Lecture 15:The Tool-Narayanaswamy-Moynihan Equation Part II and DSC

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1 Lecture 15:The Tool-Narayanaswamy-Moynihan Equation Part II and DSC March 9, 2010 Dr. Roger Loucks Alfred University Dept. of Physics and Astronomy

2 Thank you for taking me home! My eyes are completely open now! I understand the glass transition very well!!!!

3 First Let s Review p,h,t glass line p( T,0) dp dt p p eq dp dt = p p eq τ p( T, )= p eq ( T) super cooled liquid line ( equilibrium) T Temperature

4 We used these three equation. p( T,t)= p V ( T)+ p s ( T,t) p( T)= p eq ( T 0 )+α L ( T 0 )+α g ( T ) dp dt = p p eq τ d dt = T τ Tool s eq. p, H, V T 1 T 2 ( )= T 2 ( 0)= T 1

5 Initially Tool used τ = Kη = τ o e AT where η 0 and A are constants. Tool quickly realized that this did not account for the data. He postulated that η must depend on the Fictive temperature. If a liquid were cooled quickly, it would have a larger than a slower cooled liquid. This larger would correspond to a more open structure which would reduce the η. If the liquid were cooled slower, then the would be smaller and the structure is closer together and would have a larger η. To account for this, Tool (1946) assumed that a better choice of η would be η = η 0 e ( A 1 T +A 2 ) τ = τ 0 e ( A 1 T +A 2 ) where η 0 and A 1 and A 2 are constants. Tool s equation becomes d dt = T T f e A 1 T +A 2 τ 0 As clever as Tool s equation is, it can not account for the cross over experiments of Ritland. The reason for this lack of agreement is the single relaxation time.

6 Thermorheological Simplicity Define the relaxation response at a temperature T as R p ( T,t ) p T, p( T,0) p T, The range of R is 1 to 0, i.e. if t = 0, R = 1 and if t =, R = 0. The response R can typically be described by the stretched exponential R = e t b τ exp N a n e t τ n n =1 where τ exp is an experimentally determined parameter, 0 < b < 1. Stretched exponentials can be approximated by the Prony series where the a n s sum to 1. If all the τ n s have the same mathematical dependence on temperature then TRS results. To see this, rewrite the τ n R in the Prony series as τ/λ n so R becomes R = N a n e λ n t n =1 N τ = a n e λ nβ t where n =1 τ β

7 If the R for a system were measured at any value of T and then graphed as R vs. β, all of the R s at various T s would lie on top of one another, i.e. there would be one Master graph. R R For any T T 1 T 2 T 3 log t Log β Since there is one Master graph of R, define a reference temperature T r at which the Master graph would be measured. Call the time associated with this reference temperature ξ. Since the same graph would result at any temperature T, we can conclude that β = ξ τ r = t τ or more simply ξ = τ r τ t where ξ is called the reduced time.

8 We can view ξ in the following way. If a system relaxes by some amount at temperature T in a time t, ξ is the time that is needed for the system to relax the same amount at temperature T r, i.e. R( t,t)= R ξ,t r N R = a n n =1 e λ n t N τ = a n n =1 e λ nβ N = a n n =1 e λ ξ n τ r How can we extend this to temperature changes?

9 Defining a new response function M p ( t) p ( T,t 2 ) p T 2, p( T 2,0) p T 2, for T changes T 2 T 1 M has the same range as R, i.e. t = 0, M p = 1 and if t =, M p = 0. Using p( T)= p eq ( T)+α s ( T)= p( T 1 )+α L ( T T 1 )+α s ( T) And the conditions: t = 0, (0) = T 1 and ( ) = T 2 M p we becomes M p ( t)= T t f T 2 T 1 T 2

10 Narayanaswamy assumed that M p (t) obeys TRS! How? dξ = τ r [ ] dt Integrating this from 0 to t yields ξ = τ T t τ r t t dt' dt'= τ r 0 τ T t' 0 τ T t' [ ] [ ] M p in terms of the reduced time ξ M p ( t)= T t f T 2 T 1 T 2 M p ( ξ)= T ξ f T 2 T 1 T 2 ( ξ) T 2 = M p ( ξ) T While p( T,t)= p eq ( T)+α s ( t) T p T,ξ = p eq ( T)+α s ( ξ) T p( T 2,ξ)= p( T 2, ) α s TM p ( ξ)

