 mentioned about the hot wire animometer Indian suppliers, can you repeat sir? There are hot wire animometer Indian suppliers. Yeah, actually there is one of the questions for one of the participants is what are the Indian suppliers for hot wire animometer. Unfortunately there are no Indian suppliers even if they are there they are not very what to say you cannot rely on them they are not reliable. So, I would not suggest you any Indian suppliers but if you you can write an MHRD project or what is that either DST project or what is the teacher's AICTE yeah AICTE to whom am I talking to I am talking to PSG Coimbatore. PSG Coimbatore you had a I remember 10 years back you had a thick hot wire animometer I had come to PSG Coimbatore and seen the hot wire animometer very thick hot wire animometer 10 to 15 years back this was in 1998. I had seen a thick hot wire animometer which was supplied by Sunrise which was one of the companies from Bangalore but then that was too thick and it was very intrusive for the flow. So, that is why I am not suggesting any Indian supplier but you can write a DST or an AICTE proposal and get money AICTE you can write a proposal and get it either from TSI or DANTEC. So, you do not have to purchase complete kit you can purchase only single wire and with one weed stone bridge if you purchase that I think you can still manage with 7 to 8 lakhs. But you definitely need an AICTE room so that there is no dust and things like that. So, but if I try to find we will get hot wires just for measurement of velocity but not for turbulence intensity. If it is only for velocity you can go to wary.com w a a r w e.com you can go to this website you can see hot wire animometer there also and which is basically made of testo is supplying testo make hot wire animometer. So, like that there are many suppliers only for measurement of velocity but not for turbulence intensity. If you need turbulence intensity you will have to go either for TSI or DANTEC make you just press TSI or DANTEC over the way Google you will get there contact details. One of the participants question is if I have a ribbed turbulent if I have a ribbed turbulent let us say this is my pipe and these are my turbulent on both sides let us say and I whenever turbulent means to my mind always ribbed only comes may be because I have worked with ribs. So, what essentially we are saying is that in a pipe if there is a ribbed turbulent what we are saying is the heat transfer coefficient is changing locally no doubt about that the heat transfer because of the separation there will be heat transfer decrease and subsequently there will be an increase in the heat transfer if this is my flow direction. Now, the question asked is if my pipe is thick let us take two situations let me take a pipe of thickness 10 mm and let me take a pipe of thickness 1 mm and let me take this pipe as stainless steel. Why I am taking this stainless steel because there is a conduction thermal conductivity of stainless steel is slightly lower than that of copper I am not taking a copper pipe. If I take stainless steel what is happening whatever heat is heat transfer is taking place on the wall gets conducted into my wall. So, then this conduction also plays a role that is what is called as conjugate the problem what you have asked is actually what is called as conjugate conduction problem that is there is both convection and also conduction the point is we have the that is why whenever we are we are trying to measure local heat transfer coefficient. If I am trying to measure the local heat transfer coefficient I should not use should not use copper pipe I should use stainless steel pipe and that too its thickness it should be as small as possible. If I take a thick copper pipe my I cannot get the local heat transfer variation everywhere I will get constant heat transfer coefficient. So, heat transfer coefficient what I have measured is not what actually is perceived by my plate because of the fact that it has got conducted into my plate and my plate has made it average that integral of 1 by L 0 to L h L d x that averaging has been done by whom not by the calculator, but by my copper plate itself that is the reason why we should not use stainless copper pipe and that too thick pipe if I have to capture stainless steel if I have to capture local heat transfer coefficient. So, that is why we if you see most of the local heat transfer coefficient measurements will have a metal foil of the order of 0.05 mm. If you take a circular pipe of this order that is 0.05 mm and then put the thermocouples wherever you want then it is going to capture the local temperature because conduction is avoided lateral conduction is avoided otherwise lateral conduction takes place which is nothing but conjugate conduction and will affect my localness of my heat transfer coefficient. Coming back to your second question I did not understand will you please elaborate me that question what is that you try to maintain constant wall temperature by putting steam and what is that you did not get will you please elaborate for me. Sir in the when we are allowing that steam inside the annulus section and we are allowing the water inside the tube tube into heat exchanger when we are conducting the experiment for the plane tube the Nusselt number for the plane tube is about 90 whenever when I am inserting the regulator the regulator is to increase the heat transfer, but the increase in order is only about 17 percentage. So, when I am calculating the performance the value is less than 1 whether it is for steam or because of the steam sir. Ok professor actually I can understand the question, but I cannot give the answer right away, but I what I would say is I will not hand wave I will try to put this question little differently. What you are