 Hello students, myself Ganesh B. Aglavel working as an assistant professor in Department of Mechanical Engineering, Valchian Institute of Technology, Solopo. In this session, we will see second law of thermodynamics, learning outcomes. At the end of the session, students will be able to describe Kelvin Planck's statement and also the second statement is also there, we will be able to describe Clausius statement. Now, as we have studied in the once again the in the BME of first year of engineering that is first law of thermodynamics, now we have the second law of thermodynamics. And in the second law of thermodynamics, there are two statements that the first one is the Kelvin Planck and second Clausius, we will see one by one, so first statement Kelvin Planck statement, it states that Kelvin Planck states that it is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of energy by work to its surroundings while receiving energy by heat transfer from a single thermal reservoir. I will make it simple, so consider this is the thermal energy reservoir, thermal energy reservoir, try to recall the concept of thermal energy reservoir, yes. So, in thermodynamics, the thermal energy reservoirs are the systems which has infinite amount of heat and the temperature remains constant, suppose you take q amount of heat, maybe 1 lakh to lakh kilo joule, after taking that much amount of high amount of heat, the temperature of that system does not change, ok. So, that is the definition, simple definition of the thermal energy reservoir. So, as per the Kelvin Planck statement, a single thermal energy reservoir is there and this is the machine. Now, it is not possible to convert the supplied heat that is q into equal amount w into equal amount of work done, it is not possible as per the Kelvin Planck statement. Then when it will become possible, we will see. So, to make it possible what I will do, what I will do, consider there are two thermal energy reservoirs, there are two thermal energy reservoirs getting and there is one machine taking the q s amount of or q 1 amount of heat, suppose this is the q 1 while developing the w amount of work, this is the w amount of work. Whereas, there is rejection whereas, there is rejection, suppose this rejection is q 2, are you getting? So, as per the Kelvin Planck statement, if such systems are not possible practically, then to make it a practical practical system, this system should reject the some amount of heat into the another thermal energy reservoir. So, here the T 1, suppose the high temperature energy reservoir is at T 1 temperature and the low temperature energy reservoir is at T 2 temperature, where T 1 is greater than T 2, then the output is the work done. Now, this is the system which will obey the Kelvin Planck statement. So, all the IC engines are absorbing the q 1 amount of heat from the high temperature energy reservoir whereas, they are developing the work done, the work done is available at the crankshaft in terms of the brake power and the heat rejection is through the silencer for the engine. So, all IC engines follows the Kelvin Planck statement, this is the in very simple language. Now, the effectiveness of such type of engines are measured in terms of the efficiency. So, efficiency of the engines will be equal to the efficiency of the engine will be equal to. So, efficiency of the engine will be equal to the ratio of output to input that is q 1. Now, as we have seen this was the thermal energy reservoir, this was the lower thermal energy reservoir and this was your machine. So, this was q 1, this was q 2. Then I can write q w as q 1 minus q 2 which can be substituted instead of w in the numerator. So, it becomes q 1 minus q 2 by q 1. Now, we can also replace the heat by the temperature, by thermal thermodynamic scale, temperature scale. So, this becomes t 1 minus t 2 by t 1. So, for any numericals on the engine, you can use these three relations. If w is given and q is given, you could get the eta e or only heat supplied heat rejected is given, also eta efficiency can be obtained and the temperatures. But remember the Carnot efficiency is the function of temperature only, that is why the maximum possible efficiency for any engine is Carnot efficiency which is the ratio of temperature only. Now, we will move to the second statement, Clausius statement. So, Clausius states that it is impossible for any system to operate in such a way that the sole result would be an energy transfer by heat from a cooler to a hotter body. So, as per the Clausius statement, if the system is working between hot and cold thermal energy reservoirs, then the flow of energy from cold thermal energy reservoir to hot thermal energy reservoir on its own is not possible. The reverse is possible. So, what is that? As per the Clausius statement, we could transfer the energy from low temperature energy reservoir to high temperature energy reservoir by using external agency, that is external work. Now, all refrigerators and heat engines follow the Clausius statement. I hear the suppose the aim of this machine or the system is to absorb the Q2 from low temperature, then that you take in the numerator. Now, to achieve this Q2, the work done is W for the engine, the ratio of high-grade energy to the low-grade energy. Now, here if we want to find the effectiveness of such type of the machine, then we cannot use eta term. So, as per ASHRAE, there is coefficient of performance parameter is there and the system which is doing this is called refrigerator. So, all refrigerators absorb the heat from low temperature energy reservoir whereas, deliver it to the high temperature energy reservoir by consuming work done. So, this is called coefficient of performance of the refrigerator. Once again, we could replace W in terms of CM, the W Q1 is nothing but now W plus Q2. So, what I can do now? I can write Q1 minus Q2 and also I could replace the Q by temperature. So, it becomes T2 divided by T1 minus T2. Once again, for the numericals on the refrigerators, you could use only this one relation. And once again, this is known as coefficient of performance of the Karmat, which is maximum COP, theoretical maximum COP. So, in this fashion, we have seen the Kelvin Planck statement which is applicable for the engine and refrigerator works on the Clausius statement. For any further study, you can refer fundamentals of thermodynamics by Boranke Sontek and fundamentals of engineering thermodynamics by Moran and Shapiro. Thank you.