Chapter 6 Calculation of Properties of Pure...
143
1 Chapter 6 Calculation of Properties of Pure Fluids 2012/4/13
Transcript of Chapter 6 Calculation of Properties of Pure...
1
Chapter 6Calculation of Properties of Pure Fluids
2012/4/13
2
Thermodynamic properties calculations from PVT and heat capacity data
Thermodynamics properties depend on the state of the system.The state of the system can be determined by the Gibbs phase ruleMeasurable properties: the temperature, the pressure and some heat capacities such as Cp and Cv.All the derived properties such as the entropy can be calculated using the thermodynamic relations
2012/4/13
3
6.1 Some Mathematical Preliminaries
( )1 2
Thermodynamic state variable: , , , , , , , and For the single-component, one-phase system, the degree of freedom
is equal to 2 (i.e., 2 1 1 2 2).Thus, any intensive property, = , .
P T V S U H A G
F F C Pfθ θ θ
= − + = − + =
( )
( ), , ,
(A) The chain rule of differentiation.
Suppose , , then
Suppose , , , then
V T
V N T N T V
U UU f T V dU dT dVT V
U U UU f T V N dU dT dV dNT V N
∂ ∂⎛ ⎞⎛ ⎞= = +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Example: F of boiling the drinking water in a kettle = 1-2+2 = 1
4
( )
(B) The triple product rule
1
Suppose , ,
For constant, 0,
Rearrang
Z
Z Y X
Z Y
Y
X Z YY X Z
X XX f Y Z d X dY d ZY Z
X XX d X dY d ZY Z
∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∂ ∂⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
∂ ∂⎛ ⎞ ⎛ ⎞= = = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
ement and one gets Z X Y
X dY XY d Z Z
∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Already Proved it using open system in “Preface to Ch 6”.
5
6.2 The evaluation of thermodynamic partial derivatives
Define four thermodynamic properties
constant-volume heat capacity
constant-pressure heat capacity
(6.2-1)
(6.2-2)
1
VV
PP
U CT
H CT
∂⎛ ⎞ = =⎜ ⎟∂⎝ ⎠
∂⎛ ⎞ = =⎜ ⎟∂⎝ ⎠
coefficient of thermal expansion
1 isothermal compressibility
(6.2-3)
(6.2-4)
P
TT
VV T
VV P
α
κ
∂⎛ ⎞ = =⎜ ⎟∂⎝ ⎠∂⎛ ⎞− = =⎜ ⎟∂⎝ ⎠
6
( ),
then (6.2-9
Also,
a)
(6.2-9 b)
V S
V
S
dU Td S PdVU f S V
U UdU d S dVS V
UTS
UPV
= −
=
⎛ ⎞ ⎛ ⎞∂ ∂= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠
⎛ ⎞∂− = ⎜ ⎟∂⎝ ⎠
Partial derivativesFor close, reversible, no KE,PE,friction
7
( = )
(6.2-5) (6.2-9a) (6.2-9b)
(6.2-6b) (6.2-10a) (6.2-10b)
(6
.2-7b
( )
V S
SP
dU Td S PdV U UT PS V
H HT VS P
d H Td S VdP
d A S
U H PV
U A T S
dT PdV
⎛ ⎞ ⎛ ⎞∂ ∂= − =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞∂ ∂⎛ ⎞= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠
=
−
=
= +
= − −
+
−
) (6.2-11b) (6.2-11a)
(6
( = )
.2-8b) (6.2-12b) (6.2-12a
( =
)
)
V T
P T
A AS PT V
G GS
G H T S
G A P
dG SdT VdP VT P
V
⎛ ⎞∂ ∂⎛ ⎞− = − = ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
∂ ∂= − + ⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠
+⎝ ⎠
−
Basic relations
4 total derivatives
8 Partial derivatives
8
( )2 2
Maxwell relationsFor an exact function , , the following relation hold,
or
For thermodynamic relation as follows,
W
yx y x
f x y
x y y x x y y x
dU Td S PdV
θ
θ θ θ θ
=
⎡ ⎤⎡ ⎤⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞= = ⎢ ⎥⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎢ ⎥⎝ ⎠ ⎢ ⎥⎣ ⎦ ⎣ ⎦
= −
( )hen ,
And since and
One has
This is one
(6.2-13)
of
V SS V
V S
S V
U f S V
U UV S S V
U UT PS V
T PV S
=
⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂=⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
⎛ ⎞ ⎛ ⎞∂ ∂= − =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞∂ ∂= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
the Maxwell relations
Maxwell relation (commutative property)
9
Maxwell relations
Maxwell relations
; UdU Td S PdV TS
⎛ ⎞∂= − = ⎜ ∂⎝
(6.2-9a); (6.2-9b) ; (6.2 -13)
=
(6.2-6b) (6.2-10a); (6.2-10b) ;
;
V S S V
SP
U T PPV V S
H Hd H Td S VdP
U H PV
T VS P
⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂− = = −⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞∂ ∂⎛ ⎞= + = =
−
⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠
(6.2-14)
(6.2-7b) (6.2-11b); (6.2-; 11a) ; (A1 )
S P
V VT T
S
T VP S
A Ad A SdT PdV P
U A T S
S PT V V T
⎛ ⎞∂ ∂⎛ ⎞ = ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞∂ ∂⎛ ⎛ ⎞∂ ∂⎛ ⎞⎞= − − − = − = ⎜ ⎟⎜ ⎟∂ ∂⎝=⎜ ⎟ ⎜ ⎟∂ ∂⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
= +
(6.2-15)
=
(6.2-8b) (6.2-12b); (6.2-1
; 2a) ; (6.2-16)
=
) (B1P T T P
G GdG SdT V S VG
P T
H T S
G
dP S VT P
A PV
∂ ∂⎛ ⎞ ⎛ ⎞= − + − = =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟
−
+
⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
8 Partial derivatives
4 Total derivatives
4 commutative derivatives
Graphical Plot of Basic Relations
i i
i i
i i
i i
dU TdS PdV dn
dH TdS VdP dn
dA SdT PdV dn
dG SdT VdP dn
μ
μ
μ
μ
= − +
= + +
= − − +
= − + +
∑∑∑∑
Total internal energy (U) is an essential thermodynamic function. All energies are derived from it. H, A, G are additional primary thermodynamic properties. After Legendre's transformation of U, all can be expressed in canonical variables (V, T, S, P, ni).
V T
*: Independent state variables in corner.*: Dependent properties in between of state variables.*: Properties (A/U/H/G) related to variables (V/T/S/P).
S: state variable indirectly measured by Cp and Cv
, , , ,, , ,,
partial molar Gibbs energy
: those values are not partial property themself.j i j i j i j i
i i iT V n S Pi i
i T P nn V S n
G Gn
No
A H Un
t
n n
e
μ≠ ≠≠ ≠
⎛ ⎞∂= = =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝
=⎠
⎜ ⎟∂⎝ ⎠
OPEN system
11
Two more relations
; Also Divide by
;
And let the op
dU Td S PdV d H TdS VdPdT
dU dS dV d H d S dPT P T VdT dT dT dT dT dT
= − = +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
eration at constant ; And let the operation at constant ,
; V V P P
VV
V P
V PU S dV H S dPT P T VT T T T T T
U SC TT T
∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∂ ∂⎛ ⎞ ⎛= =⎜ ⎟ ⎜∂ ∂⎝ ⎠ ⎝;
(A2) = (B 2)
PV P
V P
V P
P
C CS ST T T
H SC TT T
T
∂ ∂⎞ ⎛ ⎞ ⎛ ⎞= =⎟ ⎜ ⎟ ⎜ ⎟
∂ ∂
∂ ∂⎠ ⎝
⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜
⎠
⎟∂
⎝
∂⎝ ⎠ ⎝ ⎠
⎠
12
Four important relations
=
(
A1) (A2)
(B1) (B2)
V
V VT
P
T P P
CS P SV T T T
S V S CP T T T
∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠
∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠
13
( ),S f T V=
( ),
(6.2-7)
And
(6.2-18
)
(6.2-15 )
V T
V
V
VT
V
S f T V
S Sd S dT dVT V
CST T
S PV T
C Pd S dTT T
=
∂ ∂⎛ ⎞⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠
∂ ∂⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠
∂⎛= +∂
(6.2-19)V
dV⎞⎜ ⎟⎝ ⎠
14
( ),S f T P=( ),
= and
And
(6 .2 2 - 0)
P T
P
P T P
P
P
S f T P
S SdS dT dPT P
S C S VT T P T
C VdS dT dPT T
=
∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠
( ) , ?What is the relationship of S f P V=
ideal fluid value at constant ?PV
CWhat is the S and TC
; VP
P P P P V V V V
VP
P V P V
CCS S T T S S T TV T V T V P T P T P
CCS S T Td S dV dP dV dPV P T V T P
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= = = =
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ POP Q
UIZCh 4.
4-4
15
(6.2-21
)
(6.2-
VV
V
V
V
T
dU TdS PdV T PdV
PdU C dT
C PdT dV
T P dVT
U PT PV T
T T⎡ ⎤∂⎛ ⎞+⎢ ⎥⎜ ⎟∂⎝ ⎠⎢ ⎥⎣
= − = −
⎡ ⎤∂⎛ ⎞= + −⎢ ⎥⎜ ⎟∂⎝ ⎠⎢ ⎥⎣ ⎦⎛ ⎞∂ ∂⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠
⎦
23)
(6.2-
22)
(6 )
.2-24
P
P
PP
T P
d H Td S VdP T VdP
Vd H C dT V T dP
C VdT dPT
T
H VV TP T
T= + = +
⎡ ⎤∂⎛ ⎞= + −⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂
⎡ ⎤∂⎛ ⎞−
∂⎝ ⎠ ⎝
⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦
⎠
U(T, V) H(T, P)
Can I have U in terms of changing P at constant T? Answer is NO
it is the same with 6.2-21VT P V T
T T T V T
U U P VdU dP dT C dT T P d PP T T P
U U V P VT PP V P T P
⎡ ⎤∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = + −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
16
ILLUSTRATION 6.2-1Showing That Similar Partial Derivatives with Different State Variables Held Constant Are Not Equal. Obtain expressions for difference of two derivatives (∂U/∂ T)V - (∂U/∂T)P , show that they are not equal.
Starting from Eq. 6.2-21,
Using Eq. 6.1-7 yields
Again starting with Eq. 6.2-21, we obtai
Solution
n
VV
V VV V
P
V
PdU C dT T P dVT
U PC T P CT T
U C
V
T
T
⎡ ⎤∂⎛ ⎞= + −⎢ ⎥⎜ ⎟∂⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞= + −⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠
(6.2-25)
0
VV
V PV P
P
PT PT
U U P VT PT T T T
VT
⎡ ⎤∂⎛ ⎞+ −⎢ ⎥⎜ ⎟∂⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣
∂⎛ ⎞⎜ ⎟∂
⎦
⎝ ⎠
≠
17
The Joule-Thomson coefficient
0
(6.2-26)
(6.2-2 ) 7
H
PP
P
H P
TP
Vd H C dT V T dPT
VV TTT
P C
μ
μ
∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠
⎡ ⎤∂⎛ ⎞= = + − ⎜ ⎟⎢ ⎥∂⎝ ⎠⎣ ⎦⎡ ⎤∂⎛ ⎞− ⎜ ⎟⎢ ⎥∂∂ ⎝ ⎠⎛ ⎞ ⎣ ⎦= −⎜ ⎟∂⎝ ⎠
=
18
Partial derivation of Gwith respective to T
( )( ) ( )
( )( ) ( )
2 2 2
2 2 2
22
1 1
1 1
1 1 1;
(
6.2-28)
(6.2-29
or
)
P
P
V
V
GH T ST G G HS
T T T T T T T
AU T ST A A US
T T T T T T T
d dT dT T dT
y xdydxT T
⎛ ⎞∂ −⎜ ⎟ ∂⎛ ⎞= − = − − = −⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎜ ⎟⎝ ⎠
⎛ ⎞∂ −⎜ ⎟ ∂⎛ ⎞= − = − − = −⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞= − = −−⎛ ⎞ =⎜ ⎟
⎝ ⎠⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( ) ( ) ( )
( ) ( )
22
22
2
hence 1 1
h1
(6.2-30)
P P P
V V
G G GT T TH H
T TT d dT T
A AT T U
ydx
T TT dT
x
⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = − =⎜ ⎟∂ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ ⎞⎛ ⎞∂ ∂⎜ ⎟⎜ ⎟ ⎜ ⎟= = −⎜ ⎟∂ ⎛ ⎞⎜ ⎟⎜ ⎟ − ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
( )(6.2-3ence
10)
V
AT U
dT
⎛ ⎞∂⎜ ⎟
⎜ ⎟ =⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
19
Illustration 6.2-2Showing Again That Similar Partial Derivatives with Different State
Variables Held Constant Have Different Values, CP - CV= ??
