Week 9: Lecture 45: LMTD Method

[music] [bell] [music] [bell] [music] So in the last module we were looking at classification of heat exchangers. We understood the concept of uh fouling concept of overall heat transfer coefficient etc. And now we are going to just use that in a very simple problem. Okay, everything is given. You just have to calculate. So this is a heat exchanger where uh the inner and outer diameters of the tubes are given 1.5 cm and 1.9 cm. Hi and hnot have been calculated for you. The following resistances have been given to you. Thermal conductivity of steel is given. So the job is to calculate all the resistances which is obtained from this formula here. uh a i a kn is there all quantities are known calculate the total resistance which is what this slide is going to do for you and you get ui a i you also get u not a kn which is nothing but reciprocal of the thermal resistances you want to calculate ui based on the ai so you calculated this quantity so 1 by ui ai is this 53.19 into 10us 3 which is nothing but summation of all the resistances. Now you know ai pi dl. So therefore 1 by ui is this number * pl. Therefore ui is reciprocal of that answer which is 399. Similarly, you can calculate U not also. And if it is a copper tube as opposed to a stainless steel tube with thermal conductivity is 15.1. The resistance is of the order 0.094 into 10us 3. You look at the resistances here. 2.49 into 10us 3 is the conduction wall resistance. If it were a copper tube, it is 9.4 4 into 10 -5. So of these resistances, five resistances which are there, you can neglect that if it were a copper tube. Another interesting thing to look at is look at this one 1x hi AI 26.54 13.96. Okay, so these add up to about 75% of the total resistance. 14 + uh 26 is about 40 53 is the total. So about 75 to 80% are the convective resistances. The fouling in this one is actually quite high heat. Okay. So bulk of the resistances is the convection fouling in case is actually bigger than the conduction resistance. So it's not a good idea to neglect it. Okay. So these are the inferences that you can get from such a problem. There is nothing difficult about it but physical insights are what we are looking at. Okay. So heat exchange we we already saw this heat loss by hot fluid is heat gained by cold fluid. Heat capacity rate capital CH is nothing but M dot into CP and M dot into CP capital uh CH small CC uh small CH sorry capital CH is M dot into CP of H capital CC is M dot C into CP of C. If I have hot fluid and cold fluid flowing at the same rate and they have same specific heat. Q would be MCP delta T and let me just write it instead of talking. Um M dot H CPH is CH capital C. Okay, this into T uh hot inlet minus T hot outlet equal to CC into T cold outlet minus T cold inlet. If M dot H CPH is equal to M dot CPC that means deltat T hot would be equal to delta T cold. The rise in temperature of the cold fluid would be equal to uh drop in temperature of the hot fluid. Okay. So this delta t I'm talking about this rise is equal to this drop. Okay. So this is when ch is equal to cc very special case. Okay. [snorts] Okay. Now we go to the derivation of the uh epsilon I mean LMTD how it is to be calculated. So assumptions we are looking at a tube and tube heat exchanger. We are talking of steady flow situation. Kinetic potential energy changes are negligible. Thermophysical properties are constant over the entire length of the heat exchanger. No heat loss to the surroundings and heat transfer coefficient is assumed to be constant over the entire length. It is important because if you're having a cross flow kind of heat exchanger, it would change. We are talking of H being constant. Negligible heat loss to the surrounding. That means whatever heat is lost by the hot fluid is what is gained by the cold fluid. There is no loss. Okay. Constant thermophysical properties over the range of temperatures that we are talking about. K, mu, cp, row etc are remaining constant. Okay. Okay. So very nicely systematically put here in the slides. So I'm not going to write everything. I will just explain the critical points and you can go through the derivation. We have seen this diagram already. This is representing the temperature distribution in the parallel flow heat exchanger. So deltat t inlet or deltat t1 is t hot in minus t cold in. Delta t2 or delta t exit is t hot out minus t cold out. any local delta t is t minus t cc local temperature difference. So dq is equal to u into d a s into temperature difference locally. So the das represents so if this is the tube das represents pi d into the infinitismal length dl. So pi d into dl dq heat loss by cold I mean heat loss by hot fluid