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Chapter 13 Mathematical Modeling- Thermodynamic Properties

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(1)

Optimal Design of Energy Systems

Chapter 13 Mathematical Modeling- Thermodynamic Properties

Min Soo KIM

Department of Mechanical and Aerospace Engineering

Seoul National University

(2)

Chapter 13. Mathematical Modeling

equation fitting

(conversion of data to equation form) physical relationship

complexity accuracy

(3)

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 i

i i

Y y A P D

n

Y

  

  

 

100 1 SD S G O F

  G

2

1

( )

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.

(4)

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

 

  

 

(5)

13.5 Nonlinear regression analysis

- Steepest decent method – unconstrained optimization

ex)

Chapter 13. Mathematical Modeling

ln ln

sin

y ae

bx

y a bx

y a bx

   

Y A

(6)

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

 

(7)

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

 

     

(8)

13.7 Internal energy and enthalpy

Chapter 13. Mathematical Modeling

p

h T

   s

(9)

13.8 Clausius-Clapeyron equation

Chapter 13. Mathematical Modeling

P

T

1 2

0 dG  at saturation

f g

GG

f f g g

v d Ps d Tv d Ps d T

G h Ts

dG dh Tds sdT

du pdv vdP Tds sdT vdP sdT

 

  

    

 

(10)

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

  

 

  

(11)

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

(12)

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)

13.11 Specific heats

Chapter 13. Mathematical Modeling

v

v

c u

T

     

p p

c h

T

     

, pvRT

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

  

  

   

(14)

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

(15)

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

  

       

      

   

(16)

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

(17)

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 /

vgm kg

(18)

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

(19)

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

   

       

     

(20)

13.13 Building a full set of data

<Solution>

Chapter 13. Mathematical Modeling

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