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20EI501-Process Control

Department: Electronics and Instrumentation Engineering

Batch/Year: 2021-25/III

Created by:

Ms. K.R. Chairma Lakshmi AP/EIE

Table of Contents

1. COURSE OBJECTIVES

2. PRE REQUISITES

4. SYLLABUS

SUBJECT CODE : 20EI501

SUBJECT NAME: PROCESS CONTROL

L T P C :3 2 0 4

UNIT I PROCESS MODELLING AND DYNAMICS

Need for process control – Mathematical Modeling of Processes: Level, Flow, Pressure and Thermal processes – Continuous and batch processes – Interacting and Non-Interacting system - Self regulation – Servo and regulatory operations – Lumped and Distributed parameter models – Heat exchanger – CSTR

– Linearization of nonlinear systems.

​

UNIT II FINAL CONTROL ELEMENTS

Actuators: Pneumatic and electric actuators – Control Valve Terminology - Characteristic of Control Valves: Inherent and Installed characteristics - Valve Positioner – Modeling of a Pneumatically Actuated Control Valve – Control Valve Sizing: ISA S 75.01 standard flow equations for sizing Control Valves – Cavitation and flashing – Control Valve selection

​

UNIT III CONTROL ACTIONS

Characteristic of ON-OFF, Proportional, Single speed floating, Integral and Derivative controllers P+I, P+D and P+I+D control modes – Practical forms of PID Controller – PID Implementation Issues: Bumpless, Auto/manual Mode transfer, Anti-reset windup Techniques – Direct/reverse action.

​

UNIT IV PID CONTROLLER TUNING

PID Controller Design Specifications: Criteria based on Time Response and Criteria based Frequency Response - PID Controller Tuning: Z-N and Cohen-Coon methods, Continuous cycling method and Damped oscillation method, optimization methods, Auto tuning – Cascade control – Feed-forward control

SYLLABUS

UNIT V MODEL BASED CONTROL SCHEMES

​

Smith Predictor Control Scheme - Internal Model Controller – IMC PID controller –

- Three- element Boiler drum level control - Introduction to Multi-loop Control

Schemes – Control Schemes for CSTR, and Heat Exchanger - P&ID diagram.

4. Course Outcomes

5. CO-PO/CO-PSO MAPPING

6. Lecture Plan-UNIT-III

7. ACTIVITY BASED LEARNING

Topic: Match the controller characteristics to its suitable controller

​

Activity: Think and pair

Students will recall the characteristics of the controller studied and match with the suitable name of the controller. Students can be formed as a group and be instructed to discuss the characteristics and match the correct pair.

2. Describe one example of (a) flow control, (b) pressure control, (c) liquid level control, (d) temperature control, draw the appropriate diagram and suggest a suitable controller.

​

Activity: Think and correlate the suitable controller

Students can be asked to select one example and make them to draw a process diagram and depending upon the process behavior they can be asked to suggest a suitable controller for that process.

3. Hypothetical Industries has just put you in charge of one of their batch reactors. Your task is to figure out a way to maintain a setpoint level inside of the reactor. Your boss wants to use some type regulator controller, but he is not quite sure which to use. Help your boss find the right type of controller. It is extremely important that the level inside the reactor is at the setpoint. Large fluctuation and error cannot be tolerated.

Activity: Ponder and Propose

Students can put themselves in the real situation and allowed to think about the process, its associated variables (measuring, controlling and manipulating variables) and propose a suitable controller from their own understanding. (Can be done as a group too)

8. Lecture Notes-Unit-III

Automatic control system

The automatic controller determines the value of controlled variable and compares it with the desired value to find the deviation and produces the counter action to maintain smallest deviation from desired value.

The method by which the automatic controller produces counter action is called mode of control or control action.

Fig: Closed loop System

​

Basic hardware Components

Process: The material equipment along with the physical or chemical operations which takes place (Tanks, heat exchanger and reactors)

Measuring system: Sensors like bellows, thermocouples, orifice, diaphragms, gas chromatographs etc.,

Controllers: This will determine how the manipulated variable can be changed to achieve the desired value (set point)

Final control element: A control valve or variable speed metering pump. It receives the control signal from the controller and implement it by physical adjustment.

​

Transmission lines: It is used to carry the measurement signal from sensor to

controller and the control signal from controller to final control element.

​

Process Characteristics

Process Equation: Process could be described by a set of equation called process equation.

Process load: Process load refers to set of parameters excluding the controlled variable. When all the parameters have their nominal values then the system is said to at nominal load. Whenever the parameters change from nominal values, the process load change has occurred. The controlling variable is adjusted to compensate for the load change and its effect on the dynamic variable to bring it back to set point.

​

Transient: Temporary variation of one of the load parameter.

Process lag: The time taken by the process to have a change in itself is called process lag.

Self regulation: It is the tendency of some process to adopt a specific value of the controlled variable for nominal load with no control operations.

​

Control system Parameters

​

Error: Deviation of controlled variable from the set point.

Variable range: The variation under control has a range within which control is maintained. The range can be expressed as the minimum and maximum value of the variable.

Control parameter range: Controller output range is the translation of output to the range of possible values of final control element.

Control lag: It is the time taken by the process control loop to make necessary adjustments to final control element.

Dead Time: It is the elapsed time between the instant deviation (error) and the corrective action that occurs first. It has a profound effect on the performance of control operations on a process

Cycling : The oscillation of the error about zero is cycling. The variable is cycling above and below the set point value.

​

Control actions

Direct action: A controller is said to be operating with direct action, when an increasing value of the controlled variable causes an increasing value of the controller output.

Reverse action: A controller is said to be operating with reverse action when an increasing value of the controlled variable causes decreasing value of the controller output.

​

Consider Two cases :

​

1) Level Controller LC controlling discharge control valve.

In this process I have connected a level controller to the bottom valve. For this configuration the controller needs to increase its signal (and hence the flow) when the level in the tank increases.

