 Welcome to the lecture series on process integration. This is module 4 lecture number 7. The topic of lecture is area targeting. In this lecture, we will see that how to find out the area targeting of a hen, when the hot and cold streams of the hen have unequal stream heat transfer coefficient. Area targeting is a vital component in the determination of the heat exchanger network capital cost. It plays an important role in capital energy trade of to determine the optimum delta P also. Basically, when we design a hen is in the pinch analysis for a given problem, then many heat exchanger networks can be created based on the value of delta T minimum. Then, a optimum delta T minimum is selected, which gives the tack of the hen as minimum. To do so, one should calculate the capital cost of the hen for each delta T minimum as well as the operating cost of the hen for each delta T minimum. The operating cost can be found out from the energy target and the capital cost can be found out from the area target, because once we know the area of the heat exchanger network, capital cost can easily be found out. Area targeting is done with the help of composite curves, which we have already seen how to generate a hot composite curve and how to generate a cold composite curve. Composite curves provide information about energy targets and also help in the prediction of heat transfer area of heat exchanger network. Now, in a heat exchanger network, we find different heat exchangers, some of which use to transfer heat from process hot to process cold, others are used to transfer heat from utilities. Hence, both type of heat exchangers should be included in the heat exchanger network. This can be done when utility streams forms a part of the composite curve before it is used for area target. Composite curves which include utility streams are called balanced composite curves. Utility demand for the balanced composite curve is 0, as the hot balance composite curve which is called HBCC is in heat balance with cold balance curve which is called CBCC. Now, the figure shows a composite or a balanced composite curve. The upper one shows a balanced hot composite curve, where this part of the hot composite curve shows the hot utility and this shows a balanced cold composite curve, where this part of the curve shows the cold utility. The balanced composite curve which is drawn between axis temperature and delta H, which is enthalpy can be divided into vertical enthalpy intervals from 1 to n. In the present case n is 5. Now, the first one is this, which is number 1 vertical enthalpy interval. Second is this, which contains hot composite curve up to this and cold composite curve up to this. And third one is this and so on so forth. The bounded section of hot and cold composite curves resemble with temperature profiles of a counter current heat transfer. If we see the part 1 of the vertical enthalpy interval, that is number 1 vertical enthalpy interval, this is the hot composite which moves from here to here and this is a cold composite whose temperature is the inlet temperature, this is the outlet temperature. This resembles the temperature profile of a counter current heat exchanger, if it is so and we know these two temperatures and we also know the overall heat transfer coefficient of this and we can know the delta H part, which is from here to here. So, delta H equal to Q is known. This four temperatures will give LMTD log mean temperature difference. So, Q is known, log mean temperature difference is known and if U is known, then area for this interval number 1 can be computed. Similarly, we can compute the area for interval number 2, interval number 3, interval number 4 and interval number 5. If we add these areas, then we find the overall area of the heat exchanger network. Further, if we see here, then we find that there are few points which are not available. Like if we see this number 3, that is third enthalpy interval, the temperature at A and temperature at C is known, sorry temperature at A and temperature at C are not known in the composite curve. So, these temperatures has to be found out before we go for area targeting. Similarly, here we see that where there is a change in the slope, from here we see there is a slope, there is a change in slope in this zone. So, this point is always known. Similarly, if we see the slope is changing at this point, so this temperature is known to us, but the slope is not changing from this point to this point. So, the corresponding temperature to this temperature D which is A here is not known. Similarly, there is a change in slope at B point, so B point temperature is known, but C point temperature is not known as there is no slope is changing. Unless, otherwise we know the temperature at A, D, B and C, we cannot find out the LMTD for this zone. Unless, otherwise we find out LMTD for this zone, we cannot find out the area of this zone. Even if the U is available with us. So, the first step in the balanced composite curve will be to know the temperature of those points in which temperature of either hot composite curve is not known or cold composite curve is not known. If the overall heat transfer coefficient of enthalpy interval i is U i, then area A i will be given by delta H i divided by U i