Solution Manual for Transport Phenomena in Biological Systems George A. Truskey, Fan Yuan and David F. Katz

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1 Solution Manual for Transport Phenomena in Biological Systems George A. Truskey, Fan Yuan and David F. Katz

2 Solution to Problems in Chapter, Section... The relative importance of convection and diffusion is evaluated by Peclet number, Pe = vl (S..) D ij (a) Solving for L, L = PeD ij /v. When convection is the same as diffusion, Pe =, L is.cm. (b) The distance between capillaries is -4 m, O needs to travel half of this distance, and Pe =.455. Therefore, convection is negligible compared with diffusion... Since H O = H Hb, equation (.6.4) is simplified to the following: C O = H O P O + 4C Hb S Hct (S..) P O and S are 95 mmhg and 95% for arterial blood and 8 mmhg 7% for venous blood. C Hb is. mol L - x.45 =.9 M for men, and. mol L - x.4 =.8 M for women. Based on these data, the fraction of oxygen in plasma and bound to hemoglobin is.5% and 98.5% in arterial blood, and.8% and 99.7% in venous blood for men. Corresponding values for women are.7% and 98.% in arterial blood, and.9% and 99.7% in venous blood. Most oxygen in blood is bound to hemoglobin... For CO 7% is stored in plasma and % is in red blood cell. Therefore, the total change of CO is.7(.7)+.98(.) =.8 cm per cm. For O, P O changes from 8 to mmhg after blood passes through lung artery. Using data in problem (.), the total O concentration in blood is.88 M in arterial blood and.6 M in venous blood. At standard temperature (7.5 K) and pressure ( atm =,5 Pa), mole of gas occupies,4 cm. Thus, the O concentration difference of.5 M corresponds to 5.58 cm O per cm. While larger than the difference for CO, the pressure difference driving transport is much larger for O than CO..4. The diffusion time is L /D ij = ( -4 cm) /(x -5 cm s - ) =.5 s. Therefore, diffusion is much faster than reaction and does not delay the oxygenation process..5. V = πr L and the S= πrl where R is the vessel radius and L is the length Order volume, cm surface area, cm cumulative volume, cm cumulative surface area, cm

3 .6. Order Volume (cm ) Surface Area (cm Cumulative Volume Cumulative Surface ) (cm ) Area (cm ) (a) The water content is 55% and 6% of the whole blood for men and women, respectively. Then the water flow rate through kidney is 99 L day - for men and,8 L day - for women. Then the fraction of water filtered across the glomerulus is 8.% for men and 6.67% for women. (b) renal vein flow rate = renal artery flow rate excretion rate =.9 L min - renal vein flow rate =.5 L min - (.5 L day - )/(44 min day - ) =.49 L min - (c) Na + leaving glomerulus = 5, mmole day - /8 L day - = 4 mm. Na + in renal vein = Na + in renal artery - Na + excreted (.5 L min - x 5mM 5 mm day - /(44 min day - ))/.49 L min - = 5.7 mm There is a slight increase in sodium concentration in the renal vein due to the volume reduction..8. (a) Bi = k m L/D ij = 5 x -9 cm s - x.5cm/( x - cm s - ) =.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 .9 The oxygen consumption rate is V O = 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 (.6.4) The fractional saturation S is given by Equation (.6.5). For the data given, the venous fraction saturation is.97. The arterial fractional saturation is.754 under resting conditions and.9 under exercise conditions. Men Women Rest C a =.7 M C a =.6 M Exercise C a =.9 M C a =.7 M C V =.9 M The oxygen consumption rates are Men C v =.8 M Women Rest.5 mole min -. mole min - Exercise.776 mole min mole min -.. (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 ) (S..) We want to assess the left hand side of Equation (S..) which represents the rate of oxygen removal from the lungs. From the data provided and the ideal gas equation: C alv = p alv ( 5 mm Hg) / = ( 76 mm Hg/atm ) RT.86 L atm/(mol K) ( )( K) =.54 M ( ) ( )( K) =.86 M C I = p alv RT =. atm.86 L atm/(mol K) V I = breaths/min ( )(.56.9 L) =.7 L/min males ( )(.45.4 L) =. L/min females V I = breaths/min ( ) ( ) Hb Hb C = H P! Hct + 4C S + H P Hct O O O O Since we have all terms on the left hand side of Equation (), the rate of oxygen removal from the lungs is: ( ) = (.7 L/min) (.8 mole O /L) =.4 mole O /min males V I C I C alv V I ( C I C alv ) = (. L/min) (.8 mole O /L) =.874 mole O /min females To convert to ml O /L blood, multiply to oxygen removal rate by,4 L O per mole of O. 4

