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friends, in the last lecture we initiateddiscussion on the estimation of the residence time distribution for a laminar flow reactor;let us continue with that. so, laminar flow reactor is essentially at a tube through which a fluid is flowing and let say that the volumetric flow rate with which the fluid is actuallyflowing through the reactor is v naught, and if the length of the reactor is l, then thevelocity profile of the fluid in the radial direction will be parabolic with maximum velocityat the center and all the other fluid streams which is flowing at any other r location willbe smaller than the maximum. so, this is r equal to zero and this r equal to capitalr which is the radius of the cylindrical tube. so, the maximum of velocity will be at thecenter. so, the parabolic velocity profile

as we saw in the last lecture is actuallygiven by u at any r location is two times the average velocity in a cross section orthe cutting velocity in a cross section multiplied by 1 minus r by r the whole square.and that is equal to 2 into v naught which is the volumetric flow rate with which thefluid is actually flowing through the reactor divided by the cross sectional area pi r squareinto 1 minus r by capital r the whole square. so, that is the dependence of the velocityin any radial position with respect to the position. now we said that the time that isactually taken by different fluid in different r location is going to be different, becausethe velocity with which they are moving through the reactor is different. so, therefore, thetime that they would different fluid elements

at different r location would take in orderto traverse from the entry to the exit of the reactors will be tau divided 2 into 1minus r by capital r the whole square where tau is actually given by v volume of the reactordivided by the volumetric flow rate which is essentially the space time of the reactor. so, now let us look at a particular crosssection. so, if at any radial location if we identify a small thickness of delta r;so suppose if thickness is delta r and this thickness is located at some r location. so,the inner radius will be r of this element and the outer radius will actually be r plusdelta r, and the diameter is given by 2r, where r is the radius of the tube. so, nowthe volumetric flow rate of the fluid that

actually flows through that small elementdelta r is essentially given by d v; if u is the velocity with which the fluid is actuallyflowing in that r location multiplied by 2 pi r d r. so, that is the volumetric flowrate of fluid flowing in delta r; so that is the volumetric flow rate.now what is the fraction of the total fluid that actually flows through that small elementdelta r? so, that fraction of fluid flowing through delta r; so that is actually givenby d v by v naught, where v naught is the flow rate with which the fluid actually isentering the reactor, but that is equal to u by r u by v naught into 2 pi r t r. andthat is nothing but the e of t d t which is the fraction of the fluid that is actuallygoing through the small element whose volumetric

flow rate is actually between v and deltav. and also the time that is actually spending inside the reactor is given by the time thatis between t and t plus delta t. so, therefore, the fraction of the fluid that is flowingthrough should be equal to the corresponding residence time of that particular fluid element. so, now we know that the time that the fluidelement takes to actually traverse from 0 to l; that is along the length of the reactoris actually given by t equal to tau divided by 2 into 1 minus r by capital r the wholesquare, where capital r is the diameter of the tube. now from here by differentiatingthis expression, we can find out that d t equal to 4 by tau r square multiplied by taudivided by 2 into 1 minus r by r square. so,

one needs to perform a little bit of algebrato get this expression so into r d r. so, that is the expression for d t as a functionof. so, when we take the first differential of this expression, this is the expressionthat one would get as a function of d t equal to r and d r, okay. so, we can further simplifythis by substituting the expression for the time that the fluid which is present in aparticular r location takes to travel from the inlet to the exit stream of the rector.so, that is given by 4 t square divided by tau r square into r d r. so, this is obtainedsimply by substituting this expression which is present inside these brackets with thecorresponding time. so, that is nothing but the time taken by the fluid at our locationto travel from the inlet to the exit of the

reactor. so, from here, we can find out thatr d r is actually given by tau r square by 4 t square into d t. so, now we can plug thisin to the expression for the fraction of the fluid whose volumetric flow rate is betweenv and v plus v naught v and v plus d v and the fraction that spends the residence timeof that fraction is between t and t plus d t. so, that is given by e of td t and that isequal to d v by v naught. and that is given by l d v is nothing but length of the reactorinto the area of that small element that is 2 pi r d r divided by the time that is actuallytaken by the fluid in that particular element to travel from the inlet to the outlet ofthat particular reactor of that particular

