[BLANK_AUDIO]. Okay. Welcome back. So this again I hope was pretty straightforward for you. If a turbine generates power, that's power out, and we're considering turbine, a system, the sign is going to be positive for the work transfer generated by the turbine. So we need to remember that. It always helps if we understand the system ahead of time. So if we know we're working with a turbine, or let's say, we're working with a compressor or a pump, we're going to find that those systems require work into the system. So those systems will know, should generate negative work values. And turbine should generate [INAUDIBLE] positive work value. And most start looking at, you know, expansion and compression processes today and look at which should generate positive power, which should generate negative power, or work. Okay. So this is an example. We're going to start putting things together, and we're going to start looking at how important our units are. And we're also going to slip in, because every moment is a teaching moment, an introduction to processes that are called polytropic. So we're going to do an analysis. It's going to be a two-step process, where we're going to look at air in a piston. And we're going to subject it to two steps, two processes. In the first step, we're going to compress the air in a polytropic manner. And that just is a mathematical expression, that says, hey, look, when we compress the air, it's going to obey this expression here. This expression is a polytropic expression, where we take the product of the pressure times the volume, where the volume is raised to the power of n. And that expression is equal to a constant value. We can define that constant whatever we want to call it. We can label it R. We can label it y. Whatever we want to label it. So we an initial state. We're going to use real numbers now, where air is at 100 kPa. So that's about atmospheric conditions. And it's going to be contained in a volume with a specific volume of 0.04 meters cubed per kilogram. The final specific volume at the end of the first process is 0.02 meters cubed per kilogram. And we're told that the power for the polytropic process, n, is equal to 1.3. Okay. So we're going to compress the air from one specific volume to another. We know an initial pressure. We don't know anything else about the final state other than the specific volume. The second step, in the second step, the air is subject to a constant pressure process. This is like the, thought example we did in the last [INAUDIBLE] segment. So if we have a constant pressure process, where the final specific volume is equal to the initial state, we have to determine, is that an expansion process or a compression process? So a lot of this is left unknown, a lot of this example is left unknown because I want you to think your way through it. So with thermodynamics, whenever we consider energy transfer, we're always going to use the [INAUDIBLE], use all the tools we have available. And that includes state diagrams to describe the system. So I already snuck in the state diagram for the PV, pressure volume in our last example. We're going to use that again here, which state [INAUDIBLE], which state diagram we use depends on the system that we're considering. So we've been talking about pressure and volume because we've been looking at expansion and compression work. When we go to look at, let's say, rotating machinery, we're going to look at different expressions, different form, different diagrams, because those are going to be more relevant to show us what's happening in the system. Okay. So first we have to do is sketch the process on a PV diagram. And I'm going to be even more strict here. I want you to sketch the process and label with the actual numbers. The pressures and the volumes of all three states. So we know that means pressure and volume, and states one, two, and three. So that's three pairs of information. In part B, we're going to determine the specific work, which is, again, I'm sneaking in another piece of information here. Specific values are normalized per unit mass. In this case per unit mass of air. So specific work is work for unit mass. Specific e transfer would be e transfer per unit mass. When we start talking about, let's say, kinetic energy or other term, other types of energy, we can also determine those on a per mass basis, and those are called specific terms. They're also called mass density terms. Okay. So, again, introducing a few new concepts in this example. So lets start by doing that first step, which is the PV diagram. We'll answer the first part of this problem. So here's my PV diagram, I have pressure on one access and again learning moments. We are now looking a specific volume, so that's a lowercase v. So that has units of meters cubed per kilorgram. Okay? So we always want to make sure that we understand our units and we label them correctly. And again, we talked about the many, many different units there are for pressure. in this case, we're going to use kilopascals. We could use atmosphere, bars, whatever, but we need to understand what units we're using and stick with them, generally, throughout the problem. Okay. So we have initial state, we're told, is 100 kPa and 0.04. We know we're going to compress the gas, so I'm going to [INAUDIBLE] just take some location here as state 1. And I'm putting it to the right of the axis because I know that when we compress a substance, it's going to move to a smaller volume. And we already have those numbers. We know that we moved from 0.04 meters cubed per kilogram to 0.02, so we can come here and we can say this is 0.04, and we can label some distance here at 0.02. We know the initial pressure is 100 kPa. Okay? What we don't know is what's the final pressure. If you compress a gas, ask yourself what's going to happen to the pressure. Will it increase or decrease? Okay. So hopefully you thought, it's going to increase. So, we know the pressure is going to be somewhere higher on this axis. Question is how much height. You have enough information to determine that. So let's take a look at what we were told. We know that from state 1 to state 2, the gas undergoes a polytropic process, where we have PV raised to the power n, is a constant value. I decided I don't like R. So let's go ahead and use x. That's a good number. because r could be radius, so we're going to use x. X has no special meaning in thermodynamics. Okay. So that's a constant value. What this statement tells us is that the pressure at state 1, times the volume at state 1, raised to 1.3 power is identically equal to the pressure at state 2, the volume at state 2, raised to the n power. We are going to assume this is a closed system. It's not a very bold assumption here because we have essentially no other information. So we're going to assume that this is a closed system. Why is that important? It's important because it tells us that the mass in this system is a constant. Which tells me that this expression, which is in terms of the absolute volume, is equally valid for the specific volume. So take, for example, each side of this expression and divide it by the mass raised to the n power. [SOUND] [INAUDIBLE]. [SOUND] Okay. Well this term is simply the specific volume. [SOUND] Right. So, that's important because our problems statement gives us information in specific volumes not in absolute volumes. So we look at this expression. We say, hey, look. I know state one's pressure and specific volume. I know the specific volume at state 2. Plug and chug through this problem and you can determine what