11 Episode III: Revenge of the ξ To complete the derivation, imagine that the temperature is changed from some initial value of T o to some final value T in a series of N steps, i.e. T = T 0 + T 1 + T T N = T 0 + T i N i=1 T 0 T 1 T 2 How can be extend (ξ) and p(t,ξ) to multiple temperature steps? T 3 p( T 2,ξ)= p( T 2, ) α s TM p ( ξ) t 1 t 2 t 3 p( T,ξ)= p( T, ) α s T 1 M p ( ξ ξ 1 )+... α s T N M p ( ξ ξ N ) N p( T,ξ)= p( T, ) α s T i M p ξ ξ i i=1

12 Using the chain rule and rewriting T i in terms of ξ yields T i = T ξ ξ i ξ i dt = dt dξ dξ N p( T,ξ)= p( T, ) α s T i M p ξ ξ i N p( T,ξ)= p( T, ) α s M p ξ ξ i i=1 i=1 T ξ ξ i ξ i ξ p( T,ξ)= p( T, ) α s M p ξ ξ' 0 dt dξ' dξ'

13 Likewise, the equation for fictive temperature becomes ( ξ) T 2 = M p ( ξ) T N = T T i M p ξ ξ i i=1 N = T M p ξ ξ i i=1 T ξ ξ i ξ i ξ = T M p ξ ξ' 0 dt dξ' dξ'

14 Recall in the last lecture, we stated that the fundamental flaw in Tool s equation is that it only has one relaxation time. Let s pretend that M p (ξ) is given by only one relaxation time, i.e. M p ( ξ)= e ξ τ r Substituting this M into Narayanaswamy s equation for the evolution of yields ξ = T M p ξ ξ' or 0 dt dξ' dξ'= T ξ e 0 ( ξ ξ ') τ r dt dξ' dξ' T = e ξ 0 ( ξ ξ ') τ r dt dξ' dξ' Taking the derivative with respect to ξ gives d dξ = 1 τ r e ( ξ ξ ') ξ τ r 0 dt dξ' dξ' d dξ = T τ r Tool s eq.

15 What did Narayanaswamy use for τ? τ p = τ 0 exp x H RT + ( 1 x ) H R where 0 < x < 1 Arrhenius term A dependence just like Tool! The Tool-Narayanaswamy-Moynihan equations are ξ p( T,ξ)= p( T, ) α s M p ξ ξ' 0 dt dξ' dξ' and = T M p ξ ξ' ξ 0 dt dξ' dξ' and some form for t p such as τ p = τ 0 exp x H RT + ( 1 x ) H R

DSC: Differential Scanning Calorimetry as a Black Box.

output. Ignore the details of how the apparatus works.

t required to produce the specified T vs. t A given T vs.

16 DSC: Differential Scanning Calorimetry as a Black Box. By a black box, I mean 1) what are the inputs and 2) what is the output. Ignore the details of how the apparatus works. DSC Q( t) i.e. dq dt Unknown sample T( t) i.e. dt dt The output is the Q vs. t required to produce the specified T vs. t A given T vs. t is specified The ratio of the output to input is dq dt dt dt = dq dt = C p

17 How does a DSC work? The philosophy of the device. output Q t thermocouple A given V corresponds to a T Q = I X V X dt T X sample X T unknown T known Al V Q = I Al V Al dt Q = m Al c Al T Al I X V I Al input T t Feedback mechanism. If the V doesn t correspond to the correct T, the I X is changed.

