 Good morning, today we will be talking about condensation heat transfer and in today's lecture we will just briefly touch upon what condensation heat transfer is followed by a relatively detailed discussion about film condensation wherein we will discuss the derivation in laminar film condensation on a vertical flat plate. There are certain assumptions we will be making during this derivation so in the subsequent sections we look at how the formulation changes if we remove some of these assumptions. So we'll be looking at what happens if we include the sub cooling effect, what happens if you take into account the vapor moving and what happens during flow transition. We look at some of the problems and end by talking about drop wise condensation. So condensation heat transfer is pretty much the reverse of boiling wherein when you have saturated vapor coming in contact with a cool surface it's going to transform into liquid and there can be two types you can either have a film forming on the surface or you can have droplets of liquid forming. So the essential difference as you can see is that you will have the surface wetting to be complete in the case of a film condensation and on the left hand side you can see that the surface wetting is not complete but this actually aids in the heat transfer. So essentially when you look at the heat transfer coefficients for a drop wise condensation the heat transfer is significantly much higher than the film wise condensation part and this makes it extremely preferable in your design of heat exchangers but the biggest problem is how do you sustain drop wise condensation. So essentially when you have drops forming continuously at the first go yes droplets form it has a high heat transfer coefficient but essentially what happened is all these drops coalesce and they become a film. So the stillness of drop wise condensation is the challenge and the reason why film wise condensation that is slightly lower in heat transfer coefficient essentially is because it's the resistance across any thickness of the layer any film so like your condensation heat transfer you have a resistance across that thickness similarly in your film condensation you will have a difference in the resistance so that causes a lesser heat transfer coefficient. Considering that drop wise is difficult to sustain any designer is only going to design assuming that film wise condensation occurs. Drop wise essentially because it has a shape of a bubble you have interfaces contacting the surface so that is going to promote heat transfer because of the shape of the interface rather than just a film where there's condensation heat transfer condensation and it's essentially conduction across the film so it's the shape of the interface which aids if you remember your bubble when it's growing at the wall when you have a vapor bubble you can see that there's a contact line movement which aids in the heat transfer because of the thin film that is in contact right so in similar cases when you have a drop what happens is the contact line movement aids the transfer and if you look at it this way you have a surface and there is a drop formed then the drop leaves so if you have a vertical surface a drop forms then the drop slides off and then it's again cool wall that is there right so there's going to be more another drop formed so this is ideal heat transfer but when you have a film what happens is initially the film forms on the surface but beyond this heat transfer is only across this interface so this constant replenishment of the surface access of the surface to the surrounding aids in the heat transfer so if you're going to want to design for a film condensation because that is the lower heat transfer coefficient then what are the things that you need to understand about film condensation so like I said initially the vapor is in contact with the wall but once your film forms then any more of heat transfer is going to be across this interface there's no more direct contact with the wall so your heat transfer is only a function of how much heat you can transport across this delta t and if you look at the velocity and temperature profiles over here these are more like the realistic situations where for example in your velocity profile you have the film forming and then because of the film weight it starts to flow down so you have a velocity at the interface level so your velocity profile changes from zero to a certain extent this will tend to drag some amount of vapor if the vapor is stationary the vapor can also move of its own so in this case we have assumed that the vapor is stationary but some amount of vapor is being dragged so there's going to be a drop in the velocity profile now in the case of temperature it doesn't have to be that the interface has to be in saturated temperature it can be the vapor can be slightly superheated and hence push the saturation temperature a little into the liquid layer so you will have a temperature profile like this but we would like to be able to quantify the heat transfer in other words our goal is to determine the heat transfer coefficient in this case it's quite a complicated problem and Nusselt made certain assumptions for his derivation in order to obtain something that is quantifiable like a closed form solution so he made certain assumptions first of them being that it is laminar it doesn't have to be in real in reality that it is a laminar flow so essentially you would see that in experiments it's been observed that your laminar film will slowly develop some waves or turbulences and then slowly transition into a turbulent regime so you have a laminar regime