 So, now let us I will just go again velocity profile. So, what is there is this is steady flow in pipes. So, what we said this is the Hagen Poiseuille flow, what we said is that velocity now if it is a fully developed flow, what will be there? V z is going to be a function of only R, V z is going to be a function of R and there will be no V r and V theta that is what is fully developed concept. So, using that fully developed concept we neglected V r momentum R momentum equation and theta momentum equation. If we neglect that those two equations that is you see I have written here V r equal to V theta equal to 0 because it is fully developed flow and V z is only a function of R not theta and z please note this now R momentum equation and theta momentum equation is not at all required because V r and V theta is just not there and now I have to take V z equation now you see here first term on the left hand side is 0 because V r is 0, second term on the left hand side equation is 0 because V theta is 0 and V z is not a function of z, but it is a function of R. So, del V z by del z also becomes a 0. So, whole of left hand side is 0, what is on the right hand side 0 equal to minus 1 by rho del P by del z plus nu into del squared V z by del R squared plus 1 by R del V z by del R V z is not a function of theta and V z is not a function of z that is why these two terms get become 0. Now, if I substitute this if I take this equation and do the algebra this is algebra which I do not intend to do only thing I want to emphasize is that there are two boundary conditions when R equal to R that means I am just at the pipe wall V z is equal to 0 which is no slip condition and when R equal to 0 d V z by d R equal to 0 this is the velocity profile is symmetric. So, taking these two boundary conditions I get A and B which are constant and I get this paraboloid velocity profile and which from which we can show that V z average is equal to half times the V z maximum velocity that is the center line velocity and shear stress can be calculated taking mu d V z by d R because I have got the velocity profile this is what we keep saying to get the shear stress I need to know the velocity profile only if I know the velocity profile I can do this differentiation. So, that is what we are saying and we get the friction factor. So, with this we will move on to the heat transfer I will come to all of this fluid mechanics little later after I derive this I will take you through that detour of fluid mechanics on a fast track again please be prepared for fast track but for now I am going to go as slow as I can for temperature profile and nusselt number what is that I am doing I am taking a circular pipe I am taking let me define my problem my problem is it is a circular pipe it is a circular pipe flow is in this direction and we consider steady and incompressible flow and laminar flow what are we trying to look at to get the temperature distribution as a function of R and I will try to get the temperature and from that I will get the nusselt number distribution that is the heat transfer coefficient let us see how do we do that we will first have to take the energy equation that is rho Cp del T del T plus V R see yesterday's derivation whatever we derived is it that is for Cartesian coordinates we can do the same derivation for cylindrical and spherical coordinates. So, if we do for cylindrical coordinates I get V R del T by del R plus V theta by R del T by del theta plus V into del T by del x is equal to k 1 by R del by del R del T by del R plus 1 by R square del square T by del theta square plus del square T by del x square plus q dot plus 5 V there is no volumetric heat generation. So, that is why I am making q dot equal to 0 one of the participants were asking yesterday q dot should be there you can see q dot being there in the energy equation and viscous dissipation is not there because I have assumed that the flow is incompressible if I do that now I will take another I need to take another assumption that is my flow is fully developed in what sense both in hydrodynamic sense and also in thermal sense it is thermally and hydrodynamically thermally and hydrodynamically it is fully developed. So, if I take that if I take that what do I get if I take that we have the velocity profile already you see I have not mentioned this is V z earlier what I was using instead of writing all the time V z I am just writing it as V. So, I have rho C p V del T by del x that is stream wise direction and these two I am taking steady flow. So, del T by del T is cancelled out fully developed means V r equal to 0 V theta equal to 0 these two terms go out as 0. So, I am left out on the left hand side rho C p V del T by del x what about on the right hand side temperature is not a function of theta and it is not a function of x that is what we have seen we have seen that these two are not there. So, what is that I am having 1 by r del by del r del T by del r. So, if I take these two terms rho C p V del T by del x equal to k 1 upon r del by del r del T by del r I get V del T by del x equal to k by rho C p is alpha by r del by del r del T by del r. Now, what is to be figured out here now we are doing another boundary condition what is the boundary condition I am solving it for I am solving it for constant heat flux. I am not solving this for constant wall temperature I am solving this for constant heat flux and professor Arun has already shown us that for constant heat flux boundary condition what is d T by d x he has shown that d T by d x equal to 2 q dot upon rho C p V m r V m is the average velocity q dot s is the constant heat flux this has been coming from simple energy balances and applying energy a simple energy balances and Newton's law of cooling I am applying that and I am getting this that is what I am going to invoke or use that. That means, basically what I mean to say is I know d T by d x in terms of constant heat flux and the mass flow rate let us see how do we put that. If I put that that is d T by d del T by del x equal to del T s by del x equal to del T m by del x equal to q 2 q dot s upon rho C p V m into r this is constant that is what we have seen that is the slope of the bulk mean temperature and slope of the constant slope of the wall temperature both are same that is what we said this is the basic check what we do while doing experiment if I plot temperature versus x if it is fully developed my T s is going to be linear so is my bulk temperature and the difference between the two at any location is going to be constant this is the first basic check what we do when we do experiment that is not coming then something is wrong because I can check always T s while I am doing experiments I cannot measure bulk mean temperature at every location can I know professor showed us that it has to be integrated at every location velocity and temperature at every location I need velocity and temperature they are this data is not easy to come by and that too especially I should be doing this non-intrusively if I put a thermocouple or put a hot wire anemometer it is going to intrude and disturb my velocity profile and the temperature profile what do I do I usually measure the bulk temperature at the inlet and what do I do I measure the bulk temperature at the exit how do I measure this how do I measure this if it is water it is much easy if it is water all that I need today is in fact it is also