 So, what we are saying here is that one in one case we have relation for effectiveness in terms of ratio of specific heats and NTU and in the other we are saying that NTU is a function of epsilon and C that is what if I write it for NTU what method it is what method it is if I write equation for NTU epsilon is a function of temperatures. So, if I know inlet and outlet all temperatures I know effectiveness I know C. So, NTU that means NTU approach if I write in equation for NTU actually it is LMTD approach is not it is a LMTD approach why because epsilon is having one of the outlet temperatures remember that. So, if I so what that is what professor was trying to say if you have epsilon NTU you can get LMTD approach also from LMTD how can you get to NTU approach we will see that little later on. So, now let us quickly take the figures and tell you want to have you let us take one of these figures. So, what is this figure telling me what is this figure telling me I am taking a parallel flow heat exchanger and before I go to the parallel flow heat exchanger let me take all heat exchangers in one this is the mother of all figures actually. So, this is effectiveness and NTU. So, this is effectiveness is on the y axis and NTU is on the x axis now if the this plot has been shown for C equal to 1 that is ratio of the specific heats equal to 1 that is mass flow rates into specific heat of hot equal to mass flow rate and specific heat of the cold. You see here this is the point I was trying to strike effectiveness for parallel flow is very less and effectiveness of counter flow is very high and it is counter flow you see it is increasing with the increase of NTU, but for parallel flow it becomes constant after certain NTU we do not gain much by increasing the area in parallel flow heat exchanger that is what it means in case of parallel flow. But all other heat exchangers which we are going to build are going to be in between these two here we have just plotted cross flow with both fluids and mix you plot any other heat exchanger they are going to lie in between these two that is the crux that is the crux that is the reason why in any heat transfer textbook or any heat transfer teacher is going to teachers only parallel and counter flow why because once you know counter flow you know that is the best and once you know parallel flow that is the worst. So having understood this concept now let us take up any heat exchanger what is happening here for a if c equal to 1 this is effectiveness this is NTU I have just taken parallel flow what is happening with the decrease of c what is happening with the decrease of c c is with the decrease of c min by c maximum this ratio is c min by c maximum when c min by c maximum equal to 1 it is very low when c min by c maximum equal to 0 it is very high why for c min by c maximum it is very equal to 0 it is very high what is that heat exchanger actually in which case C min by C maximum equal to 0 two phase flow heat exchanger that is. That means on one side the heat transfer coefficient is extremely high either h i or h o why because it is two phase. So, C maximum is infinite that is that is what we took know there is no temperature gradient for that case where there is no temperature gradient where the heat transfer is taking place thoroughly by latent heat then we have C min by C maximum equal to 0 that is where we have maximum heat transfer naturally two phase flow heat exchangers are more effective than the single phase heat exchanger for a given NTU if I go on decreasing C min by C maximum I am going this is like going from single phase to two phase flow heat exchanger that is the reason why effectiveness is increasing and the ultimate limit is this why because this is two phase flow heat exchanger. So, that is another reasoning what one can give and this pattern is true for any counter flow you take this is true and you take shell and tube heat exchanger this is true and you take any other heat exchanger you are going to get the same pattern. So, these are the points which we need to emphasize and and C equal to 0 effectiveness is equal to 1 minus of e to the power of minus 1 NTU this is what I have told you C equal to 0 C max tends to infinity that is what is going to get for condenser and boiler that is the case that is here what I had shown for C min by C maximum equal to 0. So, with that we will move on to if C is equal to 1 effectiveness is the lowest. So, we should the point is we should try to avoid the keep to keep the same products same products we should try to avoid there should be the difference that is because that will create more temperature difference that is other way of looking at it it will create more temperature difference that is what professor Arun was insisting if you create more temperature difference there will be more heat transfer you can create more temperature difference by having more difference in the mass flow rates and specific heat. So, that is what is being mentioned here in all of these transparencies whatever I told and these are the equations which are given which you can all these derivations are given in case and London for all the cases if someone is interested in these derivations they can go back and derive from fundamentals. But definitely for a UG parallel and counter flow one has to do and these are problems I am not going to solve any of these problems I would request you to sit down and solve these problems they are all simple plugging type problems and all that I want to tell only two more points I will tell and sign off from this is that one is can I use LMTD for NTU approach that means if I know the that is can I use LMTD approach for NTU that means for a given heat exchanger and inlet temperatures known can I compute the area using the LMTD approach can I compute yes how yes here my student says yes we can do and he says that we can do it iteratively what he is saying is that if I have a parallel flow heat exchanger I have to takes elemental small small DAS and for each elemental area you apply LMTD approach you apply you apply LMTD approach you apply LMTD approach and then you can you can go that way and compute the area at the exit at which your th0 is equal to the th0 what has been given in the problem. So LMTD piecewise if you do you can apply it for even NTU approach so earlier days when LMTD approach was found there was no computer they had to manage with simple grasp that is why case and London came up with that NTU approach in today's world both the approaches for all practical purposes they are same there is nothing different in real life heat exchanger design people simply go ahead with either LMTD or NTU based in the convenience approach is not a major issue because in today's computer either you will write a MATLAB code or an excel sheet calculation where in which you will take a small strip and go on strips and you will go on integrating them so it does not matter whether you are