Thermal Analysis · Title: Présentation PowerPoint Author: Jean-Marc SAITER Created Date: 10/22/2017 1:23:56 PM

Thermal AnalysisPr. Dr. Jean Marc Saiter

Short Courses POLYCHAR 25Kuala Lumpur

Onyx développement, Hameau du Bois Ricard, 76770 Malaunay - FranceSMS sciences et méthodes séparatives, Université de Normandie, 76821 Mont Saint Aignan

Cedex, France

1Copyright © 2017 by Jean-Marc Saiter

2

T1T2 >

ALL IS LINKED TO : ENTROPY x y

z

Thermal AnalysismetrologyTemperature, HeatDefinition ?

Copyright © 2017 by Jean-Marc Saiter

Physical property

1 Temperature

Time

Physical property

Temperature

Isothermal method dynamic method

Thermal Analysis

signal


Physical properties

Temperature Calorimetry adiabatic TDelta Temperature ATD ∆T

Power (Delta) DSC ∆P

Position, distance dilatation ∆lVolume expansion ∆V

Mass Thermogravimetry ∆m

Current Dielectric constantCDTS

magnetic

optical

Plus a fequency

Delta Temperature ATD ∆T

Position, distance dilatation ∆l

Current Dielectric constant


5

Thermodynamic

dU = TdS –P dV

dH = TdS + VdP

dU = d W = dQ if V = cst

Internal energy

Entrropy

pressure

volumeTemperature

Enthalpy

dH = d W = dQ if P = cst

Work Heat


6

From Thermodynamic

Cp - Cv= α2T/ρβΤ

Coef of thermal expansion

density

Isothermal compressibility


Very important

Heat transfert

heatmass

Heat capacity

Temperature

This is the measurementThis is what i want to know

This is a function of T and phase change

Must be a constant

If not we are blind

Calorimetry


Before any Calorimetric, measurement, we have to be sure at 100% that no mass loss will occur

during the experimental duration.


We never do a DSC measurement to see

We do a DSC measurement to get an answer to one question


Question ?

What the litterature said ?

protocole

Sample shapeSample massTemperature domain scannedHeating rateCooling rateAtmosphereKind of calorimeterKind of pans

Mass stabilityChemical stability

measurements

Data analysis

Thermogravimetry

Cheking mass constant


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Measurement by a comparison to zero

Measurement with a pre calibrated probe

Needs

Control of the probe

Control the zero

=

base line


12

DSC HEAD

Gas purge exhaustTwo owens

Low inertia

Compensation method


∆P

t

T

Electronic calibration P=UI = RI2


∆P

T

Before adjustment

After baseline adjustments

The signal is proportional to

the sample mass

the heating rate

Good balance


J/g

T

signal

J/g

T

Surface = Energy

How to do to get a good signal by means of DSC measurement

Tr

Endo

Exo

Heating mode


How to doBest signal

Small sample mass

Sample mass increases

Small heating Rate

heating Rate increases


What a good calibration is ?

The position of the signal depends upon the heating rate

If q+ increases the T signal increases

but: the melting temperature of a pure element is an invariant

q1 < q2

This is a effect of the oven geometry and heat transfert inertia

This is true for the surface.

So we may calibrate the Temperature scale according the values given in the data handbooks


Measured T

Expected T

Slope 1 is the best

T1

T2

T2

T1

Measured T

Expected T

T’2

T2

If only one element is used, i can find an infinity of line passing by this pointSo the calibration is good for T2 and absolutly not good for all the other temperatures


To be able to calibrate the Temperature scale, we need at least to know the melting temperature of 2 pure elements

Measured T

Expected T

T1

T’2

T2

T’1Only if the base line is known and exhibit a linear shape

Slope can be different from 1The measurements can be performed only between T1 and T2

OPEN THE BIG QUESTION OF THE BASE LINE QUALITY

What is true for T is true for the enthalpy


If the base line quality is bad

If the base line quality is bad

different optionsCleaning procedure

Work to optimize the base line

The data obtained will be bad too

30 k$ to 60k$

We have to linearise only in the temperature domain needs for the study

That is not possible for all the different kind of equipments available on the market


