for chemical engineering fifth year

1. Unit Three
Dynamic Behavior of First , Second and higher Order Processes
1

2. • Dynamic behavior can be seen as the deviation of the output variable from its desired value
In control, dynamic behavior arises:
– During startup and shut down of operations
– Due to disturbances
2

3. Dynamic Behavior of First-Order Processes
 In analyzing process dynamic and process control systems,
it is important to know how the process responds to
changes in the process inputs.
 Process inputs falls into two categories:
1. Inputs that can be manipulated to control the process.
2. Inputs that are not manipulated, such as disturbance
variables.
3

4.  A number of standard types of input changes are widely
used for two reasons:
1. They are representative of the types of changes that
occur in plants.
2. They are easy to analyze mathematically.
First Order Process System
• It is one whose output y(t) is modeled by the first
order of differential equation.
4

5. a) 𝒂𝟏𝒚′ + 𝒂𝒐𝒚 = 𝒃𝒇(𝒕) take the Laplace
⟹ 𝒂𝟏𝒔𝒀 𝒔 + 𝒂𝒐𝒀 𝒔 = 𝒃𝑭 𝒔 ⟹
𝒂𝟏
𝒂𝒐
𝒔𝒀 𝒔 + 𝒀 𝒔 =
𝒃
𝒂𝒐
𝑭 𝒔
⟹ 𝝉𝒔𝒀 𝒔 + 𝒀 𝒔 = 𝑲𝑭 𝒔 ⟹ 𝝉𝒔 + 𝟏 𝒀 𝒔 = 𝑲𝑭 𝒔
⟹
𝒀 𝒔
𝑭 𝒔
=
𝑲
𝝉𝒔+𝟏
= 𝑮(𝒔) ⟹ 𝑮 𝒔 =
𝑲
𝝉𝒔+𝟏
…………………………..(1)
Equation (1) is called the transfer function of first order system.
𝝉 =
𝒂𝟏
𝒂𝒐
= time constant of the process
𝑲 =
𝒃
𝒂𝒐
= static gain of the process (or simply gain)
• By block diagram
𝐺 𝑠
=
𝐾
𝜏𝑠 + 1
𝑭 𝒔 𝒀 𝒔
5

6.  The parameters of the first-order system are:
 τ (time constant)
 K (steady state gain)
 Time constant of a process
 is a measure of the time necessary for the process to adjust to a change in the
input.
 Steady state gain
 is the steady state change in output divided by the constant change in the
input.
 characterizes the sensitivity of the output to the change in input.
 The steady state gain of a transfer function can be used to calculate the
steady-state change in an output due to a steady-state change in the input.
 Both parameters depend on operating
conditions of the process.
6

7. Standard Process Inputs for
first order
Standard types
Process Inputs
1. Step Input
2. Ramp Input
4. Rectangular Pulse
3. Sinusoidal Input
5.Impulse Input
6. Random Inputs
7

8. Response of First Order System
1) Unit Step Response
 A sudden change in a process variable can be approximated
by a step change of magnitude, M: 𝑢 𝑡 =
0 , 𝑡 < 0
𝑀, 𝑡 ≥ 0
 The step change occurs at an arbitrary time denoted as t =
0.
 Example of a step change: A reactor feedstock is suddenly
switched from one supply to another, causing sudden
changes in feed concentration, flow, etc.
8

9.  If the input u(t) is a step function with amplitude M:
𝑢(𝑡) = 𝑀 ⟹ 𝑈(𝑠) =
𝑀
𝑠
⟹ 𝑌 𝑠 =
𝐾
𝜏𝑠 + 1
. 𝑈 𝑠 ⟹ 𝑌 𝑠 =
𝐾
𝜏𝑠 + 1
.
𝑀
𝑠
⟹ 𝑌 𝑠 =
𝑀𝐾
𝑠(𝜏𝑠 + 1)
⟹ 𝑌 𝑠 =
𝑐1
𝑠
+
𝑐2
𝜏𝑠 + 1
⟹ 𝑌 𝑠 =
𝑀𝐾
𝑠
−
𝑀𝐾𝜏
𝜏𝑠 + 1
⟹ 𝑌 𝑡 = 𝑀𝐾 − 𝑀𝐾𝑒−𝑡/𝜏
⟹ 𝑌 𝑡 = 𝑀𝐾(1 − 𝑒−𝑡/𝜏
) 9

