Alfred University Dept. of Physics and Astronomy

Transcription

1 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

2 First, let s review! Narayanaswamy assumed that M p (t) obeys RS. ξ = t τ t r dt' dt'= τ τ P ( t' ) r τ P [ ] [ ( t' )] M p 2 ( t) = t f 1 2 M p 2 ( ξ) = ξ f 1 2 p( 2,ξ) = p( 2, ) α s ΔM p ( ξ) ( ξ) 2 = M p ( ξ)δ

3 N = + Δ 1 + Δ Δ N = + Δ i i=1 Δ 1 Δ 2 Δ 3 t 1 t 2 t 3 p( 2,ξ) = p( 2, ) α s ΔM p ( ξ) ( ξ) 2 = M p ( ξ)δ N p(,ξ) = p(, ) α s M p ( ξ ξ i ) i=1 Δ( ξ) Δξ i Δξ i N = M p ξ ξ i i=1 Δ( ξ) Δξ i Δξ i p (,ξ ξ ) = p(, ) α s M p ( ξ ξ' ) dξ' dξ' ξ = M p ξ ξ' dξ' dξ'

4 τ p = τ exp xδh R + ( 1 x )ΔH R where < x < 1 Arrhenius term A dependence just like ool! he ool- Narayanaswamy- Moynihan equa;ons are ξ p(,ξ) = p(, ) α s M p ( ξ ξ' ) dξ' dξ' ξ and = M p ξ ξ' dξ' dξ' and some form for τ p such as τ p = τ exp xδh R + ( 1 x )ΔH R

5 DSC: Differen;al 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. Q(t ) DSC Unknown sample (t ) i.e. dt A given vs. t is specified i.e. dq dt he output is the Q vs. t required to produce the specified vs. t he ra;o of the output to input is dq dt = dq = C dt p

6 B) Linear hea;ng a glass that was linearly cooled i.e. an up scan i H 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.

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

8 A = B Y.Z. Yue Chemical Physics Leders 357 (22) ( C P C P ) c e = C C P,L P,g g Well worth reading!! G A is the area of the bird C p,l C p C p 2 C p,g B is the area of this trapezoid C p 1 c

9 Pulling all of the pieces together! ξ Is there any deeper meaning to = M p ( ξ ξ' ) dξ' dξ'? What can we use for the response M p. From experiments, M p can be fit with a stretched exponent M p ( ξ ) = exp ξ τ r b Let s subs;tute M p into the expression ξ = M p ξ ξ' dξ' dξ'= ξ b ξ ξ' exp τ p dξ' dξ'

10 Using the Prony series approxima;on to the stretched exponen;al, we obtain = exp ξ ξ' b ξ τ p dξ' dξ'= ξ N a n n =1 ξ ξ' exp τ n dξ' dξ' = a n exp ξ ξ' N ξ n =1 τ n dξ' dξ'= N a ξ e n n =1 ξ ξ ' τ n dξ' dξ' Recall that the a n s sum to 1. We can then rewrite the above equa;on as N ξ = a n e n =1 N ξ = a n e n =1 ξ ξ ' τ n ξ ξ ' τ n dξ' dξ'= dξ' dξ' N a N ξ a n n e n =1 = 1 n =1 ξ ξ ' τ n dξ' dξ'

11 = a n e N ξ n =1 ξ ξ ' τ n dξ' dξ' Does this look familiar????? Look back at the last lecture It is just Narayanaswamy s equa;on for a single τ n which reduced to ool s eq!!!!! We now have N ool equa;ons. We have come back full circle. Let s call the fic;ve temperature associated with each term in the { },n, so we now have N = a n,n n =1 What is the meaning of this equa;on? Each relaxa;on ;me τ n has its own fic;ve temperature. can be viewed as a weighted sum of the individual fic;ve temperatures for various relaxa;on process.

12 Is there anything else that we can obtain from DSC and compare with theore;cal calcula;on? Yes! We can use DSC to measure d /. We can then use NM to vs. t. If we know the cooling rate q = /dt then d = d dt dt = 1 q t d dt measure calculate How can we measure d /drom DSC? Moynihan was an expert at this!

13 We can define the fic;ve temperature in the following fashion H( ) = H eq C P,g ' Since <, H decreases by C P,g. In addi;on, we can write H( ) = H eq ( ) + C P ' where is the ini;al Further, we can write the equilibrium H eq as H eq ( ) = H eq + C P,L ' Now subs;tute H(t) and H eq () into our top expression yields H eq + C P ' = H eq + C P,L ' C P,g '

14 So we now have C P ' = C P,L ' C P,g ' If we now subtract C P,g ' from both sides we obtain C P ' C P,g ' = C P,L ' C P,g ' C P,g ' ( C P C P,g )' = C P,L ' C P,g ' C P,g ' split this integral into two pieces

15 Splikng the last integral on the right into two pieces gives ( C P C P,g )' = C P,L ' C P,g ' C P,g ' ( C P C P,g )' = C P,L ' C P,g ' C P,g ' C P,g ' switching the limits C P,g '= C P,g ' We now obtain ( C P C P,g )' = ( C P,L C P,g )' Very soon we will see how Moynihan used this expression to find.

