Spring 2001 Dr. Ken Lewis ISAT 430 2Module 3
Introduction
Three types of polymers of importance Thermoplastics Thermosets Elastomers
As a group, polymers (plastics) possess Light weight Corrosion resistance Electrical insulating resistance Thermal insulating resistance
Spring 2001 Dr. Ken Lewis ISAT 430 3Module 3
Introduction2
Applications Automobile parts Packaging materials Electrical and electronic components Household articles Utensils Tubing Foamed products Fibers Films Paints, varnishes Fiber matrix composites
Spring 2001 Dr. Ken Lewis ISAT 430 4Module 3
Introduction3
Polymer usage is surpassing most other materials, at least in volume (low density!)
Plastics are replacing Metals Glasses Woods
Spring 2001 Dr. Ken Lewis ISAT 430 5Module 3
Introduction4 – why plastics are important.
Polymers are easily shaped into unlimited designs
Many plastics are molded which is net shaping so little further processing is necessary
Heating is needed, but far less than for most metal processes.
Many times finishing or painting is not necessary.
Spring 2001 Dr. Ken Lewis ISAT 430 6Module 3
Properties Used in Processing
Enthalpy (Specific Heat)
Thermal Conductivity
Viscosity
Most of these materials are all processed under similar constraints
Both of these affect the initial plastification
and the final cooling
Affects the flow throughthe dies and spinnerets
and mold cavities
Spring 2001 Dr. Ken Lewis ISAT 430 7Module 3
Polymer Characteristics of importance
High viscosity
Material Viscosity (Pascal
Seconds)
Water 0.001
Oils 0.1 – 1.0
Dough 300
Molten Glass 100 – 10,000
Polymers 100 – 1500
Spring 2001 Dr. Ken Lewis ISAT 430 8Module 3
In order to understandPolymer processing…
We need some grounding
in viscosity and viscous flow
Spring 2001 Dr. Ken Lewis ISAT 430 10Module 3
Viscosity and Shear Rate Consider two large parallel plates separated by
a fluid.
At time t = 0 the upper plate is set in motion with a velocity v0.
As time proceeds, the fluid gains momentum and we arrive at the final steady state velocity distribution
Spring 2001 Dr. Ken Lewis ISAT 430 11Module 3
V
V
V
t < 0
t = 0
small t
large t
vx(y,t)
vx(y)
Fluid initially at rest
Upper plate set in motion
Velocity buildup inunsteady flow
Final velocity distributionin steady flow.
Y
y
x
Spring 2001 Dr. Ken Lewis ISAT 430 12Module 3
Viscosity and Shear RateConsider two very long and wide parallel plates.
One is at rest and one is moving with velocity v0.
xv y
Y
x
zy
0v
Spring 2001 Dr. Ken Lewis ISAT 430 13Module 3
Shear Rate The fluid adheres to both
walls so the velocity of the fluid
Is zero at the bottom plate
Is v0 at the top plate Is proportional to the
distance from the bottom plate
0x
yv v
Yxv y
Y
x
zy
0v
Spring 2001 Dr. Ken Lewis ISAT 430 14Module 3
Shear Rate
xv y
Y
x
zy
0v
0x
yv v
Y
We may rewrite this as:
0x
vv y
Y
The proportionality constant is just the slope of the line, or:
0 xv dvY dy
Spring 2001 Dr. Ken Lewis ISAT 430 15Module 3
Shear Rate
The slope, or the rate of change of the x-velocity in the y direction is called the shear rate. Shear rates have unit of
sec-1.
The faster the plate moves, or the closer they are together… the more stress is imposed on the fluid
0 xv dvY dy
Spring 2001 Dr. Ken Lewis ISAT 430 16Module 3
Shear Stress yx & Viscosity
To support this motion, There must be a
tangential force on the upper plate
v0
1/Y
F/A, the force per unit area is called stress.
We may rewrite this as:
xv y
Y
x
zy
0v F VA Y
xyx
dvdy
Spring 2001 Dr. Ken Lewis ISAT 430 17Module 3
Newton’s Law of Viscosity
The shear force per unit area is proportional to the local velocity gradient.
The constant of proportionality is called
the viscosity
xv y
Y
x
zy
0v
xyx
dvdy
Spring 2001 Dr. Ken Lewis ISAT 430 18Module 3
Newton’s Law of Viscosity
In the neighborhood of the moving surface, the fluid acquires a certain amount of x-momentum.
