Solution Manual For Transport Phenomena In Biological Systems, 2nd Edition

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Solution Manual forTransport Phenomena in Biological SystemsGeorge A. Truskey, Fan Yuan and David F. Katz

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2Solution to Problems inChapter 1, Section 1.101.1.The relative importance of convection and diffusion is evaluated by Peclet number,Pe=vLDij(S1.1.1)(a)Solving for L, L = PeDij/v. When convection is the same as diffusion, Pe =1, L is 0.11cm.(b)The distance between capillaries is 10-4m, O2needs to travel half of this distance, and Pe =0.0455. Therefore, convection is negligible compared with diffusion.1.2.Since HO2= HHb, equation (1.6.4) is simplified to the following:CO2=HO2PO2+ 4CHbSHct(S1.2.1)2OPandSare 95 mmHg and 95% for arterial blood and 38 mmHg70% for venous blood. CHbis0.0203 mol L-1x 0.45 = 0.0091 M for men, and 0.0203 mol L-1x 0.40 = 0.0081 M for women. Basedon these data, the fraction of oxygen in plasma and bound to hemoglobin is 1.5% and 98.5% inarterial blood, and 0.83% and 99.17% in venous blood for men. Corresponding values for women are1.7% and 98.3% in arterial blood, and 0.93% and 99.07% in venous blood. Most oxygen in blood isbound to hemoglobin.1.3.For CO270% is stored in plasma and 30% is in red blood cell. Therefore, the total change ofCO2is 2.27(0.70)+1.98(0.30) = 2.18 cm3per 100 cm3. For O2,2OPchanges from 38 to 100 mmHgafter blood passes through lung artery. Using data in problem (1.2), the total O2concentration inblood is 0.0088 M in arterial blood and 0.0063 M in venous blood. At standard temperature (273.15K) and pressure (1 atm = 101,325 Pa), 1 mole of gas occupies 22,400 cm3.Thus, the O2concentration difference of 0.0025 M corresponds to 5.58 cm3O2per 100 cm3. While larger than thedifference for CO2, the pressure difference driving transport is much larger for O2than CO2.1.4.The diffusion time is L2/Dij= (10-4cm)2/(2x10-5cm2s-1) = 0.0005 s. Therefore, diffusion ismuch 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 lengthOrdervolume, cm3surface area, cm2cumulative volume, cm3cumulative surface area, cm210.015826.270.015826.2720.0388535.320.0561.5930.0573831.440.1192.9940.0921930.230.20123.2150.1278826.640.33149.8660.2048723.280.54173.1470.2073315.560.74188.7080.2413211.030.99199.7390.310108.171.30207.89100.230463.711.53211.60110.506713.992.03215.59

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31.6.OrderVolume (cm3)SurfaceArea (cm2)Cumulative Volume(cm3)Cumulative SurfaceArea (cm2)030.5467.8630.5467.86111.1336.4941.66104.3424.1119.8245.78124.231.5010.7047.27134.943.2328.7250.51163.653.2937.6553.80201.263.5450.6757.35251.974.0470.2961.39322.284.4595.7465.84418.095.15133.7670.99551.7106.25192.3877.24744.1117.45273.5184.701018129.58403.4194.2714211311.68569.79106.019911416.21876.05122.228671522.421358.86144.642261630.572038.28175.262641742.333135.25217.593991860.2234817.76277.7142171990.057663.95367.82188120138.4212303.82506.23418521213.1819831.06719.45401522326.7231874.641046.8589023553.7554024.8116001399151.7.(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-1for men and 1,080 L day-1for women. Thenthe 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-1renal 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-1x 150mM – 150 mM day-1/(1440 min day-1))/1.249 L min-1= 150.037 mMThere is a slight increase in sodium concentration in the renal vein due to the volume reduction.1.8.(a) Bi = kmL/Dij= 5 x 10-9cm s-1x 0.0150cm/(1 x 10-10cm2s-1) = 0.75.(b) The results indicate that the resistance to LDL transport provided by the endothelium is similar tothat provided by the arterial wall.

