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SARDAR VALLABHBHAI NATIONAL INSTITUTE OF

TECHNOLOGY

SURAT

LABORATORY JOURNAL

OF

Process Control

B.Tech IV (7 th Semester)

CHEMICAL ENGINEERING DEPARTMENT

Name :_____________________________________

Class : _____________ Admn/Roll No._________

Academic Year :____________________________

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INDEX

Sr.No Experiment Title Date PageNo. Signature

1. FIRST ORDER LIQUID LEVELSYSTEM

2. SECOND ORDER CRITICALLYDAMPED SYSTEM

3. LINEARIZATION4. RESPONSE OF THERMOMETER 5. INTERACTING SYSTEM6. MANOMETER 7. HEAT EXCHANGER 8. CATALYTIC REFORMER 9. BINARY DISTILLATION

EXPERIMENT 1

FIRST ORDER LIQUID LEVEL SYSTEM

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AIM :- To study the response of first order liquid level system to step change.

APPARATUS :- Tank with input and output arrangement, Centrifugal pump , StopWatch , Measuring Scale.

PROCEDURE :-1) Start the centrifugal pump and set the flow rate into the graduated tank.2) Allow the system to reach the steady state for this flow rate. Measure the incoming

flow rate and outgoing flow rate. Ensure that input and output flow rates are samewithin the experimental accuracy. Note down the initial flow rate and height.

3) Once steady state is achieved the flow rate of liquid into the level tank is disturbed by changing the input flow rate. (Flow may be cut off). Time, t is set to zero.

4) Note down the time taken by the liquid level in the test tank to fall by onecentimeter.

5) Repeat step (4) for every centimeter fall in the liquid level in the test tank.

6) Allow sufficient time for level in the test tank to attain a new steady state under thenew reduced flow rate condition.7) Once the new steady state is achieved the final flow rate is measured and final steady

state height is noted.

THEORY: A tank of uniform cross sectional area A has a flow resistance R attached, suchas a valve, a pipe or a weir. For volumetric flow rate

q0 = h/R...(i)

This is a case of linear resistance.If qs and hs are steady state values, we define,

Q = q-qsH = h-hs ..........................(ii)

= AR (iii)

Introducing step change of H value , (H = hs1 - hs2)H(t) = H (1- e-(t/))Or H(t) = (qs1- qs2)R(1- e-(t/))

OBSERVATIONS :-1) Cross-Sectional area of tank , A =2) Initial flow rate ,qsi =3) Final flow rate,qsf =4) Initial Steady state heighthsi=5) Final Steady state heighthsf =

OBSERVATION TABLE:-

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Sr No. Time, t (s) Height (observed)h observed ( cm )

Height (theoretical)h theoretical ( cm )

123

4567810

SAMPLE CALCULATIONS:-1) Initial Resistance, R i = hsi / qsi =2) Final Resistance, R f = hsf / qsf =

3) H =hsi - hsf Or H = (qsi -qsf ) * R =4) Time constant = AR i = 5) Predicted Height or theoretical Height, htheoretical

)1(h /ltheoretica t

sie H h =

GRAPH : Plot the following curves in single graph for comparison.1. hobservedvs time2. ht vs time

RESULT :- Comment on the Nature of the Graph.

CONCLUSION :-

EXPERIMENT 2

SECOND ORDER CRITICALLY DAMPED SYSTEM

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AIM :- To study the response of a non-linear, non-interacting second order criticallydamped system.

APPARATUS :- A two tank non-interacting system , Centrifugal Pump , Measuring

Scale, Stop WatchPROCEDURE :-

1) Start the centrifugal pump and set the flow rate into the graduated tank.2) Allow the system to reach the steady state for this flow rate.3) Once steady state is achieved the flow rate of liquid into the level tank is changed,

and time, t is set to zero.4) Now, for each 5 min change in height of liquid in the second tank time is noted.5) Once the new steady state is achieved the final flow rate is measured and the pump i

switched off.

THEORY: For a critically damped second order system the response expression is given by,Y(t) = 1- (1+t/) e-t/ .(i)

The assumption made above are, A1= A2, R 1 = R 2 = R and 1 = 2 = .The response when plotted, is non oscillatory.

