Applied BioFluid Mechanics  Lee Waite and Jerry Fine
.pdf22 Chapter One
So far, in this analysis, we have made three assumptions: first, steady flow (i.e., dQ/dt 0); second, fully developed flow in a constant–cross section tube (i.e., dP/dx is constant); and third, viscosity m is constant. Now, we need to make the fourth assumption, which is the noslip condition. This means that V at the wall is zero when r equals the radius of the tube.
Therefore, set r Rtube R and V 0 to solve for C1. From Eq. (1.8), we can write
0 5 

1 dP R2 
(1.9) 










1 C1 


2m dx 
2 









or 















C1 
5 2 
1 


dP R2 
(1.10) 











2m dx 2 







Then, Eq. (1.8), which gives the velocity as a function of the radius r, becomes
V 5 
1 dP [ r2 
2 R2] 
(1.11) 

4m dx 


The fifth and final assumption for this development is that the flow is laminar and not turbulent. Otherwise, this parabolic velocity profile would not be a good representation of the velocity profile across the cross section.
Note that in Eq. (1.11), dP/dx must have a negative value; as pressure drops, that would give a positive velocity. This is consistent with the definition of positive x as a value to the right, or downstream. Note also that the maximum velocity will occur on the arterial centerline. By substituting r 0 into Eq. (1.11), the maximum velocity can be obtained and may be written as
Vmax 5 
1 dP [ 2R2 
] 
(1.12) 

4m dx 


Furthermore, velocity as a function of radius may be written in a more convenient form:
r2 
(1.13) 

V 5 Vmax c1 2 R2 d 


Review of Basic Fluid Mechanics Concepts 
23 
1.4.4Flow rate
Now that the flow profile is known and a function giving velocity as a function of radius can be written, it is possible to integrate the velocity multiplied by the differential area to find the total flow passing through the tube. Consider a ring of fluid at some distance r from the centerline of the tube. See Fig. 1.14.
The differential flow dQ passing through this ring may be designated as
dQ 5 2pVrdr 
(1.14) 
To obtain the total flow passing through the cross section, integrate
the differential flow in the limits from r 0 to r Rtube, after substituting Eq. (1.11) for V. Thus,






R 













Q 5 2p 3 Vrdr 






(1.15) 






0 







b ` 






dP 
R 



dP 
R 
a 
r4 

r2R2 
R 


Q 5 
p 

3 sr3 2 rR2ddr 5 
p 

3 
2 
(1.16) 










2m dx 0 

2m dx 0 
4 
2 
0 

This yields the expression:
Q 5 
2pR4 
dP 
i.e., 
(1.17) 
8m 
dx 
h Poiseuille’s law! 




Figure 1.14 Cross section of a tube, showing a ring of fluid dr thick, at a radius of r.
24 Chapter One
Finally, it is now possible to solve for the average velocity across the cross section and to check assumption 5, that is, the flow is laminar. The expression for the average velocity across the cross section is
Vavg 5 
Q 
5 
2R2 dP 
5 
Vmax 
(1.18) 







A 
8m dx 
2 





To verify assumption 5, use Vavg to calculate the Reynolds number and check, to be sure, it is less than 2000. If the Reynolds number is greater than 2000, the flow may be turbulent, and Poiseuille’s law no longer applies.
Remember to check whether Re 5 rVDm is less than 2000.
1.5Bernoulli Equation
For some flows, it is possible to neglect the viscosity and focus on the pressure changes along a streamline, which is a line drawn through the flow field in a direction that is always tangential to the velocity field. For such an inviscid, incompressible flow along a streamline, the Bernoulli equation can be used to investigate the relationship between pressures and velocities. This is particularly useful in situations with converging flows. The equation, named after the Swiss mathematician, physicist, and physician, Daniel Bernoulli, is written as
P1 1 12 rV12 1 rgz1 5 P2 1 12 rV22 1 rgz2
where P1 and P2 pressures at points 1 and 2 V1 and V2 velocities at points 1 and 2 z1 and z2 heights at points 1 and 2
Because the Bernoulli equation does not take frictional losses into account, it is not appropriate to apply the Bernoulli equation to the flow through long constant–cross section pipes, as described by Poiseuille’s law.
1.6Conservation of Mass
If we want to measure mass flow rate in a very simple way, we might cut open an artery and collect the blood in a beaker for a specific interval of time. After weighing the blood in the beaker to determine the mass m, we would divide by the time t, required to collect the blood, to get the flow rate Q m/t. However, this is not always a practical method of measuring the flow rate.
Instead, consider that, for any control volume, conservation of mass guarantees that the mass entering the control volume in a specific time
Review of Basic Fluid Mechanics Concepts 
25 
is equal to the sum of the mass leaving that same control volume over the same time interval plus the increase in mass inside the control volume, that is,

