Advanced Thermodynamics

Thermodynamic Properties of Fluids

Property relations for homogeneous phases

First law for a closed system dWdQnUd )(

a special reversible process

revrev dWdQnUd )(

)(nVPddWrev )(nSTddQrev

)()()( nVPdnSTdnUd

Only properties of system are involved:It can be applied to any process in a closed system (not necessarily reversible processes).The change occurs between equilibrium states.

PdVTdSdU

SandUTVP ,,,,The primary thermodynamic properties:PVUH The enthalpy:

TSUA The Helmholtz energy:TSHG The Gibbs energy:

For one mol of homogeneous fluid of constant composition:

)()()( nVPdnSTdnUd

VdPTdSdH

SdTPdVdA

SdTVdPdG Maxwell’s equations

VS SP

VT

PS SV

PT

TV VS

TP

TP PS

TV

Enthalpy, entropy and internal energy change calculations f (P,T)

dPPHdT

THdH

TP

dPPSdT

TSdS

TP

VdPTdSdH P

P

CTH

VPST

PH

TT

dPVPSTdTCdH

TP

TP PS

TV

dPTVTVdTCdH

PP

TC

TS P

P

dPTV

TdTCdS

PP

PP TST

TH

TP PS

TV

PVUH

VPVP

PU

PH

TTT

TTT PVP

PVT

PU

Determine the enthalpy and entropy changes of liquid water for a change of state from 1 bar and 25°C to 1000 bar and 50°C. The data for water are given.

H1 and S1 at 1 bar, 25°C

H2 and S2 at 1 bar, 50°C H3 and S3 at 1000 bar, 50°C

dPTVTVdTCdH

PP

dPTV

TdTCdS

PP

)(1)( 12212 PPVTTTCH P

VdPTdTCdH P 1

VTV

P

volume expansivity

)(ln 121

2 PPVTTCS P

molJH /340010

)11000)(204.18()15.323)(10513(1)15.29815.323(310.756

molJS /13.510

)11000)(204.18(1051315.29815.323ln310.75

6

Internal energy and entropy change calculations f (V,T)

dVVUdT

TUdU

TV

dVVSdT

TSdS

TV

VV

CTU

dVPVSTdTCdU

TV

VT TP

VS

dVPTPTdTCdU

VV

TC

TS V

V

dVTP

TdTCdS

VV

PVST

VU

TT

VV TST

TU

VT TP

VS

SdTVdPdG

Gibbs energy G = G (P,T)

dTRTGdG

RTRTGd 2

1

Thermodynamic property of great potential utility

SdTVdPdG

dTRTHdP

RTV

RTGd 2

TSHG

dTTRTGdP

PRTG

RTGd

PT

)/()/(

TPRTG

RTV

)/(

PTRTGT

RTH

)/(

G/RT = g (P,T)

The Gibbs energy serves as a generating function for the other thermodynamic properties.

Residual properties

• The definition for the generic residual property:

– M and Mig are the actual and ideal-gas properties, respectively.

– M is the molar value of any extensive thermodynamic properties, e.g., V, U, H, S, or G.

• The residual Gibbs energy serves as a generating function for the other residual properties:

igR MMM

dTRTHdP

RTV

RTGd

RRR

2

dTRTHdP

RTV

RTGd

RRR

2

const T dPRTV

RTGd

RR

P RR

dPRTV

RTG

0

PRT

PZRTVVV igR

PR

PdPZ

RTG

0)1(

PTRTGT

RTH

)/(

P

P

R

PdP

TZT

RTH

0

PP

P

R

PdPZ

PdP

TZT

RS

00)1(

RTG

RTH

RS RRR

const T

const TZ = PV/RT: experimental measurement .Given PVT data or an appropriate equation of state, we can evaluate HR and SR and hence all other residual properties.

