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Index

• Introduction/Objective
  • 1.1 Task 1
  • 1.2 Task 2
  • 1.3 Task 3
  • 1.4 Task 4
• Task 1
  • 2.1 Part 1
    • 2.1.1 Types of System
      • 2.1.1.1 Open Loop Systems
      • 2.1.1.2 Closed Loop System
    • 2.1.2 Functions of Sensors and Actuators
      • 2.1.2.1 Sensors
      • 2.1.2.2 Actuator
    • 2.1.3 Motor Comparison
  • 2.2 Part 2
    • 2.2.1. Gear Ratio Calculation
    • 2.2.2. Closed Loop Control of AC Servo-motor
• Task 2
  • 3.1. Part 1
  • 3.2. Part 2
• Task 3
  • 4.1. Part 1
  • 4.2. Part 2
• Task 4
  • 5.1. Part 1
  • 5.2. Part 2
• List of Firgures
• Bibliography

1. Introduction/Objective

The aim of this report is to develop to study the industrial system and the applications of various components included in the system such as electrical, electronic and mechanical system and transducers and sensors. The study is organized on the basis of the tasks defined in a systematic manner in which the preliminary study of the sensors designs and applications are done on the basis of the conventional methods of the sensors and transducers design and applications. Next, the gained knowledge and understanding is applied on the basis of the system using a simple simulation of the motor parameters and features such as voltage, current, speed etc. Further, we will see the effects of a mathematical and real time difference in the measurements systems and devices. This practical experiment is done by executing the testing of the system. On the final note, we will use all the learnings of the above tasks ad implement them in our industry and determine the performance. Following are the tasks list of the assignment.

1.1 Task 1

Part 1

• Compare open-loop and closed-loop control systems. Give an example of both the systems.
• State the basic function of each of the following as part of a process-control system:
  • Sensors
  • Actuators
• Calculate which of the following motors gives the greater load acceleration: –
  • Motor A-has a max. torque of 0.15 N-m and an inertia of 2.5 kg-m².
  • Motor B-has a max. torque of 0.25 N-m and an inertia of 5 kg-m².

Part 2

• Determine the gear ratio necessary to obtain greatest load acceleration, and then calculate the maximum load acceleration, if the load inertia is 450 kg-m², using the motor with the greater load acceleration.
• Use diagrams to help describe how it is possible to control the speed of an AC-servo motor using a closed speed control system. Explain how the error signal is formed ad discuss what ‘fault’ conditions could cause the speed to drop abnormally.

1.2 Task 2

Part 1

• For the speed control system shown in the figure below, the feedback transducer has a transfer function of 50mV per r/min and the transfer function for the motor/load is 100r/min per volt. The reference signal applied to the system is 100V.Given that the allowable speed error is 40 r/min, determine the amplifier gain (transfer function)

Part 2

• For the system in the above question the reference voltage is reduced to 80V.Determine the Corresponding reference speed.
• For the system in the above question, the load speed is adjusted to 1900 r/min. Determine the corresponding reference voltage. critically investigate the behavior of the control system to compare the above 3 approaches.

1.3 Task 3

Part 1

• For your laboratory work, you have to test a measurement system, like digital voltmeter, digital ammeters, etc. or any other measurement system available in your laboratory. The approach for this test would be practical and computer based.

Part 2

• Explain the procedure for this and talk about accuracy, resolution, tolerances, stability, sensitivity and response time for that measuring system. Critically evaluate the performance of an ideal measurement system compared to a real circuit.

1.4 Task 4

Part 1

• Write down the advantages and disadvantages of increase use of wireless and remote control in GRACO industry.

Part 2

• Analyze appropriate analytical techniques used in your industry(GRACO) and justify your recommendations to improve the performance.

