McMaster University Iraplnng InaouilJan and UIicoybiy Learning and understanding thermodynamics: a struggle against obviousness Dmitri V. Malakhov Department of Materials Science and Engineering 1 Outline Decarburization: let us warm up Maximal temperature of adiabatic combustion: do we calculate it correctly? Influence of pressure on the molar Gibbs energies: what does P do to G(x)? A choice of a reference frame in the compound energy formalism: is it unique? An advice: stay alert, be critical MSE: we are waiting for you Combustion: a source of heat Component Wt. % C 0.4 Mn 0.7 Si 1.5 Cr 0.7 V 0.2 Decarburization 4 ŕ* What causes the misfortune? , ,gas . , steel . A ,A ^gas . ^steel Mc ^ Mc or> equivalently, a^ < a( Austenitization: int region beh 1100 1000 J____________I____________I____________I____________I____________I____________I____________I____________L Y Y + cem + Y + cem a + cem i i i i i 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Wt. % of carbon erior and near-surface ave differently 100 ÍS 60 o 50 500 600 700 800 900 Temperature, C 1000 1100 6 How can decarburization be suppressed or eliminated? • Kinetics (induction heating: seconds instead of minutes) • Chemistry (heating in vacuum: an absence of mediators such as H20, C02, H2) • Thermodynamics 7 Can the adversity be avoided? gas . steel f ŕc \ v T nothing can be done air: fuel ratio, furnace design,. ~v~ nothing can be done J V ■T r steel \ T composition nothing can be done ~~T- ^^T ~, \ b nothing can be done J 8 Terminology (fuel-dependent) CH4 + 202 + 8N2 -> C02 + 2H20 + 8N2 fuel air flue gas "rich mixtures" employed "lean mixtures" __air_ excess of fuel ■ raj10 excess of air 'fuel high carbon activity stoichJmetric ^ high oxygen activity low oxygen activity ratio low carbon activity 9 Can decarburization be defeated by changing the air:fuel ratio? .01 10 I" O -8 S 10 10 10 <3 io 10 10 I i i T = 900 °C —i— -------1-------1------- 500 600 700 800 900 Temperature, C 1000 1100 7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5 airimethane ratio Thermodynamic verdict: no way! Another reason prohibiting low ratios re O) 3 10" 1- .1- .01- o " .001 0) E -4 3 10 O > 10 - -6 10 -| 10 T = 900° C 7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5 air:methane ratio 11 Temperature of adiabatic reaction between hydrogen and oxygen 1. Take 0.5 mole of 02(g) and 1 moles of H2(g) at 25°C and P = 1 atm. By definition, H° = 0. 2. Make 1 mole of H20(g) at 25°C and put AfH°298 released into a heat reservoir. 3. Maintain P = 1 atm and use all energy stored in the reservoir for heating 1 mole of gaseous H20. 4. Ask yourself a question: do I know how to calculate Tmax? 12 Of course you know, but just in case... T max jc;(r)dr= AfHm 298 "V" V--------v--------' energy in reservoir heating from 298 K to unknown 71 Kelly's expression C°(T) = a + bT + -^ + dT T max Non-linear equation Boring, but simple T. max 13 Tmax = 4620°C Tmax = 4620°C: an erroneous result 5000 4500- 4000- 3500- °- 3000 -I (D 3 2500- (D 2000 A Q_ g 1500^ \H°19%= -241826 J! total energy in reservoir T 0 50 100 150 200 Heat used for heating, kJ 250 What happens at this temperature? Output from POLY-3, equilibrium = 1, label AO , database: SSUB3 Conditions: N(H2)=1, N(02)=0.5, T=4893.29, P=1E5 DEGREES OF FREEDOM 0 Temperature 4893.29 K ( 4620.14 C), Pressure 1.000000E+05 Number of moles of components 1.50000E+00, Mass in grams 1.80148E+01 Total Gibbs energy -1.87698E+06, Enthalpy 9.34464E+05, Volume 1.18691E+00 Component Moles M-Fraction Activity