 So, today's tutorial actually we have problem 19 onwards that is lumped capacitance transient conduction three problems are there. We want all of you to solve these three problems properly. We will also solve one by one let us see how far we can get. Problem number 19. 19. The heat transfer coefficient for air flowing over a sphere is to be determined by observing the temperature time history of a sphere fabricated from pure copper. The sphere which is 12.7 mm diameter so the diameter of the sphere is 12.7 mm is at 66 degree Celsius before it is inserted into an air stream that is T i is 66 and having it temperature into an air stream having a temperature of 27 degree Celsius that is T infinity equal to 27 degree Celsius. A thermocouple on the outer surface of the sphere indicates 55 degree Celsius 69 seconds after the sphere is inserted in the air stream that is T equal to 69, T finally equal to 55 degree Celsius. Assume and then justify that the sphere behaves as a space wise isothermal object that is space wise isothermal means lumped that means it is not going to vary the temperature in terms of space and calculate the heat transfer coefficient. The material properties given are k equal to 398 Watt per meter Kelvin, C p equal to 389 Joule per kg Kelvin, rho equal to 8933 kg per meter cube. The moment you see copper by and large you can expect to be lumped but not always. So, first thing is first thing is we need to get the biot number. Biot number we cannot get directly we need to get the heat transfer coefficient based on what information that is T final and temperature given. Using that you can get the heat transfer coefficient. Let us calculate I will give you the mathematical experimental implication of this little later on. So, that is T f minus T infinity upon T I minus T infinity equal to exponential of minus h A s upon rho V C p T. So, 55 minus 27 upon 66 minus 27 equal to exponential of h h I do not know into A s by A s by V A s by V is 6 by D exponential of or first let us find out time constant minus T by tau ok T is 69. So, if I press the calculator for this I am going to get tau equal to 208 seconds. You see it is quite high compared to our 4.52 seconds because that was 0.7 mm diameter this is almost half inch 12.7 mm diameter it is exactly half inch. So, that is why it is taking more time. Now, tau equal to tau equal to rho V C p upon h A s. So, implies 208 equal to yeah V by A s is D by 6. So, tau equal to 208 ok h equal to 6 into 208 upon rho is 893 3 into diameter is 12.7 into 10 to the power of minus 3 into C p is 389. So, I get a h of 5.3 Watt per meter squared kilo. See the beauty of this problem what is so beautiful in this problem you see we can measure the heat transfer coefficient using this method. Now, you can understand when I say heat transfer coefficient much better because we have completed convection I did not deal this because of this tutorial problem in yesterday's presentation. If I had told you can measure h you would have not understood. So, I had postponed this the main reason for choosing this problem is to explain that this technique can be used for measuring the heat transfer coefficient. Now, I said that this sphere is put in a air stream. So, I can have velocity of that air stream for a given velocity for a given fluid that is air I can measure the heat transfer coefficient that is only thing I need to know is the time locking. So, that is the beauty this is how we can measure the heat transfer coefficient using simple lumped transient conduction analysis. In fact, anyway the same thing is there in our experimental portion also. Someone was asking me in the chat that where is the video for measuring the heat transfer coefficient it is very much there in the course notes folder please see there is no folder sorry in the general files which are there which are visible for everyone there is a video whose file is different I do not remember exactly dot avi I think it is dot avi extension file that is the folder I would strongly urge all of you to see that video. Now, let us check in this problem whether it is lumped or not biot number equal to h l c by k h is 35.3 and d by 6 that is 12.7 into 10 to the power of minus 3 by 6 upon 398. So, if we substitute this we are going to get 1.88 into 10 to the power of minus 4. So, which is very much less than 0.1. So, it is very much lumped. So, that is how we have gotten away with this problem, but what is the main intention of cooking this problem or choosing this problem is to elaborate that heat transfer coefficient can be measured this way. If you put it in a gas stove flame there also you can measure the heat transfer coefficient. You can think of any environment you can measure the heat transfer coefficient using this simple harmless technique. Let us move on to next problem that is problem number 20 it is little lengthier never the less we will do this because we will get to understand this what is the usage of those series solutions formulae that is a long 20 centimeter diameter d equal to 20 centimeter cylindrical shaft made of SS 304 comes out of an oven at a uniform temperature of 600 degree Celsius that is T i is 600 degree Celsius. The shaft is then allowed to cool in an environment chamber at 200 degree Celsius T infinity is 200 degree Celsius with an average heat transfer coefficient of 80 watts per meter square degree Celsius quite high heat transfer coefficient. So, determine the temperature at the center of the shaft that is theta naught or T naught comma T 45 minutes after the start of the cooling process that is T equal to 45 into 60. Also determine the heat transfer per unit length of the shaft during this process time period K equal to 14.9 K equal to 14.9 note this number SS 304 this is stainless steel this is having lesser resistance lesser thermal conductivity C p 477 joule per kg kelvin rho 7900 kg per meter cube. So, how do I go about this problem how do I go about this problem I think we will come back to this solution we will take two questions and come back by which time you guys please try the problem. Please try this problem basically it is simple analytical solution we will show the methodology you apply that and let us see how many of you would have completed. Yeah theta naught cylinder I need to know A 1 and lambda 1 which is a function of tau. So, tau is what what is tau is Fourier number Fourier number is alpha T by L square alpha is known K by rho C p and T is given 45 minutes L square you have to take the appropriate characteristic length for a this is what cylinder. So, volume by area. So, then if you substitute for that tau if you for a given tau you are going to get lambda 1 and sorry lambda 1 and A 1 in this 22 slide. So, from that you are going to get the center line temperature. So, please do this problem keep doing this problem while we take the questions from other centers and come back to you. Why say college Nagpur any questions sir. What is the difference between finite and differential analysis of fluid flow sir please explain with practical example. The question asked is what is the difference between finite control volume analysis and the differential analysis. So, I have not brought the control volume approach problems. So, I think I will I will cook a problem in which the finite control volume approach would be solved and then also similarly how to handle it differential approach, but still I would go ahead and answer this question something like this. Now, let us say I have a pipe flow I have a pipe flow and I know only the inlet velocity. Now, I think I will have to cook the problem and come back it becomes difficult for me to cook the problem directly because for finite control volume approach I need to cook a problem. So, that you can understand it whatever I will tell it will be same as what I would have told in the morning. All that I can say is infinite control volume analysis we do not need velocity distributions and temperature distributions and the pressure distributions, but in differential approach you are going to get all of that. So, I will cook a problem and show that in the module for you correct. Okay, professor has given me an example see I think in our fluid mechanics lab in a fluid mechanics lab do you have impact of jet experiment do you have impact of jet experiment? Professor. Yes sir we do. Yeah okay. So, you have impact of jet experiment in the yeah in the impact of jet experiment what comes out do we measure any velocity profile and pressure profile? No, we just apply the jet. So, there is a jet which is hitting my plate in that what do I measure I balance the I over I have the overhang okay. Professor is drawing for us in the white board. So, in the impact of jet experiment I have a flat plate and I have a jet I have a jet hitting the plate. Now, because of the jet hitting the plate I have to hold this back at the same position. So, I have I balance this with balancing mechanism and measure through that balancing simple lever mechanism I put the weight and get put the mass and get the weight what does that give me the total drag force experience or the total kinetic energy which has been or the total momentum which has been transferred from the fluid to the plate and in this experiment we demonstrate that corrode plate is better than flat plate and we say that that is why we use in pelton turbines corrode blade rather than flat plate that is what and if you use corrode plate we show theoretically that the efficiency of transferring the kinetic energy or the momentum transfer to the plate that is mechanical energy is almost 100 percent. So, what did we do? Did we get here any velocity distribution or pressure distribution? No, this is this essentially finite control volume approach here we have not taken the velocity distribution at the exit of the nozzle at all. So, this is finite control volume approach because I have I have just taken the control volume in this case either a flat plate or a corrode plate. So, this is control volume approach I think professor Arun has really given best example because we all do this experiment I think we cannot cook any better example than this I hope we have reached you. Sir, what is the significance of Prandtl number? Does it helps in selecting for media? The question asked by is by one of the professors what is the significance of Prandtl number? Does it help in choosing my fluid for any heat transfer application? See for most of the times in heat transfer the fluid choice is mostly limited by my application. Yes, in heat exchanger perhaps I have choice of fluids, but for example, electronic cooling usually I am limited by either air or dielectric fluid, but how many in regular desktop computer or laptop I cannot carry dielectric fluid I have to manage with the air. So, the fluid is fixed there that is air, but for example, it is a server and it is a huge server and I have to cool the motherboards of the server. There yes they use dielectric fluid, but of course I cannot explain this directly with Prandtl number. See all that we can say is Nusselt number is a function of Reynolds and Prandtl number as long as the Prandtl number is equal to 1 or greater than 1 it is going to carry if it is laminar flow Prandtl number to the power of half if it is a flat plate, but in internal channels it is going to be 0.3 in case of cooling and 0.4 in case of heating. So, definitely from that you can see that the heat transfer coefficient is directly proportional to the Prandtl number. If you increase the Prandtl number definitely your heat transfer coefficient is going up no doubt about that, but now let us see which all fluids have high Prandtl number. Prandtl number of air is 0.71 that is almost closer to unity Prandtl number of water is anywhere between 6 to 13.5 around room temperature it is around 6 and as you go on increasing the temperature the Prandtl number comes down sorry it comes down the Prandtl number comes down. So, it is anywhere between 3 to 6.5 up to until it becomes a steam. Now, if I take oil whose Prandtl number is very high now can I choose oil all the time? No why because the pumping power for oil is very high why because the viscosity of oil is high. So, the pumping power required for oil is very high and of course, if I use oil most of the times in most of the application it becomes messy. I may want to utmost manage with water or air, but that does not mean that I will never use oil most of the heat exchangers work with oil, but if I can reconcile with the fact that I can offer to have large pumping power I will go for oil why because Prandtl number is high. If it is no matter whether it is laminar or turbulent the heat transfer coefficient is going to increase with the increase of Prandtl number that is the answer for your question professor. We will have low conductivity or low high conductivity fluid we will have a low Prandtl number that is my question. No, no okay we should not look at k alone it is mu Cp by k together the question asked is high Prandtl number will have low conductivity how can low conductivity fluid give me high heat transfer coefficient. See the point is it is we should not look at only thermal conductivity if we take the definition of h equal to minus k del t by del y at y equal to 0 that is the heat transfer coefficient definition k alone is not contributing for the heat transfer coefficient it is del t by del y which is contributing and this del t by del y at y equal to 0 is dependent on the Prandtl number. So, we should not look at it as k alone k alone okay now that you have asked k alone I will answer this in a little different fashion we use nanoparticles in for liquids nowadays lot of research is going on this nanoparticles. So what they do is they put carbon nanotubes or alumina that is aluminum oxide small nanoparticles they put along with the water stream. So, what happens belief is that the thermal conductivity will be high because I have put a solid thermal conductivity that is aluminum thermal conductivity is 150 and water thermal conductivity is 0.6. So, I should have a thermal conductivity quite high in between these two yes thermal conductivity is high, but what have people seen hardly an increase of heat transfer coefficient by 3 to 4 percent the point is we need to look at this problem as the temperature gradient it is not the thermal conductivity alone. So, that is the answer for this question professor do not look at it as k alone the heat transfer coefficient is a function of Prandtl number Prandtl number Prandtl number I cannot emphasize more than this why because when we do dimensional similarity day after tomorrow that is on Monday from the energy equation there are two non dimensional numbers which are going to come that is Reynolds and Prandtl. So, it is not k alone it is Prandtl which is going to decide the heat transfer coefficient ok. We will get to the problem now where we left problem number 20. So the question was yeah the question was initial and final temperature was given and we were supposed to find the center line temperature of a circular cylinder. So, first thing what we need to do is we need to get the tau, tau equal to alpha t upon r naught squared let us find out alpha, alpha is k by rho C p, k is k is given to be 14.9, rho is 7900, C p is 477. So, that is equal to 3.95 into 10 to the power of minus 6 meter squared per second. Now if I substitute that in tau, tau equal to 3.95 into 10 to the power of minus 6 into 45 into 60 that is 45 minutes converting into seconds 45 into 60 upon diameter is given to be 20, 20 centimeters. So, it is 0.1 whole squared because r naught so tau equal to 1.07. So, why are we calculating tau first of all this is just to ensure that our first series solution is valid for tau greater than 0.2. So, first term is sufficient that is number one. Next is so we are not going to do this with chart we are going to directly do with the analytical solution. So, for the analytical solution I need to know the Biot number. So, let us calculate Biot number what is Biot number given by h r naught by k. So, Biot number equal to h r naught by k yeah here we have a general question by all the students perhaps most of you also had got I have got this question that is this is what this is a cylinder this is a cylinder. So, for a cylinder the characteristic length is not r naught, but remember all these relations or these solutions are derived on the basis of taking the characteristic length as r naught for cylinder and for sphere r naught and plate thickness is t. So, you do not start applying lumped thing here it is not h l c by k here Biot number by definition when we have derived when we have derived the solution it is taken as h r naught by k for both cylinder and sphere. So, this has to be kept in our mind we should not again go back and take the characteristic length l c equal to volume by surface area no. So, this is a word of caution we have to be very careful why because that is how we have defined that is all there is no other significant reason. So, Biot number equal to h is 80 r naught is 0.1 k is 14.9. So, that is Biot number equal to 0.537. So, now let us go to the chart let us go to the chart. So, the point is Biot number is 0.5 is 0.537 it is very much larger than 0.1. So, lumped is not going to hold water any more. So, we have to go to the document and get a 1 and lambda 1 what is the solution first let us write the solution theta naught t naught minus t infinity equal to t naught minus t infinity theta naught equal to t naught minus t infinity upon t i minus t infinity is equal to a 1 e to the power of lambda 1 squared into tau. Now, we need to find a 1 and lambda 1 where do we get this a 1 and lambda 1 we will get this as a function of Biot number let us go to the table and get this lambda 1 and a 1 from this table. So, that is Biot number equal to 0.537 yeah there is a standard question here I do not find 0.537 in this table I have 0.5 and 0.6 what should I do I have to interpolate can I interpolate if you strictly ask no why because interpolation is possible only when it is linear, but here definitely it is not linear it is non-linear, but I have no other option. So, that is why of course the Biot number variation here is between 0.4 to 0.5. So, may be piece wise it would become linear, but that is the assumption it, but usually it is not very much off. So, that is the reason we interpolate, but if the student ask can I interpolate because this is a non-linear function we should say yes you are right it is a non-linear function but still we will go ahead and take it as interpolation assuming that within that domain it is going to be linear. So, if I do that I am going to get a 1 as 1.122 and lambda 1 as 0.97 I guess most of you would have got this a 1 equal to 1.122 lambda 1 equal to 0.97. So, if I substitute this I would get T naught as 364 degree Celsius. In fact for this problem let us just do for the heck of it applying lumped capacitance just let us do this let us do this applying lumped capacitance and let us see how much the temperature would be different from this temperature what we have computed that is what is lumped capacitance T naught minus T infinity upon I am just brute force applying lumped capacitance that is T minus T infinity upon T i minus T infinity equal to e to the power of minus T by tau. Let me calculate tau, tau equal tau we have already calculated no this tau is different we are unfortunately we are using the same notation for two things. Now, I realize that tau is Fourier number and also time constant. So, please do not get confused when I say tau here in this equation it is time constant that is the reason most of the textbooks use for Fourier number as f o rather than tau because tau is usually chosen for time constant. So, let us now find out time constant that is Fourier number what we found as alpha T by r naught squared is actually Fourier number tau equal to rho v C p upon h a s let us calculate that tau equal to rho v C p upon h a s I hope we have rho v C p and all perhaps we do not have yeah, but still we will go ahead and make some assumptions and calculate yeah we have rho is 477 D by 6 that is D by 6 no not D by 6 this is cylinder. So, what will be the first cylinder? R naught by 2 no pi D squared by let us write pi pi by 4 D squared L divided by pi D L L gets cancelled out pi pi gets cancelled out. So, D by 4 it is that is R by 2 that is right into C p is 477 and C p is sorry rho is already told right yeah rho v C p I have told for rho it is 7900 and D is 0.2 meters upon h what is h here 80 watts per meter square Kelvin and a s is not there ok what is the tau I get let us calculate 7900 into 0.2 into 0.75 into 477 upon 80 is that right yeah 4 oh yeah 2355 2355 I get 2355.2 seconds it is quite high why because it is stainless steel ball stainless steel cylinder. So, it has to be high it is not as conductive as copper. Now, let us substitute that in this equation T minus that is what is T or T is what I need to find out T minus of T infinity is 200 T minus 200 yeah T minus 200 upon 600 minus 200 equal to e to the power of minus T that is T is 45 into 60 2700 or 45 into 60 divided by 2355.2 that is 2700 divided by 2355.2 into negative sign exponential of T minus of this would fetch me 0.3 into 400 plus 200 how much I get 327 327.1 there is a there is a question from one of the participant saying that for biot number 0.537 lambda 1 and a 1 what we have chosen as 0.97 and 1.122 are wrong let us go back and check let us go to the document. So, what is biot number 0.537 and 0.5 it is cylinder I do not know what we our participant is saying for cylinder yes it is right 0.97 and 1.122 sorry we are not wrong for a change. Probably you are looking in the column for plane wall ok please. Please see for cylinder. It appears to be 0.6533 that is the plane wall column that is plane wall column. Sorry you are seeing plane wall see the cylinder. Now, let us come back what did we get for lumped temperature. Temperature we got 327 we got the temperature as 327 let us take stock what are we doing we got biot number of 0.537 we said that what is the biot number we got 0.537 that is very high we cannot use lumped but still we went ahead and put the lumped and what is the temperature we got 327 see what is the actual temperature 364. It is a temperature difference of 364 minus 327 it is 37 degree Celsius. So, we cannot afford to assume lumped now this itself suggests that why biot number of 0.1 is so sacred. Because there are so many questions in yesterday's and day before yesterday's question answer session why biot number 0.1 only is taken. So, this is how one can check oneself whether when can I take lumped and when can I not take lumped. I think we will get back to question answer session and then solve a problem or we will get problem. Yeah this problem is through energy removal. There is second part there is second part that is heat transfer per unit length of the shaft during this time period. We will go to the document and see the relation what we need to use that is next yeah. So, here for a cylinder it is Q by Q maximum cylinder it is Q by Q maximum cylinder equal to 1 minus 1 minus 2 theta cylinder. Let us write that that is theta by theta max theta by theta max theta by theta max equal to 1 minus 2 theta naught Q by Q max sorry Q by Q max Q by Q max I am sorry Q by Q max Q by Q max equal to 1 minus 2 theta naught j 1 lambda 1 j 1 of lambda 1 that is first order lambda 1 upon lambda 1. Now, how do I get this j 1 for lambda 1 equal to 0.97? Let us go to the document for lambda 1 equal to 0.97. Let us try to get j 1. If you go to Jacobian function which are listed yeah here no sorry Bessel function that is 0.97 0.97 0.97 0.97 0.97 0.97 0.97 j 1 is supposed to be 0.43 correct j 1 equal to 0.43. So, now let us calculate before we will go ahead and calculate Q by Q maximum. Q by Q maximum equal to 1 minus of 2 theta naught we are just found theta naught just little while ago or we can find it out theta naught no we have found theta naught 0.41 0.41 into 0.43 upon 0.97. So, what do I get I am going to get this as 0.636. Now, the question is to find Q max Q max we know pi rho into pi r square L that is m c p m is density into volume rho into pi r square L c p into t i minus t infinity I have to just plug in that is 7900 into pi 7900 into pi into 0.1 square into 1 per meter length we are taking into 477 into 600 minus 200. So, I would get theta max Q max as 47350 kilo joules 47350 kilo joules if I multiply that with 0.636 I end up with actual heat transfer as 30120 kilo joules this is the total energy this is the answer. One thing we have learnt if we take recap of what we have learnt from this problem what is that we are carrying the message from this problem that is if I take lumped assumption where I am not supposed to take lumped assumption that is for biot numbers greater than 0.1 I am going to be my temperature predictions are going to be seriously in error. So, they are not going to be right. So, I should not take the lumped assumption where I am not supposed to take it that is for biot numbers greater than 0.1. So, that is what is demonstrated through this problem I think we will take one or two questions from two centers before we come to convective heat transfer questions. So, we are stopping tutorial on conduction. So, we are through with conduction tutorial and we will be moving on to convection once we come back once we come back after question answer session. So, for 5 more minutes we are taking questions from various centers. Government college Salem. Government college Salem. Sir, how is the turbulent shear stress measured? Yeah, the question asked by one of the participants is rows how is the turbulent shear stress measured? First let us take the question first let us write what is turbulent shear stress. So, what is turbulent shear stress turbulent shear stress is equal to minus rho u prime v prime bar. So, there is something called what is hot wire animometer I will come to the principle of hot wire animometer just a little while from now someone let us say use me the velocity u versus time and v versus time. So, v versus time is something like this and v versus time is something like this let us say. So, at every instant at every instant I would pick up u prime from here and at the same instant I would pick up v prime from here and then take the product of u prime v prime like that at every instance I would pick up and average that product over a long period of time and that is what I get u prime v prime then I multiply that with density and that is what is my turbulent shear stress. Now the question is how do I measure this although very nicely I plotted it how do I get this plot that is done by what is called as hot wire animometer what is hot wire animometer hot wire animometer is just a thin wire is just a thin wire how thin that wire is it is consisting of a small wire which is having prongs and this wire is one arm of the weed stone bridge we all know weed stone bridge very well. So, this wire is one arm of this weed stone bridge. So, this is the weed stone bridge and we measure the voltage output from this weed stone bridge. So, this wire is one arm of the weed stone bridge how thin is this wire it is of the order of 15 micrometer what is our hair diameter size hair diameter size is 75 micrometer. So, 75 micrometer hair itself I cannot see until I keep that under the tube light. So, how about 15 micrometer it is a difficult task if you keep the hot wire to the tube light perhaps with lot of difficulty youngsters can see, but the old people like me can definitely not see. So, 15 micrometer is the diameter of this hot wire why it is such a small diameter that is because again from heat transfer we can answer rho C p rho V C p is very less that is thermal inertia is to be less that is why it is made very small. So, that it can respond very fast its response is of the order of milliseconds because it has to capture fluctuations of the order of kilo hertz. So, it has to be time constant has to be of the order of millisecond that is the reason why it is made so small. So, this weed stone if when it is made part of the weed stone bridge it is maintained typically at 300 or 400 degree Celsius when I put this in flow when I put this in flow this hot wire gets cooled, but I will bring it back to the same temperature what is the additional power I have to put in to bring it back to the same temperature is a measure of the velocity that is why this is called as C t a that is constant temperature animometer because I am maintaining the temperature constant. So, this is a very highly specialized equipment there are to the best of my knowledge there are only 4 to 5 suppliers all over the world the major suppliers are TSI and DANTEC and each one the cheapest one causes at least 15 to 16 lakhs, but there are some cheaper versions which will cost at least 5 to 6 lakhs. So, all sudden done the coming back to your question we measure the velocity profile using that hot wire animometer and do the bookkeeping of u prime and v prime take the product and average it over considerable period of time and that is how you get the turbulent shear stress that tells us how difficult it is to get the turbulent shear stress. Now imagine you have to move point by point if you are doing in a flat plate point by point not only along the x direction, but also in the y direction we have to be thankful to all those scientists who have measured this and put these values in the literature we have to be thankful to them, but for their persistent efforts we would not have had data of friction factor or otherwise. I think I have answered your question. Government college. Government college Salem. Sir, in case of mixture of two fluids different fluids how the heat transfer coefficient h varies? The question by one of the participants is if I have mixture of two fluids how does the heat transfer coefficient vary? The answer is in the question itself why because now if let us say they are immiscible they are miscible fluids that means they can mix with each other then if the two fluids can mix with each other I have to imagine as a homogeneous fluid which will have its own density its own viscosity its own specific heat its own thermal conductivity that is its own thermo physical property. Now once I make it as a homogeneous fluid then it becomes quite easy. Now that is if it is miscible fluid that is if both the fluids can mix I think I did not state the question for the other participants. I am sorry the question asked by the participant was if two different fluids flow in a pipe for example two different fluids if two different fluids flow in a pipe how does the heat transfer coefficient get quantified? The answer is if the both the fluids are miscible that is if they can mix and form a homogeneous fluid. Now I think all of you can understand when I say homogeneous. Homogeneous means it is the fluid which is having properties homogeneous that is it does not vary with location that is density specific heat thermal conductivity viscosity are all homogeneous are all same for this imaginary fluid which is a mixture of fluid A and fluid B then it is almost like the same convective heat transfer what we are handling for miscible fluids but if it is for immiscible fluids then what is the issue here? Not only if they are density wise they are significantly different denser one will flow at the bottom and the lighter one will flow at the top. Now definitely if the density is different naturally I am sure viscosity specific heat thermal conductivities are different that means Prandtl number of the top fluid is different from the Prandtl number of the bottom fluid. So two issues so there is a boundary layer growth differently on the top wall and boundary layer growth differently on the bottom wall. So heat transfer coefficient on the top wall is different from the heat transfer coefficient on the bottom wall this is a two simplistic way of looking at it actually. So professor is drawing so the boundary layers are different the way they are growing are going to be different based on the Prandtl number but if you have to handle friction factor now if you have to handle pressure drop I have to take the interfacial shear stresses also that is not just the shear stress on the top wall and the bottom wall there is going to be a shear stress between the two fluids itself that is at the interface of these two of these two fluids and both will have different velocity. So it is going to be quite difficult to get the closed form solution but nevertheless it is a good question it is only going to tell us what are all the complexities involved in this type of problem. Heat transfer from one fluid to another. Yeah so professor has shown that at the interface there is an heat transfer between one fluid to the another fluid as well. So it is a quite a complicated problem you cannot get easy closed form solution nevertheless it is a good question again an application oriented question I would say. Is that okay? Okay sir. Yeah. Sir okay. Yes sir it is like composite wall we have to add H1 and H2. No no no. Question is is it like composite wall can I add H1 and H2? No it is not that simple we have been telling now. See the Prandtl number is different for the top wall and the bottom wall it is not H1 and H2. I have to take the local heat transfer coefficient what is drawn on the paper is just two plates but imagine a circular pipe. So it will be heat transfer coefficient along the circumference is going to be varying all over. So it is not simple H1 and H2 it is not H1 and H2. Yeah. Yeah there is professor has drawn for us 1 and 2 so H on the top wall is going to be different from the bottom wall depending on the Prandtl number difference. So it is not H1 and H2 we cannot simplify that easily that is why I have been telling this is a complicated question. Is that okay professor? Yes sir. Thank you sir. Okay thank you. Let us take up a problem. Problem number 26 the velocity profile it is a flow over a surface where in which we have been given the velocity profile and the temperature profile. Velocity profile is u of y equal to a y plus d y squared minus c y cube and temperature profile t of y is equal to d plus e y plus f y squared minus g y cube where a b c d e f g are all constants. So now obtain expressions for skin friction factor and convective heat transfer coefficient in terms of free stream velocity and free free stream temperature and appropriate profile coefficients and fluid properties that is the question. Okay so let us go to the definition. So here it is pretty straightforward actually it is simple differentiation. Let us first take up H. H equal to minus k del t by del y at y equal to 0 upon t s minus t infinity. Here the thing is we know the constant the question here is are all constants a to g are constants and we know them that is what is that when I know the profile I know the constant. So how do I get del t by del y if I differentiate this I get e plus 2 f y minus 3 g y square but now dt by dy at y equal to 0 means the second term and the third term vanish I end up getting only e. So that means if I substitute this in the top I get minus k e upon h equal to minus k e upon t s minus t infinity this is my heat transfer coefficient like this we can cook lot of problems and give it to our students so that they understand. And in fact this velocity profile and temperature profile are not out of the world they are very much they are taken polynomial that is all. So more or less our velocity profiles and temperature profiles are not going to be this complicated but still they are going to be if it is laminar they are going to be parabolic we can still give parabolic temperature profile and sorry velocity profile and ask them to find the heat transfer coefficient ok c f equal to tau s upon rho u infinity squared by 2. And what is tau s mu del u by del y at y equal to 0 upon rho u infinity squared by 2. So now if I calculate del u by del y del u by del y because we know the velocity profile del u by del y equal to a plus 2 b y minus 3 c y square. So now I want del u by del y at y equal to 0 so the next 2 terms that is second 2 terms get cancelled out I get del u by del y at y equal to 0 as a. So c f equal to mu a upon 2 mu a upon or mu a upon half rho u infinity squared that is 2 mu a upon rho u infinity squared ok let us take a trivial problem which is 24 that is stretching just to get the hang of it we will just write what is this stretching. So the first case is problem number 24 I hope everyone would have answered this that is what I would expect. So if it is stretching what is this what is this case first of all this case is linear deformation that is it is getting deformed in linear direction. So I have linear deformation is del u by del x that is this is del u by del x del u by del x and this is del v by del y del v by del y is that right. So linear deformation total linear deformation is del u by del x plus del v by del y del u by del x plus del v by del y ok. Next case what is next case I think it can be clearly seen we have already solved this but still let us do it its rotation its rotation means I should have gradients of that velocity in the other direction that is I have this is in z plane so that is del v by del x this is del v by del x del v by del x and the top one is del u by del y no I have made a mistake this is correct del u by del y del u by del y correct del u by del y because I have moved from bottom to top velocity has changed. So rotation what is the direction of del v by del x net rotation is half of actually it is half of del v by del x plus del u by del y this is the net rotation it is omega z this is equal to omega is this minus plus that is plus ok. Now let us take the last one the angular deformation I think this has been already answered so I would just state it as del v by del x plus del u by del y that is all so minus minus del v by del x minus no plus of minus no plus it is so del delta alpha plus delta beta that is my shearing strain so it is not minus so shearing strain or angular deformation equal to del v by del x plus del u by del y do not get confused it is angular deformation how much is the net angular deformation delta alpha and delta beta so del v by del x plus del u by del y is the angular deformation I think we will stop solving tutorial problems we will perhaps take one or two questions over to Amrita. Sir actually the one of the participant had a doubt regarding that tau value in your in our transient heat transfer I mean for all practical problems we say that for tau value or Fourier number greater than 0.2 we can use this one term approximation for that Fourier number value greater than 0.2 and they have the doubt that are all the practical problems for all the practical problems can we use this one term approximation or Fourier number will be greater than 2 or can we generate can we do some generalization like that. Yeah question asked by one of the participants is that for Fourier number greater than 0.2 can I apply this for all practical problems yes why not as long as the Fourier number is greater than 0.2 and it is one dimensional transient conduction problem and it is varying with time we can apply this as long as Fourier number is greater than 0.2 one term solution is still valid there is no problem. The question is are practical problems in general having tau greater than 0.2 or the question asked is are practical problems having Fourier number greater than 0.2 that cannot be answered just like that because each practical problem will have its own Fourier number but generally I think Fourier numbers are greater than 0.2 but I would not may I would not dare to make that statement so we have to check for each practical problem what is the Fourier number ourselves ok ma'am so we will meet back again on Monday morning sharp 9 o'clock energy equation energy equation thank you with full of energy we should be coming back.