 This session I would like to be devoted essentially to exercises on pages 5, 6 and 7, properties of fluids and the first law 2. And if time permits in the later half, I will take some questions on second law. But before tea break, I am not going to discuss anything second law. All I want you to do is attempt exercises from the set F 2.0 to F 2.18. I will give you some comments on these. F 2.0 is just plotting just to get comfortable with the state space of steam. 2.1, 2.2 and 2.3 are exercises which force the students to get used to steam tables, just getting used to steam tables. And I would like the students to use as much of the steam tables as possible and not the HS diagram. The HS diagram is a qualitative representation and is unless absolutely necessary, we should not use it. Then exercises 2.4 to 2.9 pertain to a model of a gas called the Van der Waals equation of state. And the equation of state is given and you are asked to determine the critical parameters in terms of A and B. Now in F 2.4, the idea is this. On a PV diagram, if this is the critical point, what happens is the critical isotherm. The isotherm pertaining to the critical temperature goes like this. At the critical point, this is Tc, this is Pc and this is Vc. So at this point, if you take the slope of the isotherm, so this is the way pressure varies with volume at constant temperature. So the slope of this isotherm is 0 at critical point. Not only that, the tangent of this isotherm is horizontal and the tangent crosses the tangent is horizontal and the tangent crosses that line itself. So it is a point of inflection. The second derivative is also 0. So second derivative of pressure with respect to volume at constant temperature is also 0 at the critical point. Use these two equations and you will be able to solve F 2.4. And after solving F 2.4, 2.5 and 2.8 etcetera, they are 2.5, 2.6 and 2.8 are straight forward, they are algebraic. 2.7 is more of an approximation, not really a problem in thermodynamics. So if you are not interested in that type of a situation, you might as well keep away from it. Now 2.10, actually 2.9 and 2.10 are 8, 9 and 1, 0 pertain to the question as to the extent to which our steam can be considered to be modellable by the Van der Waals equation of state. And it turns out that Van der Waals equation of state is not really a good model for steam. As you start solving 2.8, 2.9 and 2.10, you will notice that particular fact. But 2.10 is interesting in another aspect. You are asked to determine the work done per kg. That means we have a system containing 1 kg of ordinary water substance. Of course it is in the vapour form from 1 bar to 24 bar, 25 bar isothermally at 400 degree C and 5 bar to 25 bar isothermally at 600 degree C. Use steam table and the trapezoidal rule for integration. Remember that our steam tables provide us tabulated values at some points, but not continuously. For example, to determine the work done in compressing steam, we will have to integrate PDV, assuming it to be a quasi-static process. If you take the B part, 600 degree C isothermal compression from 5 bar to 25 bar. If you come to table 3, you will find the 600 degree C isotherm is tabulated. You will see the required range from 5 bar to 25 bar on pages 12 and 13 of the steam tables. But you will notice that the data of regarding specific volume is available only at points 5 bar, 6 bar, 7 bar, 8 bar and so on. I think it is available for a 1 bar spacing between 5 and 25. 5, 6, 7, 8, yes I think it goes uniform spacing of 1 bar. So when it comes to PDV, although the isotherm may look like this, actually what you have is the data at 5 bar, data at 6 bar, data at 7 bar, data at 8 bar. So you actually do not know anything in between. The correct thing would be, the correct visual would be something like this. I know the information at 5 bar, I know the information at 6 bar, 7 bar, 8 bar, 9 bar and so on. Say this is at 5 bar, this is at 6 bar, this is at 7 bar, this is at 8 bar and so on. Everything is at 600 degree C and we can determine these values volumes. So here we are asked to use trapezoidal rule that means the area under this curve is approximated by joining these two points by a straight line. So what is going to happen is, if you consider say the compression from 5 bar to 6 bar, just that particular part. So at 5 bar and at 6 bar, we will have these two pieces of information. At 5 bar the specific volume is, this is pressure in bar, volume here and the volume here, let me write down. At 5 bar it is 0.8041 and at 6 bar it is 0.6697 meter cube per kg. I think we are given 1 kg per kg. So we will work in the per kg mode. What we are going to do is, we are now going to assume that the variation between these two points is a straight line. This is assumed variation which leads to integration by trapezoidal rule and the area is numerically this area. If I say this is point i, state i and this is i plus 1, then this small interaction delta w from i to i plus 1, this will be given by delta w i to i plus 1 is integral p dv. So it will be considered as the average pressure into the difference in volume. So this will be average pressure will be p i plus p i plus 1 by 2. In this particular case, 5 bar plus 6 bar by 2 that is 5.5 bar multiplied by final volume v i plus 1 minus initial volume v i, the values we know. We can create a spreadsheet or write a small computer program and crank this information out. And you will also notice that v i plus 1 will be less than v i for each and every step. So individually all these numbers