Chapter 4 Dynamic Systems Higher Order Processes Prof

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Chapter 4 Dynamic Systems: Higher Order Processes Prof. Shi-Shang Jang National Tsing-Hua University Chemical

Chapter 4 Dynamic Systems: Higher Order Processes Prof. Shi-Shang Jang National Tsing-Hua University Chemical Engineering Dept. Hsin Chu, Taiwan May, 2013

4 -1 Non-interactive Systems – Thermal Tanks

4 -1 Non-interactive Systems – Thermal Tanks

4 -1 Non-interactive Systems –Thermal Tanks – Cont.

4 -1 Non-interactive Systems –Thermal Tanks – Cont.

4 -1 Non-interactive Systems

4 -1 Non-interactive Systems

4 -1 Non-interactive Systems

4 -1 Non-interactive Systems

Numerical Example

Numerical Example

4 -1 Non-interactive Systems An over-damped second order system has two negative real poles.

4 -1 Non-interactive Systems An over-damped second order system has two negative real poles. Therefore, 2 s 2+2 s+1=( 1 s+1)( 2 s+1); hence such that =

4 -2 Interactive Systems – Thermal Tanks with Recycle

4 -2 Interactive Systems – Thermal Tanks with Recycle

4 -2 Interactive Systems – Thermal Tanks with Recycle- Cont.

4 -2 Interactive Systems – Thermal Tanks with Recycle- Cont.

4 -2 Interactive Systems – Thermal Tanks with Recycle- Cont.

4 -2 Interactive Systems – Thermal Tanks with Recycle- Cont.

2 -5 Step Response of Second-Order Processes – Over-damped process m(s)=A/s =1 =3 =2

2 -5 Step Response of Second-Order Processes – Over-damped process m(s)=A/s =1 =3 =2 Inflection point

4 -2 Interactive Systems fi V 1 fo h 1 Crosssectional=A 1 f 1

4 -2 Interactive Systems fi V 1 fo h 1 Crosssectional=A 1 f 1 V 2 h 2 Crosssectional=A 2 f 2

4 -2 Interactive Systems Fi(s) F 0(s) + + + Fi(s + ) F

4 -2 Interactive Systems Fi(s) F 0(s) + + + Fi(s + ) F 0(s) H 2(s) H 1(s) H 2(s) -

Numerical Example

Numerical Example

4 -1 Second Order Systems A second order system is of the following form:

4 -1 Second Order Systems A second order system is of the following form: Another form: Kp is called process gain, is called time constant, is called damping factor. The roots of the denominator are the poles of the system.

4 -1 Second-Order Processes - Continued Definition 4 -1: A second order process is

4 -1 Second-Order Processes - Continued Definition 4 -1: A second order process is called over-damped, if >1; is called under-damped if <1; is called critical damped if =1. Property 4 -1: Consider the roots of denominator, in case of over-damped system, the poles of the system are all negative real numbers. Property 4 -2: The poles of a under-damped system are complex with negative real numbers. Property 4 -3: The pole of a critical damped system is a repeated negative real number.

State Space Approach Consider the following linear system with N differential equations K inputs

State Space Approach Consider the following linear system with N differential equations K inputs and P sensors where X is termed the state vector and M is the input vector. The following observation equation is available:

State Space Approach _ Cont. Assume that it is desirable to realize the input/output

State Space Approach _ Cont. Assume that it is desirable to realize the input/output transfer functions and neglecting the state variables.

State Space Approach _ Example Interacting Tanks

State Space Approach _ Example Interacting Tanks

State Space Approach _ Example Interacting Tanks – Cont.

State Space Approach _ Example Interacting Tanks – Cont.

State Space to Transfer Function. MATLAB >> A=[-0. 0375; 0. 0375 -0. 075] A

State Space to Transfer Function. MATLAB >> A=[-0. 0375; 0. 0375 -0. 075] A = -0. 0375 -0. 0750 >> B=[1; 0]; C=[0 1]; D=0; >> [num, den]=ss 2 tf(A, B, C, D, 1) num = 0 -0. 0000 0. 0375 den = 1. 0000 0. 1125 0. 0014 >> ss=den(3) ss = 0. 0014 >> num=num/ss num = 0 -0. 0000 26. 6667 >> den=den/ss den = 711. 1111 80. 0000 1. 0000 >> tf(num, den) Transfer function: -9. 869 e-015 s + 26. 67 ----------711. 1 s^2 + 80 s + 1

2 -5 Step Response of Second-Order Processes – Under-damped process B 2. Decay Ratio=

2 -5 Step Response of Second-Order Processes – Under-damped process B 2. Decay Ratio= 3. Rise time=tr= Response 1. Overshoot= C T A Rise time Settling time 4. Period of oscillation=T= 5. Frequency of oscillation (Natural Frequency)=1/T

Example: Temperature Regulated Reactor t=1000 s t=3060 s Feed flow rate 0. 4→ 0.

