 I do not have to harp on where condensers are used okay in two phase flow in two phase flow heat exchanger condensers are a norm so I will straight get into nacelled condensation why is it called nacelled condensation we know but before we go to nacelled condensation there are two types of there are two types of condensation one is film wise another one is drop wise so in the film wise condensation it is a very natural to imagine in a film wise condensation there is a film and in drop wise condensation the condensation is occurring in the form of drop just to give you the visual picture this is the film a liquid film is being formed here drop wise condensation so out of the two which one is the best one drop wise condensation but maintaining drop wise condensation thoroughly always difficult so you usually what to say paint or treat you treat the surface with an appropriate material or a paint such that you can maintain drop wise condensation but then this treatment not lost for eternity it is going to last only for short time short duration short means it can be few days few months okay after that again it is going to get back to film condensation so that is the reason why we always study only film wise condensation and not only we study all heat exchangers are built on the base or the condensers are designed on the basis of film wise condensation even if you do drop wise even if you heat treat it even if you treat the wall with paints what are the paints I think I have some material here which says ha steric acid if you apply it with steric acid so there is it doesn't it should not get wetted basically should not get wetted then only wetted means the fluid should not go and stick to it that that kind of treatment I should make it but then the treatment cannot last for long so whenever I design a condenser I am going to design it on the basis of film wise condensation not on drop wise condensation even if I would have treated my surface even if I would have treated my surface I would continue to design it on the basis of film wise condensation okay so with that understanding so I don't want to lose time there are this is the film there is a small film being found we are going to spend enough time on that and naturally why do we go for condensers because in two phase flows the heat transfer coefficients are very very high how high yes yes how high they can be as these drop wise condensation can have a heat transfer coefficient of 30,000 watts per meter square kelvin on the other hand film wise condensation is typically of the order of 11,700 these are just numbers just to say that drop wise condensation roughly is 2 to 3 times better than that of film wise condensation film wise condensation okay so with that background let us get into what is called as nusselt condensation it is called as nusselt condensation because nusselt was the person who derived it for the first time okay so if I take let me just go ahead if I take a flat plate I am just taking a vertical plate to give you the big picture of today condensation I take a laminar I take a plate and condensation can be as I showed in the previous picture initially it is will be laminar and then it is maybe my boundary layer is going to start from the top okay because it has to condense and fall it is quite different converse to natural convection okay but the same body force is going to come okay so then my it can it is called here wavy but we can as well call it as transition there is no harm and it can subsequently go to turbulent but I am going to focus only till here I am not going to focus anything about that okay and of course correlations can be given in the same line what we have derived for laminar to turbulent also now coming back what is this this is my boundary layer or not boundary layer I should be calling I should not be calling here boundary layer because it is a film it has got condensed why why should it condense why is film being formed on the vertical plate because my plate is now maintained at a lower temperature how lower below saturation temperature corresponding to the pressure in which it is operating here let us say we can say it is atmosphere is that okay if I can bring down my wall temperature lower than the saturation temperature of the liquid then I am going to get the condensation on my okay now having understood that so coordinates let us get the coordinates fixed so I have x I have y okay this is my x this is my y and of course this is my u and this is my v without losing time we will be coming to this little later do not worry about that so there is a velocity and there is a temperature okay so how do I start how do I start with so many so many cases done so far how do I start I have to momentum and energy so in the assumptions I am taking that laminar flow constant properties are assumed gas is assumed to be pure vapor at a uniform temperature equal to saturation temperature okay so we will there is no it is a pure vapor there is no mixture here stress I will I will emphasize this at the end there is no shear stress at the liquid vapor interface that is I am considering that the vapor outside is not moving at all it is still if there is a vapor moving there is a vapor no around if the vapor is moving around that what will happen there is interfacial shear stress there is a shear stress because one film is moving at one velocity and vapor is moving at the another velocity so you are supposed to expect a interfacial shear stress interface means it is the interface between the liquid film and the vapor okay interface inter word inter means in between intermission we use the word intermission in between phase in between the two phases that is between the phase of liquid and the vapor phase vapor phase I am taking it as 0 that is why there is no interfacial shear