Chapter 3
Fluid Properties
3.1 Introduction
A fluid is defined as a substance that deforms continuously when a shear stress is applied to
it. Figure 3.1-1 shows the fluid motion when a force F is applied to a plate on top of the fluid
causing the plate to move. As long as there is movement of the plate, the fluid continues to
flow or deform since the fluid next to the plate is under the action of a shear stress equal to
the force F divided by the surface area of the plate. When a fluid is at rest (no relative motion
between the fluid elements), there can be no shear stress.
Plate
F
Fluid
Figure 3.1-1 Fluid moves with the plate.
Both liquids and gases are fluids even though they are quite different at the molecular level.
In liquids the molecules are held close together by significant attraction forces; in gases are
relatively far apart and have very weak attraction forces. Figure 3.1-2 shows a PV diagram
for water where the isotherms are plotted with the isotherm of highest temperature on the top.
An isotherm is a curve that relates pressure to volume at a constant temperature. As
temperature and pressure increase, the differences between liquid and gas become less and
less, until the liquid and gas become identical at the critical point. For water the critical point
occurs at 374.14oC and 22.09 MPa. Because of their closer molecular spacing, liquids
normally have higher densities, viscosities, and other physical properties than gases.
Figure 3.1-2 PV diagram for water.
3-1
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 () and the shear rate (  ) or rate of shear
strain.
 =  
(3.2-1)
In this equation, the proportional constant  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

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
 =

lim
d
=
x ,  y ,  t  0
dt
t  t

t
t
 / 2  arctan[( v x
lim
d
=
x ,  y ,  t  0
dt
y  y
 v x y )t / y ]   / 2
t
arctan[( v x y y  v x y )t / y ]
lim
d
=
x ,  y ,  t  0
dt
t
For small angle , arctan() = , therefore
(v x
lim
d
=
x ,  y ,  t  0
dt
y  y
 v x y )t / y
t
3-2
=
(v x
lim
x ,  y ,  t  0
y  y
 vx y )
y
 =
dv x
d
=
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 yx denotes the viscous flux of x momentum in the y direction. In this
equation, the shear stress is defined to be the same as the momentum flux ( yx)mf. 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.
yx =  
y
h
V
x
t<0
rest
V
t=0
lower plate
moves
t>0
velocity
develops
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.
3-3
yx|y+y
y+y
y
yx|y
Figure 3.3-3 x-Momentum entering and leaving the CV = Ay
Applying the x-momentum balance on the CV yields

( Ay vx) =  yx A   yx
y
t
y  y
A
Dividing the equation by Ay and letting y  0, we obtain for constant physical properties

lim  yx
v x
=
y  0
t
Substituting yx =  
y  y
  yx
y
y
=
 yx
y
(3.2-3)
v x
into equation (3.2-3) yields a second order partial differential
y
equation (PDE)

v x
v
 2v
 2v
=  2x  x =  2x
t
t
y
y
(3.2-4)
where  = / 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

1

 n exp   n 
n 1
2
2
 ny 

 sin 
h   h 
t 
2
The solution can also be expressed in dimensionless form with vx* =
3-4
v x * y * t
,y = ,t = 2
V
h
h
vx* = 1  y* 
2

1
 n exp  n  t sin(n 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
v /V
0.6
0.7
0.8
0.9
1
x
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
3-5
2

1
 n exp  n  t sin(n y*)___

2
n 1
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
 = 
dv x
V
=
h
dy
The force to pull the lower plate at velocity V can be evaluated: F = A  yx
y 0
= A
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

Pseudo plastic
Newtonian fluid
Dilatant
Yield
stress
Shear rate
Figure 3.2-5 Behaviors of Newtonian and non-Newtonian fluids.
3-6
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.
3.3 Fully Developed Laminar Flow in Tube
We want to develop a relationship for shear stress-shear rate given volume flow rate Q and
pressure drop P across the horizontal tube 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
rz
R
Po
r
z
PL
Pz+z
Pz
z
Figure 3.3-1 Forces acting on a cylindrical fluid element within a tubey.
Consider the control volume r2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|zr2  rz2rz  P|z+zr2 = 0
Dividing the equation by the control volume yields

P z  z  P z
z
=
2
rz
r
In the limit when z  0, we obtain the differential equation for the shear stress distribution
3-7

dP
2
= rz
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 rz =  

P  PL
dP
2
= rz = o
dz
r
L
The equation is rearranged to
rz = 
P  PL r
r dP
= o
2 dz
2L
(3.3-2)
The shear stress vanishes at the centerline of the tube and achieves its highest value, w, at the
wall.
w = rz|r=R = 
P  PL R
R dP
= o
2L
2 dz
(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
(3.3-2) and (3.3-3) for
yields
dz

Therefore
dP
2
2
= rz = w
dz
r
R
rz =
P  PL r
r
w = o
2L
R
(3.3-4)
 rz

=  = constant, the fluid is a Newtonian fluid and  is called the viscosity. If rz =  


constant, the fluid is a non-Newtonian fluid and  is called the apparent viscosity. We will
follow different procedures to determine a relationship between shear stress and shear rate
depending on whether or not  and rz are known directly.
If
A)  and rz 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:
 = 
dv z
=  (rz)
dr
(3.3-5)
3-8
We can follow the following procedures to obtain a relationship between the shear rate 
and shear stress rz.
1) We calculate the volumetric flow rate from the axial velocity profile as follows.
R
Qcal = 2  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 =
Therefore
1
Qcal = 2 r2 v z
2
1 2
r
2
R
0
 2 
R
0
R
1 2  dv z 
2  dv 
r 
 dr =   0 r  z  dr
2  dr 
 dr 
3) Next, we change the integration variable from r to rz using equation (3.3-4)
rz =
R
R
r
w  r =
rz  dr =
drz
R
w
w
 rz
Qcal =  
0
Qcal =
R 3
 w3
2
 
R
 dv 
drz , (Note:  =   z  )
R  rz  
w
 dr 
w 
2
 rz

0
  rz2 drz
(3.3-6)
4) We then assume a relationship between  and rz (for example  =
 rz1 / 2
).