delta T LM i. That means, the delta H of that interval i is delta H i, U i is the overall heat transfer coefficient of ith enthalpy interval delta T LM i is the log mean temperature difference of the ith interval. And area A i is the heat transfer area in the ith enthalpy interval for vertical heat transfer of H i amount. Here, we are presuming that vertical heat transfer is taking place in enthalpy intervals. Now, in such cases generally delta H i is known, U i is known because you for to calculate U i, we should know the individual heat transfer coefficient of the stream. Those informations are known to us and in the problem this information is tabulated. However, delta T LM i has to computed and for this purpose all the temperatures exit and inlet temperature of the interval should be known. Now, there are two type of algorithms which are available for area targeting. The first algorithm is area targeting with equal heat transfer coefficient which is comparatively simple and the second algorithm is area targeting with unequal heat transfer coefficient. The topic of this lecture is area targeting with unequal heat transfer coefficient. Let us see the algorithm of area targeting when the streams have unequal heat transfer coefficient. Based on the concepts which has been discussed earlier, the area computation algorithm can be generated based on vertical heat transfer. The total area of the network based on vertical heat transfer is A which is equal to summation i equal to 1 to n where there are n enthalpy intervals and A i is the area of the enthalpy interval i. This is equal to summation i to n delta h i divided by delta T LM i into u i. Now, if the heat transfer coefficient is constant then the u i will take the shape of u and it is constant. So, it can be taken out from the summation and the summation is equal to 1 by u summation i to n delta h i by delta T LM i. However, if the u i is not constant then this formula cannot be used. Let us see some other part of the algorithm of unequal heat transfer coefficient. In this hot and cold composite curves are drawn which contains necessary information required for computing network heat transfer area. The composite curves are basically BCC curves that is balanced composite curves. The balanced composite curves are divided into vertical enthalpy intervals. Within each enthalpy interval a network design is considered which can satisfy the vertical heat transfer. And to satisfy this vertical heat transfer some assumptions has to be taken or some manipulation needs to be carried out. What is that manipulation? Taking each hot stream into the same number of branches as the number of cold streams that is f in the ith interval number of cold streams is 3 and hot stream is 2 then each cold stream in this interval is treated into two hot streams. And each hot stream of this interval is treated into three hot streams as the number of the cold streams. In this way number of hot stream in this interval becomes 6 and also the number of cold streams in this interval becomes 6. Once the number of hot streams is equal to number of cold streams in a interval three hot stream is then matched with the every cold streams. By doing so we are able to do the vertical heat transfer. Now the figure shows balanced composite curves the first one is called BHCC balanced hot composite curve and the lower one which is in blue shows the balanced cold composite curve BCC. Here we see that the hot utility has been added to this and here the cold utilities have been added to this. In this BCC it is very clear that the heat available with the hot process streams along with the hot utility is equal to the heat required by the cold utility and cold composite streams. Hence it represents the complete heat exchanger network including heaters and coolers. So that means if we find out the area using a balanced composite curve which includes balanced hot composite curve and balanced cold composite curve then whatever area we will find out it will include the heaters and coolers and process heat exchangers. Now whatever we have seen in the algorithm we will see now here. So it shows you a ith interval in the balanced composite curve there are two hot streams available and three cold streams available. So each hot stream is divided into three parts or splitted into three parts and each cold stream is splitted into two parts and in this way the heat exchange takes place. That means the first part of the hot stream one exchanges heat to the first part of the cold stream three. So this forms a exchanger network in the ith interval and facilitates the vertical heat transfer. Here we see that the hot composite hot stream rises from this temperature to this temperature which is delta th and the cold stream rises from this temperature to this temperature which is delta t c. Now in this ith interval all the hot streams rise from this temperature to this temperature that is delta th which remains constant for all the hot streams and for the all the cold streams of ith interval the cold streams temperature rise from this point to this point which is delta t c. Now if I number them that means for this heat exchanger the area is one three because one is exchanging heat with the cold stream three that is why the area is