3-1

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.

Fluid

FPlate

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.

3-2

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 () 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.

y

x

(v | - v | ) tx y+ y x y

Element at time t

Element at

time t+ t

y

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

= dt

d =

0,,

lim

tyx t

ttt

dt

d =

0,,

lim

tyx t

ytvvyxyyx

2/]/)arctan[(2/

dt

d =

0,,

lim

tyx t

ytvvyxyyx

]/)arctan[(

For small angle , arctan() = , therefore

dt

d =

0,,

lim

tyx t

ytvvyxyyx

/)( =

0,,

lim

tyx y

vvyxyyx

)(

3-3

= dt

d =

dy

dvx

The shear stress for this simple flow is also the molecular momentum flux in the y-direction

and is given as

yx = dy

dvx (3.2-2)

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.

x

y

t < 0rest

t = 0lower plate moves

t > 0velocitydevelops

t >> 0steady velocityprofile

h

V V V F

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 Ay as shown in Figure 3.3-3.

3-4

y

y+ yyx y+ y|

yx y|

Figure 3.3-3 x-Momentum entering and leaving the CV = Ay

Applying the x-momentum balance on the CV yields

t

( Ay vx) = yyx A

yyyx A

Dividing the equation by Ay and letting y 0, we obtain for constant physical properties

t

vx

= 0

lim

y y

yyxyyyx

=

y

yx

(3.2-3)

Substituting yx = y

vx

into equation (3.2-3) yields a second order partial differential

equation (PDE)

t

vx

= 2

2

y

vx

t

vx

= 2

2

y

vx

(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

vx = V(1 h

y)

V2

1

1

n nexp

2

22

h

tn

sin

h

yn

The solution can also be expressed in dimensionless form with vx* =

V

vx , y* =

h

y, t

* =

2h

t

3-5

vx* = 1 y

* 2

1

1

n nexp *22

tn sin(n y*) (3.2-5)

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.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

vx/V

y/h

t*=.01

t*=.05

t*=.1

t*=1

t*=.2

Figure 3.2-4 Dimensionless velocity profiles for flow between two parallel plates.

____ Table 3.2-1 Matlab program to plot vx* = 1 y

* 2

1

1

n nexp *22

tn sin(n 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

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.

2

2

dy

vd x = 0 (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 = dy

dvx = h

V

The force to pull the lower plate at velocity V can be evaluated: F = A0yyx = A

h

V

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.

Yieldstress

Shear rate

Ideal plastic

Real plastic

Pseudo plastic

Newtonian fluid

Dilatant

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.

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 no-

slip boundary condition at the wall.

z

z

Rr

PzPz+ z

rz

L

PoPL

Figure 3.3-1 Forces acting on a cylindrical fluid element within a tubey.

Consider the control volume r2z 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|zr2 rz2rz P|z+zr

2 = 0

Dividing the equation by the control volume yields

z

PPzzz

=

r

2 rz

In the limit when z 0, we obtain the differential equation for the shear stress distribution

3-8

dz

dP =

r

2 rz (3.3-1)

Since rz = dr

dvz , the right hand side of equation (3.3-1) is a function of r only and the left

hand side of equation (3.3-1) is a function of z only. They both must be equal to a constant

dz

dP =

r

2 rz = L

PP Lo

The equation is rearranged to

rz = 2

r

dz

dP =

L

rPP Lo

2

(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 = 2

R

dz

dP =

L

RPP Lo

2

(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

(3.3-2) and (3.3-3) for dz

dP yields

dz

dP =

r

2 rz = R

2 w

Therefore rz = R

r w =

L

rPP Lo

2

(3.3-4)

If 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.

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:

= dr

dvz = (rz) (3.3-5)

3-9

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.

Qcal = 2 R

z rv0

)( rdr

2) We express Qcal in terms of shear rate using integration by part.

d(uv) = vdu + udv v du = )(uvd u dv

Let v = vz(r) dv = dvz(r)

du = rdr u = 2

1r

2

Therefore Qcal = 22

1r

2 R

zv0

2 R

r0

2

2

1

dr

dvz dr = R

r0

2

dr

dvz dr

3) Next, we change the integration variable from r to rz using equation (3.3-4)

rz = R

r w r = w

R

rz dr =

w

R

drz

Qcal = rz

R

0

2

2

w

rz

w

R

drz , (Note: =

dr

dvz )

Qcal = 3

3

w

R

rz

0 2

rz drz (3.3-6)

4) We then assume a relationship between and rz (for example = 2/1

rz ).

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).

3-10

T

L

RiRo

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 2rL

T = r(2rL)r

Solving for the shear stress, we have

r = Lr

22T

(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 = 2

oi

LR22

T where R = (Ri + Ro)/2 (3.3-8)

The average shear rate is given by

= dr

dv io

io

RR

VV

= io

o

RR

R

= oi RR /1

(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.

3-11

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) 2 4 10 20 40

T (dyncm) 3.6105

3.8105 4.410

5 5.410

5 7.410

5

(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 = LR

22T

, and = oi RR /1

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.

Table 3.3-1 Fluid apparent viscosity at different shear rates

(rpm) T (dyncm) (1/s) r(dyne/cm2) (Poise = g/cms)

2

4

10

20

40

100

360000

380000

440000

540000

740000

10.5

20.9

52.5

105

209

524

156

165

191

234

320

14.89

7.86

3.64

2.23

1.53

0 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 74

3-12

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 + 0.827524 = 580 dyne/cm2

= 580/524 = 1.11 g/cms

For the power law model

=

r = 83.2 0.2341

= 0.69 g/cms

______ 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);

3-13

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 = tS

S1

In this expression St =

N

i

i YY1

2)( is the spread of the data around the mean Y of the

dependent variable and S =

N

i

ii yY1

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.

3-14

101

102

103

100

101

102

Shear rate (1/s)

Vis

co

sity

(P

ois

e)

DataBingham plasticPower law

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 rate-

shear stress relation (rz) = rz is substituted into equation (3.3-6) to obtain

Q = 3

3

w

R

rz

rz

0

3 drz =

3

3

w

R

w

rz

04 =

3

3

w

R

4

4

w

Q =

4

3R w =

4

3R

L

RPP Lo

2

=

L

PPR Lo

8

4

The velocity profile inside the capillary can also be obtained by integrating equation (3.3-2)

rz = dr

dvz =

L

rPP Lo

2

(3.3-2)

3-15

rz = dr

dvz =

L

rPP Lo

2

(3.3-2)

zv

zdv0

=

L

PP Lo

2

r

Rrdr

vz =

L

RPP Lo

4

2

2

1R

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

L

Q

Po

PL

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/cm

3. 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 = 2

R

dz

dP =

L

RPP Lo

2

(3.3-3)

1 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 80

3-16

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 = L

ghd

4

where d is the inside diameter of the tube.

When h = 10 cm, w1 = )10)(4(

)2.0)(10)(980)(3.1( = 63.7 dyne/cm

2

When h = 20 cm, w2 = )10)(4(

)2.0)(20)(980)(3.1( = 127.4 dyne/cm

2

If the fluid is a Newtonian fluid then

Q =

4

3R w

so that 1

2

Q

Q =

1

2

W

W

= 2

From experimental data

exp1

2

Q

Q =

12

30 = 2.5

Therefore the viscosity at high shear rate is smaller than that at lower shear rate. The

emulsion is a shear thinning liquid.