Mass 

mass 


a 

bin 
t 5 a 

bout 
t 1 smass increased 
Time 
time 
The continuity equation presents conservation of mass in a slightly different form for cases where the mass inside the control volume does not increase or decrease, and it is written as
r1A1V1 5 r2A2V2 5 constant
where r1 the fluid density at point 1
A1 the area across which the fluid enters the control volume V1 the average velocity of the fluid across A1
The variables r2, A2, and V2 are density, area, and average velocity at point 2.
For incompressible flows under circumstances where the mass inside the control volume does not increase or decrease, the density of the fluid is constant (i.e., r1 r2) and the continuity equation can be written in its more usual form:
A1V1 5 A2V2 5 Q 5 constant
where Q is the volume flow rate going into and out of the control volume. Consider the flow at a bifurcation, or a branching point, in an artery. See Fig. 1.15. If we apply the continuity equation, the resulting expression is
Q1 5 Q2 1 Q3
or
A1V1 5 A2V2 1 A3V3
Figure 1.15 A bifurcation in an artery.
26 Chapter One
1.6.1Venturi meter example
A Venturi meter measures the velocity of the flow using the pressure drop between two points on either side of a Venturi throat, as shown in Fig. 1.16.
First, suppose that the viscosity of the fluid were zero. We can apply the Bernoulli equation between points 1 and 2, as well as the continuity equation. Assume that the Venturi meter is horizontal and that z1 z2 0.
Figure 1.16 A Venturi flowmeter with pressure measuring ports at points 1 and 2.
(a) The Bernoulli equation is
P1 1 12 rV12 1 gz1 5 P2 1 12 rV22 1 gz2
(b) The continuity equation is
A1V1 5 A2V2 5 Q or V2 5 A1V1
A2
(c)Combining the equations in (a) and (b), and using the assumption, we can write
P1 2 P2 5 1 rV12caA1b2 2 1d
2 A2
(d) Therefore, solving for V1, we obtain
V1 
5 

2sP1 
2 P2d 

ãrcaA2b 
2 



2 1d 





A1 


(e)Once we have solved for V1, the ideal flow can be calculated by applying the continuity equation as
Qideal 5 V1A1
Review of Basic Fluid Mechanics Concepts 
27 
(f)In practice, the flow is not inviscid. Because of frictional losses through the Venturi, it will be necessary to calibrate the flowmeter and adjust the actual flow using the calibration constant c. The actual flow rate is expressed as
Qactual 5 cV1A1
1.7Fluid Statics
In general, fluids exert both normal and shearing forces. This section reviews a class of problems in which the fluid is at rest. A velocity gradient is necessary for the development of a shearing force. So, in the case where acceleration is equal to zero, only normal forces occur. These normal forces are also known as hydrostatic forces.
In Fig. 1.17, a point P1 in a fluid is shown at a depth of h below the surface of the fluid. The pressure exerted at a point in the fluid by the column of fluid above the point is
P1 5 gh
where P1 the pressure at point 1
g the specific weight of the fluid
h the distance between the fluid surface and the point P1
or 
P1 5 rgh 
where r the fluid density
g the acceleration due to gravity
h the distance between the fluid surface and the point P1
h
P1 
Figure 1.17 Fluid in a reservoir 

showing the depth of point P1. 
28 Chapter One
1.7.1Example problem: ﬂuid statics
Aliquid (viscosity 0.002 Ns/m2; density 1000 kg/m3) is pumped through the circular tube, as shown in Fig. 1.18. A differential manometer is connected to the tube to measure the pressure drop along the tube. When the differential reading h is 6 mm, what is the pressure difference between the points 1 and 2?
2 m
4 mm
1 
2 
h
Density of gauge fluid = 2000 kg/m3
Figure 1.18 Pipeline using a differential manometer.
Solution: Assume that the pressure at point 1 is P1. See Fig. 1.19. The pressure at point a is increased by the column of water between the points 1
and a. The specific gravity gwater of the water is 9810 N/m3. If the vertical distance between 1 and a is da, then the pressure at a is
Pa 5 P1 1 gwater da
1 
2 
h
b
Figure 1.19 Schematic of the
a 
manometer. 

