Welcome back. So, we're going to follow up on our example. We were considering a steam turbine, where we were given the entrance conditions of 69 bar and a temperature of 538 degrees Celsius. And we had the speed of the water as it entered the turbine, and remember, last time we found the exit state condition. Specifically, we needed the specific volume, and we used that to determine the mass flow rate through the system. And we also determined the velocity of the, of the water vapor as it leaves the turbine. So the question you are left with is to sketch the process on a P-v diagram. So that's what we're going to do now. So, what was important for us to recognize is that we have all the information we need to actually put numbers on a chart. So, let's recall, we're going to put our vapor dome on our Pv diagram. And I've exag-, I'm making them a little shorter, because a couple of things that we learned in this exercise was that this initial entrance state, p1, t1 was actually in the superheat region. So this was a superheat, which is up here. Right? Its at the high pressure, high temperature conditions and remember we always want to put those isotherms, those constant temperature lines on all of our diagrams here. So I'll put a couple of those on here. And we learned that the exit state that we had pressure, lower pressure and we had a saturated vapor and so we know that we can just, the easier line for us to or the easier point for us to put on the diagram is actually this saturated point. So lets take this lower temperature condition here And we will call this state two. So state one has to be at a higher pressure. So it has to be somewhere north of that state. It's in the super heat region so we don't want to move into the dome. And in fact we can say it's actually going to be somewhere up here. Maybe not to scale, but this is going to be our state one. And again the key features, is that it has to show it's in the superheat region, and we have to show that the, this specific volume is lower at the entrance state than at the exit state. So we need to make sure that there's some movement in this direction and movement in this direction relative to the final state. And then we can connect those with a line and show that's the process direction for our P-v diagram. Okay, but you had all the tools available. And remember, we used our online calculator in order to determine those specific volumes. And we, the input data are shown here. Again, pressure and temperature allows us to divine every other thermodynamic property when we're in the superheat region. And when we're in the saturation region, recall we need to have two independent properties, so they can be pressure and quality, pressure and temperature, whatever they (no period) Whatever the right combination is, it can't be pressure and temperature. Because as we see on our figure here for our two isotherms, remember T high and T low, temperature and pressure in the dome are not independent. Okay, so, [COUGH] I wanted to follow that problem up with a discussion about scaling. Because we'll often find in engineering applications, thermodynamics, and others as well, that we can learn a lot about a problem by not using the actual numbers, but looking at the relationships between the key parameters in variable form. So, last time we were looking at, we, we determined what the exit velocity was for the turbine and what we used were was the conservation of mass, which we said was a product of the density, the area and the velocity or the speed of the fluid entering it or leaving the turbine. And we were working in specific volume space, so we could write the equation that way. so we know that this is a constant for this system. We assume that it was a steady-state system, and so we have a constant value for the mass flow rate. So, this question asks us, how does the exit velocity scale as a function of these thermodynamic parameters? And all we have to do is look at the relationship between the entrance state and the exit state there. So if we're interested in V two and specifically the velocity at the exit state, then we can simply group these terms and cancel out common features. So if we assume the area is for our circle. We have pipes, you know, circular cross sectional pipes. Then what we have is a relationship between the diameters, actually the diameter squared. at the entrance state, divided by the diameter squared at the exit state. And so that's the left-hand side here, times a ratio of the densities or ratio of the specific volumes. All of that times the entrance velocity. So from these types of relationships you can see that again velocity scales as diameter's squared. for inversely proportional to the exit diameter, and directly proportional to the entrance diameter. So, these are very powerful relationships, so we didn't really need to calculate. If our problem had only asked what's the exit velocity, we didn't need to calculate the intermediate variable of the mass flow rate. All we had to do is take the state information that we, we looked up, from our state tables and the given information on the diameter of the pipes and the engine's velocity. So sometimes we can save ourselves a lot of work, mathematical work and, and that's also an opportunity for us to make errors. By looking at and being insightful, looking at the dependence of the problems and their parameters, and how they scale on key variables. So again if we increase the