EI 2406 instrumentation system design laboratory

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DEPARTMENT OF ELECTRONICS AND
INSTRUMENTATION ENGINEERING
EI 2406-INSTRUMENTATION SYSTEM DESIGN
LABORATORY
LAB MANUAL
EI2406 INSTRUMENTATION SYSTEM DESIGN LAB
List of Experiments
1. Design and Implementation of instrumentation amplifier.
2. Design and implementation of active filter.
3. Design and implementation of V/I and I/V converters.
4. Design and implementation of cold – junction compensation circuit for
thermocouple.
5. Design and implementation of signal conditioning circuits for RTD.
6. Design of orifice plate
7. Design of rotameter.

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8. Design of controlvalve (Sizing and flow –lift characteristics).
9. Design of PID controller(Using operational amplifier and microprocessor)
10.Piping and Instrumentation Diagram – Case study.
11.Preparation of documentation of instrumentation project(Process flow sheet,
instrumentation index sheet and instrument specifications sheet)
DESIGN OF INSTRUMENTATION AMPLIFIER
AIM
To design an instrumentation amplifier based on the three operational amplifier configuration
with adifferential gain of 100.
EXERCISE
1. Develop the instrumentation amplifier with differential gain of 100 and draw the input Vs
output
characteristics of the three operational amplifier based instrumentation amplifier and make a
comment on the response.
2. Compare the performance characteristics of Instrumentation amplifiers with commercial
Monolithic Instrumentation amplifier.
EQUIPMENT
1. Dual power supply – 1 No
2. Digital Multimeters – 1 No
3. Resistors – 10 No
4. Operational Amplifiers – 4 No
5. Any commercial Monolithic Instrumentation amplifier - 2 No

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DESIGN OF ACTIVE FILTERS
AIM
To design an active first order / second order Butterworth type Low – Pass /
High Pass / Band-pass filter with the following specifications.
Low pass filter : Cut – off frequency : 1 KHz
High pass filter : Cut – off frequency : 1 KHz
Band pass filter : Cut off frequency : 1 KHz < fc < 5 KHz99
EXERCISE
1. Develop an active Butterworth first order (or) second order low pass and / or high – pass,
band pass filter and determine experimentally the frequency response.
EQUIPMENT
1. Dual power supply - 1 No
2. Operational amplifiers - 2 Nos
3. Resistors - 10 Nos
4. Capacitors - 10 Nos
5. Signal generator - 1 No
6. C.R.O - No
DESIGN OF REGULATED POWER SUPPLY AND DESIGN OF V/I AND I/V
CONVERTERS
(a) AIM:-
To Design a Regulated Power Supply.
EQUIPMENT
1. Diodes IN4007
2.100 μF, 10 μF
3. IC 7805
4. Potentiometer
5. Ammeter and Voltmeter
EXERCISE
LINE REGULATION
1. Varying the Input Voltage (0 -15)V.
2. Note down the output voltage
LOAD REGULATION
1. Connect a variable Potentiometer across the output of the RPS.
2. Vary the potentiometer and note down the corresponding output current and
voltage.
(b) AIM:-
To design a voltage to current converter and a current to voltage converter and verify the
characteristics experimentally.
OBJECTIVES
1. To design a voltage to current converter (grounded load) with the following
specification
Input voltage range : (0 – 5) V
Output current range : (4-20) mA (should be independent of load)
2. To design a current to voltage converter with the following specification
Input current range : (4-20) mA