using epsilon NTU approach or LMTD approach that is another thing that is one point another point what I want to emphasize is which we did not emphasize the pumping power heat transfer heat exchanger is not designed on the basis of only heat transfer coefficient so in the problem in one of the problems I said oil I said you put a pump but pump does not come just like that free especially if I have to design if I had to decide a pump and fix a pump that pumping power will be in kilo watts or even mega watts okay. So if that is in a condenser I have seen in Tarapur nuclear power plant condenser which we usually neglect in thermodynamic condenser power is very less we neglect but that is actually around 5 to 10 mega watts in Tarapur nuclear power plant and that is as big as our mechanical engineering building that is three storey building what they mean in thermodynamics is when we neglect that pumping power of condenser is compared to the power you generate in the complete power plant which is of the order of 800 or 900 mega watts compared to 800 mega watts you can probably neglect that 3 or 5 mega watts but 3 or 5 mega watts just like that it would not come you have to build it so when you are designing a condenser you have to be worried about the pumping power also so however small it is pumping power makes lot of difference lot of influence that is why in the design of heat exchanger always the quest is for increasing the heat transfer coefficient and decreasing the pumping power and it is not easy to come by nothing comes easy you have to spend something I mean nothing comes free you have to spend something the heat transfer coefficient always increases with the increase of friction factor or the pumping power not as much as the as much as you increase the pumping power in fact the increase in heat transfer coefficient is always lesser than the pumping power you spend okay so with this I think I will we will come to end of heat exchanger and I know there are lot of questions but I need to wait for questions what I will do is next or we will take some questions now in next 15 20 minutes I had postponed yesterday in radiation see we had found that emissivity of nonconductor I am switching gears and coming to radiation that is number one number two I had promised that even in coordinators workshop that we will try to understand through electromagnetic wave theory we will try to understand and explain why emissivity is or how are they why nonconductor emissivity is greater than conductor emissivity one of the questions from Anna university also was there I am sure all of you will have this question why emissivity of a nonconductor is higher than the emissivity of a conductor number one and how does emissivity vary with temperature so it is very difficult to answer these questions until I take the recourse of electromagnetic wave theory. So here in next half an hour I do not pretend that I have understood everything of this I am just trying to see how in a very simplistic fashion take the laws which are available in light and electromagnetic I will just state those laws I am not going to derive them by stating those laws how can I compute reflectivity from reflectivity how can I compute emissivity from which I can infer absorptivity so these are the questions for I will do for conductor and a nonconductor and that too I will do it for non magnetic material if you take a magnetic material things are going to be complicated in which way they are going to be complicated I will let you know as we go along. So next half an hour I want all of you to write equations what I am writing here including you I want all of you to write the equations along with me they are not very lengthy but why I am saying you need to write with me because you will be comfortable when you write these equations along with me okay so let me get started okay so what am I saying is electromagnetic wave theory that is what is the heading you can write if you want to write in your notes what is that we are saying is there is complex index of refraction complex index of refraction refraction not reflection refraction that is m equal to n minus ik that is it is imaginary n is real part k is imaginary it is a complex number not imaginary this is a complex form this is a complex form let me where n is refractive index and k is absorptive index why it is negative because as an electromagnetic wave starts radiating through a medium it its energy is getting absorbed that is why this negative sign this negative sign is coming that is m equal to n minus ik okay now let me take two non absorbing media because I cannot take both the things both complexities in one go I will take two non absorbing media if I take an interface if I take an interface between two non absorbing media that means k is 0 I have only non absorbing non absorbing media if I take non absorbing media okay I have to move on to the next page okay so what do I get I this we have studied in plus two but still for the sake of completion this is medium one which is having refractive index n one this is medium two which is having refractive index n two and I have chosen such that n two is greater than n one okay and now if electromagnetic wave or light goes we all know that this is the angle theta one with respect to normal it undergoes refraction we all have studied this why spoon looks bent and all we have studied so we I do not have to go to those examples for you in plus two we have studied so this is theta okay this is theta two okay this is theta one okay so what is Snell's law telling us Snell's law Snell's law was found was stated in sixteen twenty one sixteen twenty one what did he say he derived from fundamentals that n one by n two equal to sin theta two by sin theta one okay where theta one is angle of incidence and theta two is angle of refraction theta one is angle of incidence and theta two is angle of refraction okay now I got how I can relate the refractive indices now next to question is from what are we looking for we are trying to look for transmissivity emissivity and absorptivity now I will try to relate or from Maxwell's relation that is by invoking Fresnel's relation as I said I am not I am only stating these relations I am not deriving them Fresnel's relation I get reflectivity as a function of theta that is tan squared theta one minus theta two upon tan squared theta one plus theta two plus sin squared theta one minus theta two upon sin squared theta one plus theta two now what is transmissivity there is no absorptivity okay so now transmissivity should be equal to one minus reflectivity okay in an on absorbing medium I can quantify now reflectivity and transmissivity which is dependent on angle of incidence and angle of refraction that gives me the refractive index that means this can be expressed in terms of refractive index also it is not very difficult for that I have to invoke Brevestor's law let me not do that I am not going to do