5 For males the value is ml O /min and for females the value is 96 ml O /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 () 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: P O P 5 ( / 6).6 S = = P O P + / 6 5 ( ) ( ).6 ( ) =.6 Likewise for the pulmonary artery: P O P 5 ( / 6).6 S = = P O P + / 6 5 ( ) ( ).6 ( ) =.6 This is substantially less than the value in the pulmonary artery under resting conditions, S =.754. The concentration in blood is: C = H P! Hct + 4C S + H P Hct For men C v = (. x 6 M mmhg )( mmhg).55 + ((. M) (.57) + (.5 x 6 M mmhg )( mmhg) ).45 =. M C a = (. x 6 M mmhg )( mmhg).55 + ((. M) (.97) + (.5 x 6 M mmhg )( mmhg) ).45 =.9 M For women C v = (. x 6 M mmhg )( mmhg).6 + ((. M) (.57) + (.5 x 6 M mmhg )( mmhg) ).4 =.75 M ( ) mmhg C a =. x 6 M mmhg ( ).6 + ((. M) (.97) + (.5 x 6 M mmhg )( mmhg) ).4 =.8 M Thus, the oxygen consumption rates are ( ) ( Hb Hb ) C = H P! Hct + 4C S + H P Hct O O O O ( ) ( Hb Hb ) O O O O 5

6 Q( C v C a ).48 mole O /min men. mole O /min women These values are about 4 times larger than the values under resting conditions. From Equation () V ( I = Q C C v a ) 5.5 L O /min men 46.8 L O /min women ( C I C alv ) For a respiration rate of breaths per minutes, the net volume inspired in each breadth is:.75 L/min for men and.56 L/min for women. In terms of the total air inspired in each breadth, it is.94 L/min for men and.7 L/min for women... CO = HR x SV where CO is the cardiac output (L min - ), SV is the stroke volume (L) and HR is the hear rate in beat min -. Stroke Volume, L Rest Exercise Athlete.8.8 Sedentary person.694. The peripheral resistance is R = p a / CO Peripheral resistance, mm Hg/(L/min) Rest Exercise Athlete 5. Sedentary person 6 W = p a dv = p a ΔV since the mean arterial pressure is assumed constant. DV corresponds to the stroke volume. Note L = cm *( m/ cm) =. m mm Hg =, Pa Sedentary person W = ( mm Hg)(. Pa/mm Hg)(.69 L)( cm /L)( m /x 6 cm ) = Work, J (N m) Rest Exercise Athlete. 4. Sedentary person Power, W (J/s) Rest Exercise Athlete. 7. Sedentary person

7 .. Although the pressure drops from 76 mm Hg to 485 mm Hg, the partial pressures are unchanged. The inspired air at,65 m is.85 mm Hg. For a mm Hg drop, the alveolar air is at 7.85 mm Hg. The oxygen consumption rate is V O = V I Assuming that the inspired air is warmed to 7 C C I = p I RT C alv = p alv RT ( ) / ( 76 mm Hg/atm ) ( )( K)..85 mm Hg =.86 L atm/(mol K) ( ) / ( 76 mm Hg/atm ) ( )( K) 7.85 mm Hg =.86 L atm/(mol K) ( C I C alv ) =.57 M =.7 M Assuming that the inspired and dead volumes are the same as at sea level ( ) = (.56 L.9 L) = 7.4 L min - V I = f V I V dead The venous blood is at a partial pressure of.98(7.85) = 7. mm Hg The corresponding saturation is.9... (65 kcal/day)*4.84 kj/kcal*(day/4 h)*( h/6 s) = 79.9 J/s Rest Athlete.4 Sedentary person.4.4. The concentrations are found as the ratio of the solute flow rate/fluid flow rate Urine, M Plasma, M Urine/Plasma Sodium Potassium Glucose Urea 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..5. 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 plasma inulin GFR = C urine inulin Q urine Solving for the glomerular filtration rate: 7

