 K ( ) = 0.00527 M C alv = p alv RT = 71.85 mm Hg ( ) / 760 mm Hg/atm ( ) 0.08206 L atm/(mol K) ( ) 310 K ( ) = 0.00372 M Assuming that the inspired and dead volumes are the same as at sea level  V I = f V I − V dead ( ) = 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 Rest Athlete 0.014 Sedentary person 0.014 1.14. The concentrations are found as the ratio of the solute flow rate/fluid flow rate Urine, M Plasma, M Urine/Plasma Sodium 0.1042 0.08444 1.233 Potassium 0.0694 0.004 17.36 Glucose 0.000347 0.00444 0.0781 Urea 0.32431 0.005183 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, C inulin plasma GFR = C inulin urine Q urine Solving for the glomerular filtration rate: 8 GFR = C inulin urine C inulin plasma Q urine = 0.125 0.001 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 mL min -1 ( ) = 125 mL min -1 9 Solution to Problems in Chapter 2, Section 2.10 2.1. Q = v • n dA ∫ = 3 2 x + 6 2 y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ dxdy = 3 2 2 x 2 + 6 2 2 yx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x = 0 2 dy y = 0 3 ∫ x = 0 2 ∫ y = 0 3 ∫ Q = 6 2 + 12 2 y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ dy y = 0 3 ∫ = 6 2 y + 6 2 y 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ y = 0 3 = 72 2 Q = 50.91 cm 3 s -1 2.2. n = 1 = a 2 + a 2 + a 2 = 3 a Rearranging, a = 1/ 3 2.3. ∇ • ρ vv ( ) = e x ∂ ∂ x + e y ∂ ∂ y + e z ∂ ∂ z ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ • ρ vv ( ) = e x ∂ ∂ x + e y ∂ ∂ y + e z ∂ ∂ z ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ • ρ e x v x v + ρ e y v y v + ρ e z v z v ( ) = ∂ ∂ x ρ v x v ( ) + ∂ ∂ y ρ v y v ( ) + ∂ ∂ z ρ v z v ( ) Differentiating term by term, ∇ • ρ vv ( ) = v ∂ ∂ x ρ v x ( ) + ∂ ∂ y ρ v y ( ) + ∂ ∂ z ρ v z ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ρ v x ∂ ∂ x v ( ) + ρ v y ∂ ∂ y v ( ) + ρ v z ∂ ∂ z v ( ) ∇ • ρ vv ( ) = v ∇ • ρ v ( ) + ρ v • ∇ v 2.4. (a) For a two-dimensional steady flow, the acceleration is: a = v x ∂ v ∂ x + v y ∂ v ∂ y For v = U o (x 2 – y 2 +x) e x - U o (2xy +y) e y , ∂ v ∂ x = U o 2 x + 1 ( ) e x - U o 2y e y ∂ v ∂ y = U o − 2 y ( ) e x - U o 2x +1 ( ) e y a = U o 2 x 2 − y 2 + x ( ) 2 x + 1 ( ) e x - 2y e y ( ) − U o 2 2 xy + y ( ) − 2 y e x - 2x+1 ( ) e y ( ) Collecting terms: a = U o 2 x 2 − y 2 + x ( ) 2 x + 1 ( ) + 2 xy + y ( ) 2 y ⎡ ⎣ ⎤ ⎦ e x − U o 2 x 2 − y 2 + x ( ) 2 y − 2 xy + y ( ) 2 x + 1 ( ) ⎡ ⎣ ⎤ ⎦ e y a = U o 2 2 x 3 + 3 x 2 − 2 xy 2 − y 2 + x + 4 xy 2 + 2 y 2 ⎡ ⎣ ⎤ ⎦ e x − U o 2 2 yx 2 − 2 y 3 + 2 xy − 4 x 2 y + 2 xy + 2 xy + y ⎡ ⎣ ⎤ ⎦ e y a = U o 2 2 x 3 + 3 x 2 + x + 2 xy 2 + y 2 ⎡ ⎣ ⎤ ⎦ e x − U o 2 − 2 yx 2 − 2 y 3 + 6 xy + y ⎡ ⎣ ⎤ ⎦ e y a = U o 2 2 x 2 + 2 y 2 + 3 x + 1 ( ) x + y 2 ⎡ ⎣ ⎤ ⎦ e x + U o 2 2 x 2 + 2 y 2 − 6 x − 1 ( ) y e y At y = 1 and x = 0 a = 2 ( ) 2 e x + e y ( ) = 4 e x + 4 e y 10 At y = 1 and x = 2 a = 2 ( ) 2 8 + 2 + 6 + 1 ( ) 2 + 1 ⎡ ⎣ ⎤ ⎦ e x + 2 ( ) 2 8 + 2 − 12 − 1 ( ) e y = 140 e x − 12 e y (b) From equation 2.2.6 Q = v • n dA ∫ = v x ∫ dydz since n = e x . Q = U o z = 0 3 ∫ y = 0 5 ∫ x 2 − y 2 + x ( ) x = 5 dydz = 3 U o 30 − y 2 ( ) dy y = 0 5 ∫ = 3 U o 30 y − y 3 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 6 150 − 125 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 650 m 3 s -1 2.5. (a) a x = e x a = v x ∂ v x ∂ x = U 0 1 − x L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 2 ∂ ∂ x U 0 1 − x L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ∂ ∂ x 1 − x L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2 L 1 − x L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 3 a x = e x a = v x ∂ v x ∂ x = U 0 2 1 − x L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 2 ∂ ∂ x 1 − x L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 2 U 0 2 L 1 − x L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 5 For values given: a x = 50 m 2 /s 2 2 m 1 − 0.5 ( ) − 5 = 25 m/s 2 ( ) / 1 / 32 ( ) = 800 m/s 2 (b) (1) The “no slip” boundary condition is not satisfied. (2) At x = L, the acceleration is undefined! 