shell. and that ratio will give what is thisdifferential volumetric flow rate multiplied by 1 by v naught. so, that is the expressionfor d v by v naught. so, now from here we can substitute r d r using the time that itactually differential time that the fluid actually takes to travel from one end to theother end of the reactor. so, that can actually be related, and so onewould get that l into pi r square divided by v naught into 1 by t into tau by 2 t squareinto d t. so, all that has been done is we have substituted r d r with the correspondingexpression we just derived a short while ago. now l into pi r square is nothing but thevolume of the reactor itself. so, l into pi r square is the volume of the reactor. andso therefore, v by v naught is nothing but

the space time of the reactor. so, that isgiven by that is equal to tau by t into tau by 2 t square into d t. so, that is nothingbut tau square by 2 t cube into d t. so, e t d t which is the residence time ofthe fluid fractional time that is actually spent fraction of the fluid that spends thattime t whose residence time is between t and t plus d t is actually given by e t d t andfor laminar flow reactor that is equal to tau square which is the square of the spacetime divided by the 2 into t cube, where t is the time that is spent by a particularfluid element at a particular location from the entry to the exits of the reactor. so,therefore, by simply comparing simply by observation we can deduce that e of t should be equalto tau square by 2 t cube, where tau is given

by volume of the reactor divided by the volumetricflow rate if we assume that the flow rate is actually constant.so, now the question is when is this particular expression valid? what is the validity ofthis expression or the question is when will the fluid start living? suppose i put a tracerat the entry of the reactor, how much time will it take for this tracer to first appearat the exit stream of the reactor; that is if i put tracer let us say pulse tracer atthe entry of the reactor, how much time will that pulse take to travel through the reactorand what will be the first time at which the fluid will actually leave the reactor at l.and this is important because e of t is essentially the age distribution of the effluent stream.so, therefore, the e of t d t is actually

valid only from the time at which the fluidis actually at which the tracer is actually seen at the exit stream or the effluent streamof the reactor. so, how do we find this? so, we can find this by observing that for a laminarflow parabolic profile, the fluid elements which is actually present at r equal to 0,it travelers at a maximum speed. so, at r equal to 0, the fluid travelers theu of r equal to u max; so that is the maximum velocity. so, now, therefore, the residencetime of the fluid element which is actually sitting at r equal to 0 or entering the reactorat the center of the reactor would actually spend the least time to travel from the inletto the exit of the reactor. so, therefore, the minimum residence time should actuallybe equal to the residence time of the fluid

stream which is actually entering at thislocation r equal to 0. so, how do we find this? so, we know that the time that is actuallytaken for the fluid elements to travels from one end of the reactor to the other end ofthe reactor is simply given by l by u. and the minimum time is actually given byl by u max; that is the maximum velocity and that is at the center of the reactor. so,from here by simply plugging in the corresponding expression, we can find that minimum timethat is actually taken for the tracer to be seen at the exit of the reactor is given byl by 2 u average; that is the cup mixing average into pi r square divided by pi r square. andthat is nothing but v by 2v naught and that is equal to half of this space phi. so, thefluid element that is actually entering the

reactor at the center of the tube would actuallytake half the space time before it actually reaches the other end of the reactor.and in fact the residence time distribution actually starts from that particular timetau by 2 the minimum time. so, therefore, the r t d function for the laminar flow reactoris actually given by e of t that is equal to 0 if time is less than tau by 2 which meansthat there is no fluid stream which is actually leaving the reactor if the time at which itis monitored is less than tau by 2, whereas at for any time greater by tau by 2, the residencetime distribution is actually given by tau square by 2 t cube for t greater than or equalto tau by 2. so, this is a nice example of how to find the residence time distributionfunction for a real reactor. so, such kind

of a method can actually be employed to findthe residence time distribution of any reactor where the dispersion is not present. so, now what about f of t the f curve forthe lamina flow reactor? so, for t greater than or equal to tau by 2 that is the timefor which e of t is actually valid and f of t is given by integral 0 to t e of t d t andthat is equal to 0 plus integral tau by 2 to t e of t d t. and now plugging in the expressionfor e of t, we can find that this is equal to integral tau by 2 to t tau square by 2t cube d t, and on integration, one would find that will be equal to 1 minus tau squareby 4 t square. so, therefore, the mean residence time which is the one of the property's ofthe residence time function is actually given

