The Processing of Polymers Spring 2001. Module 3 Spring 2001Dr. Ken Lewis ISAT 4302 Introduction...

The Processing of Polymers

Spring 2001

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


Introduction2

Applications Automobile parts Packaging materials Electrical and electronic components Household articles Utensils Tubing Foamed products Fibers Films Paints, varnishes Fiber matrix composites


Introduction3

Polymer usage is surpassing most other materials, at least in volume (low density!)

Plastics are replacing Metals Glasses Woods


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.


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


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


In order to understandPolymer processing…

We need some grounding

in viscosity and viscous flow



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


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


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


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


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


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


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


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



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


Flux is the “rate of flow per unit area”


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


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


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.


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


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


Shear Rate3

If we insert Newton’s Law:

zrz

dvdr

r

L

PP

dr

dv ambientpumpz

2


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


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


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


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)


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


So the effect of viscosity on

fluid transport

can be

IMPORTANT


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!



xyx

dvdy

or

constantyx


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 =


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


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

Velocity Gradient

Non-NewtonianPower Law Flow

NewtonianFlow

xdvdy

yx


Viscosity

Velocity Gradient

Non-NewtonianPower Law Flow

NewtonianFlow

Newtonian Fluid Viscosity (slope)

constant

Non-Newtonian Fluid Viscosity is not

constant Profound affect on

processing


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



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)


Slope = n - 1



The effect can be enormous

0.1

1

10

100

1000

10000

0.1 1 10 100 1000 10000 100000

Shear Rate (sec-1)


Slope = n - 1

Polymer B 90°CShear Rate (sec-

1)

Viscosity (Pa s)

0 1000

100 20

1000 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


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


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


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




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




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.



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


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!



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



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



66.83 10

990

P Pa

P psi

Compare this to the non shear thinned value of:

81.02 10P Pa

14,773P psi


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


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.


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

100.0

1,000.0

10,000.0

100,000.0

1,000,000.0

10,000,000.0

100,000,000.0

0 sec-1

.1 sec-1

1 sec-1

10 sec-1

100 sec-1

1000 sec-1

Effect of Temperature on Non-Newtonian Viscosity

Shear Rate

Temperature °C

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The Processing of Polymers Spring 2001. Module 3 Spring 2001Dr. Ken Lewis ISAT 4302 Introduction...

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