 I would like to talk something about two exercises in that is 13 and 18 because when it comes to students these need to be explained properly. So, OS 13 is about the situation where we have a throttling calorimeter and the situation is measure the state of wet steam that is mixture of water and steam x less than 1. So, here we know that the phase rule dictates that pressure and temperature are not independent of each other. So, we have to have one of them at most and something else. So, how do you determine the something else? It is difficult to do density measurement or it is difficult to do energy measurement. So, under certain conditions we use something called a throttling calorimeter. OS 22 is an illustration of throttling and separating calorimeter. So, you can extend this idea in OS 22. Here the idea is like this. Because the situation suppose you have a duct through which steam flows and you measure its pressure say P1 and temperature T1 and you find that the pressure and temperature turn out to be lie exactly on the saturation line. For example, the pressure could be if you look up the steam tables pressure could be 2 bar and temperature turns out to be 120.2 degree C or pressure turns out to be 5 bar temperature turns out to be 151.9 degree C. Such situations typically occur at the exit of a boiler steam generator just before it goes to the super heater or some equipment which follows the boiler drum. But we want to know what is the state is and with such a situation the state could be anywhere from saturated liquid to dry saturated steam. The throttling calorimeter is a device by which we can determine the drainage fraction of steam under the following conditions. Number one, the pressure is higher than ambient. So, that if we take a sample of steam from this pipe and connect the other end of the sampling device to air steam will start automatically flowing through it. We will have not have to use a vacuum pump for anything. And second this works when the drainage fraction of steam is very near one within a few percent of 100 percent. It may not work when the drainage fraction is as low as 0.8 or 0.7, but it is likely to work when the drainage fraction is something like 0.949596 and so on. So what we do is the following. We take a sample out of this, maybe we put a tube with some holes and some steam a representative sample is taken out. Now it is expected that P1 is greater than P ambient. So, if this is connected on the other side to the environment steam will start flowing through this and maybe a large amount of steam will flow. What we do since we want only a sample we put here what is known as a throttle valve. A valve which is only partially open or sometimes a small orifice or a series of orifices. The job of this throttle valve is to provide enough flow resistance so that even with the large pressure difference between P1 and ambient only a small amount of steam flows through this. Now as it reduces in pressure the volume increases so we provide a large area for the steam to flow out. So the inlet is here and we consider our open system to be made up of from the inlet to the sampling tube to the exit and at the exit we measure the pressure P2 which is expected to be P ambient and we measure the temperature T2. So this is the inlet state I state 1 the exit state E state 2 and there is a small amount of m dot which flows through this. The whole thing is well insulated it is not a very large piece of equipment. So we expect that it runs in a steady state there is no power output or power transfer it is well insulated and because it is a small equipment data EP is negligibly small and because of the small sample which is extracted and the large area which is provided at the exit it is expected that data E or K is also negligibly small. If you start with the steady state energy equation we get Q dot minus W dot S is m dot HE minus HI which is H2 minus H1 here plus delta EK plus delta EP. Now here Q dot is 0, W dot is 0 these two are expected to be negligible. So we end up with the result that H2 equals H1 that means from the inlet state to the exit state or for the inlet state and the exit state what we have is the same enthalpy. And if you now look at the properties of steam if you take it on say the PV diagram let us say let this be the initial pressure P1 inlet pressure and let this be the inlet state slightly wet. This is the saturated liquid line this is the dry saturated vapor line and let this state be the inlet state. If you look up our enthalpy values you will find that if I keep the enthalpy the same and if I reduce it to a lower pressure this line is likely to go like this and you will end up with a state 2 which is at pressure P2 but which is likely to be superheated. And now if you are in the superheated zone here P2 and T2 decide the state. So from here we obtain H2 and since one is wet steam P1 and T1 is not a good pair. So we use P1 and then we use H1 which from this relation turns out to be H2. If you proceed with the details of this problem you will find that in this particular case P1 is 10 bar state of steam after throttling is 0.75 