that so if we invoke Brevestor's law I can get this expression in terms of N1 and N2 for absorbing media I am going to do that now next is let me take now let us get to real life that is let me take a case where in which I have a medium that is we are interested in solid plate let me take a solid plate that is this is absorbing medium I am taking this as absorbing medium this is absorbing medium and this is air we all know it is not having any absorptivity and refractive index is also equal to one that means this is non absorbing medium air from air electromagnetic that is what happens in all our applications of radiation so that is what I have taken so here for this absorbing medium the index complex index is given by m equal to n minus ik where n is refractive index k is absorptive index so again I have theta 1 and I have theta 2 this is theta 1 now without deriving again I am going to state this rho n lambda that is reflectivity in the normal direction at a given wavelength lambda is given by n minus 1 whole squared plus k squared upon n plus 1 whole squared plus k squared okay what I am trying to say is without deriving what I am trying to say is reflectivity in an non absorbing medium was a function of refractive index in an absorbing and refractive medium it is going to be function of refractive index and absorptive index if I measure refractive index and absorptive index of every material I should be able to compute the reflectivity if it is now how do I get now if I take a solid plate I said I am taking a solid plate for a solid plate what would be the transmissivity it is opaque it is opaque if for a copper plate or a silver plate or an aluminum plate that is what I am going to do calculations for silver plate I am going to do the calculations and I am going to do for silicon carbide for magnesium oxide I am going to do for reflectivity and then show that non conductor emissivity is higher than conductor emissivity that means for a copper plate transmissivity is 0 if transmissivity for a copper plate is 0 what will be my absorptivity absorptivity will be given by 1 minus reflectivity this is for opaque medium this is for opaque medium let us remember that now I have to head for bigger equation now how do I get this n and k next question is how do I get this n and k who gives me this n and k who is going to decide this n and k or what is it this n and k dependent upon that is the question that can be obtained from electrical permittivity and magnetic permeability I am first again we go always from simple to complex I am taking a non magnetic material please note my examples what I told silver copper aluminum they are all non ferrous they are non magnetic materials that means my magnetic my magnet cannot attract them okay so those non magnetic materials will not have any magnetic permeability so in that case in that case for non magnetic material what am I up to I am trying to figure out how can I predict n and k from electrical permeability or electrical permittivity for non magnetic material for non magnetic material n squared equal to half upon little lengthy you have to be patient epsilon prime this is not emissivity this is permittivity in but primes we are using please note that it is not emissivity do not get carried away we have not reached emissivity yet it is distant plus square root of epsilon prime squared plus epsilon double prime squared what is this epsilon prime okay this is epsilon prime and this is epsilon double prime okay what is epsilon prime given by epsilon prime equal to electrical permittivity do not worry I will explain electrical permittivity also within a minute for now you take my word electrical permittivity in a given medium and electrical permittivity of that material in vacuum that is epsilon not okay that is let me write that epsilon prime epsilon is electrical permittivity electrical permittivity okay epsilon not is same electrical permittivity of the same material but measured in vacuum vacuum vacuum okay. Now epsilon double prime what is this epsilon double prime dependent on epsilon double prime is given by sigma e lambda not upon 2 by c not epsilon not in fact if you can relate this n with earlier already I have told n has this n come anywhere else earlier c equal to c not by n I had told that is refractive index the same refractive index we are talking about here okay so now you can relate why did I write c equal to c not by n there okay. So what is the sigma e sigma e is electrical conductivity mediums electrical conductivity okay that is epsilon double prime I think other things we can understand what is lambda not lambda not is the wavelength of the electromagnetic wave in vacuum and c not is the velocity of light in vacuum 2.998 into 10 to the power of 8 meters per square meter meters per second epsilon not already I have told so we know now epsilon prime and epsilon double prime still if someone does not know electrical permittivity let us go and understand that epsilon prime okay question asked this by professor Arun is what is epsilon prime and what is epsilon double prime epsilon prime is a normalized electrical permittivity that is electrical permittivity of a given material divided by electrical permittivity of the same material measured in vacuum and what is epsilon double prime it is a measure of electrical conductivity it is a measure of electrical conductivity that means electrical conductivity effects are this second term and what is this electrical permittivity is the first term okay. So refractive index is a function of electrical permittivity and electrical conductivity now let us see what is electrical permittivity okay getting bored I do not know maybe it is for me it is quite interesting because I am learning this for the first time so what is electrical permittivity this we have studied in fact in electrostatics okay when we have two charges I am trying to define what is permittivity if I have two charges and if they are the distance between the two charges is r the force Coulomb's law force equal to the force between these two charges is equal to which is having a charge q 1 and it is having a charge q 2 q 1 q 2 upon you see what all last we are invoking we went to light we went to electromagnetic and now we are getting to electrostatic we need everything whatever we study in plus two now only I am realizing whatever we study in plus two is required for our lifetime we cannot afford to forget everything is getting connected everything is one we are the ones who have divided it because for the sake of learning okay so q 1 q 2 upon 4 pi epsilon r square okay same Coulomb's law if I apply in vacuum if I apply in vacuum if I do the same thing in vacuum if not in a medium the same Coulomb's law if I apply what should I get a different force F naught but what would be different q 1 q 2 I have kept the same charges okay same charge that is 4 pi epsilon naught r square this is F by F naught gives me what what does F