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Identification of the Mathematical Models of Complex Relaxation Processes in Solids

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The approach to solving the problem of complex relaxation spectra is presented.
Presentation for the XI International Conference on Defect interaction and anelastic phenomena in solids. Tula, 2007.

Publicada em: Tecnologia, Negócios
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Identification of the Mathematical Models of Complex Relaxation Processes in Solids

  1. 1. Identification of theIdentification of theMathematical Models ofMathematical Models ofComplex Relaxation ProcessesComplex Relaxation Processesin Solidsin SolidsBakhrushin V.E.Bakhrushin V.E.University of HumanitiesUniversity of Humanities“ZISMG”, Zaporozhye, Ukraine“ZISMG”, Zaporozhye, Ukraine
  2. 2. Relaxation processes:- internal friction;- dispersion of modulus;- stress relaxation;- elastic aftereffect.Parameters:- interstitial concentrations for different states;- interstitial solubility;- local diffusion coefficients;- activation energies for jumps.
  3. 3. Identification tasks1. To choose the type of mathematical model: idealDebay peak (model of the standard linear body); thesum of ideal peaks (processes); enhanced Debaypeak; the sum of enhanced peaks; the sum of peaks +background.2. To determine the quantity of relaxation processes3. To determine the parameters of relaxation processes
  4. 4. ( )n1 1 1 1 i0 0ii 1 0iE E 1 1Q T Q exp Q coshRT R T T− − − −=   = − + −  ÷ ÷    ∑0ii 0ikTE RT lnhf=Model of spectrum at Snoeck relaxation area:10Q ,E−– background intensity and activation energy;10i 0iQ ,T−– i-th peak height and temperature– i-th peak activation energyf – sample vibration frequency
  5. 5. Parameters, which must be determined are:10i 0in,Q ,T−or10i in,Q ,E−( )0,14139 0,0032450T 12,89967 2,706674f 0,04547 0,04929f E,= + + − +An error for 300 – 800 К interval at f = 20 – 60 Hzis not more, then 1 %From Wert & Marx formula such approximation maybe obtained:( )( )m 21 1j jj 1S Q Q T min− −== − →∑( )( )21 1mj j1 2j 1 jQ Q TS min− −=−= →σ∑
  6. 6. ( )( )2exp z/z − β ψ =β π( )mz ln /= τ τ z2β = σPeak enhance may be taken into account by the modelof log-normal distribution of relaxation time:
  7. 7. maxmax idδρ =δ20,0853 0,197 0,970ρ = ϕ + ϕ +n1 1 1 i0ii 1 0iE 1 1Q Q coshR T T− − −=  = −  ÷ρ  ∑In this case:( )mlnϕ = ωτFor ρ value from N.P. Kushnareva & V.S. Petcherskydata such approximation may be obtained:
  8. 8. Graphic decomposition2 peaks without error-8-4041,4 1,7 2 2,3arcch(Q-1m/Q-1)103/T
  9. 9. 4 peaks + error-6-30361,2 1,5 1,8 2,1 2,4arcch(Q-1m/Q-1)
  10. 10. Linear least-squares method:( )n1j i i j ii 1x A cosh [B (x c )]−=ϕ = −∑1i 0i j j i 0iA Q , x 1/T , c 1/T−= = =ii i1 kB lnc hfc=
  11. 11. ( )( )( )( )( )( )njj jj 1 1njj jj 1 2njj jj 1 nxy x 0;Axy x 0;Axy x 0,A===∂ϕ − ϕ =  ∂∂ϕ − ϕ =  ∂∂ϕ − ϕ =  ∂∑∑∑LLLLLLLLLLLL
  12. 12. ( )j 1i j i ijixcosh [B (x c )] FA−∂ϕ= − =∂n n n n21 1j 2 1j 2 j n 1j nj j 1jj 1 j 1 j 1 j 1n n n n21 2 j 1j 2 2 j n 2 j nj j 2 jj 1 j 1 j 1 j 11 nj 1j 2 njA F A F F A F F y F ;A F F A F A F F y F ;. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A F F A F= = = == = = =+ + + =+ + + =+∑ ∑ ∑ ∑∑ ∑ ∑ ∑LLn n n n22 j n nj j njj 1 j 1 j 1 j 1F A F y F= = = =+ + =∑ ∑ ∑ ∑L
  13. 13. 1 11 2 12 n 1n 11 21 2 22 n 2n 211 n1 2 n2 n nn nA W A W ... A W Z ;A W A W ... A W Z ;. . . . . . . . . . . . . . . . . . . . . . . . . .A W A W ... A W Z ,+ + + = + + + = + + + =n1 1ik i kj 1 j i j k1 1 1 1W cosh B cosh Bx c x c− −=        = − − ÷ ÷  ÷  ÷  ÷ ÷        ∑n1i i ij 1 j i1 1Z y cosh Bx c−=  = −  ÷ ÷   ∑
  14. 14. The advantages of linear least-squaresmethod:- simple realization;- sufficient accuracy (up to 10 % for the main peaksheights)Method disadvantages:- linearization error;- necessity of peak temperatures preliminary definition;- possibility to obtain an ill-condition system;- supposition of uniformly precise of the data;- supposition of absolute accuracy of temperaturemeasurements;- possibility of obtaining the negative values of peakheights.
