So, we will get started good morning everybody. So, now let us get started with the most interesting and the most application oriented topic that is heat exchangers. So, we are on the penultimate day. So, let us get started with the heat exchangers. So, heat exchangers is essentially based on conduction and convection rarely we take into account radiation. So, now let us get started with the basic definition. What is what is meant by heat exchanger? Heat exchanger means many things for many people, but heat exchanger is going to mean in a only a particular way for us in this class that is what is heat exchanger? Heat exchanger I can exchange heat from point is it is exchange of heat from one fluid to the another fluid, but is there any separation between the two? Yes, all that we are saying is there is a heat transfer between one fluid with the another fluid, but there is no mixing. We are saying that there is no mixing there is a wall in between it can be pipe it can be something it can be one container another container, but there is no mixing. Where does that is no mixing? Where does mixing take place? Mixing can be in several situations for example in cooling tower. So, what are we saying? We were saying that in a heat exchanger the heat exchanger which we are considering for the design or understanding the principle is that we are not considering the mixing of one fluid to the another fluid. If there is mixing of one fluid to the another fluid for example in cooling tower the heat from the water is taken away by the air there is mixing there and other application can be direct contact condensation that is you have taken Fukushima you have heard Fukushima power plant in Fukushima power plant when there was in the nuclear reactor containment when there was a problem they were trying to get the sea water and try to pour into it why in the reactor containment the reactor containment of the Fukushima power plant was completely filled with steam and that steam had to be condensed the only way to condense this steam is to put it through spray nozzles. If you have spray nozzle if you have spray nozzles that is water spray nozzles the water is going to form a sheet and get into the steam this sheet this sheet of water which is coming from the spray nozzle is going to on this sheet the steam is going to get condensed steam is sitting at what temperature 100 degree Celsius and water is sitting at what temperature 30 degree Celsius. So the condensation of the steam on the water sheet occurs or the water droplets occur okay so we are not interested in this DCC that is direct contact condensation where in which the mixing of both the fluids is occurring we are interested or we are taking up the heat exchanger definition as when there is no mixing of fluid it is not that this direct contact condensers are not important they are important but we do not consider them in our lecture that is the point okay that DCC direct contact condensation is also a very very interesting problem by itself both fluid dynamics wise and heat transfer wise. Now let us take up the heat exchanger so what is happening in a heat exchanger in a heat exchanger I have a pipe typically this is let us say hot fluid and outer wall I have cold fluid this is I have taken a pipe in a pipe this is inner pipe and this is outer pipe okay so this is what I have drawn here so here I have drawn reverse hot and cold it does not matter here I have drawn cold and hot what are the modes of the heat transfer which can occur here conduction and convection where all it is occurring here it is occurring on the inner side let me call that as hi it is what is the what is the convective heat transfer coefficient on the inner side hi okay or the convective conductance would be hi ai okay so if it is a circular pipe what will be ai pi di pi di l okay so now what is there is a conductive resistance or conductive conductance what is that k a sorry k okay so here a would be cross sectional the cross sectional area of the pipe so or otherwise if you have to take this one by k a so what is that you take for circular you put this as log of r 2 by r 1 upon k l 2 pi 2 pi r 2 pi k l 2 pi k l k is the thermal conductivity of my solid okay similarly there will be convective resistance or convective conductance on the on the outer side outer side outer side of the inner pipe that is 1 upon h naught k naught okay that is what I have written here so there is conduction convection on the inner side and convection on the outer side so heat transfer in a heat exchanger involved convection in each fluid and conduction through the walls separating the each fluid now that is where we get to overall heat transfer coefficient that is I think we will come to this overall heat transfer coefficient little later on okay so what is that we are saying is that there are two approaches I will come back to this overall heat transfer coefficient little later on but let us get through some basic approaches one approach is LMTD approach LMTDF approach is that approach in which LMTD approach is that approach in which I know what all things I know here in LMTD approach I know m dot c that is cold side mass flow rate m dot h hot side mass flow rate Tci that is cold side temperature Thi cold side inlet temperature Tco cold side outlet temperature and Tho that is hot side outside temperature what are these let me draw that again so if this is my cold fluid this is Tci and this is Tco okay and if this is hot fluid this would be Th i and this would be Th I think we are all very comfortable with this there is no problem so that means what if I know m dot c Tci Tco what is the load of the heat exchanger q dot equal to m dot c into Cpc into Tco minus Tci which is also equal to m dot h into Cp h into Th i minus Tho is that okay this is my load of the heat exchanger that means I know the load of the heat exchanger then what is the time supposed to do in the heat exchanger what is that I am supposed to know I am supposed to find the area of the heat exchanger if I fix the diameter what should be the length that is the question I need to ask so question asked is what is the area of the heat exchanger if this is the problem I will be using LMTD approach that is logarithmic mean temperature difference approach this is well understood in our convective heat transfer lectures what is LMTD we have defined this so this is LMTD approach how do we do this using LMTD we will learn that but this is what is called as LMTD when this is known this is what we call as LMTD approach now let us see how do we how do we what is epsilon NTU approach what is epsilon NTU approach epsilon is effective effectiveness we are not a defined for now you take NTU you know UA upon m dot cp you know that but you we do not know effectiveness it is okay we will define effectiveness little later within an hour epsilon NTU approach in epsilon NTU approach what is all known I know m dot c I know m dot h I know tci I know tco I do not know tc th sorry I do not I know th i you are right th i I know this is th i I do not know th naught and tc naught and I do not know th naught is that okay so but then what do I do not know that means I do not know what is the heat load but then what do I know I know the area of the heat exchanger known when do I know area of heat exchanger if you go to alpha novel website any heat exchanger supplier website he will give this is my heat exchanger which is ready made of the shelf it is there with me this is what my standard product is so that means I know the area so what I should calculate now I should be able to calculate if I use this heat exchanger of this known area of this known diameter of this known length with this mass flow