Process Dynamics and Control : Chapter 5 Lectures

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Chapter 5

1

DYNAMIC BEHAVIOUR OF

FIRST ORDER

AND

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Chapter 5 Dynamic Behavior

Chapter 5

In analyzing process dynamic and process control systems, it is important to know how the process responds to changes in the process inputs.

A number of standard types of input changes are widely used for process modelling and control for two reasons:

1. They are representative of the types of changes that occur in plants.

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Chapter 5

3

A sudden change in a process variable can be approximated by a step change of magnitude, M:

Chapter 5

0 0 (5-4) 0 s t U M t < ⎧ ⎨ _≥ ⎩

• Special Case: If M = 1, we have a “unit step change”. We

give it the symbol, S(t).

The step change occurs at an arbitrary time denoted as t = 0. =

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Chapter 5

2. The heat input to the stirred-tank heating system in Chapter 2 is suddenly changed from 8000 to 10,000 kcal/hr by changing the electrical signal to the heater. Thus,

( )

8000 2000 , unit step 2000 , 8000 kcal/hr Q t S t S t Q t Q Q S t Q = + ′ = − = = and

Example:

1. A reactor feedstock is suddenly switched from one supply to another, causing sudden changes in feed concentration,

flow, etc.

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Chapter 5

5

Chapter 5

We can approximate a drifting disturbance by a ramp input:

( )

0 0 (5-7) at 0 R t U t t < ⎧ ⎨ _≥ ⎩

Examples of ramp changes:

1. Ramp a setpoint to a new value. (Why not make a step change?)

2. Feed composition, heat exchanger fouling, catalyst activity, ambient temperature.

=

• Industrial processes often experience “drifting

disturbances”, that is, relatively slow changes up or down for some period of time.

• The rate of change is approximately constant.

2. Ramp Input

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Chapter 5

Chapter 5 ( )

0 for 0 for 0 (5-9) 0 for RP w w t U t h t t t t < ⎧ ⎪ _{≤ <} ⎨ ⎪ _≥ ⎩ Examples:

1. Reactor feed is shut off for one hour.

2. The fuel gas supply to a furnace is briefly interrupted.

0

h

X_RP

T_w Time, t It represents a brief, sudden change in a process variable:

3. Rectangular Pulse

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Chapter 5

7

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Chapter 5

Examples:

1. 24 hour variations in cooling water temperature. 2. 60-Hz electrical noise (in the USA)

Processes are also subject to periodic, or cyclic, disturbances. They can be approximated by a sinusoidal disturbance:

( )

_{( )}

sin 0 for 0 (5-14) sin for 0 t U t A

ω

t t < ⎧ ⎨ _≥ ⎩ where: A = amplitude, ω = angular frequency

4. Sinusoidal Input

=

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Chapter 5

9

Chapter 5

Examples:

1. Electrical noise spike in a thermo-couple reading. 2. Injection of a tracer dye.

• Here,

• It represents a short, transient disturbance.

( )

.

I

U t =

δ

t

5. Impulse Input

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Chapter 5

The standard form for a first-order TF is:

where:

Consider the response of this system to a step of magnitude, M:

Substitute into (5-16) and rearrange,

First-Order System

( )

τ 1 (5-16) Y s _K U s = s +

( )

for 0

( )

M U t M t U s s = ≥ ⇒ =

( ) ( )

(5-17) τ 1 KM Y s s s = + steady-state gain τ time constant K = =

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Chapter 5

11

Chapter 5

Take L-1 _{(cf. Table 3.1),}

( )

(

₁ t/ τ

)

_(5-18) y t = KM −e− . y_∞ = KM t ___ 0 0 0.632 0.865 0.950 0.982 0.993 y y_∞ τ 2τ 3τ 4τ 5τ

Note: Large τ means a slow response. y

y_∞

τ

t

Let steady-state value of y(t). From (5-18), y_∞ =

First-Order System

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Chapter 5 Response of First Order TF for other inputs

- Response of a first order system for a ramp input

(sec 5.2.2)

- Response of a first order system for a sinusoidal input (sec 5.2.3)

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Chapter 5

13

Chapter 5

Consider a step change of magnitude M. Then U(s) = M/s and,

Integrating Process

Not all processes have a steady-state gain. For example, an “integrating process” or “integrator” has the transfer function:

( )

(

constant

)

Y s _K K U s = s =

( )

KM₂

( )

Y s y t KMt s = ⇒ =

Thus, y(t) is unbounded and a new steady-state value does not exist.

