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 = PeDij/v. When convection is the same as diffusion, Pe =1, L is 0.11cm.
(b) The distance between capillaries is 10-4 m, O2 needs 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
=HO2
P
O2
+ 4CHb 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. 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
CO2is 2.27(0.70)+1.98(0.30) = 2.18 cm3 per 100 cm3. For O2,
2
O
P
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/(2×10-5 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 surface area, cm2
1
0.0158
26.27
26.27
2
0.03885
35.32
61.59
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
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
There is a slight increase in sodium concentration in the renal vein due to the volume reduction.
1.8. (a) Bi = kmL/Dij = 5 x 10-9 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
4
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)
class that equates the oxygen removed from the inspired air with the oxygen uptake in the blood.
VI=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:
To convert to mL O2/L blood, multiply to oxygen removal rate by 22,400 L O2 per mole of O2.
( ) ( )
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 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:
Thus, the oxygen consumption rates are
6
Q Cv−Ca
( )
0.148 mole O2/min men
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.
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
=
VICI−Calv
( )
1.13. (1650 kcal/day)*4.184 kJ/kcal*(1day/24 h)*(1 h/3600 s) = 79.9 J/s
Rest
Athlete 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
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,
plasmaGFR =Cinulin
urineQurine
Solving for the glomerular filtration rate:
8
GFR =Cinulin
urine
Cinulin
plasma Qurine =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•ndA
∫=3
2x+6
2y
⎛
⎝
⎜⎞
⎠
⎟dxdy =3
2 2 x2+6
2 2 yx
⎛
⎝
⎜⎞
⎠
⎟
x=0
2
dy
y=0
3
∫
x=0
2
∫
y=0
3
∫
Q=6
2
+12
2y
⎛
⎝
⎜⎞
⎠
⎟dy
y=0
3
∫=6
2y+6
2y2
⎛
⎝
⎜⎞
⎠
⎟
y=0
3
=72
2
Q = 50.91 cm3s-1
2.2.
n=1=a2+a2+a2=3a
Rearranging,
a=1/ 3
2.3.
∇•
ρ
vv
( )
=ex
∂x+ey
∂y+ez
∂z
⎝
⎜⎞
⎠
⎟•
ρ
vv
( )
=ex
∂x+ey
∂y+ez
∂z
⎝
⎜⎞
⎠
⎟•
ρ
exvxv +
ρ
eyvyv+
ρ
ezvzv
( )
=∂
∂x
ρ
vxv
( )
+∂
∂y
ρ
vyv
( )
+∂
∂z
ρ
vzv
( )
2.4. (a) For a two-dimensional steady flow, the acceleration is:
a=vx
∂
v
∂
x
+vy
∂
v
∂
y
At y = 1 and x = 2
28+2+6+1
28+2−12 −1
(b) From equation 2.2.6
11
Solving for vmax:
vmax =2Q
π
Ri
2=2Q
2
1/ 2
⎡
⎤
2
2.7. Evaluating Equation (2.7.30) for y = -h/2 yields:
τ
w=
τ
yx ( y =−h / 2) =Δp
L
h
2
(S2.7.1)
From Equations (2.7.23) and (2.7.26),
Δp
L=8µvmax
h2 = 12µQ
wh3
(S2.7.2)
Replacing Δp/L in Equation (S2.7.1) with the expression in Equation (S2.7.2) yields
τ
w=6µQ
wh2
Solving for h:
h=6µQ
wτw
Inserting the values provided for Q, w, µ and τw yields h = 0.051 cm.
2.8. (a) Δp = ρgh = (1 g cm-3)(980 cm s-2)(2.5 x 10-4 cm) = 0.245 dyne cm-2
12
Tc= Δp
⎛
⎞
2.9. (a) To find the radius use Equation (2.4.16) and treat the pipet radius as the capillary radius Rc
= Rcap.
