Chapter 3
The Physical and Flow Properties of Blood
3.1 Introduction
Blood is a viscous fluid mixture consisting of plasma and cells. The composition of the
plasma is shown in Table 3.1-1 with protein accounts for 7-8 wt% of the plasma. Plasma is a
pale yellow fluid that consists of about 91% water and 9% other substances, such as proteins,
ions, nutrients, gases and waste products.
Solute
Na+
K+
Ca++
Mg++
Cl−
HCO3−
HPO4−−, H2PO4−−
SO4−−
Phosphocreatine
Carnosine
Amino acids
Creatine
Lactate
Adenosine triphosphate
Hexose monophosphate
Glucose
Protein
albumin
globulins
fibrinogen
Urea
Others
Total (mOsmole/L)
Table 3.1-1 Composition of the plasma
Plasma
MW
(mOsm/L)
143
4.2
1.3
0.8
108
24
2
0.5
g/ml
2
0.2
1.2
5.6
1.2
---------------------------4
4.8
69,000
35,000-1,000,000
400,000
4.5
2.5
0.3
302.8
The major proteins found in the plasma are albumin, fibrinogen, and the globulins that
consist of the alpha, beta, and gamma types. Albumin is involved in the regulation of pH and
the colloid osmotic pressure. Fibrinogen is involved in the process of blood clotting through
its conversion to long strands of fibrin. The alpha and beta globulins are involved in solute
transport, whereas the gamma globulins are the antibodies that fight infection and form the
basis of the humoral immunity. When plasma from an immune animal is injected into the
3-1
blood of a nonimmune animal, the nonimmune animal will become immune. Because this
process involved body fluids (humors), it is called humoral immunity.
Homeostasis1 is the existence and maintenance of a relative constant environment within the
body. Blood helps to maintain homeostasis due to its circulation through the blood vessels
extended throughout the body. Oxygen enters blood in the lungs and is carried to cells.
Carbon dioxide, produced by cells, is carried in the blood to the lungs, from which it is
expelled. Ingested nutrients, ions, and water are transported by the blood from the digestive
tract to cells. Waste products of cells are transported by the blood to the kidneys for
elimination. Many of the hormones and enzymes that regulate body processes are carried
from one part of the body to another by the blood. The precursor to vitamin D is produced in
the skin and transported by the blood to the liver and then to the kidneys for processing into
active vitamin D, which in turn is transported in the blood to the small intestines, where it
promotes the uptake of calcium. Lactic acid, produced by skeletal muscles during anaerobic
respiration, is carried by the blood to the liver, where it is converted into glucose.
The water, proteins, and other substances in the blood, such as ions, nutrients, waste
products, gases, and regulatory substances, are maintained within narrow limits. Normally,
water intake through the digestive tract closely matches water loss through the kidneys,
lungs, digestive tract, and skin. Therefore, plasma volume remains relatively constant.
Suspended or dissolved substances in the blood come from the liver, kidneys, intestines,
endocrine glands, and immune tissues like the spleen.
The cellular component of blood consists of three main cell types: red blood cells (RBCs) or
erythrocytes, platelets, and white blood cells (WBCs) or leukocytes. The RBCs are the most
abundant cells comprising about 95% of the cellular component of blood. The platelets and
white blood cells only comprise about 5% of the cellular component of blood, therefore their
effect on the macroscopic flow characteristics of blood is negligible. The primary functions
of RBCs are to transport oxygen from the lungs to the various tissues of the body and to
transport carbon dioxide from the tissues to the lungs. Approximately 98.5% of the oxygen
transported in the blood is by hemoglobin within the red blood cells, and the remaining 1.5%
is dissolved in the water part of the plasma. If red blood cells rupture, the hemoglobin leaks
out into the plasma and becomes nonfunctional because the shape of the molecule changes as
a result of denaturation. Carbon dioxide is transported in the blood in three major ways:
approximately 7% is transported as carbon dioxide dissolved in the plasma, approximately
23% is transported in combination with blood proteins (mostly hemoglobin), and 70% is
transported in the form of bicarbonate ions. The bicarbonate ions (HCO3−) are produced
when carbon dioxide and water combine to form carbonic acid (H2CO3), which dissociates to
form hydrogen and bicarbonate ions. Carbon dioxide is combined with water with the help of
an enzyme called carbonic anhydrase, which is located primarily within red blood cells.
Carbon monoxide, which is produced by the incomplete combustion of hydrocarbon, binds to
the iron of hemoglobin to form the relatively stable compound carboxyhemoglobin. Due to
the stable binding of carbon monoxide, hemoglobin cannot transport oxygen, and death may
occur. Carbon monoxide is found in cigarette smoke, and the blood of smokers can contains
5%-15% carboxyhemoglobin.
1
Seeley R.R, Stephens T.D., Tate P., Anatomy & Physiology, McGraw Hill, 2003, p. 640
3-2
3.2 Rheology
Rheologh is the study of the deformation and flow behavior of fluids. For a Newtonian fluid,
we have a linear relationship between shear stress (τ) and the shear rate ( γ& ) or rate of shear
strain.
τ = µγ&
(3.2-1)
In this equation, the proportional constant µ is called the viscosity of the fluid. The viscosity
is the property of a fluid to resist the rate at which deformation takes place when the fluid is
acted upon by a shear forces. As a property of the fluid, the viscosity depends upon the
temperature, pressure, and composition of the fluid, but is independent of the shear rate. Most
simple homogeneous liquids and gases are Newtonian fluid.
(vx|y+∆y - vx|y)∆t
y
∆y
Element at
time t+∆t
Element at
time t
δ
∆x
x
Figure 3.2-1 Deformation of a fluid element.
The rate of deformation of a fluid element for a simple one-dimensional flow is illustrated in
Figure 3.2-1. The flow parallel to the x-axis will deform the element if the velocity at the top
of the element is different than the velocity at the bottom. The shear rate at a point is defined
as
γ& =
δ
lim
dδ
=
∆x, ∆y , ∆t → 0
dt
t + ∆t
−δ
t
∆t
π / 2 − arctan[( v x
lim
dδ
=
∆x, ∆y , ∆t → 0
dt
y + ∆y
− v x y ) ∆t / ∆y ] − π / 2
∆t
arctan[( v x y + ∆y − v x y ) ∆t / ∆y ]
lim
dδ
=−
∆x, ∆y , ∆t → 0
dt
∆t
For small angle θ, arctan(θ) = θ, therefore
(v x
lim
dδ
=−
∆x, ∆y , ∆t → 0
dt
y + ∆y
− v x y ) ∆t / ∆y
∆t
3-3
=−
(v x
lim
∆x, ∆y , ∆t → 0
y + ∆y
− vx y )
∆y
γ& =
dv
dδ
=− x
dt
dy
The shear stress for this simple flow is also the molecular momentum flux in the y-direction
and is given as
dv x
(3.2-2)
dy
The subscript yx on τyx denotes the viscous flux of x momentum in the y direction. We may
use equation (3.2-2) to obtain an expression for shear stress as a function of the fluid velocity
and the system dimension. Consider the situation shown in Figure 3.2-2 where a fluid is
contained between two large parallel plates both of area A. The plates are separated by a
distance h. The system is initially at rest then a force F is suddenly applied to the lower plate
to set the plate into motion in the x direction at a constant velocity V. Momentum is
transferred from a region of higher velocity to a region of lower velocity. As time proceeds,
momentum is transferred in the y direction to successive layers of fluid from the plate that is
in motion in the x direction.
τyx = − µ
y
h
V
x
t<0
rest
V
t>0
velocity
develops
t=0
lower plate
moves
V
F
t >> 0
steady velocity
profile
Figure 3.2-2 Velocity profile development for a flow between two parallel plates.
The velocity profile of the fluid between the parallel plates may be obtained by applying the
momentum balance, which states that
Time rate of change
of linear momentum =
within the CV
rate of linear
momentum enters −
the CV
rate of linear
momentum exits +
the CV
sum of external
forces acting on
the CV
Since the velocity in the x direction vx is dependent on the y direction, we choose the control
volume CV to be A∆y as shown in Figure 3.3-3.
3-4
τyx|y+∆y
y+∆y
y
τyx|y
Figure 3.3-3 x-Momentum entering and leaving the CV = A∆y
Applying the x-momentum balance on the CV yields
∂
( A∆yρ vx) = τ yx A − τ yx
y
∂t
A
y + ∆y
Dividing the equation by A∆y and letting ∆y → 0, we obtain for constant physical properties
ρ
lim τ yx
∂v x
=−
∆y → 0
∂t
Substituting τyx = − µ
y + ∆y
− τ yx
y
∆y
=−
∂τ yx
∂y
(3.2-3)
∂v x
into equation (3.2-3) yields a second order partial differential
∂y
equation (PDE)
ρ
∂v
∂v x
∂ 2v
∂ 2v
= µ 2x ⇒ x = ν 2x
∂t
∂y
∂t
∂y
(3.2-4)
where ν = µ/ρ is the kinematic viscosity of the fluid. Equation (3.2-4) can be solved with the
following initial and boundary conditions:
Initial condition: t = 0, vx = 0
(3.2-4a)
Boundary conditions: y = 0, vx = V and y = h, vx = 0
(3.2-4b)
Equation (3.2-4) with the auxiliary conditions (3.4-2a,b) can be solved by the separation of
variables method with the following result
2V
y
vx = V(1 − ) −
h
π
∞
1

