 We go on with advanced reaction engineering. Today, we will be looking at some practice problems in energy balance. The problem we would like to take is an interesting exercise. Let me just explain the background to this exercise and then we will look at the problem itself. You have nitrogen N 2 O 4 giving you twice N O 2 and twice N O 2 giving you N 2 O 4. So, this is the reaction. Now, these two reactions 1 and 2 they are considered to be instantaneous. In the sense at any instant of time N 2 O 4 and N O 2 exist in equilibrium. If you look at some data on this system data looks like this N O 2 N 2 O 4. So, you have delta H F 7.96 2.23 delta G 12.26 23.4 C P specific heat 12.5 and 25 units are K cal per mole K cal per mole. This is calories per mole degree K. Since this reaction is instantaneous it is of interest to us to see whether we can use this reactions to perform some useful function. Two functions which are great interest to society is space heating and of course production of electrical energy of power. Now, space heating as you all will know is requires low temperature heat basically and generally low temperature heat is available. This is an abundance of low temperature heat particularly in the industry. Therefore, if you want to do space heating and if you can make use of this low temperature heat then you probably be able to do it economically. Now, if it comes to work of course we are limited by the first law and we are also to content with the fact that steam which is the working fluid in our engines in the power plants steam has a very large where water has a very large vapor pressure. Therefore, if you want to go to high temperatures you necessarily have to go to very high pressures as a result very high pressure equipment expensive and so on and related hazards of etcetera. So, in it is in this context we have device this exercise and to see you know how we can learn to use certain properties of substances that exist around us for some functions that we might think is of use to us that is the context. Let us put this context to some use that we are looking at by let us say as an example. We want to use for space heating application what is meant by space heating application that we just draw what might be the let us say we have a device. So, we have here a all right. So, we have a pump let us say this is this is going from N 2 O 4 going to twice N O 2 N 2 O 4 going to twice N O 2 this is a 80 degree C and this is a 20 degree C. So, material is coming in like this and going out like this. So, what do we expect is that it picks up heat at 80 degree centigrade it picks up heat. So, let us say q 1 and there are 20 gives of that heat q 2 and you have a moisture remover moisture the idea of this moisture absorber is that in case there is some moisture that gets into the system we should get rid of it because N 2 O 4 N O 2 O with corrosive gases we really do not want any moisture at all. So, in essence what we are trying to say here is that if you have a device which is able to pick up heat from a low temperature source and this N 2 O 4 reaction goes towards in the appropriate direction. And then here it this low temperature it is giving that heat and therefore, it is able to reverse that reaction. Now, we can calculate based on the thermodynamics that is given to us that K p K p at 80 C is 5.8 and K p at 20 C is 0.45. So, this is K p at 20 C is 0.45 and. So, what it means is that this reaction at 80 C the K p is 5.8 and at 20 C is K p is 4. Which means what this is an endothermic reaction that means N 2 O 4 if it has to go to N O 2 you have to supply heat. And then when you reduce the temperature the reaction reverses itself releasing that heat basically it is this instantaneous ability for this reaction to you know go both ways depending upon the temperature at which the environment is available. It is able to we are able to pick up heat and remove and supply heat to the source to the environment of our interest. So, our this whole exercise is to be to quantitatively evaluate what is the ability of a working fluid like N 2 O 4 N O 2 to work as a as a means of recovering heat from a low temperature source and supplying it to from a source at 80 and supplying it to a consumer at 20. The consumer at 20 is space heating application the source 80 degree C might be a source of I mean waste fluids that might be available in an environment may be a factory and so on. So, this is the application that we are trying to evaluate in this exercise. Now, how do we actualize this in a process after all you have