Plate Fin Heat Exchanger | Numerical Problem – Part 2 | Alison
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Module 1: Plate Fin Heat Exchanger

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Numerical Problem – Part 2

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Plate fin heat exchanger: Numerical Problem – Part 2
 
Welcome to this lecture, we are trying to solve a numerical problem relating to plate fin heat exchangers. So in the last class we have looked into the problem statement and we said that it is basically a rating problem where some of the parameters are known to us and heat exchanger specification was completely given so we have to find out what the exact temperature and we have seen that we have to calculate first of all the number of players associated with in the gas and the airstream so while trying to do that if we look into it we will find that we have alternative layer of air and the gas so if we look into this we have cast removed call ency then we have one passage designated quality than the next person is designated for air then you have a genji then you have ears so like that it will continue we have the key and so on so the basic unit we can say thatthis is typically to scream exchange will try to look for alternative hot and cold frits ring and in this case the we have also designated that this gas flame is the hot food straight and the airstream is the corflute strength so and obviously that enter heat exchanger in a we have another dimension on this site but what we have not told in the last class is that this though it is the building block like a g a g gas and air gas and air alternatively Floyd the last one is here but we will add an additional here on this side this is also a five because this is relatively older than the hot timing of course it is warmer than the ambient air so we intend to keep a kind of symmetry for this entire heat exchanger and we will talk about this requirement of the cemetery in context to the plate frame type heat exchanger later on and for time being you have to assume that we will keep an airstream at the end so that why not it’s a g I mean why not we put it like ga-ga-ga this could have been an alternative arrangement like this ain’t a strange school have been the gas stream but you can see that this gas film being the hot streak and it will need more number of insulin shins at the end or will it have insulation at the endLinda police department of the army was called the insulation this and take it extremely for it now look at we have safe then you have gas here guess so like that we have ga so if we said that there are any number of passengers for the passages for the gas if you have any number of gas then we have in PE class 1 number of passengers designated for passages designated for air so all together we have how many MP plus one and NP number of gas and air passages so we have in the number of gas passages and we have until plus one number of air passages over this entire length if we know this in pnnp + 13 + of each gas late and we know that already we have said that both the sides that having the similar type of fence and if we designate these big ass for the height of the fee and that means this is the same height this is the finished and this way designate as fear of hair do we know that be the height of the faint of their side is equals to height of the Field on The gaslight so this the BD NBA voted similar so now we can have any number of pages corresponding to the cousins we have in be too busy + n + 1 into in and there should be equal to this one metre length but it's not so there is something else we have the thickness of the plates so we have to add the plate thickness with this one and then equate with one then only would be able to find out the number of passengers in hisMP into PT plus NP + 1 into be a plus how many number of plates up there if we have two layers of the gas and answer another year so we have 123 number of screens then we have one plate to 34 number of separating plates so43 which has four if we have phone number of spheres we have five number of plates so always if we had in his class 1 + 8 + 12 + 2 x 2000mm this is given in how much it is given as the plate thickness plate thickness W2 0.8.5 mm .5 mm and then we have pa equals to BC the fan height that is equals to 2.49 mm so this is also converted into millimetre this month has been converted into millimetre and now we can estimate the number of players from this month and we will find that it is approximately coming the nearest whole number is equals to 167 so we have in number and PIN number of this many layers of the Indians passengers so we have 168 number of air passages in that case so now we can understand that we have 168 number of such air passages and 167 number of passengers arranged in a crossword clue patent and we now had to go to the next estimation I mean what we are looking for is basically the overall away I’m in this evening we are trying to calculate because that will give us if you remember in the last class we have talked about that we intend to find out the message number so we wanted to find out and that ain’t you is basically you by Simi and that we see me we want to calculate and for calculating the UAE we’re trying to evaluate all this numbers and pee and from in PV have to find out the area and so on so now if I look into it the first of all we have arrived at the total number of players so we can now try totelling the frontal area what you mean is that this is the frontal area for the airstream and the frontal a frontal if for the air is the sakali this surface area so it is equals to 1 metre x 2.36 okay well how excited are you hundred so what is the floor but this is the way it is taking place so this is the surface area so it is having .3 metre long and this is the other one dimension so this is also warning 2.3 metres square so we have .3 m squared both the frontal area for the air and gas the same as what we have the frontal area information with us now we can also try to find out the volume between the plates because why it is necessary because we know we have been given some parameters beta and thanks that has already been told about each type of fee and since we know better we can try to find out the heat transfer area from that one if we are able to estimate heat exchanger area between the plates basically this data is the door in the very first class we had talked about this is there is nothing but the heat transfer at a party with your volume and we will try it now calculate the volume between the plates are for each troop strength and the volume between the plates for it if we look for the gas side what we have is the fin height of in height is known to us this is me of Jeannie have designated it as the of g and this is one dimension the other dimension is this 1.3.3 data and its thickness is there is it busy so we have how many cc is .3 * .3 and x busy so that's the number of a single the volume of a single layer of the gas stream so how many satellites are there we have in p is equals to 167 