- 2 days ago
In today's video, we will understand
1. What is LMTD | What is LMTD Correction Factor its significance
2. Temperature Cross
3. how to Avoid Temperature Cross
1. What is LMTD | What is LMTD Correction Factor its significance
2. Temperature Cross
3. how to Avoid Temperature Cross
Category
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LearningTranscript
00:00e-learning from today's session are what is lmtd we'll see with the example what is its significance
00:08we'll see what is lmtd correction factor and why it is required what is the impact of it then we'll
00:15see what is the correlation what is the between the lmtd correction factor and temperature cross
00:20and what is the basically a temperature cross and how to avoid a temperature cross
00:25hi friends welcome to our channel concept engineering flux focus on fundamentals
00:30in today's session we'll see what is lmtd and lmtd correction factor where it is used what is its
00:37significance and then we'll try to understand what is temperature cross and what are the measures we
00:42have to take if you find the temperature cross and how to resolve the temperature cross so let's start
00:51disclaimer for our channel and the content you shown in this videos so what is lmtd if you see
01:00this graph that the temperature versus the heat flow rate so hot temperature the the hot hot fluid coming
01:09from t1 th1 to th2 and cold fuel go from in the counter current to cold uh tc1 to tc2
01:26and if you put this is an as part of the delta t of this then it it uh comes
01:32as a flag uh single line
01:33the delta t1 the delta t1 and delta t1 and delta t2 as we seen here it is th2 minus
01:41th2 minus tc1
01:46and delta t2 is th1 minus tc2 so basically lmtd is q is equal to ua into delta t2
02:02this delta t2 this delta t2 minus delta t1 which is this one upon ln of delta t2 of t1
02:13so this is
02:14this is the lmtd it's called the long mean temperature difference however the simple way of
02:20remembering the lmtd is largest delta t minus smallest delta t so this is the largest delta t always
02:29the inlet temperatures of both the inlet temperature of the hot and the outlet of the cold and the
02:36by ln of largest delta t to the smallest delta t this is the this is the way i remember
02:42the lmtd
02:46so we'll try to understand the significance of lmtd for counter current and co-current system in this for
02:54this we'll take one example suppose you have to cool the oil from 60 degree to 40 degree by using
02:59the cooling water which is inlet at 30 degree and outlet will be 35 degree in this calculation we are
03:05considering it's a purely co-current or a counter current flow there is no means if the exchanger is
03:11designed with this it is a single pass exchanger there will not be multiple passes on the tube or shell
03:16side so in that case when we calculate the lmtd the fluid hot fluid cools from 60 to 40 in
03:27counter current
03:27and cooling water temperature increase from 30 to 35 so this is your t2 which is 60 minus 35 will
03:35be 25
03:36and delta t1 and delta t1 and delta t1 will be 40 minus 30 which is 10 so 25 minus
03:4310 will be 15 and 25
03:45divided by 10 will be 2.5 so ln of 2.5 so you will get lmtd as 16.37
03:50similarly for co-current your lmtd will
03:55be 13.9 based on temperature like because it's a co-current so 40 minus 35 will come and 60
04:02minus 30 will come
04:03so 30 here so based on this lmtd we can say that the counter current exchanger
04:12need about 18 percent less surface area than the co-current because if you see your lmtd value is
04:20higher in counter current so you are getting uh you required a less 18 percent less area this is 18
04:26percent more this lmtd so you need a 18 percent less area and in this calculation we assume that
04:33the u values are remain the same and reduction in pressure drop is also negligible to just get
04:40this calculation because there is an impact of that also so if you want to take the advantage of oil
04:50in the other example if we want to take the advantage that okay we will reduce the temperature from
04:56instead of 40 to uh 60 and outlet temperature of cooling water to 40 then the count in counter current
05:05the flow will the hot fluid will go from 60 to 35 and cold fluid cooling water comes from 30
05:11to 40.
