Design and Simulation | Numerical Problem – Part 2 | Alison
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Module 1: Design and Simulation

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Design and Simulation: Numerical Problem – Part 2

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Design and Simulation: Numerical Problem – Part 2
 
Welcome to the lecture of heat exchanger fundamental and design analysis. Today we are going to solve some numerical problems, this is in continuation to our earlier discussion on the heat exchanger design problem. In the last class but we try to solve is a numerical problem based on the lmtd method so we have different we have learnt about different techniques Ltd limited interior silent excellent relations and now we want to utilise all this techniques to solve different detection the problem and depending on the knowledge of the out the values given for the Citrix problem we have to find out or apply different techniques to solve different try to solve the problem that we have discussed in the last class in the last class we tried to solve this problem where there was an industrial gas turbine heat exchanger and the heart of oil was called by the water in a concentrated heat exchanger and this was the properties which was given based on this heat exchanger problem so we have tried to estimate the overall amount of heat that is getting transferred 68524 what was the amount of heat that was very graceful and from there we try to calculate the cold exit temperature and that by we found that our assumption of the average temperature that 35 degree centigrade based on which all these properties were evaluated we have been able to find out that this is the average temperature this is the exact temperature and the average temperature is coming to b35 so it was a good case and based on that we have solved the Delta tlm the overall the log mean temperature difference and we had also based on the geometry of the cube we have been able to find out the internal heat transfer coefficient as well as the external data secure patient please look into this problem that in this problem what was given was the internal diameter and the outside diameter were both given so all the unknown parameter was the length of the cube and since the internal and the outside diameter when onto as we have been able to find out the heat transfer coefficient both internal and exit temperature we have been able to find out because our tube dimensions the diameter of the tubes when known to us and based on that we have been able to find out the overall heat transfer coefficient so from there we found the length of the exchanger and it came to be 66.5 metre it appears to be pretty long and obviously you can understand that it cannot be simple straight tube obviously we have to make a calyx like these are we have to make a call like this through enough through that it has to be a counselor to you and he will have to make a call of it and as soon as you make a call of it you will find both its internal heat transfer coefficient and the overall am in outside heat transfer coefficient of the annular space that will change and that will of course change the overall heat transfer coefficient so it will become a different problem but we are not going to solve it applied the lmpd technique to solve this particular problem look into that we want to apply the Epsilon into technique so what is known to us but has been given it was like a low in the earlier case we have been given the value the flow rate of oil and water is known to us and the entry temperature of the hot and cold weight is given so we have been able to find out this average temperature of the oil of the hot plate because we know the exact temperature of the soil and it is coming from hundred to 60 degrees centigrade and the cold fluid is coming from 30 degree centigrade and it is going out to 40 degrees centigrade so like this this is the problem statement and as usual we have assumed some exit temperature for the cold outlet so that we can evaluate all the street properties and based on that we have been able to find out the exact temperature of the cold outlet it is it has come out to be 40.2 degree centigrade is on this temperature now I have all the food is the temperature known to us and also below the flow rate and CP so based on this information if we look at we will find that we can find out the what capacity freight the capacity of the hot fluid and that will come out to be .1 x 2 131 so it is 213.1 and we also have the call freight heat capacity so that will come out to be .2 x 41 78 so that is equals to 835 what part made so we can understand that this is the same animal the minimum heat capacity this is the maximum weight capacity fluid so here we have the pulse rate as the maximum weight capacity fluid and the heart rate is the minimum capacitively so what would mean is that what it means is that the quality will find that this will come depending on its capacity it can approach this outlet Amit the inlet of the heart rate will reach to the inlet of the coal fired at the most so this is the maximum difference in temperature that can happen on this is the Delta t-maxx that is possible with minimum capacity fluid so if we now want to find out the effectiveness of these exchanges we have to write it in terms of its silent because two cube IQ Max and where we can find out it to be this is the heat transfer actual heat transfer that is taking place that is here we had assumed it for the minimum capacity fluid we have written in terms of the that is the minimum capacity fluid is the wireless the minimum capacity fluid and it has come out today find 5714 is the Epsilon of the heat exchanger effectiveness now if find another parameter that is the seeming by C-Max that is equals to see her or the ratio of the two his deposit is the minimum and the maximum and these will come out to be find 255 that is basically an issue betweentake me to the ceiling and C-Max and that will come out to be fine to 55 so here’s this is the seeming that will come here and this is the same acts that went come in the denominator so this will give you a ratio of .25 point so now we know CR and Willow Ave Salem so now we have to look for a suitable relation that will tell us the NQ so this is a counterclaima relation like this and take us to 1 by 1 – CFL this part here in this equation now we see that we know that we know the asylum so we can have an