 So, we will go ahead and get started for our final lecture, what is condensation? So, where do you come across condensation? So, condensation we have been talking yesterday, Professor Arun taught us about boiling. Now, let us worry about condensation, where does condensation take place? Condensation is there all over the place in power plants. For example, you are talking about condensers in refrigerators, power plants, steam power plants, water is boiled, if there are condensers we have seen condensers in refrigerators, we have seen condensers in condensers we draw condensers in steam power plant. If you see, if you see the condensers for example, if you see the steam power plant where do I say steam power plant, I have a boiler and from the boiler I get steam and from this steam I have a turbine, this is boiler, this is turbine, steam turbine and from the turbine you get the water coming out. Sorry, it is slightly it is not steam but it is steam water mixture which is coming out. Now, you have to get this through a condenser and from the condenser you have a small pump, this is condenser, this is the condenser and this is the pump and from this pump this is going to the boiler. So, what are we worried about, we are worried about the condenser here, this condenser is what we are looking for, what is that which has to get condensed? You see if I have a mixture, my pump has to handle the mixture of vapor and water, then my pumping work required will become very high if it is too fast. So I want to convert all the left out vapor which is coming out, which is getting out here, condense it thoroughly so that my pump is going to handle only water, only water that is the reason condenser has been put. So, if one has to design the condenser like this, I need to have the idea of the fundamentals of condenser. Of course, I am not going to go into the details of tube condensers that is condensation which is occurring on the tube, we will limit our discussion to only on a flat plate that is I will take a flat plate which is colder than the steam that is I will take a cold flat plate and surrounded by steam and that steam has to condense on my plate, what will happen to that condensation at what rate the condensation takes place and how do I define the Nusselt number that is the heat transfer coefficient and how do I handle this under laminar boundary layer case is what I am going to demonstrate today. Tube in tube condensation is pretty involved that can be taught only in heat exchanger course. So, that cannot be covered in UG heat transfer. So, enthalpy of condensation is very high and we know that the heat transfer coefficients on the condenser sides are very high, so that is why we are worried about condensation. There are two types of condensation, one is film condensation and drop condensation. Typically, whatever you see on a coke bottle which you take out from a refrigerator and keep it outside what type of condensation you see drop wise condensation you do not see a film you see a drop wise condensation, but you this is what is called as as I said there is film wise condensation and the another case is droplets there are droplets all over the place that is you see on a coke bottle which of the two is good whether film condensation is good or drop wise condensation is good this is the two things you see let us see the film what is a film what is happening in this condensation my wall temperature has to be lower than this saturation temperature of the fluid surrounding it. Let me write this so that we do not get lost what we are talking about you are saying that I have a plate you assume this plate temperature as let us say T wall equal to 25 degree Celsius that is I have taken ambient temperature and now this is surrounded by steam let us say now let us say there is a steam surrounded around it and the steam is at atmospheric pressure why I am saying atmospheric pressure because I need to know the saturation temperature what is the saturation temperature of steam at atmospheric pressure 100 degree Celsius. So I have just saturated steam which is sitting at 100 degree Celsius now because my wall is cooler than this steam surrounded by heat this steam starts getting condensed on my plate this condensation can occur in both ways that is either through film condensation or through drop wise condensation or through drop wise condensation. So in a film condensation in a film condensation what is happening a thin film a thin film is being formed a thin film is being formed what is this film made of this film is made of water this film is made of water and this film because it is made of water it acts as a resistance for the heat transfer now to occur between the plate and the steam. So film condensation you can see on the face of it itself film condensation is not so good compared to is not so effective compared to drop wise condensation drops means there are small small small films all over the place rest of the places where film is not there then it is all filled with air. So resistances are less because thermal conductivity of air is lower than the thermal conductivity of water. So the point is drop wise condensation is more effective than film wise condensation that is drop wise condensation is more effective than drop wise condensation is more effective than film wise condensation film condensation that means h of the film condensation should be greater or lower than the drop wise should be higher film sorry film wise should be lower drop wise should be higher this is drop wise and this is film is that right fine. Now, that is precisely what I am talking here in the in these transparencies there is a blanket there is a film the name itself suggests in one case there is a liquid film being formed in the other case drops are formed. So, that is why that is called as liquid film and that is what I am talking about resistance here if the in case of the film wise condensation the heat transfer coefficients are generally around 11700 watts per meter square Kelvin. If you remember in the condensation condenser problem yesterday I took 21800 watts per meter square Kelvin the heat transfer coefficients in two phase flows are enormously high enormously high. So, if you have a film of about 0.05 mm thickness you get an