= (12) = 1 +2/2H2(t) = K

OBSERVATION :-

Tank -11) Cross-Sectional area of tank , A1 =2) Initial flow rate ,qsi1 =3) Final flow rate,qsf1 =4) Final height, hsf1 =5) Resistance, R 1 = hsi / qsi =6) Time constant1 = AR 1Tank 25) Cross-Sectional area of tank , A2 =6) Initial Height , hsi2 =

7) Final height, hsf2 =8) Initial Resistance, R 2 = hsi / qsi =9) Time constant2 = AR 2

OBSERVATION TABLE:-

Sr No. Time, t (s) Height in theSecond Tank (obs.)

Height theoretical

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h observed ( cm ) h theoretical ( cm ) 1234

567891011

CALCULATIONS :-

1) = (12) =2) H for second Tank: H =hsi - hsf Or H = (qsi -qsf ) * R=3) Predicted Height or theoretical Height, htheoretical

])}/(1{1[h /2ltheoretica

t

siet H h +=

GRAPH: Plot the following curves in single graph for comparison.

1. hobservedvs time

2. ht vs time

RESULT:- Comment on the Nature of the Graph.

CONCLUSION:-

EXPERIMENT 3

LINEARIZATION

AIM : To study the response of linearization.

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APPARATUS : A trapezoidal system tank equipped with i/p & o/p arrangement withfluid circulating, centrifugal pump, measuring cylinder, stop watch.

CHEMICAL: Ethylene glycol.

PROCEDURE:

1. Start the centrifugal pump and set the flow rate into the graduated tank.2. Allow the system to reach the steady state for this flow rate. Measure the

incoming flow rate and outgoing flow rate. Ensure that input and output flow ratesare same within the experimental accuracy. Note down the initial flow rate andheight.

3. Once steady state is achieved the flow rate of liquid into the level tank isdisturbed by changing the input flow rate. (Flow may be cut off). Time, t is set tozero.

4. Note down the time taken by the liquid level in the test tank to fall by one

centimeter.5. Repeat 4) for every centimeter fall in the liquid level in the test tank.6. Allow sufficient time for level in the test tank to attain a new steady state under

the new reduced flow rate condition.7. Once the new steady state is achieved the final flow rate is measured and final

steady state height is noted.

THEORY:

For a liquid of constant density & tank of uniform cross sectional area, a material balancearound the tank gives,

q(t) q0(t) = A dh ..(2)dt

In this Non-linear tank, if the fluid is flowing through linear resistance,

q0(t) = h/R

q - q0(s) = (h-hs) = A dh ..(3)R 1 dt

At steady state q0 = q0(s)

Q = q-qsH = h-hs = AR (4)

Introducing step change H, (H = hs1 - hs2)

H(t) = (H (1- e-(t/))Or H(t) = (qs1- qs2)R(1- e-(t/))

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)/1( t sit e H hh= ................(5)

OBSERVATIONS :-

1) Cross-Sectional area of tank (at the average height of hsi and hfi ), A =2) Initial flow rate ,qsi =3) Final flow rate,qsf =4) Initial Steady state heighthsi=5) Final Steady state heighthsf =

OBSERVATION TABLE:-

Sr No. Time, t (s) Height (observed)

h observed ( cm )

Height (theoretical)

h theoretical ( cm )1234567810

CALCULATIONS :-1) Initial Resistance, R i = hsi / qsi =2) Final Resistance, R f = hsf / qsf =3) H =hsi - hsf Or H = (qsi -qsf ) * R =4) Time constant = AR i =5) Predicted Height or theoretical Height, htheoretical

)1(h /ltheoretica t

sie H h =

GRAPH : Plot the following curves in single graph for comparison.

1. hobservedvs time2. ht vs time

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RESULT :- Comment on the Nature of the Graph.

CONCLUSION :-

EXPERIMENT 4

RESPONSE OF THERMOMETER

AIM : To measure positive step change of measuring thermometer.

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APPARATUS : Thermometer, beaker, burner, stand etc. CHEMICAL: Ethylene glycol.

PROCEDURE:

Take sufficient amount of ethylene glycol or oil in a vessel. Heat it using burner upto 90C Now, dip thermometer in it. Allow temperature to increase to steady state of 90C. Nowimmediately take thermometer out of the bath and note down the time for every 5 C fall.

