CH503 –Instrumentation and Process Control Laboratory Manual … · Laboratory Manual Chemical Engineering Department School of Engineering Institute of Technology Nirma University

CH503 –Instrumentation and Process Control

Laboratory Manual

Chemical Engineering Department School of Engineering Institute of Technology

Nirma University

July 2018

CH503 - Instrumentation and Process Control Chemical Engineering Department

SoE, IT-NU

List of Experiment Sr. No. Objective

1. Study of dynamic characteristics (speed of response) of instrument (thermometer).

2. To study resistance temperature detector (RTD) and calibrate it.

3. To study thermocouple and calibrate it.

4. To study thermistor and calibrate it.

5. To study level measurement by air purge method

6. To study linear variable differential transducer (LVDT)

7. To study response of first order (thermometer) system to step input.

8. To study response of first order (liquid level) system to impulse input

9. To study response of first order (mixing process) system to step input.

10. To study response of first order (non - interacting tanks in series) system to step input.

11. To study response of second order (interacting tanks in series) system to step input.

12. To study response of second order (u-tube manometer) system to step input.

13. Virtual Lab experiments


SoE, IT-NU EXPERIMENT No: DATE: / / .

DYNAMIC CHARACTERISTICS (Speed of Response) OF INSTRUMENT

Objectives: study of dynamic characteristics (speed of response) of instrument (thermometer).

Apparatus: 250 ml glass beakers, Mercury thermometer, CCl4 thermometer & Stop watch. Materials: Lubricating oil & Water. Procedure:

1. Take 250 ml water in a glass beaker, put it on a hot plate & heat the water up to 1000C. Dip the given thermometers in the beaker and wait till the thermometer indicates 1000C. Note down the time taken by the thermometers for the change.

2. Take out the thermometers from water beaker and allow it to cool up to room temperature. Note down the time required for it.

3. Take 250 ml Oil in a glass beaker, put it on a hot plate & heat the oil up to 1200C. Dip the given thermometers in the beaker and wait till the thermometers indicates 1200C. Note down the time taken by the thermometers for the change.

4. Take out the thermometers from oil beaker and allow it to cool up to room temperature. Note down the time required for it.

5. Repeat the same procedure with water for two different temperatures (700C and 600C). 6. Repeat the above procedure with for two different temperatures (900C and 800C).

Observation Table: Room temperature: _____________ 0 C

Thermometer type: ___________

Sr.

No.

Fluid Temperature Time taken to rise

(Second)

Time taken to fall

(Second)

Mercury CCl4 Mercury CCl4

1

Water

2

3

4

Oil

5

6


SoE, IT-NU CONCLUSION:

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________

QUIZ:

1. What is dynamic characteristic of an instrument?

2. What is meaning of speed of response?

3. List dynamic characteristics of an instrument.

4. Define Hysteresis, Drift.

Marks: Signature:



RESISTANCE TEMPERATURE DETECTOR (RTD) Objective: To study of resistance temperature detector (RTD) Apparatus: Resistance temperature detector, thermometer, 250 ml glass beaker, heating mental, stop

watch, resistance measuring element Theory:

RTDs are among the most accurate, reproducible, stable, and available thermal elements. A metal

resistance element changes its resistance with temperature. Resistance thermometry is based upon the

increasing electrical resistance of conductors with increasing temperature. Pure elements have been

used for measurement of temperature by this effect and the method is one of the most accurate ones.

The relation between the resistance and change in temprrature is expressed by the following equation.

Rt = Ro [ 1 + T + T2 + T3 + ............]

Where , , ...... Are temp. Coefficient of resistance. In the narrow ranges of operation

and higher order coefficients are negligibly small so that Rt is given by

Rt = Ro [1 + T]

is positive for a metal resistance element. The commonly used metals are platinum, copper and

nickel. All RTDs require the following consideration in their manufacture. Wire wound sensors must

be supported on materials closely matching the wire in thermal expansion to minimize strain effects.

Additional assembly materials, such as cements should not introduce additional strain in the operating

temp. range. The final assembly must be stable. Only high purity materials and clean assembly method

should be used to avoid sources of contamination. The choice of material will be governed by (a) High

temperature coefficient. (b) High resistivity of the material. (c) Linearity of relation between

resistance and temperature.(d) Stability of the electrical characteristics of the material and resistance

to contamination.

Normally used metal in RTD is Pt. Generally Pt100 is a very common RTD. In this 100 indicates the

resistance of Pt wire is 100 at 00C.Certain RTD. are constructed having the resistance element

exposed, such as anemometory experiments, most are constructed so that fine wire element is coiled

and loosely supported on mica form. The coil is annealed and is then placed in a protective sheath.

Extra care is taken to fabricate the element so that the effects of mechanical shock are minimised.

After its end wire are connected, it is fixed in place with varnish or the other encapsulating material.

The entire sheath is often hermetically sealed, a fill material such as magnesium oxide or aluminium

oxide being included. A Mechanically fitting is usually attached to the sheath. This is often


SoE, IT-NU accomplished with an attached electrical terminal block or sometimes with a loading spring which

gives the tip of the sheath positive contact with its thermowell. This also prevents vibration. The low

thermal mass of these electrical resistance thermometers provides excellent response.