to actualize this or we actualize this. To actualize this what is suggested is you have two fluid beds and your fluids are moving between. So, you have fluids coming in going goes like this and then once again coming in going through like this. So, it comes out like this and goes in like this and comes out like this or other words what we are saying is that we have a hot this is at 80 and this is. So, this is coming out at 20. So, what is it that we are doing. So, we have let us say air available at 80 and we contact that with this cooling this coil where 80 degree C temperature is taken up by this gases N 2 O 4 going to N O 2 and then that is released to air this is another air stream to. So, that you know 20 if it is 20 coming in it will probably go out at 30 let us say. So, that air gets heated and that is how we realize this space heating application. So, let us model this let us model this there are two environments this model this like this there are two environments let us say environment 1 and environment 2 environment 1 is at 80 environment 2 at 20. You have fluids moving between the two why the reaction N 2 O 4 giving you 2 N O 2 takes place. In both cases it is the same reaction taking place twice N O 2. I call this environment as E and this environment as F. Therefore, you have. So, it is going like this. So, it is E A I will call it E A and E B. So, I call this as A I call this as B and then F A and F B are coming out. So, what is the system it we have an environment at 80 degree C which we call as E and we have another environment is 20 which we call as F and we have this fluids which contains N 2 O 4 N O 2 and so on. We call E A E B is what is going from E to F and as it encounters a lower temperature immediately the reaction takes place. So, that the composition changes and F A F B come out. Now, since the reactions are instantaneous what does it mean in a reaction is instantaneous it means that K P equal to P B squared divided by P A this equality is equal to P B squared the applies. We all understand this thermodynamics that if a reaction is instantaneous the compositions of the leaving stream E A and E B will be at equilibrium corresponding to 80 degree centigrade. Similarly, we can say this is K P at 80 similarly, K P at 20 also equal to P B squared divided by P A or this equality whether it is 20 and 80 should hold because the reaction is instantaneous. Therefore, the compositions of the leaving streams must be corresponding to the equilibrium compositions. Our interest of course, is to use this reaction and recover some energy. So, we need what is called as to understand what are these flows E A and E B. So, in this particular case what is mentioned is that E A it says the problem states that the nitrogen that is flowing in this stream the amount of nitrogen here whether it is E A and E B or F A and F B total amount of nitrogen is 1 mole. It is 1 mole per second is the total amount of nitrogen that is what is specified. Now, we know that if you look at E A if you look at stream E A here this stream E A this stream E I mean coming out of a equipment E it has this is this is N O N 2 O 4 and this is N O 2 therefore, the amount of nitrogen here for every mole of this contains 1 mole of nitrogen, but every mole of this contains half a mole of nitrogen. So, if E A and E B are the molar flows of A and B coming out of environment E then what is given to us is E A plus E B by 2 is 1. So, what is the problem statement problem states that we have 1 mole per second of nitrogen flowing in the circuit. That means what I am saying is that this 1 mole of nitrogen it can be partially as N 2 O 4 or partially as N O 2 depending upon the conditions of the environment, but the total nitrogen in the system is fixed which is 1 mole per second. So, 1 mole per second is circulating, but as it leaves E I mean environment E there will be certain compositions of N 2 O 4 N O 2 while it leaves F there will be different compositions N O 2 and N N 2 O 4. The reason being that the temperature at 80 degrees and 20 degrees the equilibrium compositions as determined with thermodynamics are different. So, what is Y A by definition and what is Y B all right. Now, we also know that K P I am talking about environment E now K P equal to P B square divided by P A that we know from our basics. Now, therefore this P B square by what is also can be written as P T times Y B square divided by P T Y A P T square. So, this becomes P T Y B square divided by Y A. So, essentially our equilibrium representation gives you this K P is given