average heat transfer coefficient of 11700 watts per meter square Kelvin. Now, on the other hand if it is a drop wise condensation the heat transfer coefficient can be as high as 30000 watts per meter square Kelvin you see almost more than twice more than twice, but do we design condensers on the on the basis of drop wise or film wise condensation we design our condensers on the basis of film wise condensation why because it is very difficult to maintain drop wise condensation because to maintain generally nature wants to have film wise condensation. If you want to get a drop wise condensation you have to apply some paint you have to apply some you have to maintain the surface texture. So, that the drop wise condensation is maintained, but whatever texture you maintain if you have painted let us say after six months the paint vanishes then eventually the drop wise condensation will turn into film wise condensation. So, it is better to design even if you imagine that or even if your texture is maintained you design the condenser on the basis of film condensation. So, that it is a conservative estimate initially and subsequently film wise condensation will take over. So, that is the some and substance of these two transparencies what is there here what all things are written I would request you to go through this line by line so that you understand what it is so that I do not read this. So, the main problem in drop wise condensation is sustainability there is a word here sustainability we cannot sustain drop wise condensation for long time. Eventually drop wise condensation transits into film wise condensation because of wettability because the surfaces wettability goes away. So, Teflon various things Teflon coating if you do actually you get drop wise condensation, but the coating vanishes. In fact, there are thousands of patents only on coatings which can maintain drop wise condensation as opposed to film wise condensation. So, with this let us move on to something concrete that is laminar film condensation on a vertical wall what is that I am doing. So, I am taking I am taking a vertical wall unlike you see here unlike natural convection in natural convection what was happening my boundary layer was growing from bottom to top notice the difference here. You have the condensation forming from top to bottom it is very natural why because it is to fall is not it. So, initially it is laminar subsequently it has become wavy we are not calling it as transition we are calling it as wavy and after waviness you get turbulent. So, essentially again you have laminar and turbulent. So, now what actually happens this is the wall temperature what am I doing again same conditions my wall temperature is lower than the reservoir temperature that is this reservoir the wall is kept in imagine that it is kept in a big room like this where I am sitting and this full room is filled of steam at atmospheric pressure and the steam is maintained at a saturation temperature of 100 degree Celsius. If that is the case and my wall temperature is at room temperature let us say then condensation starts occurring on the wall. So, if that is the case what will happen actually what will happen is what will happen here I have liquid this is 0 means I am close to the plate that is y x equal to 0 means this is just at the exit of the just at the plate edge of the plate. So, now what is happening at any y you take that is not a problem that is at any location just close to the wall what is there liquid film is there subsequent to that vapour is also there, but there is an interfacial velocity between liquid and vapour you see I have plotted velocity of liquid and vapour vapour will have higher velocity compared to that of liquid while film is flowing it will entrain the vapour along with it and it will carry some vapour along with it and, but vapour is going to have a higher velocity and next here you see the temperature this is the wall temperature and the liquid the temperature in the liquid film is decreasing linearly although it is not thoroughly linear we are going to assume linear and there is vapour temperature difference what am I going to say I am going to assume that there is no interfacial stress that means I am saying that vapour is going to be vapour is going to be still otherwise I need to get the interfacial shear stress that is why my life becomes very difficult if I take the interfacial shear stress under that assumption under that assumption what am I going to do I am going to redraw my figure and you see here what have I shown there is no interfacial shear stress that means at the interface between the liquid film and the vapour I am saying that there is zero shear stress this is an assumption how correct is this assumption my experimental results after computing the heat transfer coefficient will get we will get that. So, now this is zero shear stress there is zero shear stress and also I am not taking the temperature gradient which is there in the vapour region I am taking only in the only in the film you see please note this there is no interfacial shear stress this is the crux of the matter actually in this derivation there is no interfacial shear stress actually vapour wants to move along with the film or even faster than the film but we are assuming that the vapour surrounded by heat is not going to move then only if there is no movement only then there will be no slip condition at the interface then the shear stress there would be zero that is what I am saying I am assuming no slip condition here that is interfacial shear stress is zero fine. So, now I get T wall not no slip sorry that is not no slip there is no interfacial shear stress no slip means u equal to zero it is not no slip sorry it is not no slip it is just that there is no interfacial shear stress that means there is there is no the vapour is still not still vapour is also moving at the same velocity as the liquid film actually vapour is moving faster but both are moving at the same velocity when both move the no at the same velocity then there can be no shear stress. So, that is what I am telling so here this is the I am assuming the linear temperature profile if I do that and assume that my complete film is laminar as usual I am taking that my film is