10. The transient response for a step change is
10

11. First order step change characteristics
 A first-order process is self-regulating.
 The process reaches a new steady state.
 The ultimate value of the output is K for a unit step change in the input, or KM
for a step of size M.
 𝑦 → 𝐾𝑀 as 𝑡 → ∞.
 ∆ 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝐾∆ 𝑖𝑛𝑝𝑢𝑡 .
 This equation tells us by how much we should change the value of the input in
order to achieve a desired change in the output, for a process with given K.
 A small change in the input if K is large (very sensitive systems)
 A large change in the input if K is small.
11

12.  The slope of the response at t = 0 is equal to 1.
12

13.  The value of y(t) reaches 63.2 % of its ultimate value when the
time elapsed is equal to one time constant τ. 13

14.  After a time interval equal to the process time constant (t = τ), the
process response is still only 63.2% complete.
 When the time elapsed is 2τ, 3τ, 4τ, 5τ the response is 86%, 95%,
98% and 99% respectively.
 If the initial rate of change of y(t) were to be maintained, the
response would reach its final value in one time constant.
 The smaller the value of time constant, the steeper the initial
response of the System.
 Theoretically, the process output never reaches the new steady-
state value except as t →∞.
 It does approximate the final steady-state value when t= 5τ.
14

15. Example 1: Let a transfer function 𝐺 𝑠 =
15
2𝑠+4
, then calculate:
a)The static gain
b)The time constant
c)The response of the system for step change if input is 5.
Solution: 𝐺 𝑠 =
15
2𝑠+4
=
3.75
0.5𝑠+1
a)𝐾 =
15
4
= 3.75
b)𝜏 =
2
4
= 0.5
c)𝑢(𝑡) = 5 ⟹ 𝑈(𝑠) =
5
𝑠
⟹ 𝑦(𝑡) = 𝑎𝐾(1 − 𝑒−𝑡/𝜏
)
⟹ 𝑦(𝑡) = 5(3.75)(1 − 𝑒−𝑡/0.5
)
⟹ 𝑦(𝑡) = 18.75(1 − 𝑒−2𝑡) 15

16. 2. Ramp Input
 Industrial processes often experience “drifting
disturbances”, that is, relatively slow changes up or down
for some period of time with a roughly constant slope.
 The rate of change is approximately constant.
 We can approximate a drifting disturbance by a ramp
input:
𝑢 𝑡 =
0 , 𝑡 < 0
𝑎𝑡, 𝑡 ≥ 0
Examples of ramp changes:
1. Ramp a set point to a new value rather than making a
step change.
2. Feed composition, heat exchanger fouling, catalyst
16

17. 17

18. Ramp Response
If the input u(t) is a ramp function with amplitude A
𝑢(𝑡) = 𝐴𝑡 ⟹ 𝑈(𝑠) =
𝐴
𝑠2
⟹ 𝑌 𝑠 =
𝐾
𝜏𝑠 + 1
. 𝑈 𝑠 =
𝐴
𝑠2
.
𝐾
𝜏𝑠 + 1
=
𝐴𝐾
𝑠2(𝜏𝑠 + 1)
=
𝑐1
𝑠
+
𝑐2
𝑠2
+
𝑐3
𝜏𝑠 + 1
⟹ 𝑌 𝑠 =
−𝐴𝐾𝜏
𝑠
+
𝐴𝐾
𝑠2
+
𝐴𝐾𝜏2
𝜏𝑠 + 1
⟹ 𝑌 𝑠 =
−𝐴𝐾𝜏
𝑠
+
𝐴𝐾
𝑠2 +
𝐴𝐾𝜏
𝑠+1/𝜏
take inverse Laplace
⟹ 𝑦 𝑡 = −𝐴𝐾𝜏 + 𝐴𝐾𝑡 + 𝐴𝐾𝜏𝑒−𝑡/𝜏
⟹ 𝑦 𝑡 = −𝐴𝐾𝜏 + 𝐴𝐾𝑡 + 𝐴𝐾𝜏𝑒−𝑡/𝜏
as 𝑡 ≫ 𝜏 ⟹ 𝑦 𝑡 = −𝐴𝐾𝜏 + 𝐴𝐾𝑡
⟹ 𝑦 𝑡 = 𝐴𝐾(𝑡 − 𝜏)
18