2.6. (a) Using the definition of the volumetric flow rate, Q Q = v i n dA ∫ = v z rdrd θ 0 R i ∫ 0 2 π ∫ The cross-sectional area element in cylindrical coordinates is rdrd θ . Since the velocity does not vary with angular position, substitution for v z and integration in the angular direction yields: Q = v max 1 − r 2 R 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ rdrd θ 0 R i ∫ 0 2 π ∫ = 2 π v max 1 − r 2 R i 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ rdr 0 R i ∫ R i is used to denote the local radius within the stenosis. Integrating in the radial direction yields: Q = 2 π v max 1 − r 2 R i 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ rdr 0 R i ∫ = 2 π v max r 2 2 − r 4 4 R i 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ r = 0 R = π R i 2 2 v max 11 Solving for v max : v max = 2 Q π R i 2 = 2 Q π R 0 2 1 − 0.5 1 − 4 z L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1/2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 2 Outside the stenosis, R i = R 0 and: v max = 2 Q π R 0 2 (b) At z = 0, the velocity in the stensosis is v max = 2 Q π R i 2 = 2 Q π R 0 2 0.5 [ ] 2 = 8 Q π R 0 2 R i = R 0 1 − 0.5 1 − 4 z L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1/2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 0.5 R 0 The shear stress in the stenosis is: τ rz stenosis = μ ∂ v r ∂ z = μ ∂ ∂ r v max 1 − r 2 R i z = 0 ( ) 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ r = R i = − 2 μ R i z = 0 ( ) v max R i z = 0 ( ) 2 = − 32 μ Q π R 0 3 Outside the stenosis the shear stress is: τ rz = μ ∂ ∂ r v max 1 − r 2 R 0 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ r = R 0 = − 2 μ v max R 0 = − 4 μ Q π R 0 3 2.7. Evaluating Equation (2.7.30) for y = -h/2 yields: τ w = τ yx ( y = − h / 2) = Δ p L h 2 (S2.7.1) From Equations (2.7.23) and (2.7.26), Δ p L = 8 μ v max h 2 = 12 μ Q wh 3 (S2.7.2) Replacing Δ p/L in Equation (S2.7.1) with the expression in Equation (S2.7.2) yields τ w = 6 μ Q wh 2 Solving for h: h = 6 μ Q w τ w Inserting the values provided for Q, w, μ and τ w yields h = 0.051 cm. 2.8. (a) Δ p = ρ gh = (1 g cm -3 )(980 cm s -2 )(2.5 x 10 -4 cm) = 0.245 dyne cm -2 (b) Rearranging equation (2.4.16) we have 12 T c = Δ p 2 1 R p − 1 R c ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ T c = 1.838 x 10 -5 dyne cm -1 2.9. (a) To find the radius use Equation (2.4.16) and treat the pipet radius as the capillary radius R c = R cap . Δ p = 2 T c 1 R cap − 1 R c ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ For R c = 6.5 μ m T c = 0.06 mN/m= 6 x 10 -5 N/m(1 x 10 -6 m/ μ m) = 6 x 10 -11 N/ μ m Δ p = 0.2 mm Hg Since 1.0133 × 10 5 N m − 2 = 760 mm Hg 0.2 mm Hg = 26.7 N m -2 (1 m/10 6 μ m) 2 = 2.67 x 10 -11 N μ m -2 Solving for R cap 1 R pcap = 1 R c + Δ p 2 T c R cap = 1 1 R c + Δ p 2 T c R cap = 1 1 R c + Δ p 2 T c = 1 1 6.5 + 2.67 2 6 ( ) = 2.66 μ m While this result satisfies the law of Laplace, we need to assess whether the surface area is no greater than the maximum surface area of the cell, 1.4 times the surface area of a spherical cell, or 743.3 μ m 2 . The factor of 1.4 accounts for the excess surface area. Ideally, a larger cell entering a smaller capillary with look like a cylinder with hemispheres on each end. The cylinder will have length l and radius equal to the capillary. The hemispheres will have a radius equal to the capillary radius. The volume must remain constant, so V = 4 3 π R c 3 + π R c 2 L Solving for the length, 13 L = V − 4 3 π R c 3 π R c 2 = 4 3 π R 3 − R c 3 ( ) π R c 2 = 4 3 6.5 3 − 2.66 3 ( ) 2.66 2 = 48.2 μm The resulting surface area is SA = 4 π R c 2 + 2 π R c L = π 4 * 2.66 2 + 2 * 48.2 * 2.66 ( ) = 894.6 μm 2 This is larger than the surface area 530.9 μ m 2 or 1.4 times the surface area 743.3 μ m 2 . To find the radius and length, one could iteratively solve for L and surface area of use the fzero function in MATLAB. After several iterations, the result approaches a radius of 3.3 μ m and L = 29.2 μ m. If the cell had no excess area, then the cell would have no capacity to enter a capillary smaller than itself! (b) Whether or not excess area is not considered, a cell with a radius of 3.0 μ m can enter the capillary. 