Pages 1/18. Marks 2+2

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1 Pages 1/18 Subject Code: Model Answer Page No: _1_/ 18 Important Instructions to examiners: 1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. 2) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate. 3) The language errors such as grammatical, spelling errors should not be given more Importance (Not applicable for subject English and Communication Skills). 4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn. 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate s answers and model answer. 6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on candidate s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept. Q1. A) a) What is control system? What are the various blocks of feedback control system explain the function of of them? Marks A control system is a system which maintains process variable at desired value or changes it in desired manner. The closed loop control systems operates with the objective of minimizing error i.e. bring the process variable to setpoint. The closed loop controller has following elements, Error detector: The output of the error detector is deviation of process variable from setpoint i.e. Error = SP PV Controller: the controller is a decision making element of the control system. Based upon error signal controller generates its output as per the control action implemented. The control actions are ON/OFF, Proportional(P), P +I, PID Final control element: It is also called as actuator. This element brings about changes in the power input going to process as commanded by controller. Feedback: The feedback path consists of a suitable measurement circuit needed to convert the physical variable into signal required for control system. For electrical/electronic control system, the feedback circuit consists of a transducer which converts the physical variable into electrical signal.

2 Pages 2/18 Subject Code: Model Answer Page No: _2_/ 18 Q1. A) b) Compare dc and ac solenoids Marks AC solenoid AC solenoidroduces alternating magnetic field Shading ring required to produce unidirectional force Smaller force to weight ratio Pole faces make contact at the end of stroke Life of solenoid is reduced because of pole face contact Noisy operation Fast operation Inrush current high Dc Solenoid Fixed polarity Shading ring is not required It produces larger force than AC in the given space Pole face Contact can be avoided Life is much increased Quiet operation Slow operation No inrush current Q1. A) c) Select a suitable contacter for the following purposes stating reasons. i) Crane ii) Conveyer belt. Selection of contactor is based on utilization category which is as per the current handling and nature of duty. Commonly used utilization categories are from AC1-AC4 ½ for For crane application : AC contactor of AC4 category is recommended as it involves of starting, inching, reversing, and reverse current braking etc. For conveyer application: AC contactor of AC3 utilization category is preferred as it involves only start-stop of induction motors Q1. A) d) Differentiate between hydraulic and pneumatic controller (four points). 1 for

Pages 3/18 Subject Code: 12149 Model Answer Page No: _3_/ 18 Q1 B) a) What do you mean by automation? What are its advantages?

3 Pages 3/18 Subject Code: Model Answer Page No: _3_/ 18 Q1 B) a) What do you mean by automation? What are its advantages? It is the process of using automatic control systems and to reduce requirement of human effort and sensory requirement. 1) It replaces human requirements in tasks involving hard physical or monotonous work. 2) It replaces human requirement in tasks that is done in dangerous, hazardous conditions. 3) Automated system can perform tasks that are beyond human capabilities. 4) Automation enhances precision, consistency. 5) It increases productivity, enhances system efficiency. 6) It optimizes the resource consumption. 7) Automation enhances safety. 8) Automated system occupies smaller space as compared to manual operated system. 9) It enhances reliability of the system. 10) For long term use, it is much economical to use automated system. 11) It avoids human errors. Q1 B) b) Explain the constructional features, working principle of AC servomotor with neat sketch AC servo motors are is wound with two windings at 900 with respect to other. One winding is called as reference winding(fixed voltage) and the other as control wdg (variable voltage). The rotor is squirrel cage with longer length and small diameter. The rotor has high resistance to increase starting torque and linear torque speed characteristics

Pages 4/18 Subject Code: 12149 Model Answer Page No: _4_/ 18 Q2 a) Define transfer function. State the various applications of control system.

4 Pages 4/18 Subject Code: Model Answer Page No: _4_/ 18 Q2 a) Define transfer function. State the various applications of control system. Transfer function is defined as ratio of Laplace transform of the system output to Laplace transform of the input to the system. C( s) G( s) R( s) Where C(s) Laplace transform of the system output Where R(s) Laplace transform of the system output And G(s) is the transfer function. Applications of control systems: 1) Domestic applications: temperature control in refrigerator, Air conditioning, Heating 2) Industrial applications: speed control of drives, Industrial heating, chemical process control, pharmaceutical process control 3) Space application: space vehicle launching, guidance and trajectory control. In-orbit control of satellite 4) Defense applications: Missile launching, auto pilot combat vehicles etc. Q2 b) Differentiate between contactors and electromechanical relays Since a contactor is required for a higher load, a relay is always cheaper than a contactor. 1 A relay is normally used in appliances below 5KW, while a contactor is preferred when the appliance is heavier. A relay is used only in control circuit while a contactor can be used in both control and power circuits. In general contactors are little slower than relays Contactor is so designed that it can be repaired while it is not normally done in the case of relays. Q2 c) What are synchros? Explain how these can be used as an error detector Synchro transmitter: is a type of transformer used for measuring angular position of rotating object. There are three stator coils placed at 120 deg apart. The rotor is wound and excited by ac supply thr slip rings. When ac excitation is applied to rotor, the resultant current produces ac magnetic field around rotor winding. By transformer action voltage is induced in its stator coils depending upon angular position of stator coil axis w.r.t. rotor axis.

