 Good morning Mr. Hello, I have got a basic doubts about the isentropic process and adiabatic process. In the equation delta S is equal to DQ upon DT and DQ is 0 and delta S is also 0. How we can differentiate between these two properties? Can you clarify this? This is an absolutely basic question about entropy. Our definition of entropy, remember definition of entropy is always a differential definition that is Ds is DQ by T. But with the condition that this DQ interaction must be for a reversible process element or for a reversible process. This is definition. I can put it as remember again definition. Then we have shown using the inequality in the inequality part of Clausius inequality that for any process reversible or otherwise Ds is greater than or equal to DQ by T. So you can say this is definition of Ds. This is a consequence of the second law through the Clausius inequality. So that means that even when DQ is 0, DQ is 0 means only the right hand side is 0. So this will be 0 if DQ is 0 that means for an adiabatic process. But remember because of this greater than or equal to sign, this means that Ds is greater than or equal to 0 for an adiabatic process. And remember that because of this greater than sign, even when DQ is 0, Ds need not be 0. It can be greater than 0 and in fact for real life processes it will be greater than 0. Because equal to 0 means a reversible process which is very difficult to implement in practice. Over to you. Sir, in the steady flow energy equation, we have got the terms kinetic energy and potential energy which is divided by GC that is a local gravitational force. And we are assuming that GC is equal to 1 for all altitude whether it is a correct or a GC is going to change the side for the altitude is correct. Over to you sir. The question is you ask something like GC. My question is what is GC? I have never seen it and I have never used it in this thing. Our first law equation in the reasonably general form we wrote is DECV by DT is Q dot minus W dot plus m dot i H i plus V i squared by 2 plus G z i minus m dot e H e plus V e squared by 2 plus G z i. There is no GC which I see here. The G only which I see here is the local acceleration due to gravity. But that too when I write it like this G z i and G z e it is expected that the Z i and Z e are not what you say not very different from each other. So, that the local acceleration due to gravity can be considered uniform. But if you set up a open thermodynamic system say going from the surface of the earth to say you know the geosynchronous orbit something like 36000 kilometers then naturally this model of G z i G z e will not work. But write it as delta e p this is the specific gravitational potential energy at Z i this will be the specific gravitational potential energy at Z e and then use the appropriate variation of G as it goes from Z i to Z e and you will still be correct. But there is remember there is no place for anything called GC in this equation over to you. This is regarding 0 degree Celsius and 0 Kelvin or whatever we assume 0 degree Celsius is 0 energy level and from steam table we get the energy levels above 0 degree Celsius. If for convenience we have taken 273.16 as a triple point and is it ok that my idea is at 0 Kelvin 0 energy level for all the substance on the earth is 0. Is it correct? See thermodynamics defines only changes in energy. So it is never right to say for example at 0 degree C energy or thermal energy of water is 0 that is not true because we never say that energy is 0 we always talk of the change in energy. So correct way to say is at 0 degree C the thermal energy of water is taken as a reference and the reference value is set to 0 that too for algebraic convenience. For example if you take your steam tables you have values of U values of H and values of S and you come to table 3 table 2 the very first line in that. Notice that at triple point 0.01 Celsius the specific internal energy for the liquid phase U f is 0 that is defined to be marked in this table the three points temperature of 0.01 degree C then U f of 0.00 and S f of 0.000 these are defined values at the triple point of water U f and S f for the liquid phase and 0.01 degree C for all three phases because it is the triple point. So but this 0 is arbitrary it is only for convenience if you feel like it and if your favorite number is say 14.97 for some reason just add 14.97 to all values of U here all values of U not just that U f value of 0 add your favorite number to all values of U and since your H is U plus P V that number will have to be added to H also and then absolutely nothing will go wrong your steam table will differ from other people's steam table but finally you will need only the values of delta U and delta H and those will not change because in the initial state as well as final state you have added a certain number in its energy value and same thing is true of S f S f also as an independent reference. So all over the steam table if you add your favorite number to S f whether positive or negative nothing goes wrong because all problems finally require only delta S but you also asked about 0 Kelvin so what is true for water is also true for other material in fact 0 Kelvin is only a derived or a created number in our thermodynamic scale unlike 0 C which was defined at some stage as the as a fixed point on the Celsius scale 0 K is not a defined temperature it is a temperature which is extracted out of the definition over to you now over to you why see college on a course for my good morning to you my question is what is the limitation of mass flow rate of steam extraction in the regeneration of steam power plant has it got any relation with the pump characteristics the question is what is the limit on the