then there is a wavy laminar regime and then there's a turbulent regime laminar regime Nusselt was able to quantify using certain assumptions but your wavy and turbulent regime is not simple to solve hence there are pretty much only empirical treatments of that regime but it has been found out from calculations again that they facilitate more heat transfer that is a wavy and the turbulent regimes facilitate more heat transfer and that is preferable so in fact in industrial systems you'd be having designs where which will promote wavy or turbulent regimes they'll try and transition quickly to these regimes so that's preferred but today's class will only look at trying to model this laminar regime so we'll go through what Nusselt did to calculate the heat transfer coefficient so there are series of assumptions that he made so we'll start from a laminar case the coordinate system being y and x the wall temperature is t wall you're assuming so the first thing is this is a laminar regime so that is your first assumption that the flow is laminar the second one is that the properties within the medium are constant so the fluid properties do not change with temperature the next one is momentum effects are negligible we'll come back to this a little later then the vapor is stagnant essentially that this is not going to have any velocity meaning there's not going to be any shear force over here at the interface then there is going to be linear temperature variation within the film so essentially if you have a t-sat temperature then it is a linear temperature profile that's going to take care of the heat transfer then enthalpy change with subcooling is negligible we'll revisit this a little while later so essentially we have to try and get certain things from this laminar flow so the first thing that we would determine is we would like to determine is the velocity profile second thing is what is the condensation rate what kind of condensation rate are we getting so we'd like to know that third we'd like to know the thickness of the boundary layer or the thickness of the film and fourth is a heat transfer coefficient so these are the four things we'll work through right okay so first is to try and determine the velocity of the fluid so let's go to the momentum equation so the momentum equation for this case will be so this is going to be a momentum equation the first thing that you know from boundary layer approximations is that the pressure gradient in the vapor will be hydrostatic so that is rho g so rho gas or rho vapor is much better and the pressure gradient inside the liquid film is assumed to follow the same pattern okay so essentially this will get plugged in over here now going back to your assumption that momentum effects are negligible so what we're trying to say is that the flow is so slow that momentum change across a control volume due to the condensation is very less so essentially your momentum changes go to zero so in this case what happens is that you get an equation with uh so this will be rho l minus rho v into g with a negative sign so we can just transfer it like this when integrated we'll give you the velocity what are the boundary conditions so your boundary conditions essentially will be u is zero at y is zero that's a no slip boundary condition and the shear is neglected at the interface so at the interface thickness your dou u by dou y is going to go to zero so when you integrate and apply these boundary conditions you will get the velocity to be so this is the expression for velocity of the fluid inside the film the next thing that we need to find out is the condensation rate right so the condensation rate can be got by integrating so this is into dy so essentially we are taking a control volume of length dy and thickness dx so we are trying to say that rho u l dy we're integrating it from zero to the thickness of the interface we know the velocity profile we plug it in here and when we integrate we get rho l times rho l minus rho v by mu l now this is nothing but mass flow rate m dot per thickness of the plate so in this case what i mean is that if you have a plate like this and your film is flowing like this then this if taken to be b then unit width so your units will correspond to this where b is the width of the plate so that gives you your condensation rate now we have to calculate the thickness of the film if you take the control volume wall on one side whatever heat goes to the wall in the form of conduction heat transfer so you can say that this is minus k t sat minus t wall by the thickness here right so that into dx this will be balanced by whatever heat is brought in by the condensing film in across the three faces so what are those going to be here it is so in this case you're going to have m dot into hf and in this case it's going to be some dm dot into hf plus hfb that's the amount of heat it's going to bring in because it's going to have hf plus that latent heat transfer change and here it's going to be m dot plus dm dot into hf okay now let's go to the last assumption that we've made enthalpy change with subcooling is negligible now if you remember this is at t wall and this is at t sat so essentially there is some liquid here which is lesser than the saturation temperature and your enthalpy essentially should be lesser than your saturation enthalpy so hf is your saturation enthalpy but we assume that that hf is not much different from your saturation enthalpy so it can be applied across the interface so essentially we have kept hf to be the same across so when we add this equation we will get this so this is for a general case now if i'm going to do per unit width