thermodynamic properties of fluids

8/12/2019 Thermodynamic Properties of Fluids

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1

ThermodynamicProperties of FluidsDr.Sininart Chongkhong A Dao

ChE. PSU.


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2

Purpose of this Chapter To develop from the first and second laws the fundamental

property relations which underlie the mathematical structure

of thermodynamics.

Derive equations which allow calculation of enthalpy and

entropy values from PVT and heat capacity data.

Discuss diagrams and tables by which property values

are presented for convenient use.

Develop generalized correlations which provideestimated of property values in he absence of completeexperimental information.


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3

Property Relations for Homogeneous Phases

Fundamental Properties

Although this equation is derived from the special case of areversible process, it not restricted in application toreversible process.

It applies to any process in a system of constant mass thatresults in a differential change form one equilibrium state to

another. The system many consist of a single phase or several

phases; may be chemically inert or may undergo chemicalreaction.

nVPdnSTdnUd

dWdQnUd revrev

)(

(6.1)


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Define

H = Enthalpy

A = Helmholtz energy

G = Gibbs energy

PVUH

TSUA

TSHG

(2.11)

(6.2)

(6.3)


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5

Based on one mole (or to a unit mass) of a

homogeneous fluid of constant composition,

they simplified to

SdTVdPdG

SdTPdVdA

VdPTdSdH

PdVTdSdU


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6

Maxwells equaitons

TP

TV

PS

VS

PS

TV

V

S

T

PS

V

P

T

SP

VT


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7

Enthalpy and Entropy as Functions of T and P Temperature derivatives:

T

C

T

S

CTH

P

P

P

P

Pressure derivatives:

PT

PT

T

VTV

P

H

T

V

P

S


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The most useful property relations for the enthalpy

and entropy of a homogeneous phase result when

these properties are expressed as functions of Tand P (how H and S vary with T and P).

dPT

V

T

dTCdS

dPTVTVdTCdH

PP

P

p

(6.21)(6.20)


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Property Relations for Homogeneous PhasesInternal Energy as Function of P

The pressure dependence of the internal energy

is shown as

PdVTdsdU

TPT P

VP

T

VT

P

U


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10

Property Relations for Homogeneous PhasesThe Ideal Gas State

For ideal gas, expressions of dH and dS

(eq.6.20-6.21) as functions of T and P can besimplified to as follows:

(6.24)

(6.23)

P

dPR

T

dTCdS

dTCdH

P

R

T

VRTPV

ig

P

ig

ig

P

ig

P

ig

ig


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11

Property Relations for Homogeneous PhasesAlternative Forms for Liquids

Relations of liquids can be expressed in terms

of and as follows:

VTPP

U

VT

P

H

VPS

T

T

T

1


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Property Relations for Homogeneous PhasesAlternative Forms for Liquids

Enthalpy and entropy as functions of T and P

as follows:

and are weak functions of pressure forliquids, they are usually assumed constant at

appropriate average values for integration.

)....(VdP

T

dTCdS

)....(VdPTdTCdH

P

P

296

2861


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13

Example 6.1Determine the enthalpy and entropy changes of liquid water for

a change of stagefrom 1 bar 25C to 1,000 bar 50C.

T(C) P(bar) Cp(Jmol-1K-1 V(cm3mol-1) (K-1)25

25

50

50

1

1,000

1

1,000

75.305

75.314

18.071

18.012

18.234

18.174

256x10-6

366x10-6

458x10-6

568x10-6


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15

166

13

11

10513102

568458

204.182

174.18234.18

,50

310.752

314.75305.75

,1

K

molcmV

CTforand

KJmolC

barPFor

p

121

2

12212

ln

1

PPV

T

TCS

PPVTTTCH

p

p


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16

11

6

1

6

13.593.006.6

10

1000,1204.1810513

15.298

15.323ln310.75

400,3517,1883,1

10

1000,1204.1815.32310513115.29815.323310.75

KJmol

S

Jmol

H

121

2

12212

ln

1

PPV

T

TCS

PPVTTTCH

p

p

Note that the effect of P of almost 1,000 bar on H and S of liquid water

is less than that of T of only 25C.


