Ch 5 Criteria for Equilibrium Stability 1 Unit

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

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)

Limit of Intrinsic stability

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

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 ≧ 0 (431) 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

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

ΔS U, V, N < 0

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

• (1) The criterion of equilibrium • δ S = Σ δ S (k) = O (6 -6) • (2) The criterion of stability • δ 2 S =Σ δ 2 S (k) ≦ 0, if = 0 • δ 3 S =Σ δ 3 S (k) ≦ 0, if = 0, • δ j. S < 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 • (6 -7)

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

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 < 0 (6 -28)

(3) ΔS =0 , T = constant, ΔS =0 , S = constant, ΔU + PΔV < 0, Δ[U + PV ] < 0, Δ HS, P, N < 0 (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

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 Σ = ΔN j (iii) ΔN sys = ΔN HR= ΔN WR = 0

1. Enthalpy, lock thermal gate, ΔUHR = 0 ΔU Σ = Δ(Usys + UWR ) > 0 (i) ΔS Σ = ΔS sys = 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

2. Helmholtz free energy, lock the piston, ΔUWR = 0 ΔU Σ = Δ(U + UHR ) > 0 (i) ΔS Σ = ΔS + ΔSHR = 0 (ii) ΔV Σ = ΔVHR = ΔVWR = 0; V = constant (iii) ΔN Σ = 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

3. Gibbs free energy, open both the piston and thermal gate, ΔUΣ= Δ(U + UHR + UWR ) > 0 (i) ΔSΣ = ΔS + ΔSHR = ΔSWR = 0 (ii) ΔVΣ = (ΔV + ΔVWR ) = ΔVHR = 0 (iii) ΔNΣ = 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

The criterion of stability δ 2 S = (∂2 S/∂U 2)VNδU 2 + 2 (∂2 S/∂U∂V)N[i] δUδV + (∂2 S/∂V 2)UNδV 2 + 2Σ[ (∂2 S/∂U∂Ni), V, N[i]δU+(∂2 S/∂V∂Ni), U, N[i] δVi ] δNi + ΣΣ(∂2 S/∂Ni∂Nj) U, VδNiδNj < 0 δ 2 S =(SUUδU 2 + 2 SUVδUδV + SVVδV 2 + 2Σ[SU N[i]δU + SU N[i]δV]δNi + ΣΣS Ni. NjδNiδNj < 0 The constraints are δU = 0; δV = 0; δNi = 0, i= 1. 2, ……, n

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

Consider the system is perturbed to become two phase α and β, δ 2 S = δ 2 S α + δ 2 S β = {(∂2 S/∂U 2)VNδU 2 + 2 Σ(∂2 S/∂U∂V )NδUδV + (∂2 S/∂V 2)UNδV 2 + 2Σ [(∂2 S/∂U∂Ni)V, [Ni]δU + (∂2 S/∂V∂Ni)U, N[i]δVi ]δNi + 2 Σ (∂2 S/∂Ni∂Nj)U, V, N[i, j]δNiδNj }α + the similar terms of the β phase < 0 δ 2 S = { (SUUδU 2 + 2 SUVδUδV + SVVδV 2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS Ni. NjδNiδNj }α + the similar terms of the β phase < 0 (7 -1)

The equations for criterion of stability based on δ 2 S or δ 2 U δ 2 S = (N/Nβ ) { (SUUδU 2 + 2 SUVδUδV + SVVδV 2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS Ni. NjδNiδNj }α < 0 ( 7 -5) The similar procedure is used for the criterion of the internal energy, δ 2 U = (N/Nβ ) { (USSδS 2 + 2 USVδSδV + UVVδV 2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNi. NjδNiδNj }α > 0 ( 7 -6) The criterion of stability for the pure component system, for instance, δ 2 U = (N/Nβ) (USS δZ 12 + A 22 δZ 22 + G 33 δZ 3 2) α > 0 , then USS > 0 , AVV > 0 , G 33 > 0

The mathematic treatment for obtaining the criterion of stability U = y (o) = U(S, V, N 1, N 2, ……, Ni, ……. Nn) y (o) = f (x 1, x 2, ……, xi, ……. xm) m=n+2 δ 2 U = (N/Nβ ) { (USSδS 2 + 2 USVδSδV + UVVδV 2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNi. NjδNiδNj }α > 0 (7 -6) δ 2 y (o) = K ΣΣ y (o) ijδx iδx j > 0 ΣΣ y (o) ijδx iδx j = Σ ykk (k-1) δZk 2 > 0 (7 -8) = (7 -6) k = 1, 2, …. . , m (7 -9) δZ k = {δxk + (y(o)23 / y(o)11) δx 2 + (y(o)33 / y(o)11))δx 3 } k = 1, 2, ……, m δZ m= δxm for k = m ykk (k-1) Zk 2 > 0, and Zk 2 > 0 ykk (k-1) > 0, k = 1, 2, …. . , m = n + 2 y 11(o) > 0, y 22(1) > > 0, y 33(2) > 0, …………, ymm (m-1) (= y(n+2) (n+1) ) > 0

