Optimal Design of Energy Systems

Optimal Design of Energy Systems

Chapter 4 Equation Fitting

Min Soo KIM

Department of Mechanical and Aerospace Engineering

Seoul National University

Equation Development

• Performance characteristics of equipment

• Behavior of processes

• Thermodynamic properties of substances

Key elements

• To facilitate the process of system simulation

• To develop a mathematical statement for optimization

Purposes

Chapter 4. Equation fitting

4.1 Mathematical modeling

11 12 1

1 2

n

m m mn

a a a

a a a

 

 

 

 

 

 

Order of matrix

m n 

  ^A

 

3 6 A

 

 

  

 

 

 

4 5 6

A T  

  

 

Chapter 4. Equation fitting

4.2 Matrices

• Multiplying two matrices

# of columns of the left matrix = # of rows of the right matrix

𝑎

⋯ 𝑎

⋮ ⋱ ⋮

𝑎

⋯ 𝑎

×

𝑏

⋯ 𝑏

⋮ ⋱ ⋮

𝑏

⋯ 𝑏

⇒ 𝑚 × 𝑙 𝑚𝑎𝑡𝑟𝑖𝑥

Chapter 4. Equation fitting

4.2 Matrices

• Simultaneous linear equations

1 2 3

1 2

1 2 3

2 3 6

3 1

4 2 0

x x x

x x

x x x

  

 

  

1

2

3

2 1 3 6

1 3 0 1

4 2 1 0

x x x

      

      

     

      

     

Chapter 4. Equation fitting

4.2 Matrices

• Determinant (scalar)

11 12 13

21 22 23

31 32 33

a a a

D a a a

a a a

 

31 22 13 21 12 33 11 23 32

a a a a a a a a a a a a a a a a a a

  

  

cofactor of

22 33 23 32

a a a a

 

a11 ^{[( 1) ]}

i

[ ]

i j ij

submatrix formed by striking out A

th row and j th column of A

   Cofactor of 𝒂_𝒊𝒋

Chapter 4. Equation fitting

4.2 Matrices

= 𝑎₁₁𝐴₁₁ + 𝑎₂₁𝐴₂₁ + 𝑎₃₁𝐴₃₁

Example 4.1 : Matrices

Evaluate

1 0 3 4

2 1

−1 2

−1 2 1 1

0 0 2 5

(Solution)

Find row which has many zeros if possible => second row!

det

Chapter 4. Equation fitting

= 𝑎₂₁𝐴₂₁ + 𝑎₂₂𝐴₂₂ + 𝑎₂₃𝐴₂₃ + 𝑎₂₄𝐴₂₄

= 0 𝐴₂₁ + 1 −1 ²⁺²

1 −1 0

3 1 2

4 1 5

+ 2 −1 ²⁺³

1 2 0

3 −1 2

4 2 5

+ (0)𝐴₂₄

= 0 + 10 + 46 + 0 = 56

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

[ ][ ] [ ]

a a a x b

A X a a a x b B

a a a x b

     

     

        

     

     

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

a x a x a x b a x a x a x b a x a x a x b

  

  

  

 Simultaneous linear equations

 Matrix form

Chapter 4. Equation fitting

4.3 Solution of simultaneous equation

[ ] [ ] i

i

A matrix with B matrix substituted in th column

x  A

• Crammer’s rule

Chapter 4. Equation fitting

4.3 Solution of simultaneous equation

Example 4.2

Using Cramer’s rule, get 𝑥₂

2 1 −1

1 −2 2

−1 0 3

𝑥₁ 𝑥₂ 𝑥₃

= 3 9 0

(Solution)

𝑥₂ =

2 3 −1

1 9 2

−1 0 3

2 1 −1

1 −2 2

−1 0 3

= 30

−15 = −2

Coefficient matrix [A]

 

 

 

 

 

 

 

 

 

 

Triangular matrix

Back substitution

• Gaussian elimination

Chapter 4. Equation fitting

4.3 Solution of simultaneous equation

2

0 1 2

n

y  a  a x  a x   a x

• degree of the eq = highest exponent of x

• # of data point = degree + 1 → exact expression

> → best fit

Chapter 4. Equation fitting

4.4 Polynomial representations

 Points are equally spaced

 Derive 4^th degree polynomial

 n=4

2

0 1 0 2 0

3 4

3 0 4 0

( ) ( )

( ) ( )

n n

y y a x x a x x

R R

n n

a x x a x x

R R

   

       

   

       

1 0 n n 1

x x x x x _

     

2

0 1 2

n

y  a a x a x  a xn

(next page) Eq. (4.16)

4

x

Chapter 4. Equation fitting

4.6

2 3 4

1 0 1 0 1 0 1 0

1 1 2 3 4

1 2 3 4

4(x x ) 4(x x ) 4(x x ) 4(x x )

y a a a a

R R R R

a a a a

         

          

   

0 0

1 1 1 2 3 4

2 2 1 2 3 4

3 3 1 2 3 4

4 4 1 2 3 4

, ,

, 2 4 8 16

, 3 9 27 64

, 4 16 64 256

x x y y

x x y y a a a a

x x y a a a a

x x y a a a a

x x y a a a a

 

      

     

     

     

 if substitute (x1,y1)

 Substitute all the points to Equation (4.16)

Chapter 4. Equation fitting

4.6

Chapter 4. Equation fitting

4.6

2

0 1 2

y  a a xa x

1( 2)( 3) 2( 1)( 3) 3( 1)( 2)

y  c x x x  x c x  x x x c x  x x x

1, 1 1( 1 2)( 1 3)

x  x y c x x x x

2, 2 2( 2 1)( 2 3)

x  x y c x x x x

3, 3 3( 3 1)( 3 2)

x  x y c x x x x

1 1

( ) ( )