completely laminar I am not taking that my film is wavy or turbulent I am only considering laminar portion if I do that I have to take the again I have to take the help of my momentum equation x momentum equation and energy equation energy equation it is only conduction. So, within the boundary layer it is only conduction and that boundary layer I am assuming within that boundary layer my conduction is 1 D when I said linear it is only varying with x remember my coordinates let us not forget my coordinate this is x direction and this is y direction u and v this is y direction that is moving downwards is considered as positive. So, if I do that if I do that what will happen to my y momentum equation which of the equation is not important which velocity is important u or v v is important vertical velocity is important but the horizontal velocity is not at all important because it is smaller it is not at all important it is less important than that of vertical velocity. So, if I neglect the x momentum equation pressure is going to be a function of only y. So, u is very much less than v so x momentum equation vanishes and I have y momentum equation in the y momentum equation also velocity gradient in the x direction del squared v by del x squared is important or del squared v by del y squared is important x is of the order of delta y is of the order of the flat plate length. So, del squared v by del l del l squared this will become that is v upon l squared v upon delta squared. So, this is more important than the second term so I neglect this. So, then another important term I have kept in the y momentum equation what is that what is that equation I have kept in the y momentum equation what is that I have kept. So, please note notice that I have kept rho l into g rho l into g what is that liquid is falling because of the gravity that body force I have taken rho l into g. Now, tell me what is this d p by d y can you tell me what is this d p by d y that is the pressure impressed on my boundary layer liquid film is the boundary layer impressed on my boundary layer by the atmosphere. Atmosphere is what now is atmosphere moving we said atmosphere is you have taken it as still. So, that means what that is hydrostatic pressure minus rho v into g. So, I have taken rho v into g. So, if I substitute that I get rho l minus rho v into g plus mu l del squared v by del x squared is that. So, this is my momentum equation. So, if I substitute that in if I substitute that here what do I get in the on the left hand side what do I have rho l into u del v by del x plus v del v by del y equal to rho l minus rho v into g plus mu l into del squared v by del x squared. So, what is this this is inertia I told what is inertia I always remember inertia like this it is difficult to get up in the morning and come because we have inertia because we have inertia I have to take special effort to come my cycle is standing still I have to make it moving means that is cycle is having inertia I have to give some momentum. So, that it loses its inertia. So, that is inertia and what is this rho l minus rho v sinking effect sinking effect more the liquid density more the body force. So, this is the sinking effect and mu l into del squared v by del x squared whatever I am teaching now I am teaching this material from convective heat transfer by Bijan by Bijan. So, assuming inertia is negligible but not that it is not there anywhere else it is there very nicely even in jungle even in incorpora and David and by the way whatever we are studying what is this called I should not forget that this is what is called as Nusselt condensation you see again professor Nusselt has made his contribution again in condensation this is what we are studying today is called as Nusselt it was first derived by professor Nusselt whatever we are deriving now it was first derived by professor Nusselt. So, this is my if I what do I do I assume that inertia is negligible why do I do this because getting the solution of this equation is very difficult I will simply go ahead and make the inertia negligible. But it is not a valid assumption I have no reason to tell why inertia can be neglected by scale analysis I cannot say that go ahead and through the inertia because it is difficult to solve I will go ahead and assume that inertia is negligible that is right or wrong only at the end my heat transfer coefficient comparison with the experiments will tell me otherwise I have no reasoning why I should neglect this. So, if I do that now it is a balance between sinking and friction now what are the boundary conditions x equal to 0 v equal to 0 x equal to delta del v by del x equal to 0 if I integrate this equation I get mu l del v by del x plus g rho l minus rho v into x equal to c 1 and again integrate mu l v plus g into rho l minus rho v x squared by 2 equal to c 1 x plus c 2 I know I am going fast because it is simple mathematics. So, if you substitute x equal to 0 v equal to 0 you get c 2 equal to 0 and if you substitute this boundary condition you get c 1 that is g into rho l minus rho v into delta substituting this c 1 and c 2 I get velocity have I got velocity here watch carefully this equation what is this comprising of it is comprising g rho l minus rho v mu l delta squared x by delta x by delta squared do I know delta do I know film thickness no until I find the film thickness I have not got the velocity now we will go ahead and compute I do not know the film thickness that means I have not reached the answer yet taking this velocity let me start calculating the mass flow rate per unit width per unit width means the width of the plate perpendicular to my paper I am taking it as 1 later on I will multiply that with v which is the width but for now that is instead of taking m dot I am taking gamma because gamma represents the mass flow rate per unit width fine gamma of y equal to what is this on the right hand side which you see rho a v rho into rho l into v instead of area I have taken x because it is only varying with x and I have taken the width as unity 1. So, 0 to delta rho l v dx is the local mass flow rate now if I substitute the relation for v which is whole of this equation and if I integrate this equation I am going to get I am not going to do that integration because