bar. So they have used the vacuum pump for some such device to reduce the ambient pressure near the exit of the throttling calorimeter to 0.75 bar and T2 is 100 degree C. Now from this 70.75 bar and 100 degree C if you go into our steam table you will find that 75 degree C is 75 bar and 100 degree C is 0.75 bar and C is tabulated on page 11 roughly slightly above the middle 100 degree C and 0.75 bar and here H2 turns out to be 2679.4 kilojoules per kilogram and hence H1 is also 2679.4 kilojoule per kilogram. Since P1 is already given to us as 10 bar, 10 bar and H1 of 2679.4 kilojoules per kilogram decides the state and now we go to table 3 sorry table 2 page 8 where we notice the 10 bar line and if you notice that at 10 bar HF that is HF at 2 is 762.8 kilojoule per kilogram Hg at 2 is 2778.1 kilojoule per kilogram. And then you say HF2 is less than H1 is less than Hg2. Hence H1 lies between HF1 and Hg1 hence state 1 is wet steam and once you decide it is wet steam then X1 can now be calculated as H1 minus Hf1 divided by Hg1 minus Hf1 which you can calculate whatever value turns out to be. So while explaining I have solved almost all the full problem for you. Now another problem which I want to talk about is OS 18 OS 17 and OS 18 both per kilogram are attained to what we call a heat exchanger. Now we have to tell our students because when they learn this they have not studied heat transfer. So they do not know what a heat exchanger is. If you take them on a round of your campus then you can show them heat exchangers of some times which you will find in your central air conditioning plant or chiller plant and things like that or even open the bonnet of an old car. New car can also be used but the equipment is so compactly arranged and so cluttered in it that it is difficult to show the various subsystems. If you show them an old car you can show them a radiator and explain that it is a heat exchanger but in a thermodynamics class all that you tell them is that heat exchanger is a open thermodynamic system. So if this is a heat exchanger there will be two streams flowing through it without mixing with each other. So this is the transfer type of heat exchanger. Say fluid 1 rate m dot 1 enters at state 1 i and leaves at state 1 e. Fluid 2 m dot 2 enters at state 2 i and then the fluid 2 m dot 2 enters at state 2 i and leaves at state 2 e. Tell them that look this is only a schematic it is not necessary that the two fluids flow through two channels just next to each other as parallel lanes on a highway. They could cross each other one tube could be rolled on to the other they could be across each other they could be against each other. There are types of arrangements very complex once in real life and but what we should know is there are two streams not mixing with each other. So what goes in at 1 i must come out at 1 e and what goes in at 2 i goes out at 2 e. So this is an open system two inlet two exit no mixing it works in a steady state and although we claim it to be a heat exchanger overall it is an adiabatic system it is usually well insulated and no attempt is made to extract any power. So apart from q equal to 0 w dot s equals 0. Also the change in kinetic energy for any stream is negligible change in potential energy for any stream is also negligible. Now if you look at our equation the steady state equation is q dot minus w dot equal to you will have m dot 1 h 1 exit minus h 1 inlet for one stream plus m dot 2 h 2 exit minus h 2 inlet. This will what will be the reduced form of the first law but we should derive it by the general form. On the left hand side w dot is 0 w dot s is 0 q dot s is also negligible so you end up with this expression to be equal to 0 that is the heat exchanger equation. What happens inside the heat inside the heat exchanger is there is an exchange of energy in the form of heat between the two fluids and because of that as you notice here since m dot 1 and m dot 2 are both positive the flow rates one enthalpy difference will be positive one enthalpy difference will be negative depending on which it said to be the hotter fluid and which it said to be the colder fluid or cooler fluid. Here we have to determine for example in OS 18 we are given 120 kg per hour of saturated water at 8 bar enters a heat exchanger and leaves at 4 bar 200 degree c. So in OS 18 stream 1 is water, 1 inlet is saturated water at 8 bar, 1 exit is the heat exchanger 4 bar 200 degree c. So there is a in the other side the other stream is hot air enters at 600 degree c, 600 degree c 2 bar and leaves at 240 degree c 2 bar, 2 inlet is 2 bar 600 degree c and 2 exit is 2 bar. So how do you pressure difference on the air side? 240 degree c. So now you will notice that you can determine h i h e for stream 1, you can determine h i h e for stream 2, m dot 1 is given 120 kg per hour. So from this equation the only unknown remains to be calculated is m dot 2 which you can now calculate. Heat transfer rate from air to water remember here our system is the complete system, but you can devise another control volume and that control volume is containing only the one side through which say