2.Task 1

2.1 Part 1

Control system engineering, a branch of engineering which deals with the modeling, designing and development of the control systems which are a crucial part of the system of the integrated machine and industrial devices. As there are numerous kind of systems available but on a general note, they are defined as open loop and closed loop system. They are defined as follows:

2.1.1 Types of System

• Open Loop Systems
• Closed Loop Systems

Open loop systems are basically being those systems whose output is independent of the input parameters. This means that the output of the system remains constant for a particular type of input. Such systems remain undisturbed by the external disturbances. They are basically represented as follows:

Fig.1(Open Loop System)

2.1.1.2 Closed Loop Systems

The closed loop system is the one in which the response is fed back to the input system via a feedback controller. The aim of this system is to calibrate the difference as per the desired output. Thus, the closed loop system has an advantage over the open loop systems as they doesn’t have the capacity of self-calibration or correction. But, the closed loop system has one disadvantage that they are prone to the external disturbances if gets added in the signal.

Fig.2. (Closed Loop System)

The examples of closed loop are: Nervous system of human body, Aircraft autopilot system etc.

2.1.2 Functions of Sensors and Actuators

2.1.2.1 Sensors

A sensor is a transducer that converts a physical stimulus from one form into a more useful form to measure the stimulus. The function of the sensor is to convert the input into the form which can be processed by the system and based on that input the system can change the input of the process and get the desired result. They are based classified in two categories:

• Analog
• Discrete(Binary,Digital)

The examples of sensors are as photo diode, LDR, Thermocouple etc. They are the ones which measures the physical quantity such as heat, light or temperature and then converts them in the form of electrical signals readable by electrical systems.

2.1.2.2 Actuators

The actuators are basically the ones which convert a controller command signal into a change in a physical parameter. Basically they are the inverse of sensors. They bring the physical change in the system. They convert the electrical signal into a physical form such as motion. They are of various types such as:

• Linear Actuators
• Electrical Actuators
• Hydraulic Actuators
• Pneumatic Actuators

Fig.3(Actuators)

T=Jα ————————————————(1)

Where,

T=Torque

J=Moment of Inertia

Α=Load Acceleration

So, as per the given formula,

• CASE I:
  Motor A (J=2.5 kg-m², T=0.15 N-m)
  Thus from the calculation, we have Load acceleration as:
  A=T/J
  A=0.15/2.5= 0.06 (m/s²)

• CASE II
  Motor B (J=5 kg-m², T=0.25 N-m)
  Thus from the calculation, we have Load acceleration as:
  A=T/J
  A=0.25/5= 0.05 (m/s²)

Thus, from the above result we can decide that the load acceleration is more for motor A then motor B which also tells us that the lighter the body the more will be the acceleration.

2.2 Part 2

2.2.1 Gear Ratio Calculation

Now as per the above values found using the formula of torque and load acceleration, to find the gear ratio we can use the generalized ratio of the gear ratio and torque formula as:

T₂/T₁=N₂/N₁

Where

T₂/T₁= Torque Ratio
N₂/N₁= Gear Ratio

Using (1) We Can calculate the greatest load acceleration as

T₂/T₁=J₁α₁/J₂α₂

                                                                                 T₁=(T₁*J₂* α₂)/ J₁α₁              (using given value)

On Solving, we get

α₂=3.333*10^-4 m/s²                 (maximum load acceleration)

T₁= 0.14985N-m

From the provided formula of gear Ratio, we get

N₂/N₁= 1:1      (gear ratio)

2.2.2 Closed Loop Control of Ac Servo-Motor

Servo Motor are also called Control motors. They are used in feedback control systems as output actuators and does not use for continuous energy conversion. The principle of the Servomotor is similar to that of the other electromagnetic motor, but the construction and the operation are different. Their power rating varies from a fraction of a watt to a few hundred watts. The rotor inertia of the motors is low and have a high speed of response. The rotor of the Motor has the long length and smaller diameter. They operate at very low speed and sometimes even at the zero speed. The servo motor is widely used in radar and computers, robot, machine tool, tracking and guidance systems, processing controlling, etc. Following is the control system description of the AC servo motor feedback system (Closed Loop).

