 Hello and welcome to another video in the Understanding Thermodynamics video series. My name is Adrian and today we will discuss transient open systems. Open systems are also known as control volumes. Before we get to transient open systems, let us take a quick look back to what we've already discussed. In a closed system, while work can be done or performed by control mass and heat can be transferred across the system boundary, there are no mass flow across the system boundary. A typical example of a closed system is a piston-cylinder arrangement seen here in the picture. There is a certain amount of substance inside and heat may be transferred across the system boundary and work done if the volume changes. We are working with an amount of substance and amounts of heat and work and the units are kilogram and kilojoules. The energy content of our control mass is given by internal energy. In an open system, mass flow takes place across the system boundary and in a steady state open system, we say everything is constant like an aircraft engine on a long haul flight. Therefore, there is no accumulation of mass inside the system boundary and the rate of mass flow in is equal to the mass flow rate out and is expressed in kilograms per second. This is also true for energy and the rate of energy transfer is expressed in kilojoules per second or kilowatts. Now the contents for this video, we will explain the concept of transient open systems and apply the first law into example problems. So here is our first example. Consider a pot containing 1 liter of water and the water is initially at 20 degrees Celsius. The water is heated and starts to boil. 5% by mass of the water evaporates before the gas is turned off. The question now is how much heat was released by the gas? First, we need to make some assumptions. We will assume that all the heat released by the gas flame is transferred to the water. The ambient pressure is assumed to be 100 kPa. We also assume that 1 liter of water equals 1 kilogram. We also assume that the steam is saturated and at the same temperature as the boiling water. The temperature of the pot will also increase from 20 degrees Celsius to 100 degrees Celsius but in this example we are going to ignore the heat required to heat up the pot. Now let's first draw a picture. We consider only the water. The system boundary is given by the red dashed lines. Even though the volume of the system will decrease slightly due to the small amount of steam escaping and therefore some boundary work will be done, we assume that the amount of boundary work is negligibly small. You can calculate its value if you are interested in confirming the validity of our assumption. Now steam flows across the system boundary and heat is also transferred across the system boundary as shown in our picture and the state changes from 1 to 2. Now let us start our analysis. Energy conservation requires that the energy in the beginning plus the energy that flows into the system is equal to the energy at the end plus the energy that has left the system. The energy in the beginning is equal to the mass of water multiplied by the internal energy of the liquid water. The only energy that enters the system is the heat Q. Let us now consider the right hand side of the equation. The energy at the end is the mass of liquid water multiplied by the internal energy of the hot water. Energy leaves the system in the form of steam and as it is flowing its energy content is given by enthalpy. M1 is the mass of water at the beginning and is given in the problem statement. We can read the value of U1 from the tables, the internal energy of water at 20 degrees Celsius and we want to determine the value of Q. The value of M2 is given in the problem statement and the value of U2 is that of saturated liquid water at 100 kPa. We can calculate M out using the mass conservation which gives the following equation. M1 equals M2 plus M out which allows us to calculate M out. The value of enthalpy out is that of saturated steam at 100 kPa. Note that the units for energy is kilogram times kilojoule per kilogram which is equal to kilojoules. We can rearrange the equation in the previous slide to calculate Q. Now M2 is equal to 0.95. Using the steam tables we can determine the internal energy of saturated water at 100 kPa which is 417.4 kilojoules per kilogram. M out is 5% of the initial mass evaporated and the enthalpy is that of saturated steam at 100 kPa. The water is initially sub cooled but we assume its internal energy is the same as that of saturated liquid water at the same temperature of 20 degrees Celsius. We can now calculate the value of Q and we get a value of 446.4 kilojoules. See if you get the same value as me. Let's consider a second example. Example 2, Rises follows. An adiabatic piston cylinder setup contains one kilogram of water at 20 degrees Celsius and 100 kPa. Hot air flows through the pipe spiral in the water as shown in the sketch. The warm air that flows to the spiral is at 500 kPa and 100 kPa. At the outlet the temperature of the air is 380 kPa. The air releases heat to the water with the result that the water starts to boil at constant pressure. Eventually all the water is converted into superheated steam at 150 degrees Celsius and 100 kPa. The question asks how much air has flowed through the spiral to achieve this state. First again let's draw a picture. The system boundary again is given by the red dashed lines. The volume of the system increases and therefore boundary work is done pushing away the atmosphere. Air flows in and out of the system but we assume no air accumulates inside the coil. Conservation of energy means that the energy in the beginning plus energy flowing in equals the energy of the system at the end plus the energy out. In mathematical terms the energy in the beginning is equal to the mass of water times the internal energy of water at state 1 and the amount of energy associated with the air that flows across the system boundary into the system is the mass of air multiplied by the enthalpy of air at 500 Kelvin. This value can be read from the ideal gas air table. On the right hand side of the equation the energy in the system at the end of the process is equal to the mass of superheated steam multiplied by its internal energy. The energy that flowed out of the system with the air is equal to the mass of the air multiplied by its enthalpy which again can be read from the ideal gas air table. We will need to determine the boundary work. We can read the values of the specific volume of the water and superheated steam from the tables and calculate the amount of air which comes to 21.39 kilograms. Now we can also approach this problem a bit differently. If we draw our system boundaries differently different equations result but the amount of air will remain the same. See if you get the same value using the two different approaches. So in summary we consider transient open systems. The state of the system changes from state one to state two and the mass flow and heat transfer can take place across the system boundary. The volume of the system can change resulting in boundary work and the conservation of mass means that the mass in the beginning plus the mass flowing into the system must equal the mass in the system at the end of the process plus the mass flowing out of the system. This is also true for energy. Thank you very much for watching. The course notes which these videos are based on is available on my website audrunsblog.com. I'm also on Twitter if you want to connect with me, ask any questions there, you're more than welcome to do so. See you in the next video. Bye!