Η παρουσίαση φορτώνεται. Παρακαλείστε να περιμένετε

Η παρουσίαση φορτώνεται. Παρακαλείστε να περιμένετε

Ch. 5 Criteria for Equilibrium & Stability

Παρόμοιες παρουσιάσεις


Παρουσίαση με θέμα: "Ch. 5 Criteria for Equilibrium & Stability"— Μεταγράφημα παρουσίασης:

1 Ch. 5 Criteria for Equilibrium & Stability
1. Unit process: Chemical reactions 2. Unit operation: mass transfer Distillation Absorption & Adsorption Extraction & leaching Membrane separation

2 Multiple phase, multiple component systems
1. Solubility (i) complete miscible : single homogeneous phase (ii) partial miscible : multiple phases (iii) immiscible: multiple phases 2. Types of phase (i) Liquid –liquid (extraction) (ii) Vapor – liquid (distillation; gas absorption) (iii) Solid –liquid (leaching; crystallization) (iv) Solids in solids (zone melting; crystallization)

3 Limit of Intrinsic stability

4 Classification of equilibrium
In tem of potential energy Unstable equilibrium Neutral stable Metastable equilibrium Potential energy Stable equilibrium barrier

5 Postulate II & The Second Law
☆ Postulate II: In processes for which there is no net effect on the environment, all the systems with a given internal constraints will change in such a way to approach one and only one stable equilibrium state ☆ The Second Law ΔSUniv. (= ΔSisolated ) = ΔSsys + ΔSHRs = ΔSU, V, N ≧ (4-31) for all natural processes, or unstable state systems Since an isolated system, Q = 0, W = - P ΔV = 0, V = const, Further using the first law for close system, ΔU = Q + W, then ΔU = 0, or U = const If the stable processes, or stable state system, ΔSU, V, N < 0

6 Criteria for stable state system
ΔSU, V,N < 0 equilibrium ΔS = 0 S ΔS<0 ΔS>0 unstable stable One of the properties of system

7 ΔS U,V,N < 0

8 ΔS = S - So =δS + (1/2!) δ2S + (1/3!) δ3S + ….. < 0
Perturbation If minor perturbations leave the system unchanged, we define the original state as a stable equilibrium state. So(Z1o,) S S S Zio - δZi Zio + δZi Zio Zi ΔS = S - So =δS + (1/2!) δ2S + (1/3!) δ3S + ….. < 0

9 ΔS = S - So = δ S + (1/2!) δ2S + (1/3!) δ3S + ….. < (6-3) where δ S = Σ (∂S/∂Zi) δZi δ2 S = ΣΣ(∂2S / ∂Zi ∂Zj)δZiδZj δ3 S =ΣΣΣ (∂3S / ∂Zi ∂Zj ∂Zk )δZiδZjδZk (i) δU = 0 (ii) δV = 0 (iii) δNi = 0, i = 1, 2, ….., i, …., n

10 (1) The criterion of equilibrium
δ S = Σ δ S (k) = O (6-6) (2) The criterion of stability δ2S =Σ δ2S (k) ≦ 0, if = 0 δ3S =Σ δ3S (k) ≦ 0 ≦ 0, if = 0, (6-7) δjS < 0 (3) The constrains (Isolated system) (i) δU =ΣδU (k) = 0 (ii) δV =ΣδV (k) = 0 (iii) δNi =ΣδNi (k) = 0, i = 1, 2, ….., i, …., n

11 Equilibrium Criteria Derived from the combination of the first and second laws
ΔS U,V,N = ΔS system + ΔS heat resv. > (1) for irrversible spontanoeus processes ΔS heat resv = Q heat resv /To ( To = Theat resv ) and Q heat resv = - Q system = Q ΔS - Q/ To. < (2) Applying the1st law for the system, ΔU = Q + W Q = ΔU - W (3) Combine (2) and (3), ΔS – (ΔU - W )/ To < 0 ) ΔU - To.ΔS < W = - WwR Only consider P-V work, WwR = [ (- PoΔVo )], and ΔVo = -ΔVsys

12 Equilibrium criterion for natural processes
ΔU + Po ΔV – To ΔS <0 Equilibrium criterion for natural processes (1) ΔV =0 , i.e., V = const., ΔS =0 , S = const Δ US, V, N < 0 (2) ΔV =0 , V = constant, To.= T = constant ΔU - TΔS < 0 , ΔU - Δ (T S) < 0 Δ(U - T S) < 0 The Helmholtz free energy, A, is defined as A = U - TS Δ A < 0, Δ AT, V, N < (6-28)