Fig: Direct and Reverse action

2) Level Controller LC controlling inlet control valve.

In this case the controller needs to reduce the flow when the level in the tank increases.

Both configurations are equally capable of controlling the level, but they require the controller to do entirely opposite things. This is what direction of control action involves.

​

A direct acting controller is one whose output tends to increase as the

measurement signal increases.

A reverse acting controller is one whose output tends to decrease as the measurement signal increases.

​

Controller Modes

Discontinuous Modes

Two position

Single speed floating

Multiple speed floating

​

Continuous Modes

Proportional Integral

Derivative

​

Composite Control Modes

PI

PD PID

Two Position Mode [On – Off controller]

It is the simplest, cheapest and undoubtedly the most widely used type of control for both Industrial and Domestic service.

It is a position type of controller action in which the manipulated variable is quickly changes to either a maximum or minimum value depending upon whether the controlled variable is greater or lesser than setpoint.

​

The minimum value of the manipulated variable is usually zero (off) and the

maximum value is the full amount possible (on)

In general,

_⎪⎧^{0% e}_p ^{< 0}

^{P= ⎨} 100% e > 0

⎪_{⎩ p}

This relation shows that when the measured value is less than the set point, full controller output results. When it is more than the setpoint, the controller output is zero.

This relation shows that when the measured value is less than the set point, full controller output results. When it is more than the setpoint, the controller output is zero.

​

a) Electrical Two position control

Example : Room heater or water heater Liquid level control

A liquid level control is shown in fig. A float in the vessel operates an electric switch which controls power to a solenoid valve.

When the liquid level rises, the switch contacts are closed, the solenoid valve closes and the inflow is cutoff. When the liquid level falls, the switch contact are opened, the solenoid valve opens and the inflow resumes.

​

If the float has no bearing friction and the electrical contact draw no arc, the

action is sharp.

Fig: Electrical Two Position controller

​

The plot between P and e_p for two position controller is shown below

Fig: Two Position controller Neutral zone

Until an increasing error changes by Δe_p above zero, the controller output will not change state. In decreasing, it must fall Δe_p below zero before the controller changes to 0%.

The range 2Δe_p, which is referred to as the neutral zone or differential gap in two position controller causes the manipulated variable to maintain its previous value until the controlled variable has moved slightly beyond the setpoint.

​

A differential gap is caused in two position controller for the following reason.

The liquid level must rise slightly above the desired value to create sufficient buoyant force to overcome friction when the level is rising. The liquid level must fall slightly below the desired value when the level is falling so that the weight force may overcome the friction.

​

b) Pneumatic Two Position Controller

It consist of flapper nozzle and air relay or power amplifier. The back pressure (P_b) in the chamber is controlled by the position of the flapper with respect to the nozzle.

If the flapper is fully closed (X=0), the back pressure will be equal to the supply pressure (P_s). If the flapper is fully open (X is very large), the back pressure will be approximately equal to the ambient pressure. The relay is termed reverse acting because an increasing P_b there is corresponding decrease in P_o.

​

P_b = P_o = -K_x

Fig: Pneumatic Two position controller

​

Applications

​

• It is best suited to large scale systems with relatively slow process rates.

​

• In a room heating or air conditioning system, the capacity of the system is very

large in terms of air volume, but the overall effect of heater or cooler is relatively slow.

​

• Liquid bath temperature control

​

• Level control

Advantages

Only two output states i.e. ON and OFF. Simple construction

Low cost

​

Disadvantages

Response of ON-OFF controller is slow.

Not suitable for complex system

Single speed floating

Floating Mode: In this mode, the specific output of the controller is not uniquely determined by the error.

If the error is zero, the output will not change but remains (floats) at whatever setting it was when the error went to zero.

​

When the error moves off zero, the controller output again begins to change.

​

Two types

Single speed floating control

Multiple speed floating control

​

Single speed floating control

The manipulated variable changes at a constant rate in one direction when the deviation is positive and in the opposite direction when the deviation is negative.

The output of the control element changes at a fixed rate when the error exceeds

the neutral zone

(1)

dt

p

f p

^dP = ± K [e ]> Δe

^dP = rateof changeof controlleroutput with respect to time dt

​

K_f= rate constant (%/s)

If eqn (1) is integrated for the actual controller output

​

P= ± K_f t + P(0)

P(0)= controller output at t=0.

Eqn. (2) shows that the present output depends on the time history of errors that have previously occurred.

If the deviation persists, eqn. (1) shows that the controller saturates at either 100% or 0% and remains there until an error drives it towards the opposite extreme.

In this example, assume controller is reverse acting, which means the controller output decreases when the error exceeds the neutral zone. This corresponds to negative K_f in equation.

The controller starts at some output P(0). At time t, the error exceeds the neutral zone.

Therefore the controller output decreases at a constant rate until t₂. When the error again falls below the neutral zone limit at ‘t_s’, causing controller output to change.

Single speed floating control is commonly associated with systems in which the final control element is a single speed reversible motor. The output of the reversible motor is either forward, reverse or off depending on the error condition.

If the solenoid in liquid level control is replaced by reversible motor with gear reducer to move the control valve system. When the level rises the switch contact is made and the motor reducer slowly closes the control valve. As soon as the level falls, the switch contact is broken and the motor reducer reverses its direction of rotation and opens the control valve.

​

Continuous controller modes

Continuous controller modes are an extension of the discontinuous controller. The output of the continuous controller changes smoothly in response to the error or rate of change of error. This mode includes proportional, Integral and differential.

​

Proportional control mode

A smooth linear relationship exists between the controller output and the error.

In other words, proportional action is a mode of controller action in which there is a continuous linear relation between values of the deviation and manipulated variable. The action of the controller variable is repeated and amplified in the

action of the final control element.

Common names are proportional action, correspondence control, droop control and modulating control.

The adjustable parameter of the proportional mode K_p is called proportional gain or proportional sensitivity.

​

Over some range of errors about the setpoint, each value of error has unique

value of controller output.

