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Introductory Chemical Engineering Thermodynamics Chapter 7 - Departure Functions By J.R. Elliott, Jr. Internal Energy Departure Function ig V U U ig ( T,V ) = dV U ( T,V )- U

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SLIDE 1

Introductory Chemical Engineering Thermodynamics

Chapter 7 - Departure Functions

By J.R. Elliott, Jr.

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SLIDE 2

Chapter 7 - Departure Functions Slide 2

Internal Energy Departure Function

U(T,V)-U

ig(T,V) =

dV V U V U

ig T T V

      −      

∫

∞

∂ ∂ ∂ ∂

FPR tells us (dU)T = T(dS)T - P(dV)T For the ideal gas (dUig)T = T(dSig)T -P(dVig)T where (dSig)T = RdlnV and P(dV

ig)T = RT/V dV = RT dlnV

Substituting we find, (dU

ig)T = T*R dlnV - RT dlnV = 0

Therefore,

ig T

V U       ∂ ∂

=0 Returning to the FPR,

T T T

V V P V S T V U       −       =       ∂ ∂ ∂ ∂ ∂ ∂

Maxwell’s Relation

V T

T P V S       =       ∂ ∂ ∂ ∂

Finally,

∫

∞

      −       = −

V V ig

dV P T P T U U ∂ ∂ ) (

If we transform to density, the expressions we get are usually easier to integrate.

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SLIDE 3

Chapter 7 - Departure Functions Slide 3 ∴ = − ⇒ = − → ∞ → dV d V and 1

2

ρ ρ ∂ ∂ρ ρ ρ V at V , ,

∫

              − =         ⇒

ρ ρ

ρ ρ ∂ ∂ ρ ρ 1

  • d

T P R RT P RT U U

ig

        − =       − =         − ∴

∫

RT P T U P T U d T Z T RT V T U V T U

ig ig

) , ( ) , ( ) , ( ) , (

ρ ρ ρ

ρ ∂ ∂

Because Uig(T,P) - Uig(T,V) = ∫ (∂Uig/∂V)T dV = 0

  • Example. Derive the internal energy departure function for the EOS:

Z=1+4bρ/(1-bρ)-aρ/RT (∂Z/∂T)ρ = + aρ/RT2 ⇒ -T(∂Z/∂T)ρ = -aρ/RT

[ ]

RT a RT a d RT a d T Z T RT V T U V T U

ig

ρ ρ ρ ρ ρ ρ ρ ∂ ∂

ρ ρ ρ ρ

− = − = − =       − =         − ∴

∫ ∫

) , ( ) , (

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SLIDE 4

Chapter 7 - Departure Functions Slide 4

Helmholtz Energy:Departure Function

dV V A V A V T A V T A

V ig T T ig

∫

∞

              −       = − ∂ ∂ ∂ ∂ ) , ( ) , (

( )

V RT V nV RT V A nV RTd dV V RT dA PdV dA FPR

T ig T ig T T

− =       − =       ⇒ − = − = ⇒ − = ⇒ ∂ ∂ ∂ ∂

  • )

(

Also

P V A

T

− =       ∂ ∂

⇒

V RT P V A V A

ig T T

+ − =       −       ∂ ∂ ∂ ∂

Transform to ρ⇒dV = -V dρ/ρ

∫

        − =         − ⇒

ρ

ρ ρ 1 ) , ( ) , ( d Z RT V T A V T A

ig

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SLIDE 5

Chapter 7 - Departure Functions Slide 5

Gibbs energy departure function

As for the density dependent part, it is easy to see that, G = U + PV -TS = A + PV

∫

− +         − = − + − = − ⇒

ρ

ρ ρ

  • ig

id

Z d Z Z RT V T A V T A RT V T G V T G 1 1 1 ) , ( ) , ( ) , ( ) , (

Since V and P correspond to the properties of the real gas, the pressure of the ideal gas at T and V is P1 =RT/V. The change in Gibbs energy is

