Welcome back. Alright. So we already know the units per pressure and of course there's lots of choice there. Atmospherics, kilo pascal, whatever your favorites are. And of course we just reviewed the units for temperature as being kelvin, Celsius, rankine and Fahrenheit, but we're mostly going to work in kelvin because is does us, that's going to be the most accurate way for us to do our analysis. Now, in terms of density, we know that, that's a mass per unit volume. So we expect the units to be kilogram for SI system, kilogram per meter cubed. And of course, the specific volume being the inverse of the density should be those units, just reversed. So it should be meters cubed per ki-, per kilogram, again, for the SI example. Now, in terms of, we know that all of those properties are intensive properties. And specifically, you can look at the units of density. Which, again. Are now kilograms per meters cubed, or if you prefer, look at the units of specific volume. Their volume normalized by mass. So, every time we see that unit normalized in, with mass in the denominator, we expect that those are the units associated with an intensive property. Okay, that's going to become more and more important. When we, as we move forward trying to characterize the thermodynamic state of a system. Okay, so let's start taking. Let's start expanding our tools. Energy in closed systems. So first of all. I want to look at what are the different forms of energy. And I want to only consider a closed system. So one where the mass is fixed. So let's start there. Okay, so what do we know? It's a fixed system, so the mass is a constant. You know the system is capable of kinetic energy or having kinetic energy, so we have, [SOUND] the classic expression for kinetic energy, one half the mass times the velocity squared. And I just wanted to let you know that. Thermodynamics has lots of different Vs, we've already seen one, the specific volume, which is a lower case v. In this class, the nice thing about thermodynamics is that we don't have to retain spatial fidelity, so we can assume all of our systems are essentially one dimensional. So we don't really need vector notation in this class. But I just want to emphasize that this is a velocity. So I'm going to leave this vector notation to emphasize that this is a velocity. Even though, again, we're going to assume that everything is one dimensional. So, here's the velocity and of course it has units again in SI, of meters per second or feet per second if you were in British. So if this is the kinetic energy and again this is mass, we know that this has units of kilograms and we know that the product of the mass times the units of velocity squared is going to give us joules and units of joules in the case of SI units. Okay. If we look at the potential energy, the potential energy is given by the product of the mass times the gravitational constant times Z, the height of this system. So Z is equal to the height. And that has, again, units of meter if we're looking at SI units and g is gravitational constant. Okay. So our system can possess kinetic energy. It can possess potential energy. My closed system is also capable of undergoing a work transfer process. So, we talked about heat transfer. Heat transfer is an energy transfer process. Whenever we have a temperature difference, we have to have heat transfer. Work transfer is another energy transfer process also across a system boundary, so is heat transfer. Heat transfer is across a system boundary. Work transfer is not a system property, m'kay, and the work transfer depends on the process path, so I get different amount of work depending on the process path that I take. So depending on the process path, I'm going to get a different amount of work and the way that we express that path dependent behavior is through the mathematical expression for work transfer. So w is what we use to denote work transfer. And again since it's an energy transfer process, we know that it should have units of energy. So kilojoules or joules, things like that, right. And the mathematical expression that we use is an integral of the, some arbitrary force executed in a particular direction, so applied in a particular direction. And that force has to be applied over a certain distance. So we have ds here. And we move from state one to state two. So, again, the work transfer is the application of a force through a particular vector. Okay now that's kind of a really generic description of work transfer, and in the systems that we're really interested in. Right, power generation for the stationary sector for the transportation sector, there are really only a few forms of work that matter. So, we'll discuss those in greater detail and we'll move from this kind of generic description into the more specific description of the different types of work. So, the very first thing I want to use as an example is expansion and compression work. Okay, so let's write that down. There are many forms of work transfer. Right? So, we have expansion. [NOISE] Okay. There's expansion and compression work and this is the way the majority of the prime movers operate throughout the world. So, the engines that we use for the transportation sector are primarily reciprocating internal combustion engines, so we move a piston up and down and that process, that motion of expansion and compression is the type of work that drives those systems. Little pun there. In the stationary sector, the majority of the power generation is through turbine through, gas turbines and steam turbines. And those turbines spin shafts, kind of like a pinwheel, and again we'll discuss those in more details when we look at those examples. So, that's a rotational energy, so we have a force again operating in a rotational direction, so we have the work there would be described by a product of the torque and the rotational speed associated with that turbine. So that's shaft work. So we have expansion and compression work, that's the prime mover for the transportation sector, and then we have shaft work, which is the prime form of energy transfer. By work transfer in the stationary power sector. And there are all sorts of other forms of work and we can, because essentially you think about it. Any time you have a force you have potent, that's being applied over a particular