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41.9The oxygen consumption rate isVO2=Q Cv−Ca()where Q is the pulmonary blood flow and Cvand Caare the venous are arterial oxygen concentrations. The oxygen concentrations are obtainedfrom Equation (1.6.4)The fractional saturationSis given by Equation (1.6.5).For the data given, the venous fractionsaturation is 0.971.The arterial fractional saturation is 0.754 under resting conditions and 0.193under exercise conditions.MenWomenRestCa= 0.0070 MCa= 0.0063 MExerciseCa= 0.0019 MCa= 0.0017 MCV= 0.0090 MCv= 0.0080 MThe oxygen consumption rates areMenWomenRest0.0115 mole min-10.0102 mole min-1Exercise0.1776 mole min-10.1579 mole min-11.10.(a) To obtain the rate of oxygen removal from the lungs, we use the mass balance discussed inclass that equates the oxygen removed from the inspired air with the oxygen uptake in the blood.VICI−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 oxygenremoval from the lungs. From the data provided and the ideal gas equation:Calv=palvRT=105 mm Hg()/ 760 mm Hg/atm()0.08206 L atm/(mol K)()310 K()=0.00543 MCI=palvRT=0.21 1 atm()0.08206 L atm/(mol K)()310 K()=0.00826 MVI=10 breaths/min()0.56−0.19 L()=3.7 L/minmalesVI=10 breaths/min()0.45−0.41 L()=3.1 L/minfemalesSince we have all terms on the left hand side of Equation (1), the rate of oxygen removal from thelungs is:VICI−Calv()=3.7 L/min()0.00282 mole O2/L()=0.0104 mole O2/minmalesVICI−Calv()=3.1 L/min()0.00282 mole O2/L()=0.00874 mole O2/minfemalesTo convert to mL O2/L blood, multiply to oxygen removal rate by 22,400 L O2per mole of O2.()()2222OOOO1Hct4HctHbHbCHPCSHP=!++

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5For males the value is 233 mL O2/min and for females the value is 196 mL O2/min. These values area 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 representsboth 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=PO2P50()2.61+PO2P50()2.6=100 / 26()2.61+100 / 26()2.6=0.972Likewise for the pulmonary artery:S=PO2P50()2.61+PO2P50()2.6=20 / 26()2.61+20 / 26()2.6=0.3357This is substantially less than the value in the pulmonary artery under resting conditions, S = 0.754.The concentration in blood is:For menCv=1.33 x 10–6M mmHg–1()20 mmHg()0.55+0.0203 M()0.3357()+1.50 x 10–6M mmHg–1()20 mmHg()()0.45=0.0031MCa=1.33 x 10–6M mmHg–1()100 mmHg()0.55+0.0203 M()0.972()+1.50 x 10–6M mmHg–1()100 mmHg()()0.45=0.0090MFor womenCv=1.33 x 10–6M mmHg–1()20 mmHg()0.60+0.0203 M()0.3357()+1.50 x 10–6M mmHg–1()20 mmHg()()0.40=0.00275MCa=1.33 x 10–6M mmHg–1()100 mmHg()0.60+0.0203 M()0.972()+1.50 x 10–6M mmHg–1()100 mmHg()()0.40=0.0080MThus, the oxygen consumption rates are()()2222OOOO1Hct4HctHbHbCHPCSHP=!++()()2222OOOO1Hct4HctHbHbCHPCSHP=!++