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17

Property Relations for Homogeneous Phases

Internal Energy and Entropy as Function of T and V Useful property relations for T and V as independent

variables are

V

VT

V

V

T

P

P

T

PT

V

U

TC

TS


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The Partial derivatives dU and dS of homogeneous fluids

of constant composition to temperature and volume are

Alternative forms of the above equations are

dVT

P

T

dTCdS

dVPTPTdTCdU

V

V

V

V

dVT

dTCdS

dVPTdTCdU

V

V


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Property Relations for Homogeneous PhasesThe Gibbs Energy as a Generating Function

An alternative form of a fundamental property relation asdefined in dimensionless terms:

The Gibbs energy when given as a function of T and Ptherefore serves as a generating function for the otherthermodynamic properties, and implicitly representscomplete information.

P

T

P

RTGT

RT

H

P

RTG

RT

V

dTRT

HdP

RT

V

RT

Gd

2


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Residual Properties The definition for the generic residual property is:

M is the molar value of any extensive thermodynamics

property: V, U, H, S, G.

M, Mig= the actual and ideal gas properties which are atthe same temperature and pressure.

igR

MMM


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Residual gibbs energy:

G, Gig= the actual and ideal gas values of the Gibbs

energy at the same temperature and pressure.

Residual volume:

igR

GGG

1

ZP

RTV

P

RTVVVV

R

igR


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Fundamental property relation for residual properties The fundamental property relation for residual

preperties applies to fluids of constant composition.

).(

T

RT/GT

RT

H

).(P

RT/G

RT

V

).(dTRT

HdP

RT

V

RT

Gd

P

RR

T

RR

RRR

446

436

4262


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)48.6()()1(,

)46.6();44.645.6.(

0)45.6()1(

)(),43.6.(

00

0

000

0

TconstP

dPZ

P

dP

T

ZT

R

SSo

RT

G

RT

H

RT

SFrom

PdP

TZT

RTHEq

RT

Ggasidealfor

P

dPZdP

RT

V

RT

G

RT

G

TconstdPRT

V

RT

GdEqFrom

P

P

PR

RRR

P

PR

P

RP

RP

P

RR

RR


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Enthalpy and Entropy from Residual Properties

).(SPPlnRdTCSS

).(HdTCHH;onSubstituti

P

P

lnRdTCSSdTCHH

);.(and)..(EqofnIntegratio

SSSHHH;SandHtoApplied

RT

T

ig

P

ig

RT

T

ig

P

ig

T

T

ig

P

igigT

T

ig

P

igig

RigRig

516

506

246236

0

0

0

0

00

0

0

00


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25

2

2112

21

21

2

12

00

0

00

2

1

2

1

43

536

526

TT

DCTTBTA

R

C

)T/Tln(

T

dTC

C

TT

D)TTT(

CBTA

R

C

TT

dTC

C

).(SP

P

lnRT

T

lnCSS

).(H)TT(CHH

lmamlm

SP

T

T

ig

P

SP

amam

HP

T

T

ig

P

HP

R

SP

igig

R

HP

igig

The true worth of the Eq. for ideal gases is now evident.

They are important because they provide a convenient basefor the calculation of real-gas properties.


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Example 6.3

Calculate H and S of saturated isobutane vapor at 630 Kfrom the following information:

1. Table 6.1 gives compressibility-factor data

2. The vapor pressure of isobutane at 630 K 15.46 bar

3. Set H0ig= 18,115 Jmol-1and S0

ig= 295.976 Jmol-1K-1

for the ideal-gas reference state at 300 K 1 bar4. Cp

ig/R = 1.7765+33.037x10-3T (T/K)


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Solution 6.3

Eqs. (6.46) and (6.48) are used to calculate HRand SR.