mass flow extraction from steam power plant and does it have anything to do with the pump characteristic yesterday I showed you only what is known as a contact heater which requires two pumps for that single is extraction which I showed the remember that an extraction reduces the specific output and if you do a detailed analysis it turns out that where you extract and how much you extract affects the efficiency and if you do some calculations you will find that there is a certain pressure for given parameters of the cycle like boiler exit state and condenser pressure there is a certain pressure at which the extraction should be taken and a certain rate of extraction which maximizes the efficiency so it has nothing to do with pump characteristic it has something to do with the properties of steam itself and the way the cycle is laid out in the applied thermodynamic not even in the applied thermodynamic course we have a third course known as steam turbines and gas turbines in which these detailed calculations of the optimals of various types of cycles and various extraction I considered only one extraction in an actual power plant there are anywhere between for a small power plant there will be one but for a large power plant there may be as many as seven or eight extraction five and six extractions are very common and one can go through an optimization process which optimizes or maximizes the energy for this. Derivations are possible but those are very tedious derivations because the equation of state for water vapor and water is not simple over to you. Thank you sir over and out. Good morning Sivakasi over to you one question please over. Good morning sir please. Can you explain about STP and NTP over to you. Okay the question is about STP and NTP these are defined P and T for reference states and depending on who you are whether you are a physicist and the chemist and what you are doing the values will be slightly different I have seen the pressure being one atmosphere that is the one standard atmosphere which is 1.01325 bar sometimes it is one bar similarly temperature I have seen 0 degrees C 0.01 degrees C 25 degrees C 30 degrees C and 300 Kelvin when you say STP and NTP find out from that field and that textbook what is the definition but if you take if we are doing thermodynamic calculations assuming either this and this or this and this is a good approximation if you do not know that because those are the most common values I find these days over to you. Thank you sir. Amal Jyoti Kerala good morning over to you. Sir my question is air is compressed to 200 bar then cooled to atmospheric temperature if the diesel is sprayed at this point will it ignite over to you. No the question is air is compressed to 100 bar then cooled to atmospheric temperature that means you have air at 100 bar and say the ambient temperature say 300 K then you want to ignite air I do not think why should it ignite just because it is at a high pressure temperature but maybe I misunderstood the question over to you. Sir if the diesel sprayed at this point. Oh the question is whether the diesel is sprayed at that point one has to do the so you have ambient temperature and high pressure diesel is unlikely to ignite because you will just have some kinetic energy of the diesel spray you need the you know the temperature to be above the flash point and if the diesel temperature is ambient I think it is unlikely to ignite that is my feeling but I am not really sure what will happen that is my feeling over to you. Sir another question from our side my question is that we have got petrol engines and diesel engines and for heavy duty applications we are making use of diesel engines why would not we go for petrol engine over to you sir. For heavy duty applications you require not only heavy power but also heavy torque and when it comes to very large power if you determine the mean effective pressure which leads to the compactness of the engine you will find that a diesel engine has slightly or reasonably high mean effective pressure than a typical petrol engine that happens because diesel engines can work at much higher compression ratios in fact they need much higher compression ratios than petrol engines. In petrol engines we cannot increase the compression ratio because then you have a problem of pre-ignition remember that in a petrol engine we must initiate the process of combustion by a spark if we do not use a spark but compress it to a higher temperature petrol vapor will ignite by itself and that will not be under our control we need control on ignition because of this the compression ratio of petrol gets restricted to around 10 maximum whereas compression ratio for diesel can go to as high as 22-23 this gives a diesel engine a higher mean effective pressure that means they become more compact and hence for high powers like transportation and heavy equipment a diesel engine happens to be the default choice over to you. Sir we have another question sir give example for sublimation process with respect to any industry sir overs. The very common sublimation process that we come across is that of the so called dry ice carbon dioxide the triple point of carbon dioxide as a pressure higher than the ambient pressure consequently if you leave it open at ambient pressure the dry ice that is the solid carbon dioxide straight away goes into the vapor form this is the most common in fact that is why it is known as dry ice looks like ice but does not go to liquid solid goes straight away to vapor that is the most common and demonstrable form of sublimation over to you thank you sir thank you sir over and out thank you VNIT Nakpur one question from you over to you good morning sir sir I am asking this question just due to curiosity