Introductory Chemical Engineering Thermodynamics By J. Richard - - PowerPoint PPT Presentation

Chapter 8 - Phase Equilibrium in a Pure Fluid 1

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 Slide 2

• 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 3

The Clausius-Clapeyron Equation:

If starting from VLE, and maintaining VLE while changing T, dG

V =dGL

Fundamental property relation ⇒ V

V dP sat - S V dT = V L dP sat - S L dT

By definition of G: G

V = H V - TS V = H L - TS L = G L

⇒S

V - S L = (H V - H L)/T

( ) ( )

( ) ( )

( ) ( )

L V L V L vap V vap L V vap vap

Z Z RT dT H H RT RT V p V p T dT H H p dp − − = − − =

2

/ 1 / 1

Clausius-Clapeyron Equation dividing by Psat ⇒

( ) ( )

( ) ( )

( ) ( )

L V L V L sat V sat L V sat sat

Z Z RT dT H H RT RT V P V P T dT H H P dP − − = − − =

2

/ 1 / 1

( ) ( )

( )

T d Z Z R H H P n d T dT T d nP d P dP

L V L V sat sat sat sat

/ 1 1 ;

2

− − − = ⇒ − =       =

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 4

• 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 propane present? Recall that Z

V≈1 and Z L≈0 and (H V - H L )≈constant. Integrate

( )

      − − − =      

b L V sat

T T R H H atm P n 1 1 1

• For propane, ω=0.152⇒ (

)

7 . = Tr sat

P = Pc *10[-(1+0.152)] = 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 Therefore, P

sat (261K) = exp[-2372(1/261 - 1/231)] = 3.277 atm.

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 5

Short-cut Estimation of Saturation Properties

If ∆Hvap/∆Zvap was constant, then we could recover the simple low-P form. To find

• ut, plot ln(Psat ) vs. 1/T. A straight line means the slope is constant.

ARGON 1/Tr

log(Pr)

• 2
• 1

1 1.2 1.4 1.6 1.8 2

ETHANE

1/Tr log(Pr)

• 6
• 4
• 2

1 2 3

        − ∆Ζ ∆Η − =        

ref ref vap

T T R P P n 1 1

• Setting Pref = Pc and Tref =Tc ,

      − ∆ ∆ =       − ∆ ∆ − = Tr ZT R H Tc Tc T Tc ZT R H P n

c c sat r

1 1

• ⇒

( )

      − ≡       − ∆ ∆ = Tr A Tr ZT R H P

• g

sat r

1 1 1 1 303 . 2 1

• (

) ( )

7 / 3 1 7 . 1 1 log 7 . @

10

− =       − = + − ≡ = A A P T

sat r r

ω

⇒

( )

ω + 1 3 7 = A ∴ log10(Pr

sat) =

( )(

)

Tr 1 1 1 3 7 − +ω

SHORT-CUT VAPOR PRESSURE EQUATION e.g. For propane P

sat (-12°C) = Pc *10** (

)(

)

7057 . 1 1 1 3 7 − +ω

= 3.26 atm

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 6

• 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 PREOS ⇒

        − nRT H H

ig v

= -0.9952;

        − nRT H H

ig L

= -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 Clausius-Clapeyron ⇒ n(Psat ) = - ∆ Hvap /RT(ZV-ZL ) + B

( )

2726 1 3 7 3025 . 2 = + = ∆ ∆ Tc Z R H vap ω

∆Hvap = 2726*R*(ZV-ZL) = 2726*8.314*(0.67434-0.07854) = 13503 J/mol

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 7

• 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:

e.g. VLE ⇒ log (Psat/P) = 7/3(1+ω)(1-1/Tr)