THEORY: Assumption made for analysis of response of mercury bulb thermometer:(1) All the resistance to heat transfer reside in the film surrounding the bulb.(2) All the thermal capacity is in the mercury.(3) The glass wall containing the mercury does not expand or contract during the

transient response.Applying unsteady state energy balance,

h As (x-y) = mc dy ..(i)dtFor steady sate h As (xs - ys) = 0 .(ii) Now,

X = x - xsY = y - ys

From (i)-(ii)Y(s) = 1X(s) s + 1

Where = mc/h AsFor step change,

A =ysi - ysf Y(t) = A(1-e-(t/))y (Theo. Temp) = yinitial A(1- e-t/)

OBSERVATIONS:1. Room temperature (Initial temp indicated by thermometer) =

OBSERVATION TABLE:

Sr. No TemperatureExperimental

(C)

Time(sec)

Theoretical Tempy (C)

1

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23456

78910111213

SAMPLE CALCULATION:

A =ys(initial) - ys(final) = ___________(time at 63.2% Response, from graph of observed temperature vs time)

yt (Theo. Temp) =yinitial A(1- e-t/)

GRAPH: Plot the following curves in single graph for comparison. 1. Temperature experimental v/s Time2. Temperature theoretical (Predicted) v/s Time

RESULT:

1. Time constant for 63.2% response, [from Graph of Temp.(exp.) v/s t] _____________

CONCLUSION:

EXPERIMENT 5

INTERACTING SYSTEM

Aim : To study response of second order over damped system consisting of two interactingtank in series.

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Apparatus : Two interacting tank system, stop watch, scale, measuring cylinder etc.

Chemicals : Ethylene Glycol.

Theory :

Taking material balance across the tanks ,

q q1 = A1dh1 /dt .(1)

q1 q2= A2 dh2/dt.(2)

q1 = 1/R 1 (h1 -h2) ..(tank 1) .(3)

q2 = h2/ R 2(tank 2)(4)

At steady state,equation (1) & (2) becomes

q(s) q1(s) = 0.(5)

q1(s)- q2(s) = 0.(6)

Equations 1,2,5 and 6

Q - Q1 = A1dh1 /dt (7)

Q1- Q2 = A2 dh2/dt(8)

Q1 = (H1- H2)/ R 1.......................................................................................(9)

Q2 = H2/ R 2...............................................................................................(10)

H2(S)/ Q(S) = R 2/{ 1 2 s2+ (1 +2 + A1R 2) s + 1}For both tanks to be interacting,Q2(S)/ Q(S) = 1/( 2 s2+ 3 s+1)

Q2(S)/ Q(S) = 1/0.3825 + D (0.6225+ 1)

For step disturbance:Taking inverse of Laplace transform

Q2(t) = 1+ 0.17 e-t/0.38 - 1.17 e-t/2.62 H2(t) = A [1+ 0.17 e-t/0.38 - 1.17 e-t/2.62 ]

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Procedure: Start flow of liquid, measure the flow rate and attain steady state. Note down initial height and flow rate readings.When steady state is attained, give step /impulse disturbance (e.g. plug one of two inlets) Note down the reading at each 5 mm drop of liquid level in second tank, till the second

steady state is achieved.Observation:

(1) Initial flow rate =(2) Initial Steady state height in tank 1 =(3) Initial Steady state height in tank 2=(4) Final steady state height in tank 1=(5) Final steady state height in tank 2=(6) Flow rate at second steady state=

Observation Table:

Sr.No . Height Observed (cm) Time (sec) Theoretical height (cm)

Calculation:

(1 )1= A1R 1

(2)2= A2R 2

(3) =

For step disturbance:

H2(t) = A [1+ 0.17 e-t/0.38 - 1.17 e-t/2.62 ]

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So, h(t) = h initial H [1+ 0.17 e -t/0.38 - 1.17 e -t/2.62 ]

GRAPH: Plot the following curves in single graph for comparison.

1. hobservedvs time2. ht vs time

RESULT:- Comment on the Nature of the Graph.

CONCLUSION:-

EXPERIMENT 6

MANOMETER

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Aim: To find out the dynamic response of a second order under damped system to a stepchange in the input variable. Also to determine overshoot and decay ratio from the responsand compare it with theoretical values.

Apparatus :

AU-tube manometer of sufficient inside diameter, filled with mercury. A compressed air supply at 1 atm. gauge pressure. A pressure regulating valve to control the out put pressure of compressed air.