will be negative numbers and you will get a final value which is negative value and that is expected because we have the processes of compression. So work done by the system will be negative, work done on the system will be positive. Exercises from 2.11 onwards up to 2.18 are exercises which require the first law of thermodynamics and of course other things which we have studied so far. This does not require really none of them requires the second law of thermodynamics. And again let me recommend as earlier that start with the system diagram, start with the process diagram, go on completing them as the thing proceeds. Read the problem completely, write down all your thoughts. When it comes to first law, use first law only as q equals data e plus w. Use the information provided and appropriate assumptions as needed to simplify it. And because we are using steam, steam is the working substance, steam and ordinary water substance is the working substance. Relations like pv equals rt, delta u is mcv dt, delta h is mcp dt and such relations which pertain only to an ideal gas are absolutely out of question. My students know that if I see anybody writing this, I immediately stop evaluating the answer book at that stage. It is a big red line across the page and say bye bye. You do your homework completely and then let us meet again. If we enforce such discipline on, if we want to enforce such discipline on our students, then we should first enforce that discipline on us. Now, let me give you a few minutes, may be about 10 minutes and then I will be back online to take any interactions with you. So, welcome back. I am now ready to take questions. 1, 2, 6, 1, MET is Institute of Engineering, Nashik. Over to you. Good afternoon sir. I am Suyukaram. I have one question that we have solved examples of steam and some gases. So, if we consider some contaminations or impurities, so what will be the change in formally? Is there any such change? Over to you sir. Well, impurity, we use steam because it has certain very desirable properties from the thermodynamic heat transfer, economic and environmental points of view. Although we do not discuss it in thermodynamics when it comes to power plant engineering, about major part of a lecture is devoted to which is a good working fluid and it terms out that for vapour cycle steam is perhaps the best working fluid depending on the parameters of course. So, for typically coal, oil or gas fired plants, fossil fuel plants, even nuclear power plants, steam is the working fluid and hence we take utmost care to see to it that steam is not contaminated. We have demineralization plant, we have water purification plant, we have blow down systems, we have dosing systems to see to it that the purity of steam or the working fluid is maintained. However, there are processes where impure water or impure steam is not used, but may be in some water treatment plants, sewage treatment plant, such contaminated water will be the normal fluid to be handled. But there will be situations where for example water and ethylene glycol will be used, but water and ammonia mixture will be used, water and hydrochloric acid mixture will be used. This is not contamination, this is a mixture of two substances and the physical chemists have provided a series of methods by which the properties of these mixtures are determined. So, instead of a steam table and a water table, you will have a table or a set of formulae which give you the properties of those mixtures at the requisite pressure, temperature etc. There are excellent books belonging definitely to the realm of physical chemistry which talk about properties of mixtures of various kinds. So, we will use those properties. See thermodynamics ends by providing the basic relationships. Properties come from the actual behavior of the material. So, somebody has to experiment with, tabulate all those properties and we will use those properties over. Sir, one more question. Yes. Is there any minimum range of temperatures for exchange of heat? This is a question actually in heat transfer. There is no such things that if T1 and T2 differ by less than something, no heat transfer will take place. The science of heat transfer will tell you that in any mode, the larger the temperature difference, larger is the rate at which heat is transferred. So, if the temperature difference is pretty small, the rate at which heat is transferred also is pretty small. That is all we can say. Hello. Yes, go ahead. One more question, sir. Yeah, yeah, yeah. There is one more question, sir. Yeah, we have discussed about the adiabatic process, sir. Yes. And the adiabatic process may be at constant pressure or at constant volume. Yes, possible. So, yeah. So, what is exactly the significance of that process, sir? I mean adiabatic process means heat transfer does not take place or is not taking place. That is all is meant by adiabatic. So, for example, you have a constant volume system. You have a constant volume system. Its work is being done on it by stirring it. It could be an adiabatic system. Similarly, a constant pressure system. I think we have illustrations here where we have adiabatic processes at constant volume and constant pressure. So, nothing special about that. Yesterday I spent some time as the process specification can be interaction specification or it could be a path specification. Interaction specification is typically adiabatic. There is no Q type of interaction. Whereas, a path specification is one in which you say pressure is constant or volume is constant or