stress at the interface only liquid is moving but the vapor is not moving I am saying when I say interface I mean here along this vapor were to move let us say in fact there will be slight movement of the vapor in real life but if the vapor is moving then there will be a slip no then there will be slip if there is a slip then again I am going to have the professor told us it is not about stationary flat plate and flow moving on it as long as I have the relative velocity I am going to have the slip and I am going to have the shear so that is that is what it is it has to be less than no how will it condense otherwise yeah correct okay and this is start but then when the heat transfer start between the plate and when will the heat transfer start between the plate and the my film when there is a temperature gradient so that temperature gradient has to be established by keeping the wall temperature okay so now the next recourse is I will have to take up my same momentum and continuity and I am sure now by now you would have felt already that one of the momentum equations will definitely vanish which one it would be x momentum because your velocity is more in the y direction so this equation is not important although I have gone at length and try to show here all that is not required now dp this is del p by del y can be replaced this dp by dy you see here this is rho L here rho L into g and this dp by dy has to be replaced by dp by dy is the imposed pressure from the inviscid portion that is the outside the film here is that right that is nothing but again here hydrostatic pressure hydrostatic pressure dp infinity by dy which is equal to rho v into g if I substitute that I will get rho L u del v by del x plus v del v by del y equal to mu L plus del squared by del x squared y del squared v by del y squared has been thrown out there are two terms in my momentum equation del squared v by del x squared plus del squared v by del y squared I have thrown out del squared v by del y squared my film thickness is of the order of delta y is of the order of my plate height delta is substantially lower compared to that of my plate height that is why I can throw del squared v by del y squared I think now we can feel it without even thinking is not it so g into now actually all this presentation of condensation is beautifully given by professor Bajan I would suggest that please read no other book gives the way Bajan gives okay I am a standard I am a great fan of Bajan but that does not matter the way I have understood condensation by reading Bajan I have not understood that way by reading incorpore and David or Changal or any other book for the even two Facebook books for that matter when I read Nusselt condensation in Bajan because he does that through scale analysis I will tell you in a couple of minute what are we going to do so please before teaching or before our end workshop main workshop please read Bajan's portion problem with Bajan is that Bajan's book is quite expensive 9000 rupees it costs heat transfer by Bajan but convective heat transfer is cheap I just brought this is expensive 9000 may be Indian edition has come I am not aware but convective heat transfer this is the Indian edition mind what is that not Indian edition student edition restricted for Asia so this is one no this is 420 rupees so I you can read this same thing is there but in an elaborated version whatever you start here in this that much portion you read in this book okay so this is quite cheap I would strongly recommend each one every teacher should have this book because scale analysis you can understand okay okay so now coming back what is the each term what is the each term the first term on the left hand side is first two terms are the inertia first term on the right hand side is friction viscous it is friction and what about this G into rho L minus rho V here Bajan calls it calls it as instead of bio NC because it is not bio NC because it is liquid which is falling it is actually opposite to bio NC that is sinking it is coming down sinking that is what by Bajan call I think it is okay thinking is a good opposite for bio NC now now what does he do to get a closed form solution to get a closed form solution he neglects inertia it is quite difficult to understand it he neglects inertia has to be balanced by frictional force and sinking effect is that right inertia has to be balanced by friction and sinking effect but what is this contention is that sinking effect is what is happening physically let us think what is happening condensation has happened because of we are maintaining low temperature it starts falling there is a sinking effect but because of the sinking is happening there is there is resistance for the film to drop or the for the fluid to drop but it is having momentum also that momentum who has created temperature gradient temperature gradient but what he is saying is that sinking effect predominates over the inertia for a minute it is quite difficult to explain by doing order of magnitude analysis or say that okay this is larger than this so this is smaller so I can neglect it is quite difficult to do so you can even say that I will assume I am going to assume how will I check whether my assumption is right or wrong at the end whatever heat transfer coefficient I am going to compute using this assumption if I check that with my experimental data that will tell me or suggest me whether this assumption is valid or not but otherwise it is quite difficult to explain why this inertia is substantially lower compared to that of sinking effect why are we going to throw this inertia out it is a mathematical expediency means if I throw this inertia out I can get the closed form simple solution okay that is what Nusselt did that is the greatness of Nusselt okay so come on if I do if I