 So we will take couple of questions here may be 3-4 questions from various over to Mofakkam ja college Hyderabad. Actually my very popular question, the first question goes in this way, when I take the oil industry in specific, the order goes in this way the gas, first is gas, the next one is oil and the third one is water. So as it is clear in the connection the velocity decreases as we go down, so when the velocity decreases when we bring oil to take out the crude oil, how can the oil comes so easily when there is velocity decrease, this is my first question. The question is little not, it is little more application oriented, so all that I would, the question is there is oil moving over which air is moving and perhaps some other fluid question is not very audible, nevertheless there are multiple fluid flowing and now how do I handle the heat transfer coefficient or the friction factor in this sort of situation. In fact I had answered similar question in on Friday, I said that the shear stress or the temperature gradient on the side where the oil is there is quite different from on the side where the air is there, that is number one and number two there is going to be interfacial shear stress that is wherever there is an interface between oil and air there is an interface, so there is an interfacial shear stress and also temperature gradient. So I know I am not completely answered but the fundamentals what we have taught here is only for single fluid, we are not handling two phase flow, this comes under two phase flow maybe we will touch and go when we touch boiling, but I would not say that we will be touching two phase flow, here we are interested in single phase flow. I would like to take next question. Yeah next question is when the oil has come out from the well and it is sent to the refinery through the flow lines, so at the emission pressure for example it is P 1 assume that it is somewhere around 20 bar and at the exit which is at P 2 is less than the P 1. Now my question over here is when the flow is steady flow and how come there is a change in velocity when the corrosion layer is there in the flow line. The question is I am supplying the flow from high pressure 20 bar to almost atmospheric pressure where I am supplying, the question is how come this velocity, how come this pressure decreases even in the steady flow. Professor you will have to wait for internal flow but I can tell very easily that frictional losses will account for this pressure drop that is all it is but still you will understand much clearly when we handle this in internal flow. All that I can say for now is that frictional losses would compensate or are the culprit which create this pressure drop. Of course, 20 bar is too high may be 20 bar is required depending on the length if it is running for few kilometers yes because it is going through valves, bends, straight pipes so you have major losses, minor losses all put together may be it will contribute to 20 bar is that ok professor. If I want to have a steady flow the pipe decreases in the velocity what I want to do is I want to remove the corrosion because the oil contains more and more amount of sulfur. So, it corrects the pipe so easily and the thickness of the pipes it reduces. So, what I want to do when I have to remove the corrosion simultaneously the velocity should not decrease and simultaneously I have I am sure that I should maintain the same Reynolds number. See we are asking for something the question asked is one of our participant is very enthusiastic to remove corrosion and he says that he wants to utilize a high velocity and remove the corrosion. So, what he says is that when the corrosion takes place in the pipe it sediments essentially what is happening here is the area decreases but he does not want to decrease the velocity wherever the area decreases there but that is not possible it is it if corrosion takes place it is going to decrease the cross sectional area I have to satisfy continuity equation. So, my rho and a is my rho is fixed because if it is oil and a area is decreasing there is no other option but for the velocity to increase locally. So, sorry we cannot we cannot avoid corrosion corrosion is part of our life I have been told that in one of the power plants in steam power plants when they installed a steam power plant they had they had they let us say they were generating 1 mega watt but after 5 years there was a decrease in power by almost a 10 percent because of the frictional losses because of corrosion ok. Next question see this is what we are always TS minus T infinity we are using in forced convection what do we do with free convection let us cross the bridge when we reach the bridge we have not reached the bridge yet ok free convection is out of scope now. Next question please. We have done averaging for continuity equation and momentum equation and the same averaging technique can apply for energy equation. I think the question was we have done some kind of averaging for continuity and momentum equation is the same kind of averaging applicable for energy equation am I right yeah it is it is definitely valid. So, in fact that is how when turbulence was introduced you had u prime v prime bar here now you will have u prime t prime bar v prime t prime bar so on and so forth. In fact that we are just going to show that slide to you where turbulence is taken care in the energy equation just there yeah here the heat transfer rate is there rho C p v prime t prime bar that term that is essentially the transfer of energy because of the fluctuation of temperature coupled with the fluctuation of the velocity and that k t we will we will give that as we want to write it in terms of a derivative. So, we will call that minus k t d t bar by d y and this minus k t essentially is not a thermal conductivity it is not a material property it is not the fluid property it is encompassing the fluctuations associated with the velocity and the temperature. So, that is it is called as turbulent thermal conductivity that is all ok. So, I would go one step ahead and say we have not Reynolds averaged the energy equation why do not you take that as a homework and average it not why you only v prime t prime bar we will get u prime t prime bar also it depends on situation to situation whether u prime t prime bar is significant or v prime t prime bar is significant just to give the feel that v prime t prime bar is the one which overtakes over k del t by del y that we have given this like we said laminar shear stress is mu del u by del y here also laminar heat conduction we can say that it is minus k del t by minus k del t by del y and turbulent conduction is rho c p v prime t prime similar to what we did for turbulent shear stress minus rho u prime v prime bar ok. Sir, I will convert to slide number 53 second you explain that simplification process. One of the participants wants how did I simplify in this transparency ok what are we doing here let me first do the bottom one and then do the top one because now it does not matter in which order I go. So, del u v bar what is u u bar plus u prime into v v bar plus v prime whole bar that is this gradient has to be differentiated, but we know that I do not have to differentiate the gradient because we have already showed that del u v bar by del y or del u v by del y whole bar is equal to del u v bar by del y I have already shown that from one of the relations. So, what do I get that? So, this is u bar into v bar that will be there u prime v prime bar that will be there, but u u bar v prime plus v bar u prime if I am averaging what will happen to u bar u bar will come out. So, if I am averaging v prime bar we have shown already that averaging time averaging a fluctuating component that is whether it is u or v u prime bar is going to give me 0. So, I get I get this is 0 essentially because it is u bar into v prime bar v prime bar is 0. So, this term becomes 0 similarly here v bar into u u prime bar which is again 0. So, that is why these two terms become 0 is if this is understood the top thing also you should be you should have understood is that ok professor. It is a top one sir u bar square by del x equal to 2 u bar dou u by del x u bar sir that I am not able to understand. No problem one of the participants question is how did I get this 2 u bar del u bar by del x? If I differentiate x squared what do I get 2 x? So, but then u bar again is a function of x. So, I get 2 u bar del u bar by del x that is already is that ok professor over to Amrita for questions. Sir, this is regarding the universal velocity profiles go ahead go ahead. So, this is regarding the universal velocity profile for turbulent flows. So, there you have non-dimensionalized velocity and y with u tau and that requires wall shear stress to be measured. So, how do you measure this wall shear stress? And one more thing is it a constant first of all or is tau w a constant or if it is not a constant then how is it obtain it each layer? So, one of the questions is very good question professor Vivek is asking us that in the universal velocity profile in the universal velocity profile we are writing u plus as u by u tau and u tau is square root of tau wall by rho. Now, how does one measure this tau wall? So, what did we say is that in the boundary layer let us say over a flow over a flat plate I have u infinity. So, what did we say initially we have laminar sublayer in the laminar sublayer in the laminar sublayer what is tau wall? Tau wall equal to mu del u by del y this is flow over a flat plate there is no del v by del x because v equal to 0. So, del u by del y how do I get del u by del y I can put a small probe very difficult, but there are some small probes which you can insert and measure but very difficult to measure within the laminar sublayer. So, what people do is people take hot wire animometer which I had told in the last class to for measuring the turbulent stresses people take small hot wire animometers go as close to the wall as possible and measure this velocity gradient or if you cannot go as close what do you do you scale up the setup you scale up the setup such that for y plus equal to 5 my y is almost of the order of 5 to 6 mm. So, then you can take a small pitted probe and put the pitted probe and traverse the pitted probe traverse the pitted probe near the wall and get the velocity gradient that answers for laminar sublayer. Now, I have to go to turbulent boundary layer and buffer layer let me go to turbulent boundary layer in turbulent boundary layer life is easy turbulent boundary layer why life is easy I say because you can take the probe. So, but then definitely you cannot measure with simple pitted probe I will answer that little later why I cannot measure turbulent stress with simple pitted probe hot wire animometer as I said it is having less thermal inertia. So, it can capture the velocity fluctuations. So, I will measure at every location u versus time and v versus time. So, I get u prime and v prime at every instant of time and then I will multiply that u prime and v prime and time average that I will be getting u prime v prime. So, this is how one can measure either with hot wire animometer and of course, I do not want to get into laser Doppler velocity or particle image velocity it gets little complicated because I have already explained how hot wire animometer works this is how one measures u prime v prime and that is how one measures the shear stress no matter whether I am in laminar sublayer or in turbulent boundary. If these two are covered I have covered buffer layer as well is that ok Vivek. Your next question was yes sir, because what happens is in order to identify the three layers you require y plus which again request out of. Yeah, yeah the question is again professor Vivek is asking us the question that for identifying y plus you have to identify u tau yes you have to identify if I have measured the shear stress. If I have measured shear stress what do I get I have got the shear stress I can get u plus that means if I have measured shear stress means I have got u also I have measured shear stress and I have measured y as a function of y is that ok. We have got shear stress and velocity as a function of y what is u plus? u plus is u by u tau tau, where u tau is square root of tau wall by rho. Is that right? So, now what is y plus? y plus is y u tau by nu. So, I have already got u tau. So, I can get y plus. Now to answer the question how do I identify? When I am putting my hot wire animometer and I am putting my hot wire animometer, I am in laminar sub layer. There will be no u prime, v prime. There will be very less. Once in my layer u prime and v prime become significant, then very well I know that I am in turbulent boundary layer. Now I do not even have to know. Now because I even if I do not know this, if I plot the shear stresses the way I have measured in terms of either the velocity gradient or the turbulent shear stresses, if I plot them I can identify u plus and y plus. And to answer your next question are they going to be constant? No, they cannot be constant. The shear stress is going to vary with y, no matter whether I am in buffer layer. It is going to vary with x also and it is going to vary with y also. You want to ask any further question? You please go ahead. Any other question from Amrita? Sir, one more question. On Friday one of the professor was asking you about how we can change the boundary layer problem. So you and the professor Arun was telling that if in case of fixed flat plate you drill a hole either with the help of suction or blowing you can affect the boundary layer problem. But in case of that your u infinity or free stream velocity will also vary. Again free stream velocity will have an effect on the boundary layer also. Suppose if we take a open circuit wind tunnel where the close test section we can take into consideration which each section each side will act as a if we take a square section each side will act as a free stream velocity. If I am going to drill a hole at the very of the wall it will again create a my flow will not be a laminar I suppose I never tested it drilling a hole into my test section the u infinity will vary and again the u infinity will have a effect on the boundary layer. So how much far we can suggest that I am not sure I just want to know. See one of the participants question is that for playing with the boundary layer we had suggested that on the other day we had suggested that we will take we will either on that day we had told that we have a flat plate we have a flat plate and there is u infinity and we if we drill holes we will either suck or blow and that is how I can control the boundary layer. Now one of our professor is having is skeptical about this idea and he thinks that this is not practically possible for the simple fact that my u infinity is going to change all over the place. Now first of all I would like to answer for this question that this suction and blowing is very much in the real life application before I come to how to operate this in wind tunnel I will come to that little later for you specific to the wind tunnel but in real life this is very much an application where it is an application at least the place where I know it is film cooling what is film cooling if I take a gas turbine blade if I take a gas turbine blade and this is my convection passages that is through each of these passages perpendicular to this board flow is taking place what do they do to keep they they drill hole there is remember there is flow here there is flow taking place over the blade and of course my blade is rotating that is a different thing for a minute let us take this blade as stationary no issues now if I drill a hole from here what is happening what is happening when I drill a hole it is there is a air blown blown out and there is a small film forming all over my blade why so that I keep my blade protected remember this air which is jetting into the jetting outside or getting out of the blade is actually playing or meddling with the boundary layer both thermal and hydrodynamic both thermal and hydrodynamic and it is forming a film so it is not this suggestion of controlling the boundary layer is not out of the world is very much applied if you cannot feel this film cooling I give this example very generally all of us when we take out potato from cooker how do we cool it how do we peel off the potato this is what I used to do whenever I used to help my mother she used to ask me to put below the tap I just open the tap slightly and put it below the tap and so that my hand does not feel the hot temperature of the potato why because the tap water flow which is falling on my potato forms a boundary layer forms a film and that is what is protecting my skin essentially that is what here is we are doing there is a film being formed now coming to your second question of course you never drilled in your wind tunnel that says that you have experience in wind tunnel so now I can assume that and answer that yes you infinity also has to be measured every location wherever you have drilled the hole you infinity is not going to be the same though and I have to have how the you infinity varies also so it is going to be a messy affair nevertheless in fact if you see professor you want textbook for suction and blowing for flow over a flat plate for laminar boundary layer you have closed form solutions you have closed form solutions if you cannot get this you want if you are interested to get this closed form solutions put this question in moodle we will get back to you with the solutions but the summary some and substance of all this discussion is that suction and blowing is very much of very much a solution and we do play with this in most of the heat transfer applications over to Anna University's for questions hello yeah yeah the Reynolds equation which term is negligible for high high speed flow or which term is predominant one of the questions asked by one of the participants is that see in Reynolds averaging luckily I am here only in Reynolds averaging which of the terms are negligible for incompressible flow or which of the terms are important for compressible flow if you see here carefully I have not Reynolds average density also I have to if density variations are important I should have I should have written density also that is density also as rho bar plus rho prime I have not done this I have not done this the very fact that I have not done this itself means that whatever Reynolds averaging I have done is it is for incompressible flows only so so if I have incompressible flows then I have to take the density variation also then in that case if you see if you carefully see what is the continuity equation I have taken I have taken del u by del x plus del v by del y plus del w by del z the very fact that I have taken the continuity equation as this means that I have taken compressible flow so incompressible flow it is not compressible flow ma'am I would request you to take the compressible flow equations because momentum equations nothing is going to change only the continuity equation is going to change the density variation will come into picture so that is all that is I will have d rho by dt and in d rho by dt that is the total derivative remember in the total derivative I will have density fluctuations so my equations is going my equation is going to be quite lengthy and it is quite difficult to characterize in a compressible flow all the turbulent fluctuations capturing density because density means what no sensor is there which will give me the density directly I will have to measure the pressure fluctuations I will have to measure the temperature fluctuations temperature fluctuations it is quite easy to measure with a simple thermocouple of course it is not straight forward because you have to have a thermocouple as small as 25 micrometers telling is very easy as I told on the other days 75 micrometers is my hair diameter if I take a thermocouple I have to have a bead of thermocouple bead of 25 micrometers the question now is pressure, pressure measurement is going to be very difficult because transducers come very fast in fact I skipped this question what professor asked Professor Vivek had asked why I cannot measure with pitot tube I cannot measure with pitot tube the pressure fluctuations because whatever pressure I measure I have to take it through a tube and take it into my pressure transducer, pressure transducer will be fast but all my pressure fluctuations will get dampened out in my tube so my pressure tube or the tube through which my pressure fluctuations are being captured is quite difficult to put it very simple way capturing density fluctuations experimentally is quite difficult but there are always we cannot just say that they are difficult and do away with it or put it put them under the carpet we do some take larger tubes and take faster sensors and take smaller length of the tube and try to measure these pressure fluctuations but these pressure fluctuations cannot be typically faster than 100 hertz that is that is the typical typical sense typical way we circumvent these density fluctuations is that okay ma'am any questions so when we refer Reynolds analogy in text books the conditions provided are pronders number equal to one and the form drug is negligible or zero so what is the reason for the second condition okay the question asked by one of the professors is that he says that you know what he says is he says that in the text books for the Reynolds analogy to be valid parental number should be equal to one and form drag should be equal to zero yeah see form drag means actually it is the pressure gradient drag is of two types one is pressure drag and pressure drag sorry I am very impatient drag is drag is equal to pressure drag plus viscous drag before coming to this drag let us take your first question parental number equal to one see professor has already told us professor has already told us that these equations will look similar only when the momentum and the energy equation if you see if you see these two equations it is quite from this equation itself one can tell if you see these equations when these two equations look similar when or maybe instead of this maybe we can take these two equations let us this is the best two form non dimensional form this is momentum equation and this is energy equation when will these two equations be same when dp by dx is zero when dp by dx is zero and another one is when parental number equal to one so let me come back if parental number equal to one what will happen this looks one by REL del squared u star by del y star del y star squared and this also becomes one by REL del squared t star by del y star squared that answers your first part of the question why parental number should be equal to one because the momentum equation and the energy equation form will be same now another thing what is not there in the energy equation is dp star by dx star squared dp star by dx star equal to zero means what there should not be pressure variation that is what is that is what is meant by pressure drag or form drag that means there should not be any pressure variation if you take for example I will in the next class I am going to tell under what conditions this pressure drag is going to be zero that is only for flow or a flat plate pressure drag is zero if you take for a for example flow around a cylinder it is not equal to zero so flow around a cylinder we cannot apply Reynolds analogy for flow over a flat plate we can apply Reynolds analogy so this is what this is the answer for your question professor okay so let us go to shirpur or Shri Patel shirpur good afternoon sir so my question is that in case of in case of thermal boundary layer if we are going to consider boundary layer theory in case of aircraft is this theory is applicable and how so one of the questions by one of the participants is that we are assuming steady flow whether it is thermal boundary layer participant actually is asking for thermal boundary layer he is asking whether the steady flow whatever we are assuming if I take an aircraft it is valid yes let us see how does the aircraft reach steady whenever it takes off I am talking about a domestic aircraft I am not talking about a fighter aircraft if I take a domestic aircraft if I take velocity versus time when it takes off it picks up when it takes off we all have experienced today we have experienced that when it takes off actually it accelerates like anything and we are very uncomfortable because we get affected physiologically so it takes off and it picks up the velocity and typically it reaches around 700 to 800 kmph which is a Mach number of around 0.4 or 0.5 and that is where it stays and while again touching down what is that called landing while landing again the velocity decreases during cruising during cruising in the air during cruising in the air it is steady state the boundary layer in the on the wing on the wing during cruising is steady state so whether it is thermal or hydraulic yes they are also hydrodynamic thermal boundary layer is there remember the temperatures there are minus 20 degree Celsius I do not know if you have watched while traveling there are temperatures and pressures given pressure is very low compared to atmospheric and temperature is minus 20 but my plate temperature is very high so may be after sometime it will reach steady state but in terms of temperature but there is temperature gradient the point here is during cruising it would reach steady state is that okay NIT callicate you have any questions we have used for non-dimensionalization x star equal to x by L and y star equal to y by L but we are are we taking the same length in both case okay okay okay see the question asked by one of the participants is that for x star and y star we have non-dimensionalized it by one of the characteristic length L see it does not matter what dimension I take what I mean here when I say characteristic length one for x one for y I cannot take I have to take one engineering dimension which I know that is for a for example floor a flat plate it is length of the pipe length of the plate if it is flow inside a pipe it is circular diameter so these are the characteristic length one I would take for example square pipe you would take the dimension of the square so no matter whether you are in x and y when you are non-dimensionalizing you have to non-dimensionalize with one dimension that to a no dimension and that dimension can be only diameter length or the dimension of the square pipe so that answers your question both in the x momentum equation and in the energy equation we are taking a term for the body forces so we are already we are accounting for the pressure and normal stresses and what else we are expecting actually the is that the Coriolis forces or something else okay