13 (3) ΔS =0 , T = constant, ΔS =0 , S = constant,
ΔU + PΔV < 0, Δ[U + PV ] < 0, Δ H < 0, Δ HS, P, N < (6-27) (4) Po = P = constant, To.= T = constant ΔU + Δ(PV) -Δ(TS) <0 H = U + PV Δ ( U + PV -TS) < 0 The Gibbs-free energy is defined as G = U + PV –TS ΔG < 0 ΔG T, P, N < 0

14 Criterion based on H, A and G
ΔU Σ = Δ(Usys + UHR + UWR ) > 0 (i) ΔS Σ = Δ(Ssys + SHR)= ΔSWR = 0 (ii) ΔV Σ = Δ(V + VWR)= ΔVHR = 0 (iii) ΔN Σ = ΔNsys = ΔN HR= ΔN WR = 0 j

15 1. Enthalpy, lock the thermal gate, ΔUHR = 0
ΔU Σ = Δ(Usys + UWR ) > 0 (i) ΔS Σ = ΔS sys = ΔSHR = ΔSHR = 0; S = constant (ii) ΔVΣ = Δ(Vsys + VWR) = ΔVHR = 0 (iii) ΔNiΣ = ΔNisys = 0; Ni = constant, i = 1, 2,…., n Apply the first law for the work reservoir (WR) ΔU WR = - P WR ΔV WR = P ΔV (let PWR = Psys = P ) ΔU Σ = ΔU + P ΔV > 0, If P = constant ΔU Σ = ΔU +Δ(P V) > 0 ΔU Σ = Δ(U +P V) > 0 Since H = U + P V , ΔH S, P, N > 0

16 2. Helmholtz free energy, lock the piston, ΔUWR = 0
ΔU Σ = Δ(U + UHR ) > 0 (i) ΔS Σ = ΔS + ΔSHR = 0 (ii) ΔV Σ = ΔV = ΔVHR = ΔVWR = 0; V = constant (iii) ΔN Σ = ΔN = ΔNHR= ΔN WR = 0; N = constant Apply the first law for the heat reservoir (HR) ΔUHR = THR ΔSHR = - TΔS (If TWR = T = constant) ΔU Σ = ΔU + [ -Δ(TS)]= Δ(U - TS) > 0 A= U - TS ΔA T, V, N > 0

17 3. Gibbs free energy, open both the piston and the thermal gate,
ΔUΣ= Δ(U + UHR + UWR ) > 0 (i) ΔSΣ = ΔS + ΔSHR = ΔSWR = 0 (ii) ΔVΣ = (ΔV + ΔVWR ) = ΔVHR = 0 (iii) ΔNΣ = ΔN = ΔNHR = ΔNWR = 0 Apply the first law for the HR and WR ΔUHR = THR ΔSHR = - T ΔS (If TWR = T = constant) ΔUWR = - PWR ΔVWR = P ΔV (If PWR = P = constant ΔUΣ= Δ(U + P ΔV - T ΔS) = Δ(U + PV - TS) > 0 G = U + PV - TS ΔG T, P, N > 0

18

19 The criterion of stability
δ2S = (∂2S/∂U2)VNδU2 + 2 (∂2S/∂U∂V)N[i] δUδV + (∂2S/∂V2)UNδV2 + 2Σ[ (∂2S/∂U∂Ni),V,N[i]δU+(∂2S/∂V∂Ni),U, N[i] δVi ] δNi + ΣΣ(∂2S/∂Ni∂Nj) U,VδNiδNj < 0 δ2S =(SUUδU2 + 2 SUVδUδV + SVVδV2 + 2Σ[SU N[i]δU + SU N[i]δV]δNi + ΣΣS NiNjδNiδNj < 0 The constraints are δU = 0; δV = 0; δNi = 0, i= 1. 2,……, n

20 Taking Δ US, V, N < 0 as the criterion
ΔUS, V, N (= U -Uo ) = δU + (1/2!) δ2U + (1/3!) δ3U + …> 0 (1)The criterion of equilibrium δUS,V,N = δU = Σ(∂U/ ∂Zi) δZi = O (2)The criterion of stability δ2U = ΣΣ(∂2U/ ∂Zj ∂Zj)δZiδZj = ≧ 0, if = 0 δ3U = ΣΣΣ (∂3U/ ∂Zi ∂Zj ∂Zk)δZiδZjδZk≧ 0, if = 0, ……………….. If = 0 δjU > 0 The constraints (i) δS = 0 (ii) δV = 0 (iii) δNi = 0, i = 1, 2, ….., i, …., n

21 Consider the system is perturbed to become two phase α and β,
δ2S = δ2Sα + δ2S β = {(∂2S/∂U2)VNδU2 + 2 Σ(∂2S/∂U∂V )NδUδV + (∂2S/∂V2)UNδV2 + 2Σ [(∂2S/∂U∂Ni)V, [Ni]δU + (∂2S/∂V∂Ni)U, N[i]δVi ]δNi + 2 Σ (∂2S/∂Ni∂Nj)U,V,N[i,j]δNiδNj }α + the similar terms of the β phase < 0 δ2S = { (SUUδU2 + 2 SUVδUδV + SVVδV2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS NiNjδNiδNj }α + the similar terms of the β phase < 0 < (7-1)