Procedure:

1. Connect the compressed air supply to the inlet side of the pressure regulating valve.2. Keeping the ball valve,on the outlet side of the pressure regulating valve, closed adjus

the setting of the pressure regulating valve that the pressure on the outlet side comes tohe desired value.

3. Apply the regulated compressed air pressure to one side of the manometer by using thmanometric tubing provided for the pressure.4. Allow sufficient time for the manometer reading to stabilize. Note down the height o

mercury thread in one of the limbs of the manometer.5. At time T=0 with draw the air pressure and note down the height of the mercury thread

in the same limb where the reading was taken earlier .Remember that the mercury threadwill move very fast and the reading has to be taken in very short time.

6. Repeat (3) and (4) but now take the reading for the trough in the thread height if earliereading was taken for peak in the thread height.

7. Go on taking the readings of subsequent peak and though heights of mercury thread.8. Repeat steps (6) to (7) for minimum three different step sizes of pressures.

Theory:

Consider a U tube manometer. The transfer function can be developed, byconsidering a force balance on manometric fluid and applying Newtons law of motionto it. To simplify the matter, the frictional pressure drop due to motion of fluid throughU tube, will b assumed to be proportional to the velocity of the fluid and all the fluidmass is assumed to accelerate uniformly. Again it will be assumed that the cross-sectional area of manometer tube A remains constant all through.

After this an additional pressure of magnitude P is applied .As a consequence to thismanometric fluid starts rising in the right hand side and dropping by correspondingmagnitude in the left side. This rise in the height of the liquid will create back pressureand will try to resist movement of manometric fluid. Similarly viscous pressure dropwill also try to resist the movement of the fluid. Applying the Newtons Second law ofmotion gives us,

Net force on the mass= Mass acceleration.

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AP - 2hAg/gc - RAdh/dt = AL/ gc[d2 t / d h2 ] (1)

Where,A = cross sectional area of the manometer tube. [cm2]

P = Applied pressure [gm.force/cm2

]

H = Height of liquid above its steady-state level.[cm] = Density of the manometric fluid [gm.mass/cm3]R = Frictional resistance of the fluid for flow [gm.force sec/cm3]

With the above given assumption of frictional pressure drop proportional to h t velocityHagen Poiseullis equation can be written as

P = 32uL/ gc D2 = Ru (2)

Where,

L = Length of the manometric fluid in the manometer. [cms]D = Inner diameter OF U tube. [cms]

Sub. Equation (2) in equation (1) and dividing by 2g/gcand rearranging, it gives,

{L/2g}d2h /dt 2 + {16 L / gc D2 } dh/dt + h = P gc / 2g = hi (3)

Here, hi is taken as input pressure in terms of liquid column height to balance that pressure

Again notice that coefficient of d2h /dt 2 {L/2g}have dimensions of sec2 or (time2).Similarly coefficient of dh/dt have the dimensions of second or [time] .Thus the

coefficient of dh/dt can be replaced by 2

as L/2g may not be equal to 16 L / gc D2

Thus, equation (3) may be written as 2 d2h /dt 2 + 2 dh/dt + h = hi (4)

Taking Laplace transform on both the sides and noting that h as well as dh/dt are zero att=0, and rearranging, we get

H(s)/H2(s) = 1/[ 2 S 2 + 2 S + 1 ] (5)

To define equation (5) fully we require value of two parameters and and it is known assecond order system.

The response of second order system to various forcing functions can be obtained byfactorizing the quadratic term in the denominator into two linear terms that contains itsroots S1 and S2.

For less than on the roots of quadratic equation [ 2 S 2 + 2 S + 1 ] are given by

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S1 = {- / } + { 1- 2/ } = a+ibS2.= {- / } - { 1- 2/ } = a-ib (6)

When the forcing function is a unit step function, then its Laplace transform is given by

X(s) = 1/S (7)Thus, Y(s) = 1/S [ 2 S 2 + 2 S + 1 ] (8)

Taking Laplace inverse by making use of partial fractions, we get the response of thesecond order under damped system [

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RESULT :

From the graph, determine the value of overshoot and decay ratio, and compare graphicavalue with the theoretical value.

CONCLUSIONS:

EXPERIMEN 7

HEAT EXCHANGER

Aim : To study the Digital Control System for Heat Exchanger.

