 Hello my name is Adrian welcome to another video in the understanding thermodynamics video series. Today we are going to look at how we can solve for temperature pressure specific volume and quality for different scenarios. This video will look at typical examples of calculating the value of either pressure, temperature, volume, mass and quality and this will include for scenarios of ideal gases, real gases, saturated two phase liquid vapour systems and compressed liquids. So for the first question we need to calculate the mass of air in a room measuring three by four by two meters and the temperature is 25 degrees and the pressure is 87 kPa. Now there are more than one way to solve this problem but I chose this route as I wanted to demonstrate the use of the two equations that I'm going to use. The first step in solving a problem is to write down the key equation. The key equation usually contains the variable we are asked to determine. So we use the equation which states that the mass of air in the room is the total volume of the room divided by the specific volume of the air. We can calculate the volume of the room as the dimensions are given in the problem statement and the ideal gas law is valid for these conditions and can be used to calculate the specific volume. As the values of the two independent intenser variables are known the state is fixed and we can calculate the values of the required variables. Remember that the value of r the gas constant is different for every substance in this case. You can find this value in tables for ideal gases in textbooks or on the internet. Also remember when using the ideal gas law that you need to convert the temperature from degrees Celsius to Kelvin and we get an answer for specific volume of 0.983 cubic meters per kilogram and we can convert that to mass and we get an answer of 24.4 kilograms of air inside that room. Now before we get started on real gases liquids and the two-phase system problems there are two steps that you always need to do when tackling these problems. The first thing that you need to do is always determine the phase first which will then help you to use the appropriate table. Now in this question we need to determine the specific volume and quality of water at 10 kPa and 68 degrees Celsius. Now as I said we first need to determine the phase in order to know which table to use and because the values of temperature and pressure are given in the question we can use the pressure temperature diagram to determine the phase and you can see at 10 kPa and 68 degrees it falls in the superheated vapor region. If you do not have the graph with you you can always use the saturated water pressure table shown here because at 10 kPa saturation temperature is 45.81 degrees Celsius and the question tells us that the water is at 68 degrees meaning it's higher than the saturation temperature thus it is in the superheated vapor region. Now we can move on to use the superheated steam tables to determine the specific volume. So we will be using the superheated table superheated water and we will be looking at pressure of 10 kPa. So for our case where we want to know the specific volume at 68 degrees Celsius that is not shown on the tables that value is between 50 and 100 degrees which means we need to use interpolation to get the specific density at 68 degrees Celsius. Linear interpolation was introduced in a previous video so if you're unfamiliar with how to use interpolation go back to that video watch it and then come back to this problem. So through interpolation we can get the value of A which is 0.745 and now we can go and calculate the value for specific volume at 68 degrees Celsius and we get an answer of 15.612 cubic meters per kilogram. You are welcome to pause this video and see whether you can follow the logic and get the same value as me. Let's do another question. So now we need to go and determine the temperature and quality of water at 100 kPa with a specific volume of 0.1 cubic meters per kilogram. So first things first we need to determine the phase. Looking at the saturated water pressure table for 100 kPa we see that the specific volume given falls between these two values meaning that we have a two-phase mixture. This can also be shown graphically where at 100 kPa and 0.1 cubic meters per kilogram it falls inside the dome showing visually that it's in fact a two-phase mixture. So because it's two-phase mixture we immediately know that the temperature at 100 kPa is 99.61 degrees Celsius. Now we just need to calculate the quality. So for quality we can use this equation with the unknown being x. We know specific volume at saturated vapor and we know what's the specific volume at saturated liquid. We can get that from the tables and it is shown over here in the bottom. We can substitute the known values in and we can solve for x and we get x as 6.748 times 10 to the power of minus 3. You can pause this video and see if you can follow my steps and get the same answer. Now we need to determine the pressure and quality of water at 100 degrees Celsius with a specific volume of 1.672 cubic meters per kilogram. We can use the steam tables to determine if it is a two-phase mixture or not and if we go to 100 degrees Celsius and we can see that we have at saturated vapor a specific volume of 1.672 which is exactly the same as the one given to us. This means that the mixture is a saturated vapor with quality one. This can also be shown visually and you can see it touches the right side of the dome meaning that it is a saturated vapor with a quality one. So knowing that it is a saturated vapor we know the pressure at 100 degrees Celsius will be 101.42 kPa. Next question. Determine the specific volume and quality of water at 200 kPa and 100 degrees Celsius. Again the first thing you need to do is determine the phase. Now if we use our saturated water temperature table and we go down to 100 degrees Celsius we can see that the saturated pressure at 100 degrees Celsius is 101.42 kPa but the pressure given to us is 200 kPa meaning that this water is in fact subcooled water. This can be shown visually as well where at 100 degrees Celsius and 200 kPa we are in the compressed liquid or subcooled liquid region. Now while the specific volume of liquid water varies with temperature we assume that liquid water is incompressible. This means that the specific volume of liquid water at 200 kPa and 101.42 kPa will be the same as long as the temperature remains the same. Therefore for all compressed liquids we assume that the specific volume of the compressed liquid at temperature T is equal to the specific volume of the saturated liquid at the same temperature. Lastly quality is undefined because it is a subcooled liquid it is not in the two-phase region. In summary when solving for pressure temperature volume and quality you first need to determine the phase. Once you know the phase you can use the appropriate table and quality is only defined for a two-phase mixture. Hope you've enjoyed the video the course notes which these videos are based on is available on my website adreonsblog.com. I am also on Twitter my Twitter handle is at asv90 where you can ask me any questions and I'm more than happy to answer them. Thank you very much for watching and I will see you in the next video. Bye!