 In this lecture, we will look at the methodology for evaluating properties of two phase mixtures. Even before we actually look at evaluating properties of two phase mixtures, locating the state of a two phase mixture itself needs to be done. In the case of an ideal gas, locating the state given two properties was relatively straight forward if you use a PV diagram or any two independent coordinates. In this case also, two independent properties are required. So, we had established that earlier that for a simple compressible substance, two independent properties are required to fix the state of the system. But in the case of system that is actually undergoing phase change or it has two phases liquid and vapor, the process that is required to locate the state is a little bit more involved. So, we will look at that first. So, let us start with a simple experiment. This is the so-called Andrews experiment. So, basically we have a bath which is maintained at 100 degree Celsius and a piston cylinder mechanism is immersed in the bath. So, the contents of the piston cylinder mechanism or the contents of the cylinder which in this case is nothing but water. So, we start with liquid water at first. So, this is always maintained at 100 degree Celsius. So, what we are planning to do now is conduct this experiment and draw say isotherm corresponding to 100 degree Celsius on a PV diagram. So, let us say that this is the specific volume and this is the pressure. So, we would we will attempt to draw an isotherm in this in this PV diagram, isotherm corresponding to 100 degree Celsius. So, let us say that initially I know the specific volume and I know the pressure. Let us say that the pressure is a relatively high value. Let us say it is 20 mPa. Let us say that we denote the initial state like this. So, the pressure is 20 mPa and the specific volume let us say is also known. So, we locate the initial state like this using this cross. Now, while maintaining the water at 100 degree Celsius, let us say that we move the piston slowly up. So, we move the piston slowly up and we notice that as we do this the pressure decreases. But what is interesting is the pressure decreases by a lot even if we move the piston only by a small amount. So, even for small changes in the specific volume of water, we notice that the pressure changes a lot. In other words, the next point may be something like this. So, this is let us say 10 mPa and the specific volume is almost the same just slightly more than what we had before just slightly more. Now, this happens because the water inside is in a liquid state and you as you all know liquid is any liquid is highly incompressible. So, even for large changes in pressure, the change in specific volume is very, very small. So, similarly here the pressure changes even for just small pressure changes by a lot or drops by a lot even for small changes in specific volume. So, the next point comes here and we continue to do this. So, we continue to move the piston upwards increasing the volume and we continue to notice that even with small increases the pressure decreases by a lot like this. This continues until we hit a pressure of 100 kilopascal. Remember, during this process, the water is maintained at 100 degree Celsius. So, when we hit 100 kilopascal as we keep moving this piston outwards, eventually we reach a pressure of 100 kilopascal and you know that water boils at a pressure of 100 kilopascal and 100 degree Celsius. So, now if I, so this is the point at which we reach 100 kilopascal, now the water starts to boil if I move it a little bit up. So, if I now move the piston up, I notice that some of the water evaporates and occupies the space that is available. So, when I move the piston up a little bit more, some extra space is available now and the liquid water, some of it evaporates and the vapor occupies the space above the liquid and below the piston. As a result of which the water undergoes a change of phase but the pressure remains constant. The temperature is already being maintained constant as we have mentioned. Now, when the phase change is taking place because this water vapor occupies the space, the pressure remains constant. So, we go from here to here like this. If I move the cylinder a little bit more, some more water evaporates. Notice now that I have to move the piston by a finite amount for the water to evaporate. So, we keep going like this, so some more water evaporates and occupies the space and this goes on until this goes on until we reach a certain point at which all the liquid water has been converted to vapor. Now, the cylinder is occupied, initially we started with liquid water. Now we have all water vapor, the cylinder is filled entirely with water vapor. Now if I continue to move the piston outwards, decreasing the pressure, now the pressure starts to decrease because phase change is complete. Now the pressure starts to decrease and the vapor behaves just like any normal gas, well close enough to a normal gas. So, the pressure decreases and the specific volume increases. Notice that this curve is in contrast to this curve, where even small changes in specific volume resulted in large changes in pressure because it was an incompressible liquid. Here this is a compressible vapor, so the expansion process looks more like the familiar expansion of an ideal gas. So, you can notice two important things, one is let us just highlight this with red, one is this point which is the beginning of of evaporation and the other one is this point which is the ending of evaporation. In between these two points, the pressure remains constant and the temperature is already being maintained constant. So, notice that as phase change takes place, both pressure and temperature remain constant. So, when phase change takes place, pressure and temperature are not independent properties. So, they are independent here, they are independent here, but not when phase change takes place. Notice that the isotherm and the isobar corresponding to 100 kilopascal