rate and with this inlet temperature what would be the outlet temperature if I get to know the if I calculate or if I can get to know the outlet temperatures if with my epsilon anti approach I can calculate the heat load and then decide okay this heat load will be okay or not usually we will be knowing the heat load I will know I will know that okay my th i has to get down to th o of it is entering at 30 it has to get down to 40 I would be knowing that but whether that heat exchanger would suffice or will be able to cater to my requirement I will not be knowing that can be done by epsilon NTU approach. So these are the two approaches we are going to study in our heat exchanger lectures that is about this LMTDF and epsilon NTU approach there are various types of heat exchangers we need to understand two types of heat exchangers very very clearly let me state this at the cost of being repetitive one is counter flow heat exchanger another one is parallel flow heat exchanger one is parallel flow another one is counter flow heat exchanger if you take tube in tube heat exchanger that is double pipe heat exchanger parallel means both cold side TCI, TCO and THI and THO both are parallel to each other that is the cold fluid and hot fluid are parallel to each other I think most of us are very comfortable with all this but still for the sake of completeness I will have to cover all this so that is THI, THO, TCI, TCO this is parallel flow heat exchanger why I am telling I will come to an important point here this is counter flow what is that counter flow this is TCI, TCO, THI, THO now hot fluid is flowing in the opposite direction compared to that of the cold fluid so that is I have how does it look like this is the hot fluid and this is the cold fluid that is TCI, TCO, THI, THO TCI and THO what is so great about these two heat exchangers that we only study these two heat exchangers for our heat exchangers whether it is LMTD approach or epsilon NTO approach we only study these two why do we study that is what we need to emphasize to students why do we study why because parallel flow heat exchanger is going to have very low performance I am not a defined performance yet it is going to be very low that means if I design a heat exchanger for a given load it is going to have large length if I take counter flow heat exchanger if I for a given load it is going to take shortest possible length so my any heat exchanger I design under the sun any type of heat exchanger I design they are going to be its effectiveness is going to be between these two that is one is the worst fellow another one is the best okay so that is why we want to study the worst and the best and everyone else are going to be between this worst and best that is the crux of the matter otherwise people or students will keep wondering why do we worry about this parallel and why do we worry about this counter flow crux of the matter is this you take shell and tube heat exchanger you take cross flow heat exchanger which I am going to define many of you know all of this any heat exchanger you take plate fin heat exchanger plate frame heat exchanger any heat exchanger you take their effectiveness or their area is going to be between these two okay so that is what we need to emphasize that is what was my point why I drew this although it is very true okay so fine so having understood that now let us move on to that is parallel flow and that is counter flow let me just give you a feel of what are the various types of heat exchangers which are available which are generally used I will come to this beta thing little later this is the cross flow heat exchanger cross flow we all know there is a cylinder in cross flow we studied similarly there is a tube in the tube one fluid is flowing and cross flow to this tube another fluid another fluid is flowing that means one fluid is crossed that is perpendicular to the another fluid so that is why it is called cross flow heat exchanger okay actually this is the shell and tube heat exchanger if you go to market if you go to market especially refrigeration and condenser refrigeration market for him most of the times heat exchanger means it is shell and tube heat exchanger only he does not know anything else for him all that he knows is he has to take a shell and he has to fill in number of tubes if you change the load he will just change the number of tubes and make the shell and tube heat exchanger so what is so great about shell and tube heat exchanger is that it can take larger pressures you see both are of circular both are circular shape that is shell is also made of circular pipe and tubes are also made of circular pipe why circular pipe can take larger pressures if you have to operate at 150 bar 160 bar and all that stuff then shell and tube heat exchanger is a very nice design why because it can take larger pressure okay so there are multiple tubes and in the shell in the tubes the flow is taking and in the shell it is entering here and getting out it is entering here it is entering here and getting out here so what is that happening is it counter flow is this counter flow no it is somewhere between parallel and counter flow why it is coming down and it is going up and it is coming down and it is going up so here we do not know what it is it is neither parallel nor counter flow so by intuition itself you can see that shell and tube heat exchanger is going to be between parallel and counter and also here actually in the tube there is flow from one side to another side but there can be tubes such that there are multiple passes they can be like this the tubes can be like this okay that is I think I have like this let us say if there is a situation like this that means what for one case it will be parallel other case it will be counter another case parallel another case counter like that it will be hundreds and thousands of tubes I have seen heat exchangers in Thane right here I have seen heat exchangers as big as our buildings itself that is our mechanical engineering building that is three-storey floor building itself shell and tube heat exchanger why because if you want to build a heat exchanger of one megawatt or five megawatt it has to be that size the only way to increase is not to increase hi or hi there is a limitation on hi and hi there is only way to increase the surface area okay this I learnt a very hard way when I went to Thara poor nuclear power plant there also heat exchangers are as big as multi-storey buildings okay so point and heat exchangers are as small as 6 meter by 6 meter squared okay so we can build there are now people talk about micro heat exchangers also let us not get into that the point is that heat exchangers are quite large and they are going to be between parallel and counter flow that is the point I want to drive home so this is the shell and tube heat exchanger you can have all combinations gas one side liquid one side the major problem in shell and tube heat exchanger is that problem means in designing tube side there is no problem in computing the heat transfer coefficient what is the heat transfer coefficient you would use for the tube side which correlation you will use dictus bolter correlation you will use because most of the times you will design for turbulent flow most of the heat exchangers you will not design for laminar most until and unless you are operating with oil most of the heat exchangers you will use with turbulent flow so dictus bolter or nilinsky correlation you are going to use now on the shell side that is on the outer wall of this tube how will you compute the heat transfer coefficient it is a mixture of impingement and flow so people have done experiments actually the credit goes