 Hello and welcome to lecture 9 of this lecture series on introduction to aerospace propulsion. So, in the last few lectures what we have been discussing was on some of the basic concepts of thermodynamics. And in the last lecture that was lecture 8, we discussed about the first law of thermodynamics applied to closed systems that is those systems where there is no mass interaction between the system and the surroundings. So, in this lecture what we are going to discuss about is first law of thermodynamics applied to open systems that is how do you apply first law of thermodynamics to a system which has mass as well as energy interaction with the surroundings. And so in this lecture what we are going to discuss are basically the following. So, we shall be discussing about first law of thermodynamics as applied to open systems, but before we actually define it for open systems we need to look at what are meant by flow work and the energy associated with a flowing fluid. This is basically applied only for those systems where there is a mass interaction as well between the system and the surroundings. We shall then define what is meant by total energy of a flowing fluid. We shall subsequently define energy transport by mass and then we shall carry out energy analysis of steady flow systems. We shall also define what are steady flow systems and then we shall take up some examples of some steady flow engineering devices, some commonly observed engineering devices and we shall derive the steady flow energy equation for such devices. Now, before we look into the first law of thermodynamics as applied to open systems we need to understand that there are two ways or two methods in which you can approach a particular problem. One is known as the Lagrangian approach and the other is known as the Eulerian approach. And so depending upon how you would like to analyze a particular system you can use either of these approaches. So, in the case of Lagrangian approach or system approach as it is called the usual practice is that you would track a particular particle and then see what are the changes that this particular particle or a fluid element undergoes as it moves within the system boundaries. And whereas in the case of the second approach that is the Eulerian approach or it is also known as the control volume approach. We only look at the system boundaries as a whole. We are not really interested about what is actually happening to certain fluid element or a group of particles. And so we are going to use the Eulerian approach for a majority of discussions that we are going to have during this course. But in some problems we may also look at the Lagrangian approach as it might simplify the analysis in some sense. Now, so this is regarding the different approaches that you can have for analyzing a particular system. Now, besides this a particular flow process could also be steady or it could be unsteady that is the system that you are looking at may have a certain steady flow process that is being undergone by the system or it might have an unsteady flow process. So, we shall look at what we mean by steady and unsteady flow processes. So, as I mentioned we are going to talk about we are going to discuss control volume approach for majority of the discussion. And then there are two types of processes steady and unsteady. Now, in a steady flow process the rates of flow of mass and energy are constant across the system boundary. And some examples of steady flow processes are turbines, compressors, heat exchangers and so on. And we shall take up detailed analysis of some of these systems when we explicitly state the steady flow energy equation or the first law of thermodynamics as applied to these systems little later on in this particular lecture. Now, a process could also be unsteady if the rates of mass energy are not constant across the system boundaries. Some examples are like charging and discharging processes like if you are charging a tank with compressed air or some other compressed gas or you are charging a tank from a pipeline which carries a certain compressed air or certain fluid. It is basically an unsteady flow process because the rates of mass or energy are not constant across the system boundaries. So, we will not really be covering any unsteady flow process in this particular course as it is beyond the scope of the syllabus that we have. But we shall definitely be analyzing the steady flow processes and steady flow process systems in this lecture because majority of the processes that we are interested as aerospace engineers can be approximated to be a steady flow process. And that is the reason why we shall not really go into details of the unsteady flow process in this course. And so, before we take up the first law of thermodynamics for open systems, we need to understand a very fundamental law of nature known as the conservation of mass principle. If you recollect, we had already discussed about the first law of thermodynamics for closed systems wherein we stated that it is basically the conservation of energy principles stated in a different way. Well, to define the first law of thermodynamics for open systems, we also need to understand what is meant by conservation of mass. Basically conservation of mass law states that the total mass of a particular system will be equal to the difference between the total mass entering the system minus the total mass leaving the system. So, that will be equal to the net change in mass within the system. So, this is what basically the conservation of mass principle states