advanced thermodynamics note 5 thermodynamic properties of fluids lecturer: 郭修伯
TRANSCRIPT
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
Cliquidsaturatedkg
kJH l
Cvaporsaturatedkg
kJH v
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