This fluid in turn, imparts some of its momentum to the adjacent ‘layer’ of liquid causing it to remain in motion in the x direction.
Hence, x-momentum is transmitted through fluid in the y direction.
Thus, yx may be interpreted as the viscous flux of x-momentum in the y direction
xyx
dvdy
Spring 2001 Dr. Ken Lewis ISAT 430 20Module 3
Shear Flow in a Cylinder
Let’s go from plates to cylindrical flow
Flow exhibited by fluid in pipes, capillaries, etc.
The flow is purely axial
No radial components
Spring 2001 Dr. Ken Lewis ISAT 430 21Module 3
Shear Flow in a Cylinder
Fluid velocity is zero at the wall.
Fluid velocity remains constant on concentric cylindrical surfaces.
The flow is purely axial The fluid velocity reaches
a maximum at the center. This is called:
Laminar Flow
Spring 2001 Dr. Ken Lewis ISAT 430 22Module 3
Velocity Distribution in a Cylindrical Tube
There is friction, both at the wall of the tube Within the fluid itself
Thus, the fluid is: Accelerated by the
pressure gradient Retarded by the
frictional shearing stressPressure gradient
• The fluid moves under the influence of a pressure gradient.
Spring 2001 Dr. Ken Lewis ISAT 430 23Module 3
Shear Rate
The Driving Force is: 2pump ambientP P r
The Resisting Force is: 2rz rL
At equilibrium, they must balance:
2 2pump ambient rzP P r rL
Spring 2001 Dr. Ken Lewis ISAT 430 24Module 3
Shear Rate2
At equilibrium, they must balance:
2pump ambient
rz
P Pr
L
2 2pump ambient rzP P r rL
Solving for the shear stress:
So,Stress is greatest
At the wallAnd zero at the center
Spring 2001 Dr. Ken Lewis ISAT 430 25Module 3
Shear Rate3
If we insert Newton’s Law:
zrz
dvdr
r
L
PP
dr
dv ambientpumpz
2
Spring 2001 Dr. Ken Lewis ISAT 430 26Module 3
Shear Rates4
Shear rate 0 at the center (r = 0) Max at the wall (r = R)
Shear rate is an indication of the stress being seen by the fluid, and how fast it sees it!
The shear rate at the wall for a Newtonian fluid is:
r
L
PP
dr
dv ambientpumpz
2
3
32QD
Q = volumetric flow rate
D = diameter
Spring 2001 Dr. Ken Lewis ISAT 430 27Module 3
ViscositiesMaterial Viscosity (Pa s)
Water 20°C 0.001
Water 100°C 0.00028
Air 20°C 1.8 x 10-5
Air 100°C 2.1 x 10-5
Mercury 20°C 0.0016
Machine Oil 20°C 0.1
Pancake Syrup 20°C 50
Polymer A 150°C 225
Polymer A 250°C 25
Glass (SiO2) 540°C 1012
Glass (SiO2) 1095°C 103
Glass (SiO2) 1370°C 15
Spring 2001 Dr. Ken Lewis ISAT 430 28Module 3
Volumetric Newtonian Flow in a Tube
The laminar flow of a Newtonian fluid in a pipe or tube may be expressed:
4
8PR
QL
Where:
Q = the volumetric flow rate [=] m3/s or gal/min
P = the pressure drop or driving force [=] kg/m2 or Pa
R = the radius of the tube [=] m or cm
L = the length of the pipe [=] m or cm
= the Newtonian viscosity [=] Pa s
Spring 2001 Dr. Ken Lewis ISAT 430 29Module 3
The effect of viscosity on Pressure Drop
The Pressure drop across a pipe is a measure of the energy necessary to drive a fluid through the pipe.
Assume a Newtonian Fluid
Two cases: A viscosity of 0.001 Pa s (like water) A viscosity of 500 Pa s (like many polymers)
Spring 2001 Dr. Ken Lewis ISAT 430 30Module 3
The effect of viscosity on Pressure Drop
PQ 8 L
r4
r 2 cm
Q 50cm
3
sec
Let:
Then:0.001Pa s
203P Pa
0.033P psi
500Pa s
81.02 10P Pa
14,773P psi
0.5r cm 1L m
Spring 2001 Dr. Ken Lewis ISAT 430 31Module 3
So the effect of viscosity on
fluid transport
can be
IMPORTANT
Spring 2001 Dr. Ken Lewis ISAT 430 32Module 3
Viscosity
For a Newtonian fluid, the viscosity is constant. This holds for simple fluids like water, all
gases. However
For almost all polymeric fluids, the viscosity is NOT constant.