by F naught gives me epsilon by epsilon naught now you would have understood how one would measure electrical normalized electrical permittivity yeah F by F naught is okay let me correct it this way because I want epsilon by F naught by F that is this is F naught by F if I do I get epsilon by epsilon naught okay so in the previous thing what did we say there is electrical permittivity and there is electrical conductivity that is N now let me go ahead if you are impatient what are why are we doing what are we doing let us calculate let us calculate go ahead and calculate for silver reflectivity reflectivity okay so what do we do let me take silver let me take silver and let me choose a wavelength because all properties all these permittivity electrical conductive electrical conductivity electrical permittivity all our functions of wavelength remember that and temperature and temperature epsilon 6.2 and I take room temperature I take room temperature so epsilon naught equal to 3.4 per silver and epsilon prime is given by epsilon prime is given by epsilon naught minus of nu p square I am giving a new relation I am not using the old relation which I had put okay nu square plus gamma square this is coming from because this is given by different theory drood theory this is what I am presenting is a theory by name drood theory that is epsilon naught minus of nu p square what is nu p nu p is the plasma frequency plasma frequency is the frequency with which the electron vibrates at cold conditions that is plasma frequency and gamma so that is calculating the n and k and from n and k I am supposed to get the reflectivity and from there I am supposed to get the emissivity okay this I will do both for conductor and non conductor and I have taken in a conductor non magnetic material why because I want to rule out magnetic permeability so nu nu p gamma is called as oscillation damping factor you do not have to worry if you are not understanding these things there is a relation so if this is 3.4 and nu p for silver is given to be 2.22 2.22 into 10 to the power of 5 and gamma is given to be 4.3 into 10 to the power of 12 how do I get nu how do I get my nu who gives me nu nu equal to C naught by lambda is not it so C naught by lambda that is 2.998 into 10 to the power of 8 lambda do we know yes we have taken a particular frequency particular wavelength 6.2 micro meters 6.2 into 10 to the power of minus 6 I get nu equal to 4.84 into 10 to the power of 13 hertz that is nu if I substitute all of this if I substitute all of this I get nu prime as minus 2 0 0 8 4 this is nu prime okay next I need epsilon double prime this is nu prime okay now I need epsilon double prime remember in my why am I doing this I need n n is a function of epsilon prime and epsilon double prime I have calculated epsilon prime I need epsilon I have calculated epsilon single prime I need epsilon double prime epsilon double prime is given by attack frequency nu p squared gamma upon nu into gamma squared plus nu squared if you substitute that you get 185.1 and remember we had got epsilon prime as minus 2 0 8 4 what is n n squared equal to half into epsilon prime plus epsilon prime squared plus epsilon double prime whole squared if I substitute these values I get n squared as 4.1 n equal to 2.03 similarly k squared equal to half into epsilon prime plus square root of minus this is minus epsilon prime whole squared plus epsilon double prime whole squared equal to I get 2 0 8 8 that means I get a k of 45.7 that means I have got refractive indexes 2.03 and absorptive indexes 45.3 having got what is that next thing to get I have to get the reflectivity normal reflectivity I have already given this relation if I am right did I get this relation n minus 1 I have given already for an absorbing medium I gave you this relation n minus 1 whole squared plus k squared upon n plus 1 whole squared plus k squared that is what I am going to invoke now that is n minus 1 whole squared plus k squared upon n plus 1 whole squared plus k squared this is plus note this is plus and this is minus. So, if I substitute this n here and k here I am going to get rho n lambda as 0.996 you see reflectivity is so high what is absorptivity transpecivity is 0. So, absorptivity is equal to emissivity is equal to absorptivity at the same temperature when I write that it is at the same temperature it is equal to 1 minus rho n lambda that is equal to 0.004 how do I know that whether my reflectivity calculation is right or wrong experimental rho n lambda experimental is 0.98. So, our theory has now electromagnetic wave theory has now given as a calculation for reflectivity. So, what is that why is the conductor's emissivity less because typically reflectivity of a conductor is high. Now, let us do the same calculation for summary is high reflectance because of which emissivity is less for conductor that is what we understood. Now, let us do the same thing for let us do the same thing for a non conductor let me take a non conductor what is a non conductor I am taking the non conductor I am taking here is I am taking a non conductor let us say either silica let us take silica or magnesium oxide each once refractive index is 3.41 and 1.7 for a wavelength range of 3 to 10 micrometer for silica and 1 to 10 micrometer for magnesium oxide. Now, what is the reflectivity reflectivity we have already given n minus 1 whole squared plus k squared upon n plus 1 whole squared plus k squared it. So, happens that for non conductor for non conductor the k is very very less we can neglect k for non conductor that is it is absorptive index is 0 that means there is no absorption taking place in non conductor why because there are no free electrons why because there are no free electrons it is very no I think I am not answering it I really do not know why the absorptivity is why the absorptive index is less for non conductor yes let me not go overboard and answer that, but it is experimentally observed that absorptivity index of non conductor is very less if the absorptivity is taken as 0 and if I take n for silica what do I get for silica I get rho n lambda for silica I get as 0.7 and for magnesium oxide I get 0.07 what is the emissivity for silica I get it as 0.7 sorry this is 0.3 and emissivity is 0.7 that is emissivity of silica is 0.7 and emissivity of magnesium oxide is 0.93 can you see now reflectance is for non conductor it is for non conductor very, reflectance is less very less because of which emissivity is high low reflectance for non conductors is the reason for high emissivity for non conductor that is the only way had I not taken the recourse of electromagnetic wave theory I could not have answered your question why is the emissivity of a non conductor higher than the emissivity of a conductor that is the one question I have answered. So now if magnetic material is there what will happen I will just take only 5 more minutes on this issue and magnetic material is there what