to university of delaware where the credit goes to university of delaware there is a method called bell delaware method they were responsible for designing this shell and tube heat exchanger even today in the open literature whatever heat transfer coefficients and the friction factor for shell and tube heat exchanger correlations are available are all because of they just did hundreds of experiments hundreds of experiments and came up with the correlation even today cfd has not been extensively used for designing the heat exchanger if we can do tens of hundreds of phd's in designing the heat exchanger applying cfd methodology okay there is so much scope for designing heat exchangers by marrying cfd approach and experiments there is so much scope there is so much to understand so much improvement can be done still better of course it is not that industries are not doing they are doing if alpha novel does industrial research it will not reveal that research to others it is not available in the open literature that is all my point is various companies L&T L&T is doing its own in the house research there is there are various which company that is having its own research but they do not reveal us or what is that Honeywell is doing its own research but they are not going to reveal us what it is there in the open literature it is for our academia to come up with the correlation for various combinations of heat exchanger and explain it through cfd okay so I got digressed with the main thing but but there is there is juice in heat exchanger for doing research for next 20-25 years for hundreds of people that is the essence okay so that is about shall and dvd exchanger and it can be one shell two pass that is this is one pass and this is the second pass so like that I can have multiple passes this is two shell because there is a baffle plate which is differentiating this shell and this shell and there are four passes one two three four okay so like that there are there can be tens of if not hundreds tens of types of heat exchangers let us not deal with all that actually we I am sure most of the course most of the universities will float an elective called heat exchangers also for the heat exchangers basic fundamental is if you are strong in heat transfer heat exchangers course is a walk why because you are strong in correlations you know fundamentals how to apply you need to know fins you need to know heat transfer coefficient if you are strong in conduction and convection you are strong in heat exchangers because radiation is rarely taken into account okay fine now let us come back to overall heat transfer coefficient concept so what did we say here overall heat transfer coefficient is I have inner side let us say here in this case hot fluid I have taken there is outer side cold fluid I have taken that is convective resistance on the inner side is one upon h i a i okay plus let me write this because as day before yesterday one lady had asked me what is the Wilson's plot method I will answer through this question through this presentation that what is Wilson's plot method that is what is that I have is I have a pipe inner side and this is outer side okay and this is okay my drawing is wrong this is outer side and this is my thickness okay so this is inner side and this side is outer side so what is the overall resistance what is that overall resistance I am telling here so that I do not take the different notations that is one upon h i a i plus 2 pi log of r not by r i upon 2 pi k l log of r not by r i upon r not upon r i upon r i upon r i upon 2 pi k l plus one upon h not a not okay equal to one upon u a this is the overall heat transfer coefficient so usually we say either I can define u i that is if I take a i here I get u i if I take a not I will get u not but I do not usually differentiate between a i and a not I usually keep this u a only because all that we are interested in is the resistance what is this telling me net resistance for the heat transfer to take place from one fluid to the another fluid that is all we are talking about so u is meaningless unless area is specified so if let us say this answers another question if you see in refrigerator if you see in refrigerator you see that I have a problem for that let me take that problem so that we understand see I have taken I have a cooked a problem thermal conductivity of steel pipe is 15 thermal conductivity of copper is 400 d i is 1.5 centimeter that is 15 mm d not is 19 mm h not I have given as 1200 h i is 800 do not worry about this foiling resistance okay so that is actually okay let me come to this problem after I introduce what is called as foiling resistance that is in addition to this resistances I can have what is called as there can be foiling on the inner wall and there can be foiling on the outer wall that is going to create little bit of resistance and that resistance I am putting it as r fi upon a i on the inner side and rf not upon a not in the moodle I had a question how often I should change my heat exchange heat what is that how how often I should change my heater in my geyser so that foiling problem is solved we would not change the heater this is what I have seen in olden days in not in olden days during my childhood days we used to have geyser at home in the geyser what is a geyser geyser if I take a geyser this is the geyser in the geyser I put a heater you might have seen this this heater we put in the heater we put at the bottom of the geyser why do we put the heater at the bottom of the geyser I take this example I forgot to take this example in natural convection why do we put the heater at the bottom when it gets heated up the hot water moves up and the cold water which is there in the top will come down if I put the heater on the top hot water will stay only at the top and cold water will remain at the bottom so though so uniform heat transfer is not happening so that is why heater is put here that is another example which you can quote for natural convection why did I come here I came here for scaling so this what happens is on the heater calcium deposition what should you do if you can reach the heater remove that scaling if you remove that scaling by scrapping your fouling resistance has decreased that is what we usually do in case of or they apply in a shell and tuber heat exchanger if they have to reduce the fouling they take a pipe and put some acids on to it so that the scaling is washed out but believe me if I am having a heat exchanger as big as a mechanical engineering building how much time and how much effort and how much pumping power it takes to remove the scaling so that is why scaling is so important I cannot throw away the heat exchanger I have to remove it somehow okay so that is scaling scaling is an important thing I learnt this very hard way I learnt this very hard way why because we were just trying to discuss with DHEL about some supercritical boilers we are not doing that we were just discussing that at that point of time of discussion we had a engineer who was a PhD in chemical engineering that is his he had come what was his problem or what is his job in DHEL power plant his job is to maintain pH of various fluids which get into heat exchangers and get out of heat exchanger why is he worried about pH because acidity or basicity is the one which decides the folding okay so his job is to only control pH why because he has to control this scaling because he has to control this fouling okay so fouling is a very very important aspect how do I characterize this this can be characterized only through experiment it cannot be done through modeling okay so heat exchangers even today is driven more or less by experiment it