temperature is constant. Over. 1, 2, 1, 3, Saint Peter's Hyderabad, over to you. Sir, sir, you explained regarding incompressible liquid, sir. In one of your slides, you wrote some warning. That is, that is property, only one property decides the temperature. And it says warning. That is only good for volume, internal energy, entropy up to pressure of 50 bar. Sir, can you explain why that warning is given that is only good up to 50 bar? The simple, if you see the detailed steam tables, the approximation that water behaves like an incompressible fluid is true when the range of pressure is low. Since the ambient pressure is 1 bar, look at the steam tables and you will find that up to about 50 bar, the volume is essentially independent of pressure, insignificant variation with pressure. As you go beyond 50 bar, the volume does not suddenly start increasing fast. But beyond 50 bar, the volume at a fixed temperature starts changing reasonably significantly. You can use the approximation even beyond 50 bar, but it will become poorer and poorer. It is an excellent approximation up to 20, 30, 40, perhaps 50 bar. That is all I am saying. 50 bar is not some, you know, defined or derived this thing that comes out of experience and behavior. For other liquids like engine oils, this may be true to a much higher pressure, over. Sir, why the adiabatic work is independent of path? Why is adiabatic work independent of path? That is the way we have understood nature to behave and that is why it is a statement of the first law. It is not a derivation, it is an observation. The observation by experiments by Joule and others led us to believe that adiabatic work is independent of path. We proposed that to be a reasonably universal truth, put it on a pedestal, called it the first law of thermodynamics and after that we found that all the derivations which we have made based on that assumption that adiabatic work is independent of path have turned out to be true. Hence, we believe that adiabatic work is independent of path. That is all, over. 1 2 1 5 Bajaj group, Akbarpur Uttar Pradesh, over to you. First of all I would like to thank to you and to IIT Bombay for conducting this wonderful workshop. Now my query is that I would like to know something about the black hole and the laws of thermodynamics. As we have studied from the various books, these laws are universal ones. So, are they applicable for the black holes also? And if they are applicable, so precisely speaking about the second law of thermodynamics, are these black holes are considered as reservoirs over to user? When we say that the laws of thermodynamics are universal laws, again we are getting into that zone. What do you mean by universal? The thermodynamics laws particularly the laws of thermodynamics as we have defined them are for continuous systems, systems which are properly defined with definable boundaries and systems in which the quantum effects or relativistic effects are absent. We are in the so called classical Newtonian domain of physics, not the Einsteinian or the Planck type that is quantum effect or this. When you come to black hole, black hole is a relativistic phenomena. General relativity are equations of general relativity govern the behavior of black holes, not Newtonian mechanics. And hence I am not sure whether our laws of thermodynamics are directly applicable to it, but yes, there is I am told I have read some general articles. I am not an expert in cosmology, forget about black holes, but I have read some books and articles by Narlikar, Hawkins and others. It turns out that thermodynamics in some way is applicable to black holes. When can define the temperature of black holes, because they have a finite temperature they even radiate energy that is known as Hawking radiation. It is a very clear type of radiation and because it radiates energy its mass goes down and in principle it has been shown that very small size back holes because of the Hawking radiation may even vanish to 0 mass that means they may even cease to exist. But I am just making these comments based on what I have read. These are not experts comments at all and I think there was a question yesterday that on a very similar kind, so you should talk to your friends in physics particularly those who take interest and work with relativity and cosmology and they will be able to explain more things about this to you, over to you. 1118 Sri Jayachandra College Mysore, over to you. I hope you have a question on water steam and exercises based on them. Sir I am having a question like I am having a doubt in question number F 1.4 that is like yesterday we did the question. Sir the question says the problem says it is perfectly insulated system. So can I consider the situation as adiabatic also? Like if I consider it as like PV to the power of gamma equals like P1 V1 to the power of gamma equals P2 V2 to the power of gamma then I do not get any change in the volume. So I will have no work. I think I have said excuse me I have said this earlier so many times that perfectly insulated means adiabatic but adiabatic does not automatically mean PV raise to gamma is constant and I would bring again to you your attention to 1.6. If you start deriving PV raise to gamma is constant from the condition that it is adiabatic you will have to make about half a dozen assumptions of the cuff I can list the assumptions are number 1 the working fluid has to be an ideal gas with constant specific heats number 2 it has to be a quasi static process number 3 it has to be a process in which only PV work