Example: Temperature Regulated Reactor t=1000 s t=3060 s Feed flow rate 0. 4→ 0. 5 kg/s at t=0. 1. What is process gain? 2. What is transfer func? 3. What is rise time?

Textbook Reading Assignment and Homework Chapter 2, p 41 -49 Homework p 58, 2

Textbook Reading Assignment and Homework Chapter 2, p 41 -49 Homework p 58, 2 -9, 2 -15, 2 -16 Due June 8 th

Non-isothermal CSTR

Non-isothermal CSTR

Non-isothermal CSTR- Cont.

Non-isothermal CSTR- Cont.

Non-isothermal CSTR- Cont. Process Information Steady State Values

Non-isothermal CSTR- Cont. Process Information Steady State Values

Non-isothermal CSTR- Cont. Cooling flow rate 0. 8771 0. 8 Energy Balance of the

Non-isothermal CSTR- Cont. Cooling flow rate 0. 8771 0. 8 Energy Balance of the Reactor Inlet and Outlet Energy and Material fluxes of the Reactor Material Balance of the Reactor Inlet and Outlet Energy fluxes of the Jacket Energy Balance of the Jacket Rate Constant Heat Exchange between Jacket and Reactor

Transfer Functions Derived by Linearization

Transfer Functions Derived by Linearization

Linearization of the Reactor Example

Linearization of the Reactor Example

Transfer Functions Derived by Linearization – Cont It can be shown as generated as

Transfer Functions Derived by Linearization – Cont It can be shown as generated as above: [num, den]=ss 2 tf(A, B, C, D, 4) num = 0 1. 33226762955019 e-015 -2. 81748023 -1. 356898478768 den = 1. 3804 0. 3849816 0. 038454805392 >> ss=den(4) >> den=den/ss den = 26. 0045523519419 35. 8966840666205 10. 0112741717343 1 >> num=num/ss num = 0 3. 46450233194353 e-014 -73. 2673121415962 -35. 2855375273927

SIMULINK of Linear System - CSTR

SIMULINK of Linear System - CSTR

Non-isothermal CSTR- Cont. Cooling flow rate 0. 8771 0. 8 Tank temperature Time (min)

Non-isothermal CSTR- Cont. Cooling flow rate 0. 8771 0. 8 Tank temperature Time (min)

Non-isothermal CSTR- Cont. Linearized Model (page 127)

Non-isothermal CSTR- Cont. Linearized Model (page 127)

The problem of nonlinearity

The problem of nonlinearity

4 -3 Step Response of the High Order System X(s)=A/s Responses n=2 n=3 n=5

4 -3 Step Response of the High Order System X(s)=A/s Responses n=2 n=3 n=5 n=10 time

4 -3 Step Response of the High Order System. Continued Response Method of Reaction

4 -3 Step Response of the High Order System. Continued Response Method of Reaction Curve: inflection point time

4 -3 Step Response of the High Order System. Continued Real Responses Approximate time

4 -3 Step Response of the High Order System. Continued Real Responses Approximate time

4 -3 Response of Higher-Order Systems – Cont.

4 -3 Response of Higher-Order Systems – Cont.

4 -4 Other Types of Process Response Integrating Processes: Level Process

4 -4 Other Types of Process Response Integrating Processes: Level Process

4 -4 Other Types of Process Response

4 -4 Other Types of Process Response

4 -4 Other Types of Process Response

4 -4 Other Types of Process Response

4 -4 Other Types of Process Response The most general transfer function is as

4 -4 Other Types of Process Response The most general transfer function is as the following: p 1, p 2, …, pn are called the poles of the system, z 1, z 2, …, zm are the zeros of the system, Kp is the gain. Note that n m is necessary, or the system is not physically realizable.

4 -4 Poles and Zeros - Example Imaginary part Left Half Plane LHP Right

4 -4 Poles and Zeros - Example Imaginary part Left Half Plane LHP Right Half Plane RHP 0 0 Real part

4 -4 Poles and Zeros - Example Response time

4 -4 Poles and Zeros - Example Response time

4 -4 Location of the Poles and Stability in a Complex Plane Im Purdy

4 -4 Location of the Poles and Stability in a Complex Plane Im Purdy oscillatory Exponential Decay with oscillation Exponential growth with oscillation Fast Exponential growth Exponential Decay Fast Decay Slow growth Purdy oscillatory Stable (LHP) Unstable (RHP) Re

4 -4 The Stability of the linear system Definition 4 -2: A system is

4 -4 The Stability of the linear system Definition 4 -2: A system is called stable for the initial point if given any initial point y 0, such that ∣y 0∣≦ε, there exists a upper bound , such that: Definition 4 -3: A system is called asymptotic stable if given any initial point y 0, then

4 -4 The Stability of the linear system Definition 4 -4: A system is

4 -4 The Stability of the linear system Definition 4 -4: A system is called input output stable if the input is bound, then the output is bounded. (Bounded Input Bounded Output, BIBO) Property 4 -4: A linear system is asymptotic stable and BIBO if and only if all its poles have negative real parts.