05:11so this is your t2 delta t2 and this is your delta t1 so 20 and 0.5 so your
05:20lmtd comes as
05:2110.5 however when the same like if you increase the delta temperature uh of your cooling water to 40
05:30degree
05:31in co-current it can uh we want 60 to 35 hot fluid should cool cool but in co-current
05:38when you uh your
05:40temperature of cold fluid from 30 to 40 then your this delta t1 will be the minus
05:475. so it means you cannot go to the 40 in co-current
05:53and because we are talking about a single pass so it is it means
05:56we cannot have a temperature crosses in the co-current flow exchanger
06:01so this thing you need to remember we cannot have a temperature process in co-current exchangers
06:10you will see significance lmtd by reducing the inlet and outlet temperatures how to predict based on
06:17the lmtd itself that uh how much excess area we required or which which temperature is reduced will
06:24be better for our heat exchanger so suppose you again you need to cool oil from 110 degree to 30
06:30one degree c with the cooling water temperature inlet temperature of 30 degree and outlet temperature
06:36of 35 degree then we want to calculate lmtd when one uh cooling water temperature reduced by one degree
06:44so now i will reduce that cooling water temperature 29 one time and outlet one time reduced to 34.
06:51so we'll see for if you do this 34 and this is 29 then what is the impact
07:01so first the base case originally it is 110 to 31 so and 35 to 30 so just similarly the
07:10delta t2 will
07:11come 75 and delta t1 will come 1 so based on lmtd calculation 74 75 minus 1 is 74 and
07:2175 divided by 71 is
07:2375 so ln of 75 which will which will come 17.14 and when i reduced the inlet outlet temperature
07:31of cooling water
07:32to 34 then we reduce the t out here then lmtd comes almost similar that is 17.34 almost it
07:43is like
07:44uh 1 and when we reduced inlet temperature to 29 then your lmtd comes as 20.14 which is
07:57almost 17.5 percent higher than uh the for your uh with the temperature of 30 and 35 so if
08:06you reduce
08:06one degree inlet temperature of your cooling water you will get higher lmtd it means again you need a
08:1217.5 percent less surface area if you go with the one degree lower temperature so this you can predict
08:19from your lmtd calculation itself so making change change to the smaller or on the terminal like smaller
08:29means on the inlet side of the cooling water or inlet of your heat uh cooling media it always has
08:38more
08:38effect on the design than changing the largest temperature difference so if you increase if you
08:44decrease that instead of 32 delta t of your uh cold cold side to 4 degree then it will not
08:53have much
08:54impact it may mostly around 1 percent but if it is reduce the temperature at the inlet then it will
09:00have
09:00more impact around 17.5 percent so this is the significance of lmtd you can identify before getting designed
09:10now we will try to understand what is lmtm uh lmtd correction factor in earlier lecture also we have
09:15just covered the what is lmtd correction factor however in earlier uh the earlier slides what we
09:24lmtd calculated it is for the pure co-current or the counter current exchanger but majority of exchanger
09:30utilize the combination of co-current and counter current because we at least it will have multiple passes
09:34two passes maybe minimum two so so then in that cases there is a combination of co-current and
09:41counter current and to resolve this co-current and counter current combination we need a correction
09:47factor and the correction factor basically the mean temperature difference comes with f is a
09:55correction factor into the lmtd so how to get f basically tema has provided 13 chart which will be
10:04available on internet also one of the chart i am showing here so this chart will tell you to get
10:11the correction factor f which is lmtd correction factor f here which is based on the p value and the
10:22r value
10:23so p values on the x-axis is mentioned here and the r values are these curves
10:31so we will see how to get that p and r will understand p and r and then how to
10:35find the
10:36correction factors on this graph from the graph we know that there are two values needs to be found
10:43out that is r which is the ratio of heat capacity for the two two streams which is basically the
10:50m uh
10:51heat capacity ratio which is mcp or mhcp for the hot and cold fluid or you can say it is
10:57a delta t
10:58hot by delta t cold considering the th1 and th2 are the hot fluid inlet temperature respectively to tc1
11:09and tc2 are the cold fluid temperature and and c and p cp will be remains constant for both the
11:15fluid
11:19and the temperature effectiveness with respect to cold fluid which is p
11:24which is there which is on the x-axis of that uh graph if the temperature effectiveness is also the
11:32delta t cold by delta t max so we will try to understand how with this one example how to
11:40get
11:40that p and r value then it will be very clear so to understand the correction factor
