 In the fifth session, now we are going to discuss the OttoCycle in detail. Now let us see what is the outcome that you must be able to evaluate and analyze the OttoCycle that is PVN-TS diagram and most probably the expression for efficiency of the engine. Now in the last sessions we have seen that air standard cycle plays very important role in the analysis of IC engines and will go for the OttoCycle. Now OttoCycle is used for the SI engines that is the petrol engines. Now if I draw this cycle on PV diagram and TS diagram, the picture will be more clear to you. This is the TS diagram and this is the PV diagram. Now here if I go from this say 1, 2, 2, 2, 3, 3, 2, 4 and 4 to 1. 2, 1, 2, 3, 4 and 1. Then on the TS diagram this is 1, 2, 2, 2, 3, 3, 2, 4 and 4 to 1. Now we will analyze this cycle in detail. Now see what is happening here 1, 2, 2, 1, 2 and 3, 2, 4 these are PV raised to gamma is equal to constant and these two processes are V is equal to constant. So what is expected that you must know that this is the heat addition that is the QS and this is the heat rejection that is QR. So in the PV diagram this particular area will represent the work done and efficiency we know that it is equal to work done upon energy input. So efficiency term is going to be a standard term that is the QS minus QR upon QS that is standard definition that we have used. Now see here on the TS plot we have got 1, 2, 2 which is the isentropic process 2, 2, 3 which is the constant volume process and this one. Now what is the physical meaning of this? This is very important because what it indicates that we are supplying heat essentially at constant volume. Now first see at constant volume we are supplying heat and at constant volume we are rejecting heat. Then there is other process in which we supply heat at constant pressure reject heat at constant volume that is called as a diesel cycle and we have a third cycle which is a dual cycle. The problem is that with the actual applications we cannot supply heat entirely at constant volume some part of the heat gets added during constant pressure some part of the heat we have to add during the constant volume process and that cycle becomes a dual cycle. So practically speaking we do not have say theoretically perfect diesel cycle but all our engines operate which are the operating on the fuel diesel they operate essentially on dual cycle that is some part is supplied in the constant volume region some part is supplied in the constant pressure region. Now as it is air standard cycle as I am doing my analysis for per kg that is if I take m is equal to 1 kg or m dot is equal to 1 kg per second so for per kg second kg per second I can do the analysis. Now I know that when a constant volume process is taking place heat supplied heat supplied is equal to m Cv delta T this is my definition this I know from thermodynamics. Now it is m Cv now what is delta T delta T is T3 minus T2 in this situation similarly if I see q rejected it is again m Cv T4 minus T1 see specifically I use this T4 minus T1 when I say it is qr if I write heat interaction that is q it is simply m Cv is equal to final minus initial that is T1 minus T4 looking at the diagram it is clear to me that T1 is going to be less than T4 so naturally this quantity is going to be negative. So instead of using negative quantity and adding it algebraically to the heat supplied it is better that we call it rejected we know that it is minus and we write this in expression this one otherwise this is algebraic sum of the q divided by total heat supplied it is not single q it is a total heat supplied so theoretically speaking this particular is this algebraic sum this is also algebraic sum because many a time student make mistake when the cycle is shown to them it is not necessary having 4 processes suppose 6 7 processes are there means 1 2 2 3 3 2 4 4 5 5 6 6 7 7 2 8 and again 8 to 1 and in this process in some parts we supply heat some parts we reject heat so when we calculate the efficiency of an engine what is important is area under the curve is going to be always the work output that is no problem but heat supplied that we have to divide is the total amount of heat supplied in various processes and that is why it comes as here as this one now if I know this then I can derive the expression for efficiency now derivation of expression for efficiency for auto cycle so we will write the efficiency of air