 So we are into convection, we have 45 minutes for convection for today. So we will use this 45 minutes properly and let us see how much we can cover in convection and we are going to spend next 15 hours almost on convection. So this is as I keep saying in the main workshop, in the co-ordinators workshop also we realized that fundamentals of fluid mechanics are extremely essential for convective heat transfer. Without fluid mechanics we cannot just manage to understand convective heat transfer that is why we had used the word, in fact we have not used the word Professor Incropara and David have used the word fluid mechanics is indispensable, indispensable I cannot dispense fluid mechanics it is indispensable for convective heat transfer. So now let us get started with convective heat transfer without any delay. So the convective heat transfer as we said there are two types of convective heat transfer one is forced convection another one is free or people call this free convection as also natural convection why because it is naturally occurring you are not putting you are not spending anything. So for forced convection I think all of us can see why forced convection takes place because if I am under the fan, if I am under a fan actually forced convection is occurring whenever I am under fan so that is why I get I feel cooled because if I am below the fan but on the other hand in a free convection I do not have a fan or a pump so but how does that occur? It occurs because of density differences and acceleration due to gravity. If I do not have gravity if I am sitting in space natural convection cannot occur at all. So it requires not only density gradients but also the acceleration due to gravity. Who is going to create these density gradients? The temperature difference the temperature difference is going to create this density gradients. I always take this example and I like this example very much that is in summer that is what used to happen when whenever we are in hostel in our hostel whenever I used to go in late evening and I put on the fan I used to get hot air for first few minutes that is because my air my room is has got heated up because of the top wall I was in the top floor and the top wall has got heated up because of that the low the air which was getting heated up the low temperature the high temperature air has moved up because of its low density. So the all the hot air had has got stacked in the top portion of the room so when I put on my fan all that hot air is blown downward that is why I feel hot air for some time even if you would have seen this even in an AC room for some time if you just put on fan all of a sudden this is what happens in the initial phase. So that is how one can feel natural convection. So that is about forced and natural convection. So now let us get to Newton's law of cooling. So that is let me write this q equal to h A into T s minus T infinity when T s is greater than T infinity if T infinity is greater than T s you just this h A into T infinity minus T s because you know that always the direction of the heat transfer is from hotter body to the colder body no confusion on that score. So what is that we are saying in the Newton's law of cooling what is q convection this I think we do not have any doubt this is watts h is convective heat transfer coefficient we have been using this quite quite extensively but all this while in the conduction problem someone was feeding us this h but now the question in convective heat transfer is how do I get this h this is the question mark how do I get this h is the question yesterday in fact any of the questions were asked how h has to be computed from Reynolds number Prandtl number you are directly giving yes in till conduction we are not worried about how do I get this h. Now what is T s this is the wall temperature and T infinity is the fluid temperature. So h is we have noticed so far h is having unit of watt per meter squared degree Celsius now always now the very important thing I want to emphasize re-emphasize I have seen even PhD thesis is making mistake in this issue what I am going to tell that is h is not dependent let me repeat in capital letters I am writing h is not dependent h is not dependent on q and T s minus T infinity please let me restate this no matter if I increase q in any given situation T s minus T infinity will adjust such that h will remain same h do not think that from this equation directly h will increase if I increase q if for a given T s minus T infinity it is not so it can happen only in perhaps in natural convection I am stating this for forced convection for forced convection h is not depend not dependent on not I am emphasizing a lot it is not at all dependent on heat flux or heat transfer rate and delta T then what is it dependent on it is dependent on h is dependent on now we are not non dimensionalized anything I will state like a layman that is fluid properties what are the fluid properties one can think of what are the fluid properties one can think of density viscosity k that is the thermal conductivity of the fluid C p specific heat of the fluid velocity of the fluid okay so later on we will non dimensionalize all of this and say that h is going to be dependent instead of h Nusselt number will be dependent on Reynolds and Prandtl I have not introduced formally Nusselt Reynolds and Prandtl so let us not worry about that h is dependent on thermo physical properties of the fluid that is what is important okay and the fluid dynamic properties so this is to be emphasized a lot when we tell Newton's law of cooling this is what is called as Newton's law of cooling see the contributions of Newton we thought always Newton is only related to falling of apple it is not so he is not only contributed for classical mechanics he has contributed in convective heat transfer okay so now if I come back so that is what this is what is being told here q convection equal to h A s into T s minus T infinity another important thing is here A s A s is the bathing area that is surface area that is the area over which my fluid is bathing if I take this pen if I take this pen if the flow is occurring over this pen so then what is that my bathing area pi into D into L okay so that that has to be kept in mind that is the fluid is bathing my surface so whichever surface is getting bathed by my fluid is my surface area so usually we tend to take pi D squared by 4 no area always does not mean cross sectional area so here it is surface area please note this point at the cost of being repetitive I am telling this okay so after this let us move on to local and average so that is let us just take stock of okay local and average before we go to local and average I think I will introduce the definition of h first so let me introduce the definition of h see in fluid dynamics we are all very convergent when in fluid dynamics if I take a plate a flat plate if I take a flat plate if I take a flat plate and the flow is taking place over a flat plate let us say with a velocity u infinity what would be the velocity of the fluid at the wall 0 velocity of the fluid yeah this plate is stationary it is a stationary plate and fluid is moving at the velocity u infinity so the velocity of the fluid at y equal to if this is my y direction at y equal to 0 at y equal to 0 the velocity is 0 what is this condition called no slip condition this is called as no slip condition why does it not slip because of the because of which property it does not slip because of viscosity so we have understood this that is because of viscosity the fluid has the fluid or the plate offers resistance for the flow to take place and therefore naturally the fluid velocity at the tip of the wall that is just at the edge of the wall it is going to be 0 so this is what is called as no slip condition similarly in heat transfer also we have what is called as temperature jump we do not call it as no slip but this is called as no temperature jump so I think it is quite logical to understand this no temperature condition that means there is no temperature gradient that is what we mean so whatever was the velocity of the plate that was the velocity of the fluid at y equal to 0 same thing here whatever is the temperature of the plate at y equal to 0 would be the temperature of the fluid so that is what is called as no temperature jump condition instead of no slip what we are saying is at the edge of the wall at y equal to 0 t equal to t s that is the surface temperature so this is what is called as no temperature jump condition no temperature jump that is there is no jump or there is no gradient there is no sudden change in the temperature the wall temperature happens to be or that is at y equal to 0 the fluid temperature and the surface temperature are t s that is what is the no temperature jump condition now actual definition I am introducing the actual definition of h h equal to q upon t s minus t infinity is not the definition of h okay but what is the definition of h as you can see here we have we have let me write this on the white board so we have if there is a flow if I have a flat plate and if I have fluid flowing let us say at u infinity and t infinity temperature and this is t s is the temperature of the plate maintained that is constant wall surface temperature so what is happening all of us know that there is a boundary layer and there are two boundary layer one is the hydrodynamic boundary layer and another one is the thermal boundary layer for a minute let us say both of both of them are of the same size when and all we will see that later on that is this is the layer in which there are velocity gradients so that is outside this u that is the location that is the if I take this is the boundary layer thickness delta that is the thickness until which my velocity is equal to 0.99 times the u infinity okay so similarly thermal boundary also for now let us take it as wherever the gradients are not there we are saying gradients are there we are saying that is thermal boundary layer thickness so within the thermal boundary within this thermal boundary layer what is happening what is happening it is conduction there is a temperature gradient and the mode of the heat transfer is conduction mode of the heat transfer is conduction so if I come back so within the thermal boundary layer where there are temperature gradient so q dot conduction equal to q dot conduction equal to minus k del t by del y at y equal to 0 this is conduction within the thermal boundary layer now this conduction has to be equal to q convection which was which is what we had defined by Newton's law of cooling so Newton's law of cooling tells that this is equal to h into I am writing this q double dash actually q double dash equal to h into t minus of t s minus of t infinity I have taken surface temperature larger than the fluid temperature so t s minus t infinity so what is now h equal to h equal to minus k del t by del y at y equal to 0 upon t s minus t infinity now I think we can visualize that h is not dependent on the temperature gradient or the heat transfer gradient what is it dependent it is dependent on the thermal conductivity of the fluid I always try to write or most of the textbooks try to write this k as k f why because we will get confused whether I should take thermal conductivity of the solid or fluid here it is conduction within the boundary layer boundary layer is filled with fluid so that is thermal conductivity of the fluid it can be air or it can be water or it can be oil it can be any fluid which is flowing over my plate so thermal conductivity of the fluid into temperature gradient who will decide this temperature gradient the fluid velocity the fluid thermo physical properties and the boundary condition so that is what is going to decide my boundary condition if you ask the definition of h this is the definition of h definition of h is not q double dash upon t wall minus t infinity or t s minus t infinity that is the relation to measure heat transfer coefficient this is actually the definition in fact why in fact even in laser interferometry people measure del t by del y at y equal to 0 and then compute or measure the heat transfer coefficient if you are doing numerical work when you will run the code you get the temperature distribution all over the place you take del t by del y at y equal to 0 you can take it because you can measure because you have the temperature distribution all over the place that is how you get the heat transfer coefficient so basic this is the definition of heat transfer coefficient so coming back professor is saying that I need to emphasize that h l into t s minus t infinity y r we take t s minus t infinity that is the maximum temperature difference which is occurring so that is t s is the surface temperature and t infinity is the fluid temperature this is the maximum temperature difference compared to that what is the temperature or the heat transfer which has occurred within the thermal boundary layer that is what is going to decide my heat transfer coefficient actually if you really think this h h is something like resistance in fact why we are reintroducing this we have already seen that h is a resistance in conduction in our resistances we have already introduced one upon h a as the resistance so h is resistance for the heat transfer to take place higher the h lower the resistance isn't it so this is h can be thought of as resistance for heat transfer resistance for heat transfer okay so with this I think we have defined what is heat transfer coefficient now I think we will be able to appreciate the concept of local and average otherwise if I have not understood the definition of h I cannot understand the local heat transfer coefficient see always in the last problem in the preview in all the problems so far we made always we made a point to state an assumption that h is uniform that is h is not varying with location so I said even under the fin even in fin's case where we had two circular fins and there was some play 3 mm portion was left out within that 3 mm h need not be uniform and on the fins again h need not be uniform h can be varying from location to location why h would be varying or let us say in this example which is posted here there is a flow over an air of oil there is a velocity there is a fluid which is sitting at a higher temperature so the heat transfer coefficient from one location need not be the heat transfer coefficient at this location need not be same as the heat transfer coefficient at this location why because we have seen that the heat transfer coefficient is dependent on the temperature gradient so that temperature gradient is going to change from one location to another location why is this temperature gradient going to change from one location to another location because my velocity distribution is not going to be same from one location to the another location. So, my shear stresses are going to be different why because my velocities are going to be different. So, there is a change in the velocity distribution because of which there is a change in my temperature gradient that is why my h is going to change from one point to the another point. This is a stagnation point this point will have maximum heat transfer coefficient, but here if you see the heat transfer coefficient would be lower than the heat transfer coefficient experienced by this stagnation point that is which is facing the flow right across. So, that is the reason we have introduced what is called as here average heat transfer coefficient that is the total heat transfer rate is defined on the basis of average heat transfer coefficient. How do I get? I will make the small elemental pieces all over this aerofoil and integrate them I will be getting the total heat transfer rate that is q dot convection equal to q double dash convection d A s integral this is area integral. So, if I put q double dash convection equal to h into T s minus T infinity I am able to pull this out T s minus T infinity because I have taken for constant wall temperature boundary condition here. If it is constant heat flux it is not easy for me to pull out. So, T s is constant. So, T s minus T infinity I can take out of the integral if I do the area integral this is h l d A s, but if I want average heat transfer coefficient I have to divide this area integral by the surface area. So, that surface area if I divide I am going to get the average heat transfer coefficient it is usually written as h bar. So, here I think bar is missing over h there is a dash that is h bar is equal to 1 upon A s integral of h l d A s which is area integral. So, this is the area average heat transfer coefficient. So, the point to be noted here again is that the h from point to point that is from location to location for any configuration why only aerofoil if I take a cylinder for example, if I take a cylinder if I take a cylinder if the flow is in this direction the h here would be higher as compared to the h here. So, h is not going to be same all over the place. So, that is the reason we have to differentiate between h local and h average h average. So, that is the major point we need to emphasize here when we are handling locally transfer coefficient and the average heat transfer coefficient. In fact, so far problems in whatever problems we have solved for as we have handled only the average heat transfer coefficient we never bothered to differentiate between the average and the local. So, next is this is there is a small problem let us spend 2-3 minutes on this problem before we move on to the basics. So, that is this problem says that he has given us the local there is a flat plate again and the local heat transfer somehow someone has made these measurements do not ask me how they have made the measurement that we will worry about when we come to experiment. In fact, there is one very beautiful experiment where in which we are measuring the average heat transfer coefficient using a high slab there is a video I do not know how many of you have watched I would strongly recommend you when to see the video which is an experimental video which is there already uploaded. So, if I get the heat transfer coefficient the heat transfer coefficient is given to be x to the power of minus 0.1 that is the distribution of the h that is how it is varying with x. So, experimental results for the local heat transfer coefficient hx over flow over a flat plate with an extremely rough surface were found to fit the relation hx x equal to x to the power of minus 0.1 where x is the distance from the leading edge it is not form from the leading edge which is the leading edge this is the leading edge. So, this is the trailing edge. So, people usually students keep asking what is leading and what is trailing yes it is English, but still we need to know what is leading and what is trailing this is leading this is trailing because why it is leading because that is the one which is seeing the fluid first. So, that is why it is leading that is all. So, that is the leading edge from the leading edge that is x is starting develop an expression for the ratio of average heat transfer coefficient for a plate of length x to the local heat transfer coefficient hx at x. So, that is what we are trying to do now. So, this is the flat plate you have t infinity hx being varying x to the power of minus 0.1. So, now hx bar equal to 1 upon x because it is area integral I said here although it is area integral I am doing line integral why because this plate perpendicular to the board is same all over it is not varying. So, even if I multiply here into b and multiply into b as the length perpendicular that b and this b gets cancelled out. So, I get the line integral. So, that is 1 upon x yx is in the denominator because I am interested in the heat transfer coefficient for a length x 0 to x hx x dx. Fertuously we have been given the heat transfer coefficient distribution that is x to the power of minus 0.1. So, hx bar equal to 1 upon x 0 to x x to the power of minus 0.1 dx equal to 1 upon x if I integrate this I get x to the power of 0.1 0.9 upon 0.9 I get 1.11 x to the power of minus 0.1. But what is x to the power of minus 0.1 that is the local heat transfer coefficient. So, hx bar that is the average heat transfer coefficient is always 1.11 times the local heat transfer coefficient for this problem. So, that is how one can relate the average and the local heat transfer coefficient. Actually getting the local heat transfer coefficient is very difficult it is not so easy to get through experiments. We need lot of expensive devices or even lot of ingenuity in designing your experiment if you have to measure local heat transfer coefficient. Generally we tend to get average heat transfer coefficient. So, that is what is being plotted here how the heat transfer coefficient varies with x naturally from the distribution itself you can see that x to the power of minus 0.1 that means the heat transfer coefficient is supposed to decrease with the increase of the x that is what is happening here. So, next important thing is Nusselt number I think we will come to this Nusselt number definition little later because we always introduce this Nusselt number when we do the dimensional similarity of the Navier-Stokes equation and the energy equation. So, we will come to this Nusselt number definition little later because we are going to spend quite a bit of time on this transparency for now I am skipping this transparency, but you should not be worried that we are missing this out we are going to revisit this tomorrow. So, I think I will even skip this Nusselt number history. So, I will come back to this little later, but there are some basic definitions which we should be worried about or we should be very confident about before we start off the convection. So, I am just going to go on a fast track I think most of you know these things, yeah classification. So, that is internal flow and external flow that is whatever flow is occurring in a duct that is internal flow the duct can be square duct, circular duct, triangular duct, trapezoidal duct. So, that is internal flow and external flow is flow over a body is external flow. It can be again a flat plate, it can be a circular cylinder, it can be anything again we have to be very thorough with these definitions laminar and turbulent. So, this is a very important definition I am sure many of you have used these definitions very often or quite often so this figure is very self-explanatory. This experiment was designed by Reynolds apparently in University of Manchester that setup which Reynolds did this experiment is still in working condition and the UG students that is the undergraduate students use this setup even today ok that is what I have been told. So, what is the setup consisting of? This setup is having a pipe in which the flow is taking place and there is a flow taking place and I am able to control this flow rate or the velocity or the Reynolds number of course I have not yet defined what is Reynolds number I will be defining little while from now. Now what they have seen is that for certain velocities if the velocities are very low let us say it is water this experiment was essentially done in water. If I put potassium permanganate dye and water is my fluid and if the velocities are very less the dye just follows the fluid that is laminar that is laminae you might have all of us are very convergent with the term laminated sheets why because each laminated sheet is placed on the other laminated sheet. So, the fluid also flows like laminae that is they flow like layers one over the other and one layer is not talking with the other layer one layer is moving independently with the other layer they are not there is no cross talk one layer does not know that the other layer is moving above it. So, that is what that is why it is called laminar you see so thoughtfully people have named the names laminar. So, if you go to the root of the word most of the times if you are good in English we should be able to understand science also. So, I think we should open dictionaries and see the names of all names or the terminologies what we use. So, this is coming from laminae and next is turbulent I will go to turbulent before transition turbulent that is if I increase the velocity inadvertently that is I increase it very high then it happens to become mixing it the dye goes all over the place there is a cross talk from one laminated to the another laminated that is turbulent, but in between if my velocities are not so high and not so low there is a transitional region that is for some time it is laminar and for some time it is turbulent. So, that is the word transitional, but the transitional is not transient let us let us note this because my students whenever we do the fluid mechanics myself and professor Arun take this fluid mechanics lab for transitional region always they write transient region. So, this is because we do not differentiate transitional with transient, transitional is not transient why what is transitional we always use this word transitional this way we are transiting from winter to rainy season or winter to summer season or from vacation to coursework. So, it is transiting from laminar to turbulent moving it is moving from changing transition is a change from laminar to turbulent. So, that is why it is called transitional, but now how do I identify when does it become laminar I said velocity is low velocity is high how high is high and how low is low can be defined on the basis of Reynolds number I am sure all of us know Reynolds number very well Reynolds number is equal to inertia force to viscous force that is inertia force how do I understand inertia force all of us have inertia every morning I saw in today's I visited all the sites today all of the centers I did not see at 5 minutes to 9 also I did not see the participants why because all of us have inertia to come to class and get up and get ready and come. So, because we are in state of slumber we are sleeping. So, there is inertia to get up and come so that is inertia. So, is fluid particle fluid particle wants to be if it is at rest it wants to be at rest only it does not want to move I need to push it. So, that is inertia force I need to push so that it comes out of its slumber if it is moving if I have to stop again it is having inertia I have to apply that much force or I have to restrain it that much so that it changes its state. So, that is inertia force what is viscous force viscous force is because of viscosity because that is the resistance offered if it is highly viscous then it is going to offer me lot of resistance I will explain you in this way we had in our lab we had to transfer a fluid called silica gel. So, we had to do some testing in the lab so we just purchased silica gel. So, I just told my student it was a big can and we needed only little amount of silica gel just hardly 1 liter. So, I asked my student Niles Deshpande who was doing the experiment I just asked him to pour 1 liter of silica gel from that huge container to this little container which had to fill in 1 liter. I went to the class and came back he had not filled at all I asked why you did not fill it had taken so much time because it was 60000 times viscous than the water water I can pour in no time because it is less viscous ok. So, viscous is about fluidity it is having large fluidity that is why I could not pour at all. So, that is what is this it is having too much of resistance for the for itself to flow for itself to flow that is why it is viscous force. So, Reynolds number is the ratio of inertia of force to viscous force. So, that is we will we can derive this from fundamentals in form of differential form I will do that when we do tomorrow the differential form, but for now take my words that is rho V L by mu rho being the density V being the velocity L being the characteristic length you take this characteristic length as for flat plate as the length of the plate for a circular pipe diameter of the pipe and mu viscosity dynamic viscosity we should not get confused between dynamic and kinematic viscosity I if I have to write kinematic viscosity it will become V L by mu nu is kinematic viscosity kinematic viscosity I had touched upon yesterday is going to have the unit meter squared per second it is nothing, but momentum diffusivity. So, this is Reynolds number coming back here Reynolds number is laminar for all Reynolds numbers less than 2300 for transitional flow for Reynolds number of the order of it is it will transit between 2000 to 2300 is transitional flow and above 2500 it is turbulent. So, now people ask me the question is 2000 can I maintain a laminar flow at 20000 yes we can maintain laminar flow at 20000 then what is this 2300 then 2300 is no matter what disturbance I create upstream no matter whether I put a valve or ask it to go through circuitous path whatever I torture in whatever way I torture my fluid it is going to get laminarized itself for all Reynolds numbers lower than 2300 for such listing in his book says that so far for Reynolds number up to 50000 if we carefully design the experiment I can generate laminar flow. So, this 2300 is only for the utmost disturbance I can create even if I create whatever disturbance the flow is going to get laminarized. So, that is what is the that is how that 2300 has come into picture for all Reynolds numbers greater than 2300 it is going to be turbulent. So, now I think we will take the questions VIT Pune. Hello sir regarding the definition of H there are two questions in this you have mentioned dT by divided y equal to 0 but you are referring it to boundary layer where not necessarily y is always 0 1 second in denominator we have TS minus T infinity. So, H does depend on temporary difference initially in a definition you have mentioned it is not depend on Q and temporary difference. The question is we have defined minus K del T by del y upon TS minus T infinity I would still insist that H is not dependent upon TS minus T infinity for a given fluid temperature if I vary my surface temperature the temperature gradient near the wall is going to change. So, still for forced convection I am very cautious when I make this statement this is not for natural convection for forced convection as long as I am in forced convection my temperature gradient will get readjusted based on TS minus T infinity. So, I still insist that H is not dependent on TS minus T infinity. So, can we write it y equal to 0 which is starting of a plate starting of a boundary layer not a full boundary layer whereas this particular thing refers to a boundary layer you said. Will you please repeat the question? Yeah, d by T by d by y you said it referred to a boundary layer where not necessarily y is 0 but we have written y equal to 0 d by T by d by at y equal to 0. No, I did not say that at not necessarily at y equal to 0 until I have the temperature distribution within the boundary layer I will not be able to get this slope at y equal to 0. So, I need to get the temperature distribution within the boundary layer using that temperature distribution within the boundary layer I am going to compute del T by del y at y equal to 0 that is what is going to fetch me the heat transfer coefficient. Okay, thank you sir. Over to Jabalpur college. Sir, what is the lowest limit of the bias number? If I rephrase this question and if I understand it this way the question asked was what is the lower limit of bias number? If I rephrase and re-understand this question perhaps what you mean is to what lowest limit I can take my body as lumped is that right? Okay, so actually there is no lower limit what we are saying is the upper limit for all bias numbers lower than 0.1 with the convective boundary conditions we are saying that it is it can be taken as lumped for all bias numbers greater than 0.1 we cannot make the assumption that it is lumped. Does that answer your question? Yes sir, okay right sir. Over to E Road. Sir regarding this Reynolds number up to 2300 you can say that flow is going to be laminar flow then you are saying that up to 50,000 it is possible to have the laminar flow instead of turbulent flow. So what are all the factors to be reconsidered sir? Okay, the question asked is we said for transition for the laminar flow to convert itself or to transit itself into turbulent flow the Reynolds number has to be greater than 2300. That statement was made and also another statement was that we can maintain laminar flow up to a Reynolds number of 50,000. So there is a contradiction between these two statements that is the question by one of the participants. What we are saying is that we can maintain Reynolds we can maintain laminar flow for all Reynolds numbers up to 50,000 provided I take enough care in not disturbing the flow before it reaches my pipe. But if I put a valve and if I put let us say bends and if I ask it to go through tortuous paths then also even if I ask it to go through these tortuous paths then it will remain laminar only up to 2300. I cannot maintain laminar flow above 2300. If I put a valve and restrain it and ask it to go through orifices, ventures and all sorts of constrictions then it can maintain the laminar flow only up to Reynolds number 2300. That is the experimentally observed number I hope I have answered your question. If I do not have any external disturbance to the flow it might have a very high velocity but if I allow it to settle nicely and come into what you say well ordered kind of flow then it is possible that you still can have laminar. That is what professor is saying is settled away means you ask it to go through flow conditioners. That is you ask it to go through multiple straws and you ask it to go through meshes properly if you condition the flow and take it down you can maintain the laminar flow up to 50,000. That is what experimentally people have taken enough care and realized that. NIT, Tricci any questions? How do you treat the following problems? Porous powder metallurgy metals, porous ceramic materials etcetera. Heat transfer, heat transfer in the case of porous powder metallurgy materials, heat transfer in the case of porous ceramic materials. No that is beyond the scope of this course but still what we can say is that it is having both the air and also the solid material. So, what is defined as like something like void faction porosity as is defined taking the porosity you will write both for air and also the solid medium and write the governing equations. But it is quite involved I cannot answer this question directly in a straight forward manner but what I would request you is put this sort of special questions in the forum we will answer this by proper formulation. Sir you said the hatch is independent of T s and T infinity delta T value whether the physical properties are a function of T s or not. No, no see T s minus T infinity is not going to decide the heat transfer coefficient this have been telling time and again even while I stated the Newton's law of cooling. So physical properties are not going to influence T s minus T infinity what is going to adjust whether I pour water over my flat plate or oil over my flat plate or air over my flat plate for a given T s minus T infinity what is getting adjusted is my del T by del y that is the temperature gradient within the boundary layer is going to get adjusted. So what we need to appreciate is that the heat transfer coefficient is neither dependent on heat flux nor dependent on temperature gradient time and again I am telling this heat transfer coefficient is not dependent on heat flux not dependent on temperature gradient as long as I am in forced convection. So what is deciding my heat transfer coefficient is the temperature gradient within the thermal boundary layer M k triple spoon a. I am having two more questions one how will explain the difference between thermo-enemic and thermo-physical property and how to interpret for the students that is the most important job. So the police have guide us on that regard and second question in case of the redwoods viscometer we are measuring mu instead of mu why? See thermo-physical properties are the properties of the fluid like density, viscosity the question asked was one of the question asked was by one of the participants is that how do you differentiate between thermo-enemic and thermo-physical properties that is number one and if I understood correctly the second question in one of the viscometers the viscosity measured is the kinematic viscosity and it is why. So let me answer first the second question because it is simple so the it is kinematic viscosity because it is actually empirical the whatever if I am correct the which viscometer you are using I do not know but whichever viscometer you are using it will be generally a empirical relation base that is for a given container for a given amount of for a given amount of time if you collect for known amount of the volume of the fluid you will be measuring the amount of the time for collecting that much volume of the fluid and that time will be related to the viscosity. So that relation whatever you have got the relation that is in terms of kinematic viscosity which has been empirically generated so that is why it is kinematic viscosity but otherwise you can generate an empirical relation even for the dynamic viscosity. In fact we have Sable's viscometer in our lab where in which we get it for dynamic viscosity not for kinematic viscosity. So it is all how you have calibrated it not directly either kinematic viscosity or dynamic viscosity and coming to your second question what is how do I relate with the thermodynamic properties and the thermo physical properties. See I think these two are quite different thing thermo physical what are the thermo physical properties we said thermo physical properties are let me use the white board the thermo physical properties are temperature sorry thermo physical properties are density viscosity and maybe I can say kinematic viscosity also which is dynamic viscosity upon density specific heat thermal conductivity these are the thermo physical properties. But what are the thermodynamic properties generally what we define that is either enthalpy internal energy temperature pressure. So I would say that these thermo physical properties are dependent at what temperature and pressure we are talking about that is how I think the thermodynamic properties and the thermo physical properties get related. How to measure thermo conductive of liquids and gases sir. So this is a big question. So actually we are covering this measurement of thermal conductivity in experimental portion. So we have some experiments which I have put up in the power transparencies if you see the course material we have one experiment for measuring the thermal conductivity of a solid and thermal conductivity of a liquid with a thin wire. So we will address this when we are dealing with the experiment experimental portion in the sessions I do not remember where we have that session in the lab session where we are holding the video of the lab experiment I will explain this with reference to that experimental setup. Thank you sir. Over to VNIT Nagpur. Is it so is it so that sir this Newton's law of cooling is always called as Newton's law of cooling or I can say Newton's law of heating also? The question asked by one of the participants is that why is Newton's law of cooling called as Newton's law of cooling can I ask it as Newton's law of heating. I think historically it has been called as Newton's law of cooling but I do not see any reason why we cannot call it as Newton's law of heating because it is used either T s minus T infinity or T infinity minus T s. So I do not think there is anything to do with heating or cooling. Yes, if you want to hair split and call Newton's law of heating you can call there is nothing wrong in it. Actually I tried to convince my student once but he was not convinced. So there should be technical reason. No, no I think. So that is a philosophical answer I should give it to him. Yeah, yeah. So I think may be first time when Newton proposed this he wanted to cool it, cool the plate. So may be refrigerators were not there that time. So that is why they were always worried about cooling. So that is why it has come out as Newton's law of cooling. But I do not think we need to be worried about this. This is a philosophical question. Okay. One more time. Just now you have mentioned the name of the professor who has mentioned that by maintaining some ideal conditions we can maintain laminar flow up to rental number of 50,000. So may know the name of that book. Okay. See professor Schlichting, S C H, yeah I will write that. I will write that. The question asked by one of the participants is that I made mention that laminar flow can be maintained up to 50,000 in which book is this information there. It is boundary layer theory by Schlichting, S C H, L I, C H, T I, N G. If you Google you will be able to get appropriate, I think it is Pringer Verlag now. So I do not remember the edition now what is available. But whichever edition it is this information is available. Over to MA NIT Bhopal. Sir, I have one question that why we are using characteristic length in case of Nusselt number as well as in case of Bayard number. Characteristic length sir. Okay. Then we are calculating Nusselt number as well as Bayard number. Let me rephrase the question. The question asked is why are we using the characteristic length? Let me ask back are you asking what is the characteristic length I should be using or why are we using characteristic length? Why are you using characteristic? No we have defined it that way that is Nusselt equal to for example or Bayard number equal to H L by K. How did I arrive at Bayard number for example? I did not define just like that it came after non dimensionalization. Let me show you that if I come back. So if I come back to the document so let me load the transient conduction thing. Bayard number and Fourier number came through non dimensionalization of my differential equations. That is if you see this if I take the plane wall case where in which we set up the equation and the boundary condition and the initial condition. When we non dimensionalized these equations what did we come across? We came across Fourier number tau and Bayard number. So we have not introduced the characteristic length brute force. It has come on its own that is how even Nusselt number also comes out even Reynolds number also comes out when we set up the differential equation and non dimensionalized them these non dimensional numbers are going to pop up. That is how we realize that non dimensional numbers are supposed to be defined like this. This is what is called as dimensional similarity. This is what is called as dimensional similarity. So Bayard number I cannot answer you why you will see or the characteristic length has been introduced. It is coming out by non dimensionalization of the differential equations which are set up to define or to solve a given phenomena or given situation. Thank you sir. Thank you. Over to you government college Salem. Sir in solving transcendental equations in reducing PDEU to ODE you are taken eta is equal to x divided by root of 4 alpha t. How that term is taken sir? One of the participants question is that in converting the PDE to ODE in semi infinite medium how have you taken eta equal to alpha upon square root of 4 alpha t? No x upon square root of 4 alpha t. So that is the question. Let me rephrase that question again. See the question asked is in converting this ODE sorry PDE to ODE. So when we converted this PDE to ODE my ODE looked like this PDE got converted itself to d square t by d eta square minus 2 eta d t by d eta. So in doing this I have taken eta equal to x upon square root of 4 alpha t. So how did I know that eta should be equal to x upon square root of 4 alpha t? I had mentioned based on the order of magnitude analysis. That is I took an egg I took an egg and put out that egg from cooker or from hot vessel outside to the atmosphere. So I said that only the outer skin layer is experiencing the temperature difference that is the very much inner that is the yolk or the yellow portion is very much at the initial solid temperature. But only the outer skin very much outer may be near the albumin that is albumin portion that is around delta thickness is experiencing this temperature gradient. So in this temperature gradient I have done the order of magnitude analysis. You see here del square t by del x square can be written as del t by del x at x equal to x of the order of delta not equal to delta. I made the mention in the morning also I am using the notation tilde that is of the order of magnitude. It does not mean that it is equal to delta x of the order of delta minus