coincide when the phase change takes place. So, these are important points that we will elaborate on in the next slide. Now, if I repeat the experiment, if I repeat the experiment for let us say lower value of pressure, then the curve, the process curve would look something like this. If I repeat it for a lower value of temperature, it would look like this. If I repeat it for higher value of temperature, then the process curve would look something like this. This curve would look something like this, let me redraw the curve. So, the process curve corresponding to the lower value of temperature would look something like this. So, you can see that for a lower value of temperature, you can see that the phase change, the length of the phase change region or the length of the process curve corresponding to phase change is actually longer. And at a higher temperature, you can see that it is becoming shorter. In fact, one can easily visualize that as I keep increasing the temperature, there ought to be a temperature at which this phase change portion becomes almost like 0 length. There is no discernible phase change. So, we go directly from liquid to vapor. So, that isotherm would look something like this and that temperature is called the critical temperature. At the critical temperature, no phase change is discernible. We directly go from the liquid phase to the vapor phase. So, these are important trends that you should remember with respect to the PV diagram. So, let us look at a nice illustration of the same thing. So, here the red curve was the isotherm that we actually discussed in detail that corresponds to 100 degree Celsius. Notice that LM is a liquid state. So, the water was a liquid during process LM, where even small changes in specific volume resulted in large changes in pressure. Mn is the phase change portion of the process curve. When we go from m to n, it is called evaporation. When we go from n to m, it is called condensation. But m denotes either beginning or ending of phase change depending on which direction we are going. And n also beginning or ending of phase change depending on which direction we go. But at state m, we always have a liquid, which is just about to undergo or just about to finish a phase change process. So, at state point m, the water is in a state called saturated liquid. Saturated liquid, the word saturated here denotes that phase change is about to begin or about to end. Similarly, state point n is usually called saturated vapor state. Again, the word saturated denoting that phase change is about to begin or about to end. So, this is saturated liquid, this is saturated vapor. And the states here L to m are usually called compressed liquid or subcooled liquid. We will explain this terminology in a minute. But these are called compressed liquid or subcooled states here to distinguish them from the saturated liquid state. And the state points here along this segment of the isotherm. So, these are superheated states. So, we have compressed or subcooled liquid states. We have saturated liquid. So, we have saturated liquid, we have saturated vapor. In between the two states which lie in between these two are usually called the saturated mixture states. So, it is a mixture of saturated liquid plus saturated vapor. So, it is called the saturated mixture state. So, we have compressed or subcooled liquid, saturated liquid, saturated mixture, saturated vapor and then superheated vapor or superheated states. Now again you can see here that the phase change region is broader for lower temperatures and narrower for the higher temperatures and it is totally absent when the temperature becomes equal to the critical temperature which is 374 degree Celsius for water. Now we can actually connect all the saturated liquid states. So, that line AC is the locus of all saturated liquid states and line AC is called the saturated liquid line. So, it is called the saturated liquid line. Similarly, we can connect all the saturated vapor states and if you do that then you get curve Cb which is called the saturated vapor curve. So, this is a saturated liquid line and I am sorry this is a saturated vapor line. So, the saturated liquid line and the saturated vapor line delineate beginning or ending of phase change. So, to the left of the saturated liquid line or compress liquid states to the right of the saturated vapor line or to the right of the saturated vapor line we have superheated vapors and in between the two. So, this dome shaped region that we have here is called the saturated mixture region. So, all saturated mixture states lie within the dome shaped region. So, we have summarized all this information here. So, saturated liquid state all of them lie on line AC and saturated vapor state all of them lie on line BC. And the saturated liquid line is line AC that we saw before and saturated vapor line is line BC. We have already seen the critical state also. Now, we look at two new concepts saturation pressure and saturation temperature. Notice that as I already said the isotherm for 100 degree Celsius and the isobar for 100 kilopascal coincide inside the or coincide when phase change takes place or inside the two phase region. So, the isotherm corresponding to 100 degree Celsius and the isobar corresponding to 100 kPa coincide inside the two phase region. So, the pressure of 100 kilopascal is usually referred to as the saturation pressure corresponding to 100 degree Celsius. Alternatively, the temperature of 100 degree Celsius is usually referred to as the saturation temperature corresponding to 100 kilopascal. So, we look at the isotherm and isobar which coincide inside the two phase region and the pair has a special name. So, saturation pressure and saturation temperature. So, saturation pressure denoted PSAD is the pressure at which phase change takes place for the given temperature T. So, for the given temperature T, there is an isobar PSAD which coincides with this isotherm inside the two phase region that isobar the pressure corresponding to that