standard efficiency that is ASC for auto cycle if I write it so it becomes q supplied minus q rejected so it is m Cv T2 minus T1 minus m Cv T4 minus T1 divided by m Cv T2 minus T1 this is my heat supplied so I know that if I divide this m Cv m Cv will get cancelled from all this and I will get T2 minus T1 minus T4 plus T1 so in the top I will get T2 minus T1 minus T4 plus T1 divided by T2 minus T1 or if I do it systematically if I divide this I will get 1 minus T4 minus T1 upon T2 minus T1 so this is my expression now what happens when we work in thermodynamics we are interested in finding out all our answers in terms of pressure ratios in terms of compression ratios and so many other things now let us see for this particular situation if I write this expression once again that is you write here for your convenience 1 minus T4 minus T1 upon T2 minus T1 now if I see the cycle that is say PV diagram then it is clear to me that 1 to 2 2 to 3 and 3 to 4 this is my PV diagram now if I could find the values of this T4 minus T1 and T2 minus T1 in terms of compression ratio now what is the compression ratio I have got here V1 I have got here V2 which is also equal to V3 which is also equal to V4 so V1 by V2 is equal to V4 by V3 is equal to R this R is called as a compression ratio and the index that I have for here it is PV raise to gamma is equal to constant so gamma has got different value for monatomic for air so for monatomic it is 1.67 for gamma for air it is 1.4 and gamma for exhaust gas if I take exhaust gas it is 1.3 or something means final expression for efficiency that we are evaluating it may happen that it will depend upon the compression ratio and gamma and other values okay now if I see the process that is I want what I want T4 minus T1 and T2 by T1 so I have a ratio V1 by V2 and V2 V4 by V3 is equal to R so what I will do I will do for 1 to 2 if I go for process 1 to 2 now what is this process it is P1 V1 upon T1 is equal to P2 V2 upon T2 I am interested in finding out the value of T2 by T1 say for example so I will get T2 by T1 is equal to P2 by P1 P2 by P1 into V2 by V1 because it is the ratio of this and also I know that P1 V1 raise to gamma is equal to P2 V2 raise to gamma so now I want in terms of V2 by V1 because V2 by V1 is known to me and this P2 by P1 I will calculate P2 by T1 in terms of V2 by V1 now P2 by P1 P2 by P1 if I see here P2 by P1 will be V1 by V2 raise to gamma and this is V2 by V1 raise to 1 now this is actually V1 by V2 and this is V2 by V1 so if I make it reciprocal it will become V1 by V2 raise to gamma minus 1 so in short it is R raise to gamma minus 1 this is my first part similarly for process 3 to 4 if I go for 3 to 4 I will get the same thing that is say T3 P3 V3 upon T3 is equal to P4 V4 upon T4 and P4 V4 raise to gamma is equal to P3 V3 raise to gamma now I am interested in the value of T4 by T1 so T4 by T1 now if I want T4 by T1 what I will get I will get T4 by T1 or T4 by T3 sorry T4 by T3 T4 by T3 is equal to P3 P4 V4 P4 V4 upon P3 V3 and in the same way in the same way I can write this P4 by P3 into V4 by V3 P4 by P3 into V4 by V3 now I know that V4 by V3 is equal to R and this is P4 by P3 so P4 by P3 I will write in with the help of this equation P4 by P3 is equal to V3 by V4 raise to gamma so if I put this into V4 by V3 so I will get this as V3 by V4 raise to gamma minus 1 V3 by V4 raise to gamma minus 1 so I got this particular same values so by putting the values in this expression for the efficiency here I will get 1 minus 1 upon R raise to gamma minus 1 so I will get the answer as 1 minus 1 upon R raise to gamma minus 1 now here what happens if I draw the graph of efficiency versus compression ratio then I will see that this particular say graphs I get for different values of gamma is equal to 1.67 gamma is equal to 1.4 gamma is equal to 1.3 now same logic we can apply for diesel cycle we can apply for dual cycle and we can get the answers now in this cycle we have studied the effectiveness of the particular say engine which is actual engine we have to convert it into auto engine so that it is air standard efficiency we have derived and the same problem we can solve for the diesel and dual that we can see in the next sessions. Those who are interested in studying further they can see internal combustion engine by Hewitt and internal combustion engine by Gupta and for this session say once again I say thank you for patient listening.