 Welcome back to the final session today. So, we are into the discussion mode. First let me go to 1193. Ram Mege Institute, Badnira, over to you. Yes. At triple point 3 phases coexist in equilibrium. So, my question is suppose we have 10 kg of substance in a adiabatic container, there is 5 kg of ice, 3 kg of water and 2 kg of steam. After some time, will there be any changes in masses of all the 3 phases? Yes. Consider a simpler version of your question. Suppose you have an adiabatic container, now remember adiabatic means no heat transfer, nothing more, nothing less than that, okay. And suppose you have just steam and water in it or even ice and water in it, say ice and water and they are at some temperature, say 0 degrees C appropriate, 0 degrees C you would not have, say 0.01 degrees C, just ice and water, no steam. Depending on the work being done or energy interaction of some kind, it is possible that the pressure remains the same, temperature remains the same but the state changes because some water changes to ice and ice changes to water. Your query regarding triple point is more interesting because at the same pressure and temperature, you will have all the 3 phases together. So that means without changing pressure, without changing pressure, you can have interactions. What will happen is the relative masses of solid, liquid and vapor will change. So the volume will change, energy will change, entropy will change, enthalpy will change, okay. In case of water steam mixture, we define the dryness fraction. There is only one fraction we need to define because the other fraction turns out to be 1 minus dryness fraction. So the liquid fraction is 1 minus x, vapor fraction or dryness fraction is x. When you have a triple point situation, you will have to have 3 fractions, the solid fraction xs, a liquid fraction xl and a vapor fraction xv. The sum of these 3 will be 1. So 2 of them can vary independently. Over to you. Thank you sir. My next question is, in example 2.18, the process is modeled as PV is equal to constant. The process may be modeled that means the process is executed along a line on the PV diagram which says PV equals constant, yes. Is it isothermal process? Look, read the problem completely. If it is 2.18, we have some amount of steam. Now steam is not an ideal gas, okay. So a steam may execute a process or can be made to execute a process which is PV equals constant. But it does not mean it is an isothermal process. It only means that it is a quasi-static process such that at any stage if you measure pressure, measure volume, multiply them together. For all intermediate states, you will get a constant volume, constant value. That is it. And that is why you are given that the initial state is 2 bar dry saturated. So the temperature will be something like 120.2 or whatever is dictated by the steam tapers. 2 bar dry saturated, let me see, 120.2 degree C. Then it is taken to 8 bar, final state 2. So the pressure goes up by a factor of 4. So the volume would have decreased by a factor of 4. So the initial volume will be the specific volume at 2 bar liquid, sorry, 2 bar dry saturated, right. That is 0.886 meter cube per kilogram multiplied by 1.5 kg. That will be the initial volume. The final volume will turn out to be 1 fourth of that because the pressure has gone up by 4 times. Volume will reduce by 4 times. So divide that by 4 and you will get the final volume. Since you know the final pressure and final volume, you know the final state. And depending on the final state, you can determine the temperature. You can determine the energy, initial, final. The PV raise to constant will be useful when you consider work done by steam because work done by steam will be the area under the curve which can be determined using that PV equal to constant line. And then you will have to use first law to determine the heat absorbed by steam. Maybe I should explain it on this. So can you show the whiteboard please. So this is the query regarding F 2.16, sorry, 2.18. You have compression. So a cylinder piston arrangement is in order. Some pressure will be 1.5 kg steam. Initial state, 2 bar dry saturated. Final state, 8 bar. It is compressed. Anything else given? Process may be modelled as PV equals constant. So naturally this implies that it is a quasi static process. Now, process diagram. Since we are going only between 2 and 8 bar, it is not necessary for me to show the whole 2 phase dome. You can, this is pressure in bar. And let us have this as, let us work with specific volume. Everything will get scaled up by 1.5 because the mass is 1.5. 2 bar to 8 bar. 2 bar dry saturated. So this is our state 1. The process executed is PV raise to, sorry PV raise to 1 or P into V is constant. So let us assume that the PV raise to, this is P into V is constant. So this is the process. Naturally state 2 will be at 8 bar and the line dictated by PV equals constant. So this will be 2. And area under this curve when multiplied by the mass of the system assuming the scale goes down up to 0. This is the expansion curve. So now, final temperature and final state. The final state will be given by, it is given by 2 conditions. First P2 is 8 bar given. And it is also given that P2 V2 is P1 V1. Hence we can calculate or dividing throughout by mass. We get P2 V2 is P1 V1. So this gives us V2 equals V1 into P1 by P2. And if you put V1 which is dry saturated vapor at 2 bar, this is 0.886 meter cube per kg multiplied by 2 by 8 bar over bar. So this will turn out to be one-fourth of this that is 0.2215 meter cube per kg. So V2 this and P2 this is the final state. These