this EQ divided by at constant P, the
Starting
(6.2-31) he
n:
re sh
from Eq. 6.2-19,
o
w:
V
V
VP V
P V P
C Pd S dT dVT T
CS P V PC C TT T T T T
dT∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠
∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2
Now using the triple-product rule,
1
(6.2-32)
(6.2-33a )
V P
V T P
VP
P PT
VP
VT
P V TT P V
CC S P VT T T V T
CC VorT T P
∂⎛ ⎞⎜ ⎟∂⎝ ⎠
⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞ = −⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞= = − ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠
∂⎛= − ⎜ ∂⎝
2
(6.2-3 3b ) T V
PT
∂⎞ ⎛ ⎞⎟ ⎜ ⎟∂⎠ ⎝ ⎠
VP
P V
CC S ST T T T
∂ ∂⎛ ⎞ ⎛ ⎞= ≠ =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
20
( )
2
2
2
1
(6.2-33b)
(6.2-34)
Compare this with
VP
PT
VP
T
VP
T
V
V
P V VV P
CC P VT T V T
CC VT T V
CC VT T
CST T
P VC C T CT T
ακ
ακ
⎛ ⎞∂ ∂⎛ ⎞= − ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠
⎛ ⎞= − ⎜ ⎟−⎝ ⎠
= +
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠
∂ ∂⎛ ⎞ ⎛ ⎞= + = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
2
2 2
2
(6.2-35)
:
PT
V VT V T
P VV P PT
P VTV T
V P VC T C TP T
In Summary
P V P VC C T TT T V T
ακ
⎛ ⎞∂ ∂⎛ ⎞⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠
∂ ∂⎛ ⎞ ⎛ ⎞= − = +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠−
⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = = − ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
1V P TT V PP VT
⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠
1 coefficient of thermal expansion
1 isothermal compressibility
P
TT
VV T
VV P
α
κ
∂⎛ ⎞ = =⎜ ⎟∂⎝ ⎠∂⎛ ⎞− = =⎜ ⎟∂⎝ ⎠
21
Maxwell relations (commutative property)
2
2
;
Maxwell relation
VV
VV VT
VV
VT V
V
PdU C dT T P dVT
U U PC T PT V T
PT PTC P P PT
V T T T T
⎡ ⎤∂⎛ ⎞= + −⎢ ⎥⎜ ⎟∂⎝ ⎠⎢ ⎥⎣ ⎦∂ ∂ ∂⎛ ⎞⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠
⎡ ⎤⎡ ⎤∂⎛ ⎞∂ −⎢ ⎥⎢ ⎥⎜ ⎟∂⎝ ⎠ ⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞ ⎢ ⎥ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎣ ⎦= = + −⎜ ⎟⎜ ⎟ ⎜⎜ ⎟ ⎢ ⎥∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦
2
2
2
2
(6.2-36)
Similarly,
(6.2-37 )
V V
P
T P
PTT
C VTP T
⎛ ⎞∂= ⎜ ⎟⎟ ∂⎝ ⎠
⎛ ⎞∂ ∂⎛ ⎞ = − ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
22
23
U(T, V) H(T, P)
A(T, V) G(T, P)
6.2-23
6.2-24
μ =
Versus EQ(8.2-1) for multi-component system:
, , , ,, , ,,
partial molar Gibbs energy
: those values are not partial property themself.j i j i j i j i
i i iT V n S Pi i
i T P nn V S n
G Gn
No
A H Un
t
n n
e
μ≠ ≠≠ ≠
⎛ ⎞∂= = =
⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝
=⎠
⎜ ⎟∂⎝ ⎠
24
Illustration 6.2-3Use of Partial Derivative Interrelations to Obtain Useful ResultsDevelop expressions for the coefficient of thermal expansion α, the isothermal compressibility κT, the Joule-Thomson coefficient μ, and the difference CP-CV for (a) the ideal gas and (b) the gas that obeys the volumetric equation of state
where a and b are constants. (This equation of state was developed by J. D. van der Waals in 1873, and fluids the obey this equation of state are called van der Waals fluids.
2 (6.2-38a)RT aPV b V
= −−
25
[ ]
Solutiona. For
1 1 1 so that = = =
1 1
the id
so that
eal
:
1
g s
F
a
P P
TT T
P
H P P
PV RTV R V V VT P T V T V T TV V VP P V P P
for real gas
VV TTT V T
P C C
α
κ
μ α
=
∂ ∂⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
−
⎡ ⎤∂⎛ ⎞− ⎜ ⎟⎢ ⎥∂∂ ⎝ ⎠⎛ ⎞ ⎣ ⎦= = − −⎜ ⎟∂⎝ ⎠ =-
[ ]* *
* *
or the ideal gas, 11 1 0
H P P
P V
T V VT TP C C T
C C R
μ α∂⎛ ⎞ ⎡ ⎤= − − = − − =⎜ ⎟ ⎢ ⎥∂⎝ ⎠ ⎣ ⎦
= +
=
26
( )
( )
( )( ) ( ) ( ) ( ) ( ) ( )
( )
2 3
1
1 12 2 2
2
An expression for is obtained as follows:
2
2 2
1Consequently, 1 12
T
T
TT
T
H P P
P RT aV VV b
PVV
RTV a R TV a RV bV b V b V bV RVV b
T V VTP C C V a V b
V b RT V
κ
κ
κ α
μ α
−
− −
⎛ ⎞∂= − +⎜ ⎟∂ −⎝ ⎠
⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠
⎡ ⎤= − = − − =⎢ ⎥
− − −− ⎢ ⎥⎣ ⎦
∂⎛ ⎞= − − = − −⎜ ⎟∂ ⎛ −⎝ ⎠ −− ⎝
=
( ) ( )
22
2
3
and 21
P V V VT
TV R RC C C TV CV b V ba
RT V
α ακ α
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎞⎢ ⎥⎜ ⎟⎢ ⎥⎠⎣ ⎦
⎛ ⎞= + = + = +⎜ ⎟⎜ ⎟− −⎝ ⎠ −
-1 ( ) = VP
TV
α ⎛ ⎞∂⎜ ⎟∂⎝ ⎠
27
Illustration 6.2-4Showing that a Fundamental Equation of State Contains All the Information About a FluidShow that from an equation of state relating the Gibbs energy, temperature, and pressure, equations of state for all other state functions (and their derivatives as well) can be obtained by appropriate differentiation.
28
( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
Suppose we had an equation of state of the form
, and ,
, = , + , ,
, = , + , ,
,
,
P T
P
P
G GS T P V T PT P
GH T P G T
G G T P
P T S T P G T P TT
G GU T P G T P T S T P G T P T PT
PV T PP
∂ ∂⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠
∂ ∂⎛ ⎞ ⎛ ⎞=
=
− −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠−
( ) ( ) ( ), = , ,
T
T
GA T P G T P PV G T P PP
∂⎛ ⎞− = − ⎜ ⎟∂⎝ ⎠
29
( )( )
( ) ( )( )
( )( )
( )
2
2
2 2 12 2 2
2 2
2 2
2
,,
,
1
1
PP
P P
P
PV P
P TT
TT
T T
P T
GG T P TTH GC T P T
T T T
T V T G G GC T P C T TV P T T P P
G PVV P G P
GV T P
V T G P
κ
α
−
⎡ ⎤⎡ ⎤∂⎛ ⎞∂ −⎢ ⎥⎜ ⎟⎢ ⎥∂ ⎛ ⎞∂ ∂⎝ ⎠⎛ ⎞ ⎣ ⎦⎢ ⎥= = = − ⎜ ⎟⎜ ⎟ ⎢ ⎥∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
∂ ∂ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂= + = − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∂ ∂∂⎛ ⎞= − = −⎜ ⎟∂ ∂ ∂⎝ ⎠
∂∂⎛ ⎞ ∂ ∂= =⎜ ⎟∂ ∂ ∂⎝ ⎠
6.2-35
30
QUESTIONSWhy do we need two equations, a volumetric equation of state P= P(T, V) and a thermal equation of stale U = U(T, V), to define an ideal gas?2. Can you develop a single equation of state that would completely specify all the properties of an ideal gas? (See Problem 7.7.) Although fundamental equations of state are, in principle,
the most useful thermodynamic descriptions of any substance, it is unlikely that such equations will be available for all fluids of interest to engineers. In fact, frequently only heat capacity and PVT data are available; in Sec. 6.4 we consider how these more limited data are used in thermodynamic problem solving.
What is Thermodynamics database Kingdom?
It is Mollier, T-S, P-H, diagrams; data tables, monographs of all pure substances.
31
ILLUSTRATION 6.2-5Calculation of the Joule-Thomson Coefficient of Steam from Data in the Steam TablesUse the steam tables in Appendix A.III to evaluate the Joule-Thomson coefficient of steam at 600oC and 0.8 MPa.
32
( ) ( ) ( )
( ) ( )
o 3
o o
At 0.8 MP
Solution
ˆˆ
ˆ
ˆˆ
ˆ ˆC m /kg kJ/kg
500 0.4443 3480.60.5018
700 0.5601 360
a
0 36924.2
ˆ ˆˆ 700 C 500
99.
,
00
4
Cˆ7
P P
H P P
P
P
P
P
VV V TV T TTTP C C
HCT
T V H
H HHCT
μ
⎡ ⎤⎛ ⎞∂⎡ ⎤∂⎛ ⎞ −⎢ ⎥⎜ ⎟− ⎜ ⎟⎢ ⎥ ∂∂∂ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎣ ⎦ ⎣ ⎦= − = −⎜ ⎟∂⎝ ⎠
⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠
−⎛ ⎞∂= ≈⎜ ⎟∂⎝ ⎠
=
( )
( )o
kJ3924.2-3480.6kJkg 2.218
200 K kg K500 C= =
−
33
( ) ( )( )
( )
( )
3
o o 34
o
34
33
m0.5601-0.4443ˆ ˆˆ 700 C 500 C mkg 5.84 10200 K kg K700 500 C
ˆˆ
ˆ
m0.5018 600 273.15 5.84 10kg K m K3.66 10kJ kJ2.218
kg KSince 1 kJ = 1
P
P
H P
V VVT
VV TTT
P Cμ
μ
−
−
−
−⎛ ⎞∂≈ = = ×⎜ ⎟∂ −⎝ ⎠
⎡ ⎤⎛ ⎞∂−⎢ ⎥⎜ ⎟∂∂ ⎢ ⎥⎝ ⎠⎛ ⎞ ⎣ ⎦= = −⎜ ⎟∂⎝ ⎠
⎡ ⎤− + ×⎢ ⎥
⎣ ⎦= − = ×
-3 30 MPa m , and 1 MPa = 10 bar,K K3.66 0.366
MPa barμ = =
34
( )
( )
( )
o
o
Alternative methodkJˆ ; 600 C, kg
kJˆCkg
At 0.6 MPa, 500 3482.8600 3700.9
By interpolation we can find the temperature at 0.6 MPa at which
ˆ 0.6M,
0.8MPa 3699.41
P a
H
T HP
T H
H T
μ ∂⎛ ⎞= =⎜ ⎟∂⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
( )
( ) ( )( ) ( )
o
o
o
o o
kJ ˆ 600 C, kg
ˆ ˆ600 C,0.6MPa ,0.6MPa 3700.9 3699.4 600ˆ ˆ 3700.9 3482.8 600 500600 C,0.6MPa 500 C,0.6MPa
which gi
3699.41 0.8MPa
599.3023 Cves or . Therefore0.6977K
0.6977K
=0.2H
H
H H T T
TTT
H H
Pμ
Δ
= =
− − −= =
− −−
∂ −⎛ ⎞= ⎜ ⎟∂ −⎝ ⎠
= −=
K K3.49 0.349
MPa MPa bar= =
NOTE: Inversion temperature of a real fluid was occurred at Joule-Thomson coefficient = 0
35
Illustration 6.2-6Calculation of Cp for waterThe constant-pressure heat capacity of liquid water at 25 oC at 1 bar is 4.18 J/(g K). Estimate the heat capacity of liquid water
at 25 oC and 100 bar.