Pages 5/18 Subject Code: 12149 Model Answer Page No: _5_/ 18 Control transformer: The construction is similar to that of transmitter except rotor is made cylindrical to maintain uniform air gap.

The output voltage induced in secondary of control transformer is proportional to the cosine of the angle between two rotors. The voltage at the output of control transformer is given by.

5 Pages 5/18 Subject Code: Model Answer Page No: _5_/ 18 Control transformer: The construction is similar to that of transmitter except rotor is made cylindrical to maintain uniform air gap. The electrical output of synchro transmitter is given as supply to control transformer. The output voltage induced in secondary of control transformer is proportional to the cosine of the angle between two rotors. The voltage at the output of control transformer is given by. V () t K Vsin() wt o 1 r. The above equation indicates that the output voltage is proportional to difference in the input and output shaft angular positions i.e. position error Q2. d) What are the advantages of hydraulic controller 1 Q2. e) Explain the use of microprocessor and microcontroller to be worked as controller

Pages 6/18 Subject Code: 12149 Model Answer Page No: _6_/ 18 When a micro-computer /microcontroller works as a controller, the system is called as digital control system.

6 Pages 6/18 Subject Code: Model Answer Page No: _6_/ 18 When a micro-computer /microcontroller works as a controller, the system is called as digital control system. The system operates at specific instants of time known as sampling instants. It communicates with the process through ADC and DAC. The ADC is used to sample the feedback and DAC is used to send analog signals back to the process. The function of controller is handled by microprocessor/microcontroller which executes a program stored in its program memory. The program is for implementing the specific control action. Q2 f. Describe P + I and P + D action of controller P +I control action can be described by an equation CV KP error KI error dt CV0 (Kp prop gain %/%, Ki integral gain sec -1 ) In P+I action the controller output combines one-to-one relationship between error and controller output of proportional control and time accumulation of error of the integral controller. For constant error, the proportional action produces constant output and the integral controller produces increasing output as long the error persists. This feature helps in eliminating steady state error in the controller variable. the characteristic is as shown in following figure P +D control action can be described by an equation de CV KP error KD CV0 dt In P+D action the controller output combines one-to-one relationship between error and controller output of proportional control and the derivative output is dependent on rate of change of error w.r.t time. Derivative action is helpful in correcting sudden changes in setpoint or load disturbances

Pages 7/18 Subject Code: 12149 Model Answer Page No: _7_/ 18 Q3 a) Compare open loop and closed loop control system.

7 Pages 7/18 Subject Code: Model Answer Page No: _7_/ 18 Q3 a) Compare open loop and closed loop control system. Open loop control system Closed loop control system Feedback is absent Feedback is present The gain is high The gain is low There is no error detector Error detector is present Very sensitive to parameter changes Less sensitive to parameter changes Low cost High cost Examples of open loop system Examples of closed loop control: 1 1) washing machine 2) Lift or elevator 3) microwave oven 4) automatic toaster 5) sprinkler system 6) timer driven appliances 1) DC/AC position control system, 2) missile tracking system 3) Water level controller 4) Electric Iron 5) Refrigerator Q3 b) Draw any four symbols used in power and control circuits. 1 Q3 c) What are potentiometers? Explain how these can be used as an error detector A potentiometer is a resistive, variable voltage divider, with an output voltage that is a percentage of the input voltage. The amount of output voltage is proportional to the position of the wiper relative to the grounded end. By using two potentiometers, an error detector can be configured. The error output is the relative potential difference between variable terminals of the two potentiometers.