called as mixing cup temperature why because this is taken through pipe where it gets mixed up that is there is a mixing cup that is let me draw that let us say this is a pipe and now it is made to go through a mixing cup where in which it is asked to go through lot of tortuous paths and then collect that in a flask and put a thermocouple or a thermometer in this flask and get the temperature whatever temperature you get by mixing this in the mixing cup and then taking it to a flask and measuring is what is called as the mixing cup temperature or the bulk mean temperature this you can do only at the exit you cannot do anywhere at the centre anywhere in between so you get usually the bulk mean temperature at the exit and the bulk mean temperature at the so bulk mean temperature at the inlet so I am just taking one question from J N T U Hyderabad a condition wherein both constant wall temperature and constant heat flux can be maintained simultaneously if it is not possible why is it not possible over to you sir see the question is can I maintain constant heat flux and constant wall temperature simultaneously I will not answer this question through heat transfer I will answer this question through electricity I will post back professor this question to you like this can I maintain constant voltage and constant current in any transformer or any DC power supply simultaneously both it is not possible sir so now similarly it is not possible to maintain constant heat flux and constant wall temperature we understand electricity very well so that is the reason I took this analogy so no offense in this the point is it is very difficult we cannot maintain the same potential difference and as I have the same constant heat flux also the same current also it is just not possible one of the two can be maintained similarly if I maintain constant heat flux potentials get fixed that is the temperature if I maintain constant constant temperature currents get fixed that is the heat fluxes so we cannot maintain both of them together professor any other question professor just a minute sir the question is regarding the skin friction coefficient and friction factor sir what is the difference between the two and what is the physical significance of Darcy's friction factor the question asked is what is Darcy's friction factor and how is it related to skin friction factor and what is the physical significance of Darcy's friction factor okay I do not see this any of my transparencies but I think I need to derive this let me do that so what do we do for Darcy's friction factor so we derive skin friction factor or we define C f equal to tau all upon half rho u mean squared I am doing this for circular pipe because that makes it easy okay now what is friction factor defined for a circular pipe if I take if I take a pipe if I take a pipe and there is a flow taking place what how do I define this hf equal to what is that hf equal to f l v squared divided by 2 gd let me multiply on both sides rho g what do I get delta p equal to rho f l v squared divided by 2d how did this come from actually actually it came like this delta p divided by rho v squared is when I do the non-dimensional analysis which was done by Buckingham's pie theorem in fact there no no I never took this equation was done Niko Rajse perhaps done did this using Buckingham's pie theorem because he is the one who has reported the friction factor data which we use in the form of the Moody's chart that is delta p upon half rho v squared is directly proportion is a function of is is a function of l by d and friction factor if I combine this this is what it is delta p equal to rho f l v squared divided by 2d now this is what is this this is Darcy's friction factor Darcy's friction factor now how do I connect this friction factor with skin friction factor now let us do that if I take a small control volume in the pipe okay and this is my pressure p 1 and this is my pressure here p 2 and what is the shear stress tau wall which is acting opposite to the flow direction so what do I get tau wall in 2 tau wall is acting like heat transfer coefficient it is like batting area that is in inside surface area that is pi into d into l this should be equated by pressure force that is delta p I am not writing delta p equal to p 1 minus p 2 we very well know that delta p into pi d squared by 4 so pi pi gets cancelled d d gets cancelled so I am getting with tau wall equal to delta p into d by l upon 4 okay now let me substitute this tau wall now what is this tau wall okay now if I substitute this tau wall in this equation or I can substitute in this equation if I substitute this what do I get delta p equal to tau wall into l by d into 4 is that right delta p equal to now if you equate I am not going to do that if you put this here in this equation and in turn compare with that equation you are going to get 4 equal f equal to 4 c f so what is that what is that essentially friction f is also shear stress tau c f is also shear stress it is just that the way we have defined c f and the way we have defined f is different that is all that is all that is that is f equal to what did we write f equal to delta p that is f equal to delta p divided by rho v squared by 2 into l by d that is pressure drop increases with the increase of the length that is l by d that is very obvious to see that is what nicorat said did using bucking amps by theorem but people had defined c f as tau wall upon half rho u m squared usually c f definition is usually used for external flows rarely we use for internal flows but not to say that we never use it we do use but rarely we use so this is the physical significance between f and c f some one substance is that f is also shear stress c f is also shear stress basically it is the resistance offered by the flow by the plate or the surface for the flow to take place by virtue of viscosity that is no slip condition that is the answer for your question is that ok ma'am thank you sir thank you one question from Jaipur college yes sir good afternoon sir my question is in thermally developing region convective heat transfer coefficient is more or less than fully developed temperature profile. How does the heat transfer coefficient compare in the developing and the fully developed region if this is heat transfer coefficient and this is my stream wise direction let us say this is my developing region developing region and this is let us say fully developed region fully developed region the heat transfer coefficient in the developing region continually decreases and in the develop fully developed region it becomes constant ok. So, that is the answer for your question my second question is why temperature boundary layer is much more rapidly than velocity boundary layer the question asked is why is the developing length for hydrodynamically developing length is larger than thermally developing length yeah it depends in the parental number you cannot generalize like that let me tell the answer like this if it is if it is yeah here you see this is parental number greater than 1 delta is greater than delta t that means what this is delta and this is delta t this is this is delta this is delta and this is delta t. So, what what is the same for parental numbers greater than 1 developing length required for thermally developing length is much smaller than hydrodynamically developing sorry hydrodynamically developing length is much