Many times it is a function of the shear rate!
Spring 2001 Dr. Ken Lewis ISAT 430 34Module 3
Power Law Fluids
The Ostwald-de Waele Model Known as the Power Law Model
xyx
dvdy
1n
x xyx
dv dvm
dy dy
Note that for n=1, this reduces to Newton’s Law of viscosity with m =
Spring 2001 Dr. Ken Lewis ISAT 430 35Module 3
Power Law Fluids
The deviation of n from unity indicates the degree of Non-Newtonian behavior.
If n < 1, material behavior is pseudoplastic
If n> 1, material behavior is dilatant.
1n
x xyx
dv dvm
dy dy
Spring 2001 Dr. Ken Lewis ISAT 430 36Module 3
Power Law Viscosity
For most polymers, the isothermal viscosity decreases with increasing shear rate. Effect of shear on the entangled polymer chains Usually, in the literature, the viscosity is not
shown as “”, but rather “” So:
1n
x xyx
dv dvm
dy dy
xyx
dvdy
Spring 2001 Dr. Ken Lewis ISAT 430 38Module 3
Viscosity
Velocity Gradient
Non-NewtonianPower Law Flow
NewtonianFlow
Newtonian Fluid Viscosity (slope)
constant
Non-Newtonian Fluid Viscosity is not
constant Profound affect on
processing
Spring 2001 Dr. Ken Lewis ISAT 430 39Module 3
Power Law Viscosity
For a power law fluid:
nyx m
1nm
The effect of shear rate on viscosity can be enormous!
Rememberfor a
Newtonian fluidn=1
is constant
0.1
1
10
100
1000
10000
0.1 1 10 100 1000 10000 100000
Shear Rate (sec-1)
Zero Shear Viscosity
Slope = n - 1
The Effect of Shear Rate on Viscosity
Spring 2001 Dr. Ken Lewis ISAT 430 41Module 3
The Effect of Shear Rate on Viscosity
The effect can be enormous
In this case the zero shear viscosity is about 1000 Pa s.
At a shear rate of 1000 sec-1, the viscosity has dropped to about 5 Pa s
0.1
1
10
100
1000
10000
0.1 1 10 100 1000 10000 100000
Shear Rate (sec-1)
Zero Shear Viscosity
Slope = n - 1
Spring 2001 Dr. Ken Lewis ISAT 430 42Module 3
The Effect of Shear Rate on Viscosity
The effect can be enormous
0.1
1
10
100
1000
10000
0.1 1 10 100 1000 10000 100000
Shear Rate (sec-1)
Zero Shear Viscosity
Slope = n - 1
Polymer B 90°CShear Rate (sec-
1)
Viscosity (Pa s)
0 1000
100 20
1000 3
Spring 2001 Dr. Ken Lewis ISAT 430 43Module 3
Power Law Shear Rates.
It can be shown that for the flow of a power law fluid through a cylindrical pipe, the maximum shear rate is;
3
3 1n Qn r
NoteIf n = 1,
This reduces toThe Newtonian
Shear rate
3
4Qr
Spring 2001 Dr. Ken Lewis ISAT 430 44Module 3
Power Law Shear Rates2
And the volumetric flow rate Q for a Power Law fluid through a pipe can be shown to be:3 1 1
3 1 2
nn nr P
Qn mLn
NoteIf n = 1,
This reduces toThe Newtonian
Flow rate
4
8PR
QmL
Spring 2001 Dr. Ken Lewis ISAT 430 45Module 3
Properties
Polymer Tg Tm Tp nPolyethylene LDPE -100 120 160-240 65 0.35
HDPE -115 130 200-282 240 0.5
Polyvinyl chloride 80 212 160-210 80 0.3
Polystyrene 100 240 180-260 220 0.3
Nylon 6,6 55 26 260-290 100 0.75
Polycarbonate 150 230 280-310 225 0.7
Polyester (ABS) 115 180-240 210 0.25
Spring 2001 Dr. Ken Lewis ISAT 430 46Module 3
The effect of shear rate on viscosity which affects pressure drop.
