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
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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 P1 to P2: Ws = ΔH = H2 – H1 By neglecting small kinetic and potential energy changes. Thus the SHAFT work, Ws, is simply ΔH.
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
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
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
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
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 :
Example : H = H(S,P) dH = Using those above rule : Example : G = G(P,T) dG= Further derivation results in : T V etc.
Consider : H = H(P,T) dH = From other equation : dH = TdS + vdP From Maxwell equation : Hence, Putting back to the initial equation :
From the relationship : Entropy can be expressed as : S = S(T,P) total entropy change Here, the correlation between V & T should be KNOWN!
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 =
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.
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 :
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 :
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
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.
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 :
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.
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) :
For ideal gas : This is the Fundamental Property Relation for residual properties; applicable to constant composition fluids. From those equation we can derive : And
At constant T : Recall that : at constant T! Recall : Combination with the above eq. gives :
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 =
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 =
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
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.
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
0,1 bar Z 0,5 bar 2 bar 4 bar T
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)
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
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
TWO PHASE SYSTEM Postulate : during the phase transition (fusion/melting; vaporization/condensation) : dG = 0
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 :
= 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
For vaporization at low pressure : So that : Or : ClausiusClayperon equation
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)
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,
Refer to chapter 3 : Z = Z0 + ωZ1 = + ω Result in :
More direct correlation between z and (Pr,Tr) : (refer to chapter 3) Z = 1 + B0 + ω B1 This results in : Z0 Z1
Where : For a change of state of a real gas, from state-1 to state-2 : H2 = H0ig + H1 = H0ig + ΔH =
For the entropy, we obtain : ΔS = The schematic path to calculate ΔH & ΔS for real gas is :
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
Solution : The paths to estimate ΔH & ΔS can be describes as follow :
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
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
= 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
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!
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
Line of constant pressure (isobaric line) Critical point b’ Isenthalpic line a’ a’ b’ : isentropic (adiabatic – reversible) compression (S = const. ; ΔS = 0)