The range of error to cover the 0% to 100% controller output is called proportional band (P_B) because the one to one correspondence exists only for the errors in this range. The proportional band is equivalent to inverse of proportional gain.

The proportional band characterizes the range over which the error must change in order to drive the actuating signal of the controller over its full range.

​

Proportional mode can be expressed by the equation

​

P = K_p e_p + P₀

Transfer function of proportional controller K_p

p

B

K

P

₌ 100

Fig: Controller output versus error response of Proportional controller

​

Characteristics

​

If the error is zero, the output is a constant equal to P₀

If there is a error, for every 1% of error a correction of K_p percent is added to or subtract

from P₀ depending on the reverse or direct action of the controller.

There is a band of error about zero of magnitude P_B within which the output is not saturated to 0% or 100%

Proportional band is dependent on gain. A high gain means large response to error but has a narrow error band.

Fig: Offset

Offset

Whenever a change in load occurs, the proportional control mode produces a permanent ‘residual error’ in the operating point of the controlled variable. This is called as offset. It can be minimized by a large constant K_p which also reduces the proportional band.

Fig: Controller output when load varies

​

Application

It is used in processes where large load changes are unlikely or with moderate to small processes lag times.

​

Effect of proportional control on the response of a controlled process

Fig: Closed loop system

(2)

(1)

Process Y (s)=G_P (s) m(s) + G_d(s) d(s)

​

Measuring device Y_m (s)=G_m(s) Y(s)

​

e(s)=Y_sp(s) – Y_m (s)

C(s)=G_c (s) e(s)

​

Final controller element=m(s)=G_f(s) C(s)

m(s)=G_f (s) G_c (s) e(s)=G_f (s) G_c (s) [Y_sp(s) – Y_m (s)]

m(s)=G_f(s) G_c(s) e(s) =G_f (s) G_c (s) [Y_sp (s) – G_m (s) Y(s)]

Sub eqn (2) in eqn. (1)

​

Y (s)= G_P (s) G_f (s) G_c (s) [Y_sp(s) – Y_m (s)] + G_d(s) d(s)

In the above equation first term shows the effect of output when set point changes. While the second term shows the effect on output when change in load.

​

Assume G_m (s) = 1 G_f (s) = 1 G_c (s) = K_p

Consider the process transfer function as

sp

Y (s) + d(s)

G_p (s) G_f (s) G_c (s) G_d (s)

^{1 + G}_p ^{(s) G}_f ^{(S) G}_c ^{(s) G}_m ^(s) 1 + G_p (s) G_f (S) G_c (s) G_m (s)

Y(S) =

d(s)

sp

Y (s) +

p ^G_d ^(s)

G (s) K

p

^{1 + G}_p ^{(s) K} _p 1 + G_p (s) K _p

Y(S) =

K

p

p

= G (s)

τ s+1

p

p

^Kd = G (s)

τ s+1

K K

K K

p

K_d

sp

p

K K_p

s

τ_ps+1

1+

τ_p s+1 _d(s)

Y (s) +

τ_ps+1

1+

τ_ps+1

Y =

Closed loop response of first order system has following characteristics

​

It remains first order with respect to load and set point changes

The time constant has been reduced ( ) which means closed loop response has become faster than open loop response

​

The static gains have been decreased

​

Servo Problem

Fig: Output response

d(s)

p

p

sp

p

p

K K_p

s

K_d

τ s+1+ K K

Y (s) +

τ s+1+ K K

Y =

K'

p

_K'

p

τ^' s + 1

Y_sp (s) + ^d d(s)

τ^' s + 1

Y(s) =

p c

p

τ_p

1+ K K

τ^' =

K K_p

1 + KK_p

K^' =

p

d

_K'

1+ K K

₌ K_d

d

K^' & K^' → Closed

s

s p

Y (s) = ¹

d(s) = 0, i.e.step change given in input

​

K' 1

p

τ^' s + 1 s

∴Y(s) =

Take inverse Laplace transform_−t

_T '

Y (t) = K ^' (1 − e ^p )

The ultimate response after _{t →α}, never reaches the desired new setpoint. There is

always a discrepancy called offset.

​

Offset=(new set point) – (ultimate value of response)

K K _p

1

1+ K K _p 1+ K K _p

=1 – K^’= 1− =

Therefore offset becomes zero when

K _p →

Remarks

Although offset tends to zero as ^K _p ^→α, extremely large values of K_p are never are used.

1

Processes having the term in their transfer function when controlled by

proportional controller, do not exhibit offset for set point changes (Eg. Liquid level

control with pump)

​

Advantages

​

Construction is simple

These controllers has high loop gain

​

It has steady state tracking accuracy

​

It improves the disturbances signal reduction

​

It stabilizes the gain and makes the system more stable

​

Disadvantages

It cannot accommodate load change without sustained deviation. It produces the constant steady state error.

For very large gain it leads to instability of the system

It has a sluggish i.e. slow response for wide proportional band. It makes the system less sensitive to parameter variation

Integral control Mode (Reset Action Mode)

It is a mode of control action in which the value of the manipulated variable is changed at the rate proportional to the deviation.

When the controlled variable is at set point (zero deviation), the final control element remains stationary. If the deviation is doubled over a previous value, the final control element is moved twice as fast.

This mode represents a natural extension of the principle of floating control. This mode can be expressed by the following equation.

Integral time is defined as the times of change of controlled variable caused by a unit change of deviation.

P(0) = controller output at time t=0

This equation shows that the present controller output P(t) depends on the history of errors from time t=0.

​

If error doubles, the rate of controller output change also doubles.

_I e_p (1)

dp _{= K} dt

^dp = rate of change of c dt

​

K_I=Constant relating to the error (%/s)

K

1

I

I

Integral time = T = sec or min

∫

After integrating eqn. (1), the actual controller output at any time is

t

p

0

P (t) = K_I e (t) dt + P(0)

K_I=expresses the scaling between error and controller output. Large value of K_I means that a small error produces a large rate of change of P and vice versa.