22 The equations for criterion of stability based on δ2S or δ2U
δ2S = (N/Nβ ) { (SUUδU2 + 2 SUVδUδV + SVVδV2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS NiNjδNiδNj }α < ( 7-5) The similar procedure is used for the criterion of the internal energy, δ2U = (N/Nβ ) { (USSδS2 + 2 USVδSδV + UVVδV2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNiNjδNiδNj }α > ( 7-6) The criterion of stability for the pure component system, for instance, δ2U = (N/Nβ) (USS δZ 12 + A22 δZ22 + G33 δZ3 2) α > 0 , then USS > 0 , AVV > 0 , G33 > 0

23 The mathematic treatment for obtaining the criterion of stability
U = y (o) = U(S, V, N1, N2, ……, Ni, ……. Nn) y (o) = f (x1, x2, ……, xi, ……. xm) m = n + 2 δ2U = (N/Nβ ) { (USSδS2 + 2 USVδSδV + UVVδV2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNiNjδNiδNj }α > (7-6) δ2y (o) = K ΣΣ y (o) ijδx iδx j > 0 (7-8) = (7-6) ΣΣ y (o) ijδx iδx j = Σ ykk (k-1) δZk2 > k = 1, 2,….., m (7-9) δZ k = {δxk + (y(o)23 / y(o)11) δx2 + (y(o)33 / y(o)11))δx3 } k = 1, 2, ……,m δZ m= δxm for k = m ykk (k-1) Zk2 > 0, and Zk2 > 0 ykk (k-1) > 0, k = 1, 2,….., m = n + 2 y11(o) > 0, y22(1) > > 0, y33(2) > 0, …………, ymm (m-1) (= y(n+2)(n+2) (n+1) ) > 0

24 The Legendre transform, y(o) = f (x1, x2, ……, xi, ……. xm)
d y(o) = Σ {(∂ y (o) /∂xi)x[i] dxi = Σ ξi dxi And ξi = {(∂ y (o) /∂xi)x[i] = y i (o) y (j) = f (ξ1, ξ2, … ξj , xj+1, ……. xm) = y(o) - Σξi xi d y (j) = - Σxi dξi + Σξi dxi y i(j) = {(∂ y (j) /∂ xi)ξ, x[i] = ξi = y i (o) , i > j y i(j) = {(∂ y (j) /∂ ξi )ξ[j], x = - xi, i ≦ j y (m-1) = f (ξ1, ξ2, … ξj , …..ξm-1, xm) , m = n+2 y (n+1) = f (ξ1, ξ2, … ξj , …..ξn+1, xm) j j j+1

25 ymm (m-1) = y (n+2)(n+2) (n+1) = (∂2 y (n+1) /∂xn+2 2)ξ
= (∂/∂xn+2 )(∂ y (n+1)/ ∂xn+2 )ξ yx(n+2)(n+1) =(∂ y (n+1)/ ∂xn+2 ) = y(n+2)(o) = ξn+2 y (n+2)(n+2) (n+1) = (∂ ξn+2 /∂xn+2 )ξ Since ξn+2 = f(ξ1 ,ξ2 , ………., ξn+1) An intensive property,ξn+2, is determined by the other (n + 1) intensive properties, ξi , Therefore, as other (n + 1) intensive properties are fixed, ξn+2 should be constant, ymm (m-1) = y (n+2)(n+2) (n+1) = (∂ ξn+2 /∂xn+2 )ξ= 0

26 ymm (m-1) = y (n+2)(n+2) (n+1) = (∂2 y (n+1) /∂xn+2 2)ξ
= (∂/∂xn+2 )(∂ y (n+1)/ ∂xn+2 )ξ yx(n+2)(n+1) =(∂ y (n+1)/ ∂xn+2 ) = y(n+2)(o) = ξn+2 y (n+2)(n+2) (n+1) = (∂ ξn+2 /∂xn+2 )ξ ξn+2 = f(ξ1 ,ξ2 , ………., ξn+1) An intensive property,ξn+2, is determined by the other (n + 1) intensive properties, ξi , Therefore, as other (n + 1) intensive properties are fixed, ξn+2 should be constant, ymm(m-1) = y(n+2)(n+2) (n+1) = (∂ ξn+2 /∂xn+2 )ξ= 0