by 0 to infinity t of e t d t.so, this is the f curve; this is the relationship between f curve f and the time and the meanresidence time is actually given by a integral 0 to infinity t e t d t and that is becauseit is between 0 and tau by 2 e t is 0. so, this integral simply becomes tau by 2 to infinitythe limits will change and the integral will be tau square by 2 t square into d t whichis equal to tau square by 2 into minus 1 by t. and the limits are tau by 2 and infinity.so, that is the limits, and substituting the limits, we will find that that will be exactlyequal to tau which is what we instituted before that if there is no dispersion, then the meanresidence time would be equal to the space time itself irrespective of the r t d function.and, in fact, we have shown this for three

different types of reactors, the plug flowreactors, cstr and that of the laminar flow reactor. now let us look at the normalizedresidence time distribution function. so, suppose if he define theta as t by tau,tau is the space time or the mean residence time for this particular case. and so e oftheta would be 0 for theta less than 0.5, and it will be 1 by 2 theta cube for thetagreater than or equal to 0.5. and similarly, f theta will be 0 for theta less than 0.5,and it will be 1 minus 1 by 4 theta square for theta greater than or equal to 0.5. so,now if we look at the e curve and the f curve, suppose if i sketch the e curve of the normalizedresidence time distribution function, then we can observe that the e curve is going tostart at 0.5 because before 0.5 it is not

valued. so, at 0.5 it will start and so itwill start it will look something like this. and the f curve can actually be again f curvewill also start at 0.5. this is the f curve and it will start at 0.5 and then it willactually it will slowly increase and go to 1. so, that is the f curve. so, now we havelooked at the residence time distribution functions of three different reactors. so,let us now attempt to put them all together and compare how the residence time distributionfunctions are actually different for these three reactors. so, now let us compare the residence timedistribution function the f curve for cstr plug flow reactor and the laminar flow reactor.so, if we plot the normalized f curve, then

for a plug flow reactor the f curve wouldstart exactly at the space time of the reactor. so, therefore, it is exactly at 1, becauseit is a delta function, and so the f curve will actually be. so, that will be 1, thenfor a cstr that is the f curve that one would get for a cstr. now if we plot for laminarflow reactor, so it starts at 0.5 and it appears somewhere in between the cstr and the plugflow reactor. so, this is cstr the f curve for cstr, and this the f curve for plug flowreactor, and this is the f curve for laminar flow reactor.so, one can actually experimentally if one performs the one estimates the f curve andthe e curve experimentally from the methods that we described earlier that is if we usea pulse or a step input and one finds out

what is the f curve for the reactor, thenby simply making a comparison of this chart, one can actually estimate whether the realreactor is closed to what type of these three reactor that we have actually looked at sofor; that is the cstr, the laminar flow reactor and the plug flow reactor. so, such kind ofa comparison provides a method for actually diagnoses of the nature of the rtd functionfor a given real world reactor. so, that brings us to the next topic wherewe want to see how to use rtd function for diagnoses diagnostics and troubleshooting.so, rtd functions can actually be used for diagnoses of certain properties of the reactoror certain aspects of the reactor and also to troubleshoot if something undesirable isactually happening inside the reactor. so,

how do we do this? so, the rtds are actuallyused for diagnoses. so, by comparing the rtds is that are actually theoretically estimatedfor certain type of reactors and comparing that with the rtd of the real world reactorwhich may be estimated using experimental methods. by comparing that, one can actuallyfind out what is the class of the real world reactor based on the rtd function.so, rtd function can actually be used for diagnoses and not just that it can also beused rtd functions can also be used in order to model the real reactor as a combinationof ideal reactors. so rtds play a huge role in actually modeling real reactors as combinationof ideal reactors. so, the rtds the residence time distribution function, they play a crucialrole in this process as well in order to model

the real reactors as a combination of idealreactors. so, before we get into how to do the diagnostics, let us look at what are allthe common residence time distribution functions. so, he common rtds, so for a plug flow reactorthe residence time distribution is essentially a dell function. so, that is the residentialtime distribution and it is entered at the space time of the reactor and that is thee curve e of t. and if i look at cstr, so the residence time distribution is actuallyexponential that is e of t and that is the e curve for cstr. now if i take a packed bedreactor, now what has been observed; so suppose if there is a reactor and there is a fluidwhich is actually flowing through this reactor. one of the commonly observed rtd curve forsuch kind of a real world packed bed reactor,