air flows. And if you want you can have another control volume, control volume 1 is that part through which only the steam flows and you will have this q dot from 2 to 1 the rate of heat flow. Remember if we consider the whole heat exchanger as our system any internal heat transfer will not be computable. So to determine the heat transfer rate from air to water we will have to put a boundary between air and water. So we will consider either a green control volume which is the water flow control volume or the red control volume which is the air flow control volume. And then when you apply your first law to this say to the water side you will get q dot 2 to 1 equals m dot into h 1 e into h 1 i in which you can from which you can calculate q dot. The entropy production rate can also be calculated we have the expression just sum it up over 2 streams and b and you can proceed from there. I think other exercises you can tackle yourself 1 1 0 0 m e s pillai nu panvel. Sir we are talking about second law about energy you have not made any comment on that. There is various ways in which the laws of thermodynamics particularly the second law can be interpreted. I have left it at entropy differences I have left it at the inequality which comes out of the second law. The maximum I have done is defined entropy production and says any process which takes place is one in which entropy production is positive. At most it could be 0 in which case it will be a limiting ideal reversible process. People interpret it as degradation of energy but without defining what is meant by degradation of energy. People interpret it as increase in randomness increase in chaotic behavior and things like that but we do not have to interpret it anything like that it is good enough for us to interpret it as a change in entropy of adiabatic systems being always positive in real situations. That is it over to you. Thank you very much. Over and out. 1 1 1 3 s b patil in the poor. Sir my question is how to derive the reducer equation of state for under wall gas. How to derive the reduced equation of state? In fluid mechanics we are used to using dimensionless numbers. For example velocity can be made dimensionless by dividing it by either the free stream velocity or the mean velocity in the duct. We have Reynolds number and all that. So in thermodynamics when we compare different fluids and see whether their characteristics are similar we tend to use what are known as reduced variables. So instead of pressure we define pressure divided by some reference pressure, temperature divided by some reference temperature and volume divided by some reference volume. For fluids which have a critical point and which is true for most of the fluids the critical state pressure, temperature and volume at the critical state is used as the reference state. So all that you do for deriving the reduced form of the van der Waals equation is write pressure as critical pressure into reduced pressure P r, temperature as critical temperature and to reduce temperature T r and the specific volume as critical specific volume V c multiplied by reduce specific volume V r. Now that way you will end up with an equation which contains the parameters a, b and r as well as the critical constants V c, T c and P c. We have already derived relations for P c, V c and T c in terms of a, b and r. If you use those relations you will notice that a, b and r get eliminated and the equation remains only in terms of P c, V c and T c. That is the reduced equation of state or reduced form of the equation of state. The advantage is once this happens once you are able to get the reduced form of equation without any P c without any a, b or r in it then in principle you can use it for any van der Waals gas or any gas which obeys the van der Waals equation of state. Over to you. What is the significance of calculation of entropy in steam generation process? Calculation of entropy in steam generation process. I do not know, calculation of entropy and entropy differences is significant and good to look at in any process for that matter. Nothing special about steam generator process. Actually it is the entropy comes into its own and it is very important when we have adiabatic systems when like turbines, compressors, nozzles, sometimes even heat exchangers. Of course steam generator is a heat exchanger but we have a high temperature source from combustion. So there is significant amount of heat transfer, significant temperature difference also. So there will be significant entropy production as we have defined that term. Apart from that in the design of boilers the entropy generation or anything is not used. I know that. Over to you. Thank you sir. Over and out. 1272, Mangala Yatthen University, Beshwautar Pradesh. Sir, yesterday we have described one formula TDS equal to DU plus PDV. And I want to ask whether it is valid for flow system or open system or all the systems. See TDS, this is a basic property relation. It is not for any process or anything. So as a relation between properties of fluids you can always use it. It has nothing to do with whether it is an open