the question to you sir the question asked is we are already taking surface forces pressure forces and the pressure surface forces pressure forces what is the additional thing which this body force is going to give body force is a new force that is you see when I take a simple circular pipe when the flow is taking place is there Coriolis force no so there is no body force but at the same time if I take a pipe which is rotating when should I take a pipe rotating a gas turbine blade cooling passage a gas turbine blade cooling passage would be rotating so then there will be Coriolis force in case of a pipe which is rotating which will simulate a gas turbine blade cooling passage that is that is one way or otherwise there can be a passage let us say in a furnace or in a furnace when I say furnace let us say that is an induction furnace that means what there might be electromagnetic force so there are special situations where in which these body forces either centrifugal force or Coriolis force or electromagnetic force which come into picture it can be electrostatic force also so these forces generally if I take a flow in a pipe they are not important but if they are rotating or in the presence of electromagnetic force then they become important so only when they are there and they become important we take them into account in our Navier stokes equations PSG college Coimbatore any questions please sir what do you mean by cryogenic heat transfer okay there is one of the questions by the participant is what do you mean by cryogenic heat transfer cryogenic heat transfer is nothing but the heat transfer at low temperature generally what we take into account generally what we take into account is our room temperatures are higher temperatures but cryogenic temperatures are usually lower than the ambient temperature much lower than 0 degree Celsius they talk about 250 280 Kelvin so we cannot we cannot we usually do not come across those temperatures those temperatures which are much much below the room temperature but higher than our absolute 0 that is minus 273.15 degree Celsius is the cryogenic heat transfer can you give some examples are commonly used cryogens cryogenic commonly we do not the common question asked is what are the commonly used cryogenic applications okay so commonly used cryogenics is that if you take any sensor they are not commonly used of course if you take any sensor for example thermal camera or FTIR spectrometer usually we pour either liquid nitrogen so that is actually liquid nitrogen is sitting at very low temperatures I do not recollect exact temperatures but I very well know that it is very much lower than atmospheric that is at least 100 degree Celsius lower than atmospheric so the point is the in regularly for cooling for all sophisticated sensors maybe it is thermal camera or FTIR spectrometer or any other costly equipment which requires local cooling it has cryogenic temperatures involved in fact sterling cycle operated cryo cryo what is the cryo coolers are embedded in the sensor sir Nassan number correlation is non-dimensional Nassan number correlation is arrived based on number of ascensants starts from a steady incompressible laminar without discussed distribution and all but it is Nassan number correlation is available for all for many cases first it out is Navier stokes is capable to represent turbulent also then why it is recited only to laminar the question asked is actually the Nassan number has only Reynolds and Prandtl number but it starts with lot of assumptions that it is valid only for steady flow incompressible flow but Navier stokes we have used can be used is why is it Navier stokes restricted only for laminar flow the answer is Navier stokes equations is not restricted for laminar flow that is the answer first of all let me write see if you see if you see the Navier stokes equation we took the Navier stokes equation we took the Navier stokes equation and did the Reynolds averaging Reynolds averaging and these Reynolds averaged equations are valid for turbulent flows. This is one way of looking at it, but the Navier-Stokes equations in general are valid for both laminar and turbulent. So, if I have to capture very small turbulences, I have to march with time, but if I take steady state, it becomes difficult to capture turbulent. So, that is why we take the Reynolds averaged equation. So, the answer is in spite of making the assumption that it is for steady flow in for steady flow and incompressible, we are going to we are we are able to get the solutions of the Navier-Stokes equation that is what is the solutions of the Navier-Stokes equation that is friction factor as a function of Re and energy equation Nusselt number is a function of Re and Pm, but for compressible flows if you have to take then what will happen? Nusselt number will not only be a function of Reynolds and Prandtl, but also a function of a Kirch number. Why? We have seen when we non dimensionalized for compressible flows, new non dimensionalized number which comes into picture is a Kirch number. So, we cannot say that this non dimensionalization is done only for steady and incompressible. We have done for compressible also. For compressible flow, new non dimensional number which has to be taken into account is a Kirch number that we regularly do not do, but if we see compressible flows, a Kirch number comes into picture.