by P T Y B by Y A. Now, Y B and Y A can be replaced from here. So, suppose I do that we get K P equal to P T P A divided by E B square divided by E B plus E A divided by E A. Now, we are looking at environment E. So, let us do at environment E first and then do the environment F. So, K P equal to P T Y B square is coming from here E B square E A plus E B whole square denominator Y A also E A. So, it is how so 1 E B plus E A has got cancelled. So, we get this kind of equations I can simplify this further by writing P T into we notice we notice from this equation suppose I call this as some equation 1. So, you can see here from E B can be put from here as 2 E A is 2 minus 2 E A. So, we put it as 2 minus of 2 E A that is E B that is E B it comes from equation 1 divided by E B E B plus E B is what E B is let me let me write this like this 2 minus of twice E A plus E A is it. I have replaced E B plus E A like this and multiplied by E A equal to K P is it clear what we say what did we say let me just run through this whole thing once again. We have 2 environments 1 at 80, 1 at 20 the reaction N 2 O 4 to N O 2 is instantaneous. Therefore, the compositions of this mixture at 80 will be determined by its equilibria. Similarly, composition of the mixture at 20 will be equilibrium constant at 18, 20 are given. Now, we want to since both these environments are well mixed. So, we are saying that K P value which is P B square by P A which can be expressed in terms of Y B and Y A and so on. Therefore, we are able to express K P in terms of total pressure and mole fractions and mole fraction of Y A and Y B comes from basics. And we also know from the problem statement that 1 mole per second is circulating in the system. Therefore, E A plus half of E B equal to E B equal to 1. Why this word half is coming is because we notice that N 2 O 4 as 1 mole of nitrogen while N O 2 as only half mole of nitrogen. That is why when we say 1 mole of nitrogen is circulating in the system it means that E A plus half E B is 1. So, that is why we have got this equation 1. So, this equation which you got P T multiplied by all this now involves only E A. Let me just write this once again. So, our equation for so K P equal to P T times 2 minus of twice E A divided by 2 minus of twice E A plus E A within bracket of E A or it is P T times 2 minus of twice E A divided by 2 minus of E A times E A equal to K P which is equal to 5.8 at 80 degree centigrade. Let me just go through this square bi-sync here this square bi-sync here this square bi-sync here this square bi-sync here please notice notice as E B square I just forgot the square. So, it can be simplified further equal to K P equal to 4 P T into 1 minus of E A whole square divided by 2 minus of E A multiplied by E A and that K P is 5.8 at 80 degree C. Now, this quadratic I mean can be easily solved let us see how the solutions look like. So, let us expand this and simplify when you do all that it you get 11.6 E A minus of 5.8 E A. E A square equal to 4 times 1 minus of twice E A plus E A square let me just check how it looks like 1 minus. So, this is 1 minus E A square. So, we have expanding this. So, when you do that when you do that you get this term is correct and now 2 E A divided by now it is 2 multiplied by 5.8 is 11.6 that is mix make sense and then 5.8 E A that also make sense everything seems alright. So, essentially we have of course, we have taken P T total pressure equal to 1 atmosphere that is an assumption. So, this is fine this basically expanding multiplying cross multiplying and then simplifying. So, it is fine. So, we have here. So, let me so it is 4 9.6 E A square is it correct 4 9.6 E A square minus 8 E A and therefore, minus and minus 8 19.6 E A. So, I have taken this also plus 4 equal to 0. So, what is the solution? So, E A equal to 19.6 plus or minus root of B square minus E A square minus E A square minus E A square is 19.6 whole squared minus 4 times 9.6 times 4 divided by 2 into 9.6 or this 9.8 and 9.6 E A square where is 9.8. So, this gives you solution E A equal to 9.6 E A square what is E A equal to 0.23 or 1.76. This is the 2 solutions we get. Now, E B equal to 2 minus of twice E A therefore, that gives you 2 minus of 2 E A. Now, if you put this when you put 1.76 here then E B becomes negative therefore, this solution is not admissible. Therefore, the solution which is admissible is the second one 2 into 0.23 that is equal to 1.76. So, mole per second. So, E A this solution is admissible. So, many mole per second. So, what we have done is we have taken this exercise