 Now, let us see how to locate the state given P and V. Here we are given specific volume and pressure. So, what we try to do here is the following before we go through this figure, let us see what we want to do. So, let us say that this is our PV coordinate. So, 250 kPa, let us say this is the isobar corresponding to 250 kPa. So, what we would like to know is the specific volume is given. What we would like to know is, is it more than the specific volume of the saturated liquid state and less than the specific volume of the saturated vapor state or more than the specific volume of the saturated vapor state or less than the specific volume of the saturated liquid state. In other words, we want to know whether the state lies here, here or here. In order to determine that we need to know the specific volume corresponding to saturated liquid and specific volume corresponding to saturated vapor at this pressure 250 kilopascal. So, let us go to the pressure table, pressure table 250 kilopascal that is 2.5 bar which is over here. So, the specific volume corresponding to the saturated liquid is 0.001, please bear in mind that this is multiplied by 1000. So, the specific volume is 0.001 0672 and the specific volume of the saturated vapor is 0.7193. So, we may retrieve these values. So, the specific volume of the saturated liquid is now known at this corresponding to this pressure and the specific volume of the saturated vapor is known, the given value 0.5 lies in between these two. So, this is the state. So, if you are given the given P and a specific property, state is relatively easy to locate. Given P and T, you need to locate the corresponding saturation temperature or saturation pressure. Here it is very straightforward because we can retrieve these values Vf and Vg or Uf and Ug as the case may be from the directly from the table. So, since it lies between F and G, we have a two-phase mixture. Let us go through one more example, 1 MPa and specific volume 0.5 meter cube per kilogram. So, we go to the pressure table here because pressure is given. So, we go to the pressure table 1 MPa which is 10 bar. So, the specific volume of the saturated liquid is 0.001 1272 and that of the saturated vapor is 0.1945. So, we retrieve these values and then go through the same procedure again. So, we draw the isobar corresponding to 1 MPa which is this and the specific volume of the saturated liquid has been drawn from the tables and this has also been retrieved from the tables. So, the given specific volume of 0.5 is greater than Vg corresponding to this. So, which means that this is the line corresponding to 0.Vg V equal to 0.5. So, the given state R lies at the point of intersection of V equal to 0.5 and P equal to 1000. In the same way here also the given state lies at the point of intersection of V equal to 0.5 and P equal to 250. Note that these diagrams are not to scale. Please bear that in mind. These diagrams are qualitative. They are not to scale. In the same manner you can go through couple of more examples. T equal to 200 degree Celsius V equal to 0.1. You should be able to retrieve from the temperature table corresponding to 200 degree Celsius that Vf is equal to this and Vg is equal to this and the given V lies in between Vf and Vg. So, that means it is a two phase mixture region. It is in the two phase mixture region. 250 degree Celsius 0.2 meter cube. Again, you should be able to retrieve Vf and Vg from the temperature table and the given value for the specific volume is greater than Vg. So, that means the state is super heated. So, this is how we locate states for or on a PV or TV diagram for water or R134A. The procedure is identical for steam or water and R134A. For R134A also we have temperature table, pressure table and superheated table. So, the procedure is identical. Basically, since it is a simple compressible substance you need to be given two values for two independent properties. It may be P and T. It may be P and V. It may be T and V. Later on as we go through some of the examples and later part of the course, it may be P and U or P and H. The procedure remains the same or it may be T and S. Again, the procedure remains the same. So, it is either P, T and any one intensive property or P and T. So, we have covered all possibilities. So, you should be able to locate the state in the PV diagram or TV diagram using this information. So, I suggest that you review this information carefully. Go through the examples a couple of more times. Feel comfortable using the steam tables. Procedure 4. Now, we turn to the process of calculating property values. We have located the state. It is either a compressed or subcooled liquid, a two-phase mixture or a super heated state. How do we calculate property values? Now, we are not provided tables separately for compressed liquid or subcooled liquid states which is okay in a course like this. We do not really need that. We will make certain simplifications. Since it is known that it is a liquid, certain simplifications can be made and we will look at that later on. But for now, let us say that the state lies in the two-phase mixture region. For example, something like this. So, it lies in the two-phase mixture region. How do I calculate? For example, specific internal energy or specific enthalpy and so on. Any specific property. In order to do this, we introduce a new property called the dryness fraction or quality X. It is defined as