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CHE302 Process Dynamics and Control Korea University 7-1

CHE302 LECTURE VII

DYNAMIC BEHAVIORS OF

REPRESENTATIVE PROCESSES (II)

Professor Dae Ryook Yang

Fall 2001

Dept. of Chemical and Biological Engineering Korea University

INVERSE RESPONSE IN CHEMICAL PROCESSES

• Reboiler level to the change in boilup rate

– If boilup rate is increase, more liquid will

vaporized and the inventory in the reboiler will be reduced and the level will be decreased.

– However, at the beginning of the boilup rate increase, more vapor is generated and the vapor will flow upward to the stage above. In that stage, more vapor will pass through the liquid and the density of liquid will decrease so that the more liquid will spill over the weir. This results a temporary increase in the level of reboiler.

– Eventually, the density of liquid settles down and overflow will reach at another steady state.

Then the reboiler level will be decreased. Steam Bottom

Product

Seal pan Weir

Downcomer

Reboiler level

Condensate

Boilup

(2)

CHE302 Process Dynamics and Control Korea University 7-3

• Outlet Temp of the exothermic packed bed reactor to feed temp change

– An increase in feed temp will speed up the reaction rate and the reactor outlet temp will increase due to the increased reaction heat generated.

– However, at the beginning of the feed temp increase, more reaction occurs in the inlet part of the reactor and more reactants are consumed. This causes a decrease of reactant concentration in the outlet part of the reactor, and the outlet temp will decrease due to temporary drop in reactant

concentration.

– Eventually, increase in feed temp will enhance the reaction and generate more reaction heat and the outlet temp will increase slowly.

Packed-bed reactor

Feed Product

Tf, CAf T, CA

TIME DELAYS

• Fluid transportation through a pipe

– Also, called distance-velocity lag, transportation lag, dead time

• Transfer function

• Many high-order system can be approximated by a first-order plus dead-time model (FOPDT)

length of pipe volume of pipe fluidvelocity volumetric flowrate

θ = =

L v

t f(t)

θ

0 for

( ) ( ) for

y t t

x t t

θ

θ θ

 <

= − ≥ ⇒

( ) ( )

( ) Y s s

G s e X s

θ

= =

( ) 1

Ke s

G s s

θ

τ

=

+

But this transfer function cannot tell what the zeros are and what the poles are if it appears in denominator.

(3)

CHE302 Process Dynamics and Control Korea University 7-5

• Polynomial approximations to

– Taylor series approximation

– 1/1 Pade approximation

– 2/2 Pade approximation

e

θs

2 2 3 3 4 4 5 5

1 2! 3! 4! 5!

s s s s s

eθ = −θs+θθ +θθ +L

2 2 3 3

1 1

2! 3! 1

s s s bs

e s

as

θ θ θ θ

= − + − + ≈ + L +

2 2

1 (+ −a θ)s+(0.5θ)s + ≈ +L 1 bs , (0.5 2 ) 0

/ 2, / 2

a b a

a b

θ θ θ

θ θ

− = =

⇒ = = − 1

1 ( /2) ( ) 1 ( / 2) G s s

s θ θ

= +

2 2 3 3 4 4 2

2

1 1

2! 3! 4! 1

s s s s bs ds

e s

as cs

θ θ θ θ θ

= − + − + − ≈ + +

L + +

2 2 2 3 2

1 (+ −a θ)s+(0.5θ aθ +c s) +(aθ / 2θ / 6cθ)+ ≈ + +L 1 bs ds

2 2

2 2 2

1 ( /2) ( /12)

( ) 1 ( /2) ( /12)

s s

G s s s

θ θ

θ θ

− +

= + +

2

2 3 4 3 2

2

, (0.5 ) ,

( / 2 / 6 ) 0, ( /24 / 6 /2) 0

/2, / 2, /12

a b a c d

a c a c

a b c d

θ θ θ

θ θ θ θ θ θ

θ θ θ

− = − + =

= + =

⇒ = = − = =

• Taylor series approximation

– Increase the order of numerator (physical realizability) – AR and phase angle are different from the exact