is equal to heat gain by cold fluid. M dot cph dth temperature decreases along the coordinate direction. Therefore you have to introduce a minus sign to make Q positive. Temperature increases in the coordinate direction for cold fluid. M dot C CPC into DTC. So that's fine. So from these I will get DTH and DTC. Why? Because I want to relate these two now with the overall heat transfer coefficient. The first equation has TH and TC. So I want to get those things from here. So dth is delta q divided by m dot cph. Same thing here. So dth minus dt c is nothing but d of this one which is equal to I'm substituting for this. This is dq with a minus sign as it is m dot. xcph minus of minus becomes plus here. Okay. So minus dq uh dth minus dt c dt is positive. Therefore it will be substituted as it is. So dq dot is u into th minus d s from here I am going to substitute for dq here. Whatever I got as dq here will be now used here and then separation of variables will happen. So essentially this is what is done. Let's go back and take a look. So this quantity dq I'm going to write in terms of the temperature differences and put it here. D by t minus tc. Okay minus tc is of interest to me. Okay. So that and this quantity d of th minus tc are being written together. So I will substitute and separate the variables. So this is nothing but u a s into 1 by this one. So I what did I do? I substituted this last quantity. This is this one. I substituted this equation. the last one here and then took the th minus tc to the denominator here okay and then integrated that's what has been done so integrate from I inlet to outlet u into das s times this whole bracket as it is so u is das is here the whole bracket as it is remains go through the mass this is nothing but log of th minus tc evaluated at inlet and exit U into DA S * this particular quantity. This whole square bracket is a constant. U is a constant for a given heat exchanger. What is U? U depends on mass flow rates. U depends on NASA number. U depends on heat transfer coefficient. Right? From there we get so HI knot is constant. Conduction resistance is constant. Everything is constant. So U is a constant. Integration of DAS S from inlet to outlet would be the surface area of the heat exchanger AS. So therefore this equation now will get manipulated. Q dot is equal to this. Now I will I want everything in terms of temperature and I want Q. So I will write 1 / M.H CPH as TCO minus TCI by Q 1 / M dot C etc. this way. Substitute all this. Bring this to this side. Q is equal to minus UAS into this quantity. Absorb the minus sign. Q is equal to UAS into this particular bracket. This particular bracket is called as a log mean temperature difference. Notice what it is. THO minus TCO minus THI minus TCI. Go back to this diagram. Delta tsub2 minus delta t_sub_1 divided by log of deltat ts2 divided by delta t1. This is delta t2 exit deltat t1 delta ts2 delta t1 delta ts2 minus deltat t1 divided by log of delta t2 by delta t1. This is your lmd. Okay. So this LMTD is what you need to calculate and from that once you calculate that particular quantity you will be able to calculate the heat transfer rate it. Okay. So please remember that I draw this diagram and whatever is the inlet delta T that is what is delta T1. Now if you go back to go to the counterflow heat exchanger the derivation is exactly the same except that the minus sign for dq will appear in both the cases. Apart from that there is no change. Delta t1 here is going to be the t h in minus t cold out. Delta t2 is th out minus t cold in. Okay that is going to change. You will regroup everything. you will get the same answer. Okay. So, whatever be the configuration you draw the temperature versus location diagram and whatever is the delta T at this particular inlet, we shouldn't call it inlet and outlet because hot fluid and cold fluid would have different inlet and existent counterflow. Whatever is at x=0 that delta t the other one at x= l that delta t subtract those two divided by log of this by that okay so it's a very straightforward concept nothing great here and of course we said when these two are constant when the mass flow product of m dot cp is constant for hot and cold ch is equal to cc delta t1 equal to delta t2 the lines become parallel to each other and your log mean temperature difference. If you look at this case 0 by 0 form would come because these two numbers are identical. 