that the total mass which enters the system minus the total mass that leaves the system should be equal to the net change in mass within the system. That means that there is no creation or generation of mass within the system boundaries and it is basically the mass contained in the system is the net difference of the mass that is entering the system and the mass leaving the system. So, stated in equation we would have m subscript in minus m subscript out should be equal to delta m C v where C v stands for the control volume. So, the mass in minus mass out is equal to delta m that is net change in mass of the control volume or in rate form rate of mass flow in minus rate of mass flow out is equal to d m C v by d t. And therefore, you can calculate the total mass within the control volume m C v as the integral of rho times d v by rho is density and v is the volume. And in the rate form the rate of change of mass within the control volume is d m C v by d t which is d by d t of rho d v. Now, this is for a control volume. So, basically the conservation of mass states that there is a certain if there is a change in the mass of the system that is attributed to the mass that is coming in and the mass that is going out. Now, to understand the flow process we also need to understand or we should be familiar with the concept of what is known as the flow work or flow energy. Now, if you are looking at a certain flow process that there is a certain mass entering a system and a certain mass leaving the system then it means that there is certain amount of work that is required to push a certain mass into a control volume or out of the control volume. And this mass or this work that is required to either push in the mass or push out the mass of a control volume is known as the flow work or flow energy. And this is the major difference between a closed system and an open system and that makes a lot of difference in the analysis of closed and open systems. That is because of the certain amount of work that is required for pushing in a certain mass of fluid into a system or pushing out a certain mass out of the system. So, there is a certain work required. Now, to define or to understand this better let us consider a certain system which has a certain volume v let us say and the pressure acting on that particular control volume or particular fluid volume is p through an area a. So, we know that the work done for any particular process is equal to force times the distance. And to understand this better let us consider that there is an imaginary piston which is pushing this mass flow into the control volume. And so the piston moves by let us say a distance of L and so this requires a certain force which is let us say f. So, f times L will give you the work required and we also know that force is nothing but the pressure times the area. And since we already know the pressure or we have defined the pressure as p and area is a. So, p times a will give you the force acting on the piston multiplied by L will be equal to work. And so f into a into L will be the work. And since a times L is volume we get the net work done as equal to p times the volume p v. So, basically the flow energy will come out to be the product of the pressure times the volume. This was some way obvious to you as well because in the last lecture or in the two lectures before today's lecture we had discussed about work and heat transfer. And one of the major forms of work was displacement work. So, this is basically displacement work but we are defining it in a different way here. And that is basically known as the flow work or flow energy. So, if you want to look at it in a different way if you are as I mentioned we consider a certain fluid element of volume v. And if the fluid pressure acting is p through a cross sectional area a and L let us say is the distance through which an imaginary piston must move moved. So, the work required or work done in pushing this fluid element across the system boundary would be f times L where f is pressure times the area. And therefore, it is p a times L and therefore, it is equal to p v. So, work done during this process is p v. So, I have illustrated this in a small example here. So, what I was mentioning is about this fluid element which you can see here has a certain volume v pressure acting is p and a mass m. And this is the control volume which has been indicated by dotted lines. And the piston that is shown here is an imaginary piston. This is just to indicate that there is a certain force which has to be acted upon this fluid element to push it into the control volume. So, this is the force acting f it is acting through an area a and as the piston moves by a distance of L this much amount of fluid which has a volume of v pressure p and mass m goes into the control volume. And similarly, there is a certain amount of mass flow leaving the system. So, the work done for this process is this force acting on the imaginary piston multiplied by the distance L through which it was displaced. So, force times L and the force is pressure times area. So, p a into L and a into L is volume area times the length is volume. So, therefore, the work done for this process or the flow energy or the flow work is equal to p times v pressure times the volume. Now, we shall now look at what is the total energy. Now, that you have understood what is flow energy. We will now look at what is the total energy that is associated with a fluid that is flowing that is either let us say entering a system or leaving a control volume. So, it has a certain