that is actually it look as below. so, twopeaks have been observed; this is one of the commonly observed type of rtd function, wheretwo peaks are observed, and typically, the first peak if there are two peaks and thefirst peak which actually appears before the space time of the reactor indicates that theremay be channeling in the reactor, channeling or the bypassing inside the reactor.so, that is the first peak which appears before the space time of the reactor and the secondone the second peak which actually appears after the space time of the reactor; thatindicates that there may be dead zones which may be present inside the reactor which doesnot serve any useful purpose inside the reactor. so, now if we attempt to depict this in thepacked bed reactor, so there may be channels

which may be present inside the reactor throughwhich the fluid which is actually going through the packed bed reactor will easily escapeand leave the reactor. and because there is a channel with which the fluid easily escapes,the time that they spend inside the reactor should actually be is actually smaller thanthe space time of the reactor. and that is the reason why the first peakcorresponds to the channeling of the fluid steam inside the reactor. on the other hand,if there are dead zones which are actually present inside where the reactor is virtuallyinaccessible, then there will be some of these fluids which are actual present; they willspend too long time inside these dead zones before they leave the reactor. and that iswhy it appears as a second peak particularly

the tail part of the distribution curve. so,another commonly observed rtd function is that of a tank reactor a stir tank rector.so, suppose if we have a tank and its well stirred and let say this is the inlet to thetank and this is the outlet to the tank and ma be there is bypassing through this particulartank and maybe there are some dead zone which are present here. so, this is the dead zone,and this is the bypassing. so, if such kind of a situation is there in a tank reactorthat can actually be observed in the rtd curve in the rtd distribution function. so, thetypical so the first peak which appears very close to time t equal to 0 is because of thischanneling or because of this bypassing. and this bypassing can occur because of the placementof the entry and the exit fluids stream of

the tank reactor; that these fluids streamssimply quickly escapes and leaves the reactor, and that can actually be captured by thissharp peak which is actually present at time close to 0.that is at the initial stages and then the long tails which is actually, so this is correspondsto the channel bypassing, and then the long tail which is actually present here is because of the dead zone. so, the launch tail indicates the presence of a dead zone inside the tank reactor, and this long tail is because this dead zone is actually not available for the fluids to actually go and they are notexchanging material with the location inside the tank which is well mixed. and, therefore,whatever residual fluid which is present here, they will take a very long time before theyactually appear at the effluent stream of

the reactor. therefore, this dead long tailvirtually corresponds to the dead zones that may be present inside the reactor. so, thiskind of an approach the detection of the residence time distribution curve can actually providea lot of information about what is actually happening inside the reactor, and this commonrtd that has been explained just now is a good example of that. so, let us now look at the operation of cstr.so, let us consider a cstr, and let us look at what are all the various types of operationsof cstr, how it can be operated, and what are the distribution curves for each of thesesituations? now this is very important to understand because if there is a problem withthe cstr and if it falls in one of these operational

modes, it helps in diagnosing what is theproblem with the actual reactor what is the problem with the functioning of the actualreactor, and then the methods to correct it can actually be implemented or actually canbe deciphered later or can actually be thought of and strategies can actually be improvisedlater. so, there are three modes of operation. suppose,if there is real reactor whose volume is known and let us say a volumetric flow rate withwhich the fluid is flowing through the reactor is known, then one could actually look atwhat is called the normal operation where it behaves like an ideal cstr, where all locationsin the reactor is actually available for the reaction which means there are no dead zoneswhere the reaction does not happen. and also

it is assumed that there is perfect mixingin the reactor, and there is no bypass of fluid which means that all fluid that comesin actually undergoes reactions, spends the sufficient amount to time inside the reactorand then they leave the reactor. so, as a second mode of operation is cstrwith bypassing. so, if we understand how the residence time curve of a cstr with bypassingis going to appear, the shape of the curve can actually provide a clue as to if we understandwhat it is, then that actually be used as a diagnostic tool to find out if there isbypassing in the reactor. and then the third operation is with dead zone. so, now for awell mixed cstr, suppose if i put a tracer suppose if i actually insert a tracer intothe cstr, then one can write a mole balance

for the tracer and the mole balance will be.suppose there is a tracer and the mole for the tracer will be v into d c by d t; thatis equal to minus v naught into c suppose if it is a pulse tracer minus v naught intoc and that should be equal to d c by d t minus 1 by tau into c. so, that is the mole balancefor the tracer, and by integrating this expression, one would find that c of t is equal to theinitial total concentration of the traces c t naught into exponential of minas t bytau and we know that the rtd function for thisreactor is actually given by the rtd function is given by 1 by tau that is the rtd function and then the ft curve which is the f curve is actually given by 1 minus exponential of minas t by tau. so,this we have already seen. now suppose if