smaller than the thermally developing length. So, it and for parental number less than 1 opposite that is hydrodynamically developing length will be much larger than thermally developing length. So, why do you say that the thermal developing length is always smaller than hydrodynamically developing length it is not true it is not ok. I think we will stop question answers and now it is time that we will go ahead with our discussion ok. So, we will get started with where we left. So, we said that we have steady flow incompressible flow and fully developed thermally and hydrodynamically and also constant heat flux boundary condition is taken and it is a circular pipe flow and we showed that this is this is what the equation turns down to and d t by d x is given by q q q 2 q dot upon rho v m c p r this is known constant heat flux I know the average velocity I know the radius of the pipe I know c p that means d t by d x I know why do why should I know that because I want to substitute here for delta t by del x. So, that is known now now do I know v yes I know the velocity profile I have derived that in terms of central line velocity or in terms of average velocity. So, u v equal to v maximum into 1 minus r by r whole square. So, if I have to present this v maximum in terms of average velocity it would be becoming 2 into average velocity. Now, if I substitute these two here in this equation what do I get v equal to v maximum into 1 minus r by r whole square that is what I am doing in the next step. So, that is let me do this as slow as I can I do not want to run rush through here I have taken v del t by del x equal to alpha by r del by del r del t by del r please watch my words I am going as slow as I can v equal to v maximum into 1 minus r by r whole square and del t by del x equal to q q dot s upon rho c p v m r then what is that I am looking for temperature as a function of r I am looking for. If I substitute this here and v max I am substituting for 2 into v average because this v average can be cancelled out with this v average into 1 minus r by r whole square equal to alpha I am writing this as k by rho c p r into d by d r d t by d r now you see I am saying the temperature is only a function of r now rho c p rho c p gets cancelled out v average v average gets cancelled out I get 2 into 2 4 q dot s upon k r that means this k I have shifted to the left hand side k r into 1 minus r by r whole square equal to 1 by r d by d r of r d t by d r now I am going to multiply this r to the left hand side that is I am pushing this r which is there in the denominator of the right hand side to the left hand side if I do that I will have to multiply by r that is 4 q dot s upon k r equal to r minus r q by r square equal to d by d r of r d t by d r that is if I integrate this now c 1 plus 4 q dot s upon k r into r square by 2 minus r to the power of 4 upon 4 r square equal to r d t by d r now what is the boundary condition let us try to find c 1 r equal to 0 temperature profile is symmetric so d t by d r equal to 0 so I get if I substitute r equal to 0 and d t by d r equal to 0 I get c 1 equal to 0 so now so I can remove this c 1 and write this equation again and now divide it by r so 4 q dot s upon k r if I divide it by r I get r by 2 minus r q by 4 r square equal to d t by d r again I integrate this I get c 2 plus 4 q dot s upon k r into r square by 4 minus r to the power of 4 upon 16 r square equal to t now the onus is or the responsibility is to get this c 2 I take r equal to r equal to t at r equal to r that is at the wall t equal to t s do I know this t s do I know this t s no what is my boundary condition it is constant heat flux I do not know t s but still no matter whatever is the boundary condition that is even if it is constant heat flux it is going to have a temperature t s what is the temperature I do not know let it be as t s I will take r equal to r s t equal to t s because I need to know the potential if I do not know the potential it is just current I cannot solve the problem so I need to know the potential so I am assuming the potential as t equal to t s so c 2 plus 4 q dot s upon k r into r square by 4 minus r to the power of 4 upon 16 r square equal to t s which will give me c 2 equal to t s minus of 3 q s dot r upon 4 if I substitute this in this equation that is substitute c 2 in this equation which is what I have done I get t equal to t s minus 3 q s dot r upon 4 k plus 4 q s dot upon k r into r square by 4 minus r to the power of 4 upon 16 r square let me simplify this so t equal to t s minus q s dot r upon k into 3 by 4 minus r square by capital R square minus plus r to the power of 4 upon 4 r to the power of 4 now I am let us keep this temperature as it is but remember I do not know t s I need to find t m why do I need to find t m q double dash equal to h into t s minus t m so if I get t m then I will be able to get my h q double dash I know let us see how do we get this t s minus t m to get that first I will try to get t m so for getting t m I have t here why did I do this because in t m I need a temperature profile and also the velocity profile so bulk temperature we have derived this already bulk temperature is given by 2 upon v average r square integral of 0 to r t of r v of r r t r so by putting this in this equation I will be able to get r by in the numerator and denominator has got cancelled so I get 2 upon v average r square if I substitute for t of r whole of this equation and for v r v substitute 2 v average into 1 minus r by r full square this is the paraboloid which we have derived little earlier into r dr if I integrate this I think I will not do that integration it is straight forward integration if I do this integration I am going to get t m equal to t s into t s minus 11 24 q s dot r by k I have put all the steps I have not skip even a single step so we have tried to incorporate all the steps please go through those steps you will get t m minus t s or t s minus t m equal to 11 by 24 q s dot r by k what is h h equal to q s dot upon t s minus t m for t s minus t m I have already found this if I substitute this I will get 24 by n but I have defined nusselt number as h d by k instead of h r by k so I multiply by 2 both sides I get 48 by 11 that is nusselt number on the basis of diameter of the pipe I get 4.3636 now very interesting result we need to stop and think nusselt number in laminar flow in internal flows is constant is constant it is independent of both Reynolds number and Franklin number so how do you explain this I am sure you will be asking me questions so before you ask I am going to preempt you the only plausible not possible plausible that is probable answer for this is that because the flow is taking place in pipe as laminae one lamina is not acting with the another lamina so there is no increase of the nusselt number with the increase of the Reynolds number so there is no interaction with the with one layer to the another layer that is all one can think of and this has been experimentally checked and found that for constant heat flux boundary condition no matter what Reynolds number you do as long as the flow is laminar that is the Reynolds number is still around 2000 less than 2000 to 2300 you continue to get the nusselt number as 4.3636 in fact doing experiments for laminar flow is much difficult than turbulent flow because heat loss computations are difficult if you do the same thing for constant wall temperature