two items together P and V. Remember P and V will always give us a final state because it is not a combination Pt. If it is Pt then we have to be careful. Now notice that at 8 bar look up the table page 8. In my case it is page 8. If you are using a different table it will be something different. Now this is at 8 bar we have Vf equals 0.001115 meter cube per kg. Vg is 0.240 meter cube per kg. Since Vf is less than V2 of course this is you should I would write Vf2 and Vg2. Since Vf2 is less than V2 less than Vg2 hence or I should say since therefore state 2 is wet steam. And since it is wet steam now you can calculate X2 is V2 minus Vg sorry Vf2 divided by Vg2 minus Vf2. This gives you the value of X2. And from now P2 V2 has given us X2 and now P2 X2 will allow us to calculate U2 T2. Since it is wet steam the final temperature is the saturation temperature 170.4. The initial temperature we did not calculate, but since it is P2 is 2 bar and it is saturated sorry P1 is 2 bar and since it is given X equal to 1 this implies that T1 is 120.2 degrees C. And it is obvious to you now that because it is steam PV equals constant does not represent an isothermal process. So the initial temperature is 120.2 degrees C, the final temperature is 170.4 degrees C. There is a very significant temperature difference. So now with this we are able to determine final temperature final state. Now change in energy delta E first you will have to assume to be equal to delta U assume because there is no other component of energy mentioned. So this now becomes M into U2 minus U1 and U1 can be read off U2 can be calculated from here. M is given U2 is calculated U1 can be read off from the initial state. And now we have to determine W work done by steam. First you have to write W is W expansion. This is assumed because there is no mention of a stirrer or any electrical work. Since it is a quasi-static process this will be integral 1 to 2 of PDV quasi-static. So we can evaluate it. This will be true anyway, but we can evaluate it because it is quasi-static. And because the process is P1 V1 is PV is constant. So P can be replaced by P1 V1 divided by V. P1 V1 can be taken out of the integral sign. So it will be logarithm of V2 by V1. We know P1, we know V1 that is M into the specific volume at 1. And V2 by V1 is known as P1 by P2 which is 4. So this can be calculated. And after this is calculated heat absorbed by steam. Mind you notice that this will be a negative number because V2 is smaller than V1. And then Q should always be calculated as delta E plus W. No other choice. We have determined delta E, we have determined W. So you can determine Q. This is the procedure. Over. Yes sir, what is supercritical fluid? Okay, supercritical fluid is the name given to the state of a fluid where the temperature is higher than the critical temperature. But generally we can, there is no proper definition. Generally we can say any state in which either the pressure or the temperature is higher than their corresponding critical values, we can say it is a supercritical fluid. 1056, Malla Reddy College of Engineering and Technology. Sikandarabhad. My question is triple point is assumption and postulate. So how do we plot in this PV diagram as in real platform. So what is your question regarding triple point? But I did not understand what you said after that. I will repeat sir, triple point is assumptions or postulate. No triple point, wait, wait, wait, wait, I will stop you there. Triple point is not assumed. The triple point is not postulated. Triple point exists and you can have a simple apparatus which you can create using a good glass blower, a vacuum pump and freezing unit and pure water, so called conductivity water or as pure water. And you can create the triple point apparatus in your lab and measure the triple point. If you go on Google or YouTube and say triple point of water, you will get enough information including excellent videos of how a triple point apparatus is created and triple point temperature is noted. I should not say measured because triple point temperature is only to be noted. It is defined over to 1, 0, 1, 2. My question is related to charging of cylinder while charging the evacuated cylinder by external gas line which is having sufficiently high temperature when the charging process completes. That means the temperature of the pipeline is equals to the temperature of the cylinder. The gas inside the cylinder is having a temperature significantly higher than that of the main gas pipeline. I want to know the reason for this. The reason, actually it is interesting you are all jumping the gun. This is a problem pertaining to open systems and we will tackle it at that time but I will bring your attention to exercise where I have that exercise somewhere here. In open thermodynamic system there is a exercise OS14 on page 15. This is very similar to what you are saying and it will be clear to you why the temperature rises. It is a very simplified version of what you say. An evacuated bottle is filled by allowing air to fill in. When we come to open system studies, we will come to that problem and then you can solve many other problems like this. The next visit is 1107. Yes sir, please explain the rudimentary system and one way work transfer. First thing we appreciate that depending on the type of system some work modes are one way some work modes are two way. Simple illustration is if you stir a liquid, if you take a fluid and stir it that is a one way work mode because by stirring you can do work on the fluid. If the fluid is initially in equilibrium and you