Procedure:

1. Clean the glass beaker with water and fill it with water up to half of the volume.

2. Put the glass beaker on the heating mental, start heater and raise the temperature of water

and wait till it reach at steady state temperature.

3. Join the RTD with resistance measuring element and insert RTD into the glass beaker.

4. For each five degree temperature rise in water note down the temperature indicated by

temperature indicator and resistance by resistance measuring element simultaneously in the

observation table.

Observation table: Sr.

No.

Temperature,

deg. C.

Resistance,

ohm.

Sr.

No.

Temperature,

deg. C.

Resistance,

ohm.

1. 10.

2. 11.

3. 12.

4. 13.

5. 14.

6. 15.

7. 16.

8. 17.

9. 18.

Calculations : Rt = Ro [ 1 + T]

Where, = temp. Coefficient of resistance

Rt = resistance at any temperature

Ro = intercept of the graph

[ 1 + T] = Rt/ Ro

T = [Rt/ Ro ] -1

= [ Rt/ Ro ] -1/ T


SoE, IT-NU Graph: Draw plot of resistance verses temperature and find value of and R0 from graph. R0 is

intercept on Y axis at T = 0. Slop of the graph is R0.

Results:

1. Experimental value of Co-efficient () = _______________

2. Experimental value of resistance (Ro) = _______________Ω

3. Graphical value of Co-efficient () = _______________

4. Graphical value of resistance (Ro) = _______________Ω

Conclusion : ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ Quiz:

1. Explain meaning of PT 100 RTD.

2. Write brief note about different types of RTD.

3. State advantages & disadvantages of RTD.

Marks: Signature:



STUDY OF THERMOCOUPLE Objectives: To study thermocouple and calibrate it. Apparatus: Thermocouple (J type), thermometer, 250 ml glass beaker, heating mental, stop watch,

milivoltmeter. Materials: Water. Theory:

The salient features of the thermocouple are:

Two dissimilar conductors in the form of rod or wire electrically insulated except at the hot

junction.

A refractory sheath to afford protection from injurious furnace gases. Another metal sheath

may also be used to prevent mechanical damage.

Cold junction temperature Control.

An instrument for measuring e.m.f., either a milivoltmeter or for more accurate measurement

a potentiometer system.

Compensating leads to allow the measuring instrument to be sited at considerable distance

from the thermocouple without the necessity for using expensive couple materials as extension

leads.

Thermocouples are low-cost, small convenient, wide range, reasonably stable, accurate and fast.

They are used for measuring high temp. as well as low temp. Thermocouple provides an accurate

and reliable indication of temperature for many kinds of industrial application. Imparting heat to

the junction of two dissimilar metals causes a small continuous e.m.f to be generated. Seebeck

discovered these phenomena in 1821; the thermal e.m.f is produced in a thermocouple. Electric

circuit when the temp. at the two junctions are different. Based on this the thermocouples are

constructed. An ordinary thermocouple consists of two different kinds of wires, each of which

must be made of a homogeneous metal or alloy. The wires are fastened together at one end to form

a measuring junction. The free ends of the two wires are connected to the measuring instrument

to form a closed path in which current flow. After the thermocouple wires connect to the measuring

instrument, the junction is designated as reference junction and it is maintained at constant temp.

It is most important that each section of wire in a given circuit be homogeneous, since, with no

change in the composition or physical properties along its length, the circuit emf depends only

upon the metals employed and the junction temp., circuit emf are independent of both length and


SoE, IT-NU

dia. of wires as well as the method by which the junction is made, it may be welded, soldered,

twisted or an intimate contact

Desirable properties of Thermocouple

1) Relative large thermal emf.

2) Precision of calibration.

3) Resistance to corrosion and oxidation.

4) Linear relation of emf to temp. so that scale is more easily read.

In order to protect them the thermocouple wire, it is usually covered by thermal insulation (e.g.

ceramic) and a sheath for mechanical protection (e.g. thermo-well)

Procedure:

1. Clean the 250 ml glass beaker with water and fill the beaker with water up to half of the

volume.

2. Put the glass beaker on the heater, start heater and raise the temperature of water and wait for

the steady state (850 c) to reach.

3. Join the thermocouple with mill voltmeter and insert thermocouple into the glass beaker.

4. For each 50C temperature rise in water note down the temperature indicated by thermometer

and mV indicated by milivoltmeter simultaneously table.

Observation Table:

Sr.

No.

Temperature,

deg. C.

Milivolt meter

reading(mv)

Sr.

No.

Temperature,

deg. C.

Milivolt meter

reading(mv)

1. 10.

2. 11.

3. 12.

4. 13.

5. 14.

6. 15.

7. 16.

8. 17.

9. 18.

Graph: Draw a graph of mV versus Temp. Compare the graph with the standard curve of J type

thermocouple.


SoE, IT-NU CONCLUSION :

________________________________________________________________________________

________________________________________________________________________________

_______________________________________________________________________

QUIZ:

1. Explain principle of thermocouple.

2. Give example of industrial applications of thermocouple.

3. Give name of five thermocouples with their temperature range.

4. Write about see back effect.

Marks Signature with date


SoE, IT-NU EXPERIMENT NO.: DATE:

CALIBRATION OF THERMISTOR Objectives: To study thermistor and calibrate it.

