1 -Thermodynamics Properties of Fluids

7/23/2019 1 -Thermodynamics Properties of Fluids

1/41

Thermodynamic Propeties of

Fluids

Chemical Engineering

Thermodynamics II


2/41

In this lecture we will talk about

Finding property relations for homogenous

phases in terms of easily measurable variables

Define residual properties and derive residual

relationships

Use of thermodynamic diagrams and tables


3/41

Property relations for homogenous

phases

dWdQnUd )(

revrev dWdQnUd )(

)(nVPddWrev

)(nSTddQrev

)()()( nVPdnSTdnUd

First law for a closed system

a special reversible process

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.


4/41


phases

SandUTVP ,,,,The primary thermodynamic properties:

PVUH Enthalpy:

TSUA Helmholtz energy:

TSHG Gibbs energy:

)()()( nVPdnSTdnUd

For n number ofmoles, sub d(nU)

dPnVnVPdnUdnHd )()()()(

dPnVnSTdnHd )()()(

dTnSdPnVnGd )()()(

dTnSnVPdnAd )()()(

PdVTdSdU

VdPTdSdH

SdTPdVdA

SdTVdPdG

For one mol of homogeneous fluid of constant composition:


5/41


phasesRemember that for Exact differential equations

= +

PdVTdSdU

VdPTdSdH

SdTPdVdA

SdTVdPdG

VS S

P

V

T

PS S

V

P

T

TV V

S

T

P

TP P

S

T

V

=2

=

Maxwells equations


6/41

Enthalpy and Entropy as function of T

and P

dPP

HdT

T

HdH

TP

VdPTdSdH

dPT

VTVdTCdH

PP

TP P

S

T

V

PT T

VTV

P

H

Maxwells

equations

P

P

CTH

VP

ST

P

H

TT


7/41

Enthalpy and Entropy as function of T

and P

dPP

SdT

T

SdS

TP

T

C

T

S P

P

dPT

V

T

dTCdS

P

P

TP P

S

T

V

VdPTdSdH

PP T

ST

T

H

Maxwells equation


8/41

Example 1

Determine the enthalpy and entropy changes of liquid water for achange of state from 1 bar and 25

C to 1000 bar and 50

C. The data

for water are given. (see data on example 6.1, p204)

dPT

VTVdTCdH

P

P

dPT

V

T

dTCdS

P

P

)(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 /13.510

)11000)(204.18(10513

15.298

15.323

ln310.75

6


9/41

Example 2Estimate the change in enthalpy and entropy when liquid ammonia at 270K is

compressed from its saturation pressure of 381 kPa to 1,200 kPa. For saturated

liquid ammonia at 270K, Vl=1.551x10-3m3kg-1 and=2.095x10-3K-1

VdPTdTCdH P 1 VdPT

dTCdS P

Constant T: VdPTdH 1 VdPdS

)3811200)(1055.1)](270)(10095.2(1[ 13113 kPakPakgmKKH

)3811200)(1055.1)(10095.2( 13113 kPakPakgmKS

)(1 12 PPVTH

)( 12 PPVS

kg

J

H 7.551

Kkg

JS 661.2

Ans.

Ans.


10/41

Internal energy and Entropy as

functions of T and V

dVV

UdT

T

UdU

TV

dV

V

SdT

T

SdS

TV

dVP

T

PTdTCdU

V

V

V

V

C

T

U

PV

ST

V

U

TT

PdVTdSdU

dVPV

STdTCdU

T

V

VT T

P

V

S

Maxwells

equations:

T

C

T

S V

V

dVT

P

T

dTCdS

V

V

VV T

ST

T

U

VT T

P

V

S

PdVTdSdU

Temperature and volume may serve as more

convenieny variables than do temperatureand pressure.