The relationship between P rate of change and error is represented graphically as below. Rate of output change depends on gain and error

Characteristics

If the error is zero, the output stays fixed at a value to what it was when the error went to zero.

c

p

(First order system) ; G =

G =

m f

I

p

p

; G = G = 1

K

K

τ s

τ s +1

Y (s)

s p

p f c m

G_p G_f G_c

1+ G G G G

Y (s) =

If the error is not zero, the output will begin to increase or decrease at the rate of K_I percent / second for every 1% of error.

The transfer function of integral control is ^KI (or) ₁

s T_I S

Applic_;ations

The integral mode eliminates the offset. This mode continuously looks at the total past history of the error by continuously integrating the area under the error curve and eliminates the offset.

​

Effect of Integral control action

​

Consider

decreases.

Therefore the response moves in general from sluggish overdamped to faster but oscillatory underdamped behavior.

​

^{If τ}I ^{decreases,ξ decreases then also underdamped behavior is achieved.}

From above two cases, increasing K_c and decreasing ^{τ I} , the speed of closed loop response improved with high deviation and longer oscillation.

Fig: output response

Advantages

​

It reduces steady state error i.e. effect of offset.

It provides high controlled output at a particular time after the error generated is for high value of KI.

​

It responds to the continued existence of deviation.

​

Disadvantages

​

It is never used alone.

​

It makes the system unstable for oscillatory response.

​

It introduces hunting in the system response about its steady state condition.

​

Derivative control mode

The controller output depends on the rate of change of error. The other terms for derivative control are rate response, load component and anticipatory response.

​

Drawback

This mode cannot be used alone because when the error is zero or constant, the controller has no output. For a noisy response with almost zero error, it can compute large derivatives and thus yield large control action.

The analytic expression is

K_D = T_D = rate or derivative time in min.

It is the time interval by which the rate action advances the effect of the proportional control action.

If a derivative mode is set for time T_d, it will generate a correction action immediately which the proportional mode would have generated T_d time later.

Application: This mode becomes necessary as the size of processing equipment increased. In large processes, the derivative mode is to predict process errors before they have evolved and take corrective action in advance

Fig: Derivative Controller response

For a given rate of change of error there is a unique value of controller output. The extent of controller output depends on the rate at which this error is changed and not on the value of the error.

d

P = K

K_D =derivative gain constant (% s/%)

dt

With the presence of derivative term, the controller anticipates what the error will be in the immediate future and applies control action which is proportional to the current rate of change of error. So called anticipatory control

^dep =rate of change of error (% / s)

Characteristics

​

If the error is zero and constant in time, the mode provides no output

If the error changes with time, it contributes an output of K_D % for every 1% / sec rate of change or error. Transfer function of derivative control = K_D s (or) T_D s

​

Effect of derivative control function

K

p

p

G_m = G_f = 1 G_c = K_D S

G (s) =

τ s + 1

(s)

K

K

s p

p

⎠

⎝

⎛ ⎞

1 + ^{⎜ ⎟} K _p T_D s

⎜ ⎟ _{K p TD s}

⎛ ⎞

Y (s) = ^{⎝ p ⎠} Y

⎜

τ s +1

⎟

^⎜τ s +1^⎟

The derivate control does not change the order of the response. (Process order is one and closed loop order is one)

The effective time constant of the closed loop response is i.e. larger than . This means the response of controlled process is slower than that of original first order process. As K_p increases, the effective time constant increases and the response becomes progressively slower.

​

Advantages

​

It can overcome the overshoot and severe cycling.

It has a rapid response to counter the effect of rapidly changing errors. It responds to the changes of the speed and direction to the deviation.

​

It does not affect the steady state error directly, but anticipates the error.

​

It increases the stability of the system by initiating an early corrective action

​

Disadvantages

It cannot be usedalone, since it cannot give any output for zero or constant error.

​

It is ineffective for slowly changing error and hence causes the drift.

It amplifies the noise signal and causes a saturation effect on the system. It does not eliminate the steady state error. (offset)

Fig: Output response for step input signal

Y_{s p} (s)

p p D

K K _p T_D s

Y (s) = ₍

τ + K K T )s + 1

Composite control Modes

The following combinations are commonly used in industrial processes Proportional – Integral control (PI)

Proportional – Derivative control (PD) Proportional – Integral –Derivative control (PID)

​

Proportional Integral Control

This control mode results from a combination of the proportional mode and the integral mode.

​

This mode is also called as proportional plus reset action controller.

​

The analytic expression for PI controller

t

P = K _p e_p + K _p K_{I ∫}e_p dt + P_t (0)

0

P_t (0)= Integral term value at t=0 (initial value)

​

Advantage :

One to one correspondence of the proportional mode is available and the integral mode eliminates the inherent offset.

t

K _p

P = K _p e_p + _{T ∫}e_p dt + P_t (0)

Where T_I is the integ^I ral⁰ time constant or reset time in minutes. The reset time is

an adjustable parameter and is sometimes referred to as minutes per repeat. Usually it varies in range 0.1 ≤τ_I ≤ 50 min . Reciprocal of T_I, is known as reset rate.

Consider that the error changes by a step of magnitude ‘e’. Initially the controller output is K_p e (the contribution of integral term is zero). After a period of T_I minutes the contribution of integral term is

That is, the integral control action has repeated the response of the proportional action. This repetition takes place every T_I minutes and has its name reset time.

Fig: PI controller output for step input

Reset time is the time needed by the controller to repeat the initial proportional action change in its output. The integral action causes the controller output to change as long as an error exists in the process output. Therefore such as controller can eliminate even small errors.

e (T_I ) K _p e

I

K

p

T

K

p

T

T_I

∫

p

e dt =

I 0

Characteristics

When the error is zero, the controller output is fixed at the value that the integral term had when error went to zero. Output = P_t (0).

If the error is not zero, proportional term contributes correction and the integral term begins t decrease or increase the accumulated value depending on the sign of the error and the direct or reverse action.

