 Before we move on to other correlation I would like to take some questions I think before we take questions professor Arun wants to add something. One question which was there in Moodle was in the data handbook there are multiple correlations which is the one that I use for a given problem. There is no unique answer for that see there might be families of correlation for flow over a flat plate couple of them three or four of them somebody might have done for some fluid etcetera etcetera etcetera. What you have to do is look at the range of conditions of applicability of that correlation see if it is done if it is valid for the range of properties that you are having and then use it. And what should a good correlation is the one which should be valid for a wide range of fluids and overall what you say property values etcetera. Now if I do an experiment for with say R 1 2 3 or something and develop a correlation explicitly for that I cannot use it for flow with water or anything like that. So and please keep in mind that empirical correlations have a certain amount of uncertainty involved in them caused by the experimental uncertainty also ok. So these correlations are valid with plus minus 20 percent 15 percent 10 percent whatever it may be. So you get an answer with one particular correlation another correlation is also equally valid you get a slightly different answer does not matter ok. So this is something we need to emphasize and students will come and say this answer is wrong that answer is correct both answers could be equally correct or equally wrong ok. Erode any questions there is a question mark. In case of transport of crude oil from one country to another country the pipelines are taken underneath of the sea. In that case there is a chance for the pipeline to go through the high specs. So in that case the thermal boundary layer what what happens to thermal boundary layer because the pumping power will be high in case of if the thermal boundary layer is more and sometimes there is a chance for blockage of the crude oil flow. So what is your suggestion. See the question asked by one of the participants is that the crude oil transportation from one country to another country is generally taken through sea and sometimes it goes through icebergs that word I did not hear clearly I hope it is icebergs basically what pressure wants to impress upon us is that it is going through lot of temperature gradients. So whether the whether the crude oil is going to get affect the movement of the crude oil is going to get affected. See the answer I would think as like this see the boundary layer see there is going to be temperature gradient no doubt about that and instead of heating it is cooling the oil and in fact like in semi infinite medium assumption we had said that how below we should be taking yes if the temperature freezes my oil if it is below the freezing point definitely nothing can be done but I do not think it is going to be frozen. So yeah see professor says that seawater is generally warm at the base so it is not going to freeze or solidify the oil which is there inside that is that might be one of the reasons maybe we can just go back and check what would be the temperatures at the sea bed. So the point is I do not think they would lay the pipes intentionally where the oil is going to get frozen. So that is the answer I would think temperature is generally greater than 4 degree Celsius. So I do not think it is going to freeze my oil at all the freezing point of the oil would be much lower. I hope I have answered your question professor. So what happens to the thermal boundary layer and hydrodynamic boundary layer in such cases? The question asked is what happens to thermal boundary layer and the hydrodynamic boundary layer. Thermal boundary layer or the hydrodynamic boundary layer is like this let me it is going to be like whatever conventional of course this is little premature question but still I would go ahead and answer this because now that we have been asked so this is a pipe. So now parental number of the oil I would think it is much much greater than 1 so that means delta is much much greater than delta t. So thermal boundary layer thickness is very much lower than delta so that means delta is thicker that is this is delta and delta t is thinner this is delta t. So of course we are not introduced this concept because you asked me so we are telling this this is the hydrodynamic developing length and this is hydrodynamic developing length and this is thermal developing length. So the thermal developing length in case of oil will be higher than that of developing length the hydrodynamic developing length but I do not think all this is of any importance because my pipeline is much much much much larger developing length if the is typically of the order of 100 times the pipe diameter but my pipeline size is running to few kilometers. So developing length concept has no relevance much relevant but Nusselt number and friction factor are as good as which are derived for any other Newtonian fluid as long as we can assume crude oil as Newtonian. In such cases the pumping power will be extremely high then there will not be any flow to take place now. So question asked is in such cases for high Prandtl number pumping power will be very high yes pumping power will be very high because viscosity is very high for pumping power it does not know C p and all k C p it does not know for friction factor what is important is the viscosity. Viscosity is very high pumping power will be very high in fact this is one of the cases where we say classic example laminar flow will exist usually the flow is not turbulent in pipeline supplies it is laminar flow why because viscosity is very high I need to give that kind of pumping power if I have to supply the oil I cannot say that I cannot pump it has to be pumped it is laminar and the pumping power will be significantly larger than that of water or air. Ok professor. Will there be any provision to keep the oil well below the firing point of the oil in order to make the flow easier. See I really do not know the question is will there be any provision to keep below I really do not know how exactly the crude oil pipes are kept whatever I am answering is based on fundamental I would request you professor now that you have asked so many questions can you take this as a homework and answer this question over Moodle how exactly crude oil pipes are supplied if you do not answer definitely we will put it across but I would request you to having taken your three questions I think you can put the answer I think it is fair enough to ask you to put the answer for this question on the Moodle yourself I think ok. Ok Panvel. Hello Nasell number equal to say R e raise to m P r raise to n these are the constants of correlations. So, what are the different methods to get it what are the different methods to find out these constants how to obtain these value of constants. Yeah see one of the participants question is that we said that N u equal to A N u equal to C R e to the power of m and P r to the power of n how do I get this C m and n yeah it is a very good question anyway we are going to answer just little while from now we have solved it for laminar flat plate case, but it is not always possible to get the closed form solution if I cannot get the closed form solutions with mathematics for example, for turbulent flow one would take the recourse of experiments we would find the experiment we would conduct experiments and say that through experiments at least we know the functional form how to represent the results. So, I will find C m and n through experiments the turbulent shear stress is deducting from the daughter shear how it is. See one of the participants question is do we deduct the turbulent shear stress from the total stress no we do not deduct see as I have been saying since morning we have laminar sub layer and we have buffer layer and we have turbulent boundary layer I am not deducting in the turbulent boundary layer what is there is only minus rho u prime v prime bar as the turbulent shear stress and in the in the buffer layer both of this minus rho v prime u prime bar and mu del u by del y they are of the same order both are of the same order now in the laminar sub layer it is mu del u by del y we are not deducting in the laminar sub layer for example, u prime is 0 and v prime is 0. So, there is no question of having rho u prime v prime bar. So, this is not there this is 0. So, we are not deducting anything from anywhere it is just whatever is measured is just being reported I think what professor Arun says is that may be your question is why is it negative sign no minus rho u prime v prime is actually positive that is a good question always I used to forget this, but I used to give this as a homework for student, but actually it is not it is like this if u prime is generally if it is positive v prime is generally v prime negative. Why because where does the energy come from wherever if it is lost in one direction it has to be compensated in the other direction. So, if u prime is positive v prime is negative. So, minus this minus is essentially to make this positive if one is positive and the other is negative means product is negative minus of this product is again negative sorry the product of all of this is positive. So, total thing is positive. So, we are not deducting this minus sign is essentially to make it positive any other question sir what about moving plate boundary part about what boundary layer for moving plate the question is by one of the participants what will happen for a if I have a flat plate in addition to the velocity if this is moving if this is moving no issue the question here is always it is the relative velocity which generates the boundary layer. So, if it is moving the velocity at the edge of the boundary layer will be equal to the velocity of the plate, but there is as long as there is a relative velocity between u infinity and the plate velocity there is going to be boundary layer that is what is their covetive flow. In fact we are going to deal with this when we come to internal flow, but we are not going to solve for moving plate and this is essentially what we study in various covetive flow. Point is for moving plate also boundary layer is going to be there and that boundary layer depends on the relative velocity between the fluid velocity and the plate velocity. I have a question regarding the energy equation which we derived in the morning the left side of the energy equation is convection and the right side contains the diffusion the pressure work and the dissipation. So, is it appropriate to call the left side as convection or should we call it should we be calling it as advection instead of convection. One of the participants question is that if we see the energy equation in the energy equation we have we have this term what we have called this as convection this is conduction pressure work and viscous dissipation. One of the participants question is why do not we call this as advection rather than convection, but if you really see it is a combination of both advection and convection see why because why I say convection because convection is nothing but conduction within the boundary layer del t by del y and del t by del x I am taking within the thermal boundary layer, but why is it there it is because of the convection velocities u and v. So, it is very difficult to differentiate both the advection and the convection component or advection and conduction within the boundary layer. So, that is the reason why we usually call this as convection knowing very well or in the back of our minds that it is the conduction within the thermal boundary layer that is all the answer is for this. Baramati any questions please. Sir my question is about the Eckert number what is the limit for the velocity when you consider Eckert number above the certain velocity value. See the question is what in the Eckert number now that we are anyway here only in the Eckert number what is the velocity limit above which we should consider or neglect Eckert number actually it is there here itself Eckert number equal to Eckert equal to v square divided by C p into T s minus T infinity. Now of course it depends on the fluid also, but if I assume that my fluid is air C p for fluid is 1050. So, for the Eckert to become 1 let us say for Eckert to become 1 what should happen one of you should help me. So, that is if Eckert number is 1 v squared equal to 1050 into delta t or I should I should have my delta t such that my temperature gradient should be such that my v v v squared should be of the order of delta t or this number v squared and delta t should be such that my Eckert number should be around 1 when will that happen when the velocity is very very high. Because I cannot write delta t just like that because it depends on the velocity I cannot write I cannot fix delta t and go on increasing v or I cannot fix v and go on decreasing delta t. It is they are coupled based on my velocity the temperature gradient is high what they say is that if my velocity is supersonic or hypersonic delta t is of the order of T s itself is of the order of T s itself then perhaps then it becomes that is if the temperature increases so high so high virtue of kinetic energy that is all almost all the kinetic energy is transformed into thermal energy. So, that is what happened with Kalpana Chawla its crash was essentially because one of the tile had gone up come out the insulating material on the what is that rocket had come out and the temperature had picked up because velocity. So, that is why the there was the trash or whatever accident had occurred my next question is sir one more question is there sir is it a incompressible fluid is in ideal or it is in existence. The question is is incompressible flow ideal or existing incompressible flow what is incompressible flow what is incompressible flow that cannot be compressed are there fluids which cannot be compressed are very difficult to compress yes water is a incompressible flow very difficult to compress water. So, if you take the velocity of sound in water it is very high it is in thousands actually. So, but if the velocity of sound in air is 330 meters per second both are linked. So, point is water can be assumed or water can be considered why assumed it can be considered as incompressible flow. So, incompressible flow is not an ideal situation, but inviscid is an ideal situation that means there is no fluid which is thoroughly inviscid. So, this is an ideal situation, but this is not a ideal situation. So, one more question is there I have not understood the statement when a certain number is 1 heat transfer across the layer is only by conduction. I think we will take this question when let me answer this question now that this question we had expected this question. The question asked is Nusselt number equal to 1 means conduction is equal to convection. See we need to understand conduction upon convection. See we need to understand one thing here we have written h l by k actually it is not conduction over the length l or actually what is that we are equating it is conduction within the boundary layer conduction is equated with the. So, what should I take here actually I should take h delta t by k h delta t by k, but do I know delta t today only we derived for flow over a flat plate for Prandtl numbers less than 1 delta t is of the order of r e to the power of minus half and p r to the power of minus 1 by 3, but for all situations for a moment delta t is of delta t is of the order of r e to the power of minus half and p r to the power of minus 1 by 3 that is I cannot get this type of solution for all cases for that cases I do not know delta t delta t is unknown. So, we can definitely say this one this thing is right when only when I say h delta by k in that form only one has a meaning that is conduction within the thermal boundary layer is equal to convection if I write h l c by k that equal to 1 has no meaning actually, but I write h l by k because I do not know delta t I give all correlations in terms of characteristic length when I write in terms of characteristic length h l by k equal to 1 need not mean that conduction is equal to convection it is equal to 1 only when conduction within the thermal boundary layer is equal to conduction convection any other questions. Sir one more question in temperature gradient d t by d y at y is equal to 0 is this definition or this means is it exactly on the surface part or on the layer which is in no slip condition. See one of the questions is is this minus k delta t by del y is h equal to h equal to minus k delta t by del y at y equal to 0 is this only on the first layer or subsequent layer. Now let me answer this way delta t by del y at y is equal to 0 is it not getting affected because of the velocity distribution in the temperature distribution within the thermal boundary layer. Yes for argument for argument sake we can say that it is conduction in the first layer why because there is velocity in the upper in the subsequent layer, but I think we can conveniently say that actually it is conduction within the thermal boundary layer, but this del t by del y is affected because of the velocity distribution and the temperature distribution within the thermal boundary layer. How do we get the temperature distribution after the solution of the velocity distribution. So u has been solved if you go mathematically continuity x y momentum equation solution gave you that u and v which when you put in the energy equation the u d t by d x plus v d t by d y equal to alpha d square t by d y square that gave you the temperature distribution. So this both u and v have come from the velocity distribution which was obtained from the momentum and the continuity equation. So k d t by d y is definitely affected by the velocity above the layer. So I cannot have k d t by d y at y equal to 0 independent of what is happening above it. So it is definitely affected. Everything is interlinked. Sir having two questions for you one is when you are talking about the Nusselt number is function of x star r e and Pantel number is it a better idea that when x star is increasing that your r e is also increasing over the flat plate and if you. The question asked by one of the participants is that we said that N u equal to is a function of x star and r e and P r e is with the increase of x is r e also increasing. Yes if you take N u equal to what is our relation point and let me take the relation because I do not remember 0.332. So that is 0.332 r e to the power of half P r to the power of 1 by 3 P r to the power of 1 by 3 N u x r e x. So that means what is r e x rho u infinity x by mu. So as x increases that means x increases for a given fluid and for a given velocity definitely r e x is increasing definitely r e x is increasing. So there is no doubt about that r e x is increasing what is the second question please. So in case of the flat plate suppose we are calculating say if you are having a 1 meter length if you are calculating the r e over the plate say at different locations 0.1, 0.2, 0.3, 0.4. So we need to calculate x or r e based on the L. The question asked is as we move along the length let us say for a flat plate of 1 meter if I calculate x equal to 0.1, x equal to 0.2 and x equal to 0.3, x equal to 0.4 will I calculate Nusselt number at x equal to 0.1 on the basis of L equal to 1 meter or x equal to 0.1 meter. Right from the beginning we have been talking about this local heat transfer coefficient and average value. This Reynolds number being affected by the length gives you a local value of the Nusselt number local value of the heat transfer coefficient. So at x is equal to 0.1 I am going to get the value of heat transfer coefficient at x equal to 0.1, x is equal to 0.4 it will be a different heat transfer coefficient based on the Reynolds number at that location. So if I have this this is a common x book problems if I have electronic chips which are being cooled by force convection the chip at the first chip from the leading edge versus the 10th chip versus the 100th chip. So each of this the heat transfer coefficient is going to be different it is going to be decreasing with respect to increase in x and that quantity is a local quantity. Now if you are asked to calculate what is the total heat dissipation from the plate or something then it is over the entire length that case you will take a average h value in which case your Reynolds number will be based on the length of the plate not the local distance your Nusselt number will be based on the length of the plate not the local Nusselt number. So whatever is the subscript that you use for Nusselt number x that is the same as subscript that you use for Reynolds number if that is x this is x if Reynolds number is l this is l one other word of caution if you have a plate of 1 meter and there are 10 chips the 10th chip which is at x is equal to l Reynolds number local will be r e l but Nusselt number for that will still be a local because you want to calculate heat transfer at that location it is again play with English only at a given location if you are asked it is a local quantity over a given length if you are asked that is an average quantity ok. So heat transfer at x equal to 1 meter means how much heat is lost at that location heat transfer over 1 meter length of the plate would mean what is the heat transfer over the full length of the plate there you will use the h bar value ok. V N 19 Agpur. Sir what is the effect of pressure on thermo physical properties? This is a good question the question is what is the effect of pressure on thermo physical property there is going to be effect of pressure but the effect of temperature is more significant than effect of pressure but definitely pressure is also going to have an effect we cannot say that there is no effect of pressure but for example thermal conductivity if we take thermal conductivity or viscosity for that matter if we take both of them are more affected by temperature rather than pressure. So usually that is the reason in most of the textbooks also they do not list out the effect of pressure on thermo physical properties because it is influence on the thermo physical properties is significantly lower compared to that of temperature. Amal Jyoti College. Hello sir we have actually three questions one is that sir in the derivation of the energy equation you have not considered the heat generation term so why it is so and second question is that in suppose you are going to measure the velocity in case of a laminar sorry in case of buffer layer we are placing a hot wire anemometer so the will the anemometer affect anything to the flow conditions so will it cause a very serious problem to the flow condition. Now third question is that why it is that the Reynolds transport theorem is called as a transport theorem. Okay first let me list out question by question one first question is I before I forget first question is why volumetric heat generation is not used first question is why volumetric heat generation why volumetric heat generation not considered in energy equation number one not considered in energy equation and second question is in the buffer layer in the buffer layer in the buffer layer in the buffer layer if you if I introduce hot wire anemometer if it will it affect velocity that is the second question and in Reynolds transport theorem in Reynolds transport theorem why is Reynolds transport theorem called as Reynolds transport theorem. Okay let me take up randomly in a buffer layer yes whenever you intrude a whenever you intrude a probe into a flow definitely the probe is going to disturb the flow that is why I said you scale up your plate you scale up your plate such that your velocity bound that is the buffer layer thickness is very large. So that is if you take a huge plate let us say of 1 meter and 2 meters wide then you would perhaps get a boundary layer let us say of let us say of 10 centimeter then in that 10 centimeter if you introduce a probe of 3 mm then the intrusion levels are low that is one way of looking at it otherwise there are today non-intrusive way of measuring things that is laser Doppler anemometer or particle image velocity particle now particle image velocity so these are non-intrusive non-intrusive how do how does it work how does it work in a non-intrusive for example particle image velocity I am just going to hand wave it is quite difficult to make you understand but what is done is it is a pulsed laser that is one laser to the second laser it is going to be there is a time difference of nanosecond that is 10 to the power of minus minus 9 seconds so 10 to the power of minus 9 seconds delta t I have a pulsed laser now if I introduced a particle and I have tracked this particle let us say I take this particle and I take picture between 2 nanoseconds that is sorry between 1 nanosecond and if I were to track this nanoparticle within 1 nanosecond a distance delta x same particle mind you this is a and a but this a is after time t so I have got delta x by delta t I get velocity this is non-intrusive so it is we are not introducing any probe or anything so people I have measured with particle image velocity symmetry and laser Doppler animometer where in which they are non-intrusive and people have found that in fact when velocity distribution was figured out they were all done with either Prandtl probe or hot wire animometer but later on these universal velocity distribution has been confirmed even through non-intrusive measurement one thing we should be able to appreciate is that any measurement is going to be having some uncertainty no measurement is a perfect measurement so every measurement is going to have problem near wall nevertheless people all over the world in different places have measured and find that universal velocity distribution that is in the buffer layer is what we have got so that is the answer for the second question third question why is it called Reynolds transport theorem see whatever loss we have studied for example conservation of mass mass of the system is conserved that is system system is not moving here but what is happening if I take my fan as my control volume air is coming in and getting out what am I tracking I am tracking a fluid particle which is moving in and out of a control volume so whatever I apply to a control volume into which my fluid particle is going in and going out so that so that is Reynolds transport theorem why transport because my fluid particle is getting transported why Reynolds may be perhaps I this I do not know why may be Reynolds only has derived this I do not know but perhaps either he has derived or in his honor it has been named as Reynolds transport theorem one of the easiest or best examples of this is that if you have a balloon in which you are trying to blow and there is a leak so your mass is coming into the control volume the control volume is changing shape and at the same time mass is leaving the control volume. So, this Reynolds transport theorem basically looks at this whole thing now in m out and the change of shape of the control volume because you have transformed from a system to a control volume whose dimensions can also change that is also there is that ok. I have left out another question q dot not considered ok. So, if q dot is there that can be taken as a source term in the energy equation see why for example, in combustion if there is combustion there is net chemical energy release we add that as a chemical energy release that is source term on the right hand side of our energy equation. So, that is how even volumetric heat generation also can be accounted there is we have not taken volumetric heat generation because we want to be as generic as possible that is all the same way we took the energy generation term in the heat diffusion equation if you see here one important thing what I have missed out is if you see the energy equation and if I put u equal to v equal to 0 what is that I get I get the heat diffusion equation only. So, that is essentially the answer for your question to take the volumetric heat generation is that ok. So, for Reynolds transport theorem can we apply to mass diffusion case or is it only for fluids or can we apply to any larger areas further areas. Will you repeat can you apply to this Reynolds transport equation can we apply only to the fluid can we apply to fluid cases heat generation cases heat cases or in any other cases so suppose there is a mass diffusion case that is mass transfer cases also can we apply this type of equation. See the question is can we apply the Reynolds transport theorem for mass diffusion also yes Reynolds transport theorem can be applied for mass diffusion also if you are handling combustion you have to keep track of each species. So, for each species you are going to write Reynolds transport theorem only that is again conservation of mass and conservation of momentum. So, Reynolds transport theorem can be applied for mass diffusion also what we need to realize is that Reynolds transport theorem can be applied as long as my fluid as long as my fluid particle is moving ok. So, let us start off with the tutorial problems. So, we will take up one problem I know there will be lot of questions today I would encourage you to ask us as many questions as possible. So, we will solve less of problems, but more of questions today till 5 30. So, let us take up problem number 27. So, what is problem number 27 saying? So, I need to experimental measurements of the convection heat transfer coefficient for a square bar in cross flow for a square bar in cross flow air is v velocity v and l is characteristic length l is 0.5 meters. So, now in the first case h 1 bar equal to remember bar means average heat transfer coefficient 50 watts per meter square Kelvin and when v 1 is 20 meters per second and h 2 bar h 2 bar equal to 40 watts per meter square Kelvin when v 2 is 15 meters per second ok. I think it is natural heat transfer coefficient has decreased with the decrease of velocity ok. Now, the question is a Nusselt number is of this functional form a nu bar equal to C r e to the power of m, P r to the power of n. What will be the where C m and n are constants without telling we have been thinking that since morning I mean we have been telling that since morning. What will be the convection heat transfer coefficient for a similar bar with l equal to 1 meter when v equal to 15 meters per second that is the question asked. So, how do I go about this problem? First of all what is that I need to find that is when v equal to 15 meters per second and l equal to 1 meter and bar is similar to this there is no doubt about that. So, now assumptions I think we do not have to write we will just state steady state by now we know steady state and we are taking average heat transfer coefficient that is a nu bar or h bar and properties are assumed to be constant. Now, for external forced convection flow r e l equal to v l by nu that is the first thing nu is mu by rho ok. So, now what is that we are saying h bar l by k equal to h bar l by k equal to C r e to the power of m P r to the power of l. So, h bar equal to k by l C r e l to the power of m P r to the power of l we can now we will keep it like that r e l to the power of m P r to the power of m. Now, h 1 bar by h 2 bar let us take we have two cases given h 1 bar by h 2 bar h 1 bar by h 2 bar equal to k by l 1 C we do not have to C is constant for both r e l 1 to the power of m P r to the power of m upon k by l 2 C r e l 2 to the power of m P r to the power of m. So, k f k f gets cancelled C C gets cancelled P r P r gets cancelled. So, I am getting l 2 by l 1 l 2 by l 1 into r e l to the power of r e l by r e l 1 upon r e l 2 to the power of m. So, what is h 1 bar given h 1 bar is given to be 50 and l 2 equal to l 1 no. So, l 2 is equal to l 1 l 2 is equal to l 1 let us expand h 1 bar is 50 by 40 h 1 bar is 50 50 upon 40 equal to l 2 l 1 is same. So, l 2 equal to l 1 they will get cancelled out r e l 1 we can write v 1 by v 2 because viscosity is same for both the fluids. So, it will get cancelled out and l 1 equal to l 2. So, that is also getting cancelled out. So, we are left out with only v 1 by v 2. So, what do I get now? So, I get v 1 by m I get that is 20 by 15 to the power of m that is m equal to 0.7825 0.78 yeah this is how one can if you know if you somehow experimentally measure the heat transfer coefficient we can get this power m that is I think this is the this is we are not reached yet. Our question now asked is what will be the value of now we can write h bar equal to I think we will write what is h bar when v equal to 15 meters per second and l equal to 1 meter let us write this. So, that is l equal to 1 meter per second. So, now let us write h bar upon h 1 h bar b for velocity equal to 15 meters per second for 15 meters per second velocity equal to and l equal to 1 meter. So, h bar nu for velocity equal to 15 meters per second l equal to 1 meter. So, h bar nu upon h 1 bar equal to k by l 1 k l 1 it is there we can just take that functional form l 1 by l l 1 by l l nu into r e nu upon r e l 1 to the power of m which is 0.7825. So, now h bar nu equal to what is h 1 bar 50 into l 1 is 0.5 where did I get this 0.5 from see the question in the question it is given as l equal to 0.5 this 50 is for 0.5 in the diagram in the problem. So, r e nu is v v l by v 1 v nu v nu l nu upon v 1 l 1 to the power of m. So, now h nu equal to h nu equal to h nu equal to 15 into 0.5 by 1 into 15 into 1 meter 1 upon 20 into 0.5 to the power of 0.7825. So, h bar is equal to 34.3. So, second part we are not doing because the same procedure you can apply all that is as these length is different and velocity is different. So, answer is h bar equal to 59.06. So, we are not going to solve this problem second part because it is a similar exercise question is what the results would the results be same if the side of the bar rather than it is a diagonal were used as the characteristic length. If the side of the bar is used as the characteristic length no only thing is that we can use side of the bar as the characteristic length no issue only thing is that my constant would have changed otherwise my constant would not have changed that is all that is all c m n will be n is independent of the characteristic length why because n is parental to the power of n which is a fluid property, but c and m depend on the characteristic length. So, the way I choose the characteristic length it will depend on the characteristic length. So, we will take another we will start taking questions again back. So, we can go to may be K. K. Vag you have any questions. Yes sir why Reynolds number critical Reynolds number is taken as 5 into 10 s to 5 for this laminar flow a flat plate and why it was 2000 for a pipe. See the question asked is for flow over a flat plate for flow over a flat plate we are taking critical Reynolds number as 5 into 10 to the power of 5, but for 5 into 10 to the power of 5, but for pipe R e pipe we take as 2300 why because these two are different why they appear because we have taken different characteristic lengths in both the cases R e is equal to rho u infinity l c by mu l c for flat plate we have taken l c equal to characteristic length which is the length of the flat plate for the pipe we generally take diameter, but what we should take what we should take we had specified this in Nusselt number also H delta t by K that is what we told. Similarly, for Reynolds number also we should take rho u infinity delta hydrodynamic boundary layer upon mu. If I take rho u infinity delta by mu no matter what class of problem I choose that is external flow, internal flow, natural convection, jet flow or any other flow you can think of they are all going to be of the order of 20 to 30. So, because we are taking different characteristic lengths not taking appropriate hydrodynamic boundary layer thickness we get different critical Reynolds number, but why do we do this knowingly why because hydrodynamic boundary layer thickness is not known. What we know is the characteristic length that is the engineering parameter that is either the diameter or the flat plate length that is the reason why we take flat plate length or the pipe diameter because of which I get the critical Reynolds number different. So, there is another question this is related to why we are considering only the fluid properties in deciding the flow whether it is a laminar or turbulent why not the surface properties. The question asked is for deciding whether the flow is laminar or turbulent or transitional why we are taking only the fluid properties why not the solid properties. Now who is offering resistance for the flow to take place because of the fluid viscosity it is not whether it flows on a copper plate or it flows on a stainless steel plate or it flows on an aluminum plate who is creating resistance it is not the solid property material by virtue of which it is being resisted it is because of viscosity. If there is no viscosity then my fluid would have simply slipped over the plate whether it is stainless steel or whatever it is because of viscosity only it is getting no slip condition we are coming across because of which only we have hydrodynamic boundary layer or the thermal boundary layer that is no slip and the no temperature jump condition. So, there is no bearing or there is no connection whatsoever between the solid properties and the boundary layer boundary layer is only because of the viscidity or the viscosity of the fluid. One more last question sir it is not expected to have a hydrodynamic boundary layer but the thermal boundary layer is expected for the heat transfer hydrodynamic boundary layer is not expected thermal boundary layer whether it is desirable sir for heat transfer. See there are the participant is asking me two questions whether hydrodynamic boundary layer is expected or thermal boundary layer is expected I do not know what do we mean by expected but what I feel by the question is that when I say expected we are not supposed to have the boundary layer that is when I have a thermal boundary layer I am not supposed to have the hydrodynamic boundary layer that is not possible it is expected both are expected why because there is resistance for the flow and there is resistance for the heat transfer to take place. So, because of the conduction in the thermal boundary layer there is del t by del y and because of the resistance offered by the flow there is within the boundary layer that is hydrodynamic boundary layer there is velocity. Now is thermal boundary layer a desirable thing no not at all a desirable thing because that is the resistance what is f for fluid dynamics that is Nusselt number for heat transfer what is delta for hydrodynamics that is for fluid flow that is delta t for heat transfer. So, both delta and delta t are expected but are unwanted but they are there they are there nothing can be done. So, we have to overcome them then the only way to overcome them is either to have a fan or a pump. So, that is why our pumping power otherwise this pump should have never been invented if there were not to be any boundary layer it would have been just going by slipping they will just go by because of the slippage it just by you give the head then they will just flow once you generate the flow they will continue to flow there is no question of any resistance or getting the flow stopped somewhere this is because the resistance is there inherently for the heat transfer or for the fluid flow to take place boundary layers are unwanted but they are there and they are going to limit us the amount of the heat transfer or amount of the what is that fluid flow why today laptop or anything miniaturization is not possible let us come to application essentially because of thermal analysis because we are not able to cool the chip the heat fluxes are going up but my chip temperature is expected to stay at 30 to 40 degree Celsius otherwise it is not going to behave the way it is designed to behave the heat fluxes are going up but my temperature is expected to be at 50 degree Celsius people generally take this example if you take the sun's temperature sun's heat fluxes are of the order of 50's or 100's kilo watts per meter square what is the sun's temperature 5500 Kelvin but same amount of the heat fluxes we are expecting on my laptop but what is the temperature I am expecting to maintain it is 50 degree Celsius not an easy task that is why we are not able to make my laptop as small as possible that is why their life of the laptop for example is much lower than that of desktop because I am able to cool the desktop even why desktop also if you do not put properly in an AC room if you just put in an outside its life gets affected because of cooling okay ma 90 Bhopal have a question how can we correlate the bondy layer thickness developed over the fin and the effect of this thickness of the layer be on the distance between the two adjacent fin okay I think can we postpone this question why because we are not told the concept of developing flow so if we complete internal flows we will be able to the question asked is if on the fins how does the boundary layer start and stop and in between fins how does it grow actually we have to understand the concept of developing flow so I think we will take up this question after we cover the internal flows all that I can say is if you know if you are very good at internal flows this is the answer it is continuously developing flow it is never a fully developed flow but after sometime it may be reached periodically fully developed flow you will not be able to understand what I have told you will understand I will after I complete internal flow any other question please quickly see generally I just want to make one clarification morning we had one question we have not taken unsteady terms in the non dimensionalization of both the momentum and the energy equation even if you take the non dimensionalization for example del u by del t if you take so what will you get u instead of u you will get u infinity u star you get u infinity outside del u star u infinity outside and del by del t for t you take it as t star equal to t upon l upon u infinity so you get u infinity squared by l so that is yeah so if you if you see non dimensionalization that is del t by del t if you will get rho c p rho c p u infinity rho c p u infinity by l so you are going to get del t star by del t star so you are not going to get any other new term that is all I want to add here see you are going to get you want to write that is del t star rho c p del t by del t equal to rho c p del t star t star sorry t infinity minus t s into del t star upon del t star where t star equal to t star is equal to t upon t upon u infinity by l no l upon u infinity characteristic length l upon u infinity yes l upon u infinity l upon u infinity let me so yeah so l upon v so now if I substitute this in the top for t star t star so into t star into l by v that is in the denominator l upon v so I get this is equal to rho v c p upon l rho v c p rho v c p upon l t infinity minus t s del t star by del t star we are dividing throughout by this term so what am I going to get I am going to get del t star by del t star nothing new so no new non dimensional number is going to come either in the momentum equation or in the energy equation for unsteady flows as well that is the that is the point I wanted to professor Sevetkar had asked this question in the morning I had not given such a convincing answer but this is the convincing way to look at this so whether I take unsteady flows or steady flows I am not going to get a new non dimensional number both in momentum and energy say lump will thermal boundary layer separate like boundary layer separation in case of hydrodynamic boundary layer separation like boundary layer hydrodynamic boundary layer separation will thermal boundary layer also separate the question asked is will thermal boundary layer separate like hydrodynamic boundary layer yes if there is sudden obstruction if I am asking it to go through ribs, if I just draw ribs let us say, let us say I take flow over a flat plate with two ribs let us say two ribs. So, here velocity boundary layer is separating and then I have again regrowth of the boundary layer. This is same thing is happening both for not only for velocity boundary layer, but also for thermal boundary layer. So, thermal boundary layer also separates wherever velocity boundary layer separates that is how I think we can look at it ok. Sir, my question is that suppose there is a flow in a flat plate and there is a flow of a that is gases flow over our flat plate have high velocity gases and temperature of the gases also high. Suppose we injecting a nozzle through nozzle of water what is effect of that flow in that plate. The question is the question is there is a flat plate over a flat plate high temperature gases are flowing. Now, all of a sudden or gradually I have injected water mist or just water over this. So, we are what we are asking we are asking for two phase flow what are we studying we are studying single phase flow. So, when in a two phase flow it is actually interaction between the gases and the liquid. So, that we are not covering in any of the fundamentals. So, but anyway to answer the question what is happening is then there is heat transfer between the gas and the plate and water and the plate and gas and water also. So, it is the interaction between the gas plate water plate and gas water. So, it is not going to be easy task to answer this question and there is no closed form answer for this. So, but then it is a two phase flow problem all that I can say is it is a two phase flow problem wherein which I need to take the interfacial shear stresses and interfacial heat transfer coefficients also. So, you are saying gas is very high may be my water will get evaporated also. So, I really do not know and if my water and the plate temperature is very less may be my water will get condensed also sorry. So, either ways whichever has got evaporated back again it will get condensed. So, it is a very difficult task it all depends on the numbers what I choose. So, it is essentially the sum and substance of this question is that it is two phase flow problem. Sir, can we apply the N S equation over here? N S equation and then energy equations. Question is can I apply the energy equation over here? Yes, but I have to apply the energy equation separately for gas separately for water and then couple them through that is what I have been saying interfacial shear stress, interfacial shear stress these two get connected through interfacial shear stress and interfacial heat transfer coefficients. That is what happens in casting that is what happens in melting. So, wherever two phases are there interfacial shear stresses and interfacial heat transfer coefficients come into picture, but energy equation, momentum equation, continuity equation separately for both the phases can be taken.