  15. 15. The method of gradient descent(linearized least square method)Main differences:- an expression of ideal peak is linearized by Taylorseries expansion in the neighborhood of some point(initial estimate) with abandonment of only linearterms;- the possibility to choose the different types ofobjective function (cancellation of supposition that thedata have the same errors)[L. Crer et. al, 1969; M.S. Ahmad et. al, 1971;O.N. Razumov et. al., 1974; A.I. Efimov et. al., 1982.]
  16. 16. ( )( ) ( )1 1 1mj j2j 1 j kQ Q T Q T0, k 1,2,...qa− − −=− ∂= =σ ∂∑{ }1 1 10 01 0n 1 n 01 0na E,Q ,Q ,...,Q ,E ,...,E ,T ,...,T− − −=q 3n 2= +( )( )21 1mj j1 2j 1 jQ Q TS min− −=−= →σ∑Fromwe can obtain:at general case, q 3n= without background,q 2n= for ideal Debay peaks.
  17. 17. After linearization we obtain:( ) ( )m0 1k k k mkm1a a a M Z ,−=∆ = − = ∑l( ) ( )1 1mk 2j 1 j kQ T Q T1M ,a a− −= ∂ ∂=  ÷ ÷σ ∂ ∂ ∑ll( )( ) ( )1 1 1mj j2j 1 jQ Q T Q TZ ,a− − −=− ∂=σ ∂∑llwhere:derivatives are determined in 0ka
  18. 18. Adjusted values:0k k ka a a , 0 1.= + γ∆ < γ ≤From the definition of ka∆ follows, that it correspondswith the general formula of gradient search methods:0k k 1a a gradS .= −βGradient methods realize an iteration procedure, inwhich such stopping conditions may be used:p p 1k ka a ;−− < ε p p 11 1S S ;−− < δ 1gradS ;< ξ , , 0.ε δ ξ >
  19. 19. Problems and disadvantages:- poor convergence at the case of large number ofpeaks;- possibility of iteration stopping at the critical point,which is not the point of minimum;- possibility of getting into a loop, when the objectivefunctional S is ravine;- absence of realization at standard libraries of the mostpopular software packages;- М matrix must be positively defined at the every stepof iterations
  20. 20. ( ) ( )(k 1) (k) 1 (k) (k)H ,+ −= −X X X G XQuasi-Newton algorithm{ }1 1 10 01 0n 1 n 01 0na E,Q ,Q ,...,Q ,E ,...,E ,T ,...,T− − −=2 2 221 1 2 1 q2 2 222 1 2 2 q2 2 22q 1 q 2 qS S S...a a a a aS S S...a a a a aH... ... ... ...S S S...a a a a a ∂ ∂ ∂ ÷∂ ∂ ∂ ∂ ∂ ÷ ÷∂ ∂ ∂ ÷∂ ∂ ∂ ∂ ∂=  ÷ ÷ ÷ ÷∂ ∂ ∂ ÷∂ ∂ ∂ ∂ ∂ 12qSaSaG...Sa∂  ÷∂ ÷∂ ÷ ÷∂= ÷ ÷ ÷∂ ÷ ÷∂ 
  21. 21. ijijh , i j;c0, i j. == ≠n1 T1 k k kk 1P z ,−== ∑ v v° 1 1 11H C PC ,− − −=zk are eigenvalues and vk are eigenvectors of matrix:Grinshtadt technique:1 1P C HC ,− −=It is necessary to provide the positive definiteness ofHesse matrix or to find an approximation of Н-1
  22. 22. andF is a Fisher criterion value for the correspondingnumbers of degrees of freedom and significance level,∆2- sum of errors squares (relative errors) ofexperimental points.Adequacy criteria for spectrum models:2SF≤∆2FS∆≤( )0,052SF F 2,0...2,5> ≈ ⇒∆number of model parameters must be increased;2FS∆> ⇒number of model parameters must be decreased.
  23. 23. Quasi-unimodelity (an absence of physicallydifferent minimums) of objective functional, that is allminimums of objective functional correspond to thesame physical model of a spectrum.Deviation from quasi-unimodality may be causedwith:- the presence of excess peaks in the model;- absence of some essential peak in the model;- presence at the real spectrum of some collateralpeak, which height is close to measurement error.
