EASWARI ENGINEERING COLLEGE, CHENNAI-89 Department of Electronics and Instrumentation Engineering Important Two marks Questions and Answers – Process Control [EI2352] –Section B UNIT -1 PART – A 1.
List the • • • • • • •
advantages of automatic process control. Improvement in quality of the product Increase in quantity of the product Improvement in uniformity of the products Saving in power and energy Saving in process material Saving in process equipment Decrease in human drudgery
2.
What is the need for process control? • To optimize the performance • To increase the productivity • To suppress the influence of external disturbances • To make it cost effective
3.
What are the major components of process control system? 1. Plant (Process) 2. Measuring System 3. Controller 4. Final Control Element
4.
What is meant by degree of freedom of a process? Give an example. The Degrees of freedom of a processing system are the independent variables that must be specified in order to define the process completely and is defined by the following relationship: Degrees of freedom = number of variables – number of defining equations The degrees of freedom of a process represents the maximum number of independently acting automatic controllers that can be placed on that process. In a heat exchanger system, heated fluid flow rate, heating fluid flow rate, heating fluid inlet temperature, heating fluid outlet temperature, heated fluid inlet temperature and heated fluid outlet temperature are the six variables available. They can be defined by first law of thermodynamic equation. Hence the degrees of freedom of this system is 6 -1 = 5.
5.
Define time constant of a process. Time constant of a process is given by the product of the process capacitance and process resistance. It is the time taken by the step response of the process to reach 63.2% of the steady state value.
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Dr.S.Nagarajan, Professor / EIE, Easwari Engineering College, Chennai-89
EASWARI ENGINEERING COLLEGE, CHENNAI-89 Department of Electronics and Instrumentation Engineering Important Two marks Questions and Answers – Process Control [EI2352] –Section B
6.
A tank operating at 3 m head and 25 lpm outflow through a valve has a cross sectional area of 1.2 m2. Calculate the time constant. Head = H = 3 m Outflow = Q = 25 lpm = 25 x 0.001 m3 / min Cross sectional area = A = 1.2 m2 Resistance = R = H / Q = 3 / (25 x .001) m / (m3/min) = 120 min / m2 Capacitance = C = Cross sectional area = A = 1.2 m2 Time Constant = Resistance X Capacitance = RC = 120 X 1.2 = 144 min
7.
What is the difference between interacting and non-interacting systems? Give an example. In non-interacting systems, any dynamic changes in system 1 will affect the dynamic behavior of system 2. But the dynamic changes in system 2 will not affect the dynamic behavior of the system 1 In interacting systems, any dynamic changes in any of the systems will affect the dynamic behavior of the other system. Example: Non-interacting: level tanks not connected through pipe links
Interacting: level tanks connected through pipe links
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Dr.S.Nagarajan, Professor / EIE, Easwari Engineering College, Chennai-89
EASWARI ENGINEERING COLLEGE, CHENNAI-89 Department of Electronics and Instrumentation Engineering Important Two marks Questions and Answers – Process Control [EI2352] –Section B 8.
Distinguish between servo and regulator operations. In servo problem, the feedback controller is designed to act in such a way to keep the process variable close to the changing setpoint under the condition of no disturbances and no load changes. In regulator problem, the feedback controller is designed to eliminate the impact of load changes or disturbances with set point remaining the same. Block diagram for Servo problem:
Block diagram for regulator Problem:
9.
What is a batch process? A process in which sequence of operation is carried out to produce a quantity of a product ( the batch) and in which the sequence is then repeated to produce further batches is called batch or sequential process.
10.
Distinguish between a batch process and a continuous Batch Process 1. Discrete loads of raw mateterials are fed 1. into the system 2. Each batch of raw materials can be 2. processed differently in the various pieces of equipment. The same equipment might be used to produce different products or product grades. 3. Movement of batch from one equipment to 3. another can occur only when the operation is complete and the next equipment is ready to receive the material. 4. Sequential logic is a part of the normal 4. operating mode. 5. Special failure routines will be executed in 5. the event of failure of processing phases.
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process. Continuous Process Feed of materials is continuous Raw materials are processed in identical fashion.
Materials flow steadily from one equipment to another.
Sequential logic is used only during start-up and shut-down. The process is generally shut-down when failure occurs.
Dr.S.Nagarajan, Professor / EIE, Easwari Engineering College, Chennai-89
EASWARI ENGINEERING COLLEGE, CHENNAI-89 Department of Electronics and Instrumentation Engineering Important Two marks Questions and Answers – Process Control [EI2352] –Section B UNIT -2 1.
What is referred to as ‘Reverse action’ in the controller operation? A controller is said to be operating in REVERSE ACTION when an increasing value of the controlled variable causes a decreasing value of the controller output. Example: Temperature control of a furnace with fuel as energy - If the temperature increases, the controller output should decrease to close the valve for decreasing the fuel input to bring down the temperature.
2.
What is meant by differential gap? In any practical implementation of the two position controller, there is an overlap as error (eP) increases through zero or decreases through zero. In this span, no change in controller output occurs and this region of 2∆eP is referred to as neutral zone or differential gap.
3.
Why do you require differential gap adjustment? Differential gap is required in on-off controllers to prevent excessive cycling i.e. the oscillation of error about zero.
4.
Define proportional band. Proportional band is defined by the equation
PB =
ଵ K౦
where KP is the proportional gain. The controller
output is not saturated at 0% or 100% within this band of error about zero. 5.
Define Offset. How it can be minimized in P-controller? Offset is the permanent residual error in the operating point of the controlled variable produced by the proportional controller when a change in load occurs. This can be minimized by setting a larger proportional gain for the proportional controller.
6.
How does the use of Integral Controller affect the stability of the process? In the integral controller, the value initially begins to change rapidly. As the final control element is started operating, the error decreases and it is slowed down. If the process lags are too large, the error can oscillate about zero or even be cyclic, thereby affects the stability of the process.
7.
Why the derivative control mode is not recommended alone? In derivative control mode, the controller output is zero when the error is constant and thus the controller takes no action for the error. Hence derivative control mode is not used alone.
8.
Write the transfer function of proportional integral control. The controller output of the Proportional Integral Controller is given by ୲
P = K P EP + K P K I න EP dt + PI (0) where P KP
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= Controller output, = Proportional Gain,
Dr.S.Nagarajan, Professor / EIE, Easwari Engineering College, Chennai-89
EASWARI ENGINEERING COLLEGE, CHENNAI-89 Department of Electronics and Instrumentation Engineering Important Two marks Questions and Answers – Process Control [EI2352] –Section B KI EP PI(0) On applying Laplace transform,
= Integral gain, = Error and = Integral term value at t=0 (initial value)
PS K P EP S K P K I
EP S 1 K P K P K I EP S S S
KPS KPKI PS EP S S PS KPS KPKI EP S S 9.
Design electronic P controller with PB = 50%. PB= 50% 100 100 KP 2% PB 50 R% In the circuit K P 2 R Taking R1=1 K, R2 =2R1 = 2K Electronic P controller:
10.
Design an electronic controller for Kp=5 and Ti=10 sec. Ti = 10 sec R2C = 10 sec Assuming C = 100 µF, R2 = 10 / C = 100 K KP 5
R% 5 R As R2=100 K, R1 =100 / 5= 20 K
In the circuit K P
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Dr.S.Nagarajan, Professor / EIE, Easwari Engineering College, Chennai-89