isobar is called the saturation pressure. Similarly, saturation temperature temperature at which phase change takes place for the given pressure P. So, for the given pressure P, we look at the isotherm which coincides with the isobar corresponding to the given pressure inside the two phase region and that temperature is called the saturation temperature. Now, compressed or sub cooled region lies to the left of the saturated liquid line. The important thing about these states is there for a given pressure and temperature. Notice that pressure and temperature or independent in the sub cooled or compressed liquid region or in the superheated region. So, P and T are independent here, P and T are independent here, but P and T are not independent inside the dome region. So, so the given P and T is such that P is greater than the saturation pressure corresponding to the temperature or the temperature given temperature is less than the saturation temperature corresponding to the given pressure which is why it is either called a compressed liquid state because P is greater than P sat. So, it is compressed beyond the saturation pressure. So, it is called a compressed liquid state or since T is less than the corresponding saturation temperature, it is called a sub cooled state. So, it is either called a compressed liquid state or a sub cooled liquid state that is where the name comes from. Superheated region as we already mentioned lies to the right of the saturated vapor line and the dome shaped region encloses all two phase mixture states. So, these are some important concepts that you should remember. Now, what we will try to do is having this diagram in mind, we will try to do two things. Number one, how do we locate a given state in a diagram like this and how do we calculate properties corresponding to the given state that is what we are going to do next. Now, as I said two independent properties are required to fix a state of a simple compressible substance. It can be PV as we saw before or it can also be TV. So, you can see that isobars and a TV diagram look like this. There is also a portion of isobar here which I have not indicated for the sake of clarity. So, isobars actually continue like this. So, you should know PV diagram what an isotherm looks like or how an isotherm looks on the PV diagram, TV diagram and how an isobar looks on the PV on the TV diagram. So, you should be able to use both PV and TV coordinates comfortably and later on we will also be using so-called TS coordinates where S is the specific entropy. So, that is also introduced later on and that is very helpful for illustrating certain types of ideas. So, let us now move on to locating states. So, we are given the pressure 250 kilopascal temperature 200 degree Celsius and we have to locate the state. Since we are given pressure and temperature, we presume that these are independent and so the state is either a compressed or subcooled liquid state or a superheated state. That is what we understand from this. Now, how do we determine where which side the state given state falls in. We need tabulated data for this purpose and or steam tables for this purpose and let us see what this looks like. Here is the property table for water. We also have a similar table for refrigerant R134A. So, as you can see here it is, this is the temperature table. So, we basically have three tables associated with water. One is the temperature table, so-called temperature table. Next one is the so-called pressure table and then we have the superheated table. We do not have separate set of tables for compressed liquid states. The reason will become clear. For this sort of codes compressed liquid tables are not required. Now, table A if you see looks like this. So, this is called the temperature table because the first column here is temperature and it goes in equal increments of temperature. So, you can see that starting from 0.01 degree Celsius, it goes all the way up to the critical temperature which is 374.12. Now, what does it give? So, for each of you take any temperature, for the given temperature, you can see that the second column gives a saturation pressure corresponding to the temperature and we have pairs of columns. This is specific volume. This is specific internal energy, specific enthalpy and specific entropy. Quantities here are denoted with a subscript F or subscript G. In the early days, subscript F referred to fluid and subscript G referred to gas. We now understand subscript F to be saturated liquid and G to be saturated vapor. So, VF here is a specific volume corresponding to the saturated liquid at that temperature and VG is a specific volume corresponding to the saturated vapor at that temperature. That is 10 degrees Celsius is what we have highlighted and similarly for specific internal energy specific enthalpy and so on. Now, if I go to pressure table, it gives the same information but instead of having equal increments of temperature, we now have pressure in the first column and this goes in equal increments of pressure more or less. Now, the second column, so if I take any pressure, the second column here gives the saturation temperature corresponding to that pressure and again, specific volume of saturated liquid, specific volume of saturated vapor and so on. So, both is present the same data. One gives it in increments of uniform increments of temperature, other one gives in uniform increments of pressure so that the data in the two tables interleave like this and give you a comprehensive coverage of the entire two phase region. Now, the given value if you recall, we were asked to look at the state corresponding to 250 kPa and 200 degrees Celsius. So, let us first use the pressure table and then see what we get. So, from the pressure table corresponding to 250 kPa, which is 2.5 bars. So, 2.5 bars. So, we go into the table at 2.5 bar, the saturation temperature is 127.4 degrees Celsius and that is what we have written here. So, saturation temperature corresponding to 200 kPa is 127.4 degrees