Plate Fin Heat Exchanger | Numerical Problem – Part 3 | Alison
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Module 1: Plate Fin Heat Exchanger

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Numerical Problem – Part 3

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Plate Fin Heat Exchanger: Numerical Problem – Part 3
 
Welcome to this lecture we are trying to solve the numerical problem related to plate fi type heat exchangers this fitting problem out of the performance evaluation of the heat exchanger. So until the last class we have tried to estimate the heat transfer surface area and also we have evaluated the mass flow rate corresponding to their site and the gas side if we look at we will find that we have estimated the mass flow rate of air and that is equals to 2.0 kg per second and then we had the corresponding value for the gas side was 1.66 kg per second so this must flow rate of air and the muffler rate of the gas when we multiply it with the corresponding the cbcp valueswe don’t know exactly what is the expected temperature of the air or what is the exact temperature of this gas so that’s what we’re supposed to find out and as we have seen in the earlier case is also we have to make an rough estimation of this gas exit or the air exit so that we can take the fruit properties or we can get the fluid properties at the mean temperature so should we go for any wild guess of the should we go for some kind of all reasonable estimate of the make an estimate of the gas that will be close to the Earth understand thatrcpa not very different they’re not really didn’t I mean dependent on the temperature in that case we can say that this energy or the gas side is the minimum capacity fluid or seeming will correspond to the gas and accordingly we have to we can try to find out in the field properties based on the assumption that leaked we assume that this is heat exchanger effectiveness is .75 so sorry this gospel type heat exchanger this is the assumption we have we are starting with this exemption will lead to an estimate of the exit temperature on the gas side as well as for the inside and once we know theproperty and then once we know the properties to even try to calculate the Reynolds number and the other I mean other properties so if we are you miss Allen 2.75 will be able to negotiate them with the with the headaches the temperature is it temperature Epsilon is equals to .75 and since we have resumed similar specific heat for the air and the gas we will know that the gas is the minimum capacity fluid so we can say that tal of the city city gas of the inlet – steel gas at the exit is the minimum capacity and the SES PNC of entry - 3 years you will find that Epsilon that is equals to .75 and 30 is equals to mine is t a n into mass flow rate of the Earth divided by the PCR minus t a r that is the maximum difference in temperature x the minimum capacity and we have assumed the CP of the air and the city of India and so this is here in this relation will find that we know the entry fee into a gas entry entry and Amy Bayer AG ratio is already known to us so we will be able to find out there exit temperature this will come out to be 635.8 degree certificate so we show an idea about the air and gas exact temperature so we can now calculate the mean temperature and the gas 5 mins temperature will now come out worship will now come out to btm that is equals to 900 + 375 / 2 laser arithmetic mean .5 day you sent today and for the airside we have the main temperature as 200 it was its entry temperature and 635.8 is the estimated temperature based on the edge salon because 2.75 so that gives you 417.9 degree centigrade is the average temperature in this temperature in temperature we would be able to calculate the fluid properties of the emit different field properties like what does the word properties we need the new Winnie the CP we need we need the pin number and the number of years because of the permit is what we will now get ready for ENT for gas we have meal because to 40.1 into 10 to the power – 6 Pascal second is the unit and then we have this is the means that we have CP that is equals to 1.122 kilo joule per kg Kelvin then we have CPR bls CPR class so CPR where is equals to 1.073 kilos Casey Calvin and then we have paid for this one is 2.694 sources of the values known for the gas and the air so based on this early we would be able to now calculate the gas velocity what is gas velocity for density or mass velocity for the gas side so this is equals to MG don’t buy a free-flow area for metres Katie services kg per metre square per second this unit is KG per metre square second and simile they can also calculate mass velocity for the air and the most velocity for air will come out as m dot / 800 already have able to calculate the mass flow rate that is because 22 kg per second and this is just nothing but point 10 .1089 metre square so that gives you 18.365 kg per metre square second so these are the mass velocity of air and the gas side already known to us so we know about The gaslight mass velocity and the inside is philosophy we would be able to calculate the hydraulic diameter corresponding to The gaslight RDC this will come out to be gcdh by me you and this is the view of the gas so this will come out to me if we put all these parameters this is equals to 15.342 * .00154 / view of the gas has been taken as 40.1 into 10 to the power – 6 so this will come out to be 589 similarly if we travel to estimate the RV for the inside we will find that this is a move by MC th by meal of the year side and if you put all the values it will come out to be 842 so now we have the knowledge about the Reynolds number side and back inside the corresponding to these values we will try to find out when you go to the nurse that damn well we will be able to locate the this is nothing but the standard number of x phone number to the park two-thirds basically this is the chair and this is theand this is the house number so we have the house number here and please note that this is x 10 to the – 3 corresponding to the at 5 at 9 and 8:42 will have to find out the key factors so the gas will have a shape factor given as corresponding to point 017BA part 2 3rd so since we know this Jay we have already estimated ccpp all these balloons unknown so4j corresponding toCUDA 17 and we can try to find out the heat transfer coefficient