Plot (Z/T)P/P and (Z-1)/P vs. P

From the compressibility-factor data at 360 K (Z-1)/P

The slope of a plot of Z vs. T (Z/T)P/P

Data for the required plots are shown in Table 6.2.

P

dPZ

P

dP

T

Z P

P

P

)1(00


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11

1

11

4

0

14

0

7345314868970

38412360314894930

3148

689702596094930486

94930103726360466

259601103726

KJmol...S

Jmol.,..H

KJmol.RFor

...R

S

),.(.EqBy

.).)((RT

H),.(.EqBy

.P

dP)Z(K.

P

dP

T

Z

R

R

R

R

P

P

P


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31

11

1

11

11

01

01

10

3

3

676286734541153148300

36016105576295

559821284123003604110511518

516506

16105314864912

41105314867912

09329300360

300360

3302

360300

2

100373377651

100373377651

KJmol...ln.)ln(..S

Jmol.,.,)(.,H

).(and)..(EqointSubstitute

KJmol.).(.C

KJmol.).(.C

K.)/ln()T/Tln(

TTT

KTT

T

T..BTA

R

C

T..BTAR

C

S

ig

P

H

ig

P

lm

am

lmlm

S

ig

P

amam

H

ig

P


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Residual Properties by Equations of StateResidual Properties from the Virial Equation of State The two-term virial eq.gives Z-1 = BP/RT.

)56.6(),47.6.(int

)55.6(/

),44.6.(

)54.6(,

dTdB

RP

RSEqoonSubstituti

dT

dB

T

B

R

P

T

RTGT

RT

HEqBy

RT

BP

RT

GSo

R

P

RR

R


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In application is a more convenient variable than V,

PV = ZRT is written in the alternative form.

).(d

)Z(d

T

ZTZln

R

S),..(EqFrom

).(Zd

T

ZT

RT

H);.(and)..(EqoftionDifferenti

T

)RT/G(

T

P

P

Z

RT

H),.(and)..(EqFrom

).(ZlnZd

)Z(RT

G);..(EqointSubstitue

Z

dZd

P

dP)dZZd(RTdP),.(RTZP

R

R

RR

R

6061476

5961586576

1426406

58611496

576

00

0

2

0


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The three-term virial equation.

)63.6(2

1ln

)62.6(

2

1

)61.6(ln2

32

).60.6()58.6.(int1

2

2

2

2

dT

dC

T

C

dT

dB

T

BTZ

R

S

dT

dC

T

C

dT

dB

T

BT

RT

H

ZCBRT

G

throughEqosubstituedisCBZ

R

R

R

Application of these equations, useful for gases up tomoderate pressure, requiresdata for both the second andthird virial coefficients.

R id l P ti b C bi E ti f St t


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Residual Properties by Cubic Equations of State

I

dT

dq

T

Z,qI)bln(;simplifyTo

)b)(b(

)b(d

dT

dq

T

Z)b)(b(

)b(dq

b

)b(d

b

b

);.(),.(

)b)(b(b

dTdq

TZ

).()b)(b(

bq

b

bZ

)b)(b(

bqbZ

)...(Eq,/VeRTy)..(Eq

dd1)-(Z

d

d1)-(Z

Eqs.ofintegralsThe

bygivenqassubstitutandbdevides

0

0

0

00

0 0

1

11

111

606586

11

646111

1

111

1

5131423


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The generic equation of state presents two cases.

Zb

bI:

)b.(

Z

ZlnI

bZwhenceRT

PZ

)a.(b

b

lnI:

1

6561

656

1

11

IICase

RT

bP

Z.offavorineliminatedisWhen

ICase


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37

)68.6(ln

)(ln)ln(

)67.6(1ln

)(ln1

)66.6()ln(1

)66.6()1ln(1

qITd

TdZ

R

S

qITdTdZ

RTH

bqIZZZRT

G

aqIZbZ

RT

G

r

r

R

r

r

R

R

R


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Find values for the HRand SRfor n-butane gas at 500 K

50 bar as given by the Redlich/Kwong Eequation.

