introductory chemical engineering...

Introductory ChemicalEngineering Thermodynamics Chapter 7 - Departure Functions

By J.R. Elliott, Jr.

Chapter 7 - Departure Functions Slide 2

Internal Energy Departure Function

U(T,V)-Uig(T,V) = dV

V

U

V

Uig

TT

V

−

∫

∞ ∂∂

∂∂

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

ig)T = RT/V dV = RT dlnV

Substituting we find, (dUig

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

Therefore,

ig

TV

U

∂∂

=0

Returning to the FPR, TTT V

VP

V

ST

V

U

−

=

∂∂

∂∂

∂∂

Maxwell’s Relation VT T

P

V

S

=

∂∂

∂∂

Finally, ∫∞

−

=−

V

V

ig dVPT

PTUU

∂∂

)(

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


∴ = − ⇒ =−

→ ∞ →dV dV

and12ρ

ρ ∂ ∂ρρ

ρ V

at V , 0,

∫

−=

⇒

ρ

ρ ρρ

∂∂

ρρ0

1- d

T

P

RRT

P

RT

UU ig

−=

−=

−∴ ∫ RT

PTUPTUd

T

ZT

RT

VTUVTU igig ),(),(),(),(

0

ρ

ρ ρρ

∂∂

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

ad

RT

ad

T

ZT

RT

VTUVTU ig ρρρρρ

ρρ

∂∂ ρρ

ρ

ρ

−=−=−=

−=

−∴ ∫∫ 000

),(),(


Helmholtz Energy:Departure Function

dVV

A

V

AVTAVTA

V ig

TT

ig ∫∞

−

=−

∂∂

∂∂

),(),(

( )V

RT

V

nVRT

V

AnVRTddV

V

RTdAPdVdAFPR

T

ig

T

igTT

−=

−=

⇒−=−=⇒−=⇒

∂∂

∂∂ l

l)(

Also P

V

A

T

−=

∂∂

⇒ V

RTP

V

A

V

Aig

TT

+−=

−

∂∂

∂∂

Transform to ρ⇒dV = -V dρ/ρ

∫

−=

−⇒ρ

ρρ0

1),(),(d

Z

RT

VTAVTA ig


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

∫ −+

−=−+−=−⇒ρ

ρρo

igid

ZdZ

ZRT

VTAVTA

RT

VTGVTG1

11

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

Since V and P correspond to the properties of the real gas, the pressure of the ideal gas atT and V is P1 =RT/V. The change in Gibbs energy isG T p G T V RT n P P RT n PV RT RT n Zig ig( , ) ( , ) )− = = ( / ( / ) = ( )l l l1

⇒ = − − −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

RTZ

ig( , , ) ( , )ln( )

−−

)ln(11),(),(

ZZdZ

RT

PTGPTG

o

ig

−−+

−=−

⇒ ∫ρ

ρρ


Summary of density dependent formulas fordeparture functions from equations of state.

( )∫ −+

−=− ρ

ρρ

∂∂

o

ig

Zd

T

ZT

RT

HH1

( ) ( )( ) ( )∫ −−+−=− ρ

ρρo

ig

nZZdZ

RT

PTGPTGl1

1,,

( )ρρ

∂∂ρ

d

T

ZT

RT

UU

o

ig

∫

−− ( ) ( )( ) ( )

∫−=− ρ

ρρo

ig

dZ

RT

VTAVTA 1,,

( ) ( )( ) ( )∫

−−

−=− ρ

ρρ

∂∂

o p

ig dZ

T

ZT

R

VTSVTS1

,, ( ) ( )( ) ( )∫ +

−−

−=− ρ

ρρ

∂∂

o p

ig

nZd

ZT

ZT

R

PTSPTSl1

,,


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 to25 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-8Tc = 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


II. (190,25)ig →(105,5)igHig(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/moleSig(190,25) - Sig (105,5) = A ln(T2/T1)+B(∆T) +C∆(T2)/2+D∆(T3)/3 - Rln(P2/P1) ;∆Sig = -4.224 ln(463.15/378.15) + 0.3063(85) + (-1.586E-4)(4632-3782)/2 + + 3.215E-8(4633-3783)/3 - 8.314 ln 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 theaccuracy with which the chart can be read. Entropy is especially difficult because thetemperature and pressure effects tend to cancel and we end up with the small differencebetween large numbers. In reality, the Peng-Robinson equation is only accurate to about10% on enthalpy if compared to a highly accurate equation.