Remember the problem of finding the pressure drop necessary to push a fluid through a pipe at a desired flow rate.
Two cases The Newtonian fluid (water) with a viscosity of 0.001 Pa s.
The polymer with a zero shear viscosity of 500 Pa s. Let the power law exponent n = 0.55
And remember the conditions:
3
50 0.5 L=1msec
cmQ r cm
Spring 2001 Dr. Ken Lewis ISAT 430 47Module 3
The effect of shear rate on viscosity which affects pressure drop.
Remember the problem of finding the pressure drop necessary to push a fluid through a pipe at a desired flow rate.
In the first case, the results are the same since the fluid is Newtonian and the viscosity is constant….
0.001
203
0.33
Pa s
P Pa
P psi
Spring 2001 Dr. Ken Lewis ISAT 430 48Module 3
The effect of shear rate on viscosity which affects pressure drop.
Remember the problem of finding the pressure drop necessary to push a fluid through a pipe at a desired flow rate.
In the second case The fluid is non-Newtonian This means that the apparent viscosity will be a function of the
shear rate Thus, we must first find the shear rate at the above conditions, Then using our power law relationships find the apparent
viscosity at that shear rate Finally using the power law equation, calculate the pressure
drop that will occur.
Spring 2001 Dr. Ken Lewis ISAT 430 49Module 3
The effect of shear rate on viscosity which affects pressure drop.
We know:
3
3 1n Qn r
3
33
503 1 3 0.55 1 sec0.55 0.5
cmn Qn r cm
1613.4sec
Spring 2001 Dr. Ken Lewis ISAT 430 50Module 3
The effect of shear rate on viscosity which affects pressure drop
1613.4sec
And from the equation
for a power law viscosity
1nm
Remember:m is the zero shear viscosity
500 Pa sAnd
n = 0.55
0.55 11 0.55 1500 613.5secn
m Pa s
27.8Pa s Look at theDifference!
Spring 2001 Dr. Ken Lewis ISAT 430 51Module 3
The effect of shear rate on viscosity which affects pressure drop
3 1 1
3 1 2
nn nr P
Qn mLn
We know the power law equation for the volumetric flow rate, Q
Rearranging and solving for P
3 1
3 1
2
n
n
n
nQ
nP m L
r
Spring 2001 Dr. Ken Lewis ISAT 430 52Module 3
The effect of shear rate on viscosity which affects pressure drop
Rearranging and solving for P3 1
3 1
2
n
n
n
nQ
nP m L
r
Inserting and solving:
0.553
0.554.82 2
50 4.82sec 2 500 1
10
cmm
P Pa s mcmrcm
3 1 3 0.55 14.82
0.55
n
n
Spring 2001 Dr. Ken Lewis ISAT 430 53Module 3
The effect of shear rate on viscosity which affects pressure drop
66.83 10
990
P Pa
P psi
Compare this to the non shear thinned value of:
81.02 10P Pa
14,773P psi
Spring 2001 Dr. Ken Lewis ISAT 430 54Module 3
Velocity profiles of Newtonian and Non-Newtonian Fluids
1 0.5 0 0.5 1
1
1
2
3
4
54.854
3.429 105
Newton n
NonNew n
10.98 w n
Note the difference. The Newtonian profile is
parabolic The Power Law fluid is blunted.
Why? Remember the viscosity for the
PL fluid is a function of shear rate.
Shear rate is highest at / near the wall
The shear gets dissipated and the central part of the PL flow is called “plug flow”.
Ramifications – the fluid there is stagnant
Spring 2001 Dr. Ken Lewis ISAT 430 55Module 3
Velocity profiles of Newtonian and Non-Newtonian Fluids
1 0.5 0 0.5 1
1
1
2
3
4
54.854
3.429 105
Newton n
NonNew n
10.98 w n
Note that the PL flow rate for the same pressure drop is higher
The shear rates are higher and the viscosity becomes lower.
Spring 2001 Dr. Ken Lewis ISAT 430 56Module 3
Effect of Temperature on Viscosity
Usually models using a form of the Arrhenius equation
E
RTAeShear Rate
(sec-1)
Activation energy E (kcal/mole)
0 12.8
10-1 11.4
10 10.3
101 8.5
102 7.2
103 6.1