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6Q Cv−Ca()0.148 mole O2/minmen0.132 mole O2/minwomenThese values are about 14 times larger than the values under resting conditions.From Equation (1)VI=QCv−Ca()CI−Calv()52.5 L O2/min men46.8 L O2/minwomenFor a respiration rate of 30 breaths per minutes, the net volume inspired in each breadth is: 1.75L/min for men and 1.56 L/min for women. In terms of the total air inspired in each breadth, it is 1.94L/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 HRis the hear rate in beat min-1.Stroke Volume, LRestExerciseAthlete0.08330.238Sedentary person0.06940.2The peripheral resistance isR=pa/COPeripheral resistance, mm Hg/(L/min)RestExerciseAthlete205.2Sedentary person206W=padV∫=paΔVsince the mean arterial pressure is assumed constant. DV corresponds to thestroke volume.Note 1 L = 1000 cm3*(1 m/100 cm)3= 0.001 m3100 mm Hg = 13,333 PaSedentary personW = (100 mm Hg)(133.3 Pa/mm Hg)(0.069 L)(1000 cm3/L)(1 m3/1x106cm3) =Work, J (N m)RestExerciseAthlete1.114.12Sedentary person0.9254.00Power, W (J/s)RestExerciseAthlete1.117.22Sedentary person0.9248.33

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71.12.Although the pressure drops from 760 mm Hg to 485 mm Hg, the partial pressures areunchanged. The inspired air at 3,650 m is 101.85 mm Hg. For a 30 mm Hg drop, the alveolar air isat 71.85 mm Hg.The oxygen consumption rate isVO2=VICI−Calv()Assuming that the inspired air is warmed to 37 CCI=pIRT=101..85 mm Hg()/ 760 mm Hg/atm()0.08206 L atm/(mol K)()310 K()=0.00527 MCalv=palvRT=71.85 mm Hg()/ 760 mm Hg/atm()0.08206 L atm/(mol K)()310 K()=0.00372 MAssuming that the inspired and dead volumes are the same as at sea levelVI=f VI−Vdead()=20 0.56 L−0.19 L()=7.4 L min-1The venous blood is at a partial pressure of 0.98(71.85) = 70.32 mm HgThe corresponding saturation is 0.930.1.13.(1650 kcal/day)*4.184 kJ/kcal*(1day/24 h)*(1 h/3600 s) = 79.9 J/sRestAthlete0.014Sedentary person0.0141.14.The concentrations are found as the ratio of the solute flow rate/fluid flow rateUrine, MPlasma, MUrine/PlasmaSodium0.10420.084441.233Potassium0.06940.00417.36Glucose0.0003470.004440.0781Urea0.324310.00518362.57The results indicate that urine concentrates sodium to a small extent, potassium to a higher level andurea to very high levels. Glucose is at a lower concentration in urine than plasma, suggesting that itstransport across the glomerulus is restricted.1.15.Assuming that inulin is not reabsorbed by the kidneys and returned to the blood, then the massflow rate of inulin across the glomerulus must equal the mass flow rate in urine. The mass flow rateis the product of the mass concentration (mass/volume) multiplied by the flow rate (volume/time).Thus,CinulinplasmaGFR=CinulinurineQurineSolving for the glomerular filtration rate:

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8GFR=CinulinurineCinulinplasmaQurine=0.1250.001⎛⎝⎜⎞⎠⎟1 mL min-1() =125 mL min-1