throw this my equation looks simple so mu del squared v by del x squared plus g into rho l minus rho v equal to 0 what are the boundary conditions at x equal to 0 v is 0 and at x equal to delta what is this del v by del x equal to 0 interfacial shear stress is 0 okay so now if I integrate this I get sorry mu l del v by del x plus g rho l minus rho v into x equal to c 1 further integration I will get mu l v plus x squared by 2 c 1 x plus c 2 if I apply first boundary condition c 2 is 0 if I apply second boundary condition I am going to get v v equal to g into rho l minus rho v so now you get the velocity did I get the velocity here do I know everything on the right hand side do I know everything on the right hand side of the velocity boundary layer not boundary layer thickness film thickness is not known until I know the film thickness I do not know the velocity let me keep that I have to figure out the film thickness how okay I will come to that little later only thing I will just try to do here is mass flow rate here mass flow rate no one is going to give us the mass flow rate is being generated because of the temperature gradient so what do I get mass flow rate here Vajan uses tau so I am also using tau tau of y equal to rho l v dx 0 to delta that is per unit width I am taking perpendicular to go the plate width I am taking as 1 so rho l v into dx integrated over 0 to delta if I integrate that I am going to end up with g rho l rho l minus rho v into delta cube by 3 million if I have to get total mass flow rate I will have to multiply this gamma y into b so that is what I have done and this is the mass flow rate but I cannot be very happy because still my unknown is delta delta is not known so all that is saying that flow rate is proportional to sinking at the vertical if the density gradient is high if rho l minus rho v is large my mass flow rate has to very easy to physically feel and viscosity is sitting in the denominator if viscosity rises my mass flow rate has to I think this is very obvious from our force and convection so this is the physical field now how do I get next question is to get the delta what am I going to do within my delta is that I am going to assume that it is going to be purely conduction because I have thrown is it logical or not why because I have thrown all my inertia term so if I so it is conduction pure conduction conduction in which direction x direction x direction so if it is then if it is 1D conduction means what should be the governing equation d square t by dx square get back to my e diffusion equation del square t by del y square del square t by del z square transient term all that are going to vanish I am going to end up with del square t by del x square equal to 0 if I integrate this I am going to get a linear equation that is del t by del x equal to c 1 and x equal to 0 t equal to t wall ct equal to so what are the boundary conditions at x equal to delta t equal to t sat and at x equal to 0 t equal to t wall if I put those boundary condition I am going to get t equal to t sat minus t wall into x by delta plus t wall so this is the linear temperature profile we all know if I assume conduction that is what I am supposed to get between t wall and the t sat okay now this is again standard I do not have to what is my h definition h equal to k l into del t by del x at the wall that is at x equal to 0 upon t sat minus t wall remember one new thing I have introduced for q double dash this we know q double dash equal to k l del t by del x at wall but what is the definition here I am using h into earlier for q double dash was equal to h into t wall minus t infinity or t infinity minus t wall here I am taking t sat minus t wall please note that difference t sat minus t wall if I substitute that here temperature gradient is going to be k l into t sat minus t wall within it is a linear temperature profile so I can write this t sat minus t wall upon delta so t sat minus t wall t sat minus t wall get cancelled out k l by delta till I have not got delta I have put mass flow rate in terms of delta heat transfer coefficient in terms of delta now our pursuit is to get delta let us see whether how do we get that so next is okay you will have to follow my steps so that is mass flow rate equal to this is what we have derived see if I try to find dm dot by dy how does this mass flow rate vary with y that is the vertical ordinate so you get 3 delta square d delta by dx is that right is that right rate of condensation of vapor rate of condensation of vapor over the vertical distance is it understood or not why okay let us spend time this is something critical so why am I written 3 delta square divided into d delta by dx rate of condensation of vapor over a distance y or is it mistake it is a mistake it is d delta by because I am differentiating with respect to y it has to be d delta by dy it is d delta by dy okay so then in the next what is this where is this heat transfer going where is this heat transfer going we are forgetting latent heat so dq dot whatever heat transfer is taking place within the film it has to be balanced through rate of heat transfer from the vapor to the plate through liquid film has to be equal to the heat released as the vapor as it is getting so that is what I have done dq dot equal to dm dot hfg dm dot by dm dot hfg is equal dm dot equal to dq dot is also equal to dm dot hfg is the latent heat that is this but this is also equal to what is this what is this conduction k l into b dy into t sat minus t wall upon delta so if I compare this equation with this equation I am going to get sorry yeah if I compare a and b unfortunately they are in two different locations I can I am going to get one huge equation if I solve with this with respect to this is y it is y again copy paste problem y if I integrate that I am