19. Example 3: Let 𝐺 𝑠 =
27
9𝑠+3
, then calculate:
a)The static gain
b)The time constant
c)The response of the system for ramp change if input is 6.
Solution: 𝐺 𝑠 =
27
9𝑠+3
=
9
3𝑠+1
a)𝐾 =
27
3
= 9
b)𝜏 =
9
3
= 3
c)𝑢(𝑡) = 6 ⟹ 𝑈(𝑠) =
6
𝑠
⟹ 𝑦(𝑡) = 𝑎𝐾(𝑡 − 𝜏) ⟹ 𝑦 𝑡 = 6 9 𝑡 − 3
⟹ 𝑦(𝑡) = 54(𝑡 − 3) 19

20. 4. Sinusoidal Input
 Processes are subject to periodic, or cyclic, disturbances.
 They can be approximated by a sinusoidal disturbance:
𝑢 𝑡 =
0 , 𝑡 < 0
𝐴𝑠𝑖𝑛(𝜔𝑡), 𝑡 ≥ 0
where: A = amplitude, 𝜔 = angular frequency
 Sinusoidal output wave has the same frequency as that of input sinusoid.
 Amplitude Ratio between the output wave and input wave is: 𝐾/ 𝜔2𝜏2 + 1
 Output wave lags behind the input wave with a phase difference: tan−1
(−𝜔𝜏)
20

21. Response of a First-Order System to a Sinusoidal Input
 Consider a simple first order process,
 Let the sinusoidal input u(t) = A sin ωt perturb the system.
 Then the output of the process will be
 Computing the constants taking inverse Laplace Transform
of the above equation we obtain,
21
As t ∞ the exponential term approach to zero

22. Dynamic Behavior of Second-Order Processes
A second-order system is one whose output, y(t), is described by a
second-order differential equation.
The following equation describes a second-order linear system:
where K = Process steady-state gain
t = Process time constant
x = Damping coefficient 22

23.  The very large majority of the second- or higher-order systems encountered
in a chemical plant come from multi-capacity processes, i.e. processes that
consist of two or more first-order systems in series.
 The standard form of second order TF is:
Dynamic response for second-order system
1. Step responses
For a step change of magnitude M, U(s) = M/s
𝑌 𝑠 =
𝐾𝑀
𝑠(𝜏2𝑠2 + 2𝜏𝑠 + 1)
23

24.  The two poles of the second-order transfer function are given by the
roots of the characteristic polynomial.
 The form of the response of y(t) will depend on the location of the
two poles in the complex plane.
24
Characteristic Behavior of Second-Order Transfer Functions
Case Damping
factor
Nature of the response
(roots of characteristic equation:)
Characteristic behavior
1  > 1 real, distinct roots Over damped
2  = 1 real, equal roots Critically damped
3  < 1 complex & conjugate roots Underdamped

25. Effect of the damping coefficient
•The value of  completely determines the degree of
oscillation in a process response after a perturbation.
a) > 1 (Over-damped)
Two distinct real roots
Sluggish response
No oscillation
𝑦 𝑡
= 𝑀{1
25

26. b)  = 1 (Critically-damped)
• The transition between over damped and under damped.
 Two equal real roots
 Faster than over damped
 No oscillation
𝑦 𝑡 = 𝐾𝑀[1 − (1 +
𝑡
𝜏
)𝑒−𝑡/𝜏
]
 Responses exhibit a higher degree of oscillation and overshoot
(y/KM> 1) as approaches to zero.
 Large values of yield a sluggish (slow) response.
 The fastest response without overshoot is obtained for the
critically damped case (= 1). 26

27. Step response of critically and over damped second
order processes
27

28. c) 0<<1 (Under-damped)
 Two complex conjugate roots
 Fastest response
 Oscillating response (the damping is attenuated as  decreases)
𝑦 𝑡 = 𝐾𝑀{1 − 𝑒−
𝜀𝑡
𝜏 cos
1 − 𝜀2
𝜏
𝑡 +
𝜀
1 − 𝜀2
sin
1 − 𝜀2
𝜏
𝑡 }
d)  ≤ 0 (Not-damped): gives unstable solution
 Unstable system (the oscillation amplitude grows indefinitely)
28

29. 0 2 4 6 8 10 12 14 16 18 20
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0.6
0.4
 = 0.2
t / t
y
/
(KA)
Step response of under damped second
order processes 29

