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MECHANICAL DEGRADATION

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Presentation on theme: "MECHANICAL DEGRADATION"— Presentation transcript:

1 MECHANICAL DEGRADATION
The polymer chain is ruptured by mechanical means. The effect is to reduce the polymer molecular mass. This is sometimes done deliberately. Example:- MASTICATION of natural rubber but if it occurs in service the polymer properties will be adversely affected. Usually the chain break is made permanent by oxygen attack and mechanical deg. Is more accurately described as MECHANOCHEMICAL DEGREDATION

2 MECHANICAL DEGRADATION
1) Shear of chemical bonds Solid state Results perhaps in relaxation Y X i.e. break made permanent e.g. by chemical reaction most likely with O2 Correctly call this type of degradation as mechanochemical degradation Limiting RMM:- under a given set of conditions related to reduction in viscosity and ability of chains to more rather than be broken

3 MECHANICAL DEGRADATION IN SOLUTION
1) Ultrasonics cavitation bubble forms to collapses chain subjected to effectively a shock wave which shears the chain. Air bubble Intensive shear force

4 2) Turbulant flow PolymersSolution Sprayed out high shear Mechanical degradation Capillary

5 3) Violent shaking or stirring
It will be the highest molecular masses which will be affected., and hence whilst the average molecular mass will fall the molecular mass distribution will change. a) average value falls b) distribution narrows In making up solutions for molecular mass characterization the addition of antioxidants may not be permitted. I.e. the value of Mn would be lowered.

6 Mechanical Degradation in the Solid State
Example Mastication:- Deliberate break down of chains to reduce viscosity to aid dispersion of compounding ingredients. In use of N.R. i.e. RMM up to 5X106 With synthetic polymer RMM controlled in polymerization reaction in order to avoid reed to masticate polymer.

7 EFFICIENCY OF MASTICATION
For a given set of shearing conditions efficiency of mastication depends on:- 1) Shearing force. 2) Molecular mass of original polymer. 3) Temperature. 4) Environment.

8 SHEARING FORCE Greater the shearing force the greater the reduction in molecular mass.

9 MOLECULAR MASS The greater the molecular mass the greater the reduction in molecular mass. Always falls to a limiting value. At low temperature Polymer molecules have high viscosity and therefore, chains are subjected to maximum shearing force (high efficiency). As the temperature increases viscosity falls and the mobility of chains increases and avoid the shearing force (efficiency decreases). At high temperature Degradation is caused by thermal oxidative breakdown Mastication continually exposing new surfaces for attack (efficiency increases).

10 TEMPERATURE At low temperature
Polymer molecules have high viscosity and therefore, chains are subjected to maximum shearing force (high efficiency). As the temperature increases viscosity falls and the mobility of chains increases and avoid the shearing force (efficiency decreases). TEMPERATURE Temperature Efficiency At high temperature Degradation is caused by thermal oxidative breakdown Mastication continually exposing new surfaces for attack (efficiency increases).

11 ENVIRONMENT The shearing forces ruptures the chain. Under N2 +
In air The shearing forces ruptures the chain. Efficiency Mastication time Under N2 + benzoquinone Under nitrogen (N2), chain recombination can occur (low efficiency) Under N2 In air (O2 present), the radicls formed will immediately react with O2, followed by rearrangement reaction which make the break permanent (high effiiciency) Adding benzoquinone under N2 results in increasing the efficiency because benzoquinone acts as inhibitor and stabilises the radicals formed (prevents chain recombination).

12 In hot mastication, thermal oxidative degradation is occuring.
Binzoquinone functions as an antioxidant (efficiency decreases as benzoquinone concentration increased). Under N2 Efficiency Benzoquinone concentration In air at 140 oC (hot)

13 BASIC MECHANISM 1- Shearing force breaks chain Example:-
Natural rubber CH CH3 CH2 – C = CH – CH2 - CH2 – C = CH – CH CH CH3 CH2 – C = CH – CH2 + CH2 – C = CH – CH i.e R R

14 + 2- Under nitrogen, low temperature 3- In presence of oxygen.
R R R – R i.e. recombination 3- In presence of oxygen. R O R – O – O Stable product Permanent break 4- In presence of radical acceptor e.g. benzoquinone RO R + O

15 Thank You See You Next Lecture


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