it is quite difficult to do it is an iterative solution we do not really do for undergraduate for post graduate this we give as an assignment so please do not give this as an assignment for UG guys if it is for post graduate students definitely go ahead and give this as an assignment you get 3.66 only one question which comes to our mind is for constant heat flux boundary condition I am getting a nusselt number of nusselt number of 4.366 but for constant wall temperature for constant heat flux I am getting nusselt number of 4.36 for constant wall temperature I am getting a nusselt number of 3.66 why is the nusselt number for constant wall temperature lower than that of constant heat that can be explained through temperature distribution if for the constant heat flux case how does the temperature vary with location so this is the temperature of the wall and this is the bulk fluid temperature and here in case of constant wall temperature this is for constant heat flux for constant wall temperature this is wall temperature and my fluid temperature what is happening the delta T in constant heat flux case is constant throughout but for constant wall temperature case the delta T gajendra you are sleeping constant wall temperature if you what is happening the temperature difference is decreasing constant for constant wall temperature as I move along the temperature difference between the wall and the fluid is decreasing heat transfer is essentially because the temperature difference if the temperature difference is decreasing the heat transfer is not so effective. So that is why the heat transfer coefficient in case of constant heat flux case is higher than that of constant wall temperature case so everything has a reason if we can try to understand. So now that is about constant wall wall temperature like that one can as I said in the morning in shah and London for various configuration you can get both friction factor and nusselt numbers of course this is for laminar flow both for constant wall temperature and constant heat flux no matter whatever is the configuration you will see that for constant heat flux case the nusselt numbers are higher than that for constant wall temperature case. So now there are various correlation for a developing flow you will see that there is L by D sitting and for property another important thing here is that in few correlations you would see property variation that is how there is a property variation from inlet to outlet my bulk temperature is changing from inlet to outlet. So my properties are also going to vary from inlet to outlet so how do I account usually the property variation is accounted by this factor that is mu B by mu wall or mu surface to the power of 0.14 mu B is taken as the viscosity at t i plus t e by 2 that is inlet temperature and outlet temperature average of the two. So it is not an unreasonable assumption because at least if it is constant heat flux the wall temperature is varying linearly. Now before I come to this problem before I come to this problem I just want to deal with turbulent flow in tube in turbulent flow in pipes in pipes what is happening there is a standard correlation what is called as Dittus Volcker correlation. So this is a mu equal to let me write it because we need to I think as engineers we need to remember this I do not think we can afford to forget this that is nusselt number equal to nusselt number is equal to 0.022. 3 r e to the power of 0.8 and p r to the power of n where n equal to 0.4 for heating and 0.3 for cooling. See one important thing one way what we need to say is that is cooling of the fluid not the tube cooling of the fluid through the tube. So that is heating and cooling it is with reference to the fluid not with reference to the tube and the properties are taken at T i plus T e by 2 that is properties of rho mu C p k in r e and p r are taken as the average temperature of inlet and outlet temperature. How is this equation come from this equation has come thoroughly through experiments and this is what is called as Dittus Volcker. If I remember correctly this correlation was published in 1950 and even today it works and he took around hundreds of experimental data and he collected all that data which was existing till that time and put an empirical correlation and of course he knew that he knew equal to C r e to the power of m and p r to the power of n and he also knew that n is roughly around 0.3. So that is what he did for heating and cooling is suggesting differently. This is 0.4 0.3 essentially takes care of property variations which we do not do in this that is mu b by mu walt to the power of n that is what he is doing here. So and r e to the power of 0.8 is an empirical correlation. So whenever you do turbulent flow you always check whether you are getting r e to the power of 0.8. Another important thing is that I had not answered for in external flows that is in turbulent flows the heat transfer coefficient is independent of heat transfer coefficient is independent of boundary condition. It does not matter whether I do with constant wall temperature and constant heat flux constant wall temperature or constant heat flux. So in both the cases I am going to get the same number why because it is of turbulence it is so random it is so much mixing that it does not know what is the boundary condition. So whatever they have measured for constant wall temperature and constant heat flux the heat transfer coefficient for the in both the cases are within the experimental uncertainty they are matching. So that is why it is stated that the heat transfer coefficient is independent of boundary condition as long as my flow is turbulent. So I think with this we have stated the correlation both in tubes and pipes sorry both for laminar and turbulent. So there are various correlations I am not going to touch each one one by one another important correlations which is usually used is Nielinski's correlation. Dittus Bolter works very well for gases and Nielinski works very well for liquids. I have learned this very hard way whenever we were measuring for water we never got the matching well for water with Dittus Bolter. But somehow I landed up with this Nielinski's correlation which was there in our lab in a old thesis I landed up in this correlation and then I got my correlation my experimental data right. What is that is Nielinski around 1970's or 1975 over and above the data of what Dittus Bolter collected he collected again for next 25 years data and fit the correlation and gave the correlation in terms of friction factor. In fact this correlation can be derived from fundamental this form can be derived from fundamental what is not derived is only this 12.7 12.7 is just tweaked based on the experimental data. Actually this comes from universal velocity profile and the universal temperature profile it is beyond the scope of our course we will not be able to teach all that but if we take a convective heat and mass transfer course we need to derive how this comes from. So this is semi empirical we can say it is not purely empirical although Dittus is purely empirical but Nielinski is semi empirical and for liquid metal again for you have different correlations for liquid metals and again you can see it is r e to the power of 0.85 although we are making a small nitpicking here you see that there is a slight variation not very much variation you can say constant wall