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3442 Industrial Instruments 2 Chapter 9 Controller Principles

Princess Sumaya Univ. Electronic Engineering Dept. 3442 Industrial Instruments 2 Chapter 9 Controller Principles. Dr. Bassam Kahhaleh. 9: Controller Principles. Process Characteristics Process Equation A process-control loop regulates some dynamic variable in a process. Example :

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3442 Industrial Instruments 2 Chapter 9 Controller Principles

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  1. Princess Sumaya Univ.Electronic Engineering Dept. 3442Industrial Instruments 2Chapter 9Controller Principles Dr. Bassam Kahhaleh

  2. 3442 - Industrial Instruments 2 9: Controller Principles Process Characteristics • Process Equation A process-control loop regulates some dynamicvariable in a process. Example: The control of liquid temperature in a tank. The controlled variable is the liquid temperature TL TL is a function: TL = F(QA, QB, QS, TA, TS, TO)

  3. 3442 - Industrial Instruments 2 9: Controller Principles Process Characteristics • Process Load • Identify a set of values for the process parameters that results in the controlledvariable having the setpoint value. • This set = nominalset. Process load = all parameter set – the controlledvariable

  4. 3442 - Industrial Instruments 2 9: Controller Principles Process Characteristics • Process Change • Process Load Change A parameter changes value from its nominal value causes the controlled value to change from its setpoint. • Transient Change A temporary change of a parameter value.

  5. 3442 - Industrial Instruments 2 9: Controller Principles Process Characteristics • Process Lag The time it takes for the process to respond after a process load or transient change occurs, to ensure that the controlled variable returns to the setpoint.

  6. 3442 - Industrial Instruments 2 9: Controller Principles Process Characteristics • Self Regulation The tendency of some processes to adopt a specific value of the controlled variable for nominal load with no control operations.

  7. 3442 - Industrial Instruments 2 9: Controller Principles Control System Parameters • Error e = r – b. • Percentage

  8. 3442 - Industrial Instruments 2 9: Controller Principles p: percentage Control System Parameters • Error • Control Parameter u: actual

  9. Speed (RPM) 600 140 I (mA) 4 20 3442 - Industrial Instruments 2 9: Controller Principles Example A controller outputs a 4 – 20 mA signal to control motor speed from 140 – 600 RPM with a linear dependence. Calculate: Control System Parameters • Current corresponding to 310 RPM • The value of (a) in percent. 310

  10. Speed (RPM) 600 310 140 I (mA) 4 20 3442 - Industrial Instruments 2 9: Controller Principles Example Sp = m I + So 140 = 4 m + So 600 = 20 m + So  m = 28.75 rpm/mA So = 25 rpm 310 = 28.75 I + 25  I = 9.91 mA Control System Parameters

  11. 3442 - Industrial Instruments 2 9: Controller Principles Control System Parameters • Control Lag The time it takes for the final control element to adopt a new value (as required by the process-control loop output) in response to a sudden change in the controlled variable. • Dead Time The elapsed time between the instant a deviation (error) occurs and when the corrective action first occurs. • Cycling The cycling of the variable above and below the setpoint value.

  12. 3442 - Industrial Instruments 2 9: Controller Principles Control System Parameters • Controller Modes • Continuous / Discontinuous Smooth variation of the control parameter versus ON / OFF. • Reverse / Direct Action An increasing value of the controlled variable causes an decreasing / increasing value of the controller output.

  13. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Two-Position Mode Neutral Zone

  14. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Two-Position Mode Example A liquid-level control system linearly converts a displacement of 2 – 3 m into a 4 – 20 mA control signal. A relay serves as the two-position controller to open or close an inlet valve. The relay closes at 12 mA and opens at 10 mA. Find: The relation between displacement level and current The neutral zone

  15. I (mA) 20 4 2 3 Displacement (m) 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Two-Position Mode Example H = K I + HO 2 = K (4) + HO 3 = K (20) + HO  K = 0.0625 m / mA HO = 1.75 m HH = 0.0625 * 12 + 1.75 = 2.5 m HL = 0.0625 * 10 + 1.75 = 2.375 The neutral zone = HH – HL = 2.5 – 2.375 = 0.125 m

  16. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Two-Position Mode Example As a water tank loses heat, the temperature drops by 2 K per minute. When a heater is on, the system gains temperature at 4 K per minute. A two-position controller has a 0.5 min control lag and a neutral zone of ± 4 % of the setpoint about a setpoint of 323 K. Plot the tank temperature versus time.

  17. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Two-Position Mode Example Neutral zone = 310 – 336 K Temp. gain = 4 K per minute Setpoint = 323 K Temp. loss = 2 K per minute Control lag = ½ minute

  18. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Multiposition Mode

  19. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Multiposition Mode Example

  20. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Floating-Control Mode If the error is zero, the output does not change but remains (floats) at whatever setting it was when the error went to zero.

  21. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Floating-Control Mode • Single Speed

  22. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Floating-Control Mode • Single Speed

  23. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Floating-Control Mode Example Suppose a process error lies within the neutral zone with p = 25%. At t = 0, the error falls below the neutral zone. If K = +2% per second, find the time when the output saturates. Solution 100 % = (2 %/s * t) + 25 % t = 37.5 s

  24. 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes • Floating-Control Mode • Multispeed

  25. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Proportional Control Mode Direct (- K) & Reverse (+ K) Action

  26. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Proportional Control Mode % per % Proportional Band:

  27. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Proportional Control Mode Example Valve A: 10 m3/h per percent. PO = 50 % KP = 10 % per % Valve B changes from 500 m3/h to 600 m3/h Calculate: The new controller output The new offset error

  28. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Proportional Control Mode Solution QA must go up to 600 m3/h QA = 10 m3/h * P P = 60 % 60 = 10 eP + 50 eP = 1 % Example Valve A: 10 m3/h per percent. PO = 50 % KP = 10 % per % Valve B changes from 500 m3/h to 600 m3/h Calculate: The new controller output The new offset error

  29. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Integral Control Mode Integral Time: TI = 1 / KI % per sec % sec

  30. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Integral Control Mode Example An integral controller is used for speed control with a setpoint of 12 rpm within a range of 10 – 15 rpm. The controller output is 22% initially. The constant KI = - 0.15 % per second per % error. If the speed jumps to 13.5 rpm, calculate the controller output after 2 s for a constant ep.

  31. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Integral Control Mode Example setpoint = 12 rpm (range of 10 – 15 rpm) P(O) = 22% KI = – 0.15 speed = 13.5 rpm constant ep 2 seconds time Solution eP = (12–13.5)/(15–10)*100 %. eP = – 30 % P(t) = KIeP t + P(0) P(t) = (– 0.15) * (– 30) * 2 + 22 P = 31 %

  32. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Derivative Control Mode Not used alone because it provides no output when the error is constant. sec

  33. 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes • Derivative Control Mode Example

  34. 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes • Proportional – Integral Control (PI)

  35. 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes • PI Example KP = 5 KI = 1 sec-1 P(0) = 20% Plot a graph of the controller output as a function of time

  36. 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes • PI Example KP = 5 KI = 1 sec-1 P(0) = 20%

  37. 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes • Proportional – Derivative Control (PD)

  38. 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes • PD Example KP = 5 KD = 0.5 sec Po = 20%

  39. 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes • Three – Mode Controller (PID)

  40. 3442 - Industrial Instruments 2 9: Controller Principles End of Chapter 9

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