 Welcome back, I am UX Hathay, Assistant Professor, Department of Mechanical Engineering, Vulture Institute of Technology. Now this is fifth session, already in last four sessions we have studied a lot about basics of thermodynamics. Now we are studying applications and as usual before going to discuss we will see what are the outcomes, the expected outcomes of this. So at the end of this session you will be able to derive the expression for efficiency of heat engine, you will be able to analyze various statements of second law of thermodynamics and you will apply Carnot theorem. Now see the first of all let us recall what we have studied, we have seen in the first video, second video and third video onwards the heat, work which are forms of energies, the first statement of first law of thermodynamics, the conservation principle and the first law are equivalent and also we have seen that first law has got a limitation that is if you do the process in one direction there is no guarantee that process will be there in the reverse direction. Now today we are going to see the heat engine, how it operates and how to find out the efficiency of heat engine. Now let us see if I draw a PV diagram, PV diagram and let us say this is my one process say path A from 1 to 2. Now looking at this process 1 to 2, immediately the question comes to your mind that this is volume V1, this is volume V2. Now what is happening to the volume, volume is getting decreased, now decrease in volume will never be automatic, for the decrease of volume we require certain work. So we have to spend work means when I go from V1 to V2 there is a reduction in volume, there is a reduction in volume and reduction in volume requires work, it requires work means this process 1 to 2 will consume work, okay there is no problem for it but now suppose I come from process 2 to 1 by process say path B, then immediately we know that now for this process there is increase in volume, now as there is increase in volume we say that the net work done by this particular cycle, now see the world what I am using cycle is the area of the PV diagram. Now this is a loop and if this loop is existing and if I have a cycle which is in the clockwise direction see the direction I will get work output. So we have decided that clockwise is work cycle is work producing, it is work producing machine, it is work producing machine. Now when I go from 1 to 2 and I get from 2 to 3 there are two quantities which are important one is the Q1 and second is Q2, now whether I supply heat or heat is rejected that will decide how much amount of work I am going to get. Now looking at the diagram you can see that 1 to 2 is a compression this can be obtained either with the help of work or heat but you know that out of work and heat work is superior. So as far as possible if you use heat and that work can be done that particular task can be completed by heat it is well and good. If we have heat and work option but work that particular process cannot be carried out say with the help of heat then only use the work. So without say loss of generality we assume that Q1 is heat supplied, Q1 is heat supplied and Q2 is expansion so it is heat rejected, heat rejected. Now the definition of a thermal efficiency in general is work done upon heat supplied. Now what is work done Q1 minus Q2 because finally whatever amount of heat I have supplied and whatever amount of heat I am getting rejected so Q1 minus Q2 is the work done upon heat supplied is Q1. So I will get 1 minus Q2 by Q1 as the efficiency. Now this is very fundamental, now if you go to another say basic principle about the say efficiency you will find that when efficiency I write as 1 minus Q2 by Q1 if it is equal to 1 its meaning is Q2 is 0. Now if I am not rejecting heat then only I can get the conversion of entire amount of heat supplied into work which is against the first law of thermodynamics means we cannot have perpetual machine of first kind. Now this is about the heat engine. Now go to the important topic that is second law of thermodynamics, second law. See historically say first second law was developed first because it was responsible person was Karnot who was an engineer. Now the second law is important from the point of view of say visibility of a process. So there are various versions of the process that if I say that if I want to go from the point 1 to point 2, point 1 to point 2 the problem is whether this process would occur or not and if I come from 2 to 1 again so the feasibility of this cycle is judged by second law that is one thing. And second law is a common day experience that whenever heat transfers it is from high temperature to low temperature whenever there is a flow of fluid it is from high pressure to low pressure whenever there is a concentration difference then high concentration to low concentration there is a mixing means there is a potential difference what is important there is a potential difference for the process to occur and always it is from high potential to low potential. So high to low is a natural and we cannot come from low to high say without expanding energy. So looking at this particular phenomenon there are various statements given by various scientists one is a clausius statement. Now this clausius statement says that if you want to transfer heat there are two bodies one is at T1 and second is at T2 and assume that say T1 is less than T2 and if I want to transfer heat from T1 to T2 now just imagine you have two buckets in which one water is having 30 degrees another bucket water is having at what say 60 degree temperature you connect a copper rod and expect that heat is being transferred from low temperature to high temperature will it happen in nature never because we know that there is a temperature difference and it will not happen on its own. So clausius statement said that it is not possible operating in a cycle which is important word cycle in a single process it is possible that we can convert entire amount of heat into work. I give simple example if I have got an isothermal process if I have got isothermal process say isothermal expansion or isothermal compression whatever it may be now here what happens if I take isothermal expansion means I have to supply heat and expansion gives me work first law says that q-w is equal to delta u delta u is equal to 0 for isothermal so q is equal to w so a student make be confused that for an isothermal expansion I get q is equal to w means whatever amount of heat I have supplied I have convert into work then it is a violation of the clausius statement or violation of the conversion of entire amount of heat into work but remember all these laws second laws statements are specific to the cycle means if you operate anything in a cycle you will never get transfer heat from low temperature to high temperature without expenditure of energy so this is the clausius statement. Now what is the consequence of this you know that in a refrigerator in a refrigerator this is temperature say T1 this is temperature T2 T1 is less than T2 I have to transfer this heat or remove the heat from this say amount of q and transfer it to the higher temperature reservoir then for this I must spend certain amount of work now when this work is shown then it says that clausius says that it is possible because what you are doing now you are removing heat from a temperature reservoir which is at low temperature to higher temperature by consuming some amount of work so it is not against second law of thermodynamics and you are operating in a cycle similarly suppose I have got a reservoir at some temperature T1 I remove heat and say some amount of q and I can completely convert it into work so what is this this is engine so whatever amount of heat you have taken you are converted into work so this is a kelvin plank statement so kelvin plank statement says that in a cycle you cannot remove heat from a single reservoir and convert it entirely into work without rejecting it to the sink so naturally the sink is mandatory for the cycle to operate when we do this some amount of heat will be taken some amount of heat will be rejected and we will get the heat transfer now third point is the cannot cycle if you draw it on the PV diagram you will find that it consists of four cycles so this is PV is equal to constant and this is also PV is equal to constant this is PV raise to gamma is equal to constant and this is PV raise to gamma is equal to constant so this is isothermal expansion followed by adiabatic expansion isothermal compression followed by adiabatic compression and you know that isothermal and adiabatic it is a very slow process it is very fast process so coupling of slow and fast is impossible if you find out the expression for the efficiency of a cannot cycle using the same that is heat supplied upon say efficiency is equal to work done upon heat supplied you will get 1 minus T2 by T1 where T1 is greater than T2 and there is a limit that if T2 by T2 and T1 are in absolute degree and we cannot have T2 equal to T1 then efficiency 0 there is no work done if T2 is equal to 0 that is absolute 0 we can get efficiency is equal to 1 which is not possible and that is perpetual motion machine of second kind is impossible now the detail about the machine and its application will see in the next videos those who are interested in studying further they can refer to the some standard books available say references are thermodynamics by P. K. Naal and thermodynamics by Eunice Sengel thank you for patient hearing.