∑ n exp  − n π
n =1
2
2
νt 
 nπy 
 sin 

h   h 
2
The solution can also be expressed in dimensionless form with vx* =
3-5
v x * y * νt
,y = ,t = 2
V
h
h
vx * = 1 − y* −
2
π
∞
1
∑ n exp (− n π t )sin(nπ y*)
2
2 *
(3.2-5)
n =1
Table 3.2-1 lists the Matlab program to plot dimensionless velocity profiles at various
dimensionless times. The results are shown in Figure 3.2-4.
1
0.9
0.8
0.7
0.6
y/h
t*=1
t*=.2
0.5
t*=.1
0.4
t*=.05
0.3
t*=.01
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
vx/V
0.6
0.7
0.8
0.9
1
Figure 3.2-4 Dimensionless velocity profiles for flow between two parallel plates.
*
*
____ Table 3.2-1 Matlab program to plot vx = 1 − y −
yoh=0.05:.05:.95; np=length(yoh);u=yoh;
thetav=[.01 .05 .1 .2 1]; nt=length(thetav);
n=1:20; ns=n.*n;
hold on
for k=1:nt
theta=thetav(k);
for i=1:np
y=yoh(i);
sum1=(1.0./n).*exp(-ns*pi*pi*theta)*sin(n*pi*y)';
u(i)=(1-y)-2*sum1/pi;
end
yp=[0 yoh 1];up=[1 u 0]; plot(up,yp)
end
xlabel('v_x/V');ylabel('y/h'); grid on
3-6
2
π
∞
1
∑ n exp (− n π t )sin(nπ y*)___
n =1
2
2 *
As time approaches infinity, the system reaches steady state and the summation terms in
equation (3.2-5) become zero. The steady state velocity profile is then
vx * = 1 − y*
(3.2-6)
The steady state solution can also be obtained directly from equation (3.2-4) by setting the
temporal derivative equal to zeo.
d 2vx
=0
dy 2
(3.2-7)
Integrating equation (3.2-7) twice, we obtain
vx = Ay + B
The two constants of integration are evaluated from the boundary conditions:
y = 0, vx = V and y = h, vx = 0
Therefore
B = V and A = − V/h
Hence
vx = V(1 − y/h) ⇒ vx* = 1 − y*
The shear rate at any position y in the fluid is given as
γ& = −
dv x
V
=
dy
h
The force to pull the lower plate at velocity V can be evaluated: F = A τ yx
y =0
= Aµ
V
h
Fluids are classified as Newtonian or non-Newtonian, depending upon the relation between
shear stress and shear rate. In Newtonian fluids the relation is linear while in non-Newtonian
fluids, the shear stress is not a linear function of shear rate as shown in Figure 3.2-5.
Ideal plastic
Real plastic
τ
Pseudo plastic
Newtonian fluid
Dilatant
Yield
stress
Shear rate
Figure 3.2-5 Behaviors of Newtonian and non-Newtonian fluids.
3-7
The slope of the Newtonian fluid line is the viscosity. For the non-Newtonian fluids, the
slope is not constant therefore its value at a given shear rate is called the apparent viscosity.
The apparent viscosity of a dilatant fluid increases with shear rate while the apparent
viscosity of a pseudo platic decreases with shear rate. The ideal or Bingham plastic has a
linear shear stress-shear rate relation for stresses greater than the yield stress. Real plastic or
Carson fluid also flows with stresses greater than the yield stress. The apparent viscosity
however decreases with shear rate and at some point the Carson fluid behaves as a
Newtonian fluid. Heterogeneous fluids that contain a particulate phase that forms aggregates
at low rates of shear require a yield stress. Blood is a heterogeneous fluid with the
particulates consisting primarily of the red blood cells. Therefore blood follows the curve
shown for real plastic. At low shear rates, red blood cells clump together to form aggregates.
This behavior results in high value of apparent viscosity. However, at shear rate higher than
100/s, red blood cells do not clump together, therefore blood behaves as a Newtonian fluid
with an apparent viscosity of about 3 cP. The properties of blood change rapidly if removed
from the system and so it is extremely difficult to perform experiments on it under laboratory
conditions.
The heart delivers blood in short busts during contraction into the aorta, therefore the flow in
the artery is pulsatile and not uniform. The arteries are elastic and a typical cross section may
change significantly with time due to the pulsating nature of the flow of blood. Thus it may
be unreasonable to treat the arteries as rigid tubes. As blood arrives at the arterioles and
capillaries, the variation in blood pressure becomes negligible as shown in Figure 1.1-3. The
flow in the arterioles and capillaries can be treated as flow in a rigid pipe. It should be noted
that Poiseuille who developed the theory of pipe flow was in fact a physician, and his interest
was the study of blood flow.
Figure 1.1-3 Pressure variation in the systemic circulation.
3-8
3.3 Fully Developed Laminar Flow in Capillary
We want to develop a relationship for shear stress-shear rate given volume flow rate Q and
pressure drop ∆P across the horizontal capillary as shown in Figure 3.3-1. We use cylindrical
coordinates with the following assumptions: the length of the tube (L) is much larger than the
tube radius (R) (i.e. L/R > 100) to eliminate entrance effect; steady incompressible and
isothermal flow; one-dimensional flow in the z direction only, therefore vz = vz(r); and noslip boundary condition at the wall.
L
τrz
R
Po
z
r
Pz+∆z
Pz
PL
∆z
Figure 3.3-1 Forces acting on a cylindrical fluid element within a capillary.
Consider the control volume πr2∆z shown in Figure 3.5-1. For steady flow, the summation of