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9Solution to Problems inChapter 2, Section 2.102.1.Q=v•ndA∫=32x+62y⎛⎝⎜⎞⎠⎟dxdy=322x2+622yx⎛⎝⎜⎞⎠⎟x=02dyy=03∫x=02∫y=03∫Q=62+122y⎛⎝⎜⎞⎠⎟dyy=03∫=62y+62y2⎛⎝⎜⎞⎠⎟y=03=722Q = 50.91 cm3s-12.2.n=1=a2+a2+a2=3aRearranging,a=1/32.3.∇•ρvv()=ex∂∂x+ey∂∂y+ez∂∂z⎛⎝⎜⎞⎠⎟•ρvv()=ex∂∂x+ey∂∂y+ez∂∂z⎛⎝⎜⎞⎠⎟•ρexvxv +ρeyvyv+ρezvzv()=∂∂xρvxv()+∂∂yρvyv()+∂∂zρvzv()Differentiating term by term,∇•ρvv()=v∂∂xρvx()+∂∂yρvy()+∂∂zρvz()⎛⎝⎜⎞⎠⎟+ρvx∂∂xv( )+ρvy∂∂yv( )+ρvz∂∂zv( )∇•ρvv()=v∇•ρv()+ρv•∇v2.4.(a) For a two-dimensional steady flow, the acceleration is:a=vx∂v∂x+vy∂v∂yForv= Uo(x2– y2+x)ex- Uo(2xy+y)ey,∂v∂x=Uo2x+1()ex-Uo2yey∂v∂y=Uo−2y()ex-Uo2x +1()eya=Uo2x2−y2+x()2x+1()ex- 2yey()−Uo22xy+y()−2yex-2x+1()ey()Collecting terms:a=Uo2x2−y2+x()2x+1()+2xy+y()2y⎡⎣⎤⎦ex−Uo2x2−y2+x()2y−2xy+y()2x+1()⎡⎣⎤⎦eya=Uo22x3+3x2−2xy2−y2+x+4xy2+2y2⎡⎣⎤⎦ex−Uo22yx2−2y3+2xy−4x2y+2xy+2xy+y⎡⎣⎤⎦eya=Uo22x3+3x2+x+2xy2+y2⎡⎣⎤⎦ex−Uo2−2yx2−2y3+6xy+y⎡⎣⎤⎦eya=Uo22x2+2y2+3x+1()x+y2⎡⎣⎤⎦ex+Uo22x2+2y2−6x−1()yeyAt y = 1 and x = 0a=2()2ex+ey()=4ex+4ey

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10At y = 1 and x = 2a=2()28+2+6+1()2+1⎡⎣⎤⎦ex+2()28+2−12−1()ey=140ex−12ey(b) From equation 2.2.6Q=v•ndA∫=vx∫dydzsincen=ex.Q=Uoz=03∫y=05∫x2−y2+x()x=5dydz=3Uo30−y2()dyy=05∫=3Uo30y−y33⎛⎝⎜⎞⎠⎟=6 150−1253⎛⎝⎜⎞⎠⎟=650 m3s-12.5.(a)ax=exa=vx∂vx∂x=U01−xL⎛⎝⎜⎞⎠⎟−2∂∂x U01−xL⎛⎝⎜⎞⎠⎟−2⎡⎣⎢⎢⎤⎦⎥⎥∂∂x1−xL⎛⎝⎜⎞⎠⎟−2⎡⎣⎢⎢⎤⎦⎥⎥=2L1−xL⎛⎝⎜⎞⎠⎟−3ax=exa=vx∂vx∂x=U021−xL⎛⎝⎜⎞⎠⎟−2∂∂x1−xL⎛⎝⎜⎞⎠⎟−2⎡⎣⎢⎢⎤⎦⎥⎥=2U02L1−xL⎛⎝⎜⎞⎠⎟−5For values given:ax=50 m2/s22 m1−0.5()−5=25 m/s2()/ 1 / 32()=800 m/s2(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, QQ=vindA∫=vzrdrdθ0Ri∫02π∫The cross-sectional area element in cylindrical coordinates is rdrdθ. Since the velocity does not varywith angular position, substitution for vzand integration in the angular direction yields:Q=vmax1−r2R2⎛⎝⎜⎞⎠⎟rdrdθ0Ri∫02π∫=2πvmax1−r2Ri2⎛⎝⎜⎞⎠⎟rdr0Ri∫Riis used to denote the local radius within the stenosis. Integrating in the radial direction yields:Q=2πvmax1−r2Ri2⎛⎝⎜⎞⎠⎟rdr0Ri∫=2πvmaxr22−r44Ri2⎛⎝⎜⎞⎠⎟r=0R=πRi22vmax