11:47we are cooling oil from 105 degree to 95 degree in a shell and tube heat exchanger with the cooling
11:55water which is at 30 degree inlet temperature and 45 degrees outlet temperature and the exchanger is
12:01having one shell and the four tube passes then we want to calculate the mean temperature difference
12:08so the temperature is 105 to 95 it's hot fluid is getting cool and cool fluid getting heated from
12:1630 degree to 45 degrees so delta is delta t2 will be 60 and delta t1 will be the 65
12:23the stream delta t because to calculate p and r we need a stream delta which is 105 to means
12:28hot
12:29uh inlet to hot outlet is 10 degree and cold outlet delta t is 15 degree and the maximum stream
12:37delta t that
12:38is t max will be your hot inlet hot inlet 105 to your cold inlet which is 30 degree that
12:47is coming 75
12:50so when you want to calculate r then r is uh hot hot delta t to the cold delta t
12:58so
12:5910 by 15 which is coming around 0.67 and then p which is the temperature effectiveness which is
13:06cold delta t by the maximum t max so which is 75 so comes on 0.2 and based on
13:13this if you go back to
13:14the graph and see you will get the value of a chart from the 0.99 actually it is at
13:20end of that curve
13:21comes here somewhere so that's why it is just 0.99 very close to uh the the like the one
13:28why it is
13:30because your temperature differences in the hot fluid does are very low so when you calculate lmtd
13:36your lmtd comes is 60 62.48 and uh to calculate the mean temperature difference for the four passes
13:43you can uh for this exchanger it will be f into lmtd so f is your 0.99 and your
13:49lmtd is 62.48 so which
13:52comes around 61.8 so however this calculation we we done without knowing that which side is cold fluid
14:03either it is on tube side or shell side we not knowing and we did this calculation however that
14:08when you do it this in the tema tema because it it's required in uh htri that which side is
14:15cold
14:15fluid so there is a different formula to calculate p and r when it goes to the tema so let's
14:21see that
14:21formula random td correction factor as per tema it's it see p and r terms uh they do in uh
14:32in normal
14:33calculation it is like based on the temperature difference but it is uh in terms of shell side and
14:38tube side fluids for the tema so how will they do it's r is the delta t of shell side
14:44by delta t of
14:45your tube side so if you see uh tube side uh shell side we have a cooling water then it
14:53is 15 degree
14:53here and then tube side will have the exchanger so it is 10 degrees so 1.5 and we are
14:59assuming this
14:59as an example and if the p delta t on the tube side is 10 for the p the temperature
15:07effectiveness and
15:07delta t max is 75 so based on these values also if you go in the chart and see the
15:13f factor comes
15:150.99 by doing on the both ways we are getting the correction factor same in all the cases but
15:22most
15:22effective way of doing is this only because it's followed by the tema as well so lmtd calculated in
15:31each case was the lmtd for counter flow and this is always higher than the value of co-current flow
15:37when both the arrangement are present in exchanger it must be higher value which is which is the
15:44corrected with the correction factor which is less than one so what is what we are finding out and the
15:50only exception to this when the one fluid undergoes the isothermal process isothermal process means
15:56there is no temperature difference there is no delta t on one of the side so then it will go
16:02f as a 1
16:05f is uh in this case f is very high value this is because the both fluid had relatively small
16:12temperature changes so basically the your hot side fluid have very low temperature difference which
16:18is the 10 degree that's why it is very close to the uh 0.99 or you can say to
16:25the unity or the
16:26one and that uh that uh the smaller the temperature changes the process closer to the isothermal
16:34process so it means you will have a very low temperature difference between your inlet and
16:40outlet for the both the shells and tube sides if it's isothermal process so one more example if you
16:47want to cool uh the the oil from 100 degree to 40 degree and the shell side shell and tube
16:53exchanger with the
16:54cooling water temperature of 30 to 45 degree then the temperature uh that is 105 to 45 it will go
17:04and cooling side it will go from 30 to 45 so your temperature difference now here it is because you
17:09are cooling more up to 45 similar to the exactly as the outlet temperature of your cooling water the
17:15delta t2 will be 60 and delta t1 will be the 15 your stream delta t the hot stream delta
17:22t will be
17:2260 degree your cold stream delta t will be 15 degree and your uh t max will be 75 which
17:30is hot inlet minus hot
17:34cold inlet so for this r will be 60 by 15 which is 4 and p is 15 by 75
17:42which is 0.2 and based on that
17:43the same graph you will come you will get the f as 0.8 so f uh had a value
17:52of 0.8 the 0.8 is always taken
17:54as the limiting value when it is used in hand calculation and it it occurs the temperature