P

P

ig

P

PH

igP

ig

P

P

ig

P

P

T

T

igP

igRig

PdP

TZRTTTDCBATTMCPHRH

PdP

TZRTTTCH

PdP

TZRTDCBATTICPHRH

PdP

TZRTdTCHHHH

0

2000

0

200

0

200

0

20

),,,;,(

),,,;,(

0

PP

P

ig

PP

PS

igP

ig

PP

P

ig

PP

P

T

T

igP

igRig

PdPZ

PdP

TZTR

PPR

TTDCBATTMCPSRS

PdPZ

PdP

TZTR

PPR

TTCS

PdPZ

PdP

TZTR

PPRDCBATTICPSRS

PdPZ

PdP

TZTR

PPR

TdTCSSSS

0000

00

0000

0

000

00

000

0

)1(lnln),,,;,(

)1(lnln

)1(ln),,,;,(

)1(ln0

Calculate the enthalpy and entropy of saturated isobutane vapor at 360 K from the following information: (1) compressibility-factor for isobutane vapor; (2) the vapor pressure of isobutane at 360 K is 15.41 bar; (3) at 300K and 1 bar, (4) the ideal-gas heat capacity of isobutane vapor:

molJH ig /181150 KmolJS ig /976.2950

TRC igP

310037.337765.1/

P

P

R

PdP

TZT

RTH

0

PP

P

R

PdPZ

PdP

TZT

RS

00)1(

Graphical integration requires plots of and (Z-1)/P vs. P.PTZ

P

/

9493.0)1037.26)(360( 4

0

P

P

R

PdP

TZT

RTH

6897.0)02596.0(9493.0)1(00

PP

P

R

PdPZ

PdP

TZT

RS

molJH R /3.2841

KmolJS R /734.5

mol

JHEICPHRHH Rig 5.21598)0.0,0.0,3037.33,7765.1;360,300(0

Kmol

JSREICPSRSS Rig

676.286

141.15ln)0.0,0.0,3037.33,7765.1;360,300(0

Residual properties by equation of state

• If Z = f (P,T):

• The calculation of residual properties for gases and vapors through use of the virial equations and cubic equation of state.

P

P

R

PdP

TZT

RTH

0

PP

P

R

PdPZ

PdP

TZT

RS

00)1(

Use of the virial equation of stateRTBPZ 1

PR

PdPZ

RTG

0)1(

two-term virial equation

RTBP

RTG R

P

P

R

PdP

TZT

RTH

0

P

P

R

PdP

RTBP

TT

RTH

0)1(

P

P

R

PdP

TB

dTdB

TRPT

RTH

0 2

1

RTBPZ 1

dTdB

TB

RP

RTH R

RTG

RTH

RS RRR

dTdB

RP

RTS R

21 CBZ

PR

PdPZ

RTG

0)1(

Three-term virial equation

ZCBRTG R

ln232 2

P

P

R

PdP

TZT

RTH

0

2

21

dTdC

TC

dTdB

TBT

RTH R

RTG

RTH

RS RRR

RTZP

Application up to moderate pressure

Use of the cubic equation of state

• The generic cubic equation of state:

))(()(

bVbVTa

bVRTP

)1)(1(11

bbbq

bZ

bRTTaq )(

qIbdZ )1ln()1(0

IdTdqd

TZ

0

Tconstbb

bdI

0 )1)(1()(

qITdTdZ

RTH

r

rR

1

ln)(ln1

qITdTdZ

RS

r

rR

1

ln)(lnln

RTbP

Find values for the residual enthalpy HR and the residual entropy SR for n-butane gas at 500 K and 50 bar as given by Redlich/Kwong equation.

176.11.425

500rT 317.1

96.3750

rP 09703.0r

r

TP 8689.3)(

r

r

TTq

))((1

ZZ

ZqZ01

685.0Z

qITdTdZ

RTH

r

rR

1

ln)(ln1

qITdTdZ

RS

r

rR

1

ln)(lnln

Tconstbb

bdI

0 )1)(1()(

13247.0ln1

ZZI

molJH R 4505

KmolJS R

546.6

Two-phase systems

• Whenever a phase transition at constant temperature and pressure occurs,– The molar or specific volume, internal energy,

enthalpy, and entropy changes abruptly. The exception is the molar or specific Gibbs energy, which for a pure species does not change during a phase transition.

– For two phases α and β of a pure species coexisting at equilibrium:

GG

where Gα and Gβ are the molar or specific Gibbs energies of the individual phases

GG dGdG SdTVdPdG

dTSdPVdTSdPV satsat

VS

VVSS

dTdP sat

VdPTdSdH STH The latent heat of phase transition

VTH

dTdP sat

The Clapeyron equation

Ideal gas, and Vl << Vg

satv

PRTVV

)/1(ln

TdPdRH

satlv

The Clausius/Clapeyron equation

VTH

dTdP sat

The Clapeyron equation: a vital connection between the properties of different phases.