0 divided by log of 1 which is again 0 0 by 0 it doesn't it is undefined but physically driving temperature difference is actually nothing but the log mean temperature difference. Same thing because it is constant. If you do mathematically do L orital rule it will come out to be delta T1 or delta T2. Okay. It is a special case very important case typically asked in GATE exams and masters PhD interviews etc. When the specific heats product of M.CP are the same for both hot and fluid cold fluid. What is the delta t? Delta t log mean is nothing but the local delta t which is constant. Okay. [snorts] So limiting case of counterflow heat exchanger is when I told you remember uh we'll we'll do this counterflow heat exchanger we can have the following thing. The hot fluid will do this. The cold fluid can do this. TI, TOCI, TCO. Right? Or I could have cold fluid doing this and the hot fluid doing this. Which one does what? So that we can see very simply by looking at the products. M dot h cph t hot inlet minus t hot outlet is equal to m dot cold cp cold cold t cold out minus t cold in. So if I write deltat t hot divided by deltat t cold which is nothing but m dot cpc divided by m dot h cph. If m dot cpc is greater than m.h cph that means if this is greater than one delta t would be greater than delta tc which means I would get the fall in the hot fluid temperature. Fall in the hot fluid temperature would be larger than the rise in the cold fluid temperature. So this graph is for the case when CC is greater than CH that is implies capital C greater than capital CH right. The other one obviously is when CH is greater than CC. Can check this also very easily. Now what is important here this exit temperature the one which is having the smaller this is the smaller one in the limiting case can reach the inlet temperature of the cold fluid. So, CH less than CC as L tends to infinity THO will tend to TCI. Am I right? This has a very very important implication in the heat exchanger design. We will define this as effectiveness later on. Similarly, look at this as CC less than CH. CC is less than CH. In the limiting case, as L tends to infinity, T cold out will tend to th. So what is important? Look at these two cases. the one with minimum, the smaller of the two, the C minimum fluid. If you look here, the C minimum fluid is cold. If you look here, the C minimum fluid is the hot fluid. The hot fluid exit is hot fluid exit goes to cold fluid inlet. Here C minimum is cold. So that cold fluid exit is going to hot fluid inlet. Okay, this limiting cases are very very important. Okay, we will see that implication when we do epsilon ntu method. That is what is uh exemplified here. Now what what what is the other thing that we are trying to say here is when I have uh given heat load say for example I'll put it here while writing Q is equal to U A deltat T log mean given heat exchanger we saw we told this also already so if I have parallel flow heat exchanger the deltat t at every location is going to be progressively reduced so given Q and area deltat T log mean for parallel flow less than delta T for counterflow given Q and area given area and temperatures distribution Q counterflow will be larger than Q parallel flow. Okay, the temperatures will adjust such that the log mean temperature difference would be higher and therefore you would get heat transfer for counterflow would be always greater than the heat transfer for parallel flow. It comes from this simple thing because each infinitismal addition to the length doesn't give you the same effect as the previous one. Therefore, law of diminishing returns kind of thing, you don't have uh if I go to a greater and greater length, I am not reaping the same benefits. That's what is told here. So, LMTD counterflow for specified inlet temperatures is greater than LMTD parallel flow. Okay. So, these is this is a calculation where four numbers are given and you are asked to calculate the LMTD. You can check this very easily. Look at the difference. 41° centigrade for counterflow, 27° for parallel flow. So, your dimensions being given this is about 1 and a half times. So the counterflow heat exchanger is smaller. Parall flow heat exchanger would be 1 and a half times larger than the counterflow. If you fix the diameter, the length would be 1 and 1/2 times larger, which means the pressure drop would be one and a half times larger. Okay, so the implications are there everywhere. Okay, not just in heat transfer. In the fluid mechanics part also the implications are there. Okay. So this is a simple numerical example where you are asked to calculate surface area given everything. It's a counterflow heat exchanger. You have diameters inlet and outlet diameters given and the temperatures are given the mass flow rate etc. is given you are asked to calculate the surface area. So how do I do that? So diameter diat t you can this data is given to you and the mass flow rates are given inlet and one exit temperature