8 GFR = C urine inulin C Q =.5 plasma urine inulin. ml min - ( ) = 5 ml min - 8

9 Solution to Problems in Chapter, Section... Q = v nda = y= x= 6 Q = + y dy = y= Q = 5.9 cm s -.. n = = a + a + a = a Rearranging, a = / ρvv.. ( ) = e x x + e y y + e z x + 6 y dxdy = x + 6 yx y= 6 y + 6 y z ρvv ( ) = e x = ( x ρv v) + x ( y ρv v y ) + ( z ρv v) z y= = 7 x= x + e y y + e z z ρe v v + ρe v v + ρe v v x x y y z z dy ( ) Differentiating term by term, ( ρvv) = v x ρv x ρvv ( ) + ( ) + z ρv z y ρv y ( ) = v ( ρv) + ρv v ( ) + ρv x.4. (a) For a two-dimensional steady flow, the acceleration is: v a = v x x + v v y y For v = U o (x y +x)e x - U o (xy +y)e y, v x = U o x + ( )e x - U o ye y a = U o ( x y + x) ( x + )e x - ye y Collecting terms: a = U o x y + x a = U o a = U o ( ) x + v y = U o y ( )e x - U o x + ( )e y ( ) U o xy + y ( ) + xy + y ( )y e U x o ( x v ) + ρv y ( y v ) + ρv z ( z v ) ( ( )e y ) ( ) ye x - x+ ( x y + x)y xy + y ( )( x + ) x + x xy y + x + 4xy + y e x U o yx y + xy 4x y + xy + xy + y e y x + x + x + xy + y e x U o yx y + 6xy + y e y ( ) x + y a = U o x + y + x + At y = and x = a = ( ) ( e x + e y ) = 4e x + 4e y ( ) ye y e + U x o x + y 6x e y 9

10 At y = and x = a = ( ) ( ) + e + ( x ( ) 8 + )e y = 4e x e y (b) From equation..6 Q = since n = e x. v nda = v x dydz Q = 5 y= z= U o 5 ( x y + x) dydz = U o y x=5 ( )dy = U o y y y= 5 = 6 5 = 65 m s - v.5. (a) a x = e x a = v x x x = U x L x x L a x = e x a = v x v x x = U For values given: = L x L x L x x U x L x L a x = 5 m /s ( m.5) 5 = 5 m/s = U L ( ) / / x L ( ) = 8 m/s 5 (b) () The no slip boundary condition is not satisfied. () At x = L, the acceleration is undefined!.6. (a) Using the definition of the volumetric flow rate, Q π Q = vinda = v z rdrdθ R i The cross-sectional area element in cylindrical coordinates is rdrdθ. Since the velocity does not vary with angular position, substitution for v z and integration in the angular direction yields: π R i Q = v max r R i R rdrdθ = πv max r R i rdr R i is used to denote the local radius within the stenosis. Integrating in the radial direction yields: R i Q = πv max r R i rdr r = πv max r 4 4R i R r = = π R i v max

11 Solving for v max : v max = Q π R i = Outside the stenosis, R i = R and: v max = Q π R (b) At z =, the velocity in the stensosis is v max = Q π R = Q i π R.5 R i = R.5 4 z / L =.5R The shear stress in the stenosis is: Q π R.5 4 z L [ ] = 8Q π R / τ rz stenosis = µ v r z = µ r v max r ( ) R i z = = µr i ( z = )v max r = Ri R i z = = µq ( ) π R Outside the stenosis the shear stress is: τ rz = µ r v max r R r = R = µv max = 4µQ R π R.7. Evaluating Equation (.7.) for y = -h/ yields: τ w = τ yx (y = h / ) = Δp L From Equations (.7.) and (.7.6), Δp h (S.7.) L = 8µv max = µq (S.7.) h wh Replacing Δp/L in Equation (S.7.) with the expression in Equation (S.7.) yields τ w = 6µQ wh Solving for h: h = 6µQ wτ w Inserting the values provided for Q, w, µ and τ w yields h =.5 cm..8. (a) Δp = ρgh = ( g cm - )(98 cm s - )(.5 x -4 cm) =.45 dyne cm - (b) Rearranging equation (.4.6) we have