the viscous and pressure forces acting on the control volume must be equal to zero.
P|zπr2 − τrz2πr∆z − P|z+∆zπr2 = 0
Dividing the equation by the control volume yields
−
P z + ∆z − P z
∆z
=
2
τrz
r
In the limit when ∆z → 0, we obtain the differential equation for the shear stress distribution
−
2
dP
= τrz
dz
r
(3.3-1)
dv z
, the right hand side of equation (3.3-1) is a function of r only and the left
dr
hand side of equation (3.3-1) is a function of z only. They both must be equal to a constant
Since τrz = − µ
−
P − PL
2
dP
= τrz = o
dz
r
L
The equation is rearranged to
τrz = −
(P − PL )r
r dP
= o
2 dz
2L
(3.3-2)
The shear stress vanishes at the centerline of the capillary and achieves its highest value, τw,
at the wall.
3-9
τw = τrz|r=R = −
(P − PL )R
R dP
= o
2 dz
2L
(3.3-3)
Equations (3.3-2) and (3.3-3) are valid for both Newtonian and non-Newtonian fluids since
we has not specified a relationship between the shear stress and shear rate. Solving equations
dP
yields
(3.3-2) and (3.3-3) for
dz
−
Therefore
2
2
dP
= τrz = τw
dz
r
R
τrz =
(P − PL )r
r
τw = o
R
2L
(3.3-4)
τ rz
τ
= µ = constant, the fluid is a Newtonian fluid and µ is called the viscosity. If rz = η ≠
γ&
γ&
constant, the fluid is a non-Newtonian fluid and η is called the apparent viscosity. We will
If
follow different procedures to determine a relationship between shear stress and shear rate
depending on whether or not γ& and τrz are known directly.
A) γ& and τrz are not known directly
We want to find a general relationship between the shear rate and some function of the shear
stress in terms of the measurable quantities Qmeas., Po − PL, L, and R. That is:
γ& = −
dv z
= γ& (τrz)
dr
(3.3-5)
We can follow the following procedures to obtain a relationship between the shear rate γ&
and shear stress τrz.
1) We calculate the volumetric flow rate from the axial velocity profile as follows.
R
Qcal = 2π ∫ v z ( r ) rdr
0
2) We express Qcal in terms of shear rate using integration by part.
d(uv) = vdu + udv ⇒
∫ v du = ∫ d (uv ) − ∫ u dv
Let v = vz(r) ⇒ dv = dvz(r)
du = rdr ⇒ u =
1 2
r
2
3-10
Therefore
1 2
r vz
2
Qcal = 2π
R
0
− 2π ∫
R
0
R
1 2  dv z 
2  dv 
r 
 dr = − π ∫0 r  z  dr
2  dr 
 dr 
3) Next, we change the integration variable from r to τrz using equation (3.3-4)
τrz =
R
R
r
τw ⇒ r =
dτrz
τrz ⇒ dr =
R
τw
τw
τ rz
Qcal = π ∫
0
πR 3
Qcal =
τ w3
2
τ 
R
 dv 
R  rz  γ&
dτrz , (Note: γ& = −  z  )
 dr 
τw  τw
2
τ rz
∫
0
γ& τ rz2 dτrz
(3.3-6)
4) We then assume a relationship between γ& and τrz (for example γ& =
τ rz1 / 2
).
µ
Equation (3.3-6) is then integrated to obtain Qcal. We will accept the assumed
expression between γ& and τrz if Qcal ≈ Qmeas. Otherwise step (4) is repeated.
B) γ& and τrz are known directly
The shear stress and shear rate can be determined using a cup-and-bob or Couette
viscometer. As the name implies, the Couette viscometer consists of two concentric cylinders
as shown in Figure 3.3-2. The fluid is in the annular gap between the outer cylinder (cup) and
the inner cylinder (bob).
T
Ri
Ro
Ω
L
Figure 3.3-2 Couette viscometer.
The outer cylinder is rotated at a fixed angular velocity (Ω). The shearing force is transmitted
to the fluid, causing it to deform or flow. The inner cylinder is kept stationary by a torque (T)
3-11
that can be measured by a torsion spring. The shear stress at any position r within the gap (Ri
≤ r ≤ Ro) is determined by a balance of moments on a cylindrical surface 2πrL
T = τrθ(2πrL)r
Solving for the shear stress, we have
τrθ =
T
2πr 2 L
(3.3-7)
Setting r = Ri gives the stress on the bob surface (τi), and setting r = Ro gives the stress on
the cub surface (τo). If the gap is small [i.e., (Ro − Ri)/Ro ≤ 0.02], the flow in the annular gap
can be approximated by the flow between two parallel plates. In this case, an average shear
stress should be used
τrθ =
τi +τo
2
≈
T
where R = (Ri + Ro)/2
2πR 2 L
(3.3-8)
The average shear rate is given by
γ& =
dvθ
V − Vi
Ro Ω
Ω
≈ o
=
=
dr
Ro − Ri
Ro − Ri
1 − Ri / Ro
(3.3-9)
Equations (3.3-8, 9) provide the experimental values for the shear stress and the shear rate
that can be fitted by a non-Newtonian fluid model.
Example 3.3-1.0 ---------------------------------------------------------------------------------The viscosity of a fluid sample is measured in a cup-and-bob viscometer. The bob is 15 cm
long with a diameter of 9.8, and the cup has a diameter of 10 cm. The cup rotates, and the
torque is measured on the bob. The following data were obtained:
Ω(rpm)
T (dyn⋅cm)
2
3.6×105
4
3.8×105
10
4.4×105
20
5.4×105
40
7.4×105
(a) Determine the viscosity of the sample.
(b) Fit the data with the following model equations
τ = τo + µ∞ γ& (Bingham Plastic Model)
and
τ = m γ& n (Power Law Model)
(c) Determine the viscosity of this sample at a cup speed of 100 rpm in the viscometer using
the above models.
0
Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 74
3-12
Solution ------------------------------------------------------------------------------------------
Since (Ro − Ri)/Ro = (10 − 9.8)/10 = 0.02, we can use the equation (3.3-8, 9) to determine the
shear stress and the shear rate
τrθ =
T
Ω
, and γ& =
2
2πR L
1 − Ri / Ro
The apparent viscosity η is determined from