• ymm (m-1) = y (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+1) = (∂ ξn+2 /∂xn+2 )ξ • Since ξn+2 = f(ξ 1 , ξ 2 , ………. , 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+1) = (∂ ξn+2 /∂xn+2 )ξ= 0

• ymm (m-1) = y (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+1) = (∂ ξn+2 /∂xn+2 )ξ ξn+2 = f(ξ 1 , ξ 2 , ………. , 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+1) = (∂ ξn+2 /∂xn+2 )ξ= 0

y 11(o) > 0, y 22(1) > 0, … , y(m-1) (m-2) > 0, ymm (m-1) (= y(n+2) (n+1) ) = 0 The number of criterion becomes y 11(o) > 0, y 22(1) > 0, … , y(m-1) (m-2) > 0, ykk(k-1) > 0, k = 1, 2, ……. . , m-1 y (k-2) = f(ξ 1 , …, ξ(k-2) , x(k-1) , ……. . , xm ) y (k-1) = f(ξ 1 , …………, 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-2) (1) ykk(k-2) > 0, y(k-1) (k-2) > 0 (2) Decrease y(k-1) (k-2), increase [yk(k-1) (k-2) ]2/ y(k-1) (k-2) , As continue to decrease y(k-1) (k-2) , [yk(k-1) (k-2) ]2/ y(k-1) (k-2) will be increased, and making{ ykk(k-2) - [yk(k-1) (k-2) ]2/ y(k-1) (k-2) } approach to 0. Further. decrease y(k-1) (k-2), { ykk(k-2) - [yk(k-1) (k-2) ]2/ y(k-1) (k-2) } becomes smaller than negative, i. e. , ykk(k-1) < 0

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

Membrane Equilibrium

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)

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)

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)

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)

Derivation for Equations (7 -5), (7 -6) and (7 -9) δ 2 S = (N/Nβ ) { (SUUδU 2 + 2 SUVδUδV + SVVδV 2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS Ni. NjδNiδNj }α < 0 ( 7 -5) The similar procedure is used for the criterion of the internal energy, δ 2 U = (N/Nβ ) { (USSδS 2 + 2 USVδSδV + UVVδV 2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNi. NjδNiδNj }α > 0 ( 7 -6) The mathematic expression, δ 2 y (o) = K ΣΣ y (o) ijδx iδx j > 0 ΣΣ y (o) ijδx iδx j = Σ ykk (k-1) δZk 2 > 0 k = 1, 2, …. . , m

Consider the system is perturbed to become two phase α and β, δ 2 S = δ 2 S α + δ 2 S β = {(∂2 S/∂U 2)VNδU 2 + 2 Σ(∂2 S/∂U∂V )NδUδV + (∂2 S/∂V 2)UNδV 2 + 2Σ [(∂2 S/∂U∂Ni)V, [Ni]δU + (∂2 S/∂V∂Ni)U, N[i]δVi ]δNi + 2 Σ (∂2 S/∂Ni∂Nj)U, V, N[i, j]δNiδNj }α + the similar terms of the β phase < 0 δ 2 S = { (SUUδU 2 + 2 SUVδUδV + SVVδV 2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS Ni. NjδNiδNj }α + the similar terms of the β phase < 0 (7 -1)

δ 2 S = (N/Nβ ) { (SUUδU 2 + 2 SUVδUδV + SVVδV 2 + 2 SUNδUδN + 2 SVNδVδN + S NNδN 2}α < 0 ( 7 -5) The result is extended to the equations for criterion of stability of multiple components system δ 2 S = (N/Nβ ) { (SUUδU 2 + 2 SUVδUδV + SVVδV 2 + 2ΣSUN[i]δUδNi + 2ΣSVN[i]δVδNi ] + ΣΣS Ni. NjδNiδNj }α < 0 ( 7 -5) The similar procedure is used for the criterion of the internal energy, δ 2 U = (N/Nβ ) { (USSδS 2 + 2 USVδSδV + UVVδV 2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNi. NjδNiδNj }α < 0 ( 7 -6)