4 -1 6 -
3
2
SolutionData:
ˆ1 1 2.56 10 K , ˆ
ˆand
9.6 10 K (
1.003 cm /
g v n)i e
g
P P P
V VV V TT T
V
αα − −∂⎛ ⎞ =⎛ ⎞∂ ∂⎛ ⎞= = = × ×⎜⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎟∂⎝ ⎠
=
36
2
2
2
2
2
2
From Eq. 6.2-37, we haveˆ ˆ
= or equilivalent =
ˆ ˆˆ or ˆ=
ˆ =
ˆ
PP
P P
T PT
P P
P
T P
C C VT TP P T
VT
V VT
V VVT T
C VTP T
Tα
α
αα
⎛ ⎞ ⎛ ⎞∂ ∂ ∂⎛ ⎞ − −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞∂ ∂=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞∂ ∂⎛ ⎞ − +⎜ ⎟⎜ ⎟
⎛ ⎞∂⎜ ⎟∂⎝ ⎠
⎛ ⎞∂ ∂⎛ ⎞+⎜ ⎟ ⎜
∂ ∂⎝ ⎠ ⎝ ⎠
⎟∂ ∂⎝ ⎠⎝ ⎠
2
38 6 -2
33 4
3
ˆ ˆ ˆ
cm298.15 K 1.003 6.5536 10 9.6 10 Kg
cm 1 J J2.89 10 2.890 10g K 10 bar cm g K bar
P PTV T V V
Tα αα
− −
− −
⎡ ⎤ ⎡ ⎤∂⎛ ⎞ = − +⎢ ⎥⎜ ⎟ ⎢ ⎥∂⎝ ⎠ ⎣ ⎦⎢ ⎥⎣ ⎦
⎡ ⎤= − × × +
∂
×⎣ ⎦
= − × ×
⎛ ⎞⎜ ⎟
= −
⎠
×
∂⎝
37
( )( ) ( )
( ) ( )
( )
o o 4
o
, we have
ˆˆ ˆ, ,
J Jˆ ˆ25 C,100bar 25 C,1bar 2.890 10 99bar= 0.0286g K bar g K
ˆTherefore, 25 C,10
ˆAssuming that is reasonably constant with pressu
0bar 0.024 18
e
.
r
PP P
P T
T
P P
P
CC T P P C T P PP
C C
C
C P
−
⎛ ⎞∂+ Δ − = Δ⎜ ⎟
∂⎝ ⎠
− = − × −
= −
∂
×
∂
J86g K
and we see that the heat capacity of water at these conditions (and indeed that of most liquidsand solids) only very weakly dependent on
4.1514
press is .ure
=
38
Gibbs-Helmholtz relation
( )( )
( )
2 2 2
2
1 1
( 6.2- 39
1
)
P
P
P
dG SdT VdPG ST
G G G H TS HT ST T T T T T T
G HTT T
dT
= − +
∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠
⎛ ⎞∂ ∂ −⎛ ⎞⎜ ⎟ = − = − − = −⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎜ ⎟⎝ ⎠
⎛ ⎞∂⎜ ⎟ = −⎜ ⎟∂⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
( )( )
22
1;
(6.2-41/
0)
P
dT dT T dT T
GT HT
⎛ ⎞= − = − ⎜ ⎟⎝ ⎠
⎛ ⎞∂⎜ ⎟ =⎜ ⎟∂⎜ ⎟⎝ ⎠
G-H relationship can be represented by dimensionless as well
39
6.4 THE EVALUATION OF CHANGES IN THE THERMODYNAMIC PROPERTIES OF REAL SUBSTANCES ACCOMPANYING A CHANGE OF STATE
The Necessary DataIn order to use the energy and entropy balances for any real substance for which thermodynamic tables are not available, we must be able to compute the changes in its internal energy, enthalpy, and entropy for any change of state. The equations of Sec. 6.2 provide the basis for such computations. However, before we discuss these calculations it is worthwhile to consider the minimum amount of information needed and the form in which this information is likely to be available.(A) Volumetric Equation-of State Information(B) Heat Capacity Data
40
(A) Volumetric Equation-of State Information
The van der Waals equation of state (1873)The Redlich-Kwong equation of state (1949)The Peng-Robinson equation of state (1976)The general cubic equations of stateThe Virial equation of state (1901)The BWR (Benedict, Webb, and Rubin, 1940) equation of stateThe Bender equation of state (1970)
41
The van der Waals equation of state
( )2
2
(6.2-38
was proposed by
a)
J
(6.2-
.
38
b)
aP V b RTVRT aP
V b V
⎛ ⎞+ − =⎜ ⎟
⎝ ⎠
= −−
D. van der Waals in to describe the volumetric or PVT behavior of both vapors and liquids, work for which he was awarded the Nobel Prize in Physics in 1910. The constants and can be dete
1
r e
8 3
n
7
mia b deither by fitting this equation to experimental data or, more commonly, from critical-point data. Values for the parameters for several gases appear in Table 6.4-1. The van der Waals equation of stat not very accurate
the first equation capablee is and is mainly of his
of predicting the transittori
ion c
interest in that betweenvapor and
it waliq
s uid.
42
43
The Redlich-Kwong equation of state (1949)
( )1/ 2 (6.4-1
the temperature dependence of was introduced in the attractive
)
term.
RT aPV b T V V b
a
= −− +
What is the meaning of the first term of RHS?
44
The Peng-Robinson equation of state (1976)
( )( ) ( )
( ) ( )2 2
0.45724
(6.
4-2)
0.07780
(6
.7-1)c
c
c
c
a TRTPV b V V b b V b
R Ta T TP
RTbP
α
= −− + + −
=
=
1 1
(6.7-2)
(6.7-3)
0.37464 1.54226c
TT
α κ
κ
⎛ ⎞= + −⎜ ⎟⎜ ⎟
⎝ ⎠= + 20.26992 (6 - .7 4)ω ω−
45
The general cubic equations of state
( )( )( )2
where each of the five parameters , , , , and can depend on temperat
(
ure. Inpractice however, generally only
3
6.4- )VRTP
V b V b V V
b
η θδ ε
θ δ ε η
−= −
− − + +
is taken to be a function of temperature, and it isadjusted to give the correct boiling temperature as a function of pressure.Table 6.4-2 gives the parameters of Eq. 6.4-3 for some commonequations o
θ
the hundreds of this classf state from among that have been publi shed.
46
47
The Virial equation of state (1901)( ) ( )
( ) ( )
21
where and are the temperature-dependent second and third virial coeffi-cients. A
(6.
lthough higher-order te
4-5)
rm
B T C TPVRT V V
B T C T
= + + +L
( ) ( )( ) ( )
only for the second virial coeffics can be defined in a similar fashion, data generally
are available two body interactions
three body interactionsWhen truncated at the (T) te
ie
rm, the
t
v
n .
iri
B T f
C T fB
=
=
al equation of state can be used only at low densities; as a general rule, it be used at pressshould not above 10ures for most flu
barids.
Leiden-Z form
{ }2
2
2
Berlin-Z form and Leiden-Z form:
Berlin- and Leiden- Residual volume as follows:
'( ) '[ ] ' '[ ]
1 '[ ] '[ ]
1
R IG
R IG
RT RT RTV V V B RT C P B RT C PRTP P P
RT RT RTV V V BP P PV
PVZ B P C PRT
Z B Cρ ρ
⎧ ⎫⎡ ⎤Δ = − = − + + = − −⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭
⎡ ⎤Δ = − = − + +⎢ ⎥
⎣ ⎦
= = + +
= + +
2 2
RT RT RTC B CPVPV PV
⎧ ⎫ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥= − −⎢ ⎥⎨ ⎬ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭
{ },0 0
, 20 0 0V
1
At zero pressure, Berlin- and Leiden- residual volume of a real gas:
lim( ) lim ' '[ ] '
lim( ) lim lim
RT xp p
V V
RT xp p p
V VZ
V B RT C PRT B RT
RT RTV B CPV PV
→ →→∞ →∞
→ → →→∞ →∞ →∞
=
⎧ ⎫⎪ ⎪Δ = − − = −⎨ ⎬⎪ ⎪⎩ ⎭
⎧ ⎫⎡ ⎤⎡ ⎤ −⎪ ⎪Δ = − − =⎨ ⎬⎢ ⎥⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭
B BZ
⎧ ⎫ = −⎨ ⎬⎩ ⎭
48
The BWR (Benedict, Webb, and Rubin, 1940) equation of state
( )
23
25 23
1 11
1 exp / (6.4-6a
wherc the eight constants , , , , , , , and are specific
)
PV A C aB bRT RT RT V RT V
a VRT VRTV V
a b A B C
α β γ γ
α β γ
⎛ ⎞ ⎛ ⎞= + − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞+ + + −⎜ ⎟
⎝ ⎠to each fluid and
arc obtained by fitting the equation of state to a variety experimental da of ta.
49
The Bender equation of state (1970)
( )2202 3 4 5 2 2
2 3 41 2 3 4 5
26 7 8
9 10
11 12
133 4 5
14 15 153 4 5
17 18 19
1 exp /
/ / / /
/ ///
/
/ / /
/ / /
(6.4-6b)
with
T B C D E F HP R G a VV V V V V V V V
B a a T a T a T a T
C a a T a TD a a TE a a TF a T
G a T a T a T
H a T a T a T
⎡ ⎤⎛ ⎞= + + + + + + + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
= − − − −
= + += += +=
= + +
= + +
50
Heat Capacity Data
( ) ( )2 2
1 1
2
2
2
2
1 2
2, ,
2 1 2, ,
At each temperature we can integrate Eq. 6.2-37 to obtain
At constant ,
Integrate from to ,
, ,
P
T P
PP
P T P
P P PP T P TP
C VTP T
VT dC T dPT
P P
VdC C P T C P T T dPT
⎛ ⎞∂ ∂⎛ ⎞ = − ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠
⎛ ⎞∂= − = − ⎜ ⎟∂⎝ ⎠
∫
( ) ( )
( ) ( )
2
1
2,
2 1 2,
2
2 1 2
, ,
Similarly,
, ,
(6.4-7)
(
6.4-8)
T
P T
P P P TP
V VV
VC P T C P T T dPT
PC V T C V T T dVT
⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠
⎛ ⎞∂= + ⎜ ⎟∂⎝ ⎠
∫
∫
2
1
,
,(6.4-9 )
V T
V T∫
51
( ) ( )
( ) ( )
1
2,*
*
20,
2
2
For an ideal gas, 0, ,
,
,
(6.4-10)
(6.4-11
P T
P P TP
V
P
VV
P V
VC P T T T dPT
PC V T T T V
C
C dT
=
→ → ∞
⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠
⎛ ⎞∂= + ⎜ ⎟∂⎝ ⎠
∫,
,)
V T
V T=∞∫
Which one is easy for calculation of 6.4-10 or 6.4-11?
It is the latter because a cubic volumetric EOS is generally explicit in pressure not in volume!
Using 6.4-10 and 6.4-11, and then 6.4-8 and 6.4-9 can be employed to get Cp and Cv values at the state 2.
52
Ideal gas heat capacity
* 2 3P
*PSee Appendix A.II for for some compounds.
C a bT cT dT
C
= + + + +L
Note:Ideal gas relationships are not fade out since you need them for calculations from given state-1 to unknown state-2!