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Output voltage range : (0-5) V
100
EXERCISE
1. Determine experimentally the characteristics of voltage and current converter an plot
output current versus input voltage and comment on the response.
2. Determine experimentally the characteristics of current to voltage converter and plot
output voltageVs input current and comment on the response.
EQUIPMENT
1. Resistors - 10 No
2. Operational amplifiers - 5 No
3. Transistor (NPN / PNP) - 2 No
5. Digital Multimeters - 2 No
6. Loop analyzer - 1 No
DESIGN OF LINEARIZING CIRCUIT AND COLD-JUNCTION COMPENSATION
CIRCUIT FOR THERMOCOUPLE
AIM
To design a cold – junction compensation circuit for thermocouple.
OBJECTIVE
To design a automatic reference correction circuit for thermocouple.(A solid – state
temperature sensor or RTD can be used for the cold function measurement)
EXERCISE
1. Develop the circuit for reference junction compensation.
2. Keep the hot junction temperature at say 4000C.
3. Vary the cold – junction temperature from 30 – 900C and observe the output voltage
for with and without cold-junction compensation.
4. Plot the output voltage versus cold-junction temperature and comment on the
response.
EQUIPMENT
1. Thermocouple - 1 Nos
2. Operational amplifier - 3 Nos
4. AD – 590 or RTD - 1 Nos
5. Resistors - 10 Nos
7. Multimeters - 1 No
DESIGN OF SIGNAL CONDITIONING CIRCUIT FOR STRAIN GAUGES AND RTD
(a) Aim:
To design Signal Conditioning Circuit for Strain Gauge.
Specification as follows
1. Input Range 0 to 1 Kg
2. Output Voltage 0 to 5 V
3. Device -Bourdon Strain Gauge (350 Ohm)
Equipment
1. Bonded Strain Gauge

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101
2. Loads (100 gm to 1 Kg)
3. Operational Amplifier
4. RPS
5. Resistors
Exercise:
Develop Signal Conditioning Circuits for different loads and plot output voltage versus Load.
Comment on Linearity
(b) Aim
To design a signal conditioning circuit to RTD. The specification are as follows
Temperature Range : 300 C – 1000C (Approximately)
Output voltage : 0 – 5 V DC
Sensor : RTD (Pt 100)
Current through RTD : Not to exceed 10mA
Equipment
1. RTD (Pt 100) - 1 No
2. Resistors - ?
3. Operational amplifiers - 4 Nos
4. Temperature bath - 1 No
5. Multimeter - 1 No
6. Trim Pot - 3 Nos
Exercise
1. Develop the signal conditioning circuit and plot the output voltage versus temperature
and comment on the linearity.
DESIGN OF ORIFICE PLATE AND ROTAMETER
Designof Orifice Plate
Aim:
To Design an Orifice Plate for the given Specification.
Equipment
1. Pump and Reservoir
2. Pipeline with Orifice plate
3. Collecting Tank
Exercise:
1. Convert Electrical Signal to Differential Pressure
2. Determine the interval data
3. Calculate D/d
4. Calculate sizing factor
Designof Rotameter
Aim:
To Design a Rotameter for given Specification
Equipment
1. Pump and Reservoir
2. Pipeline with Orifice plate
3. Collecting Tank

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Exercise
1. Swithch On the Motor
2. Adjust the Rotameter to read the required flow rate.
3. Start the Timer
4, After 5 Min Note the Head in the tank.
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5. Drain the tank.
6. Repeat the Procedure and Calculate Cd in each case
CONTROL VALVE SIZING
Aim:
To design a Control Valve and Study the flow lift Characteristics
Equipment:
1. Linear Control Valve
2. On/OFF Control Valve
3. Air Regulator
4. Rotameter
5. Pump
Exercise
1. By varying the inlet pressure note down the stem moment value and the flow
rate.
2. Draw the Graph for pressure Vs Flow rate, Stem Moment Vs Flow rate
DESIGN OF PID CONTROLLER
Designof PID Controller using Op-Amp
Aim:
1.To the study the response of P,PI,PD ,PID Controllers using Op-Amp
Equipment
1. Signal Generator
2. IC 741
3. Resistors and Capacitors
4. CRO
5. Bread Board
Exercise
1. Design a Analog PID Controller for various values of Kp, Ki, Kd
2. Apply the error Signal from signal Generator (Square, Sine)
3. Note down the response from the CRO.
DESIGN OF PID CONTROLLER USING MICROPROCESSOR
Aim:
To the study the response of P, PI, PD ,PID Controllers using Microprocessor.
Equipment
1. Signal Generator
2. Microprocessor based kit with ADC and DAC Section
3. CRO
Exercise:

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1. Enter the PID Algorithm in Microprocessor
2. Give the Error Signal to ADC Section of Microprocessor Kit.
3. Execute the Microprocessor Program
4. Note down the output response of PID Controller in the DAC Section
Microprocessor Kit
INDEX
SL.NO Date Name of the experiments Marks Staff Signature

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PIN Diagram:
Ex No: Date:
INSTRUMENTATION AMPLIFIER

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Aim:
To design an instrumentation amplifier.
Apparatus required:
SL.NO ITEM SPECIFICATION QUANTITY
1
2
3
4
5
IC
Resistors
Regulated power supply
Digital Multimeter
Bread board
741
1KΩ
(0 – 15)V
---
---
3
6
2
1
1
Procedure:
1. Find’ R’ to get required output voltage.
2. Give the connections as per circuit diagram.
3. Vary the input voltage.
4. Observe the output voltage and compare them with the theoretical values.
Formula:
V0 = - (R2/R1) (1+2R/Rg) Vd
Theory:
Instrumentation amplifier is an amplifier with high input impedance, very low output
impedance, low offset and low drifts voltage. This configuration is better than inverting or
non-inverting amplifier because it has minimum non-linearity, stable voltage gain and high
common mode rejection ratio (CMRR > 100 dB.). This type of amplifier is used in
instrumentation field where the output voltages are very low such as in thermocouples, strain
gauges and biological probes.
Tabular column:

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SL.NO Rg V1 V2 Vd= (V1-V2) V0
(Theoretical)
V0
(practical)
1
2
3
4
5
Calculation:
In this circuit, consider all resistors to be of equal value except for Rg. The negative
feedback of the upper-left op-amp causes the voltage at point Va (top of Rg) to be equal to V1.
Likewise, the voltage at point Vb (bottom of Rg) is held to a value equal to V2. This establishes a
voltage drop across Rg equal to the voltage difference between V1 and V2. That voltage drop

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causes a current through Rg, and since the feedback loops of the two input op-amps draw no
current, that same amount of current through Rg must be going through the two "R" resistors
above and below it. This produces a voltage drop between points V1’ and V2’ equal to:
V1’ – V2’ = (V2 – V1) [1+(2R/Rg)].
The differential amplifier in the circuit diagram takes the voltage drop between V1’ and
V2’, and amplifies it by a gain of 1 (assuming again that all "R" resistors are of equal value). It
possesses extremely high input impedances on V1 and V2 inputs (because they connect straight
into the non inverting inputs of their respective op-amps). It also provides the same for the input
through adjustable gain resistor. Hence the overall voltage gain in the instrumentation amplifier
is,
Av = [1+ (2R/Rg)].
The important features of an instrumentation amplifier are
1) high accuracy
2) high CMRR
3) high gain stability with low temperature coefficient
4) low DC offset
5) low impedance.
Result:
Circuit Diagram:
1. Low Pass Filter:

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Low pass Filter Vi=
SL.NO FREQUENCY(Hz) O/P VOLTAGE(Vo) GAIN=20log(Vo/Vin)
Ex No: Date:
DESIGN OF ACTIVE FILTERS

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Aim:
To study and design an active filter using operational amplifier circuit.
Apparatus Required:
1
2
3
4
5
6
IC
Resistors
Capacitor
Bread board
Function Generator & CRO
741
10K,1.59K,5.86K
0.1uF
(0 – 15)V
---
---
1
2,2,2
2
1
1
1
Procedure:
1. Connect the circuit diagram for Low pass and High pass filter as shown.
2. Now input signal (sine wave) at particular frequency is applied.
3. The different frequencies is applied to the circuit is noted and corresponding output
voltage is noted.
4. Graph is drawn for frequency Vs Output voltage.
Theory:
Low pass Filter:
V0 =
Magnitude of Response,
2. High Pass Filter:

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High pass filter Vi=
SL.NO FREQUENCY(Hz) O/P VOLTAGE(Vo) GAIN=20log(Vo/Vin)

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Recalling that magnitude of a complex number is square root of sum of squares of real and
imaginary parts. There are also phase shifts associated with the transfer function Vo/Vi
through out. This is obviously a low pass i.e low frequency signals are passed and high
frequencies are blocked. If ω<< 1/RC then ωRC<< 1, then magnitude of gain is approximately
unity and unity output equals input. If ω >>1/RC then gain goes to zero.
Vo/Vi= |Ho| ω o / (ω 2+ ω o
2)1/2
At low frequencies w<<1/R1C, the circuit will act as an amplifier with gain R1/R2= Ho.
High pass filter:
Using analysis techniques similar to those used for LPF it can be shown that
Vo/Vi= |Ho| ω o / (ω 2+ ω o
2)1/2
Which general form for first order LPF. At low frequencies the capacitor acts as a short
so gain of amplifier goes to Ho= -R1/R2 At low frequencies (w<<wo) the capacitor is an open
and gain of circuit is Ho. For this circuit wo= 1/R2C. Therefore this circuit is a high pass filter.
Result:
Circuit Diagram:

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Tabular Column:
SL.NO Temperature in oC Output Voltage(V)
Ex No: Date:

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DESIGN AND IMPLEMENTATION OF COLD – JUNCTION
COMPENSATION CIRCUIT FOR THERMOCOUPLE
Aim:
To design a cold junction compensation circuit for thermocouple.
Apparatus required:
1
2
3
4
6
7
8
IC
Resistors
Capacitor
AD590
Trim Pot
Bread board
741
1KΩ
0.47μF
(0 – 15)V
---
10K
1
3
2
1
1
1
1
Procedure:
1. First measure the thermocouple output with respect to room temperature in mV and
this is called as Vo.
2. With respect to Vo design Rg, R1, R2.
3. Measure the output of AD590 and reset the gain value such that it meets the Vo of the
table value.
Theory:
When temperature near ambient are to be measured with the thermocouple and it is
inconvenient to use a fixed reference junction, therefore the compensating circuit must be
employed in the measuring system. An arrangement for automatic compensation is shown. A
temperature sensitive bridge is included in a thermocouple circuit, such that variations in ambient
temperature are compensated by the changes in resistance.

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Tabular column:
SL.NO Temperature in oC Resistance of RTD in Ω Output Voltage(V)
Ex No: Date:

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DESIGN AND IMPLEMENTATION OF SIGNAL
CONDITIONING CIRCUIT FOR RTD
Aim:
To design signal conditioning circuit using RTD.
Apparatus required:
1
2
3
4
5
6
IC
Resistors
Oven and thermometer
RTD
Bread board
741
1KΩ,10kΩ,47kΩ
(0 – 15)V
---
-----
1
3,4,1
1
1
1
1
Procedure:
1. Connections are made as per circuit diagram.
2. Balance the bridge initially so that the output voltage occurs.
3. Increase the temperature of the oven and note down the output voltage.
4. Calculate the gain to get the required output voltage.
5. Observe the output voltage of the amplifier and compare it with theoretical value.
Theory:
Cold junction compensation:
Signal conditioning circuits normally employ amplifiers, temperature compensation
devices etc.

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Design:
RTD:
RT = Ro (1+α∆T)
Assume R0 = ---Ω, α = 0.0039/oC, T = --- oC.