– Low accuracy

• Pade approximation

– Does not change the order of transfer function – Only phase angle is different

– Higher accuracy

– Oscillatory behavior (complex poles and zeros)

Pure delay FOPDT

(4)

CHE302 Process Dynamics and Control Korea University 7-7

APPROXIMATION OF HIGHER-ORDER SYSTEMS

• Approximation by FOPDT or SOPDT models

1 2

( ) ( 1)( 1) ( n 1)

G s K

s s s

τ τ τ

= + + L +

1 2 1

If (τ ? τ ,L,τ n), ( is theτ "dominant time constant")

2 1

( ) where

( 1)

s

n

G s Ke

s

θ θ τ τ

τ

= + +

+ L

1 2 3 1 2

If ,τ τ ?(τ ,L,τ n), ( and areτ τ the "dominant timeconstants")

3

1 2

( ) where

( 1)( 1)

s

n

G s Ke

s s

θ θ τ τ

τ τ

= + +

+ + L

• nth-order system with equal time constants

– Step response ( )

1

n n

G s K

n s

= τ

 + 

 

 

1 /

0

( / )

( ) 1

!

n i nt

i

y t KM e nt

i

τ τ

=

=

/ 1/

0

lim ( ) lim lim

lim(1 )

( 1) ( 1)

s

n s s

n n n n n s x

x

K K K

G s Ke

s x

n s n

τ τ τ

τ τ τ

→∞ →∞ →∞

= = = =

+

+ +

1/

By deinition, lim(10 ) x 2.7182818285...

e x x

+ =

@

(5)

CHE302 Process Dynamics and Control Korea University 7-9

FITTING DATA TO EMPIRICAL MODELS

Fitting FOPDT model using step test

With a step response 1. Obtain gain

2. Estimated time delay by inspection

Visual inspection

Use of tangent line at inflection point

3. Find time constant

Graphical method

Rearrange data: the slope is the reciprocal of time constant

Sundaresan and Krishnaswamy (1977)

( ) (0) ( ) (0)

y y

K u u

= ∞ −

∞ −

( ) ( ( ) (0)) ( )

ln ( ) (0)

t KM y t y y yi ti

e KM y y

θ τ θ

τ

− − ∞ −

= ∞ − = −

35.3% 85.3% 85.3% 35.3%

1.3t 0.29t 0.67(t t )

θ = τ =

(0) /

y =KM τ

• Fitting SOPDT model using step test

– Harriot’s Method for overdamped systems

– Smith’s method

• Use t20% and t60%.

• Read parameters from graph

• Nonlinear/Linear Regression

1 2 t73% /1.3

τ τ+ =

73%/2.6 1 1 2

Calculate and read /(yt τ τ τ+ ) from graph

If graph is out of range, it is not applicable and possibly higher- order or underdamped.

Use of optimization to minimize the error between the data and calculated model value.

2 , , ,

1

min ( ( ) ) where ( ) ( , , , , )

N

i i i i

K i

y t y y t f t K

τ ζ θ τ ζ θ

=

=

(6)

CHE302 Process Dynamics and Control Korea University 7-11

INTERACTING AND NONINTERACTING PROCESSES

• Non interacting process

– The first tank affects the second tank but second tank does not affect the first tank and so on. This is called “noninteracting”.