Theory : The Digital Control System is the recent development in ChemicalEngineering Branch. In this practical, the stream is taken from the 7th tray and it is at higher

Type of Value Overshoot Decay ratioGraphical valuetheoretical value

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temperature and the temperature has to be lowered so we are using heat exchanger .The hofluid flows into the tube side and the cold fluid (water) flows into the shell side. When thehot fluid comes out its temperature decreases and the temperature of cold fluid increases Now vary the flow rate of hot fluid, cold water and note down various parameters that var proportionally respectively.

Also vary the temperature of hot stream, cold stream, and take down the readings.

The logic behind this is that all the flow rate varies due to the proportional control attachedto it .For having the temperature of the stream leaving the heat exchanger and entering thfractionator and at higher.Process Description :

The objective in training a student on this system is to explore the control system tied upwith non-linear process behavior.The purpose of Heat Exchanger in this model is to control the temperature profile of FCCU

fractionator. This is accomplished by cooling the TPA (Top Pump Around),entering the 4th

tray to 105 Deg C. The TPA comes from the 8th tray of FCCU fractionator at150-200 DegC. The temperature controller regulates the flow through heat exchanger & by-pass, therebcontrolling the temperature of the mixed stream to maintain desired set point (105 Deg C)

HEAT EXCHANGER PROCESS DESCRIPTION:

The model uses standard Shell & Tube Heat Exchanger, with water on tube side & TPAentering the shell side .The temperature of the TPA entering the shell is controlled by theflow of cold water whose temperature varies from 15 to 50 Deg C, depending upon thmalfunction value. The temperature of the TA, entering the shell, ranges from 150 Deg Cdepending upon the malfunction value.

SIGNIFICANT OPERATING VARIABLES:

The most important controller is Temperature Controller that is used to control the TPAreturn temperature to desired set point. Flow controllers for TPA entering Shell Side & coldwater entering tube side are also provided for controlling the plant in fully automatic mode

Training Exercises:

The exercises begin with the program operating at the normal condition, following aDesign Start. If any difficulties arise in the midst of work, reinitialize the program andcontinue without getting bogged down in by manipulation of the controls.

Exercise 1: Temperature Control

Action: Decrease the set point of temperature controller TIC-01Result: The controllers output decreases, controller output increases temp will settle alower value than design.

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Discussion: The decrease in set-point shall demand more cold TPA from shell side of heaexchanger as compared to the TPA through by-pass which causes the temperaturecontroller to increase the cold TPA through by-pass.

Exercise 2 : FIC-01Action : Increase setting of FIC-01Result : Output of FIC-01 increases thereby increasing flow of cold water through tubeside of heat exchanger.

Exercise 3 : FIC-02Action : Increase setting of FIC-02.Result : Output of FIC-02 increases thereby increasing flow of TPA.

EXPERIMENT 8

CATALYTIC REFORMER

Aim : To study the Digital Control System with the Catalytic Reformer.

Theory :The Catalytic Reformer Plant converts feed stocks into high Octane gasoline. A by- product of the process is hydrogen. To avoid catalytic poisoning, the feed is firstdesulfurised to extremely low limits of sulfur, using some of the hydrogen produced by threforming reactions.

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Catalytic Reforming involves both cracking and isomerisation .Straight-run gasoline andlight naphthas usually have very low octane numbers. By sending this fraction to areforming unit and giving them a light crack their octane number may be increased .Thiupgrading is probably due primarily to isomerisation. The plant is divided into mainsections as regards chemical reactions. The first of these is the desulphurization section

and the second is the reforming section. The Simulator will be considering reformingsection only. Reforming section includes one reactor, one furnace and one feed effluenheat exchanger.(whereas plant has three sets of these equipment ) .

The plant is divided into main sections as regards chemical reactions. The first of these ithe desulphurization section, and the second is the reforming section. The Simulator will bconsidering reforming section only. Reforming section includes one reactor, one furnaceand one feed effluent heat exchanger (where as plant has three sets of these equipment).

In simulator, feed flow is simulated by simply opening feed valve and opening feed valvand recycle gas. Composition of feed flow and recycle gas can be changed from Instructo

console. The reactor pressure is assumed to be constant at 31kg/cm2

G .

Training Exercises:

FIRST SET EXERCISES:

Ex.1. Reactor Inlet Temperature.