Equation (3.3-6) is then integrated to obtain Qcal. We will accept the assumed
expression between  and rz if Qcal  Qmeas. Otherwise step (4) is repeated.
B)  and rz 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).
3-9
T
Ri
Ro

L
Figure 3.3-2 Couette viscometer.
The outer cylinder is rotated at a fixed angular velocity (). The shearing force is transmitted
to the fluid, causing it to deform or flow. The inner cylinder is kept stationary by a torque (T)
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 2rL
T = r(2rL)r
Solving for the shear stress, we have
r =
T
2r 2 L
(3.3-7)
Setting r = Ri gives the stress on the bob surface (i), and setting r = Ro gives the stress on
the cub surface (o). 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
r =
i o
2

T
where R = (Ri + Ro)/2
2R 2 L
(3.3-8)
The average shear rate is given by
 =
dv
V  Vi
Ro 

 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.
3-10
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:
(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
 = o +   (Bingham Plastic Model)
and
 = m  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.
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
r =

T
, and  =
2
2R L
1  Ri / Ro
The apparent viscosity  is determined from
=
 r

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.
(rpm)
2
4
10
20
40
100
0
Table 3.3-1 Fluid apparent viscosity at different shear rates
 (1/s)
r(dyne/cm2)
(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
Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 74
3-11
For the Bingham plastic model, we obtain
 (dyne/cm2) = o +   = 147 + 0.827  (1/s)
For the power law model, we obtain
(dyne/cm2) = m  n = 83.2  0.234
At 100 rpm or  = 524 s-1, for the Bingham plastic model
 = 147 + 0.827524 = 580 dyne/cm2
 = 580/524 = 1.11 g/cms
For the power law model
=
 r
= 83.2  0.2341 = 0.69 g/cms

______ 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);
3-12
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
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
 Y ) 2 is the spread of the data around the mean Y of the
i 1
N
dependent variable and S =
 (Y
i 1
i
 yi ) 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.
3-13
10
2
Viscosity (Poise)
Data
Bingham plastic
Power law
10
1
0
10
1
10
2
10
Shear rate (1/s)
10
3
Figure 3.3-3 Behavior of non-Newtonian fluid.
---------------------------------------------------------------------------------------------------
3.4 The Hagan-Poiseuille Equation
We now consider the case of a Newtonian fluid flowing through a capillary. The shear rateshear stress relation  (rz) =
Q=
Q=
R 3
 w 3
 rz

0
 rz
is substituted into equation (3.3-6) to obtain

 rz3 drz =
R 3  rz
 w 3 4
w
=
0
R 3  w4
 w 3 4
R 3
R 3 Po  PL R R 4 Po  PL 
w =
=
2L
8L
4
4
The velocity profile inside the capillary can also be obtained by integrating equation (3.3-2)
rz =  
P  PL r
dv z
= o
dr
2L
(3.3-2)
3-14
rz =  

vz
0
P  PL r
dv z
= o
dr
2L
dv z = 
vz =
Po  PL 
2L
Po  PL R 2
4L
(3.3-2)
r
 rdr
R
  r 2 
1    
  R  
Example 3.4-1. 1 ---------------------------------------------------------------------------------You are asked to measure the viscosity of an emulsion, so you use a tube flow viscometer
similar to that shown below, with the container open to the atmosphere.
h
Po
L
PL
Q
The length of the tube is 10 cm, its diameter is 2 mm, and the diameter of the container is 3
in. When the level of the sample is 10 cm above the bottom of the container the emulsion
drains through the tube at a rate of 12 cm3/min, and when the level is 20 cm the flow rate is
30 cm3/min. The emulsion density is 1.3 g/cm3. What can you tell from the data about the
viscous properties of the emulsion?
Solution -----------------------------------------------------------------------------------------Equation (3.3-3) provides a relation between the wall shear stress and the pressure drop
across the tube
w = rz|r=R = 
1
P  PL R
R dP
= o
2L
2 dz
Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 80
3-15
(3.3-3)
This equation is valid for any Newtonian or non-Newtonian fluid. Po is essentially the
pressure at the bottom of the container and PL is the ambient pressure Patm. Therefore
Po  PL = gh
The wall shear stress is then given by
w =
ghd
4L
where d is the inside diameter of the tube.
When h = 10 cm, w1 =
(1.3)( 980)(10)( 0.2)
= 63.7 dyne/cm2
(4)(10)
When h = 20 cm, w2 =
(1.3)( 980)( 20)( 0.2)
= 127.4 dyne/cm2
( 4)(10)
If the fluid is a Newtonian fluid then
Q=
so that
R 3
w
4

Q2
= W2 = 2
Q1
W1
From experimental data
 Q2 
30
  =
= 2.5
 Q1  exp 12
Therefore the viscosity at high shear rate is smaller than that at lower shear rate. The
emulsion is a shear thinning liquid.
3-16