 Hello and welcome to another video of understanding thermodynamics. In this video we will discuss open systems, sometimes also referred as control volumes. But first let's take a look at the bigger picture. So we started with properties. Properties are used to describe the characteristics of substances and initially we only discussed pressure, temperature and specific volume. But later saw that internal energy is also a property. Then we encountered the concept of conservation of energy which states that energy cannot be created or destroyed. This concept is formalized in the first law of thermodynamics. The forms of energy we considered were heat, boundary work and internal energy. Also is the center of gravity of the control mass shown here by the little red dot. Moves during the process we may also need to take kinetic and potential energy of the control mass into consideration. And lastly we applied the first law to closed systems. In a closed system there are no mass transfer across the system boundary and therefore the name control mass. Now in this video we will begin the discussion of open systems or control volumes. In an open system mass transfer across the system boundary takes place and here you can see a photograph of a jet engine on a test stand. Air and fuel flow into the jet engine and the combustion products flow out of the engine at high velocity. So the contents of this video then includes an introduction to the concept of open systems. We will explain the concept of flow energy and enthalpy and then we will explain how to calculate the value of enthalpy for ideal and perfect gases. When we inflate the tire of a bicycle we need to pressurize the air in the pump in order to force it to flow through the connecting tube into the tire. We perform work when pressurizing the air. If we assume the pressure inside the pump stays constant the work necessary to force the air into the tube is equal to the pressure inside the pump multiplied by the volume of air that flows into the tire. Work is a form of energy and this energy is called flow energy. Therefore the energy required to cause one kilogram of fluid to flow is equal to P multiplied by the specific volume of the fluid and is called flow energy. It is convenient to combine these two types of energy and internal energy and flow energy into one variable enthalpy. Enthalpy is an intensive variable. Let us see how the enthalpy of an ideal gas is calculated. So the equation used to calculate enthalpy is given by the equation shown. For an ideal gas the term in square brackets reduces to zero and the enthalpy of an ideal gas is only dependent on temperature. Let me show you why. For an ideal gas specific volume equals RT divided by P. Therefore the differential dV dt equals R divided by P and T multiplied by dV dt equals TR divided by P which is also equal to the specific volume and that's specific volume, subtract specific volume equals zero. From this we then obtain an equation that enabled us to calculate a value for enthalpy relative to a reference value at a reference temperature. In general the specific heat of an ideal gas is not a simple equation and the integration is not trivial. We do not want to perform the integration every time we need a value of enthalpy. So therefore the integration is performed for common gases under the assumption of ideal gas behavior and the values are tabulated. Now let's consider air. The value of enthalpy of an ideal gas such as air as a function of temperature is found in the tables. As in the case of internal energy the enthalpy of air is a property and if we know the value of two independent properties we can calculate the value of the other unknown properties. It is important to note that in the case of an ideal gas the values of temperature, internal energy and enthalpy are not independent and the value of only one can be specified independently. Let us do an example. The enthalpy of air at 200 kPa is 241.1 kJ per kilogram. Now determine the values of temperature and specific volume. We can solve this problem by using the table. We see that the enthalpy value supplied to us is between two of the tabulated values and we will have to interpolate to find the temperature of the air. See if you find the same value as me. We can then use the ideal gas law to determine the specific volume. Note that temperature, enthalpy and internal energy are not independent of each other. Once you specify the value of one the others are fixed. Now let's consider a special ideal gas. If the specific heat of an ideal gas does not depend on temperature we call such a gas a perfect gas. We often assume that helium and other mono atomic gases are perfect gases. For a perfect gas it is easy to perform the integration of the equation for the calculation of enthalpy. The specific heat can be taken out from under the integral and the following equation results. Under the perfect gas assumption it is important to note that we now can only calculate the change in enthalpy. If we were asked to calculate the value of enthalpy for helium at 300 Kelvin and 200 kilopascals we would not have been able to do it. Now you need to watch out. A common mistake is to assume perfect gas behavior for steam and use the value of Cp for steam found in tables to calculate the change and even enthalpy itself for steam. Now take note of the following. First, the value of Cp in the tables is under the assumption of ideal gas behavior. Only under very specific conditions which is low pressure and high temperature the assumption of ideal gas behavior is valid. Secondly, in general water can undergo a phase change and ideal gas does not change phase. The specific heat of liquid water differs to that of steam and Cp does not take latent heat into consideration. Thirdly, the tabulated value of Cp is usually for a specific temperature. The specific heat for steam is not independent of temperature. So if you have not checked that the ideal gas assumption is valid under all circumstances for your situation and that the temperature does not vary too much do not use Cp to calculate the enthalpy of steam. And at no stage in this video playlist will we be assuming ideal gas behavior for steam. Okay, to summarize. Energy is necessary to enable a fluid flow. We call this flow energy. Pressure multiplied by specific volume. Internal energy and flow energy are usually combined to form a property we will call enthalpy. Enthalpy is a property and we now have five properties pressure, temperature, specific volume, internal energy and enthalpy. In general this means that if we know the value of enthalpy and the value of another independent variable that the state is fixed and we can calculate the values of all the other properties. Enthalpy of an ideal gas is only a function of temperature and is not affected by pressure. This means that temperature, internal energy and enthalpy are not independent variables. The values of enthalpy of gases under the assumption of ideal gas behavior relative to a certain reference point can be found in tables. If we assume perfect gas behavior it is possible to easily calculate the change in enthalpy. As a perfect gas is a special type of ideal gas the enthalpy of a perfect gas is also not affected by pressure. Thank you very much for watching. The course notes which this video is based on is available on my website audienceblog.com. I'm also available on Twitter. If you've got any questions, please do not hesitate to ask them. And as always, see you in the next video. Bye.