Pages 8/18 Subject Code: 12149 Model Answer Page No: _8_/ 18 Q3 d) Explain in brief pneumatic flapper-nozzle system with the help of diagram Flapper nozzle amplifier directly converts displacement

8 Pages 8/18 Subject Code: Model Answer Page No: _8_/ 18 Q3 d) Explain in brief pneumatic flapper-nozzle system with the help of diagram Flapper nozzle amplifier directly converts displacement into pressure signal. At one end of flapper nozzle amplifier, there is as supply of pressurized air( 20PSI) and the other end there is a small opening as a nozzle. A flapper is positioned against the nozzle opening. The pressurised air passes thr the restriction(orifice) before releasing thr nozzle. When flapper completely closes the nozzle, air cant ecscape thro nozzle and a back pressure is developed in the output port. When flapper moves away from the nozzle and the pressure in the output port is dropped( atmospheric pressure). Thus a small movement of flapper brings about large changes in the output pressure. Q3 e) Draw a pneumatic PID controller and derive its transfer function

9 Pages 9/18 Subject Code: Model Answer Page No: _9_/ 18 Q3 f) Write a program for I/O connections Any suitable answer Q4 A) a) Name various kinds of pilot devices normally employed in electric circuits to initiate control action 1) Push button 1 2) Illuminated pushbutton Emergency stop push buttons 3) Selector switch 4) Limit switch 5) Proximity switch 6) Pilot light 7) Buzzer, annunciator Q4A) b) Compare permanent magnet and variable reluctance stepper motor 1 PM stepper motor Design is complex VR stepper motor Design is simple 30-3 deg /step 1.8 deg/step Quieter operation Full,half, microstepping Use of permanent magnet rotor Noisy operation Only full step Non magnetic soft iron rotor

10 Pages 10/18 Subject Code: Model Answer Page No: _10_/ 18 Q4. A) c) Explain the operation of OPAMP in inverting and non-inverting mode The non-inverting op-amp is used for amplifying signals in instrumentation. It is a negative feedback amplifier with gain larger than unity. The gain of the non-inverting op-amp is Inverting op-amp- It is a negative feedback amplifier. The negative sign in the gain expression indicates that the output is negated. The gain of the inverting amplifier is Q4 A) d). What are the advantages and disadvantages of PLC Advantages 1) High reliability and long life 2) Flexibility in programming 3) Small footprint 4) Multitasking and multifunctional capability 5) Accurate and repeatable time measurement 6) Flexible installation and easy expansion 7) Easy troubleshooting Disadvatages 1) PLCs are not suitable for handling large data and complex numerical calculations 2) Lack of standardisation makes it difficult to change PLC 3) Limited Graphical user interface 4) Cost increases with analog interface

Pages 11/18 Subject Code: 12149 Model Answer Page No: _11_/ 18 4. B) a) Describe on-off, P and I action of controller. 3+3 P +I action described earlier Q4.

11 Pages 11/18 Subject Code: Model Answer Page No: _11_/ B) a) Describe on-off, P and I action of controller. 3+3 P +I action described earlier Q4. B) b) Draw a neat diagram of SCADA system and explain the function of RTU 3+3 Prog. Auxiliary memory CPU C.P.U. Communicatio Display and control console RTU RTU RTU Simple SCADA system with single computer RTU remote terminal unit. In SCADA systems, an RTU is a device installed at a remote location that collects data, codes the data into a format that is transmittable and transmits the data back to a central station, or master.

Pages 12/18 Subject Code: 12149 Model Answer Page No: _12_/ 18 Q5 a) Name five basic types of static-switching device and indicate what one is equivalent to in electromagnetic systems of control.

12 Pages 12/18 Subject Code: Model Answer Page No: _12_/ 18 Q5 a) Name five basic types of static-switching device and indicate what one is equivalent to in electromagnetic systems of control. 1 Controlled power semiconductor switch like thyristor, triac is analogous to electromagnetic contactor, Any suitable answer Q5 b) Explain the lead-lag network Lag compensation works as low pass filter and improves steady state performance. Noise is easily filtered because of lowpass action. Lead compensation provides high pass characteristics. It improves dynamic response and increased bandwidth. Lag-Lead network combines advantages of both.

Pages 13/18 Subject Code: 12149 Model Answer Page No: _13_/ 18 Q5 c) Compare PLC and PC 1 Q5 d) Explain the basic blocks like CPU and I/Os of PLC The CPU consists of a microprocessor, memory chip and

Types of memory are volatile and nonvolatile memory.