is done and nothing else although this is a perfectly insulated system one condition is satisfied it works with hydrogen we are told treat hydrogen as an ideal gas with constant gamma. So other two conditions are also satisfied but two more conditions only PV work and quasi static process it is stirred at constant pressure that is given so we may take that as a hint and assume that it is a quasi static process but the final condition is that it should only be a PV type of work and here although there is PV type of work involved because the volume changes constant pressure process but we are also told that it is stirred the moment it is stirred we have a component of work other than PV in the moment we have component of work other than PV work adiabatic does not mean PV raise to gamma PV raise to gamma is constant. So that is what I want you and I have been warning again and again that do not jump to conclusions adiabatic means Q equals 0 that is it anything else from Q equal to 0 if it is to be derived derive it make appropriate assumption over. One more question sir yes go for any gases why C p is greater than C v sir you take any gases the C p value is more than C v value what is the reason behind that sir. Okay I think you will need to study kinetic theory for this but the simple reason is the following one qualitative reason not a physical reason see if you heat a gas or put energy to the gas to raise its temperature suppose you do it at constant volume if you do it at constant volume and supply it energy to raise the temperature you only have to increase the internal energy okay you take the same gas and start at constant pressure what happens as you try to energize it to raise its temperature maintain the pressure at the same value the volume has to expand as the volume increases work is done work is done means energy is required or energy is expended okay. So the energy which is provided say in the form of Q will have to be used not only to increase U delta U plus also to do the PDV type of work and hence between two temperatures if you have a constant volume process and constant pressure process although delta U is the same in the second case there will be PDV work done hence additional energy in the form of it needs to be provided and hence you will find Cp is higher than Cv for materials which hardly do any expansion work for example incompressible fluid the Cp and Cv are almost the same we will be showing that when we do property relation okay sir one more question sir sir according to the first law of thermodynamics for a closed system for a cyclic process Q is equal to W or W is equal to Q. It means that 100% of work can be converted into 100% of it for example 2 unit of work is equal to 2 unit of Q if you take an example is it possible to convert 2 units of work to 4 units of it 2 units of work to 4 units of it no that will violate first law in a cyclic process you mean no. Yes sir for a cyclic process in a closed system is it possible because Q is W is W is 2 units Q has to be 2 units sir then I will continue the question sir regarding this only okay but according to the COP of a refrigeration system it is the ratio of refrigeration effect divided by work input means Q by W there you will get the more than 1 how it is this question is okay when for a cyclic process you are making a simple algebraic mistake for a cyclic process when you say Q equals W it is the net work and the net heat transfer when you are considering refrigeration for the refrigeration effect and for calculating COP you only consider the heat extracted from the cold system or if it is a heat pump if you consider only the heat provided to the system you take any cycle refrigeration power W net equals Q net there is no option to that that is first law your refrigeration cycle calculate W net calculate Q net and the 2 have to be equal 1 0 0 2 Amal Jyothi college Kottayam over to you. Sir I have a doubt regarding evaporation and boiling what is the difference basically before you have mentioned evaporation the only difference being evaporation is a little slower. Yeah see evaporation boiling steam generation these are different words used for the same thermodynamic process in real life the same process has different connotations and different wordings associated with it even in real life for example the our movement can be a stroll it can be a walk it can be a brisk walk it can be a dash it can be a run in a similar way evaporation boiling are all a process in which a liquid gets converted into its vapor but if you really want to see the difference suppose I take a drop of water on my hand and keep it open it does not boil I will say that it evaporates because it converts itself into vapor in a very slow fashion but if the same thing I put on a same drop of water I put on a tawa or a hot plate it will flash into steam it will say it is simply boiled off in fact technically we will say it will simply flash into steam in no time you put a water on the boil vessel on the stove initially you will find some small vapor rising from the surface because the water has not become hot we call that evaporation just colloquial words. When the whole thing becomes hot you will find the hot surface the surface also above the bottom surface above 100 degrees C bubbles will start rising from that there will be vigorous mixing we call it boiling thermodynamically both are the same processes liquid getting converted to vapor in the detail there is some difference but thermodynamically there is hardly any difference. Amal Jyothi I am