Equipments: Thermistor, glass beaker, thermometer, heating mentle stopwatch, resistance

measuring devise.

Theory Thermistors are semi conductors made from specific mixtures of pure oxides of nickel,

manganese, copper, cobalt, tin, uranium, zinc, iron, magnesium, titanium and other metals inserted

at temperatures above 982 °C. Their distinguishing characteristics are a high temp. coefficient and

the fact that their resistance is a function of absolute temperature. Their temp. coefficient is usually

negative, although it can be positive as well. In some cases with negative coefficient type,

thermistor resistance decreases at the rate of over 3 % for each degree F temp. rise. Thermistors

have the desirable characteristics of small size, narrow span, fast response and a very high

sensitivity. Thermistors do not need cold junction compensation. Thermistors are available in a

great variety of configurations, they are inexpensive. They are not affected by polarity and their

stability increases with age. They are the most sensitive temp. detectors available. There are two

types of thermistors. The types of thermistors depend on the value of b. If b is negative than the

thermistor is of negative type. If the b is positive, then the type of thermistor is positive. The

negative is more common. Positive thermistors are used within a limited range of temp. (50 to 225

°C.) only. The negative temp. coefficient resistance thermistor has a resistance-temp. relation,

given by the empirical equation,

where, Rt = Ro exp [ - b (1/T0 - 1/T) ]

Rt = Resistance at T K.

Ro = Resistance at To K.

b = Thermistor constant is dependent on material and manufacturing techniques, K.

Most familiar is the bead type, which is usually glass-coated. However thermistors can be made

into washers, discs, or rods. They can also be encapsulated in plastic, cemented, soldered in bolts,

encased in glass tube or needles or a variety of other forms. These assemblies serve to support the

sensor protect against damage to the wires. Main advantages are its small size, fast response,

variety of configurations possible, its stability and its low cost. Its main disadvantage is very non-

linear temp. vs resistance curve.


SoE, IT-NU Procedure:

1. Clean the glass beaker with water and fill it with water up to half of the volume.

2. Put the glass beaker on the heater, start heater and raise the temperature of water and wait

steady state to reach.

3. Join the thermistor with resistance measuring element and insert thermistor into the glass

beaker.

4. For each five degree temp. rise in water note down the temp. by temp. indicator and resistance-

by-resistance measuring element simultaneously in the observation table.

Observations: Initial temperature T0 , ________ o C = _______K Any temperature, TX , K = note in Observation Table Observation Table:

Positive Thermistor Sr. No.

Temperature o C

Temperature K

Resistance, ohm.

Rt, (+Ve type)

ln Rt ( 1/T0 ) – ( 1/Tx )

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15

Negative Thermistor Sr. No.

Temperature o C

Temperature K

Resistance, ohm.

Rt, (-Ve type)

ln Rt ( 1/T0 ) – ( 1/Tx )

1 2 3 4 5 6 7 8


SoE, IT-NU

9 10 11 12 13 14 15

Graph:

Plot graph of ln Rt versus (1/T0) – (1/Tx ) Draw straight line on max points of the graph. Intercept of graph on y axis is ln R0 Slope of graph = b = temperature coefficient for the thermistor. CALCULATION: Find out the value of temperature coefficient, b for the thermistor and compare its value with the practical value.

Use following equation: ln Rt = ln Ro + b [(1/T0 ) – ( 1/Tx ) ]

RESULTS:

1. Experimental value of Co-efficient (b) = _______________

2. Experimental value of resistance (Ro) = _______________Ω

3. Graphical value of Co-efficient (b) = _______________

4. Graphical value of resistance (Ro) = _______________Ω

CONCLUSION: ________________________________________________________________________________________________________________________________________________________________________________________________________________________ QUIZ:

1. What is principle of thermistor ?

2. What are different shapes of thermistor sensors?

3. Write advantages & disadvantages of thermistor .



SoE, IT-NU EXPERIMENT NO: DATE: / / .

LEVEL MEASUREMENT BY AIR PURGE METHOD

Objective: to study method of level measurement working on hydrostatic principle (Air purge

method).

Apparatus: Level measuring instrument, Scale

Materials: Water

Theory:

Level Measuring instrument is of six types visual Gauge glass, Float type, Float activated, Level and

shaft mechanism, Magnetic float, Head measurement. Air purge method falls under head

measurement category. If the density or specific gravity of liquid in the open type vessel is known the

only pressure measurement can lead for measurement of liquid level. This is the indirect method of

liquid level measurement device. This system is satisfactory for all liquids. The only limitation is the

clogging of bubble pipe in a sinusoidal system such as chemical slurries etc. A known length (Say lm)

and diameter (1/2 -1 inch) is lowered into the vessel. A pressure gauge or a manometer of suitable

range is connected separately to the upper end of a bubbler system. It operates by building up a

pressure in the feed line until the gas escapes and flow stabilizes at a rate determined by valve in the

feed line. The pressure in bubbler pipe necessary to cause flow is just equal to the pressure exerted by

liquid head above the tip of feed line. So by knowing specific gravity of the liquid, its level can be

measured in the open vessel.