G T p G T V RT n P P RT n PV RT RT n Z

ig ig

( , ) ( , ) ) − = = ( / ( / ) = ( )

  • 1

⇒ = − − − G T P G T P RT G T P V G T V RT G T P G T V RT

ig ig ig ig

( , ) - ( , ) ( , , ) ( , ) ( , ) ( , ) =

G T P V G T V RT Z

ig

( , , ) ( , ) ln( ) − −

) ln( 1 1 ) , ( ) , ( Z Z d Z RT P T G P T G

  • ig

− − +         − = − ⇒

∫

ρ

ρ ρ

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SLIDE 6

Chapter 7 - Departure Functions Slide 6

Summary of density dependent formulas for departure functions from equations of state.

( ) ∫

− +       − = −

ρ

ρ ρ ∂ ∂

  • ig

Z d T Z T RT H H 1

( ) ( )

( )

( )

∫

− − + − = −

ρ

ρ ρ

  • ig

nZ Z d Z RT P T G P T G

  • 1

1 , ,

( )

ρ ρ ∂ ∂

ρ

d T Z T RT U U

  • ig

∫

      − −

( ) ( )

( )

( )

∫

− = −

ρ

ρ ρ

  • ig

d Z RT V T A V T A 1 , ,

( ) ( )

( )

( )

∫

        − −       − = −

ρ

ρ ρ ∂ ∂

  • p

ig

d Z T Z T R V T S V T S 1 , ,

( ) ( )

( )

( )

∫

+         − −       − = −

ρ

ρ ρ ∂ ∂

  • p

ig

nZ d Z T Z T R P T S P T S

  • 1

, ,

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SLIDE 7

Chapter 7 - Departure Functions Slide 7 Example 7.1. Use of PREOS to get enthalpy and entropy departures. Propane gas undergoes a change of state from an initial condition of 5 bar and 105°C to 25 bar and 190°C. Compute the change in enthalpy and entropy. For propane : A=-4.224; B=0.3063; C= -1.586E-4; D=3.215E-8 Tc = 369.8 K; Pc = 42.49 bar.; ω=0.152 Solution: Path, for H(190,25) - H(105,5) [H(190,25) - Hig(190,25)]+[Hig(190,25) - Hig(105,5)]+[Hig(105,5) - H(105,5)] Similarly for S(190,25) - S(105,5) [S(190,25 - Sig(190,25)]+[Sig(190,25) - Sig(105,5)]+[Sig(105,5) - S(105.5)]

  • I. Departure Function + II. Ideal gas + III. Departure function
  • I. (190,25) → (190,25)ig

190 + 273.15 = 463.15K & 25 bar ⇒ Tr = 1.25135; Pr = 0.58837 PREOS ⇒ Z=0.8891 ⇒ (H-Hig) = (-0.3869) 8.314*463.15 = -1490 J/mol (S-Sig ) = (-0.2757) 8.314 = -2.2918 J/mol-K

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SLIDE 8

Chapter 7 - Departure Functions Slide 8

  • II. (190,25)ig →(105,5)ig

Hig(190,25) - Hig(105,5) = ∫Cp dT = -4.224(463-378) + 0.3063(4632-3782)/2+ (-1.586E-4) (4633-3783)/3 +(3.215E-8)(4634-3784)/4 = 8405 J/mole Sig(190,25) - Sig (105,5) = A n(T2/T1)+B(∆T) +C∆(T2)/2+D∆(T3)/3 - Rn(P2/P1) ; ∆Sig = -4.224 n(463.15/378.15) + 0.3063(85) + (-1.586E-4)(4632-3782)/2 + + 3.215E-8(4633-3783)/3 - 8.314 n 5 = 6.613 J/mol-K