4 -4 Stability - Example Response G 4 G 2 G 1 time m(s)=1

4 -4 Stability - Example Response G 4 G 2 G 1 time m(s)=1 G 3

Open Loop Unstable Process. Chemical Reactor (text page 139)

Open Loop Unstable Process. Chemical Reactor (text page 139)

Open Loop Unstable Process. Chemical Reactor

Open Loop Unstable Process. Chemical Reactor

Homework Text p 148 4 -4, 4 -5, 4 -7, 4 -8, 4 -10,

Homework Text p 148 4 -4, 4 -5, 4 -7, 4 -8, 4 -10, 4 -11, 4 -12 Due April

Supplemental Material Development of Empirical Models from Process Data

Supplemental Material Development of Empirical Models from Process Data

S-1 Introduction An empirical model is a model that is developed from experience and

S-1 Introduction An empirical model is a model that is developed from experience and their parameters are found based on experimental tests. The most frequent implemented empirical models are first order, second order and/or with time delays. The input changes is basically a step or an impulse.

S-2 First Order without Time Delay Systems Using Step Input Consider a first order

S-2 First Order without Time Delay Systems Using Step Input Consider a first order system with a output signal y(t) and input signal m(t), then:

First Order with Time Delay Systems Using Step Input Consider a first order system

First Order with Time Delay Systems Using Step Input Consider a first order system with a output signal y(t) and input signal m(t), then:

Example: A Typical Experiment Time (second) Y(temperature, o. C, 70100 o. C) Y(temperature, m.

Example: A Typical Experiment Time (second) Y(temperature, o. C, 70100 o. C) Y(temperature, m. A, 4 -20 m. A) Y (temperature, %) 0. ln(1 -Y) 0 70 4 1 71. 74 4. 928 0. 058 -0. 0598 2 76. 51 7. 472 0. 217 -0. 2446 3 80. 8 9. 76 0. 360 -0. 4463 4 84. 64 11. 808 0. 488 -0. 6694 5 88 13. 6 0. 600 -0. 9163 6 90. 76 15. 072 0. 692 -1. 1777 7 93. 16 16. 352 0. 772 -1. 4784 8 94. 99 17. 328 0. 833 -1. 7898 9 96. 64 18. 208 0. 888 -2. 1893 10 97. 75 18. 8 0. 925 -2. 5903 0

Graphical Fitting Methods Fit 1: Method of 63. 2% Response Fit 2: Method of

Graphical Fitting Methods Fit 1: Method of 63. 2% Response Fit 2: Method of initial slope Fit 3: Method of Log plot

Example: An Experiment Plot Fit 1 G 1(s) = 1 5. 5 s +1

Example: An Experiment Plot Fit 1 G 1(s) = 1 5. 5 s +1 Fit 2 G 2 (s ) = e - 0. 5 s 6. 8 s + 1

Method of log plot Consider a First-Order Plus Dead Time Model

Method of log plot Consider a First-Order Plus Dead Time Model

Method of log plot - Continued 0 0 5 10 15 -0. 5 -1

Method of log plot - Continued 0 0 5 10 15 -0. 5 -1 Series 1 -1. 5 Linear(Series 0) -2 -2. 5 -3 R 2 = 0. 9861 Fit 3

Method of log plot - Continued Fit 3 real Fit 1 Fit 2

Method of log plot - Continued Fit 3 real Fit 1 Fit 2

S-3 Over-damped Second Order Systems Using Step Input

S-3 Over-damped Second Order Systems Using Step Input

Smith’s Method for Second Order Systems Step 1: Get time delay by observing the

Smith’s Method for Second Order Systems Step 1: Get time delay by observing the response curve. Step 2: Find time t 20 such that y/y =0. 2, find t 60 such that y/y =0. 6 Step 3: Get t 20/t 60, then and From the right figure.

Example t 60=5 t 20=1. 9 t 20/t 60=0. 38 From the figure t

Example t 60=5 t 20=1. 9 t 20/t 60=0. 38 From the figure t 60/ =2. 4 =5/2. 4 =2. 1 =1. 2 2=4. 32, 2 =5. 04