27 y11(o) > 0, y22(1) > 0, … , y(m-1)(m-1) (m-2) > 0, ymm (m-1) (= y(n+2)(n+2) (n+1) ) = 0
The number of criterion becomes y11(o) > 0, y22(1) > 0, … , y(m-1)(m-1) (m-2) > 0, ykk(k-1) > 0, k = 1, 2, …….., m-1 y (k-2) = f(ξ1 ,…,ξ(k-2) , x(k-1) ,…….., xm ) y (k-1) = f(ξ1 ,…………,ξ(k-1) , xk ,…….., xm ) By the step down procedure, ykk(k-1) = ykk(k-2) - [yk(k-1) (k-2) ]2/ y(k-1)(k-1) (k-2) (1) ykk(k-2) > 0, y(k-1)(k-1) (k-2) > 0 (2) Decrease y(k-1)(k-1) (k-2), increase [yk(k-1) (k-2) ]2/ y(k-1)(k-1) (k-2) , As continue to decrease y(k-1)(k-1) (k-2) , [yk(k-1) (k-2) ]2/ y(k-1)(k-1) (k-2) will be increased, and making{ ykk(k-2) - [yk(k-1) (k-2) ]2/ y(k-1)(k-1) (k-2) } approach to 0. Further. decrease y(k-1)(k-1) (k-2), { ykk(k-2) - [yk(k-1) (k-2) ]2/ y(k-1)(k-1) (k-2) } becomes smaller than negative, i.e., ykk(k-1) < 0

28 y(m-1)(m-1) (m-2) > 0 (7-15)
It indicates the term,, ykk(k-1) becomes negative before y(k-1)(k-1) (k-2) keeps positive does, in other words, only if y(k-1)(k-1) (k-2) > 0, ykk(k-1) should be positive. Consequently, the necessary and sufficient condition criterion of stability, y(m-1)(m-1) (m-2) > (7-15) The limit of stability or spinnodal condition is y(m-1)(m-1) (m-2) = (7-16) y(m-1)(m-1) (m-2) = £i /Π y(r +1-)(r+1) (r) 0 ≦i ≦ m -2 (7-20) For a stable system, £i > (7-17) At the limit of stability, £i= (7-18) m = 3 r = i

29 £i = y (i+1) (i+1)(i), y (i+1) (i+2)(i), …….. …. y (i+1) (m-1)(i) y (i+2) (i+1)(i), y (i+2) (i+2)(i), …….. …. y (i+2) (m-1)(i) y (i+3) (i+1)(i), y (i+3 )(i+2)(i), …….. …. y (i+3) (m-1) (i) ……………………………………. y (m-1) (i+1)(i), y (m-1) (i+2)(i), …….. …. y (m-1) (m-1) (i) 0 ≦i ≦ m -2 £3 = y(m-1)(m-1) (m-2) = y44(3) = £2 /y33(2) = £1 /y33(2) y22(1) = £o /y33(2) y22(1) y11(o)

30

31 Multiple Phases multiple components equilibrium
A multi-phases (Π), multi-component (n) system δS = Σ δS (s) = Σ {[1/T (s)] δ U(s) + [P (s)/T (s)] δ V(s) –Σ [μi (s)/T (s)] δNi (s) }= 0 δU = 0 = Σ δU (s) δV = 0 = Σ δV (s) δNi= 0 = Σ δ Ni(s) ,i = 1, 2,…., j,…., n δS = Σ [1/T(k) - 1/T(1) ] δ U(k) + [P (k)/T (k) - P (1)/T(1) ] δ V(k) – Σ [μi (k)/T (k) - μi(1)/T(1) ] δNi(k) = 0 T(k) = T(1) P(k) = P(1) k = 1, 2,…., Π μi (k) =μi(1) , i=1,2,…., n Π n Π Π Π Π n k ≠1

32 Chemical Reaction equilibria
ν1C1 + ν2C2 + ….. + νiCi =0 Σ νjCj = 0 δN1/ν1 = δN2/ν2 = ………= δNi/νi = δξ δNj/νj = δξ , j = 1, 2,…i Σ δNj (s) = νj δξ For the inert components, Σ δNj (s) = 0, j = i + 1, …n δS = Σ [1/T(k) - 1/T (1) ] δ U(k) + [P (k)/T (k) - P (1)/T(1) ] δ V(k) – (1/T (1))Σνj μj δξ - Σ[μi (k)/T (k) - μi(1)/T(1) ] δNi(k) = 0 T(k) = T (1); P(k) = P (1); Σνj μj = 0; μi (k) =μi(1) , i = i+1,….., n

33

34 Membrane Equilibrium

35 T(1) = T(2) ; P(1) = P(2) ; μA (1) = μA(2) ; μB (1) = μB(2)
The internal wall is movable, diathermal, and permeable to both A and B δU = 0 = δU (1) + δU (2) δV = 0 = δV (1) + δV (2) δNA = 0 = δ NA(1) + δ NA(2) δNB = 0 = δ NB(1) + δ NB(2) T(1) = T(2) ; P(1) = P(2) ; μA (1) = μA(2) ; μB (1) = μB(2)