is more experimental driven rather than numerical of course CFD is important and should be pitched in so that we can we can afford to do less experiment okay so this is about inner fouling and outer fouling let us go to this example where in which I have put the inner fouling outer fouling and heat transfer coefficients are there if I put this let me take up the resistance what is this resistance now log r naught by r i upon 2 pi k l what is this resistance this resistance is 1.675 into 10 to the power of minus 3 if I take is that right no it is 2.491 into 10 to the power of minus 3 if I take stainless steel pipe on the other hand if I take copper tube what is this resistance coming down to 0.094 you see it has come down by 10 times it has come down by 10 times that is the reason why you see in refrigerators all copper tubes all over copper tubes because conductive resistance log r naught by r i upon 2 pi k l if you increase this k you can decrease the conductive resistance so that is how conductive resistance if you take a highly insulating material and the thickness less if you take the thickness less log r naught by r i also will become less so if you take thin tube and highly conductive material conductive resistance is less then you can neglect if you that is what is being shown through this problem if I can neglect the conductive resistance so what is my overall and the thickness is small then what will happen 1 upon what will happen to this equation 1 upon if a i is approximately equal to a naught and conductive resistance is approximately equal to 0 I will get 1 upon u equal to 1 upon h i plus 1 upon h naught okay so now very interesting thing that is Wilson plot we will be coming that is if I that is what I am saying is what I am saying is if I take a tube in tube heat exchanger and I have inner side h i and outer side h o if m dot let us say this is hot side and this is cold side let us say m dot h and m dot c what is that I learned 1 upon u equal to 1 upon h i plus 1 upon h naught okay so what do I get I think I think I have to postpone this explanation because I have not okay delta t l m t d I have used this so I can introduce so if I know t c i if I measure t c i t c o t h i t h o please listen to me carefully I am measuring a method I am explaining a method to measure heat transfer coefficient itself one of the questions yesterday's was what are the various methods of measuring heat transfer coefficient this is one of the methods to measure heat transfer coefficient if I know t c i t c naught t c t h i and t h o we know that q dot equal to in convection we have studied q dot equal to u a into delta t l m t d what is this delta t l m t d what is this if I have parallel flow heat exchanger sorry I am drawing a small figure this is t h i t h o t c i t c o until I see this figure I cannot write delta t l m t d that is my problem so t h i minus t c i minus of t h o minus t c o upon log of t h i minus t c i upon t h o upon t c o so what is that I know if I know if I have measured all these temperatures inlet and outlet inlet and outlet cold side and outside I know delta t l m t d and do I know q dot do I know q dot m dot c p delta t either in the hot side or the cold side you take you get the q dot so what is now I will get u a I will get so 1 by u a and if a i equal to a naught if I take a thin tube 1 by u equal to 1 by h i plus 1 by h naught let us say on the hot side I will keep my mass flow rate so high so high that h naught is very high h naught is very high if the mass flow rate m dot h is very high h naught is very high mean 1 by h naught will become negligibly small negligibly approximately 0 let us say then what will happen 1 by u equal to 1 by this implies that 1 by u equal to 1 by h i that means what I am able to measure the inner side heat transfer coefficient of course this is average heat transfer coefficient it is not local it is average heat transfer coefficient this is what is called as wilson's plot method I will not go to what is the wilson plot and plot it and show that but if someone is interested there are papers on wilson's plot method I can always give you those references but this is essentially what is wilson's plot method of measuring the heat transfer coefficient you see if you are strong in fundamentals you can generate your own methods that is what people do now you want a micro tube micro finned tube whose heat transfer coefficient you want to measure you put a micro finned tube inner side and outer side you keep h naught very high then you have to you can measure the h i with this we will move on to what is called as of course there are if you put a fins on the inner side or outer side you have to take into account the fin efficiency for resistance ok so with this we will move on to there are various there are various cases which are listed for which the overall heat transfer coefficient are listed out I am not going to go and visit each one and explain you what is what but this is going to give you the if it if the gas is there you can see that the overall loss coefficient is less if it is liquid the overall loss coefficient is high with this I do not think I am not going to spend too much of time on each case and say that this is high this is low I am going to now get started with delta t l m t d approach that is l m t d approach and I am not going to write this definitely I am not going to derive this on step by step because I am sure each one of us are quite comfortable with all of this so in next 20 minutes I intend to complete this delta t l m t d approach and solve a problem on this ok so what is this delta t l m t d approach in the delta t l m t d approach we said that inlet mass flow rate that is mass flow rate on the cold side and the hot side we know inlet temperatures we know T c i T c o T h i T h o we know that is what we said as delta t l m t d approach yeah this is delta t that this is l m t d approach we know m dot c m dot h T c i T c o T h i T h o so under these circumstances let me do the analysis what am I doing before I do that I take a pair I am going to do this derivation for parallel flow heat exchanger the same derivation is there for counter flow heat exchanger in this node same lines you can follow there is no difference what is that this is the parallel flow heat exchanger and I am taking T h in as the inlet temperature of the hot and T h o is the outlet temperature of the hot side T c in is the inlet temperature of the cold side and T c o is the outlet temperature of the cold side now at every location there is a temperature difference between hot and cold and this temperature difference is decreasing with the increase of length. So now you see delta t delta t is decreasing now let me take a small elemental area that is a length d x that is what is this this length here is d x or elemental area d a s if I take d a s as the elemental area what is happening in this elemental area there is a temperature drop on the hot side d T h and there is a temperature increase on the cold side d T c and there is a heat transfer from hot side to cold side delta q dot this delta q dot equal to m dot c c p c d T c which is equal to minus m dot c p h d T h why is this minus sign why is this minus sign because temperature is decreasing with the increase of length now before we get into this analysis I need to state the assumptions under which this analysis I am doing what are the assumptions I am assuming that steady flow has reached okay heat exchangers take infinite time I should not use the word infinite take long long time before they reach the steady state if you take a power plant the power plant takes almost couple of hours to reach steady state okay and if there