put in a stirrer or a spoon, ask the liquid to stir it, the liquid will not do it. That is a one way mode of heat transfer. Whereas a fluid in a cylinder piston arrangement if you reduce the pressure on the piston, external pressure on the piston slightly that will expand. If you increase the external piston on the piston slightly it will compress. So a expansion compression process is a two way work mode. Now if you have appreciated this then you will also be able to appreciate that there are systems which have more than one two way work mode possible. An electrolyte for example, can expand and contract and can also be charged and discharged. So electrolytes, electromagnetic substances, the fluids used in the magneto-hydrodynamics these are all complex systems because they have more than one work mode. You have compression expansion work possible, you have electrical work possible, perhaps you have magnetic work possible. Now going to the other extreme it is possible to have system which are inherently incapable of doing any two way mode of work or we can create systems by constraining them in such a way that no two way work mode is possible. For example, we have seen that a gas in a cylinder piston arrangement is a simple compressible system and it has one two way work mode that of expansion and compression. But if I fix the piston at one place, seal it with araldite or MC then whatever be the change in external pressure the gas will not get compressed or gas will not try to expand. So this way that becomes a system in which there is no two way work mode possible and hence it becomes a rudimentary system. There are simpler systems which are basically rudimentary for example, your mercury in glass thermometer. You can say solid is to some extent compressible but in the range of temperature and pressures which we use the mercury in glass thermometer or the clinical thermometer is a rudimentary system because there is no two way work mode possible. You cannot bend or unbend it, you cannot twist or untwist it, you cannot charge or discharge it, you cannot expand or compress it and hence it is a rudimentary system and hence that single the advantage of a rudimentary system is the state is defined by a single property and we can directly map that property to define a scale of temperature or we can define a scale of temperature using that single property. We do not have to worry about what happens to other properties when it interacts thermally with other systems over to you. When you get out of this debo mode F 2.17, what is the problem? How to solve it? How we will move in this column? How we will move we have to find out the volume, total volume? Look, if you read it we have a rigid insulated vessel, rigid means no change in volume expansion work is 0. Insulated vessel means adiabatic situation any heat interaction will be 0. Then it is divided into two chambers by an adiabatic partition wall. One chamber contains 1 kg of wet steam at 4 bar and some dryness fraction. Other chamber contains 0.5 kg of dry saturated steam at 2 bar. If the partition between the chambers is removed and the fluids on both sides allowed to mix, determine the specific volume, specific internal energy, pressure and dryness fraction in the final state. The situation is this F 2.17. We have a rigid insulated vessel. I will just show it by dotted line a partition. Our system is the total internal of this rigid insulated vessel. So rigid means delta V is 0. This implies W expansion is 0. Insulated means Q is 0. Now let us say this is the A part, this is the B part. The initial state 1 consists of one part A is 1 kg 4 bar dryness fraction 0.7. So using this we can determine all properties. In the final state the partition is removed and everything is allowed to mix. After mixing you get a, of course, there is a B part. MB is 0.5 kg, PB is 2 bar, XB is 1. From here we can again determine all properties. Now final state 2, what do we know about the final state? First thing because it is rigid, we can say V2 equals V1. And since V1 is going to be made up of two parts, V1 is going to be VA plus VB. And this will be because MA and MA VA plus MB VB. VA and VB you would have calculated from the states. This is one property of the second property is mass of 2 is going to be mass of 1 which is MA plus MB. We know both of them, sorry I am using lower case M for mass, so let me continue using that. So the final M2 is going to be M1 because it is a rigid vessel, nothing is coming in, nothing is going out which will be MA plus MB which will be 1 plus 0.5 kg, 1.5 kg. So I know the final volume, I know the final mass, so I know the final specific volume. That is one aspect of state 2. The second aspect of state 2 is I do not know what the pressure is, I do not know what the temperature is. But the first law includes energy, so let us see whether I can use the first law to get a fix on the energy of the system. The first law says Q equals delta E plus W, note that you always start from this equation, Q is given to be 0. For delta E we will assume delta E is delta U because there is no mention of any movement for any other aspect. W we will assume that this is equal to W expansion because there is no indication of a stirrer or electrical work and since the volume is rigid this turns out to be 0 as we have already noted here. And that gives me delta U is 0 and that means M into U2 minus U1 is 0 and that means U2 equals U1 or even without going through this perhaps it is better to write this as U2 equals U1 which is MAU A plus MDU B and all information about initial states is known, so