one three and the q exchanged in this heat exchanger is q one three and the overall heat transfer coefficient is u one three and the delta t log mean temperature difference is delta t l m one three. Similarly for this heat exchanger which is between stream number two and cold stream number five this is area is two five heat transfer or the capacity of heat exchanger is q two five u overall heat transfer coefficient is u two five and delta t l m is delta t l m two five. Now if we adopt this nomenclature and we know that area is equal to q divided by u delta t l m when u is the overall heat transfer coefficient and delta t l m is the log mean temperature difference then the area calculated will be a i is equal to a one three a one four a one five a two three a two four and a two five because there are six heat exchangers in that ith interval and if I add up the areas of all the six heat exchangers which were a one three a one four a one five a two three a two four and a two five then this will give me the complete area of the enthalpy interval i. Now by substituting the values of a one three to a two five in terms of q u and delta t l m we can write down that a i is equal to q one three divided by delta t l m one three u one three plus q one four divided by delta t l m one four u one four plus that q one five divided by delta t l m one five u one five plus and so on so forth. This should be remembered that the values of t h i by t c o and t h o by t c i for all hot and cold stream in the interval will be the same therefore delta t l m should be the same for all heat exchangers in a particular interval so this shows how the delta t l m of the ith interval enthalpy interval can be computed. So, this is delta t h i minus t c o minus t h o minus t c i divided by this l n t h i minus t c o t h o minus t c i. Let us extend the algorithm of the unequal heat transfer coefficient for area targeting in the ith interval we have seen that the delta t l m that is log mean temperature difference of all streams in the ith interval are equal. Hence we can say that delta t l m one three delta t l m one four delta t l m one five delta t l m two three and delta t l m two four and delta t l m two five which are the individual streams delta t l m's are equal to delta t l m i. Further the overall heat transfer coefficients which were u one three u one four u one five u two three u two four and u two five can be now expressed in terms of the individual heat transfer coefficient of the streams. For example in this we see that the stream one this part of the stream one is exchanging heat with this part of the stream three and if the overall heat transfer coefficient of this heat exchanger is one u one three then one by u one three is equal to one by h one plus one by h three because this has two streams this heat exchanger stream one and stream three. Even if we split a stream into many parts the heat transfer coefficient of the stream does not change that means for stream one if this part has got h one heat transfer coefficient this part will also have h one heat transfer coefficient and this part will have also h one heat transfer coefficient. Similarly for cold stream three if this part has got heat transfer coefficient h three this part will be also have heat transfer coefficient h three. So, I can write down one by u one three is equal to one by h one plus one by h three similarly one by u one four is equal to one by h one plus one by h four and so on so forth I can express all the overall heat transfer coefficient in this manner where h one and h two are film side heat transfer coefficient for hot stream one and two h three h four h five are the heat transfer coefficient of the cold stream three four and five respectively. Now, if I take the delta T l m i common so one by delta T l m i will be equal to q one three by h one plus q one three by h three plus q one four by h one plus q one four by h four q sorry q one four by h four similarly I can express this then finally I can find out by solving these equations that the total heat transfer area of the ith interval will be given by one by delta T l m i summation j one two m q j by h j of i. Now, here the i stands for number of enthalpy intervals and j stands for the number of streams hot stream cold streams and utility streams in the ith enthalpy interval. There are n number of enthalpy intervals and there are m number of streams in an enthalpy interval the value of m will change with i and delta T l m i is the l m T d for the ith interval and q j is the enthalpy change for the j th stream in the ith interval h j is the heat transfer coefficient of the j th stream in the ith interval. Now, the summation j q j by h j can be written by summation q j h q by h j h plus summation on cold streams number of cold streams j c q by h j c that means I am now separating the hot streams and cold streams and doing summation individually for hot streams changing the heat of hot streams and change in the heat divided by the heat transfer coefficient for the cold streams. So, q j by h j can be written in this fashion when integrations are done for the number of hot streams and number of cold streams separately. Now, we know that the q can be expressed in terms of the formula q equal to m c p delta