Example 7.6. Use of Referenced PREOS to get enthalpy and entropyPropane gas undergoes a change of state from an initial condition of 5 bar and 105°C to25 bar and 190°C. Compute the change in enthalpy and entropy by using a commonreference state of 230K and 0.1MPa.For propane : A = -4.224; B = 0.3063; C = -1.586E-4; D = 3.215E-8Tc = 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 theequivalence of this procedure to the way steam tables are computed. Furthermore, thecomputation of H2 - Href or S2 - Sref is entirely equivalent to the procedure given inExample 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/mole3. 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-K4. Repeat at 378.15 K and 0.5 MPa, H1 - Href = 29586 J/mole; S1 - Sref = 104.13 J/mole-K5. Subtract ⇒ ∆H = 36901-26756 = 7315 and ∆S = 109.15-104.13 = 5.02 J/mole-K


Example 7.3 Enthalpy departure for PREOSObtain a general expression for the enthalpy departure function of the PREOS.Solution: In the previous example we were able to obtain both pressure-explicit anddensity-explicit equations. Therefore, we could solve the problem two different ways.For the PREOS, we can only solve one way.

( ) ( )2221

/

1

1

ρρρ

ρ bb

RTa

bZ

−+−

−=

( )

+−

−+=

−

dT

da

TT

a

bb

RT

T

ZT

1

21

/222 ρρ

ρ∂∂

ρ

( )[ ]Pc

TcRaTcTaa cc

222

45724.0;/11 ≡−+= κ

where κ ω ω= 0.37464 +1.54226 - 0.26993 2

−

−+= − 2/1

2

1112 T

TcTc

Ta

dT

dac

κκ ⇒ ( )[ ]( )TcTTcTadT

daT c //11 κκ −+−=

( )[ ]( )

−+−−−

−+=

− TcTTcT

bRT

a

bRT

a

bb

b

T

ZT c //11

21 22κκ

ρρρ

∂∂

ρ ( ) ( )rTFbb

b2221 ρρ

ρ−+

≡

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


( )( ) ( ) ( )

∫ ∫ =−+

=

−

ρ ρ

ρρ

ρρρ

ρρ

∂∂b

o

b

o

rT b

bdTF

bb

b

b

bd

T

ZT

2221

( ) ( )( )

−+++=

+−++

+−

ρρ

ρρ

ρ

b

bn

TrF

b

bn

TrFb

211

211

8

)(

1)21(

1)21(

21

21

80

ll

Note ( ) ( ) ( )

( )

−+++=

−⇒== ∫ BZ

BZn

TrF

b

bd

T

ZTb

RT

pRT

bp

Z

Bb

T 21

21

80

ρ

ρρ

∂∂ρ

ρ

l

( ) ( )( ) ( )[ ]

−+−−

−++++−=−

rrc

ig

TTbRT

a

bRT

a

BZ

BZnZ

nRT

HH κκ 1121

21

8

11 l

= ( )( )

+

ακ Tr

B

A

B

B1

82-1+Z

2+1+Zn-1-Z l


Example 7.4 Gibbs Departure for PREOS.Obtain a general expression for the Gibbs energy departure function of the PR-EOS.

( ) ( )2221

/

1

1

ρρρ

ρ bb

RTa

bZ

−+−

−=

( ) ( ) ( )2222 21

/

121

/

1

1

1

11

ρρρ

ρρ

ρρρ

ρρ

ρ bb

RTa

b

b

bb

RTa

b

b

bZ

−+−

−=

−+−

−−−

−=−

( ) ( )( ) ( )∫ −−+−=− ρ

ρρo

id

nZZdZ

nRT

pTGpTGl1

1,,

=ln(1-bρ) – lnZ + Z – 1 + ( )( )

ρρ

b

b

bRT

a

2-1+1

2+1+1n

8l

=ln(Z-B)-( )( ) 82-1+Z

2+1+Zn

B

A

B

B

l


Example 7.7 Liquefaction revisitedReevaluate the liquefaction of methane consideredpreviously using the methane chart by performingthe analogous calculations with the PR EOS.Natural gas, assumed here to be pure methane, isliquefied in a simple Linde process. Compression isto 60 bar and precooling is to 300K. The separator ismaintained at a pressure of 1 bar and unliquefied gasat this pressure leaves the cooler at 295 K. Whatfraction of the gas is liquefied in the process?

Precool

Heat Exch

ThrottleFlashDrum

Compressor

1

23

4

5

6

7

8

Tc =190.6; Pc =4.60MPa; ω=.008; Cp ≈ Cp (200K) ≈ 28.45 J/mol-KSolution: 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).

( ) ( ) ( ) refsatLidididid HHHHHHHH +−+−+−= 1

1111

1111

30060

30060

30060

30060

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

satLidididid HHHHHHHHH +−+−+−== 11111

1111

2951

2951

2951

29518

= 0 + 28.45(295-111) + 8255 + 4538 = 18028J/moleH6 ≡ 4538 E-Bal ⇒ H3 = qH8 + (1-q)H6 ⇒ q=0.9304 ⇒ 6.96% liquefiedThis compares to 7.13% when we read the chart to the best of our ability.

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