η=
τ rθ
γ&
Table 3.3-1 lists the results from the calculation. Table 3.3-2 lists the Matlab program to fit
data with the Bingham plastic and power law model.
Ω(rpm)
2
4
10
20
40
100
Table 3.3-1 Fluid apparent viscosity at different shear rates
γ& (1/s)
τrθ(dyne/cm2)
η(Poise = g/cm⋅s)
T (dyn⋅cm)
360000
10.5
156
14.89
380000
20.9
165
7.86
440000
52.5
191
3.64
540000
105
234
2.23
740000
209
320
1.53
524
For the Bingham plastic model, we obtain
τ (dyne/cm2) = τo + µ∞ γ& = 147 + 0.827 γ& (1/s)
For the power law model, we obtain
τ(dyne/cm2) = m γ& n = 83.2 γ& 0.234
At 100 rpm or γ& = 524 s-1, for the Bingham plastic model
τ = 147 + 0.827×524 = 580 dyne/cm2
η = 580/524 = 1.11 g/cm⋅s
For the power law model
η=
τ rθ
= 83.2 γ& 0.234−1 = 0.69 g/cm⋅s
γ&
3-13
______ Table 3.3-2 Matlab program to fit shear stress and shear rate data ______
% Example 3.3-1
%
rpm=[2 4 10 20 40];
Torque=[36 38 44 54 74]*1e4;
ndata=length(rpm);
Ri=9.8/2;Ro=10/2;L=15;
omega=rpm*2*pi/60;
shear_rate=omega/(1-Ri/Ro)
Rave=(Ri+Ro)/2;
stress=Torque/(2*pi*Rave^2*L)
vis=stress./shear_rate
co=polyfit(shear_rate,stress,1); tao=co(2)
vis_inf=co(1)
x=log(shear_rate);y=log(stress);
co=polyfit(x,y,1); n=co(1)
m=exp(co(2))
drate=(shear_rate(ndata)-shear_rate(1))/25;
s_rate=shear_rate(1):drate:shear_rate(ndata);
tao1=tao+vis_inf*s_rate;vis1=tao1./s_rate;
vis2=m*s_rate.^(n-1);
loglog(shear_rate,vis,'d',s_rate,vis1,s_rate,vis2,':')
xlabel('Shear rate (1/s)');ylabel('Viscosity (Poise)')
legend('Data','Bingham plastic','Power law')
grid on
% Evaluate the correlation coefficient
vis_ave=mean(vis);
St=(vis-vis_ave)*(vis-vis_ave)';
tao1=tao+vis_inf*shear_rate;vis1=tao1./shear_rate;
vis2=m*shear_rate.^(n-1);
S1=(vis-vis1)*(vis-vis1)';r1=sqrt(1-S1/St);
S2=(vis-vis2)*(vis-vis2)';r2=sqrt(1-S2/St);
fprintf('Correlation coefficient for Bingham plastic = %8.4f\n',r1)
fprintf('Correlation coefficient for Power law = %8.4f\n',r2)
>> e3d1
shear_rate =
10.4720 20.9440 52.3599 104.7198 209.4395
stress =
155.8910 164.5516 190.5334 233.8365 320.4426
vis =
14.8865 7.8568 3.6389 2.2330 1.5300
tao =
147.2304
vis_inf =
0.8270
n=
0.2337
m=
83.1932
Correlation coefficient for Bingham plastic = 1.0000
Correlation coefficient for Power law = 0.9938
3-14
A crude measure of the how well the data is fitted by an expression is given by the
correlation coefficient r, which is defined as
r = 1−
S
St
N
In this expression St =
∑ (Y
i
i =1
− Y ) 2 is the spread of the data around the mean Y of the
N
dependent variable and S =
∑ (Y
i =1
i
− y i ) 2 is the sum of the square of the difference between
the data (Yi) and the calculated value (yi).
Figure 3.3-3 shows a plot of viscosity versus flow rate for the Bingham plastic and the Power
law models. The Bingham plastic model fits the data better as evident by its higher
correlation coefficient (1.0) in comparison with that (0.9938) of the Power law model.
10
2
Viscosity (Poise)
Data
Bingham plastic
Power law
10
1
0
10 1
10
2
10
Shear rate (1/s)
Figure 3.3-3 Behavior of non-Newtonian fluid.
---------------------------------------------------------------------------------------------------
3-15
3
10
3.4 The Hagan-Poiseuille Equation
We now consider the case of a Newtonian fluid flowing through a capillary. The shear rateshear stress relation γ& (τrz) =
Q=
Q=
πR 3
µτ w 3
τ rz
∫
0
τ rz
is substituted into equation (3.3-6) to obtain
µ
τ rz3 dτrz =
πR 3 τ rz
µτ w 3 4
τw
=
0
πR 3 τ w4
µτ w 3 4
πR 3
πR 3 (Po − PL )R πR 4 (Po − PL )
τw =
=
4µ
4µ
2L
8µL
The velocity profile inside the capillary can also be obtained by integrating equation (3.3-2)
τrz = − µ
∫
vz
0
(P − PL )r
dv z
= o
dr
2L
dv z = −
vz =
(Po − PL )
2L
(Po − PL )R 2
4 µL
(3.3-2)
r
∫ rdr
R
  r 2 
1 −   
  R  
3.5 The Velocity Profile for a Casson Fluid in Tube Flow
Above a shear rate of 100/sec, blood is a Newtonian fluid with a constant viscosity of about 3
cP. Over the whole range of shear rate, blood may be described by the following empirical
equation known as the Casson equation.
τ1/2 = τ 1y/ 2 + s γ& 1/2
(3.5-1)
In this equation, τy is the yield stress and s is a constant, both of which must be determined
from experimental data. The force applied to stagnant blood must provide a stress greater
than the yield stress before the blood will flow. The reduced average velocity U is defined as
the ratio of the average velocity and the tube diameter
U=
Vave
4Q
=
D
πD 3
(3.5-2)
3-16
The reduced average velocity U has the same dimension as the shear rate (sec-1) and a plot of
τw versus U will show the similar behavior as a plot of shear stress versus shear rate. We can
(τ rz1 / 2 − τ 1y/ 2 ) 2
into equation (3.3-6)
substitute γ& =
s2
Q=
πR 3
τ w3
τw
∫
0
γ& τ rz2 dτrz
(3.3-6)
to obtain
4Q
1
U =
=
3
3
πD
2τ w
τw
∫τ τ
y
2
rz
(τ rz1 / 2 − τ 1y/ 2 ) 2
s2
dτrz
The lower limit is changed to the yield stress τy since the fluid will not move until τ > τy.
U =
U =
U =
1
τw
2s τ w
2
3
1
2 s 2τ w
1
2s 2
3
∫τ τ
y
2
rz
(τ rz − 2τ rz1 / 2τ 1y / 2 + τ y ) dτrz
 τ w4 4 7 / 2 1 / 2 τ w3   τ y4 4 7 / 2 1 / 2 τ y3 
 − τ w τ y + τ y  −  − τ y τ y + τ y 
3   4 7
3 
 4 7
τ w 4 1 / 2 1 / 2 1 τ y4 τ y 
+ 
 − τw τ y −
3
4
7
84
3
τ
w