U1 is known, so U2 is known and then you get U2 is U2 by M2, so state 2 gets defined in terms of V2, U2 and of course M2. But it is V2 and U2 which will give you the final fix up to Vt, V2 and U2 you can come straight away volume specific internal energy. When it comes to pressure and drainage fraction this V2 and U2 is an odd thing to look at, if you look at their values you would perhaps get a hint from your steam tables that the pressure is going to be somewhere between 2 bar and 4 bar in that range and it is going to be wet steam. Then what you will have to do is by trial and error determine P2 and X2 by trial and error and you proceed like this assume some P2 it may not be right using P2 and V2 but not using U2 you calculate one drainage fraction let me call it X2 prime using the same value of P2 use P2 and U2 using these two you can calculate another value of X2 prime, X2 prime equals X2 double prime or is within a very small fraction like 3 0s 1 remember our typical Xs are calculated as 4 decimal places if it is the various difference is only in the 4th place it is okay. Then P2 is the final pressure otherwise try a different guess of P2 that is the process or what you can do is do the following guess various values of P and plot the drainage fraction calculated by the two methods may be if you select different values of P it is possible that X2 prime this is P2 X2 prime goes like this whereas X2 double prime goes like this at some place the two graphs will intersect then this is the value of final X2 and this is the value of final P you can determine for a few sets of points then draw smooth curves through them and wherever they intersect you have a solution I hope this satisfies you over to you I am now attempting 1 1 2 2 Sir as we can analyze the temperature pressure work in a practical manner Sir can we understand the term enthalpy in practical way Sir or enthalpy is just a mathematical form of internal energy and work done or is there any physical significance of this term enthalpy because in good old books on heat and power and engines including perhaps Da low and Rangham and such enthalpy is mentioned as the total forget that enthalpy has no significance except that it is a short form for U plus PV and do not even bring into picture the idea that PV represents work or flow work of any kind PV here represents only pressure volume product okay it may come out of our first law of thermodynamics for open systems that PV I know comes out of the flow work term but H when it is defined it is defined only as a short form of U plus PV there is no physical significance for H we use it because that form U plus PV comes up so often that we find it convenient to create a short form for it that is it over to you. Sir one more question Sir as we have already discussed that work is a path function so it follows the inexact differential Sir is the inexact differential has any significance or it is just a mathematical term that if it follows the equation mdx plus ndy equal to constant 0 or constant then yes you are right we will come to this exact and inexact differentials tomorrow when we do property relations but an exact differential is exactly what you said that exact differential is one which when integrated from 0.1 to 0.2 in a coordinate system the integration turns out to be independent of the path an exact differential can be shown as the differential of some complicated function of x, y, z whereas an inexact differential cannot be shown so of course there are theories of then integrating factors and all that but we do not have to bring them in just now the basic difference between exact and inexact differentials if we understand it that is good enough for us. Hello Sir one more question yes can all process in thermal energy reservoir can be considered as quasi-static or not okay see we are interested only in two aspects of thermal reservoir number one its temperature which is always maintained constant number two later on we may need to estimate or determine the entropy difference or entropy change of a reservoir okay that we will determine as the heat absorbed by the reservoir divided by the temperature divided by its temperature okay and because of that we do not really have to worry about whether all processes inside a thermal reservoir are quasi-static or not okay that that question just does not need to be answered over. Sir one more question can we use same plant as heat pump in winter and as a refrigerator in summer from a thermodynamic point of view yes but in an actual practice what happens is in the summer the plant will work between an outside temperature of say 45 degree C and inner temperature of say 25 degree C so the range of operation is between 25 and 45 in winter you may have the inner temperature or the higher temperature as 25 degree C outer temperature as say 0 degree C so the range of temperatures in which that heat pump operates is different so you will have to worry whether the oils will do their job properly whether the working fluid will have the appropriate characteristic remember that our air conditioner we have modern air conditioners but the old window air conditioners they used R 22 as the refrigerators why because they worked in outside temperature of about 40 45 inner temperature of about 20 25 right refrigerators the ones which we have in our house and which are used to make ice and keep cool things chilled they work with ambient temperatures of about say 35 40 may be 45 degree C but they work with inner temperatures of 0 minus 5 minus 10 degree C see the range is slightly wider but we suddenly