Procedure:

1. Fill the tank with water.

2. Start the compression & adjust the flow such that a known no. of bubbles (i.e. one by one)

are coming out of bubbler pipe.

3. Take the readings of water level in the tank and level difference in manometer.

Observation Table:

Sr.

No.

Liquid level H

cm.

Level Difference

(H – H0)

( cm of water column )

P= Hg

gm / cm.s2

1.

2.


SoE, IT-NU

3.

4.

5.

6.

7.

8.

9.

10.

CALCULATIONS :

P= Hg

Graph :

Plot graph of P Vs level.

CONCLUSION:

________________________________________________________________

________________________________________________________________

________________________________________________________________

QUIZ:

1. Describe different methods for direct level measurement.

2. Write brief description of hydrostatic methods used for level measurement.

3. Explain different methods used for level measurement of solids.



SoE, IT-NU EXPERIMENT No.: DATE:

STUDY OF Linear Variable Differential Transducer (LVDT)

Objective: to study working of Linear Variable Differential Transducer (LVDT)

Apparatus: Experimental Kit, C.R.O., Frequency Counter

Theory: The most widely used an inductive transducer; to translate the linear motion into electrical signal

is the L.V.D.T.

The basic construction of L.V.D.T. is given in figure1. The transformer consists of single primary

winding P1 and two secondary winding S1 & S2 wound on cylindrical former.

The secondary windings have equal number of turns and are identically placed on either side of

primary windings.

The primary winding is connected to alternating current source.

A movable soft iron core is placed inside the former. The displacement to be measured is applied

to an arm attached to the soft iron core. In practice the core is made nickel iron alloy, which is

slotted, longitudinally to reduce eddy current loses.

The two secondaries of L.V.D.T. are connected in series opposition as shown in figure 2. thus o/p

voltage of transducer is difference of two voltages.

E0 = E1 – E2

When core is at null position flux linking with both secondaries are same hence equal e.m.f.

produced in them. So that output voltage E0 is zero at null position.

If core is moved to left of null position, more flux link with winding S1 & less with S2. So that

output voltage E1 of secondary winding S1 is more magnitude of output voltage is thus E1 – E2 &

output voltage is in phase with E1.

Similarly, if core is moved to right to null position, the flux linking with secondary winding S2

becomes longer than that linking with winding S1. This result in E2 is larger than E1. output voltage

in this case is E0 = E2 – E1 is in phase with E2.

As the core is moved in one direction from null position the differential voltage i.e. the difference

of two secondary voltage will increase while maintaining an in phase relationship with voltage

from the input source. In other direction from null position, the differential voltage will also

increase but 180 out of phase with the voltage from source. By comparing the magnitude and

phase of the output voltage with that of source, the amount and the directions of core and hence

the displacement may be determined.


SoE, IT-NU

The amount of voltage in either secondary winding is proportional to movement of core. Hence

we have an indication of linear motion. By noting which output voltage increasing or decreasing.

We can determine the direction of motion.

The output voltage signal can also applied to a recorder or to a controller that can restore moving

system to its normal position.

The output voltage of an L.V.D.T. is a linear function of core displacement within limited range

of motion say about 5 mm from null position.

Procedure:

1. Connect mains cord to 230V supply switch on unit and see that the supply LED glows.

2. Observe mains output of signal generation on C.R.O. and adjust the frequency to 1KHz which

is excitation frequency of L.V.D.T.

3. Now connect signal generator is set for excitation signal requirement of L.V.D.T. Connect

signal generator output as excitation signal input of L.V.D.T.

4. Connect output of L.V.D.T. to A.C. milivoltmeter.

5. Keep micrometer fully clockwise position observe & record output.

6. Now rotate micrometer anti clock wise, one position at a time & for each position observe &

record the output.

7. Plot graph of output voltage V/s input displacement.

Observations Table: Excitation Frequency:- ___________ KHz.

Sr. No. Displacement in mm Output Voltage (mV) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.


SoE, IT-NU RESULT: CONCLUSION: ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ Quiz:

1. What is an LVDT? 2. Why use an LVDT? 3. How does an LVDT work? 4. Give details of LVDTs and their support electronics. 5. What types of industries and applications use LVDTs?




RESPONSE OF FIRST ORDER (THERMOMETER) SYSTEM TO STEP INPUT.