you want to compare these three cases thatis the three modes of operation, then we can now slowly try we can now attempt to findthe rtd curve for these three operations. so, suppose let us start with the perfectso let start with the first case of perfect cstr, so we tagged we used the symbol p forthe perfect cstr or operation of the cstr in a perfect mode that is there is no bypassand there is no dead zone which means that this cstr is actually an ideal cstr.and the volume and the volumetric flow rate are basically the measurable quantities ofa real reactor; let us say we know of the volume and the volumetric flow rate with whichthe fluid is actually flowing inside the cstr. so, we know of the residence time distributioncurve and e of t versus t and so this starts

at 1 by tau and then it actually decreaseswith time. and then we know the corresponding f curve that is an exponential increasing function and then it goes all the way up to 1. so, this is the f curve and the e curve for acstr. so, now if the space time of the reactor tau is that is very large, then the decayof this exponential curve and the corresponding concentration curve is actually going to beextremely slow which means that is the space time is large, then the tracer actually spendsa lot more time inside the reactor. and, therefore, the decay of this e curveas a function of time is going to be very slow. on the other hand, if the space timeis very small then the amount of time that the tracer spends inside the reactor is goingto be very small. and, therefore, the e of

t and the time curve is going to have a sharpslope at small times which means that it is going to decay faster. so, now let us lookat the second case of bypassing cstr operation under bypassing conditions. so, suppose we look at suppose we considera cstr where bypassing is known that bypassing is present. so, we refer to that as bp. ifv b is the volumetric flow rate of the fluid which is actually bypassing the reactor; sothat is the bypass volumetric flow rate. so, v b is the bypass volumetric flow rate andlet us assume that v s b is the volumetric flow rate which is the actually going throughthe system volume. so that is that enters the system volume. so, therefore, v naughtwhich is the volumetric flow rate with which

the fluid is actually entering the reactorshould be equal to v b plus v s b. now if we assume that v s is the volume ofthe tank, then we know that v s b is actually less than v naught because it is only a fractionof the total volumetric flow rate with which the fluid is actually being pumped that goesinto the reactor. so, therefore, clearly tau s b should actually be greater than the spacetime of the reactor which means that the amount of time that this fraction of the fluid whichis are not being bypassed the amount of time that it spend inside the reactor is actuallylarger than the actual space time of the reactor itself based on the overall volumetric flowrate. so, remember that tau is actually definedas v by v naught where v and v s is the volume

of the reactor and v naught is the volumetricflow rate with which the fluid is actually flowing through the reactor. so, therefore,because this tau s b is greater than tau, the ct the concentration curve and the e curve,they are going to decay very slowly. so, as we observed a few minutes ago as we actuallydiscussed a few minutes ago because the resident time of the fluid stream which is actuallygoing through the system volume is actually larger than the space time. the decay of theconcentration and the e curve is going to be slower when compared with the case of aperfect operation that is there it is no bypassing, okay. so, in a similar fashion, now we canactually look at what is the possible residence time distribution under this condition. sowhat is possible?

so, various possible residence time distributionshave been considered and one of the possible distribution is that it will be v b dividedby v naught into delta t minus 0; that means this is the component or fraction which isactually bypassing the reactor and leaving the fluid stream very soon after it actuallyenters the reactor that plus v s b square dived by v into v naught that multiplied byexponential of minus t by tau s b. so, that is a possible residence time distributionfunction that actually describes the residence time distribution of a cstr which is actuallyoperated along with a bypassing of same part of the fluid stream that enters the reactor.so, now the system can actually be depicted in the following way. so, suppose if thisis the cstr and this is the inlet stream and

if there is a bypassing of the fluid streamin the cstr, then this can actually be depicted in the flowing cartoon. so, suppose if hereis a tank and if let us say that the inlet fluid stream is actually split into two partswhere there is a bypass component v b which actually goes and directly joins the exitstream and only fraction of the inlet fluid stream actually which is v s b the volumetricflow rate v s b actually enters the cstr and participates in the same volume which is available,otherwise, and leaves the reactors. so, this is the s p and it leaves the reactor,otherwise, same volume; the amount of the volumetric flow rate of the tracer that actuallyenters the reactor is v s b and then v s b is what this it actually leaves if we assumethat bypass is essentially taking some of