back on stream go ahead. Thank you sir my doubt is sir any regarding temperature is there any difference between boiling and evaporation because evaporation can happen even at room temperature. Yeah see we see evaporation happening even at room temperature because the air surrounding the water contains water vapor at a much lower partial pressure. So, for example, water is at 30 degrees C and if you look up the saturation pressure of water at 30 degrees C which is 0.04246 bar if the water vapor in the air is less than of pressure partial pressure of water vapor in the air is less than 0.04246 bar then water will definitely evaporate because the air has the capacity to absorb water. But if the water vapor in the air is so high that its partial pressure is 0.04246 or is higher then it will not evaporate that is why in coastal towns like Mumbai for example our sweat does not evaporate so fast because the humidity in the atmosphere is large but in dry places like Pune or Nagpur or Delhi we do not seem to be sweating but that is not true whatever sweat comes out of our pores almost immediately evaporates because the air is dry it has a large capacity for absorbing water vapor. You will study more of this next week when you study psychrometric processes over. Sir when I when we say boiling point is 100 degree Celsius at one atmosphere does it imply implicate that the relative humidity is 100 percent. No when we talk of boiling point we are not considering any relative humidity boiling point means we are considering an equilibrium in a closed system containing just liquid and its vapor no air is present. So boiling point of water at one atmosphere is 100 degree Celsius means if you have a closed system containing water and its vapor in equilibrium with each other and if you bring the pressure to one atmosphere for the two phase equilibrium to remain the temperature will be 100 degree Celsius. This is long explanation of what we mean by boiling point of water at one atmosphere is 100 degree Celsius. Thank you sir we have one more question. Yes sir how can we convert water at room temperature and atmosphere pressure into water at saturated compressed liquid. You have water at higher pressure you have water at one atmosphere pressure and room temperature and room temperature say 30 degree Celsius and 30 degree centigrade and what you want to do. I want to convert into 5 bar saturated liquid there. 5 bar saturated liquid first use a pump to pump it pump the pressure to 5 bar and then heat it up using a flame or petrol or diesel or an electric heater till it comes to saturation temperature. I want to do it in a boiler how can I take the water to 5 bar saturated liquid. No a boiler a boiler a boiler if it is a flow system a boiler by itself will not take it from 1 bar to 5 bar. If you want to do it in a boiler then it has to be a closed boiler like a pressure cooker. So for example in a pressure cooker that is what you do suppose in a pressure cooker you put water you are putting it at 1 bar and 30 degree Celsius but with that weight on it it pressurizes as it gets heated up and it comes to 2 bar and saturation temperature at 2 bar. It does not go beyond 2 bar because of that valve moment it tries to go beyond 2 bar the pressure is released by allowing some material to go out. What is the real physical process taking in a pressure cooker suppose I have got a 1 bar atmospheric pressure and 30 degree material inside the ok it is the physical process the physical process is constant volume heating till the weight lifts itself and allows some seems to go out till that point it is a constant volume heating process. So that is I have first I will be taking this to 1 bar saturated steam I will be taking it. No, no, no you are not taking in between you are not going along the saturation line you are going along a constant volume line. If you want 1 bar saturated liquid then you should not leave the weight on. If you want 1 bar saturated liquid remove the weight so that the pressure in the pressure cooker remains always 1 bar or 1 atmosphere. Then when it comes to the boil it will then what you are doing is you are doing constant pressure heating because initially the pressure is 1 bar initial temperature is 30 degree C. So it will come at constant pressure from 1 bar 30 degree C to 1 bar nearly 100 degree C. Sir I have got one more question what is the how can I make a sub cooled liquid in industrial situations? Sub cooled liquid just cool it to the required temperature. Keeping the pressure constant. Yes keeping the pressure constant or you have saturated liquid increase the pressure by putting it in a pump or something either way you will reach sub cooled liquid. In a pressure cooker I am selecting the weight such that the pressure remains at 5 bar. At the instant of reaching that pressure will I have some quantity of saturated liquid at 5 bar inside the pressure cooker. Yes if I should not call it a pressure cooker I should call it a pressure vessel because pressure cooker by itself will not be safe enough to go to 5 bar. But suppose you have a pressure vessel the weight or the safety valve with set at 5 bar and initially say you have at 1 bar 30 degree C. As you go on heating it the pressure will rise the temperature will rise and it will eventually reach 5 bar. It will as it gets heated up it will become 5 bar saturation and the moment pressure tries to go beyond 5 bar by any means whether it has reached saturation temperature or not the safety valve will open the pressure will get released. So the purpose of the pressure release valve is to see to it that the pressure does not exceed beyond the preset value. But if you heat it long enough it will reach 5 bar and in fact in hospitals and many other labs you have what is known as an autoclave used for sterilization it goes to typically sometimes they use 5 bar but rarely it is a different kind of pressure vessel. 