because that's what i'm going to deal with most of the time so if this is b i'm going to do per unit width i'm going to divide all of this by b so in that case minus m dot by b into hf minus dm dot by b into hf i'll arrive at an expression which goes like k tau t sat minus t wall into dx is equal to dm dot by b into now this requires an m dot by b which we had just determined so i'm going to take this m dot by b expression and i'm going to differentiate it so i'm differentiating it plug it back in there and then i can integrate it which just gives me this expression so i'm just substituting this back in here and i'm integrating so i know that for the getting the value of c the thickness of the boundary layer is zero when my plate is just beginning at this point over here so that is the condition i'm going to apply over there and this will give me for the thickness of the boundary layer so this gives you the thickness of the boundary layer now the final portion is pretty simple so you need to find out the heat flux right so essentially that gives me h is equal to heat transfer coefficient by t sat minus t wall and we know that this heat transfer is through conduction so you have k by tau so essentially we know the thickness so we just plug this in here so your heat transfer coefficient expression would be something of this sort so you've got an expression for the velocity you've got a mass flow rate and you've got a boundary layer thickness if you look at the boundary layer thickness there is an x term involved there that essentially is the length along the plate so your thickness varies right so at what distance along the plate do you want the boundary layer thickness that can be obtained and that x denotes that coordinate distance your Nusselt number essentially which is h d by k has two forms the local Nusselt number which is your hx any local distance along the plate and then there is an average Nusselt number which essentially will get from your average so if you plug in the length l it'll give you the average so if you instead of this x over here if you have l here it'll give you your average heat transfer coefficient and an average Nusselt number some amount of value is plugged in because you take into account you'll see in subsequent cases there are different cases that they take into account and there are some coefficients that are added to it people have tried to non-dimensionalize it which gives you in terms of the Reynolds number and the length scale the Reynolds number is the Reynolds number of the film for m dot by mu lb, b being the width of the plate and these are the kind of relations that you get for the non-dimensionalized version if you remember we said that let's ignore the sub-cooling of the film what did we do we just assume that across that control volume the enthalpy is equivalent to the enthalpy of the fluid so all across there was this enthalpy of the fluid was assumed to be hf which is at saturation temperature so how do we account for the sub-cooling effect so essentially you're going to have some amount so if you take this to be dx then the amount of heat added per unit length can be for your phase change which is hfg times dm dot by b times dx plus some amount of heat is gone to sensibly heated to the saturation level so you're going to have d by dx of so you're going to have rho l c p l u l delta t so delta t will be t sat to any temperature from t any temperature t to t sat and along the thickness so essentially you're going to have this is nothing but the symbol there so hfg dm dot by b times dx plus d by dx of this so essentially per unit length this is the kind of change in your heat transfer right so now you know your velocity let me just rewrite my velocity there you've just derived your velocity profile to be rho l minus rho v times g mu l so this can be plugged in over here into equation one this is equation two and now we need to find out this value over here so it's some temperature relative to t sat so we know that it's a linear temperature profile variation so essentially t sat minus t by t sat by t w so you know that your boundary layer thickness it's going to be delta so your y you can get your temperature at any position so this is your third relation and then when you plug it in i also want to be able to uh so the idea behind this is that i want to express this in terms of some heat transfer coefficient by dx i want to bring it to this form form of the first term over here so essentially i need m dot by b expression so what i'm going to do is i know my mass transfer rate expression that i derived previously now if you look at this in this term you have rho l c p l u l will give me rho l minus rho v by mu l so there's some form of this in here so i'm going to express all of this in terms of that so essentially step number one plug this in step number two use this expression over here and try to express your mass transfer m dot by b in this find that expression over there so if you do that you essentially will come to this form dq by dx is equal to h f g by dou sigma by 2x so this is your first term the second term will take the form so what we do here is we call a h f g prime which is nothing but one plus so i'm going to take all this h f g common outside and i'll have one plus c p l so that i call as h f g prime so essentially i'm defining the h f g prime which is a substituent heat transfer coefficient when you try and include your subcooling effect so that is what you've got over here heat transfer coefficient for subcooling effect people have taken into account convection as well and that's the correlation that you see at the bottom 1 plus 0.68 