​

Integral term cannot become negative so it will saturate at zero.

Transfer function of PI controller =

​

Application

It is used in systems with frequent or large load changes.

​

Disadvantage

During startup of a batch process, the integral action causes a considerable overshoot of the error and output before settling to the operation point.

​

Integral windup (or) reset windup

The integral term of a PI controller causes its output to continue changing as long as there is non zer error. Often the error cannot be eliminated quickly, and given enough time they produce larger and larger values for the integral term, which in turn keeps increasing the control action until it is saturated (valve fully open / close). This condition is called integral windup and occurs during manual operational changes like shut down, changeover etc. When the process is returned to automatic operation, the control action will remain saturation leading to large overshoots.

If there is a sudden large change in setpoint, the error between setpoint, the error

⎦

⎣

_s ⎥

⎡ ^K_I ⎤

K _{p ⎢}1 +

between setpoint and process output will suddenly shoot up and the integrator output due to this error will build up with time. As a result the controller output may exceed saturation limit of actuator. This windup unless prevented may cause continuous oscillation of process The characteristics of actuator, the output varies linearly with input till the input is within certain range beyond that output becomes constant either at maximum or minimum values of output.

Fig: Actuator response

There are number of ways that have been developed to minimize problems with reset wind up. Two methods are

1. First order feedback 2. Reset back calculating

Fig: Standard PI control without windup compensation.

​

First order Feedback

Fig: First order feedback

This method uses a switch to break the integral action, whenever the actuator goes to saturation.

When the switch is closed, the controller acts as a PI controller. On the other hand when the switch is open, it is simple P controller.

​

When the switch is closed, Transfer function becomes

Transfer function of PI controller when switch is closed=

In the linear ranghe of actuator, the switch is closed and behaves like PI controller. When the actuator is at saturation, the switch is opened and acts as a P control.

Reset back calculating

Fig: Reset back calculating

Slope of actuator in linear range is unity. As a result when the actuator is operating in the linear range the error e_A is zero, the controller acts as PI controller. But when the actuator is in saturation mode, e_A is negative for a positive error(e). This will reduce the integral action in overall control loop.

⎥

T s

^⎡ 1 ^⎤

K 1 +

⎣ I ⎦

p ⎢

Advantages

It provides better stability to the system.

​

It provides simplicity and directness.

It fully eliminates the steady state

error i.e. offset.

It has good transient response.

​

It stabilizes the controller gain.

​

Disadvantages

It takes the longer time to stabilize the controller gain than proportional controller action.

​

It suffers from only oscillation induced by the integral overshoot.

It requires excessive stabilization, when the process has many energy elements or

dead time.

​

Proportional Derivative control Mode

A derivative control action may be added to proportional control action and the combination is termed as proportional derivative control action

The analytic expression

It cannot eliminate offset of proportional controllers. It can handle fast process load changes.

Transfer function is K_p (1+K_Ds)

o

p p p D

de_p

dt

• P

P = K e + K K

Advantages

It allows the rise of narrower proportional band with its lesser offset.

It increases the controller gain during the error change. It can compensate the rapidly changing error.

It can handle the fast process load change.

​

It can compensate some of the lag in a process

​

Disadvantages

​

It cannot eliminate the offset of proportional controller.

Proportional Integral Derivative control Mode

It is also called as three mode controller. In industrial practice it is called as proportional plus reset plus rate controller. The combination of Proportional integral and derivative modes is one of the most powerful but complex controller mode operations. This system can be used virtually for any process condition.

This mode eliminates the offset of proportional mode and still provides fast response.

The three adjustment parameters are proportional gain, integral time and derivative time.

The PID controller can result in better control than one or two mode controllers. In practice, the control advantage can be difficult to achieve because of difficulty in selecting the proper tuning parameters.

​

Application

Used in process industries to control slow variables such as temperature, pH and other analytical variables.

de_p dt

t

^I 0

+ P_t (0)

P = K e

p p

The analytic equation ₊ ^K_p

T

∫

p

e dt + K K

p D

Advantages

It reduces the overshoot which often occurs when integral control action is added to proportional control action.

It counteracts the lag characteristics introduced by the integral control action. It approaches the tendencies towards oscillations.

It senses the rate of movement away from the set point and gives corrective action earlier than only with P or PI

It is more effective for control process with many energy storage element than P+I control action used alone.

It eliminates the offset i.e. steady state error introduced by proportional control action.

​

It stabilizes the gain of the controller

​

Electronic PID controller

The characterized equation for PID mode is

Where

P= controller output in percent of full scale

e_p = Process error in percent of maximum K_p = Proportional gain

​

K_I = Integral gain

K_D = Derivative gain

P_t(0) = Initial controller integral output

(1)

de_p

dt

t

P = K e

p p

• K_p

T

∫

^I 0

e dt + K K + P_t (0)

p p D

The zero term of proportional mode is not necessary because the integral term accommodates for offset and nominal setting.

​

This mode can be provided by op-amp circuits.

Where R₃ has been chosen from 2πR₃C_D<<T for stability

​

Comparing (1) and (2)

Fig: Electronic PID controller

Discontinuous mode controller

​

Electronic On-OFF controller or Two position controller

​

The detection of an error signal is accomplished in electronic controllers by taking

out

dv

e

dt

p D

t

+ V (0)

Therefore, Output = - V = G V + G G V dt + G G

∫

0

out p e p I e

the difference between voltages. One voltage is generated by the process signal current passed through resistor.

The second voltage represents the set point which is usually general by a voltage divider using a constant voltage as a source.

​

Two position controller

​

A two position controller can be implemented by a great variety electronic and

electromechanical designs.

Bimetal strip and mercury switch combination is employed for position controller applications.

The bimetal strips bends, when temperature decreases and reaches a when the mercury slides down to close an electrical contact. The inertia of mercury tends to keep the system in that position the temperature increases to the value above the set point temperature. This provides the required neutral zone to present excessive cycling system.