will happen if there is a magnetic material m continues to be n minus i k no doubt about that but n and k are going to be functions of both electrical permittivity and magnetic permittivity magnetic permeability n squared minus k squared equal to epsilon mu c naught squared n k equal to sigma e mu lambda naught c naught upon 4 pi what is mu what is mu mu is magnetic permeability magnetic permeability that is mu and this is epsilon please epsilon epsilon now it is electrical permittivity and sigma e is what is sigma e what is sigma e electrical conductivity we have told that and c naught is the velocity of the electromagnetic wave in vacuum and lambda naught is the wave length in vacuum. So what it means is that n and k are dependent both on electrical permittivity magnetic permeability and electrical conductivity 3 parameters n and k are functions of functions of electrical permittivity magnetic permittivity and permeability and electrical conductivity. So these 3 properties decide my n and k which in turn decide my reflectivity which in turn decides my emissivity that is all we have shown I know I have not derived any relations all basic laws from optics and electromagnetic wave theory I have straight lifted and presented here this is a very what to say very humble attempt just to explain how emissivity is vary or how difficult it is to explain the emissivity variations with emissivity with wave length with temperature and with material ok. So what I would suggest to you is why do not you go ahead and calculate why do not you go ahead in the same lines what I have done why do not you please calculate yourself so that if you do the calculation once you will be comfortable why do not you calculate once for two other materials that is homework what I am giving on this is it is not here and forget I want you to calculate that is for aluminum for aluminum and copper nu p is 3.07 plasma frequency copper is 2.25 hertz gamma oscillation damping factor 31.24.55 epsilon experimental is no this is reflectivity reflectivity is reflectivity is 0.99 for aluminum experimental and 0.97 when you calculate you will get the values closer to this ok. So why do not you and you calculate this at the same lambda 6.2 micrometer and room temperature of course temperature is not coming to picture anywhere but all this nu p gamma rho are all going to be nu p and gamma are going to be functions of temperature and wave length ok. This is the homework which you are going to take for calculating nu ok. One last word before I sign off on this electromagnetic wave theory business is how is emissivity going to be a function of temperature emissivity is found to be function of temperature in this format 0.776 T upon sigma e to the power of half minus of bracket open 0.309 minus of here I am writing 0.089 log of T by sigma e bracket closed 0.077. T by sigma e minus of 0.0175 T upon sigma e to the power of 3 by T ok typically here yeah ok. So now it becomes complex what is that nothing complex all that we are saying is emissivity should increase with temperature at one point that is one thing at the same time there is a dependence on what is this sigma e electrical conductivity temperature with the increase of temperature electrical conductivity decreases at least for metal ok. So emissivity increases with the increase of temperature at one point and electrical conductivity decreases with the increase of the temperature. So it is the interplay between these two for different material which decides whether emissivity increases or decreases with the temperature in most in most of our graphs in emissivity with temperature somewhere it was increasing somewhere it was decreasing that is because of the interplay between the temperature and electrical conductivity. This is a whatever I have taught now is not complete answer for all the questions. This is an humble attempt just to give the feel how difficult it is to explain the emissivity absorptivity transmissivity reflectivity property variations with temperature and material that is all it is just an humble attempt do not take this what to say to the T means it will differ for different material it is restricted to the assumptions which for which they are derived for each law is derived for that is what with that I think we will move on to next half an hour is for question answers we will spend time on question answers. VN-90 Nagpur in the topic heat exchanger the you have considered one solution for in the numerical the hot fluid is flowing through the tube and cold fluid is flowing over the tube and if the situation is reversed the cold fluid is flowing through the tube and over the tube the hot fluid is flowing what change in the heat transfer will be occurring and by what means that is occurring. See you can sit down and calculate no see if I change flip hot and cold said what will it change what will it change yes if I take the problem in principle you can sit down and do that I can say that you please do it yourself but I will not do that let me attempt to answer that question let me go to the numerical and think allowed the question asked is in the numerical which is solved one fluid cold fluid I do not remember the cold fluid is flowing one side and hot fluid is flowing on the other side if I flip them that is in this problem in this problem hot fluid is flowing on the annulus that is the shell side and cold fluid is flowing in the tube side what change I will get what change I will get if I flip these two fluids properties no no property change properties would be same and Reynolds number should be different Reynolds number should be different what will happen nothing will happen in principle you will more or less get the same numbers that is h i again will tend to become laminar in the tube side and h o on the outer side will be turbulent only so in principle you will be ending up of course the length would be slightly different this length would be slightly different because h i and h o will get slightly differed but in principle length will be of the same order is that ok professor if you put hot fluid on the outside then your insulation has to be far better correct right that is that is why you tend to put wherever possible cold fluid on the outside because heat is going from hot to cold now if I put hot fluid on the outside it has to go into the cold and also it will escape out radially so it will go in two dimensions two directions outward radially inward radially it is not cold really it is 30 so it is near ambient condition ok sir helical coil is arranged in a shell through the coil hot water is flowing and over the coil cold water is flowing that is in the shell ok what change in the heat transfer will be occurring if the situation of flow of the two fluids is reversed ok I will not question is I have a helical coil immersed in shell and tube heat exchanger shell and tube on shell side one fluid flows cold fluid on tube side hot fluid flows now