1

What is a polymer and the repeat unit?

Polymers are very large, longchain molecules.

Some are in the region of 10000atoms long

H2

C*HC *

Close inspection of the longchain shows that it consists of aunit that is repeated many times.A convenient way to show apolymer is to write out therepeating unit.

For PS, the repeat unit is…

Morphology - amorphousAmorphous – means no order or regularity to the material. There are no

crystallites present.

Simulation of 50 chains.

Like spaghetti….

Many entanglements

Computer simulation of a singlepolymer chain - random walk.

The chain adopts a coiled shapeto minimise the overall energy ofthe structure

An amorphous polymer can beeither a liquid or a glassdepending on the temperature

2

The glass to liquid transition

The temperature at which the glassbegins to transform to a liquid istermed the glass transitiontemperature, Tg.

Above Tg, the polymer is a liquidand the chain mobility is relativelyhigh.

Below Tg, chain mobility is virtuallyzero. The polymer behaves like asolid. The structure of the liquid islocked or “frozen” in at the Tg.

Tg

Temperature

H

Volume, V

Length, L

Cross-linked polymers

Polymer chains can be physically bonded together, or cross-linked. These links can be short chain lengths that ‘tie’polymer chains together.

No cross-links(thermoplastic)

Cross-linked(thermoset)

Cross-link

3

An example of a cross-linked system - rubber

Start with a tree -Hevea brasiliensis.A source of naturalrubber - cis-polyisoprene. Ifsulphur is addedand the mixture isheated - cross-linking occurs(vulcanisation).

*

H2C C

HC

H2C *

CH3

sulphur

Rubber - mechanical properties

Cis-butadiene is amorphous and adopts a random coil structure.Application of a load causes the coiled backbone to extend. However,plastic deformation is prevented due to the presence of cross-links.

Cross-links prevent the chains from ‘slipping’ past each other. Removethe load and the chains will re-coil elastically.

4

Semi-crystallinepolymers.

Cooling from theliquid state canresult in theformation ofcrystals(spherulites)

This photographshows theformation ofspherulites withtime

Formation of a spherulite

Repeated branching develops the ‘wheat sheaf’intermediate structure. Further branching makesthe crystallite spherical in shape - spherulite

5

Microstructure

SpheruliteExtensive folding and close packing of the polymer chainin sheets called lamella

What happens when you heat up a polymerfrom a low temperature?

• Non-crystallisablepolymer

• A crystallisablepolymer that isinitially amorphous

• A semi-crystallinepolymer

• A single Tg

• A Tg, crystallisationand melting

• Melting, but with asmall Tg

6

DSCDifferential Scanning Calorimetry

Require a 5mg sample ofthe polymer.

Require 2 aluminiumcontainers (‘pans’)

Decide the temperaturerange and heating rate

Thermal analysis - DSC (differential scanning calorimetry)

‘Difference in heat flow to asample and reference cell is

monitored with time ortemperature’

Sample + pan pan

If the sample takes in orgives out heat, there isan imbalance in thetemperature of the cells

The DSCcompensates for thisdifference intemperature byaltering the powerapplied to the sampleheater

DSC measures this difference with temperature

7

What does a DSC trace look like?

300 350 400 450 500 550 600 650

-8

-4

0

4

8

Re

lative

He

at

Flo

w

Temperature / K

C C

Glass transitiontemperature (Tg)

Crystallisation exotherm

Melting

Endo

Semi-crystalline

DMTADMTA involves adynamic bendingtest in which thetemperature of thesample is varied

DMTA allows the dynamicstiffness and damping in amaterial to be measured

8

A typical DMTA trace

E1

Class test

You have joined the R&D department of a large golfequipment manufacturer. Your first task is to evaluate arange of materials that may be used in golf ball cores.

Two tests have been carried out, DSC and DMTA. Youare supplied with the variation of relative heat flow(RHF) and Storage modulus (E1) with temperature andalso a measure of the damping in the material (tan δ).You must identify the best material for use in a golfball core by eliminating unsuitable materials

9

Example

RH

F

T-50°C 50°C

Tan δ=0.1THIS MATERIAL HAS VERYLOW DAMPING WHICH ISIDEAL FOR A GOLF BALLCORE, BUT IT EXHIBITS AMELTING PEAK IN THETEMPERATURE RANGE OFINTEREST. THEREFORE THEMATERIAL IS UNSUITABLE

Short concise answers are required – please print for clarity. Use a(single) separate sheet to identify each transition on the traces andexplain why each material is/is not suitable. Conclude by statingwhich material (s) is/are suitable.