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RTD:
RTD stands for resistance temperature detector. RTD has a positive temperature
coefficient. The principle of RTD is as there is a change in the temperature, the resistance
changes, this change in resistance is measured. As RTD has a positive temperature coefficient as
temperature increases, resistance increases, the resistance at any temperature that can be
calculated using the formula.
RT= Ro[1+α (t – to)]
RT is resistance at T
Ro is resistance at 0oC, room temperature.
α Temperature coefficient of material 0.04.
Thus the value of resistance can be calculated.
Result:
Circuit Diagram:

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Tabular column:
S.No. Input Current (mA) Output Voltage (Volts)
Ex No: Date:

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DESIGN OF CURRENT TO VOLTAGE CONVERTER
Aim:
To design and construct a current to voltage converter
Apparatus required:
1
2
3
4
5
6
7
RPS
Operational Amplifier
Resistors
Trim Pot
Voltmeter
Ammeter
Bread Board & Connecting Wires
----
IC 741
100Ω,1KΩ,10KΩ
10KΩ
0 – 30V
0 – 30mA
----
1
2
2,4,5
2
1
1
Sufficient nos.
Procedure:
1. Give the connections as per the circuit diagram.
2. Vary the input current (4-20 mA) in steps of 1.5 mA.
3. Observe the output voltage and tabulate.
4. Plot the graph between input current and output voltage.
Theory:
In electronics, a transimpedance amplifier is an amplifier that converts current to voltage.
Its input ideally has zero impedance, and the input signal is a current. Its output may have low
impedance, may be matched to a driven transmission line; the output signal is measured as a
voltage. Because the output is a voltage and the input is a current, the gain, or ratio of output
to input, is expressed in units of ohms. Inverting amplifier configuration of an op-amp
becomes a transimpedance amplifier when Rin is 0 ohms.
Result:
Circuit Diagram:

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Tabular column:
S.No. Input Voltage (Volts) Output Current (mA)
Ex No: Date:

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DESIGN OF VOLTAGE TO CURRENT CONVERTER
Aim:
To design and construct a voltage to current converter.
Apparatus required:
1
2
3
4
5
6
7
RPS
Resistors
Trim Pot
Transistor
Ammeter
----
IC 741
560Ω,1KΩ 10KΩ
10k
CL100
0 – 30mA
----
1
2
2,7
1
1
1
Sufficient nos.
Procedure:
1. Give the connections as per the circuit diagram.
2. Vary the input voltage (0-5 V) in steps of 0.5 volts.
3. Observe the output current and tabulate.
4. Plot the graph between input voltage and output current.
Theory:
Voltage to current converter is also known as transconductance amplifier. The inverting
input voltage (Vi = I R) is given. The input voltage is converted into an output current of
V/R. The same current flows through the signal source and load.
Result:

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Ex No: Date:
CALIBRATION OF ROTAMETER
Aim:
To calibrate a rotameter for flow rate measurement.
Apparatus required:
1
2
3
4
Pump and reservoir
Pipeline with rotameter
Collecting tank
Stopwatch
----
----
---
---
---
---
---
1
Procedure:
 Switch on the motor
 Adjust the rotameter to read the required flow rate
 Start the timer.
 After 5 mins note down the head in the tank
 Drain the tank,repeat the procedure Td in each case.
Theory:
The rotameter's operation is based on the variable area principle: fluid flow raises a float
in a tapered tube, increasing the area for passage of the fluid. The greater the flow, the higher
the float is raised. The height of the float is directly proportional to the flow rate. With liquids,
the float is raised by a combination of the buoyancy of the liquid and the velocity head of the
fluid. With gases, buoyancy is negligible, and the float responds to the velocity head alone.
The float moves up or down in the tube in proportion to the fluid flow rate and the annular
area between the float and the tube wall. The float reaches a stable position in the tube when
the upward force exerted by the flowing fluid equals the downward gravitational force exerted
by the weight of the float. A change in flow rate upsets this balance of forces. The float then
moves up or down, changing the annular area until it again reaches a position where the forces
are in equilibrium. To satisfy the force equation, the rotameter float assumes a distinct
position for every constant flow rate. However, it is important to note that because the float
position is gravity dependent, rotameters must be vertically oriented and mounted.

