Design and Simulation | Numerical Problem – Part 3 | Alison
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Module 1: Design and Simulation

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Design and Simulation: Numerical Problem – Part 3

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Design and Simulation: Numerical Problem – Part 3





Welcome to this lecture on the design and simulation of heat exchangers. In this lecture we will try to solve some numerical problems in the last class we have solved some numerical problem and we will try to solve the same numerical problem with the different approach. In this problem it is a cross-flow he texting you back exhaust gas is the pool by the water so the parameters are already known to us we know the Lord entry and exit temperature of the gas and the water Street and since the inlet and exit temperature son know we also know the properties and that remained constant it is not changing with the length of the heater single also the overall heat transfer coefficient based on the heat transfer surface area and that is equals to you weighed 100 watt per metre square and what we need to find out is the hot side or the gas side surface area now before this in a book the flights are on mixed so before going into the details of this problem are trying to solve this problem we just want to hear mention what is meant by do it on me because this one time and they can't come that sum of the flu make some pizza on me before you go to the next the q is having this floor I’m in the teams are separated and destroyed this is this train is coming from a particular find getting distributed to different size mixing with the straight on this road is not messing with this fluids separated by this place now for this straight straight this is also coming from some other pipe and gettingthis configuration fine that this new extreme remain on makes the because they’re not mixing with each other they are not able to mix with each other when I see this we have now removed the fence so there is possibility that this works pain after passing through this one there is a possibility that they will be mixing up so this sports team is mixed and this Field Trip is on mixed so depending on the situation we may have bought therelation of the excellent relations will change and that you have to keep it in mind this problem talks about the books on mixed condition now in this celebration of that problem with the crossbow attack which makes we try to solve buy some method and now we are trying to solve it by if lmtd relation because this since this is a conspiracy we know that we can correlate this heat exchanger heat transfer by overall heat transfer coefficient and the area product x the factor factor and the data log mean temperature in this case was known to us both the entry and exit empiricists so if we look at this temperature profile we will find it is across the exchanges so the temperature profiles and not actually known to us but we assume it to be as it’s this is our counter-current exchange where the horse race trainer is entering from this the end and moving out at hundred degree celsius well as the pulse rate is entering a 35-degree and moving out at 125 degree certificates held this being a countercurrent exchanger we would have easily determine the log mean temperature difference by se dil Tak table of the larger 1 – 30 of the smaller / log me in on the 30 of July 30 all thisturn on the coffee and the exit and temperature of the corn field strength so the total height is known in the situation we also know that delta Kellogg me and so that we can easily find out to 111 .7 degree celsius this is thi 3334 is basically one 25th is an outlet that is equals to 103 CI that is equals to 35 degree centigrade so I was talking with this hot in late is 340 called exit is 125 so accordingly if we put this value we will find the 13 of May as 111 .07 degree satellite so now if we go look into this calculation of the affected we had to look for the eighth title on the correction factors for this configuration graduate schemes remain on mixed so in our configuration we have been told that but the phrase ships are not mixing with each other so accordingly we have to identify and we find that will book the footsteps search on mixed conditions this is the configuration of the all this is the chart which will tell us the correction factor as a function of age and we visit the two parameters what is basically the hot in Lake – 3 exit and this cold2 / - 98 the small PA is basically the cold fluid and this is the heart inlet and exit call Linette and cold except for this we have written to zero as TCI here and cold inlet is written as sorry TCI this is nothing but you see I am this is tcu and this is t h I – 300 we find that the director parameter sun is out and the other parameter is PT is tcu – kci / thl let – kcr so cold it is 35 if we look at the sun125 – TCI is 35 ÷ 3/2 in LED is 300 – TCI is 35 so accordingly will find that because 2.34 similarly the art value is the hot entry the inlet is 300 – ½ x 8 is equals to 130 have tissue that is equals to 125 – PCI is 35 and will find it to be early cuz 2.22 2.22 so now we have to find out for articles to 2.22 we have for all the cost of diesel the Angel Houston 2.4.6 like that so here’s some had this is for articles 22 andso we have no almost all the Bellamy close know we know that you the amount of fake getting transferred we already have an idea about the Delta t log mean temperature we have the correction factor so accordingly if we put all these parameters in this equation we cannot calculate the area because already we have this overall heat transfer coefficient also us equals 200 what time is Calvin that’s also given accordingly we can find out this area to be 39.1 m square in our area calculation we got similar almost similar value of 39.1 or 39.7 m square and like this crazy is mainly because of the I mean we have estimatedinaccuracies involved with that man and that inconsistency in this area measurement or area calculation is because of that will try to song the same problem but they will change it to or I mean from a design problem to simulation problem like in the ankle calculation trying to evaluate the heat transfer surface area in this problemthis is like in exchange for it is already known to us it is in service but somehow we have put according to that I will design calculation we have estimated the area to be in 39.1 of 39.7 accordingly we have put manufacturer TextMe across the latest single