The measured wall stress τw is obtained from τw = −
(3.3-7)
(P − PL )R and the measured
R dP
= o
2 dz
2L
4Q
. Equation (3.3-7) is then fitted to the data providing values
πD 3
for the two parameters: τy = 0.0289 dynes/cm2 and s = 0.229 (dynes-sec/cm2)1/2. The
physiological values of U are in the range of 5 to 1,000 s-1. The flow in a small vessel is
often observed to be intermittent. It can stop completely and then start again1
U is obtained from U =
The axial velocity profile for a Casson fluid is shown in Figure 3.5-1. For this fluid to flow,
the wall shear stress, τw, must be greater than the yield stress, τy. Since the value of the shear
stress reduces with r, there will be a critical radius, rcrtitical, where the local shear stress will
equal τy. From the tube centerline to this critical radius, the core fluid will have a flat velocity
profile with value vz = vcore. From the critical radius to the pipe wall the velocity will decrease
with r.
1
Merrill, E. W., A. M. Benis, E. R. Gillialand, T. K. Sherwood, and E. W. Salzman. 1965. Pressure flow
relations of human blood in hollow fibers at low flow rates. J. Appl. Physiol. 20:954-967.
3-17
At r = rcritical
τy = τrz(r)
rcritical
R
Vcore
Figure 3.5-1 Velocity profile of a Casson fluid.
We can determine the velocity profile by integrating the expression for the shear stress from
r = rcritical to r = R.
τrz1/2 = τ 1y/ 2 + s γ& 1/2 ⇒ γ& = (τrz1/2 − τ 1y/ 2 )2/s2
(3.5-1)
From equation (3.3-2)
τrz = −
a=
Therefore
γ& = −
vz
0
(3.3-2)
(Po − PL ) = τ w
where
∫
(P − PL )r = ar
r dP
= o
2 dz
2L
2L
R
dv z
1
1
= 2 (a1/2 r1/2 − τ 1y / 2 )2 = 2 (ar − 2 a1/2r1/2 τ 1y / 2 + τy)
dr
s
s
dv z = −
1
s2
∫
r
R
(ar − 2 a1/2r1/2 τ 1y / 2 + τy)dr
Integrating the equation and putting in the limits we obtain the velocity profile for the
interval rcritical ≤ r ≤ R
Rτ w
vz(r) =
2s 2
   r  2  8
 1 −    −
   R   3
At r = rcritical, τrz = τy = −
Since τw = −
τ y 
 
