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

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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 = vL
Dij
(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 HO2 = H Hb , equation (1.6.4) is simplified to the following:

CO 2 = H O 2 PO 2 + 4C HbS Hct (S1.2.1)
2OP 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 cm3 per 100 cm3 . For O2 , 2OP 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 CO 2 , the pressure difference driving transport is much larger for O 2 than CO 2 .
1.4. The diffusion time is L2 /D ij = (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 = πR 2 L 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)
Cumulative Surface
Area (cm2)
0 30.54 67.86 30.54 67.86
1 11.13 36.49 41.66 104.34
2 4.11 19.82 45.78 124.2
3 1.50 10.70 47.27 134.9
4 3.23 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 4.04 70.29 61.39 322.2
8 4.45 95.74 65.84 418.0
9 5.15 133.76 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 16.21 876.05 122.2 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 = km L/Dij = 5 x 10-9 cm s-1 x 0.0150cm/(1 x 10-10 cm2 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 VO2
= Q C v − C a( ) where Q is the pulmonary blood flow and Cv
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 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.
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 = palv
RT = 105 mm Hg( ) / 760 mm Hg/atm( )
0.08206 L atm/(mol K)( ) 310 K( ) = 0.00543 M
C I = palv
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 O2 /L blood, multiply to oxygen removal rate by 22,400 L O2 per mole of O 2 .
( ) ( )2 2 2 2O O O O1 Hct 4 HctHb HbC H P C S H P= ! + +

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