# Chapter 13 Mathematical Modeling- Thermodynamic Properties

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## Optimal Design of Energy Systems

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### Chapter 13. Mathematical Modeling

equation fitting

(conversion of data to equation form) physical relationship

complexity accuracy

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### 13.3 Criteria for fidelity of representation

(1) Sum of the deviations squared

(2) Average percent absolute deviation

(3) Goodness of fit

2 1

n

i i

i

2

1

n i i

i i

2

1

n

i m ean

i

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

x

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### 13.5 Nonlinear regression analysis

- Steepest decent method – unconstrained optimization

ex)

bx

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### 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)

) , )

,

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)

p

h T

   s

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### 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

(10)

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)

### 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

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### Chapter 13. Mathematical Modeling

Tds du Pdv Tds dh vdP

sdT df Pdv sdT dg vdP

   

     

s v s p

p T v T

(13)

v

v

p p

If ideal gas :

2 2

2

p p

p

p

T

T p

2

2

p T

(14)

2

2

1/ 2

Van der Waals

Redlich-Kwong

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p T

p

T P

p T

p

T P

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### 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)

<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

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### 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)

<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)

<Solution>

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## References

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