1073 Bannary Amman Institute of Technology erode over to you. Sir in problem number F point 2.11 F point F 211 go ahead. Yes sir the critical condition can we take the critical condition as a 374.15 degree Celsius and the pressure as a 220.1. Yes critical condition is at the critical point. Is it correct means sir the oven temperature is 349.9 given sir but the steam condition is 374.715. Yes the steam the initial conditions is the critical condition and steam is inside a rigid container. The rigid container is placed inside an oven which is at 349.9 degree C and allowed to reach thermal equilibrium that means the final temperature of our system will be 349.9 degree C. Then how can we find the x value in the dryness fraction? See you have the initial if you look up the white board I think they will be projecting the white board now. F 2.11 we are let us take now instead of a PV or even PV diagram will do PV diagram. This is the critical condition the temperature at the critical conditions is 373. whatever is the value 374.15 and critical point this is the initial condition state 1. Let this be the 350 degree C or 349.9 degree C isotherm. It is in a sealed container rigid sealed container and that means during the process the volume does not change. So this is the initial state the final state will be at the same volume but at 349.9 degree C. So this is state 2 and we have now the condition v 2 equals v 1. Since v 1 is known we know v 2 and v 2 is known t 2 is known. So we know the final state from which we can determine x 2. Over to you. Actually in Van der Waals equation the real gases the A and B constants is there constant value or is it depends on some of the factors? Just the way r varies r depends on what the gas is. A and B also depend on what the gas is. A and B tend to be fitted for a particular gas in a particular zone of pressure and temperature. Sometimes they are fitted to match the critical point. Sometimes they are fitted to match a particular isotherm but they are not universal constants. They are not universal A and B. They are gas specific or fluid specific values of A and B. Over to you. I want to know we are studying thermodynamics always under the equilibrium conditions. Is there any textbook referring that non-equilibrium condition thermodynamics sir? See we are working on thermodynamics but we are not considering equilibrium conditions. We are not considering equilibrium conditions. We are considering processes. So some non-equilibrium effects are included particularly in non-quasi-static processes. However the states which we have initial final and all the states with which we work are states of equilibrium. So you are right in a way we are essentially working with states of systems which are equilibrium states. And hence there are some people who say that what we study as thermodynamics is not really dynamics it is thermostatics and there is some sense in that. If you really want to study non-equilibrium thermodynamics there are books on that and there is a there was he expired a few years ago. Illya Prijo Jean a physical chemist and whole life significant contributions to non-equilibrium thermodynamics. P R I G O G I N E look up his work and there are many other people for example Ansagar has done some classical work in non-equilibrium thermodynamics. But most of the work on non-equilibrium thermodynamics has remained in research papers there are very few standard text books of non-equilibrium thermodynamics. Over to you. One more question sir. Sir I want to know which text book is preferable for B students in a self-financing engineering colleges. Text book for engineering thermodynamics you have mentioned six books see here. So I prefer Van Weiland sir and I do not know about this second text book Shapiro fundamentals of engineering thermodynamics comparing with this Van Weiland what I want to know this second book also sir. Okay Van Weiland is a reasonably good book but I would prefer Moran and Shapiro because the treatment out there is more rigorous and more or less follows our scheme of things. Moran and Shapiro edition is also available in Indian print or an Asian print. I recommend that before recommending it to students you pick up a copy or get your library to buy a copy compare Van Weiland and Moran and Shapiro and then you can take your own decision. But I if you just look at it from the student friendliness point of view I think Van Weiland may be slightly better only slightly better. But finally you know the type of students you have their abilities and their perseverance. So you take a decision by comparing the two books yourself over to you. 1099 Mehdi Capps Institute pick number Indore over to you. Yeah I have a very general question sir in most of the books time is not considered as a thermodynamic variable but we have used in process diagram time as a thermodynamic variable any comments. No in the process diagram particularly when you have voltage against time torque against time. Time is not used as a thermodynamic variable. It is because things are specified as a function of time over a particular period of time. So we plot it. There is no harm in plotting voltage against current but then you