– Each tank can be modeled as

– For the whole system

cAi, qi

cAn, q

V1

V2

Vn

cA1, q

cA2, q

… . ( 1)

( )

( ) 1

Aj j

A j j

C s K

C s =τ s +

1

( )

( ) ( 1)

n An j

Ai j j

C s K

C s = = τ s

+

• Interacting process

– Many chemical processes exhibit interacting nature.

qi

q1 q2

R1 R2

V1 h1 V2 h2

1 1 2

1

1 ( )

q h h

= R

1

1 i 1

A dh q q

dt = − 2 dh2 1 2

A q q

dt = −

2 2

2

q 1 h

=R

1

1 1 2

1

1 ( )

i

A dh q h h

dt = − R 2 2 1 2 2

1 2

1 1

( )

A dh h h h

dt = R R

1 1 1( ) 1( ) 2( ) 1 i( )

A R sH s% +H s% H s% = R Q s%

2 1 2 2 2 2 1 2

2 2 1

1 2 1 2 1 2 1 2 1 2

( ) /( )

( ) ( ) ( )

( ) /( ) 1

A R R R H s R R R

sH s H s H s

R R R R H s A R R R R s

+ = = +

+ + + +

% % % %%

2 2

2

1 2 1 2 1 1 2 2 1 2

( )

( ) 1

i( )

H s R

A A R R s A R A R A R s Q s =

+ + + +

%%

2 2

2 2

( )

2 1

i( )

H s R

s s

Q s =τ ζτ

+ +

%% 1 2 1 2 1 1 12 22 1 21 2

( )

where ,

2

A R A R A R A A R R

A A R R

τ = ζ = + +

Since arithmetic mean geometric mean, >1 (overdamped)ζ

(7)

CHE302 Process Dynamics and Control Korea University 7-13

STAGED SYSTEM

• Three-stage absorber

– Assumptions

• Constant liquid holdup (H)

• Perfect mixing on each stage

• Neglecting holdup of gas

• Equilibrium:

– Definition:

• The stage liquid residence time

• The stripping factor

• The gas-to-liquid ratio

i i

y =ax +b

1 1

( ) ( )

i

i i i i

H dx G y y L x x

dt = − +

/ τ = H L /

σ = aG L / K =G L

1 ( ) 1

i

i i i

H dx aGx L aG x Lx

dt = − + + +

– With

1

1 2

2

1 2 3

3

2 3

( ) (1 )

(1 )

(1 )

f

f

dx K y b x x

dt

dx x x x

dt

dx x x x

dt

τ σ

τ σ σ

τ σ σ

= − − + +

= − + +

= − + +

1 2

1 2 3

2 3

0 ( ) (1 )

0 (1 )

0 (1 )

f

f

K y b x x

x x x

x x x

σ

σ σ

σ σ

= − − + +

= − + +

= − + +

( )

( )

( )

2 1

2 2

2

2 3

( ) /(1 )

(1 ) ( ) /(1 )

(1 ) (1 ) ( ) /(1 )

f f

f f

f f

i i

x x K y b

x x K y b

x x K y b

y ax b

σ σ

σ σ σ σ

σ σ σ

= + +

= + + + + +

= + + +

= +

1

1 2

2

1 2 3

3

2 3

(1 )

(1 )

(1 )

f

f

dx Ky x x

dt

dx x x x

dt

dx x x x

dt

τ σ

τ σ σ

τ σ σ

= − + +

= − + +

= − + +

% % % %

% % % %

% % % %

1 2

2 1 3

3 2

( 1 ) ( ) ( ) ( )

( 1 ) ( ) ( ) ( ) 0

( 1 ) ( ) ( ) ( )

( ) ( )

f

f

i i

s X s X s KY s

s X s X s X s

s X s X s X s

Y s aX s

τ σ

τ σ σ

τ σ σ

+ + =

+ + =

+ + =

=

2

2

2 2

2

1 2 3 2

2

2 2

2 (1 )

1 1 1

( ) 1 ( )

1 2 (1 )

1 1

1 1 1

f

s s

X s K Y s

s s s

τ τ σ

σ σ σ σ

σ σ

σ σ σ τ τ τ σ

σ σ σ

+

+ +

+ + + +

+ +

= + + + + +  + + ++ + ( ) 0

X%f s =

(8)