Objective: to study the response of first order system to step input. Apparatus: 250 ml glass beakers, Mercury thermometer, Stop watch. Materials: Water. Theory: Consider a thermometer bulb positioned in a flowing stream of fluid for which the temperature ti varies with time and tt is the temperature indicated by the thermometer. The following assumptions are made: 1. All the resistance to heat transfer lies in the film surrounding the bulb. The resistance offered

by glass and mercury is negligible. 2. All the thermal capacity is in the mercury and the temp. is uniform throughout. 3. The glass wall containing mercury does not expand or contract during transient response. Applying unsteady state energy balance, input-output = accumulation. h a(ti - tt) - 0 = mcpdtt /dt ------------------------------(1) asurface area of bulb for h.t cpheat capacity of mercury. t time hfilm coefficient for heat transfer At steady state, h a(ti S - ttS) = m cp d ttS/dt = 0 ------------------------------------(2) (1) - (2) gives, h a[(ti - ti S) - (tt - ttS)] = m cp d(tt - ttS)/dt ------------------------------(3) Let Ti = ti - ti S and T = tt - ttS h a(Ti -y) = m cp dy/dt ----------------------------------------- (4) If we let m*cp/h*a = , eq. (4) becomes, Ti - T = dT /dt -------------------------------------(5) Taking laplace transform of eq. (5) we get, Ti(s) - T(s) = [ s T(s) - T(0)]


SoE, IT-NU Ti(s) - T(s) = s T(s) T(s)/ Ti(s) = 1/(s + 1) -----------------------(6) The parameter is called as time const. of the system and has the unit of time. The expression of eq.

(6) is called the transfer function of the system. Transient response to a step function, T(t) = A u(t) Ti(s) = A / s Substituting in eq. (6) we get, T(s) = (A / s) *1/(s + 1) Solving by partial fractions, (A/)/[s*(s+1/)] = c1/s + c2/( s+1/c) c1 = a and c2 = -A y(s) = A/s - A/(s +1/) Taking inverse lap lace, y(t) = A(1 - exp(-t/) ) -------------------(7) eq.(7) is called as the response for 1st order system for a step input. Procedure:

1. Heat the water in 250 ml glass beaker up to approximately 1000C. 2. Dip the given thermometers in the beaker and note down the time required by the thermometer

for every 20C raise in temperature from room temperature to till it attend the final steady state temperature.

3. Now take out the thermometer from the beaker and note down the initial temperature indicated by the thermometer.

4. Note down the time required for every 20C decrement from initial temperature indicated by the thermometer bulb to till it attend the final steady state temperature.

Observation Table:

System 1: Thermometer placed from air to boiling water

Observation No. Observations 1 Initial Steady State temperature(Ti ) = _____0C 2 Final Steady State temperature (Tf ) = _____0C

3 Amplitude A = Tf - Ti = _____0C


SoE, IT-NU

Sr. No. Temperature indicated by

thermometer (Tt) ( 0C )

Time in

sec

Percent change from steady state =

[y(t)/A]*100 = 100*

if

it

TTTT

1.

2.

3.

4.

5.

6.

7.

8.

9.

System 2: Thermometer placed from boiling water to Atm. Air

Sr. No. Temperature indicated by

thermometer (Tt) ( 0C )

Time

in sec

Percent change from steady state =

[y(t)/A]*100 = 100*

if

it

TTTT

1.

2.

3.

4.

5.

6.

7.

8.

9.

Observation No. Observations 1 Initial Steady State temperature(Ti ) = _____0C

2 Final Steady State temperature (Tf ) = _____0C

3 Amplitude A = Tf - Ti = _____0C


SoE, IT-NU CALCULATIONS : Amplitude A = Tf – Ti

y(t) = Tt- Ti

y(t) = A(1 – e (-t/))

Where = time constant

Find (practically) from y(t) = A(1 – e (-t/) ).

Find (graphically) on x- axis at 0.632 on y axis.

Graph :

Plot graph of [y(t)/A] Vs Time (for both systems)

RESULTS:

Sr. No. Systems (practically) (graphically)

1 System –1

2 System –2

CONCLUSION: ________________________________________________________________________________________________________________________________________________________________________________________________________________________ QUIZ:

1. Explain meaning of First Order System.

2. Describe Time Constant and its significance.

3. Show graphical representation of response of first order system to step input.

4. List the errors related with thermometers.

5. Time constant of which system is found higher than other system in experiment? Why?




RESPONSE OF FIRST ORDER (LIQUID LEVEL) SYSTEM TO IMPULSE INPUT

Objective: To study the response of first order liquid level system to Impulse input.

Apparatus: Tank with linear resistance outlet, Beakers.

Materials: Water

Theory:

The Impulse response of a first order system can be developed as below

Magnitude of Impulse = A

X(S) = A

Y(S)/X(S) = 1/(s +1)

Y(S) = A/(s +1) = (A/)/ (s+1/)

Take Inverse Laplace Transformation

Y(t) = A (e-t/) /

Procedure:

(1) Arrange the Apparatus for liquid level system.

(2) Start the pump and allow the flow of water through control valve.

(3) Maintain steady state flow at least six times and each time note down level of tank

(4) Find out the resistance and time constant of tank.

(5) Now maintain one steady state flow rate and note down level in tank.

(6) Give an impulse input to flow of tank by adding water by beakers very quickly.

(7) Note down time taken for addition and quantity of water added.

(8) Start stop watch and note down variation in level with time of tank.

(9) Finally note down steady state flow rate and level of tank.

Observations:

Initial level in tank before Impulse Input is given = __________mm, (h1, 0)

Highest level in tank after Impulse Input is given = __________mm, (h2, h)

Final level in tank after Impulse Input is given = __________mm, (h2, 0)

Amplitude A = (h2, max) - (h2,final) = __________mm

Area of Tank 1, A1 = __________ m2


SoE, IT-NU Observation Table - 01

Sr.