τ w 
1/ 2
  r 3/ 2 
τ y 
1 −    + 2  
  R  
τ w 
rcritical dP
2 dz
τ
r
R dP
, we have critical = y
2 dz
R
τw
3-18
r 

1 −  
R 

(3.5-3)
The core velocity is obtained by substituting
Rτ w
vcore =
2s 2
τ
r
r
= critical = y into equation (3.5-3)
R
R
τw
   τ  2  8  τ 1 / 2
y
y
 1 −    −  
   τ w   3  τ w 
  τ y 3/ 2 
τ y 
1 −    + 2  
  τ w  
τ w 
 τ y 
1 −   (3.5-4)
 τ w 
Example 3.5-1. ---------------------------------------------------------------------------------Consider blood flows in an arteriole with a diameter of 0.01 cm and an average velocity of
0.05 cm/s. Blood is a Casson fluid (τrz1/2 = τ 1y / 2 + s γ& 1/2) with yield stress τy = 0.0289
dynes/cm2 and s = 0.229 (dynes-sec/cm2)1/2. Determine the wall shear stress τw and plot the
velocity profile.
Solution ------------------------------------------------------------------------------------------
We can solve for the wall shear stress from equation (3.3-7)
U =
where U =
4
1 τ w 4 1 / 2 1 / 2 1 τ y τ y 
−
−
+ 
τ
τ

