Fixed Bed Regenerator: Numerical Problem – Part 1





Welcome to this lecture, we are talking about the fixed bed regenerator, today we will try to solve some numerical problem we have already stated about the problem in the last class and this is about the heat transfer and pressure drop estimation for packed bed regenerator with the spheres, spherical shorts or it was something like this it has been made of some late shot and its diameter is specified specified I’m in the porocity has already been specified and it was inside the pipe of diameter 154 mm and length 159 15 mm so the gas which is used in this is nitrogendesign why we talk about the recent assimilation not design will find that these two parameters will be essential to god it will be necessary to estimate the design or the or it’ll be necessary to simulate the originator so if we now first concentrate on this packed but with the sphere and with this particular problem so we have been told about Sutton correlations that we have to use for this practice and it is the heat transfer is expressed in terms of the cold ones affected and then we have also the friction factor given for different riddles number reasons for our 11004 rzr 1000 so we can understand that which has three evaluate the Reynolds number and then only you will be able tospectacles spectacles spectacles face so we have to find out what is the mass velocity and that will be based on the free floor area or the fertile area and velocity so then we have the heater is the surface area that will also be necessary particularly and it may be different for the practice field it is if it is air made out of the word screen self hear this is the finally this is the expression to be used for the pressure drops particularly for this problem if we now and look into we will find that we have the order value of the equivalent diameter and then see the mass velocity now we’d like to calculate the first the DVR the equivalent diameter so the parametersthis is the equivalent diameter so once we know the equivalent diameter we will come to the mouse LEGO city part where we know the mass velocity vs .80 kg per second then we have the city that has been given us .38 and the frontal area of yesterday you can understand that a frontal will be four 5 squared by 4 so this is what is that and this is of course the this is like qhs field with this kind of face and the front of area is obviously the surface area which is nothing but why disturb iPhone so if we now calculate the free-flow area which will be the central area x the city so this is by dispersed by 4 we have been given this internal diameter and if we put this valuable find she will become 11.11.13 08 kg per metre square into second so this is about the chief hello so we have the value of this equivalent diameter we have the value of g and what we need to know is the which the city part of the freight so that we have already been told about the fluid the fluid is nitrogen an average pressure of 200 200 and it’s temperature is pressure is already known so and back temperature and pressure we have to evaluate the velocity mint properties of nitrogen and we will come out to be we will put its value its 12.95 if we know she is equals to 11.30 wait x equivalent diameter was 6.54 into 10 to the power – 4 and then we have the meal equals to 12.94 7 into 10 ^ -6able to decide whether we will use the correlation we have to use for the friction factor and we already know that the correlation for the cold wind chill factor so now if we go back to our discussion now we will come back to the heat transfer coefficient in the form of contract and the school when you factor is basically nothing but point to 32 l e to the power point 3.33 x and y to the power 23 so she will also put the other parameters and then we’ll see how we go about it ok56 so this is nothing but the Stanton number and PIN number to the card 2010 number is agcp so we already have an idea about the severity of taint now we also know from the field properties of nitrogen the value of sleepy so we can calculate the heat transfer coefficient that is what we are looking for from this relation so this is already known 5034 256 and then that is the cause to 7 into 10 to the power 3 coolpad classical what’s the value of 6 and 1/5 and we have also the value of a number p is equals to .7 to 5 sixand accordingly we will find that the heat transfer coefficient will come out to be 50 2.7 watt per metre squared coming so this is what is needed for this feature is that that’s what we are looking for then we will go for the friction factor because we have seen that the number is less than thousand so we will put it in this relation and from there we will be able to estimate the friction factor so the friction factor is coming to be 2.346 if you put the Audi is equals to 571 that’s what we have estimated for The this tablet and once we know the friction factor we can put all this time is it are here for the friction factor already have known as it has already been openedOlympus Pointe 914 metres and then she value is known the equivalent diameter we have calculated the density sensor the nitrogen is flying at a temperature of 8 at an average temperature of 200 Kelvin to atmospheric pressure the density of the gas is also known and does density is equals to 3.28 and Casey park medical store install the silence in this relation will find that this will come out to 6117 3 kPa to pa and that will be equivalent to 61.1 or 200 kPa so we can estimate the pressure drops for this the fact that for a floor out of the gym product of .8 points are the product was given .08 kg per second we find that the estimated pressure drop is 61.2 kPa 9 that we have to order pending on the regenerated matrix material we had to calculate the pressure dropped and having the heat transfer coefficient as it is necessary for the design on the simulationso now you will go back to our discussion on the simulation of the regenerator problem you have taken it from the balancer Chrysler transfer book and it’s about