12 T c =.88 x -5 dyne cm - T c = Δp R p R c.9. (a) To find the radius use Equation (.4.6) and treat the pipet radius as the capillary radius R c = R cap. Δp = T c R cap R c For R c = 6.5 µm T c =.6 mn/m= 6 x -5 N/m( x -6 m/µm) = 6 x - N/µm Δp =. mm Hg Since. 5 N m = 76 mm Hg. mm Hg = 6.7 N m - ( m/ 6 µm) =.67 x - N µm - Solving for R cap = + Δp R pcap R c T c R cap = + Δp R c T c R cap = + Δp = R c T c ( ) =.66 µm 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,.4 times the surface area of a spherical cell, or 74. µm. The factor of.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 π R c + π R c L Solving for the length,

13 L = V 4 π R c = π R c 4 π R ( R c ) = 4 π R c ( ) = 48. µm.66 The resulting surface area is SA = 4π R c + π R c L = π ( 4 *.66 + * 48. *.66) = µm This is larger than the surface area 5.9 µm or.4 times the surface area 74. µm. 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. µm and L = 9. µm. 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. µm can enter the capillary... A momentum balance is applied on a differential volume element, πrδrδy, as shown in the figure below. p r πrδy p r +Δr π ( r + Δr)Δy + τ yr y +Δy πrδr τ yr y πrδr = (S..) Divide each term by πrδrδy and take the limit as Δr and Δy go to zero results in the following expression: d( rp) = dτ yr r dr dy (S..) 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. Solving for the shear stress, τ yr = C y + C. Substituting Newton s law of viscosity and integrating yields: v r = C y µ + C µ y + C (S..) Applying the boundary conditions that v r = at y=±h/, = C h 8µ + C h µ + C (S..4a) = C h C h 8 µ + C (S..4b) Adding Equations (S..4a) and (S..4b) and solving for C,

14 C = C h (S..5) 8 Inserting Equation (S..5) into Equation (S..4a) yields C =. Thus the velocity is: v r = C y µ h (S..6) 8 The volumetric flow rate is: Q = Q = πrc µ h / v nda = πrv r dy = πrc y= h / µ y 6 h y 8 h / y= h / = πrc h µ h / y h dy (S..7) y= h / 8 4 = πrc h 8 6µ (S..8) Solving for C and inserting into equation (S..6) v r = 6Q y πrh h 8 The shear stress can thus be written as; τ w = τ yr y= h / = µ dv r = 6Q dr y= h / πrh y = µq y= h / πrh (S..9) (S..).. Flow rate per fiber, Q f = Q/5 =.8 ml/6 s =. ml/s Average velocity per fiber: <v f > = Q f /πr f = (. ml/s)/(.459*(. cm) ) <v f > = 4 cm/s Re = ρ<v f >D f /µ =.5*4*./. = 9.7. L e =.58DRe =.58*(. cm)(9.7) =.4 cm << L = cm... (a) The momentum balance is the same as that used for the case of pressure-driven flow in a cylindrical tube in Section.7.. dp dz = d(rτ rz ) (S..) r dr (b) The velocity profile is sketched below: Integrating the momentum balance and substituting Newton s law of viscosity, τ rz = Δp L r + C r = µ dv z (S..) dr 4

15 Note that the shear stress and shear rate are a maximum at r = than zero, then r will have a maximum in the fluid. C Δp / L. Assuming that C is greater (c) Integrating Equation (S..) yields: v z = Δp 4µL r + C µ ln(r) + C (S..) Applying the boundary conditions V = Δp 4µL R C + C µ ln(r C ) + C (S..4a) Subtracting = Δp 4µL R + C µ ln(r) + C (S..4b) ( ) + C V = Δp 4µL R C R Solving for C : µv C = ln R C R + Δp 4L µ ln R C R ( R C R ) ln R C R (S..5) (S..6a) Using this result to find C C = Δp V 4µL R ln R R C + Δp 4µL ( R C R ) ln R ln(r) R C (S..6b) The resulting expression for the velocity profile is v z = ΔpR 4µL r R + V + Δp ln r ( 4µL R C R R ) ln R R C (S..7) (d) The shear stress is: Δp τ zr = µ dv z dr = rδp µv + ( L + 4L R C R ) ln R r (S..8) R C (e) At r = R, the shear stress is: 5

University of Washington Department of Chemistry Chemistry 453 Winter Quarter 2013

Lecture 1 3/13/13 University of Washington Department of Chemistry Chemistry 53 Winter Quarter 013 A. Definition of Viscosity Viscosity refers to the resistance of fluids to flow. Consider a flowing liquid