all of the steps are very clearly given here and substitute 0 to delta limits I get gamma of y that is the mass flow rate per unit width equal to rho l into g into rho l minus rho v upon mu l delta cube by 3 I have still not solved the problem because delta is unknown now we have completed momentum equation now I will take what is the next equation energy equation but that is this heat transfer I have multiplied with width b. So, this is the mass flow rate so now remember if the sinking effect is more that is if g into rho l minus rho v is more means my mass flow rate is high that is what I understand from this equation and if the viscosity of the liquid is high means my mass flow rate condensation flow rate is less. So, this is coming out from this equation still we do not know what is delta now heat transfer energy equation is simply d square t by dx square equal to 0 because I am assuming that it is simple conduction that means I am neglecting the convective terms that is very important if I do that if I integrate that I get t equal to c 1 x plus c 2 what are the boundary conditions at x equal to delta what is the temperature at x equal to delta what is the temperature saturation temperature because steam is there and x equal to 0 wall temperature I am maintaining constant wall temperature c 2 equal to t wall if I substitute this boundary condition at any location x I get t equal to it is not x it is there is a mistake here it is y it is not x is it x yeah it is x because it is boundary layer is in the x direction not y sorry it is x t equal to t sat minus t wall into x by delta plus t wall if I do that if I do that and now what is that I need to substitute see I do not think I need to do if I take that convection equal to conduction in the boundary layer what is convection defined in two phase in this case not two phase flow this is single phase only h into t sat minus t wall equal to k l into d t by dx at the wall and do I know d t by dx at the wall do I know d t by dx at the wall that is I take it as t sat minus t wall upon delta that is the temperature difference within my film and that if I equate it to that is k h equal to minus k l k l d t by dx at the wall upon t sat minus t wall this is the definition of h if I substitute that I get h equal to what is that h equal to k l by delta is that okay did I go fast from here it is fine is that okay if you are not got tell me I will equate it fine so fine so now still I have not got the heat transfer coefficient I have represented remember I have represented mass flow rate in terms of delta and heat transfer coefficient in terms of delta now how do I get this delta I have to invoke sensible heat equal to latent heat of condensation I have not invoked that condition okay that is I am just doing this let me finish off that and invoke see this is my mass flow rate now I would need this d m dot by d y little later that is why I am writing this and keeping if I differentiate this what do I get b rho l g all of this are constant only delta is a function of y my film thickness is a function of y if I differentiate this I get 3 delta squared by 3 d delta by y this is the rate of condensation with respect to y now now what do I do what is the condition I am invoking rate of heat transfer from the vapor to the plate through the liquid film is equal to heat released as the vapor is condensed is that right vapor condensation rate has to be equal to the rate of the heat transfer gained by my plate through conduction or convection isn't it so d m dot h of g is equal to k l into area that is b into d y into t sat minus t wall upon delta is that okay so if I do that d m dot by d y if this d y if I pull out I get d m dot by d y I just now derived what is d m dot by d y in terms of d delta by d y if I substitute that what am I doing what am I trying to do in this problem let us stop one thing what am I trying to do I want to find the heat transfer coefficient and the rate of condensation and the boundary layer thickness okay so I get if I substitute for that d m dot by d y here whatever equation I have got into all of this so if I equate to this if I equate this and now what is that I should do I now get delta q equal to d delta into all of this what can I do next what is that I have to integrate this if I integrate this I get delta to the power of 4 and all of this as a function of y what is delta at y equal to 0 it is 0 so I get c equal to 0 so I get delta I get a very nice equation for delta do I know everything on the right hand side of this delta mu l I know k l thermal conductivity I know t sat I know t wall I know rho l I know g I know rho v I know h f g I know now film thickness I have found very nicely in terms of thermo physical properties and temperatures okay if I get delta I have written already equations for mass condensation rate and h in terms of delta if I substitute this delta there I am going to get the condensation rate and the heat transfer coefficient I said h equal to k l by delta if I substitute that if I substitute that if I substitute that delta here and I am integrating this for complete length why because I want to get the average heat transfer coefficient integrating 0 to l y to the power of minus 1 by 4 upon 1 by l because all of this is constant they are not varying with y so if I do that I get average heat transfer coefficient equal to 0.943 into all of this okay similarly condensation rate equal to all of this into delta cube by 3 I get mass flow rate equal to all of this okay if I again rearrange this I get this now the only question is that at what temperature I should be computing all these properties it is but natural let I have to take it as film temperature film average film temperature is T wall plus T sat by 2 that is precisely what written here T film equal to T sat plus T wall by 2 now I took in all of this analysis let me just take a recap of what I have done what did we do we assume that my boundary layer is my film is laminar I should not be using the word boundary layer although boundary layer concepts are what we are using my film is laminar that is number 1 number 2 this steam which is around it I am assumed it as saturated number 2 and number 3 what is the assumption I have made there is no interfacial shear stress between