system or closed system. Now if we are at it, I would like you to you and all others who are listening to understand and appreciate this because this quite often leads to some confusion. Now consider for simplicity a closed system and a small process. Now in that small process we will have DQ equals DE plus DW. Now let us assume that DE equals DU and let us assume that it is a simple compressible process. Now in that case I can write DQ, now remember this is first law. DQ equals DU, now simple compressible system I can write this PDV plus DW other where DW other is other type of work say electrical or stirring or any other non 2W type of work. The only two way work is PDV because it is a simple compressible system. Now this also is first law. Now what does the second law say? The second law says that DS is greater than or equal to DQ by T. Since T is always positive I can write DQ is less than or equal to TDS. Now that means I have combined this and this and I will get TDS is greater than or equal to DQ and DQ is DU plus PDV plus DW other. Now for this equation I have used first law and I have used first law. Now remember this equation I will write it again on the top of the page TDS is greater than or equal to, TDS is greater than or equal to DU plus PDV plus DW other and this is a consequence of first and second laws. Now I will use the property relation. Now which we have derived yesterday the property relation says TDS equals DU plus PDV and why can I use this because this is a property relation for a simple compressible system. So, now I have one equation which is an inequality for TDS after first and second law and another equation which is an equality for TDS using the property relation. Now the next thing which I have to do is replace in this equation for TDS DU plus PDV and what do I get? I get DU plus PDV is greater than or equal to DU plus PDV plus DW other and now DU plus PDV will cancel out from either side and I will get for a simple compressible system DW other is less than or equal to 0 for a and what does it mean? That means if you have a fluid system like a liquid or a vapor or whatever then if you do work for a simple compressible try to do work by a mode other than the PDV mode because this is DW other that work interaction will always be such that you can do work on the system you cannot extract work from the system and this is the thermodynamic consistent demonstration of the fact that you can put a stirrer in a fluid and start stirring it doing work on the system, but you cannot put a stirrer in a fluid which is already in equilibrium and ask the fluid stir the stirrer and do work on it. It cannot do that because the laws of thermodynamics prevented this is what we have shown. So this is another justification that work interactions which do not depend on any property of the material will be one way work interactions and this is the one way direction that is demonstrated by the laws of thermodynamics over to you. Okay, thank you sir over to you sir. Chikodi 1089 KLE college of engineering and technology over to you. Actually we want to know that how that a plug flow has been characterized yourself. Now the energy of the exit plug will be equal to the mass in that that is mass in that much volume multiplied it by its specific energy. The specific energy for the inlet plug is EI the specific energy for the exit plug is EE and if you go back a bit you will see that the mass in the inlet plug is rho I AI VI delta T the mass in the exit plug is rho EE VI delta T. So all that has been done is we have taken these two expressions and use those mass in the inlet plug and exit plug whatever you see in the brackets here is mass in the inlet plug what you see in the brackets here is mass in the exit plug and all that we have done is multiplied the mass in the inlet plug with the specific energy at inlet mass in the exit plug with specific energy at the exit over to you. As we know that two temperatures were taken for the heat engine is there any concept of how the temperature distribution effect can be considered for the efficiency calculations. The crucial thing here is to realize that it is the corollary of Carnot which is more important that if you have two temperature levels then any reversible heat engine will have the same efficiency. The beauty of this is this does not depend on any detail of how that reversible engine works. So it has absolutely no relation with the temperature distribution from T1 to T2 which may be occurring inside the engine. We just do not have to worry about it and that is the strength of Carnot theorem and it is only on that strength that we can define a thermodynamic scale of temperature over to you. If the water is heated in infinite number of stages by bringing the water in contact with infinite number of reservoirs each succeeding reservoir being at a higher temperature than the next receding one what will be the change in entropy of the universe. This is I think an exercise either in Sears book or the Zimansky book. You can show that as you increase the number of reservoirs each increasing by smaller and smaller delta T the you