Tubular Heat Exchanger: Shell and Tube Design - Part 2
Welcome to this lecture this is in contunuation to our earlier discussion where we were trying to design the shell and tube heat exchanger. Where we have been asked to design a Shell and tube heat exchanger for heating up the wall water with the condensed water and trying to design it and we have estimated the overall heat transfer there we found that this heat transfer area corresponding to this is 0.20 point 05 M12 and this is 15.01 metre square so that means this ratio AF by AC was 1.34 and this design is within our limit where it was expected that oversized design should be less than 35% so based on this surface area as we have said that we will be working with this 20.25 m square heat transfer surface area know what we need to design is how many number of cubes will be able to provide this kind of heat transfer surface area so fun that we now go to the next slide read this surface area is you have as you can understand that we had estimated this external heater on for surface area and imagine that we have in the number of cubes each and a surface area of 50 so a number of cubes each of length l each of surface area 50 so that gives you the total number of heat transfer surface area as it’s not all because this entry from I didn’t think the main Square stage whether we are arranging them in the triangular page with 45 degree are 60 degree or 30 degree so accordingly thisthis number this one so we had to assume this year is equals to 1 and PT is already given PT that is basically it’s like this if we look into it it’s like this we have for such tubes the OD of each tube is digital given by 19 mm and thisCT and PT by 20 this is given as 1.25 here we have the clearance this is nothing but cleared and see he will be using this town later on this will also be necessary for the clear dances you know so this is what we have been given and no sorry so so based on that this information since we have a single artist you single too fast and we have square feet we have one and we have PT 935 accordingly if we put this one and the value of a 1 also accordingly we will be able to estimate the diameter of the shell so this is this gate field the diameter if you put all these values we will find that this constant pt67 is coming and this way we’ll be able to find out because the soil is known ctp is known ichiro already have estimated it to be 20.0 find the speech ratio is also known as Syria is also known as known so there is the diameter of the shell can be estimated so this diameter of the shell is comingdiameter of the ship so this is also it snow already it is 1.25 we already have said about it now if you look into the previous equation we find that we also try to find out what is that at because this is our power target I mean how many number of teams finally we have to decide so if we do that now we will try to calculate what is the number of cubes so already you have decided the DS so he will also find that this it is related to the TS and if we calculate this one we will find if we put all the values we will find that it is coming approx 117 numbers so we have to take a round of value because it it is calculation will come as 116.48 but we cannot have a fractional number of too so we will be taking the next higher value as 117 so LC can understand that we should have a diameter of typically of point 294 metres and we have to accommodate something like 117 number of such tubes so this is a summary of a rough estimate that we have obtained for this election year we have safepoint 294 will not go we’ll go for nearly about .3 metre diameter of the shellAudi of the team already has been specified to be 19 metre for the cube and it’s it is 16 m 16016 mm and we have also been told about the Battle spacing and already we know about the square feet sowe can all try to find out what would be the actual dimension of the heat exchanger support that what we need to do is that we are having two different techniques one is the current method popularly known and another one is the bell delivery method so we will not talk about this method in this class are in this lecture and we will just briefly talk about the current method where it will be her this is the estimations are based upon the equivalent diameter where we can find the equivalent diameter this is for the speech and we find this is depending on the pitch pitch ratio BT square and then 5 and this is the digital the external diameter of the YouTube so we can estimate is this equivalent diameter please mind that this is not the hydraulic diameter this is we have talked in the earlier close this is based upon the heat transfer perimeter I will wait at perimeter of the heat transfer area so now if we look into the this is the land across the area and this is the sell-side mass flow velocity this will also appear in our calculation of This is how we calculate I’m sorry okay so this is the third time this year as I told you earlier that this is the clearance this is the battle spacing and PT is the beach the beach so the transfer speech and the same side mass velocity this is the last one divided by the band across the area is from here we have to put this value to get the last load is velocity on the safe side let us try to get those values once we are able to get those values we will be able to estimate thetransfer coefficient so then we should also look for this is about the heat transfer coefficient we also need for the pressure drop calculation and this will depend on that mass velocity of the cell site and also it will depend on in me the number of battles which will be used so this is the batten spacing this is the length and NB is Elvira -1 and here if you can look into it just under the lip tattoos in this expression for the pressure drop on the safe side this will be in the class one so that if we have anything on lot of bubbles the fluid will be passing in the plus one number of times over the cubes to banks and we get the pressure drop the related to that this way and also we have this for years as a ratio of the park this is evaluated at the bank and uses that the stillness temperature and this will come over here this this is basically the same size diameter and the friction factors so it will be an expression like this for this is this is that it for the Reynolds number in the range of 400 to 1 into 10 to the power 6 and as you can understand that this is not very I’m in this takes care of the number of battles in it and we had to use a slightly different kind of correlation I’m in for the sell-side so this is how we will get the friction factor and evaluate we can ever relate the friction factor first of all we have to calculate the Reynolds numberRDS the sunshine then we can calculate the friction factor then we can try to calculate the pressure drop on the sales side and we had to see whether it is within the allowed limit so now we can get back to our original discussion so I'm sorry previous snow if we are able to get all this information I’m in theand now we have to rectify or rather we had to go for the letting problem as you can understand that we have done a rough estimation of the heat exchanger now we have to apply the current method or I mean where we have been told about the heat transfer coefficient and the pressure drop correlations has already been given and now we have to select some kind of heat exchanger are there is a guideline by the tubular exchanger manufacturer manufacturers association that is Tamar so there you will find that for accommodating 117 number of tubes as that that is the number of tubes we have estimated so we have to get a guideline from 1000 to accommodate that number of tubes of 170 and obviously I will find that exact number of 117 number of tubes will not be available to the nearest possible number of tubes which is close to 117 but it should be more than 170 so accordingly we will find that there is a guideline and for such design we find that the same diameter and since it’s the it’s the given in terms of a single pass double pass or triple pass and son so here in this case since it was optional that we can choose a single cell or double salt a mineral salt and double to type so to pass weekend think of somethingbut this is 15 and ¼ inch diameter of the CDs and the number of tubes now according to this this is 124the nearest possible one that we can accommodate that number of tubes has already been standardized by Tamar standard so according to that one we find that 15 ¼ inch number of a mean diameter shelf will be able to accommodate 124 number of cubes in one squareis also defined by then and here’s the batten spacing and we have also been specified about the to order that is 19 and the 16 mm is the ID of the cube so and the battle spacing is remaining the same pitch PT it is now slightly changing 2.0 254 mmusing number of passes with two passes was single but now we will have number of two passes and pickles to 20 cordingley this is based upon the preliminary design calculation that we have done now we have gone to the table standard today if I ever design and now we have come to this conclusion that we will be using something like intercourse to 124 Renaissance square feet 1 mm 1 inch thread pitch and the number of passes of course our two accordingly we would be able to calculate all the pressure drop and the heat transfer coefficient as we have suggested in the previous slides that how to estimate the heat transfer coefficient depending on the DL you we have to estimate the devaluation we have to estimate the as then we have to estimate the chairs from there we will be able to calculate the RDS and from there we should estimate the overall heat transfer coefficient and the pressure drop by patient and then we will be able to justify whether that liquid pressure drop and the heat transfer is achieved or not and then we will find that if it is within the 11 limit we will conclude that the design is perfect otherwise we have to choose a different configuration for the standard TEMA standard and do the calculation again, thank you