do we have any practical application of Ericsson engine and Sterling engine anywhere as on today over to you no practical application of Ericsson engines Sterling engines are being worked on in fact they were working Sterling engines but when the IC engines petrol and diesel came up they lost out some research some development is going on on Sterling engines the advantage is that they will be external combustion engines but Sterling refrigerators are very common in the gas liquefaction field over to you I think Kalina cycle is it same as Rankine cycle with more number of heat exchangers or is it a separate standard cycle the question from K. K. Wag Nashik was about Kalina cycle whether it is a Rankine cycle with modifications or whether it is another standard cycle what I know of Kalina cycle is Kalina cycle is not a pure fluid cycle it uses mixture as a working fluid and because of that there are additional heat exchangers separators mixers involved if you a very very crude way of looking at Kalina cycle is remember we talked of the vapor compression refrigeration cycle and you can consider Rankine cycle and vapor compression refrigeration cycle to be roughly or crudely inverses of each other with appropriate modifications similarly you have the vapor absorption refrigeration cycle and if you say vapor compression refrigeration cycle and vapor absorption refrigeration cycle on the refrigeration side then on the power side you can roughly say the corresponding cycles could be Rankine cycle and the Kalina cycle the Kalina cycle is nowhere near being considered a standard cycle now it is a proposed cycle and successfully implemented commercial power plants based on Kalina cycle I do not think exist now it is still under development and experiment stage over to you. Hello sir one more question sir sir as we have seen absolute temperature and absolute temperature in degree Kelvin and degree Celsius we have seen the relation is there anything special about absolute pressure it is atmospheric pressure plus gauge pressure how this idea may be coming over to you sir absolute temperature is actually a misnomer what we have defined is the thermodynamic temperature or ideal gas temperature or anything like that on that remember for the Kelvin temperature triple point of water is defined as some value 273.16 Kelvin that is the only temperature defined on Kelvin scale so based on that 0 Kelvin turns out to be a derived temperature if you have a system at 0 K you will have to measure and demonstrate that the temperature is at 0 K 0 K is not a fixed point on the Kelvin scale there is only one fixed point and that is the triple point of water that is not true of pressure nothing in the system so no internal pressure on the walls of the system that is something which can be created and that is our fixed point for pressure and that is 0 pressure so pressure which is measured with respect to vacuum is the so called absolute pressure we have simply been using the word pressure for it but when you measure pressure using a manometer or some such thing usually what you end up measuring pressure by default is the pressure difference between your system and the surroundings and since the surroundings are usually the ambient conditions what you end up measuring is the pressure difference between the system and the ambient as indicated by the instrument usually called the pressure gauge so pressure as indicated by the gauge became known as the gauge pressure and that is how the term gauge pressure came add to the gauge pressure the local ambient pressure and you have the real thermodynamic pressure that we are talking about over to you sir I have a problem on work interaction piu can you go to the exercises on work interaction piu number piu sir exercises over to you sir yeah work interaction piu this is the quasi static and isothermal compression or pressurization of a metal block from 0 bar that means very low pressure to 1000 bar I do not have the thing just now with us but I remember this was discussed as the keyword here is in the third line the last line on page 2 that we have to assume that the density and isothermal bulk modulus of the metal to remain constant and between remain and constant within bracket you have the world almost that is the keyword because you have to assume some variation but at the appropriate stage neglect that variation because if you say that the density is going to remain absolutely constant then the answer is simply 0 there is no change in volume but we have a small change in volume but that change in volume in spite of the pressure going from 0 to 1000 bar it is so small that we might as well neglect it so using the bulk modulus you get a relation between the pressure and volume and the volume during the process is considered almost constant and is given by the mass of the block divided by the density I think if you look at the archives of this course which should be upon the moodle if they are not already there otherwise ask a question on moodle and you will get the answer within a few days over to you thank you sir over and out sir Balchand college sangly over to you hello sir my question is are there any power plants in India where the working boilers are at critical pressure or to you sir as far as I know today we do not have any installed supercritical pressure power plant in India but I would not be surprised if in the next four five years we start developing some in Europe we have such plants not very large scale but users critical supercritical power plant supercritical boilers and corresponding turbines etcetera are reasonably well established for the last maybe 30 years 35 years so there should be any difficulty it is only a question of scale and economics at that location