  • III. (105,5)→(105,5)ig

105 + 273 = 378.15 & 5 bar → Tr = 1.02258; Pr = 0.11767 Z = 0.9574 ⇒ (H-Hig ) = (-0.1274) 8.314*378.15 = -400 J/mol (S-Sig ) = (-0.0852) 8.314 = -0.7081 J/mol-K ∆Htot = -1490 + 8405 + 400 = 7315 J/mol (Note: Chart ⇒ (1265 - 1095)*44 = 7480 J/mol) ∆Stot = -2.292 + 6.613 + 0.708 = 5.029 J/mol-K (Note: Chart ⇒ (1.52-1.50)*44*4.184=3.7 J/mol-K) Moral: The difference between the chart and the Peng-Robinson equation is significant, but could be because of error in the Peng-Robinson equation or sensitivity to the accuracy with which the chart can be read. Entropy is especially difficult because the temperature and pressure effects tend to cancel and we end up with the small difference between large numbers. In reality, the Peng-Robinson equation is only accurate to about 10% on enthalpy if compared to a highly accurate equation.

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SLIDE 9

Chapter 7 - Departure Functions Slide 9 Example 7.6. Use of Referenced PREOS to get enthalpy and entropy Propane gas undergoes a change of state from an initial condition of 5 bar and 105°C to 25 bar and 190°C. Compute the change in enthalpy and entropy by using a common reference state of 230K and 0.1MPa. For propane : A = -4.224; B = 0.3063; C = -1.586E-4; D = 3.215E-8 Tc = 369.8 K; Pc = 42.49 bar.; ω=0.152 Solution: In this example, we are directed to use a reference state such that, for enthalpy: H2 - H1 = (H2 - Href) - (H1 - Href), and for entropy: S2 - S1 = (S2 - Sref) - (S1 - Sref). Note the equivalence of this procedure to the way steam tables are computed. Furthermore, the computation of H2 - Href or S2 - Sref is entirely equivalent to the procedure given in Example 7.1.

  • 1. REF: Enter the values of Tc , Pc , ω, A, B, C, D and define the T, P, and root of

interest.

  • 2. Press PVTF to enter the pressure and temperature and choose the root of interest.

E.g. at 463.15 K and 2.5 MPa, V = 1369 cm3/mole

  • 3. Press UHSG to compute internal energy, enthalpy, entropy, and Gibbs free energy.

E.g. at 463.15 K and 2.5 MPa, H2 - Href = 36901 J/mole; S2 - Sref = 109.15 J/mole-K

  • 4. Repeat at 378.15 K and 0.5 MPa, H1 - Href = 29586 J/mole; S1 - Sref = 104.13 J/mole-K
  • 5. Subtract ⇒ ∆H = 36901-26756 = 7315 and ∆S = 109.15-104.13 = 5.02 J/mole-K
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SLIDE 10

Chapter 7 - Departure Functions Slide 10

Example 7.3 Enthalpy departure for PREOS

Obtain a general expression for the enthalpy departure function of the PREOS. Solution: In the previous example we were able to obtain both pressure-explicit and density-explicit equations. Therefore, we could solve the problem two different ways. For the PREOS, we can only solve one way.

( ) (

)

2 2

2 1 / 1 1 ρ ρ ρ ρ b b RT a b Z − + − − =

( )

            + − − + =       − dT da T T a b b R T T Z T 1 2 1 /

2 2 2ρ

ρ ρ ∂ ∂

ρ

( ) [ ]

Pc Tc R a Tc T a a

c c 2 2 2

45724 . ; / 1 1 ≡ − + = κ where κ ω ω = 0.37464 +1.54226

  • 0.26993

2

      −                 − + =

− 2 / 1

2 1 1 1 2 T Tc Tc T a dT da

c

κ κ ⇒

( ) [ ]( )

Tc T Tc T a dT da T

c

/ / 1 1 κ κ − + − =

( ) [ ]( )

     − + − − − − + =       − Tc T Tc T bRT a bRT a b b b T Z T

c

/ / 1 1 2 1

2 2

κ κ ρ ρ ρ ∂ ∂

ρ

( ) (

)

r

T F b b b

2 2

2 1 ρ ρ ρ − + ≡

F(Tr) is shorthand. Also B ≡ bP/RT ⇒ bρ = B/Z and A ≡ aP/R2T2 ⇒ a/bRT = A/B

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SLIDE 11

Chapter 7 - Departure Functions Slide 11

( )