is change in the load it will again take much more time to reach in the morning there is heavy load in the afternoon it is part load and like that the load keeps on in the night again the load changes load increases. So my power plant also get has to modulate itself for this changes in the load electrical load so point is it takes time for a heat exchanger to reach steady state and there are no kinetic and potential energy changes in a heat exchanger there are lots of bends and valves and it is going through lot of things but we are going to assume that velocity is going to be constant and whatever height difference is there I am not going to consider that that is potential energy changes are neglected thermo physical properties are constant especially Cp and viscosity and rho I am assuming constant this is not always true especially when we are taking liquids because the viscosity at 30 degree Celsius is not same as the viscosity at 90 degree Celsius for water so it is going to be different but for now I am going to go ahead and make a relaxation and take the assumption that thermo physical properties are constant over entire length another important thing is no heat loss to the surrounding that is whatever heat lost by hot water is equal to what heat gained by cold fluid but not always true because there is loss to the ambient there is loss to the ambient conditions. So that loss that is why we were talking about yesterday insulation aluminum foil and things like that and heat transfer coefficient is constant over entire length that means we are assuming that the flow is thermally fully developed and hydro dynamically fully developed if you recollect we said that the heat transfer coefficient will decrease initially and then become constant in the fully developed region heat exchangers are usually running in few meters of length that initial developing length heat transfer coefficient I need not have to worry even if I worry that heat transfer coefficient is higher if I design it on the fully developed length concept my design is going to be a conservative design. So I do not have to worry about this developing length in heat exchanger so with this assumptions in the back of my mind so let me equate this energy balance what was I saying delta q dot equal to m dot c c p c d t c equal to minus m dot h c p h d t h so d t c equal to delta q dot upon m dot c c p c and d t h equal to minus delta q dot upon m dot h c p h. So now what is delta q dot delta q dot equal to u into that is heat transfer coefficient what heat transfer coefficient overall heat transfer coefficient this is our newton's law of cooling nothing different u into d as into t h minus t c so now what is d that is equation 2 in 1 this is equation 2 and this is equation 1 if I subtract 2 minus 1 what am I getting 2 minus 1 d t h minus d t c that is d t h minus d t c if I do that what will I get d t h minus d t c equal to minus delta q dot upon m dot h c p h minus of delta q dot upon m dot c c p c same thing I am doing here but delta q dot minus delta q dot I can take out and d differential I am writing in terms of temperature difference d into t h minus t c equal to minus delta q dot into 1 upon m dot h c p h plus 1 upon m dot c c c c. So if I substitute delta q dot what is this delta q dot we just now said u d as into d t h minus d t c so that is if I substitute that delta q dot in 3 here that is if you are not able to get me delta q dot equal to u into t h minus t s into d as that is for delta q dot I am substituting delta q dot equal to u t h minus t c into d as that is u t h minus t s t c into d as that t h minus t c I am pushing it to the left hand side so left hand side I have d into t h minus t c upon t h minus t c. Let me just get started from LMTD approach in the LMTD approach we know the mass flow rates cold side hot side inlet temperatures and outlet temperature that is t c i and t h naught under these assumptions under these assumptions we said that we need to compute the what is the aim what is the analysis what is that we are trying to do in this in this derivation we are trying to relate the temperature with the overall heat transfer coefficient and the load so that I can get the area what am I up to in LMTD approach given the temperatures given the mass flow rate we need to compute the area that is the point that is what we are looking for so if I take a counter flow heat exchanger in which it is going from t h in to t h out t c into t c out and I have delta q dot equal to m dot c c p c d t c sorry before this we said that there is a small elemental area d as within this d as on the hot side there is a temperature drop of d th that is this and there is a temperature increase on the cold side d t c so that heat transfer within this elemental area is delta q dot equal to m dot c c p c d t c equal to minus m dot h c p h d th this minus sign is because of decrease of the temperature slope is negative here slope is positive so d t c equal to that is from this delta q dot upon m dot c c p c and d d t h equal to minus delta q dot upon m dot h c p c p c at the same time we know that delta q dot equal to u d as into t h minus c c what is this u overall heat transfer coefficient we have defined just a little while ago so you can see that I will deduct 2 minus 1 and substitute in that 2 minus 1 for delta q dot from this equation okay so equation 2 minus equation 1 gives me minus delta q dot into 1 upon m dot h plus c p h 1 upon m dot h c p h plus 1 upon m dot c c p c equal to d t h minus d t c that is d of th minus d t c for this delta q dot I am substituting u d as into t h minus d c that t h minus d c I have push up to the left hand side so now I am integrating it from inlet to outlet inlet to outlet throughout the length so if I integrate this this is nothing but log d theta by theta that is log of theta not by theta i that is th not minus d c not upon t h i minus d c i equal to minus of u as into why I am doing how why can I pull out this u outside what is the assumption I have invoked in pulling this outside I have stated this assumption what is that heat transfer coefficient is constant is uniform not constant I have to use the word constant with respect to time uniform with respect to space heat transfer coefficient is uniform because of which my u is 1 upon u as equal to what 1 upon h i a i plus 1 upon h not a not plus log of r 2 r 1 upon 2 pi l k as long as my h and I are constant I am going to get my u as as constant so I can pull out this u and d as integration I get as I get this how why can I pull out this because c p h and c p c I have assumed them as constant with respect to space not time uniform with respect to space or they are constant with respect to temperature you are right constant with respect to temperature because temperature is varying log of th not minus d c not upon log of th minus th i minus d c i equal to minus u as upon this 1 upon m dot h c p h plus 1 upon m dot c c p c what is this m dot h c p h I know why load equal to q dot equal to m dot h c p h into th i minus th not if I substitute that here q dot in the numerator q dot will be in the denominator and th i minus th not will be in the numerator similarly q dot will be in the denominator d c not minus d c i will be in the numerator let us see that equation that which we have if I rearrange this minus q dot I will push to the left hand side and this log of whole thing I will push to the right hand side so q dot equal to minus u a not u as into I am writing that just that th i minus d c i minus th not minus t c not upon th not minus log of this so minus if I absorb I will get u as into th not minus t c not minus th i minus t c i upon log of this temperature so that is