 myself, S. M. Watt, I stand professor from mechanical engineering department of Olsen Institute of Technology, Solapur. Today I am going to explain how to calculate the efficiency of the Otto cycle. You know that basically in the IC engine there are different cycles are to be there. Otto cycle, diesel cycle, Carnot cycles are to be there. So, these are the theoretical cycles. So, one of the theoretical cycle is Otto cycle. Basically it is a air standard cycle. Air is used as a working media and here to assume some data that it is a working substance, it obeys all the gas laws and when it obeys all the gas laws it is a perfect gas ok. And heat added and heat rejected. So, there will not be heat losses from system to the surrounding and surrounding to the system. So, one of that air standard cycle is Otto cycle. Today I am going to explain. The basic learning outcome of this session will able to derive the expression for the Otto cycle for calculating the efficiency of that engine, efficiency of that plant, efficiency of the machine. So, here I am showing this graph of this air standard cycle. Basically here are four cycles are four process are there or cycle. I am showing here process this 1 to 2, 2 to 3, 3 to 4 and again comes to initial condition 1. Initially the process started at from 0 to 1. Here we have to neglect it initially starting coefficients. Once the process started it moves to 1 to 2, 2 to 3, 3 to 4 and 4 to 1 again moves to 1 to 2. So, here there are two adiabatic process are there and two constant volume process are there. Air is used as a working media. So, here process 1 to 2 this process 1 to 2 it is a adiabatic process. Which adiabatic process? It is a what we call compression process. So, this compression is a isentropic process. So, process 1 to 2 is a adiabatic process. Process 1 to 2 is a adiabatic. Whatever the air you have to suck in the 0 to 1 it get compressed. So, it moves from volume V 1 to the V 2. Here it is V 1 to the V 2 and due to which temperature is to be increases. This is a process 1 to 2. So, it obeys the gas law P into V raise to gamma is equal to constant. Then process 2 to 3 after 1 to 2 it moves here 2 to 3 it is a constant volume heat addition process. So, heat is added Q in is there and this volume V 2 is equal to V 3. So, it is a constant volume heat addition process. So, in this process heat process 1 to 2 or it is called as the Q s is m C v. C v means it is a constant volume process m C v T 3 minus T 2. So, temperature rises from T 2 to the T 3 and at the point of 3 it is a highest temperature is to be there. And whatever the heat addition is to be going on here that is a constant volume. So, that is why this process 2 to 3 is called as a constant volume heat addition process. Process 1 to 2 is a adiabatic process adiabatic compression process or isentropic process to be there and process 2 to 3 is a constant volume heat addition process. Then third process, process 3 to 4 here is a expansion process. After heat addition at that point burning is going on in that IC engine heat is added or temperature is such a high conditions that. So, when the explosion is going on you get the work output from that system and the volume is raised from 3 to 4 V 3 to the V 4 and this process again it is a adiabatic process it obeys adiabatic law. So, process 3 to 4 is a expansion process or here you get the work output from that system. So, process 3 to 4 is a adiabatic expansion process. It is a adiabatic expansion process and it obeys all the gas laws that is adiabatic process PV raise to gamma is equal to constant. That 1 to 2 is adiabatic process that is a compression process then 2 to 3 is a constant volume heat addition process 3 to 4 is a adiabatic expansion process. It is also called as a you get the work output from the system and it pushes that piston to go downward and piston receives a reciprocating motion and that reciprocating motion is transmitted to the connecting rod and from that connecting rod to the crank 1 and from that crank to the flywheel or a pulley and here you get a rotational mechanical energy and last one after the completion of 3 to 4 process less process is 4 to 1. So, process 4 to 1 constant volume heat rejection