)(TfP

TBAP sat ln For the entire temperature range from the triple point

to the critical point

CTBAP sat

ln The Antoine equation, for a specific temperature range

1

ln635.1 DCBAP sat

rThe Wagner equation, over a wide temperature range.

rT1

Two-phase liquid/vapor systems

• When a system consists of saturated-liquid and saturated-vapor phases coexisting in equilibrium, the total value of any extensive property of the two-phase system is the sum of the total properties of the phases:– M represents V, U, H, S, etc.– e.g., vvvv VxVxV )1(

vvlv MxMxM )1(

Thermodynamic diagrams and tables

• A thermodynamic diagram represents the temperature, pressure, volume, enthalpy, and entropy of a substance on a single plot. Common diagrams are:– TS diagram– PH diagram (ln P vs. H)– HS diagram (Mollier diagram)

• In many instances, thermodynamic properties are reported in tables. The steam tables are the most thorough compilation of properties for a single material.

Fig. 6.2

Fig. 6.3

Fig. 6.4

Superheated steam originally at P1 and T1 expands through a nozzle to an exhaust pressure P2. Assuming the process is reversible and adiabatic, determine the downstream state of the steam and ΔH for the following conditions:(a) P1 = 1000 kPa, T1 = 250°C, and P2 = 200 kPa.(b) P1 = 150 psia, T1 = 500°F, and P2 = 50 psia.

Since the process is both reversible and adiabatic, the entropy change of the steam is zero.

(a) From the steam stable and the use of interpolation, At P1 = 1000 kPa, T1 = 250°C:

H1 = 2942.9 kJ/kg, S1 = 6.9252 kJ/kg.KAt P2 = 200 kPa, S2 = S1= 6.9252 kJ/kg.K < Ssaturated vapor, the final state is in the two-phase liquid-vapor region:

vvlv SxSxS 22222 )1( vv xx 22 1268.7)1(5301.19252.6

9640.02 vx

0.2627)3.2706)(964.0()7.504)(964.01(2 HkgkJHHH 9.31512

(a) From the steam stable and the use of interpolation, At P1 = 150 psia, T1 = 500°F:

H1 = 1274.3 Btu/lbm, S1 = 1.6602 Btu/lbm.RAt P2 = 50 psia, S2 = S1= 1.6602 Btu/lbm.K > Ssaturated vapor, the final state is in the superheat region:

FT 28.2832

lbmBtuHHH 0.9912

lbmBtuH /3.11752

A 1.5 m3 tank contains 500 kg of liquid water in equilibrium with pure water vapor, which fills the remainder of the tank. The temperature and pressure are 100°C, and 101.33 kPa. From a water line at a constant temperature of 70°C and a constant pressure somewhat above 101.33 kPa, 750 kg of liquid is bled into the tank. If the temperature and pressure in the tank are not to change as a result of the process, how much energy as heat must be transferred to the tank?

mHQdt

mUd cv )(Energy balance:dt

dmHQdt

mUd cvcv )(

cvcv mHmUQ )(

PVUH

cvcvcv mHPmVmHQ )()(

cvcvcv mHHmHmQ )()( 1122

At the end of the process, the tank still contains saturated liquid and saturated vapor in equilibrium at 100°C and 101.33 kPa.

cvcvcv mHHmHmQ )()( 1122 CliquidsaturatedkgkJH 70@0.293

CliquidsaturatedkgkJH l

cv100@1.419

CvaporsaturatedkgkJH v

cv100@0.2676

From the steam table, the specific volumes of saturated liquid and saturated vapor at 100°C are 0.001044 and 1.673 m3/kg , respectively.