replace R 22 by R 12 and now with R 134 A or something like that okay so the properties of the fluids required and hence the surrounding materials which are required they change sometimes significantly even with small changes in temperature so yes thermodynamically in principle you can use but the same thing whether for example you are in Udaipur may be the winters are cold summers are hot so may be your question is in summer I have a air conditioner can I turn it around and use it as a heater in winter my only question is try and see what happens it will work but whether it will be good for the health of that and whether it will be a very effective and efficient heat pump you have to turn it around over to you 1277 technology education and research integrated institutions Kurukshetra over to you as we take a boiler and can you tell me what type of interaction is there and what type of system if we take if we are only interesting in the heat transfer from the hot gases and in the water what is the question about I want to you want to consider what what type of system is that if we are only interested in the heat transfer from the hot gases and in the water what type of system is there your boiler the both sides are open thermodynamic systems hot gas on one side and the steam being generated on the other side it was rudimentary or we can say it was a simple or complex no which you are considering a boiler right what is the system that you are considering boiler is simple boiler simple boiler in which water flows and becomes steam and in through which you put fuel and air and hot gases come out of the other side that type of boiler yeah then it is an open system there is no question of it being rudimentary open system means you have perhaps expansion work we have flow work which is some sort of expansion work you have a pump which is absorbing some type of work there is no question rudimentary system etc are specific to close thermodynamic system sir there is the same question I have I have the doubt if it is a close boiler in which there the water is on the upper side and the fire we are firing the boiler from the bottom side and we are collecting the steam on the upper side of the water space that is the steam space then what type of work interaction is there in that there is only flow work as the steam goes out if you say that the vessel is rigid there will not be any expansion work if there is no stirrer or pump there will not be any electrical or stirrer work the only work interaction will be the expansion work and of course heat transfer but that is not a work interaction heat transfer from the hot gases to the boiler boiling water when we come to open systems we will have enough number of such problems let us just now discuss close systems. But in case of thermometrics also a close system in which we apply heat to the bulb thermometer bulb and the mercury inside the bulb it got expanded and we call it a rudimentary system see remember there is also one way of interaction no when I say a mercury in glass thermometer is a rudimentary system my system includes everything my system boundary is the outer boundary of the glass casing if you are looking at just the mercury which is expanding then that is not a thermometer that is a system you can say it is a cylinder piston arrangement with mercury in it which is expanding but if I create a capillary in which on one side you have mercury and above that you have vacuum and the capillary it is sealed in a glass envelope and my system is whatever is included in that glass outer surface of the glass then it is a rudimentary system the moment you change the boundaries of a system you are defining a new system and the new system may not have all the characteristics of the earlier system one has to be that is why I am emphasizing that location of the boundary is very important when a slight change in the boundary changes the situation sometimes very significant over to this. It is a closed boundary anti mercury is inside that yes but agree but your closed boundary is different from my closed boundary which is the outer surface of the glass envelope and when I define a thermometer and claim that it is a rudimentary system I have defined my system to be the one inside the outer surface of that glass envelope when you say mercury and it is in a closed boundary but that is a different boundary you are talking of a different system that system may not be a rudimentary system. Sir there is one more question also yes at low pressure and if we keep the temperature constant the water will start boils at low suppose the temperature is for 50 at low pressure sir generally we calculate that if the pressure is 1 atmospheric and the temperature is 100 degree centigrade the water start boiling but if we take the pressure low and keeping the temperature constant still then the amount of we can collect the amount of papers why do not we apply that formula in boiler. No you are looking at a different process now here you are looking at a process where you are keeping the temperature the same let us again go to the pressure temperature diagram and let us start from the triple point and go to the critical point. Let us say this is the pressure which is 1 bar this will be something like 99 point or this is let me say 1 atmosphere so this is say 100 degree C and if you have water at this temperature it will not boil because the pressure is above its saturation pressure but if you reduce the pressure then you are changing the state of the system like this are going along this line. So if you go on reducing the pressure