this, what the pressure is at the end state for [INAUDIBLE], the end state of step 1 or the process, the process from 1 to 2. So, that pressure, if you go ahead and plug in the numbers here, will be 246 kilopascal. And make sure, you know, you all remember how to do this type of solution but it should be pretty straightforward. So what we know now is that the volume at state two is 0.02, the specific volume is 0.02 and the pressure is 246 kPa and so I can now label state 2. A polytropic process looks sort of like it shouldn't be double-valued down here. So, just expand my doc here. looks like a curve. and probably can't see the level of curvature unless you have a large space that you're covering. So this is a Polytropic process. [SOUND]. Okay. And I can label it equally either absolute volume or specific volume. Cool, we've got from step one to step two. Now we need to get from step two to step three. What do we know? Well, we know that the air is subject to a constant pressure process, so we know it's going to be somewhere along a constant pressure line. And what we need to decide is will my line move to the right or will my line move to the left to get to state three? Well we're told the final specific volume is equal to the initial state, and that's state 1. [SOUND] So the volume at state 3 is equal to the specific volume at state 1. Okay. There we go, right? We move across. And that's the process diagram. Two, two processes. But this is the complete process diagram from one to two and two to three. So, hopefully that was pretty straight forward. And you can see how you're going to have to be very careful about interpreting the information that, that's given to you. Now, in thermodynamics classes, we, we do a lot of pre-chewing your food for you. So we give you the information that you need. In the real world, often you'll have either an underdefined or an overdefined system. So that's the hardest part of these problems is understanding what information to use, and what information can you neglect or ignore in your analysis. So again in this [INAUDIBLE] in this class, we'll really set things up, so that you can really focus on the analysis. But understand, just writing the problem statement is quite a challenge in the real world, often. Okay. Having said that, now we have our process diagram. Let's go ahead and determine the specific work for each step in the overall process. So I'm going to move on to the next screen here, so we have enough space to work with. And all we need to do is remember that, okay, this is expansion and compression process. We already saw that there's a compression step, that's one to two, and there's an expansion step, that's two to three. We need to treat each of these steps individually, and then we can add them together. To determine the work transfer for this process what we need to remember is that there are two steps. And what we can't do is bundle the steps together. Now we know that we have expansion and compression work. So that's what we're trying to find. And remember, we're asked to find the specific work. So the work normalized by the massive air in the systems. So we'll denote that as a lower case w. And that's equal to the expansion and compression work. And again, if we were to just calculate the work from one to three, that would be incorrect. This is a two-step process. Work transfer is path dependent. So in reality, the total work is given by the sum of the contribution from the two steps. So from one to two, and from two to three. We sum the expansion work transfer. Expansion and compression work transfer contributions. Okay. So let's look at each of these steps one at a time. So we'll start with the work transfer from one to two. And remember, we need to understand how the pressure varies as a function of specific volume. And in that first part of that process, we're told that we have a polytropic compression process. And we know that that means that the pressure and the volume vary using this proportionality. Pressure times volume to the n equals a constant. So if we solve for pressure we know that, we have some constant divided by volume to the n. We can take this expression and plug it into the first, the work transferred term for the first part of this two step process. So we do that here, so we have a constant volume to the n and that n is known differential of a specific volume. Okay. So let's go ahead and plug in the value for n, which is 1.3. Go ahead and do that integration and what we end up with is an expression that looks like the, let's see, constant. We have a negative sign upfront, because remember, our work our sign convention for work transfer. And we know that the compression process, compression is work in. Okay. 1 divided by volume to the 0.3 power and we're going to evaluate that from, again, state 1 to state 2. And if we go ahead and say hey this constant can be determined by using either the information at state 1 or state 2, whichever one is fully defined. And in this problem of course P1 and the pressure in the specific volume at state 1 are fully defined. So we can go ahead and substitute that in here to the 1.3 power, divided by 0.3. And we have 1 divided by [SOUND] the specific volume raised to the 0.3 power at state 2 minus [SOUND] this parallel expression for the state 1 condition. And then all of these values are known. We plug in our numbers and we get. an answer for the work from one to two, the specific work transfer of minus 3.08 kilojoules per kilogram. And again this is less than zero which confirms our, confirms our understanding that work in is negative. That's our sign, our sign convention. [SOUND]. Super. So now we go onto step two to three. And that's, of course, much easier because, again, we need to understand how the pressure varies as a function of specific volume. But that was a constant pressure process, so we can just pull the pressure outside the integral and we get [SOUND] nice simple expression for those specific work transfer for that second part of that process. And again we know all of these volumes, we know the pressure. Again this is the pressure we can either use the pressure at state 2 or at state 3. We'll go ahead and call that P2. plug in those numbers and we get 4.92 kilojoules per kilogram for the specific work transfer from two to three. This number is positive, it's an expansion process. So, again, this confirms our sign convention, which says, expansion is work out. Expansion, work out. And so that is a positive value. Now we're asked to find the total work for both steps, so all we do is add these two together. And that 4.92 plus the minus point, 3.08. And what we find is that the net work for these two step process, two, two processes and sequence is positive. So we have a net work out of the system and kilojoules per kilogram units. So net work out. So we've learned a lot in this process and some of it implicit and some of it explicit. We've used our sign convention, that's great. We see, we've used our definition of expansion and compression work. Okay. So we've seen some actual numbers here. We saw that the system has some intrinsic signs associated with it, whether or not it's work in for a compression step or if it's work out for an expansion step. And we've also seen that you cannot evaluate this system, you can't answer the question which is the overall work for the process, by just going from state 1 to 3. So you've done this example for a polytropic compression process, now what I want you to do. s draw on a PV diagram, what would a polytropic expansion process look like?