CHE302 Process Dynamics and Control Korea University 7-15

– Step increase in gas feed composition

2 2 2

2

2 2

( ) 1 ( )

1 2 (1 )

1 1 1

X s K Y sf

s s

σ

σ τ τ σ

σ σ

= + + + ++ +

2

3 2 3 2

2

2 2

( ) 1 ( )

1 2 (1 )

1 1

1 1 1

X s K Y sf

s s s

σ

σ σ σ τ τ τ σ

σ σ σ

= + + + + +  + + ++ +

3 3

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

( )

( ) ( ) ( ) ( ) ( ) ( )

f f f f f f

Y s aX s Y s aX s Y s aX s

Y s = Y s Y s = Y s Y s = Y s aK =σ

3rdorder+two zeros 3rdorder

2ndorder

• Analysis using transfer function

– Easy to develop

– Types of response can be recognized very conveniently

– Parameter effects can be analyzed through lumped parameter – Linearization required when the system is nonlinear

• If the effect of flow rate change, the previous example becomes nonlinear.

• Due to linearization, new model has to be obtained when the operating condition changes widely.

– For the nonlinear system analysis, numerical integration should be considered

• Utilize differential equation solvers – Non-stiff case: explicit method – Stiff case: implicit methods – ODE

– DAE – PDE

(9)

CHE302 Process Dynamics and Control Korea University 7-17

DISTRIBUTED PARAMETER SYSTEMS

• Lumped parameter system (ODE)

– Dependent variables depend only on time, not on spatial location

– Perfect mixing assumption eliminates the dependency on – spatial coordinates and the analysis will be conducted in

averaging sense

– All balances are valid around the boundary

• Distributed parameter system (PDE)

– Dependent variables depend on time and also spatial location – Perfect mixing is not valid. (double pipe heat exchanger,

packed bed reactor

DPS EXAMPLE

• Double pipe heat exchanger

– Assumptions

• Plug flow (no radial or angular variations in TL)

• Neglect axial conduction

• Heat transfer resistance of the inside tube metal is neglected

• Steam jacket is well mixed

SL

AL

(10)

CHE302 Process Dynamics and Control Korea University 7-19

• For a control volume, from the energy balance for the liquid

(ALis a heat transfer area per unit length)

• For a control volume, from the energy balance for the wall ( , )

( ( , ) ) ( ( , ) )

( ( , ) ( , ))

L

L L L L L L L ref L L L L ref

L L W L

dT t z

C S z v C S T t z T v C S T t z z T

dt

h A z T t z T t z

ρ = ρ ρ + ∆ −

+

( , ) ( ( , ) ( , ))

( ( , ) ( , ))

L L L L L

W L

L L L

dT t z v T t z z T t z h A

T t z T t z

dt z ρ C S

+ ∆ −

= − +

1 ( ) where

L L L L L

W L HL

HL L L

dT T C S

v T T

dt z h A

τ ρ τ

= − + =

As 0∆ →z

( ) ( )

W S S L L

S W W L

W W W L L L

dT h A h A

T T T T

dt = ρ C S ρ C S

1 1

( ) ( )

W

S W W L

SW W L

dT T T T T

dt =τ τ

• Discretization

– PDE => ODE

– Eliminate z-dependency – Replace

• Number of point N will decide the accuracy of solution

– Finite difference approximation (FDM)

– Other methods: method of characteristics, FEM, OC, etc.

( , ) ( ) for 0, ,

( , ) ( ) where (0) 0, , ( )

j

L j L

j

LW j W

T t z T t j N

T t z T t z z N L

= =

= = =

L

L

( , ) ( , )

(forward difference approximation) ( , ) ( , )

(backward difference approximation)

( , ) ( , )

(central difference approximation) 2

L L L

L L L

L L L

T T t z z T t z

z z

T T t z T t z z

z z

T T t z z T t z z

z z

+ ∆ −

=

= − ∆

= + ∆ − − ∆

(11)

CHE302 Process Dynamics and Control Korea University 7-21

• Using FDM with centered difference approx.

