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Module 1: Surface Condensers

    Study Reminders
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    Surface Condensers – Part 3





    Welcome if you recall that we are in between solving a problem on surface condenser. So we will without much introduction, we will proceed and a let's go to the next slide, so here you see the example which we are continuing from heat balance we have got two very important quantities that is the Steam flow rate and do s, this is one thing we have got and we have got him not see total heat transfer total rate of heat transfer that once we have got any from the linq pool and velocity we have got anything that is the total number of teams here we have assumed that only you know there will be a bank but oneroof there will be 70 tubes to this is your 70 this is your first tube so there will be like this and we need to know how many cubes will be on a Scion a call I’m not on here or on a column how many teams will be there so there will be 70 columns on a call decided anything we need to indeed you will have some sort of a turbulent flow from the experience one can tell that the coolant flow will be turbulent of course from Reynolds number we will get this kind of eating sorry we have probably gone forward little bit so let us go back and then let us try to do it should be turbulent flow and it will be given by a transfer coefficient wire which will be given by national number from the number kind of relationship so we know we know we know we have calculator selected a cool and velocity of 2 m per second to renounce number we get this number and we used to kind of their number next slide so probably nusselt number we get the transfer coefficient for this is been sighted transfer coefficient now we know the condensing symptoms temp temperature show in the heat exchanger the inlet side that means English for be cool and so in Lakeside we know what is the temperature difference outlets Arduino funky pawan genelia from there we can calculate log mean temperature difference if we see let’s say this is your new page sure this is your water and this is then we have got this distance is first there will be conviction conviction then there will be inside fowlie I am indicating by AFI then there will be conduction due to you all tube wall then there will be again falling I am calling it if you and then there will be condensation should very well-known thermal circuit and here we will have your Steam ok here will have your stick so what have to stay there are 12345 resistance but you see this convective resistance a child have to calculate an issue you have to calculate in calculation already I have described for hr calculation this temperature difference which we will call as delta TW2 this is this is needed if you go back to the station you will find that the temperature is now the local is temperature today’s temperature and then he will be different so it is a very difficult calculation for what we will doconnected like this this side frida’s entering his life through it is going out so whatwhat will be do we calculate at this point that is the outside heat transfer coefficient we will calculate a show at this point and we will take an average but doing this we have to go for an iterative method this is very important that even for doing this we have to go for an iterative method and let us see how this iterative method works next we need to estimate the sales side heat transfer coefficient that is the tube outside convective heat transfer coefficient due to condensation as this depends on local heat flux iteration is necessary for this week love the resistance’s into sell-side condensacion which is unknown and which I had told that it is the function of the 830 w and the remaining resistances which are known so overall heat transfer coefficient is now one resistance which has club all the different kind of resistances and another resistance it is the condensing registration registered on the team then it will be given by this kind of a formula so what are there howling resistance outsideinduction the distance of the tube we will use some sort of a mean diameter instead of using the logarithmic relationship for conduction so we can go without taking any kind of all these things because the properties of liquidwill it rain this coming from NASA relationship and again nothing relationship has to be modified this is your muscles average age on YouTube this is due to and Asianand that means the total thing is nothing and then this is nothing and this is in addition by condensation and here you see that delta t has been delta TW is unknown which is there should this is one thing again we have to so which is unknown is this delta this is you and in here in 70 already we have assumed so ultimately there we get a is equal to this particular which is a function of this is that what is the temperaturewhat is the local temperature difference between the two streams and it is this sum of all the other registered 6qw Prime is the local heatplus given by q is equal to 2 into 10 Tak ki saree materials that we get by this kind ofstudy study duration at the inlet and outlet so why we have to go for iteration because we have to calculate you we have to calculate we have to calculate you in it and we have to calculate you outlet and from there we have to calculate average yesterday so it is there to order tube wall producers in contention or what we can show