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8/8/2019 Chapter 3 Thermodynamics Properties of Fliuds (Part 3)
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Chapter 3
Thermodynamics Properties
of Fluids
Chemical EngineeringThermodynamics
8/8/2019 Chapter 3 Thermodynamics Properties of Fliuds (Part 3)
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8/8/2019 Chapter 3 Thermodynamics Properties of Fliuds (Part 3)
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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
[!
8/8/2019 Chapter 3 Thermodynamics Properties of Fliuds (Part 3)
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Figure 6.5 : The Lee/Kesler correlation for
c
R
RTH
0
as function of Trand Pr
8/8/2019 Chapter 3 Thermodynamics Properties of Fliuds (Part 3)
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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
[
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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 !
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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)
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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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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