Plate Fin Heat Exchanger: Pressure Drop
 
Welcome to this lecture, we are talking about the plate fin type heat exchangers and as you know that pressure drop is an equally important parameter in the case of any heat exchanger. So how do we estimate the pressure drop in plate fin type heat exchangers that we are going to study in this lecture. And if we look into a typical play Queen tie fighters in your business being taken from kaiser London’s compact heat exchanger you can also have a look into the fundamentals of heat exchangers by shy and calculate the area will have the detail description of the delegation of these equations of the pressure drop will only look at the final expression of the pressure drop in this lecture sit here if you look at we have this is a cross-sectional team will find that some of the passengers are some of the messages that open and to fetch the flower templates and this other places through which the floor is taking place and it will again come out so this is what is thethis is the actual reason that it is the handle and this is the other exact reason so this is what the fluid is moving out this is the entry so as you can understand that this flow when it is coming and getting to the surface of this cold it will stain sunlit that is some kind of induction in the heat transfer floor area where it was going through this area now it has some reduced area because some of the passengers the black remember that we have talked about the when we talked about the construction of the plant winter wheat exchanges we have alternative hot and fluidly and when does fluid is flowing through this direction if we look into this you know the 3D view so if it is a playful type exchange rate cross-flow time and this is the sidebar for this fluid flow and similarly are for this side you can see that this is the sidebar so this is when this flower is taking place from this the end you will find that it is this side but is blocked for this floor so the truck and take place only if we take this is as the frontal area we will find that almost all in part of a 50% or you know some portion of the floor area is blocked for this freight similarly but this flight is flying from design to this and it finds that only if this passage is the available for it but it was coming from this and it was having this much frontal area but only this much for a free-flow area it is serving of those the fame sell their fans will also block the passages and so we can say that there is some kind of contraction in the floor area and on the other hand what will happen on the exit and find that the floor is suddenly in getting expanded so I said no we can say that we have three components in this pressure drop when is the liquid pressure drop between this is the temperature drop between a and b and this is if we divided it into a b and this is between a and b and this is between 9 and 2 so we have some kind of don’t be a big man and a and then we have some kind of pressure drop in the cold which will consist of now we will come into that part and there is another day Turkey actually share this this will be a negative one and this will be minus b by b22 so we have the three components in this pressure drop the one is due to sterling construction then we have the oil pressure drop when we have sudden expansion and that it will give some pressure rise in this weekend so here we have pressure drop here we have pressure drop and here there is some kind of pressurizedno pressure but you will come to town this pressure between in the cold but it is between a and b lies between a and b have the special it will consist of two what we call it the pressure drop and I have also found this type of pressure drop but the skin friction and the bang bang associated and that will be included with in that fraction friction factor if so will not separately the look into the farm truck or the skin fractions that will be taken inside the friction factor f and we have our insurance is the hydraulic radius and the room 102 was the density at the entrance and the exit and then we have the exact effective and over again find two more times one is the sigma and other one is that KY this guy takes care of the exhibitor effect of the coefficient of coefficient of thermal expansion so we have more than four times in the pressure drop the correlation or delta byp100 we were looking into this part we have not taken care of the flu acceleration of the entrance effect of the exit effect of this poor acceleration we have not encountered I mean in our earlier equation only we were talking about this delta p by p what is the cause to this but if we want to have a accurate estimation of the pressure drop then we will have to take account of all this entrance exit effect but you will find that these efforts that this morning and evening as compared to The corporation the entrance exit if it’s a bit small so we will look into the this as we had told listen devices that kkck finals so if we know about the signal value and if you have an idea about the Hardy values that we can try to find out the key and kacy from this Block and try to estimate how much is that a pressure drop due to sudden contraction and sudden expansion but as I told you that this constitutes or give a very small percentage of the total pressure drops okay so the next time we will go to numerical problem here we try to the numerically estimate the pressure drop this example I have taken from the shine shaker lakes that the fundamentals of heat exchangers and there are certain parameters are the specifications given and we will just ignore them down and then try to calculate the pressure drop or try to estimate the pressure drop across the path for estimating the pressure drop what we need to know what is the pressure I’m in the same specification of the geometric pacificationthe genetics Pacific ocean I mean that one of the few times I got the frame and details and then what we need to know is the frontal area or the length and the width or the height of the heat exchanger then only will be able to estimate the pressure drop the general idea is that we need first of all the geometric specification of the heat exchanger then we need to look for the lead to look for the operating conditions Sofia the first this particular point the finger and CP is given in the field density is given as finding city what is meant by Fenton city we have already talked about it and this is having 615 fence per metre fence per metre so if we have 615 615 principle that means we have this kind of thing say this is the separating play the river separating two separating plates and we have