 In the third session of the performance of IC engines, we are going to discuss Carnot cycle in detail from the thermodynamics point of view. I am Professor Vivek Sathe, assistant professor, Department of Mechanical Engineering, Walsh and Institute of Technology, Solapur. Now let us see what is expected outcome of this session. First thing is after this we will be able to analyze Carnot cycle thermodynamically. Now this is very important because some parts are procedural and some parts are conceptual. When we say that thermodynamic analysis of Carnot cycle means what is expected from us that we must know that when a 2T engine is there the thermodynamic efficiency already we have studied that for a 2T engine efficiency is equal to 1 minus TL upon TH that we know. Now the question is if I give you some example say for example I have here 600 Kelvin source 300 Kelvin sink I have an engine producing some work it is taking heat from the reservoir it is rejecting heat to the reservoir I take this as 1000 say 100 and I say that this is equal to 50 rejected is equal to 50. Now if somebody claims that this is my engine which works between this and this so whether this will operate or not. So what is thermodynamic restriction by the corollaries of second law we have studied that any engine which is reversible operating between two temperature reservoirs will have the same efficiency. Now see this is very important concept if two reservoirs are given and I have any reversible engine suppose I have these two reservoirs say TH and TL if I have one engine say irreversible 1 then irreversible 2 then say irreversible 3 and so on they will produce certain amount of work W1W2W3 so cannot says that you take any engine you take any engine its efficiency is same what it means this engine may take 100 joule this engine may take 2000 joules this engine may take 1 mega joule it may reject some different values but maximum efficiency that it can achieve is restricted by the formula 1 minus TL by TH. Now in this case that particular efficiency comes out to be 50 percent as already we know. So heat reject supplied is equal to heat rejected in this situation because 50 percent shows that particular part and we know that efficiency is equal to net work done upon heat supplied. So what is the work done finally it is efficiency into heat supplied so it is 0.5 into say 100 it comes out to be 50. In second situation as I told you that I have here instead of this one I have 60 then this becomes 40. Now in this chain scenario my efficiency is still 0.5 there is no change in efficiency because efficiency is maximum but now work done is equal to efficiency into heat supplied that is point that is 50 but what work I have shown over here 60. So from the first law of thermodynamics by the concept of first law this analysis is valid because 100 is equal to 60 plus 40. So as long as I say that 100 is equal to 60 plus 40 it is algebraically correct by first law it is correct by second law it says that it is not possible why it is not possible because there is a restriction that any engine operating between two temperature reservoirs has a maximum efficiency of they cannot cycle that is why we are saying like this. But now today we are going to discuss this as a part of say using the glaciers inequality. Now glaciers inequality you remember is given by a simple definition that is integral dq by t is less than or equal to 0. Now when I say dq by t less than or equal to 0 there is no restriction of this particular say process there may be one process two process three processes where equivalent it is equivalent to q i by t summation of q i by t is equal to or is less than or equal to 0. Now take the same example the previous example and whenever we say that this is 100 this is 600 this is 300 and we will take the second case in which Carnot has said that efficiency is not correct and that is why by Carnot theorem it is not possible. Now I take it the proof with the help of your glaciers inequality what glaciers inequality says if this is the engine then engine is my system engine is my system. So heat supplied to the engine is 100 joule at what temperature it is supplied 600 it is supplied to the engine at what temperature 600 plus now 40 is the heat rejected by the engine and at what temperature it is rejected 300 right. Now what has happened so if I go by this calculations you will find that this is 1 by 6 minus 4 by 3. So answer will be 1 minus you can multiply by 2 here 8 upon 6 so it becomes minus say heat rejected by the system it is 1 minus 4 by 6 so it is minus 7 by 6. Now for the system that I have studied heat supplied and heat rejected so heat supplied is positive heat rejected is negative the temperature at which it is supplied is 600 the temperature at which it is rejected is 300 and the answer I am getting is minus 7 by 6. Now the question is whether this system will prove the glaciers inequality or not that is our question or whether we are going somewhere wrong in the calculations that is let us check it. Now if I treat this as a system heat supplied is 100 at what temperature it is supplied 600 Kelvin plus heat rejected is minus 40 at what temperature it is rejected this 300. So I can get this particular say 4 by 30 sorry 4 by 30 then my calculation will become 6 LCM of 6 and 30 is 30. So I have to multiply here by 5 and this is 4 so my answer will be 1 by 6. So when I get answer plus 1 by 6 it says that it is not following the glaciers inequality and hence this engine is not feasible. Now what has happened this is valid for 2T engine but suppose I take the situation in which I have an engine I do not know whether it will work or not but I will just take any arbitrary values and we will see whether it will work or not say this is 600 then this is say 900 it is taking heat from this say 300 joule it is taking from this say 400 joule and it is rejecting so 700 is taken suppose it is resulting 500 joule. So theoretically speaking it must produce say 200 joule of work and this temperature is say we are taking say around 300 around 300 this is my engine. Now if I ask you whether this engine will is feasible or not the first question is the feasibility the feasibility of this engine. Now can you do analysis of this engine with the help of Carnot efficiency there is a technical issue if I say this part only that is 600 to 300 and say that it is operating between 600 and 300 I will find out what is the amount of heat supplied what is the amount of heat rejected but I do not know out of this 400 500 is rejected is not possible this 400 plus 300 700 is feasible but out of this 500 what is the component of 400 I do not know. So it becomes a very complicated problem to handle thermodynamically. So clashes inequality is a very handy tool that we can have. Now we will see that we will see summation of qi upon ti i equal to 1 to n if it is less than or equal to 0 it is valid engine otherwise not. So I will see now as a system 600 is the amount of sorry 400 is the amount of heat supplied at temperature 600 then 300 is the amount of heat supplied at a temperature of 900 and 500 is the amount of heat rejected at a temperature of 300. Now I will see the exact things what is happening here in this situation so I will get 4 by 6 plus 3 by 9 minus 5 by 3. So I will get this 3 plus 2 5 5 minus 5 is 0 by sorry 3 this is 3 so 2 plus 3 3 minus 5 is minus 2 so minus 2 by say 3 so luckily this is working. Now see this is a very interesting thing many times it happens in thermodynamics that we have many sources we have many sinks not necessary this one sink is here there may be another sink which we are supplying say at 200 Kelvin some amount of heat and we have to check the feasibility of the engine. Now once you have a tool of Carnot efficiency and the clashes inequality we can do the analysis of your thermodynamic cycle very easily. So I think now you have got in nutshell the concept of Carnot engine the concept of clashes inequality why we apply to T concept only for Carnot efficiency calculation and when we want to find out the feasibility forget about the efficiency we want to find out the feasibility we have to check for clashes inequality and if you want to find out the efficiency of this engine there is no other choice but we have to conduct the trials that we have to give how much amount of heat we have supplied how much amount of work we have obtained physically and then we have to get the actual output from the system. Now those who are interested in studying further they can refer these two books IC engines by Heywood and Gupta and in the next sessions we will be discussing about the air standard efficiency as a case study we will study only auto cycle and diesel and dual cycle calculations you can do an auto on your own. So thank you for patient listening for this particular session.