amount of additional energy. If you recall we had defined total energy for closed systems as the sum of internal energy plus kinetic energy plus the potential energy. Now, in the case of fluid which is flowing there is an additional work or energy associated with it and that is given by the flow work or flow energy which is the product of pressure times the volume. So, the total energy of a fluid which is flowing will be the sum of the internal energy u plus the kinetic energy plus the potential energy and the flow energy that is p v. So, the total energy which we normally represent for flow processes by theta will be equal to u plus k plus p plus p v which is the flow energy. Now, in one of the earlier lectures I had defined a term which is known as a combination property as enthalpy and enthalpy we had defined as the sum of internal energy u and the flow energy p v. So, the total energy is now equal to the sum of the enthalpy, the kinetic energy and the potential energy. So, for a flowing fluid there are the total energy consists of 4 different terms it is the sum of internal energy u, kinetic energy, potential energy and the flow energy. Whereas, for closed system the total energy consists of 3 terms which is just the internal energy, kinetic energy and the potential energy. So, there is no flow energy associated with closed systems and so it makes sense to define this term combination property as enthalpy because you enthalpy actually takes care of the internal energy as well as the flow energy. So, you do not have to worry about these 2 terms separately and enthalpy has already taken into account for the internal energy and the flow energy. So, that is the convenience of defining this combination property of enthalpy. So, the total energy of an open system or a flow process will be the sum of enthalpy which is in turn equal to u plus p v and enthalpy plus kinetic energy plus the potential energy p e. So, the total energy associated with the flowing fluid is normally associated as h plus k e plus p e and so if you were to define it in terms of an equation the total energy of a flowing fluid is e is equal to non flowing fluid rather total energy associated with the non flowing fluid is u plus k e plus p e and which is in turn equal to u plus v square by 2 that is kinetic energy per unit mass plus g time z which is the potential energy per unit mass. For a flowing fluid on the other hand theta is equal to theta is the total energy is equal to h plus k e plus p e and which is in turn equal to u plus p v which is equal to h plus v square by 2 plus g z. So, you can see that the total energy consists of 3 parts for a non flowing fluid and for a flowing fluid the total energy consists of 4 components or 4 terms and so that is the basic difference between a flowing fluid and a non flowing fluid. Now, I mentioned that for a mass for a process which involves a fluid element which is crossing the system boundary or if there is a certain flow that is taking place then we can define or the total energy per unit mass is actually equal to is equal to the mass which is of the flowing fluid multiplied by the energy itself. Therefore, the energy transport associated with the mass will be equal to m times of theta that is mass times the total energy. So, we have already defined total energy for the flowing process which was the sum of h plus kinetic energy plus potential energy. So, the amount of energy transport let us say e subscript mass is equal to mass times theta and where theta is the flow energy equal to h plus v square plus g z and if you were to look at this in the rate form the rate of energy transport e dot mass is equal to m dot times theta where m dot is the mass flow rate. So, the amount of energy transport e mass is in kilo joules and the rate of energy transport will be in kilo watts because you are looking at mass flow rate and so you would get kilo joules per second which is equal to kilo watts. So, the amount of energy transport here will be the product of the mass and the total energy associated with this particular process and the amount of or the rate of energy transport will be equal to the product of the mass flow rate times the total energy. So, you would get m dot multiplied by h plus v square by 2 plus g z where h is the enthalpy for the process v square by 2 is the kinetic energy and g z is the potential energy per unit mass and so this multiplied by the mass flow rate would give you the total energy associated or energy transport associated with that mass in kilo watts and so the amount of energy that a certain fluid flow process carries depends on as we discussed four terms which consists of internal energy which is you the flow work or flow energy which is p v the kinetic energy and the potential energy. So, this is the total energy associated with a flow process this multiplied by the mass will give you the total energy transport associated with this particular mass flow process. So, it is important for us to understand these flow processes because many of the engineering systems that we shall be discussing soon are basically involving flow processes and a certain mass associated with these flow processes and so we need to understand the importance of flow processes and also how do you apply the first law of thermodynamics to different flow processes and we will take up some examples towards the end of this lecture on how you can derive first law of thermodynamics or so the so called energy equation for different steady flow processes like turbines and nozzles etcetera which we will take up towards