THERMODYNAMICS PROPERTIES OF FLUIDS

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CHAPTER VI. THERMODYNAMICS PROPERTIES OF FLUIDS. Numerical values for thermodynamic properties are essential to the calculation of HEAT and WORK . Example : work requirement for a compressor to operate ADIABATICALLY , to rise the pressure of gas from P 1 to P 2 : Ws = Δ H = H 2 – H 1

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## THERMODYNAMICS PROPERTIES OF FLUIDS

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1. CHAPTER VI THERMODYNAMICS PROPERTIES OF FLUIDS

2. Numerical values for thermodynamic properties are essential to the calculation of HEAT and WORK. • Example : work requirement for a compressor to operate ADIABATICALLY, to rise the pressure of gas from P1 to P2: Ws = ΔH = H2 – H1 By neglecting small kinetic and potential energy changes. Thus the SHAFT work, Ws, is simply ΔH.

3. PROPERTY RELATIONS for HOMOGENOUS PHASES First law of thermodynamics, for a CLOSED system, containing n moles : d(nU) = dQ + dW For REVERSIBLE process : d(nU) = dQrev + dWrev Where : dWrev = p.d(nV) dQrev = T.d(nS) …(a) Where : U = molar internal energy; energy/mole S = molar entropy; entropy/mole V = molar volume; volume/mole

4. Eq. (a) combines 1st law and 2nd law  contains only PROPERTIES of THE SYSTEM (in the form of STATE function). Eq. (a) derived for REVERSIBLE process, but the application is not restricted just for reversible process; since the eq. (a) already expressed in STATE VARIABLES. Recall : H U + PV Define : 1. Helmholtz energy A U – TS 2. Gibbs energy G H – TS

5. From enthalpy : nH = nU + P(nV) d(nH) = d(nU) +P.d(nV)+(nV).dP …(b) Equation (a) : (a)  (b), results in : Similarly : d(nA) = d(nU) – T.d(nS) – nS.dT …(c) (a)  (c), results in : Analogue with that : d(nG) = (nV)dP – (nS).dT But remember : all equations are written for the ENTIRE MASS and CLOSED system

6. Application to a unit mass (or one mole) of homogenous fluid with CONSTANT COMPOSITION : dU = T.dS – p.dV dH = T.dS + V.dP dA = -p.dV – S.dT dG = V.dP – S.dT From those equations, one can derive that : MAXWELL Correlations

7. Note : Those math expression can be derived from : if F = F(x,y), then the total differential, dF = or, dF = M.dx + N.dy Where, M = and N = Further differentiation :

8. Example : H = H(S,P) dH = Using those above rule : Example : G = G(P,T) dG= Further derivation results in : T V etc.

9. Consider : H = H(P,T) dH = From other equation : dH = TdS + vdP From Maxwell equation : Hence, Putting back to the initial equation :

10. From the relationship :  Entropy can be expressed as : S = S(T,P)  total entropy change Here, the correlation between V & T should be KNOWN!

11. Example : For an ideal gas, the PVT behavior is expressed by : PVig = RT So that, Putting back to general dH & dS equations: dHig = Cpig.dT +  dHig = Cpig.dT dSig =  dSig =

12. Alternative form : (volume expansion coeff.) From Maxwell equation, we get :  and,  The dependence of H and S on pressure, can then be expressed as : dH = Cp.dT + V(1-βT)dP And, dS = Since β and V are WEAK function of P; they can be assumed constant at appropriate average values.

13. The value of β usually applied only to liquids. • For LIQUID, far from the critical point : P has little effect on S, H, and U. • For an incompressible fluid, usually assumed that β0. In this case :   However, , since :

14. Example 6-1 Determine the enthalpy and entropy changes for liquid water for a change of state from 1 bar and 25oC to 1000 bar and 50oC. The following data for water are available :

15. Solution : ΔH = Cp.dT + V(1 – βT)dP Cpavg(T2 – T1) + Vavg (1 – βavg.T2)(P2 – P1) And ΔS = Cp - β.V.dP Cpavgln - βavg.Vavg.(P2 – P1) The path of integration : H1;S1 at 1 bar 25oC At 1 bar. 2 H2;S2 At 1000 bar, 50oC At 25oC 1 bar 50oC

16. Note : • ΔH & ΔS is state functions; the path of integration can be arbritrary. • Cp weak function of T V & β weak function of P Hence, At P = 1 bar ; Cpavg = ½ (73,305 + 75,314) = 75,310 J/mole/K At t = 50oC ; Vavg = ½ (18,240 + 17,535) = 17,888 cm3/mole βavg = ½ (458 + 568) x 10-6 = 513 x 10-6 K-1 Substitution gives : ΔH = 75,310(323,15–298,15) + = 3,374 J/mole ΔS = 75,310 ln = 5,14 J/mole/K Can use arithmetic avg.