ACTION: The student changes the temperature set point of temperature controller TIC 104RESULT: Temperature of combined feed brings to its new set point.DISCUSSION: Temperature controllers regulate the fuel gas flow.

Ex.2.Reformer Feed Control.

ACTION: Student varies the set point FIC-101.

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RESULT: Feed rate changes to new set point value of FIC-101.DISCUSSION: Notice finally now this change is reflected to the throughput to reactor i.e.flow to the reaction section changes.

Ex.3. Recycle Gas Flow Rate.

ACTION: Student varies the setting of HC 101.RESULT: Circulation of hydrogen to the reactors is changed.DISCUSSION: Note indicator FI-102, Discuss why a change in the flow rate is not particularly upsetting the reformer flash drum pressure control system.

SECOND SET EXERCISES :EX.1. Reformer Charge Rate.

ACTION: Student varies the set point FIC-101.

RESULT: Essentially all flow rate in the plant change.DISCUSSION : Eventually, the fresh feed rate settle out to a value, which satisfies demand,set on FIC-101.Furnace firing rates change, to satisfy the reacting requirement representefor above reason by setting FIC-`101.More or less hydrogen is produce. So the reformeflash drum pressure controller reacts. The recycle gas compressor system stay at set valuethus changing some basic operating ratios in the plant.

Note also how much time is required for steady conditions become established after achange of FIC-101 set point. This is a special significant feature of the plant and deservespecial notice.

EX.2. Reformer reaction Temperature Control.ACTION: The student changes the set point of TIC-104.RESULT : All temperatures in the reformer section change.DISCUSSION : Take note how all the temperatures change, how much time it takes for conditions to settle out, 7 briefly note the fact that hydrogen production rate is change.

EX.3 . Recycle gas control.ACTION: Student varies the setting of HC-101RESULT: The circulation of hydrogen to reactor is changed.DISCUSSION: Note the indicator FI-102 , discuss why a change this flow rate is not particularly upsetting to the reformer flash drum pressure control system.

Major Equation Used in Models:

The catalytic reactor model is considering the following reactions :

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1. P N + H22. N A+ H23. P + H2 LHC4. N + H2 LHC

P = Paraffines N=NaphthenesA=AromaticH2 = HydrogenLHC = Light Hydrocarbon

EXPERIMENT 9

BINARY DISTILLATION

Aim : To study the Digital Control System for Binary Distillation.

Theory : In Binary Distillation, Distillation of Benzene and Toluene occurs Benzene

comes out from the top and toluene comes from the bottom and we have to note down threflux flow rate that varies proportionally.The steam flow rate is varied and accordingly reflux, distillate, and residue rate is variedThen various readings for varying top head pressure is changed and result on the othe parameter is seen. Then we will close the discharge valve and then we will note down theffect.

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The DISTILLATION TOWER OPERATIONS Program has to do with conventional bubble cap plate distillation tower. These are so named because they have a number ohorizontal plates or trays, which are usually equally spaced in the upright tower shellThese bubble plates provide time for mixing and interchange of vapor and liquid flowrequired for carrying out distillation separation. The relatively large number of trays use

generally produces tall and upright structure, and is the reason for the terms column otower.

The temperature of the boiling liquid is fixed by ht column pressure and liquid compositioand heat balance is always maintained. For the continuous balanced operation, the sensiblheat of the liquid and vapor flows into the plate plus the latent heat of the condensingvapour must equal the sensible heat of the liquid and the vapor flow out, plus the heat losand the heat of mixing. Since the temperature change from one plate to another is usuallysmall, the heat balance effectively reduces the balance between the heat given up bycondensing vapor and the heat requires to vaporize the liquid leaving the plate. And heat ovaporization and condensation have similar values; the vapour rate out of the plate is nearl

equal to the vapor rate into the plate.Liquid enters the bottom part of the tower from the bottom plate down flow pipe. The morvapour produced, the richer the bottom liquid becomes, in high material. Therefore foconstant column pressure, the bottom plate temperature can be used to adjust the fraction ohigh boiling material in the bottom product flow. Increasing the bottom plate temperaturewill increase the bottom fraction of boilers.

The reflux rate effectively determines the time that the feed materials remain in the columnA large reflux rate means high portion of the overhead vapors are returned to the columgiving the material a longer time exposure to the separation operation, making it richer inlow boiling material. A smaller reflux flow will produce a less pure overhead product.