54
Evaluation of ΔH, ΔU, and ΔS( ) ( )
( ) ( ) 2 2
1 1
1 1 2 2
,
2 2 1 1 ,
To compute the change in enthalpy in going from the state , to the state , , we start from
(2,1) , , (6.4-12)
(2,1)
T P
T P
PVd H
T P T P
H H T P H T P
d V T
d
C TT
H
∂⎛ ⎞=
Δ = −
+ −∂
=
⎜⎝
∫
( )
( )
( )
( )
1
1 1
2
2
1
2
2
, 0
,
,
, 0
, 0 *P, 0
Three steps calculation procedure (solid l
(2,1)
(6.2-22
ine
)
)
T P
T PP
T
T P
P
P
T PP
P
T
VV T d
dP
PT
VV
H C dT
T dPT
=
=
=
= ⎡ ⎤∂⎛ ⎞−⎢ ⎥⎜ ⎟∂Δ = +
⎝ ⎠⎣ ⎦
⎡ ⎤∂⎛ ⎞−⎢ ⎥⎜ ⎟∂⎝
⎡
⎠⎣
⎤⎢ ⎥⎟
+
⎣ ⎦
⎦
⎠
∫
∫
∫
(6.4 -13)
Introducing Residual functions (Departure functions)
“Evaluation of ΔH, ΔU, and ΔS” requires the IDEAL GAS calculation and the residual function calculation. Then Thermodynamics database kingdom can be built up.
Figure 6.4-1 Integration path in the P-T plane.
Cancel U&L limits and de-Integrate reduced to 6.2-22!
55
Isobaric heating followed by isothermal compression(dash line in Fig 6.4-1)
1 2 2 2
1 1 1 2
2 2 1 2 2
1 1 1 2
1
, ,
, ,
2 ,*P 20 ,
2
20
(6.4-14
:
a)
P T P T
PP T P TP
T T P P T
T T P TP
P
VH C dT V T dPT
V VC dT T dP dT V T dPT T
Since
VT dPT
⎡ ⎤∂⎛ ⎞Δ = + −⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦⎧ ⎫ ⎡ ⎤⎛ ⎞∂ ∂⎪ ⎪ ⎛ ⎞= − + −⎨ ⎬⎜ ⎟ ⎢ ⎥⎜ ⎟∂ ∂⎝ ⎠⎪ ⎪⎝ ⎠ ⎣ ⎦⎩ ⎭
⎧ ⎫⎛ ⎞∂⎪ ⎪⎨ ⎬⎜ ⎟∂⎪ ⎝ ⎠⎩
∫ ∫
∫ ∫ ∫ ∫
∫
suuuuuuuuuur
2 1 2
1 1
1 2
1
1 2 1 1
2 1
2
20
0
, ,
0, 0,
=
T P T
T TP
P T
TP P
P T P T
P T P TP P
VdT T dT dPT
VT V dT dPT T
V VT V dP T V dPT T= =
⎧ ⎫⎛ ⎞∂⎪ ⎪= ⎨ ⎬⎜ ⎟∂⎪ ⎪ ⎪⎝ ⎠⎭ ⎩ ⎭⎧ ⎫⎧ ⎫⎡ ⎤∂ ∂⎪ ⎪ ⎪ ⎪⎛ ⎞ −⎨ ⎨ ⎬ ⎬⎢ ⎥⎜ ⎟∂ ∂⎝ ⎠⎪ ⎪⎣ ⎦⎪ ⎪⎩ ⎭⎩ ⎭
⎡ ⎤ ⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞= − − −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦
∫ ∫ ∫
∫ ∫
∫ ∫ (6.4 -15)
6.2-37
6.4-15 + 6.4-14a = 6.4-13
( )
( )2 2
2
,
, 0Sum =
(6.4 13)
T P
T PP
VV T dPT
in EQ
=
⎡ ⎤∂⎛ ⎞−⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦−
∫
56
Entropy change between two states
( )
( )
( )
( )
( )
( )
( )
( )
1 2 2
1 1 2
2
1
2
1
1 2 2
1 21
, 0 ,
, , 0
*, 0 P, 0
*, 0
,,
, ,V, 0
For and
(6.4-16
are state variables,
a)
T P T P
T P T PP
T P
T PP
T T P
T P
V T V
T V T VV V
V VdP dPT T
P
CS dT
PdV dVT
T
T VCS dTT T
=
=
=
=
=
=
=∞
=∞
∂ ∂⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
∂ ∂⎛
Δ = + −
Δ = + +⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
∫∫ ∫
∫∫∫ (6.4 -17a)
Figure 6.4-2 Integration path in V-T plane.
Again, dV is easy for EOS calculation since EOS is explicit in pressure not in volume!
Cancel U&L limits and de-Integrate reduced to 6.2-20.
6.2-19
6.2-20
57
Internal energy change between two states (T, V) from state 1 to state 2
( )
( )
( )
( )
1
1 1 1
2
2
2
2
,
,
, 0 *
,
V,
,
0
(6
For and are state variables,
. - 8) 4 1
T V
T VV
T P
T V
T VV
T P
T V
U C dTPT P dVT
PT P dVT
=∞
=
=
∞
=
⎡ ⎤∂⎛ ⎞ −⎢ ⎥⎜ ⎟∂⎝ ⎠⎢ ⎥⎣ ⎦
⎡ ⎤∂⎛ ⎞ −⎢ ⎥⎜ ⎟∂
Δ = +
+⎝ ⎠⎢ ⎥⎣ ⎦
∫
∫
∫6.2-22
58
Ideal gas results
( ) ( )
( ) ( )
2
1
IG IG IG IGIG
0,IG IG *2 2 1 1 P0,
IG IG *2 2 1 1 VV ,
; 0 = ;
, ,
U ,
(6.4-14b
V U ,V
)
P P V V
P T
P T
V V R V P P R PV T T PT T P T T T V T
H T P H T P C dT
T T C dT
=
=
=∞
∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − = − = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
− =
− =
∫
( ) ( )
( ) ( )
( ) ( )
2
2
1
1 2 2 2
1 1 1 2
1 2 2 2
11 1
1
1 2
V= ,
*0, ,IG IG P2 2 1 1 , 0,
*, ,IG IG
*IG IG P
V2 2 1 1 ,
22 2 1 1
1
G2
,
I
(6.4-16, ,
, ,
, ,
b)
ln
T
T
P T T P T
P T T P T
V T T V T
V T T V T
T
T
CR RS T P S T P dP dT dPP T P
CR RS T V S T V dV dT dVV T
C PS T P S T P dT RT P
S T
V
∞
=
=
=∞
=∞
− = − + −
− = +
− = −
+
∫
∫ ∫
∫ ∫
∫
∫
∫
( ) ( ) 2
1
*IG V 2
2 1 11
(6.4-17, , ln b ) T
T
C VV S T V dT RT V
− = +∫
Combine and cancel Infinity terms
What will be the ? V
, ,
; V
IGIGP
T x T x
CCP
∂∂ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
59
Enthalpy departure function (T, P)( ) ( ) ( ) ( )
( )
( )
( )
( )
( ) ( ) ( ) ( )
1 2 2
1
2 1
1 2
2 1
IG IG2 2 1 1 2 2 1 1
IG IG IG IG2 2 1 1,
, 0 ,
, , 0
,
I
, , , ,
(6.4-1 ), , 9
T P T P
T P T PP
T T P
P
P
H T P H T P H T P H T P
V VV T dP V T dPT T
H H
H H H T P H T P H H
=
=
− = −
⎡ ⎤ ⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞+ − + −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤− −+ ⎦ −= −⎣
−
∫ ∫
( ) ,G
, 0,(6.4-22)
Re in tegration of the following term at constant T :
T P
T PT
PP
PP
call
Vd H C dT V T dPT
VV
ed
T dPT=
⎡ ⎤∂⎛ ⎞= + −⎢ ⎥⎜ ⎟
⎡ ⎤∂⎛ ⎞= −⎢ ⎥
−
⎜ ⎟∂⎝
∂⎝ ⎦
⎣ ⎦
⎠
⎠
⎣
∫
(6.2-22)
6.4-14a + b
Residual functions (Departure functions)
6.4-13
60
Entropy departure function (T, P)( ) ( ) ( ) ( )
( )
( )
( )
( )
( ) ( ) ( ) ( )
1 2 2
1
2 1
2
2
1
1
IG IG2 2 1 1 2 2 1 1
IG IG IG IG2 2 1 1, ,
, 0 ,
, , 0
I
, , , ,
(6., , 4 -20)
T P T P
T P T
T P T P
PP P
S T P S T P S T P S T P
V R V R
S S S T P S T P S S
dP dPT P T P
S S
=
=
− = −
⎡ ⎤ ⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− − − −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦
= ⎡ ⎤− + −− −⎣ ⎦
−
∫ ∫
( ) ,G
, 0, (6.4-23)
Re in tegration of the following term at constant T ::
T P
T PTP
P
P
P
caC Vd S dT d
V R dPT
lled
PT T
P=
⎡ ⎤∂⎛ ⎞= − −⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦
∂⎛ ⎞= − ⎜
−
⎟∂⎝ ⎠
∫
(6.2-20)
6.4-16a + b
Residual functions (Departure functions)
61
Entropy departure function (T, V)( ) ( ) ( ) ( )
( )
( )
( )
( )
( ) ( ) ( ) ( )
1 21 2
1
2 12
21
1
IG IG2 1 2 12 1 2 1
, ,
, ,
IG IIG IG
, ,
G2 12 1
, , , ,
+
(6.4-2 , , 1
T V T V
T V T VV
T V T V
V
S T V S T V S T V S T V
P R P RdV dVT V T V
S T V S STS SVS
=∞
=∞
− = −
⎡ ⎤ ⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− + −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤= + − −− ⎣ ⎦ −
∫ ∫
( ) ( ) ( ),IG IG
,,
)
(6.4-24)
Re in tegration of the following term at cons
, ,
tant T ::
T V
T VT V
V
V
V
C
P RS S S T V S
Pd S dT dVT T
called
T V dVT V=∞
⎡ ⎤∂⎛ ⎞− = − = −⎢ ⎥⎜ ⎟∂⎝
∂⎛ ⎞= + ⎜ ⎟∂⎝
⎢⎣
−
⎠ ⎥⎦
⎠
∫
(6.2-19)
6.4-17a + b
Residual functions (Departure functions)
62
Converting H(T, P) to H(T, V)data measurement ＜＝＞ EOS calculation
( ) ( );
The triple produc
1 (6.4-25)
d
t rule among , and
1 ; at const ant , P VT P
d PV PdV VdP
P T V
PdP d PV dVV V
V PT
V P T TT V P
= +
⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞ = −⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠
= −
∂ ∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠⎝ ⎠⎝ ⎠
( )
( ) ( )
( )
IG
, 0 0 0
0 0 0 0
0
(6.4-
1
26)
T P
P P P
P P PP P
P P P P
P P P PV V
P
P
V
V VV T dP VdP T dPT T
P P PV d PV dV T dV
H
d PV PdV T dVV V T T
Pd PV
P
H
TT
dVT
= = =
= = = =
=
⎡ ⎤ ⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− = −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ ∂ ∂⎛ ⎞ ⎛ ⎞⎡ ⎤= − + = − +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎣ ⎦∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦
∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠
⎛ ⎞⎜ ⎟∂
−
⎝ ⎠
= ∫ ∫ ∫
∫ ∫ ∫ ∫
∫
( ) ( )
( ) ( ),IG
,,
0
0
1
1 (6.4-2 )
1
7
P
PV
z V V
T P V
T P VT PP V
z V VV V
V PH H V T dP RT Z T P dVT
P dV
P Pd ZRT T P dV RT Z T P dVT T
T
=
= =∞
= ∞
=∞
=
⎡ ⎤⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− = − = − + −⎢ ⎥⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎡ ⎤−⎢ ⎥
⎢ ⎥⎣ ⎦⎡ ⎤ ⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞= + − = − + −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝
⎢ ⎥⎣
⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎦
⎦ ⎣ ⎦
⎣
∫ ∫
∫
∫ ∫ ∫
63
Converting from S(T, P) to S(T, V)
( ) ( );
The triple product rule among
d
, and
1 ; at constant , P V P VT
V PP dVT
d PV PdV R R RdP d PV dVdP
P T V
V P T TT
VP P V
V
V
TP
= +
⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞ = −⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎠⎝ ⎠
= −
( )
( ) ( )
( )
IG
0 0 0
1
,
ln
ln
(6.4-26)
P P P
P P PP P
V V V V
V V V VV V
z
zV
T P
R V R VdP dP dPP T P T
R R P P Rd PV dV dV R d PV dVPV V T T V
P RR d
S
RTT V
S
Z
= = =
=∞ =∞ =∞ =∞
=
⎡ ⎤ ⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− = −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤⎡ ⎤ ∂ ∂⎛ ⎞ ⎛ ⎞= − + = + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥ ∂ ∂⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦
⎡ ∂⎛ ⎞= + −⎜ ⎟∂⎝ ⎠⎣
− = ∫ ∫ ∫
∫ ∫ ∫ ∫
∫
( ) ,IG
, 0,ln (6.4-28)
ln
T P V
T P VT PP V
V V
V VV
R V P RS S dP R z
P RdV R z dVT
dV
V
P T T V
=∞ =∞
= =∞
⎡ ⎤⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞− = − = + −⎢ ⎥⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⎢ ⎥⎣
⎤ ⎡ ⎤∂⎛ ⎞= + −⎢ ⎥ ⎢ ⎥⎜ ⎟∂⎝ ⎠⎢ ⎥ ⎢ ⎥⎦ ⎣
⎦ ⎣
⎦
⎦
∫
∫
∫
∫
Easy for data measurement Easy for cubic EOS calculation
64
Enthalpy and entropy changes for Peng-Robinson equation of state (PR-EOS)
( ) ( )( )( )
( ) ( )( )( )
IG
,
IG
,
1 21 ln (6
2 2 1 2.4-2
1 2ln ln
2 2 1 2
where
9)
(6.4-
3
=
)
/
0
Z
T P
T P
daT a Z BdTH H RT Zb Z B
daZ BdTS S R Z B
b Z B
PV
⎛ ⎞ − ⎡ ⎤⎜ ⎟ + +⎝ ⎠ ⎢ ⎥− = − +⎢ ⎥+ −⎣ ⎦
⎡ ⎤+ +⎢ ⎥− = − +⎢ ⎥+ −⎣ ⎦
( ) ( )
2 2
2 2
2
and / ; A /
0.45724 ; 0.07780 ; 1 1
0.37464 1.54226 0.26992
c c
c c c
RT B Pb RT Pa R T
R T RT Ta T T bP P T
α α κ
κ ω ω
= =
⎛ ⎞= = = + −⎜ ⎟⎜ ⎟
⎝ ⎠= + −
6.4-28
6.4-27
Recalled “Evaluation of ΔH, ΔU, and ΔS” requires the IDEAL GAS calculation and the residual function calculation. Then Thermodynamics database kingdom can be built up.