36 Case (a) The internal boundary is permeable only to B, diathermal and movable
δNA = 0 = δ NA(1) = δ NA(2) δNB = 0 = δ NB(1) + δ NB(2) δS = [1/T(1) - 1/T(2) ] δ U(1) + [P (1)/T (1) - P (2)/T(2) ] δ V(1) – [μB (1)/T (1) - μB(2)/T(2) ] δNB (1) = 0 T(1) = T(2) P(1) = P(2) μB (1) = μB(2)

37 Case (b) The internal boundary is rigid, diathermal, and permeable to both A and B,
δU = 0 = δU (1) + δU (2) δV = 0 = δV (1) = δV (2) δNA = 0 = δ NA(1) + δ NA(2) δNB = 0 = δ NB(1) + δ NB(2) δS = [1/T(1) - 1/T(2) ] δ U(1) - [μA (1)/T (1) - μA(2)/T(2) ] δNA(1) – [μB (1)/T (1) - μB(2)/T(2) ] δNB (1) = 0 T(1) = T(2) μA (1) = μA(2) μB (1) = μB(2)

38 Case (b) The internal boundary is movable, adiabatic, and permeable to both A and B
δU = 0 = δU (1) = δU (2) δV = 0 = δV (1) + δV (2) δNA = 0 = δ NA(1) + δ NA(2) δNB = 0 = δ NB(1) + δ NB(2) However, mass interchange between the subsystems, can vary the energy of each compartment; thus in reality, no additional restrains, and at equilibrium, T(1) = T(2) ; P(1) = P(2) ; μA (1) = μA(2) ; μB (1) = μB(2)

39

40 Derivation for Equations (7-5), (7-6) and (7-9)
δ2S = (N/Nβ ) { (SUUδU2 + 2 SUVδUδV + SVVδV2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS NiNjδNiδNj }α < ( 7-5) The similar procedure is used for the criterion of the internal energy, δ2U = (N/Nβ ) { (USSδS2 + 2 USVδSδV + UVVδV2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNiNjδNiδNj }α > ( 7-6) The mathematic expression, δ2y (o) = K ΣΣ y (o) ijδx iδx j > 0 ΣΣ y (o) ijδx iδx j = Σ ykk (k-1) δZk2 > k = 1, 2,….., m

41 Consider the system is perturbed to become two phase α and β,
δ2S = δ2Sα + δ2S β = {(∂2S/∂U2)VNδU2 + 2 Σ(∂2S/∂U∂V )NδUδV + (∂2S/∂V2)UNδV2 + 2Σ [(∂2S/∂U∂Ni)V, [Ni]δU + (∂2S/∂V∂Ni)U, N[i]δVi ]δNi + 2 Σ (∂2S/∂Ni∂Nj)U,V,N[i,j]δNiδNj }α + the similar terms of the β phase < 0 δ2S = { (SUUδU2 + 2 SUVδUδV + SVVδV2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS NiNjδNiδNj }α + the similar terms of the β phase < 0 < (7-1)

42 (1)ΣδU = δUα + δUβ = 0; δUα = - δUβ ; (δUα)2 = (δUβ)2
ΣδV = δVα + δVβ = 0; δVα = - δVβ ; (δVα)2 = (δVβ)2 ; ΣδN = δNα + δNβ = 0; δNα = - δNβ; (δNα)2 = (δNβ)2 δUαδVα = ( - δUβ)( -δVβ ) δNα δVα =( - δNβ)( -δVβ ) δUα δNα =( - δUβ)( -δNβ ) (2) SUU = (∂2S/∂U2)VN= [(∂S/∂U)(∂S/∂U)VN ]VN = [(∂S/∂U )(1/T )]VN = (-1/T 2) (∂T/∂U)VN = (-1/T 2N) (∂T/∂U)VN SUUα + SUUβ = [( -1/T 2N)α + ( -1/T 2N)β](∂T/∂U )VNα Since (∂T/∂U ) VNα = (∂T/∂U ) VNβ; Tα = Tβ = (-1/T 2 )[Nα + Nβ) /NαNβ](∂T/∂U )VNα = (-1/T 2 )[N/ Nβ][(∂T/∂(NU )]VNα = (-1/T 2 )[N/ Nβ](∂T/∂U )VNα = [N/ Nβ] SUUα