11/41

Gibbs energy as a Generating Function

SdTVdPdG

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

)/(

),( TPRT

G

RT

G

Total differential equation as a function of P, T

RT

G

RT

H

R

S

RT

PV

RT

H

RT

U

G(P,T) serves as a generating

function for the other

thermodynamic properties


12/41

Residual properties

The definition for the generic residual property:

Mand Migare 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 generatingfunction for the other residual properties:

igR MMM

dTRT

HdP

RT

V

RT

Gd

RRR

2


13/41

Residual properties

dTRTHdP

RTV

RTGd

RRR

2

dPRT

V

RT

Gd

RR

P RR

dPRT

V

RT

G0

P

RT

P

ZRTVVV

igR

P

R

P

dPZ

RT

G

0)1(

PT

RTGT

RT

H

)/(

P

P

R

P

dP

T

ZT

RT

H

0

PP

P

R

P

dPZ

P

dP

T

ZT

R

S

00 )1(

RT

G

RT

H

R

S RRR

Z = PV/RT: experimental measurement .Given

PVT data or an appropriate equation of state, we

can evaluate HRand SRand hence all other residual

properties.

;


14/41

Residual properties by equation of

state

RTBPZ 1

P

R

P

dPZ

RT

G

0)1(

two-term virial equation:

RT

BP

RT

GR

P

P

R

P

dP

T

ZT

RT

H

0

P

P

R

P

dP

RT

BP

TT

RT

H

0)1(

P

P

R

P

dP

T

B

dT

dB

TR

PT

RT

H

0

2

1

dT

dB

T

B

R

P

RT

HR

dTdB

RP

RTS

R


15/41

Residual properties by equation of

state

P

R

P

dPZ

RT

G

0)1(

three-term virial equation:

P

P

R

P

dP

T

ZT

RT

H

0

01Z

d

T

ZT

RT

HR

21 CBZ RTZP

ZCBRT

GR

ln2

32

2

2

2

1 dT

dC

T

C

dT

dB

T

BTRT

HR

RT

G

RT

H

R

S RRR

2

2

1ln

dT

dC

T

C

dT

dB

T

BTZ

RT

HR


16/41

Residual properties of the cubic

equation of state

))((

)(

bVbV

Ta

bV

RTP

)1)(1(1

1

bb

bq

bZ

qIbd

Z )1ln()1(0

IdT

dqd

T

Z

0

Tconstbb

bdI

0 )1)(1(

)(

RT

bP

))((

)(

)( bVbV

T

bVRT

RT

RT

P

Z

ZI ln

1

Zb

b

I 1

;Case I:

Case II: ;


17/41

Residual properties of the cubic

equation of state

qIbd

Z )1ln()1(0

IdT

dqd

T

Z

0

qITd

TdZ

RT

H

r

r

R

1

ln

)(ln1

qI

Td

TdZ

R

S

r

r

R

1

ln

)(lnln

Van der Waals equation of state

Zb

bI

1Case II:

qIZZRT

G )ln(1 Virial equation of state;


18/41

Example 3Find values for the residual enthalpy HRand 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

0

1

685.0Z)0973.0)((

0973.0)09703.0)(8689.3(09703.01

ZZ

ZZ

13247.0ln1

Z

ZICase II:


19/41

Example 3Find values for the residual enthalpy HRand the residual entropy SR

for n-butane gas at 500 K and 50 bar as given by Redlich/Kwong

equation.

685.0Z

13247.0ln1

Z

ZICase II:

qITdTdZ

RTH

r

r

R

1

ln)(ln1

qITd

TdZ

R

S

r

r

R

1

ln

)(lnln

molJHR 4505

Kmol

JSR

546.6

2/1ln/)(lnln2

1)(ln)( 2/1 rrr TdTdTTTT

Ans.

Ans.


20/41

Two-phase systems

Whenever a phase transition at constant

temperature and pressure occurs,

The molar or specific V, U, H, and S changes abruptly. The

exception is the molar or specific Gibbs energy, which for apure species does not change during a phase transition.

