chapter 3 thermodynamics properties of fliuds (part 3)

8/8/2019 Chapter 3 Thermodynamics Properties of Fliuds (Part 3)

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Chapter 3

Thermodynamics Properties

of Fluids

Chemical EngineeringThermodynamics


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3.7 Generalized property

correlations for gases

0

x

x!

P

P

R

P

dP

T

ZT

RT

HP = PcPr T = Tc Tr

dP = Pc

dPr

dT = Tc

dTr

The resulting equations are :

r

r

r

r

c

dr

r

xx

0

2

r

rP

r

rP

Pr

r

R

P

dPZ

P

dP

T

ZT

R

S rr

r

x

x!

001

xxP P

P

R

PdPZ

PdP

TZT

RS

0 01


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The correlation forZ is based on equation:

Z = Z0 + Z1

Differentiation yields :

rrr PrPrPr

T

Z

T

Z

T

Z

x

x

x

x

x

x 10

[

Substitution for Z andrP

rT

Z

x

x:

r

rP

Pr

r

r

rP

Pr

r

c

R

PdP

TZT

PdP

TZT

RTH r

r

r

r

xx xx! 01

2

0

0

2 [

r

rP

Pr

r

r

rP

Pr

r

R

P

dPZ

T

ZT

P

dPZ

T

ZT

R

S r

r

r

r

-

xx

-

xx

!0

11

0

00

1 [


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The first integrals on the right sides ofthese two equations may be evaluates

numerically or graphically for various

values of Tr and Pr from the data forZ0

given in Tables E.1 and E.3.

The integrals which follow in each

equation may be similarly evaluatedfrom the data forZ1 given in Tables E.2

and E.4 .


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Alternatively, the evaluated may be

based on an equation of state; Lee and

Kesler used a modified form of the

Benedict/Webb/Rubin equation of state

to extend their generalized correlation to

residual properties .

c

R

c

R

c

R

RT

H

RT

H

RT

H10

[!

R

S

R

S

R

S RRR10

[!


7/36

Figure 6.5 : The Lee/Kesler correlation for

c

R

RTH

0

as function of Trand Pr


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Calculated values of the quantities

RS

RS

RTH

RTH

RR

c

R

c

R 1010

,,,

as determined by Lee and Kesler are

given as functions ofTrand Pr in TableE.5 until Table E.12.


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-

r

r

r

rr

c

R

dT

dBTB

dT

dBTBP

RT

H1

10

0 [

The generalized second virial-coefficient

correlation valid at low pressures formsthe basis for analytical correlations of the

residual properties.

! rr

r

R

dT

dB

dT

dB

PR

S 10

[


10/36

2.5

1 722.0

rr TdT

dB!

where

6.10 422.0083.0

rTB !

6.2

0

675.0rr TdT

dB !

2.4

1 172.0139.0rT

B !


11/36

The generalized correlations forHRand

SR , together with ideal-gas heatcapacities, allow calculation of enthalpy

and entropy values of gases at any

temperature and pressure.

!2

0202

T

T

Rig

P

ig HdTCHH

!1

0101

T

T

Rig

P

ig HdTCHH


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The enthalpy change for the process,

H = H2 H1 , is difference between these

two equations :

RT

T

Rig

P HHdTCH 122

1

!( Similarly , for entropy :

!(2

112

1

2lnT

T

RRig

P SSP

PR

T

dTCS


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Alternative form :

RRH

ig

P HHTTCH 1212 !(

RR

S

ig

P SSP

PR

T

TCH 12

1

2

1

2 lnln !(


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Example 6.9

Estimate V, U, H,and S for 1-butene

vapor at 200C and 70 bar ifHand S are

set equal to zero for saturated liquid at

0C. Assume that the only data availableare:

Tc= 420.0K Pc = 40.43 bar = 0.19Tn = 266.9 K (normal boiling point)


15/36

Solution 6.9The volume of 1 -butene vapor at 200C

and 70 bar is calculated directly from the

equation V= ZRT/P,where Zis given by

Eq. (3.57) with values of Z and Z1

interpolated in Tables E.3 and E.4. For the

reduced conditions;

127.10.420

15.273200!

!rT 731.1

43.40

70!!rP


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The compressibility factor, Z is:

Z =Z0+ Z1= 0.485 + ( 0.19l )( 0.142 )= 0.512

Whence,

138.28770

15.47314.83512.0 !!! molcmP

ZRTV


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ForHand S , use a calculation path like

that ofFig. 6.7, leading from an initial stateof saturated liquid 1-butene at 0C, where

Hand S are zero, to the final state of

interest. In this case, an initial vaporization

step is required, leading to the four-steppath. The steps are:

(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

).


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Step (a): Vaporization of saturated liquid

1-butene at 0C. The vapor pressure must

be estimated, since it is not given. One

method is based on the equation:

T

BAPsat !ln

The vapor-pressure curve contains both the

normal boiling point, for which Psat= 1.0133bar at 266.9 K. and the critical point, for

which Psat= 40.43 bar at 420.0 K.


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9.2660133.1ln

BA !

0.42043.40ln

BA !

Simultaneous solution of these two

equations yields:

A= 10.1260 B = 2699.11

Using A and B values, for 0C (273.15 K),

Psat = 1.2771 bar, a result used in steps (b)

and (c).


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Here, the latent heat of vaporization isrequired. Equation (4.12) provides an

estimate at the normal boiling point, where

Trn = 266.9/420.0 = 0.636:

979.9

636.0930.0

013.143.40ln092.1

930.0

013.1ln092.1!