L-1

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Chapter 5

Consider a liquid storage tank with a pump on the exit line:

Common Physical Example:

- Assume:

1. Constant cross-sectional area, A. 2.

- Mass balance:

- Eq. (1) – Eq. (2), take L, assume steady state initially,

- For (constant q),

( )

q ≠ f h (1) 0 ( 2 ) i i d h A q q q q d t = − ⇒ = −

( )

1 i

( )

H s Q s Q s As ′ = ⎡_⎣ ′ − ′ ⎤_⎦

( )

0 ′ =

( )

1 H s Q s As ′ = ′

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Chapter 5

15

Chapter 5

• Standard form:

Second-Order Systems

( )

_τ2 2 _2ζτ ₁ (5-40) Y s _K U s = _s + _s +

which has three model parameters: steady-state gain

τ "time constant" [=] time

ζ damping coefficient (dimensionless)

K

• Equivalent form: natural frequency 1 τ n

ω

⎛ ₌ ⎞ ⎜ ⎟ ⎝ ⎠

( )

2 2 _2ζ 2 n n n Y s _K U s _s _s

ω

= + + = = = = Dr. M. A. A. Shoukat Choudhury

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Chapter 5 When??

Composed of two first order subsystems (G₁ and G₂)

Or, Y₁(s)

)

1 )(

1 (

1

1 )

(

)

(

)

(

)

(

2 1 2 1 2 2 1 1 2 1 2

+

=

+

×

+

=

s

K

s

K

s

K

s

G

s

G

s

U

s

Y

τ

Y₂ (s) Y₂ (s)

Second-Order Systems

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Chapter 5

17 2 1

τ

=

1

2 K

=

G(s)

₂ ₂

+

s

ζτ

τ

2 1 2 1

2 =

τ

+

τ

ζ

Roots/Poles:

τ

ζ

±

2

−

1 −

• Standard form

The Characteristic Polynomial

Second-Order Systems

2 2

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Chapter 5

• The type of behavior that occurs depends on the numerical value of damping coefficient, :ζ

It is convenient to consider three types of behavior:

Damping Coefficient

Type of Response Roots of Charact. Polynomial

Overdamped Real and ≠ Critically damped Real and = Underdamped Complex

conjugates

ζ 1> ζ 1= 0 ζ 1≤ <

• Note: The characteristic polynomial is the denominator of the transfer function:

2 2

τ s + 2ζτs +1

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Chapter 5

19

Where,

Case 1: when Over damped Response

For a unit step input, the response will be:

)

)(

(

)

(

2 1

s

P

s

K

s

y

p

−

=

Second-Order Systems

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Chapter 5

Case 2: when Critically damped Response

Case 3: when Underdamped Response

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Chapter 5

21

Chapter 5

Dr. M. A. A. Shoukat Choudhury

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Chapter 5

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Chapter 5

23

Chapter 5

1. Responses exhibiting oscillation and overshoot (y/KM > 1) are obtained only for values of less than one.

2. Large values of yield a sluggish (slow) response.

3. The fastest response without overshoot is obtained for the critically damped case

Several general remarks can be made concerning the responses show in Figs. 5.8 and 5.9:

ζ ζ

(

ζ 1 .=

)

Dr. M. A. A. Shoukat Choudhury

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Chapter 5

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Chapter 5

25

Chapter 5

1. Rise Time: is the time the process output takes to first reach the new steady-state value.

2. Time to First Peak: is the time required for the output to reach its first maximum value.

3. Settling Time: is defined as the time required for the

process output to reach and remain inside a band whose width is equal to ±5% of the total change in y. The term 95%

response time sometimes is used to refer to this case. Also, values of ±1% sometimes are used.

4. Overshoot: OS = a/b (% overshoot is 100a/b).

5. Decay Ratio: DR = c/a (where c is the height of the second peak).

6. Period of Oscillation: P is the time between two successive peaks or two successive valleys of the response.

r t p t s t Dr. M. A. A. Shoukat Choudhury

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Chapter 5

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Chapter 5 Exercise Problem

27

Problem 5.14: A step change from 15 to 31 psi in actual pressure results in the measured response from a pressure indicator as shown below, (a)Assuming second order dynamics, calculate all important parameters

and write an approximate transfer function in the form

1 2 K = ) ( ) ( 2 2 ' ' + + s s s P s R ζτ τ

Where, R’ is the instrument output deviation (mm), P’ is the actual pressure deviation (psi)

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