For two phases and of a pure species coexisting at

equilibrium: GG

where Gand Gare the molar or specific Gibbs energies of the

individual phases


21/41

Two-phase systems

dGdG SdTVdPdG

dTSdPVdTSdPV satsat

VS

VVSS

dTdP

sat

STH The latent heat of phase transition

VT

H

dT

dPsat

The Clapeyron equation

Ideal gas, and Vl


22/41

Two-phase liquid/vapor systems

When a system consists of saturated-liquid andsaturated-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:

Where M represents V, U, H, S, etc.

e.g for volume:vvvv VxVxV )1(

vvlv MxMxM )1(


23/41

Thermodynamic diagrams and tables

A thermodynamic diagram represents the temperature,pressure, volume, enthalpy, and entropy of a substance on asingle plot. Common diagrams are:

In many instances, thermodynamic properties are reported in

tables. The steam tables are the most thorough compilation of

properties for a single material.

TS diagram

PH diagram (ln P vs. H)

HS diagram (Mollier diagram)


24/41

Example diagrams

PH Diagram TS Diagram


25/41

Example diagrams

PH Diagram

Liquid water below its boiling point is

superheated to steam:

Line 12 : Heated at constant P to

saturated temperature

Line 23 : Vaporized at constant T, P

Line 34 : Superheated at constant P


26/41

Exampl 4Superheated 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 H for the following

conditions:

(a) P1= 1000 kPa, T1= 250C, and P2= 200 kPa.

(b) P1= 150 psia, T1= 500F, 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= 250C:

H1= 2942.9 kJ/kg, S1= 6.9252 kJ/kg.K

At P2= 200 kPa,

S2= S1= 6.9252 kJ/kg.K < Ssaturated vapor,

the final state is in the two-phase liquid-vapor region


27/41




conditions:

(a) P1= 1000 kPa, T1= 250C, and P2= 200 kPa.

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 H

kg

kJHHH 9.31512

vvlv HxHxH 22222 )1(

Ans.


28/41




conditions:

(b) P1= 150 psia, T1= 500F, and P2= 50 psia.

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

H1= 1274.3 Btu/lbm, S1= 1.6602 Btu/lbm.R

At 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

Ans.


29/41

Example 5A 1.5 m3tank contains 500 kg of liquid water in equilibrium with pure water

vapor, which fills the remainder of the tank. The temperature and pressureare 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


30/41




70





At the end of the process, the tank still contains saturated liquid and

saturated vapor in equilibrium at 100C and 101.33 kPa.

Cvaporsaturatedkg

kJHvcv

[email protected]

Cliquidsaturatedkg

kJHlcv

[email protected]

kJH [email protected]

From the steam table, the specific volumes of saturated liquid

and saturated vapor at 100C are 0.001044 and 1.673 m3/kg ,

respectively.


31/41




70





kgmv

7772.0673.1

)001044.0)(500(5.11

vl mmm 222 7507772.0500

lv mm 22 001044.0673.15.1

kJHmHmHm vvllcv 524458)( 222222

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

kgml

65.12502 kgmv

116.02


32/41




70





kJHmHmHm vvllcv 211616)0.2676(673.1

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

vl mmm222

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 Ans.


33/41

Generalized property correlation for

gasses

rcPPP rcTTT

P

P

R

P

dP

T

ZT

RT

H

0

PP

P

R

P

dPZ

P

dP

T

ZT

R

S

00)1(

r

r

P

r

r

Pr

r

c

R

P

dP

T

ZT

RT

H

0

2

rr

r

P

r

rP

r

r

Pr

r

R

P

dPZ

P

dP

T

ZT

R

S

00)1(

10 ZZZ

r

r

r

r

P

r

r

Pr

r

P

r

r

Pr

r

c

R

P

dP

T

ZT

P

dP

T

ZT

RT

H

0

12

0

02

rr

r

rr

r

P

r

rP

r

r

Pr

r

P

r

rP

r

r

Pr

r

R

P

dPZ

P

dP

T

ZT

P

dPZ

P

dP

T

ZT

R

S

0

1

0

1

0

0

0

0

)1()1(

;