what change it would occur if I flip this whatever change I explained it to you for the example whatever I quoted same effect would be there here also so there is no change so I would request you to sit down and do the calculation for the numerical what I have done you will appreciate the answer for this problem I know you are not happy with my answer but that is the answer other thing is that probably fouling effects will come in ok so when fouling effects come in you will have to think of how to clean them off so based on that consideration also you will put which fluid goes in which fluid goes out but that is secondary there is basically there is not much of a difference in terms of concept in applying things ok J N T U Hyderabad sir in gas turbine cooling technique what is meant by film cooling and what is meant by imbendment cooling that is my first question and next question is out of these two which one will give the better heat transfer coefficient actually this question has been asked in the question is in gas turbine blade cooling what is jet impingement cooling what is film cooling professor I have answered this in great detail in moodle in fact jet impingement cooling gives you the higher heat transfer coefficient compared to film cooling there is no doubt about it because jet is impinging directly film cooling is like cooling a potato I have taken this example already we cool the potato while peeling the potato by keeping the potato under water by making a film of cold water around the potato that is what we do in gas turbine blade also out of these two jet impingement heat transfer is the most effective one so much more detailed way I have answered this in moodle already there are around 5 to 6 replies for this question ok professor any other questions please folding factor how what is the maximum range of folding factor in heat exchanger how it affect with respect to heat transfer or heat transfer coefficient ok question is what is folding factor how does it vary from heat exchanger to heat exchanger and how does it affect the heat exchanger or the heat exchanger heat transfer coefficient ok so first let us take the second question that is how does the heat transfer coefficient get affected by the folding yes surface is getting roughened so heat transfer coefficient is going to get we tend to think that heat transfer coefficient may increase but my pumping power is same so my velocity may decrease with the passage of time because of the because of the folding because of my pumping power has decreased but at the same time cross sectional area also has decreased so velocity might increase also generally the velocity increase and the pumping power are not going to affect so much but the heat transfer coefficient there might be a slight increase or we can conveniently say that heat transfer coefficient may not get affected that much but what gets affected is the folding resistance there is an additional resistance which is coming into picture one more interface is been created on the wall so there is an additional resistance because of which my load will come down my load will come down that is my overall heat transfer coefficient will come down my resistance will go up because of which my load if it is designed for 100 megawatt it may work only for 98 megawatt the second second year or third year okay so how does one measure the I mean how do we get the folding resistance it is done by measurement only they do some accelerated experiments that is you increase the concentrations you increase the concentrations of those salt and do an accelerated experiment that means by increasing the concentrations of the materials or the chemicals which are responsible for fouling you increase the concentration so that whatever fouling would have occurred in one year you would have perhaps in the lab done in one day okay that scaling you do in terms of time by increasing the concentrations of the chemicals responsible for fouling and that is how one measures the fouling resistance otherwise in real life industry one cannot go and measure fouling resistance is just not possible okay. If we make elliptical holes will there be any heat transfer the change the question is in my LPG burner instead of circular holes will I make elliptical holes okay that is one PhD okay no problem but usually how does the holes here my student is sitting Samir who is doing the experiments on jet impingement or with various shapes of holes usually people are going for non-circular holes only when noise is an issue especially for compressible jets that is if we take a non-circular hole especially elliptical one elliptical one creates lesser noise so acoustics is the one which is driving the which is driving as for choosing even in a combustor what gas turbine blade combustor the elliptical hole or even if you if you have to go for high speed jet flame jets in my burner which is there which is being used at home they are not compressible jets they are incompressible okay they are not compressible that means Mach numbers are not exceeding Mach number greater than 0.3 Reynolds numbers if we sit down and calculate they may be around 5000 to 10000 they are turbulent but they are not compressible so I do not see great advantage by going for elliptical holes or burner geometry if we but if we are talking about high compressible jets which is the case in case of gas turbine blade people are not only going for elliptical but they are also trying non-circular that is triangular rectangular and other weird shapes but not from thermal aspect but from acoustics aspect okay. One more question sir why radiation is completely neglected in design of a heat exchanger I think radiation also plays a little role in a heat transfer over to you sir. One of the very good question is radiation is usually in the classrooms is completely neglected in the heat exchangers radiation is little important that is the question yes radiation is very important we cannot neglect especially when I am operating when I am handling high temperature heat exchanger but in the resistance within tube in tube if I take where is radiation becoming important radiation is becoming important while computing the heat loss from the outer side of the heat exchanger but between the two fluids where the wall is there what is the mode of the heat transfer one can think of it is conduction and convection now fluids are non participating medium that is why there is no radiation but where is radiation is there radiation is there on the outer wall where it is talking to the atmosphere that is which is deciding the heat loss that is wherever we are assuming that there is no heat loss in my heat exchanger that assumption has to be relaxed and taken into account but otherwise in computing the areas I do not think we need to take into account the radiation at all but radiation is important but it is only in computing the losses I think we are moving away from the center