w
y
2s 2  4 7
84 τ w3
3
(3.3-7)
U ave
0.05
= 5 sec-1
=
D
0.01
The unknown wall shear stress τw can be solved by Newton’s method using the Matlab
program listed in Table 3.5-1.
______ Table 3.5-1 Matlab program to solve for the wall shear stress and plot vz ______
%
f='(taow/4-4*(taow*taoy)^.5/7-taoy^4/(84*taow^3)+taoy/3)/(2*s^2)';
taoy=0.0289;% dynes/cm2
s=0.229; % dynes-sec/cm2
taow=.4; % dynes/cm2
dtao = .01;
U=.05;
% cm/s
D=0.01; % cm
Ubar=U/D;
for i=1:20
ft=Ubar-eval(f);
taow=taow+dtao;
ft2=Ubar-eval(f);df=(ft2-ft)/dtao;
etao=ft/df;
taow=taow-dtao-etao;
if abs(etao)<1e-4, break, end
end
fprintf('taow(dynes/cm2) = %9.5f\n',taow)
rc=taoy/taow;
dr=(1-rc)/20;
3-19
r=rc:dr:1;
R=D/2;
vz=R*taow*((1-r.^2)-8*(rc.^.5)*(1-r.^1.5)/3+2*rc*(1-r))/(2*s*s);
rp=[0 r];vzp=[vz(1) vz];
plot(vzp,rp)
ylabel('r/R (R = 0.005 cm)');xlabel('vz(cm/s)');
grid on
>> tubeflow
taow(dynes/cm2) = 2.69727
The velocity profile is plotted in Figure 3.5-2 where the critical radius is given by
τ
rcritical
0.0289
= 0.0107
= y =
R
τw
2.69727
The velocity profile is evaluated from equation (3.5-3)
Rτ w
vz(r) =
2s 2
   r  2  8
 1 −    −
   R   3
τ y 
 
τ w 
1/ 2
  r 3/ 2 
τ y 
1 −    + 2  
  R  
τ w 
r 

1 −  
R 

(3.5-3)
It should be noted that the velocity profile is plotted against the dimensionless distance
r
.
R
Since the wall stress is much larger than the yield stress, the core velocity (vcore = 0.0958
cm/s) is limited to a small region near the centerline.
3-20
1
0.9
0.8
r/R (R = 0.005 cm)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.01
0.02
0.03
0.04
0.05
vz(cm/s)
0.06
0.07
0.08
0.09
0.1
Figure 3.5-2 Velocity profile of the Casson fluid in problem 3.5-1.
3.6 Generalized Mechanical Energy Balance Equation
For laminar flow of a fluid in a cylindrical tube of radius R and length L, the HaganPoiseuille equation provides a relationship between volumetric flow rate and pressure drop
across the tube as follows.
Q=
πR 3
πR 3 (Po − PL )R πR 4 (Po − PL )
τw =
=
4µ
4µ
2L
8µL
3-21
P2
V2
P1
V1
Z1
Z2
Figure 3.6-1 A general piping system.
For a general piping system shown in Figure 3.6-1, we need the generalized relationship,
equation (3.6-1), that can account for the effect of pressure drop on incompressible fluid
flow, changes in elevation, tube cross section, changes in fluid velocity, sudden contractions
or expansions, and friction loss through pipe and fittings such as valves and flow meters.
P1
ρ
+ gZ1 +
α1V12
2
+ η Wp =
P2
ρ
+ gZ2 +
α 2V22
2
+ hfriction
(3.6-1)
Each term in this equation has units of energy per unit fluid mass flow rate or (length/time)2.
P = pressure
ρ = fluid density
g = acceleration of gravity
Z = elevation relative to a reference surface
V = average fluid velocity
α = kinetic energy correction factor
α = 2 for laminar flow
α = 1 for turbulent flow
Wp = work done per unit mass flow rate
η = pump efficiency (η < 1)
hfriction = friction loss due to piping and fitting
The kinetic correction factor is due to the fact that the velocity profile is not uniform over the
cross-sectional area of flow. For uniform flow, the rate of kinetic energy entering a C.V. is
given as
1

E& k =  ρV 2  VA
2

The kinetic energy per unit mass flow rate is then
3-22
E& k
1
= V2
ρVA 2
For turbulent flow, the velocity profile is almost flat, therefore
E& k
1
≈ V2 ⇒ α = 1 for turbulent flow
ρVA turbulent 2
Uniform
velocity profile
vz
V
Laminar
velocity profile
Figure 3.6-2 Laminar velocity profile in a pipe.
The velocity for laminar flow in a pipe is given as
  r 2 
r
vz = 2V 1 −    = 2V (1 − δ2), where δ =
R
  R  
The rate of kinetic energy entering a C.V. is
E& k =
Therefore
∫
R
0
1 2
 ρv z  2πrdr
2