43 (3) SVV = (∂2S/∂V2)UN (∂S/∂V) UN = - (∂U/∂V) SN / (∂U/∂S)VN = - (-P)/T SVV = (∂2S/∂V2)UN = [(∂ (P/T) /∂V]UN = [T( ∂P/∂V )UN - P( ∂T/∂V )UN]/T2 = (1/N) [T( ∂P/∂V )U - P( ∂T/∂V )U]/T2 SVVα + SVVβ = {(1/N) [T( ∂P/∂V )U - P( ∂T/∂V )U]/T2}α+ {(1/N) [T( ∂P/∂V )U P( ∂T/∂V )U]/T2}β At equilibrium, (∂P/∂V)Uα = (-∂P/∂V)Uβ and (∂T/∂V)Uα = (∂T/∂V)Uβ SVVα + SVVβ = {(1/Nα + 1/Nβ) [T( ∂P/∂V )U - P( ∂T/∂V )U]/T2}α = [(Nα+ Nβ)/NαNβ] [T( ∂P/∂V )U - P( ∂T/∂V )U]/T2}α = (N/Nβ {[T[(∂P/∂(NV)]UN – P[∂T/∂(NV )U]/T2}α = (N/Nβ ) SVVα

44 (4) SUV = (∂2S/ ∂U∂V) N = [(∂/∂U ) (∂S /∂V)UN]VN
(∂S/∂V) UN = P/T SUV = [(∂ (P/T) /∂U]UN = [T( ∂P/∂U )VN - P( ∂T/∂U )VN]/T2 = (1/N) [T( ∂P/∂U )V - P( ∂T/∂U )V]/T2 SUVα + SUVβ = {(1/N) [T( ∂P/∂U )V - P( ∂T/∂U )V]/T2}α+ {(1/N) [T( ∂P/∂U)V P( ∂T/∂U )V]/T2}β At equilibrium, (∂P/∂U)Vα = (∂P/∂U)Vβ and (∂T/∂U)Vα = (∂T/∂U)Vβ SUVα + SUVβ = {(1/Nα + 1/Nβ) [T( ∂P/∂U)V - P( ∂T/∂U)V]/T2}α = [N/NαNβ] [T( ∂P/∂U)V - P( ∂T/∂U)V]/T2}α = (N/Nβ {[T[(∂P/∂(NU)]VN – P[∂T/∂(NU )VN]/T2}α = (N/Nβ ) SUVα

45 (SNNα + UNNβ) = {(1/N) [ - T ( ∂μ/∂x) UV -μ(∂ T/∂x ) UV/T2]}α +
(5) xα = Nα/(Nα + Nβ ) = Nα/N ; xβ = Nβ/N SNN = (∂2S/∂N2)UV= [(∂/∂N) (∂S/∂N)]UV (∂S/∂N)UV = - (∂U/∂N)SV /(∂U/∂S)NV = - μ/T (Triple-product rule) [(∂/∂N) (∂S/∂N)]UV = [(∂ (-μ/T) /∂N)] UV = (1/N) [(∂ (-μ/T) /∂x)]UV = (1/N) [(-∂ (μ/T) /∂x)]UV = (1/N) [- T ( ∂μ/∂x) UV -μ(∂ T/∂x ) UV/T2] (SNNα + UNNβ) = {(1/N) [ - T ( ∂μ/∂x) UV -μ(∂ T/∂x ) UV/T2]}α + {(1/N)[ - T ( ∂μ/∂x) UV -μ(∂ T/∂x ) UV/T2]}β (∂ μ/∂x)UVα = (∂ μ/∂x)UVβ ; (∂T/∂x)vnα = (∂T/∂x)UVβ; Tα = Tβ (SNNα + UNNβ) = [ (1/Nα) + (1/Nβ)] [ - T ( ∂μ/∂x) UV -μ(∂ T/∂x ) UV/T2]}α = (N/NβNα) {[ - T ( ∂μ/∂x) UV -μ(∂ T/∂x ) UV/T2]}α = (N/Nβ){ [ - T ( ∂μ/(N∂x)) UV -μ(∂ T/∂(Nx)]UV/T2]}α = (N/Nβ ) SUVα

46 δ2S = (N/Nβ ) { (SUUδU2 + 2 SUVδUδV + SVVδV2 + 2SUNδUδN +
2 SVNδVδN + S NNδN2}α < ( 7-5) The result is extended to the equations for criterion of stability of multiple components system δ2S = (N/Nβ ) { (SUUδU2 + 2 SUVδUδV + SVVδV2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS NiNjδNiδNj }α < ( 7-5) The similar procedure is used for the criterion of the internal energy, δ2U = (N/Nβ ) { (USSδS2 + 2 USVδSδV + UVVδV2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNiNjδNiδNj }α < ( 7-6)

47 Derivation for the mathematic expression for the criterion of stability
δ2U = (N/Nβ ) { (USSδS2 + 2 USVδSδV + UVVδV2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNiNjδNiδNj }α < ( 7-6) U = U(S, V, N1, N2, ……, Ni, ……. Nn) Let y (o) = U, x1 = S, x2= V, x3= N1, ……….., xm= Nn y (o) = f (x1, x2, ……, xi, ……. xm) m = n + 2 δ2y (o) = K ΣΣ y (o) ijδx iδx j > 0 δ2y (o) = K Σ ykk (k-1) δZk2 > k = 1, 2,….., m (7-9) δZ k = {δxk + (y(o)23 / y(o)11) δx2 + (y(o)33 / y(o)11))δx3 } k = 1, 2, ……,m δZ m= δxm for k = m