65
ILLIJSTRATION 6.4-1Construction of a Thermodynamic Properties Chart using the PR EOSAs an introduction to the problem of constructing a charter (able of the thermodynamic properties of a real fluid, develop a thermodynamic properties chart for oxygen over the temperature range of -100 oC to + 150 oC and a pressure range of 1 to 100 bar. In particular, calculate the compressibility factor (Z), specific volume (V), molar enthalpy (H), and molar entropy (S) as a function of temperature and pressure. Also, prepare a pressure-volume plot, a pressure-enthalpy plot, and a temperature-entropy plot for oxygen.
( ) ( )( )
* 2 5 2 9 3P
IG
o
IGo o
reference state of oxygen to be the ideal gas state at 25 C and 1 bar
J 25.64 1.519 10 0.7151 10 1.311 10mol K
Choose the ,
25 C, 1 bar 0; 25 C, 1 bar 0
0.45724
C T T T
H T P S T P
a T
− − −⎛ ⎞ = + × − × + ×⎜ ⎟⎝ ⎠
= = = = = =
= ( )2 2
2
; 0.07780
1 1 ; 0.37464 1.54226 0.26992 =0.4069
=154.6 K; =5.046 MPa
c c
c c
c
c c
R T RTT bP P
TT
T P
α
α κ κ ω ω
=
⎛ ⎞= + − = + −⎜ ⎟⎜ ⎟
⎝ ⎠
Building up Thermodynamics database for Oxygen!
Must be given !
66
( ) ( )
( )( ) ( )
2 2
2 2
Solution.
For each temperature, calculate 0.45724 and , then at each pressure
solve the equation
for t
I. Volume
/ ; A /
he volume. or equiva
c
c
B
R Ta T T bP
a TRTPV b V V b
Pb RT P RV b
a Tb
α=
= −−
= =+ + −
( ) ( ) ( )3 2 2 2 31 3 2 2 0
lently (and preferably), so
;
lve the equation for Z
:
Mention: Illustration-6.7-1 + problem 6.7-PRPROP programs +NBS-soft
Z B Z A B B B Z AB B B
big root Vapor small root Liquid
+ − + + − − − + − + + =
− − − −
( ) ( ) ( ) ( ) ( ) ( )( )
( ) ( ) ( ) ( ) ( ) ( )
( )
IG IG IG IGo o
IG *P298.15K,
IG IG IG IGo o
*IG P
298.15K,
II. Enthalpy
III. En
, 25 C,1 bar , , , 25 C,1 bar
, 25 C,1 bar , , , 25 C,1 bar
ln1 b
tropy
ar
ware
T
T P
T
T P
H T P H H T P H T P H T P H
H H C dT
S T P S S T P S T P S T P S
C PS S dT RT
− = − + −
= − +
− = − + −
⎛ ⎞= − + − ⎜ ⎟⎝ ⎠
∫
∫
67
Given P, T values ==== Calculated Z, V, H, S values, Compared it with Table 7.5-1 covered from -100 to -200 C, which used PR-EOS as well.
How accurate of these data compared to experimental data?PR-EOS (or other EOS) is applicable for non-polar compounds (hydrocarbons), but not for polar compounds (waters). Keep in mind that steam table was established by 59 EOS constants correlated with many experimental data, which can be applied for power plant boilers and turbines.
68
69
Figure 6.4-3 Pressure-volume diagram for oxygen calculated Using Peng-Robinson equation of state.
V
Near straight line,Ideal gas behavior!
70
Figure 6.4-4 Pressure-Enthalpy diagram for oxygencalculated using Peng-Robinson equation of state.
H
Reference at25 C and 1 bar (not 0 bar)Check EQ 4.4-3.
71
Figure 6.4-5 Temperature-entropy diagram for oxygencalculated using Peng-Robinson equation of state.
SUsing these mentioned methods, Can you be able to obtain all thermodynamics properties for new coolant-R134a or for new heating media-Dowtherm at V&L phases?
Reference at25 C and 1 bar (not 0 bar)Check EQ 4.4-3.
72
For liquids or solids
pressure
1
1
are weak function of or . Thus,tempertufor
liquid o
Both useful coefficients as follows:
these two values may be con
r
s
solid p
idered a
h
s
ases.
con ntresta
TT
P
VV P
VV T
κ
α
∂⎛ ⎞− =⎜ ⎟∂⎝ ⎠∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠
Mentioned that pressure have little effect on L&S properties!
73
74
For Volume of liquids or solids
( ) ( )
( ) ( )( )[ ]
, ,
, ,
, 1 (6.4-31)
T
T
T
VV T P P V T P PP
V T P V T P P
V T P P
κ
κ
∂⎛ ⎞+ Δ = + Δ⎜ ⎟∂⎝ ⎠= − Δ
= − Δ
From definition
75
ILLUSTRATION 6.4-2Effect of Pressure on the Volume of Liquids and SolidsCompute the change in the molar volume of copper, sodium chloride, ethanol, and mercury for the very large pressure change of going from 1 bar to 1000 bar at 20oC.
( ) ( )[ ]
( )
[ ]
( )
o 6 -1Cu
oNaCl
SOLUTION, , 1
Using the equation, we havecc20 C, 1000 bar 7.12 1 0.77 10 bar 999 bar
molcc cc 7.12 1 0.000769 7.115
mol mol
20 C, 1000 bar 2
TV T P P V T P P
V
V
κ
−
+ Δ = − Δ
⎡ ⎤= × − × ×⎣ ⎦
= × − =
=
( )
( )
6 -1
o 6 -1EtOH
o 6 -1Hg
cc cc7.02 1 4.15 10 bar 999 bar 26.91 mol molcc cc20 C, 1000 bar 58.04 1 109 10 bar 999 bar
mol molcc cc20 C, 1000 bar 14.81 1 3.80 10 bar 999 bar 14.75
mol m
52.
o
04
l
V
V
−
−
−
⎡ ⎤× − × × =⎣ ⎦
⎡ ⎤= × − × × =⎣ ⎦
⎡ ⎤= × − × × =⎣ ⎦
76
COMMENT
We see that the change in volume with a factor of 1000 change in pressure is very small for the two solids and the liquid metal considered, but that there is an 8.9 percent change in volume for the more compressible liquid ethanol.
77
Enthalpy
[ ]
( ) ( )
(6.
At constant temperature
Therefore, assuming that and are essentially constant with pressure give
4
s,
-32)
1,
PP
P
Vd H C dT V T dPT
Vd H V T dP V TV dPT
V
H T H T VP P P
α
α
⎡ ⎤∂⎛ ⎞= + −⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦
⎡ ⎤∂⎛ ⎞= − = −⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦
+ Δ − = −[ ] (6.4-33 ) T Pα Δ
From definition
78
ILLUSTRATION 6.4-3Effect of Pressure on the Enthalpy of Liquids and SolidsCompute the change in enthalpy of the four substances considered in the previous example on going from 1 bar to 1000 bar at 20oC .
( ) ( ) [ ]
( ) ( )
[ ]
( ) ( )
o o 4 -1Cu Cu
o oNaCl NaCl
, ,
cc20 C,1000 bar 20 C,1 bar 7.12 1 293 K 0.492 10 K 999 barmol
cc cc bar J=7.12 1 0.0144 999 bar=7010.3 701 mol mol mol
cc20 C,1000 bar 20 C,1 bar 27.02 1 29
1
3 mol
H T P P H T P
H
V T
H
H H
Pα
−
+ Δ − =
⎡ ⎤− = − × × ×⎣ ⎦
− × =
− = −
− Δ
( ) ( )
( ) ( )
4 -1
o o 4 -1EtOH EtOH
o o 4 -1Hg Hg
JK 1.21 10 K 999 bar=2604 mol
cc J20 C,1000 bar 20 C,1 bar 58.04 1 293 K 11.2 10 K 999 bar=3919 mol mol
cc20 C,1000 bar 20 C,1 bar 14.81 1 293 K 1.81 10 K 999 bar=mol
H H
H H
−
−
−
⎡ ⎤× × ×⎣ ⎦
⎡ ⎤− = − × × ×⎣ ⎦
⎡ ⎤− = − × × ×⎣ ⎦J1401
mol
79
COMMENT
Note that in each case the enthalpy change due to the pressure change is large and cannot be neglected.
80
Internal energy
( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) [ ]
, ,
[ , , ] , ,
, , ,
(6.4-3 )
4
, , 1
U T P P U T P
H T P P P P V T P H T P PV T P
H T P P H T P V T P P
V P VT P V P
H T
VT P
VP TP P PH T
α
α
α − Δ
+ Δ −
⎡ ⎤= + Δ − + Δ − −⎣ ⎦
⎡ ⎤= + Δ − − Δ⎣ ⎦
= Δ
−
− Δ − Δ =
+ = ΔΔ −
Recalled: [ ]P
Vd H V T dP V TV dPT
α⎡ ⎤∂⎛ ⎞= − = −⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦
81
ILLUSTRATION 6.4-4Effect of Pressure on the Internal Energy of Liquids and SolidsCompute the change in internal energy of the four substances considered in the previous illustrations on going from 1 bar to 1000 bar at 20oC.