over to you thank you sir another question is regarding the test three it is on the di-thermic movable piston sir will you explain the system formulation for the same there are a number of questions in the third test that is the test which we conducted today and I do not know which one you are talking about but remember one thing when you have a system and you have a di-thermic movable piston and you have something else here may be a movable piston or an immovable partition whatever now consider this as your part a of the system consider this as part b of the system and let us say that for simplicity that is an inherent assumption that the piston in between which is di-thermic and movable so let us neglect the thermal capacity and any presence of it except that it keeps the two sides separate the moment we say it is movable we will assume it to be leak proof friction less and this implies that at any stage p a is equal to p b similarly di-thermic means t a equals t b and that means if a and b in a few problems at least I know a and b was the same ideal gas like air which is approximated like an ideal gas so when you have a di-thermic movable piston since both sides are at the same pressure and both sides are at the same temperature you can consider it to be one single system if both gases are the same if the gases are different or if the fluids are different these conditions t a equals t b and p a equals p b will work as constraints and you will have to use the characteristic of a and characteristic of b to get the solution over to you. Sir one more thing when we insert a stirrer inside the system b how the system will change? When you insert a stirrer either here or here the same interaction that you are doing work this is our standard work direction because it is stirrer the magnitude would be a negative it is like inserting a stirrer in this big system if this is a fixed piston then a and b together are doing a constant volume process if this is a movable piston and say constraint to constant pressure then a and b together would be doing a constant pressure process in either case during the process if you assume a quasi static execution of the process temperature of a and temperature of b will be equal pressure of a and pressure of b will be equal. So, just consider it to be almost one homogenous system may be two parts with two different characteristics if the two gases are different and you should be able to reach the solution over to you. Okay sir thank you over and out. G. H. Raishoni college Nagpur good afternoon over to you. Sir my question is about IC engine sir there are two type of engine external IC external engine and external combustion engine and internal combustion engine. So, my question is what is interminant external combustion engine and what is continuous external combustion engine? Over to you sir. During my lecture yesterday morning on the cycles I have explained what is meant by an internal combustion engine and an external combustion engine. This classification pertains only to power plants only those power plants which have a fuel which undergoes the process of combustion and releases thermal energy which is absorbed in the form of heat at say the q 1 or q dot 1 of an engine and there is a heat transfer process involved between whichever medium in which in which in which the heat is released to the working fluid. If this happens we say this is an external combustion engine. So, in any external combustion engine you will have definitely two parts one is the engine and it is working fluid and the other one is fuel plus air releasing heat which will be transferred across this boundary. So, because of this the working fluid of the engine does not get contaminated because of the products of combustion it does not come in contact with the combustion products at all. So, the composition of the working fluid remains the same and it can be reused till it either leaks out or deteriorates because of some other reason. In the internal combustion engine part of the working fluid is fuel itself and you start off with a combination of fuel plus requisite amount of air or more than sufficient amount of air. The process of combustion and heat release takes place within the fluid or within the working fluid that is in the engine no external heat transfer process is involved and because of this the composition changes and you will have to exhaust this working fluid because you cannot use it again. So, such machines internal combustion machines will all the way also be open cycle machines that also was explained yesterday. So, that is about internal combustion and external combustion. We have in internal combustion engines again the gas turbine types based on Brayton cycle and the IC engine types based on the auto cycle or the diesel cycle. I think your second question was continuous combustion engine and intermittent combustion engine. If you see the gas turbine engine it is a multi component or multi unit engine. So, implementation of the Brayton cycle with typically 3 units being an open cycle you have a compressor and a combustion chamber and a turbine that will result in the minimal realization of a gas turbine engine. There is no cooler because the exhaust products are simply thrown out into the environment and you can because it is a continuous flow of air turbine is not a reciprocating device. It works in a it can be made to work in a steady state. So, you have steady combustion. So, you have a so called steady means continuous combustion, but you take an IC engine and you take say 1 cylinder of an IC engine which is say 1 unit of an IC engine. There are a number of stroke typically 4 strokes and at some point during the cycle you will have to initiate combustion and when the