 Now in the second session, we will study in detail exactly how to evaluate the performance of efficiency of cannot heat engine and Professor Vivek Sathe of Department of Mechanical Engineering Vulture Institute of Technology, SOLUB. Now before we start, we must know what is the outcome of this session. Now here we are learning to evaluate the efficiency of a cannot heat engine which already you know from the basics of thermodynamics but we will now highlight certain important issues of it and then we will proceed further. Okay, in the last session we have seen that cannot propose the cycle which consisted of two isothermals and two isentropes. So on the TS diagram, we have shown that 1, 2, 3, 4, it is 1, 2, 2, 3, 2, 4 and 1 this one. Okay. Now the concept of one T engine, one T engine, now what is the meaning of one T? One T means engine interacting only with one reservoir means if I take one reservoir T1 and then it is say producing certain amount of work that is W is not valid thermodynamically. So there must be minimum two, minimum two T for the engines. What is the meaning of minimum two T means there must be reservoir one which from which heat is being taken, there is another reservoir to which heat is being rejected then we can have an engine. So naturally the question comes, if I ask you is it possible to have an engine with three T that is three temperature reservoirs, why not we can. So generally we say that N T engine, N is greater than 1 is the valid definition of heat engine means if I have N reservoirs, so how it will look like, very simple. I have got one reservoir T1, I have got second reservoir T1, T2, I have got second third reservoir T3, I have got another two reservoirs say this is T1, this is T2, this is T3, this is T4 and this is T5. I arbitrarily I can give some temperatures but my question is it can take heat from this, it can take heat from this then it can reject heat to this, it can reject heat to this, it can take heat from this and so on. So finally this engine is such that it is taking heat from so many reservoirs. Now theoretically say conceptually it is very difficult to identify such an engine where we can supply heat from those four sources but it is feasible. Just I give an assignment or an idea that you just think over it that how we can have we can have a three T engine or how we can have a four T engine and you come across such numericals when we say we are going to solve the problems and you will get an idea that it is possible but there is a difference, cannot efficiency that we are going to devise or say valid today is valid only for a two T engine. So first thing is, cannot engine or cannot efficiency is valid only for two T engine. Then whether engine works or not, how we can check with the cannot efficiency equation. So there is another say very good concept in thermodynamics which is known as glaciers inequality. Now what is glaciers inequality? Glaciers inequality says that dq by T is less than or equal to 0. Now dq by T if it is a process is reversible we call this as an entropy. See do not make a confusion regarding the concept of entropy s and dq by T. One thing is clear as long as heat transfer is taking place at reversible condition the dq by T becomes entropy, otherwise it is not. So if my process is reversible I can say that dq by T is entropy and dq by T is less than or equal to 0. To make it more comprehensive cannot the glaciers added that you can have this one dq i i equal to 1 to n. Now what has happened after what is the engine, engine is nothing but it is heat and work interaction is it not. So when heat and work interactions are there can we restrict it to only two no. One is not valid because it is thermodynamically not acceptable then two is a cannot engine and for two we have got a cannot efficiency but that two we can again say discuss here in detail that sometimes the numbers say that yes first law is valid then cannot efficiency also giving some number but whether this particular engine works or not if you want to check it either you have to find out the cannot efficiency of the engine and we have to check the efficiency of our engine and if that efficiency is less than cannot engine we will say that yes that engine will work. But instead of that if you use the glaciers inequality we will get a better result if you sum all the heat interactions with the reservoirs dq by T and add them together and if their net sum is less than 0 or equal to 0 if it is equal to 0 it is reversible engine if it is say less than 0 it is irreversible engine then the system will definitely work that part we will see in detail when we discuss about this engine. Now first of all we will now go for finding out the expression for efficiency of the cannot engine that is our today's task once again I draw here TS now what I do instead of writing the numbers that is say 1 2 3 4 that I will write as for the processes 1 2 3 and 4 I will write this temperature as TH and this temperature as TL now what has happened this is the lowest temperature this is the highest temperature I know that this is the heat supplied to the engine and this is the heat rejected that in the first slide we have already seen it now in this process what happens what is the efficiency of an engine it is the work done upon heat supplied this is a classic definition this is the classic definition of engine now what is work it is a heat supplied minus heat rejected upon heat supplied heat supplied minus heat rejected upon heat supplied now this will give me idea 1 minus q rejected upon you supplied okay now cannot cycle so on the TS diagram I can have heat supplied in the form of area under the curve now if I ask you what is the heat rejected it is area under this curve it is area under this curve so this will represent the heat rejected this shaded area will represent the heat and if I want to see that what is the heat supplied so heat supplied is this for this process and this green area will represent the heat supplied now I will come to the basic point now if I say heat rejected so it is a perfect rectangle so area of the rectangle I know so it is equal to 1 minus it is the lowest temperature into s3 minus s1 divided by th into s3 it is s4 minus s1 and this is s3 minus s2 this is my expression now you know that s3 s4 minus s1 and s3 minus s2 they are same they are same because it is a rectangle so I can cancel them and what I get is equal to 1 minus tl upon th so 1 minus tl upon th is the expression for efficiency of a Carnot engine then there are certain corollaries that already you have studied that this Carnot efficiency is the maximum efficiency if you have any heat engine operating between two reservoirs now the question comes if I have an engine operating between th and tl and its efficiency is greater than say Carnot efficiency will it be feasible or not so we will discuss that part here now let us assume that th highest temperature is equal to say 600 lowest temperature is equal to 300 of course I am talking in terms of Kelvin the numbers I have taken for convenience so that I can get some good number figure now if I calculate the efficiency of this engine it will be tl is 300 this is 600 so my efficiency is going to be 50% okay means if I have got 500 600 and 300 my efficiency is 50% what is the meaning of this if I have got this t1 this t2 at 600 this is at 300 and suppose suppose I take 100 joules at from this temperature I can reject 50 joules to the lowest reservoir and I can get the work of 50 now this is a valid expression now whether this engine is reversible or not that we don't know because Carnot assume that it is a reversible engine we are done it but in case if it is not reversible then what will happen to this 50 what is your guess the guess is 50 number that we have got is the maximum limit for the work that your engine will give now suppose I say that it rejects 60 it rejects 60 and use 40 as a work output is it a valid engine yes it is a valid engine but suppose it gives 40 as a rejection and 60 as a output it is not valid now how it is valid not been valid Carnot engine or Carnot efficiency expression only restricts the maximum efficiency but who will tell you positively that yes this engine will not work so this is a class that is classes inequality so that part we will discuss in the next session in detail okay now just to review in the first two sessions that is session one and session two now we have got a confidence that for IC engines Carnot engine is a standard engine it is a 2t engine it works on four processes all are internally reversible as well as externally reversible to isothermals and to adiabatics then the efficiency that we can calculate using the TS plot because area under the curve represents the heat supplied and the expression we get is a very standard expression 1 minus TL by TH which will give the maximum efficiency for engine operating between two reservoirs any engine operating having efficiency greater than that is not feasible but cannot tell it by some solid reasons like entropy and other things it is a fact nobody has disproved it so this part you have learnt in detail in the next session we will see the classes inequality will take some interesting changes in the temperatures of the TL and TH and see what happens to the efficiency of an engine and so that we can say see what is the scope for improving further those who are interested in studying further they can refer the IC engine by Janheur and internal combustion engine by Gupta so thank you for this particular say part of the session we'll meet next time next session