 Hello and welcome to another Understanding Thermodynamics video. In the previous video we introduced the concept of flow energy and noted that flow energy and internal energy is combined to reflect the energy of a flowing fluid. We also discussed enthalpy and ideal gases and in this video we are focusing on real gases, liquids and solids. So we will cover the calculation of enthalpy for real gases two-phase vapor liquid equilibrium systems and then sub-cooled liquids and solids. The entropy of steam as a function of pressure has this familiar dome shape that we're getting used to. This line is the saturated vapor line, this line over here is the saturated liquid line. The area inside the dome is the two-phase region. This area is the superheated vapor and gas and over here we have the sub-cooled and compressed liquid region. Let's do a few examples. Let's consider steam at 100 kilopascals with an enthalpy of 2000 kilojoules per kilogram. We need to determine the outstanding properties. We first need to determine the phase. Is it superheated, two-phase or a compressed liquid? When we plot the state of our system on the pressure enthalpy diagram as the one you've seen in the previous slide, it is clear that it lies inside the dome and is therefore a two-phase mixture. We can also determine the phase by looking at the tabulated values on the saturated water pressure table and at 100 kilopascals, the enthalpy of saturated liquid is 417.51 kilojoules per kilogram and the enthalpy of saturated vapor is 2,675 kilojoules per kilogram. The enthalpy of our system is between these two values and therefore is a mixture of vapor and liquid. For a two-phase mixture, the temperature is the saturation temperature at 100 kilopascals, which is 99.61 degree Celsius. In order to determine the value of specific volume, we need the value of x also known as the quality. Let's see how this is done. So the enthalpy of a two-phase mixture is given by the following equation. From this, we can determine the quality of our system using the known values and we get a quality of 0.701. We can use a similar equation that we just used for enthalpy to calculate specific volume. The specific volume values for the saturated vapor and liquid can be read from the table in the previous slide as well. See if you get the same answer as me. Let's do another example. Consider steam at 150 degrees Celsius with an enthalpy 2,780.2 kilojoules per kilogram. Now we need to determine the outstanding properties. First, we need to determine the phase. We can find the enthalpy of saturated liquid water at 100 kilopascals in the saturated liquid water temperature table. At 150 degrees Celsius, the enthalpy of our system is larger than that of the saturated vapor. And we can therefore conclude that our system is a superheated vapor and we need to consult the superheated tables and find the pressure there. You will see that at 10 kilopascals and 150 degrees, the enthalpy of steam is 2,783 kilojoules per kilogram, which is too large. We will also see that at 50 kilopascals and 150 degrees, our enthalpy is exactly the same as that shown on the table. And we can say that the pressure is thus 50 kilopascals. In general, it may be necessary to interpolate between these different pressure values to find the value that you're looking for. Let's do another example. Consider steam at 1,000 kilopascals and 50 degrees. Determine the outstanding properties. Now first again, we need to determine the phase. On the saturated water pressure table shown, we can see that the saturation temperature of water at 1,000 kilopascals is 179.88 degrees Celsius. From the saturated water temperature table, which is not shown here, we can find the saturation pressure of water at 50 degrees Celsius is 12.35 kilopascals. This is also evident in the pressure temperature phase diagram where we can see we are in the compressed or sub-cooled liquid region. For liquid water, we assume the pressure does not affect the enthalpy of water and that the enthalpy of liquid water at temperature T and pressure P is equal to the enthalpy of saturated liquid at temperature T and in this case, that's 50 degrees Celsius. And we can read the value of enthalpy from the table. Also note that the values of internal energy and enthalpy is almost the same. This is because the product of pressure and specific volume is very small due to the small value of specific volume. And we usually assume that for liquids, enthalpy and internal energy have the same magnitude. The implication of this assumption will become clear in the next slide. When working with sub-cooled liquids for instance to massive water and solids, it is often convenient and sufficiently accurate to assume that delta U equals delta H equals Cp delta T. The values for solids and liquids can be found in tables such as the ones shown here. And for liquid water, we use a value of 4.18 for the specific heat. So in summary, the diagram of enthalpy as a function of pressure or temperature has that familiar dome shape. We can draw a free hand sketch of this dome or use the tables to determine the phase. And for a sub-cooled liquid, we assume that enthalpy at a certain temperature and pressure equals the saturated liquid enthalpy at that given temperature. For solids and liquids, the value of specific volume is small and we usually assume that numerical values of enthalpy and internal energy are the same. The course notes which these videos are based on is available on my website, adreonsblog.com. I'm also on Twitter. If you've got any questions, you're more than welcome to post them there and I'll answer them. Thank you very much for watching and I will see you in the next video. Bye.