2.10. A momentum balance is applied on a differential volume element, 2 π r Δ r Δ y, as shown in the figure below. p r 2 π r Δ y − p r + Δ r 2 π r + Δ r ( ) Δ y + τ yr y + Δ y 2 π r Δ r − τ yr y 2 π r Δ r = 0 (S2.10.1) Divide each term by 2 π r Δ r Δ y and take the limit as Δ r and Δ y go to zero results in the following expression: 1 r d rp ( ) dr = d τ yr dy (S2.10.2) 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 C 1 . Solving for the shear stress, τ yr = C 1 y + C 2 . Substituting Newton’s law of viscosity and integrating yields: v r = C 1 y 2 2 μ + C 2 μ y + C 3 (S2.10.3) Applying the boundary conditions that v r = 0 at y=±h/2, 0 = C 1 h 2 8 μ + C 2 μ h 2 + C 3 (S2.10.4a) 0 = C 1 h 2 8 − C 2 μ h 2 + C 3 (S2.10.4b) Adding Equations (S2.10.4a) and (S2.10.4b) and solving for C 3 , 14 C 3 = − C 1 h 2 8 (S2.10.5) Inserting Equation (S2.10.5) into Equation (S2.10.4a) yields C 2 = 0. Thus the velocity is: v r = C 1 μ y 2 2 − h 2 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (S2.10.6) The volumetric flow rate is: Q = v • n dA ∫ = 2 π rv r dy y = − h / 2 h / 2 ∫ = 2 π rC 1 μ y 2 2 − h 2 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ dy y = − h / 2 h / 2 ∫ (S2.10.7) Q = 2 π rC 1 μ y 3 6 − h 2 y 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ y = − h / 2 h / 2 = 2 π rC 1 h 3 μ 1 24 − 1 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − π rC 1 h 3 6 μ (S2.10.8) Solving for C 1 and inserting into equation (S2.10.6) v r = − 6Q π rh 3 y 2 2 − h 2 8 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (S2.10.9) The shear stress can thus be written as; τ w = τ yr y = − h / 2 = μ dv r dr y = − h / 2 = − 6Q π rh 3 y y = − h / 2 = 3 μ Q π rh 2 (S2.10.10) 2.11. Flow rate per fiber, Q f = Q/250 = 0.8 mL/60 s = 0.01333 mL/s Average velocity per fiber: <v f > = Q f / π R f 2 = (0.01333 mL/s)/(3.14159*(0.01 cm) 2 ) <v f > = 42 cm/s Re = ρ <v f >D f / μ = 1.05*42*0.02/0.03 = 29.7. L e = 0.058DRe = 0.058*(0.02 cm)(29.7) = 0.034 cm << L = 30 cm. 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 dz = 1 r d(r τ rz ) dr (S2.12.1) (b) The velocity profile is sketched below: Integrating the momentum balance and substituting Newton’s law of viscosity, τ rz = − Δ p 2L r + C 1 r = μ d v z dr (S2.12.2) 15 Note that the shear stress and shear rate are a maximum at r = 2C 1 Δ p / L . Assuming that C 1 is greater than zero, then r will have a maximum in the fluid. (c) Integrating Equation (S2.12.2) yields: v z = − Δ p 4 μ L r 2 + C 1 μ ln(r) + C 2 (S2.12.3) Applying the boundary conditions V = − Δ p 4 μ L R C 2 + C 1 μ ln(R C ) + C 2 (S2.12.4a) 0 = − Δ p 4 μ L R 2 + C 1 μ ln(R) + C 2 (S2.12.4b) Subtracting V = − Δ p 4 μ L R C 2 − R 2 ( ) + C 1 μ ln R C R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (S2.12.5) Solving for C 1 : C 1 = μ V ln R C R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + Δ p 4 L R C 2 − R 2 ( ) ln R C R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (S2.12.6a) Using this result to find C 2 C 2 = Δ p 4 μ L R 2 − V ln R R C ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + Δ p 4 μ L R C 2 − R 2 ( ) ln R R C ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ln( R ) (S2.12.6b) The resulting expression for the velocity profile is v z = Δ pR 2 4 μ L 1 − r 2 R 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + V + Δ p 4 μ L R C 2 − R 2 ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln r R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln R R C ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (S2.12.7) (d) The shear stress is: τ zr = μ dv z dr = − r Δ p 2 L + μ V + Δ p 4 L R C 2 − R 2 ( ) ln R R C ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ 1 r ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (S2.12.8) (e) At r = R, the shear stress is: