 For now what I will do is I will get started with tutorial problems today because we need to solve those tutorial problems. Let us see how do we do that. So I have already a request from few participants that I need to solve 33, problem number 33 and problem number 37. So let me take problem number 33. So let me see how I can coordinate with myself. Today I am missing professor Arun. He has gone for PhD interviews. So let us get started with tutorial problem 33. What is this problem number 33 stating is there is a flat plate, there is a flat plate and its length is 3 meters and U infinity 1 which is flowing from left to right on the top is U infinity 1 is 60 meters per second and T infinity 1 is 200 degree Celsius, very high temperature, hot gases are flowing and on the bottom side there is movement of the air from right to left that is U infinity here is 10 meters per second and T infinity here is 20 degree Celsius. So there is a heat transfer from hot gas to the flat plate and there is cooling at the same time by the cold air which is sitting at the bottom. Now what are the properties at T infinity 1? So what is that I should do? See actually what is that I should do? I have to take the properties at average film temperature. Do I know the wall temperature here? I do not know the wall temperature. So what do I do? I will go ahead with by taking the properties at the T infinity 1 because I do not have any other option. So if I do that or here in this problem they have given as properties because to exclude as this confusion but where did those properties come from at this temperature only? That is Prandtl number equal to 0.7, K is 0.03, nu kinematic viscosity is 20.92 into 10 to the power of minus 6 meters per second meter squared per second. Similarly for the bottom stream it is Prandtl number equal to 0.707, K equal to 0.0263 watt per meter Kelvin kinematic viscosity is 15.89 into 10 to the power of minus 6 meter squared per second meter. Rest is straight forward. First thing is what I should do in any in any forces convection problem what is that I should do? Reynolds number. So what is the question asked? Assumptions are steady state and let me write this is important. Assumption, steady state is not an important assumption but there is an important assumption other than that that is the reason why I am writing steady state and second thing is that I am assuming that plate is very thin that is why perhaps everyone has asked the problem number 33 why because we need to make this assumption. Plate is very thin when I say why am I making this plate as very thin? You see what is happening? There is a boundary layer developed here like this and on the bottom how is the boundary layer developed? Like this that means there is convective resistance here that is 1 upon h 1 a 1 upon h 1 a and there is convective resistance here that is 1 upon h 2 a and there is conductive resistance to neglect that conductive resistance I am taking that is L by K a that is thickness by K a that thickness I am making it as minimal as possible so that the conductive resistance is 0. So plate is thin conductive resistance is negligible I am not saying 0 I am saying negligible ok is negligible ok. So what is the question asked? Question asked is I will solve only for one part second part is logical evaluate the heat flux through the plate at x equal to 1 meter measured from the left edge ok let us do that x equal to 1 meter so that is Reynolds number this let me call as fluid 1 and let me call this as fluid 2. So Re 1 equal to how do I calculate that? U infinity 1 x upon nu what is U infinity 1 here? 60 into x is how much? What is x? 1 meter nu is 20.92 into 10 to the power of minus 6. So what is that I get? I get 28.68 into 10 to the power of 5 why do I write in terms of 5? 10 to the power of 5 to make out whether it is a laminar or turbulent definitely it is turbulent. So this is turbulent flow turbulent flow so because it is turbulent I have to take the correlation for Nusselt number nu 1 equal to 0.0296 Re x to the power of 0.8 PR to the power of 1 by 3. So what is PR? I know PR 0.7 I know Re I will get Nusselt number as 3852 and from that I get h 1 x upon k k I know already 0.03 and x is what is x? 1 meter I will get h 1 as what is h 1? 115.57 watts per meter squared kilo that is h 1. Similarly I have to do for h 2 let us do it for h 2 so that is I will have to move on to the next slide h 2 again for h 2 what do I calculate? What do I calculate? Re 2 Re 2 equal to U infinity 2 x upon nu what is U infinity 2 here? 10 what is x here? It is 2 meters it is not 1 meter from the left I said but for the bottom one what is that? It is 2 meters from the right because fluid is moving in the from right to left we should not get confused this problem is specifically cook to confuse the reader. If you are clear in our boundary layer I will not get confused. So 10 into 3 minus 1 that is 2 upon nu is 15.89 into 10 to the power of minus 6. So what is the Reynolds number I get? I get a Reynolds number of can anyone help me out? I do not see your Reynolds number quickly. Yeah you are right 12.59 into 10 to the power of 6 watts per sorry Reynolds number has no unit I should not be writing 10 to the power of 5 what is this? It is laminar or turbulent it is turbulent. So again I will use nu 1 nu 2 equal to 0.0296 re x to the power of 0.8 pr to the power of 1 by 3 I get an assault number of 199.968 which implies that here remember h L by k equal to approximately this that is 1 triple 9.968 what is this L? I am supposed to take here again 2 meters no confusion 2 meters thermal conductivity is given that is 0.0263 if I substitute this I will be getting an h of 26.3 watts per meter squared Kelvin. Compare h 1 with h 2 compare h 1 with h 2 h 1 is 115.57 h 2 is 26.3 why is the heat transfer coefficient on the top wall higher than the heat transfer coefficient of the bottom wall? Higher velocity not higher temperature higher velocity higher Reynolds number because of which my nusselt number has gone up it is more forced convection on the top wall than the bottom wall temperature gradient has nothing to do temperature gradient will not affect I told that. So here my students says that because temperature gradient is more it is why I said again and again heat transfer coefficient is not a function of delta t and heat flux please do not think that it is because it is 200 that is why my heat transfer coefficient is high nothing to do I have seen phd thesis going wrong in this aspect that is why I am emphasizing this I have rejected phd thesis based on this fundamental concept going wrong. So please remember that heat transfer coefficient in forced convection is not a function of heat flux not a function of temperature difference because here Reynolds number is high on the top wall heat transfer coefficient is higher. So h 2 is 26.3 now heat transfer rate how do I get heat flux? Heat flux equal to how do I get that? t infinity 1 minus t infinity 2 upon 1 by h 1 where is a 1 gone I have embedded I am writing it for q double dash plus 1 upon h 2 what is t infinity 1 200 what is t infinity 2 20 what is h 1 115.57 what is h 2 26.3 I will be getting a heat flux of 3856.4 Watt per Watt per Watt per meter square remember I have not written L upon thickness upon k a because I have neglected the thickness is very small please remember this. So from where to where the heat transfer is taking place actually from the hot gas my plate is also getting heated up and at the same time my plate is getting cooled by the by the cold fluid. So hot fluid is transferring the heat transfer is taking place from the hot fluid to the cold fluid through the plate through the plate. So with this I think there is a second part so what is it saying is that is the direction of the flow important how would it affect the heat transfer that is what we are saying is that in the next part if I take in both the directions are same that is if I take if I take flat plate and instead of having reverse if I take hot air in the same direction as that of cold air what will happen what will happen the boundary layer will increase the gut feeling would be the heat transfer rate here will be lesser lesser the first thing is that but it is not right why I will tell you why because R e 1 and h 1 is not going to change what will I get R e 1 is same I am not going to write R e 1 h 1 is 115.71 but what is R e 2 going to be same or different x is 1 meter now it is no longer 2 meter. So R e 2 will become now 10 into 1 upon 15.89 into 10 to the power of minus 6 I get 6.29 into 10 to the power of 5 still it is turbulent if I put that I get h 2 of 30.197 what has happened to my heat transfer coefficient actually heat transfer coefficient has increased heat transfer coefficient is increased so because of that I will get q double dash as 4309.79 heat transfer coefficient is increased means what my resistance has decreased so my heat transfer rate should increase so that is what I have got earlier what was the heat transfer rate I got 3856 now I have got 4309 because heat transfer coefficient at this location has increased so this is how you can look at this problem in several ways several different ways and try to understand the heat transfer coefficient concept. So once you have understood this rest all problems more or less in fact once you understand the boundary layer concept solving problems in convection is easier but understanding is difficult. So what I will do is for next 10, 15 minutes I will not take questions now I will introduce one another concept which I had skipped that is what is called as scale analysis if you see here see I had taught you one scale analysis I will solve 37 problem do not worry I will solve only one more problem before we sign off today I want to introduce one important concept that is scale analysis and one of the professors had very earnestly asked yesterday and I had postponed because I myself was stuck and I said that you keep working as a homework because I myself had stuck and I perhaps have now understood that. So let me say what is that it is about scale analysis so we did scale analysis for Prandtl number less than 1 that is we did it for thin or thick thin boundary layer so let me see where it is here in the scale analysis. So we said that if we do the order of magnitude analysis I will get delta is of the order of L r e L to the power of minus half and C f is r e L to the power of minus half this is very well understood and we did it for Prandtl number less than 1 ok Prandtl number less than 1 and we got delta t by L as r e L to the power of minus half and p r to the power of minus half. Now nusselt number I get r e to the power of half and p r to the power of half now what will happen here Prandtl number is less than 1 means my hydrodynamic boundary layer is thinner than the thermal boundary layer that is delta by delta by delta t is very much less than 1. So delta is very much less than delta t here we went away went ahead and took U infinity is same throughout my thermal boundary layer ok that was an easy thing and very simplified thing. But now life is not easy when I come to when I come to thin thermal boundary layer that is let us do that I am going to do this because whole of this transparency whatever is there is wrong except this U by U infinity equal to delta t by delta ok except that everything is not right ok. So let me take thick boundary layer that is I am taking Prandtl number greater than 1 that is for water and oil I want you to derive along with me because this is an involved concept. So delta by delta t is greater than 1 what does this mean delta is hydrodynamic boundary layer is much much thicker than the thermal boundary layer. Now if I take a flat plate and draw the hydrodynamic boundary layer it will look something like this that is this is my hydrodynamic boundary layer it is U infinity and this is my hydrodynamic boundary layer thickness delta. Now what is my thermal boundary layer thickness ok. So thermal boundary layer thickness is smaller than that of hydrodynamic boundary layer thickness that is this is delta t ok. So with this let us go back and see how does U infinity vary that is U varies how does U vary how does U vary let me take U is varying linearly that is why I have plotted linearly you see when I am doing scale analysis I have plotted all straight lines if you have noted carefully why because I am assuming linear relationship but when I was plotting boundary layer I was plotting very nice parabolic nature but I am not doing that now. So now if I do that that is U equal to that is U equal to M y U equal to M y plus C ok. So what is that what are the boundary conditions when y is equal to 0 U is equal to 0 let me write that when y is equal to 0 U is equal to 0. Now when y is equal to delta U is equal to U infinity when y is equal to delta U is equal to U infinity in fact I should not be using equal to I am just going actually I should be writing of the order of so if I put this what will become constant C equal to 0 and what will I get M as U upon what will I get that U upon U infinity equal to U upon that is M turns out to be U infinity by delta. So let me not do this because it is straight forward U equal to M y because my emphasis is somewhere else U equal to M y U equal to M y what is M U infinity by delta. So U equal to M y is U infinity y by delta. So now U by U infinity equal to y by delta what is my y what is my y which is of interest to me thermal boundary layer thickness I am not interested what is happening above the thermal boundary layer thickness. So U by U infinity equal to delta T by delta that is it is not equal to U by U infinity is of the order of if you think that I am going very fast what is that I get let me do this if I am going very fast. So what is that I get this implies that C equal to 0 C equal to 0 now U equal to at y equal to delta U equal to U infinity equal to M into delta. So what is M what is M M equal to U infinity by delta. So now U equal to U infinity by delta into y what is what is that y which I am interested in here U is of the order of I should not be writing equal to why I am not interested about delta T because my heat transfer is taking place within that delta T. So y is delta T. So I get U is of the order same thing U is of the order of U infinity into delta T by delta let us keep this in the back of our mind. Now what is my energy equation U delta T by delta x plus V delta T by delta x T by del y is equal to alpha del square T by del x squared we always neglect and take del y square because delta T by l square is much lower than delta T by delta T square here ok. So what will I get what will I get for U what is that I should be substituting for U I just now form U infinity into delta T by delta into what is this del T what is the scale of del T delta T and what is del x what is the scale of del x l comma what is V what is V V is we have to be careful here we will go to continuity equation what is continuity equation telling del U by del x plus del V by del y equal to 0 ok. So what is U what is the scale of U where is that U I want that is at delta T let that U here be U only let me call that as U only I am not wanting U anywhere else I want U at delta T ok. So U at delta T is what U at delta T is what U infinity delta T by delta what is the scale of x what is the scale of x 1 upon l del x is l x l is of the order of V what is the scale