Tubular Heat Exchanger | Heat Transfer Co-efficient | Alison
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Module 1: Tubular Heat Exchanger

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Heat Transfer Co-efficient

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Heat Transfer Co-efficient
 
Welcome we are now trying to evaluate the heat transfer coefficient and pressure drop in tubular heat exchanger. Let us  quickly look into the configuration or what we have learnt in the previous classes. So we were basically trying to find out the heat transfer coefficient on the sell side because already we have talked about the tube side heat transfer coefficient another case for the internal floor and now we're trying to look for the external heat transfer coefficient of the fluid flow when the fluid flow is taking over the tube and we now want to find out the heat transfer coefficient for the configuration so this is particularly relevant for the fluid flow over the tubes in the cell site so there we have tried to find out the if you remember in case of internal floor we have defined the number by cdh by by DHS the hydraulic diameter and she was the mass velocity given by the mass flow rate party unit free-flow area and that’s all the rules number was defined for the floor when the floor is taking place true that you insidewhy talking about the external flow or flow over bank of tubes evil looking for the maximum velocity when it is occurring at the minimum total area of tissue area and now we can understand that this has been done because we wanted to calculate the least number based on that maximum velocity and we’re defining the Reynolds number by Rosie xth buy new 12th is the external diameter of the Earth and looking for the maximum velocity in case of an ellipse quicklyfactor is given by this relation and this is very fast this Romans number we have also learnt that for our greater than to introduce the law for the friction factor values like this we have also told that The corrs on chill factor which is nothing but the Stanton number into products from the number to the part 2 3rd is equals to this relation and this is dallas-fort the turbulent flow turbulent internal flow through a circular queue I will quickly go to the other two configurations that we have stopped about whilst talking about the external so now we’re talking about external floor and this is the inline arrangement of the queues where he is the velocity and we have estimated the zmax to be the frontal area / the a mean that is equals to this is the minimum area a1 and finally it will be given by is still buy is 2 – 10 in velocity so this is for the inline element of the to know if we look into the other elements that is possible I mean that is in the standard condition we have seen that the tubes are in this packet condition and if we connect the central between this will form a kind triangle this is what is the280 – be less than St – DST can be related by ordinary but she will be able to relate the stsd this is the diagonal pitch and it can be related to the longitudinal and transverse pitch by this relation so this isis there by dej loaf ft by 20 ft is the transfer switch the zero is the outside diameter of the tube and a cell is a longitudinal teach and delivers outside diameter this ratio that even on this side and this is value no esta by DJ Loa swelling on the side so we have c&n for each configuration this is valid phone number of rows more than that means when the floor is taking place like this in a we have in number of n equals to 1 in equals to 2 so like this we have number of rows and along the direction of the flow and it has to be more than at least 10 or more than 10 so what happens if the number of tubes are less than 10 then obviously there will be some kind of correction factor that we have to take into account so we’ll come to that part later but as you can understand that these are some discrete numbers that I mean it is not necessary that the a sell-by gojira will come as 1.125 on St by The Zero will come as 1.25 or 1.5 so we may finally wrote depending on the situation of the problem we will find that we had may have to interpolate some of the numbers so I will now go toso here we have some numbers and here we have similar numbers for yesterday by busy this is the wind chill factor related to the Reynolds number this number is based on aditi Max versus rdt Max we have already learnt that order t-maxx is related to the maximum velocity and we have learnt how to do since roti Max in 2D or the external diameter or 20 / meal and we have already learnt how to calculate the minutes so this is the only Asian given for the two bands was there that uses identity in the staggered condition so we will look into as we are talking that this is something like some desperate position that a cell biology dosa for example is this number is .6 and .9 know if by chance if something is coming at .85 or Sydney it is not defined here so let us look into this position say between 1.25 and 1.5 if someone is finding that his number is 1.4 she may find it difficult to find out exactly what will be the number for the sea and an ISO as an alternative we have I know that I’m in setup correlations given in this heat exchanger selection brittingham Kamal designs for 5 get off we find that here this is defined in as they pass functionand bearing for relations for the inline tube arrangements so depending on the Reynolds number of mass we will be able to find out what are the heat transfer appropriate nursing number this is related to I’m in this relation is given by in terms of the at least 10 configuration the tube surround installation the different responding number b and donell number wbl different boot is the prime number but this is evaluated at a bulk mean temperature this is evaluated at the wall temperature so so that’s the difference we have in this two numbers and this LED also again this is evaluated at the Vulcan temperatureso if you’ll also find another time that is seeing here also we find it number seeing and in the area coalition also you will find that we have the decency and that she is scared of the number of legs if it is less than 10 it will be some type that but if it is more or more than 10 or equal to 10 then it becomes 21 so that's so if we know we have the correlations available for the cube science I mean sorry that’s inside the tubes are arranged in either in inline configurations are planning to integrate conditions so based on this information as we have said earlier that we had to not take account of this seeing how to estimate in CN will