30. 30

31. The ff terms are used to describe the dynamics of underdamped processes:
1. Rise Time (t) is the time the process output takes to first reach the new
steady-state value.
2. Time to First Peak (tp) is the time required for the output to reach its first
maximum value.
3. Overshoot(OS) = a/c (% overshoot is 100 a/c).
Overshoot is the distance between the first peak and the new steady state.
4. Decay Ratio(DR) = b/a (where c is the height of the second peak).
 DR is a measure of how rapidly the oscillations decreasing.
 The ratio of two successive peaks is a constant quantity and is equal to the
decay ratio for that underdamped process.
5. Settling time(Ts): is the time required by the response to reach the Steady state
31

32. 5) Period of Oscillation(P) is the time between two successive peaks.
Summary: For particular case of an under damped second-order processes
the ff analytical expiration developed for some of these characteristics.
• 𝑂𝑆 = exp −
𝜋𝜏
1−
2
• 𝐷𝑅 = (𝑂𝑆)2
= exp −
2𝜋𝜀
1−
2
• 𝜔 =
2𝜋
𝑇
• 𝑇 =
2𝜋
𝜔
=
2𝜋𝜏
1−
2
• 𝑡𝑝 =
𝜋𝜏
1−
2 32

33. Example on second order system
1) Let a transfer function 𝐺 𝑠 =
9
𝑠2+2𝑠+9
; determine:
a) Natural period of oscillation
b) Static gain
c) Damping coefficient
d) Overshoot
e) Decay ratio
33

34. Solution: 𝐺 𝑠 =
9
𝑠2+2𝑠+9
a) 𝜏2=
1
9
= 0.3 𝑠𝑒𝑐
b)𝐾 =
9
9
= 1
c) 2𝜏 =
2
9
⟹  = 0.33
d)𝑂𝑆 = exp −
𝜋
1−2
= exp −
0.33𝜋
1 − (0.33)2
= exp −1.27 = 0.28 = 28%
34

35. 2) For the system
Determine the :
a) Standard form of second order
b) Natural period of oscillation
c) Static gain
d) Nature of the response
F(𝑠) 1
4𝑠 + 8
2
𝑠 + 3
Y(𝑠)
35

36. Solution: 𝐺 𝑠 =
1
4𝑠+8
.
2
𝑠+3
=
2
4𝑠2+20𝑠+24
a)𝜏2
=
4
24
= 0.167 ⟹ 𝜏 = 0.4 sec
b)𝐾 =
2
24
= 0.08
c) 2𝜏 =
20
24
⟹  = 0.33 ⟹ 0.8 = 0.83
⟹  = 1.02
a) The nature of response is over-damped.
36

37. 3) Consider a second order of a step change of magnitude 8 into the
system and transfer function G(s) =
90
40𝑠2+16𝑠+10
, then determine the:
37

38. Solution: G(s) =
90
40𝑠2+16𝑠+10
=
9
4𝑠2+1.6𝑠+1
a)𝐾 =
90
10
= 9
b) 𝜏2 =
40
10
= 4 ⇒ τ = 2min
c)2𝜁𝜏 =
16
10
= 1.6 ⇒ 4ζ = 1.6 ⇒ ζ= 0.4
d) ζ= 0.4 ⟹ underdamped
e)𝑦 𝑡 = 𝐾𝑀{1 − 𝑒−
𝜁𝑡
𝜏 cos
1−𝜁2
𝜏
𝑡 +
𝜁
1−𝜁2
sin
1−𝜁2
𝜏
𝑡 }
⟹ 𝑦 𝑡
= 9(8) 1 − 𝑒
−0.4𝑡
2 cos
1 − 0.4 2
2
𝑡 +
0.4
1 − 0.4 2
sin
1 − 0.4 2
2
𝑡
⟹ 𝑦 𝑡 = 72 1 − 𝑒−0.2𝑡 cos 0.46𝑡 + 0.43 sin 0.46𝑡
38

39. g) DR = 𝑂𝑆 2 = 0.0625 = 6.25%
h) 𝑃 =
2𝜋𝜏
1−𝜁2
=
2𝜋(2)
1−(0.4)2
= 14 𝑚𝑖𝑛
i) 𝑓 =
1
𝑃
=
1
14
= 0.07
𝑐𝑦𝑐𝑙𝑒
𝑚𝑖𝑛
j) 𝜔 = 2𝜋𝑓 = 2𝜋 0.07 = 0.45 𝑟𝑎𝑑/𝑚𝑖𝑛
39

40. 40
𝟐.

41. 41