temperature and constant heat flux are same. So this is Dittus Bolter correlation actually always there is a problem when you are measuring the heat transfer coefficient between 2300 and 10000 it is a very difficult zone. So we always whenever you are designing a heat exchanger you ensure that your Reynolds number is greater than 10000 although we say that it is it transits the flow transits from laminar to turbulent around 2300 for an engineer it is not 2300 for an engineer it is 10000 you ensure that your flow is greater than 10000 so that you are safe safely you have ensured that the flow is indeed turbulent. So that is how for a fluid mechanism he will say no my Reynolds number greater than 2300 is just transiting to turbulent but for a heat transfer expert who is doing heat exchanger design he will ensure that Reynolds number is greater than 10000 for doubly being sure that the flow is indeed turbulent even these correlations are valid for Reynolds number greater than 10000 for simple reason that for Reynolds number less than 10000 uncertainties in measurements of h is quite high. So that is the reason why we do not suggest this correlation for Reynolds number less than 10000 and of course for rough surface 1 upon square root of f you can see that this is epsilon by D sitting this is Colebrook correlation this is not this is semi empirical I will just take 5 minutes to tell how that correlation has come I am not going to derive that because for that is simple fluid mechanics but nevertheless I will just do it on a fast track how one can get simple friction factor see all of us use for friction factor in pipe what is the friction factor correlation we use generally we use what is called as blazes friction factor again blazes okay blazes is there everywhere blazes is like God to us f is equal to 0.3164 re to the power of minus 0.25 okay this is blazes friction factor now this is coming purely from empirical relation simple curve fit now what did Prandtl do is that we have velocity distribution. So if I take the flow rate pi r squared u average and I know the velocity if I take velocity profile and what is my velocity profile I have the universal velocity profile u plus equal to y plus if I take that that is u plus equal to a log y plus plus b with a equal to 2.5 and b equal to 5.5 and do the transformation of all this in terms of u plus and y plus I am not doing the algebra I am sorry about that I cannot do that if you do this algebra but the full algebra is here all that I am doing is transform this u and y and r in terms of u plus y plus and r plus the definition of u plus is u by u tau and y plus is y u tau by nu and why I am doing this because I know u plus in terms of y plus if I do that and substitute and integrate that and if I do that integration I am going to get the correlation as 1 upon square root of f equal to 2.035 log of re square root of f minus 0.9129. So this was derived by Prandtl what did Prandtl do he did not blindly apply f equal to 0.3164 re to the power of minus 1 by 4 he went back and saw that with this data with this correlation whatever here with this relation what he has derived from fundamental universal velocity profile he took the Nicaragic experimental data which is available if you see here he took the Nicaragic's data and he sees that he sees that the blazes correlation does not work for higher Reynolds numbers. So he tweaked this 2.035 and 0.9129 just to 2 and 0.8 and he finds that for full Nicaragic's whole data this equation fits very well. It is not about empiricism why I am emphasizing this because why is it so universal that universal profile is universal because of this relation you could derive the friction factor from fundamental. So that is the beauty of this and now Prandtl was convinced that universal velocity profile is applicable and can be used for deriving an engineering parameter actually for Nusselt number also one can derive this I am not doing that I am just saying that for if I use universal velocity profile I can derive friction factor as a function of Reynolds number again it is proved for friction factor all that I need to know is Reynolds number friction factor is a function of Reynolds number everything is getting converging whatever we are studying is the whole lot of effort of 100 years. So that is why everything seems to be falling in place that is 1 plus 1 equal to 2. So this is what is being told here and for rough surface also we can do the same thing I am not doing please go back and do this you can get if you do the again for universal velocity profile for rough surface because for rough surface I am only in turbulent boundary layer I have to take again a plus log of y plus but instead of y plus I take r upon epsilon if you do that if you do that you will get 1 upon square root of f equal to 2 log epsilon by d 3.7 r e square root of f. So this is what is called as Cole-Brugst correlation. So I think we have solved all of the friction factors this is the velocity profile for turbulent flow you go through this and this says that this is parabolic and for turbulent flow this is at profile and you take care of all other things that is this n that is power law for circular for turbulent profile u by u average is equal to y by r to the power of 1 by n and that n is a function of Reynolds number that is what is being shown here. So with this I think we will take questions and so I do not think we will solve any of these problems because they are because they are simple plug-in problem they are elaborately put all that we say is first find out r e pick up the right correlation friction factor and pick up the right correlation for Russell number and get the heat transfer coefficient and get the heat transfer rate. I think we will take more questions rather than solving these mundane problems that is plug-in problem PSG Coimbatore. Sir in one of the graphs which shows the variation of H with theta for flow over cylinder the Nusselt number is high at the stagnation point where the flow is almost at rest. So I do not know how to interpret can you please explain. The question is for external flow over a cylinder the Nusselt number is taken is high at the initial stagnation point and then drops how do you explain that is that the correct way of interpreting yeah. So, this what is Nusselt what is stagnation point stagnation point refers to the point of 0 velocity the flow has stagnated. There is at that point the entire what you say entire kinetic energy has come to 0 after that what is happening the flow the boundary layer is being formed over the cylinder because of this boundary layer formation we are we are trying to if you are trying to talk from a heat transfer point of view the boundary layer is like a region which is offering a resistance to heat transfer. So that point stagnation point where the fluid has been brought to rest that point is the region of highest heat transfer. Then what happens the boundary layer develops the resistance to heat transfer increases slowly and then after this so called transition to turbulence etcetera there is a further increase. So, the primary or rather I should say the decrease in the heat transfer from the initial stagnation point onwards is due to the development of the boundary layer at that point. Thank you sir one more question this is related to the variation of hedge in the entry region the hedge decreases. In the entry region and I would