17. RESIDUAL PROPERTIES Consider Gibbs free energy : dG = VdP – SdT [by definition : G = H – TS] Can be expressed as : G = G(P,T) The alternative form is : (pure math derivation) Results in :

18. From the above equation, it can be derived : ; and Other relationships for ; etc. can be derived with the same methods. If the relation : is known, all the other thermodynamics properties can be evaluated! (by simple mathematical operations).  GENERATING FUNCTION for other thermodynamics properties.

19. However : • No convenient experimental method for determining the values of G (or G/RT). • The equation of Gibbs are of LITTLE PRACTICAL use. Define : the RESIDUAL GIBSS energy, GR G – Gig In an analogues way, can be defined residual volume : VR = V – Vig = V - ; and since : V = Then, VR = (z – 1) Back to the Gibbs equation (the modified one) :

20. For ideal gas : This is the Fundamental Property Relation for residual properties; applicable to constant composition fluids. From those equation we can derive : And

21. At constant T :  Recall that : at constant T! Recall : Combination with the above eq. gives : 

22. Similarly, as : GR = HR – TSR , we obtain : The calculation of Enthalpy & Entropy, is then : H = Hig + HR and S = Sig + SR Where, Hig = Higo + ; and Sig = Sigo + The enthalpy and entropy eq. can be simply expressed as : H = S =

23. Where :  for : Cpig = A + BT + CT2 + = R [ A + B.Tam + (4.Tam2 – T0.T) + ] where Tam ½ (T + T0) = R [A+B.Tlm+Tam.Tlm (C+)] where Tlm =

24. Note : The calculation of H & S for REAL gas is generally more convenient using RESIDUAL PROPERTIES, i.e. : H = Hig + HR and S = Sig + SR Where : ; However for LIQUID, the original equations mostly more appropriate, i.e. : dH = CpdT + [v – T] dP = CpdT + v (1 – βT) dP And, dS = Cp = Cp - βv dP

25. Example 6-2 Calculate the enthalpy and entropy of saturated isobutane vapor at 360 K. use the following information : • The vapor pressure of isobutane at 360 K is 15,41 bar. • Set Hoig = 18.115 J/mole & Soig = 295,976 J/mole for ideal gas reference state at 300 K &1 bar.

26. Solution : There are 2 steps for calculating H & S of real gas, i.e. : H = Hig + HR and S = Sig + SR • Calculation of RESIDUAL PROPERTIES : and The value of should be calculated FIRST! But how?  given by the SLOPE of a plot of z vs T, at constant pressure. Numerical / graphical Numerical / graphical

27. (refer to table 6.1 for z data!)

28. 0,1 bar Z 0,5 bar 2 bar 4 bar T

29. The result of : vs P at 360 K, see table 6.2 From table 6.2  plot vs P vs P Calculate the INTEGRAND (numerical? Graphical? Up to you)

30. 2. Calculate the ideal properties : Hig & Sig Hig = Higo + = 18.115 + = 24.439,8 J/mole [R=8,314 J/mole/K] Sig= Sigo+ = 295,967 + 8,314 = 280,942 J/mole/K Integration results : = 26,37x10-4 K-1 = -0,2596

31. Hence :  HR = (-0,9493) x 8,314 x 360 = -2.841,3 J/mole  SR = (-0,6897)(8,314) = -5,734 J/mole/K H = Hig + HR = 24.439,8 – 2.841,3 = 21598,5 J/mole S = Sig +SR = 292,411 – 5,374 = 286,676 J/mole/K

32. TWO PHASE SYSTEM Postulate : during the phase transition (fusion/melting; vaporization/condensation) : dG = 0

33. Hence, For two phases α and β of a pure species co-existing at equilibrium : Gα = Gβ Gα= molar Gibbs energy of phase α Gβ= molar Gibbs energy of phase β And hence : dGα = dGβ Recall that : dG = VdP – SdT So that; V α dPsat– S α dT = V β dPsat– S β dT Results in :