of area is equals to 40 m square and the overall heat transfer coefficient is still around 100 watt per metre square and as he know we have changed the pump fluid flow rate the gas flow rate from earlier value to 1.5 kg per second and the entry temperature has also changed from 300 to 250 diesel rate now we want to find out how is the heat exchanger formance or how is the exact temperature of the side or the heart rate and the gold flake so basically it is no longer design problem rather it is assimilation problem or the heater performance problem so here working snow is the heated seat configuration is know and what we intend to calculate is the inlet sorry the exact temperature of the fruit strips now if we look so here as I told you that already we need to are already some of the bad amateur so it’s known to us intra temperatureso as I told you that the heat exchanger configuration is already known that means that he trusts or surface area and the heat transfer coefficient is already given the fluid in wet conditions are low the entry temperature and we are assuming that include streams are a human or a minute is already decided that they will be on remaining on mixed and we want to find the out the fruit properties because this will also be necessary to calculate the heat transfer and play assume that the three properties are constant this is this is manoj home to be specifically as because it is 100 kilojoule per joule per kg Kelvin and for water it is for 197 accordingly we can find out the current rate heat capacity so that is the mass flow rate of the corporate and the specific heat of the corporate services for 197 * 1 kg per second so that will come out to before 1970 ID for the heart rate the capacity is image dog and she is so this is 1.5 kg per second and this is the specific heat so it gives us 1500 watt per metre 1500 watt per kg so we can understand that this help fruit strain is the simile and this is the school first train is the C-Max on maximum capacity so accordingly we can ride school 1st Street is that right there is a cost to cement by Sea MAX we can calculate this will become so many calls to 1500 what was Calvin and this is same access 4197 so this will give us see our equals 2.357 now we have the seemingly with us we have the same valuators we already said that the election configuration is known so basically we’re trying to find out what is the n t and q is equals to you way by simian and for the hot fruit strain it is already known the way it is by see me what is the wage you which has been given us call Alisa Lewis has been givenshark for the Epsilon into as it has been shown in the earlier case I mean when both the first teams are remaining on mixed the Epsilon and two relations is can be planted from this relation but now you say we have this ain’t you value known to us listen to is 2.67 now just now we have calculated it to be 2.67 and from 10 until value ideally we should be able to put this value from this relation in this relation and we should be able to find out the value of Epsilon herethank you it is somewhere this is too and this is uncle because 23 in between this is 2.5 so 2.678 will be on the same here so if we go up and also we have the CN value known to us this year yesterday in the previous lives we have calculated in the previous like in the previous success line we have calculator cl equals 2.357 so now if I go to this point 357 if you look at this graph this is 4.25 and this is because to park but it’s finally kcal equals to zero this is equals to .25 and 2 CC equals 2.5 so it will be in which went somewhere and we can find out so it’ll be something like this and here is corresponding value we can see that it will be about this so it will be about .8 to 83 or 84 I mean this is obviously an approximation the approximate value wheel of Time from the graph wireless if we put directly this relation in this relation of that the value of paint you and Siri will directly of fender so once we move the apps I know we can try to find out temperature of the hot and cold so what is the hot entry this is 250 degree centigrade and the cool fluid is remaining saying that that is equals to 35 degrees centigrade so we can and the semen also we have sent this to be 1500 Watts so I can’t even try to find out this q-max and that will come out to be 3.2 to 25 so when we know the q-max we can find the out the actual height getting transferred from this relation because they know the definition of cube is equals to cube iqmax on this is what is known to us so we know only DSL only have calculated we know the q-max we can find out the actual he transferred so that we can find out to be 2.71 because into 10 to the power 5 assuming a value of Epsilon equals 2.84 so when Oxford MS I’m in both the heat transfer now CC x the difference in temperature and c h x this is nothing that c h x the difference in temperature and this is CeCe x the delcatty say this is c h x r t h in this situation what we know is the skew already we have calculated so the scheme is now we know the high temperature record for the most part x cph is already known and accordingly we will find that the exact temperature is 69.3 and the cold will be belong.com.au qt.com ccpc find out so but we also I’m in the same problem when we have designed the heating yet we made her situation that some of the parameters of some of the operating conditions will change or with the time in a deliberately we are supposed to change the operating condition and how the heat exchanger performance is adding with I mean with that change in the operating condition is it’s an interesting palomita so that design problem now will become simulation problem and as we have seen that we had designed the heat exchanger where the operating parameters has been changed and the flow rate of the heart rate has been changed to 1.5 kg per second and now that has changed the exit temperature to 69 point 3 degree centigrade where is the cold inlet condition and I'm in exact temperature has now come tosimulations are at the design problem we had to take help of these apps idling too hard we have to take help of that lmtt or it will be an evil entity and we have to look for the appropriate equation to solve that correlation factor if all we have to find out the appropriate that we have to take careof the application and for calculating the correlation factor and accordingly find out the heat transfer or we have to design on different design parameters. Thank you 




 


 

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