Potential Ref.stat H2 1.0000E+00 6.6667E-01 1.7999E-12 -1.1003E+06 SER 02 5.0000E-01 3.3333E-01 2.6171E-17 -1.5534E+06 SER GAS Status ENTERED Driving force 0.0000E+00 Moles 1.5000E+00, Mass 1.8015E+01, Volume fraction 1.0000E+00 Mole fractions: H2 6.66667E-01 02 3.33333E-01 Constitution: H 6.47172E-01 HlOl 1.20372E-02 H102 8.06924E-07 O 3.24793E-01 02 2.81954E-03 03 5.47919E-09 H2 1.28658E-02 H201 3.11755E-04 H202 1.44469E-09 A dramatic difference Output from POLY-3, equilibrium = 1, label AO , database: SSUB3 Conditions: P=1E5, N(H2)=1, N(02)=0.5, H=0 DEGREES OF FREEDOM 0 Temperature 3077.91 K ( 2804.76 C), Pressure 1.000000E+05 Number of moles of components 1.50000E+00, Mass in grams 1.80148E+01 Total Gibbs energy -1.01149E+06, Enthalpy 4.40821E-10, Volume 3.10200E-01 Component Moles M-Fraction Activity Potential Ref.stat H2 1.0000E+00 6.6667E-01 1.2037E-10 -5.8452E+05 SER 02 5.0000E-01 3.3333E-01 3.2235E-15 -8.5394E+05 SER GAS Status ENTERED Driving force 0.0000E+00 Moles 1.5000E+00, Mass 1.8015E+01, Volume fraction 1.0000E+00 Mole fractions: H2 6.66667E-01 02 3.33333E-01 Constitution: H201 5.85041E-01 H 7. .70546E-02 H102 4.50911E-05 H2 1.48661E-01 02 5. .06614E-02 H202 2.49697E-06 H101 1.05477E-01 0 3. . 30571E-02 03 1.90507E-08 T = 4620° C -> T = 2805°C max max ^^^ ^ v_________________________j \_________________________j V V only H20 is considered all species are considered A less obvious fault in our calculations f ^2 eas "\ d a c oratio ? = 0 y ratio=10 7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5 ainmethane ratio Who said that 2:1 was the best ratio? 2805.00- O o 0) (0 (Ď Q_ E Q) 2804.95- 2804.90- 2804.85- 0.475 0.480 0.485 0.490 0.495 0.500 n ex 202 + 8N2 + nCH4 2040 2030 -L 2029 2020- 2010- o °- 2000-CD 3 1990- co CĎ 1980 Q. ÜJ 1970^ 1960-1950-1940 2014------♦--- 15°C I C = 1.058 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 n CH, Fuel & oxidizer? Yes, but ratio as well! 20 taught many thermodynamics-related courses '■•iAiiir j^ Shewmon "Transformations in metals" 0 NB 1 figure 4-10. Free-energy diagram and phase diagram indicating change in solubility of j8, and eutectoid temperature when ß is present as fine spheres (labeled ß'). Porter, Easterling "Phase transformations in metals and alloys" Fig. 5.20 The Gibbs-Thomson effect, (a) Free energy curves at T{. (b) Corresponding phase diagram. Hillert "Applications of Gibbs energy-composition diagrams" Fig. 27. Change in compositions when a pressure is applied to one of the phases in a two-phase equilibrium. 0 *b" *b Rationale \8PyT V always positive u If P increases, then G moves upward 25 Liquid/BCC eq in the Fe-Li 1800 1600- 1400 ^ 1200 CD i_ a> E 1000 CD H 800 600- 400 system 0.4 0.6 Mole fraction of Li 7795 T= 1000 K, reference states are pure iquid Fe and pure liquid Li 0) o g 03 O) (Ď c (D C/) _Q _Q b 3-2-1-0--1- a;g >BCC r A-PL í. -3--4 AfC ^^ *" -5--6-_7 ^ *-* ^^r ^^^^^ O -9-10-11 * s s ^ * s s x?cc=0.465 Ll i j ^ !xL=0.727 -. Ll i i i ^ 0 0 0 1 0 2 0 3 0 4 0.5 0.6 0.7 0.8 0.9 1. 0 Mole fraction of Li 7= 1000 K, reference states are pure BCC Fe and pure BCC Li 0) o g 03 O) (Ď c (D C/) _Q _Q b 7 n 6-5-4- A-PL O 2-1-0-1-2-3-4- V f ^BCC / Af ----- — _ _ __ vJ -^^•fc. 6-7 i ~" ~~ xBCC=0.465| !x'=0.727 ~~ ~" 0 0 0 1 0 • 2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Mole fraction of Li T= 1 000 K, "standard element references" (SGTE) 0 o g H—» CO o - o >> O) s— CD CD C/) _Q _Q b i---------1---------1---------1—*n---------1---------r*—i---------r .0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Mole fraction of Li Not G per se, but A,G! rdG} \8PjT V always positive f d^G\ f V dP Jt A{V ? 