48 Let δZ 12 = {δS2 + 2 (USV/USS )δSδV + 2 (USN /USS ) δSδN + 2
Consider a single component system, δ2U = δ2Uα + δ2U β = (N/Nβ) {USSδS2 + 2USVδSδV USNδS δN + 2 UVNδVδN + UNNδN2}α < 0 Use the square method for the equation “y” δ2U = USS {δS2 + 2 (USV/USS )δSδV + (UVV/USS ) δV2 + 2 (USN /USS )δS δN + 2 (UVN/USS )δVδN + (UNN/USS )δN2} = USS {δS2 + 2 (USV/USS )δSδV + 2 (USN /USS ) δSδN + 2 (USVUSN/USS2 ) δVδN + (USV/USS )2δV 2 + (UNN /USS ) 2δN2} + [UVVδV2– (USV2/USS) δV 2] + [2 UVNδVδN - 2 (USVUSN/USS2 )δVδN ] +[ UNNδN2 – (UNN2 /USS ) δN2 ] = USS {δS + (USV/USS )δV + (USN /USS)δN}2 + ( UVV- USV2/USS)δV 2 + 2 (UVN - USVUSN/USS2)δVδN + (UNN - USN2 /USS ) δN2 Let δZ 12 = {δS2 + 2 (USV/USS )δSδV + 2 (USN /USS ) δSδN + 2 (USVUSN/USS2 )δVδN + (USV/USS )2δV 2 + (UNN /USS ) 2δN2} ={δS + (USV/USS ) δV + (UNN /USS )δN} 2 -

49 δZ 1 = {δS + (USV/USS ) δV + (UNN /USS )δN}
δ2U = USSδZ 12 + (UVV- USV2/USS) {δV [(UVN - USVUSN/USS2 )/ (UVV USV2/USS) ]δVδN + [( USV - USN2 /USS ) /(UVV- USV2/USS) ] 2 δN2 } + {(UNN - USN2 /USS ) - [( USV - USN2 /USS )2 /(UVV- USV2/USS) ] }δN2 Let δZ 22 = {δV [(UVN - USVUSN/USS2 )/ (UVV- USV2/USS) ]δVδN + [( USV - USN2 /USS )/(UVV- USV2/USS) ] 2 δN2 } = {δV + [( USV - USN2 /USS )/(UVV- USV2/USS) }2 δZ 2 = {δV + [( USV - USN2 /USS )/(UVV- USV2/USS) } δZ3 = δN δ2U = USSδZ 12 + (UVV- USV2/USS) δZ 22 + {(UNN - USN2 /USS ) - [( USV - USN2 /USS )2 /(UVV- USV2/USS) ] } δZ3 2

50 Use the Legendre transform and Table 5-3
U = U(S, V, N) y(o) = y( x1, x2, x2) A = A(T, V, N) y(1) = y(ξ1, x2, x2) G = G(T, P, N) y(2) = y(ξ1, ξ2, x2) y(o)11 = USS , y(o)22 = UVV , y(o)12 = USV , y(o)13 = USN , y(o)23 = UVN , y(o)33 = UNN y(1)11 = ATT, y(1)22 = AVV, y(1)12 = ATV, y(1)13 = ATN, y(1)23 = AVN , y(1)33 = ANN y(2)33 = GNN {(Table y(1)ij = y(o)ij - y(o)1iy(o)1j/ y(o)11 , i >1, j > 1} y(1)ii = y(o)ii - y(o)1i2/ y(o)11 , i >1 UVV - (USV2/USS) = y(o)22 - y(o)12 2 / y(o)11 = y(1)22 = AVV Ψ= {[UNN – (USN2 /USS )] – [(UVN - USVUSN/USS)] /(UVV – (USV2/USS)] } = [y(o)33 – y(o)132 / y(o)11] –[y(o)23 – y(o)12 y(o)13 / y(o)11] / [y(o)22 – y(o)122 / y(o)11]

51 = [y(o)33 – y(o)132 / y(o)11] – [y(o)23 – y(o)12 y(o)13 / y(o)11] / [y(o)22 – y(o)122 / y(o)11]
= y(1)33 – y(1)23 / y(1)22 = y(2)33 = GNN δZ 1 = {δS + (USV/USS ) δV + (UNN /USS )δN} = {δx1 + (y(o)23 / y(o)11) δx2 + (y(o)33 / y(o)11))δx3 } δZ 2 = {δV + [( USV - USN2 /USS )/(UVV- USV2/USS) } = {δx2 + [(y(o)12 - USN2 / y(o)11)/(UVV- USV2/USS) } δZ k = {δxk + (y(o)23 / y(o)11) δx2 + (y(o)33 / y(o)11))δx3 } k = 1, 2, ……,m δZ m= δxm for k = m The criterion stability for pure component system δ2U = (N/Nβ)(USSδZ 12 + A22δZ 22 + G33δZ3 2) > 0 , then USS > 0 , AVV > 0 , G33 > 0