to 80 degrees and 20 degrees. We recognize that the reaction is instantaneous. Therefore, the composition is described by the equilibrium at 80 degrees and based on that and then the condition that 1 mole is circulating through the system which means that E A plus half of E B equal to 1. So, we put that condition and then we sort of went through the material balance to find out what is the value that is appropriate. So, this is as far as E environment E is concerned. So, we have to repeat this for environment F because what is the situation in environment F. So, this is environment environment F. What is the situation environment you have k p equal to 0.45 equal to p t times p b square sorry p b square sorry p b square divided by p a. So, this is 0.45. So, 0.45 equal to p t multiplied by y b square divided by y a and now that is equal to y b we can substitute for y b in terms of F a and F b and so on. So, let us do that we have done that already. So, we do not have to do it again. So, our k p equal to 0.45 equal to k p we look like we can see do not write the whole thing once again or equations. Now, we will go 4 p t within brackets of 1 minus of F a whole square divided by 2 minus of 2 F a multiplied by F a. So, it will be identical what we got before there it is 4 p t 1 minus e a square here it will be 1 minus a F a square denominated 2 minus 2 e a 2 minus 2 F a. So, it is the same kind of it is similar. Now, then we can cross multiply and simplify and so on when you do that we get 0.9 F a minus 0.45 F a square equal to 4 times 1 minus of twice F a plus F a square. Let us see where you have got it right. So, 0.9 so this is 2 multiplied by 0.9 F a all right. So, it is once again 0.9 F a 0.42 into 0.45 why is this 0.45 F a 2 minus of 2 F a we missed plus F a is there. So, that F a I had missed out. So, this is that is the mistake. So, 2 minus of F a so this. So, it simplifies as 4.45 F a square minus of 8.9 F a plus 4 equal to 0. Let me see if this is correct. So, 428 F a if you take it to this side 8 minus of 0.9 have you got 0.9 F a. So, it is 0.8 minus of 8.9 is that 8.9 correct this term is correct this term is taken this term is taken. Now, 0.45 F a square plus 4 F a square this is fine and then plus 4. So, this is the equation. So, let us look at the solution to this. So, the differential equation this is once again we are in environment environment F. So, our equation is 4.45 F a square minus of 8.9 F a plus 4 equal to 0. Therefore, F a equal to minus b plus or minus b square minus of 4 into 4.45 into 4 divided by 2 into 4.45. So, this is the solution and it turns out that it looks something like this 2.83 divided by 2 multiplied by 4.45. So, that gives us 2 solutions 0.68 and 1.317 mole per second and we also know that F b equal to 2 minus 2 F a. So, this is the equation this is F a therefore, F b equal to 2 minus of 2 times F a if it is 0.68 it comes to 1.32. So, 0.68 is that right. So, it is F b. So, it is 2 is 1.36.64 mole per second and this solution is not admissible because 2 times becomes it violates the principle. So, we have what we have done is that by setting out the equations under equilibrium we have been able to determine the value of E a and E b. Let us just look at our system once again. So, that we understand what we are doing. What are we saying now is based on the thermodynamics of the system that we have we have F a and F b this these numbers are this is 0.64 F a is 0.64 sorry F a is 0.68 F b is 0.64 and E a 0.23 and E b is 0.64 and E b is 0.64 and E b is 0.23 and E b 1.76. So, you can add up the 2 E a plus E b. So, if you divide this by 2 it is 0.88 and it comes out to about 1 some calculation mistake therefore, it does not add exactly to 1. So, what we are saying is that once you impose the condition that they are in equilibrium that instantaneous at 80 and 20 and assuming that they are all well mixed vessels and therefore, the composition the exit is same as composition inside the equipment and so on. We get value for E a E b F a and F b. Now, the question that is of interest to us is to find out what is the amount of energy that we can get out of this. So, to do this what we have to do is we have to do the energy balance that is to that. So, we have the energy balance what is your energy balance we say that the enthalpy at 80 equal to H a E a plus H b E b. That means, the enthalpy of component a multiplied flow rate of component a enthalpy of component b multiple flow rate of component b. Now, we have to just calculate what is H a at 80 equal to H a at 25 plus C p into t