you will first read reach a saturated liquid point at that temperature then it will start evaporating and then for all you know if you start continuing your process further it will even become superheated steam at low pressure this is possible to do over to you Ph.G college Coimture 1136 over to you. Sir what is the significance of the super critical fluid sir because we are now we are in the industries we are using super critical boilers in nuclear power plants. So what is the qualitative significance of that super critical fluid sir? First a correction I do not know of any nuclear power plant we use a super critical steam but fossil fuel fire that is coal fired or gas fired power plants do use super critical steam. When it comes to power plant the temperatures even in old power plants have been super critical that is above 373 degree C a steam power plant is claimed to be a super critical power plant only when the pressure is super critical not the temperature. Temperatures of 500, 550, nearly 600 degree C have been used for the last 20, 30 years but the pressures above 220 bar those are rare we are started using them only recently. And there is nothing special about super critical boilers tomorrow we will realize that if you have a power plant the efficiency of the power plant tends to be higher if the upper temperature at which it is absorbed increases and hence increasing the temperature with a fluid like water will naturally lead to higher temperatures and correspondingly higher pressures. So we end up being in the realm of super critical steam power plant otherwise there is nothing special about being super critical if we go to high enough pressures and high enough temperatures to gain efficiency we automatically end up in the super critical zone of water that is it over. Sir my next question sir in the Gibbs phase rule sir in the degrees of freedom so if it is coming as F is equal to 2 is there any thumb rule to select the properties either I can go for pressure temperature or pressure volume. So just it is coming as F is equal to 2 yes is there any thumb rule to select the properties. There is no thumb rule to select the properties when F equals to you are essentially free to select any two properties you feel like they could even be pressure and temperature. Well I am not an expert in phase rule but I recommend you a book by Findlay F I N D L A Y the title of the book is the phase rule you will find everything about the phase rule right from its basic derivation in that book you will find more details discussed in that book it is a classic you should find it in your library may be it is in the physical chemistry or chemical thermodynamic section over 1167 BVM F. Now I have I have one suggestion that at the end of each exercise you may give some answer so that whenever we solve it we can have we can have the confidence of the answer. Well your suggestion in welcome in fact till a few years ago I used to do that and if I take a print out of of course this is got modified but I used to have answers at the end of the exercises particularly those which require or which end up in a numerical answer but then I found out that the students are interested only in getting that answer and not following the procedure. So after that I decided that answers are meaningless in fact recently I have started setting up problems in which students have to explain in detail how that answer is obtained without obtaining the answer itself. I will may be there are a few problems here I will bring them to your attention later go ahead. Another question in in question F 2.12 there is a mixing of two stream and the how to find out the work done in this case? Ok 2 kg of saturated liquid water at 12 bar is mixed with 1 kg of superheated steam at 12 bar 300 C. Mixing is adiabatic and at constant pressure I hope you do not take any objection to that adiabatic and then constant pressure. Determine the changes in volume and internal energy the dryness fraction in the final state and work done here you proceed like this. This is this is a problem which can be solved either by modeling it as a closed system or by modeling it as an open system the final answer will turn out to be the same. So what we have is we have a chamber the constant pressure process. So we can have a cylinder piston arrangement and we can say that initially the system is partitioned into two parts part A and part B. So the initial state state 1 is made up of A plus B where A is 2 kg 12 bar pressure saturated liquid that is x is 0 and the B part is 1 kg again 12 bar 300 degree C. So properties of A and properties of B are all determinable. So all the properties of the initial state are known. Note in particular that the pressure of each part is 12 bar and hence the external pressure should also be 12 bar to keep it in equilibrium. We can now assume that the mixing takes place by say assuming that the partition is removed and let the final state of the complete system be 2. Now on the P V diagram you are going to have the whole process taking place at some constant pressure. Since the initial pressure is 12 bar we expect the final pressure to be also 12 bar and since it said that at constant pressure we expect the whole process to take place on this 12 bar line. Now what is going to happen is our initial state 1 now notice that this is the total volume in meter cube and not specific volume. If you write sketch in terms of specific volumes this is P and specific volume and this is the 12 bar line the A part is going to be saturated liquid at 12 bar the B part is going to be super heated steam