( ) ( )
( ) ( )
( ) ( )
( )
o oCu Cu
4 -1
o oNaCl NaCl
o oEtOH EtOH
, ,
20 C,1000 bar 20 C,1 bar
cc 1 J J7.12 293.15 K 0.492 10 K 999 bar 10.3 mol 10 cc bar mol
J20 C,1000 bar 20 C,1 bar 95.7 mol
20 C,1000 bar 20 C,1
U T P P U T P
U U VT P
U U
U U
VT Pα
α
−
+ Δ − =
− = − Δ
= − × × × × × = −
− = −
−
− Δ
( )
( ) ( )o oHg Hg
Jbar mol
J20 C,1000 bar 20 C,1 ba
1915
r 78.5 mol
.5
U U
=
− = −
−
82
CommentWith the exception of ethanol, the changes in internal energy are small, much smaller than the changes in enthalpy for these substance. Indeed, for moderate changes in pressure, rather than the large pressure change considered here, the change in internal energy with pressure can be neglected.
83
Entropy
( ) ( )
At constant temperature:
(6.4-35
, ,
)
(6. 4-36)P
Vd S dPT
T P P S T PS V Pα
∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠+ Δ − = − Δ
From definition
84
ILLUSTRATION 6.4-5Effect of Pressure on the Entropy of Liquids and SolidsCompute the change in enthalpy of the four substances considered in the previous example on going from 1 bar to 1000 bar at 20oC .
( ) ( )( ) ( )
( ) ( )
o oCu Cu
4 -1
o oNaCl NaCl
oEtOH
, ,
20 C,1000 bar 20 C,1 bar
cc 1 J 0.492 10 K 7.12 999 barmol 10 cc bar
J= 0.035 701 mol K
J20 C,1000 bar
(
20 C,1 bar 0.327 mol K
2
6.4-36
0
)
0 C,1 0
T P P S T P V P
S
S
S
S S
S
α
−
+ Δ − = − Δ
−
⎡ ⎤= − × × ×⎣ ⎦
−
− = −
( ) ( )
( ) ( )
oEtOH
o oHg Hg
J0 bar 20 C,1 bar mol KJ20 C,1000 bar 20 C,1 bar 0.268
mol K
6.534S
S S
− =
= −
−
−
85
CommentsAgain, with the exception of ethanol, the changes in entropy are quite small. For moderate changes in pressure the entropy change of a condensed phase with pressure is negligible.
86
The illustrations shows that unless the pressure change is large, and the condensed phase is quite compressible (as is the case for ethanol here), the changes in volume, internal energy, and entropy with changing pressure are quite small.
However, the change in enthalpy can be significant and must be accounted for. Indeed, it is this enthalpy difference that is important in computing the work required to pump a liquid from a lower pressure to a higher pressure.
Short Summary
87
ILLUSTRATION 6.4-6Energy Required to Pressurize a LiquidCompute the minimum work required and the heat that must be removed to isothermally pump liquid ethanolfrom 20oC and 1 bar to 1000 bar in a continuous process.
How much work required in compressing air in Chapter 4?Recalled 4.5-1 (3.4-4)- from 1 to 10 bar needs 7834 J/mol.
88
1 2 1 2
1 21 2
SOLUTlONConsider the pump and its contents to be the systemThe steady-state balance is
0 or =
The steady-state e balance is
=
nergy
mass
0= + + ;
dN N N N N Ndt
dU W QN H N H Q WdT N N
= = + = − −
+ = − +
& & & & &
&&&& & &
& & ( )
( )
( ) ( )
2 1
gen
1 2 2 11 2
o o2 1
The steady-state balance (with = 0 for minimum work) is
=0= + + or
From the previous illustratio
, we ha
ent
veJ20 C, 1000 bar 20 C, 1 bar 6.534
rop
mol
n
s
y
H H
S
dS Q QN S N S T S SdT T N
S S S S
−
= −
− = − = −
&
& && &
&
( ) ( )
( ) ( )
( )
o o2 1
2 1
2 1
KJ20 C, 1000 bar 20 C, 1 bar 3919
molJ JTherefore, 293.15 K 6.534 1915
mol K molJ 1915 3919 5834
mol
H H H H
Q T S SNW Q H HN N
− = − =
= − = × − = −
= − + − = + =
&
&
&&
& &
89
Comment
( ) ( ) [ ]
( ) ( )
( )o
the effect of thermal expansion, that is, , then from Eq.If
6.4-33 , , 1
and from Eq. 6.4-36,, , 0
Then, for the problem here,
20
set
we n
C,1000
eglecte
ba
d = 0
r 20
H T P P H T P V T P V P
S T P P S T P V P
H H
αα
α
+ Δ − = − Δ = Δ
+ Δ − = − Δ =
− ( )o C,1 bar
cc 1 J J58.40 999 bar 5834 mol 10 cc mol mol
= × × =
90
( ) ( )o oand 20 C,1000 bar 20 C,1 bar 0
Jso that 0 and 5834 mol
Note that the total work is the same as before for this isothermal pressurization (why?), but inthis case no heat would have to be re
S S
Q WN N
− =
= =& &
& &
moved to keep the liquid at constant temperature.To pressurize liquid , which is closer to being incompressible than ethanol, over thesame conditions one obtains (without making the assum
merpt
curyion
Much less work is n
that a = 0)J J2
eeded for this
93.15 K 0.286 78.6 mol K mol
J J 78.6 1401 1478.6 mol mol
since very little work is being done less compressible f to
c
luid (H
g)
hange
QNWN
α
⎛ ⎞= × − = −⎜ ⎟⎝ ⎠
= + =
&
&
&
&
the volume of the liquid; the work is mainly going to change the fluid pressure.
Works required in compressing fluids in descending order of Mercury, Ethanol, more compressible gases.
91
ILLUSTRATION 6.4-7Computing the Difference between CP and CV valuesCompute the difference between the constant-pressure and constant-volume heat capacities at 20oC for the four condensed phases considered in the previous illustrations.
( ) ( )
( )
( )
( )
2
P V
24
P V 2 6Cu
P V NaCl
P V EtOH
P V Hg
SOLUTIONFrom Eq. 6.2-35 we have
0.492 10cc bar 1 J J293.15 K 7.12 0.656 mol K 0.77 10 10 bar cc mol K
J2.795mol K
J19.7mol K
J3.742mol K
T
C C TV
C C
C C
C C
C C
ακ
−
−
− =
×− = × × × =
×
− =
− =
− =
92
COMMENTNote that the difference between the constant-pressure and constant-volume heat capacities is significant.
In fact (CP-CV) of ethanol can be quite large if the fluid has a large coefficient of thermal expansion, as is the case for ethanol.
For comparison, the difference between CP and CV for an ideal gas isR =8.314 J/(molK).
93
ILLUSTRATION 6.4-8
o
4
The coefficient of thermal expansion of liquid water at 25 C was given in Illustration 6.2-6 to be
ˆ1 2.56 10 Kˆ
Use this information an
Effect of Pressure on the Enthalpy of Liquid Water
P
VTV
α −⎛ ⎞∂= = ×⎜ ⎟
∂⎝ ⎠
o 3
o
o
d the fact that ˆ(25 C, saturated liquid at 3.169 kPa) = 0.001003 m /kg and ˆ (25 C, saturated liquid at 3.169 kPa) = 104.89 kJ/kg
to determine the enthalpy of liquid water at (a) 25 C and 1 bar and
V
H
o(b) 25 C and 10 bar.
94
ILLUSTRATION 6.4-8
( ) ( ) [ ]( )o o
33 4
3
3
SolutionUsing Eq. 6.4-33, on a mass basis we haveˆ ˆ ˆ25 C, 25 C,3.169 kPa 1 3.169 kPa
kJ m=104.89 1.003 10 1 298.15 2.56 10 96.83kPakg kgkJ Pa m kJ kJ=104.89 0.0897 1 104.98kg kg kPa m k
1 bar 100H H V Tα
− −
= + − −
⎡ ⎤+ × − × × ×⎣ ⎦
+ × =
( )( )
o
33 4
gˆ 25 C,
kJ m104.89 1.003 10 1 298.15 2
10 bar
1000.56 10 3.169 kPakg kg
kJ=104.89+0.92=105.81kg
H
− −⎡ ⎤= + × − × × × −⎣ ⎦
95
COMMENTWe see from this that for modest pressure changes, the liquid enthalpy changes little with pressure. Therefore, little error is incurred if, at fixed temperature, one uses the saturated liquidenthalpy in the steam tables at higher pressures.
However, this will lead to errors near the critical point (C.P.), where the density of the liquid changes rapidly with small changes in both temperature and pressure.
Near or above C.P.:When pressure increases, it becomes worse!
96
6.5 AN EXAMPLE INVOLVING THE CHANGE OF STATE OF A REAL GAS
Given low-density heat capacity data and volumetric equation-of-state information for a fluid, we can use the procedures developed in this chapter to calculate the change in the thermodynamic properties of the fluid accompanying any change in its state (state 1 to state 2).
Thus, in principle, we can use the mass, energy, and entropy balances to solve energy flow problems for pure fluids, such as those heating media(Dowtherm) and cooling media(Refrigerant).
Thermodynamic properties of pure fluids may obtained from
1. Thermodynamic properties plots.2. An equation of state such as van der Waals equation.
97
ILLUSTRATION 6.5-1Comparing Solutions to a Problem Assuming the (a) Gas Is Ideal, (b) Using a Thermodynamic Properties Chart, and (c) Assuming That the Gas Can Be Described by the van der Waals Equation of State
Nitrogen gas is being withdrawn from a 0.15 m3 cylinder at the rate of 10 mol/min. The cylinder initially contains the gas at a pressure of 100 barand 170 K. The cylinder is well insulated, and there is a negligible heat transfer between the cylinder walls and the gas. How many moles of gas will be in the cylinder at any time? What will the temperature and pressure of the gas in the cylinder be after 50 minutes?
a. Assume that nitrogen is an ideal gas.b. Use the nitrogen properties chart of Fig. 3.3-3.c. Assume that nitrogen is a van der Waals fluid.
Data: For parts (a) and (c) use [ ]* 3J/(mol K) 27.2 4.2 10 (K)PC T−= + ×
98
( ) ( )
( )
SolutionThe mass and energy balances for the contents of the cylinder are
or (a)0
and
dN N N t N t Ntdt
d NUN H
dt
= = = +
=
& &
&
that portion of the gas that always remains in the cylinder
where = 10 mol/min., the result of wFollowing Illustrati riting an entropy balance on 4.5-
for is
0
(b)
2N
dSdt
−
=
&
( ) ( )
( ) ( ) ( )3
cyl
or 0
Also, from = , we have 0.15 m0
0 0Equati
(c
ons a, b,
)
(d)
S t S t
V N VV
N tV t V t
= =
= = == =
c. and d apply to both ideal and nonideal gases.
Illustration 4.5-3 (3.4-5)
Referred to Illustration 4.5-3 (3.4-5)-closed, adiabatic, reversible system assumption.
99
( ) ( )( )
( )
( )
35
4 3
( )a. Using the ideal gas equation of state
m8.314 10 bar 170 K0 mol K0 =0 100 bar
1.4134 10 m / molso that
0 1061.3moland
1061.3 10 mol
Computation of N t
RT tV t
P t
N t
N t t
−
−
× ×== =
=
= ×
= =
= −
100
( ) ( )
( )
( )
( ) ( )
3
3 35
3
35
b. Using Figure 3.3-3mˆ ˆ170 K, 100 bar 170 K, 10 MPa 0.0035kg
g m 1 kg m170 K, 10 MPa 28.014 0.0035 9.80 10mol kg 1000 g mol
so that0.15m0 1529.8 mol
m9.80 10mol
and 1529.8 10 m
V T P V T P
V T P
N t
N t t
−
−
= = = = = ≈
= = = × × = ×
= = =×
= − ol
101
( ) ( )
2
725
c. Using the van der Waals equation of state
The van der Waals equation is
With the consiants given in Table 6.4-1, we have8.314 170 0.1368100 bar=1 10 Pa0 3.864 10 0
Solving t
RT aPV b V
V t V t−
= −−
×× = −
= − × ⎡ ⎤=⎣ ⎦
( ) ( )
( ) ( )
( )
35
3
his cubic equation, we obtainm0 9.435 10 and 0 1589.8molmol
and 1589.8 10 mol
Note also that since the volume of the cylinder is constant (0.15 m ), and as is known,we can compute the
V t N t
N t t
N t
−= = × = =
= −
( ) ( )3
molar volume of the gas at any later time from
0.15m= (e)
In particular, we can use this equation to compute ( = 50 min
V tN t
V t ). These results, togetherwith other information gathered so far, and some results from the following sections, arelisted in Table 6.5-1.