30 Let us make a phase y from pure components _______________(l-jc)Aa+jcBp____________________ ------------------v------------------ mechanical mixture AstG = (1 - x) AtiG°Aa^ + xAtiG Oß^y B (1-jí)Ay + jíB1 AldG = RT[(1 - x)In(l - x) + x\nx] 1 Ideal y solution Real y solution Reference states are pure liquid components rOL->L , a id/^L , a ex/^L AGL = (1 - x) AttG^L + x AttG^ + A^G> A^ ~v— =0 _> ^_ "V— =0 rOL—>a . a id/^^a . *ex/oa AGa = (1 - x)AtiG°AL^a + jcAttG^ wa + A^ + AexGc r3AGL^ V ap = 0 Jt 3AGa^ dP + x dAtiG°BL^^ Jt v \ dp )T y dp = (i-x)(vz-v};)+x(vz-vs)<0 Jt J ^_ usually negative usually negative P m Why "usually"? Because there are rare exceptions such as H20, Bi, Sb, cast iron 32 What does this mean?! rL AG does not change its position AGa shifts downward by [(1-x)(fal -V*) + x(v^ -V*) xP The solid phase a is stabilized by pressure applied Now reference states are pure solid components rOa->-L rOa—>L , a id/^íL , a ex/^fL agl = (i - x) AtfiT^+xKGb™+a^ + a^g; <0a->a <0a^a , a id/^a , *ex/oa AGa = (1 - jc) AtrG^a^a + x A„G^a + zTCT + A°*G' ~" ^žfT "Tne) ŕdACŕ} = 0 rdAGL^ v ^ jT (l-,)|^ľ^ 1 ap T V w* Jt ={i-x){vi-n)+x{vz-n)>Q _> ^_ ------------v------------ ------------v------------ usually positive usually positive P m What's going on?! AG™ does not change its position AGL shifts upward by f(l - x) (FA - FA) + x(FBL - FBa) xP The liquid phase is destabilized by pressure applied 35 Reference states are pure liquid components AGL does not change its position, AGa shifts downward by Xi-x)(vZ-vz)+x(vi-vs) xP AGa does not change its position, AGL shifts upward by --------v-------- our result (1- x)(vl -v:) + X(V^ -V*j xP A A In terms of relative positions, these 3 situations are identical > f AGa shifts upward by [(1 - x) FA + xFBa ] x P AGL shifts upward by [(l - jc) FA + xFBL] x P Reference states are pure solid components AGa does not change its position, AGL shifts upward by Xi-x)(vZ-vz)+x(vi-vš) xP AGL does not change its position, AGa shifts downward by --------v-------- our result (1- x)(vl -v:) + X(V^ -V*j xP A A In terms of relative positions, these 3 situations are identical V AGa shifts upward by [(1 - x) FA + xFBa ] x P AGL shifts upward by [(l - jc) FA + xFBL] x P Shewmon "Transformations in metals" 4-6 Consider a system in which four phases exist with a G(7VB) diagram as shown in Fig. 4-13. Show the phase diagram that results if the free energy of the a, ß, and y phases decreases relative to that of the liquid as the temperature is decreased. Do this by first showing the G(NB) diagram for several lower temperatures. G I I A B figure 4-13 Home task: not G, but AfG! / v dG dT \ Jp Always negative í d AfG } V dT Jp ? 39 Retrograde solubility 40 Making an excusable mistake Steel contains 0.4 wt.% of C 4,=12.011g/mole AFe = 55.847 g/mole xc^ 0.0183 Fe C 1 c0.9817 ^0.0183 conf = -i?(0.98171n 0.9817+ 0.01831n 0.0183)« 0.761 K x mole ~ss~ a wrong result a-Fe (ferrite) and 5-Fe -""OJ-x-^""-Od # metal-Atom ° atom in octahedral interstice w metal-Atom o atom in tetrahedrai interstice y-Fe (austenite) # metal atom # metal atom o atom in octahedral interstice o atom in tetrahedral interstice 43 Correcting BCC^Fe^C^Va^ 1 3(1- y) _3-3y xc — l+3(l-y) 4-3y I xc= 0.0183 V A V 0.0062 ^a0.9938/ v--------------------v--------------------' mixing cite 3 I S%£ = 0.926 instead of 0.761 the mistake xc — FCC^Fe^Va,), I i-j _i-y l + (l-y) 2-y i xc =0.0183 V A V 00187 ^a0.9813J ~SS~ mixmg cite \ FCC conf s> = 0.758 instead of 0.76 1 Sublattice model (CEF) (A,B)fl (C,D)c , Ü + C = \--------For the sake > ^ w x ~v„ . , oi simplicity V / V / ^ . , of simplicity beneiicial (A,.,,B7)(C1„D..) (Na;_,,K;)(Cl„,Br;) --------v------------- -------------v-------- cation site anion cite (Sro2+,Bar,La-_Ŕ_c,Va°)(Ti;;,Va;)0 -----v--------------------------------/v--------------v----- A-cite B-cite (\-y)A + yB + (\-z)C + zD^(Al_y,By)(Cl_z,Dz) Another way to synthesize the phase r(l-^)A + .yB + (l-z)C + zD--------- * a AC + ßAD + yBC + SBD (A^B^C^DJ 46 Playing field (A,B)a(C,D). Compounds 3.K.3. end-members Fig. 1. Representation of composition in a quaternary system where the components mix with each other, two and two. Hillerťs Suggestion was based on a powerful KISS principle aAC + ßAD + /BC + SBD -> (a^B^C^DJ (1-^)(1-z)AC + (1-^)zAD +^(l-z)BC + ^zBD^(A1_3;,B,)(C1_z,Dz) A (l-^)(l-z) + (l-^)z = l)^ - yý>ýz y^/ýz = 1 - y 48 Reference surface, not reference line or plane or hyperplane *GA0Dd- -°GBbDd Fig. 2. Suggested surface of reference for the free energy in a quaternary system where the components mix with each other two and two. What's about uniqueness? aAC + ßAD + yBC + ÖBD -> (Ax_y,By){Cx_z,T)z) \f\ 1 0 0^ 2 0 0 11 3 10 10 4^0 1 0 1, l-3 = (0 1 -1 0) (0 1 -1 0) + 2 = (0 1 0 1) v----------------v----------------' 4 rank = 3 4-3 = 1 Hillerťs choice was the simplest and most convenient one A a + ß = \-y B r+s = y C a + Y = \ — z D ß + S = z Tikhonov regularization <\ lVx^ vi V A _> i_ '1 \X2j í \\ \b Ax-b -^min x -(JA) X But what if A is ill-conditioned or singular? Ax-b + ľx —»min, ľ = a x I regularization identity parameter matrix x(a) = (,4T,4+rTr)~Vb How does it work? ix1 + x2 = 1 0.5 0.4- 0.3 0.2 0.1 0.0 What is so special about x, = x = 0.5? 1 2 small a u n distorted system x(tf) = (^+rTr)"Vb large a distorted system 1E-4 1E-3 0.01 0.1 1 10 100 Regularization parameter a r < V 2 2 = 1 = 1 > mm Let us choose a particular solution possessing the minimal norm aAC + ßAD + yBC + SBD^(Al_y,By)(Cl_z,Dz) a + ß = \-y r+s = y a + y-\—z < ß + S = z a2 + ß1 + y1 + S2 -> min a > 0, ß > 0, y > 0, S > 0 53 „reference ^J/foriTIUla Unit CD O o ^"^1^- O >r ' '., v'-. - \, -. \ \ '. \\\-\ -"■ / V: V\ Y-v*-'v A-'A-\- p A/7Vý\\-AV\---v,\tX ° / ■.-'?'' '■J / - J-vk o / ■ .::'ŕ#;: *■ y,;'" ■/'v ^ w'x>vA'^iifiiiii v p / ' -'íi- 11 n cn/v 'í ,\/v í,-V • ''y 'J^^^^^^S o mNú'H)ií^^^^m cn / , v.,: / '< / / ?. ' i^^^^^^^^« m o ľ'í tJ'i */\;*^^^^^^^^^^ c o /Y/ - .\ í J^^^^^^^^^^^^^ o bo / ,?-., / ^^^^^^^^^^^^^^ ^^^^m / .*-*.■-'.•■'•."•'-"-'.•-"•-'-'/-''•','.',-.'•'.'-'.'■"-'■•'.''■.'/• ^^b ■ ■ / ■■ '*' ' l j^^^^Q^^^^^^^^^^^^^^^^^^^^& Q. „reference ^J/formula Unit j^ a, O) Nl CO K) Q) (Ž) O ^ Comparison .0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 N ~r—i------r—i------n—i------n—i------r 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 y y Hillert Euclid An interesting "side effect" aAC + yffAD + /BC + ^BD->(A1_y,By)(C1_z,Dz) AD Are all end-members always needed? A Z AD AC BC «AC + /?AD + rBC+^BD^(A1_j,B,)(C1_z,Dz) Only three compounds are required < A a + ß = \-y B r = y C a + y = \- z D ß = z (1-j-z)AC + zAD + jBC^(A1_j,B,)(C1_z,Dz) Non-negativity condition (l-y-z)AC + zAD + yBC^(Al_y,By)(Cl_z,Dz) 1 - y - z > 0 =^> y + z < 1 A Z AD Another particular solution a + ß = \-y r+s = y a + y = 1- z ß + S = z a2 + ß2 + y2 + S2 —» min a>0,ß>0,y>0,S>0 < a + ß = \-y y + ô = y a + y — \ — z ß + ö = z min {a, /?, y, S) —» min a>0,ß>0,y>0,ö>0 http://mse.mcmaster.ca/ 61