minus t r. So, this is from the data which is given is 22.3 plus 25 multiplied by 20 multiplied by 80 minus 25 is that clear because the environment is at 80. So, this is the amount this becomes. So, H a at 80 equal to 22.3 plus 25 into 80 minus 25 is 55. So, it becomes 3605. Is this does it make sense what is 25 25 is what what is the units of just let us say the numbers because many of these things units have to be properly taken care. It is calories per mole while all these are in kilo calories. So, 22 they change 22 because and this is calories per mole. So, it is 0.125. So, it is divided by 1000. So, this is not an important quantity. So, it is something like 22.3 kilo calorie per mole. H a at 80 similarly, we have to find out. So, what is the total this is this is the H component as such this component at 22.3 H component b at this is for component b at 80 degrees is H b at 25 plus C p t minus of t r. It is a small quantity will neglect this not a very important quantity. So, H b at 25 we can calculate from the data that is given to us H b at 25 the data is given to us here 23.1 and 2.23 sorry 2.23 and. So, that comes out to be from the data it turns out to be 7.9 it is 7.9 kilo calories per mole. So, what we have done we have calculated what is the enthalpy of component a what is the enthalpy of component b. Therefore, we can now calculate what is the energy that is associated with the let us do that what is the energy associated with the stream what is the energy associated with the stream fairly straight forward energy associated. So, it means total amount of energy at the environment 80 is E a H a plus E b H b what is E a V a is 0.23 multiplied by 22.3 and then E b is 1.54 multiplied by it is 7.96. So, many kilo calories per second is this clear what we are saying is we know the molar flows our system please do not forget our system our system is we just get hold of the we can draw it once again this is our system is this is this is E and this is F this is going like this is coming like this this stream E a is 0.23 this is 0.23 E a is 0.23 and E b is 1.54. So, if you divide this by by 2 it is 7.7. So, 0.77 from 0.23 is 1. So, it sort of satisfies the material. So, this is as far as at 80 degree stream is concerned this is 20. So, we want to do the same thing at what happens at 20 we repeat that at 20. So, we calculate what is the energy associated with streams that are leaving the. So, for the 20. So, H so the total amount of energy at environment at 20 is F a H a plus F b H b this we know. So, H a at 20 we have to calculate that is equal to 2.23 plus that C p term C p t minus of t r which we will neglect. And similarly, we have H b at 20 equal to 7.96 minus C p that term delta t term will neglect. So, it is 7.96. So, our H t at 20 equal to E sorry F a H a plus F b H b. So, F a is how much F a is 0.68 H a is 2.23 and F b is F b is 0.64 multiplied by H b and the H b is 7.96. So, many kilo cal per second. So, so this is rough for roughly 2 by 3. So, that is about this is about 0.74 this is 8 4.5 0.1 that is about 5.84 k cal per second. This is energy balance H a H b just a minute we want to calculate H t what is H t at 80. Let us just quickly calculate this that is about 1 4th 1 4th of this would be about 5.4 5.5 and this is 8 is about 12 that is about 17.5 kilo cal per second. So, what we are saying now is the energy that is the energy that is in this stream is 17.5 units. The energy that is in this leaving stream is 5.84 units. What are we saying? What we are saying is that energy that is leaving the 80 degree centigrade environment is 17.5 units. What is leaving is 5 point therefore, the energy that is delivered that means what is taken up minus what is this energy that is delivered would be therefore, energy released into environment f equal to 17.5 minus 17.5 how much is this 5.84 this is 5.84 minus 5.84 5.84 that turns out to be 11.5 is about 11.64 kilo cal per second. Now, if I calculate this nitrogen circulating is 1 mole per second or 14 grams per second. Therefore, the energy that is delivered to the stream per gram of nitrogen. So, specific energy delivery delivered is 11.64 divided by 0.014 kg. So, that is about 0.014 is about 70 77. So, it is about 780 kilo cal per kg nitrogen. We understand what are we saying? This is an important figure. So, we have been able to transfer 11.64 kilo calories per second into environment f. So, much of space heating has been provided. To do this we had to circulate 1 mole per second of nitrogen of 28 I am sorry is 28. So, it is 390 kilo calories per kg