like to know what is the reason for the decrease in hedge. See what is happening again answer incidentally happens to be same the question is in the developing region heat transfer coefficient is decreasing with the increase of developing length. So, why is it so again the boundary layer thickness is when we started off there is no boundary layer thickness that is why my heat transfer coefficient is maximum. Now as I am moving towards the center of the pipe my boundary layer thickness is increasing because of which my heat transfer coefficient continuously decreases and then it becomes constant why because my boundary layer thickness is also constant. So, that is another way of looking at it and explaining that why heat transfer coefficient continuously decreases and then it becomes constant. Baramati any questions please. First about thermal and hydraulic boundary layer in a case of stationary plate, but if we talk about the moving plate for example, submarine moving in a sea then there is a temperature and velocity gradient then how is the formation of thermal and hydraulic boundary layer there. Very good question actually I have not touched this the question is the question asked by one of the participants is that what happens to my boundary layer when I am moving that is when my plate is moving the participant has taken the example that my submarine is moving my submarine is moving and, but my fluid is still see the answer is as follows when Wright brothers did his experiment he did not actually in a wind tunnel whenever we do experiments in a wind tunnel whenever we do experiment I put my aerofoil fixed and my flow is flowing over it my flow is flowing over it it is same as my aircraft moving in a still air what is important is the relative velocity, but not which one is moving and which one is stationary the boundary layer growth does not depend on the which one is moving that is whether the solid is moving or the fluid is moving here in the submarine case for example, my submarine is moving and, but my fluid is still and in the aero plane also plane is moving, but the fluid is still if I have to simulate these two cases in my lab I will put my submarine still and make it make the water flow over the submarine, but the boundary layer growth in my lab whatever I get and the boundary layer growth what I get in the submarine they are all same. So, someone substance is that what is important is the relative velocity if the relative velocity is same the boundary layer growth is going to be same then the heat transfer coefficients are going to be same and then the friction factors are also going to be same same thing for thermal boundary layer thickness also over to you. What is the effect of cryogenic liquid on the velocity profile and temperature profile? The question is what is the effect of cryogenic liquid on the liquid temperature profile and the velocity profile cryogenic liquid is also not out of the world liquid nitrogen only if it is in liquid phase it acts like a liquid if it is in gaseous phase it acts like a gas. So, it is as good as any other fluid as long as we know it is Prandtl number that is all. Sir one more question. Good afternoon sir while morning we discussing the constant wall heat flux condition temperature profiles were drawn at two locations x 1 and x 2. So, whether what happens to surface pressure whether it increases or remains constant how it increases surface temperature constant wall heat flux condition. See constant wall heat the question is for constant wall heat flux condition we had drawn two temperature profiles at two different axial locations. The question participant is asking is what happens to the surface temperature does it increase? The answer is we are supplying heat constant wall heat flux is being maintained that means watt per meter square is fixed. Therefore, the fluid temperature will also have to increase this is logically clear. Now, what we are saying is because heat flux is constant the wall temperature will also increase. In fact, if you look at this projection I have shown a small arrow for T s 1 a shorter length whereas, T s 2 on the next diagram is quite big. Fully developed flow heat flux is constant, but what we are saying is the wall temperature will increase why should it increase because I have inherently added more heat to the fluid. We are able to understand bulk fluid temperature increasing with the same logic wall surface temperature will also increase. Sir, one more question please. Sir, this is regarding Nusselt number, Nusselt number in laminar flow with increase in aspect ratio that is a by b ratio Nusselt number is increasing can you explain this? Let us start you have shown that higher is a by b higher is Nusselt number. The question is with in internal flows for higher aspect ratio channel there is an increase in the Nusselt number compared to that of the low aspect ratio channel. So, what you are saying is actually as a by b is increasing that is a is becoming wider than b that means b is becoming smaller and smaller compared to that of a. The question is here in this figure as a by b is increasing that is as my a by b is increasing my Nusselt number whether it is constant wall temperature or constant heat flux the Nusselt numbers are increasing. Why because the interference between the one wall to the another wall that is the boundary layer interaction is increasing in case of higher aspect ratio in case of higher aspect ratio. Let me draw let me let us draw this and see yeah. So, if I have lower aspect ratio as opposed to higher this is the first one is the higher aspect ratio and sorry first one is the low aspect ratio and the second one is the high aspect ratio. In this the interaction in the second one the interaction of the boundary layer is more rigorous because of which the Nusselt numbers in case of the in case of high aspect ratio the Nusselt numbers are Nusselt numbers are higher that is how one can think or one wall if you have to talk it in a very layman one wall is talking with the another wall more rigorously compared to the first case that is the square case. R C Patel Shripur my question is that in case of bulk mean sorry mean film temperature and bulk temperature. If there are inlet and outlet temperature is given for the water or any fluid and the surface temperature of the plate or tube is given at that time we are going to consider mean film temperature. So, what is the reason that we are going to take mean film temperature as the temperature at inlet and outlet both are given means where we use bulk mean temperature and where we use mean film temperature. The question asked is where do we use bulk mean temperature and where do we use mean fluid temperature mean fluid temperature when he says mean fluid temperature he means that it is the average of inlet and outlet temperature of the fluid. See if we see carefully in our correlation usually the mean film temperature that is T B plus sorry T infinity plus T S by 2 concept is used in external flows. But for most of the correlations if not all for most of the correlations in internal flows we use inlet to outlet temperature average that is the bulk mean temperature that is inlet. So, to summarize for we take T i plus T e by 2 for internal flows and we take T s plus T infinity by 2 for external flows for most of the correlations is that ok. Is there in internal fluid also we are going to consider mean