102
Summary in table
Which calculated values can not be accepted?
It demonstrated that calculated density of fluids is usually poor by EOS model, especially by ideal gas calculation.
103
Since we know the specific volume after 50 minutes, we need to determine only onefurther state property to have the final state of the system completely specified. In principle, either the energy or entropy balance could be used to find this property. The entropybalance is more convenient to use, especially for the nonideal gas calculations. Thus, allthe calculations here are based on the fact (se ( ) ( )
* *V V P
a. Using the ideal gas equa
e Eq. c) that 0 50 min
Computation of ( = 50min) and ( = 50 min)
From Eq. 4.4-1,
Now for the ideal gas
27.2 8
tion of
.
state
V
V
S t S t
T t P t
C Pd S dT dVT T
C C C R
= = =
∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠
= = − = −( )
( )
( )
4
4
3 3
32.672 10
170 1.413 10
43
4
J314 4.2 10 18.9 4.2 10mol K
Also, since = constant, or = 0, we have
18.9 4.2 108.314 0
2.672 10or 18.9ln 4.2 10 170 8.314ln 0170 1.413 1
The0
T V
T V
T T
S S
T dVdTT V
T T
−
−
− −
−= ×
= = ×
−−
−
+ × = + ×
Δ
+ ×+ =
×+ × − + =
×
∫ ∫
( ) ( )( )
, so that solution to this equation is 130.3 K
50 min50min 40.5 bar
50 min RT t
P tV t
== = =
=
Referred to Illustration 4.5-3 (3.4-5)-closed, adiabatic, reversible system assumption.
104
( )3 3
4 3m 1mol g m50min 1.457 10 1000 5.20 10mol 28.014g kg kg
To find the final temper
b. Using Fig. 3.3-3
Fig. 3.3-3ature and pressure using , we locate the initial point( = 170 K, = 10 MPa) and follow
V t
T P
− −= = × × × = ×
-3 3
a line of constant entropy (dashed line) throughˆthis point to the intersection with a line of constant specific volume = 5.2 10 m /kg.
This intersection gives the pressure and temperature of the V ×
( )( )
= 50 min state. We find
This construction is shown in the following diagr
50min 1
am, which is a portion of Fig.
33K
50min 3.9MPa = 39
3.3-3.
bar
T t
t
i
P
= ≈
= ≈
105
106
( )
c. Using the van der Waals equation of stateHere we start with Eq. 6.2-19,
and note that for the van der Waals gas
The integration path to be followed isi. 0 ,
V
V
V
C Pd S dT dVT T
P RT V b
T t V t
∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠
∂⎛ ⎞ =⎜ ⎟∂ −⎝ ⎠
= ( ) ( )( ) ( ) ( )( ) ( ) ( )
( ) ( )
( )
( ) ( )
*V V
*50min 50minV0 170K
0 0 ,
ii. 0 , 50min ,
iii. 50 min , 50 min , 50 min
so that 50 min 0 0V T t V t
V t T V
T t V
T t V T t V C C
T t V T t V t
S t S t
CdV dVR dT RV b T V b
=∞ = =
= = =∞
= → = = ∞
= = ∞ → = = ∞ =
= = ∞ → = =
= − = =
= + +− −∫ ∫ ∫
T=Constant
V=Constant
T=Constant
6.4-21 for ΔS calculation
107
( )
( ) ( )
( )( ) ( )
350 min 50 min
0 170K
35
53
5
18.9 4.2 10or 0=
mThus. using b = 3.864 10 , we havemol
13.77-3.864 100=8.314 ln 18.9 ln 4.2 10 170
9.437 3.864 10 170
The solution to the equ
V t T t
V t T
dV TR dTV b T
T T
−= =
= =
−
−−
−
+ ×+
−
×
⎡ ⎤×+ + × −⎢ ⎥− ×⎣ ⎦
∫ ∫
( )
( )
35
ation is 50min 133.1KNow, using the van der Waals equation of state with
m = 133.1 K and = 13.76 10 ,mol
we find 50min 39.5 bar
T t
T V
P t
−
= =
×
= =
108
Summary in table
Which one can not be accepted (V or T or P)?
109
COMMENTS(1) one cannot assume that a high-pressure gas is idealand expect to make useful predictions. E.G. predicting specific volume (density) using ideal gas is poor.
(2) thermodynamics calculations involving real fluids, the use of a previously prepared thermodynamic properties chart for the fluid is the most rapid way to proceed.
(3) the use of a simple volumetric equation of state for the fluid- accurately or (4) The following corresponding-states correlations-approximately is tedious and possible.
110
6.6 THE PRINCIPLE OF CORRESPONDING STATES
111
112
6.6 THE PRINCIPLE OF CORRESPONDING STATES
we consider here the principle of corresponding states, which allows one to predict some predict some thermodynamic properties of fluids from thermodynamic properties of fluids from generalized property correlationsgeneralized property correlations based on available experimental data for similar fluids.
113
Two types of information needed in thermodynamic calculations
(1) the ideal gas heat capacityThe ideal gas heat capacity is determined solely by the intra-molecular structure (e.g., bond lengths, vibration frequencies, configuration of constituent atoms) of only a single molecule, as there is no intermolecular interaction energy in an ideal gas.
114
Two types of information needed in thermodynamic calculations
(2) the volumetric equation of a stateIt has also been found that the volumetric equations of state obeyed by all members of a class are similar in the sense that if a given equation of state (e.g., Peng-Robinson or Benedict-Webb-Rubin) fits the volumetric data of this class (e.g. low-polar paraffins), the same equation of state, with different parameter values, is likely to fit the data for other molecular species in the same class (e.g. middle-polar paraffins).
115
The van der Waals equation of stateas an example
It is useful to review the van der Waals generalized correlation scheme since, although not very accurate, it does indicate both the structure and correlative parameters used in modern fluid property correlations.
116
A plot of P versus V for various temperatures for a van der Waals fluid is given in Fig. 6.6-1.
A brief list of measured critical temperatures, critical pressures Pc and critical volumes Vc (i.e., the pressure and volume of the highest-temperature liquid) for various fluids is given in Table 6.6-1.
117
Figure 6.6-1 The pressure volume behavior of the van der Waals equationof state
118
119
Parameters (a, b) of EOS for each substance determined from characteristics of critical point (C.P.) and inflection point (I.P.) of the P-V diagram
What will be the (a, b) values for the mixture used in EOS?
The mixing rule for the mixture should be considered!
)x(b),x(a
( ) ( ) (1 ) 0 at
( ) 2 reduced to 2
m m
i j ii jj ij iji j
m mii jj
i j i ii i iii j
a x x x a a k k i molecule j molecule
b bb x x x x b i x b
= − = − = −
+⎛ ⎞= = Σ = = Σ⎜ ⎟
⎝ ⎠
∑∑
∑ ∑
An quadratic mixing rule can be used for mixture parameters of :
120
Critical-point characteristics2
2
2
. . : 0 and . .: 0 at and
(6.6-1)
c c
c cT T
cc
c
RT
P PC P I P P T
aPV b
V V
RTP
V
V
⎛ ⎞⎛ ⎞∂ ∂= =⎜ ⎟⎜ ⎟∂ ∂
−
⎠ ⎝ ⎠
=
=
⎝
−
( )
2
2 3
2
(6.6- 2a )
(6.6-2b)
evaluated at 20 , ,
c
c
cc c cT c
ab V
RTP aT P VV VV b
P
−−
⎛ ⎞∂= = − +⎜ ⎟∂ −⎝ ⎠
∂∂ ( )2 3 4
evaluated at 2 60 , ,
Thus, three equations interrelate the two unknow
(6.6-2c)
ns and .
c
cc c cT c
RT aT P VV VV b
a b
⎛ ⎞= = − −⎜ ⎟
−⎝ ⎠
121
Critical-point characteristicsSolving for a and b:Eqs. 6.6-2b divide by 6.6-2c to ensure that the critical-point conditions of Eqs. 6.6-1 are satisfied, we obtain
9 8c cV RTa =
3
By direct substitution into 6.6-2a, the compressibility at the critical point is3 0.375
(6.6-
8
3a)
c
ccc
c
Vb
P VZRT
=
= = =
2
Using these results in Eq. 6.6-3c yields
=
(6.6-3c)
(6 27
.6-3b)caPb
122
2 2
2
6.6-3 ,6.6 3 :27 and 64 8
or in terms of the critical pressure and volume
(6.6-4
3
a)
and
c c
c c
cc
From b cR T RTa bP P
Va P V b
−
= =
= =
2
3
substitute 6.6-4b into 6.6-1 and divide it by Pc(Vc):
3 3 1 8
(6.6-4b
)c
c
cc c
r rc c
VP V TP V V T
T PT PT P
⎡ ⎤ ⎡ ⎤⎛ ⎞⎛ ⎞+ − =⎢ ⎥ ⎢ ⎥⎜ ⎟⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎣ ⎦
= =
[ ]
( )
3 3 1 8
(6.6-5)
(6.6-6)
, 8 Corresponding
r
rc
rr r
r rr
P V T
VV
f P V T
V
V =
⎡ ⎤+ −⎥
=
=⎢⎣ ⎦
states principle
123
6 -5 3
7
ILLUSTRATION 6.6-1Simple Example of the Use of Corresponging states Assume that oxygen ( =154.6 K, = 5.046 10 Pa, and = 7.32 10 m /mol) and
water ( = 647.3 K, = 2.205 10 Pa, and =cc c
cc c
T P V
T P V
× ×
× -5 3 5.6 10 m /mol) can be considered.
a. Find the value of the reduced volume both fluids would have at = 3/2 and = 3.b. Find the temperature, pressure,
van de
and v
r Waals fluid
olume of each gas
s
r rT P
×
o 62 2
at = 3/2 and = 3.
c. If O and H O are both at a temperature of 200 C and a pressure of 2.5 10 Pa, find their
specific volumes.
SOLUTIONa. Using E = 3/2 and = 3 i q. 6.6n , one finds6 h- t at
r r
r r r
T P
T P V
×
= 1 for both fluids.
124
2
35
O
ILLUSTRATION 6.6-1Solutionb. Since = 1 at these conditions [see part (a)], the specific volume of each fluid is equal to the critical volume of that fluid. So
m7.32 10 mol
rV
V −= ×2
2 2 2 2
2 2 2 2
35
H O
7 7O c,O H O c,H O
O c,O H O c,H O
6
6
m 5.6 10mol
1.514 10 Pa 6.615 10 Pa
232.2 K 971 K
c. For oxygen we have2.5 10 0
3 3
.495; 473.5.046
1.5
10
1.5
r r
V
P P P P
T T T T
P T
−= ×
= × = × = × = ×
= × = = × =
×= = =
×3
3
6
7
33
2 /154.6 3.061; 16.5
mand 1.208 10mol
2.5 10Similarly, for water, 0.113; 473.2 / 647.3 0.703 and 15.95 2.205 10
m8.932 10mol
r
r r r
V
V
P T V
V
−
−
= =
= ×
×= = = = =
×
= ×
EQ 6.6-6
125
The limitations of the van der Waals EOS
(1) Zc = 3/8= 0.375, But for most fluids, Zc= 0.23~0.31. It is bad prediction near the critical region using the vdW EOS.
(2) Two parameters, Pc, and Tc may noteffectively characterize the behaviors of non-spherical molecules.
(3) Third parameter (Zc or ω) that accounts for un-symmetric molecules, needed to be defined andthree-parameters equations of state were applied for real fluids.