distance you have work transfer. So there's work transfer due to surface tension, due to magnetic fields, due to all sorts of different, you know, chemical type of force applications. But the ones that we're going to care about most, the primary examples that you'll see around you are expansion and compression work, and shaft work. Okay. So, we're, I promised you we're going to look at expansion and compression work. And let me set up a little example for you that's kind of the classic example of a piston in a compression process. So here's my piston. It's sitting in a cylinder. So here's the piston. And there's some pressure that's applied whether it's atmosphere pressure or maybe I have higher than atmosphere pressure. On the backside of this piston, but there's some pressure, P, so that's P to denote pressure. We know that that expansion and compression work is given by the differential element, so this is a differential element here of the work, of the pressure times the area. Because that's going to give me the force element there. Times the differential motion DX that this force executes or applies to the piston. So this is the force on the piston. And so if this is the piston at state one. We can imagine and here's my. Sorry, that's supposed to be a dotted line there so here's my piston in state one. We can imagine if this pressure increases on this side of the piston or for example if let's say this gas cools then the pist, then the density of this gas is going to shrink and the Piston is going to compress, is going to move in this direction. So we can imagine a piston at state two. I'll try and draw that in a different color so we can highlight that. [BLANK_AUDIO] So we can see that there's been a volume change in the gas or in the fluid that's within my little chamber here from state one to state two. We can see that the volume at state 1 is larger than the volume at state 2. This is a closed system, right? I see no mass transfer across the system boundary, so this is going to be constant. So this is a closed system. And, if the mass is constant and the volume changes, then we know that the specific volume has to change too, right? Because the specific is simply the absolute volume divided by the mass in the system. So, we have, let's go ahead and remind ourselves of all that good stuff. So, if this is the volume, and then, if this is the specific volume [SOUND]. [BLANK AUDIO} Okay so the, if the lower case v that denotes my specific volume, m is my mass, then we know the relationship between these two Is simply that the specific volume is the absolute volume divided by the mass, like that. Okay? And we can see that the volume at state one is greater than the volume at state two. And we know that because mass is a constant, the specific volume at state one is greater than the specific volume at state two. Okay. Let's keep going. So we come back over here and we say, okay this is a force on the piston. We're going to execute that through some differential distance, Dx. So if we have kind of like, if this is X=0, and if we were to push this differential element, or differential step, that would be Dx, right so that would be sum DX. But we want to actually evaluate this system over the entire process from state 1 to state 2. So we are going to integrate this expression on both sides from state 1 to state 2. And that's going to give us a system of where if we take area times dx. We know that that's going to be essentially the same as the pressure times, not essentially, is identically equal to the area. The area times dx is identically equal to the differential volume dV. So the area here, sorry I should have defined that earlier, this is the area of the piston surface. Okay so the product of the area times it' differential element dx is simply the differential volume. So that would be that little diffrential piece here. And that movement the differential amount volume right here. And so the work from one to two is simply the integral expression of the product of the pressure times its differential volume DV. Now, because the piston moves downward from state one to state two we need to know how the pressure within the system varies as a function of volume. Okay. [SOUND] In order for us to evaluate this integral, we need to understand how the pressure varies as a function of volume. If the differential change in volume is positive. So if the final state is greater than the initial state then we know that we have an expansion process. [SOUND] And that the differential work element is going to be, that value would be greater than zero. If this differential change in volume is negative [SOUND], that's a compression process. [BLANK AUDIO]. And we know that the differential work will be less than zero. Now this example that I've shown here in the figure from one to two. We know that that is a compression process and we know that that, again, the volume shrinks as we. At the final state. The final state is less than initial state in terms of the volume. And so we have a compression process. So we would expect this expression, this work to be negative based on this sign convention here. So we've introduced a lot of different concepts here. The work transfer as a form of energy transfer again it's a force. Being applied through a particular distance. Then specifically we've introduced expansion and compression work. We said okay well for expansion and compression work we can set up this generic piston and we know that, that piston applies a certain pressure across an area through that little vector distance. And that this area times this differential distance is going to give me a differential volume and we came up with this general description for expansion and compression work and we talked about the sign convention associated with it. So what I want you do before we start the next unit is I want you to think about for a constant pressure process. What would the pressure look like? What would a diagram of this process look like on a PV diagram? So let me set that up for you. So I want you to sketch for me on a pressure volume diagram what would a constant pressure compression process look like? So just pick some arbitrary state one. Okay and draw for me a compression process to state two using a constant pressure path. And that's where we'll start next time. Thank you.