del t by del x at x tending to r of the order of 0. What is the temperature gradient at delta? It is almost initial solid temperature there is no gradient. But there is gradient at x at of the order of 0 what is that temperature gradient? T i minus T naught upon delta that takes care if I substitute that in del square t by del x square I will end up getting T i minus T naught upon delta square. Now let us handle del t by del t within a time span of t temperature has undergone change from T i to T naught. So that is T naught minus T i upon T minus 0. So if I equate these two that is minus T i minus T naught upon delta square is of the order of 1 by alpha T naught minus T i by T I get if I absorb this negative sign numerators get cancelled out I get delta square is of the order of alpha t. So delta upon square root of alpha t is what I have defined as eta. This is how I get to know that eta is supposed to be x upon square root of alpha t. As I said 4 is just a mathematical convenience introduction even if you remove 4 you can convert the PDE to ODE. 4 is just to cancel out few terms and make the make or appear make appear the equation look little more elegant. Otherwise this 4 has no significance. So eta if I take equal to x upon square root of alpha t I can convert this PDE to ODE. I hope I have answered your question. Yes sir one more question sir. Sir transcendental equations having many roots how we select one or the other root what is the procedure or what is the condition you impose selecting one or the other root sir. Of the so many roots available. See the question asked is by one of the participants is by the question asked by one of the participants is that is that see we have a transcendental equation lambda n. Let us just take the plane wall case we have the solution as theta equal to a n e to the power of minus lambda square tau cos lambda n x by l. I said for tau greater than 0.2 for tau greater than 0.2 it would suffice to take only the first term solution. Let me go slow let me go slow because it looks like we have not got what we have told. You see theta of x comma t if we carefully see this equation it is a series solution that is as we wrote in the morning professor Arun wrote this a 1 e to the power of minus lambda 1 square tau cos lambda 1 x by l plus a 2 e to the power of minus lambda 2 square tau cos lambda 2 x by l plus I can go on putting the other terms. But what we are saying is that for all Fourier numbers greater than 0.2 first term is would suffice to represent the complete temperature distribution. So, if I take only the first term my equation which I had written as a series solution I need to take only the first term. So, if I take the first term how will it look like. So, t minus t infinity upon t i minus t infinity equal to a 1 e to the power of minus lambda 1 square tau cos lambda 1 x by l. So, how do I get this lambda 1 what is this lambda 1 lambda 1 is the root of this transcendental equation that is lambda n tan lambda n equal to biot number. So, if you see here I should have written here actually these are not g task this is lambda 1 lambda 2 see for any given biot number I have lambda 1 lambda 2 lambda 3 lambda 4 I have to I am supposed to take only the first root the first root if you see this actually this is lambda 1 is the first root for any biot number I have to just choose the first root if I plug in that first root which is lambda 1 here and the Fourier number I will be and get the a 1 again of course, for every biot number that is there in the next here for every biot number I have lambda 1 and a 1. So, if I take the first root that would suffice and that would give me that would fetch me the complete temperature distribution at any location and at any point of time is that ok professor. Yes sir, but one more question sir you are talking all the time about plane walls cylindrical walls and then a spherical walls which is made of what material sir whether I can we extend that to anisotropic material also whether are you about isotropic material ok. The question asked by one of the participants is that whatever solutions we have given for transient conduction so far is all for is it applicable for any k yes as long as my thermal conductivity is homogeneous and isotropic that is it is not having any preference with the direction yes we can apply if it is having preference with the direction then I have to take the whole my set of equation itself will change no see what is what is the equation I have taken let us look at the equation if I go back let me use the white board. See what is my equation telling let me write the heat diffusion equation in a most generic form that is I said del by del x k del t by del x plus del by del y k del t by del y plus del by del z k del t by del z plus q dot equal to rho c p del t by del t this was my equation now if I have to take even one dimensional conduction and no heat generation let us say then what will happen there is no temperature variation with y and there is no temperature variation with z. So, what will happen so I will get and there is no volumetric heat generation let us say I get del by del x of k del t by del x equal to rho c p del t by del t. So, now the question is is k constant yes k throughout in the x direction has to be constant then only I can pull this out and write k del squared t by del x squared equal to rho c p del t by del t. Let us say k is not a constant, but is a function of x let us say in the yesterday's one of the problems we took k equal to k naught plus a x. So, I have to plug in that distribution here and then solve this equation. So, that is what makes a life very difficult of course, when we give closed form solution we always want to make our life easy and not just for the sake of ease we can explain few principles very easily when I simplify the problem. So, if k is a function of x I need to plug in that function here in this k. So, if I plug in that function and then solve the equation I should be getting the solution, but if it is thoroughly anisotropic and if it is varying with y and z my equation is going to be complicated and I cannot perhaps get the closed form solution. Then I have to take the recourse of numerical method is that ok professor. Now, I think I have taken three questions I will move on to the next section. Thank you sir, well received sir. Yeah, I think we will stop question answer session here ok bye bye.