what we get q dot u as what is delta t 1 delta t 1 is the temperature difference at the exit and delta t 2 is the temperature difference at the inlet and log of delta t 1 upon delta t so q dot equal to u as delta t l m t d I can calculate u 1 upon u as in which I do not know the area I will know the u I know q dot because I know the load delta t l m t d I know I can get the a s let us quickly solve a problem which will elaborate this procedure what is the problem is it is having a parallel flow heat exchanger no problem what we are saying is before we move on let me take counter flow heat exchanger in a counter flow heat exchanger we can do the same analysis in the same lines I would request you all of you to sit down and do this derivation without seeing the derivation which is given here which is given here seeing the derivation of parallel flow heat exchanger apply the same principles and you will get again for counter flow heat exchanger also q equal to u a s delta t l m t d that is temperature difference temperature difference here would be T h i minus T c o and other one would be T h o minus T c this is delta t 1 or this is delta t 2 which order you take it does not matter actually okay delta t 1 minus delta t 2 upon log delta t 1 upon delta t 2 or delta t 2 minus delta t 1 upon log delta t 2 upon delta t 1 you do not have to worry what matters is the temperature difference okay now I said I made a statement in the morning that counter flow heat exchanger is always better than parallel flow heat exchanger why if you if you just check I have stated here l m t d counter flow heat exchanger is greater than l m t d of parallel flow heat exchanger if you just check that I have just taken random some temperatures T h i 110 T h o 60 T c i 30 T c o 55 if I take this what will be my delta t l m t d for counter flow 41.24 what will be it for parallel flow 27.05 counter flow whatever circus you do counter flow will be higher now the question is why now the question is why I think answer is there already the way I will give you the clue the way we explained it for constant heat flux and constant wall temperature on the same lines I can explain for a counter flow what is the temperature how is the temperature difference looking like almost the temperature difference is constant for the parallel flow how is the temperature difference looking like continuously the temperature difference is decreasing with the increase of the length. So the heat load if I calculate delta q 0 right from the beginning to the end as I move at the farthest end my heat transfer load has come down has come down because of which my delta t l m t d this has precipitated or this has manifested itself in delta t l m t d delta t l m t d that is the reason for counter flow heat exchanger is greater than that of parallel flow heat exchanger. Now I will quickly take a problem just to give you the feel of area of heat transfer coefficient professor Arun wants to chip in and add something this concept of driving temperature difference becoming lower in case of parallel flow we have seen a similar concept right at in the fin stage also longer the length of fin the driving temperature difference for heat transfer decreases that means you put twice the length but you are not going to get more and more you are not going to get twice the utility that is one thing transient conduction lumped capacitance also when you did you know the decay was fast initially afterwards the heat transfer became slower. So what is it telling me all these things you look at fin or you look at transient conduction or you look at constant wall temperature condition or an application which is heat exchanger it is a delta t which is going to matter and in all the four cases if you see the initial delta t especially fin transient conduction lumped analysis as well as constant wall temperature case the nature of the temperature distribution is same one is horizontal other is exponentially increasing or exponentially decreasing. So that the delta t is exponentially decreasing always here also in parallel flow the delta t is going to decrease of course not exponentially exponentially will be a special case the point to notice that. So actually very nicely professor Arun connected very different things the summary is temperature difference temperature difference temperature difference because that is the driving potential. So now let us take a problem the problem is I have oil and water and water is entering at 30 degree Celsius and I do not know at what temperature it is getting out although I have written it 40.2 and I will be able to calculate that and the mass flow rate is at 0.2 kg per second and I have an oil which is being flowing at 0.1 kg per second and entering at 100 and getting out at 60 degree Celsius first thing is in the problem I need to calculate the properties at what temperature I need to calculate the properties at the average of the inlet and outlet that is the best thing I think by now you know why we take always the average temperatures I do not have to elaborate that similarly for oil and water I have taken Cp µk Prandtl note Prandtl is 501 Cp µk Prandtl of water is 4.85 now what is the load of the heat exchanger how did I get this T mean as 35 first this I need to calculate the load of the heat exchanger that is I know on the oil side all the things inlet outlet mass flow rate all that I need is Cp if I take the Cp of the oil if I calculate m dot h Cp h th i minus th 0 I get 8.5 kilo watts if I plug in that 8524 kilo watts and put Cp Cp and mass flow rate of water I get TCO TCO as 40.2 that is how I have got 40.2 now having got the load what is that we are trying to solve in this problem area what is the area of this heat exchanger so what will I do I will choose a diameter most of the times you have to choose the diameter no one will give the area because this is a tube in tube heat exchanger I have to choose the diameter what is the diameter I have chosen I have chosen 1 inch pipe 25 mm I have chosen if I choose 1 inch pipe first thing again delta TLM TD I will compute and keep that is not an issue 43.2 degree Celsius I will compute and keep because I know all the temperature I do not have to show you that first thing is I have to compute h i h o what for I have to try to compute u a s so we know that 1 upon u a s equal to 1 upon h i a i plus 1 upon h 0 a 0 plus log of r 2 by r 1 upon 2 pi l k so I am neglecting the conductive resistance I am trying to calculate only h i and h o and I am neglecting the thickness also okay so if I do that diameter is 25 first thing is on the water side I have to compute water side is tube side or shell side tube side so tube side I have to compute Reynolds number 14,050 that means it is turbulent so I can use detus-bolter correlation if I plug in detus-bolter correlation I get h i of 2250 now on the oil side what is the oil side Reynolds number this is very important point here for Reynolds number I have to take velocity for velocity for Reynolds number I have to take hydraulic diameter that is what is the hydraulic diameter of an annulus d0-di that is 4 into area upon perimeter right 4 into wetted area by perimeter so you get d0-di so 4 rho v dh by mu that is Reynolds number you get you see the Reynolds number Reynolds number is very small why because it is oil viscosity is very high 3.25 into 10 to the power of minus 2 so with this Reynolds number now whatever be the parental number I know the Nusselt number now which Nusselt number I should be taking 5.56 should I take constant wall temperature or constant heat which one should I take I should take constant wall temperature why because I want to take mostly temperatures as I said in