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How to do to extract the good temperature and good the enthalpy from a DSC curves


Ideal signal

Endo

WHY ?

glass Crystal

Like liquid liquid

Glass transition Cold crystallization

melting

Are not material constantsq+ and q- play a game


H, V, S

T

liquid

crystal

crystallization

∆CP

TS

α

During cooling


H, V, S

T

liquid

crystal∆CP

Tf

α

fusion

During heating


H, V, S

T

liquid

crystal

crystallization

α

During cooling

High cooling rate

glass

Tg

∆CP


TTg

H or V or S

TTg

Power,

∆P#∆Cp(Tg)

∆ Tg

By Calorimetry during heating

TRUE : ONLY if q+ = q- and no ageingGlass transition TgDuring heating


28

Tg mesurement

T

glass

H or V or S

Slope = Cp (L)

Slope = Cp (G)

Tg


29

T

H or V or S

TTg

Power,

Tx

Sx S

Tg mesurement

Tx= Tg when Sx=S


T

H or V or S

TTg remains the same

Power,

∆Cp(Tg) remains the same

If q+ ≠ q- but the same q-

q+>q-

q+<q-

∆ (Tg) is modified


T

H or V or S

TTg

Power,

q+>q-

Tx

Sx

S

If q+ ≠ q- but the same q-

Tx= Tg when Sx=S


TTg

H or V or S

Tf1Tf2Ta

T

∆H

Tg mesurement

Ageing effects

Tx= Tf when Sx=S


Effect of particle size of aged powder Selenium on measuring the glass transition. As seen, aged Selenium shows a large peak on glass transition.

Tg mesurement Ageing effects and particles sizes

interesting


H, V, S

T

liquid

crystal

crystallization

α

glass

Tg

Like liquid

Glass transition

No melting


Endo

glass Crystal

Like liquid liquid

Glass transition Cold crystallization

melting


H, V, S

T

liquid

crystal

glass

Tg

Cold crystallization

melting


Endo

Crystal

Like liquid liquid

Glass transition

melting

Melt hereWhat it was

Cold crystallized here3 processesNumber 1 nucleationNumber 2 growthNumber 3 : diffusion

TC

Crystallization enthalpy



H, V, S

T

liquid

crystal

crystallization

α

glass

Tg

Glass transition


H, V, S

T

liquid

crystal

crystallization

α

glass

Tg

Glass transition



Cp mesurement

More difficult

Requires an excellent knowledge of the apparatus performances

Requires an excellent calibration procedure

Requires a long experimental time to be good

Why ?

Because we want an absolute value for Cp,

and we are only able to make a comparison


43

∆P = d ∆W/dt = f(T)

= d∆H/dt

=d∆H.dT/dt.dT =q+.d∆H/dT

Heating rate

DSC curve

= q+ CpNO That is WRONG

∆P (Tg) =q+ ((Cpl_Cpref1)-(Cpg-Cpref2)

Not known

(Cpl-Cpg) at Tg = ∆Cp(Tg) = ∆P/q+

if Cp ref1 = Cpref2

TAKE CARE


44

Journal of Materials Education, vol. 30, 2008, pages 51-95.

Quantitative and Transient DSC MeasurementsI. Heat Capacity and Glass Transition

Jean-Marc Saiter, Mehrdad Negahban

Philippe dos Santos ClaroPierre Delabare

Marie-Rose Garda


∆CP at Tg

Tg

TcTm∆HC

∆CP at TC is very difficult to determine

∆HmTmelting is often difficult to get for mixtures and for polymersOften taken at the maximum of the main peak,

To resume

Heating rateCooling rate

AgeAnnealing temperature


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Thermal Analysis · Title: Présentation PowerPoint Author: Jean-Marc SAITER Created Date: 10/22/2017 1:23:56 PM