Optimal Design of Energy Systems

Optimal Design of Energy Systems

Chapter 4 Equation Fitting

Min Soo KIM

Department of Mechanical and Aerospace EngineeringSeoul National University

EquationDevelopment

• 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

T

A Transpose of a matrix [A] => interchanging rows & columns

ex>

1 4

2 5

3 6

A

1 2 3

4 5 6

TA


4.2 Matrices

• Multiplying two matrices

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

𝑎11 ⋯ 𝑎1𝑛⋮ ⋱ ⋮

𝑎𝑚1 ⋯ 𝑎𝑚𝑛

×𝑏11 ⋯ 𝑏1𝑙⋮ ⋱ ⋮𝑏𝑛1 ⋯ 𝑏𝑛𝑙

⇒ 𝑚× 𝑙 𝑚𝑎𝑡𝑟𝑖𝑥


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


4.2 Matrices

• Determinant (scalar)

11 12 13

21 22 23

31 32 33

a a a

D a a a

a a a

11 22 33 12 23 31 13 21 32

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 32a a a a

11a [( 1) ]i

[ ]

i j

ij

submatrix formed

by striking outA

th row and j th

column of A

Cofactor of 𝒂𝒊𝒋


4.2 Matrices

= 𝑎11𝐴11 + 𝑎21𝐴21 + 𝑎31𝐴31

Example 4.1 : Matrices

Evaluate

1034

21−12

−1211

0025

(Solution)

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

det


= 𝑎21𝐴21 + 𝑎22𝐴22 + 𝑎23𝐴23 + 𝑎24𝐴24

= 0 𝐴21 + 1 −1 2+21 −1 03 1 24 1 5

+ 2 −1 2+31 2 03 −1 24 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


4.3 Solution of simultaneous equation

[ ] [ ] ii

A matrix with B matrix substituted in th columnx

A

• Crammer’s rule



Example 4.2

Using Cramer’s rule, get 𝑥2

2 1 −11 −2 2−1 0 3

𝑥1𝑥2𝑥3

=390

(Solution)

𝑥2 =

2 3 −11 9 2−1 0 32 1 −11 −2 2−1 0 3

=30

−15= −2

Coefficient matrix [A]

Triangular matrix

Back substitution

• Gaussian elimination



2

0 1 2

n

ny a a x a x a x

• degree of the eq = highest exponent of x

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

> → best fit


4.4 Polynomial representations

Points are equally spaced

Derive 4th degree polynomial

n=4

2

0 1 0 2 0

3 4

3 0 4 0

( ) ( )

( ) ( )

n ny y a x x a x x

R R

n na x x a x x

R R

1 0 1n nx x x x x

2

0 1 2

n

ny a a x a x a x

(next page)

Eq. (4.16)

4

x


4.6 Simplifications when the independent variable is uniformly spaced

2 3 4

1 0 1 0 1 0 1 01 1 2 3 4

1 2 3 4

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





2

0 1 2y 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

( ) ( )

( ) ( )

nnj i

i

i j i j i i

x x ommiting x xy y

x x ommiting x x


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


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 thewater. 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 thewet-bulb temperature (WBT) and the range R


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


(Solution)

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

⇒ 𝐶 = 15.247 + 0.32637𝑊𝐵𝑇 + 0.007380𝑊𝐵𝑇2

Set 2nd 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


4.9 Exponential forms

ln ln ln

my bx

y b m x

Log-log plot (y=bxm)

If y approaches some value b, as orx x

Estimate b

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

Fitting (y vs. xm)

Correct value of b

Curve y=b+axm

my b ax


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


4.10 Best fit : Method of least squares

The sum of the squares of the deviation is a minimum

y a bx

2( ) 0i i

za bx y

a

2

1

( ) minm

i i

i

z a bx y

2( ) 0i i i

za bx y x

b

i ima b x y 2

i i i ia x b x x y


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 inthe sense of least-squares deviation to the data points (1, 4.9), (3, 11.2), (4,13.7), and (6, 20.1)

(Solution)


𝑥𝑖 𝑦𝑖 𝑥𝑖2 𝑥𝑖𝑦𝑖

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 = 44𝑎0 + 14𝑎1 = 49.914𝑎0 + 62𝑎1 = 213.9

i ima b x y 2

i i i ia x b x x y

𝑦 = 1.908 + 3.019𝑥

4.10 Best fit : Method of Least Squares

• Method of least squares for2y a bx cx

2

2 3

22 3 4

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 + 𝑏𝑥𝑖 + 𝑐𝑥12 − 𝑦𝑖)

2 → 𝑚𝑖𝑛


4.11 Method of Least Squares Applied to Nonpolynomial Forms

• Method of least squares

→ apply to equation with constant coefficients

cf) sin 2 cy ax bx


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

, 1xcy ab where b c

(1) (2) (3)


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Optimal Design of Energy Systems