film temperature sometimes in some numericals I have seen. See in some numericals sometimes we take mean film temperature not the mean fluid temperature that is specific to that correlation. But that is specific to the correlation in which he might have the experimentalist might have generated the correlation using the mean film temperature rather than taking the mean fluid temperature. So whoever is generating the correlation is going to specify for me whether it is going to be average bulk temperature or average film temperature. Another thing would be in an application where exit bulk fluid temperature is not known we can say that in the limiting condition the bulk fluid temperature at exit would be equal to the surface temperature. So, that also you can say that T s plus T inlet by 2 where this T s is approximated to T mean at exit in the limiting condition. So, that we also you can look at it. Sir one of the participant having a doubt regarding the convection or conduction in case of grease a lubricating fluid which we are using in vehicles that is grease. In case of that fluid which type of or which mode of heat transfer we have to consider. Ok grease is a sticky fluid the question asked is if grease is there in our vehicles which we are using how should I handle this grease what will be the mode of the heat transfer grease is actually a what to say a wax grease is wax that means it is acting like a sort of solid until it reaches the melting point temperature. Once it starts melting then it starts acting like a liquid. So, in fact fluid dynamically we can think of this as a thixotropic fluid that means some initial stress is required to melt it once it has gotten melted it becomes like a liquid. So, it is not going to be Newtonian, but once wax becomes like a liquid then we can perhaps assume it as a Newtonian fluid and whatever fundamentals we have studied so far can be applied to the wax which has melted into a liquid. But till wax melting to liquid it is a melting problem that is whatever heat is generated it has to be consumed for melting that is latent heat of fusion or latent heat of melting we have to utilize that for that and subsequent to that it will become convective heat transfer where in which generally of course it depends whether where you are using forced convection or natural convection. The question asked was by Anna University was that when a spacecraft moves into atmosphere or when it leaves the space how does the boundary layer grow? Before moving the space actually if I see in the question what it is told is that the please comment the development of hydrodynamic boundary layer and thermal boundary layer when the spacecraft enters the atmosphere from space. So, it is it has gotten entered into the atmosphere. So, what is happening is when it is entering we need to appreciate that it is entering at a supersonic not supersonic hypersonic flow that is Mach number is going to be Mach number is going to be very high that is as high as 5 times that means velocity Mach number is velocity of sound velocity of a fluid or the velocity of my rocket in my case to the velocity of sound in air equal to 5. So, V upon 330 if I take as the gamma RT taking ambient temperature conditions of course in space the temperature is going to be different. So, I am taking 3 330 into 5 what is that I am going to get it is going to be around 1500 plus 150 that is 1650 meters per second. So, it is not a small speed it is not a small speed when this kind of velocity is there when this kind of velocity is there there is for high speed flows the q double dash equal to h into T wall minus T infinity would not work. The boundary layer will be growing the same way it is going to grow for let us say if this is my spacecraft when it is going to come the boundary layer is going to grow the same way. But the only difference is that this boundary layer is compressible boundary layer it is not incompressible boundary layer compressible boundary layer what all we have studied so, far it is all incompressible boundary layer what is it which is going to make it different for compressible boundary layer is that q double dash is no longer T wall minus T infinity, we define q double dash as h into T wall minus T adiabatic wall temperature, wherein which adiabatic wall temperature is also not known. That is we cannot take T infinity as the reference, the fluid which is near the wall is also going to get heated up and I cannot take T infinity as my reference, I have to take some other reference temperature, that reference temperature is what is called as T adiabatic wall temperature. So point is the boundary layer, both thermal and the hydrodynamic boundary layer will grow the same way as it grew for flat plate and all, but the only difference is that this boundary layer is compressible boundary layer and the heat transfer coefficient definition that is the Newton's law of cooling is not going to be q double dash equal to h into T wall minus T infinity, but it is going to be q double dash equal to h into T wall minus T adiabatic wall. So, but then both there are two unknowns that is h is an unknown and also adiabatic wall temperature is unknown, both have to be measured experimentally, how to measure them I think let us let us not take up that right now. So, we will move on to next question n 83 3 Trichurapalli or n 83 3 Sir while dealing with convection phenomenon we always come across flow over flat plate, flow over cylinder, flow over sphere, but in real life situation we will come across so much complex shapes and bodies. So, how for these kind of complex shapes how to derive a heat transfer equation as well as Nusselt number equation and how to model this kind of systems over to you Sir. See the question asked is the real life objects are not going to be flat plates or spheres or cylinders, how do I handle flow over real life objects. For example, human body if I take flow over a human body human body itself is a is an object I can have a boundary layer on my body that is what happens in essentially in cold time I keep myself my hands what is that I keep I keep decreasing my surface area by folding my hands when I am whenever I am feeling cold essentially what I am trying to do is I am trying to decrease the heat transfer rate by decreasing the surface area. So, heat transfer coefficient is same cooling heat transfer coefficient same T s minus T body is also T body minus T infinity is also same, but what I am trying to do is a s I am decreasing coming back to your question how do we handle for non non symmetric objects or real life objects. For example, a human body I will idealize it as a flow around a cylinder of course, if I have to really make nitpicking and I have to start modeling shoulders and all again same equations equations are not going to be different same conservation of mass conservation of momentum and conservation of energy is valid only thing is that I will have to take the recourse of CFD close to form solutions are not possible either I have to do the experiments scaled down or I have to take the recourse of computational fluid dynamics to solve this in computational fluid dynamics also you define you divide the complete control volume into small control volume and apply the energy balance momentum balance and the mass balance for each of these control volume. Sir when constant heat flux the Nusselt number is 4.36 when it is a constant wall