the end of the lecture. And I mentioned in the beginning of this lecture that the flow process can be of different types and one of them is known as a steady flow process and the other is an unsteady flow process and I mentioned that steady flow processes are those wherein the mass and energy do not change across the system boundaries. So, what is the importance of understanding or analyzing steady flow processes well there is a lot of importance in understanding steady flow processes because several engineering devices or systems can be approximated can be well approximated as steady flow systems some examples are turbine compressors nozzles etcetera. And so during a steady flow process the basic assumption is that no intensive or extensive property within a control volume change with time. So, that the element of time does not really appear in a steady flow process that is if there is a d by d t term in the steady flow equation well in the energy equation that will become 0 because rate of change of any of these properties any intensive property or any extensive property the rate of change of that with time will be equal to 0. And so that is the basic definition of a steady flow process and so as I mentioned there are many engineering devices which we can approximate the many engineering devices which we can approximate as steady flow devices like turbines compressors nozzles etcetera. And during a steady flow process we may discuss that the none of these properties like extensive or an intensive properties were actually change with time. And therefore, the boundary work associated with a steady flow system will be 0 because the volume of the control volume is fixed or constant and so boundary work associated with this process will be 0. And the other property with the steady flow system is that the total mass or energy entering the control volume must be equal to the total mass or energy leaving the control volume. So, there is no accumulation of mass or energy within the control volume if that were the case it would become an unsteady flow process. And I mentioned that we are not going to discuss about unsteady flow process in this lecture we shall be talking only about steady flow processes during this lecture. So, what are the properties of steady flow process well the basic properties of a steady flow process is that or that none of the properties whether they are extensive properties or intensive properties they do not change within the control volume with time. Of course, the properties can change within the control volume itself, but not within there is no rate of change of the control the properties within the control volume with time. And it is also a fact that no properties will actually change at the boundaries of the control volume with time. And that the rate of change of mass or energy with time is or across the control surface is actually a constant if the process is to be approximated as a steady flow process. And the different thermodynamic properties like pressure, temperature etcetera these have fixed values at a particular location and they do not change with time. That is if you look at a system and if it has to be approximated as a steady flow system or a steady flow process then the different thermodynamic properties of the system will have fixed values at different locations within the system and they do not change with time. So, these are some basic properties that need to be satisfied if a particular process has to be approximated or if a process has to qualify as a steady flow device or a steady flow process. And we shall very shortly define the first law of thermodynamics for steady flow devices and then we shall apply the first law for some examples of steady flow engineering devices. So, we need to keep these properties in mind if we are going to approximate a particular process or a system as a steady flow process. So, this is an example of just an illustration of showing what are the properties associated with steady flow systems. For a steady flow system which has certain mass entering and a certain mass leaving the system the net mass of the control volume is a constant. The net energy associated with this control volume is also a constant. Now, this is the first on the left hand side what you see is an example of a single entry and single exit system. You could also have a system which has multiple entries and multiple exits. The basic definition remains the same that the net mass within the control volume if it is constant and the net energy of the control volume if it remains a constant then we can approximate this process as a steady flow process. So, under steady flow conditions the fluid properties at the inlet or exit remain a constant and they do not change with time. So, the element of time in the steady flow energy equation would be the particular term would become equal to 0 because we are assuming that the properties do not change with time. So, let us now look at how we can derive the energy equation for a steady flow process. So, for steady flow system the amount of energy entering a control volume in all its forms that it could be energy transfer by heat, it could be energy transfer by work or it could be energy transfer by virtue of the mass flow itself. So, whatever energy is entering a control volume should be equal to the energy leaving the control volume because we have seen that energy of a control volume whether it is single entry or if it is multi entry the net energy of the system within the control volume is a constant. So, the total energy entering