 An exercise you should go through is ask them the following things and that take out your exercise sheet page 5 in the third part of the sheet 3.1. Find the state of a cloth system containing 1 kg of water substance as mentioned below. So you have given pressure, temperature and you have to determine whether it is saturated liquid, dry saturated vapor, sub cooled liquid or superheated vapor. The nomenclature is important. So again you sketch the PT diagram. This is liquid, this is vapor. On this saturated point, what we have here is saturated liquid. What we have on this end is dry saturated vapor. A vapor which is anywhere here is known as superheated vapor and a liquid which is anywhere here is known as either sub cooled liquid or compressed liquid. Sub cooled because if you go along isobar, this would be the saturated liquid. You have reduced the temperature, so it is sub cooled or you could have considered a liquid here that would be saturated liquid. You have compressed it and brought it to that point. So sub cooled liquid or compressed liquid mean the same thing. Similarly superheated vapor means if at the same pressure, if it were here it would have been saturated vapor. You have heated it further, increased it temperature, you have called it superheated vapor. You should tell them because they are always amenable to mathematical relation, whether it constant. Can we call, if the temperature is less than the saturation temperature, it is compressible liquid. Compressed liquid, not compressible. Yes, you can call it. Any liquid which is below saturation, purely in the liquid zone. Even in isobar it does not matter. It is a name given to the state. Fortunately for superheated vapor, we have only a unique name superheated vapor. We do not call it expanded liquid. But for liquid for some reason, we call it either sub cooled liquid or compressed liquid, whichever. So again you sketch because a question will be asked. Given a pressure temperature, how do you determine whether it is a liquid or whether it is a vapor? So you draw a line like this and say given a pressure, this is the saturation temperature or maybe I should let me redraw this. We are given a point Pt in state space. We have to determine whether it is sub cooled liquid, saturated liquid vapor or superheated vapor. What we do is, we will lay out our liquid line and determine whether on which side it lies. So one way to look at it is calculate given P0 T0. So this is P0 T0. One way is obtain P sat at T0 at the given temperature of obtain what is the saturation value. If P0 is greater than P sat at T0, what do you say? Sub cooled or compressed liquid. If P0 is less than P sat at T0, that means the point is somewhere here. You will call it superheated vapor. On the other hand, we may look at it the other way. Given P0 T0, on the previous case we calculated P sat at T0. In the second option we will calculate T sat at P sat at T0. So for the given pressure, we will determine what the T sat is at P0. And then it is possible that T0 is greater than T sat at P0. T0 is less than T sat at P0. What do you say this? T0 is greater than T sat. It is superheated. If this is so, then it is sub cooled. Then the question arises that we have only talked about greater than or less than, greater than or less than. What happens if it turns out that T0 equals T sat at P0? Or P0 equals P sat at T0. In that case, point lies exactly on the saturation line. This is the case. In this case, the situation could be I have a reasonable amount of liquid with vapor. That is perfectly okay. Or I could have a small amount of liquid, few droplets, most of it vapor. Or I could have a few bubbles of vapor and most of it liquid. But in all cases, there will be a large number of cases from a speck of liquid to almost everything, only a speck of vapor to any proportion in between. We have a mixture of liquid and vapor in equilibrium. That means that of saturated liquid plus dry sat vapor in some proportion. Since it is a two phase situation, P and T or T and P is not a full or proper combination. Yes, sir. In some case, there will be a possibility of only liquid and vapor also possible. Yes, then there is no problem. If it is only liquid, it is the saturated liquid property as listed in the table. If you are sure there is no vapor or even the boiling point is same for the particular pressure. Yes. And similarly, if you have only vapor, no liquid or just a trace of liquid negligible, then it is dry saturated vapor. Now in this case, because two phases are together, phase rule dictates that pressure and temperature are not independent. That is why we have the saturation line. And hence, if you say one atmosphere 100 degree C, we do not know what it is. It could be any reasonable ratio between liquid and vapor. So in that case, we define for mixture saturated liquid plus dry sat vapor. This is given the name technically of wet vapor. Why not dry liquid? I do not know. That is tradition. It is wet vapor. And since we cannot show it on the P T diagram properly, because our point exactly lies here, we go to a diagram which is P and something else, say V. So if this is the pressure, we extend it in this and say that look, this is our P naught. This will be our T naught exactly on the saturation line. So this is P naught. This is the F point representing saturated liquid. This is the G point representing dry saturated vapor. Our state will lie anywhere between these two points. That is the range. It could lie for example here or it could lie here or here anywhere. So instead of P and T which is not acceptable, the acceptable things are P and something other than T. It could be V or U or H or even S or it could