 I am Mr. Vivek Sathay, Assistant Professor, Mechanical Engineering Department, Vulture Institute of Technology. Today we are going to say study the performance of IC engines in detail in the coming five set of videos or more than that. Now let us see what are our learning outcomes. At the end of this session, you will be able to evaluate Carnot efficiency of heat engine. Now, very important thing is whenever we start the study of heat engines, Carnot engine is very important. As you have studied the second law of thermodynamics and you know it very well that Carnot has defined the maximum efficiency limit for any heat engine that is available in the world. Okay, now let me ask you a simple question. You have studied second law of thermodynamics, say second law and you know that any heat engine that operates between two temperature reservoirs must reject some heat to the sink. If it is not rejecting heat to the sink, what is the situation? I have this temperature reservoir say T1. My engine is taking say heat from the source and it is producing certain amount of work. Now this is the situation where Q is the amount of heat taken from a temperature reservoir T1 and W is the amount of work produced. Now as far as first law of thermodynamics is concerned, W is equal to Q is valid because first law says that energy is conserved in any process. So if you see here, if Q is 100 joule, W is 100 joule, is there any objection for writing W is equal to Q? Absolutely not because you know mathematics, you can say take the simple calculations and you say that is whatever amount of heat is taken from T1 is the same amount of heat rejected to the atmosphere, sorry it is producing to the form of work. So that work is converted into heat but this is not possible as suggested by Carnot and before Carnot we must analyze the concept of glaciers in equality. Now you know that whenever this rejects something, suppose this is the heat supplied and this is the heat rejected to a reservoir at some temperature T2. Now this becomes a valid assumption that some amount of heat is taken, some amount of heat is rejected, whatever amount of heat rejected suppose it is say 20 then this may be 20 is equal to 80. Now still there is a problem, if I say that 100 joules are taken from a temperature reservoir T1, 20 joules are rejected to the temperature at reservoir T2 and 80 joules is the work produced. Now again you can say that first law is valid but does this engine work? That is the main question. Many a time students say that okay there is a suction, there is a heat supplied, there is a heat rejection and some work is produced. Arithmetic says that yes it is a matching numbers, 100 minus 20 is equal to 80 and it must work and here comes the concept of Carnot engine. Now Carnot designed an engine, actually it is a hypothetical engine and he assumed that all the processes involved are internally reversible. Now what is the meaning of reversible process? It is totally reversible means the process is internally reversible as well as externally reversible. Meaning is I can show the diagram on PV and TS plot as complete lines so there is no breaks that is the first point. And second thing is most advantage of assuming that it is a reversible process is I can use TS diagram for finding out the energy calculations because you know from your knowledge of thermodynamics that whenever I draw a PV diagram, then area under the process 1 to 2 represents work done it may be plus or it may be minus depending upon the process. Now if you look at this 1 to 2 can you tell me whether it is a work absorbing or work producing process just think over it yes what is happening volume is given volume has gone to V2. Now tell me when it moves from 1 to 2 does contraction of volume from V1 to V2 will occur automatically have you seen anywhere in nature in the process that from somewhere the air has gone to some other process as other part where in the pressure has increased automatically impossible. So is some driving force required some work is to be spent and so that we can get the volume V2. So some work is required here means this is a power absorbing cycle. Now if I want to produce the power then I have to use another process which is known as expansion process and the work produced by expansion must be more than work required for the compression. If I say that if I go from 1 to 2 and come by 2 to 1 if I go from 1 to 2 and come from the same process from 2 to 1 will it produce work yes process is complete I gone from 1 to 2 I come from 2 to 1 but is there any work done as far as my PV diagram is concerned there is no work done because area of the PV diagram covered by the loop is equal to well done. So is it a loop no it is a line it is not covering any area so I cannot say that there is a work done either on the system or by the system. So now the question is whenever I have this say TS diagram for example and if I say I have any process like this it is not a standard process because standard process is either constant temperature constant entropy constant volume constant pressure or some other parameter that I kept constant. Now the question is area under this diagram area under this diagram you know by definition what is S it is DQ by T. So if I multiply this T into S what I get DQ so it represents the energy interaction. Now this represents energy interaction only when this process is reversible so cannot took advantage of this particular concept while developing the heat engine. So we suggested that there are 4 processes 2 adiabatics and 2 isothermal they are connected alternately that is adiabatic is connected to isothermal isothermal to adiabatic adiabatic to isothermal and he said that all the processes are reversible. So if I want to draw the Carnot cycle on the TS diagram on the TS diagram it is a perfect rectangle it is a perfect rectangle so I go from 1 to 2 2 to 3 3 to 4 and 4 to 1. Now if I ask you what is the process 1 to 2 immediately your question is your answer is it is isentropic because you know that S1 is equal to S2 S1 is equal to S2 there is entropy remains constant and if it is reversible then it must be reversible adiabatic isentropic that you part you know that these 3 combination it is a reversible it is adiabatic then it must be isentropic. So it is a reversible adiabatic or we call it is an isentropic process then what is happening it is a comparation or expansion yes what is happening to temperature temperature is being increase increase from T1 to T2. So when there is increase in temperature we say that it is a comparation process. Now 2 to 3 what is the process 2 to 3 it is isothermal because on the TS diagram it looks like a horizontal line. Now my question is whether 2 to 3 can be properly called as isothermal expansion or it should be called as a isothermal heat supply. This is a very important concept because many a time student make mistake here they do not understand that whenever we have an isothermal expansion process why we want to supply heat. Now tell me what is the meaning of isothermal? Isothermal is a process where dt is equal to 0 what is the meaning that from point to point when I move from one location to second location infinitesimally small location there should not be any change in temperature. So temperature should remain constant. So if there is a drop in temperature or if there is a rise in temperature by some means I have to make the temperature constant. Now assume that I am at 0.2 and there is some temperature say T2 when I move to the right what is happening to entropy yes increasing. Now increase in entropy is directly related to increase in volume do you agree with this? Yes because what is the concept of entropy it is a measure of disorder. So whenever I have a system at some temperature T2 and some entropy and if I reduce the temperature if I reduce the temperature yes then I can get it to the right side and volume will increase. So volume increases now increase in volume automatically suggests that there is a increase in entropy and at the same time I know that expansion results in drop in temperature. So temperature will fall now when temperature falls what happens process is no longer isothermal. So to keep the temperature constant I have to supply temperature. So that is why this process is heat supplied. In the same way when I come from 4 to 1 it is the heat rejected because the process 4 I have some entropy here entropy decreases. So automatically there is a increase in temperature that increase in temperature I must decrease because I want constant temperature and that is why it is called as heat rejected process. So you understood that 1 to 2 is a compression then 3 to 4 is expansion and 2 to 3 and 3 to 4 4 to 1 are heat rejection and say heat supply and heat rejection processes. So cannot use this as a basic cycle for his analysis. So in today's session you have learnt that for any heat engine which that is we are studying IC engines in detail cannot place very important role cannot heat engine and cannot assume that it consists of 4 processes all are reversible internally as well as externally. Then it is why divisory cycle that we can show on a TS plot as a perfect rectangle and in the next session we will see how to find out the efficiency of a cannot cycle and some interesting facts about the efficiency. If those who are interested in studying further they can refer the books by John Haywood or internal combustion engine by Gupta. So thank you for this session we will see in the next session how the efficiency of the cannot engine is evaluated. Thank you.