the vapor and the liquid film which is there under these assumptions I invoked the momentum equation and in the momentum equation x momentum equation we threw because u is very much lower than x and I invoked only the y momentum equation and I invoked y momentum equation through this apart and substituted for dp by dy rho v into g and I get 3 forces which are very important inertia force sinking effect and friction and applied the boundary condition and I got the velocity profile in terms of delta next I assume that only of course this is mass flow rate that is the mass condensation rate and I assume that it is conduction which is happening within my film not convection terms are not so important I have assumed and then I got the heat transfer coefficient in terms of film thickness and invoking that latent heat that is the condensation which is occurring because the latent heat is equal to heat conducted or the heat converted within my boundary layer that is heat transfer from vapor to the plate through liquid film by convection is equal to heat released as the vapor is condensed that is m dot h h h f g d m dot h f g equal to this conduction within the boundary layer if I put that I get the boundary layer film thickness that film thickness if I substitute in h I get the heat transfer coefficient from this heat transfer coefficient I can even plot nusselt number that is not an issue and from that I get condensation there are two unknowns in this problem usually film velocity is known I mean in a flow in our boundary layer problems velocity is known free stream velocity here nothing is known like that because condensation rate is not known that is unknown and heat transfer coefficient is unknown both we have found if there is sub cooling that is if my steam is or if my wall temperature what is this sub cooling what is this sub cooling my wall temperature is much lower than the saturation temperature to take into account of this sub cooling I have people I have given just a correction factor that is 0.68 Cpl into T sat minus T wall. So, they have given which we are not considered earlier that is sensible heat we are not considered is not it there is latent heat you see in our assumption we just took the latent heat, but there is actually there is sensible heat also between the wall and the fluid. So, that sensible heat is considered in terms of 0.68 Cpl T sat minus T wall if you take that how who gave this 0.68 0.68 came through analytical analytical calculations and experiments. So, we will have to present that as corrected H F G prime instead of taking H F G you better take it as H F G prime just H F G instead of just H F G you take sub cooled portion also, but that sub cooled portion you multiply this 0.68 and you take that as corrected latent heat then it works well in the same relation which we derived in the above case that is same relation is valid same relation is valid only instead of H F G you will have to take H F G prime that is what Nassel said and then we said there is a Jacob number that is Cpl into sensible upon latent is defined as what is called as Jacob number H F G prime equal to H F G into 1 plus 0.68 Jacob number you are going to get that is that is just an introduction how to tell what is what is the amount Jacob number represents sensible heat upon latent heat nothing great or nothing physically coming out from that. Now, if someone wants it we can even define what is called as Reynolds number because we are all very much concerned about we represents all flow in terms of Reynolds number. So, to take into account of that fact we define Reynolds number equal to rho u m d h upon mu L what is d h that is hydraulic diameter let us see if I take a plate perimeter is L area is L into delta because neglecting neglecting this small portion I have taken only L. So, hydraulic diameter equal to 4 a by p you get 4 delta the point is in all cases you are going to get this as you are going to get this as 4 delta only whether you take a cylinder horizontal cylinder vertical cylinder horizontal you take a vertical plate you are going to get this as 4 delta. So, I am not going to tell you exactly why p equal to L I have taken and a I have taken L into delta you figure it out yourself and you will understand what I am talking about. So, you get if I substitute that as hydraulic diameter and from mass flow rate if I substitute u m for u m I am going to get Reynolds number as sorry Reynolds number as 4 g delta cube by 3 mu L square but still delta if you know delta if you substitute that delta you will get the r e because we have now got delta already. So, now h average can also be represented in terms of Reynolds number 1.47 k L r e to the power of minus 1 by 3 g upon mu L square to the power of 1 by 3. So, now that is about laminar now we have to worry about wavy and also turbulent. For wavy people have done experiments and they have found that this is how Kututlath J Kututlath J way back in 1963 has it is sort of semi empirical he has done this and find that heat transfer in wavy channel is going to be something like this that is r e k L upon 1.08 r e to the power of 1.22 minus 5.2 into g by mu L this is kinematic viscosity mu L square to the power of 1 by 3. Now when does flow become wavy that is the question I have not answered yet when the Reynolds number is greater than 30 what is my Reynolds number definition now Reynolds number equal to 4 rho L squared g delta cube by 3 mu L squared and r e equal to if you substitute the delta for vertical plate in terms of the heat transfer coefficient you can get. Let me not put that way you know delta already in terms of thermo physical properties and wall temperature difference if you know that you can compute the Reynolds number can I say that can I say that yes. So, that is what I have done here. So, for all Reynolds numbers less than 30 my flow is laminar my flow is laminar greater than 30 it is wavy. Now this is the equation which is valid both for this is this is the equation which is suggested by Chen it is not only I will yeah this is suggested by Chen other equation this was Kututlath says and this is Chen's correlation I have forgotten to write that this is Chen this is Chen's correlation now for turbulent flow turbulent flow Laboud snow has given