because we have taken three questions we will go to MA NIT Bhopal. First is how the temperature distribution varies with length in case of cross flow heat exchanger. This is a very specific. We have demonstrated for parallel flow and counter flow heat exchanger. That is what I said see the question is it is neither a cross flow sorry neither a parallel flow nor a counter flow. So, it is very difficult to plot the temperature plot let me go ahead and one point I have missed I need to understand one point and tell you is this plot always like this see I take parallel and counter flow heat exchanger always like this I always write a plot always like this and for counter flow I always draw like this is it going to be always like this can my profile for a parallel flow be like this and be like this can it be like this yes it can be can counter flow heat exchanger can it be like this can it be like this instead of concave convex can it be combination of concave convex convex concave concave can it be yes it can be then, but who is deciding this convexity or concavity it is the property variation how does C P vary with temperature how does it vary that is what is going to decide there is nothing sacred about these two plots there is nothing sacred about that they need not be that way at all they need not be that way at all who is going to decide this convexity and concavity is the functional property C P how is it going to vary with temperature as an exercise take any fluid and take the variation of C P with temperature plot that with temperature compute take the heat load and compute this with the heat transfer coefficient in an excel sheet you can calculate that you plot that you will find this C P C P functionality as temperature and you will see that all sorts of concavitys and convexities come into picture. Now, coming to your question how in cross flow it is going to vary it is neither going to be cross flow or I mean parallel flow or counter flow it is very difficult for me to sit down and plot I have to sit down and do the experiments and plot the question what you have asked is an open ended question it is not having a closed form answer we have to sit down and do the experiments in particular for the cross flow heat exchanger what I have designed and fabricated and what I am testing. So, I cannot answer this question in a specific form that this is indeed the answer there is no unique answer for this question over to you. Next question is sir why efficiency has not been taken as a performance parameter in case of heat exchanger. Why efficiency is not taken into account we have taken only effectiveness if you want we can define a effectiveness as efficiency only you know efficiency is effectiveness only you know that is the maximum heat transfer you could do and what is the may heat transfer you have gained is effectiveness that is efficiency only thing is that instead of taking effective instead of calling it as efficiency we are calling it as efficient that is all. Thank you sir. NIT 3G any questions quickly. Sir while determining the boiler efficiency basically boiler is a heat exchanger. So, how to calculate the radiation loss in a boiler sir over to you sir. As it was question is how to calculate the efficiency of how to calculate the radiation losses in a boiler. So, between the fluids there is no heat loss due to radiation. So, it is only from the outer surface of the fluid to the outer wall of the boiler to the environment. So, if you know the temperature roughly of the outer wall of the boiler you can calculate the H A into T wall minus T infinity plus sigma epsilon A into T wall to the power of 4 minus T infinity to the power of 4. That means I need to know the heat transfer coefficient which is natural convection and emissivity of the surface. There was one tutorial problem of a plate suspended in air where you are asked to find the cooling rate where radiation and convection was exactly the same thing. Now, you understand reflectivity why it should be there because emissivity has to be low and radiation loss has to be less that is why we put aluminum foil. Sir while designing the heat exchangers so, how to incorporate this radiative heat transfer coefficient in the formulae. Where? Outuser. Ok question is how to incorporate radiative heat transfer in heat exchanger. Let us say everyone is asking this question I thought I have answered this question, but let us draw that let us draw where is the energy going from where to where let us keep track of that. If where is the energy going I have outer fluid I have inner fluid and this is my thickness of my wall this is my counter flow. So, now where is the heat transfer going from from the hot fluid to the cold fluid. Let me assume that this is hot and this is cold which is what we usually do this is now where is the heat transfer taking place and what are the modes of the heat transfer. We have conduction convection radiation conduction and convection we have explained already. Can radiation take place? Between the fluid. Between the fluid where is the radiation taking place between the fluid from this fluid to this fluid is there any radiative heat transfer no from this wall to the fluid there is no because fluid is not participating. So, there is no radiative heat transfer between between the fluid, but the radiative heat transfer is there from outer wall to the outer atmosphere that is all it is that is all it is. So, there is so it is valid to assume not to consider radiation in any of our analysis is that ok. Thank you sir. How do we fix the tube fluid and the shell fluid? How does one fix the tube fluid and the shell fluid? So, this is there is no one unique way to say that this is the fluid always has to be in shell and this is the fluid always has to be in tube there is no unique way. See typically tube side heat transfer coefficients are larger than the shell side heat transfer coefficient and tube side they are more or less uniform and shell side you have lots of non uniformity. So, you will use typically that fluid which will give you higher heat transfer coefficient on the shell side which will give the higher heat transfer coefficient and whichever fluid gives the lower heat transfer coefficient is put in the tube side. When I say which fluid which gives the lower heat transfer coefficient will decide me that is coming from Nusselt number Nusselt equal to r e to the power of something p r to the power of something whichever is having higher p r usually will have higher k will have higher k usually will have higher k. So, higher parental number usually are put on the shell side on the tube side lower parental number fluids are put for the simple reason that tube side heat transfer coefficient is higher than the shell side heat transfer coefficient, but not always this is just generally some rule sometimes shell side also can be high if my shell side fluid flow rate is higher than the tube flow rate then