is given. So you can calculate the fourth temperature by using heat loss by oil is heat gained by water. So the fourth temperature is actually calculated. So.1 kg 60 and 100 is given to you. 2 kg and 30° is given to you. M dot cp delta t for hot fluid is m dotcp delta t for cold fluid. From that you would get this 40.2. Okay. So essentially all the four temperatures are known. So that step is not given here explicitly but you you'll be able to do it. Properties are given. You're asked to find the surface area. Draw the diagram for temperature distribution. This is a counterflow heat exchanger. So you you draw the diagram for counterflow heat exchanger temperature distribution. This is your delta t1. This is your delta t2. This is th out, tci, thi, tcco. You know these numbers. Calculate deltat t log mean. Once you calculate delta t log mean just go back to this formula calculate the heat transfer coefficients for this geometry calculate the h you know the diameters calculate the thermal resistances calculate all the resistances you get your total resistance you get your total UA Q is calculated by this simple thing M dot CP delta T of that is the heat that needs to be transferred. So mcp delta t this is a number that you know which is equal to u aa delta t log min delta t log mean we can calculate u aa is what we would get from this you know all the resistances okay so from that you can get UA equal to reciprocal of the thermal resistances from that you'll be able to calculate the area everything else would be known problem is solved very nicely here water side reol's number 4 m dot by pi d mu nasult number heat transfer coefficient 2250 okay oil side is annular flow hydraulic diameter is d minus diol's number is row dh into bulk mean velocity bulk mean velocity is m dot by row into flow area so m dot by row into flow area is by4 d² - di² calculate this quantity This renol's number would come about 56 for di by d equal to.56. You calculate nasult number which is from the table. Okay, there's a table which you would see in any heat transfer textbook for annulus. Given Renaol's number di by d calculate nasult number or read out the nasult number which is nothing but hn dh by k from that calculate hnot is 38.5 38.4 4 2250 here 38.4 four here this is the dominant resistance okay if you have to put fins it will be on this side okay so it's a thin wall tube if you want to take so you get this as overall heat transfer coefficient I would diameters di are basically the outside diameter of the shell and the of the outer pipe I mean inner diameter of the outer pipe and the so what is 45 and what is 25? I will tell you this is 45 and this is 25. Nothing is given about the thickness. Okay. So we are assuming it has a thin wall pipe. This is the outer tube diameter. This is the inner tube diameter. This is used to calculate the hydraulic diameter. So our conduction is very small because nothing is given about inner pipe material or inner pipe thickness. That's why I'm able to take this as thinwalled pipe. So overall heat transfer coefficient therefore is calculated as 1 / hi + 1 / h. It comes close to 38.4 as you can see 37.76. So Q we calculated from first step M dot oil CP oil delta T oil 8524 that is equal to U into pi DL into log mean temperature difference which is 43 D in inner one you took so you get the L 66.54 m so point to note if the thickness of the inner pipe was given then you would have to worry about whether this is outer area or inner area then I would have had to add the conduction resistance here but now because it is a thin wall pipe there is only one p that I have to deal with that is the pipe dimension there is no outside dimension of pipe inside dimension of pipe it's only one number h outside was the do h inside is very large compared to H outside. Therefore, H outside is the dominant resistance. Okay. Okay. So, special cases as we can see we saw this even in the internal flow. Uh when I have condensation happening that is the fluid is condensing it gives heat and the cold fluid is going to have a rise in temperature. The hot fluid corresponds to a fluid of infinite M dot CP because delta T is zero. Similarly, evaporator or boiling fluid is this one where CC is tending to infinity because the cold fluid temperature is not rising. Q is nothing but M. HFG. HFG is the latent heat of evaporation or condensation which is the same depending on how you look at it. So that's what is given here. So [clears throat] what is this LMTDF approach? So LMTDF approach is essentially what is used when the heat exchangers are strictly not parall flow or counterflow and tube heat exchanger where the path is the fluid is going this way. So in certain locations