Fin Heat Exchangers | Finned Tube Heat Exchangers – Part 3 | Alison
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Module 1: Fin Heat Exchangers

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Fin Plates in Fin Tube Heat Exchangers – Part 3

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Analysis of Fin Plates in Fin Tube Heat Exchangers – Part 3
Hello everyone, we were discussing fin tube heat exchangers in the last few lectures we have seen how to calculate the efficiency for fin tube exchangers and how to calculate different geometrical parameters particularly area of fin tube heat exchangers and ultimately we have calculatedput the transfer coefficient and the pressure drop that we depend on the humidity of the earth's surface Dewberry mint number of rows number of columns andis needed so that’s why we will take up some problem or example typical example of seem to be texting nothing has to be get my heart only the method of selection or rather the logic we follow to solve them that has to be understood so that another problem comes we will be able to see this is what I like and certain general meeting is there for general it is I am pointing out as we proceed with this course let me do one thing I like to I like to eat certain overviews so that you will be able to understand what we’re going to do so let’s say this is the you are ais 2 q and flu or gas or ear which is the gas in many cases is in this direction is normal to discuss the transfer fee let me discuss the prospect of it should be in the air passes through these different cross-sections of area so when the air is passing through this portion sure it is passing through some sort of restricted pace and going it is passing through this then it has got more cross-sectional area here we go through experiencing or else there will experience some sort of acceleration and deceleration should we need some amount of pressure drop Los arcos it is passing past solid bodies which are properties and these tubes are having pains etc so there will be some sort of lost due to that which is called friction loss but it could be both your friction loss and your loss due to form that generally it is done and one can write the letter p the pressure drop that is equal to that is equal to do 22 component the acceleration plus the friction contribution of acceleration and friction as and then Max1st in which is due to acceleration and another is k which is code uncode due to friction so I told that friction and farm dragon all these losses due to discuss nature of the food that will plug into kfn then k a is equal to 1 + 6 square again I will get another relationship 48 and then see what is this sigma sigma is some sort of geometrical through this is the minimum floor area some of these things are being used are here and divided by the total frontal areathis is the height of beef in this is d with the defenders do it the defin and the space between two Fields these things will come here to rkf depends not only on the tree of the Dewberry but it also depends on other factor function of Reynolds numberbest on maximum fluid velocity and tube diameter and is recommended for are in the range of 10 to the power 3 to take 35 before you better not be amazed this and then it will damage if it's a club we come into picture basically it is dependent on Reynolds number diameter of the tube diameter of the team’s etc and then I feel tubes instability within the range of 5 into 10 to the power 5 into 10 to the 4th we have got some other day and at I have defined in our area class so you see you had to go because this formula are available at least somebody is some sort of a cube layout for which formula is not there then of course it is it is a matter to be a wreck and in that case one has two again contact experiment to get this kind of correlations otherwise if we pick our selection from the available geometry future used in the industry then this kind of formula are available andof analysis and that is what I am going to describe here and tfk a depends on sigma we have shown in this particular how to calculate sigma and we have also given some probably correlations some existing relations 49 dissidia that is what and with this we can calculate the liquid quantities now pick up a problem and with the problem we will see how we can solve this and how we can use the things which we have launched a heat exchanger consists of four walls of raiford use thermal conductivity of the steel tube that has been given and in equilateral triangle are treated with fitted with corbels or dumb enough to use to reduce the bypass the you have rule for rectangular cross section of a rectangular cross section c and the following dimensions link has been giving root diameter has been given and then the then the tip diameter has been given then actually this fastlane one should take as the length of the tube and second in this is a typographical mistake over here so this is the length of the field or height of the pjw is the w is the width of the scene and is is the gap between two firms then p1 is the peach one page in the direction of flow and b2 sorry p1 is the pitch normal to the flow direction and p2 is the pitch in the direction of let me explain in which is .5 metre that is the length of the qdr is the root diameter of the base of the diameter of the humidity is the outer diameter of the field from here you can calculate using this to you can calculate the fan height for this is giving define hi this is giving the finger and this is giving the film with w is more we feel it is is the gap between two friends and p1 is the pitch you please consult our earlier diagram there immediately 151 actually is it direction normal to give direction of clothes and p2 is in the direction of the tubes are heated on the inside by condensing paper that maintains a uniform to all surface temperature of 343 k and cold on the outside by cross-flow up here initially at 288 k flowing at the rate of .914 kg per second what is the total rate of heat transfer and pressure drop across the heat exchanger there are a few things which I like to which I’d like to mention so let’s say it this is this is tubes are these tubes arethe tubes are arranged in equilateral triangular Arabic tubes are arranged like this so this is basically if you join the centre of the dogs then they are in equilateral triangle and then you can know that this angle ABC angle included angle b 60 behalf of the jungle with this is how do you say so this is kind of a struggling think of now you see what happened that what happens that here we are having more number of views in the next column we are having less number of let’s