the volumetric flow rate and directly joiningit with the effluent stream. so, this kind of a distribution curve essentially capturesthis representation. so, now if i attempt to sketch the e curve; so right at t is equalto 0, there is going to be a fall in the e curve. and this is because some fraction actuallygets bypassed and directly goes and joints the effluent stream, and therefore, thereis going to be a sharp fall in the e curve that is the fraction that is actually leaving.and then after which there is going to be a exponential fall in the e curve. so, thisfirst part corresponds to the bypass and the second part corresponds to this exponentialterm, and this location here is essentially given by v s p square divided by v into vnaught. so, that is this location from where

the exponential fall in the e curve actuallystarts. now the corresponding f curve will be, so if this is the corresponding f curve,then we will see that the f curve actually has a jump right at t equal to zero. so, thisis one; this is the f curve. so, there is a jump right at t equal to zero and the jumpactually occurs up to v b divided by v naught which corresponds to the bypass the fractionof the inlet volumetric flow rates the fraction of the inlet stream which actually gets bypassedand leaves the effluent stream immediately. and that is reflected in the f curve and alsoin the e curve. so, the important message from here is that an e curve and f curve ifactually measured experimentally can indicate whether there is a bypass in such kind ofa system. so, the third mode of operation

is what happens if there is dead volume whichis present inside the tank reactor inside the cstr. so, if there is a dead volume inside, so letas assume that it is a cstr; there is no bypassing. it is assumed that there is a no bypassinginside the reactor and then there is some dead volume. let us say that the volume ofthis dead zone is essentially v d, and this is dead volume essentially is one is the volumewhere the fluid stream does not reach that location and so it is the presence of thislocation is of no use for the performance of the reactor. so, the overall volume isactually equal to v d plus v s d where s d is the available volume which is actuallyaccessible by the fluid stream which is flowing

into the reactor.so, now this can actually be depicted as so there may be some zone below which is actually a dead zone where the fluid stream actuallydoes not access this location and then there may be an exit stream effluent stream throughwhich the fluid that enters actually leaves the reactor. so, the e curve for this particularsituation would actually look like it starts at 1 by tau s d. so, remember that the accessiblevolume is v s d which is smaller than the actual volume of the reactor, and for thesame flow rate if it was conducted under perfect conditions that there no dead volume, thenthe tau s d which is the space time of the reactor for the fluid stream which is actuallyaccessing the non dead volume space in the reactor.so, that will actually be smaller than that

of the actual space time of the reactor vby v naught. and as a result, the exponential curve the c curve and the e curve is goingto decay faster than if it were to be conducted under normal perfect operation and the correspondingf curve would be. so, the f curve will also will be correspondingly steeper and it willactually slowly increase and go to 1. so, now if we put all these three together, letus make a comparison of the rtd functions for these three modes of operations. so, suppose let us draw the e curve. now fora perfect operation, the e curve starts at 1 by tau which is the tau is the space timeof the reactor and then it is exponential decay as a function of time; this is the ecurve. now is there was bypassing in the reactor,

then right at t is equal to zero, there willbe a sharp fall in the e curve and then followed by an exponential decay with respect to time.now supposing if there was dead volume inside the reactor supposing if this correspondsto the perfect operation, this corresponds to bypassing; now if there were to be decayif there were to be dead volume inside the reactor, we just observed we just noted afew moments ago that the decay of the e curve is going to be significantly faster.and, therefore, the curve starts above 1 by tau because tau s d so this starts at 1 bytau s d. and we said that tau s d is actually smaller that of tau, and, therefore, 1 bytau s d is going to be larger than 1 by tau and then it start's from here and the decayis actually faster than that of the perfect

operation, because tau s d is actually smallerthan the space time of the reactor. so, now if there is a real data for a tank reactor,then one can actually look at one can compare the actual rtd function measured experimentallywith the rtd functions present in this graph, and that can actually provide a clue as tocompare that with the perfect operation that can provide a clue whether there is a bypassor if there is a dead volume present inside the reactor.now similarly we can actually plot the f curve. so, for a perfect operation that is the kindof behavior that is the perfect operation and then for bypassing there is a jump rightat t equal to 0; that is the kind of behavior for bypassing and the bypassing the curve starts exactly at v b by v naught and then for dead volume case,