Δp=2Tc
1
Rcap
−1
Rc
⎛
⎝
⎜⎞
⎠
⎟
For Rc = 6.5 µm
Tc = 0.06 mN/m= 6 x 10-5 N/m(1 x 10-6 m/µm) = 6 x 10–11 N/µm
Δp = 0.2 mm Hg
Since
1.0133 ×105N m−2=760 mm Hg
0.2 mm Hg = 26.7 N m-2(1 m/106 µm)2 = 2.67 x 10–11 N µm-2
Solving for Rcap
1
Rpcap
=1
Rc
+Δp
2Tc
Rcap =1
1
Rc
+Δp
2Tc
Rcap =1
1
Rc
+Δp
2Tc
=1
1
6.5 +2.67
2 6
( )
=2.66 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, 1.4 times the surface area of a spherical cell, or 743.3
µm2. 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
Solving for the length,
13
L=
V−4
3
π
Rc
3
π
Rc
2=
4
3
π
R3−Rc
3
( )
π
Rc
2=4
3
6.53−2.663
( )
2.662=48.2 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 3.3 µm and L =
29.2 µm.
If the cell had no excess area, then the cell would have no capacity to enter a capillary smaller than
itself!
2.10. A momentum balance is applied on a differential volume element, 2πrΔrΔy, as shown in the
figure below.
14
C3=−C1h2
8
(S2.10.5)
2.11. Flow rate per fiber, Qf = Q/250 = 0.8 mL/60 s = 0.01333 mL/s
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.
15
Note that the shear stress and shear rate are a maximum at
r=2C1
Δp / L
. Assuming that C1 is greater
(c) Integrating Equation (S2.12.2) yields:
16
τ
zr =µ
dvz
dr =−RΔp
2L+
µ
V+Δp
4LRC
2−R2
( )
ln R
RC
⎛
⎝
⎜⎞
⎠
⎟
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
1
R
⎛
⎝
⎜⎞
⎠
⎟
(S2.12.9)
For the values provided
τ
zr r=R=−0.17 cm
( )
100 dyne/cm 3
( )
2
+0.03 g/cm/s
( )
+25 dyne/cm 3
( )
0.15 cm
( )
2−0.17 cm
( )
2
( )
( )
1
0.17 cm
( )
⎛
⎝
⎜⎞
⎠
⎟
ln 0.17
0.15
⎛
⎝
⎜⎞
⎠
⎟
τ
zr r=R
=−16.0 dyne/cm2
This compares with a shear stress of -8.5 dyne/cm2 in the absence of the catheter.
2.13. For a Newtonian fluid
2.14. A schematic of a volume element is shown below.
17
The momentum balance is:
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)
V1=
τ
1−
τ
0
( )
H
µ
or
τ
1=
µ
0V1
H+
τ
0
(S2.15.1a,b)
18
τ
yx =mdvx
dy
n−1dvx
dy
=mdvx
dy
⎛
⎝
⎜
⎞
⎠
⎟
n
(S2.15.7)
Rearranging yields an expression for the velocity gradient
dvx
dy
=
τ
yx
m
⎛
⎝
⎜
⎞
⎠
⎟
1 / n
(S2.15.8)
Integrating, and evaluating the boundary condition that vx = 0 at y = 0.
vx=
τ
yx
m
⎛
⎝
⎜
⎞
⎠
⎟
1 / n
y
(S2.15.9)
Evaluating the velocity at y = H for the conditions given,
4V1=2
τ
1
m
⎛
⎝
⎜ ⎞
⎠
⎟
1 / n
H
(S2.15.10a)
16V1=4
τ
1
m
⎛
⎝
⎜ ⎞
⎠
⎟
1 / n
H
(S2.15.10b)
Dividing equation (S2.15.10b) by equation (S2.15.10a) yields:
4=2
( )
1 / n
(S2.15.11)
Solving, yields n = 0.5.
2.16. For a Bingham plastic the momentum balance is unchanged from equation (2.7.57)
The yield stress τo can be determined from the torque required to begin rotation of the outer cylinder,
T = 2πR2Lτo. 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:
19
22
2.17. For a power law fluid the shear stress is related to the shear rate as:
n−1
2.18. Letting t0 = 0, the shear rate function is:
γ
x(t)=
γ
0/
ε
, for −
ε
<t<0
0, for t<−
ε
or t >0
⎧
⎨
⎩
20
Thus, equation (2.5.15a) becomes:
‘)‘(
00dtttG
yx ∫−−=
εε
γ
τ
2.19. Since the apparent viscosity depends upon the shear rate, the fluid is not Newtonian. For a
power law fluid,
1–n
m =
γη
app
. Taking the logarithm of each side yields.
ln
η
app
( )
= ln(m)+(n-1)ln(
γ
)
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