in all its forms heat work or mass because these are the three modes of energy interactions which a system can have with the surroundings. So, the net energy entering in all these forms should be equal to the net energy leaving the system. So, energy balance for a steady flow system would be equal to energy in minus energy out the net is equal to rate of change of energy of a system, but in the case of steady flow processes the rate of change of energy is equal to 0 because it is a steady flow process and therefore, energy in will be equal to energy out that is E in is equal to the rate of change of energy leaving the control volume. Now, the energy in as I mentioned could be in different forms it could be in the form of heat work or mass and, but whatever be the form of energy interaction which the system has with the surroundings the net rate of change of energy transfer by heat mass or work should be equal to rate of change of energy transfer out of the system by heat work or mass. And this is basically the energy balance or the energy equation for a steady flow process. We can write this more explicitly and if you look at the steady flow process the energy equation in expressed in more explicit form is as I mentioned energy transfer can take place in three modes heat work and flow or mass flow and. So, those are the three modes which have been written here. So, q in dot which is rate of input of heat minus w dot in which is rate of work input plus the mass flow input which is equal to sigma in m dot theta and the sigma is for accounting the different multi input flow rates into the system. And since E in is equal to E out this should be equal to q dot out which is rate of heat transfer out of the system minus w out dot which is rate of work out of the system plus or work output of the system plus sigma out of m dot theta. And this third term as we just now discussed is the total energy associated with the flow process. And so that should be equal to q in dot which is heat transfer in minus w in dot which is work input plus sigma in times mass flow rate multiplied by the enthalpy the sum of enthalpy plus kinetic energy plus the potential energy. And this is applicable for each inlet this should be equal to q dot out which is heat transfer out of the system minus w dot out which is work done by a system plus m dot out multiplied by the flow energy which is h plus v square by 2 plus g z. So, this defines the general energy equation for steady flow processes which consists of three terms. Therefore, in a steady flow energy equation you have three distinct terms one is the heat transfer term the work done term and the flow work term. So, q in minus w in plus mass flow rate times the flow work or flow energy should be equal to the same terms leaving out of the system. And if the system consists of multiple entries and multiple exits that will affect the third term that is the flow energy term you have to calculate the flow energy for each of these entries and each of this exits from the system. So, steady flow energy equation in general will consist of these three different terms which have to be applied at the entry as well as the exit. You can now compare this steady flow energy equation with the energy equation which we had discussed in the last lecture which was applicable for closed systems or closed processes. So, you might recall that we did not have this flow energy term in the first law of thermodynamics as applied to closed systems. So, for the first law as applied to closed systems was q minus w is equal to delta e or delta u and so there was no flow energy in a closed system or closed process. So, energy equation we can write the generalized form of energy equation which can be expressed as q dot minus w dot is equal to the some difference between the mass flows entering out of the system while mass flows leaving the system or the flow energy associated with the mass leaving the system minus the energy associated with mass entering the system. And so q dot here would refer to the net heat transfer into the system and w dot would refer to the net work transfer from the system. So, q dot minus w dot is equal to sigma out m dot times the flow energy term that is h plus v square by 2 plus g z minus sigma in for the each inlet m dot times h plus v square by 2 plus g z. And here q dot refers to the net heat input into the system and w dot is the net work output of from the system. This again is an assumption as I mentioned in the last lecture that we normally assume that for a given process there is heat input into the system and work output from the system. And if after our any calculation that you are carrying out these numbers come out to be negative it just means that the net heat was not input, but there was a net heat output from a system or if w comes out to be negative it means that there was net work done on the system. Because we normally have systems which generate work and which require heat input and which is why it is a common practice to assume net heat input to the system and net work output from the system. Now, if you look at this particular equation for a single entry system which is how most of the engineering devices would be that you might for example, a turbine or a compressor or nozzles and diffusers these are all devices which have single entry and single exit. The system becomes well the equation energy equation becomes simpler in the sense that you have only one particular term for the flow work. And so the energy equation gets modified as q minus w is equal to the difference in