the curve actually increases rapidly and thenit reaches 1. and so this is for the dead volume case, and this is for the perfect operationcase. so, either e curve or the f curve can actually be used to detect what is the diagnosisif there is any problem in the operation of that particular cstr.so, therefore, the recipe is that if the volume of the reactor and the volumetric flow ratesare known. so, volumetric flow rare can actually be measured and v is the volume of the tank.so, one can actually compare e t and f t f curve of ideal reactor. so, ideal cstr andthat is the perfect operation, and you can compare that with experimentally measurede and f curve. so, i put a subscript e for experimentally measured. so, one can actuallycompare the experimentally measured rtd functions

with the rtd functions of the ideal reactor,and that can be used to diagnose the presence of the bypassing or the presence of dead volumeinside the cstr. so, next let us look at these three operations for a tubular reactor. so, let us assume that it is a plug flow reactor.so, the first case is perfect operation. so, let us tag that with a p. so, here is thetube, and there is a fluid which is actually flowing at a volumetric flow rate of v naughtand the volume of the plug flow reactor is v. so, now the rtd curve we know that it isa delta function centered at the space time of the reactor. so, that is e of t, and thenthe f curve is essentially given by it starts at tau and reaches 1. it is a step functionin the f versus t plane. so, that we already

know. now what happens in the bypassing case? so, suppose if there is a bypass in the reactorsuppose if there is bypassing in the reactor, so if i tag that with b p, then that can actuallybe depicted as suppose if v naught is the volumetric flow rate with which the fluidis supposes to enter the reactor and if this is the reactor and that is the final effluentstream volumetric flow rate, then a fraction of the fluid is actually bypassed and so wecan represent that using by actually taking some part of the feed and the directly connectingit to the effluent stream. so, that is the depiction of the bypassing in the reactorand v s b is the volumetric flow rate with which the fluid it is actually flowing throughthe reactor.

now the residence time distribution for thiskind of a system can actually be written as will actually have two peaks. so, the firstpeak will actually appear very close to time t equal to zero, and this is because of thebypassing of the fluid and then there will be another peak which will appear much laterand that is because of the fluid that is actually flowing through the plug flow reactor. andthe residence time the space time of the reactor will actually appear somewhere in between.so, this is the e curve, and so the first peak is due to bypassing. the first curveis actually due to bypassing, and the second curve is actually due to the material throughthe reactor. that is because of the material that is actually flowing through the reactor.now why is there a delay in these two peaks

or why is there a delay in the residence timefor the material that is flowing through the reactor. the delay is because the tau s bwhich is the space time based on the fluid that is actually flowing through the reactoris actually larger than the space time of the reactor based on the overall volumetricflow rate that is actually expected that is the actually flowing through the reactor.now, one can actually sketch an f curve for the same f of t. so, that starts at v b by v naught that goes to one and this is tau and this is tau s b. so, remember that this secondpeak is actually centered at the space time based on the volumetric flow rate of the fluidthat is actually flowing through that is the actual volumetric flow rate which is accessibleto all parts of the reactor.

so, now let us look at the third case of plugflow reactor with dead volume; we tag it with d v. and suppose if the volume of the fluid volume of the reactor which is actually not accessible to the fluid is given by v d, then the total volume isv d plus s d. so, this is the dead volume. so, that is the dead volume and typicallythis happens because there will be recirculation of the fluid at the entry locations in thereactor and that causes the inaccessibility of those regions for the fluid stream andthat can virtually be called as the dead volume inside the reactor. so, that can actuallybe depicted as. so, this is the volume of the reactor v naught v s t that is the deadvolume which is removed from the reactor and that is the volumetric flow rate with whichthe fluid is actually entering and leaving

the stream.so, now in this case the tau s d which is the space time based on the volume of thereactor which is actually accessible for the fluid stream that is given by v s t by v naughtand that will be less than v by v naught because v s t is smaller than the volume smaller thanthe total volume of the reactor which is and v by v naught is nothing but the space time.so, in this case, the tracer will actually leave the reactor early because some partof the reactor is actually inaccessible and so the space time is actually the actual spacetime the space time based on the volume which is available in the reactor is actually smallerthan the actual space time of the reactor. and so the e curve would actually look likethere will be one delta function and that

will actually be centered at tau s d and thespace time will be much later that will be the e curve. and similarly, the f curve willbe, so that is the e curve and the f curve for a plug flow reactor with a dead volume.so, what we have seen in the this lecture is looking at the different rtd functionsfor different operation of a cstr and different operations of a plug flow reactor, and wewill continue from here in the next lecture. thank you.