charge is consumed the fuel in the charge is consumed the combustion will well terminate itself. And next time again you have to have a spark or a new fuel to be injected and the combustion will be initiated again. So, you can call that this is an in what you say intermittent combustion engine. These are the only descriptions of various engines whether to give it the potential or put it on a pedestal of a classification or not that is to be decided over to you. But, how we refined the intermittent and continuous engine for external combustion. This is you have explained this is about the internal combustion engine. Open gas turbine is the example of continuous internal combustion engine and reciprocating system also example of the interminate IC engine. But, what about the external combustion engine? Because the steam engine is the example of interminate external combustion engine and steam turbine is the example of continuous external combustion engine. So, how we will define the internal interminate external combustion engine and continuous external combustion engine. How to use? See I doubt whether the steam locomotive can be considered an intermittent external combustion engine because the process of combustion is uniform out there although the it is a reciprocating cyclic device. So, the process of steam consumption may not be uniform, but the boiler has enough you know buffer in terms of mass and thermal energy. So, as not to be worried about that intermediate extraction of thermal energy in the form of steam going to the if you start a or if you work a Britain cycle or a gas turbine cycle with just you know switching on and switching off the combustion periodically based on some VIM or some requirement then it will become an external interminate intermitted combustion engine. But, I do not really see whether there will be any application for such thing. You switch on a gas turbine when you need it run it continuously and switch it off when you do not need it any more over to you. Thank you sir. One more question sir about work done. Sir in PV plot during the calculation of work done we take V volume on the x axis and pressure on the y axis. Ok sir and during the calculation of work done for non-flow process we take perpendicular strip on the x axis and during the non during the flow process we take perpendicular strip on the y axis why during the derivation of work done over to you sir. If you see the in a closed system which you may say non-flow work I will call it W expansion. W expansion is remember any work is forced into displacement. So, that can get converted to integral p dV over the process. It is expected that it is quasi-static so that we can integrate it. If you show it on a figure it is p V 1 to 2 over that quasi-static process like this. Now, when you consider an open system in open system also there will be a W expansion if the system expands or contracts like that of a football bladder or a car tire. But apart from that you also have a flow work and the flow work is again remember like any work flow work is calculated as F into the displacement in its direction. When we did the derivation we had an outlet to the open system and we said F this is the exit state and then we said let the fluid which was at this exit state at t plus delta t occupy this location. For algebraic simplicity we had said that let us assume local equilibrium at the inlet and exit ports and one dimensional flow at the inlet and exit port. And here we had said that we are taking a cut section perpendicular to the direction of flow. So, this is our d s this d s turned out to be velocity at the exit into delta t. And what was the force acting? Consider that there is a piston here which is sucking the fluid out or the fluid in the control volume is forcing the piston away. So, the force on the piston will be pressure at the exit plane multiplied by area at the exit plane. So, our F d s gives us P e A e B e delta t. All that you do is multiply this by the specific volume at exit and divide it with specific volume of exit. Of course, as I talked I said multiply first and divide later, but I think as I wrote I said divide first and multiply later does not matter. Then you notice that this is your rate of flow out. So, that is m dot e and that is why this component of the flow works turns out to be m dot e P e B e into delta t and similarly at the inlet you will have m dot i P i V i into delta t and that is out there the force is force and displacement are in the opposite direction. So, you end up with a negative sign. So, a very similar derivation will give you P i V i into m dot i delta t. So, this is our flow work term in the time delta t. I think that should explain the difference to you if any, but remember that there is no real difference, the difference is in application in detail of it. Otherwise, it is always some force into some displacement over to you. PSG, Coimbatore over to you. So, good afternoon sir. This is regarding assignment sir. Assignment will be posted in the Moodle and we have 18 participant. How many number of questions will be posted sir? The question is about the assignment. I will put, try to put up the assignment by Sunday evening on the Moodle. So, each and every participant will be able to download it. There is hardly any teamwork involved. Each participant will have to, there are there is essentially one problem, which is common to all. Otherwise, the other problems are some things for the participants themselves to set. I will give exercises from the exercise sheet and I would like the, I would provide hints to how to combine them or how to extend them, because all the participants are supposed to be teachers. And so apart from solving problems, they should also develop the