The pressure at b can now be calculated from the pressure at a. Point b is at a higher position than point a. Therefore, the pressure at b is less than the pressure at a. The pressure at b is
Pb 5 Pa 2 ggauge h
5 P1 1 gwater da 2 ggauge h
Review of Basic Fluid Mechanics Concepts 
29 
where ggauge is the specific weight of the gauge fluid and h is the height of point b with respect to point a.
The pressure at 2 is also lower than the pressure at b and is calculated from
P2 5 Pb 2 sda 2 hdgwater
5 P1 1 gwater da 2 ggauge h 2 sda 2 hdgwater
Finally, the pressure difference between 1 and 2 is written as
P1 2 P2 5 sggauge 2 gwaterdh
or
P 5 a2000s9.81d mN3 2 1000s9.81d mN3b 10006 m 5 58.9 mN2
1.8 The Womersley Number : a Frequency Parameter for Pulsatile Flow
The Womersley number, or alpha parameter, is another dimensionless parameter that has been used in the study of fluid mechanics. This parameter represents a ratio of transient to viscous forces, just as the Reynolds number represented a ratio of inertial to viscous forces. A characteristic frequency represents the time dependence of the parameter. The Womersley number may be written as
v 
or a 5 rÅ 
vr 
a 5 rÅ n 
m 
where r the vessel radius
v the fundamental frequency7 r the density of the fluid
m the viscosity of the fluidthe kinematic viscosity
In higherfrequency flows, the flow profile is blunter near the centerline of the vessel since the inertia becomes more important than viscous forces. Near the wall, where V is close to zero, viscous forces are still important.
7 The fundamental frequency is typically the heart rate. The units must be rad/s for dimensional consistency.
30 Chapter One
Some typical values of the parameter a for various species are
human aorta 
a 20 
canine aorta 
a 14 
feline aorta 
a 8 
rat aorta 
a 3 
1.8.1 Example problem: Womersley number
The heart rate of a 400kg horse is approximately 36 beats per minute (bpm), while the heart rate of a 3kg rabbit is approximately 210 bpm (Li, 1996). Compare the Womersley number in the horse aorta to that in the rabbit aorta.
a 5 r Åvn or a 5 r Åvmr
Li also points out that the size of red blood cells from various mammals does not vary much in diameter. No mammal has a red blood cell larger than 10 mm in diameter. Windberger et al. also published the values of viscosity for nine different animals, including horse and rabbit. For this example problem, we will assume that the blood density is approximately the same across the species. The published values of viscosity were 0.0052 Ns/m2 for the horse and 0.0040 Ns/m2 for the rabbit.
Allometric studies of mammals show that various characteristics, like aortic diameter, grow in a certain relationship with the size of the animal. From Li, we can use the published allometric relationship for the aorta size of various mammals to estimate the diameter of the aorta. Thus, we have (Li, 1996, Equation 3.7, p. 27)
D 5 0.48 W 0.34
where D is the aortic diameter, given in centimeters, and W is the animal weight, given in kilograms. Therefore, for this problem, we will use an aortic diameter of approximately 3.7 cm for the horse, 2.0 cm for the man, and 0.7 cm for the rabbit. Then,











p 











36a 

brad/s 1060 kg/m3 



ahorse 5 rÅ 



mã 



vr 
5 
3.7 
30 
5 32 

m 

100 


0.0052 Ns/m2 













p 













210a 


brad/s 1060 kg/m3 

arabbit 5 rÅ 
vr 

0.7 

mã 



30 



5 








5 17 

m 
100 



0.004 Ns/m2 

From the results, one can see that transient forces become relatively more important than viscous forces as the animal size increases.
Review of Basic Fluid Mechanics Concepts 
31 
Review Problems
1.Show that for the Poiseuille flow in a tube of radius R, the wall shearing stress can be obtained from the relationship:
4m Q tw 5 pR3
for a Newtonian fluid of viscosity m. The volume rate of flow is Q.
2.Determine the wall shearing stress for a fluid, having a viscosity of 3.5 cP,
flowing with an average velocity of 9 cm/s in a 3mmdiameter tube. What is the corresponding Reynolds number? (The fluid density r 1.06 g/cm3.)
3.In a 5mmdiameter vessel, what is the value of the flow rate that causes a wall shear stress of 0.84 N/m2? Would the corresponding flow be laminar or turbulent?
4.It has been suggested that a power law:
t 5 2b advdr bn
be used to characterize the relationship between the shear stress and the velocity gradient for blood. The quantity b is a constant, and the exponent n is an odd integer. Use this relationship, and derive the corresponding velocity distribution for flow in a tube. Use all assumptions made in deriving Poiseuille’s law (except for the shear stress relationship).
Plot several velocity distributions ( / max versus r/R) for n 1, 3, 5, . . ., to show how the profile changes with n.
5.Aliquid (viscosity 0.004 Ns/m2; density 1050 kg/m3) is pumped through the circular tube, as shown in Fig. A. Adifferential manometer is connected to the tube, as shown, to measure the pressure drop along the tube. When the differential reading h is 7 mm, what is the mean velocity in the tube?
2 m
4 mm
h
Density of gauge fluid = 2000 kg/m3
Figure A