trying to say is that when you say Nusselt number you are if I understand correctly you are using a tube in tube heat exchanger hopefully in one you are allowing this steam and in the another one you are allowing something else. So, I do not know where the ribbed tabulators are there and how are we calculating the local heat transfer coefficient my request is that will you please will you please generate a document one page document where in which you put the figure in which what are the you put figure and with inlet and outlet conditions and give me the temperatures and what are the heat transfer coefficients you are computing. If you give me that I can definitely come back I can definitely come back on this question. So, we will go to Amal Jyothi college Kerala. This is a doubt regarding then order of magnitude analysis and the non-dimensionalization. When we apply this order of magnitude analysis and the non-dimensionalization show same equation what is the difference in the effect? Question asked is what is the difference between the order of magnitude analysis and the non-dimensionalization procedure. What does non-dimensionalization procedure give me? We get that Nusselt number is a function of Reynolds number and Prandtl number that is Reynolds number Nu equal to C Re to the power of m Pr to the power of n C Re to the power of m Pr to the power of n. What did we get by order of magnitude analysis for lower Prandtl numbers? I got m and n, but I did not get c that c has to come from experiments or from analytical solution. m as 1 by 2 and n as 1 by 3 we got it from order of magnitude analysis. It is like order of magnitude analysis gives me gives me the order of magnitude of like for example, Nusselt number increases with the increase of the Reynolds number how to the power of half. It will increase with the increase of the Prandtl number how to the increase of 1 by 3 that feel we will get from order of magnitude analysis. But what non-dimensional numbers are important to look for that is coming from my non-dimensional analysis. So, that is the difference between non-dimensional analysis and order of magnitude analysis. Another thing in non-dimensional analysis is sorry in order of magnitude analysis is you will get to know which terms are contributing to the effect. So, you can you can eliminate throw away terms from the governing equation to simplify it. So, that you can get a reasonably approximate not wrong reasonably approximate solution. So, that throwing away of terms if I throw terms which are important then I have thrown the problem out, but if I if I can neglect terms for example, inside the boundary layer or sorry outside the boundary layer I am saying that the viscous forces are negligibly small. It does it is not that viscosity is gone to 0 there viscous forces are smaller compared to the other dominant forces. Therefore, I can neglect that and my equation becomes simplified. Therefore, I can get a close form solution in some cases for the simplified equation that is what we get by order of magnitude analysis. So, we we can estimate the values relative importance of the terms. Any other questions from the centre VN 90 Nagpur? Sir can you comment on the flow of fluid in helical coil? I have seen one of the questions in even in the moodle the same question has been put. Can you comment on flow in a helical coil? See what is what is an additional non-dimensional number which comes for helical coil? Helical coil also nusselt number and friction factor is going to be function of Reynolds and Prandtl number, but what is the other effect which is coming to picture the curvature? That is my stream lines are asked to go through corrode lines that is curvature. So, what is called as a dean number that is if I remember correctly Reynolds into curvature that is R by D Reynolds into R by D there is a non-dimensional number called dean number. Dean number is what is taken into account that is dean also consists of again Reynolds number only dean number equal to Reynolds into R by D that is radius of curvature upon diameter of the pipe diameter of the pipe. Why because this R by D that is the radius of the curvature and the diameter of the pipe represents the how much torture my stream line is undergoing while it is going through helical pipe that torture is quantified by this R by D, but if you ask me whether R by D is it to the power of 1 or to the power of 2 or to the power of 3 it depends for different curvatures. So, if you see typical correlation for an helical coil you would see that nu equal to C into R e to the power of m P R to the power of n R by D to the power of may be x. So, that is how you would represent the Nusselt number in case of helical coil why because the only additional thing which is coming compared to the straight pipe is the curvature which is torture of or the tortuousness my stream line or the fluid particle is going through is that ok. Sir, can we take the friction in helical coil as F is equal to 64 by R e to the R e at D? Question asked is for an helical coil can I take friction factor as 64 by R e I cannot take why friction factor is equal to 64 by R e is for straight pipe, but for circular pipe for a helical pipe I friction factor is a function of R e and again R by D. So, in fact there are papers in fact there is a large body of work done by Jamia Milia Islamia institute I do not remember the professor's name is having in fluid dynamic research paper I cannot upload that paper, but I can give you that reference of that paper in which he gives standard correlations which are experimentally found for different helical coils. So, where in which he gives friction factor as a function of Reynolds number and R by D. So, that is that is the answer for your question when I say Nusselt number that is valid for