going to get this equation okay but delta is equal to 0 at y equal to what is delta at y equal to 0 y has started at the tip of my plate 0 film is 0 so c is equal to 0 I am going to get delta as a function of delta has got I have obtained delta means I have got now both mass flow rate and also the heat transfer coefficient velocity of course the heat transfer coefficient that is what I have done in the following steps which I am not going to follow because delta is there I have told that heat transfer coefficient is equal to k by delta delta we have just now found I will tell you for that delta if I have to get the average I will introduce what sir Arun has taught as 1 by l 0 to l h y dy I am going to get the heat transfer coefficient till I have not answered the question whether that assumption was right or wrong so people have found that even in spite of that assumption people have found that this equation whatever heat transfer coefficient I compute using this equation and the experimentally measured heat transfer coefficient that is why that assumption is a no I will come to that I have not come to that see another thing what is that agreed till that now people after this people go on modifying various things I have now assumed saturation temperature but what if there is sub cooling what if there is sub cooling okay if there is sub cooling so they just rosin hoe rosin hoe takes that h f g instead of taking h f g here he suggests h f g prime which is equal to h f g plus 0.68 into Cpl T sat minus T wall what is this this is latent heat this is sensible heat and this is latent he came up with a number called Jacob number which is sensible heat upon latent heat and that ratio you see here he suggested this 0.68 okay and when I take this and I take the properties at T sat plus T wall by 2 so this all that he said is if there is sub cooling replace h f g in the previous equation with h f g prime okay so that is where h f g prime so this is the Jacob number definition and this is what h f g prime is going to be basically the heat released is going to have both the latent heat and also the sensible heat so that yeah one second one second one second one second so then now the question is people people do this represent this data in terms of Reynolds number because we have this always inclination to go back to Reynolds number but do we know Reynolds number here like natural convection do we know natural convection Reynolds number no here also we do not know Reynolds number but we have computed mass flow rate now have we computed mass flow rate we had shown that gamma y equal to g rho l blah blah blah into delta q delta we have computed delta we have computed here if I plug in that delta I am supposed to get my mass flow rate and that mass flow rate is if I substitute that that delta q if I substitute delta I do not substitute anything that huge equation people usually do not substitute people try to put that in terms of heat transfer what is delta k by h so that is what because we just showed know h equal to k by delta so same thing I have rewritten and k delta at x equal to l is k l upon h at x equal to l so if I substitute that I get Reynolds number as 4g okay 4g delta same thing if I substitute that and I substitute this local h turns out to be 3 by 4 times the h average if I substitute that I get h average in terms of Reynolds in terms of Reynolds number now just to show that here I have taken hydraulic diameter as 4 delta all the time you will see that you will see that I have taken 4 delta here hydraulic diameter I have taken it as 4 delta that is true I just put this figure I just copy pasted from Schengel I just you can go through this because this is just area upon perimeter it turns out that for any configuration it turns out that it is equal to 4 delta is that okay so now with this so I guess what is let us summarize what is that we have studied so far what did we do we took a vertical plate and we kept that vertical plate temperature below that of the saturation temperature so then what is happening the condensation is happening on my plate now we are interested in the film thickness we are interested in the film thickness we wrote the momentum equation and the energy equation which momentum equation we three vote x momentum equation in the y momentum equation we had we ended up with inertia force viscous force and the sinking force or sinking effect and we went ahead and made a very bold assumption that inertia effects are very less so we threw that out so it so happened that now it becomes viscous force becomes equal to sinking effect using that we could get a relation for delta no we could get what is that we got using that we got the velocity but in terms of delta and we went ahead and made the assumption because we have thrown the inertial term we went ahead and assumed that inertia term in my convection term in my energy equations are also neglected so I end up with conduction equation conduction I am assuming that again it is only 1D conduction if I do that conduction it turns out that h is equal to k by delta again am I unknown is delta so for getting delta what is that I invoked I balance it the equation whatever heat transferred by conduction is equal to latent heat latent heat without that information I would not have got delta that is very key okay I think it is always a good thing to summarize in terms of algorithm after your derivation is over the way we did it am going fast but on the board what did we do we took this equation through this you just to put sort of algorithm it gives big picture because here I did not stop anywhere in between and asked what are we doing why are we doing because I was fast okay so I got the delta then I got the velocity distribution heat transfer coefficient I got the velocity the same thing now