will only get a point that this is voltage this is current. There will be no indication of how long that current was allowed to pass and that is why we use the time axis. Time that way is not a thermodynamic variable. In tomorrow's further discussion on the second law of thermodynamics we will just have before and after that is the only part of time that we will have in we will involve ourselves with it. So there is no way by which a thermodynamics discusses what is meant by one second or what happens in one second. Thermodynamics does not look at rate processes. If at all it looks at rate processes that is simply by differentiating our equation with respect to time. For example, we have q equals delta e plus w as our statement of first law. All that we will do is differentiate it with respect to time and we will say q dot equals d e by d t plus w dot that is it. That is the way we will include time but there is no thermodynamic process by which time will be included in our scheme of things. Except second law will say what is possible before and after that is it. One more question sir. In enthalpy's expression we are writing h is equal to u plus p v. In most of the time p v has been expressed as the flow work. Is this the correct way to express p v the flow work? Look p v by itself does not represent flow work. Day after tomorrow when we derive the expression for the first law in open systems we will see how the flow work terms flow work how the flow work terms there are two of them typically but could even be every stream was one flow work associated with it. That turns out to be of the form p v but that does not mean p v is flow work. The flow work can be shown equal to p v but if you just take p v product it is just p v product unless it is in the context of a flow into or out of an open system you cannot say that it is flow work and h is shown to be equal to that is not a proper statement to make h is defined to be equal to u plus p v. We have called the one process that is a sublimation. Do you have any specific name for a process which is opposite to the sublimation? Any specific name sir? No, it is there is no specific name it is simply sometimes it is called freezing sometimes it is called solidification. So there is no you know pukka opposite of sublimation available anywhere. There are unfortunately there are situations where the same process has more than one name and there are also situations where more than one process share the same name. So depending on the context we have to decide what is what over and over and out. So you have one more question. Go ahead last question from you. Yes? In exercise number 2.11 we have used the word o 1 whether it is correct? Yes, o 1 is a chamber which is maintained at 349.9 degrees C. It will maintain the temperature of 349.9 degrees centigrade always. Yes. Then there will not be any change in temperature of the mixture at the equilibrium. See initially the our system is 2 kg of steam in a rigid container initially at critical conditions and critical conditions the temperature will be something like 373.374.15 degree C. I have just now explained the whole thing to another centre. I do not know whether your centre received it or not at that time. This is the process which was shown. Initially the system is at the critical point 1 and 349.9 isotherm is this and the process starts from the critical point and comes to the 349.9 degrees C. Because it is a rigid container the initial volume and final volume are the same. All these volumes are V C and this pressure initial pressure is P C. Final pressure will be saturation pressure at 349.9 degrees C. But then where the heat is rejected? The heat is rejected from the system to whatever is the surrounding in the oven over and out before tea time I will take one more centre 1173 Tyagarajar College of Engineering Madurai over to you. Sir we have learnt the properties of steam and water. Is there any difference in the properties of deuterium DTO the heavy water we are using in the power plants? Yeah. Can we use the same property relations? No. See heavy water DTO is an excellent nuclear fluid used in nuclear power plants and perhaps at a few other places some reactors some nuclear power plants and some other reactors. Properties of heavy water are slightly different. For example the boiling point is I think differs by a few degrees C so does the critical point. The properties are not drastically different but they are slightly different. Now and those properties including its equivalent of steam tables and Mollier diagram they are all available. But only thing is since only a few people use it they are not published in a 24 rupee steam table. Properties their formulation they are all available in the literature you can make a search and find out. If you ask me a question what are the properties of absolutely pure H2O I would say that look they are unlikely to be any significantly different from this because the fraction of DTO in this is only 7 parts in 10 raise to 5. So that is a very small fraction of 1 percent so it is unlikely that you will see any difference here. And industrially or chemically or for any reason it is just not needed for us to have H2O purer than the naturally available H2O DTO combination. There is absolutely nothing to gain by reducing the DTO concentration in ordinary water to below 70 ppm. So maybe it will be easier for us to find all these properties. Almost pure DTO but it will be more difficult for us to find almost pure H2O. Thank you over and out and I think it is time for tea.