ability of creating problems or setting up exercises. What you see is the development, which began with my teachers. I have inherited those problems, modified them further and every year based on the question students ask and the mistake they do, I modify the problem, I add to the problems and create an updated set. So, now that I have effectively something like 750 or 800 students with me, why shouldn't I depend on them to create some more exercises for me? Part of the assignment would essentially be that. Over to you. Sir, after solving that problem, we have to upload it in Moodle sir. What is that? Yes, you will have to upload them on the Moodle and along with the assignment, the details will be posted. But one thing is certain, which I have already announced that I only want pdf files. I do not want any doc files or docac files or some swf files or xyz files. I want only pdf files. That does not mean that I am forcing you to use some package. You may write down nicely by hand. Absolutely no problem. Just scan it up. Every scanner will be able to create a pdf file for you. Upload that file. Over to you. NIT Calicut, I hope you have overcome the audio problem. Over to you. Sir, regarding the break problem, we have shown the break drum and can you please explain by considering the three boundaries, the working reaction and the heat interaction. The question is about the break power. So, I think a simple way of doing things perhaps is to show. I am just showing a block diagram. This is not a reality diagram. Let us say this is the working fluid part of any engine. So, what you have? You have streams going in. Typically, the streams going in would be flow of fluid, flow of air. Streams coming out would be m dot exhaust. Now, there may be a direct interaction between the action. For example, the exhaust may be coming out through an exhaust duct, which may be directly exposed to the surroundings. So, you will have a q dot loss. Let me put it loss 1, but the working fluid provides a work output, which you can say w dot indicated power to the mechanism and structure of the machine. There will be some heat transfer, because usually our working fluid internal combustion engine is supposed to be adiabatic. It is our assumption. So, you can say this is some q dot loss to from the working fluid to the mechanism and structure. Now, the mechanism and structure may itself have m dot say cooling water again the same m dot cooling water or sometime cooling air. And the structure itself may have some heat loss to the surroundings q dot loss 3. The structure may absorb this much power, but the structure will then deliver w dot break power. Now, out of this, the things which are easy to measure is this w dot break power, because you can use a machine dynamometer of any kind to measure that. Then you can say heat loss to cooling water. Actually, it is nothing but m dot this m dot cooling water into enthalpy at water at this point B minus enthalpy of water at this point A. If you have another air stream include that if you have an oil stream going in and coming out include that add all those streams which are going through the structures. And that will be what we call heat loss, but it is actually enthalpy outflow energy outflow, because of the flowing fluid. These losses and these losses which are directly from the structure are difficult to measure, because the mechanism is combined convection and radiation. And we generally do not measure or we do not have the ability to measure in any detail the surface temperature distribution and the surface characteristics. So, we make a reasonable guess and try to estimate these losses this one as well as this one. And because these remain estimates and because whatever you measure will have certain uncertainties with it, your first law will not exactly be satisfied. And that small dissatisfaction we say unaccounted losses, but so long as they are consistent with the magnitudes and our uncertainties, then that is ok. Over to you. Here, this is one of those exercises which I put in to find out who the culprit is, who through that sack of sand. And actually you will notice that there are four questions here, 0, 1, 2 and 3. I am following the tradition of thermodynamics starting counting from 0. And here I have said please help the prof by computing the quantities needed to answer 1, 2 and 3. And end of the question was finally guess the answer to 0. Since here my students that is all of you participants are at least a few tens of kilometers away from you. I know that the answer is no one of you, but when I put it to my students in a examination many students wrote the answer to the 0th part as definitely not me sir. But anyway let me come back to the solution of the problem. You have some mass in this particular case a sack of sand which is thrown from some height h onto the ground. So let us say this is the m of the system and it goes down a height h. So finally it comes down here. Now read the explanation. You may assume that the height of the building is 30 meters it is given. Also it may be a good idea to assume that the initial and final temperature of the bag was 25 degrees C which was also the temperature of the environment. So T initial was 25 degrees C, T final was also 25 degrees C and T environment was also 25 degrees C. The problem has really been simplified. Even though frictional heating may have raised the temperature of sand as it fell and smashed to the ground heat transfer to the environment could have cooled it to the ambient temperature. Now although it has been asked in the order 1, 2, 3, delta S of the bag of sand, delta S of the surroundings and what was the amount of entropy produced that by our standard definition would be delta S of the bag plus