convection in the heat exchangers closer boundary condition is constant wall temperature although in not through a strictest sense but I take constant wall temperature another reason is that it is lower heat transfer coefficient if it is lower my estimate of area would be that means my resistance estimated is higher that means my design of the heat exchanger would be a conservative one I would have estimated larger area then if I build that heat exchanger it would perhaps take more area than what I have designed it for so with that I get h0 as 38.4 so now 1 upon 1 upon ui 1 upon h i ai plus 1 upon h0 a0 because thickness is very small thickness is very small I have neglected I have taken that ai equal to a0 otherwise what I should have taken ai is pi di l and a0 should be pi d0 l pi d0 l okay see here we should not take a1 a1 a1 a1 a1 a1 a1 a1 as pi by 4 into d0 squared minus d0 di squared okay or I should not take usually my students ask me the questions sir I should take the outer inner diameter of the outer pipe no what is the convective resistance which we are computing I am computing the convective resistance on the outer wall of the inner pipe remember that okay so that is why my a0 should be pi d0 l all that I am saying is that in this problem my simplification is that d0 is closer to di because thickness is very small I have neglected that so I get u equal to 37.76 you see the oil side is oil side heat transfer coefficient is very less oil side it is resistance is high resistance is high so overall heat transfer coefficient is closer to that of oil in fact it is it is almost same it is 38.84 this is 37.76 okay so I get q dot equal to u as delta t l m td I get a length of 66.54 66.54 meters how many foot 66 3 meter 3 feet is 1 meter so roughly so 66 x 3 66 x 3 is around 200 feet 200 feet 200 feet means it is almost 20 storey building 10 storeys 1050 is one storey building so 20 storey building is the length of the heat exchanger is that right or wrong 200 feet 200 feet divided by 10 is 20 20 storey building is not an easy task so I do not think I can build an heat exchanger that too for a load of how much 8.5 kilowatts if it is mega watts I can imagine 8.5 kilowatts this length is too much what should I do what should I do I should decrease the resistance on the oil side why it has come down because it is laminar I make it turbulent if I have to make it turbulent I have to increase my pumping power pumping and oil is not an easy task so if I make it turbulent and get the heat transfer coefficient as much as 2002 50 as much as 2002 50 how many times it would have gone up 2002 50 divided by 38 60 times see my length will come down to nearly 1 meter 1 1.2 meter that is the beauty if we had not understood this convective resistance concept I would not have designed I know where is the pinch I can go there and take care of that and design my heat exchanger one meter means it makes sense it is just this length only thing is that I need a pump for the oil that decision that I have to go for a pump came from which is affecting my size of the heat exchanger okay so this is about LMTD approach and there is the you see here there is this is just nothing but constant there is a constant okay what is what is this transparency saying is this is what always we ask the question for our students what is delta TLMTD when m dot c CPC equal to m dot h CPH in most of the interviews in most of the entrance test you will have this question why because what is the delta TLMTD I do not have to do any mathematics for this first I can see physically what is that delta TLMTD it is temperature difference between hot side and cold side why did we define delta TLMTD because it is continuously varying from inlet to outlet but if m dot h CPH and m dot c CPC are same are equal what is this temperature difference going to be it is going to be same throughout then what should be the logical temperature difference in u a delta T a it has to be the temperature difference at any location it does not matter okay but if I substitute in my delta TLMTD I will be in trouble because I will get 0 by 0 so I am in a problem if you apply L hospitals rule for the delta TLMTD and take limits you will get appropriate temperature difference that is this delta TLMTD so it is point is you should not get carried away by mathematics if you are strong in physics that is if you are strong in understanding physical significances you do not have to take the record you do not have to get carried away by mathematics that is the point which we want to tell and you can ask this question for students you can ask this question and ask them to show through L hospitals rule that it is indeed delta T1 equal to delta T2 this is our pet question for our quiz always not always every time we cannot ask the same question okay for condenser and boiler if I have it if I have a two phase flow heat exchanger condenser in a condenser what is happening the steam side this is you are having the constant temperature because the heat transfer is by latent heat here on the boiling side this fluid is at a constant temperature so that means m dot hcph is tending to infinity because delta T here is tending to 0 so there but still I have m dot hcph delta T so what does this that is this q dot equal to m dot h hfg what does this mean still what I mean is by saying this by using this approach by taking this q dot equal to m dot h hfg you can still go ahead and use delta TLMTD approach and design your condenser and boiler all that which is going to be different in condenser and boiler is on one side it is going to be two phase flow so that is whichever side condensation is taking place whichever side boiling is taking place you do not know the heat transfer coefficient that heat transfer coefficient has to come from two phase flow that is all that we are not taught we have taught you only for single phase so if you take on the boiling side or the condenser side appropriate heat transfer coefficient you can still design the condenser and boiler using the LMTD approach that is what is this transparency telling me so the load comes from m dot hfg so with this I think we are through with no we are not through with delta TLMTD approach where there is one another little thing that is correction factor I said all heat exchangers are going to be between parallel flow and counter flow and we know that our heat exchanger whichever heat exchanger is going to be it is going to be performing little lower than the counter flow heat exchanger how much lower is taken care by a correction factor called f f is given by obtained by charge so these are the charge let us say this is a shell and tube heat exchanger shell and tube heat exchanger these are temperatures T1, T2, T1, T2 so this is a relation called R R equal to T1 minus T2 upon T2 minus T1 this is the ratio of the mass flow rates and specific heat product another ratio P T2 minus T1 upon T2 minus T1 if I because if I know the temperatures I can calculate P and I can calculate R for a given P and for a given R I get the correction factor you see this correction factor logically has to be between 0 and 1 now where did this chart come from who gave is it experimental general perception is that charts are usually generated through experiments not so not so this is derived from fundamentals in fact when we teach heat exchangers course we have derived this from fundamentals if you want these derivation you can refer to heat exchangers process heat transfer by no have it if you want the derivation of this you can refer to this text book that is process heat transfer by sorry not by Kern by have it if you Google book search you will get the addition and things like that