A method using op-amp implementation of ON-OFF control with adjustable neutral zone is given in figure. Here the controller input signal is assumed to be a voltage level with an ON voltage of V_H and an OFF voltage V_L and the output is compared V_0.

The comparator output switches states when the voltage on its input Vin equal to the set point value Vsp. Analysis of this circuit shows that the high (ON) switch voltage isV_H = V_SP

Low (OFF) switch voltage is V_L= V_SP – (R₁/R₂ ) V_o Width of the neutral zone between V_L& V_H can be adjusted by variation of R₂

Fig: Electronic On Off controller

Fig: Controller output response

​

Electronic Proportional control

​

V_out= G_pV_e+V_o

G_p= R₂/R₁

​

V_e = error voltage

V_o = output voltage

Fig: Electronic Proportional controller

Electronic Integral controller

Fig: Electronic Derivative controller

V_e= error voltage

V= output voltage

Fig: Electronic Integral controller

V_out = G_{I ∫} V_e dt + V_out (0)

V_e= error voltage ^{GI =} _RC

V_out(0)= initial output voltage

Electronic Derivative controller

1

o

dV_e dt

out D

G_D = R ₂C

• V

V = G

Electronic PI controller

Fig: Electronic PI controller

Fig: Electronic PD controller

e

1

out e

R R

1

G_I = G

V_e= error voltage

​

V_out(0)= initial output voltage

Electronic PD controller

1 ^{⎝ ⎠}

V = ^{R 2} V

₊ ^R ₂ ⎛ ¹ ⎞

⎜ _RC ⎟_∫ ^{V dt +}

⎟

⎝ 1 3 ⎠

p

3

e

⎝ 1 3 ⎠ ⎝ 1 3 ⎠

out

^⎛ R ₂ ^⎞

R + R

G = ^⎜

G_D = R ₃C

• V_o

dV_e dt

^⎟V + ^⎜

^⎟(R C)

R + R R + R

^⎛ R ₂ ^{⎞ ⎛} R ₂ ^⎞

V = ^⎜

Auto/ Manual Transfer

Bumpless transfer is either a manual or automatic transfer procedure used when switching a PID controller from auto to manual or vice versa. Its aim is to keep the controllers output the same when switching auto/manual, that is if the controller is at 50% output in auto it should retain that 50% output as you switch it to manual. If you switch from manual to auto the same should apply. Most modern PID controllers have bumpless transfer built in, including PLC and DCS PID controllers.

The term bumpless transfer refers to the process being controlled, meaning that the process is not disturbed when switching the PID controller from Auto to manual or from manual to auto.

“Bumpless transfer” is a term used to describe the desired action a controller (PID, for example) is expected to maintain when transferring from manual to auto mode, or vice-versa.

In auto mode, the controller calculates a deviation between the desired setpoint and the actual value of the process variable, and adjusts the signal to the output to reduce the value of the deviation (or error).

However, in manual mode, the operator decides what this output signal will be, bypassing the auto mode settings, and this value could be different than what the controller has calculated.

In this case, when the operator switches from manual to auto, the controller applies its calculated output value to the process, and since there is a good chance the auto and manual outputs do not match, there will be a “bump” in the process: valves will open/close suddenly, pumps will increase/ decrease in speed abruptly… until the controller adapts the output values again. This is bad for the process (sudden jumps in actual value shows lack of overall control) and can affect adversely the process actuators (valves and pumps and motors do not like being forced to change control values quickly; repeated fast changes can even damage the actuators).

To produce a bumpless transfer, the PID controller will force the manual mode output value to the output in auto mode (no bump) and then ramp slowly the auto output to the value calculated by the controller in auto. This way passage from manual to auto is gradual, the process shows no abrupt changes and the process actuators are not affected.

​

Bumpless Tuning

In this case, without a bumpless tuning feature, the bump occurs due to changing controller gain or derivative settings during controller tuning.

It is because of the bump in controller output caused by changing tuning settings that it is good practice to place a controller in manual mode momentarily while making tuning changes. Because most controllers have bumpless transfer, it eliminates the bump when switching the controller back to auto after making the tuning changes in manual mode. However, this becomes a problem if the tuning settings are changed, as in the case of gain scheduling.

Bumpless tuning can be achieved without the need to place the controller in manual mode by calculating how much the controller output will jump due to the new proportional and derivative settings, and subtracting an equal quantity from the integral term, so that the sum of the three terms (the controller output) remains unchanged. Similar to bumpless transfer, bumpless tuning can also be achieved by using a velocity algorithm.

​

Practical forms of PID controller

​

Pneumatic PID controller

Pneumatic signals are the signal in which the gas is used as the process medium and the general range of pressure is 3-15 PSI for actuating the final control element.

The main parts of pneumatic PID controllers are

setpoint bellows

integral bellows with integral or reset orifice

derivate orifice restriction

​

flapper nozzle arrangement

Four bellows are connected to a force beam. The measured process variable is converted to air pressure and connected to the measured value bellow. Similarly the air pressure corresponding to the setpoint signal is applied to the setpoint bellow. The error corresponding to the measured value and setpoint generates a force on the top side of the force beam. The bottom side of the force beam is connected to the two bellows, proportional and integral bellows and a flapper nozzle amplifier. The output air pressure is dependent on the gap between the flapper and nozzle. An air relay enhances the air handling capacity. The output pressure is directly fed back to the feedback bellows (proportional and integral bellows) through a restrictor. The opening of the restrictor decides the integral and derivative action to be applied.

Fig: Pneumatic PID controller

PROBLEMS

​

1. The temperature in a certain process has a range of 300 to 440 K and a set point of 384 K. Find the percent of span error when the temperature is 379 K.