can show that the change in entropy of the universe reduces from a large positive number it will come to a small positive number but for it to reach 0 you will have to make the number of reservoirs really to infinite. Any finite number will have a small but finite entropy production making the process irreversible but your idea is right that is what we will need but you will need an infinitely large number of reservoirs even reservoirs differing by 1 degree centigrade will not do 0.1 degree centigrade will not do you say 10 raise to minus 9 degree centigrade even that will not do because that is still a positive temperature difference over to you. Why we ignore the pressure variation with elevation for a storage tank in a gas the if the density of the gas is not very large then compared to the mean pressure the pressure variation will be small and since the pressure variation is small the variation in other properties also like is also likely to be small. So the results which you obtain regarding what happens to the gas and the way it behaves will not significantly differ if you replace that varying pressure by a mean pressure at an appropriate value. The purpose of doing all these approximations is not to say that that does not matter but to say that the variation may be of some consequence but if I neglect that and base all my calculations on the mean pressure the results which I obtain are good enough to do all the things which are needed to be done for example the gas drain rate or the gas heating rate or whatever but there will be situations where perhaps you will have to consider that variation in which case there is no choice you cannot neglect that over to you yes. Is it possible to attain to zero temperature zero Kelvin temperature experimentally. No because if you attain and create a system at zero Kelvin then that means you can run a reversible 2 T heat engine between some other temperature say ambient and zero Kelvin and that will have an efficiency of 100 percent so that itself will violate the Kelvin Planck statement. So zero Kelvin is a temperature which we can write down on paper we can discuss it in a workshop like this but which will not be attained. Can we call that entropy will measure the irreversibility. Not entropy, entropy produced or entropy production is a measure of irreversibility and next week when we considered combined first and second laws we will be studying what is traditionally called exergy analysis or availability analysis where we will provide some significance to the quantitative or the numerical value of entropy production. Today we can only say that any real process is one in which the entropy production has to be positive in the limit it can be zero but that would be an ideal reversible process and we are unlikely to be able to execute such a process in practice. But what is the difference between a process in which the entropy production is say 1 kilo joule per Kelvin and another in which it is say 2 kilo joules per Kelvin that we will be able to realize only after studying the combined first and second laws next Wednesday. Over to you. Thank you sir, over and on. 1150, Godavath institution's Atigre over to you. Sir my question is what is the value of drainage fraction at critical point. Okay see drainage fraction is defined only when there is a distinction between the liquid phase and the vapor phase so that density of liquid and density of vapor have two distinct values and so are other properties. At the critical point there is no distinction between the liquid phase and the vapor phase that is the lowest pressure at which it occurs. Since there is no distinction in the liquid phase and vapor phase there is no meaning in defining a drainage fraction at the critical point over to you. Sir can you measure the entropy? By measure you mean a direct measurement of entropy then the answer is no entropy or entropy difference between two states can always be derived but cannot be measured. Similarly even energy difference between two states can always be derived but it cannot be directly measured. Sir as we are having static sign dynamics in engineering mechanics like that is there the study like thermostatics as we are saying thermostatic valve and what about thermostatics sir the study of engineering branch is there. Okay actually since we have we lay significant emphasis on equilibrium may be things pertaining to equilibrium states that is property relations etc. you could consider as thermostatics in our course where a change of state relation between processes interactions and end states that can be considered thermodynamics and if you want to bring it the time rate by considering some other extraneous properties like conductivity and all that in that case perhaps the science of heat transfer which we will be studying or students will be studying slightly later that also could be included in so called thermodynamics over to you. Sir yesterday it was a discussion can we says the opposite word of sublimation as the throttling because during grain we will get the ice blocks and that is the instantaneous process and also in a jet