34. = entropy change = volume change “which occurs when a unit amount of pure chemical is transferred from phase αto β, at eq. T,P” Recall that : dH = TdS + VdP At constant T,P (equilib.) : ΔHαβ = T.ΔSαβ Recall the CLAYPERON equation : From phase transition : liquid vapor ; = latent heat of vaporization

35. For vaporization at low pressure : So that : Or : ClausiusClayperon equation

36. Clausius – Clayperon : α slope of plot of lnPsatvs (1/T) • Experimentally : ≠ f(T) at low pressure (P ↓) Clausius – Clayperon eqn. applies. • However at high pressures (P ↑) ; f(T). ↓ if T↑ (from triple point to critical point). Semi empirical equations for Psatvs T : (derived based on Clausius – Clayperon) • Ln Psat = A - [Clausius – Clayperon] • Ln Psat = A - [Antoine Equation] (more accurate ; most popular one) • Ln Psat = A - + D ln T + F.T6 [Riedel eqn.] where A,B,D,F = constants (wide temperature range)

37. THERMO. PROPERTIES OF GASES : GENERALIZED CORRELATION Recall reduced properties : and Define : reduced pressure, Pr = P/Pc  dP = Pc.dPr Reduced temp., Tr = T/Tc  dT = Tc.dTr Putting back to the above eqns. : and,

38. Refer to chapter 3 : Z = Z0 + ωZ1 = + ω Result in :

39. More direct correlation between z and (Pr,Tr) : (refer to chapter 3) Z = 1 + B0 + ω B1 This results in : Z0 Z1

40. Where : For a change of state of a real gas, from state-1 to state-2 : H2 = H0ig + H1 = H0ig + ΔH =

41. For the entropy, we obtain : ΔS = The schematic path to calculate ΔH & ΔS for real gas is :

42. Example : Estimate ΔH, ΔS, and ΔU to vaporize 1-butene from its original saturated – liquid condition at 0oC, 1,273 bar to 200oC 70 bar (vapor). The data available : Tc = 419,6 K ; Pc = 40,2 bar ; ω = 0,187 (T in K) ΔHv at 0oC 1,273 bar = 21753 J/mole

43. Solution : The paths to estimate ΔH & ΔS can be describes as follow :

44. Path : (a) : vaporization at T1, P1 (b) : transition REAL GAS  IDEAL GAS at T1, P1 (c) : change from (T1, P1) to (T2, P2) in the IDEAL GAS state (d) : transition IDEAL GAS  REAL GAS at T2, P2 Step (a) : ΔHa = ΔHv = 21753 J/mole ΔSa = ΔSlv = = = 79,64 J/mole/K Step (b) : Tr = = 0,651 ; Pr = = 0,0317 B0 = 0,083 - = - 0,756

45. B1 = 0,139 - = -0,904 = -0,0978 • HbR = (-0,0978)(8,314)(419,6) = -341 J/mole SbR = (8,314)(-0,105) = -0,87 J/mole/K Step (c) : = 20564 J/mole

46. = 22,16 J/mole/K Step (d) : Identical to step (b), with Tr = =1,13 and Pr = = 1,74 Results : ΔHbR= -8582 J/mole ΔSbR= -14,38 J/mole/K

47. Hence : (ΔH)real = ΔHa +ΔHig + (HdR – HbR) = 21753 + 20564 + [-8582 – (-341)] = 34076 J/mole (ΔS)real= ΔSa+ΔSig+ (SdR– SbR) = 79,64 + 22,16 + (-14,38 – (-0,87)) = 88,29 J/mole/K (ΔU)real = (ΔH)real – ΔPV = (ΔH)real – (P2V2 – P1V1) (note : V2 = Z2RT2/P2 ; Z2 = Z0 + ω Z1 etc.) Neglected!

48. Isothermic line Isentropic line THERMODYNAMICS DIAGRAM 1 2 : Heating liquid 2 3 : vaporization process 3  4 : heating vapor 1= sub cooled liquid 4 = superheated vapor Critical point

49. Line of constant pressure (isobaric line) Critical point b’ Isenthalpic line a’ a’  b’ : isentropic (adiabatic – reversible) compression (S = const. ; ΔS = 0)

50. Pressure enthalpy diagram for ammonia