​

Solution:

​

The percent error is

𝑟 − 𝑏

𝑃

𝑒 =

^𝑏𝑚𝑎𝑥−𝑏

𝑚 𝑖 𝑛

∗ 100

384 − 379

^{𝑒𝑃 =} 440 − 300 ^{∗ 100}

​

𝒆_𝑷 = 𝟑. 𝟔%

2. A liquid-level control system linearly converts a displacement of 2 to 3 m into a 4- to 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:

(a) the relation between displacement level and current, and (b) the neutral zone or displacement gap in meters.

Solution:

​

a. The relation between level and a current is a linear equation such as

​

H = KI + H₀

PROBLEMS

b. The relay closes at 12 mA, which is a high level, H_H of

H_H = (0.0625 m/mA)(12 mA) + 1.75 m

H_H = 2.5 m

​

The low level H_L, occurs at 10 mA, which is

H_L = (0.0625 m/mA)(10 mA) + 1.75 m

H_L = 2.375 m

​

Thus, the neutral zone is H_H – H_L = (2.5 – 2.375), or 0.125 m.

​

​

3. 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

​

The relation between controller output and time is

P = KPt + p(0)

​

When P=100,

​

100% = (2%/sec)(t) + 25%

that, when solved for t, yields

​

t = 37.5sec

PROBLEMS

​

4. Consider the proportional-mode level-control system shown in fig. Value A is linear, with a flow scale factor of 10 m3 / h per percent controller output. The controller output is nominally 50% with a constant of KP = 10% per %. A load change occurs when flow through valve B changes from 500 m3 / h to 600 m3 / h Calculate the new controller output and offset error.

Solution:

Valve A must move to a new position of 600 m3/h flow or the tank will empty. This can be accomplished by a 60% new controller output because

​

Q_A=(10m^3/h)/% X 60%=600m^3/h

​

as required. Because this is a proportional controller, we have

​

P = K_P.e_p + P0

​

with the nominal condition P_O = 50%. Thus,

​

e_p=(P - P_O) / K_P X 100= (60-50) / 10%

​

e_p=1%

​

so a 1% offset error occurred because of the load change.

PROBLEMS

​

5. An integral controller is used for speed control with a setpoint of 12 rpm within a range of 10 to 15 rpm. The controller output is 22% initially. The constant K_I =-0.15% controller output per second per percentage error. If the speed jumps to 13.5 rpm, calculate the controller output after 2 s for a constant e_p.

​

Solution:

𝑒_𝑃 = −30%

​

The rate of controller output change is then given by,

𝑑𝑝

_𝑑𝑡 = 𝐾_𝐼 𝑒_𝑃 = −0.15𝑆⁻¹ −30%

𝑑𝑡

^𝑑𝑝 = 4.5%/𝑠

The controller output for constant error can be found

from

PROBLEMS

​

6. For the given the error of figure, plot a graph of a proportional-integral controller output as a function of time. K_P = 5, K_I = 1.0S-1, and P_I(0) = 20%

The error can be expressed in 4 regions

​

Region – 1

​

t<0, In this region e_p=0

Region – 2

​

1 ≤ t ≤ 1, In this region e_p= t

Region – 3

​

1 ≤ t ≤ 3, In this region e_p= 1

Region – 4

​

t>3, In this region e_p= 0

The PI controller is expressed as

_𝑡2 1

PROBLEMS

For region 3 , the controller outputs can found as

𝑃₃ = 5 + 10 + 22.5 = 37.5%

​

For region 4 , the controller outputs can found as

𝑃₄ = 0 + 0 + 32.5 = 32.5%

7. A temperature controller controls temperature from 100° to 200°C. A sensor provides an output of 2 to 8 V for this temperature range. The controller output drives a heater with an output of 0 to 5 volts. What circuit gain is needed if the controller is to be used with a proportional gain of 4%/%?

​

Solution:

​

The range of measurement voltage will be ∆Vm = 8 V - 2 V = 6 V. The output

PROBLEMS

​

8. A temperature-control system inputs the controlled variable as a range from 0 to 4 V. The output is a heater requiring 0 to 8 V. A PID is to be used with K_P = 2.4%/%, K_I = 9% / (%/min), K_D=0.7 % / (%/min). The period of the fastest expected change is estimated to be 8 s. Develop the PID circuit.

​

Solution:

​

The input range is 4 V, and the output range is 8 V. Let us figure the circuit gains.

​

For the proportional mode, a 1% error means a voltage change of (0.01).(4V) = 0.04V. This

should cause an output change of 2.4% or (0.024)(8V) = 0.192V . Thus,

​

G_P = (0.192V / 0.04V) = 4.8

For the integral term, an error of 1% should cause the output to change by 9%/min, which is (9/60) =0.15%/s Thus,

G_I = (0.0015S^-1)(8V) / (0.04V) G_I = 0.3S^-1

For the derivative term, an error change of 1% per min or (0.04 V/60) = 6.67 X 10^-4 V/s

should cause an output change of 0.7% or (0.007)(8V) = 0.056V. Thus,

G_D = (0.056 V / 6.67 X 10^-4 V/s) G_D = 84s

​

These results provide the following relations:

R₂/ R₁ = 4.8

1 / (R_IC_I) = 0.3S^-1 R_DC_D = 84s

From the fastest period specification, we form the relationship

2𝜋𝑅₃𝐶_𝐷 = 0.1 8𝑠 = 0.8

​

which gives seven unknowns and four equations. We can pick three quantities. Let us try R₁

= 10KΩ, C_I = C_D = 10µF. This gives

​

R₂ = 4.8R₁ = 48 KΩ

R_I = 1 / (0.3C_I) = 333 KΩ R_D = 84 / C_D = 8.4 MΩ

R₃ = 0.8 / 2𝜋𝐶_𝐷= 12.7 KΩ

8.4 MΩ seem too large for practical consideration. Let us change C_D to 100µF. Now we get

10. ASSIGNMENTS

ASSIGNMENTS

1.A velocity control system has a range of 220 to 460 mm/s. If the setpoint is 327 mm/s and the measured value is 294 mm/s, calculate the error as percentage of span.