plane where air particles are connoted directly into the ice particles is it so with this opposite word of sublimation is throttling is it right sir. No throttling is an entirely different phenomena throttling is a simple fluid dynamical process in which in a small zone like a capillary tube or a small orifice or a slightly opened valve the resistance to flow is so large that even a small amount of flow requires a large amount of pressure drop that is all throttling the effect of throttling is to be calculated thermodynamically as we have done in one of those exercises and you will realize that it has nothing to do with the sublimation process in solid state physics and in material science sublimation has an opposite but they use the same word condensation. So they say that carbon dioxide vapor condenses directly into solid carbon dioxide. So technically the word condensation can be used to denote a liquid of a vapor as well as the solid of a vapor whereas a liquid becomes solid it is called freezing gas becoming liquid is known as condensation in a similar fashion gas becoming solid is also known as condensation over to you. Okay sir when we are defining the temperature in the textbook we will see the meaning it is a degree of hotness or coldness but when we are defining to the electronic level for example when we are defining the current we are saying that charge transfer from one electron to another electron then how we can define this temperature with respect to the atom level considering the electronics and the kinetic energy like that sir. Okay in our scheme of thermodynamics temperature was only the a label given to isotherm labels given to various isotherms were defined as temperature we later on quantified it using the second law of thermodynamics Carnot theorem etc. etc. If you want to give a different interpretation to temperature you will have to go through kinetic theory where kinetic theory derives relations where for a conglomeration of particles which we in the continuum domain say behave like a gas you can derive an equation of state where you can show that the volume multiplied by the mean momentum of the particles equals some constant into the mean kinetic energy of the particle okay and then we compare this with the our equation of state from macroscopic thermodynamics and say that look this is very similar to pv equals RT provided we assume that p represents the average momentum of particles and t represents the average kinetic energy of the particles. So that is one way you can interpret temperature as kinetic energy of particle actually even kinetic theory does not really say that temperature is the mean kinetic energy of particles you can say temperature can be used to represent the kinetic energy of particles that means for a given set of particles if they are in a state where they have a higher kinetic energy they will have a higher temperature if measured on the macroscopic scales and vice versa over to you. Sir while we are considering the plasma state of steam or water is it the ideal state we can can we say that plasma state of water is the ideal state. No there is see there is no such no state as an ideal state an ideal thing is something for us to define ideal gas we call it an ideal gas because we want to call it an ideal gas the gas does not itself know that it is ideal or otherwise and for a fluid like water there is no such thing as an ideal state we can only have a reference state that is it and even the reference state is for us to define over to you. Sir for the turbine and the compressor e raise it is decreasing and increasing we are finding out the efficiency now you are giving the example of the duct how you derived the efficiency for the duct. No there is no such thing as an efficiency for the duct the idea of isentropic efficiency is only for three components a nozzle a turbine and a compressor that is it for no other thing something like an isentropic efficiency is defined and remember that that isentropic efficiency it is something which we define thermodynamics ends by telling you that entropy for such an adiabatic system should be such entropy change will be such that at the exit it will be higher than that at the inlet except in the limiting case in which case it will be equal that is where thermodynamic after that it is for us to define using those ideas over to you. Thank you very much center 1001 MA90 Bhopal over to you. Sir yesterday you are discussing the property relations yes in that you are discussing L-Mode function and Gibbs function right so first thing what is the significance of these two functions and second one is can you give some examples in the real life where we find these actually functions. Okay the utility of these functions is mainly from the point of view of thermodynamics and particularly of physical chemistry they have been defined because they have been found useful the simplest example of this is enthalpy has been defined because it has been found useful and we have that u plus pv term coming up quite often particularly when we have open thermodynamic system actually there is nothing wrong if the