T. Now, in a ith interval the delta h for all the hot streams are equal similarly delta c for the old all the cold streams are equal hence q j by h j summation j can be written as delta T h for ith interval summation j h that means I am summing over the number of hot streams m c p divided by h for g h plus delta T c for ith enthalpy interval into summation over the cold streams j c m c p by h j c. Within an enthalpy interval all hot streams undergo the same temperature difference that is delta T h and all the cold streams also go undergo the same temperature difference that is delta T c. Now, if we put this into the equation then all the above equations which has been discussed up till now will form a algorithm in which I have considered the different heat transfer coefficient that means the algorithm is applicable to find out the area target under a condition when the stream heat transfer coefficients are different. To illustrate the algorithm let us take up a problem the stream data of the problem is given here you will notice that with the stream data we have added a new column which shows the stream heat transfer coefficient of utilities as well as process streams like for h 1 stream the heat transfer coefficient is 0.01 for h 2 stream its heat transfer coefficient is 0.04 for h 3 stream it is 0.5 and for st which stands for steam its heat transfer coefficient is 0.05 and for cold water its heat transfer coefficient is 0 2 for this problem we have taken delta T minimum is equal to 10 degree centigrade. Now, our requirement is to find out the minimum heat transfer area please note this I am going to calculate the minimum heat transfer area not maximum heat transfer area of the hen considering unequal heat transfer coefficient using bath algorithm. So, whatever the algorithm we have shown you it is called bath algorithm. Now, once the stream table is available we can go for problem table algorithm PTA to find out minimum hot and cold utilities. If we do this we find that the minimum hot utility requirement is 1064.5 to kilowatt minimum cold utility requirement is 855.84 kilowatt. Now, based on these values of hot and cold utilities we will try to find out what will be the value of Cp capital Cp for cold utility as well as hot utility. So, the Cp of cold utility will be Q Cu minimum that is the amount of cold utility requirement divided by T out minus T in of the cold utility and this will give rise to 855.84 divided by 60 minus 20 and thus the Cp of the cold utility comes to be 21.396 kilowatt per degree centigrade. For hot utilities specially when we are using steam we have to find out what is the temperature of the input stream and what is the temperature of the out. We know that steam condenses at a certain temperature and is a isothermal condensation, but however if you see practically steam has to move from one point to another point that means there is a temperature or there is a pressure gradient which will allow the steam to flow from one point to another point. This clearly indicates that the steam will not condense at a constant pressure, but it will condense at a differential pressure that differential pressure may be very less. So, here we assume that the temperature of the steam input steam is 300 as given in the table, but the T out of the steam is 299 which is 1 degree less than the T in. When we are doing this we are not committing any mistake because we are in thermal balance and the Cp of the hot utility which we calculate will take care of this. In this case the Cp of the hot utility calculate is 1064.52 kilowatt. Once the Cps of this are known we can plot the hot and cold temperatures in ascending order. So, we are going to compute the balance hot composite curve and that is why we will only take the data from the steam table which is related to the hot streams that is h 1, h 2 and h 3 and the steam which is the hot utility. So, when we consider the supply and target temperature of the streams we find that 77 is the minimum temperature of the hot streams including hot utility and 343 is the highest temperature. Now, we write down the corresponding Cp values for the hot streams and hot utility according to the temperature intervals. So, h 1 stream moves from 159 degree centigrade to 77 degree centigrade and its Cp value is 22.85. Hot stream h 2 moves from 267 to 80 degree centigrade its Cp value is 2.04 and h 3 which is the hot stream 3 is moving from 243 to degree centigrade to 90 degree centigrade and its Cp value is 5.38 and the steam which is from 300 to 290 degree centigrade 99 degree centigrade its Cp value is 1064.52. Now, we will compute the cumulative Cp values in different temperature intervals. Now, in this temperature interval 80 to 77 degree only h 1 is present and the Cp value of h 1 is 22.85 and that is why in this temperature interval the cumulative Cp value is 22.85. However, from 90 to 80 degree temperature interval there are two streams h 1 and h 2 and the Cp value the summation of the Cp value will be 22.85 plus 2.04. So, it becomes 24.89. Similarly, in this interval 90 to 159 there are three hot streams present. So, the summation Cp h b will be the summation of all these three quantities. So, we see here Cp value is 30.27 here from 267 to 159 only two hot streams are