in a given sample or in the system, the given system, let us say the total amount of mass in the system is m. And it is a mixture of liquid water and water vapor. So, the amount of vapor, let it be mg. So, the dryness fraction is defined as a ratio of the mass of vapor divided by the total mass contained in the system or mg divided by m liquid plus m vapor, mf plus mg. Now, the specific volume of the mixture itself is nothing but the total volume occupied by the substance divided by the mass. The total volume is nothing but the sum of the volume. So, this is equal to sum of volume occupied by liquid plus sum of volume occupied by the vapor. And this may be written as mf times vf where vf is the specific volume of the liquid and mg times vg where vg is the specific volume of the vapor. So, if you simplify this expression, you can write this as, so the specific volume may be written as vf plus the dryness fraction x times vg minus vf. Alternatively, you may also, if you want better interpretation probably is possible if I write this as 1 minus x times vf plus x times vg. So, you may interpret it like this, the specific volume for any two phase mixture is the weighted sum of the specific volume of the saturated vapor and the specific volume of the saturated liquid. The weights being 1 minus x for saturated liquid and x for the saturated vapor. Both these formulae are acceptable and we probably will be using more of this although this allows for a nicer interpretation in the sense that we can see that at any two phase mixture state, any intensive property is a weighted sum of the properties of the saturated liquid in the saturated vapor state. The weights being 1 minus x for the saturated liquid state and x for the saturated vapor state. So, here is how we have generalized this, any intensive property, I am sorry any intensive or specific property in the two phase region may be written as phi f plus this or alternatively as we just wrote it may be written as the weighted sum of phi f and phi g, weights being x and 1 minus x for the saturated vapor and saturated liquid respectively. So, the phi can be specific volume, specific internal energy or specific enthalpy. So, now we know how to calculate property values when the state lies in the two phase mixture region. So, we can easily apply first law now and delta u needs to be calculated, we can use this expression to calculate delta u after getting u f and u g from the steam tables either pressure or temperature table depending on the information that is given. Now, if the state is superheated like for instance here, if it lies in the superheated region, it is relatively straightforward, we can simply use the tables that are provided. So, here we can go to the superheated table. So, you can see that table C corresponds to superheated state. So, once we have two properties which are given, which are used to fix the state of the system, then we can either P or T and V or P and V then all the values specific volume or all specific properties may be retrieved directly from here. So, if you are given T and V, then we go into the corresponding superheated table and retrieve all the intensive property values that we want, it is very straightforward. Now, what do we do if the state lies in the compressed liquid region? That is what we are going to see next. So, if it is in the two phase mixture region, we retrieve the values corresponding to saturated liquid state and saturated vapor state and calculate any other intensive property or specific property as a weighted sum of these two with corresponding weights from minus x and x. If it is a superheated state, we can directly retrieve the property values from the table. What happens if it is a compressed liquid or subcooled liquid? So, here we exploit the fact that the specific volume changes very little with pressure as the liquid is incompressible. We already saw this before that the specific volume changes very little. It is almost the same as the specific volume of the saturated liquid. So, if I look at this diagram, then you can see that the specific volume corresponding to say this temperature or this temperature is more or less the same, it does not really change all that much. So, it is an approximation that we may use and the internal energy also for a liquid, internal energy changes very little with pressure. There is no dependence of the internal energy on pressure. So, we can make use of these two approximations and write for a compressed liquid state that the specific volume at any temperature and pressure is almost the same as the specific volume of the saturated liquid corresponding to that temperature. Notice that we take the value corresponding to that temperature and the specific internal energy is approximated as the specific internal energy of the saturated liquid at the corresponding temperature. The effect of pressure is negligible. So, we ignore the dependence on pressure and take u to be u of t, p to be just u of t. Notice that it would be incorrect, although numbers may be alright, it would be incorrect to write h of t comma p as h f of t. That is strictly speaking not correct, although this is used in many textbooks, strictly speaking that is not correct because by definition, h of t comma p is equal to u plus p v. And based on the two approximations we have made, I may replace u with u f of t and p remains as it is and v is replaced with v f of t. So, h of t comma