the enthalpies at the exit and the entry plus the difference in the kinetic energies plus the difference in potential energies at the inlet and exit. So, if you were to write down the steady flow energy equation for single entry device then the equation would look like q minus w is equal to q dot minus w is equal to m dot times h 2 minus h 1 plus v 2 square minus v 1 square by 2 plus g times z 2 minus z 1. So, this h 2 minus h 1 is the net enthalpy v 2 square minus v 1 square by 2 is the net kinetic energy g times z 2 minus z 1 is the net potential energy. So, the same equation per unit mass would be q dot minus w dot is equal to h 2 minus h 1 plus v 2 square minus v 1 square by 2 plus g times z 2 minus z 1. So, this is the general energy equation for steady flow processes in some text books you might see that this equation is referred to as the steady flow energy equation. And so steady flow energy equation for single entry and single exit devices is q dot minus w dot is delta h plus delta k e plus delta p e. We shall now discuss about some common steady flow devices which we see in daily life some of these commonly used steady flow energy devices are nozzles and diffusers or compressors and turbines, throttling devices, mixing chambers, heat exchangers etcetera. So, we shall now derive the steady flow energy equation for some of these engineering devices which are commonly used. And we shall see how the steady flow energy equation the general form of steady flow energy equation can be used by applying appropriate boundary conditions for these devices like nozzles diffusers or compressors turbines or certain mixing chambers throttling devices etcetera. So, the first such device which we shall discuss are nozzles and diffusers. So, a nozzle is a device which increases the velocity of a fluid at the expense of pressure and diffuser on the other hand is a device which increases the pressure of a fluid by slowing it down. So, a nozzle and a diffuser in some sense are devices which are opposite to that of each other in terms of their basic function. And so the cross sectional area of a nozzle decreases in the flow direction for subsonic flows and it increases for supersonic flows. Well you might wonder what are subsonic and supersonic flows at the moment we will just define them as those flows which have a Mach number less than 1 are known as subsonic flows and those flows which have a Mach number greater than 1 are known as supersonic flows. And I think we will define this little later on in the course when we discuss about compressible flows. And so nozzle in a subsonic flow and nozzle has decreasing area in the direction of the flow and increasing area in direction of the flow in supersonic flows. Reverse of this is true for diffuser because I mentioned that nozzles and diffusers are two devices which have opposite functions and so their geometry also in some sense will be opposite to that of each other. So, we will derive the energy equation for or we will simplify the energy basic energy equation for let us say a nozzle. It will be very similar to that for a diffuser as well with appropriate boundary conditions applied. Now, we know that in the case of an energy equation we have three different terms the heat transfer work and the mass on the flow energy term. Some of these terms will become 0 as you apply it for different engineering devices. So, in this particular example we are discussing that is for nozzle and diffuser. I mentioned that a nozzle is a device which has increase in velocity along the flow direction for subsonic flows. So, v 1 let us say is the velocity at the inlet v 2 is the velocity at outlet. So, nozzle increases the velocity at the expense of pressure and therefore v 2 is usually greater than v 1. For a diffuser it is the opposite if v 1 is the inlet velocity v 2 is the exit velocity v 2 will be less than v 1. You can also notice that a nozzle and diffuser have opposite shapes that is a nozzle in the reverse way would look like a diffuser. So, nozzles and diffusers are those devices which can cause large changes in fluid velocities and therefore there is a large change in the kinetic energy of the fluid as they pass through a nozzle and a diffuser. So, the energy equation the general form of energy equation was E in is equal to E out for steady flow systems and in the case of nozzle and diffuser there is no net change in heat transfer. Therefore, q dot can be approximated to be 0 there is no work done by a nozzle or a diffuser. Therefore, w dot is 0 and we can also approximate the change in potential energy to be 0 as long as the nozzle is horizontal or the diffuser is horizontal even if they are not the nozzles and diffusers do not have a long length and in that sense the net change in potential energy across the nozzle or diffuser can also be assumed to be 0. So, if you apply all these boundary conditions on the energy equation applying energy net a heat transfer as 0 work done as 0 and delta p is equal to 0 then the energy equation reduces to m dot into h 1 plus v 1 square by 2 is equal to m dot times h 2 plus v 2 square by 2. Since mass entering in and mass entering leaving the system as the same m dot will get cancelled out. Therefore, the energy equation for a nozzle or a diffuser would be h 2 is equal to h 1 minus v 2 square by minus v 1 square by 2. So, or you could also write it as delta h is equal to the delta k e across the system. So, net change in the enthalpy is equal to in some sense the net change in kinetic energy there is no net change in potential energy heat transfer is 