delta S of the surrounding. And finally what was the amount of lost work that would be the entropy produced into the temperature of the surroundings. So this is our T naught of the combined first and second law analysis. One way to look at the problem is the initial state of the sand is height of 30 meters h i or z i is 30 meters, T i is 25 degrees C. The final state of the sand is 30 meters z final is 0 meters, T final is also 25 degrees C. The interaction there is enough hint that there is a heat transfer to the surrounding but since it is a free fall and it is smashed on to the ground. If you assume that the ground is solid and does not deform under the drop of the sand then we can say that the work interaction is 0. In any way the first law will have to be delta E is Q minus W. W is 0 because there is no agency here to absorb any work and so the first law reduces to delta E equals Q. You can say that the system is this or system is the sand sack initially out here finally out here. So turning this around Q equals delta E and of all the components because T i is the same as T f. So there will be no change in the thermal energy of the sand. Initially we assume that it was just drop from there. So initial velocity is 0, final velocity anyway is 0. So there is no delta E k the only thing is delta E p. So that will be g into z final minus z initial. So this gives you minus g into h because h is the difference in the height z f is 0, z i is h meter. So this will be minus z h in appropriate units and the direction of Q is this or if you want you can show it into this sand bag. Now this is the heat absorbed by the system and because there is no change in the temperature of the sand the changes only in the energy part of the sand. The delta S of the system is also 0. Now this is from the sand bag as a system looking at it from the environment point of view heat absorbed by the environment is minus heat whatever is absorbed by the system equal and opposite amount of heat will be absorbed by the surrounding because it is an interaction. So this comes to be plus g h and by our definition of heat entropy change of the surroundings. So this will be Q naught by T naught that will be g h by T naught and the lost work W lost is delta S naught into T naught because sorry one more step in between delta S universe is delta S plus delta S naught which turns out to be g h by T naught and then this delta S then W lost is delta S universe into T naught is g into h. Actually there is nothing surprising in this answer and we could have come to that answer straight away by considering that look if instead of just throwing the sack of the sand downwards over a height h I could have set up a winch like thing or a chain pulley arrangement and I would have just balanced it with an equal weight assuming there is no mechanical advantage in the chain pulley arrangement balanced it by an equal weight on the other side and just tap the sand bag and sand bag would have come down by a height h and on the other hand the weight on the other side would have risen by an height h giving us the work output g h that is the work output which we could have obtained but did not obtain so that is the lost work. In fact you can start from that argument and from the lost work you can say hence the entropy produced could be this much. I hope that explains it over to you. Thank you very much sir. College of Engineering Pune over to you. Thanks a lot for this workshop we learnt many new concepts and new approaches and we all are waiting for you for your book on thermodynamics and I have one question to ask how reversible adiabatic work interaction does not depend on the path. We have seen this while defining the energy. So, how to explain it? Thank you over to you sir. See the reversible adiabatic work between two states that means you have an isentropic process between the two states it is a unique process and it is a reversible process that means for our fluid systems only the expansion type of work could be done over a unique process and that is why it is unique. Actually this comes out of the fact that the d q by t for a reversible process is unique that is the derivation from the second law of thermodynamics and from that this follows. We define entropy delta s equals 0 means isentropic that means you have an adiabatic process and reversible adiabatic process just the way I have argued now will give you a unique amount of work over to you. Sir we have started before we have not done anything from first law or second law we have formed the definition of energy and for that we have used this. So, how to explain it before we have done first law and second law? How to explain it before without taking help of second first and second law? Madam if you do not take help of first and second law how do you make an explanation using the science of thermodynamic because the thermodynamic argument will definitely invoke first law and second law whenever needed so over to you. Okay thank you sir over to you. Anirama Amdabad over to you. Hello sir please give hint to solve the problem number PR 8 property relation from property relation PR 8 over to you. We are almost at the end of the discussion session. So, I will quickly tell you that PR 8 was discussed in the class sometime after property relations were derived. So, let me ask the people here whatever I have written here and on the tablet will be uploaded on moodle right when how long will it take. So, I am told that in a reasonable amount of time of about 2 or 3 weeks mid July you will have the whatever I have written down here either on the iPad here or on the tablet will be there upon moodle. So, even this discussion will be there on moodle. So, now I will say over and out. Thank you very much.