p is u f of t plus p times v f of t. So, this is consistent with the two approximations that we have made here. So, you should always use h as u f plus p times v f. Because if you say that this is equal to h f of t, then the p here actually becomes equal to p saturation because you are taking the saturated liquid data. So, this is strictly speaking not correct because the pressure is known and that is different from this pressure. Remember it is a compressed liquid state. That means p is greater than p set. We already know that we have already seen that. So, that means that effect will be there. It may be a small effect, but it is still there. So, from a, you know, from a consistency point of view, h of t comma p for a compressed liquid should be used like this and not as h f of t. Although the values themselves may not differ by too much, it is important to understand this distinction and then do things accordingly. So, now we know how to locate the state and how to calculate the properties once the state has been located, whether it is in the compressed liquid or sub cooled region or whether it is in the two phase region or whether it is in the superheated region, we know how to calculate the property. So, we have now completed calculation of properties for pure substances. We saw how to calculate properties for ideal gases, ideal gas mixtures and now for two phase mixtures. Now we close this module with an interesting discussion on the connection between what we saw so far those two phase region and ideal gas region. So, we were drawing the TV diagram for water so far. So, one of the questions that arises is can we not take the state in the superheated region? So, if a state lies in the superheated region, can we not take this to be an ideal gas? Why not take this to be an ideal gas with molecular weight 18? Can we not use p v equal to r t or can we not use u equal to c v times t in the superheated region? Why use the table? The answer to that question is what we will discuss next. So, qualitatively steam will behave as an ideal gas in this region, I have indicated here qualitatively. So, it behaves as an ideal gas in this region only. In other regions it will not behave as an ideal gas so p v will not be equal to r t. For any two phase mixture it could be for instance nitrogen, it could be oxygen, it could be CO2. For any two phase mixture the ideal gas region lies to the right like this and the values of the pressure and temperature which bound this region will vary from one gas to another. For N2 these values may be different, for O2 these values will be different, for CO2 these values will be different, but qualitatively this is where we have the ideal gas region in comparison to the saturated liquid and saturated vapor line. So, that is where it lies. But the values themselves will change depending on the gas that we are looking at which is somewhat inconvenient, but quite interestingly it turns out that if you calculate what is called a reduced pressure and reduced temperature which is nothing but the given pressure divided by the corresponding critical pressure. So, if you are dealing with water you will take the given pressure divided by the critical pressure for water, if it is O2 you take the actual pressure divided by the critical pressure for O2. So, instead of using T and B if you re-plot using reduced temperature and reduced pressure then we get a very nice diagram. So, you plot PV over RT on the y-axis and P over P critical which is called the reduced pressure and this is parameterized by the reduced temperature which is T over T critical. Then irrespective of whether this is O2 or N2 or CO2 or helium for all gases we get a single curve for each value of T reduced or T over T critical. That is the nice thing about this particular plot. We do not have to worry about the nature of the gas and now if you want to explore or ask the question where the ideal gas region lies in this diagram notice that this rectangular region is indicative of ideal gas behavior to within plus minus 10 percent. So, PV over RT is either 1.1 or 0.9. So, approximately we can say that this is the ideal gas region. So, the ideal gas region generally is valid for all values of pressure including very low values, but valid only for high values of temperature. Notice that here the temperature is 2 times the critical temperature here it is 2.5 times the critical temperature. So, it is valid for high values of temperature but for all values of pressure. So, this gives us sort of a orientation of the ideal gas region relative to the two-phase region because we are discussing it, we have we discuss ideal gases and ideal gas mixtures and then we discuss two-phase mixtures. So, it generally gives an impression that the two are sort of totally different or disconnected from each other that is not the case. This one shows that the two actually all part of the same phase space TV or PV ideal gas region lies here like this in this region and the two-phase region lies here. So, this is the two-phase region they are connected and depending on where we are we can use the ideal gas equations equation of state and the closed form for calculating properties or we need to use the tables for calculating properties. So, what we will do in the next lecture is work out examples where we have different situations, we identify an appropriate system, we apply first law and carry out a first law analysis of systems.