can be extended to other configuration you can do the same thing for cylinder same analysis you can do there is no harm only thing is that little bit that constant will change you can give that as exercise for student okay and for turbulent again there are correlation you cannot derive for turbulent there are correlation before end workshop I will put those okay I have seen that they are there in jungle there I have seen that but whatever be the configuration whatever be the configuration I want to make one statement that whatever be the configuration my heat transfer coefficient ultimately is going to look in this form only all that it is going to change is okay this is for vertical now for wavy portion also there is a correlation given okay how do you decide whether I am in laminar or wavy is based on Reynolds number here for all Reynolds numbers less than 30 it is laminar for all Reynolds numbers less than 30 it is going to be laminar so how do how do I know that it is coming through experimental okay so there is a correlation given for wavy laminar and for turbulent this is the correlation okay so these are all obtained by experiment of course there are some theories also one can work out for at least for vertical plates but for complicated configuration it becomes quite difficult so that is about the I would stop here for condensation but I have just filled in because based on your request whatever I had I have just adjusted that so that at least the crux of the matter I can drive home we can build upon this skeleton before the main workshop now open for question yes ma'am what is your question. In the initial term sir but when you neglect the inertial I mean when you have heated shakers actual heated shakers in which flow is going to face the particular value of Reynolds number Reynolds number then it means that inertial term how will you neglect you have some 1800 also Reynolds number you showed for laminar. We are mixing up please we are mixing up please that is right it is like asking a question why I am rephrasing this question because I got your question it is like asking I am going to design a heat exchanger but I am you are suggesting me a heat transfer coefficient for natural convection but in my heat exchanger there is going to be a forced flow how can I apply the heat transfer coefficient which has been derived for natural convection for forced convection can I use this for heat exchanger in which flow is taking place can I use this for that no absolutely no there is no flow here it is like natural convection it is like natural convection that is condensation in tubes is a different ball game with forced flow there are correlations actually in fact there is a beautiful correlation called correlation it so turns out that Nusselt number upon a new Dittus bolter I take this example again because Dittus bolter is the mother every everywhere a new condensation upon a new Dittus bolter is equal to function of quality thermodynamic quality it turns out like that it shas correlation it works beautifully well that is the correlation I am supposed to use and how it has been obtained it has come out through experiment. So for various because when it is condensing my quality is going to increase so as it increases I am going to fix the length so please do not get confused it is like to make myself understand I am I am not trying to use natural convective heat transfer coefficient correlation to force recognition okay yes sir zero shear stress because I know that there is vapor I know that there is vapor moving to make my life easy I am taking the interfacial shear stress as zero but I do not think it is a very invalid assumption because interfacial vapor is not really going to move at a very high velocity okay vapor is having velocity means all this will fall that is what if there is a flow taking place then it gets into forced flow this is like as I keep saying I always imagine like natural convection and force recognition here my my air which is there outside is still still still still that is it yeah yeah see you please imagine natural convection and forced convection that is where ma'am also got confused so it is natural convection it is like natural convection only thing is I have a film form is that okay any more questions I can spend five more minutes yeah adding the same topic okay by by all shear stress we mean that either you can shear stress means what let us get back to basic question what is shear stress velocity gradients so far we usually get tuned in our minds because shear stress can occur because my plate is moving that is what I am saying that's why I am neglecting that interfacial shear stress that shear stress only I am neglecting and actually that vapor is not going to be still in real life it is going to be slight slight velocity is going to be there okay let's go ahead and assume that vapor is still that shear stress only I am taking it as 0 I am taking it as 0 otherwise I cannot get the closed form solution how in which other way I can answer you why I can take this shear stress as 0 I can't get it any other way because that is right that is shear stress only that is why I am saying it why I am saying what is that written there 0 shear what is that it is written there 0 shear del u by del y is equal to 0 so we are assuming that assuming this is an assumption but that's a good question maybe I will think back I have not thought much because this just came out of the blue after maybe I have thought this maybe after 4 or 5 years I am teaching this but that's fine we will keep this in the back of the mind but this I can go ahead and say that this is an assumption because it works that way I can get away but still that's not a convincing answer we will see how best we can answer it later on is that okay any more questions fine.