at 12 bar 300 degree C. They are going to mix and maybe this part will move something like this this part will move something like this and we will end up with a final state somewhere on this line. But this line gives a not really a correct picture because A and B do not individually move they mix together and then if this is our state 1 the final state may be to the left of it to the right of it that will tell us but finally it goes to somewhere here or somewhere here the state 2 may be here or the state 2 may be here and wherever it goes this will be suppose it goes from 1 to 2 like this this will be the amount of expansion work done. If it goes like this this will be the amount of expansion work done. This is the visual depiction 2 will be somewhere here but you understand that A and B do not individually come to 2 they mix and the final mixture mixed system goes from 1 to 2. Of course here writing 1 like this also is not very correct because it is made up of 2 sub parts with same pressure but 2 distinct states otherwise. Now, let us analyze it and see what is the set of assumptions that we have to make it is a closed system. So, the first law will simply be Q equals delta E plus W Q is given to be 0 given adiabatic delta E let us assume it to be delta U then there is nothing else mentioned. There is no stirrer etcetera mentioned. So, let us assume this is W equals W expansion also assumed. So, that means the first law ends up being delta U plus W expansion is 0. Now, let us make an assumption that the process is quasi static because it is mentioned that it takes place at constant pressure and hence W expansion which is integral p dv will turn out to be from here to here we have the assumptions quasi static and p constant and again using p constant we write this as delta U plus delta pv is 0 and that gives us because enthalpy is defined delta H is 0. This means H 2 equals H 1 and H 2 equals H 2 will then be M 2 H 2 equal to H 1 is made up of two parts M a H a plus M b H b. We know everything on the right hand side on the left hand side M 2 by so called conservation of mass will turn out to be M a plus M b. So, in this equation we know everything except H 2 which can now be calculated. Since we know p 2 p 2 and H 2 give us the final state. Compare H 2 with the saturated liquid enthalpy and dry saturated vapor enthalpy if it lies in between it is wet steam if it lies higher than dry saturated vapor enthalpy it is superheated vapor. I think the answer turns out to be wet steam and once you get the final state W expansion is calculated as p delta V and all those details and then come on over to you. I have one question if there any apparatus practical apparatus to visualize the people point. Yes, I am surprised that a question which was asked less than 45 minutes ago is being asked again. I have informed someone that go to Google or YouTube put in triple point of water apparatus and you will get enough information and enough videos excellent video showing how a triple point of water apparatus is created. So, that you can sense the and you can calibrate your thermocouples or thermometers by that all that and it is not a very complicated apparatus all you need is a good vacuum machine vacuum pump a good glass blower excellent quality of glass and clear absolutely pure water and within a week or so with a few trial and few trials and errors you will be able to create an apparatus to demonstrate the triple point of water. In fact it is an excellent apparatus to have in a thermodynamics lab or in a physics lab or in a physical chemistry lab it will be an excellent calibration piece over to you. Second thing sir that in a break particularly drum and a single show that you shown yesterday one drum and one single show. Yes. So, pulley or drum is running with some well RPM. So, it will have some kinetic energy one half i omega square and break is absorbing show is absorbing that energy. So, that is converted into heat. So, here we can consider this two as a two separate system. So, can we consider the interface as a rudimenting system? No, see interface is a boundary interface by itself is not a rudimentary system because interface itself is not a system. A system is made up of whatever is enclosed inside the boundaries or interfaces that it encloses. So, can we consider shoe as a rudimenting system? Yes, a shoe a solid can be considered as a rudimentary system. If you consider that it does not expand contract. Yes sir, but the question is there that if we measure the temperature by thermometer it do not have anywhere, but here shoe is wearing continuously. See the moment you make that complication of shoe is wearing continuously then you are treating the shoe as an open thermodynamic system because some mass is going out of the shoe and our idea of a rudimentary system pertains only to close thermodynamic system. Rudimentary and other aspects are not applicable to open thermodynamic system because open thermodynamic systems are artificially defined systems for the purpose of handling flow into and out of systems over to you. 1115 Shivaji University, Kolhapur. Sir, my question is regarding the relation of temperature and pressure. Sir, as you quoted the example of pressure cooker. So, let us take that example if we cook some food in that pressure cooker at the standard atmospheric pressure and same pressure cooker if we cook that food at some higher altitude. So, what will be the difference? Which in which case the food will be cooked fastly? See pressure cookers are used for in cooking they are used for two purposes. First one is if you go if