Thermal Analysis. Short Courses POLYCHAR 25 Kuala Lumpur. Copyright 2017 by Jean-Marc Saiter

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1 Thermal Analysis Pr. Dr. Jean Marc Saiter Onyx développement, Hameau du Bois Ricard, Malaunay - France SMS sciences et méthodes séparatives, Université de Normandie, Mont Saint Aignan Cedex, France Short Courses POLYCHAR 25 Kuala Lumpur 1

2 Thermal Analysis Definition? Temperature, Heat metrology z T 2 >T 1 ALL IS LINKED TO : ENTROPY x y 2

3 Thermal Analysis Physical property Physical property 1 Temperature signal Time Temperature Isothermal method dynamic method 3

4 Physical properties Temperature Calorimetry adiabatic T Delta Temperature ATD T Power (Delta) DSC P Mass Thermogravimetry m magnetic optical Position, distance dilatation l Volume expansion V Current Dielectric constant CDTS Plus a fequency 4

5 Thermodynamic Internal energy Temperature du = TdS P dv Entrropy pressure volume Enthalpy dh = TdS + VdP Work Heat du = d W = dq if V = cst dh = d W = dq if P = cst 5

6 From Thermodynamic Isothermal compressibility Cp - Cv= α 2 T/ρβ Τ density Coef of thermal expansion 6

7 Calorimetry Very important Heat transfert Heat capacity heat mass Temperature This is a function of T and phase change This is what i want to know This is the measurement Must be a constant If not we are blind 7

8 Before any Calorimetric, measurement, we have to be sure at 100% that no mass loss will occur during the experimental duration. 8

9 We never do a DSC measurement to see We do a DSC measurement to get an answer to one question 9

10 Question? What the litterature said? Sample shape Sample mass Temperature domain scanned Heating rate Cooling rate Atmosphere Kind of calorimeter Kind of pans protocole measurements Mass stability Chemical stability Thermogravimetry Cheking mass constant Data analysis 10

11 Needs Measurement with a pre calibrated probe Control of the probe Measurement by a comparison to zero Control the zero = base line 11

12 Two owens Gas purge exhaust DSC HEAD Low inertia Compensation method 12

13 Electronic calibration P=UI = RI 2 P T t 13

14 P Before adjustment After baseline adjustments The signal is proportional to the sample mass the heating rate T Good balance 14

15 How to do to get a good signal by means of DSC measurement Endo J/g signal J/g Tr T Surface = Energy Exo T Heating mode 15

16 How to do Best signal Small sample mass Small heating Rate Sample mass increases heating Rate increases 16

17 What a good calibration is? The position of the signal depends upon the heating rate If q+ increases the T signal increases This is a effect of the oven geometry and heat transfert inertia but: q1 < the melting temperature of a pure element is an invariant q2 So we may calibrate the Temperature scale according the values given in the data handbooks This is true for the surface. 17

18 Expected T T1 T2 Slope 1 is the best T1 T2 Measured T T 2 Expected T If only one element is used, i can find an infinity of line passing by this point So the calibration is good for T2 and absolutly not good for all the other temperatures T2 Measured T 18

19 To be able to calibrate the Temperature scale, we need at least to know the melting temperature of 2 pure elements Expected T Slope can be different from 1 The measurements can be performed only between T1 and T2 T 2 T 1 Only if the base line is known and exhibit a linear shape T1 T2 Measured T What is true for T is true for the enthalpy OPEN THE BIG QUESTION OF THE BASE LINE QUALITY 19

30 k$ to 60k$ We have to linearise only in the temperature domain needs for the

20 If the base line quality is bad The data obtained will be bad too If the base line quality is bad different options Cleaning procedure Work to optimize the base line 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 20

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22 How to do to extract the good temperature and good the enthalpy from a DSC curves 22

23 Ideal signal Endo Glass transition WHY? Cold crystallization melting Like liquid liquid glass Crystal Are not material constants q+ and q- play a game 23

24 H, V, S During cooling CP TS liquid crystallization crystal α T 24

25 H, V, S During heating liquid fusion crystal α Tf CP T 25

26 H, V, S During cooling High cooling rate liquid CP crystallization glass α crystal Tg T 26

27 Glass transition Tg During heating TRUE : ONLY if q+ = q- and no ageing H or V or S By Calorimetry during heating Power, Tg T P# Cp(Tg) Tg Tg T 27

28 Tg mesurement H or V or S Slope = Cp (L) Slope = Cp (G) glass Tg T 28

29 Tg mesurement H or V or S Tx= Tg when Sx=S T Power, S x S Tx Tg T 29

30 If q+ q- but the same q- H or V or S q+>q- q+<q- T Power, Cp(Tg) remains the same (Tg) is modified Tg remains the same T 30

31 If q+ q- but the same q- H or V or S q+>q- Tx= Tg when Power, Sx=S S T S x Tx Tg T 31

32 Tg mesurement Ageing effects H or V or S T a T f2 T f1 Tg T H Tx= Tf when Sx=S T 32

33 Tg mesurement Ageing effects and particles sizes interesting Effect of particle size of aged powder Selenium on measuring the glass transition. As seen, aged Selenium shows a large peak on glass transition. 33

34 H, V, S No melting Glass transition Like liquid liquid crystallization glass α crystal Tg T 34

35 melting Endo Glass transition Cold crystallization Like liquid liquid glass Crystal 35

36 H, V, S liquid Cold crystallization glass melting crystal Tg T 36

37 melting Endo Glass transition TC Like liquid liquid Crystal 3 processes Number 1 nucleation Number 2 growth Number 3 : diffusion Crystallization enthalpy Melt here What it was Cold crystallized here 37

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39 H, V, S Glass transition liquid crystallization glass α crystal Tg T 39

40 H, V, S Glass transition liquid crystallization glass α crystal Tg T 40

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42 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 42

43 TAKE CARE P = d W/dt = f(t) = d H/dt DSC curve Heating rate =d H.dT/dt.dT =q +.d H/dT = q+ Cp NO That is WRONG P (Tg) =q + ((Cpl_C pref1 )-(Cpg-C pref2 ) Not known (Cpl-Cpg) at Tg = Cp (Tg) = P/q + if Cp ref1 = Cp ref2 43

44 Journal of Materials Education, vol. 30, 2008, pages Quantitative and Transient DSC Measurements I. Heat Capacity and Glass Transition Jean-Marc Saiter, Mehrdad Negahban Philippe dos Santos Claro Pierre Delabare Marie-Rose Garda 44

45 To resume CP at Tg Tg Tc Tm HC CP at TC is very difficult to determine Hm Tmelting is often difficult to get for mixtures and for polymers Often taken at the maximum of the main peak, Heating rate Cooling rate Age Annealing temperature 45

DSC Methods to Quantify Physical Aging and Mobility in Amorphous Systems: Assessing Molecular Mobility

DSC Methods to Quantify Physical Aging and Mobility in Amorphous Systems: Assessing Molecular Mobility R. B. Cassel, Ph.D. TA Instruments, 109 Lukens Drive, New Castle, DE 19720, USA ABSTRACT The specific