estimate of the excellent so this is already known to us so this so you a we can also write it to me you into it I ain’t well that is the area we’re looking for through which he transferred is taking place and what are the parameters we know we know the overall heat transfer coefficient you over early transfer conduct means we no see means no idea is not end until you are ready we have been able to estimate so we can now so we have these values are you known him in his nose and we can now try to estimate there and you will find that link is coming as same as default 66.5 m so here in this little be able to find that book the excellent you are a limited approach is giving the same estimate of the overall heat transfer it may appear in practical to have so long heat exchangers but we find that this land is necessary to if we have it where she’ll Stevens kill heat exchanger of given d I n g o so this yeah that is across that I treated you and it is an active crossover and no more details have been given and in this exchange is this hot exhaust gas is heating up water and we know this floorbecause we know what the inlet and exit so we know the average temperature and we can even evaluate the food properties at the average temperature so it has already been given the specific heat of the gas side through it is 100000 joule per kg calorie and for the waterside it is 419 South Hill park education and the entry and the exact temperature are like this the whole class is interning at 318 cool 200 where is the hotel is getting one in the top 35 to 125 deluxe so with this information if we now look into it the overall heat transfer coefficient based on the gas side surface area is also known to us and that has been given to me 100 watt per metre square so this is already known to us as is and understand that we have not been given the details of this thing to transfer exchanger what is the cubes and diameter what is the number of things to be used so instead of telling all those details to find out the internal heat transfer of the external transfer coefficient of heat transfer coefficient has already so what we are supposed to find out is this is like this we have that in the heart coming here this is entering at 303 get this is coming out to be accurate and this is installing 5 degree centigrade is what is coming in and this is going out at 1.5% what we need to find outyou have the specific heat known and we need to find out the gas site but any support this problem let us try to see how we can process it so first of all if we look at the poultry what is the cold feet the cold fluid is is water water is the cold fluid and its properties announced road it is also known what is the flow rate insidecc is equals to 1 kg per second what is the CPC CPC is 4197 joule per kg and if we try to find out the CC the cold capacity that is MC dot x CTC that will come out to be one in 24197 Wattpad so this is one parameter then for the walk through it now if we try to find out for the hot plate but will find this NH dot is not given you do not know so we do not know this image top but how to find out ch then we know ch equals to image.com this is what we need to find out so far that we will use another relation well we know that c c x c out Linus eci is equals to c h thn – 38 out from this relation Outlook that the CEO is not this year is known all this temperature sun and also share of 10CC so we can find out CH2 BCC that is equals to 4197 and x PCO PCOS how much 13125 – 35 / 300 – PCOS 111 o&cc that is equals to 4197 sisters will come out to beexplain the difference in temperature between England and Ireland temperature so obviously the fluid which is experiencing more difference in temperature of larger change in temperature so that will have the minimum capacity so that means we have CR temperature of larger change in temperature so that will have the minimum capacity so that means we have cement buy cement because 21889 / 4197 and that if we give it will come out to me .45 so now we know we can also find out the queue disk you will come out to me and I see that is the actual kid getting transferred CPC TCL – TCM that is equals to 1 kg per second and then we have 4197 then we have 125 – 35 so this will come out to be 77 into 10 to the power 5can be obtained so that is equals to be hot in minus PC in so this comes out to be we all the cement cement we have obtained to be 1889 Watford Mater hospital and this is it hot in is 300 – PCI that is equals to 35 degrees centigrade so so we can now find out what is that effectiveness so this is cube IQ Max and it will come out to be .75 so when we know the heat transfer coefficient then I know that the exchange now we can use the application that we can estimate the ECU so once we know that you we will be able to find out and q is given us and you is you by Simi and we have already been told about the you and seemingly has already calculated and if we can estimate the NQ then you’ll be able to find out that now we already have the knowledge of Epsilon 3.75 so now we need to find out the appropriate qualification for the heat exchanger and will come there we have to obtain the issue now we know these exchanges to be across from it and there was another information that has been given for the exchanges that are on mixed fruits bothand again inside this one that sent you to the PowerPoint 78 so this equation cannot be rearranged to calculate the NQ from the excellent value so but already we have calculated the end we have calculated the island we have the hair salon salon on for us we have the CR value we have the sea minimum value also with us but we intend to calculate the nth you but this relation will dynamically give you the Epsilon not the NQ so we cannot rearrange this equation also so we have to take help of the equation or the graphical solution where we will find that the graphical relation for the excellent assistant you this is absalom was assigned to this can be obtained for different seeming by C-Max RC perseid values ranging from 0 to 1 and this is for Ciara call Cheryl this is 4 cl equals 2.25 year equals 2.5 and plus we have CRT cuz 2.45 so corresponding to the articles 2.45 and it’s allergy cause 2.75 side equals 2.75 minute finding that the accused be someone here that is nearly about .21 so if we consider this 2.21 2.1 so we can calculate the surface area and that will come as 39.7 meter square. Thank you.




 


 

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