so instead of 3 by 8 that's 0.68 if you take into account convection so with this we can go and look at a problem saturated steam at one atmosphere condenses on the outer surface of a 100 mm pipe which is one meter long having a uniform surface temperature of 94 degrees Celsius estimate the total condensation rate and the heat transfer rate to the pipe so you're asked to estimate the total condensation rate and the heat transfer rate to the pipe so essentially it's the first two your heat transfer rate is total total heat transfer rate is calculated using the total heat transfer coefficient and then your condensation rate is given by q by h l v right so you need your totally heat transfer coefficient remember the total nusselt number that you had calculated average nusselt number so you will have to use that first and then you will have to also use something which will give you the h l v that we just looked at previously so so your average nusselt number is given by 0.943 rho l minus rho l minus rho v g h l v l cube by u l k l delta t so we know that it's a tube with surface temperature 94 degrees it's atmospheric so your saturation is 100 degrees and pretty much your properties are known so if you and the length of the tube is one meter long now to be able to calculate your average nusselt number you need your heat transfer coefficient for this case we are assuming that it's convection subcooling effect everything we are taking into account so we know we have just calculated h l v is equal to 1 plus 0.68 cpl t sat minus t wall by chelvi so i know all my inputs over here i calculate h l v dash or prime and that comes up to 2274 kilojoule per kg and this when you plug it back in there will give you the nusselt number from which we can calculate the average heat transfer coefficient because nusselt number you know is so your average heat transfer coefficient is 7360 watt per meter square kelvin okay so we also need to find out m dot so m dot is q by h dash l v it's pretty much substitution you know a you know delta t you know h dash l v you know q so you can plug in h l v and get m dot so now we've looked at the proper list of assumptions wherein you assume there's no subcooling effect and the vapor does not move now you pulled out the subcooling effect assumption the second more critical assumption is vapor is stationary so what happens if how do you account for moving vapor in this equation so going back to your momentum equation now in this case your vapor is moving right so in a sense there's going to be an additional gradient equivalent imposed on your flow so essentially it's dpl by dx is going to be a summation of rho v g plus dpv by dx due to the motion it can be equated to some kind of gradient that is imposed so what you're going to do is we're going to impose artificial density which is essentially rho v star so this will go in over here this still goes to zero so essentially you will have that's the expression rho v star and you will have an equation which goes like mu rho square u by rho y square is equal to um so you'll have rho yeah rho v star minus rho l g into mu so if i integrate it again twice the same procedure you will get that as a velocity expression now that how does that come about so you have two boundary conditions one is u is zero at y is zero which is at the plate but your dou u by dou y is no longer zero some constant shear force so at y is equal to 10 and this gives you the second term essentially similar procedure for your mass flow rate so you can work it out and you can see if you can get that expression now when you come to these expressions get complicated because there's a second term involved and that's your boundary layer thickness all of this cannot be solved they have to be solved numerically essentially in an iterative iterative process people have tried to make that simpler slightly but by non-dimensionalizing it by those terms that you see x star delta star tau star and that gives rise to that expression over there now this all this has to be solved numerically so Butterworth essentially suggested a simpler option wherein he said let my heat transfer coefficient be a sum of two cases one where no vapor is moving so my heat transfer coefficient is only gravity driven and the first one where the motion of the vapor is much stronger that gravity is almost negligible so if i decouple the two effects and he proposes so for the second one the gravity case we have already derived that that is your nozzle case and the first one with experiments he correlates and comes up with the heat transfer coefficient so this one essentially takes into account vapor motion subcooling convection everything i'll have to actually go check but i would assume it's the normal roovi because your he goes through all the process of roovi star etc and then comes up with roovi yes yes right yes okay so you mean to say that essentially it's a constant no this we are considering it to be so with so if you have variation it will be a function of x but a constant function like dpv by dx is equal to some constant okay so now we talked about laminar motion and i mentioned that there's a wavy laminar and a turbulent case so the transition happens at they've proposed it to be at Reynolds number at 30 and Reynolds number 1800 so for m dot mu l by b if it's greater than 30 then you'll have to model for wavy laminar and if it's greater than 1800 is turbulent so there are no derivations per se but empirical correlations so for the wavy laminar condensate case your heat transfer coefficient is given by this correlation over here and your Reynolds number you can determine using this formula and