• Initial conditions and boundary conditions

• Inputs: Steam T (T

S

(t)) and Feed T (T

F

(t))

• Outputs: T

L

(t) and T

W

(t) at each location in z

• 2 PDE’s become 2N ODE’s

0 0 0

0 0 0

0

, , ,

0

, , ,

0

0

(0, ) ( ) (0) (0), , (0) ( ) (I.C. for )

(0, ) ( ) (0) (0), , (0) ( ) (I.C. for )

( ,0) ( ) ( ) ( ) (B.C. for )

( , 0) ( ) ( ) (

N

L L t L L t L L t N L

N

W W t W W t W W t N W

L F L F L

W A W A

T z T z T T T T z T

T z T z T T T T z T

T t T t T t T t T

T t T t T t T

= = =

= = =

= =

= =

L L

) (B.C. for W)

t T

1 2 0

1 1

2 3 1

2 2

1

1 ( )

2

1 ( )

2

1 ( )

L L L

W L

HL

L L L

W L

HL

N N N

N N

L L L

W L

HL

dT T T

v T T

dt z

dT T T

v T T

dt z

dT T T

v T T

dt z

τ τ

τ

= − +

= − +

= − +

M

1

1 1 1

2

2 2 2

1 1

( ) ( )

1 1

( ) ( )

1 1

( ) ( )

W

S W W L

SL HL

W

S W W L

SL HL

N

N N N

W

S W W L

SL H L

dT T T T T

dt

dT T T T T

dt

dT T T T T

dt

τ τ

τ τ

τ τ

=

=

=

M

• Transfer function

– FDM usually generates tri-diagonal matrix.

– Quite high order system if N is big.

– May not be convenient to analyze.

1

1 1 1

1 1 2 2 2

2

1

1 1

1 0 0

1 0

( )

0 0

0 0 1 0

L

L W F

L

W L

W F

N N N

W

N L L

s b

T

T K

c s b

T T

K T t

c

b

T T

c s

τ

τ

τ

 +      

 +      

      

   =  + 

      

      

 + 

 

L

O M

O O M M M

M O O O

L

1 1

1

2 2

2

1 0 0

0 1

0 ( )

0 0 1

W W L S

W W L S

L S

N N

WN W L S

s T T K

s T T K

K T t

s T T K

τ

τ

τ

 +       

       

 +   =  + 

       

       

 +       

     

L

O M

M O O M M M

L

Tri-diagonal matrix

(12)

CHE302 Process Dynamics and Control Korea University 7-23

MIMO systems

• Multi-Input Multi-Output System

– MISO (n=1) and SIMO (m=1) are possible.

– 2x2 system

1 11 12 1 1

2 21 22 2 2

1 2

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

m m

n n n nm m

Y s G s G s G s U s

Y s G s G s G s U s

Y s G s G s G s U s

    

    

  =  

    

    

    

    

L

M M O M M

L

Transfer function matrix

G11 G21 G12 G22 U1(s)

U2(s)

Y1(s)

Y2(s) + +

+ +

• From SS model to TF model for MIMO system

– With deviation variables after linearization

– Generally for (pxm) system, A is (nxn), B is (nxm), C is (pxn) and D is (pxm) matrices, respectively.

– Then, G(s) is (pxm) transfer function matrix.

1 1 1

2 2 2

1 1 1

2 2 2

x x u

x x u

y x u

y x u

     

= + ⇒

     

     

     

= + ⇒

     

     

A B x = Ax + Bu

C D y = Cx + Du

&

&

&

1

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

s s s s s s s

s s s s s s

= + ⇒ − =

 

= + ⇒  − 

X AX BU I A X BU

Y CX DU Y = C I A B + D U

( ) s  ( s )

1

G = C I  − A B + D

참조

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