Solution

Tr= 500/425.1 = 1.176, Pr= 50/37.96 = 1.317

From Table 3.1:

8689.3176.108664.0

42748.0);54.3.(

09703.0176.1

317.108664.0);53.3.(

2/3

r

r

r

r

T

TqEq

T

PEq

Ex. 6.4


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39

Z


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40

11

1

546.6)78735.0(314.8

505,40838.1500314.8,

78735.013247.0)8689.3(5.0)09703.06850.0ln(:)68.6.(

0838.1)13247.0)(8689.3)(15.0(16850.0:)67.6.(

:.21ln/)(ln,ln

21)(ln

13247.0ln

68500

)09703.0(

09703.009703.08689.309703.01

1:)52.3.(

KJmolS

JmolHThus

R

SEq

RT

HEq

ThenTdTdTTWith

Z

ZI

Then:..ZyieldsEq.thisofSolution

ZZ

Z

ZZ

ZqZEq

R

R

R

R

rrrr


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41

These results are compared with those of othercalculation in Table 6.3.


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42

TWO-PHASE SYSTEMS

).(ZR

H

)T/(d

Plndor

).(ZRT

HdTPlndZ

PRTVBut

).(VT

H

dT

dP;vaportoliquidfromtransitionPhase

equationClapeyronThe:).(VTH

dTdPT/HS,Thus

)transitionphaseofheatlatentThe(STH);..(EqofnIntegratio

V

S

VV

SS

dT

dP,tarrangemenRe

dTSdPVdTSdPVdGdG,GG

l

lsat

l

lsat

l

sat

l

l

lsat

sat

sat

satsat

7461

736

726

716

86

2

The Clapeyron eq.

for pure-speciesvaporization


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43

Temperature Dependence of the VaporPressure of Liquids

r

.

r

sat

rr

sat

sat

Twhere

).(

DCBA

)T(Pln;ToffunctionA

B.App,.BTableingivenaretstanconsAntoine

).(CT

BAPln:.eqAntoineThe

T

BAPln

1

7761

2

766

6351

C di St t C l ti


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44

Corresponding-States Correlationsfor Vapor Pressure

:

:

)81.6()(ln

)(lnln

)80.6(43577.0ln4721.136875.15

2518.15)(ln

)79.6(169347.0ln28862.109648.6

92714.5)(ln

)78.6()(ln)(ln)(ln

:/

1

0

61

60

10

sat

r

r

rr

rr

sat

r

rr

r

rr

rr

r

rr

rrrrr

sat

r

n

n

n

nn

P

T

where

TP

TPP

TTT

TP

TTT

TPwhere

TPTPTP

ncorrelatioKeslerLee

The reduced normal boiling pointThe reduced vapor pressure corresponding to 1 atm

Ex. 6.6


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45

Ex. 6.6Determine the vapor pressure for liquid n-hexane at 0, 30,

60 and 90C: (a) With constants from App. B.2.

(b) From the Lee/Kesler correlation for Pr

sat

Solution

(a)

(b) Eq.(6.78);

From Table B.1,

From Eq.(6.81) =0.298

The average difference from the Antoine values is about 1.5%.

317.224

04.26968193.13ln

tPsat

03350.025.30

01325.1,6736.06.507

9.341

satrr nn PT

t/C Psat/kPa(Antoine)

Psat/kPa(Lee/Kesler)

t/C Psat/kPa(Antoine)

Psat/kPa(Lee/Kesler)

0

606.052

76.465.835

76.1230

9024.98

189.024.49

190.0


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46


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48

Two-Phase Liquid/Vapor System

)82.6(

:.,,,,

)82.6()1(

:

1)1(

:

)(

bMxMM

formealternativAnetcSHUVMwhere

aMxMxM

equationgenericThe

xxVxVxV

fractionmassxVxVxV

molesnnnVnVnnV

ll

lv

vllv

ll

vlll

THERMODYNAMIC DIAGRAMS


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49

THERMODYNAMIC DIAGRAMS


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50

GENERALIZED PROPERTYCORRELATION FOR GASES

)84.6()1(

)83.6(

:)48.6()46.6.(int

,

00

0

2

r

r

P

r

r

P

P

r

r

R

r

r

P

P

r

r

c

R

rcrcrcrc

PdPZ

PdP

TZT

RS

P

dP

T

ZT

RT

H

andEqsosubstitue

dTTdTTTTdPPdPPPP

r

r

r

r

r

10 ZZZ


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51

)86.6(

1:)84.6.(

)85.6(

:)83.6.(

10

11

0

01

0

10

0

12

0

02

10

R

S

R

S

R

S

P

dPZ

T

ZT

P

dPZ

T

ZT

R

SEq

RT

H

RT

H

RT

H

P

dP

T

ZT

P

dP

T

ZT

RT

HEq

T

Z

T

Z

T

ZZZZ

RRR

r

r

Pr

r

P

r

r

Pr

r

PR

c

R

c

R

c

R

r

rP

Pr

r

r

rP

Pr

r

c

R

PrPrPr

r

r

r

r

r

r

r

r

rrr

Table E.5 - E.12


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52

Analytical correlation of the residual properties at low pressure


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53

Analytical correlation of the residual properties at low pressure

)88.6(

)87.6(

,);56.6()55.6.(

,

sec

10

11

00

1010

rr

r

R

r

r

r

rr

c

R

r

r

R

r

rr

c

R

rrrc

c

dT

Bd

dT

BdP

R

S

dT

BdTB

dT

BdTBP

RT

H

dT

BdP

R

S

dT

BdTBP

RT

HandEqs

dT

dB

dT

dB

dT

BdBB

RT

BPB

formsncorrelatiotcoefficienvirialonddgeneralizeThe


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54

)90.6(

722.0

)89.6(675.0

)66.3(172.0

139.0

)65.3(422.0

083.0

2.5

1

6.2

0

2.4

1

6.1

0

rr

rr

r

r

TdT

dB

TdT

dB

TB

TB


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55

HRand SRwith ideal-gas heat capacities

RT

ig

P

igRT

ig

P

ig HdTCHHHdTCHH 10

012

0

02

12

)92.6(ln,

)91.6(

12

1

2

12

2

1

2

1

RR

T

T

ig

P

RR

T

T

igP

SSP

PRdTCSSimilarly

HHdTCH

For a change from state 1 to 2:

The enthalpy change for the process, H = H2H1

Alternative form

)94.6(lnln

)93.6()(

12

1

2

1

2

1212

RR

S

ig

P

RR

H

ig

P

SSP

PR

T

TCS

HHTTCH


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56

A three-step calculational path


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57

A three step calculational path Step 11ig: A hypothetical process that transforms a real

gas into an ideal gas at T1and P1.

Step 1ig2ig: Changes in the ideal-gas state from (T1,P1)

to (T2,P2).

Step 2ig2: Another hypothetical process that transform

the ideal gas back into a real gas at T2and P2.

RigRig SSSHHH 111111

)96.6(ln

)95.6(

1

2

12

12

2

1

2

1

P

PR

T

dTCSSS

dTCHHH

T

T

ig

P

igigig

T

T

ig

P

igigig

RigRig SSSHHH222222

E 6 9


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58

Ex. 6.9

Estimate V, U, H and S for 1-butane vapor at 200C, 70 bar

if H and S are set equal to zero for saturated liquid at 0C.Assume: Tc=420.0 K, Pc=40.43 bar, Tn=266.9 K, =0.191

Cpig/R=1.967+31.630x10-3T-9.837x10-6T2 (T/K)