of this is for the gas side and you will find that he will come out to be true 60.83 what per metre squared Kelvin similarly we need to put the value of the heat transfer coefficient corresponding to a site where we have the J-Lo given as .013 for rent if we put all those values you will it will come out with 36.82 1 watt per metre squared Calvin so now we have the heat transfer coefficient for the assigned and the gas side and two wheel using the similar kind of heat exchanger services we have different heat transfer coefficient as the mass flow rates or the volume it recorded seven different and that corresponds to different re-file you and that corresponds to different heat transfer coefficient for the heater on Armenia side out the outside so now we have the knowledge about the heat transfer coefficient next day so we need to find out what is the overall fuel efficiency since the frames are connected between the plates we need to also know what is the efficiency so that our if we can put that one by the way basically if you remember that in the last few years we have told that this is the heat transfer coefficient x the heat transfer area x the overall heat transfer coefficient plus this will be won by a c and then AC and heater for Lucy and also we have the wall heat transfer paper I’m in the resistance estimated for the work so far this one we will go late later and already we have evaluated HIV have already evaluated a c&h this all this parameter is unknown and we now need to find out this the overall heat transfer coefficient for the gas and yes I saw that what we need to do is that we need to calculate the MC that is if we remember that this is not the Florent this is related to the field where that in the field efficiency will be termed as 10 hyperbolic m into l ÷ well and hear this I'm will now try to calculate because we knowhe is the heat transfer coefficient because they limited ke is the thermal conductivity of the material and a is the cross-sectional he transferred or sectional area30 so if you put these values for this particular situation we will find that this is going to be 28 by care of the frame material and then we have the Delta that is the thickness of the faint and then we have one class delta by LS thatKrista lenz length of the train and then we have a whole root over so this is what is the height I mean this implemented for this ETA on the feeder fishing seat and we have a book for the gas side and the gas side and the inside we can calculate from this correlation so that will give her this NTR to be 634 634.94 and maa will come out to be 615.37 hours so she has had the support LA and tell us that has been taken as a b by 2 minus delta that is equals to 249 by 2 - .1 0-2 and that comes out to be 1.143 millimetre now the question is why did we take this by 2 minus delta if you remember that in some of the classes we have talked about the half an ideal ization there’s that half of the field is as it attracted to this plate and half of the field is that as attested to this the upper plate andplay sight into this in details when we talk about the Manchester play phonetic selective form into you do I have that this happened idealization works are good for a man to step it is new cases and we have this earlier cuz to answer BAC is equals to 3 by 2 minus Delta it will be .86 57 so from there we can now tell plate heater overall for the gas side it will come out to be 1 – 1 – ETA a&e services 8880 and there comes that kyf – ya so this is nothing but this parameter is also given the in our calculation arm infinity specification so this comes out to be point doubleso you can evaluate it using this correlation for this it will be basically 1 – 1 – ETA she sorry ETA 840 x a fye and that will come out as metre 04 theor we can go to calculator the other blue so this item is basically the guilty dog live / care of the wall / ½ the wall and those delta w is the thickness of the frame so that is equals to .5 into 10 to the power – 3 m and k w is already given 18 what party did Alvin and the CW is the nothing but in Dublin is equals to how many number of players were there in one into l2 x 2 n p + 2 number of players we had so this comes out to be 30.2 for my services by substantial and this is 30.24 and this comes out to be too many metre square so all the blue will come out as 9.186 into 10 to the power – 7inside and if you put all the values we will find that from here away is coming to B12 985Peter is already 12985 if it is this many you where it is there so now we have to calculate the ratio between the CD is equals to indore city of the gas and say is because to him not a city of the year for the year and we have this to be 1863 what were given this comes out to b2146 what park Alvin and them the NTU comes to be Eva by seeming so this is just nothing but 6.97 and the ratio between these two cryc Mitzi mean by C-Max that is equals to .868 so we had long these values for the and cube and sphere and then based on this relation are we will be able to calculate what is called the Epsilon and we know this relation for the across political and from this correlation if we estimate this Epsilon will come out to be at 3 to 8 and since we know these are excellent value they cannot calculate the exact temperature and also we can calculate that the overall heat transfer secured BF silent x the TGI – plin2 ceiling this is the way this is from the basic definition of the heater effectiveness and if you put all these values so this is supposed to come nearly 2.8 328 * this is the absolute value to this is equals to 900 – 271 whatever value for the seeminglyand you will find that they are not matching with the assumption that we had made as you can understand that we have estimated as or as you’ve done a 50 + 0.75 so already we have deviated from there so not based on this new value of the heat exchanger effectiveness we will have two sets of the exact temperature and based on that earlier we can now again making another destination of the contextual effectiveness and then we can make an estimate of the total heat transfer and then we had to find out that when we reasonably close value are predicted then we can stop our reitaration so that’s all we have to go ahead with the design of the simulation of the heat exchanger problem. Thank you for your attention.




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