Chapter 5 Thermodynamic Properties of Real Fluids It has already been demonstrated through the first and second laws, that the work and heat interactions between a system and its surroundings may be related to the state variables such as internal energy, enthalpy and entropy. So far we have illustrated the calculations of energy and entropy primarily for pure (component) ideal gas systems. However, in practice this is an exception rather than a rule as one has to deal with not only gases removed from ideal gas state, but also with liquids and solids. In addition, mixtures rather than pure components are far more common in chemical process plants. Therefore, computation of work and heat interactions for system comprised of real fluids requires more complex thermodynamic formulations. This chapter is devoted to development of such relations that can help calculate energy requirements for given changes of state for real systems. As in the case of ideal gases the goal is to correlate the energy and entropy changes for real fluids in terms of their volumetric and other easily measurable macroscopic properties. 5.1 Thermodynamic Property Relations for Single Phase Systems Apart from internal energy and enthalpy, two other ones that are particularly useful in depiction of thermodynamic equilibrium are Helmholtz free energy (A) and Gibbs free energy (G). We defer expanding upon the concept of these two types of energies to chapter 6; however, we state their definition at this point as they are instrumental in the development of property correlations for real fluids. • Specific Helmholtz free energy: A= U − TS ..(5.1) • Specific Gibbs free energy:G= H − TS ..(5.2) For a reversible process in a closed system the first law gives: dU= dQ + dW Or: dU= TdS − PdV ..(5.3) Using H= U + PV and taking a total differential of both sides: 1 | Page dH=++ dU PdV VdP ..(5.4) Putting eqn. 5.3 in 5.4 we get: dH= TdS + VdP ..(5.5) In the same manner as above one may easily show that the following two relations obtain: dA=−− SdT PdV ..(5.6) dG= VdP − SdT ..(5.7) Equations 5.3 to 5.7 comprise the fundamental energy relations for thermodynamic systems where there is a single phase with constant composition. In principle, they may be integrated to compute the energy changes for a system transiting from one equilibrium state to another. 5.2 Maxwell Relations All the four types of energy relations above satisfy the mathematical condition of being continuous variables, as they are themselves functions of state variables. One can thus apply of the criterion of exact differential for these functions. For a function of the form P= PXY(,) one can write the following total differential: ∂∂PP dP=+=+ dX dY MdX NdY ..(5.8) ∂∂XYYX ∂∂PP Where: M= and N = ..(5.9) ∂∂XYYX ∂M ∂22 P ∂∂ NP Further, = and = ..(5.10) ∂YXY ∂∂ YX ∂ X ∂∂ XY ∂∂MN It follows: = ..(5.11) ∂∂YX XY Applying eqn. 5.11 to 5.3, 5.5, 5.6 and 5.7 one may derive the following relationships termed Maxwell relations: ∂∂TP = − ..(5.12) ∂∂VSSV ∂∂TV = ..(5.13) ∂∂PS SP 2 | Page ∂∂PS = ..(5.14) ∂∂TV VT ∂∂VS = − ..(5.15) ∂∂TPPT 5.4 Relations for Enthalpy, Entropy and Internal Energy One may conveniently employ the general energy relations and Maxwell equations to obtain expressions for change in enthalpy and entropy and internal energy for any process, which in turn may be used for computing the associated heat and work interactions. Let H= HTP(, ) ∂∂HH Then: dH= dT + dP ∂∂TPPT ∂H But = CP ∂T P ∂H Thus: dH= CP dT + dP ..(5.16) ∂P T ∂∂HS Using dH=+⇒ TdS VdP =TV + ..(5.17) ∂∂PP TT ∂∂SV From Maxwell relations as in eqn. 5.15:= − ..(5.18) ∂∂PTTP Thus using eqns. 5.17 and 5.18 in 5.16 we get: ∂V dH= CP dT +− V T dP ..(5.20) ∂T P In the same manner starting from the general function: U= UTV(, )and S= ST (, P ) and applying appropriate Maxwell relations one may derive the following general expressions for differential changes in internal energy and entropy. ∂P dU=+− CV dT T P dV ..(5.21) ∂T V 3 | Page dT∂ V dS= CP − dP ..(5.22) TT∂ P dT∂ P Or, alternately: dS= CV + dV ..(5.23) TT∂ V Thus, eqns. 5.20 to 5.23 provide convenient general relations for computing enthalpy, internal energy and entropy changes as function of volumetric properties and specific heats. If a fluid is described by a suitable EOS, these equations may be conveniently integrated to obtain analytical expressions for energy and entropy changes. --------------------------------------------------------------------------------------------------------------------- Example 5.1 Derive an expression for enthalpy change of a gas during an isothermal process assuming using the −= following EOS: P() V b RT (Click here for solution) --------------------------------------------------------------------------------------------------------------------- 5.5 Residual Property Relations An alternate method of computing energy and entropy changes for real gases involves the definition of residual property. The specific residual property M R is defined as follows: MR = MTP(, ) − Mig (, TP ) ..