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11Solving for vmax:vmax=2QπRi2=2QπR021−0.5 1−4zL⎛⎝⎜⎞⎠⎟2⎛⎝⎜⎞⎠⎟1/2⎡⎣⎢⎢⎤⎦⎥⎥2Outside the stenosis, Ri= R0and:vmax=2QπR02(b) At z = 0, the velocity in the stensosis isvmax=2QπRi2=2QπR020.5[]2=8QπR02Ri=R01−0.5 1−4zL⎛⎝⎜⎞⎠⎟2⎛⎝⎜⎞⎠⎟1/2⎡⎣⎢⎢⎤⎦⎥⎥=0.5R0The shear stress in the stenosis is:τrz stenosis=μ∂vr∂z=μ∂∂rvmax1−r2Riz=0()2⎛⎝⎜⎞⎠⎟⎡⎣⎢⎢⎤⎦⎥⎥r=Ri=−2μRiz=0()vmaxRiz=0()2=−32μQπR03Outside the stenosis the shear stress is:τrz=μ∂∂rvmax1−r2R02⎛⎝⎜⎞⎠⎟⎡⎣⎢⎤⎦⎥r=R0=−2μvmaxR0=−4μQπR032.7.Evaluating Equation (2.7.30) for y = -h/2 yields:τw=τyx( y=−h / 2)=ΔpLh2(S2.7.1)From Equations (2.7.23) and (2.7.26),ΔpL=8μvmaxh2=12μQwh3(S2.7.2)ReplacingΔp/L in Equation (S2.7.1) with the expression in Equation (S2.7.2) yieldsτw=6μQwh2Solving for h:h=6μQwτwInserting the values provided for Q, w,μandτwyields h = 0.051 cm.2.8.(a)Δp =ρgh = (1 g cm-3)(980 cm s-2)(2.5 x 10-4cm) = 0.245 dyne cm-2(b) Rearranging equation (2.4.16) we have

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12Tc=Δp21Rp−1Rc⎛⎝⎜⎜⎞⎠⎟⎟Tc= 1.838 x 10-5dyne cm-12.9.(a) To find the radius use Equation (2.4.16) and treat the pipet radius as the capillary radius Rc= Rcap.Δp=2Tc1Rcap−1Rc⎛⎝⎜⎞⎠⎟For Rc= 6.5μmTc= 0.06 mN/m= 6 x 10-5N/m(1 x 10-6m/μm) = 6 x 10-11N/μmΔp = 0.2 mm HgSince1.0133×105N m−2=760 mm Hg0.2 mm Hg = 26.7 N m-2(1 m/106μm)2= 2.67 x 10-11Nμm-2Solving for Rcap1Rpcap=1Rc+Δp2TcRcap=11Rc+Δp2TcRcap=11Rc+Δp2Tc=116.5+2.672 6( )=2.66μmWhile this result satisfies the law of Laplace, we need to assess whether the surface area is no greaterthan the maximum surface area of the cell, 1.4 times the surface area of a spherical cell, or 743.3μm2. The factor of 1.4 accounts for the excess surface area. Ideally, a larger cell entering a smallercapillary with look like a cylinder with hemispheres on each end. The cylinder will have length land radius equal to the capillary. The hemispheres will have a radius equal to the capillary radius.The volume must remain constant, soV=43πRc3+πRc2LSolving for the length,

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13L=V−43πRc3πRc2=43πR3−Rc3()πRc2=436.53−2.663()2.662=48.2 μmThe resulting surface area isSA=4πRc2+2πRcL=π4 * 2.662+2 * 48.2 * 2.66()=894.6 μm2This is larger than the surface area 530.9μm2or 1.4 times the surface area 743.3μm2.To find the radius and length, one could iteratively solve for L and surface area of use the fzerofunction in MATLAB. After several iterations, the result approaches a radius of 3.3μm and L =29.2μm.If the cell had no excess area, then the cell would have no capacity to enter a capillary smaller thanitself!(b) Whether or not excess area is not considered, a cell with a radius of 3.0μm can enter thecapillary.2.10.A momentum balance is applied on a differential volume element, 2πrΔrΔy, as shown in thefigure below.pr2πrΔy−pr+Δr2πr+Δr()Δy+τyr y+Δy2πrΔr−τyr y2πrΔr=0(S2.10.1)Divide each term by 2πrΔrΔy and take the limit asΔr andΔy go to zero results in the followingexpression:1rd rp()dr=dτyrdy(S2.10.2)Note that if the gap distance h is much smaller than the radial distance, then curvature is notsignificant. Each side is equal to a constant C1. Solving for the shear stress,τyr= C1y + C2.Substituting Newton’s law of viscosity and integrating yields:vr=C1y22μ+C2μy+C3(S2.10.3)Applying the boundary conditions that vr= 0 at y=±h/2,0=C1h28μ+C2μh2+C3(S2.10.4a)0=C1h28−C2μh2+C3(S2.10.4b)Adding Equations (S2.10.4a) and (S2.10.4b) and solving for C3,