the correlation this correlations I have taken from Bejan as I said Reynolds number greater than 1800 between 30 and 1800 we have wavy wavy flow wavy film. So, above 1800 I have turbulent and again you can see that it is a function of Reynolds number and g by mu l square. So, that is what you get and if it is inclined all that you need to do is inclined equal to h vertical into cos theta. What is this equation? This equation is valid both for wavy and turbulent I have forgotten to write here again this is this equation is valid for both let me write this. So, that we do not go too fast and let us solve a problem that is I am going to write only this equation h l into h l bar by k l h l bar by k l that is average heat transfer coefficient upon thermal conductivity of liquid into mu l squared by g mu l squared by g to the power of 1 by 3 equal to Reynolds number to the power of minus 0.44 r e l to the power of minus 0.44 plus 5.82 into 10 to the power of minus 6 5.22 into 10 to the power of minus 6 r e l to the power of 0.88 p r to the power of 1 by 3 whole to the power of whole to the power of half whole to the power of half. So, that is the correlation for both wavy and turbulent that means it is going to cover what all domains Reynolds number equal to 80 to whatever number greater than 80 it is Reynolds number greater than 30 that is not 80 anything greater than 30. Let me rewrite this Reynolds number greater than 30 this equation is valid let us see there is a one interesting graph here let us spend time on that graph and then we will move on to a problem which is an interesting problem again this is you see here this is heat transfer coefficient into mu l squared by g to the power of 1 by 3 upon k l actually very interesting here why I say interesting because mu l squared by g all of you sit down and get me the unit of mu l squared by g to the power of 1 by 3 what is mu l squared by g to the power of 1 by 3 Amir says that he gets mu l squared by g to the power of 1 by 3 he is getting meter let us see do we get meter or not mu l squared by g to the power of 1 by 3 what is mu what is the unit of mu meter squared per second that is meter squared per second whole squared upon what is the unit of g meters per second squared to the power of 1 by 3. So, what do I get meter to the power of 4 upon meter into what is this I get second squared upon second squared to the power of 1 by 3 second squared second squared get cancelled out I get meter cube to the power of 1 by 3 that means meter to the power of 3 by 3 that is meter. So, that means it is so ingenious you see what is the unit I have taken here h l by k that is Nusselt number that is Nusselt number mu l squared by g to the power of 1 by 3 that is how the Nusselt number has been chosen and this is Reynolds number. So, you can see that until laminar case wavy for Reynolds number 30 this dashed line is the heat transfer coefficient computed using the laminar derivation whatever we did. So, that is what it is and from here to here it is wavy laminar 30 to 1800 subsequent to that it is different for different differential numbers. So, that is what we found and quickly within 5 minutes we shall solve a problem. So, we have saturated steam at atmospheric pressure condenses on a 2 meter high and 3 meter wide vertical plate it is and that is maintained at 80 degree Celsius by circulating cooling water through the other side determine the rate of heat transfer by condensation to the plate and the rate at which the condensate drips of the plate at the bottom very interesting problem. So, if you take T sat and T wall 180 divided by 2 I get film temperature as 90 at this film temperature if I take all the properties of the liquid and calculate h of g prime I get 2 3 1 4 into 10 to the power of 3 joule per kg. Now, I substitute now question is do I know whether it is laminar or turbulent I do not know what is the way out first I will calculate heat transfer coefficient why because my Reynolds number is depend I can calculate Reynolds number only after I calculate the condensation rate I can calculate condensation rate only after I get the heat transfer coefficient is that right let me do let me go slow here it is quite we have to think a little bit what we are doing I will go ahead and assume that my flow is laminar that is my film is laminar if I do that and this relation whatever I have derived this is for laminar case if I substitute all thermo physical properties and the temperatures I am going to get h l bar as 2656.2 watt per meter square kilo now I need to calculate the condensation rate I do not have to remember any relation which I derived very easily q dot equal to m dot h s f h of g. So, q dot equal to h l a s into T sat minus T wall that gives me so much load this is my load heat transfer rate heat transfer rate equal to m dot into h f g please do not take it as h s f h f g if it is h f g I get m dot equal to 0.1329 kg per second. So, Reynolds number equal to 4 into m dot by b mu l this is what we have defined 4 gamma upon mu l. So, now I get Reynolds number as 562.5 what is this suggesting I am in wavy for wavy and turbulent I have already got a relation which I wrote just now on my white board this is the relation we wrote. So, that relation if I put I am going to get a heat transfer coefficient of 7691.4 and from that I get the heat transfer rate and from that I get the condensation rate and from that I get the Reynolds number which is 4221 now I know for sure that my condensation is actually decided that it is turbulent it is neither wavy nor laminar. So, the Reynolds this Reynolds number confirms that condensation is wavy laminar domain. So, this is about condensation and full credit I mean whatever we have learnt it is all because of the hard work of professor Nusselt this is what is called as Nusselt condensation you see Nusselt is there everywhere throughout convection professor Nusselt is there without his efforts there could not have been convective heat transfer that is why appropriately we have given Nusselt number because we do not want to forget him anywhere. So, with this we are saying that we are stopping condensation we are not saying that we have completed condensation we are stopping condensation simply because we have not covered tube in