Advanced Thermodynamics

Note 5Thermodynamic Properties of Fluids

Lecturer: 郭修伯

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 S

P

V

T

PS S

V

P

T

TV V

S

T

P

TP P

S

T

V

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

dPP

HdT

T

HdH

TP

dPP

SdT

T

SdS

TP

VdPTdSdH P

P

CT

H

VP

ST

P

H

TT

dPVP

STdTCdH

TP

TP P

S

T

V

dPT

VTVdTCdH

PP

T

C

T

S P

P

dPT

V

T

dTCdS

PP

PP T

ST

T

H

TP P

S

T

V

PVUH

VP

VP

P

U

P

H

TTT

TTT P

VP

P

VT

P

U

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

dPT

VTVdTCdH

PP

dPT

V

T

dTCdS

PP

)(1)( 12212 PPVTTTCH P

VdPTdTCdH P 1

VT

V

P

volume expansivity

)(ln 121

2 PPVT

TCS P

molJH /3400

10

)11000)(204.18()15.323)(10513(1)15.29815.323(310.75

6

molJS /13.5

10

)11000)(204.18(10513

15.298

15.323ln310.75

6

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

dVV

UdT

T

UdU

TV

dVV

SdT

T

SdS

TV

VV

CT

U

dVPV

STdTCdU

TV

VT T

P

V

S

dVPT

PTdTCdU

VV

T

C

T

S V

V

dVT

P

T

dTCdS

VV

PV

ST

V

U

TT

VV T

ST

T

U

VT T

P

V

S

SdTVdPdG

Gibbs energy G = G (P,T)

dTRT

GdG

RTRT

Gd

2

1

Thermodynamic property of great potential utility

SdTVdPdG

dTRT

HdP

RT

V

RT

Gd

2

TSHG

dTT

RTGdP

P

RTG

RT

Gd

PT

)/()/(

TP

RTG

RT

V

)/(

PT

RTGT

RT

H

)/(

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

dTRT

HdP

RT

V

RT

Gd

RRR

2

dTRT

HdP

RT

V

RT

Gd

RRR

2

const T dPRT

V

RT

Gd

RR

P RR

dPRT

V

RT

G0

P

RT

P

ZRTVVV igR

PR

P

dPZ

RT

G0

)1(PT

RTGT

RT

H

)/(

P

P

R

P

dP

T

ZT

RT

H0

PP

P

R

P

dPZ

P

dP

T

ZT

R

S00

)1(RT

G

RT

H

R

S 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

P

dP

T

ZRTTTDCBATTMCPHRH

P

dP

T

ZRTTTCH

P

dP

T

ZRTDCBATTICPHRH

P

dP

T

ZRTdTCHHHH

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

P

dPZ

P

dP

T

ZTR

P

PR

T

TDCBATTMCPSRS

P

dPZ

P

dP

T

ZTR

P

PR

T

TCS

P

dPZ

P

dP

T

ZTR

P

PRDCBATTICPSRS

P

dPZ

P

dP

T

ZTR

P

PR

T

dTCSSSS

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

P

dP

T

ZT

RT

H0

PP

P

R

P

dPZ

P

dP

T

ZT

R

S00

)1(

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

Z

P

/

9493.0)1037.26)(360( 4

0

P

P

R

P

dP

T

ZT

RT

H

6897.0)02596.0(9493.0)1(00

PP

P

R

P

dPZ

P

dP

T

ZT

R

S

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

1

41.15ln)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

P

dP

T

ZT

RT

H0

PP

P

R

P

dPZ

P

dP

T

ZT

R

S00

)1(

Use of the virial equation of state

RT

BPZ 1

PR

P

dPZ

RT

G0

)1(two-term virial equation

RT

BP

RT

GR

P

P

R

P

dP

T

ZT

RT

H0

P

P

R

P

dP

RT

BP

TT

RT

H0

)1(

P

P

R

P

dP

T

B

dT

dB

TR

PT

RT

H0 2

1

RT

BPZ 1

dT

dB

T

B

R

P

RT

H R

RT

G

RT

H

R

S RRR

dT

dB

R

P

RT

S R

21 CBZ

PR

P

dPZ

RT

G0

)1(

Three-term virial equation

ZCBRT

GR

ln2

32 2

P

P

R

P

dP

T

ZT

RT

H0

2

2

1 dT

dC

T

C

dT

dB

T

BT

RT

H R

RT

G

RT

H

R

S RRR

RTZP

Application up to moderate pressure

Use of the cubic equation of state

• The generic cubic equation of state:

))((

)(

bVbV

Ta

bV

RTP

)1)(1(1

1

bb

bq

bZ

bRT

Taq

)(

qIbd

Z )1ln()1(0

IdT

dqd

T

Z

0

Tconstbb

bdI

0 )1)(1(

)(

qITd

TdZ

RT

H

r

rR

1

ln

)(ln1

qITd

TdZ

R

S

r

rR

1

ln

)(lnln

RT

bP

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

96.37

50rP 09703.0

r

r

T

P 8689.3)(

r

r

T

Tq

))((1

ZZ

ZqZ

01 685.0Z

qITd

TdZ

RT

H

r

rR

1

ln

)(ln1

qITd

TdZ

R

S

r

rR

1

ln

)(lnln

Tconstbb

bdI

0 )1)(1(

)(13247.0ln

1

Z

ZI

mol

JH R 4505

Kmol

JS 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

V

S

VV

SS

dT

dP sat

VdPTdSdH STH The latent heat of phase transition

VT

H

dT

dP sat

The Clapeyron equation

Ideal gas, and Vl << Vg

satv

P

RTVV

)/1(

ln

Td

PdRH

satlv

The Clausius/Clapeyron equation

VT

H

dT

dP sat

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

)(TfP

T

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

to the critical point

CT

BAP sat

ln The Antoine equation, for a specific temperature range

1

ln635.1 DCBA

P satr

The 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 Hkg

kJHHH 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

lbm

BtuHHH 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

dmHQ

dt

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 Cliquidsaturatedkg

kJH 70@0.293

Cliquidsaturatedkg

kJH l

cv100@1.419

Cvaporsaturatedkg

kJH 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

P

dP

T

ZT

RT

H0

PP

P

R

P

dPZ

P

dP

T

ZT

R

S00

)1(

r

r

P

r

r

Prr

c

R

P

dP

T

ZT

RT

H0

2

rr

r

P

r

rP

r

r

Prr

R

P

dPZ

P

dP

T

ZT

R

S00

)1(

10 ZZZ

r

r

r

r

P

r

r

Prr

P

r

r

Prr

c

R

P

dP

T

ZT

P

dP

T

ZT

RT

H0

12

0

02

rr

r

rr

r

P

r

rP

r

r

Prr

P

r

rP

r

r

Prr

R

P

dPZ

P

dP

T

ZT

P

dPZ

P

dP

T

ZT

R

S0

1

0

1

0

0

0

0

)1()1(

),,(

10

OMEGAPRTRHRBRT

H

RT

H

RT

H

c

R

c

R

c

R

),,(

10

OMEGAPRTRSRBR

S

R

S

R

S RRR

Table E5 ~ E12

RT

T

igP

ig HdTCHH 202

2

0

RT

T

igP

ig HdTCHH 101

1

0

RRT

T

igP HHdTCH 12

2

1

RT

T

igP

ig SP

PR

T

dTCSS 2

0

202

2

0

ln

RT

T

igP

ig SP

PR

T

dTCSS 1

0

101

1

0

ln

RRT

T

igP SS

P

PR

T

dTCS 12

1

22

1

ln

RR

H

igP HHTTCH 1212 )(

RR

S

igP SS

P

PCS 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

mol

cm

P

ZRTV

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)

T

BAP 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

T

P

RT

HThe latent heat at normal boiling point:

The latent heat at 273.15 K:

38.0

1

1

nr

rlvn

lv

T

T

H

H

mol

JH lv 21810

STH Kmol

JS lv

84.79

Step (b) 650.0rT 0316.0rP

0985.0),,(

10

OMEGAPRTRHRB

RT

H

RT

H

RT

H

c

R

c

R

c

R

1063.0),,(

10

OMEGAPRTRSRB

R

S

R

S

R

S RRR

mol

JH R 3441

Kmol

JS R

88.01

Step (c)

mol

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

Kmol

J

EEICPSS ig

18.22

2771.1

70ln314.8)0.0,6837.9,3630.31,967.1;15.473,15.273(314.8

Step (d) 127.1rT 731.1rP

430.2

10

c

R

c

R

c

R

RT

H

RT

H

RT

H

705.1

10

R

S

R

S

R

S RRR

mol

JH R 84852

Kmol

JS R

18.142

mol

JHH 34233848520564)344(21810

Kmol

JSS

72.8818.1418.22)88.0(84.79

Total

mol

JPVHU 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

mol

cm

P

ZRTV

3

7.196 762.1

10

pc

R

pc

R

pc

R

RT

H

RT

H

RT

H 029.1

10

R

S

R

S

R

S RRR

molJH R /4937 KmolJS R /56.8