it is a quite an expense CR CRC press CRC press so it is quite an expensive book but it is a good book if you are teaching heat exchangers this is the book one at least we follow myself and professor around follow this book yeah it is a good addition to your libraries it is not possible to purchase personally please add this to your library okay so now point is none of these charts for any of the heat exchanger are obtained experimentally they are all derived that means they are all subjected to the assumptions what we make generally steady state no kinetic energy potential energy changes and then properties are constant all those assumptions are still valid that means if I build a heat exchanger based on these chart my real life heat exchanger is not going to be just the way my chart is telling it is going to be little different from what I have designed but ballpark within 10 to 15 percent perhaps it is going to work okay from my design design criteria so that F has been listed for various configurations I am not going to show that for all of these configurations they are there for cross flow heat exchanger and this R will carry on this definition of R is same for subsequent one okay so if you go ahead and use these chart there are problems which I am not going to solve it for you they are quite straight forward simple plug-in problems please do that one is for condenser another is for this F that is for applying this correction factor F there is this okay multi pass heat exchanger so for applying correction factor F how we get and there is cross flow heat exchanger please solve this problem just recap is that we go for LMTD approach only when we know the mass flow rate inlet temperatures outlet temperatures and you want to compute the heat transfer area you use the LMTD approach now over to professor Arun for epsilon NTU approach.


Welcome to this course. I am Anna Sparopto, a system professor in network security at the University of Tanda, and I will be your main lecture for this module on Internet Security, Attack and Defense. This module has been developed by a team of researchers at the University of Tanda, and it is part of the Internet Security course. This module focuses on attack defense techniques, as we see them in use in network nowadays. If you are interested in security protocols and web hacking, you might want to follow also the other modules of this MOOC. Each lecture in this module has three main ingredients. The first is attacks. In our lecture, we will cover a selection of attacks that we encounter in current networks. The second ingredient is defense. We will introduce and reason about defense and mitigation techniques that can be applied to real attacks. The last ingredient is measurements. Network measurements is the approach we will take in this course to observe and characterize attacks. This means that we will give example of real attacks that have been measured in production networks. Also, you will familiarize yourself with real network traces and learn how to analyze them. After this course, you will be able to understand, work with and apply concepts in the field of attacks, defense, and measurements. In particular, you will be able to describe the enabling mechanism of current attacks and discuss possible countermeasure. We will also acquire practical experience in our attacks work and how you can characterize them. Finally, you will be able to design possible defense strategy based on the countermeasure you will see in the course. The module is organized as follows. After the introductory week, we will have seven weeks of lectures. We will start with an introduction to network measurements with the goal of providing an overview of the topic and familiarizing you with trace analysis. After that, we will dive into a set of attacks. First, we will present an overview of attacks such as scans, worms, and botnets. Then, in week four and five, we will talk about denial of service attacks and DNS security. In weeks six and seven, we will focus on defense by looking at intrusion detection, firewalls, and nets. Finally, we will conclude this course with a lecture on advanced persistent ref. In this overview, we have highlighted the main focus of the lecture, that is either attack, defense, or measurements. In truth, each lecture contains elements of each of these three ingredients. So you can expect to hear about example of measured attacks and mitigation measures in several places in this module. Each lecture provides the following material. In the video lectures, you can find the relevant information about the topic of the lecture. These videos are a good starting point into the topic, but they are not exhaustive percent. We always encourage our students to deepen their knowledge with additional material, which we will propose as self-study. Each lecture, we also propose a set of quizzes that will help you to memorize important concepts. Finally, each week, with the exception of this first lecture, we'll have a homework and its own exercise to be handed in before the next lecture.


Stevenson falls to Mary Washington today. Glenn Clark with head coach Jackie Boswell. Coach, obviously, you're going up against a really good team today. As far as what we saw, how much of it was them, how much of it was you, just what went on, I guess? Well, I thought we came out and we played really scared. We put these teams on the schedule to challenge us. I don't think we were physical with them. I think we let them just come in here and do whatever they wanted. And that was a little bit disappointing, I thought. I thought we could have been a little bit tougher than that. I think at times we tried to do that. But I think consistent has been the word we've been trying to use all year. We can't really get consistent at anything right now. We have a good defensive game. Then we don't have a good offensive game. We have a good offensive game. Don't have a good defensive game. Come out here, want to be challenged, want to compete and be physical, can't do that. So it's just lack of consistency. And we talk about how young they are, but they come in here every day. We're at 57, 59 practices or whatever. Got to get a little tougher. So we'll see where we bounce back on the conference, because that's the most important thing right now. Obviously, a quick turnaround for that.  What you're getting from this team in practice, do you feel like they are growing, even if the results aren't there, do you feel like this team is growing and improving from day to day? I do. I do. We talked about it all the right the other day. We cut down on the turnovers, which is something that was bothering us again. We just got to be able to grow a little bit faster, I guess, than what they're growing right now to get to where we want to be. I don't think we should be at the bottom of the conference, but it doesn't matter what I think. It only matters what we can go out there and improve. All right, so just for a second, if I could. It's been a tough stretch here, but I want to highlight maybe some of the things some of your players are doing away from the floor. And I heard that Kalea has really been putting in an amazing effort during this winter break. Yeah, I mean, she does community service for people that need it. And she's just a really good person all the way around. So every day, 9 to 12, while people are sleeping in on their winter breaks, out there doing community service. So shout out to Kalea for that. It is really remarkable. It's very cool. All right, Coach, best of luck on the road on Wednesday. We'll see you back here on Saturday, all right? Great, thank you. She's Jackie Boswell. I'm Glenn Clark. Go Mustang Sports.tv.