17:59of the hot fluid and the cold fluid leaving the exchanger have the same value so if you get the
18:05f value as 0.8 then you will find that you can analyze that that your inlet and uh outlet
18:11temperature of
18:11cooling water and the outlet temperature of your hot side is almost same or it is same so then only
18:18you will get the f as pointed so based on these uh small uh numbers or while doing lmtd you
18:27can able
18:27to find out that what is your f factor or based on the f factor you can understand that okay
18:32both the
18:32side of my exchanger are going to come the same
18:37so understand the temperature cross when the temperature of hot hot fluid outlet temperature
18:44is very close to the cold fluid temperature or you can say hot fluid outlet temperature is lower than
18:49your cold fluid outlet temperature then temperature cross happen so in normal exchanger where the
18:55temperature cross is not there where the T1 is your inlet temperature and Tout is your outlet temperature
19:00which is above the and the t1 tc1 is your inlet temperature for the cold fluid and tc out is
19:09your outlet temperatures and there is a significance gap or approach available between the t cold out
19:16and the t hot out then there is a no temperature cross this is a good design of exchanger but
19:23suppose if this is the t1 t hot in and this is t hot out and this is t1 t
19:31cold in and t cold out
19:33so now here the temperature of your hot fluid hot out is lower than your temperature of your cold out
19:43so this is the this is called the temperature cross so if the the t t cold out minus t
19:52hot out is
19:53greater than 5 to 10 percent so this is greater than 5 to 10 percent then you should understand
19:57that there should be more than one exchanger is required and the guideline is that if f is less
20:04than 0.8 then it must it under it should understand that there is an additional area is required so
20:10you can go for there is a how to avoid this temperature cross we have multiple options
20:16so suppose this is an example where the tube side 150 is inlet and 100 is outlet cooling and
20:24hot fluid the cold fluid is getting hot from 80 degree to 105 degree so and then you can try
20:30to
20:31calculate f the correction factor and you can just check that one exchanger is sufficient or not
20:37also you can calculate whether the one exchanger is required or not and overall heat overall f and
20:48how to calculate it just try this at this example based on the earlier examples how to avoid a
20:56temperature cross in exchanger so first thing is arrange a multiple smaller heat exchanger if you see in
21:02the graph these these are there are four exchangers you can to get the same delta t from this to
21:10this
21:10because here this is the temperature cross coming here this is the temperature cross this section
21:17so you can put some multiple exchangers in that you will not find any temperature there will be the gap
21:24between the temperature in and out for the each exchanger this is the one of the way the other way
21:31increase the number of tube passes or using a specific shell like f f shell if you take you
21:38will have a partition in that so this will also this partition will can be act as like a two
21:43shells
21:46enhance the turbulence add baffles use coagulant tubes and increase the overall heat transfer
21:52coefficient if you increase the heat transfer coefficient by adding baffles if you have other
21:56you do not have any other constraint then you can go for this as well if the process allows increasing
22:03the mass flow rate this is the one option then the lower heat capacity rate will change the
22:08temperature slope profiles and and the stretch of the profile can be apart so if you have a more flow
22:15rate then definitely your your this cold side temperature will not go up it will comes like
22:21this if you have more flow available so these are the few options available to avoid the temperature cross
22:30so in the temperature cross when we when we put a multiple exchanger this x is a inlet inlet and
22:38outlet
22:38temperatures the duty and it is not necessary duty shall be equal this exchanger duties can be different like
22:47if you can see these are the duties these duties can be different even delta t uh also not necessarily
22:55to be same like delta t between the two exchangers or this exchanger this exchanger and this exchanger shouldn't
23:01may not require necessarily the same f factor is also not required to be equal
23:07so you can understand that closer the heating curve more exchangers are required
23:13and deviation from the true counter current loss of efficiency and which required an extra area if
23:19you dividing uh deviating from the counter current then you need an extra area so this is the learning
23:25from today's uh session thank you so thank you very much write your question and comment i will be happy
23:36to
23:36answer it you can reach us on conceptengineering2025 at gmail.com links and links for the other sessions are given
23:48in
23:48description
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