available. So, summation Cp will be 2.04 plus 5.38 which comes out to be 7.42 and this temperature interval only h 3 is working. So, Cp value will be 5.38 and this temperature interval h 3 plus this steam is working. So, the Cp value will be combination of these two figures comes out to be 1069.9. In this temperature interval 343 to 300 only one stream is working h 3 and hence its Cp value is 5.38. Then after computing this we will find out the heat required for each interval q h b which is equal to delta T into summation of the Cp values. So, this will be delta T here 80 and 77 it is 3 into 22.85. So, this is equal to 80 minus 77 into 22.85 is 68.55. So, this goes here similarly we can fill up the heat required for each interval and we should remember this is all we are doing for hot stream and hot utility and we are computing data for BHCC and once this data is ready we will plot balanced hot composite curve. Now, we plot the cumulative heat required for each interval. So, the first interval will be 68.55 the second interval will be this plus this and for the third interval this will be this plus this plus this. So, we have filled up the cumulative heat required for each interval. Now, we will see how to compute the data for balanced cold composite curve. For balanced cold composite curve we will select from the stream table the data for cold streams that is C1 and C2 and cold utility. The temperature levels are written here as from 20, 26, 60, 118, 127 and 265 in the increasing order. The cold utility CW goes from 20 degree centigrade to 60 degree centigrade it is Cp capital Cp is 21.396 C1 moves from 26 to 127 degree centigrade it is MCP value is 9.33 and the C2 goes from 60 degree to 265 it is MCP value is 60 to 265. Now, we compute this summation of Cp values in different temperature intervals. Now, in the first interval it is 26 to 20 and only the cold utility is working. So, here 21.396 will be written. So, it is 29, 1.396. However, for this 60 to 26 temperature interval two streams are working C1 and CW. So, this value will be a summation of this and this value. Similarly, in the third temperature interval this is 118 to 60 two cold streams are working. So, here the value will be summation of this and this and for the last temperature interval this is 19.61 as only this stream is working. So, once this is available to us we can compute Qcb which is delta T into summation of the Cp's of the interval. Here the summation is 21.396 the temperature difference is 6 degree that is 26 minus 20. So, 6 into 21.396 will give you 128.376. Similarly, for this temperature levels interval from 60 to 26. So, delta T is 60 minus 26 into this Cp value 30.726 this comes out to be 104.684. Similarly, we have filled up this. Now, we will fill up the cumulative Cb values this is the for the first interval this value is 128.376 for the second interval this is the summation of this value and this value which is 1173.06 for this interval this is the summation of this value this value and this value. So, it comes out to be 1714.2. So, similarly we fill up this column also. Now, the data is ready for balanced composite curve we have created data for the cold balanced cold composite curve as well as balanced hot composite curve. So, this is the hot composite curve data and this is the cold composite curve data this is available for us. Now, we can blot them. So, this is my cold composite curve balanced cold composite curve this is my balanced hot composite curve. Now, these are the points where this line is changing slope here the line is changing slope here the line is of the cold balanced cold composite curve is changing slope here the balance hot composite curve is changing slope here the balanced cold is changing the slope so on so forth. Now, what we see that though the temperature here is known the corresponding temperature at the cold balanced cold composite curve is not known. Similarly, here the temperature of the balanced cold composite is known but corresponding temperature at the hot balanced composite curve which is this point is not known. Now, if I consider this to be enthalpy intervals and I want to find out the area of this enthalpy interval I should know all the four temperatures out of these four temperatures only two temperatures are known and rest two are not known. However, one thing we get it here it is a straight line relationship. So, we can use the technique of linear interpolation to find out the temperature here. So, we will see that that how this technique should be used here we see that this can be divided into 11 enthalpy intervals and here we see there are some temperature which is not known like T C 12 is not known T C 10 which is corresponding to this is not known T C 9 is also not known but we know the C P summation C P value of this interval. So, based on the C P values we can calculate the temperatures using the linear interpolation method. So, these values are not known. So, we have to compute these values cumulative Q H P and cumulative Q C B are massed by omitting cumulative enthalpies common to both values and the entries are then stored in ascending order this identifies all the points where composite curve has a vertex. Now, from these two tables of which one is for balanced hot composite curve and