13 Pages 13/18 Subject Code: Model Answer Page No: _13_/ 18 Q5 c) Compare PLC and PC 1 Q5 d) Explain the basic blocks like CPU and I/Os of PLC The CPU consists of a microprocessor, memory chip and other integrated circuits to control logic, monitoring and communications. The User program and operating system program is stored in program memory of PLC. Types of memory are volatile and nonvolatile memory. I/O section of PLC IO modules enables physical connection of field devices to PLC system Input modules convert electrical signals coming in from input field devices such as pushbuttons, to electrical signals that the PLC can understand. Output modules take information coming from the PLC and convert it to electrical signals the output field devices can understand, such as a motor starter, or a hydraulic solenoid valve. The filed signals are opto isolated from PLC internal logic circuit

14 Pages 14/18 Subject Code: Model Answer Page No: _14_/ 18 Q5 e) What is ladder diagram? How it is different than the normal circuit diagram A Ladder diagram is schematic representation of logical sequence of PLC operation. Ladder diagrams have two supply line and individual rungs which inputs on left side and outputs on right side. The basic symbols used in ladder diagrams are ( ) Output (\) No-output [ ] NO contact [\] NC contact A typical ladder diagram is shown here Supply rails A normal circuit diagram is representation of physical elements and their interconnection and a ladder diargram depicts logical sequence of various events and relation between input and outputs Q5 f) Write a program for starting of induction motor Ladder diagram of DOL starting of induction motor 1) When start push button pressed, M will be energised, motor will be ON. 2) When M is energised LAMP1 glows(on pilot lamp). 3) When in OFF condition LAMP2 glows ( OFF pilot lamp). 4) In case of Overload OL will trip and motor will stop. 5) When STOP button pressed M will be de-energised and motor will stop.

Pages 15/18 Subject Code: 12149 Model Answer Page No: _15_/ 18 Q6 a) Explain the procedure of tuning P + I + D controller Tuning of PID controller referes

The selection of these parameters is based on study of the step response of the process.

Following steps are followed, 1) Apply step input to the process in steady state and in open loop and record process vaeriable Vs.

15 Pages 15/18 Subject Code: Model Answer Page No: _15_/ 18 Q6 a) Explain the procedure of tuning P + I + D controller Tuning of PID controller referes to selction of Kp,KI,KD for getting optimum performance of the PID controller for the given process. The selection of these parameters is based on study of the step response of the process. By using Zeigler- Nichols empirical tuning rules exact values of PID parameters can be selected. Following steps are followed, 1) Apply step input to the process in steady state and in open loop and record process vaeriable Vs. time till the PV settles to a new value. 2) Estmate the values of dead time(d), time constant(t) and the process gain(k) from the step response of the system. 3) Based on D,T,K estimate the values of Kp,KI,KD by using Zeigler Nichols tuning formulae. Q6 b) Draw a neat diagram of PLC and explain the function of input module. Input module already explained

Pages 16/18 Subject Code: 12149 Model Answer Page No: _16_/ 18 Q6 c) Explain the bus system and power supply system of PLC.

16 Pages 16/18 Subject Code: Model Answer Page No: _16_/ 18 Q6 c) Explain the bus system and power supply system of PLC. The CPU uses the data bus for sending data between the different elements The address bus to send the addresses of locations for accessing stored data, The control bus for signals relating to internal control actions, The system bus is used for communications between the I/O ports and the I/O unit Power supply section of PLC: Quality of power supply is very important for its reliable operation. Its function is to provide internal power to the digital circuit of PLC. The power supply unit has to regulate the supplied dc voltage levels and inform CPU about fault condition. The PLC power supply is SMPS which consists of dc to dc converter. The ac voltage input the PLC is always taken at the secondary of a step down transformer which is called as control transformer. This protects the PLC from the effects of distortion, noise and EMI problems.

Pages 17/18 Subject Code: 12149 Model Answer Page No: _17_/ 18 Q6 d) Give the features of anyone standard PLC available in market Features: 1 1) No of digital inputs, 2) No of digital outputs, 3)

17 Pages 17/18 Subject Code: Model Answer Page No: _17_/ 18 Q6 d) Give the features of anyone standard PLC available in market Features: 1 1) No of digital inputs, 2) No of digital outputs, 3) Size of program memory 4) Size of data memory 5) Power Supply voltage 6) Scan interval 7) No of analog input outputs and bit resolution Any relevant answer may be given credit Q6 e) Write a program for stepper motor control Four digital outputs Q1-Q4 are taken to drive a stepper motor. Each output remains high for fixed time in sequential manner

18 Pages 18/18 Subject Code: Model Answer Page No: _18_/ 18 Q6 f) Draw a neat diag ram of Distributed Control System (DCS). he DCS model is different: instead of centralizing all system processing in one controller, it spreads out or distributes processing tasks to smaller, local systems. The intelligent processors at the I/O level handle many functions locally, leaving the central processor free to supervise the system as a whole. Central controller Supervisory fucntion Local controller Local controller Local controller Local controller Process 1 Process 2 Process 3 Process 4

EEET 2204 Industrial Automation

EEET 2204 Industrial Automation EEET 224 Industrial Automation EEET 224 Industrial Automation by Dr Peter Graszkiewicz. Discrete-State Control.. Basic Input and Output Devices 3-phase power supply N/C N/O contactor terminals fuses contactor