No.

Flow rate

LPH

Flow rate

Q, m3/s

Tank-1 Height

h1,m

1 1500

2 2000

3 2500

4 3000

Graph:

1. Q h1

CALCULATIONS:

From graph Q h1 slope =________& R1 = 1/slope= ________s/m2

1 = A1* R1 = _______s

Observation Table - 02

Sr. No. Tank-1 Height,

h2,x,mm

Time in

sec

Height diff. y(t)=

h2,x - h2,0

Percent change from steady

state = [y(t)/A]*100

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

CALCULATIONS : Amplitude A = (h2, max) - (h2,final)

y(t) = h2,x - h2,0

y(t) = A (e-t/) /


SoE, IT-NU

Where = time constant


Find (theoratically) by = A*R

Find (practically) by y(t) = A (e-t/) / (by trial & error/ goal seek in excel)

Graph :

Plot graph of y(t)[= (h2,x - h2,0)] Vs Time

RESULTS:

(Experimentally) = _____________ sec.

(Graphically)= ___________ sec.

(Practically)= ___________ sec.

CONCLUSION:

_______________________________________________________________________________________________________

_______________________________________________________________________________________________________

______________________________________________________________________

QUIZ:

1. Explain meaning of liquid level system.

2. Explain difference between step input and impulse input.

3. How we can increase or decrease time constant of liquid level system?




RESPONSE OF FIRST ORDER

(MIXING PROCESS) SYSTEM TO STEP INPUT. Objective: to study the response of first order mixing process to step input.

Apparatus: Vessel with overflow connection, Titration set, Stopwatch, Measuring cylinder

Materials: 0.5 N NaOH, 0.5 N HCl

Theory:

The transfer function of the first order system can be represented as: Y(s) / X(s) = 1/ (1+ s) For mixing process, = V/q Where, = Time constant, V= Volume of mixture q= Volumetric flowrate If a step input of A is given, response curve will follow y ( t ) - y ( s ) = A ( 1 - e - t / ) Where, y (t)= Conc. of overflow at time t. y(s) = Steady state conc. Steady state conc. y(s) = 4 gm/lit of NaOH solution at time t=0 and when fresh water is passed in to the vessel then step input being negative. A= y() - y(s) = 0- 4 = -4 gm/lit.

Procedure:

1. Fill up the vessel with ____ N NaOH solution up to the overflow condition 2. Run the stirrer 3. Pass tape water in the vessel at know flowrate and start stopwatch. 4. After each 3 min. interval collect the sample in bottles (atleast 6 bottles) 5. Find normality of each sample by titrating it against ____ N HCl solution.

Observations:

1. Vessel volume up to overflow V = __________m3

2. Tap water flow rate q = __________m3 / sec

3. The time constant =V/q = __________sec

4. Initial steady state concentration y(i) = __________ gm/lit.

5. Final concentration y() = 0 gm/lit.

6. Step input Amplitude, A= y() - y(s) = __________ gm/lit.

7. Volume of sample taken for titration V1 =_________ ml.

8. Normality of HCl N2 = ___________ N


SoE, IT-NU Observation Table:

Sample

No.

Time,

sec

Burette

reading

(V2)

Normality

N1= N2* V2

V1

Concentration. ,

y(t) = N1*40

Y(t)

=y(t)-y(i) Y(t) /A

1. 0

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

CALCULATION:

Amplitude A = y() - y(i)

Y(t) = y(t) - y(i)

Y(t) = A(1 – e (-t/) ); Where = time constant


Graph: plot graph of Y(t) /A t and find out of time constant.

RESULTS:

(Experimentally) = _____________ sec.

(Graphically)= ___________ sec.


SoE, IT-NU CONCLUSION:

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

_______________________________

QUIZ:

1. Why mixing process is considered as a first order system?

2. What are the factors affecting time constant of mixing process?

3. Write brief description of mixing process.




RESPONSE OF MULTI CAPACITY SYSTEM (NON - INTERACTING TANKS IN SERIES) TO STEP INPUT

Objective: To study the response of first order system in series using two tank liquid level system

(Non-interacting system) to step input.

Apparatus: Non-Interacting system Apparatus, stopwatch, scale etc.

Materials: Water

Theory:

Assumption:

(1) The liquid to be of constant density.

(2) The tanks to have uniform cross-sectional area.

(3) The flow resistance to be linear.

A balance on tank 1 gives

q – q1 =A1dtdh1 ……………. (1)


q1 – q2 =A2dt

dh2 …………………… (2)

The flow-head relationships for the two linear resistances are given by the expressions

q1 =1

1

Rh …………… (3)

q2 =2

2

Rh ………….. (4)

Combining equations (1) and (3) and introducing deviation variables give the transfer function for

tank 1

1

1

1

1

sSQSQ

…………… (5)

Where Q1 = q1 – q1s , Q= q – qs and 1 = R1A1

In the same manner, we can combine equation (2) and (4) to obtain the transfer function for tank 2,

thus

12

2

1

2

sR

SQSH

………… (6)


SoE, IT-NU

Where H2 = h2 – h2s and 2 = R2A2.