Solution

13

10

8.28770

)15.473)(14.83(512.0

512.0)142.0(191.0485.0

;4.3.)57.3.(

731.143.40

70127.1

0.420

15.273200

molcmP

ZRTV

ZZZ

EandETableandEq

PT rr


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59


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60

Step (a): Vaporization of saturated liquid 1-butane at 0C


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61

Step (a):Vaporization of saturated liquid 1 butane at 0 C

The vapor pressure curve contains both

The latent heat of vaporization, where Trn=266.9/420=0.636:

)75.6(ln T

BAPsat

11.699,2126.10,0.420

43.40lnint;

9.2660133.1lnint;

BAWhence

BApocriticalthe

BApoboilingnormalthe

1137,229.266314.8979.9

979.9636.0930.0

)013.143.40(ln092.1930.0

)013.1(ln092.1

JmolH

TP

RTH

lv

n

r

c

n

lvn

n


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62

11

1

380

380

8479

15273

81021

810213680

350013722

1341

1

65004201527315273

KJmol..

,

T

HS

Jmol,.

.),(H

)....(T

T

H

HFrom

./.TK.atheatlatentThe

lv

lv

.

lv

.

r

r

lv

n

lv

r

n

Step (b): Transformation of saturated vapor into an ideal


63/70

63

Step (b):Transformation of saturated vapor into an idealgas at (T1, P1).

Tr= 0.650 and Pr= 1.2771/40.43 = 0.0316

)88.6(

)87.6(

10

11

00

rr

r

R

r

r

r

rr

c

R

dT

Bd

dT

Bd

PR

S

dT

BdTB

dT

BdTBP

RT

H

)90.6(722.0

)89.6(675.0

)66.3(172.0

139.0

)65.3(

422.0

083.0

2.5

1

6.2

0

2.4

1

6.1

0

rr

rr

r

r

TdT

dB

TdT

dB

TB

TB

11

1

1

1

88.0)314.8)(1063.0(

344)420)(314.8)(0985.0(,

KJmolS

JmolHSo

R

R

Step (c): Changes in the ideal gas state


64/70

64

Step (c):Changes in the ideal gas state

Tam= 373.15 K, Tlm= 364.04 K,

A = 1.967, B = 31.630x10-3, C = -9.837x10-6

Hig= 20,564 J mol-1

Sig= 22.18 J mol-1K-1

1

2

1

2

1

2

12

1212

2

1

2

1

966

956

P

PlnR

T

TlnC

P

PlnR

T

dTCSSS:)..(Eq

)TT(CdTCHHH:)..(Eq

S

ig

P

T

T

ig

P

igigig

H

ig

P

T

T

ig

P

igigig

Step (d): Transformation from the ideal gas to real gas


65/70

65

Step (d): Transformation from the ideal gasto real gasstate at T2and P2.

Tr= 1.127 Pr= 1.731 At the higher P; Eqs.(6.85) and (6.86) with interpolated

values from Table E.7, E.8, E.11 and E.12.

1

13

11

1

11

2

1

2

2

2

218,3210

)8.287)(70(233,34

18.1418.22)88.0(84.79

233,34485,8564,20)344(810,21

18.14)314.8)(705.1(

485,8)0.420)(314.8)(430.2(

705.1)726.0)(191.0(566.1

430.2)713.0)(191.0(294.2

JmolbarJcm

PVHU

KJmolSS

JmolHH

KJmolS

JmolH

R

S

RT

H

R

R

R

C

R


66/70

Ex 6 10


67/70

67

Ex. 6.10

Estimate V, HR, and SRfor an equimolar mixture of

carbon dioxide(1) and propane(2) at 450 K and 140 bar by

the Lee/Kesler correlations.

Solution

From Table B.1,

41.215.58

140335.1

0.337

450,

15.58)48.42)(5.0()83.73)(5.0(

0.337)8.369)(5.0()2.304)(5.0(

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

2211

2211

2211

prpr

ccpc

ccpc

PTWhence

barPyPyP

KTyTyT

yy


68/70

68


69/70


70/70

thermodynamic properties of fluids

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