chapter 7 chang

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1 Development of Empirical Dynamic Models from Step Response Data Some processes too complicated to model using physical principles material, energy balances flow dynamics physical properties (often unknown) thermodynamics

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Page 1: Chapter 7 Chang

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Development of Empirical Dynamic Models from Step Response Data

Some processes too complicated to model using physical principles

• material, energy balances

• flow dynamics

• physical properties (often unknown)

• thermodynamics

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Black Box Models

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Step Input

• Step response is the easiest to use but may upset the plant manager

• Other methods– impulse - dye injection, tracer

– random - PRBS (pseudo random binary sequences)

– sinusoidal - theoretical approach

– frequency response - modest usage (incl. pulse testing)

– on-line (under FB control)

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Fitting of 1st-Order Model

/

0

1

1

0.632

1 1

t

t

K MG s U s

s s

y t KM e

y KM

dy

KM dt

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(θ = 0)

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FOPDT and SOPDT Models

2 2

First-Order-Plus-Dead-Time (FOPDT) Model

1

Second-Order-Plus-Dead-Time (SOPDT) Model

2 1

s

s

KeG s

s

KeG s

s s

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For a 1st order model, we note the following characteristics in step response:

1. The response attains 63.2% of its final response at one time constant (t = ).

2. The line drawn tangent to the response at maximum slope (t = ) intersects the 100% line at (t = ).

There are 3 generally accepted graphical techniques for determining the first-order system parameters and .

( )1

sKeG s

s

Fitting of FOPDT ModelC

hap

ter

7

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Method 1: Sundaresan & Krishnaswany (1978)

1. Find K from stead-state response.2. Normalize step response by dividing all data with KM (t =

0, y = 0; t →∞, y = 1)

3. Use 35.3% and 85.3 % response times (t1 and t2), i.e.

4. Calculate

= 1.3 t1 – 0.29 t2

= 0.67 (t2 – t1)

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1

2

0.353

0.853

y t KM

y t KM

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Method 2: Numerical Fitting

(1) Find and in ( ) 1

to fit data of vs.

(2) Find and in ln

to fit data of ln vs.

t

y t KM e

y t

KM y t t

KM

KM y tt

KM

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Method 3: Fitting an Integrator Model to Step Response Data

In Chapter 5 we considered the response of a first-order process to a step change in input of magnitude M:

/1 1 ty t KM e

For short times, t < , the exponential term can be approximated by

/ τ 1τ

t te

so that the approximate response is:

1 1 1τ

t KMy t KM t

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is virtually indistinguishable from the step response of the integrating element

22 (7-23)

KG s

s

In the time domain, the step response of an integrator is

2 2 (7-24)y t K Mt

Hence an approximate way of modeling a first-order process is to find the single parameter

2 (7-25)τ

KK

that matches the early ramp-like response to a step change in input.

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Figure 7.10. Comparison of step responses for a FOPTD model (solid line) and the approximate integrator plus time delay model (dashed line).

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Fitting 2nd-Order Models

1 21 1

MU s

sK

G ss s

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Harriot’s Method

1.3

0.73

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0.39

0.26

1 2

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Harriot’s Method

0.73

0.73

0.731 2

0.5 1 2

0.5 0.5

1) Determine experimentally to satisfy

0.73

2) Calculate 1.3

3) Calculate 0.5

4) Determine from experimental data

4) From Figure 7.6, det

t

y t KM

t

t

y y t

1 2.ermine and then calculate

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Smith’s Method

60

20

20

60

1) Determine t and t experimentally so that

0.6

0.2

2) Fig 7.7 ,

y t KM

y t KM

t

t

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1.3=

1.79= 8.260

t

84.0

81.3

2

1

1 2 Sum of squares

S 3.81 0.84 0.0757 NLR (θ=0) 2.99 1.92 0.000028

FOPTD (θ = 0.7) 4.60 - 0.0760

Smith’s Method

20% response: t20 = 1.8560% response: t60 = 5.0t20 / t60 = 0.37

from graph

Solving,

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