Derivation for the mathematic expression for the criterion of stability δ 2 U = (N/Nβ ) { (USSδS 2 + 2 USVδSδV + UVVδV 2 + 2ΣUSN[i]δSδNi + 2ΣUVN[i]δVδNi ] + ΣΣUNi. NjδNiδNj }α < 0 ( 7 -6) U = U(S, V, N 1, N 2, ……, Ni, ……. Nn) Let y (o) = U, x 1 = S, x 2= V, x 3= N 1, ………. . , xm= Nn y (o) = f (x 1, x 2, ……, xi, ……. xm) m=n+2 δ 2 y (o) = K ΣΣ y (o) ijδx iδx j > 0 δ 2 y (o) = K Σ ykk (k-1) δZk 2 > 0 k = 1, 2, …. . , m (7 -9) δZ k = {δxk + (y(o)23 / y(o)11) δx 2 + (y(o)33 / y(o)11))δx 3 } k = 1, 2, ……, m δZ m= δxm for k = m

Consider a single component system, δ 2 U = δ 2 Uα + δ 2 U β = (N/Nβ) {USSδS 2 + 2 USVδSδV +2 + 2 USNδS δN + 2 UVNδVδN + UNNδN 2}α < 0 Use the square method for the equation “y” δ 2 U = USS {δS 2 + 2 (USV/USS )δSδV + (UVV/USS ) δV 2 + 2 (USN /USS )δS δN + 2 (UVN/USS )δVδN + (UNN/USS )δN 2} = USS {δS 2 + 2 (USV/USS )δSδV + 2 (USN /USS ) δSδN + 2 (USVUSN/USS 2 ) δVδN + (USV/USS )2δV 2 + (UNN /USS ) 2δN 2} + [UVVδV 2– (USV 2/USS) δV 2] + [2 UVNδVδN - 2 (USVUSN/USS 2 )δVδN ] +[ UNNδN 2 – (UNN 2 /USS ) δN 2 ] = USS {δS + (USV/USS )δV + (USN /USS)δN}2 + ( UVV- USV 2/USS)δV 2 + 2 (UVN - USVUSN/USS 2)δVδN + (UNN - USN 2 /USS ) δN 2 Let δZ 12 = {δS 2 + 2 (USV/USS )δSδV + 2 (USN /USS ) δSδN + 2 (USVUSN/USS 2 )δVδN + (USV/USS )2δV 2 + (UNN /USS ) 2δN 2} ={δS + (USV/USS ) δV + (UNN /USS )δN} 2 -

δZ 1 = {δS + (USV/USS ) δV + (UNN /USS )δN} δ 2 U = USSδZ 12 + (UVV- USV 2/USS) {δV 2 + 2 [(UVN - USVUSN/USS 2 )/ (UVVUSV 2/USS) ]δVδN + [( USV - USN 2 /USS ) /(UVV- USV 2/USS) ] 2 δN 2 } + {(UNN - USN 2 /USS ) - [( USV - USN 2 /USS )2 /(UVV- USV 2/USS) ] }δN 2 Let δZ 22 = {δV 2 + 2 [(UVN - USVUSN/USS 2 )/ (UVV- USV 2/USS) ]δVδN + [( USV - USN 2 /USS )/(UVV- USV 2/USS) ] 2 δN 2 } = {δV + [( USV - USN 2 /USS )/(UVV- USV 2/USS) }2 δZ 2 = {δV + [( USV - USN 2 /USS )/(UVV- USV 2/USS) } δZ 3 = δN δ 2 U = USSδZ 12 + (UVV- USV 2/USS) δZ 22 + {(UNN - USN 2 /USS ) - [( USV - USN 2 /USS )2 /(UVV- USV 2/USS) ] } δZ 3 2

Use the Legendre transform and Table 5 -3 U = U(S, V, N) y(o) = y( x 1, x 2) A = A(T, V, N) y(1) = y(ξ 1, x 2) G = G(T, P, N) y(2) = y(ξ 1, ξ 2, x 2) 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 5 -3 -3 y(1)ij = y(o)ij - y(o)1 iy(o)1 j/ y(o)11 , i >1, j > 1} y(1)ii = y(o)ii - y(o)1 i 2/ y(o)11 , i >1 UVV - (USV 2/USS) = y(o)22 - y(o)12 2 / y(o)11 = y(1)22 = AVV Ψ= {[UNN – (USN 2 /USS )] – [(UVN - USVUSN/USS)] /(UVV – (USV 2/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]