(5.24) Where, MTP(, ) is the specific property of a real gas at a given T & P, and Mig (, TP )is the value of the same property if the gas were to behave ideally at the same T & P. Thus for example: VR = VV − ig Using the generalized compressibility factor ‘Z’: VR =( ZRT /)( P − RT /) P VR =( Z − 1) RT / P ..(5.25) The residual properties are usually used for gases only. Using such a property for a liquid (or solid) is inconvenient as then it would also include the property change of vapourization (and solidification) which generally are large in magnitude. This detracts from the advantage of working with the residual property as a measure of small corrections to ideal gas behaviour. Thus the use of residual functions is restricted to prediction of real gas behaviour only. To exploit the 4 | Page concept of residual properties we take advantage of the Gibbs free energy as it can be used as a generating function for other thermodynamic properties. Derivation of Residual functions: We start from the generic equation:G= H − TS ..(5.2) And dG= VdP − SdT ..(5.7) Taking the total differential for the functionG/ RT : GG1 = − d dG2 dT ..(5.26) RT RT RT Substituting eqn. 5.2 and 5.7 in 5.26 we get: GV H = − d dP2 dT ..(5.27) RT RT RT One may write the same equation specifically for an ideal gas, whence: GVig ig Hig = − d dP2 dT ..(5.28) RT RT RT Subtracting eqn. 5.28 from 5.27: GVRR H R = − d dP2 dT ..(5.29) RT RT RT Thus we may write the following further generative relations: VRR∂(/ G RT ) = ..(5.30) RT∂ P T HRR∂(/ G RT ) = −T ..(5.31) RT ∂T P SR HG RR And further: = − ..(5.32) RT RT RT RRR R P GGG G From eqn. 5.29: d = − ; but = 0 ∫0 RT RTPP RT =0 RT P=0 5 | Page RR GVP Thus: = dP ..(5.33) RT∫0 RT R G P dP Putting eqn. 5.25 in 5.33: =(Z − 1) ..(5.34) RT ∫0 P Differentiating eqn. 5.34 w.r.t T in accordance with 5.31 gives: R HP ∂ Z dP = −T ..(5.35) ∫0 RT∂ TP P Finally using eqns 5.32, 5.34, and 5.35: R SPP∂ Z dP dP =−TZ −−( 1) ..(5.36) ∫∫00 R∂ TPP P The last two equations may be expressed in alternative forms in terms of reduced temperature and pressure: R P HZ2 r ∂ dPr = −T ..(5.37) r ∫0 RTC ∂TPrr Pr R PP SZrr∂ dPrrdP =−TZ −−( 1) ..(5.38) r ∫∫00 R∂ TPrr Pr Pr --------------------------------------------------------------------------------------------------------------------- Example 5.2 Derive an expression for enthalpy change of a gas during an isothermal process assuming using the following EOS: Z = 1 + APr / Tr (Click here for solution) --------------------------------------------------------------------------------------------------------------------- 5.6 Residual Property Calculation from EOS From Virial EOS: Using 5.35 and 2.12 one obtains: HR P B dB = − ..(5.39) RT R T dT 6 | Page Next, on substituting eqns. 2.13 – 2.15 in 5.37 the following relations result: H R 1.097 0.894 = −+ω − Pr 0.083 1.6 0.139 4.2 ..(5.40) RTCrT T r Similarly using 5.36 and 2.12 we get: SR P dB = − ..(5.40) R R dT Finally employing eqns. 2.13 – 2.15 in 5.38: S R 0.675 0.722 =−+ω Pr 1.6 4.2 ..(5.41) RTrr T From Cubic EOS: One may use the following form of the cubic EOS presented in chapter 2: RT a P = − ..(2.21) V− b V22 ++ ubV wb Or equivalently: ZZZ32+α + βγ +=0 ..(2.23) While the eqns. 5.35 and 5.36 are useful for volume explicit EOS, they are unsuitable for cubic EOS which are pressure explicit. For the latter type of EOS one may show that the appropriate equations for residual enthalpy and entropy are (see S.I. Sandler, Chemical, Biochemical and Engineering Thermodynamics, ch. 6, 4th Edition, Wiley India, 2006): R HP1 V ∂ =Z −+1 T− P dV ..(5.42) ∫V =∞ RT RT∂ T V R S 1 V ∂PR =+−ln Z dV ..(5.43) ∫V =∞ R R∂ TVV The above relations may be applied to the various cubic EOSs to obtain the necessary residual property relations. The final results are shown below. From RK-EOS: 7 | Page HR 3a ZB+ =(Z −− 1) ln ..(5.44) RT 2bRT Z SR a ZB+ =ln(ZB −− ) ln ..(5.47) R 2bRT Z 0.42748RTT22α ( ) 0.08664RT where,; a = C RK r b = C PPCC −1/2 α RK()TT r= r For SRK-EOS: ∂a R Ta− H 1 ∂T ZB+ =(Z −+ 1) − ln (5.48) RT RT bRT Z SR 1 ∂+a ZB =ln(ZB −− ) ln ..(5.49) R bRT∂ T Z ∂aa =− (0.48+− 1.574ωω 0.1762 ) ..(5.51) ∂ T αSRK()T r TT C 0.42748RTT22α ( ) 0.08664RT where,; a = C SRK r b = C PPCC 2 1/2 2 and,αSRK ( T r )=++− [1 (0.48 1.574ωω 0.176 )(1 −Tr )] For PR-EOS: ∂a Ta− HR ∂T ZB++(1 2 ) =(Z −+ 1) ln ..(5.52) RT 2 2bRT Z+− B(1 2 ) SR 1 ∂a ZB ++(1 2 ) =ln(ZB −− ) ln ..(5.53) RT2 2bRT∂ Z+− B(1 2 ) ∂aa =− (0.37464+− 1.5422ωω 0.629922 ) ..(5.54) ∂ T α PR()T r TT C 8 | Page 0.45724RTT22α ( ) 0.07779RT where,; a = C PR r b = C PPCC and,α ( T )=++− [1 (0.37464 1.5422ωω 0.629922 )(1 −T 1/2 )] 2 PR r r As part of computing the residual enthalpy or entropy (and internal energy) at any set of temperature and pressure, the molar volume or correspondingly Zvap needs to be first computed employing the usual algorithm for solving cubic EOS described in section 2.3.3.