0 work done is also 0. So, this would be the basic energy equation for nozzles and diffusers. Let us now look at the second set of devices turbines and compressors and I think you probably are already aware that pumps compressors and fans are devices which are used to increase the pressure of a fluid and therefore, they require work input turbines on the other hand generate work. And so, pumps and turbines in some sense are like nozzles and diffusers they are opposite in function in some way or the other. And in the case of if you look at the energy equation the heat transfer kinetic energy potential energy may or may not be 0 it depends upon the type of device that you are looking at, but usually it is a practice to assume heat transfer across the system boundary to be 0. If you consider that the turbine boundaries or casing is well insulated or compressor boundaries are insulated then the process can be considered as adiabatic. And so, heat transfer is 0 kinetic energy and potential energy may or may not be 0 kinetic energy is usually not really assumed to be 0, because there is a change in velocities across the system boundary. So, they may or may not be 0 potential energy again depending upon the type of system we may assume it to be 0 or it sometimes is not taken as 0. And we shall in fact when we solve some example problems later on in a later lecture we will see how much is the error that is introduced if you were to neglect kinetic energies and potential energies for such systems. So, if you look at the energy equation for let us say turbine which is what we are going to derive today. In the case of turbine we are going to assume that the changes in kinetic energy and potential energy are close to equal to 0 and so is q equal to 0, because we are going to assume that the process is adiabatic in some sense. So, let us look at a turbine a schematic of a turbine here sketch what you see here is a turbine which has a certain insulation which means that there is no heat transfer across the system boundaries across the control surface. And the turbine generates a network output and there is a certain mass flow entering the system mass flow leaving the system. And so the turbine generates a network across the system boundaries. So, if you look at the energy equation for a turbine that will be equal to since q dot is 0 we get m dot multiplied by h 1 plus v 1 square by 2 plus g z 1 is equal to m dot is equal to work output that is w dot out plus m dot times h 2 plus v 2 square by 2 plus g z 2. Now, if you assume kinetic energy and potential energy to be negligible then the network output is equal to m dot times h 1 minus h 2 that is work output from the turbine is mass flow times the difference in enthalpies. So, difference between the enthalpy times the mass flow will give you what is the work output that this particular turbine is giving. You can modify the same equation the general energy equation for a compressor as well and depending upon whether you neglect kinetic energy and potential energy you can derive a very similar expression of work input required for compressors or pumps or fans. So, the equation will be similar to this, but just that the signs of h 1 and h 2 would be different because the enthalpy leaving a compressor or a fan or a pump would be higher than the enthalpy coming in because there is work done on the system. So, you can modify the equation appropriately for compressors or pumps. The equation for compressor and pump obviously will be the same because they are thermodynamically same devices. The next device that we are going to discuss about is a throttling device. Well a throttling device is one which basically these are devices which cause flow restrictions leading to significant pressure drop in the fluid. Some examples are capillary tubes valves or porous plug etcetera and unlike turbines wherein you there is a pressure drop across a turbine, but also a work output. In the case of throttling devices they produce a pressure drop without involving any work and this pressure drop is often accompanied by a large drop in temperature as well. And this is the reason why throttling devices are very commonly used in refrigeration and air conditioning systems wherein throttling device forms one of the components of refrigeration cycle. So, throttling devices do not generate any work output and they obviously do not have any heat transfer as well and we will derive the energy equation for throttling devices. So, some examples of throttling devices are shown here. One example is a valve as you adjust the valve there is a large change in pressure across the valve and so it is that is considered to be a throttling device. A porous plug is one wherein you have a porous substance within a pipe and as mass flow flows through the porous plug that leads to significant drop in pressure. Capillary tube is another example of a throttling device thermodynamically all the three devices are the same. So, for throttling devices the net heat transfer is 0 can be assumed to be 0 there is no work output therefore, W is 0 kinetic energy and potential energy can again be assumed to be close to 0. Therefore, the energy equation will basically reduce to H 2 is equal to H 1 that is the enthalpy across a throttling device is the same. This means that throttling processes are isenthalpic processes that is these are processes with wherein enthalpy is a constant. Now, since H 2 is equal to H 1 we can also write them in terms of their components that is internal energy and flow energy. Therefore, u 1 plus p 1 v 1 is equal