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Tabular column:
Sl.No Flow rate
(LPH)
Time(sec) Head(cm) Qact(m3
/s)
*10-5
Qth(m3
/s)
*10-5
Cd % Error
Calculation:

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Rotameter (Constant pre drop variable area meter):
Rotameter consists of the vertical tube with tapered tube in which float assumes vertical
position corresponding to each flow rate through the tube for the given flow rate. Float
remains stationary since the vertical forces of differential pressure, gravity, viscosity and
buoyancy are balanced. Downward force is constant, since the pressure drop across the float is
constant.
Formula:
Qact =A2h/t (m3 /sec)
Qth =flow rate (lph)×10-3 × 3600 (m3 /sec)
Ʈd = Qact / Qth
% error = (Qth- Qact ) / Qth
Result:

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ProcessSetup:
Tabular column:
Height ( h) = 100mm
Sl.No Flow(LPH) Head(*13.6)mm
of wc
Time
taken
Theretical
Discharge
mm3/sec
Actual
Discharge
mm3/sec
Cd

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Ex No: Date:
DISCHARGE CO-EFFICIENT OF ORIFICE PLATE
Aim:
To determine the co-efficient of discharge of an Orifice Plate by measuring differential
pressure.
Apparatus required:
1
2
3
4
Pump and reservoir
Pipeline with orifice set up
Collecting tank
Stop watch
----
----
---
---
---
---
---
1
Procedure:
1. Switch ON the unit.
2. Set Rotameter to a particular flow(lph) using pump speed regulator knob
3. Note down the pressure gauge G1.
4. HV1 valve should be partially open.
5. Close HV2 and note down the time taken for a 100 mm ( h) increase in height in the measuring tank.
6. Tabulate the readings and calculate co-efficient of discharge of Orifice Plate.
7. Repeat the same procedures for different flow rate and calculate ‘Cd’.
8. The mean value ‘Cd’ thus gives the co-efficient of discharge of the Orifice Plate.
Theory:
Orifice meter is used to measure the flow rate of a fluid through a pipe. It is based on the
Bernoulli’s principle and of course it is one of the practical application of Bernoulli’s
equation.
Orifice Plate is a simple in setup, cheaper in cost. It consist of a flat circular plate with circular
sharp edged hole in the center referred as Orifice. In general design of Orifice, the Orifice
diameter in between 0.4 to 0.8 of pipe diameter.
Normally the differential pressure across b and c is measured using a Gauge. Since the
pressure
changes at (Vena contracta) ‘c’ is going to the very negligible. We consider that to be of

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atmospheric pressure. Hence the gauge indication is assumed to be differential pressure.

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Normally the differential pressure across b and c is measured using a Gauge. Since the
pressurechanges at (Vena contracta) ‘c’ is going to the very negligible. We consider that to be
ofatmospheric pressure. Hence the gauge indication is assumed to be differential pressure.
Coefficient of discharge (Cd): It is defined as the ratio of the actual discharge from an Orifice
to the theoretical discharge from the Orifice. Mathematically
Coefficient of discharge =
Act
The
Q
Q ,
QAct = Volume / time taken for collecting in to the tank  3
sec
m
QThe =  31 2
2 2
1 2
2
.
sec
a a gh m
a a


Cd = Coefficient of discharge,
a1 = cross sectional area of inlet (π d1
2/4)
a2 = cross sectional area of outlet( π d 2
2/4)
d1= diameter of inlet section in meter,
d2= diameter of throat section in meter
1 2( )
12.6
100
h h
h

  ,
1 2h h = difference of mercury level in the manometer.
G = acceleration due to gravity = 9.81 m/sec2
Diameter of Orifice = 8.5 mm(d2)
Diameter of Pipe = 13 mm(d1)
Diameter of Measuring Tank = 240 mm
Volume= Area of measuring tank height
Result:

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ProcessSetup:
Tabular column:
Quick opening(ON/OFF) control valve
Pressure drop across control valve (Δp)=
Actuarator pressure(Psi) Stem position (%) Rotameter flow (LPH) Cv = Q√G/∆P
Eequal percentage control valve

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Ex No: Date:
DESIGN OF CONTROL VALVE
(SIZING AND FLOW –LIFT CHARACTERISTICS)
Aim:
1. To study the characteristics of quick opening(ON/OFF) control valve
2. To study the characteristics of equal percentage control valve
3. To study the characteristics of linear control valve without positioned
Apparatus required:
1
2
3
Quickopening(ON/OFF)Control valve
Equal percentage Control valve
Linear Control valve
-
-
-
1
1
1
Procedure:
1. Before conducting the experiment, make sure the availability of water in reservoir
tank. Fill clean and soft water in the reservoir
2. Connect air supply pipe to the regulator. Confirm there is no loose connection.
3. control valve positioned should be in “bypass” mode.
4. Hand valve settings for ON/OFF [equal percentage]{linear}control valve
characteristics study;HV2[HV3]{HV4}(the regulating valve, which is provided at the
inlet of the control valve)and HV5[HV6]{HV7} should be fully open. Regulating
valves of other control valves should be fully closed.
5. Initially, set the output pressure of air regulator to 15psi by varying the knob. The
quick opening valve is fully open.
6. Keep partially open the vent valve(HV8),when air regulate or lifts to its maximum
range.
7. Switch on the unit.
8. Set maximum flow in the rotameter by adjusting the bypass valve(HV1) and Inlet
regulating valve(HV2)[HV3]{HV4}.
9. Maintain the pressure drop across the control valve in pressure gauge(G1)
(eg:1/1.5/2psi)remains constant by varying the hand valve(HV2)[HV3]{HV4}.Note
the pressure drop across the valve at fully open(G1).
10. Never disturb the hand valve (HV2)[HV3]{HV4],once its adjusted for particular
opening.

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11. Observe flow and inlet pressure variations. Note down the air regulator
pressure(G2).rotameter flow, and stem position in control valve.
Linear control valve without positioned
ModelGraph:

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12. Decrease the pressure in air regulator to 12 Psi,at same time ,pressure across the
control valve slightly increases, adjust hand valve(HV1)to maintain predefined
pressure in G1.
13. Note the flow in rotameter and stem position in control valve, air regulator pressure.
14. Slowly decrease/increase the air pressure regulator for achieving different stem
positions till the valve Is fully closed/open.
15. Tabulate the rotameter flow, air regulator pressure and stem position.
16. Plot the graph between rotameter flow in the y-axis and stem position in x-axis.
17. Calculate the control valve co-efficient from the table.
Theory:
In most of the industrial process control systems control valve is the final control
element.. The control valve consists of two major components, viz. Actuator and Valve. The
actuator is made up of flexible diaphragm; spring and spring tension adjustments, plate, stem
and lock nut, housing. The valve is made up of body, plug, stem, and pressure tight
connection.
The function of a control value is to vary the flow of fluid through the value by means of a
change of pressure to the valve top. The relation between the flow through the valve and the
valve stem position (or lift) is called the valve characteristic. There are three main types of
valve characteristics. The types of valve characteristics can be defined in terms of the
sensitivity of the valve, which is simply the fraction change in flow to the fractional change in
stem position for fixed upstream and downstream pressures. Mathematically
Sensitivity = dm / dx
In terms of valve characteristics, valve can be classified into three types:
1. Linear, 2. Increasing sensitivity, 3. Decreasing sensitivity.