say this is my control volume or this is thethe buses to this then here we get larger vacant space sure if air passes through this let’s see here is passing through this and here he will get larger we can space so most of the year that will try to bypass to these cushions to this problem will be therethe distribution will not be lopsided and we have selected the true struggle and inventive and we will get a good distribution of air through this staggered araminta effective this is one thing which one has to appreciate so here it is told that fitted with call Wells or coming up to see what is dummy half dude that I have explained the tube is heated from inside with the help of a condensing stick if it is heated from the inside by condensing steam generally in condensation heat transfer coefficient is pretty high so what has been done in this problem that we have neglected the resistance inside the tube due to condition because that resistance is very small compared to other assistance through this is also one thing you have to remember and in many cases one can also regular season is dense the metallic tube because metal has got a wire transfer conductivity conductivity and a new thermal resistance compared to the resistance offered by the gas which is outside so basically they’re only registering series but the distance on their site that becomes the deciding factor ok and fruit properties have been give give the deer density has been give me address for city has been given and CP has been giving k thermal conductivity of air that has been in and then random number that has also with this latest go to the next linehow they have how old is popular have been arrived at and what you can do you can also solve this problem of your own though it has been given to taking this problem statement if x is the outside we get at 8.30 .824 m square 4 minutes number this is getting the emacs queries number this is 1.61 and into 10 to the power 4 and you can do all the relations I had mentioned earlier so it falls within the regulations so let us go to the next line nextlight the average heat transfer coefficient now can be calculated show my cell number we can calculate the average national number this kind of formula I have given earlier so this is your licence number and then there are two factors okay to give two factors how do we calculate these two factors quotation for high fin tubes we have used correlation for—to blue fin tubes is this formula and I can’t use is this formula so we have to decide which one we can use and already we have given this formula so your family are what are the terminologies used in this phone then we haveand then from there we have got the average nusselt number from there we have got the transfer coefficient which is 127.31 per metre square building now they smell you having one can one does not have to get hit by hard but hundred hundred 2000 f250 or what per metre square kill me that is the transfer coefficient for air in false conviction at least this much one can remember and this kind of feeling of a number is very important when we we are designing second transfer request then she has to be calculated last year also I have shown how to calculate Philippines in the same formula we are getting defensive point this is also one thing that we should definitely is not on the higher side then we have not selected if improperly we have not selected I feel justified reduced area by the unit or we can get something that is FNB transfer device for a condensation is taking place and it is being cold outside with gear oil changewhere one side temperature remains constant other side temperature and then we will discuss about this kind of heat exchanger indicates maybe from the next lecture on what and then there is only one really stands because all the thermal resistance as are negligible compared to the quayside resistance the formula is equal to this complex or not so complex this has come to a number of steps I would request all of you already you are coming here with lmtd already you are familiar with the how to do the temperature variation estimation so please try to get this expression so once you get this expression then it is very easy delta-t is 45.6 k what is this 10:30 this is the Ducati experienced by here which is passing outside the condenser coil so now your door is equal to 8 830 that is this is the least amount of transfer calculator calculator calculator calculator10 bucks a square that is needed for pressure calculation emacs we have calculated so this is what we can get ka acceleration coefficient I have defined as a plastic mask where and then I can get it as one plus sorry 1.15 stand as it is 1 + something which is a positive on today it will be more than one and it depends on Sigma then kfkf calculation is not straight forward depends on what kind of relationship we picked up a correlation suitable for this based on Reynolds number and the geometrical parameter so we have calculated if then delta p is equal to k a s k a plastic this this is UK a fuller into k a y we have to see how many rows are there so that’s why I was four or number of rows we have to give so that is what we have been and then hofbrau in Maxwell House roadyou did not send what we have done we have first come first describe we have first describe the the method of pressure drop calculation in a finned tube raw and column variable number of females and then we had taken up a problem which is a finned tube condenser outside air is flowing for that we have calculated the overall heat transfer to deliver transfer other and then we have calculating the pressure drop this should give you some idea that how to do calculation for a free ticket so we will we will try to make some sort of a conclusion that that we have started with quit mentation of the transfer and then we defined also what are compact heat exchangers I had told that things are very useful method for augmenting a transfer and they’re very extensively used heat exchanger so that’s why we have spent some time on YouTube with extender differentfinally we have taken up a problem where we could be total amount of heat transfer and depression so there are many such problems and the books which I have described pairing with which we have even referred as a reference so those books also you can see and if any queries are there you are welcome to put up those queries but it is important that whatever I have given the  derivation, you can check those are very simple geometrical deriviations and sometimes it is derivation from LMTD, so please do this as a derivation of your own. thank you.

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