tube condensation I am sure there will be a question what will happen in case of tube in tube condensation. Tube in tube condensation again Nusselt number will be dependent on Dittus-Bolter correlation, but the other parameter which comes into effect is the thermodynamic quality a nu by nu naught equal to a function of quality that is what is called as Shah's correlation. So, let us not worry about all of that I think I want to spend just one more minute five more minutes on the natural convection experiment because there were so many questions whether I should be taking thickness or not Sameer has gone ahead and did that calculation. So, we get some interesting results what is interesting let us see. So, we said that let me just take the natural convection case see in the natural convection what did we see we melted the ice got melted and we latent heat was taken equal to convection and what was the condensation rate I am supposed to take somewhere around three because this is where somewhere around three because this is where it is reaching steady state. Now if I take and question was always the question was I have an ice block like this I have an ice block like this and boundary layer is being formed on both what about the boundary layer formed on the side walls why are you not taking into account you are very right Sameer has gone ahead and done that I will quickly do that it would not take much time not more than two minutes m dot equal to I have already posted this in model but still three into ten to the power of minus five kg per second q dot equal to m dot h s f this time that is 331.4 into three into ten to the power of minus five I get ten watts this is already there radiation loss I had taken only the area that is equal to sigma a t infinity to the power of four minus t s to the power of four t infinity that is sigma is 5.67 into ten to the power of minus eight into area I had taken earlier two into height 0.16 into width 0.105 now I have to take thickness also into account that is plus two into height 0.16 into thickness is 0.01 into temperature that is 298 to the power of four minus 273 to the power of four I get 4.865 watts earlier my radiation loss was what was my radiation loss yeah radiation loss was earlier 4.44 now my radiation loss has gone up to 4.865 watts now let me compute convective heat transfer coefficient that is now q equal to q actual equal to ten minus 4.865 watts I get 5.135 so q dot actual equal to h into area into t wall minus sorry t infinity minus t wall what is q dot actual 5.135 h I do not know area is again same two into 0.16 into 0.105 remember we had just taken this area I had not taken the side area that is thickness plus two into 0.16 into 0.01 this is the ten mm thickness 0.01 into 298 minus 273 I get h equal to 5.584 watts per meter squared Kelvin what is an assault number I get 35.7 okay let us remember that I am getting a nusselt number of 35.7 now nusselt experimental 35.7 nusselt correlation one which was stated earlier is 33 now if I take n u equal to 0.59 r a h to the power of minus 1 by 4 if I substitute that I am going to get it as what is that I get 34.8 now what has happened you see one of the participant thanks to him he has corrected my numbers that is experimental I am getting 35.7 one correlation is giving me 33 another correlation is giving me 34.8 beautiful is not it it is a simple experiment just taking rupees 200 we are able to demonstrate natural convection heat transfer coefficient I want to take an example of professor C V Raman's C V Raman's noble prize experiment took not more than 100 rupees the point is if you want to demonstrate something you do not need money all that you need is the thought process okay if you think properly and if you design your experiment properly you can demonstrate very nicely the fundamental principles that is the point I want to make but still we should not be very excited that my match my reading is very much matching we should keep in the back of our minds this 33 is having at least plus or minus 20 10 percent that means this is 33 plus or minus 3.3 and this is 34.8 plus or minus 3.5 and this is 35.7 plus or minus 3.6 this is there in the back of our mind this only we are excited because this if I add 33 plus 3.3 I get 36.3 so my 35.7 is within that 36.3 again 34.8 plus 3.5 is again within the band of 35.6 all that I am saying is my number is within that band within that uncertainty band that is all I am saying so that is the so thanks to the participant this is the this is the greatness of discussion this is because we discussed and we calibrated ourselves that is what we learn through observation had we not done the observation this much discussion would not have occurred so that is the greatness of experiments if you do experiments we get observation if you observe something you tend to question that and try to understand that okay that is the essence of experiments with this I think we are through with the teaching session we will open okay the professor Arun is suggesting that two three centers had missed the experimental portion two experiments which had told he says that before we break for not break we transit into question answer session let us quickly go through those two experiments yes that is what I had we had promised also yes we will take 5 to 7 minutes if you further have any questions in spite of my presentation you please you please come back on the moodle with questions so now let me tell let me open that okay this is this is little involved first let me take easier one and then go to the little tougher one okay so now unsteady heat conduction in a sphere see why I have designed this experiment and why we are even proposing in our lab in IIT this because all experiments what I what we interfaced when we with our coordinators we learned that most of the experiments what we are doing are all steady state experiments they are not transient to drive home the transient conduction problem we are we have designed this problem again I have taken this book this experiment from the book which I suggested yesterday Professor Granger's book okay this is not my experiment this is Professor Granger's experiment not Professor Granger I do not recollect which professor has suggested one of the professors