to u 2 plus p 2 v 2 that is the sum of internal energy plus the flow energy is a constant across a throttling device. Now, this means that you can have different values of internal energy at the inlet and exit, but that is compensated by a corresponding change in the flow energy. Let us say you have a drop in internal energy across the throttling device this has to be compensated by an increase in the flow energy term across the throttling device. This is required because the net change in enthalpy across a throttling device will be the same that is H 2 will be equal to H 1 and therefore, the sum total of u 1 plus p 1 should be conserved. And so across a throttling device the change in internal energy plus the p 1 v 1 should be compensated by corresponding changes in internal energy and the flow energy after the throttling device. So, to illustrate it if you look at one case where p 2 v 2 is greater than p 1 v 1 this has to mean that u 2 should be less than u 1. So, that there is H 1 equal to H 2 conserved. Therefore, this also means that if flow energy increases the temperature has to decrease because u is a function of temperature and the reverse is also true that is if flow energy decreases you may also have increase in internal energy and therefore, temperature. Now, for an ideal gas enthalpy is only a function of temperature and therefore, for an ideal gas or for a system which involves only ideal gases the enthalpy well the temperature has to remain a constant during a throttling process because for an ideal gas enthalpy is only a function of temperature. So, for ideal gases enthalpy is only a function of temperature which means that enthalpy has to be a constant and therefore, temperature also has to remain a constant across a throttling device. Just an example here as I mentioned if across this device let us say at the inlet of the device you had u 1 is equal to 87. Some kilo joules p 1 v 1 as something else total enthalpy is 88.56 after the throttling device let us say u 1 has reduced and p 1 v 1 therefore, has to correspondingly increase. So, that the net enthalpy is the same. So, during a throttling process enthalpy of a fluid has to remain constant, but internal energy and flow energies are inter convertible. So, you can convert internal energy into flow energy partially and so on and vice versa across a throttling device. The last device that we are going to discuss is a mixing chamber. Mixing chamber is a section wherein there is a mixing process taking place. One some examples are the mixing of hot and cold water at a T junction of a shower for example. So, here you have two different masses coming in M 1 and M 2 and leaving out as M 3. For energy equation therefore, will reduce to M 1 H 1 plus M 2 H 2 is equal to M 3 H v. Since we assume again that heat transfer is 0 work is 0 changes in kinetic energy and potential energy are 0. So, if we combine the energy and mass balances because M 3 is equal to M 1 plus M 2 energy equation reduces to M 1 dot H 1 plus M 2 dot H 2 is equal to M 1 dot plus M 2 dot times H 3. So, this would be the energy equation for a mixing chamber process that is the net enthalpy leading the system is equal to sum of the enthalpies entering the system. Now, so that brings us to the end of this lecture wherein we discussed about the first law as applied to flow processes. So, let us take a look at what we had discussed during this lecture. We discussed about the first law of thermodynamics applied for open systems or flow processes. We then discussed about flow work and the energy associated with a flowing fluid and then what is the total energy that is available for a flowing fluid. Then as a consequence of mass flow rate what is the energy associated with mass flow process. We then discussed the energy equation for steady flow processes. We derived the energy equation for steady flow processes and also we discussed about application of these energy equation, the basic energy equation for certain steady flow devices like nozzles and diffusers, turbines and compressors, then mixing chambers or heat exchangers and also we discussed about how we can apply energy equation for throttling processes. So, the idea was to help you in understanding what is the process involved in making certain justifiable assumptions and applying first law thermodynamics for different flow processes. So, if you come across a different type of flow processes based on what we had discussed here you could probably be able to extend or simplify the general energy equation to any other open system or any other flow processes. Now, in the next lecture what we are going to discuss about our second law of thermodynamics. We shall define what are known as thermal energy reservoirs. We shall then state the Kelvin-Planck statement of the second law of thermodynamics. We shall subsequently discuss about refrigerators and heat pumps and as a consequence what is the Clausius statement of thermodynamics second law of thermodynamics. We shall then prove that the Kelvin-Planck statement of second law and the Clausius statement of second law are equivalent and towards the end of the next lecture we shall again discuss about certain devices which violate the second law of thermodynamics which are known as perpetual motion machines of the second kind. We had already discussed this for the first law which were known as perpetual motion machines of the first kind. Those devices which violate second law are known as perpetual motion machines of the second kind. So, that is something we shall discuss during the next lecture.