whole world decides to forget about enthalpy and works only about only with u and pv but then every time we will have to we will not tabulate enthalpy but we will have to work with u and pv in a similar fashion the two functions Helmholtz function A and Gibbs function G have been derived because they have been found useful in doing detailed thermodynamic calculations and calculations in physical chemistry. We are not spending much time on property relations in fact given to given the choice I would spend another 5 to 6 hours with my students on property relations demonstrating to them what is the utility of A all for a hint in property relation exercises yes the last two exercises PR 20 and PR 21 we look if you look up PR 20 it asks you to demonstrate that if the Helmholtz function or specific Helmholtz function of a system is specified as a function of temperature and volume then all properties including any change of phase saturation lines can be derived and in fact this is the way our steam tables are specified if you go to the link provided on the Moodle page through the NIST steam tables you will find that the specification of properties of steam is only in terms of one single function A as a function of T and V from there everything else is defined or derived including critical point liquid vapour saturation point from triple point onwards everything else including specific it can be extracted out of it that is the power of the Helmholtz function but in our day to day calculations we do not come across that so often Gibbs function is useful because we have showed yesterday that the work which we can extract from a system which executes a process which is isobaric come isothermal would be limited or the maximum work which can extract will be the decrease in the Gibbs function. So this is the this is related to the chemical potential of a reacting mixture in fact if you study physical chemistry you will soon be made to realize that chemical potential is nothing but specific Gibbs function with respect to the change in one particular component what is known as the partial molar Gibbs function that is the chemical potential. So chemical equilibria rates of reaction all those things they depend on Gibbs function since we are not delving any deep into combustion and chemical reactions we will not be appreciating the importance of Gibbs function in this series of lecture. However it is my duty to expose you that there is something called Gibbs function which we should be aware of over to you. Type is the parameter in steady flow energy equation like q dot term is joule per second then how thermodynamics differs from subject heat transfer. See here you will notice that this first law of thermodynamics is derived just by differentiating the first law with time q dot it is still related only to energy changes or energy change rates and work rates it is not related to any temperature difference or temperature gradient it is the job of the science of heat transfer to relate the difference in temperatures to the rate at which heat is transferred. For example yesterday we solved a problem I think the very first exercise is in the second law it says there is a 150 watt of steady heat flow from one system to another thermodynamics does not tell you how that 150 watt is arranged that is the job of heat transfer over to you. Can you tell me the difference between ideal gases and perfect gases too? I do not really know because when I was in school the idea of ideal gas and the idea of perfect gas was the same. Some textbooks called it ideal gas some textbooks particularly in physical chemistry used to call it a perfect gas. So in physical chemistry you have a perfect solution and the perfect crystal so maybe they like the word perfect. Later on I found that some textbook define an ideal gas the way we do and then what we call or what I have been calling ideal gas with constant specific heats they tend to define as a perfect gas. So there is no global definition I get a feeling ideal gas is a more common nomenclature and if you want to use perfect gas to mean something else or the same thing say so and use it that is it. Unfortunately the scientific community has not standardized nomenclatures and symbols except a few things like pressure, velocity, internal energy, entropy and enthalpy. Even availability, exergy these are not standardized over to you 1118 Sri Jayachama Rajendra College Mysore when you say open system and control volume are same can we consider control volume area to be a closed system. If so how both open system and control volume are similar see open system and control volume are synonyms they can be applied to any system through which a material can flow in or flow out particularly a fluid but it could be any other material if during a process some mass flows in or mass flows out or both we call it either a control volume or an open system. The traditional method was control mass and control volume the modern method is closed system and open system I think that should explain it by the way I have been frantically been informed that t is ready and we have to break.