R
R
r r
E& k = πρ ∫ v z3 rdr = πρR2 ∫ v z3   d  
0
0
R R
1
1
E& k = πρR2 ∫ 8 V3(1 − δ2)3δdδ = 8πρR2V3 ∫ (1 − δ 2 ) 3 δdδ
0
0
1
E& k = 8πρR2V3  8  = πρR2V3 = ρA V3
Therefore
E& k
2
= V2, and α = 2 for laminar flow
ρVA la min ar 2
3-23
The friction loss is given by the following equation
hfriction = 4 ∑
i
where
fi =
τw
1
ρV 2
2
LV 2
fi i i +
2 Di
∑
j
V j2
2
Kfitting,j
(3.6-2)
= friction factor in tube segment i with length Li and diameter Di.
Vi = average velocity within tube segment i.
Kfitting = friction loss factor or loss coefficient for pipe fittings, some typical values are
given in Table 3.6-1. The velocity Vj in the summation is for the fluid just
downstream of the contraction, expansion, or fitting.
Table 3.6-1 Friction loss factor for various pipe fittings.
Fitting
Kfitting
7.5
Globe valve, wide open
3.8
Angle valve, wide open
0.15
Gate valve, wide open
A1
A2
4.4
Gate valve, half open
0.7
Standard 90o elbow
0.35
Standard 45o elbow
Sudden contraction
1.5
Tee, through side outlet
0.4
Tee, straight through
Sudden contraction

A 
0.4 1 − 2 
(turbulent flow)
A1 

2
A1
A2
Sudden expansion
 A2



−
1
(turbulent flow)
A

 1

Sudden expansion
The friction factor for laminar flow (Re =
f=
ρVD
< 2000) is given by
µ
16
Re
(3.6-3)
The friction factor for turbulent flow (Re > 4000) can be estimated by
f = {− 1.737 ln[0.269
ε
D
−
2.185
14 -2
ε
ln (0.269 +
)]}
Re
D
Re
3-24
(3.6-4)
In this equation ε is the surface pipe roughness and D is the inside pipe diameter.
Representative values for surface roughness are given in Table 3.6-2.
Table 3.6-2 Surface roughness
Surface
ε (ft)
ε (mm)
Concrete
0.001-0.01
0.3-3.0
Cast iron
0.00085
0.25
Galvanized iron
0.0005
0.15
Commercial steel
0.00015
0.046
Drawn tubing
0.000005
0.0015
If the fluid flows through a noncircular duct, then the equivalent diameter, Deq, can be used in
equations (3.6-2, 3, 4). The equivalent diameter is defined as
Deq = 4rH = 4
where
Across
Pwet
rH = hydraulic radius
Across = cross sectional area of the flow
Pwet = wetted perimeter of the duct
D o Di
Figure 3.6-3 Flow through an annular tube.
For the flow through an annular tube, the equivalent diameter is given as
Deq = 4
π ( Do2 − Di2 ) / 4
= Do − Di
π ( Do + Di )
Example 3.6-1. ----------------------------------------------------------------------------------
Water is pumped from the upper reservoir to the lower reservoir through the piping system
shown. Determine the power required for the pump if the water flow rate is 60 kg/s. The
fittings from pipe D1 to pipe D2 and from pipe D2 to pipe D3 can be considered to be standard
90o elbows. Data:
h1 = 10 m, h2 = 3 m, L1 = 50 m, L2 = 300 m, L3 = 2 m, D1 = 0.2 m, D2 = 0.5 m, D3 = 0.03 m,
water viscosity = 1 cP = 10-3 kg/m⋅s, ρ = 1000 kg/m3. The pipe roughness is 0.05 mm. The
pump efficiency is 75%.
3-25
(1)
h1
D1, L1
h2
(2)
D3 , L 3
Globe valve
D2, L2
Solution ------------------------------------------------------------------------------------------
Applying the mechanical energy balance between (1) and (2) we have
P1
ρ
+ gZ1 +
α1V12
2
+ ηWp =
P2
ρ
+ gZ2 +
α 2V22
2
+ hfriction
Let the reference level be at (2), the end of pipe 3, the energy equation becomes
Patm
ρ
+ g(h1 + L1 − L3) + 0 + ηWp =
g(h1 + L1 − L3) + ηWp = gh2 +
A(m2)
3.14×10-2
1.96×10-1
7.07×10-4
D(m)
.2
.5
.03
hfriction = 4 ∑
i
4 ∑ fi
i
∑
j
V j2
2
LV 2
fi i i +
2 Di
V(m/s)
1.91
0.306
84.9
∑
j
V j2
2
α 3V32
2
Patm + ρgh2
ρ
+0+
α 3V32
2
+ hfriction
+ hfriction
Re
3.82×105
1.53×105
2.55×106
ε/D
2.50×10-4
1.00×10-4
0.0017
f
0.00406
0.00431
0.00600
Kfitting,j
LiVi 2
50 × 1.912
300 × 0.306 2
2 × 84.9 2
]
+ 4.31×
+ 6×
= 2× 10-3[4.06×
2 Di
0.2
0.2
0.2
= 5.77× 103 m2/s2
Kfitting,j = 0.5×1.912×0.4
sudden contraction, Kfitting = 0.4
+ 0.5×0.3062×0.7
standard 90o elbow, Kfitting = 0.7
3-26
∑
j
V j2
2
+ 0.5×0.3062×7.5
open globe valve, Kfitting = 7.5
+ 0.5×84.92×0.7
standard 90o elbow, Kfitting = 0.7
Kfitting,j = 2.52× 103 m2/s2
hfriction = 5.77×103 + 2.52×103 = 8.29×103 m2/s2
Therefore
g(h1 + L1 − L3) + ηWp = gh2 +
α 3V32
2
+ hfriction
9.81(10 + 50 − 2) + 0.75 Wp = 9.81×3 +
Wp = 1.51×104 m2/s2
84.9 2
+ 8.29×103
2
The power required for the pump is
W& p = m& Wp = 60×1.51×104 = 9.08×105 W = 1220 hp
Note: 1 hp = 746 W
---------------------------------------------------------------------------------------------------
3-27