say for example in certain locations the fluid would behave parallel flow fluid will behave counterflow. So this is your set of pipes and this is your this one uh baffle here. The fluid is going to do this. This is the second baffle in this part. If this is flowing this way in this part it is behaving counterflow. Then there would be a path where it is cross flow. Cross flow is basically doing perpendicular to the direction of flow. Okay. This is cross flow and parallel and counterflow in the same arrangement. So it is neither parallel nor counter strictly. So you have this concept of F which is like a correction factor. So select the type of heat exchanger. Determine any unknown inlet temperature. This is a design procedure. Calculate log mean temperature difference and the correction factor. How do you get the correction factor? We will see. obtain the overall heat transfer coefficient and the heat transfer area. So what is new is step three this intermediate part. Okay. So the formula gets changed to UA delta T log mean into F that F is a correction factor which takes into account deviation of the given heat exchanger away from counterflow. If F is equal to 1, it's a perfect counterflow heat exchanger. Otherwise anything less than one it's a it is deviating away from counterflow and f is obtained by gra charts which essentially are given to you in the textbooks or heat exchanger handbooks. So this is one shell single shell and 2 4 6 any multiples of two. So two pass this is this is two pass here or you could have four pass or six pass. You calculate this capital R which is a ratio of temperature differences and calculate this P which is again ratio of certain other temperature differences with P on the X-axis and R as the parameter read out the correction factor. Okay. So like this you have for two shell and 4 8 or 12 any multiple of four passes then you can do this for single pass cross flow both fluids unmixed what do I mean by both fluids unmixed the fluid flowing from left to right is flowing through tubes fluid flowing from top to bottom is also flowing through tubes they don't mix with each other or they don't mix amongst themselves also that is why it is both fluids unmixed mixed. Again you have another graph for it. Similarly single pass cross flow with one fluid mix other unmix. Fluid flowing from left to right is unmixed. They are flowing individually. Whereas fluid flowing from top to bottom is completely mixing with each other. Another set of graphs for it. Okay. So what we have is essentially a numerical problem here where you you have all the information given and you are to calculate the uh f factor and calculate the delta t log mean and the size. So dimensions have to be calculated. So you have TCI all the four temperatures are known mass flow rate of the cold fluid that you obtained by energy balance 7.62 62 and go to R, get your P, get the following factor. I mean, get the factor F, not the falling factor, get the log mean temperature difference and delta T actual is F into delta T log mean. If it is a shell and tube type, if it is unmix cross flow type, this is like this. And you can see how these numbers are going to change significantly for various heat exchanges. Okay. So this is a cooked up problem. So you're just it is just given to show how much these things are going to affect 17.3 from 21 19.9 from 21.64. Okay, we'll just complete in a couple of uh slides. So f is a function of p for a given r and tends to fall very steeply indicating considerable sensitivity temperature definition used for p. uh [clears throat] so in multi-pass etc you may have local reversal of heat transfer and that is possible so locally the cold fluid may become so hot that it will give heat to the hot fluid for achieving 8 or more f multiple shells are used in series this is like a design thing uh whenever cold fluid reaches a higher temperature locally than the hot fluid temperature cross occurs zone of reverse heat transfer will occur and that means that's not a good thing from design point of view. Okay, this there are ways to calculate this etc. So if you do incremental analysis you will be able to find that out. Thumb rule f is greater than.8 Eight typically is what is used. Number of shells in series are increased then the result would come closer and closer to counterflow heat exchanger which means F would be larger and larger. Okay, this is the design procedure which we have anyways seen. The intermediate step number three is where this F comes in. Everything else is the same. In the next module we will look at the other method for design of heat exchanger which is the effectiveness NTU method. Thank you. [music] >> [music and bell]