shell side heat transfer coefficient also can be high. That is why I said in any heat exchanger there is no unique answer for all these questions which tube should be which fluid should be which side what should be the diameter what should be the length there is and which heat exchanger I will use under which circumstances it is very difficult to answer. If you have to take typically tube in tube heat exchanger we stop at 10 kilo watts after 10 10 no sorry 6 mega watts sorry what is that number 20 kilo watts after 20 kilo watts only you will take tube in tube heat exchanger after that the length becomes too big you cannot handle. So, you will have to go for shell and tube heat exchanger. Sir assuming the constant heat flux and the hot plate side how do we analyze the heat exchangers? How does one analyze a heat exchanger assuming a constant heat flux range take the appropriate heat transfer coefficient for the constant heat flux range if it is turbulent it does not matter whether it is constant heat flux or constant wall temperature in turbulent we have stated in convective heat transfer no matter what is the boundary condition my heat transfer coefficient is to be is going to be independent of the boundary condition. So, it simply does not matter whether it is constant heat flux or constant wall temperature if my boundary condition if my flow is turbulent. One more question from the centre quickly. Hello sir, my question on convection does not depend on Q and temperature difference in force convection it depends on only three point difference will you change the heat convection? So what is the question I understand in forced convection we said that heat transfer coefficient is independent of heat flux and temperature difference is it true in free convection I emphasize at this point in free convection you see what is in free convection heat transfer coefficient is dependent on what? Grashof number and Prandtl number what is Grashof number dependent on? Grashof number is dependent on G beta delta T L cube divided by alpha nu what is sitting in Grashof number delta T that means heat transfer coefficient is dependent on delta T in case of natural convection if it is constant heat flux I will define Grashof number in terms of constant heat flux. So, in free convection heat transfer coefficient E is dependent on delta T or heat flux. How does a thermal camera different from pyrometer? How does a thermal camera different from pyrometer? I will tell this in a very simple manner there are complicated answers for this I am not going to give all complicated answers thermal pyrometer and thermal camera are same let us take it like that. There are some optical pyrometers let us not worry about them let us not complicate the answer there are other pyrometers which work on optics let us not get into that generally thermal camera is considered as optical proof thermal camera is pyrometer. Pyrometer means what? Pyro means temperature meter means measure measurement of temperature anything which measures temperature is a pyrometer. So, thermal camera is a pyrometer for that matter thermocouple is also a pyrometer, thermometer is also a pyrometer enough ok V I T Pune. Good afternoon sir my question is related with heat transfer augmentation whenever we are augmenting the heat with the help of using some mechanical inserts there is always increase in heat transfer coefficient but at the cost of pressure drop. Pressure drop we are measuring in terms of friction factor in the due course of time you have explained the one term that is called as enhancement ratio or enhancement factor to justify the use of that particular insert in that application. So, that factor is something Nasez number upon Nasez number without insert divided by F upon F 0 raise to 1 by 3 can you illustrate how that 1 by 3 factor is actually coming. Actually it is little involved actually I would request the question as this for enhancement factor for enhancement it is said that you take N u by N u naught upon F by F naught to the power of 1 by 3 it is a quite a lengthy derivation all that I would say is for both the cases for smooth and rough all that what we are saying is constant pumping power constant pumping power for the constant pumping power it turns out that V to the power of 3 V to the power of 3 turns out to be F by F naught to the power of 1 by 3 please put this question in moodle I will answer this I will answer this I will answer this through moodle it cannot be answered that straight forward manner it will take at least ten minutes time please bear with me please put this in moodle I will answer this. Okay, thank you sir. Yeah, Nirmai University. Sir in a baffled heat shell and tube heat exchanger how we can calculate heat transfer coefficient. Okay, see I told a question is how does one calculate the heat transfer coefficient in in the baffled shell and tube heat exchanger professor I told this in the morning there is an elaborate procedure in bell delaware method twenty steps are there twenty to thirty steps are there they are all given in heavy but if you still insist that you want we have that nodes in bell delaware method please put this question in moodle we will upload that nodes okay it is a very elaborate twenty it will run to twenty thirty pages if I have to write that here and I do not remember also okay so please put it across that question definitely I will put that put up that in moodle okay please refer to have it more than my notes have it is the right textbook okay. One more question is there. Hello sir in case of domestic LPG stows there must be some radiation losses so can we put say D say C is to avoid those radiation losses no I said C will it be fruitful question is will there be radiation losses in case of LPG flame see I have told LPG flame does not have emissivity so emissivity is not there means there is no radiation from the flame to myself okay so there is no radiation losses at the most you might be feeling little bit of heat because there is a vessel sitting on it the vessel has got heated up and vessel it is having emissivity and that is radiating heat to us if I put a small tumbler and heat it I will not feel heat if I put a big boiler for heating water then I will feel the heat not because from the radiation from the flame but because of the radiation from the plate so that is the answer for your question okay. Yes sir I am talking about the radiation from the vessel itself can we reduce the radiation losses from the vessel so that the heat transfer can be enhanced. You heat transfer can be how to reduce the heat loss from the vessel to the human being yes answer is in putting thermal shield if one puts an asbestos thermal shield around it I do not see the heat transfer because my asbestos sheet takes long long time before it reaches the high temperature by the time my cooking is over okay sir thank you over I think we are coming to the end of the session and we are stopping this.