time regenerator fix battery generator it is made of 7.182 kg metal spheres and its material properties unknown so it’s both the specific heat and thermal conductivity of this material has already been given the paucity of this is also known and it is made of packed with face and heating cooling periods are equal and the total time period is for second so the hostility is too II and the cold period is also have two second the fridge showing through this system is is helium so it's city is also given and thedifference for patients with did not calculate it has already been specified you can understand that depending on the situation or like here already the flow rates have already been given and we need to calculate in as we have specified just we have to use similar expressions are to calculate the heat transfer coefficient on this end transfer coefficient on the cold side and for the hot streak and the core strength and it’s hidden surface area is also known as you can understand but producer producing regard to the heat exchanger design process where art design of assimilation process are particularly this is where talking about the simulation where we find that the simulation is about the situationwe need to talk and we have been told about the other talented so understand that you have to find out is the generator and what will be the process to do that so we’ll discuss about it so we need to find out what is the capacity and for that we need to find out the value of the castle and already it has already been told the city of India has already been given so we can understand that we will be able to find out the sea mean so the heat transfer coefficient and the heat transfer surface area is also given so that you can also be calculated so won’t you we'll be able to calculate the stop here wants to knowdepending on the value of the CR if it is not the one we had to find out an inclusive and 10-q and equivalent cm each other that the situation so depending on that we will be using this chart for the given is the function of and 21 cm and that's a no on the basis of aid to effective values and see if effective so we will try to find out if silent and once we find out that the salon we know that we have to calculate effective exercise depending on that he will be able to find out the correlation will be using this correlation to calculate the regenerated effectiveness of that is to be followed will go one-by-one in this case two two bananas what we do is pass the parcel go back tothis region of CP is already given so first of all we will find out in the ch so that is the cost to image and say it and that comes out to be pt04 is the mass flow and then we have the corresponding 5.20 the specific heat that is this many kilos should park Casey Calvin and we have this is Casey park this is KG per second so if we multiply these two parameters this comes out to be point 2080 10kw but this is kilowatt per Kelvin and we can also calculate the CC so this will become empty and dark and then CC then what is CCNA data slightly less and wheat is 0.3 62.03 6 * 5.2 kilo joule per kg Kelvin and then you have for this one the spelling kg per second so this comes out to be point 1872 kW power Kelvin so you can understand that this is a female and the cooperation with the same and Floyd and this is the same acts do it so we can all try to find out the CR that is equals to c mean by C-Max and that will come out to be pt-90 so this is not equals to 1 so we know that the capacity weight ratio cmu college where to find out we have to also find out the NQ equivalent so we will find out the appropriate relation to calculate the CV and a few effective and then will proceed to the calculation of the 1 by 8 + 1 HHH but you seem so this is the value ofFreeman and thus is nothing but this is away is 198 cache2 the part this one so in this case we have seen that both the hcsc and a h at the same as the AC is equals to 8 h is equals to 8320 and ac is equals to a h is equals to .950 so basically these two terms are getting angry and the NTU becomes save one by what is the that 1.18 7.2 this kilowatt per Kelvin we have converted into Watford Calvin and then we have this to buy this 8320 * .95 so this whole to the power minus one will give you the value of into vehicles to 21.11 so once you know that you will you will be able to calculate the effective value and that into effective value will become who likes to use that expression that we have already talked about so the nqf ekta value comes out to be 20 okay and then we have also now we need to as we have said that matrix capacity weight ratio that cm that is the calls to it is 1.882 * .840 and that is the total time for and then you have to / 0.18 72 so then we have this will come out to be 2.111 the equivalent cm then you will find that it is coming nearly to be 2.0 and then if we now have the value of this effective and q and the effective value of c em that is equals to 20 and 2 and the compounding value for the excellent that will become from there we will get an Epsilon is equals to 1935 so we now have an estimate of the rough estimate of the effectiveness because we have used that chart which corresponds to see Eric was 21 but since this year is not equals to one for us we had to find out this effective and you for responding to this effective see Amy and listen to effectively and that we are getting the silent one so as we have said earlier that based on the value will be calculating the perimeter x and the perimeter is comes out to be 1.5 184 and you already have sold it about the expression of x and from there if we’re not put the value of that one and the corresponding it's all done with the band minus tax services nothing but if we put the value of x in this expression they will find that this is 9727 so that originally that is 97.27% effective in its ineffectiveness is basically .0273 or 2% 2.7 3%. So this is how we estimate the effectiveness of the regenerator or simulate the performance of a regenarator in a nutshell. Thank you for your attention.