52

53

54

55

56

57 Example 7-3 U = y(o) = U(S, V, NA, NB, NC) ( m = 5)
The criterion of stability is ykk(k-1) > 0, k = 1, 2, …….., (m-1) = 1, 2, 3, 4 y11(o) > 0, y22(1) > 0, y33(2) > 0, y44(3) > 0 A = y(1) = A(T, V, NA, NB, NC) G = y(2) = G(T, P, NA, NB, NC) G’ = y(3) = G’(T, V, μA, NB, NC) y11(o) = USS > 0, y22(1) = AVV > 0, y33(2) = GPP > 0, y44(3) = G μA μA > 0 Use the step down procedure y44(3) = y44(2) - [y34(2)] 2/ y33(2) = {y33(2) y44(2) - [y34(2)] 2 }/ y33(2) = y33(2) y34(2) y34(2) y44(2) y33(2)

58 F’ : (1-δjk) yjk(j-q) –δjk = (1-δ34) y34(1) –δ34 = y34(1)
y22(1) y23(1) y24(1) y32(1) y33(1) y34(1) y42(1) y43(1) y44(1) y22(1) y23(1) y32(1) y33(1) D2 1 = y44(3) = D’: q -1 term, (1-δjk) y(j-p)k(j-q) , p = (q-1),(q -2), .1 D’: 1 term, (1-δ34) y24(1) = y24(1) F’ : (1-δjk) yjk(j-q) –δjk = (1-δ34) y34(1) –δ34 = y34(1) J’: q -1 term, (1-δij) yi (j-p) (j-q) , p = (q-1),(q -2),..1 1 term, (1-δ43) y42 (1) = y42 (1) K’: (1-δij)yij (j-q) –δji (1-δ34) y34(1) –δ34 = y34(1) = y43(1) L’: q -1 term, (1-δji) (1-δjk) yik(j-q) , 1 term, (1-δ34) (1-δ34) y44(1) = y44(1) y22(1) y23(1) y32(1) y33(1) Table 5-5, yik(j) , i, k > j j = 3, j – q = 1 , q = 2 D’ F’ D21 y44(3) = J’ K’ L’ D21

59 D : j-1 term, (1-δjk) ymk(o) , m = 1.2…,m
y11(o) y12(o) y13(o) y14(o) y21(o) y22(o) y23(0) y24(o) y31(o) y32(o) y33(0) y34(o) y41(o) y42(o) y43(0) y44(o) y22(1) y23(1) y24(1) y32(1) y33(1) y34(1) y42(1) y43(1) y44(1) y44(3) = C: j -1 term, (1-δji) ymi(o) , m = 1.2…,m 2 terms, (1-δ34) y14(o) = y14(o) , y24(o) D : j-1 term, (1-δjk) ymk(o) , m = 1.2…,m 2 terms, (1-δ34) y14(o) = y14(o) , y24(o) E : (1-δji)yji (o) –δji (1-δ34) y34(o) –δ34 = y34(o) F : 1 term, (1-δjk) yjk(o) –δjk = y34(o) G : (1-δji)(1–δjk) yik(o) = y44(o) Table 5-4, yik(j) j < i, k D F Dj(o) yik(j) = C E G Dj(o)

60 y11(o) y12(o) y13(o) y14(o) y21(o) y22(o) y23(0) y24(o)
y22(1) y23(1) y24(1) y32(1) y33(1) y34(1) y42(1) y43(1) y44(1) y33(2) y34(2) y34(2) y44(2) = = y44(3) = y22(1) y23(1) y32(1) y33(1) y33(2) y22(o) y23(o) y24(o) y32(o) y33(o) y34(o) y42(o) y43(o) y44(o) {y33(1) – (y23(1))2/ y22(1)] } y33(2) y22(1) y22(1) y23(1) y23(1) y22(1) y22(1) y33(2) = y33(2) y22(1) y11(o) y11(o) y12(o) y13(o) y21(o) y22(o) y23(o) y31(o) y32(o) y33(o) y11(o) y12o) y21(o) y22(o) y11(o) y12 (o) y21(o) y22(o) = y22(1) = y12(o)


Κατέβασμα ppt "Ch. 5 Criteria for Equilibrium & Stability"

Παρόμοιες παρουσιάσεις


Διαφημίσεις Google