Celsius. Now, how do we use this information to locate the state? This is what we do. So, let us do it. So, let me just sketch it roughly here and then we will look at this diagram. So, let us say that this is your PV diagram and let us say that the given pressure 250 kilo Pascal, the isobar corresponding to 250 kilo Pascal is like this. So, what we try to do now is try to get the isotherm which coincides with this isobar inside the two phase region that is the saturation temperature corresponding to 250 kPa. We have located from the steam tables that the saturation temperature corresponding to 250 kPa is 127.4. So, let us now draw the isotherm. So, this is 127.4 degrees Celsius. What is that? It coincides with the isobar inside the two phase region. Now, the given temperature is 200 degrees Celsius. So, we know that the isotherm corresponding to 200 degrees Celsius will be above this isotherm. So, let us draw that qualitatively. We have already seen that the, so this is 200 degrees C. So, the given state 250 kilo Pascal and 200 degrees Celsius lies at the point of intersection of these two here. So, that means that state lies in the superheated region. That is what is illustrated in this diagram here. So, we drew the isobar corresponding to 250 kPa. Then we identified the isotherm which coincides with this isobar in the two phase region that is the saturation temperature corresponding to 250 kPa. We retrieved it to be 127.4. So, now we know that the isotherm corresponding to 200 degrees Celsius lies above this. So, the point of intersection of 250 kPa, 200 degrees Celsius is state A and that is superheated. That is one way of locating the state. The alternative way is to use the temperature table. Let us see how that is done. We have given that P is 250 and kPa and T is 200 degrees Celsius. Let us now go to the temperature table. So, corresponding to 200 degrees Celsius, we go inside the table with the 200 degrees Celsius. The saturation pressure is 15.55 bar. So, let us now use this information to locate the state. So, from the temperature table, we retrieve the saturation pressure to be 15.55 kPa. So, how do we use this information now? We do the following. So, let us say that this is our PVE coordinate. So, we have the isotherm at 200 degrees Celsius. Let us say this is our isotherm at 200 degrees Celsius. So, the isobar that coincides with this isotherm is the saturation pressure corresponding to 200 degrees Celsius that is 15.55 that is 15.55 kPa. So, the given pressure 250 kPa is less than this value. So, that means isobar corresponding to 250 kPa will look like this. So, the point of intersection of 250 kPa at 200 degrees Celsius lies over here and that is in the superheated region and that is shown in this diagram. So, you can see that. So, this is the isotherm corresponding to 200 degrees Celsius. The saturation pressure is 15.55. This is the isobar corresponding to 250. Remember, this sketch is not to scale. It is qualitative and the point of intersection of this isobar 250 kPa and the isotherm 200 degrees Celsius lies in the superheated state. Now, let us move on to the next example. So, here we are given the pressure to be and we have been given the pressure to be 1000 kPa and the temperature to be 140 degrees Celsius. So, now let us go to the pressure table 1000 kPa and 140 degrees Celsius 1000 kilopascal corresponds to 10 bar. So, we go to the pressure table 10 bar and the saturation temperature corresponding to 10 bar is 179.9 degrees Celsius. So, let us go through this process one more time Pv 1000 kilopascal. So, this is the isobar corresponding to 1000 kilopascal. We just retrieved the saturation temperature corresponding to 1000 kilopascal to be 179.9. So, that is the isotherm that coincides with this isobar inside the two phase region. The given temperature is 140 degrees Celsius. So, the isotherm corresponding to 140 will be below this isotherm. So, the isotherm corresponding to this 140 degrees Celsius will be like this. So, this is 140 degrees Celsius. So, the point of intersection. So, the point of intersection of 100 kilopascal and 140 degrees Celsius lies here which is in the subcooled or compressed liquid state. So, that is illustrated in this diagram here. So, we can see that here. So, this is 1000 kilopascal. This is the isobar corresponding to 1000 kilopascal. This is the isotherm corresponding to the saturation temperature for 1000 kilopascal 179.9. The given isotherm lies below this. So, it looks like this and the point of intersection of 140 degrees Celsius and 1000 lies in the compressed liquid or subcooled liquid region. We can repeat this from the temperature table. So, if you go to the temperature table, we can retrieve the saturation pressure corresponding to 140 degrees Celsius as 361.2. Let us just quickly take a look at that. So, let us go to the temperature tables. So, 140 degrees Celsius is here and the saturation pressure is 361.2 kilopascal, 361.2 kilopascal. So, you can see that. So, this is the isotherm corresponding to 140 degrees Celsius and the saturation pressure is 361.2. The given pressure 1000 kilopascal, the isobar corresponding to that lies above 361.2 looks like this. So, the point of intersection of the given isotherm 140 degrees Celsius and the isobar 1000 kilopascal lies in the compressed liquid region. So, this is how we use the steam tables to locate the state. Whether the state is in the superheated region or compressed liquid region or in the mixed region has to be determined using the steam table. Once we have determined that, property evaluation comes next. The table has to be used even for locating the state. But the clue is if you are given P and T, then you may presume that P and T are independent, which means that the given state will lie in the either in the compressed or subcooled liquid region or in the superheated region. That is important thing. So, here we have been given P and T. So, we know to begin with that it must be either here or here. The state must lie in one of these two places.