friction factor also. Sir, the helical coil is being placed in a shell. So, when the counter flow arrangement is expected the constant wall temperature and constant heat flux temperature conditions are not to be assumed perfect. So, in that case how to evaluate the performance? The question asked by one of the participants is that if I have a helical coil inserted in a shell I have a shell and in that I have a helical coil this is my shell let us say and this is the inlet of the shell and this is the outlet of the shell and now I have a helical coil through which the flow is getting in and getting out very crude figure sorry for that this is the helical coil. Now, the question asked is how to because the wall temperature neither on the outer side nor on the inner side are constant wall temperature nor constant heat flux. So, how do I which correlation I should be taking for heat transfer coefficient? It is a very good question. Most of the heat exchangers the closest boundary condition between these two boundary conditions is constant wall temperature boundary condition, but if you really need pick and ask no it is not really constant wall temperature yes I would agree with you it is all not neither constant wall temperature nor constant heat flux, but then which correlation you would use? I would use constant wall temperature boundary condition because we will see later part of the day now immediately the heat transfer coefficient for the constant wall temperature is generally lower than that observed for constant heat flux. So, I would like to design my heat exchanger on a conservative basis which is constant wall temperature. So, answer is I will assume constant wall temperature and take the heat transfer coefficient both on the inner side and the outer side and compute the heat transfer coefficient go ahead and design my heat exchanger fully believing that or thinking that my heat exchanger would take more load than what I have designed that is what I mean when I say conservative estimate because I am under predicting my heat actually real heat is going to be slightly higher because it is not truly constant wall temperature that is the answer for your question professor. Thank you sir over to you M K triple S Pune Yes sir sir there is a slight confusion today morning while deliberating with the flow through tube professor Prabhu he explained that the entry length is in case of laminar flow it is dependent on Reynolds number at the same time the entry length in case of turbulent flow he has also deliberated that it is equal to L e by D is equal to 4.4 into Reynolds number is to 1 by 6 means it is also dependent on Reynolds number, but while discussion with the Arun Kumar professor Arun Kumar he said that it is independent of Reynolds number and the slides also shows that it is equal to 10 D in case of turbulent flow even most of the books also say that the entry length is independent of Reynolds number. So, what is the correlation that is to be used and whether it is dependent on Reynolds number in case of turbulent flow or not dependent on Reynolds number please sir over to you. So the question asked by one of the participants is that the what is the let me put it let us not get into Prabhu and Arun. So, what my point is what is the developing length for hydrodynamically fully developed flow and what is the developing length for hydrodynamically and thermally fully developed flow. Let us take an adiabatic case that means there is no heat transfer involved in that case what I mentioned is you only write nice. So, in that case for hydrodynamically fully developed flow L by D as you rightly said that is remember this I am writing for adiabatic flow no heat transfer adiabatic flow. If it is adiabatic flow there is no heat transfer L by D is equal to 0.06 Re and L by D for turbulent flow is 4.4 into Re D to the power of 1 by 4 this is only for adiabatic flow and this is for hydrodynamically developing flow that is for the velocity profile becoming invariant with this stream wise direction. Let me repeat this that is my velocity profile is going to be a function of r only, but not going to be a function of x and theta that is about adiabatic flow. Now, if I take an heated case may be it is constant heat flux may be it is constant water temperature no problem if it is a constant if it is a heated case for thermally fully developed what is the developing length required L by D equal to that is L by D equal to 0.05 Re D for laminar and for turbulent it is typically 10 D. Now, when I say this 0.05 Re D for laminar you note that say there is a subscript H that only means that it is heated flow it means this is hydrodynamic L H laminar is 0.05 Re D, but what about L H for thermally laminar it is 0.05 Re into P R into D. So, that means usually when we are trying to reach a thermally fully developed flow we usually give enough developing length upstream for the flow to become hydrodynamically fully developed. So, thumb rule is for let us say for turbulent flow thumb rule is you put 25 D upstream to your pipe before heating why because to ensure that the flow is hydrodynamically developed having ensured that my flow is hydrodynamically fully developed. Now, if I start heating it would take further length of 10 times the pipe diameter. So, that it would become thermally fully developed that means temperature is whatever condition professor stated that is we will get thermally fully developed that means the heat transfer coefficient would become invariant with stream wise distance. I think now that clears the question the answer for your question is that okay. Yes sir, thank you. Thank you sir, all to you. Sir one more question. Yeah. We got many courtesans available in handbook, but the courtesans are not given with the name who has given it. So, where do we get this literature who has proposed it, how they have conducted it, how they have got it everything. So, Okay the question asked by one of the participants is that there are several correlations listed in data handbooks I do not know which data handbook he is referring to, but in general in the data handbooks none of the correlations are given against a name, but most of the correlations for example let us say we got yesterday C F that is Blaise's correlation that we know and most of the correlations for example which we are going to study today for turbulent flow 0.023 RE to the power of 0.8 PR to the power of 1 by 3 that is Dittus-Bolter correlation, Nielinsky correlation. So, most of the correlations because there are tens of correlations there are tens if not hundreds of situations where in which there are different correlations. So, you cannot go on giving name most of fundamental ones are given name to the best of my knowledge we know Blaise's correlation, we know Dittus-Bolter correlation, Nielinsky's correlation. So, but if you want to know where is that correlation come from definitely yes you can get that each one has a journal paper published after at least he should have published a journal paper otherwise that correlation would not have come whatever correlations we see in a text book or in a data handbook are all taken from a journal paper. If you definitely put that situation like if you are saying flow around a cylinder if you put flow around a cylinder you will you will land up with Churchill's correlation. So, Churchill paper you will get if you read the Churchill paper you will understand how he has done the experiment how he has got the averaging what properties he has taken how good is that correlation with reference to the experimental data one thing we need to remember especially for turbulent flow there are hundreds of correlations, but with a pinch of a salt each correlation is going to have an uncertainty at least 10 to 20 percent if someone tells me that this heat transfer coefficient measurement is as good as 5 percent it is very unlikely. General thumb rule is for uncertainty for friction factor is around 6 percent and for heat transfer coefficient is around 15 percent it can be more than 15 percent we will be very happy if anyone quotes uncertainty in heat transfer coefficient more than 15 percent if someone says it is only 1 percent 2 percent most likely is under coating is uncertainty the point is none of the correlations are accurate to the t they are limited to the accuracy of the experiments on based on experimental data based on which that correlation has been generated. If you want any specific correlation to know how it has come you please put it in the moodle we will try to answer based on whatever information we have erode. Sir my question is if I am blowing a one bar pressure over here flat plate I am getting some value of some new that is kinematic viscosity frontal number and Cp and some k thermal conductivity. If I am replacing my blowing air to a five bar what is the changes in new frontal number Cp and k just I am increasing the pressure. See the question is I have a flat plate case I have a flow over a flat plate I have taken flat plate in the first case I am having I am blowing air hopefully from a compressor which is releasing air at one bar and temperature 30 degree Celsius let us say that is to say that it is ambient in the second case I am having same compressor but this time I am supplying it at five bar but temperature is continued to be 30 degree Celsius. What is the first change on the properties? On the properties if it is air it is very insensitive to the that is mu Cp in fact in the moodle also yesterday there was a question I have answered this rho mu not rho mu Cp k mu Cp k are almost independent of pressure almost but rho we know rho is going to be P by RT. So, rho will be five times more than the rho what you would have found in one bar now this apart the thermo physical properties I hope you are somehow controlling the velocity same if you do not control in the five bar case the velocity will be very high how high we equal to square root of delta P times higher depending on the nozzle what you are using so that if you have a coefficient of discharge or something the property sum and substance is that thermo physical properties mu Cp k are almost independent of pressure density if it is if it is air it is going to vary with the ideal gas law P by RT and that is it otherwise everything else is same MES Pillai-Panvel. Hello sir as you said in thermally fully developed region heat transfer coefficient H remains constant but is there in any impact of increase in velocity of fluid on heat transfer coefficient H over to you sir. The question asked is in the thermally developing region that is if I have if I were to plot Nusselt number or H versus X by D what is happening here Nusselt number or the heat transfer coefficient decreases in the developing region and in the this is developing this is developing region and this is fully developed region in the fully developed region the heat transfer coefficient is high I mean heat transfer coefficient is constant. Now how does H vary with velocity H varies typically if it is a flat plate it varies with the to the power of half it is laminar if it is laminar and H varies to the power of 0.8 or e to the power of 0.8 0.8 if it is turbulent the point is with the increase of the velocity definitely the heat transfer coefficient is going to increase no matter whether we are handling laminar flow or turbulent. And that is obvious right if you are trying to cool something if you have a fan it is going to cool faster so heat transfer coefficient is what is increasing. So I think we will stop the question answer session for a while now.