Development - Process Control - Lecture Slides, Slides of Process Control

This lecture is from Process Control course. Some key points for this lecture are: Development, Material, Energy Balances, Flow Dynamics, Physical Properties, Thermodynamics, Step Input, Dye Injection, Random, Sinusoidal

Typology: Slides

2012/2013

Uploaded on 03/18/2013

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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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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

2

Black Box Models

4

Chapter 7

5

Fitting of 1st-Order Model

( ) ( )

( ) (^) ( )

( )

/

0

t

t

K M

G s U s

s s

y t KM e

y KM

dy

KM dt

τ

τ

τ

τ

=

7

FOPDT and SOPDT Models

( )

( ) (^2 )

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

s

s

Ke
G s
s
Ke
G s
s s

θ

θ

8

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

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

( ) 1

s Ke G s s

θ

Fitting of FOPDT Model

Chapter 7

10

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 (t 1 and t 2 ), i.e.
  4. Calculate θ = 1.3 t 1 – 0.29 t τ = 0.67 (t 2 – t1 )

Chapter 7

1 2

y t KM y t KM

=

11

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 t
t
KM

θ

13

Chapter 7

is virtually indistinguishable from the step response of the integrating element

2 ( )^2 (7-23)

K G s s

=

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

y 2 ( ) t = K Mt 2 (7-24)

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

(^2) τ (7-25)

K K =

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

14

Chapter 7

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

16

Chapter 7

Harriot’s Method

1.

0.

17

0.

0.

τ 1 ≥τ 2

19

Smith’s Method

( )

( )

60

20 20 60

1) Determine t and t experimentally so that
2) Fig 7.7 ,
y t KM
y t KM
t
t

ζ τ

20

Chapter 7