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

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Process Characteristics Process Equation A process-control loop regulates some dynamic variable 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)

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Process Characteristics Process Load Identify a set of values for the process parameters that results in the controlled variable having the setpoint value. This set = nominal set. Process load = all parameter set – the controlled variable

9: Controller Principles Princess Sumaya University 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.

9: Controller Principles Princess Sumaya University 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.

9: Controller Principles Princess Sumaya University 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.

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

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Control System Parameters Error Control Parameter p: percentage u: actual

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

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Control System Parameters 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 4 20 I (mA) Speed (RPM) 140 600 310

9: Controller Principles Princess Sumaya University 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.

9: Controller Principles Princess Sumaya University 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.

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes Two-Position Mode Neutral Zone

9: Controller Principles Princess Sumaya University 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

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes Two-Position Mode 2 3 I (mA) Displacement (m) 4 20 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

9: Controller Principles Princess Sumaya University 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.

9: Controller Principles Princess Sumaya University 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

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes Multiposition Mode

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes Multiposition Mode Example

9: Controller Principles Princess Sumaya University 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.

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes Floating-Control Mode Single Speed

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes Floating-Control Mode Single Speed

9: Controller Principles Princess Sumaya University 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

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Discontinuous Controller Modes Floating-Control Mode Multispeed

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes Proportional Control Mode Direct (- K) & Reverse (+ K) Action

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes Proportional Control Mode % per % Proportional Band:

9: Controller Principles Princess Sumaya University 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

9: Controller Principles Princess Sumaya University 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 Solution QA must go up to 600 m3/h QA = 10 m3/h * P P = 60 % 60 = 10 eP + 50 eP = 1 %

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes Integral Control Mode Integral Time: TI = 1 / KI % per sec % sec

9: Controller Principles Princess Sumaya University 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.

9: Controller Principles Princess Sumaya University 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) = KI eP t + P(0) P(t) = (– 0.15) * (– 30) * 2 + 22 P = 31 %

9: Controller Principles Princess Sumaya University 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

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Continuous Controller Modes Derivative Control Mode Example

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes Proportional – Integral Control (PI)

9: Controller Principles Princess Sumaya University 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

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

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes Proportional – Derivative Control (PD)

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes PD Example KP = 5 KD = 0.5 sec Po = 20%

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles Composite Control Modes Three – Mode Controller (PID)

9: Controller Principles Princess Sumaya University 3442 - Industrial Instruments 2 9: Controller Principles End of Chapter 9