temperature the Nusselt number is 3.66 why it is lesser in constant wall temperature. Okay I had answered this the question asked is for constant heat flux boundary condition we have an assault number of for a constant heat flux boundary condition we have an assault number of for constant heat flux boundary condition we have a Nusselt number of 4.36 for constant wall temperature case we have a Nusselt number of 3.66 why is that for constant wall temperature lower than that of constant heat flux. In fact I had answered this question in my lecture itself perhaps this point has got missed no problem T versus X if I put so for constant heat flux what do we get this is T s and this is T m okay this is bulk temperature this is bulk temperature and this is surface temperature. So note that the difference between the surface temperature and the bulk temperature is constant is not varying with space on the other hand for a constant wall temperature boundary condition case that is T wall equal to constant what is happening and this is the bulk temperature what is happening this is actually exponential function pressure around derived this that is logarithmic temperature. So that is this temperature difference is decreasing with the increase of X so delta T is decreasing as the temperature gradient is decreasing the heat transfer amount of the heat transfer also should decrease. So basically the net delta T lm of constant wall temperature will be lower than the delta T lm for constant heat flux for delta T lm in constant heat flux is T s minus T m so there is no variation so that is the reason why for constant heat flux boundary condition we get higher Nusselt number compared to that of constant wall temperature case okay over to you Amal Jyothi Kerala any questions sir actually we have two questions one is that the surface roughness how do we measure the surface roughness is it the height of the maximum ridge or valley something like that and second question is that the entry length suppose we calculate for the laminar flow I put the Reynolds number as 2000 and diameter of the pipe as almost 1 inch now I got the answer as at least 3 meters that means that suppose you are going to fix a wall or something in between that now will it affect the flow conditions or laboratory do you want to actually consider the entry length and fix the apparatus according to that okay so now the first question was how do we measure there are two questions asked by one of the professors that is how do we measure the first question is surface roughness surface roughness how does one measure this and the second question is if I put a value by developing length is going to get affected how do I handle this let me answer first question for surface roughness what people usually do is they measure what is called as they measure the surface roughness using either tally surf meter that is there is a or a perthometer this is also called as perthometer if you discuss this with manufacturing guys they will be knowing better than me but nevertheless there is a stylus if I have a rough surface this is my rough surface let us say there is a stylus which moves on this surface when it moves on the surface it captures the movement and based on this movement again I will take like the way we handled turbulence I will take average surface roughness maximum surface roughness minimum surface roughness and of course we are taking not the maximum but we take the average surface roughness that is the answer for the first question. Second question is that if we take L by D equal to what was that 0.06 RE for laminar flow and if I take Reynolds number equal to 1000 meter and diameter equal to 25 mm what do I get L equal to 0.06 into 25 into 10 to the power of minus 3 into 1000 that means 10 to the power of minus 3 and 1000 gets cancelled out 25 into 0.06 256 is 150 256 is 150 and I get how much length no 1.5 meter I get 1.5 meters yes I need to put 1.5 meters as the developing length let us say this is my developing length now let us say you are putting an orifice plate or something see for an orifice plate what is the necessary condition we need to generate fully developed flow okay for develop for getting fully developed flow what do you ensure you need to ensure that there is 1.5 meters given now for example in between there is a value then it is going to disturb the flow again after the value I need to put again 1.5 so flow of course there are some flow conditioners and all let us not get into that the point is that if you have a value or any disturbance bend 90 degree 180 degree that is if you have a 90 degree bend like this or 180 degree bend like this or 180 degree perpendicular again it takes a bend out of plane that is 90 degree 2 90 degree bends one out of plane so then in all of these cases you have to downstream this disturbance you have to have again developing length of 1.5 meter given otherwise the flow is not fully developed okay. Sir in case of roughness value do we actually measure the height of the petrobrans which is from the surface or do we take it as the average of the maximum height or the minimum height do we take it as average or we take it as the maximum height okay the question asked is for after measuring the surface roughness do I take the average roughness or things like that see if I were to plot I will take the tally surf meter or plateau meter and I will plot height versus protuberance height from the base versus location all over the place all over the place that is this is my flat plate on which protuberances are there and on which I have moved my tally surf meter so if I put my height my height is also going to be something like this so how do I how do I represent this I represent this by again h bar that is h average and then again h prime so one is the rms value another one is the average value so to answer your question both average height and the rms value are to be represented is that okay thank you sir over and over VJTI Mumbai any questions Sir I would like to know about the laminar sublayer in case of internal flow okay the question asked is what is laminar sublayer in case of internal flow laminar sublayer in case of internal flow and the laminar sublayer in case of external flow these two are not different that is what I said and that is why it is called as universal profile so in case of this is this is the universal profile this is the viscous sublayer we said that the viscous sublayer is same whether it is internal flow or external flow viscous sublayer is same so you are going to get y plus equal to that is u plus equal to for all u plus equal to y plus that is for all y plus less than 5 u plus is equal to y plus that is what is y plus y plus is y u tau by nu what is u tau u tau is square root of tau all by rho so if you are not understanding let me answer this little differently because I think none of us are understanding what exactly is this let me take I think I will generate this problem or I will generate this problem and put up this in moodle that is if I take the friction factor from because I do not have calculator that is why I will not be able to do this so I will take the friction factor from the Moody's chart and then calculate the tau all and from tau all we will calculate back the y so that is how one can check whether I have broken the sublayer or not the point is laminar sublayer whether it is internal or external fluid is just the same.