Having the transfer function for each tank, we can obtain the overall transfer function H2(s)/Q(s)

by multiplying equation (5) & (6) to eliminate Q1(s):

11

1

2

2

1

2

sR

ssQsH

……….. (7)

The overall transfer function of equation (7) is the product of two first order transfer functions,

each one of which is the transfer function of a single tank operating independently of the other.

Inversion of equation (7) by means of partial fraction expansion gives

)]11(1[)( 21

1221

2122

tt

eeRAtH

Procedure:

1. Arrange the Apparatus for non-interacting system.

2. Start the pump and allow the flow of water through control valve.

3. Maintain steady state flow at least six times and each time note down levels of tank-1 and

tank-2.

4. Find out the resistance and time constant of each tank.

5. Now maintain one steady state flow rate and note down level in tank-2.

6. Give a step change to flow of tank-1 by regulating the control valve quickly (i.e. Open control

valve quickly by three of four turns).

7. Start stop watch and note down variation (increase) in level with time of tank-2.

8. Finally note down steady state flow rate and levels of both tanks.

Observations:

Sr.No Observations

1. Length of tank-1(L1)= _____________ m

2. Width of tank-1(W1)= = _____________ m

3. Cross sectional area of tank-1 =L1 *W1 = _____________ m2

4. Length of tank-2(L2)= _____________ m

5. Width of tank-2(W2)= = _____________ m

6. Cross sectional area of tank-2 =L2 *W2 = _____________ m2



Sr.

No.

Flow rate

Q, m3/s

Tank-1 Height

h1,m

Tank-2 Height

h2,m

Observation Table - 02

Initial level in tank-2 before step change is given =__________ m, (h2,0)

Final level in tank-2 after step change is given = ___________ m, (h2,f)

Amplitude A = (h2,f) - (h2,0) = __________m

Sr.

No.

Time

t

Tank-2 Height,

h2,x Practical Height

Tank-2 Height Deviation,

H2,x m = h2,x - h2,0

Graph:

2. Q h1

3. Q h2

4. H2 (t) / (h2 - h2,0) t (Practically)

5. H2 (t) / (h2 - h2,0) t (Theoretically)

CALCULATIONS:

From graph Q h1 slope =________& R1 = 1/slope= ________s/m2

From graph Q h2 slope = _______ & R2 = 1/slope= ________s/m2

1 = A1* R1 = _______s

2 = A2 * R2 = _______ s


SoE, IT-NU

Sr.

No.

Time

t

Practically Theoretically

Y(t) = H2 (t)

= h2,x - h2,0

Y(t)/ A =

H2 (t)/(h2,f- h2,0) Y(t) = H2 (t)

Y(t)/ A =

H2 (t)/(h2,f- h2,0)

RESULTS:

CONCLUSION:

________________________________________________________________________________

________________________________________________________________

QUIZ:

1. Explain meaning of non-interacting tanks in series.

2. Draw three non interacting tanks in series.

3. If we increase no. on tanks in series, what effect we could see on response? Explain with

graphical representation.




RESPONSE OF MULTI CAPACITY SYSTEM (INTERACTING TANKS IN SERIES) TO STEP INPUT.

Aim: To study the response of first order system in series using two tank liquid level system (interacting system) Apparatus: Interacting system Apparatus, stopwatch, scale etc. Chemicals: Water Theory: Assumption: (1) The liquid to be of constant density. (2) The tanks to have uniform cross-sectional area. (3) The flow resistance to be linear. A balance on tank 1 gives

q – q1 =A1dtdh1 ……………. (1)


q1 – q2 =A2dt

dh2 …………………… (2)

The flow-head relationships for the two linear resistances are given by the expressions

q1 = 211

1 hhR

…………… (3)

q2 =2

2

Rh ………….. (4)

At steady state, equations (1) and (2) can be written

qs – q1s = 0 ………………… (5)

q1s - q2s = 0 ………………… (6)

Subtracting equations (5) from equation (1) and equation (6) from equation (2) and introducing

deviation variables give

dtdHAQQ 1

11 …………… (7)

dtdHAQQ 2

221 ………… (8)


SoE, IT-NU Expressing equation (3) and (4) in terms of deviation variables gives

Q1 = 1

21

RHH ……………. (9)

Q2 = 2

2

RH ……………. (10)

Transforming equations (7) through (10) gives

Q(s) – Q1(s) = A1sH1(s) ………….. (11)

Q1(s) – Q2(s) = A2sH2(s) ………….(12)

R1Q1(s) = H1(s) – H2(s) ……………. (13)

R2Q2(s) = H2(s) ……………… (14)

The analysis has produced four algebraic equations containing five unknowns : (Q, Q1, Q2, H1 and

H2). These equations may be combined to eliminate Q1, Q2, and H1 and arrive at the desire transfer

function:

12121

221

22

sRAsR

sQsH

……….. (15)

Procedure: Arrange the Apparatus for interacting system. Start the pump and allow the flow of water through control valve. Maintain steady state flow at least six times and each time note down levels of tank-1 and tank-2 Find out the resistance and time constant of each tank. Now maintain one steady state flow rate and note down level in tank-2 Give a step change to flow of tank-1 by regulating the control valve quickly (i.e Open control

valve quickly by three of four turns) Start stop watch and note down variation (increase) in level with time of tank-2 Finally note down steady state flow rate and levels of both tanks Observations: Cross sectional area of tank-1 = m2 Cross sectional area of tank-2 = m2 Table-01 Sr. No.