Fig.4. (AC Servo-Motor)

In the technical terms, the X/R Ratio which is also known as impedance ratio, decides the working range of motor as per shown below graph: –

Fig.5. (X/R ratio graph)

The Torque speed characteristic of a normal induction motor is highly nonlinear and has a positive slope for some portion of the curve (see Ref.2).  This is not desirable for control applications. as the positive slope makes the systems unstable. The torque speed characteristic of an ac servo motor is fairly linear and has negative slope throughout. The rotor construction is usually squirrel cage or drag cup type for an ac servo motor. The diameter is small compared to the length of the rotor which reduces inertia of the moving parts (see Ref.4). Thus it has good accelerating characteristic and good dynamic response. The supplies to the two windings of ac servo motor are not balanced as in the case of a normal induction motor. The control voltage varies both in magnitude and phase with respect to the constant reference vulture applied to the reference winding.

The direction of rotation of the motor depends on the phase (± 90°) of the control voltage with respect to the reference voltage. For different rms values of control voltage the torque speed characteristics are shown in Fig.

Fig.6. (Torque Speed graph)

The torque varies approximately linearly with respect to speed and also controls voltage. The torque speed characteristics can be linearized at the operating point and the transfer function of the motor can be obtained. The error signal of AC servo motor is the maximum permissible error in the speed of the AC servo motor (see Ref.3).

The generalized transfer function of the system is:

ω(s)/V_c(s)={K_m/ (Τ_m(s)+1)}

where,

ω(s)=speed in s Domain

V_c(s)=Control winding voltage in s Domain

K_m= Motor Gain

Τ_m(s)=Motor Torque is s Domain

Thus, from the above method we can find the overall system gain of the AC servo Motor.

The disturbances and the error signals, the error signal arises:

• Due to transients.
• Due to phase imbalance.
• Due to armature reaction.
• Due to Harmonics.

Fault Conditions:

As the motor is an electromechanical device, the possibility of the error can’t be ignored. Thus, let us discuss about the possible fault conditions which can be raised:

• Power Failure:
  As power failure seems to be a very common failure possibility, but its importance is not ignored when employed in a very precise application such in medical operations etc. To avoid the error assurance of an auxiliary supply is prevalent.
• Transients:
  The transients are the spikes which are generated in the supply due to the presence of electronics devices which can create imbalance and affect the motor operation. the use of filters and compensators can solve the issue.
• Armature reaction:
  As we know that the motor has the phenomenon of armature reaction which can cause the quadrature and direct phase desynchronization in the supply leading to the armature reaction which directly lowers the speed of the motor and will also increase the loss. The remedy for this is to align the magnetic and electrical axis appropriately.

3. Task 2

3.1 Part 1

As per the mentioned task and the given figure,

Fig.7. (Problem System)

We can derive the system by the use of control system method. So, From the figure, we can assume that,

Fig.8. (Solution Figure)

This resembles to the general feedback system So, with this, we can use the rule for finding the amplifier gain:

 So the general feedback formula is given as:

C(s)/R(s)=A(s)G(s)/(1+(A(s)G(s)H(s)))

On substituting and calculating the value, of the above mentioned block gains we get:

A(s) = 0.01

3.2 Part 2

• On the condition of providing, the input voltage (R(s)) as 80 volts and On applying the same formula as in (1), we get:

  C(s)=0.08 r/min

• Speed is adjusted to 1900 r/min.

  R(s)=190.105 volts

This shows that the when the input systems value is reduced, then the output systems value is also reduced which shows the principle of linearity. This means that with the input, the output follows the principle of superposition and homogeneity (See Ref.1).

4. Task 3

4.1 Part 1

The objective of this task is to perform the testing of measurement system such as digital voltmeter etc. Thus, we will perform a test of a digital voltmeter. The test will be performed in a simulation which measures a voltmeter reading in performing a measurement of the resistance voltage drop.

So, the parameters on which we will perform the test of the instrument are:

• Accuracy
• Resolution
• Tolerances
• Stability
• Sensitivity
• Response Time

So to perform each of the tasks, let us attempt by using a table of test results

Following is the simple figure of the circuit used for testing:

Fig.9. (Testing Schematic)

 

Fig.10(Graph of Current Reading)

 

Fig.11(Graph of Power Dissipation Reading)

4.2 Part 2

Procedure:

1. Set the voltage to 12 volts and the voltmeter resistance to 100M ohms and note down the ammeter and wattmeter reading.
2. Next, increase the voltage by 2 and repeat above step till 20 volts.
3. Then, change the value to 100,10 and finally 1 and repeat from step 1.and note down the readings of ammeter and wattmeter
4. Now calculate the theoretical values of current and power using the ohms’ law and power dissipation formula.
5. Then plot the curves of current and power you will get the deviation visible, which shows the difference from the actual and theoretical value.