be T and anything other than P, V, U, H, S. So what we do is we define. Now I may be comfortable with P and U. Somebody else may be comfortable with T and U. Some third person may be compatible with P and H. One of P, T will do. The other cannot be T or P or if you want, somebody will say I am interested in V and U. I want neither P nor T. That also is okay. But thermodynamically, since we have a good feel for pressure and temperature, if both pressure and temperature together are useless, at least one of them we would like to use. So traditionally, although U and V are acceptable pair of properties to define a state, we will select one of P or T and the other one either as V, U, H, S. But again we have an arbitrariness. So that is reduced by defining what is known as a dryness fraction. This is the way to introduce dryness fraction. You say that in our previous thing, let us say that the vapor has a mass mV. The liquid has a mass mL. So that the total system mass is mV plus mL. And hence, we say that this state will be defined as mL of liquid plus mV of vapor. And the dryness fraction, symbol X, usual symbol X is defined as mass of vapor divided by the total mass of the system. It is a ratio, it is a dimensionless number and it goes from 0 to 1. So a state here, the saturated liquid state. If I use G, that is okay. You can use L and G because we are already familiar with F and G, F for liquid and G for gas. Let us go back, then we will modify this. If you are comfortable with, this is mg, liquid is mF. So the total is mF plus mg and here we will again redefine this. So this is mg, this is mF plus mg. Symbolism you can decide. If you are textbook forces, you can decide as a particular symbol, use it. Now what is this point? The point where you have liquid and hardly any vapor, but saturated liquid. This will be represented by X equal to 0 because mg is 0, everything is mF. This point, G point, F equal to 1 because everything is vapor, only a speck of liquid. Just to see to it that it is an equilibrium vector. So this will be X equal to 1 and in between points will vary from X equal to 0 to X equal to 1. So you tell them that the states in this zone of the state space are not directly listed here, but the limits are listed. The saturated liquid properties are listed, trisaturated and vapor properties are listed. So using those, now how do you calculate properties for wet vapor? For wet vapor, tell them that now the volume will be, since you say that all properties are extensive properties, volume of our system will be volume of the liquid part, but volume of the vapor part. Grinus fraction is a thermodynamic property. Yes, it is a property related to the state of the system. Is it an extensive property or intensive property? It is an intensive property. We will come to that now. See the volume is a, it is, see I have sketched here vapor in the upper part, liquid in the lower part. That is because we always assume there is gravity. So if there is a bottle here, today my water bottle is missing and this water bottle is useless. You cannot see a liquid there. So that the vapor rises above, liquid settles below. So you cannot say that the system is uniform everywhere, but we will consider it to be homogenous. We will neglect the effect of gravity and we will say everywhere in the system, any small volume you take, there will be equal part, uniformly spread parts of liquid and vapor. That is an assumption. Otherwise if you look at it, out here, grinus fraction is 1. Out here, grinus fraction is 0. Then you will say the system itself by definition is not in equilibrium because I do not have a unique grinus fraction. We will neglect this gravity, but we will show it as liquid vapor separate. But our, will be droplets of liquid in a cloud of vapor or bubbles of vapor distributed in a liquid uniformly. So homogenous is the assumption, which we will make. In which case x is uniquely defined, it is a property. Which one? Yes, in presence of gravity it is an approximation. Yes, it is an approximation. But it is a very good approximation because all that gravity does is, you know fog. You create a fog, initially it will stay slowly because of gravity all over dune and fog and dust will lie down. It is a slow effect. So it is in most cases it is negligible. So for wet vapor, any property, say volume, being extensive that will be the volume of the vapor part and volume of the vapor part and the liquid part. So if you define the specific volume for each part and the whole system, this is the specific volume of the whole system. This will be mass of the liquid part, specific volume of the liquid part plus mass of the vapor part, specific volume of the vapor part. And if you divide throughout by Mf by Mg, you will get v equals, Mf divided by Mf minus Fg will be 1 minus x vf plus x vg. Or if you solve for x, x will be v minus vf vg plus vf vg plus vf vg plus vg minus vf. And tell them that all this is valid if you replace v and v by u and specific internal energy h and specific enthalpy and s and specific entropy. We have not defined it yet but since the values are tabulated, they might as well get used to those values. But tell them that this is not valid for density. Do not write density of the system is 1 minus x into density of liquid plus 1 into x into density of vapor because that is the reciprocal of specific volume. Anything is with per kg of the system mass, this is valid. Density is not per kg of the system mass, it is per unit volume of the system mass. And of course you may write x equal to v minus vg, all this equals u minus uf ug minus. And then bring them to the, bring them bring their notice, bring them to notice this that when