  24. 24. Absence of model residuals serial correlation(Darbin & Watson criterion):( )m 2j j 1j 2m2tj 1e ed ,e−==−=∑∑( )1 1j j je Q Q T− −= − - model residuals.d 2≈ - serial correlation is absent;d 0→d 4≈- positive serial correlation;- negative serial correlation(there are excess peaks).
  25. 25. 0510500 600 700 800Q-1·103T, KGiven data (4 peaks + error)
  26. 26. Initial approach (4 peaks) σ=0,1
  27. 27. 04812500 600 700 800 T, KQ-1·103The result of decompositionS=0,30F=0,74
  28. 28. -0,25-0,10,050,2500 600 700 800∆Q-1·103T, KResidualsd=2,18
  29. 29. 04812500 600 700 800 T, KQ-1·103Initial approach 1 (3 peaks) σ=0,1
  30. 30. 04812500 600 700 800 T, KQ-1·103The result of decomposition 1 (3 peaks)S=20,52F=50,31
  31. 31. -2-101500 600 700 800∆Q-1·103T, KResidualsd=0,33
  32. 32. 04812500 600 700 800 T, KQ-1·103Initial approach 2 (3 peaks) σ=0,1
  33. 33. 04812500 600 700 800 T, KQ-1·103The result of decomposition 2 (3 peaks)S=13,58F=33,28
  34. 34. -2-101500 600 700 800∆Q-1·103T, KResidualsd=0,54
  35. 35. 04812500 600 700 800Q-1·103T, KInitial approach 1 (5 peaks) σ=0,1
  36. 36. Initial approach 2 (5 peaks)04812500 600 700 800Q-1·103T, Kσ=0,1
  37. 37. Given 3_1 3_2 4 5_1 5_2T1 570 571,3 578,6 569,4 569,4 569,4T2 620 632,5 619,8 619,8 619,8T3 690 698,1 685,5 690,2 690,2 690,2T4 750 742,4 749,2 749,2 749,2T5 499,6 718,2Q1 6 6,3 7,4 5,9 5,9 5,9Q2 3 3,5 3,1 3,1 3,1Q3 12 13,0 12,2 12,0 12,0 12,0Q4 3 3,6 2,9 2,9 2,9Q5 0,0 0,0σ=0,1
  38. 38. 04812500 600 700 800 T, KQ-1·103σ=0,3Initial approach 1 (4 peaks)
  39. 39. 04812500 600 700 800 T, KQ-1·103Initial approach 2 (4 peaks) σ=0,3
  40. 40. 04812500 600 700 800 T, KQ-1·103Initial approach 3 (4 peaks) σ=0,3
  41. 41. 04812500 600 700 800 T, KQ-1·103The result of decompositions=3,01F=0,76
  42. 42. -0,5-0,20,10,4500 600 700 800∆Q-1·103T, Kd=1,20Residuals
  43. 43. 04812500 600 700 800 T, KQ-1·103Initial approach 1 (3 peaks) σ=0,3
  44. 44. 04812500 600 700 800 T, KQ-1·103Initial approach 2 (3 peaks) σ=0,3
  45. 45. 04812500 600 700 800 T, KQ-1·103Initial approach 3 (3 peaks) σ=0,3
  46. 46. The result of decomposition04812500 600 700 800 T, KQ-1·103s=15,87F=3,99
  47. 47. -2,5-10,5500 600 700 800∆Q-1·103T, KResidualsd=0,57
  48. 48. 04812500 600 700 800Q-1·103T, KInitial approach 1 (5 peaks) σ=0,3
  49. 49. 04812500 600 700 800Q-1·103T, KThe result of decompositions=2,92F=0,74
  50. 50. -0,5-0,20,10,40,7500 600 700 800∆Q-1·103T, KResidualsd=1,21
  51. 51. 04812500 600 700 800Q-1·103T, KInitial approach 2 (5 peaks) σ=0,3
  52. 52. 04812500 600 700 800Q-1·103T, KThe result of decompositions=1,41F=0,35
  53. 53. -0,5-0,20,10,40,7500 600 700 800∆Q-1·103T, KResidualsd=2,47
  54. 54. Given 3 4 5_1 5_2T1 570 578,7 569,8 571,2 569,7T2 620 621,8 622,8 621,3T3 690 686,5 691,2 691,2 690,6T4 750 749,2 757,1 757,2 751,7T5 537,5 851,0Q1 6 7,2 5,9 5,7 5,8Q2 3 3,0 2,9 2,9Q3 12 12,4 12,2 12,2 12,0Q4 3 3,6 2,9 2,9 2,9Q5 0,3 0,6σ=0,3
  55. 55. 01020450 550 650 750Q-1·103T, K0,180,360,420,60[N], at.%:Nb – 2 at.% W – N (3 peaks, 1 result)