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For the linear type valve characteristics, the sensitivity is constant and the characteristic
curve is a straight line (e.g. linear valve). For increasing sensitivity type, the sensitivity
increases with flow. (e.g. Equal percentage or Logarithmic valve). In practice, the ideal
characteristics for linear and equal percentage valves are only approximated by commercially
available valves. These discrepancies cause no difficulty because the inherent characteristics
are changed considerably when the valve is installed in a line having resistance to flow, a
situation that usually prevails in practice.
Equal percentage control valve:
Flow changes by a constant percentage of its instantaneous value for each unit of valve lift.
Quick opening control valve:
Flow increases rapidly with initial travel reaching near its maximum at a low lift.
Calculation:

42
CONTROL VALVE CO-EFFICIENT:
The number of us gallons of water/min that flow through a fully open valve with a ∆p of 1 psi.
Cv = Q√G/∆P
Where,
Q= flow rate in GPM(1GPM=227.1247LPH)

43
G=specific gravity of water(=1)
∆p=pressure drop across the control vlve in psi
Determine the control valve gain,
Result:
Circuit Diagram:
Tabulation:
Controller PB or Kc
τi (sec) τd (sec) Amplitude
Time period Frequency(Hz)

44
P
I
D
PID
Ex No: Date:
DESIGN OF PID CONTROLLER USING OPERATIONAL AMPLIFIER
Aim:
To design and construct a PID controller using operational amplifier.
Apparatus Required:
1
2
3
4
5
6
7
RPS
Function Generator
CRO
Resistors
Capacitors
1MHz
30 MHz
IC 741
470Ω,1KΩ,10KΩ
0.1µF
-
1
1
1
3
1,4,7
2
Required
Procedure:
1. Connections are made as per the circuit diagram.
2. Give a square wave input of 4V peak to peak at 1KHz using function generator.
3. The output is observed in a digital storage oscilloscope (DSO).
4. Draw the graph between Magnitude and Frequency of the output voltage.

45
Theory:
Two position control results in a continuously oscillating response. These oscillations will
affect the performance of final control element. This can be avoided by replacing two position
controller by continuous mode controllers such as proportional, Proportional + Integral and
Proportional + derivative. The proportional controller produces an output signal that is
proportional to error e. this may be expressed as P = Kc e + Ps
Where P = Output signal from controller,
Kc = Gain or sensitivity, e = Error = Set point – measured value, Ps = Constant.
Proportional band (Pb) is defined as the error response based as a percentage of the range
of measured variable to move the valve from fully open to fully close. The relation between
proportional band in % and Kc = 100/ [Pb (%)]

46
Proportional Integral control mode is described by the relation P = Kc c / TI e dt + Ps
Where TI = Integral time
The Proportional Integral Derivative control mode is given by the expression
P = Kc e + Kc TD de/dt + Kc / TI e dt + Ps
Formula used:
Vout = (R2/R1) Ve + (R2/R1) 1/ RI CI Ve dt + (R2/R1) RD CD dVe/dt + Vout(0)
Gp = R2/R1; GD = RDCD ;GI = 1/ RI CI

48
Ex No: Date:
PIPING AND INSTRUMENTATION DIAGRAM – CASE STUDY
Aim:
To draw the PI diagram for the flow, pressure and temperature process.
Symbols Used:
Process signal lines
Instrument air supply
Instrument electric power supply
Un defined signal
Pneumatic signal
Electric signal
Hydraulic signal
Filled thermal element capillary tube
Guided EM signal
Fiber optic cable guided same signal
Unguided EM signal
Unguided sonic signal
Alternative radio communication link
System with software link
Field is locally mounted
Normally accessible to an operator
Signal connector
Central or mail control room
Central or main control room and not
Secondary or local control room.

50
Ex No: Date:
PREPARATION OF DOCUMENTATION OF INSTRUMENTATION
PROJECT (PROCESS FLOW SHEET, INSTRUMENTATION INDEX
SHEET AND INSTRUMENT SPECIFICATIONS SHEET) – CASE
STUDY

EI 2406 instrumentation system design laboratory

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