for your turbulent case the heat transfer coefficient is given over here so with these correlations let's look at one problem which is for a vertical plate now an important thing to look at over here is that you have a tube case and you have a plate case so when can you apply these correlations can you apply it to both cases so essentially if your laminar the the thickness of the film is much lesser than your tube diameter then whatever you're deriving for a plate will be applicable for a tube so in our previous example case it was for condensation on a tube but your film thickness if you notice would be much lesser than your diameter of the tube and hence you can apply what it sees it as a plate essentially that's what we mean if so the radius is very high the condensation thing sees it as a plate and you can apply the correlation so in this case we are looking at a vertical plate 500 mm high and 200 mm wide is to be used to condense saturated steam at one atmosphere first of all find out the surface temperature of the plate in order to achieve a condensation rate of 25 kilogram per hour so essentially in this case you have not been given a surface temperature so let's look at so the plate is 500 mm and a width of 200 mm and it is giving me 25 kilograms per hour so i'm required to find the surface temperature so the first expression we know here is that whatever condensate you're forming m dot h l v can be related to your heat transfer h a t sat minus t wall right so this is your first expression now in this we do not know this we do not know this and we do not know this now this we have an expression for which is that 1.1 plus 0.68 again here we do not know this so we can't find out what this is so we do not know t wall and that means we'll have to make an initial guess and it becomes an iterative process so the first case your saturation temperature is 100 degree celsius so let me say i assume t wall is 74 degree celsius so first plug that into equation 2 which gives you h dash l v to be 2331 kilo joule per kg so that gives you 2331 now in light of what you've been learning about wavy laminar laminar and turbulent we'll have to first check if this is laminar case wavy laminar or turbulent case so going back we'll have to find out the Reynolds number so Reynolds number for a general case this is a correlation obviously but for a general case is 4 m dot by mu l b so that's something we're going to apply here 4 m dot by mu l b and that comes up to 4 29th so if you look at the previous case 4 29 means it lies in the wavy laminar regime so essentially you'll have to use the heat transfer coefficient from the wavy laminar case so that is nothing but so this gives me a value of 7312 watt per meter square Kelvin now this i can plug it back into equation and once i plug it back into equation 1 i can find out t wall which gives me in this case 78 so to what i have to do is basically do an iterative process so what am i doing first plugging in t wall some assumption determining h l dash l v trying to find out the average heat transfer coefficient to be able to plug in these two values getting a t wall essentially both have to match so i iteratively solve till both match and you would find that actually if you iteratively solve it'll match at 0.78.2 degree celsius so that is your surface temperature now i would like you to actually go and think about what will be the heat transfer coefficient if the dimensions of the plate become opposite essentially it becomes a shorter plate but with wider width what will happen to your heat transfer coefficient second thing if the plate were to be inclined at 45 degree celsius and if you remember the first problem we worked out where we use the laminar formulae go back and look at whether it actually lies in the laminar case or it falls into the wavy or the turbulent case to just go to the final part which is drop wise condensation essentially it is only handled numerically because it's a highly complex phenomenon and how people look at it is well we can simplistically look at it as a series of resistances across the bubble so you have your wall resistance conduction then you have across the liquid some amount of resistance for the heat transfer then because a bubble has to form there is a metastable state you need some kind of super heat so that is the capillary resistance there's some heat transfer resistance across the interface and then there is heat transfer resistance at the vapor so people quantify each of these resistances this is for the bubble and then surrounding the bubble there's going to be just vapor right so that's the convection resistance directly so that's a parallel process so people try and quantify this or numerically model it even using dns like i showed you for the boiling so they do use a similar approach calculate drop wise condensation but a universal coefficient heat transfer coefficient formula very simplistic it's been produced proposed by griffith people use it as a first-hand calculation and the heat transfer coefficient is given by this formulation so we've looked at boiling heat transfer we looked at the bubble dynamics we looked at pool boiling pool boiling curves we looked at two-phase flows homogeneous separated flow we looked at the pressure drop then we looked at flow boiling sub-cool boiling and so on and so forth and then finally we looked at condensation where we looked at film condensation and drop wise condensation we derived in detail the laminar film boiling case we looked at velocity profile the mass flow rate the thickness of the boundary layer and the heat transfer coefficient so with that today's lecture ends thank you