diameter by a factor two, you would increase the velocity dependent on if it's the entrance. You'd increase the velocity by a factor of four. And if you increase the diameter of the exit by 2, you'd decrease the velocity by a factor of 4. And those types of scales are very powerful for us to interpret and apply, for our problems. Okay, so those are some good examples to get us started. Let's get back to what we started on for this topic, which was the control volume analysis for conservation of energy. So we applied, you saw our example here of the, we applied the conservation of mass, and, in general, remember our, our general form, we said, in minus out equals stored. [SOUND] Okay. So this is the form of the conservation of mass. The conservation of energy has the exact same analog. It's completely parallel to the conservation of mass. So In let's consider all of the terms that could come into the system. We could have heat transfer into the system, which is going to be positive. We could have work transfer into the system which would be negative. So I'm going to put a negative sign in front of this Just like we did with the conservation of mass to represent the fact that, the sign convention, remember. We talked about the sign convention. So the sign convention says work In is negative. That negative negative is a positive, and it'll be an appropriate balance in this system. We also know now, that for control volume systems, there's mass flow into and potentially out of the control volume. And that mass carries with it, energy. So we know that all the different mass flow rates into the system carry internal energy plus kinetic energy. Plus potential energy. And if we have potential kinetic mass to leave the system. We're going to have mass flow rate that has associated with it. Internal energy, kinetic energy, and potential energy. And I look a this equation, I say, hey these are max flow rates. So this is equate, these terms are at a rate basis. So I'd better come back over here and correct my heat transfer and work transfer terms to make sure that they represent heat transfer rate and work transfer rates. Now, so these are my ins and outs, right? I've now identified all the different ways that energy can be transferred into and out of an arbitrary control surface or control volume. And we also need to allow the system to potentially accumulate Energy, so we're going to represent that term just like we did up here, for the conservation of mass, as a control volume, as an unsteady control volume term. And that energy withing the control volume is an uppercase E, so we know that that is an extensive property. And, again, has the same contributions that we see here, in terms of the internal energy, the kinetic energy, and that's the mass in the control volume, plus the mass of the control volume times g, the gravitational constant. Times the potential energy associated with the control volume. So same, same terms that we had before. So again, whenever you write any of these governing principles. These equations, like the conservation of mass or the conservation of energy. They have to have exactly parallel forms. If you're using a control mass. Then you need to have a control mass version of the conservation of energy. If you have a control volume for the continuity equation. This is also referred to as the continuity equation then you have to have a control volume form of the expression for energy. So its a self checking system that we have. It corrects itself, hopefully, if you identify oh, that's not consistent. You make the correction. Okay, so that's our generic form of the conservation of energy. Now I want to take a moment and look at a work term in particular. The work term we know has two parts to it. There's the part that we discussed which earlier, which is there's shack work, there's compression and expansion type work. All these different types of work, electrical work. So this term, we're going to say is due to Let's see. Well, we'll label this just arbitrary control volume work here. And we'll call this shaft work as an example. Let's say, for example, this should be shaft work. This can be, magnetic, et cetera. Okay, but, because every fluid that enters or leaves a control line is at a pressure that's greater than zero, it's a non-zero pressure, that pressure, when applied to an area, is a force. So pressure times area, we would call that force. So that means that if we remember our arbitrary definition for work, every time a fluid moves into or out of a control volume, because that fluid is at a non-zero pressure, it exerts what we refer to as flow work. So we're going to take a look at that term very carefully here. So we, remember we will recall that the arbitrary definition for work is A force that's executed through a specific distance. Right? So this would be actually, del work is force, is the product of the force times this differential differential element, dx. So if we consider the flow work, this pressure times an area is a force. And it again executes that force, that force is applied to this differential, dx element. Remember, everything you do here is, is essentially one-dimensional. We're not dealing with complex dimensions in this class. but that would be a next step that you could easily think about how could you build more fidelity into your analysis. Well, let's say you were looking at flow over the Hoover dam. Well, that's probably a, non, it probably has some