126
As has been already indicated, the accuracy of the van der Waals equation is not
very good. This may be verified by comparing the results of the previous illustration
with experimental data, or by comparing the compressibility factor Z = / forthe van der Waals equation of stale with experimental data for a variety of gases.In particular, at the critical point (see Eq. 6.6-3c); van der Waals = c
PV RT
Z / 0.375 3 / 8
(Table
6.6
w
-
hi
1)
le f
, so
or most
that th
fluids the criti
e van der Waals
cal compressibility is in the
fails to pre
range 0.23 to 0.
dict accurateequation critical-point behav
31
ior.
cc
c
c
Z
P V RT = =
(however, a great improvement over the ideal gas equation of state, whichpredicts that = 1 for all conditions.)The fact that the critical compressibility of the van der Waals not fluid is equal toth
Z
different values for the van der Waals parame-ters are obtain
at for most real fluids also means that , depending on whether Eqs. 6.6-3a, Eqs. 6.6-4a, or
Eqs. 6.6-ed for an
4b are usy
e o
d ne
to f
luid
rela In practice, thecritical temperature or critical pressure, so that Eqs. 6.6-4a and critical-point data ar
te these parameters to thee
most frequently used to o
critical pro
btain the van
pertie
de
s.
r Waa Indeed, the entries inTables and 6.6-1 are related in this way. Thus, if the parameters in Table 6.4-1are used in the van der Waals equation, the critical temper
ls paramet
ature and
ers.6
pres.4-1
sure will becorrectly predicted, but will be by the fa
where is the real fluid critical compressibili
the ctor
van der Waals 3
critical volume too hig
.
h
8ty
c
cc
cZ
ZZZ
=
127
128
( )Z= / , where the functional relationsh
(6.6-7)ip between , and Z is d
Two parameters corresponding states
r r
r r
PV RT Z P TT P
=
etermined from experimental data, or from a very accurate equation of state. That such a procedure has some merit is evident from , where the compressibility data for different fluiF dig. 6. sha
6-2ve been made to almost superimpose by plotting each as a function of its reduced
temperature and pressure.Two parameters corresponding states equations are good for nonpolar spherical molecules,
however, onl
( )c
c
y fair representations for nonspherical molecules.
Z= / , , Alternatively, fluid characteristics other than can be used as the additional p
Three parameters corresponding states equation
r rPV RT Z P T ZZ
=
c
since formany substances the cri
arameterin the generalization of
tical density, and hence , isthe simple corresponding
known with limited accu-states principle, in fact,
, if
at a
racy
ll, there is a rZ
c.eason in avoiding the use of Z
129
( )vap10 0.7 /
Vapor pressure measurement is required.
1.0 log
Three-parameter correspondi
Acentric Factor defined by professor Pitzer (Founder of Swiss Phamaceutical
ng st
Company
es
)
at
:
r cP T Pω ⎡ ⎤=⎣ ⎦= − −
( ), ,proposed:
r rT PZ Z ω=
The third parameter was found unanimously as
130
Figure 6.6-3 Z = f(Tr, Pr) at Zc = 0.27
Two-parameter C.S. calculation
131
( )
( )
( )
( )r
,IG
, 0,
r r
2
22, ,IGr, 0 , 0,
r r
IG
started from 6.2-22 at constant T
,
r r
r r
T P
T PT PP
P P
T P T P rcT P T PT P
P P
VH H V T dPT
RTV Z T PP
V RT ZV TT P T
TRT Z ZH H dP RT dPP T P T
H H
=
= =
⎡ ⎤∂⎛ ⎞− = −⎢ ⎥⎜ ⎟∂⎝ ⎠⎣ ⎦
=
∂ ∂⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞∂ ∂⎛ ⎞− = − = − ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
−
∫
∫ ∫
( )( )
r
2,,r, 0
r
IG
,r r
r
i.e. Fig. 6.6.-4 (6.6-8)
,
r r
r r
T PT P rT P
c P
T P
c
H H
T ZR
f
dPT P
PT
T
T
=
⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠
−=
∫
Enthalpy departure (residual)
Z = PV / nRT
Divided by Tc and Pc
Z = C CC
C
P VRT
132
Figure 6.6-4 HR = f(Tr, Pr) at Zc = 0.27
Two-parameter C.S. calculation
133
Enthalpy difference (H2-H1)
( ) ( ) ( ) ( ) ( ) ( ){ }( ) ( )
0from Fig. 6
IG IG0 0
.6-4
IG*P
, , 0 , , 0 , ,
, , (6.
Enthalpy change obtianed from corresponding s
6 9)
t
-
a
T
cTc
H T P H T P H T P H T P H T P H T P
H T P H T PC dT T
T
− = = − = + −
⎡ ⎤−= + ⎢ ⎥
⎢ ⎥⎣ ⎦∫
( ) ( ) ( ) ( ) ( ) ( )
r2
2
1
2 r1 1, ,
IG IG*
2 2 1 P
, , , ,, ,
tes
r rT
T
c cTc PcP T
H T P H T P H T P H T PH T P H T P C dT T T
T T
⎡ ⎤ ⎡ ⎤− −− = + −⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫
(6.6-10)6.4-19
134
Entropy departure (residual)/Entropy difference (S2-S1)
( )
( ) ( )
r r
r rr
,IG r
IGr r,
r, 0,
r r r
1 started from 6.2-20 (Prob. 6.6) (6.6-11)
, Fig. 6.
6-5T P
T P
T PT PP
TZ ZS S R d
S
PP P T
Entropy diffe
S T
r n
f
e
P
=
⎡ ⎤⎛ ⎞− ∂⎢ ⎥− = − + ⎜ ⎟∂⎢ ⎥⎝ ⎠⎣ ⎦
− =
∫
( ) ( ) { }2 2
1 1 r2 2 r1 1
*IG IGP
2 2from Fig. 6.6-, 5
1 ,
state:
, ,
r rT P T P
T P
T P
ce obtained from CorrespondingC dPS T P S T P dT R S S S ST P
⎡ ⎤ ⎡ ⎤− = − + − −⎣ ⎦ ⎦− ⎣∫ ∫ (6. 6-12)6.4-20
P
P
C VdS dT dPT T
∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠
Z = C CC
C
P VRT
135
Figure 6.6-5 SR = f(Tr, Pr) at Zc = 0.27
Two-parameter C.S. calculation
136
Rework of ILLUSTRATION 6.5-1
Comparing Solutions to a Problem Assuming the (a) Gas Is Ideal, (b)Using a Thermodynamic Properties Chart, and (c) Assuming That the Gas Can Be Described by the van der Waals Equation of State
Nitrogen gas is being withdrawn from a 0.15 m3 cylinder at the rate of 10 mol/min. The cylinder initially contains the gas at a pressure of 100 bar and 170 K. The cylinder is well insulated, and there is a negligible heat transfer between the cylinder walls and the gas. How many moles of gas will be in the cylinder at any time? What will the temperature and pressure of the gas in the cylinder be after 50 minutes?a. Assume that nitrogen is an ideal gas.b. Use the nitrogen properties chart of Fig. 3.3-3.c. Assume that nitrogen is a van der Waals fluid.Data: For parts (a) and (c) use
[ ]* 3J/(mol K) 27.2 4.2 10 (K)PC T−= + ×
137
ILLUSTRATION 6.6-2Using Corresponding States to Solve a Real Gas ProblemRework Illustration 6.5-1, assuming that nitrogen obeys the generalized correlations of Figs. 6.6-3, 6.6-4, and 6.6-5.
r r
SOLUTIONFrom Table 6.6-1 we have for nitrogen = 126.2 K and = 33.94 bar, and from the initialconditions of the problem,
170 1001.347 2.94
From
6126.2 33.94
, Z = 0.741Fig. 6.6-3 , so
c cT P
T P
V t
= = = =
=( ) ( )( ) ( )
( ) ( )
( ) ( )
3IG 4
3 34
0 m0 0 1.047 100 mol
Therefore, following Illustration 6.5-1,0 1432.7 mol and 1432.7 10 mol
0.15 m m50min 1.6082 101432.7 50 mol mol
RT tZ ZV t
P t
N t N t t
V t
−
−
== = = = ×
=
= = = −
= = = ×−
138
( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
IG IG0 0
IG IG50 min 50 min
IG50 min
SOLUTION ( )To compute the temperature and pressure at the end of the 50 minutes, we
conuse
0 50min
0 ,
50 min ,
50 min
i ue
0
t d
,
n
t t
t t
t
S t S t
S t S T P S S
S t S T P S S
S t S t S T P
= =
= =
=
= = =
= = + −
= = + −
= − = = − ( )
( ) ( )( ) ( ) ( )
( ) ( )
IG0
IG IG
50 min 0
IG IG50 min 0
3
,
+
50min, , 27.2 ln
17050 min
4.2 10 50min 170 8.314ln100
t
t t
t t
S T P
S S S S
T tS T P S T P
P tT t
=
= =
= =
−
− − −
=− =
=⎡ ⎤+ × = − −⎣ ⎦
6.4-16b
139
IG IGt=0
r r
Both of the ( ) terms are obtained from the corresponding-states charts, ( ) iseasily evaluated, since the initial state is known; that is, = 1.347 and = 2.946, so that,
fr Figom .
S S S ST P
− −
IG I
r
Gt=
0
r
( ) = 2.08 cal/(mol K) = 8.70 J/(mol K). To compute ( )
at = 50 minutes is m
6.6-5
because neither nor is ore difficult . The procedure to be
followed is1.
known
( = 50 min) i
S S S S
t
V t
T P
− − − −
s known, so ( = 50 min). (A reasonable first guess is the ideal gas solution obtained earlier.)2. Use ( = 50 min) and ( = 50 min) to compute, by trial a
guess
nd er
a value o
ror, ( = 50 mi
f
n
t
V t T t P t
T
( )r rc
IGt=50 min
) from
, ,
3. Use the values of and from steps 1 and 2 to compute ( - )4. Determine whether ( = 50 min) = ( = 0) is satisfied with the trial
c
RT T P RTP Z Z T PV T P V
P T S SS t S t
⎛ ⎞= =⎜ ⎟
⎝ ⎠
Our solu
values
tion aft
of and .
er a number ofIf not, guess another value of ( = 50 min) and go back to step 2.
is ( = 50 min) = 136 K ( = 50 min)
tri
= 41 bar
als
TP T t
T tP t
( ) ( ) 0S t S t= =Fig. 6.6-5
140
COMMENT
Because of the inaccuracy in reading numerical values from the corresponding-states graphs, this solution cannot be considered to be of high accuracy.
It should be pointed out that although the principle of corresponding states and Eqs. 6.6-7(residual UR), 6.6-8 (HR), and 6.6-11(SR) appear simple, the application of these equations can become tedious, as is evident from this illustration. Also, the use of generalized correlations will lead to results that are not as accurate as those obtained using tabulations of the thermodynamic properties for the fluid of interest.
Therefore, the corresponding-states principle is used in calculations only when reliable thermodynamic data are not available.
141
6.7 GENERALIZED EQUATIONS OF STATE
2
2 2
(6.2-38b)
(1) van der Waals equation of state
27 and 64 8
(2) Peng-Robinson equation
(6.6-4a)
o
c c
c c
RT aPV b V
R T RTa bP P
= −−
= =
( )( ) ( )
( ) ( )2 2
f state
0.45724
0.07780
(6.4-2)
(6.7-1)
(6 .7 - 2
1 1
)
-
c
c
c
c
c
a TRTPV b V V b b V b
R Ta T TP
RTbP
TT
α
α κ
= −− + + −
=
=
⎛= + ⎜
⎝2
0.37464 1.5
(6.7-3)
(6.4226 0.26992 7-4)
κ ω ω
⎞⎟⎜ ⎟⎠
= + −
Better and Accurate
Get these EQs usingC.P. and I.P. criteria.
Can you get these EQs as 6.6-6 using C.P. and I.P. criteria?
2
2
. .: 0
. .: 0
c
c
T
T
PC PV
PI PV
⎛ ⎞∂=⎜ ⎟∂⎝ ⎠
⎛ ⎞∂=⎜ ⎟
∂⎝ ⎠
142
For the thermodynamic properties calculations for process design, the equations of state method is normally applied and accurate results can be obtained.
However, a digit computer will normally be used when using the complicated equation of states approach.
Introducing the Illustration 6.7-1 program solved by MathCad. i.e. Reworked 6.5-1.
143
Homework for Chap 6
Problems6.3, 6.5, 6.7a&b, 6.8, 6.16, 6.46a