now find that there is a graph when we can coordinate to seeing this is seen as a function of number of tubes so we see that if the number of tubes are one but we have second kind of see and value for different order values also if the order value is betweendecline in sport the staggered condition and the continuous line is for the inline condition so if we have the tubes arranged in inline condition then we should look for this car is there in the standard condition depending on the Reynolds number with either we look for discounts or will look for discount if the Reynolds number is more than 20 separate we find that installation what I mean is this graph will tell you what will be the correction factor corresponding to number of rows equals to 2468 10 and if it’s more than 10 you can see it was almost reaching to value if it is said 40 number of rows it is almost 1 if it is about 13 it is .99 it is about this point and so on sonumbers 9 depending on the number of tubes if it is less than 10 or 12 then we can accordingly take care of that correction factors that we have to incorporate within that correlation that has been given in this earlier slice so now we will try to quickly look into a small problem that we can try to solve here it is saying that atmospheric air is flowing across a bank of struggled tubes and the number of tubes at 8 so they’re in the garage tonight Rose in the along the flow direction the diameter of the tube is given the longitudinal spacing and the trunk processing is also given the extreme velocity is specified and it is flowing at a condition of 20 degree central it so if the surface temperature of the team is maintained at 90° c then we’re supposed to find out the average heat transfer coefficient so let us try to solve this problem so what are the things that has been given to us we already know that we have some number of tubes how many number of players are there is like this we had 123 45 6 and we have to700 flow direction is this the velocity v is 6 metres per second the diameter of the tube is already specified its diameter is 1 mm 1 cm and the longitudinal spacing that means the centre to centre this distance is how much 1.5 cm we have the transverse spacing that is this to this is 2.54 it is not to the skull and it is looking the smaller dimension is looking more than this one but it if it if it is properly or even find that this is widely spaced ok and 47 is that this way is flowing at the temperature be in Finnerty is equals to 20 degrees centigrade and the samples of these tubes are maintained at a temperature of 180 degree centigrade so what we know now try to find out is an average heat transfer coefficient for this configuration so basically if you look into this configuration you may find that in children tube heat exchangers we may have this kind of arrangement of The cubes and ovens with the fluid flow is taking place and we’re interested to find out the heat transfer coefficient for this configuration configuration of a specified RV supply use the heat transfer coefficient but in reality we may have to calculate it and we now try how to calculate this kind of heat transfer coefficient so now what is the starting point we have already been given 1.5 CST is 2.54 cm and the metre is equals to 1 cm and first of all we had to find out the fluid properties refinement the fluid properties at 20 degrees centigrade and the density is 1.2 double zero 32045 kg per metre cube then we have the CP this is equals to 1.005 kilo joule per kg Kelvin then we had the viscosity 1.82 into 10 to the power – 5 Newton second per metre square then we have the permit number equals 2.713 and the thermal what can you tell me and also this is pill number p and evaluated at the bulk mean temperature and we are the blue that is the wall temperature that is 180 degree central get this comes out to be point 685 soap these are the values we know the number of rows r8 and the velocity b is equals to 6 metre per second so with this information now we try to calculate the different parameters what are the different parameters that you have to find out first of all we calculate the diagonal pitch that is equals to root over aircel squared plus St by 20 square and square root so this comes out to is 1.95 cm then we have is d minus d that is equals to .95 m you can understand why we’re trying to find out is the minus key because we do not know exactly at what location that minimum of a floor area is there or we are other trying to find out why the maximum velocity is occurring so then we’ll try to find out is t minus d by 2 and that comes out to be pt770 we see that is d – 30 less than st-line SD by two is 2.95 and this is 577 so this condition is not true so this is not I made less than so we can understand that the minimum area is occurring at a 1 and we have to evaluate the velocity D-Max is equals to s t y s t minus t and p so this will come out to be is tiwa Savage ft is 2.54 and 2.54 – 1 into 6 this will come out to be 9.9 metre per second so once we know the velocity maximum maximum velocity then we can calculate the re Max and that will be going to roll into remax in 2D / meal and this becomes 9.9 * .01 is a diameter and 1.2045 is the density of the air ÷ 31.82 into 10 to the power – 5 so this gives you the reverse number six fivelook for the appropriate if you go back find the appropriate correlation for this condition and even find that the number because 2.35 into PT 98.9 at coming for seeing because we have any calls 28 and corresponding to any calls to wait we have to look into that chart and there will have the correction factor points 98 then we have Rd to the power .6 this is corresponding to an earlier cuz 26549 so there will find that the organisation is it to the 4.6 then we have been calibrated and the Walkman temperature that we come at point p o p p o p to the power point 36 and then PRC by prw that comes out to be .25 and then you have is the buy is longitude known as 0.2 so this when you evaluate it comes out to 66.30 once we know the missing number this can be related to HD by k and we have the value of k we have the value of b we can find out the heat transfer coefficient that will come out as 170.4 watt per metre square Kelvin so engineers that value which may otherwise supply different transfer calculations now you have learnt how to calculate it from different correlations and the expression of the heat transfer coefficient we will now be able to find out. By that I mean different or other parameters which are necessary. Thank you for your attention.




 


 

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