pretty decent spatial variation, so it's probably not one-dimensional flow. Again, depending on the fidelity that we gotta achieve in our calculations. Okay, so we have the force times the differential element dx right here. And, what we're, but all of our equations here are on a rate basis, so let's take the flow work here, so this is my flow work, and we say, well, if you look at the rate basis for this expression on the right-hand side, we can see, oh, there's a dx dt term, well, that's simply the velocity of the flow. Again, assuming one-dimensional flow, so that simplifies to something that looks like this. Okay. So we're going to take that expression and we're going to come back over here and look at what happens to the work term once we have those expressions added. So, we know, again with our arbitrary control volume, that we have potential for flow to both enter and to leave the system. So I have the flow work, as the work of the, fluid as it leaves the control volume. And here's the flow work associated with the fluid as it enters the control volume. And again now applying my sign convention, that positive work is work out of a system and negative work is work into a system. So that's taking into consideration the, the sign convention for this class. So we also need to remember that hey, wait a minute, the, mass flow rate is simply the product of the mass times the specific volume times. Excuse me, is equal to the area times the speed. Okay, remember this is also, we can also write it in terms of densities. So let's take this information, here, but actually I should say this information here, and substitute it back in for these two terms into our original expression for the flow work. And what we now have is that the product of the area times the velocity can now be given by the mass flow rate times the pressure, times the specific volume for the conditions at the exit, minus the mass flow rate times the pressure at the inlet times the specific volume at the inlet. Okay. So now we have, oops, sorry, I dropped the pen here. So this is the, let's just fix this. [LAUGH] Okay. This is the flow work here only. So we're going to take these two components to work. The shaft work term. This kind of arbitrary control volume work term, and the flow work term, and we're going to put that back into our expression we had on the previous slide, for the overall, conservation of energy for control volume. So recall that, expression [COUGH]. And I should have summation terms here. We'll squeeze those in. Because we know we can have multiple in, entrances and exits for a control volume. I'll fix my subscript as I go here. now we're going to have [SOUND]. Okay. So, a lot of writing you can see to have this complete form of the conservation of energy. But once we get to this, you're going to have, hopefully have a aha moment. Okay. So, we look at these terms here, and we can see that I can take the flow work terms and I can group them into the expressions that I have for the energy transfer due to the mass flow rate into and out of the system. And what I hope, you'll hopefully see is that we have this common term that looks like this. And if you recall, we talked about state properties and we introduced a number of different properties in a previous unit. And one of those properties we introduced we said would be, by definition, the enthalpy flow fluid, and this is why. We defined enthalpy as essentially a shortcut in our accounting for the thermodynamic properties in controlled volumes. So, if we look at these expressions here, what you see, is that every single control volume, when there's flow moving into or out of it, which is the definition of a control volume. You have flow transfer or mass transfer across the, control surface. We have flow work. So if know we have to account for the flow work every single time, we decide well, let's just bundle that in with the internal energy of a fluid. Because it's also dependent on those thermodynamic properties, the density of the fluid, the pressure of the fluid, and so for the control volume conservation of energy, what we end up with is an expression that looks like this. [SOUND] That's a 2. Make that a little clearer. Oh, all that and I missed the big finale. That's the h. Okay, so this is the governing equation for the conservation of energy for a control volume. It's the most important form of the conservation of energy expression. So if we have control volumes, we work with enthalpies. If we have control masses, we work with internal energy. Okay. So you're going to see this equation again and again in our expressions here. So now here it is, repeated in a much prettier form here. conservation of energy for control volume. Now, two of the most common simplifications we will use in this class, and most people will use in most applications, are to assume steady-state conditions, and to assume that a system or device might be adiabatic. So what I want you to do is to pause for a moment and think about what are the mathematical implications of those two assumptions. They aren't linked, they're just two different assumptions. Often we will make them at the same time, but because something's steady state doesn't mean it's adiabatic, so I don't want to confuse you in that regard. But they're two very common assumptions, and what I want to do is think about how would you simplify this expression using those two assumptions, and that's what we'll start with next time.