this is for balanced cold composite curve we will note down the enthalpy cumulative Q H B and Q C B values. So, these value and these values are massed together and they are sorted in ascending order if two values match we take one out of it we have already computed all the data required for the plotting of balanced hot composite curve and cold composite curve. Now, the job left is to find out the temperatures of those areas or those points where for balanced hot composite curve the temperature is not known and those areas where the for the balanced composite curve for cold the temperatures are not known we will see here. Now, here it is T H I is given and we clearly see there are some enthalpy intervals where temperatures are not known for hot similarly for cold there are some intervals where cold temperature is not known. Now, we will fill up the summation of C P H B and or summation of C P cold in those areas where either its value is available for hot or its value is available for cold like for example, at 80 degree centigrade of hot steam temperature the value of C summation of this C P H B is available which has been taken from here. Similarly, for 26 degree centigrade of the cold the value of summation C P C B is taken here as 21.396 that means wherever the temperatures are available we have taken the values of the summation of C P H B or C P C B wherever possible. Now, let us see that what are the temperature which are not available to us like this temperature for the hot is unknown this temperature for the hot is unknown this temperature of the hot is unknown this temperature is unknown however, these temperatures are available similarly for the balanced cold composite curve this temperature is unknown this temperature is unknown and these are the temperature which are unknown. So, now our job is to find out these unknown temperatures and the method which we will be employing for finding out the unknown temperatures is the linear interpolation because the segments of the balanced hot composite curve or the balanced cold composite curves are straight lines and hence we can very effectively use the linear interpolation technique to find out the temperatures. Now, the temperature which are not known are given in this color for the hot similarly we will find out the temperature which are not known given in the balanced cold composite curve in the green color. So, let us see how to find it out now this is the equation using which we will be finding out the unknown temperatures the explanation is given here, but it is better to go through the method to understand it. Now, to start with we will try to find out the unknown temperature here which is T H 3. Now, T H 3 is given by using these temperatures and cumulative enthalpies and summation of delta C p values. Now, this shows that I want these temperatures. So, the below temperature is given which is 90 I have written it 90 then corresponding to this temperature the cumulative Q is 317.45 taken 317.45 and we have deducted this value here 128.376 then we have divided this with the summation C p values which is available here. So, it gives 82.40 as the unknown value. So, this moves to unknown value here. Now similarly another temperature which is T H 5 is also not known. So, we calculate T H 5 here. So, if you see the T H 5 then we will be using this 4 values to compute T H 5. Now, if you see if you want to find out this temperature the next unknown temperature is this 159 because these two temperatures are also unknown. So, there is no information available about the temperature in this two temperature intervals. So, the known temperature is here 159. So, we take 159 minus the enthalpy here is 2406.08 we have taken this then minus the enthalpy which is known here is this the unknown level is this. So, we have taken this then divided by the M C p value summation M C p value which is available here on this row. So, this is 118.27. So, moves to that and fills it up 118.27. Similarly, these two unknown values are also computed based on linear interpolation technique. Now, we have completed the unknown temperature values for the computation of area target for hot balanced hot composite curve. The same thing has to be repeated for the balanced cold composite curve. So, the same type of equation is used here to compute the unknown temperatures of the balanced cold composite curve this is a unknown temperature which is T C 2 unknown temperature T C 4 unknown temperature T C 8, T C 9, T C 10 and T C 11. So, the first attempt will be taken to compute the T C 2 value. So, T C 2 is equal to 26 minus. So, these are the temperatures cumulative enthalpy and summation C p values which will be using. So, this is equal to 26 because this is the unknown value next value is 26. So, 26 minus this value 128.376 minus this value divided by the summation of C p values available here. So, it is 23.20. So, it moves to this place and it becomes 23.20 value. Similarly, the unknown values here are computed and transferred. So, the second so these are the values which will be used now 60 minus we are computing this value unknown value that is T C 4. So, this is 60 minus this cumulative value then minus this value then divided by in this row the value of summation C p. So, the calculated