process. So, these are the 4 processes. So, how to calculate this now you have to write that efficiency of that plant is equal to heat supplied minus heat rejected divided by heat supplied. What is the heat supplied? Heat supplied is equal to where it is heat supplied. Heat supplied has process 2 to 3 that is M Q S is equal to M C V into bracket T 3 minus T 2 and heat rejected that is in Q R that is in a process 4 to 1, 2 to 3 is a heat addition and 4 to 1 is a heat rejection process. So, Q R is equal to M C V T 4 minus T 1. So, we have to put this value in above equation. So, efficiency is equal to heat supplied is M C V T 3 minus T 2 M C V into bracket T 3 minus T 2 minus heat rejected M C V into bracket T 4 minus T 1 the whole divided by heat supplied that is M C V T 3 minus T 2 or efficiency is equal to 1 minus T 4 minus T 1 upon T 3 minus T 2. So, this is the first equation you have to write it here, but here in this case as it is adiabatic process V 1 by V 2 is called as a compression ratio V 1 by V 2 is equal to compression ratio R C is equal to V 1 by V 2 and expansion ratio is equal to V 4 by V 3 here expansion ratio is V 4 by V 3 and compression ratio is V 1 by V 2. Now, as this is a isentropic process are to be there process 1, 2, 2 and 3, 2, 4. So, that is why you have to write T 2 by T 1 is equal to V 1 by V 2 raise to gamma minus 1 or T 4 by T 3 T 4 by T 3 is equal to V 3 by V 4 raise to gamma minus 1 T 2 by T 1 is equal to compression V 1 by V 2 raise to gamma minus 1 and T 4 by T 3 is equal to V 3 by V 4 raise to gamma minus 1, but here in this case it is a both 1, 2, 4 is a constant volume process and 2, 2, 3 is a constant volume process. So, V 2 is equal to V 3 and V 1 is equal to V 4 V 2 is equal to V 3 and V 1 is equal to V 4 is a constant volume processes. So, put this value here T 2 by T 1 is equal to R raise to gamma minus 1 and T 4 by T 3 is equal to as V 3 is equal to V 3 is equal to here V 2 and V 4 is equal to V 1 you have to put the value V 3 is equal to V 2 and V 4 is equal to V 1 raise to gamma minus 1 or it is equal to 1 by R raise to gamma minus 1. Therefore, you have to write T 2 by T 1 is equal to it is a 1 by R raise to gamma minus 1 you have to write T 3 by T 4 is equal to R raise to gamma minus 1. So, T 2 by T 1 is equal to T 3 by T 1 T 4 is equal to R raise to gamma minus 1. So, put this value in the efficiency what is the formula for efficiency. So, efficiency is equal to here you have to write 1 minus T 4 minus T 1 divided by T 3 minus T 2 put the value. So, efficiency is equal to 1 minus put the value of T 4 what is the value of T 4 value of T 4 is equal to T 3 upon R raise to gamma minus 1 T 3 upon R raise to gamma minus 1 minus put the value of T 1 put the value of T 1 is T 2 by R raise to gamma minus 1 T 2 by R raise to gamma minus 1 this whole divided by T 3 by T 2. So, efficiency is equal to now 1 minus this T 3 get cancelled 1 T 3 minus T 2 1 minus 1 upon R raise to gamma minus 1. So, it is a value could calculate the efficiency of the IC engine that is for a auto cycle. Now pause the video think over that which value we have to require to calculate this one or when you have to increase the efficiency of that plant because you know all these derivations that is for 1 2 2 is a adiabatic process that is 2 adiabatic process 1 2 2 and 3 2 4 and another 2 or a constant volume 2 2 3 and 4 to 1 then how to increase the efficiency of that plant yes to increase the efficiency of that plant to increase the efficiency of that plant you have to increase the value of R R means a compression ratio R means the value of compression ratio. So, when you have to go for a graph of this one how to increase the efficiency of that plant. So, efficiency versus compression ratio you take the value 20 percent efficiency 40 percent 60 percent 80 percent the gamma you have to regularly use a 1.4 that is a idle 1. So, it is a for a and gamma for 1.8 and gamma for 1.2. So, this value when the value of R is to be increased the efficiency of that plant going on increase 1. So, from this derivation you have to know how to increase the efficiency of that plant or how to calculate the efficiency of the auto cycle. So, taking the references by basic mechanical engineering by SINGLE.