1. Chapter 4 Dynamic Systems: Higher Order Processes Prof. Shi-Shang Jang National Tsing-Hua University Chemical Engineering Dept. Hsin Chu, Taiwan April, 2014

2. 4-1 Second Order Systems A second order system is of the following form: Another form: K p is called process gain,  is called time constant,  is called damping factor. The roots of the denominator are the poles of the system.

3. 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.

4. 2-5 Step Response of Second-Order Processes – Over- damped process m(s)=A/s Inflection point  =3  =2  =1 

5. 4-1 Non-interactive Systems – Thermal Tanks

6. 4-1 Non-interactive Systems –Thermal Tanks – Cont.

8. 4-1 Non-interactive Systems

10. Numerical Example

11. 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 =

12. 4-2 Interactive Systems – Thermal Tanks with Recycle

13. 4-2 Interactive Systems – Thermal Tanks with Recycle- Cont.

16. 4-2 Interactive Systems f1f1 Cross- sectional=A 2 h2h2 V2V2 f2f2 h1h1 V1V1 fifi Cross- sectional=A 1 fofo

17. 4-2 Interactive Systems H 1 (s) H 2 (s) + + F 0 (s) F i (s) + - F 0 (s) F i (s) + -

18. Numerical Example

19. 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:

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

21. State Space Approach _ Example Interacting Tanks

22. State Space Approach _ Example Interacting Tanks – Cont.

23. State Space to Transfer Function- MATLAB >> A=[-0.0375 0.0375;0.0375 -0.075] A = -0.0375 0.0375 0.0375 -0.0750 >> B=[1;0];C=[0 1];D=0; >> [num,den]=ss2tf(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: -9.869e-015 s + 26.67 --------------------- 711.1 s^2 + 80 s + 1

24. State Space Approach _ Example Recycling Heating Tanks

25. State Space Approach _ Example Recycling Heating Tanks (Cont.)

26. State Space Approach _ Example Recycling Heating Tanks (MATLAB) >> A=[-0.45 0.05;0.225 -0.275];B=[0.4 0;0 0.05];C=[0 1];D=[0 0]; >> [a,b]=ss2tf(A,B,C,D,1) a = 0 -0.0000 0.0900 b = 1.0000 0.7250 0.1125 >> [a,b]=ss2tf(A,B,C,D,2) a = 0 0.0500 0.0225 b = 1.0000 0.7250 0.1125

27. 4-3 Generation The most general transfer function is as the following: p 1, p 2,…,p n are called the poles of the system, z 1, z 2,…,z m are the zeros of the system, K p is the gain. Note that n  m is necessary, or the system is not physically realizable.

28. 4-4 Poles and Zeros - Example Imaginary part Real part 0 0 Left Half Plane LHP Right Half Plane RHP

29. 4-4 Poles and Zeros - Example time Response

30. 4-4 Location of the Poles and Stability in a Complex Plane Re Im Purdy oscillatory Fast Decay Slow Decay Exponential Decay with oscillation Slow growth Fast Exponential growth Exponential growth with oscillation Stable (LHP) Unstable (RHP)

31. 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

32. 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.

33. 4-4 Stability - Example time Response G4G4 G2G2 G3G3 G1G1 m(s)=1

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

35. Example: Temperature Regulated Reactor t=3060st=1000s Feed flow rate 0.4 → 0.5kg/s at t=0. 1.What is process gain? 2.What is transfer func? 3.What is rise time?

36. 4-5 Over-damped Second Order Systems Using Step Input

37. 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.

38. Example: A Typical Experiment Time (second) Y(temperature, o C,70- 100 o C) Y(temperature,mA,4-20mA)Y (temperature, %)ln(1-Y) 0 704 0. 0 1 71.744.928 0.058 -0.0598 2 76.517.472 0.217 -0.2446 3 80.89.76 0.360 -0.4463 4 84.6411.808 0.488 -0.6694 5 8813.6 0.600 -0.9163 6 90.7615.072 0.692 -1.1777 7 93.1616.352 0.772 -1.4784 8 94.9917.328 0.833 -1.7898 9 96.6418.208 0.888 -2.1893 10 97.7518.8 0.925 -2.5903

39. Example t 20 =1.9 t 60 =5 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

40. Textbook Reading Assignment and Homework Chapter 2, p41-49 Homework p58, 2-9, 2-15, 2-16 Due April 25th

41. Non-isothermal CSTR

42. Non-isothermal CSTR- Cont.

43. Process Information Steady State Values

44. Non-isothermal CSTR- Cont. Cooling flow rate 0.8771  0.8 Energy Balance of the Reactor Material Balance of the Reactor Energy Balance of the Jacket Inlet and Outlet Energy and Material fluxes of the Reactor Inlet and Outlet Energy fluxes of the Jacket Rate Constant Heat Exchange between Jacket and Reactor

45. Transfer Functions Derived by Linearization

46. Linearization of the Reactor Example

47. It can be shown as generated as above: Transfer Functions Derived by Linearization – Cont. [num, den]=ss2tf(A,B,C,D,4) num = 0 1.33226762955019e-015 -2.81748023 -1.356898478768 den = 1 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 3.46450233194353e-014 -73.2673121415962 -35.2855375273927

48. SIMULINK of Linear System - CSTR

49. Non-isothermal CSTR- Cont. Cooling flow rate 0.8771  0.8 Time (min) Tank temperature

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

51. The problem of nonlinearity

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

53. Open Loop Unstable Process- Chemical Reactor

54. 4-3 Step Response of the High Order System X(s)=A/s n=2 n=3 n=5 n=10 time Responses

55. 4-3 Step Response of the High Order System- Continued Method of Reaction Curve: time Response inflection point 

56. 4-3 Step Response of the High Order System- Continued time Responses Real Approximate

57. 4-3 Response of Higher-Order Systems – Cont.

58. 4-4 Other Types of Process Response Integrating Processes: Level Process

59. 4-4 Other Types of Process Response

61. Homework Text p148 4-4, 4-5, 4-7, 4-8, 4-10, 4-11,4-12 Due April

62. Supplemental Material Development of Empirical Models from Process Data

63. 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.

64. 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:

65. 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:

66. Example: A Typical Experiment (  U=1m 3 /min) Time (second) Y(temperature, o C,70- 100 o C) Y(temperature,mA,4-20mA)Y (temperature, %)ln(1-Y) 0 704 0. 0 1 71.744.928 0.058 -0.0598 2 76.517.472 0.217 -0.2446 3 80.89.76 0.360 -0.4463 4 84.6411.808 0.488 -0.6694 5 8813.6 0.600 -0.9163 6 90.7615.072 0.692 -1.1777 7 93.1616.352 0.772 -1.4784 8 94.9917.328 0.833 -1.7898 9 96.6418.208 0.888 -2.1893 10 97.7518.8 0.925 -2.5903

67. Graphical Fitting Methods Fit 1: Method of 63.2% Response Fit 2: Method of initial slope Fit 3: Method of Log plot

68. Example: An Experiment Plot Fit 1 Fit 2 15.5 1 )( 1   s s G  18.6 5.0 2    s e sG s

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

70. Method of log plot - Continued Fit 3

71. Method of log plot - Continued 070 171.7470 276.5173.0517 380.879.852 484.6484.9364 58888.7377 690.7691.5797 793.1693.7045 894.9995.2932 996.6496.481 1097.7597.369

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

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

74. 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.

75. Example t 20 =1.9 t 60 =5 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