( ) ( ) (

)

∫ ∫

= − + =       −

ρ ρ

ρ ρ ρ ρ ρ ρ ρ ∂ ∂

b

  • b
  • r

T

b b d T F b b b b b d T Z T

2 2

2 1

( )

( ) ( )

      − + + + =                 + − + +         + − ρ ρ ρ ρ

ρ

b b n Tr F b b n Tr F

b

2 1 1 2 1 1 8 ) ( 1 ) 2 1 ( 1 ) 2 1 ( 2 1 2 1 8

  • Note

( ) ( )

( ) ( ) 

     − + + + =       − ⇒ = =

∫

B Z B Z n Tr F b b d T Z T b RT p RT bp Z B

b T

2 1 2 1 8

ρ

ρ ρ ∂ ∂ ρ ρ

  • (

)

( ) ( ) ( ) [ ]

      − + − −       − + + + + − = −

r r c ig

T T bRT a bRT a B Z B Z n Z nRT H H κ κ 1 1 2 1 2 1 8 1 1

  • =

( ) ( )

      +       α κ Tr B A B B 1 8 2

  • 1

+ Z 2 + 1 + Z n

  • 1
  • Z
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SLIDE 12

Chapter 7 - Departure Functions Slide 12

Example 7.4 Gibbs Departure for PREOS.

Obtain a general expression for the Gibbs energy departure function of the PR-EOS.

( ) (

)

2 2

2 1 / 1 1 ρ ρ ρ ρ b b RT a b Z − + − − =

( )

( ) (

)

2 2 2 2

2 1 / 1 2 1 / 1 1 1 1 1 ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ b b RT a b b b b RT a b b b Z − + − − = − + − − − − − = −

( ) ( )

( )

( )

∫

− − + − = −

ρ

ρ ρ

  • id

nZ Z d Z nRT p T G p T G

  • 1

1 , , =ln(1-bρ) – lnZ + Z – 1 +

( ) ( )

      ρ ρ b b bRT a 2

  • 1

+ 1 2 + 1 + 1 n 8

  • =ln(Z-B)-

( ) ( )

8 2

  • 1

+ Z 2 + 1 + Z n B A B B      

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SLIDE 13

Chapter 7 - Departure Functions Slide 13

Example 7.7 Liquefaction revisited

Reevaluate the liquefaction of methane considered previously using the methane chart by performing the analogous calculations with the PR EOS. Natural gas, assumed here to be pure methane, is liquefied in a simple Linde process. Compression is to 60 bar and precooling is to 300K. The separator is maintained at a pressure of 1 bar and unliquefied gas at this pressure leaves the cooler at 295 K. What fraction of the gas is liquefied in the process?

Precool Heat Exch Throttle Flash Drum Compressor 1 2 3 4 5 6 7 8

Tc =190.6; Pc =4.60MPa; ω=.008; Cp ≈ Cp (200K) ≈ 28.45 J/mol-K Solution: To facilitate comparison to chart, set the reference enthalpies equal. Let: Href=H

satL(1bar) = 4538J/mol (283.6 J/g as given on chart, T sat(1bar)=111.0K).

( ) ( ) ( )

ref satL id id id id

H H H H H H H H + − + − + − =

1 111 1 111 1 300 60 300 60 300 60 300 60

= -0.4334(8.314)300+28.45(300-111)+8.9453(8.314)111 + 4538 = 17089 J/mole

( ) ( ) ( )

ref satL id id id id

H H H H H H H H H + − + − + − = =

1 111 1 111 1 295 1 295 1 295 1 295 1 8

= 0 + 28.45(295-111) + 8255 + 4538 = 18028J/mole H6 ≡ 4538 E-Bal ⇒ H3 = qH8 + (1-q)H6 ⇒ q=0.9304 ⇒ 6.96% liquefied This compares to 7.13% when we read the chart to the best of our ability.