you write this this will be h minus hf hg minus hf. So this hg minus hf turns up and that is one of the reasons why hg minus hf is tabulated. And later on when we work with open systems and turbines, compressors h and s are the important properties for us and that is why hf minus hg, hg minus hf and hg minus hf are tabulated. Now that takes care of the saturation line and the wet vapor. Then we say that we have to look at people point, critical point. Suppose a system state lies here that is superheated vapor or superheated steam. It is single phase so P and T is an okay combination and with this combination the properties are tabulated in table 3. Come to page 11, we have table 3. Now unlike tables 1 and 2 where the first column was the independent variable and all other columns had the information, this is a two dimensional table. Tables 1 and 2 are one dimensional. This is a two dimensional table because properties are listed as a function of pressure and also as a function of temperature. So for example if you want the properties at 2 bar and 200 degrees C, you should look at the row for 2 bar which is a 4 high row and 200 degrees C. So this is the block in which the properties are listed at 2 bar and 200 degrees C. The 4 properties listed are specific volume, specific energy, specific enthalpy and specific entropy. Make them read out the values loudly. Now show them around and show them that it is a big table going from page 11 right up to page 16 actually really up to page 17. Slowly introduce them to various points in this table, read out a few values but then there are certain clear things about this table. For example there are some blocks which are blank. Why are they blank? Question will come up. Then what is this column and what are these values below the pressure? So explain to them that below these values for example 2 atmosphere 120.2 degrees C is the saturation temperature at 2 atmosphere because the saturation temperature is 120.2 degrees C at 2 bar. When it is 2 bar 100 degrees C or 2 bar 50 degrees C, it is not superheated vapor. It will be subcooled liquid. The authors of this have decided to tabulate properties of only superheated steam. Instead of that blocking it off they could have tabulated properties of subcooled water because these are not points where steam does not exist. Ordinary water substance exist but in a subcooled liquid form. So in this table wherever a state would be of subcooled liquid that particular block has been left blank. And this table is saturated dry saturated vapor properties at that pressure and if need be can be used for interpolation. For example at 2 bar 150 degrees C we have this tabulation 200 degrees C we have this tabulation. So suppose we want at 180 degrees C we will have to interpolate between 2 values. One value at 150, one value at 200 degrees C. But suppose I want at 130 degrees C then I know at 130 it is vapor, superheated vapor. But I do not have a tabulation below 150. So what we do is we took take the properties at 120.2 which is the dry saturated vapor property which are listed here. So between 120.2 degrees C this value and 150 degrees C these values we interpolate. At the first course level we will generally not set up problems where significant amount of interpolation will be needed. Problems will be simple. But some cranky student will come up with some idea. What if I have to have property? So you have to explain this to him. Again you can but an interpolation between 2 points is always better than extrapolation. Because what lies beyond we do not really know and that is why even from a numerical analysis point of view extrapolation is not as safe as interpolation. That is all. Now when it comes to this range of this table the range of this table is from saturation temperature upwards and pressures up to critical points. This is the range of table 3. I have to change back because now I am showing the table. If you see as you go further you will find end of this table which is on page 16. You have gone up to 220 bar which is just below critical pressure and you have gone up to 700 degrees C which is much beyond critical temperature. Again at 220 bar saturation temperature is 373.8 so the tabulation begins at 375. At 350 it will not be superheated vapor so that is left blank. Again you will notice that on this page up to page 15 he has tabulated values of V, U, H and S. But just to save space on this page the values of only V, H and S are tabulated. Not only that the values of V are multiplied by 100 before tabulation. That is because if you look up here the values of V are 0.0 something 0, 2, 0, 3 always 0.0. So to reduce the space required he has got rid of U because generally H will be required more in calculations for open system and V is plotted as V into 10 raise to 2. So the specific volume at say 100 bar and 400 degree C is 2.641 divided by 100 meter cube per kilogram. Bring this to their notice particularly if you are using this table or any such table. Again dropping of U is of no great importance because it can always be calculated as H minus T V. And then you tell them that supercritical steam nothing special about supercritical steam. It is just this zone. This is in table 4. It is essentially an extension of table 3. You will notice that the pressure start from 230 just beyond 220 and temperatures go from 350 to 800. And again V into 10 raise to 2 just for simplicity of tabulation and U is dropped but H is provided again for providing more information on the same page. Now that brings us to. Sir. Yes sir. Is it sufficient to represent a steam with U and V? Yes. In terms of specific volume. U and V yes. The only thing is calculating pressure