 Hello. Welcome to another video in the understanding thermodynamic series. My name is Adrian and we are busy with internal energy. In a previous video, we considered the internal energy of gases and today we will look at the internal energy of two phase mixtures. In this video, we will quickly discuss the concept of internal energy again. We will see how we fix the state from the supplied variables given in the problems and then we will determine the unknown properties that's left. So in the previous video, we have said that internal energy is the sum of the total of all microscopic forms of energy. One of the microscopic forms of energy is the kinetic energies in the individual particles. The kinetic energies of the particles in a solid is lower than the kinetic energies of the particles in a gas. And for an ideal gas, the internal energy only consists of the kinetic energies of the gas particles. It is therefore independent of the distance between gas particles and the internal energy of an ideal gas is only a function of temperature and not of pressure. Now the second component of internal energy is the energy associated with inter particle interactions and there may be energy necessary to bring particles closer together or to increase the distance between particles. Now when such inter particle interactions exist as is the case in a real gas pressure does have an effect on internal energy and internal energy is then dependent on temperature and pressure which is the case with a real gas. Internal energy is an intensive variable and it can be used to fix the state. Now let's consider the determination of the internal energy when the values of temperature and pressure are known. Now for the first question it asks us to determine the internal energy of water at 50 kPa and 100 degrees Celsius. Now we first need to determine the phase before we can continue. As discussed previously in order to determine the phase we draw a freehand sketch of the temperature pressure phase diagram or you can also use the saturation table and from there we can determine that the saturation temperature at 50 kPa is 81.32 degrees Celsius. So in this case we are in the superheated region because 100 degrees Celsius that's been given to us is greater than the saturation temperature of 81.32 degrees Celsius. As such the quality of the steam is also undefined. We can now determine the internal energy from the superheated water tables as was done in a previous video and we get an answer of 2,511.5 kJ per kilogram. Now let's look at a compressed liquid example. The question asks us to determine the internal energy of water at 20 degrees Celsius and 50 kPa. Now if we look at the sketch from our previous question we can see that at 20 degrees Celsius and 50 kPa we are in the compressed liquid region and as such the phase is compressed liquid. Now if we assume the liquid water is incompressible increasing the pressure will not bring the water molecules closer together and therefore for an incompressible substance such as water internal energy is independent of pressure. Therefore we assume that the internal energy at a temperature T and a pressure P is equal to the internal energy of the saturated liquid at the same temperature. Now for our example the internal energy can therefore be read from the saturated water table and at 20 degrees Celsius we get an internal energy of 83.913 kJ per kilogram. Now let's have a look at the internal energy and temperature phase diagram. If we heat a compressed or sub cooled liquid at a constant pressure let's say 5000 kPa the internal energy will increase until we reach the saturated liquid line. Adding more heat will cause water molecules to break free from the inter particle bonds keeping them in the liquid state and as we heat the water more more vapor will form and we move into the two phase region until all the liquid has evaporated and a saturated vapor results. Adding more heat will result in a super heated gas. If the value of the internal energy was greater than that of the saturated vapor we knew it was super heated. Now let's do an example showing this. The question asks us to determine the temperature and specific volume of water at 1000 kPa with an internal energy of 2000 kJ per kilogram. Now if we look at the saturated warped water pressure table for 1000 kPa you will see for the internal energy 2000 lays between the values of 761 and 2582 kJ per kilogram which means that the state is a two phase. The temperature is therefore equal to the saturation temperature at 100 kPa and that is 179.88 degrees Celsius. Now we still need to determine the specific volume and for a two phase mixture we know that we can determine the specific volume with this equation given using the quality of the steam. While we can read the values of the specific volume of the saturated vapor and the saturated liquid from the tables we first need to determine the value of x. Now we know for internal energy for a two phase mixture we can use this equation shown and we can get the saturated values from the tables and we know the internal energy of the mixture and therefore we can substitute the known values in and the only unknown is the quality. We can get a quality then of 0.6800. Now we can use the value of the quality that we've calculated and substitute it back into our first equation for specific volume and we get an answer for specific volume of 0.1325 cubic meters per kilogram. Now let's look at the temperature internal energy phase diagram. Again you will see the compressed liquid region, the two phase region and the superheated vapor design. Now let's do an example. So this example asks us to calculate the pressure internal energy and specific volume of water at 220 degrees Celsius with an x or the quality of 0.1. Now the fact that they've given us x of 0.1 we can immediately say that it is a two phase mixture. Because we have a two phase mixture we can say that the pressure is therefore equal to the saturation pressure at 200 degrees Celsius and can be read from the saturated water table and we can see at 220 degrees Celsius we get a value for the saturation pressure of 2319.16 kilo Pascal. Finally we can use the equations that we've used previously for the two phase mixture to calculate the values for specific volume and internal energy using the saturated values at 220 degrees Celsius. See whether you can get the same answers that I've got shown here on the slide. So in summary the specific internal energy pressure and in specific internal energy temperature phase diagrams have the familiar dome shaped curve and when it comes to two phase mixtures we always need to determine the phase first by either using the relevant freehand sketch or steam tables. And after we've determined the phase we can then go and use the correct steam table either the superheated or the saturated steam tables. And that's it. Of course notes which these videos are based on is available on my website oddreanceblog.com. I'm also on Twitter if you want to connect with me and you've got any questions please do not hesitate to ask them. Thank you very much for watching and I will see you in the next video. Bye.