Introductory Chemical Engineering Thermodynamics By J. Richard Elliott and Carl T. Lira Chapter 8 Phase Equilibrium in a Pure Fluid Chapter 8 - Phase Equilibrium in a Pure Fluid 1 Example. Pressure in a Variable Volume Cylinder A piston-cylinder contains some propane liquid and some vapor at -12°C. The piston is forced down a specified distance and the pressure is increased. The vessel is allowed to cool back to its original temperature. What is the final pressure in this vessel? Solution This is a trick question. If there is still vapor left, then the pressure will go back to where it started. If not, then we are compressing a liquid and the pressure could become quite high. We would need to compute how high based on an equation of state. First, let's examine the case of no change in pressure. Since the temperature and pressure from beginning to end are constant, the change in Gibbs energy of the system from beginning to end must be zero. ⇒ For the whole system G = MLGL+ MV GV dG = MLdGL +MV dGV + GL dML +GV dMV ⇒ Since T and p are constant, dGL =dGV =0. But dML = -dMV 0 = GL -GV or GL =GV . This is a very significant result! If the pressure comes back to the original pressure then GL = GV . If GL≠GV , then the pressure does not come back to the original pressure. But we know that when the pressure does not come back it must be because there is only liquid. In other words, GL= GV is a constraint for phase equilibrium. If we specify one more constraint (eg. T), then all of our other state properties can be calculated. Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 2 The Clausius-Clapeyron Equation: V If starting from VLE, and maintaining VLE while changing T, dG =dGL V sat V L sat L Fundamental property relation ⇒ V dP - S dT = V dP - S dT V V V L L L By definition of G: G = H - TS = H - TS = G V L V L ⇒S - S = (H - H )/T vap ()V − L ()()V − L dp = H H dT 1/ RT = H H dT vap vap V vap L 2 V L Clausius-Clapeyron Equation p T ()p V − p V ()1/ RT RT ()Z − Z sat dividing by P ⇒ sat ()V − L ()()V − L dP = H H dT 1/ RT = H H dT P sat T ()P satV V − P satV L ()1/ RT RT 2 ()Z V − Z L dP sat 1 − dT − ()H V − H L = dlnP sat ; d = ⇒ dln P sat = d()1/ T P sat T T 2 R()Z V − Z L Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 3 Example. Clausius-Clapeyron near or below the boiling point What is the value of the pressure in a piston-cylinder at -12°C with vapor and liquid V L V L propane present? Recall that Z ≈1 and Z ≈0 and (H - H )≈constant. Integrate P sat ()H V − H L 1 1 ln = − − 1atm R T Tb ω ⇒ (P sat ) [-(1+0.152)] For propane, =0.152 Tr =0.7 = Pc *10 = 3.034 atm. at Tr=0.7⇒ T = 259K The CRC lists the boiling temperature of propane as -42°C=231K. ln(3.034/1) = - ∆H/R(1/259 - 1/231) ) -∆H/R = -2372 sat Therefore, P (261K) = exp[-2372(1/261 - 1/231)] = 3.277 atm. Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 4 Short-cut Estimation of Saturation Properties If ∆Hvap/∆Zvap was constant, then we could recover the simple low-P form. To find sat out, plot ln(P ) vs. 1/T. A straight line means the slope is constant. ARGON ETHANE Pvap ∆Η 1 1 0 0 ln = − − P R∆Ζ T T 1 1.2 1.4 1.6 1.8 2 -2 123 ref ref -1 -4 -2 -6 log(Pr) log(Pr) 1/Tr 1/Tr Setting Pref = Pc and Tref =Tc , ∆H Tc Tc ∆H 1 sat 1 ∆H 1 1 ln P sat = − − = 1− ⇒ log()P = 1− ≡ A1− r ∆ ∆ r ∆ R ZTc T Tc R ZTc Tr 2.303 R ZT Tr Tr 1 7 @T = 0.7 log P sat ≡ −()ω + 1 = A − 1 = A()− 3 / 7 ⇒ A = ()1+ω r 10 r .7 3 sat 7 ∴ log (P ) = ()1+ ω ()1− 1 SHORT-CUT VAPOR PRESSURE EQUATION 10 r 3 Tr sat 7 e.g. For propane P (-12°C) = P *10** ()1+ω ()1− 1 = 3.26 atm c 3 0.7057 Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 5 Example. General Application of the Clausius-Clapeyron equation. Liquid butane is pumped to a vaporizer as a saturated liquid under a pressure of 1.88 MPa. The butane leaves the exchanger as a wet vapor of 90 percent quality and at practically the same pressure as it entered. From the following information, estimate the heat load on the vaporizer per gram of butane entering. Tc =425.2 K; Pc =3.797 MPa; ω=0.193; Solution: Use VP eqn. to find the T at which this takes place ⇒ Tsat = 383 K v − ig L ig H H H − H ⇒ PREOS nRT = -0.9952; nRT = -5.2567; Therefore, Hv = (-0.9952+5.2567)8.314*383 = 13568 J/mol 90% quality→ Q = 0.9*13568 = 12211 J/mol * 1mol/58g = 211 J/g Note: Alternatively, we could have used the short-cut Clausius-Clapeyron another way sat Clausius-Clapeyron ⇒ ln(P ) = - ∆ Hvap /RT(ZV-ZL ) + B ∆H 7 vap = 2.3025 ()1+ ω Tc = 2726 R∆Z 3 V L ∆Hvap = 2726*R*(Z -Z ) = 2726*8.314*(0.67434-0.07854) = 13503 J/mol Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 6 II. Generalized Fluid Properties Overview of UNIT II. 1. Calculus of State Functions a) FPR b) Maxwell’s Relations c) P=P(T,V) ⇒ dU(T,V), dS(T,V) ⇒ dH(T,P), dS(T,P) 2. Equations of state: e.g. P = RT/(V-b) - a/V2RT 3. Departure Functions: e.g. (U-Uig)/RT = ∫-T(∂Z/∂T)dρ/ρ e.g. (U-Uig)/RT = -aρ/RT = -aP/ZR2T2 4. Phase Equilibrium in a Pure Fluid: sat 7 e.g. VLE ⇒ log (P /P) = /3(1+ω)(1-1/Tr) Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 7.