and temperature will be difficult because the tabulation is in terms of pressure and temperature. But is there any direct method or only by trial and error we can get the other? Only by trial and error and by interpolation. I think this kind of problem find very difficult to find out the other properties. Yes. Yes. U and V. I know it is difficult to find out but when you do detailed turbine calculations you will come to a stage where the state at some point is known in terms of H and S, enthalpy and entropy not in terms of pressure or temperature. So, we have to work backward to pressure and temperature particularly in simulation or off design calculation of turbine performance where you have to predict in between turbine stages what is the pressure and temperature. What you get from your governing equations enthalpy and entropy from first law and second law. So, from enthalpy and entropy you have to go back to p and p. It is difficult but by interpolation you have to do it. There are equations for example, you may tell the students. End of question ask in the anonymous exam. That U and V is given. In fact, we have a problem in our tutorial sheet but that is in terms of that finally ends up in wet vapor. Maybe it is some similar problem. Then we solve problems we will see today. Yes sir. Sir, in the table 4 that is meant for supercritical steam. Page 17. Whereas in the page 16 the critical temperature for the pressure of 220 bar is 373.8. Page 16 the last one the critical temperature for the pressure of 220. It is not critical temperature at 220 bar the saturation temperature is 373. Whereas in the table 4 it is meant for supercritical steam. So, the pressure given is 230 for that there is no need of that 350 degree centigrade column. Which 350 degree centigrade. Now, this column does not exist because at 230 there is no saturation temperature. We are beyond the critical pressure. Sir that is why no that even for 220 the saturation temperature is 373. Right. So, for 230 should be more than that probably more than 373. No, no, no, no. Let me come to this. Let me show the let me show a part of the state space which is near the critical point. This is the critical point and this is the line, but I am not showing the lower part. This is pressure, this is temperature. Now, this is let me write this. This is 373.15 degree C. This is 221.2 bar. Now, in table 3 you have a point of 220 and 375 that is you have a point this point. And we have points beyond it up to 700 degree C. Now, when it comes to the next column, next row in which is the first row in table 4 that is this was 220. Now, this is 230 bar. At 230 bar we can have here at this is 375 which was that 375. You have a point you have 375 here. But what you say is we also have 350, but that is beyond the critical pressure. It is in the fluid part. We cannot say it is in the liquid part because out there the distinction between liquid and vapor has vanished. What you are doing is you are extrapolating this and say maybe it is crossing that line, but there is no point in extrapolating that line. At 230 bar actually you should call it supercritical vapor or supercritical fluid. He has not plotted at 230 bar below 350. He could have plotted at 300, 200, 100 that is perfectly all right. It is supercritical fluid and for those who are interested you can tell them that there is an international committee in international society for the properties of steam which collects data on properties of steam and fits equations to it because you can write computer programs to do all the property calculations. Those equations are if you write an equation it occupies a page and a half. In the total set of equations contain roughly 300 or 400 constants. Compared to that PV equals RT 1 constant. If you use CPCV maybe one more constant or a few more constants. Compared to that for steam there are a few hundred constants. So managing that equation is impossible. Compared to them that our ideal gas equation is absolutely nothing. It is very easy. That is why we use equations for an ideal gas. For steam we use tables. But if you really need detail you can program those equations and use them. There is nothing wrong with that. Hello sir. Yes sir. In steam table you have said that HFG is the difference between HG and HF. Right. Can we define that term as a latent heat? That is traditionally known as latent heat. Then what is the base of generation of this steam tables? How it is generated? Property data. There are hundreds of teams which have measured saturation temperature, they have measured enthalpies, they have measured specific heats. From that all these steam tables are generated. Those who are interested I have a, there is a site on which all this information is available. You look up I forget. International conference on the properties. I will get you that link tomorrow. You just put properties of steam and you will get all those links in Google. Sir, I have a doubt. Just as we increase the pressure, density of the vapour goes on increasing. PV diagram from PV diagram. From the PV diagram, as you increase the pressure, density of vapour goes on decreasing. Specific. Specific volume goes on decreasing, density goes on increasing. And for liquid the reverse. So, I can understand the volume of vapour, but what about liquid? What is wrong in that? As we increase the pressure from PV diagram, we can see specific volume of vapour goes on decreasing. So, density is increasing and on the left hand side. For the liquid last thing that as the pressure increases. It is a specific volume. Why should the specific volume increase? Density decrease. Yes. Because the idea is as pressure increases density should