has suggested and has been written in Granger's book the same thing I have put if you want details you can go back to that book and read this what I am doing is I am taking a small sphere made of any insulating material or it can be copper also when to go for copper and when to go for insulating material I will tell little while from now you put a thermocouple you take a sphere you take a sphere what is that I need is just that you make a sphere or a cylinder either of the two I am pertaining my explanation with reference to sphere with perspex or Teflon let us say and I put a thermocouple and take it out now I take this sphere and put it in a hot water bath now my sphere will start picking up temperature now I have measured what is the temperature I have measured temperature versus time okay now we have done this what is this this is under I have put it in a hot water bath analysis we have done for this remember this is not lumped this is not lumped I am taking that this is one-dimensional conduction so why have I put at the center because I am tracking the center line temperature T0 if you remember the definition to T0- theta 0 that is T0- T infinity equal to let me write that T0- T infinity upon TI- T infinity equal to e to the power of you remember the relation e to the power of minus lambda 1 squared tau my minus lambda 1 squared tau what is tau e to the power of minus lambda 1 squared tau is Fourier number that is alpha t lambda 1 squared alpha t A1 will be there for minus lambda 1 squared alpha t by r0 squared r0 squared is that right so now if I take T0 is the temperature as I am marching with time TI is the initial temperature this is the fluid temperature A1 is here yes is right A1 is okay so now if I plot what do I do if I plot this what am I doing let me just take stock it is yeah A1 is there if I plot this if I take log of this log of T0- T infinity upon TI- T infinity equal to what will I get log of A1 minus of lambda 1 squared alpha t by r0 squared what is this of the form this is y this is c and this is my m y equal to minus this is x minus mx plus c if I plot that it is going to be something like this okay so what is this what is this constant I get what is this that is nothing but my c when x equal to 0 that is when t equal to 0 whatever y I get this is not directly right and this is log of this is what I measure I have to plot one more plot where in which I plot log of this let me call this as theta log of theta versus e then whatever plot I get it is going to be linear and this is not right this is this is nothing but my log of A1 which is constant from this what do I get A1 I get if I go to my table if I go to my table I have to open transient conduction no chance there is no other option for me if I go to transient conduction let me go to the table yes is this no is this yes so now thanks to zooming concept now now let me focus only on cylinder okay so now what do I get sphere sphere or we can do for cylinder also for a sphere what do I get this is my constant A1 I have got having got A1 if I get A1 I can get my for a given A1 there is one unique lambda 1 and biot number so if I put that back here I get lambda 1 and I get biot number if I get lambda 1 what is that slope slope is equal to lambda 1 squared alpha by R0 squared I know my M because from this curve I get the slope and I get lambda 1 from the table I know R0 squared I can calculate thermal diffusivity of my material beautiful isn't it so that is what it is the demonstration now I said I will come back which material you are going to choose if you say that oh I need data logger to get the temperature measurement transient I need no not really I do not agree with you why because if you take perspex it is going to respond very slow you can take with the multimeter also there is no problem if you take perspex or Teflon or any insulating medium you make a sphere with that insulating medium there is no problem if you have data logger your rich percent data cards are available even for 25 and 30,000 rupees Indian make the analog data cards so if you purchase them it is not going if then you can do with copper also then you can do with copper also then there is an interesting thing if you take copper if you take copper you can make it lumped also if you make lumped then you can get direct then it takes a different note okay so you take it cannot become lumped if you take a bigger copper cylinder okay so point is there are multi facets for a given experiment it is not just one way of doing things if you start making building an experiment and start doing experiment definitely you are going to get infinite ideas not only from yourself from your students also when they ask questions why not do this why not do that definitely it will just open several windows several windows for thought process okay so that is one of the experiments and then the next experiment is this I am going to zip and really fast okay so actually what it is is this gives thermal conductivity and thermal diffusivity in one shot but this is little involved it is not that is straightforward but still we can build it it is not at all when I say involved it is not impossible it is little difficult that is all what we are saying is that I have an insulating medium let us say which is filled here it can be liquid also actually I have a wire heater wire which is put here and there is a thermocouple embedded at different location if I substitute if I apply heat if I apply heat that is this Q if I track this temperature of the thermocouple at it presented at a different location I get that temperature distribution as t minus t0 equal to Q1 upon 4 pi k log t plus log of 4 alpha by r squared minus 0.5772 so if I plot this again and do the this is the equation I get I have not derived I said this derivation is in Karla and Giger this is a line source what is applied okay if I have a line source and this is the temperature distribution I can again apply y equal to mx plus c funda and get both k and alpha with this input I am going I think I know I have not done justice to this experiment but still I have given the glimpse of it how do we attack it okay so with this we will take up questions so professor Arun will keep starting taking questions I will just come back in a minute.