this kind of thing and like this we have submit Wayne how frequently they are appearing so that’s what is the film frequency so from this sent to this time this is what is the pin spacing out this is the famous person on this will be having how many number of Fields we have in one inch are in one metre so we have 615 number of films in 1 m and thereby we can tell that one thing how much is the distance between this thing to this thing so that’s what we get an idea about the fringe spacing so here we have the field density others are some great things for 15 so we have six 15 millimetre 4mm we have this one135 so we have a 6.35 mm plate spacing then we have the fafsa to plan so I can now understand that this is an upset at least has been given us offset stripping and for offset stripping we have one if we look at this as a note this is the top few if we say this is like we have the same arrangement and this is what we call it as the lens length or the queen of succulent this is equals to 3.18 millimetre and a floor-length so the effluent in one this is Eleanor sorry this is a to This is the end to this l2 is equals to .6 metre so the air is flowing in this direction and we have it is traversing a length of .6 later and we have to find out in the pressure dropped due to this air flow through point six metre after April and then we have the hydraulic diameter this passage is given as a point not to 2383 this is what is the hydraulic diameter this has already been given and the film metal thickness that is a delta that is because 2.15 mate sorry millimetre and as you can understand that if we are using think I feel the pressure drop is going to be hired and we have also been given the free-flow area as well as in the the fountain area I mean we have already learnt about that the minimum free flow any AFL or sharing this once we have defined it as if it was a sea that is equals to .1177 then m square and the free-flow area by the frontal area that is sigma is equals to .437 we have these values and given for the construction of the chill metrical properties now we need to look into the operating conditions so in the operating conditions we find that the following trick flow rate is given point six metre cube per second this is the volume at Rick flavoured weed out of a fidget I’m sorry if you thought of a is given as .6 m cube per second it has already been estimated as the number 786 otherwise you know we had to calculate this one from the geometry and the feeling FanFiction factors that if has already been estimated for us .0683 this is the kind of friction factoris A4 bigger central get well as the principal said this is because to 194.5 degrees centigrade .5 degree centigrade and the gas constant r the gas constant for air has been given us to seven .04 joule per kg class so based on this information now we have to calculate the pressure drop the first time that we need to do is that as you can understand if you look into our pressure drop expression we have both kkc then signal then we had the different densities grow one rupee and one is already given so if you know then try to I'm in Howard and we had all the parameters known to us we had to calculate then the one by one if you look into this we have no one we had to estimate we have to estimate the road to then we have to estimate row m then we need to look at the g of the mass velocity then we need to have an locate to the KY and then we have Casey and the sigma so not as you can understand that we have already been told about this sigma it has already been given kck we have to estimate based on this signal values so depending on the same of Alice and the Harley value we have to estimate their sake and kissing and then we have to estimate room one and room two and paste on this room 102 we will estimate this one by arrow and as one by root 1 + 1 by root 2 is equals to 1 by 3 and so from there we will calculate the room and then we have to estimate trichy we have been given about the mass flow rate of the parliamentary chlorate is given from the volume metric chloride will multiply it with the density appropriate density to get the mass flow rate and we also know we have already been told about the free flow area so we can estimate the value of g and then we can calculate the total pressure drops if you look out one-by-one into does parametres now if we will find that role of the inlet can be estimated as a function of time and effort as you need to be an ideal gas we can use the app NRT record relations services Pierre in the by RTA in let us or you have to convert this into the Calvin and accordingly this will come out as 1.3827 kg per metre cube so this’ll be the density of the air and the entry then we can similarly calculate the density of air at the exit and we will have to ask you some kind of pressure drops but we do not have the pressure drop idea but it’s the oven we need an assumption at this point and we will assume that the pressure drop is not substantial so that you know we willgetting really large pressure drops book the value of 110 kilopascals and these are well you already have told the temperature is already known so accordingly we will be able to find that this is coming as .8195 kg per metre cube for the density of the exit 84 Winston this to value of roulette Andrew except we can then calculate the drawer and this will come out as 1.0291 kg per metre cube so this’ll be the main density and then we have the GG can be estimated from the valuator product x royal and the entry and then we have the free flow area as we have estimated that already it is given so Franklin we would be able to calculate the value of the last velocity 7.086 kg per metre square per second so that is the image and photo ID number given we can try to estimate the calculation I meant we can estimatefind out that that Kelly is coming at points 3-3 and discussing is giving us pointy ones so then you put the the song the spirit is now known to us so accordingly we can try to find out that delta p buy one and this will come out as PT 01536 so we can find that the pressure drop is 1.69 kilo Pascal and that’s not really a very I mean it’s not on it’s the only about 1.5% of that total in the pressure so that the assumption that has been made to her in our calculation that the exit pressure is not really very different from the inlet one is valid and we can now I’m in that assumption for we don’t need any separate iterative calculation to the repeat this one out to estimate the exitpressure drop. So this is how we calculate the pressure drop in a plate fin type heat exchanger, so we have already talked about it. Thank you for your attention.