19 The Empty Al pan acts as the reference sample Sealed Al pans containing the samples

20 Pans go in here

22 Before we explain how to measure and T g using a DSC, lets first examine some typical C p vs T results A) C p vs. or a linear cooled liquid i.e. a down scan T i α L = α V +α S = dh dt = C L P C P L t H C P g T T α g = α V = dh dt = C g p < C P L

23 B) Linear heating a glass that was linearly cooled i.e. an up scan T i H T t C P L C P g As the glass is relaxing toward the super cooled equilibrium line, heat is given off i.e. H is decreasing so this region is exothermic. T

24 C) A liquid cooled by a down quench T i = inal t C P L H C P g inal T i = inal T i =

25 D) A linear up scan on an annealed glass T i t C P L H C P g T T g

26 What information does a C p vs T graph provide? Recall that C p = dh dt C p dt = dh T 2 C p T 1 T 1 T 2 dt = dh = H 2 H 1 = H If the system is a glass T 1 and a liquid at T 2, then H = H L -H g. How can we use this to find and T g? For example, how do you measure the and T g of a quenched glass?

27 Yuanzheng Yue s Enthalpy-Matching Method 1) Make a glass and quench it. The cooling rate and are unknown. What is the of this quenched glass? 2) Place a sample of the quenched glass into a DSC and heat the sample up to the liquid state at some fixed linear rate say 20 0 C/min or 10 0 C/min. Call the C p for this first upscan C p1. 3) Cool the liquid at the the same linear rate, i.e. say 20 0 C/min, to room temperature. 4) Reheat the cooled glass sample using at the same linear rate of 20 0 C/min back up to the liquid state. Call the C p for this second upscan C p2. The graph for C p2 will not have a severe of a dip since the glass has relaxed. 5) Graph of C p1 and C p2 vs. T curves.

28 H upscan 1 upscan 2 t

Applying Yue s technique is easy in practice. We ll set it up in steps. Why the technique works requires more effort. I ll explain what to do first before I give the explanation.

29 Applying Yue s technique is easy in practice. We ll set it up in steps. Why the technique works requires more effort. I ll explain what to do first before I give the explanation. First, calculate the integral C p2 is insensitive to since the glass was brought to the liquid state on the first upscan. T g 2 1 A C p C P T c 2 C p dt Clearly, this is just the area between the second and first upscans. C p C p 1 This dip will deepen if is higher since the glass will relax more on the first upscan. T c T e

30 Second, calculate the integral B C p,l C P,g T g dt The C P,g is the C P curve for the glass. It is found by extrapolating the C P2 curve before T g. To extrapolate the C P,g curve use the following fit C p,g = a + bt + c T 2 + d T 0.5 where the a, b, c and d are constants that must be determined experimentally T G C p C p 2 C p,l T G is given by the following method. Find the inflection point on the C P2 curve and draw a tangent. The T at which it intercepts C P,g is T g. C p,g B is the area of this trapezoid C p 1 T c

31 It turns out that integrals A and B are equal A = B 2 1 ( C P C P ) T c T e dt = C C P,L P,g T g dt To find, change the upper limit in the right integral until the two integrals are equal. When they equal, that value is! WHY????

32 To understand why these two integrals are equal, let s examine each integral separately. Start with B. B C P,L C P,g dt Recall from previous lectures that α p - α g = α s where α P = C P T g and α g = C g in our case. The structure/configuartion of the liquid that is quenched will change from to the T g. Past T g the relax times are too large for any appreciatable relaxation B C P,S dt = H structure to occur. Above the liquid is still in equilibrium. T g Further, H = E + p V and most of H comes from E since V is small compared so H structure E structure

33 Now let s consider the left integral A. T e 2 1 A C p C P T c dt Below T c both C p1 and C P2 are identical. Recall that the slopes of p vs T graphs for low T were all identical! Above T e, both C p1 and C P2 are identical since they are in the both liquids. The vibrational contributional to C p1 and C p2 are identical at a given T. Therefore, the vibrational contributions cancel and all that is left is the contribution from structural changes. Note that if the upper limit of this integral was extended to, the integral would not since C p1 = C P2 in the liquid region. Therefore, A is also equal to H structure. A = B Yue is very clever! This is an active area of work!!!!!!!

Alfred University Dept. of Physics and Astronomy

Lecture 16:he ool- Narayanaswamy- Moynihan Equa;on Part II and DSC March 11, 21 Dr. Roger Loucks Alfred University Dept. of Physics and Astronomy loucks@alfred.edu First, let s review! Narayanaswamy assumed