you are a mountaineer and go up a hill the pressure will be so low and pressure will be lower than ambient and hence your water will boil at a temperature lower than 100 degree C and when it boils at a temperature lower than 100 degree C your cooking will not be proper. Things may not get cooked at all because you need to reach a certain temperature for proper cooking of some food and even if you have reached that temperature since that temperature is low the rate of cooking will be very low. You will have to spend a lot of time wasting a lot of fuel and hence we use a pressure cooker to increase the pressure of the medium and the food which is being cooked to a higher value than the ambient. Now in a mountain you can use it for increasing the pressure from say 0.7 bar or 0.7 atmosphere to 1 atmosphere but there is no harm in increasing it further and doing the job quicker and in that case you can do it even at sea level or at normal pressure and that is what we do in pressure cookers. Although the food cooks faster people who say that the taste of food which is cooked in a pressure cooker and the taste of food which is cooked without pressurizing it is different but that is I think we are getting into more details because although the cooking reactions speed up as the temperature rises the other biochemical reactions of biochemical degradation which are also biochemical reactions they also get speeded up and hence perhaps there is an optimal temperature beyond which you should not go for a given quality of food over to you. Thanks a lot sir my second question is regarding sublimation state, as you stated solid will be directly converted to a liquid, sorry vapour vapour, is there any real example for that sublimation states? Yes you take solid carbon dioxide the so called dry ice at atmospheric pressure it is below it is triple point so solid carbon dioxide simply evaporates into vapour carbon dioxide and even some solids like camphor or you know those the Ododil type of solid room or cabinet fresheners you keep them they do not become liquid they simply slowly evaporate into vapour itself 1111 RVS college Coimptor. Some sub cold liquid to liquid just like Malia chart, steam tables can you give some table or chart to evaluate the properties of sub cold liquid, hello sir. Okay I am getting you but I am disconnecting you because there is a lot of echoing the question asked from RVS college Bangalore was that RVS college Coimptor was that are there tables charts or some information sub cold liquid and the answer is yes if you go to detailed steam tables as I showed today morning the steam tables published by national institute of technology available from download for download from the net. I think if you put I will try to put the link up on Moodle today late evening but if you just put NISD steam tables in Google you will find that link. In this detailed tabulation at every 5 degrees C or so of properties of sub cool liquid of ordinary water substance are tabulated. If you want to use a computer program then there is an international formulation on properties of water and steam a very detailed formulation program is not available but the equations are available you can use those equations to create your own steam tables over and over and out 1 2 8 1 S V E R I college of engineering ponder poor over to you. Hello sir I do have two questions my first question is we derive the equations just for example we are having D U is equal to M C V D T for that equation is it necessary the process must be quasi static or in general also it is applicable. See if you write D U equals M C V D T so long as you are talking of differentials this is true but D U will be M C V D T plus some other thing so the assumptions involved in this is U is a function only of T and it is a differential process for a differential process the question of quasi static does not arise that is you are considering two small neighboring states these are the only two conditions under which these are the two conditions under which this is valid and since U is a property delta U will be M integral of C V D T or you can sometimes write this as M some average C V into delta T these two are also valid under the conditions following U equals either U equals F of T or states 1 and 2 are at the same temperature and since we are talking of a change of state we do not have to worry about the actual process which links them two links the two states is quasi static or not over yes. So, I have 13 years of experience in teaching and multinational software companies. So, why I am saying this how what is the impact of this APF management analysis with respect to transient thermal analysis so what is the budding engineers and what are the students how we give a correct and accurate teaching with the students in respect of APA with respect in relation to transient thermal analysis because structural analysis that is different so we are dealing with engineering thermodynamics so which is going to be deal with transient thermal analysis and steady state thermal analysis which is a very advanced boost for modern industry. Thermodynamics does not consider any fields so in thermodynamics there is no transient thermal analysis. So, tools like finite difference techniques or finite element techniques are not directly applicable to basic problems in thermodynamics they are applicable to problems in fluid mechanics feed transfer and definitely to solid mechanics. So, I do not think I have I do not think your question is actually relevant in this course I think it is well beyond 530 so we stop here today.