number of such as their services what is the total amount of volume up between the plates on the gas side so this comes out or to be 167 x BG 2.3.3 so that will come out as .03742 metre square similarly we can try to estimate the value of a point between the plates volume between the plates for the air and obviously you can understand that this is the year we have one more number of layers in this one so we have 168 x b into a and * .3 .3 so it will be slightly higher than this volume and sorry this is basically of volume it is meticulous we have this number will come out as .03765 m cube so this is the volume between the plates for the gas and this is the volume between the plates for the air side and this is the volume between the plates for the gas side ppg and CP of air so now we have the information about the volume between the plates so we can try to estimate the closest service area associated with the gas side of the inside so if you look into that part we will be able to calculate the heat transfer surface area because Nino busy that has been given us to 25454 metres square so that she calls to transfer surface area that’s save the gas side and the volume between the plate that we have now calculators just now we have calculated for this is because 2.03742 metre cube and metre cube is equals to x x father father side we can write this thing type of feelings there so it is again to 254 and that is because to this is metres per metre cube and that is because two of the inside / we have calculated the volume between the plates the and that’s slightly different for this one this is 765 so this will correspond to a of air and on this site will have a slightly higher heat transfer surface area associated with this this will come out to be at 4.8 63 m square and if we calculate the same figure for this one they should have been 2384 point 345estimated by new from the configuration of the heat exchanger so once we know the heat transfer surface area one of our requirements is the estimation of the rails number if we have to calculate the Reynolds number Reynolds number if you have to calculate then what we need to know is the free-flow area and the mass flow rateso we have been told about the parliamentary product we have now been able to calculate the heat transfer service area now we need to calculate the free-flow area and we will now estimate it from the information given on the basis of the information given related to the hydraulic diameter of the hydraulic diameter for this heat exchanger has already been given for the East passages has been known to us and if you know the hydraulic diameter from that hydraulic diameter with a can try to since that hydraulic diameter give the relation between the three floor area and the total heat transfer surface area should be able to estimate the floor area from the knowledge of the hydraulic radius and the height of hydraulic diameter and the heater is the surface area now that the hydraulic diameter definition we have written it like this the hydraulic diameter the height is equal to 4 times the freeflow idea this was the free-flow area and / the wetted perimeter so we have the free flow area that was the definition we have adopted for the hydraulic diameter and this is a weighted perimeter know if we multiply both decide the denominator and numerator and we will find that this is just nothing but fall into a 0 and 2 l / this is p and that p x is nothing but the heat transfer surface area so we have this information of this hydraulic diameter we have the information about the sickness that area we know the length and what we need to calculate is that a zero from that relations so now if we go back to our dimension so this is parked outside and gas side we have calculated we have been able to calculate the heat transfer area and the transfer area for the agi is basically a gas side was at 4.30 an estimated to be at 4.45 square and now if we look at the hydraulic diameter d h that has already been given for the heat exchanger and we know it to the point 01548 that equals to 4 times 80 and then we have the length of it is because to this is the length this is the dimension and that is equals to .3 metre and / that he transferred area that is equals to 84.345 so from there we would be able to calculate ratio and this is a hero for the west side so we have the 80 hydraulic diameter * 84.345 and then we have ÷ 4 and 3 so this will come out to the point 1082 Square that’s the amount of heat transfer associated with the gas right now if we look into the similar heat transfer area for their side what we need to find out is basically in the similar relation only thing is that this parameter d h x the heat transfer surface area / 1412 so this is just nothing but now if we put this parameter circle come out the 10.1 089 metre square that’s the clitoris behaviour associated with the CSI obviously since there is one number of I mean one more layered associated with the year the transfer service area will be more on the inside so now knowledge about the heat transfer area wethe floor area we can also calculate the word is called the white understudy the sigma g we have alsoand similarly the sigma for floor area of the air side / a frontal area for the inside so this will come out to be .361 and will become .363 5363 visit the values for the signal and the signal will come there I’m in terms of the pressure drop relations and other correlations this becomes is full so we will now go on to the next parameters that we can estimate on the basis of this one so that’s already in her and I did the floor area now what we need to find out now we can try to find outflipside so if you didn’t bet the gas was gas pressure was 160 kPa and the inside was 200 kPa was the corresponding pressure and if we noif we try to find out the density of air at 200 kPa so we can use the music as a a1c related to be ideal gas and the corresponding pressure relation so we can use the PV NRT relation the ideal gas law and then we have this road g is equals to PTR / RTC of and from there we know you’re able to talk is one we will get value of 4751 kg per metre cube as the density of the gas and the inlet similarly role of a and the inlet can be considered as a pa / 80 AI and business come out to be one point formetre cube per second and the volumetric flow rate of the gas given as 13 points 3.494 q per second so we can now calculate the mass flow rate of the gas side 2 * 6 + 8 * this density music that will give you this one as the mass flow rate NG is equals to 1.66 m per second and similarly on this side in the mass flow rate of air will come out to be 1.3 5/8 * 1.4 726 this to see if we multiply they will find it to be nearly 2 kg per second so the mass flow rate of air is more or as compared to the mass flow rate of the gas so based on this we will be able to go to the next step for calculating the minimum fluid and then what is the minimum  capacity fluid and based on that we will have to make the other estimations. Thank you for your attention.




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