  56. 56. -3,5-2,5-1,5-0,50,51,52,5450 550 650 750∆Q-1·103T, Kd: 0,69; 0,521,94; 1,14Residuals
  57. 57. 01020450 550 650 750Q-1·103T, K0,180,360,420,60[N], at.%:Nb – 2 at.% W – N (3 peaks, 2 result)
  58. 58. -3,5-2,5-1,5-0,50,51,52,5450 550 650 750∆Q-1·103T, Kd: 0,85; 1,331,88; 1,27Residuals
  59. 59. 01020450 550 650 750Q-1·103T, K[N], at.%: 0,180,360,420,60Nb – 2 at.% W – N (4 peaks)
  60. 60. -2,2-1,2-0,20,81,82,8450 550 650 750∆Q-1·103 d: 1,83; 2,102,84; 1,67Residuals
  61. 61. 01020450 550 650 750 T, KQ-1·103[N], at.%: 0,180,360,420,60Nb – 2 at.% W – N (5 peaks, 1 result)
  62. 62. Residuals-2-1012450 550 650 750∆Q-1·103T, Kd: 1,60; 2,363,09; 1,83
  63. 63. 01020450 550 650 750 T, KQ-1·103[N], at.%: 0,180,360,420,60Nb – 2 at.% W – N (5 peaks, 2 result)
  64. 64. -2-1012450 550 650 750∆Q-1·103T, Kd: 2,54; 2,362,96; 2,25Residuals
  65. 65. T1 T2 T3 T4 Т5 S F3_1 539,1 657,2 685,2 70,1 2,03_2 647,1 673,6 748,7 86,5 1,64 537,9 650,9 674,7 749,0 43,3 3,35_1 528,8 656,9 676,4 748,0 593,0 35,7 4,05_2 535,5 641,4 674,4 745,5 665,5 34,1 4,2Q1 Q2 Q3 Q4 Q53_1 2,2 18,2 13,43_2 9,3 21,3 2,74 2,1 10,8 19,5 2,65_1 1,5 13,6 15,9 2,6 2,15_2 2,0 7,0 17,9 2,6 5,8
  66. 66. Nb – 12 at.% W Nb – 6at.%W4peaks5 peaks 4peaks1 set 2 setE1, kJ/mol 102,2 86E2, kJ/mol 109,5 111,5 110,1 109,8E3, kJ/mol 116,3 116,9 116,5 116,5E4, kJ/mol 128,9 129,6 129,1 128,3E5, kJ/mol 1456 145,9 145,9 145,9
  67. 67. 00,30,60,9500 600 700 800Nb – 2 at.% Hf – 0,32 at.% N (3 peaks):E = 1,47; 1,61; 1,76 kJ/mol;d = 0,90; F = 1,73.
  68. 68. 00,30,60,9500 600 700 800Nb – 2 at.% Hf – 0,32 at.% N (4 peaks):E = 1,29; 1,48; 1,62; 1,79 kJ/mol;d = 1,26; F = 2,64.
  69. 69. 00,30,60,9500 600 700 800Nb – 2 at.% Hf – 0,32 at.% N (5 peaks, 1 result):E = 1,44; 1,54; 1,63; 1,77; 1,91 kJ/mol;d = 1,06; F = 3,21.
  70. 70. 00,30,60,9500 600 700 800Nb – 2 at.% Hf – 0,32 at.% N (5 peaks, 2 result):E = 1,26; 1,46; 1,57; 1,64; 1,80 kJ/mol;d = 1,55; F = 3,58.
  71. 71. nN ii 10ii 2 2 2ii0iiM(T) M M (T);MM (T) ;1 4 fE 1 1expR T T.2 f= = − ∆∆∆ =+ π τ    −  ÷   τ = π∑The temperature dependence of dynamic elasticmodulus in a case of n processes, which satisfy themodel of standard linear body, may be determinedfrom a system:MN – non-relaxed modulus.
  72. 72. Model parameters, which must be identified, are:0i 0iM ,T .∆( ) ( )m 21 exp j jj 1S M T M T min,= = − → ∑We are to solve such minimization problem:( )exp jM Twhere are experimental data for modulus at Tj.(*)
  73. 73. Functional (*) has a great number of minimums,so the result of minimization strongly depends on initialassumption.Adequate model may be obtained by using as T0iinitial values the results of relaxation spectrumdecomposition and setting initial values as0iM∆10i N 0iM 2M Q .−∆ =T0i values after minimization are very close with theinitial ones, and values change essentially. Butthere is a correlation (r = 0,90 – 0,97) between partialSnoek peaks heights and results for :0iM∆0i10i HM2,00 0,15.Q M−∆= ±0iM∆
  74. 74. Nb – 12 at.% W – N[N], at.%:  - 0,11;■ – 0,16;▲ – 0,22;♦ - 0,31

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