Optimal Design of Energy Systems
Chapter 13 Mathematical Modeling- Thermodynamic Properties
Min Soo KIM
Department of Mechanical and Aerospace Engineering
Seoul National University
Chapter 13. Mathematical Modeling
equation fitting
(conversion of data to equation form) physical relationship
complexity accuracy
13.3 Criteria for fidelity of representation
(1) Sum of the deviations squared(2) Average percent absolute deviation
(3) Goodness of fit
Chapter 13. Mathematical Modeling
2 1
( )
n
i i
i
SD S y Y
2
1
100
n i ii i
Y y A P D
n
Y
100 1 SD S G O F
G
21
( )
n
i m ean
i
G Y y
2 1
1 ( )
n
i i
i
R M S y Y
n
1
1 ( )
n
i i
i
B ia s y Y
n
1
1 n
i i
i
A A D y Y
n
eqn. exp.
13.4 Linear regression analysis
- Parameters appear in linear form not the variable
Chapter 13. Mathematical Modeling
ln
sin cos 2
x
y a bx
y a be c x
y a x b x
13.5 Nonlinear regression analysis
- Steepest decent method – unconstrained optimization
ex)
Chapter 13. Mathematical Modeling
ln ln
sin
y ae
bxy a bx
y a bx
Y A
13.6 Thermodynamic properties
Chapter 13. Mathematical Modeling
Property equations
1. Equations for vapor
2. The specific heat at zero pressure 3. Relation for saturated conditions 4. The density of liquid
o p
f
P v T
c
P T
13.7 Internal energy and enthalpy
Chapter 13. Mathematical Modeling
) , )
,
rev rev
v s
q du w
q Tds w Pdv
Tds du Pdv
du Tds Pdv
u u
T P
s v
,
p s
dh Tds vdP
h h
T v
s P
13.7 Internal energy and enthalpy
Chapter 13. Mathematical Modeling
p
h T
s
13.8 Clausius-Clapeyron equation
Chapter 13. Mathematical Modeling
P
T
1 2
0 dG at saturation
f g
G G
f f g g
v d P s d T v d P s d T
G h Ts
dG dh Tds sdT
du pdv vdP Tds sdT vdP sdT
13.8 Clausius-Clapeyron equation
Chapter 13. Mathematical Modeling
2
ln ln
fg
fg
dP Ph
dT RT
P h C
RT P A B
T
( )
( , )
g f fg
g f g f
fg
g f g
g
s s h
dP
dT v v T v v
h RT
v v v
Tv P
Chapter 13. Mathematical Modeling
13.8 Clausius-Clapeyron equation
( )
(2593.1 211.4)
(1 ) 0.625
(50.5 273.15)(11.775 0.0010124)
fg
g f
P T h
T v v
K kPa
<Solution>
- calculation with Clausius-Clapeyron equation
12.961 12.335 0.626 kPa - calculation from steam table
13.10 Maxwell Relations
Chapter 13. Mathematical Modeling
Tds du Pdv Tds dh vdP
sdT df Pdv sdT dg vdP
( , ) ( , ) 1 P v T s
s v s p
p T v T
T p T v
v s p s
v s p s
T p T v
13.11 Specific heats
Chapter 13. Mathematical Modeling
v
v
c u
T
p pc h
T
, pv RT
If ideal gas :
2 2
2
1
1
p p
p
p
T
T p
h s
T T T
c
c s s v
T p p T T p T
2
2
0
p T
c v
p T T
13.12 P-v-T equations
Chapter 13. Mathematical Modeling
2
2
1/ 2
( ) ( ) 1
( )
pv RT pv ZRT
RT B T C T
p v v v
RT a p v b v
RT a
p v b T v v b
Van der Waals
Redlich-Kwong
13.13 Building a full set of data
Chapter 13. Mathematical Modeling
p T
p
T P
h h
dh dT dp
T p
s v
c v T v T
p T
p T
p
T P
s s
ds dT dp
T p
c s v
T p T
13.13 Building a full set of data
<Example 13.6> Starting with a base enthalpy of h=0 kJ/kg for saturated liquid water at 0℃, the enthalpy of superheated vapor at 4000 kPa and 500℃ using the following equations: (1) Redlich-Kwong equation of state (a=43.951, b=0.001171), (2) cpo from Table 13.3 and (3) the p-T equation for saturation conditions at low pressure,
.
Chapter 13. Mathematical Modeling
ln P 19.335 5416 / T
13.13 Building a full set of data
<Solution>
Chapter 13. Mathematical Modeling
( )
0.0434 273.15 206.54 2 501.5 /
0 0 2501.5 2501.5 /
fg fg
g f g
fg
o g
h h
dp
d T T v v T v
h kJ kg
h at C kJ kg
2
0 , 0.6108
5416 0 .0 4 4 3 4 /
at oC p kP a
dp p kP a K
d T T
From the Redlich-Kwong equation, Clapeyron equation :
206.54 3 /
vg m kg
13.13 Building a full set of data
<Solution> From Table 13.3
Chapter 13. Mathematical Modeling
50 0o & 0.6 108 2 5 0 1 .5 9 8 6 .9 5 3 4 8 8 .5 /
C kP a
h kJ kg
Integration from 0 to 500℃ at 0.6108 kPa
0
0
0
1.8703 0 1.9603 250
2.1319 500
o p
o p
o p
c at C
c at C
c at C
6 2
0 1 .8 7 0 3 0 .0 0 0 1 9 6 8 0 .6 5 2 8 1 0
cp t t
Equation fitting (t is in ℃)
500
0 0 986.95
o
o
C
C cp dt
13.13 Building a full set of data
<Solution>
Chapter 13. Mathematical Modeling
7 7 3 .1 5 ( ) 3
0 .0 1 0 / 0.2
c a
b p
v v
v T v v m kg
T
From the Redlich-Kwong equation
499.9 0.085963245748
500.0 0.085975713311
500.1 0.085988180541
o
a o
b o
c
t C v
t C v
t C v
5 0 0 & 4 0 0 0
3 4 8 8 .5 ( 0 .0 1 0 ) ( 4 0 0 0 0 .6 1) 3 4 4 8 .5 /
oC kP a p
P
h d h c d T v T v d p
T
kJ kg
13.13 Building a full set of data
<Solution>