nitrogen. So, what this working fluid which is N2O4 NO2 system is able to do theoretically based on whatever numbers we have generated is that it is able to pick up and deliver 11.64 kilo cal per second for 1 mole per second of nitrogen circulating 1 mole is 28 grams per second. So, I have divided this by 28 028. So, let me just do this once again. The numbers are in your mind. These are important numbers which may have some implications later. So, energy delivered equal to 11.64 divided by 0.028 that comes to about 390 kilo calories per kg. So, we are able to deliver 390 kilo calories per kg of nitrogen circulating. Now, these numbers are very important. We come to that in a minute. Now, the context to looking at this should not be forgotten. The context is a following N2O4 NO2 is a system in which there is no phase change. That means everything is in the gas phase number 1. Now, therefore, we can look at fairly high temperatures may be 120, 140 depending upon the kind of low temperature heat that might be available in your environment. So, if your low temperature heat available is 120, 130, 140, whatever then clearly if you are looking at steam as a working fluid then we are looking at much higher pressures. So, essentially it is the high pressure that is to be handled with steam which makes the use of steam a little expensive because of course, steam is not corrosive, it is got great properties and so on. But, you have to go to higher pressures. Here that feature is not there. And second thing which is equally important with that we are able to transfer is fairly large quantity of you know energy is not small. As you can see here it is 390 kilo calories per kilogram. Now, we can do a similar kind of calculations for work also and I want to do that I will do that right away just to go this calculations. Of course, this is not very difficult to do. So, I will not give you the details. So, we have here if you do the same calculation for work you find that work output measured as g t that is g t at 80 minus g t at 20. That means how do you find g t? G t is calculated as e a g a plus e b g b. Similarly, g t at 20 is calculated as f a g a, sorry f a g a plus f b g b. Now, g a and g b to calculate this is the g and g b you should know the standard heat of free energy of form it all that is given. So, I put all those numbers and got this. So, when you put in all your numbers it looks something like this that work output equal to about 17 kilo cal per second. And nitrogen circulation is 1 mole per second or 28 grams per second. Therefore, you have to calculate for specific output equal to 17 kilo cal divided by 0.028. Now, 17.028 is something like how much is it 84.3 that is 30 is about 510 kilo cal per kilogram of nitrogen. So, if indeed we want to drive an engine between 80 and 20 you get about 510 kilo cal per gram of nitrogen. This is the difference between if you are looking at steam to do the same kind of work we find that the amount of work you will get is very very small. Because of variety of reasons one of course, that at 80 and 20 the energy that is contained in steam is not large enough and the temperature difference is also not sufficient and so on. So, just to cut this long story short what we have tried to say here is that if you are using this working fluid which is NO 2 N 2 O 4 working fluid. Then we are able to derive pick up heat from a low temperature source may be 80 degrees may be 100 may be 130 and since it is there is no phase change involved in this system you do not have to go to higher pressures and therefore, it is a huge advantage from the point of view of recovery of heat at low temperatures. So, here is an instance of a chemically reacting system being used to pick up heat and deliver heat or to actually pick up the energy and deliver work. This is not a very common system that you will see in daily life N 2 O 4 NO 2 highly corrosive and therefore, it is not used commercially that, but the point of discussing this here is not so much that is not commercially used, but to point out the fact that if you can think of making a synthetic substance which is inert which is got this kind of properties which can be used as a working fluid and therefore, we do not have to look at steam and there will be a great advantage as far as commercial and then daily life is concerned and it would be a great product if you can think of a product like this. That is the context of looking at this example to illustrate how we can use a chemically reacting system to give work or to give heat as might be the case.