kJHmHmHm vvllcv 211616)0.2676(

673.1)001044.0)(500(5.1)1.419(500)( 111111

vl mmm 222 750673.1

)001044.0)(500(5.1500

lv mm 22 001044.0673.15.1

kJHmHmHm vvllcv 524458)( 222222

kJmHHmHmQ cvcvcv 93092)0.293)(750(211616524458)()( 1122

rcPPP rcTTT

P

P

R

PdP

TZT

RTH

0

PP

P

R

PdPZ

PdP

TZT

RS

00)1(

r

r

P

r

r

Prr

c

R

PdP

TZT

RTH

0

2

rr

r

P

r

rP

r

r

Prr

R

PdPZ

PdP

TZT

RS

00)1(

10 ZZZ

r

r

r

r

P

r

r

Prr

P

r

r

Prr

c

R

PdP

TZT

PdP

TZT

RTH

0

12

0

02

rr

r

rr

r

P

r

rP

r

r

Prr

P

r

rP

r

r

Prr

R

PdPZ

PdP

TZT

PdPZ

PdP

TZT

RS

0

1

0

1

0

0

0

0

)1()1(

),,(10

OMEGAPRTRHRBRTH

RTH

RTH

c

R

c

R

c

R

),,(10

OMEGAPRTRSRBR

SR

SR

S RRR

Table E5 ~ E12

RT

T

igP

ig HdTCHH 2022

0

RT

T

igP

ig HdTCHH 1011

0

RRT

T

igP HHdTCH 12

2

1

RT

T

igP

ig SPPR

TdTCSS 2

0

202

2

0

ln

RT

T

igP

ig SPPR

TdTCSS 1

0

101

1

0

ln

RRT

T

igP SS

PPR

TdTCS 12

1

22

1

ln

RR

H

igP HHTTCH 1212 )(

RR

S

igP SS

PPCS 12

1

2ln

Estimate V, U, H, and S for 1-butene vapor at 200°C and 70 bar if H and S are set equal to zero for saturated liquid at 0°C. Assume that only data available are:

KTc 420 barPc 43.40 191.0 .)(9.266 ptboilingnomalKTn 263 10837.910630.31967.1/ TTRC ig

P

127.1rT 731.1rP

512.0)142.0)(191.0(485.0

10

ZZZ

molcm

PZRTV

3

8.287

Assuming four steps:(a) Vaporization at T1 and P1 = Psat

(b) Transition to the ideal-gas state at (T1, P1)(c) Change to (T2, P2) in the ideal-gas state(d) Transition to the actual final state at (T2, P2)

Fig 6.7

Fig 6.7

Step (a)

TBAP sat ln

9.2660133.1ln BA

42043.40ln BA

For = 273.15K,Psat = 1.2771 bar

979.9930.0

)013.1(ln092.1

nr

c

n

lvn

TP

RTH

The latent heat at normal boiling point:

The latent heat at 273.15 K:38.0

11

nr

rlvn

lv

TT

HH

molJH lv 21810

STH Kmol

JS lv

84.79

Step (b) 650.0rT 0316.0rP

0985.0),,(10

OMEGAPRTRHRB

RTH

RTH

RTH

c

R

c

R

c

R

1063.0),,(10

OMEGAPRTRSRB

RS

RS

RS RRR

molJH R 3441

KmolJS R

88.01

Step (c)

molJEEICPHH ig 20564)0.0,6837.9,3630.31,967.1;15.473,15.273(314.8

KmolJ

EEICPSS ig

18.22

2771.170ln314.8)0.0,6837.9,3630.31,967.1;15.473,15.273(314.8

Step (d) 127.1rT 731.1rP

430.210

c

R

c

R

c

R

RTH

RTH

RTH

705.110

RS

RS

RS RRR

molJH R 84852

KmolJS R

18.142

molJHH 34233848520564)344(21810

KmolJSS

72.8818.1418.22)88.0(84.79

Total

molJPVHU 3221810/)8.287)(70(34233

Gas mixtures• Simple linear mixing rules

i

iiy i

ciiPc TyT i

ciiPc PyP

Estimate V, HR, and SR for an equimolar mixture of carbon dioxide and propane at 450K and 140 bar

KTyTi

ciiPc 337)8.369)(5.0()2.304)(5.0(

barPyPi

ciiPc 15.58)48.42)(5.0()83.73)(5.0( 41.2prP

335.1prT 697.00 Z

205.01 Z

188.0)152.0)(5.0()224.0)(5.0( i

iiy 736.010 ZZZ

molcm

PZRTV

3

7.196 762.110

pc

R

pc

R

pc

R

RTH

RTH

RTH 029.1

10

RS

RS

RS RRR

molJH R /4937 KmolJS R /56.8