= [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} = {δx 1 + (y(o)23 / y(o)11) δx 2 + (y(o)33 / y(o)11))δx 3 } δZ 2 = {δV + [( USV - USN 2 /USS )/(UVV- USV 2/USS) } = {δx 2 + [(y(o)12 - USN 2 / y(o)11)/(UVV- USV 2/USS) } δZ k = {δxk + (y(o)23 / y(o)11) δx 2 + (y(o)33 / y(o)11))δx 3 } k = 1, 2, ……, m δZ m= δxm for k = m The criterion stability for pure component system δ 2 U = (N/Nβ)(USSδZ 12 + A 22δZ 22 + G 33δZ 3 2) > 0 , then USS > 0 , AVV > 0 , G 33 > 0

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 y 11(o) > 0, y 22(1) > 0, y 33(2) > 0, y 44(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) y 11(o) = USS > 0, y 22(1) = AVV > 0, y 33(2) = GPP > 0, y 44(3) = G μA μA > 0 Use the step down procedure y 33(2) y 34(2) y 44(3) = y 44(2) - [y 34(2)] 2/ y 33(2) = {y 33(2) y 44(2) - [y 34(2)] 2 }/ y 33(2) = y 34(2) y 44(2) y 33(2)

y 22(1) y 23(1) y 24(1) y 44(3) = y 32 (1) y 42 (1) y 33 (1) y 34 (1) y 43 (1) y 44 (1) y 32(1) y 33(1) D’: 1 term, (1 -δ 34) y 24(1) = y 24(1) D’ D 2 J’ y 32(1) y 33(1) D’: q -1 term, (1 -δjk) y(j-p)k(j-q) , p = (q-1), (q -2), . 1 j = 3, j – q = 1 , q = 2 y 44 y 22(1) y 23(1) Table 5 -5, yik(j) , i, k > j (3) = D 2 1= 1 F’ K’ L’ D 2 1 F’ : (1 -δjk) yjk(j-q) –δjk = (1 -δ 34) y 34(1) –δ 34 = y 34(1) J’: q -1 term, (1 -δij) yi (j-p) (j-q) , p = (q-1), (q -2), . . 1 1 term, (1 -δ 43) y 42 (1) = y 42 (1) K’: (1 -δij)yij (j-q) –δji (1 -δ 34) y 34(1) –δ 34 = y 34(1) = y 43(1) L’: q -1 term, (1 -δji) (1 -δjk) yik(j-q) , 1 term, (1 -δ 34) y 44(1) = y 44(1)

y 11(o) y 12(o) y 13(o) y 14(o) y 44 (3) = y 21 (o) y 22 (o) y 23 (0) y 24 y 31 (o) y 32 (o) y 33 (0) y 41 (o) y 42 (o) y 43 (0) Table 5 -4, yik(j) j < i, k (o) y 22(1) y 23(1) y 24(1) y 34 (o) y 32(1) y 33(1) y 34(1) y 44 (o) y 42(1) y 43(1) y 44(1) C: j -1 term, (1 -δji) ymi(o) , m = 1. 2…, m 2 terms, (1 -δ 34) y 14(o) = y 14(o) , y 24(o) yik(j) = D Dj(o) C E F G Dj(o) D : j-1 term, (1 -δjk) ymk(o) , m = 1. 2…, m 2 terms, (1 -δ 34) y 14(o) = y 14(o) , y 24(o) E : (1 -δji)yji (o) –δji (1 -δ 34) y 34(o) –δ 34 = y 34(o) F : 1 term, (1 -δjk) yjk(o) –δjk = y 34(o) G : (1 -δji)(1–δjk) yik(o) = y 44(o)

(1) y 11(o) y 12(o) y 13(o) y 14(o) y 32(1) y 33(1) y 34(1) y 21(o) y 22(o) y 23(0) y 24(o) y 42(1) y 43(1) y 44(1) y 31(o) y 32(o) y 33(0) y 34(o) y 22 y 33(2) y 34(2) y 23 (1) y 24 = = y 34(2) y 44 y(2)33(2) y 44(3) = (1) y 41(o) y 42(o) y 43(0) y 44(o) y 22(1) y 23(1) y 22(o) y 23(o) y 24(o) y 32(1) y 33(1) {y 33(1) – (y 23(1))2/ y 22(1)] } y 32(o) y 33(o) y 34(o) y 33(2) y 22(1) y 33(2) = y (1) 23 = y 22(1) y 23(1) (o) yy 22 11 12 13 y 21(o) y 22(o) y 23(o) y 31 (o) y 42(o) y 43(o) y 44(o) y 32 (o) y 33(2) y 22(1) y 11(o) y y 21 y 22 (o) 12 o) (o) y 22 (1) = y 11(o) y 12 (o) y 21 (o) y 22 (o) y 12(o)