34/41


gasses

c

R

c

R

c

R

RT

H

RT

H

RT

H 10 )()(

rr

r

rr

r

P

r

rP

r

r

Pr

r

P

r

rP

r

r

Pr

r

R

P

dPZ

P

dP

T

ZT

P

dPZ

P

dP

T

ZT

R

S

0

1

0

1

0

0

0

0

)1()1(

c

R

RT

H 0)(

c

R

RT

H 1)(

r

r

r

r

P

r

r

Pr

r

P

r

r

Pr

r

c

R

P

dP

T

ZT

P

dP

T

ZT

RT

H

0

12

0

02

c

R

c

R

c

R

RT

S

RT

S

RT

S 10 )()(

Lee/Kesler correlation


35/41


gasses

10 BBRT

BPB

C

C

Generalized second-virial coefficient mixing rule

rrr dT

dB

dT

dB

dT

Bd 10

dT

dB

R

P

RT

SR

dT

dB

T

B

R

P

RT

HR

r

r

r

r

c

R

dT

dBTB

dT

dBBP

RT

H 1

10

0

rr

r

R

dT

dB

dT

dBP

R

S 10

6.1

0 422.0083.0

rTB 2.4

1 172.0139.0

rTB


36/41


gasses

Rig

HHH

r

r

r

r

c

R

dT

dBTB

dT

dBBP

RT

H 1

10

0

rr

r

R

dT

dB

dT

dBP

R

S 10

RRT

T

ig

P

HHdtCH12

2

1

Rig SSS RR

T

T

ig

P SSP

PR

T

dTCS 12

1

2ln2

1

RR

H

ig

P HHTTCH 1212 )(

RR

S

ig

P SSP

PR

T

TCS 12

1

2

1

2 lnln


37/41

ExampleEstimate U, H, and S for 1-butene vapor at 200C and 70 bar if H and Sare set

equal to zero for saturated liquid at 0C. Assume the only data available are:

KTc 0.420 KTn 9.266 191.0barPc 43.40

263 10837.910630.31967.1 TTR

CigP

(a) Vaporization at T1and P1=Psat

(b) Transition to ideal gas state at

T1and P1

(c) Change to T2 and P2in the ideal

gas state(d) Transition to the actual final

state at T2 and P2


38/41

Example 6Estimate U, H, and S for 1-butene vapor at 200C and 70 bar if H and Sare set


KTc 0.420 KTn 9.266 191.0barPc 43.40

263 10837.910630.31967.1 TTR

CigP

(a) Vaporization at T1and P1=Psat

T

BAPsat ln

From normal boiling point and critical point P and T

9.2660133.1ln

BA

42043.40ln

BA

1260.10A

11.2699Bat 0C,

15.273

11.26991260.10ln satP barPsat 2771.1


39/41



KTc 0.420 KTn 9.266 191.0barPc 43.40

263 10837.910630.31967.1 TTR

CigP

nr

c

n

n

T

P

RT

H

930.0

)013.1(ln092.138.0

1

1

nr

r

lv

n

lv

T

T

H

H

nr

c

n

n

T

P

RT

H

930.0

)013.1(ln092.1)9.266)(314.8(

636.0930.0

)013.143.40(ln092.1

nH

molJ/137,2238.0

1

1

nr

r

lv

n

lv

T

T

H

HmolJHlv /810,21

lvlv STH KmolJSlv 15.273/810,21 1 KmolJSlv /83.78


40/41



KTc 0.420 KTn 9.266 191.0barPc 43.40

263 10837.910630.31967.1 TTR

CigP

molJ/485,8c

R

c

R

c

R

RT

H

RT

H

RT

H 10 )()(

c

R

c

R

c

R

RT

S

RT

S

RT

S 10 )()(

)9.266)(314.8()713.0)(191.0()294.2(2 RH

KmolJ /18.14

)9.266)(314.8()726.0)(191.0()566.1(2 RS

(d) Transition to the actual final state at T2

and P2

Use Lee/Kesler correlations


41/41



KTc 0.420 KTn 9.266 191.0barPc 43.40

263 10837.910630.31967.1 TTR

CigP

Finally,

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

KmolJS /72.8818.1418.22)88.0(84.78

PVHU

molJJbarcm

/218,3210

)8.287)(70(233,34

13

1 -Thermodynamics Properties of Fluids

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