Flow rate Q, m3/s

Tank-1 Height h1,m

Tank-2 Height h2,m

h1 - h2, m



Initial level in tank-2 before step change is given =__________ m, (h2,0)

Final level in tank-2 after step change is given = ___________ m, (h2,f)

Amplitude A = (h2,f) - (h2,0) = __________m

Sr.

No.

Time

t

Tank-2 Height,

h2,x Practical Height

Tank-2 Height Deviation,

Y(t) = H2,x m = h2,x - h2,0

Graphs: Q h1 - h2

Q h2

H2 (t) / (h2- h2,0) t (Practically)

H2 (t) / (h2- h2,0) t (Theoretically)

Calculations: From graph Q (h1-h2) slope = & R1 = 1/slope= s/m2 From graph Q h2 slope = & R2 = 1/slope= s/m2 1 = A1* R1 = ________ s 2 = A2 * R2 = _______ s Substitute values of 1, 2 and A1R2 in equation no. 15 and derive time equation by taking inverse

laplace transforms to calculate theoretical H2. Write equation derived for theoretical H2:____________________________________ Table-03

Sr.

No.

Time

t

Practically Theoretically

Y(t) = H2 (t)

= h2,x - h2,0

Y(t)/ A =

H2 (t)/(h2,f- h2,0) Y(t) = H2 (t)

Y(t)/ A =

H2 (t)/(h2,f- h2,0)


SoE, IT-NU Results: Conclusion:

Interacting system data:

Tank-1 Tank-2 Length, cm Width, cm




RESPONSE OF SECOND ORDER (U-TUBE MANOMETER) SYSTEM TO STEP INPUT.

Objective: To study the step response of a U-tube manometer and to find the value of time lag () and damping coefficient ( ) for the system. Apparatus: U-tube manometer , manometeric liquid, stop watch. Theory: Consider a simple U-tube manometer. Pressure drop of magnitude P = P1- P2 is imposed suddenly on the two legs of the manometer. Let us do the force balance on plane “C” of the manometer. (Force due to P1 on leg1 ) - (Force due P2 on leg 2) - (Gravitational Force due to level diff.) - (Drag Force due to friction) = (Mass of liquid in the tube ) * (acceleration )

:. P1A1-P2A2 - gA2*2h -force due to friction = m dvdt* -----------------(1)

Where, P2,P1=pressures at the top of legs 2 & 1 respectively. A2,A1= Cross sectional areas of leg 2 & 1 respectively (e.g. A1=A2=A) =density of fluid. m=mass of fluid = AL v= avg. velocity of fluid in the tube. h=deviation of liq.level from initial plane to steady state. L=length of liq. in manometer. Poiseuille’s eq. can be used to relate the fluid friction force to flow velocity.

A dhdt

R PL

4

8 ------------------------------------------------ ( 2 )

Where, R=radius of pipe through which liq. follows. = viscosity of the flowing liquid. Now force due to fluid friction can be given by:

dtdh

RALRP 2

2 8 3

dtdhv 4a

2

2

dthd

dtdv

4b

from equations (3) & (4); equation (1) becomes,

2

2

2

82dt

hdALdtdh

RALgAhPA


SoE, IT-NU Defining above equation for steady state and subtracting from above equation we get:

2

2

2 242

dtHdAL

dtdH

RALgAHPA

Where H is a deviation variable and is given as follows:

H h hs 2 dividing both the sides by gA

gPH

dtdH

gRL

gL

dtHd

22

2 42

___________________ (5)

Let 2

2

gL

6

2

42gR

L 7

gK P

1

PKHdt

dHdt

HdP 22

22 ________________________(8)

Taking Laplace transform, transfer function is obtained as

12)()(

22

ssK

sPsH P

9

Eq. (6) & (7) indicate inherent second order dynamic characteristic of manometer. Now if unit step function is given Output for <1 i.e. for under damped system, output for the system will be given as follows:

101

tan1

sin1

11)(2

12

2

t

eKtHt

P

Procedure: (1) Fill the manometer with the manometric fluid. (2) Give the -ve step input to the manometer by suddenly applying a pressure drop. At different heights record the time until liquid get steady state. (3) Plot graph of deviation variable H Vs t (4) From the plot calculate overshoot and decay ratio. (5) From Overshoot calculate , from the formulae:

21 ExpOVERSHOOT

Also calculate from equation ( 6 ) and (7) and comment.


SoE, IT-NU Observation: Final steady height: Initial steady state height:

Sr. No. Height of liquid in manometer H, cm

Time t (sec)

Graphs: H t Results: Properties Theoretical Practical Properties Theoretical Practical Decay ratio Rise time Over shoot Frequency

Conclusion:


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CH503 –Instrumentation and Process Control Laboratory Manual … · Laboratory Manual Chemical Engineering Department School of Engineering Institute of Technology Nirma University