this is the cue ball let me hear it then show it so basically it is like this there is a dinner calling while there is an outer fouling layer and there is a condensate layer let’s say this is your condenser plate is your delta TW ok so this is against temperature and there is another this is your we have to understand this is notcalculate ho from equation to album we have three question we have noted as one two and three let me go back and show these equations that will be important for you should this is Lori question 3 this is your equation one and this is your equations to show equation 1 2 and 3 are very important so let us go back calculate age from equation to calculate you from equation one above and recalculate best rtw from equation solution and continue this is what we have to do this is what we have to do and then let’s go back sorry let’s go for what move forward and then in the next slide table I should be good this should be a temperature difference between the two strings show at the inlet is 25.8 and the outlet and for the initial case of Bill w9.47 and then again we land up with some overall heat transfer coefficient 1575 that means 1005 and 75 and we continue this we get ultimately 9.44 degree Celsius is still 30% similarly vacuum 6 degrees Celsius as delta t w at the outlet we do the same kind of iteration and ultimately we get the ultimately we get the heat transfer coefficient overall heat transfer coefficient at 16:32 so I’ll let me tell you what we are up to please not the corrections the these these delta t w should be days that if this then that we should be there today they are at the inlet and outlet the temperature difference between these teams with all these things what we do we try to calculate the overall heat transfer coefficient at the inlet and outlet going new transfer coefficient at the inlet and outlet wetransfer and thenlet me go back to the previous page so that you can see once again so he let Sadie transfer coefficient we have got 1575 what time is sq KM and outlets I do transfer coefficient we have got 1632 what time it is where can we show with this to we calculate the average heat transfer coefficient every heat transfer coefficient is this everything transfer coefficient is the transfer now we can calculate the amount of heat transfer which is known which is known because how do I know we have already calculated if you recall it so we know how much time we have to condense and bring it to the saturated liquid condition then we know that steam at what condition it is entering the condenser from there we know QQ we know everyday transfer coefficient we know logarithmic temperature difference we know and they’re only unknown that is there that is a poo because everyday transfer coefficient has been calculated based on outside area of the cube and EO we can calculate this is the value of a should this side area rnd is still number also have calculated number of use so this week I’ve got an outside tube diameter also we know so if we know outside you’ve damaged then we can calculate the tube length needed to get this outside area of The cube latestEminem this is the only problem anything to be discussed to be appreciated so I mean just can’t suddenly very very briefly I will recalculate what we have done maybe next time when we will again start with this problem we will repeat the same thing but at the end you should understand how we have done the design so basically the steam conditions have been given sucking minimumwe had assumed everything you know diameter of the tube outer diameter of the tube all the material properties because materials tube materials we have issue and Steam and water proper property and then what we have got we got shot in which basic quantities of the heat exchanger one is total rate of heat transfer another is the Steam flow rate another is the water flow rate with these things and we have also assumed or taken as design data what should be the velocity of water through the tube so from there we have got the number of that there will be a bank and in the tube bank only column there will be 70 number we could calculate the inside heat transfer coefficient really but outside the transfer coefficient we had to calculate knowingly some sort of a temperature difference for each iteration is needed and we have gone for iteration and ultimately we have determined the outside heat transfer coefficient every help drive southsider transportation average overall heat transfer coefficient and with all these things then we are able to calculate the outside area of all the tubes or cumulative outside area of the tubes needed to condense the amount of steel which we need to contact so then from there we could calculate the number of tubes already have calculated now we can calculate the length of the tubes now probably we can check whether the pressure drop is alright or not now we can have some idea what would be the same diameter now we can go for some sort of litigation or some sort of exploration if we select other kind of velocity what will be the effect on the design of the heat exchanger show all this exercise we can do with this I with this load I ain’t here but the problem discussion on this problem does not in here we would like to take the benefit of the problem which is which comprehensive which is becoming more or less a practical problem to understand other aspects of a surface condenser when it is having shell and tube kind of configuration. Thank you.