Thus, it is evident from the data that there is a slight deviation from the ideal characteristics which is about of 1% tolerance. The sensitivity of the system is practically calculated by the simulation time and  the time taken by the meter to make the value stable which is incidentally equals to the stability. The final conclusion tells us that there is a manufacturing tolerance error in the device which declines the accuracy of the device by ±1%(See Ref.5).

5. Task 4

5.1 Part 1

As an industrial technician in the GRACO industry, I discovered various industry specific modifications possible in the system where many systems can be easily controlled, supervised and automated via the application of wireless monitoring and controlling systems such as pneumatic systems with Wi-Fi systems control features (See Ref.6). Further modification has advantages and disadvantages which are as follows:

Features	Advantage	Disadvantage/Limitation
WI-Fi based supervisory system	Better control and administration	Network Dependent
Remote pressure sensor	Accurate measurement	Needs periodic maintenance
Wi-Fi based temperature sensor	Better temperature management and power saving	Needs frequent and dedicated monitoring
Electrical system sensors	Central control system possible and better control	Additional support for motor breakdown conditions
Electronics systems	Low power system with power saving and efficient work output	Needs to be taken care separately
Mechanical systems sensors	More output and more stability	Hard to modify

 

5.2 Part 2

The above recommendations are made on the basis of the various analytical techniques and the methodologies for the same. The various systems for the analysis of the above systems are defined as follows:

• Wi-Fi Based Supervisory System:
  Number of manpower deployed in the task and the number of reports of breakdown issues in the daily log.
• Remote Pressure Sensor:
  The analytical technique is used for this purpose is based upon the dynamic performance of the actuating systems and their various peaks and maximum output time periods and the level of its core application.
• Wi-Fi Based Temperature Sensor:
  The Temperature sensor’s use in analyzed in the manufacturing sections temperatures variation and the maximum peak temperatures in the hot summer and during its most prolonged working schedule on continuous nature.
• Electrical, Electronic and Mechanical Sensor:
  This requirement is recommended in the basis of the losses and the power consumption and their operation statistics in which the system needs to be operated as well as it requires human interference which not only hinders the work but also the additional supply to keep the machine continuously power ON will also add to the additional powerloss of the industry which either can be solved by using the solar panels as an auxiliary supply of power or else we can use the electronics sensors for this application in order to save the power and for a better economic operation of the system.

6. List of Figures

1. Fig.1 (Open Loop System)_____________________________________ 5
2. Fig.2 (Closed Loop System)____________________________________ 6
3. Fig.3 (Actuators) ___________________________________________7
4. Fig.4 (AC Servo-Motor)______________________________________ 9
5. Fig.5 (X/R Ratio Graph)_____________________________________ 9
6. Fig.6 (Torque Speed Graph)__________________________________ 10
7. Fig.7 (Problem System)_____________________________________ 12
8. Fig.8 (Solution Figure)_____________________________________ 12
9. Fig.9 (Testing Schematic)___________________ ________________15
10. Fig.10 (Graph of Current Reading) ____________________________15
11. Fig.12 (Graph of Power Dissipation Reading) ____________________16

7. Bibliography

1. S,Norman,Control System Engineering 7^th edition,Wiley Publications,2015,New York,USA
2. Bhunia,S.C.,Pal Srimanta,Engineering Mathematics,Oxford Publications,2015,Oxford,UK
3. Kothari,Nagrath, Power Systems Engineering,Tata McGraw Hill,2015,New Delhi ,India.
4. https://www/mathworks.com/plot
5. https://wirelesscomm.com/industry+based+wifi
6. https://www.wikipedia.org/industrial+systems

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