16 But wait there s more!!!!!!! Recall the fundamental theorem of calculus x = f x F x a dx where a is a constant df dx = f x What happens if F(x) is a composite func;on, i.e. F(g(x))? g ( x ) F( g( x) ) = f x a dx Need to use the chain rule ( ) df g x dx ( ) = df g x dg x dg dx = f g x ( ) dg dx

17 Apply the fundamental theorem of calculus for a composite func;on to our expression ( C P C P,g )' = ( C P,L C P,g )' [ C p ( ) C P,g ( ) ] = C p,l [ C P,g ( )] d d = C p [ C P,g ( ) ] [ C p,l ( ) C P,g ( )] Using DSC you can measure every term on the right side Calculate this with NM eq.

18 How did Moynihan use this expression to find? ( C P C P,g )' = ( C P,L C P,g )' Consider the C p graph for a liquid that is cooled through the glass transi;on and then reheated through the glass transi;on. he H vs. graphs and C p vs graphs are 2 t C P L H C P g

19 Moynihan s Method orange ( C P C P,g )' Area bounded by C p and C P,g C P L ( C P,L C P,g )' purple Area of the trapezoid C P g In Moynihan s method, approaches a lower limit of g

20 In prac;ce how do you solve the NM equa;ons ξ p(,ξ) = p(, ) α s M p ( ξ ξ' ) dξ' dξ' ξ and = M p ξ ξ' dξ' dξ' Assume some form for τ P. ypically τ P = τ exp xδh R + ( 1 x )ΔH R Recall that the reduced ;me is given by ξ = t τ t r dt' dt'= τ τ P ( t' ) r τ P [ ] [ ( t' )] Assume that M p can be approximated by a stretched exponen;al M p ξ = exp ξ τ r b

21 Rewrite the reduced ;me ξ in terms of and the hea;ng/cooling q = /dt t ξ = τ r dt' τ P ξ = τ r t τ P dt' ' ' ' = τ r ' qτ P ' If we have a func;on of ξ ξ as we do in M p then ' ' '' ξ ξ'= τ r τ qτ P ( ' ) r qτ P '' ' ' ' ' '' ξ ξ'= τ r + τ qτ P ( ' ) r τ ' qτ P ( ' ) r qτ P '' ξ ξ'= τ r ' '' qτ P '' split the integral

22 We now break up the (t) into N sec;on as = + Δ i he NM eq. for can now be wriden as N i=1 ξ = M p ξ ξ' dξ' dξ' = M p ( ξ ξ' ) = exp ξ ξ' b τ r = Δ i exp ξ ξ' b N N i=1 τ f = Δ i exp i=1 ' r '' qτ P '' b ξ ξ'= τ r ' '' qτ P ''

23 Finally N = Δ i exp i=1 ' '' qτ P '' b N N Δ = Δ i exp j i=1 j =i q j τ P, j b he τ p is tricky since it depends on τ P = τ exp xδh R + ( 1 x )ΔH R

24 Use the following cute trick with τ p If we break into temperature steps that are small, it would not be unreasonable to assume that τ p at temperature step i is very close in value to τ p at temperature step i- 1 So instead of wri;ng t p at temperature step i as τ P,i = τ exp xδh R i ( + 1 x )ΔH R,i We can write τ p,i as τ P,i = τ exp xδh R i ( + 1 x )ΔH R,i 1 his is fine since we need to know the ini;al condi;on of i.e. (), = a given. We need to take smaller and smaller Δ un;l this approxima;on as no effect.

25 So what do we need to actually do a NM calcula;on? We need 4 parameters: b for the stretched exponen;al M p ( ξ ) = exp ξ τ r b t dt' where ξ = τ r t' τ P [ ] and τ o, x, and ΔH for τ P,i = τ exp xδh R i ( + 1 x )ΔH R,i 1 We also need the thermal path, (t), and the ini;al value (). N N Δ hen use Excel or some other program to iterate = Δ i exp j i=1 j =i q j τ P, j b

26 Repeat this procedure for ξ p(,ξ) = p(, ) α s M p ( ξ ξ' ) dξ' dξ' Finally an applica;on!!

27 It is well known that the index of refraction of glasses, n, varies with the cooling rate. Recall the Ritland and Napolitano and Spinner experiments. Further, it has been empirically determined that n depends on the prior cooling rate in the following fashion. n d ( h X ) = n d ( h ) + m nd ln h x h where h X and h are two different cooling rates and m nd is typically a negative constant. Can NM shed any insight into this expression?

28 What assumptions did they make? Over the visible range, the index of refraction will have a strong density dependence. Assume that the density is a linear function of the fictive temperature. Further, assume that there is only one univerisal for the enthalpy, density and n. n( λ) = n( λ) ref + n ( λ ),ref How can we calculate? Ref: U. Fotheringham et al. Refractive Index Drop Observed After Molding of Optical Elements: A Quantitative Understanding Based on the ool- Narayanswamy-Moynihan Model, J. Am. Ceram. Soc., [3] (28) Ref: U. Fotheringham et al. Evaluation of the Calorimetric Glass ransition of Glasses and Glass Ceramics with Respect to Structural Relaxation and Dimensional Stability, hermochimica Acta, 461 [1-2] (27)

29 Use NM ( t) = ( t) ς dς' exp[ ( ς ς' ) b ]dς' where ς = t dt' τ t' and [ ] = τ exp H R τ ( t), t x t + 1 x ( t) x =.789 he parameters used for one glass are t = 1.68 x 1-46 s H/k = b =.656 he boundary condition they use it = above the glass transition.

30 Some results! wo different glasses. Each glass was taken through two different cooling rates. glass 1 glass 2 F A comparison of DSC with NM. Excellent agreement!!!!

31 more results n d n d n d n d

32 hank You! Any questions?