temperature is 32.15 it moves to here. Similarly, these 4 values are also computed and filled up. Now, we are able to compute all the unknown temperatures of the balanced hot composite curve and all the corresponding temperatures in the balanced cold composite curve and this is required to find out the LMTD for all the enthalpy intervals of the balanced composite curve. So, this work is now finished. So, we will go for computation of area. So, this values are filled up here and we will use this table to compute the area target. Now, for computation of the area target we will have summation C p divided by H values for hot and summation C p by H cold values for cold. Now, these are our streams and these are the heat transfer coefficient available for different streams like for H 1 the heat transfer coefficient is 0.1 for H 2 it is 0.04 for H 3 it is 0.5 and for the steam it is 0.05 for cold water this is 0.2 and so on so forth. So, we have to fill up these two columns. Now, if I see for this interval only one hot stream is working that is H 1. So, the summation C p divided by H can be computed as for interval 1 C p H for interval 1 is this value. Now, divided by the H that is heat transfer coefficient of this stream H 1 which is 0.01. So, we divide this by 0.01 and find out the value 228.5. Now, for the second this goes to here 225 228.5 and for this interval we have two streams which are working. So, we have already computed for this stream. So, we will use this result and then we will newly calculated for these streams that means 2.04 divided by 0.04 and we then add them up this with the value which we compute for this stream the total will be written here. So, let us see for stream interval 2 for enthalpy interval 2 this value will be summation of this value will be equal to we plus it this value divided by the heat transfer coefficient of the second stream and we add this two we get this two 79.5. So, that moves here similarly for this three here we have two streams working for this fourth we have three hot streams working here and for here we have two hot streams that means one hot streams H 3 and the utility stream is working. So, we can find out the summation C P by H values for hot in those intervals and we can fill it up. Similarly, we can go for the cold one if I see the interval number 1 then only the cold water stream is working its C P value is 21.396 and the H value is 0.2. So, this value becomes 106.98 which is 21.396 divided by 0.2. So, this value I get now in the second interval the same cold water stream is there that is cold utility stream. So, the value is the same 106.98 because it is not changing whereas in the third interval there are two cold streams. So, my value will be 21.396 divided by 0.2 plus 9.93 divided by 0.01 when I add them together I get a value 1039.98. So, similarly this column has been filled. Now, we are interested in finding out what is the Q by H value because Q by H value is needed for the computation of H. So, Q by H is computed as the differential in temperature. Now, what it is there it is 80 minus 77 these two 80 minus 77 into the value 228 which is this plus then 23.204 minus 20 this is these are delta t values into summation C P values. So, this is multiplied 106.98 and when we add these two terms we get 1028264 in this Q by H term the Q for the cold as well as Q for the hot is considered and that is why we are multiple adding two terms. So, this result goes here similarly for the next we can compute and we can fill up this Q by H values. Now, the next is to compute LMTD because we know the temperatures the extreme temperatures extreme mean and extreme out temperatures. So, four temperatures are needed to compute LMTD value. So, we see that we find out the LMTD value for interval 1. So, for this the LMTD value is this this is 77 minus 20 77 minus 20 then 80 minus 23.2 then ln this divided by this which comes out to be 56.90. So, we are using this four values to compute this LMTD similarly for this we can use this and this and this and this values to compute LMTD and we can fill it the LMTD values here. Then area can be calculated because area is summation Q by H divided by the LMTD for each interval. So, for interval 1 we have Q by H values for this and LMTD value for this. So, we can divide this value 1028.264 divided by 56.90. So, we get this value 18.07. So, we similarly we fill it this column and when we add this I get the area because now I have calculated the area of each interval and when I add this all I get the total area of the heat exchanger network which includes heaters and coolers. So, my final result is total area which is 4154.659 meter square. Now, this area is the minimum area after this lectures we will take the area targeting part 2 in which we will discuss if the heat transfer coefficient of all the streams are same then how to compute the minimum area using the bath algorithm and then we will discuss that in which condition bath algorithm will give minimum area and in which condition there will be a departure of minimum area that means it will not give minimum area and it will give more area and then we will discuss what are the other alternatives available for area targeting which will lead to minimum area. These are the references and thank you.