( ) ( )

n n

j i

i

i j i j i i

x x ommiting x x

y y

x x ommiting x x

 

 



 

Chapter 4. Equation fitting

4.7 Lagrange interpolation

2

1 : 1 1 1 1

S  P a b Q c Q

2

2 : 2 2 2 2

S  P a b Q c Q

2

3 : 3 3 3 3

S  P a b Q c Q

2

0 1 2

2

0 1 2

2

0 1 2

( ) ( ) ( )

a S A A S A S b S B B S B S c S C C S C S

  

  

  

( ) ( ) ( ) 2

P a S b S Q c S Q

   

Chapter 4. Equation fitting

4.8 Function of two variables

Example 4.3 : Function of two variables

The range is the difference between the inlet and outlet temperatures of the water. In table below, for example, when the wet-bulb temperature is 20℃

and the range is 10℃, inlet and outlet temperature are 35.9℃ and 25.9℃

each. Express the outlet temperature 𝒕 in Table below as a function of the wet-bulb temperature (WBT) and the range R

Chapter 4. Equation fitting

Wet-bulb temperature, ℃

Range, ℃ 20 23 26

10 25.9 27.5 29.4

16 27.0 28.4 30.2

22 28.4 29.6 31.3

(Solution)

For 3 WBTs, get parabola that represents (R,t) i) For WBT = 20℃

(R,t) : (10,25.9), (16,27.0), (22,28.4)

⇒ 𝑡 = 24.733 + 0.075006𝑅 + 0.004146𝑅² ii) For WBT = 23℃

⇒ 𝑡 = 26.667 + 0.041659𝑅 + 0.0041469𝑅² iii) For WBT = 26℃

⇒ 𝑡 = 28.733 + 0.024999𝑅 + 0.0041467𝑅²

Example 4.3 : Function of two variables

Chapter 4. Equation fitting

(Solution)

Set 2^nd degree equation (constant terms(C) , WBT) ∶ 24.733,20 , 26.667,23 , 28.733,26

⇒ 𝐶 = 15.247 + 0.32637𝑊𝐵𝑇 + 0.007380𝑊𝐵𝑇²

Set 2^nd degree equation (coefficient of 𝑅 , WBT) and (coefficient of 𝑅² , WBT) as well

𝑡 = 15.247 + 0.32637𝑊𝐵𝑇 + 0.007380𝑊𝐵𝑇² + 0.72375 − 0.050978𝑊𝐵𝑇 + 0.000927𝑊𝐵𝑇² 𝑅 +

(0.004147 + 0𝑊𝐵𝑇 + 0𝑊𝐵𝑇²)𝑅²

Chapter 4. Equation fitting

4.9 Exponential forms

ln ln ln

y bxm

y b m x



 



Log-log plot (y=bx^m)

If y approaches some value b, as orx   x  

Estimate b

Calculate m with log-log plot (y-b vs. x) Fitting (y vs. x^m)

Correct value of b

Curve y=b+ax^m

 y  b ax^m

Chapter 4. Equation fitting

4.9 Exponential forms

Misuse of least square method Example of least square method

 Misuses of least square method (a) Questionable correlation

(b) Applying too low degree

Chapter 4. Equation fitting

4.10 Best fit : Method of least squares

The sum of the squares of the deviation is a minimum

y  a bx

2( _{i i}) 0

z a bx y

a

    





2 1

( ) min

m

i i

i

z a bx y







2( _{i i}) _i 0

z a bx y x

b

    





i i

ma  b  x   y

2

i i i i

a  x  b  x   x y

Chapter 4. Equation fitting

4.10 Best fit : Method of least squares

• Method of least square for

Example 4.4 : Best fit : Method of least squares

Determine 𝑎₀ and 𝑎₁ in the equation 𝑦 = 𝑎₀ + 𝑎₁𝑥 to provide a best fit in the sense of least-squares deviation to the data points (1, 4.9), (3, 11.2), (4, 13.7), and (6, 20.1)

(Solution)

Chapter 4. Equation fitting

𝑥_𝑖 𝑦_𝑖 𝑥_𝑖² 𝑥_𝑖𝑦_𝑖

1 4.9 1 4.9

3 11.2 9 33.6

4 13.7 16 54.8

6 20.1 36 120.6

∑ 14 49.9 62 213.9

𝑚 = 4

4𝑎₀ + 14𝑎₁ = 49.9 14𝑎₀ + 62𝑎₁ = 213.9

i i

mab





2

i i i i

a







𝑦 = 1.908 + 3.019𝑥

4.10 Best fit : Method of Least Squares

• Method of least squares for ^{y  a bxcx2}

2

2 3

2 3 4 2

1 _{i i i}

i i i i i

i i i i i

a b x c x y

a x b x c x y x

a x b x c x y x

  

  

  

   

   

   

෍

𝑖=1 𝑚

(𝑎 + 𝑏𝑥_𝑖 + 𝑐𝑥₁² − 𝑦_𝑖)² → 𝑚𝑖𝑛

Chapter 4. Equation fitting

4.11 Method of Least Squares Applied to Nonpolynomial Forms

• Method of least squares

→ apply to equation with constant coefficients cf)

y  sin 2 ax  bx

Chapter 4. Equation fitting

4.12 The art of equation fitting

• Choice of the form of the equation

Polynomials with negative exponent⁽¹⁾ Exponential eq.⁽²⁾

Gompertz eq⁽³⁾ combination

, 1

cx

y  ab where b c 

(1) (2) (3)

Chapter 4. Equation fitting