2.A controlling variable is a motor speed that varies from 800 to 1750 rpm. If the speed is controlled by a 25- to 50-V dc signal, calculate (a) the speed produced by an input of 38 V, and (b) the speed calculated as a percent of span.

3.A floating controller with a rate gain of 6%/min and P(0) =50% has a +5gal/min deadband. Plot the controller output for an input given by figure below. The setpoint is 60 gal/min.

4. A PD controller has K_P = 2.0, K_D=2s, and P_O=40%. Plot the controller output for the error input given in the figure.

5. A liquid-level system converts a 4–10-m level into a 4- to 20-mA current. Design a three-mode controller that outputs 0–5 V with a 50% PB, 0.03-min reset time, and 0.05-min derivative time. Fastest expected change time is 0.8 min.

11. PART-A

12. PART-B

13. SUPPORTIVE ONLINE CERTIFICATION COURSES

• Online Course: SWAYAM

​

Course Name: Chemical Process control

Course Instructor: Prof. Sujit Jogwar, IIT Bombay Duration: 8 weeks

​

AICTE approved FDP course

​

• Online course:Coursera

Course Name: Sensor Manufacturing and Process Control

​

University: University of Colorado Boulder

Course Instructor: James Zweighaft, Jay Mendelson Duration: 5 weeks

​

• Online course:Udemy

​

Course Name: Introduction to process control and Instrumentation

Course Instructor: WR Training, Petroleum Petrochemical & Chemical Engineering

​

Duration: 10 sections,75 lectures

14. REAL TIME APPLICATIONS IN DAY TO DAY LIFE AND TO INDUSTRY

A good example of temperature control using PID would be an application where the controller takes an input from a temperature sensor and has an output that is connected to a control element such as a heater or fan. The controller is usually just one part of a temperature control system, and the whole system should be analyzed and considered in selecting the proper controller.

​

PID Controller Problem Example

Almost every process control application would benefit from PID control. Here are several PID controller problem examples:

Heat treatment of metals: "Ramp & Soak" sequences need precise control to ensure desired metallurgical properties are achieved.

Drying/evaporating solvents from painted surfaces: Over-temperature conditions can damage substrates while low temperatures can result in product damage and poor appearance.

Curing rubber: Precise temperature control ensures complete cure is achieved without adversely affecting material properties.

Baking: Commercial ovens must follow tightly prescribed heating and cooling sequences to ensure the necessary reactions take place.

15. CONTENT BEYOND SYLLABUS

​

Topic: Linear fractional order controllers; A survey in the frequency domain

Today, there is a great tendency toward using fractional calculus to solve engineering problems. The control is one of the fields in which fractional calculus has attracted a lot of attention. On the one hand, fractional order dynamic models simulate characteristics of real dynamic systems better than integer order models. On the other hand, Fractional Order (FO) controllers outperform Integer Order (IO) controllers in many cases. FO-controllers have been studied in both time an frequency domain. The latter one is the fundamental tool for industry to design FO-controllers. The scope of this paper is to review research which has been carried out on FO-controllers in the frequency domain. In this review paper, the concept of fractional calculus and their applications in the control problems are introduced. In addition, basic definitions of the fractional order differentiation and integration are presented. Then, four common types of FO-controllers are briefly presented and after that their representative tuning methods are introduced. Furthermore, some useful continuous and discrete approximation methods of FO- controllers and their digital and analogue implementation methods are elaborated. Then, some Matlab toolboxes which facilitate utilizing FO calculus in the control field are presented. Finally, advantages and disadvantages of using FO calculus in the control area are discussed. To wrap up, this paper helps beginners to get started rapidly and learn how to select, tune, approximate, discretize, and implement FO-controllers in the frequency domain.

Reference: Linear fractional order controllers; A survey in the frequency domain Article in Annual Reviews in Control · April 2019

16. ASSESSMENT SCHEDULE (PROPOSED DATE & ACTUAL DATE)

17. Prescribed Text Books & Reference Books

​

TEXT BOOKS:

1. Seborg, D.E., Edgar, T.F. and Mellichamp, D.A., “Process Dynamics and Control”, WileyJohn and Sons, 2nd Edition, 2003.
2. Bequette, B.W., “Process Control Modeling, Design and Simulation”, Prentice Hall of India, 2004.
3. Stephanopoulos, G., “Chemical Process Control - An Introduction to Theory and Practice”, Prentice Hall of India, 2005.

​

REFERENCES:

1. Coughanowr, D.R., “Process SystemsAnalysis and Control”, McGraw - Hill International Edition,2004.
2. Curtis D. Johnson, “Process Control Instrumentation Technology”, 8th Edition, Pearson, 2006.
3. Considine, D.M., Process Instruments and Controls Handbook, Second Edition, McGraw, 1999.
4. Bela.G.Liptak., “Process Control and Optimization”., Instrument Engineers’ Handbook., volume 2, CRC press and ISA, 2005.
5. Ramesh C. Panda., T.Thyagarajan., “An Introduction to Process Modelling Identification and Control for Engineers” Narosa Publishing house Pvt. Ltd, 2017.

18. MINI PROJECT SUGGESTIONS

​

Project 1: Examine the output response for a two tank interacting system using on-off controller

Aim: To obtain the output response for a two tank interacting system when step input is applied with on-off controller.

​

Project 2: Examine the output response for a flow process

using P controller

Aim: To obtain the output response for a flow process when step input is applied with P controller.

Project 3: Examine the output response for a level process using PI controller

Aim: To obtain the output response for a level process when step input is applied with PI controller.

Project 4: Examine the output response for a temperature process using PD controller

Aim: To obtain the output response for a temperature process when step input is applied with PD controller.

Project 5: Examine the output response for a temperature process using PID controller

Aim: To obtain the output response for a temperature process when step input is

applied with PID controller.

Project 6: Examine the output response of second order system using PID controller in Matlab

Aim: To obtain the output response of a second order system when step input is applied with PID controller using Matlab

Thank you

Disclaimer:

​

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