decrease. Yes. But here remember that as pressure increases temperature is also increasing. And only when temperature is constant that general idea that when pressure increases density should increase is true. Let us say what you are looking at is a combined effect of both pressure and temperature. Both for liquid as well as for that. Now that brings us to a situation where we know how to take care of states which are saturated liquid, dry saturated vapour, superheated vapour and even supercritical fluid. Unfortunately the data here and here is not fully tabulated. I say fully tabulated because in the recent editions if you come to page 18, table 5, we have a very crude tabulation of compressed liquid water. Can you understand this table? Actually this is 6 tables in one and at 6 different pressures of 50 bar, 100 bar, 150 bar, 200, 300 and 500. As a function of temperature the properties are listed. The properties are V, U, H and S. Effect of pressure, but unfortunately it shows for a fixed pressure the effect of temperature. To see the effect of pressure you take for example the 60 degree C line, 60 degree C and then you will see at 50 bar, 100 bar and 150 bar what is the effect on pressure. You will notice that take the volume at 60 C at 50 bar it is 0.001015, then 1013 and then 1011. It decreases but decreases only slightly. Similarly you take internal energy 250.23, 249.36, 248.51. It decreases but decreases only slightly. Similarly look at entropy 0.8285, 0.8258, 0.8232. Decreases but decreases only slightly. You can interpolate between this but unfortunately the spacing is too crude 50, 100, 150. Quite often we will have problems of pumping from say 1 bar to 10 bar, 1 bar to 8 bar. Even 50 bar is too large a problem. We do not have that information. So if we cannot use table 5 and apparently this is one of those rare steam tables which has this tabulation. You might have your own steam tables. I had asked them to bring open them and you will find no information about this at all. So what do we do for subcooled liquid? For subcooled liquid we do the following. If data is not available, see this table for example gives some idea. There are big tables which provide a detail as much detail as superheated vapor. Maybe if I will bring one such table so that you can see it tomorrow. Subcooled liquid we use an approximation and we assume the subcooled liquid to be an incompressible fluid. We use it as a model and the idea is like this. I will show extend this. This is pressure, this is temperature. We are somewhere here subcooled liquid zone. Our pressure is higher than the saturation pressure. Our temperature is lower than the saturation temperature. Let us say this is given by P naught T naught. What we say is we compare it with this point. We take this is saturated liquid at T naught. And let us say that the saturation pressure there is P sat at T naught. This is the state in which we are interested. The saturated liquid state is a state which is tabulated. Data is available. What we say is we consider these two states. The state naught and this state let me call it the F state. What is the difference between state 0 and state F? Temperature is the same. Pressure is different. But if we assume that the subcooled liquid can be modeled as an incompressible liquid then that assumption tells us that the effect of pressure on three properties specific volume internal energy and entropy is negligible. So we say that V at this point will be V at F and same thing for U and S. In particular we can expand this by saying V at P naught T naught when it is subcooled liquid will be approximated by V F at T naught. Similarly, U at P naught T naught when we are sure that it is subcooled liquid will be approximated as U F at T naught. There is no effect of pressure. Similarly, S at P naught T naught will be approximated as S F at T naught. However, H will not be approximated. H at P naught T naught will have to be calculated as U at P naught T naught which is approximated as U F at P naught plus P naught into V at P naught T naught. So this will become U F at T naught no effect of pressure. This will become V F at T naught no effect of pressure. But this P naught directly exists there and hence enthalpy will be a function of pressure. No no no this will become we have I said H at P naught T naught is U at P naught T naught. This is simply expansion plus P naught because it is P V, V at P naught T naught and by our approximation this is U F at T naught plus P naught into V F at T naught which is not H F at T naught. H F at T naught will be saturation pressure here. This is H this is H naught this is H. But this is not equal to H F at T naught. Notice not equal to H F at T naught and many other place only it is an accident of the equation of state of an ideal gas and Joule's law that H happens to be a function only of temperature. That is a rare case any other gas any other liquid H will be generally a function of pressure. If you want you can go further write this as U F at T naught plus this will be P sat into V F at T naught. This will be P sat at T naught plus P naught minus P sat at T naught and then you can write this as if you want H F at T naught plus. But I think for the students we should leave it at that. But remember that going back to this you should tell them that if I start from this point with this approximation if I go from this point upwards along this line constant temperature line with this approximation my values of V U and H will remain unchanged. But my value of H will go on increasing linearly with P because that P into V F at T naught that exists. With this our property of skin tables property calculations is over.