of y what is the scale of y delta T. So what is the scale of V now U infinity delta T squared upon delta l is that ok. Mejendra you are with me ok. So if I substitute that here U infinity delta T square upon delta l what is the scale of del T delta T what is the scale of del y which thickness I am interested delta or delta T in my energy equation delta T this should be of the order of alpha delta T upon delta T square. Now what will happen to this 1 delta T will get cancelled out with 1 delta T will get cancelled out with the another delta T. How do these two terms compare with each other how do these two terms compare with each other they are same what does that tell to me what does that tell to me it tells that U del T by del x is of the same order of V del T by del y I cannot throw any one of the two and they are same they are of the same order. So let me take only one of the two let me take see it is like telling 1 rupee plus 1 rupee is 2 rupees it is of the order of rupees only it is not if you take 10 rupee plus 11 rupee it is still 10s of rupees only not 100s is that right. So that is what we are saying so U infinity del T by this delta T delta T I can neglect del T by delta squared l is of the order of alpha upon 1 by delta T square is that okay I have to move on to the next transparency let me do that so if I move on what will I get for del T if I move this del T squared from right hand side to the left hand side what will I get delta T cube is of the order of yes alpha l into delta squared upon U infinity something missing delta squared I have put delta T cube is of the order of alpha l by U infinity into not delta squared where did this delta squared come from it is delta only sorry here I have made a mistake in the previous thing please note that this is not delta squared while transferring I have transferred it wrongly it is delta only I do not know whether I can do that erasing now yeah I am able to do that luckily so this is delta this is delta okay so I get delta this is right this is delta this is delta alpha l delta by U infinity so what is let me do some rearrangement what is delta alpha l by U infinity what is delta l r e l to the power of minus half this we have derived earlier then what is that let me divide both sides by l squared delta T cube upon l squared is of the order of what do I get alpha delta by let me multiply and divide it by nu that is what do I get U infinity U infinity this l squared only I have pushed it this side I am not dividing this l squared only I have pushed it let me multiply and divide by l okay let me multiply and divide it by l U infinity by U infinity l by nu I have done several things here let me tell you one by one l this l squared I have pushed it to the left hand side and I have multiplied and divided it by nu and I have multiplied and divided it by l okay now what is this and also there is another term no r e l to the power of minus half right so this l let me push to the left hand side what will I get delta T cube upon l cube what is nu by alpha what is nu by alpha no no Prandtl number Sameer one upon Prandtl number okay what is U infinity l by nu yes one by r e very good one by r e into r e l to the power of minus half so delta T cube upon l cube is of the order of one by Prandtl r e l to the power of if I push this one minus one by half to the denominator what will I get three by two so if I write delta T by l is of the order of P r to the power of minus one by three r e to the power of r e to the power of minus half and I will not do nusselt number equal to h delta T by k if I substitute that I will get directly nusselt is of the order of r e to the power of half P r to the power of one by three is that okay so elegant it is so all questions asked on this are now answered are now answered so I would take one or two questions for ten fifteen minutes and then solve the last problem so over for discussion nirma university amdabad any questions on the material which I have taught today and the concept and the experiments which we did yeah today you have discussed about the wavelength and the range of the red color being used for the traffic signals because of the longer wavelength now the question I am yeah here the red wavelength will also have the low energy content because its wavelength is longer so will it not affect the visualization at a longer distance because of intensity of the radiation emitted by the red wavelength will be as compared to other wave no the question asked by one of the participants is that red wavelength is having higher wavelength as opposed to that of the green and larger wavelength means smaller energy see wavelength does not decide energy I told in the morning what did we say wavelength it is decided by sorry the energy is decided by whom it is decided by the frequency so frequency decides the frequency decides the energy not the wavelength wavelength is decided by the medium okay so please do not get confused with the energy and the wavelength and the frequency frequency decides the energy wavelength is decided by the medium in which my electromagnetic wave is my electromagnetic wave is travelling ok.