!

!

(

nr

c

n

lv

n

T

P

RT

H

Whence,Hlvn= ( 9 .979 )( 8.314 )( 266.9 )= 22,137 J mol-1


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The latent heat at 273.15 K, orTr =

273.15/420.0 = 0.650,is given by Eq.(4.13):

Hlv= (22,137)(0.350/0.364)0.38

= 21,810 J mol-1

By Eq. (6.70),

1184.79

15.273810,21

!!(!( KJmolT

HSlv

lv

38.0

1

2

1

2

1

1

!

(

(

r

r

T

T

H

H

EFEF

STH (!(


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Step (b): Transformation of saturated-

vapor 1 -butene into an ideal gas at the

initial conditions (T1, P1). Since the

pressure is relatively low, the values of

HR1 and SR

1 are estimated by Eqs. (6.87)

and (6.88) for the reduced conditions, Tr=0.650 and Pr = 1.2771/40.43 = 0.0316.

-

!

rr

rrr

c

R

dT

dBTB

dT

dBTBP

RT

H1

10

0 [

!

rr

r

R

dT

dB

dT

dBP

R

S 10[


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From computer:

HRB (0.650,0.031 6,0.191) = -0.0985SRB (0.650,0.0316,0.191) = -0.1063

HR1= (-0.0985)(8.314)(420.0)

= -344 J mol-1

SR1=(-0.l063)(8.314)

= -0.88 J mol-1 K-1

As indicated in Fig. 6.7, the property changes

for this step are -HR1 and -SR

1 because the

change is from the real to the ideal-gas state.


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Step (c): Changes in the ideal-gas state

from (273.15 K, 1.2771 bar) to (473.15 K.

70 bar). Here, Hig and Sig are given by

Eqs. (6.95) and (6.96):

!!(2

112

T

T

ig

P

igigig dTCHHH

!!(2

11

212 ln

T

T

ig

P

igigig

P

PRT

dTCSSS


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1118.222771.1

70ln314.8474.55 !!( KJmolSig

From computer,

8.314 x ICPH(273.15,473.15;1.967,31 .630E-3,-

9.837E-6,0.0)

= 20,564 Jmol-1

8.314 x ICPS(273.15,473.15;1.967,31 .630E-3,-

9.837E-6,0.0)

= 55.474 Jmol-1 K-1

1

Jmol56

4,20

!(

ig

H


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Step (d): Transformation of 1-butene from

the ideal-gas state to the real-gas state at

T2and P2. The final reduced conditions

are: Tr= 1.127 Pr= 1.731

At the higher pressure of this step, HR2 and

SR2 are found by Eqs. (6.85) and (6.86),together with the Lee/Kesler correlation.

With interpolated values from Tables E.7,

E.8, E.11, and E.l2, these equations give:

430.2713.0191.0294.22 !!c

R

RT

H

705.1726.0191.0566.12 !!c

R

RT

S


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Whence,

1

2 485,80.420314.8430.2

!!Jmol

H

R

112 18.14314.8705.1 !! KJmolSR

The sums of the enthalpy and entropy

changes for the four steps give the totalchanges for the process leading from the

initial reference state (where Hand S are set

equal to zero) to the final state:H = H = 21,810- (-344) + 20,564 - 8,485= 34,233 Jmol-1

S = S = 79.84- (-0.88) + 22.18 - 14.18

= 88.72J

mol-

1 K-1


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The internal energy is: 1

13218,32

10

8.28770233,34

!!! Jmol

barJcmPVHU

These results are in far better agreementwith experimental values than would have

been the case had we assumed

1-butene vapor an ideal gas.


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31/36

Example 6.10

Estimate V, HR,and SR for an equimolar

mixture of carbon dioxide(1) and

propane(2) at 450 K and 140 bar by the

Lee/Kesler correlations.


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Solution 6.10

The pseudocritical parameters are found

by Eqs. (6.97) until (6.99) with criticalconstants from Table B.1:

|i

iiy [[ |i

icipcTyT |

iicipc

PyP

KTyTyT ccpc 0.3378.3695.02.3045.02211 !!! barPyPyP ccpc 15.5848.425.083.735.02211 !!!

188.0152.05.0224.05.02211 !!! [[[ yy


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Whence.

335.10.337

450!!prT 41.2

15.58

140!!prP

Values ofZ0and Z1 from Tables E.3 and

E.4 at these reduced conditions are:

Z0= 0.697 and Z1= 0.205

Eq. (3.57) yields:

Z= Z0+ Z1= 0.697 + (0.188)(0.205) = 0.736

10 ZZZ [!


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Whence,

137.196140

45014.83736.0 !!! molcm

P

ZRTV

From Tables E.7 and E.8,

730.1

0

!

pc

R

RT

H

169.0

1

!

pc

R

RT

H

Substitution into Eq. (6.85) gives:

762.1169.0188.0730.1 !!pc

R

RT

H

c

R

c

R

c

R

RT

H

RT

H

RT

H10

[!


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Whence,

HR= (8.314)(337.0)(-1.762) = -4,937 J mo1-1

By Tables E.11 and E.12 and Eq. (6.86),

SR/R= -0.967 + (0.188)( -0.330) = - 1.029

Whence,SR= (8.314)(-l.029) = -8.56J mol-1 K-1

R

S

R

S

R

SRRR 10

[!


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The End

chapter 3 thermodynamics properties of fliuds (part 3)

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