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14C3=−C1h28(S2.10.5)Inserting Equation (S2.10.5) into Equation (S2.10.4a) yields C2= 0. Thus the velocity is:vr=C1μy22−h28⎛⎝⎜⎞⎠⎟(S2.10.6)The volumetric flow rate is:Q=v•ndA∫=2πrvrdyy=−h / 2h / 2∫=2πrC1μy22−h28⎛⎝⎜⎞⎠⎟dyy=−h / 2h / 2∫(S2.10.7)Q=2πrC1μy36−h2y8⎛⎝⎜⎞⎠⎟y=−h / 2h / 2=2πrC1h3μ124−18⎛⎝⎜⎞⎠⎟=− πrC1h36μ(S2.10.8)Solving for C1and inserting into equation (S2.10.6)vr=−6Qπrh3y22−h28⎛⎝⎜⎞⎠⎟(S2.10.9)The shear stress can thus be written as;τw=τyr y=−h / 2= μdvrdry=−h / 2=−6Qπrh3yy=−h / 2=3μQπrh2(S2.10.10)2.11.Flow rate per fiber, Qf= Q/250 = 0.8 mL/60 s = 0.01333 mL/sAverage velocity per fiber:<vf> = Qf/πRf 2= (0.01333 mL/s)/(3.14159*(0.01 cm)2)<vf> = 42 cm/sRe =ρ<vf>Df/μ= 1.05*42*0.02/0.03 = 29.7.Le= 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 acylindrical tube in Section 2.7.3.dpdz=1rd(rτrz)dr(S2.12.1)(b) The velocity profile is sketched below:Integrating the momentum balance and substituting Newton’s law of viscosity,τrz=− Δp2L r+C1r= μdvzdr(S2.12.2)

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15Note that the shear stress and shear rate are a maximum atr=2C1Δp / L. Assuming that C1is greaterthan zero, then r will have a maximum in the fluid.(c) Integrating Equation (S2.12.2) yields:vz=− Δp4μL r2+C1μln(r)+C2(S2.12.3)Applying the boundary conditionsV=− Δp4μL RC2+C1μln(RC)+C2(S2.12.4a)0=− Δp4μL R2+C1μln(R)+C2(S2.12.4b)SubtractingV=− Δp4μLRC2−R2()+C1μln RCR⎛⎝⎜⎞⎠⎟(S2.12.5)Solving for C1:C1=μVlnRCR⎛⎝⎜⎞⎠⎟+Δp4LRC2−R2()lnRCR⎛⎝⎜⎞⎠⎟(S2.12.6a)Using this result to find C2C2=Δp4μL R2−VlnRRC⎛⎝⎜⎞⎠⎟+Δp4μLRC2−R2()lnRRC⎛⎝⎜⎞⎠⎟⎛⎝⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟ln(R)(S2.12.6b)The resulting expression for the velocity profile isvz=ΔpR24μL1−r2R2⎛⎝⎜⎞⎠⎟+V+Δp4μLRC2−R2()⎛⎝⎜⎞⎠⎟lnrR⎛⎝⎜⎞⎠⎟lnRRC⎛⎝⎜⎞⎠⎟(S2.12.7)(d) The shear stress is:τzr=μdvzdr=−rΔp2L+μV+Δp4LRC2−R2()lnRRC⎛⎝⎜⎞⎠⎟⎛⎝⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟1r⎛⎝⎜⎞⎠⎟(S2.12.8)(e) At r = R, the shear stress is:

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