 A piston-cylinder arrangement contains steam and an electric resistance heater. The cylinder is perfectly insulated to ensure minimal heat transfer with its surroundings and rests on a table in a large room where the ambient conditions are 20°C and 1 bar. The steam in the cylinder is initially at 200°C and 150 kPa and occupies a volume of 0.25 cubic meters. Now you plug the heater into a 120V wall outlet and measure the current running through the heater to be 5 amps. The heater heats up the steam causing the steam to isobarically expand. You stop the heater once the gas has expanded to 0.5 cubic meters. Complete the following. Determine the mass of the steam at the beginning of the heating process, the temperature of the steam at the end of the heating process, and the time required for this process to occur. I will begin with a system diagram. I'm defining two state points here. One, the beginning of the process, I will call state one. The other, the end of the process, I will call state two. And my goal is going to be to pick out two independent, intensive properties for both state points that will fix both state points, at which point I can look up whatever I need in our steam tables. So at state one, I know that we have an initial temperature and pressure of 200°C and 150 kPa. Those two independent, intensive properties will allow me to look up whatever I want. We know the initial volume as well, 0.25 cubic meters. And the end condition is defined as being a volume of 0.5 cubic meters. Now what else do I know about state two? Think about it for a second. Maybe pause the video if you want. Many of you are probably saying, well it's probably reasonable to assume that the mass is constant, so mass two would equal mass one, and that's a reasonable assumption to make. We could assume we had a closed system, or that we can treat this as a closed system. That would mean that our mass was not allowed to cross the system boundary, meaning m2 is equal to m1, but that is not two independent intensive properties. For one thing, volume isn't even intensive, so there must be more to it. What else do we know about state two? Has to do with the fact that the problem told us it was isobarically expanding. Isobaric means constant pressure, which means p2 is equal to p1. So what are the two independent intensive properties which fully define state two? They are pressure and specific volume. I know, it's not what you expected, right? But John, I hear you asking, we don't even know the specific volume of state two, but we have enough information to be able to describe it, because we can look up anything we want at state one, including but not limited to specific volume one. So we could use specific volume one to calculate the mass of state one, which would be the same as state two, and then we could use the volume and mass at state two to calculate the specific volume of state two, and then we could use the pressure and the specific volume at state two to determine whatever it was that we wanted. Now that leads me to my next question. What is it that we want? Well, the problem first asks for the mass of the steam at the beginning of the process. So for part a, I'm looking for mass one, which is going to be total volume one divided by specific volume one, which means that we are going to want to look up specific volume one, which I already have written down. For part b, we are trying to determine the temperature at the end of the process, which once we have that state point fixed, we can just look up. So b will be done when we look up t2. Then c asks us to determine the time required for this process to occur. So for that, we are going to have to relate work and power, meaning I know the electrical work is applied at a certain rate because I have a voltage and a current, and we are ignoring power factor as a concept right now. So I can multiply the voltage in the current, 120 volts and five amps, to come up with the amount of electrical power applied to my system. That would be 120 times five, which is 600. So 600 joules of electrical work is applied per second. If I could determine how much work was actually occurring, then I have myself a duration. Now, do we know anything that would allow us to deduce what the electrical work is? Well, do we have a way to describe how energy is related across a process? It is our energy balance. And I'm going to perform my energy balance on our mass of steam. So first question, do we have a steady process or a transient process? We have a transient process. We know that because stuff is changing with respect to time. Like for example, the volume of the steam changes over the course of time. And in fact, the problem asks us for a duration, which implies that time is important. Therefore, we can't neglect the effects of time. Transient means we have to include the left-hand side of our energy balance here. So this could be delta u, could be delta ke, and it could be delta pe. And then I already am assuming this is a closed system, and energy can cross the boundaries of a closed system as heat transfer or work. Therefore, my full energy balance for this problem is delta u plus delta ke plus delta pe is equal to q in plus work in minus the quantity q out plus work out. My next step is to begin to eliminate terms. So I will start from left to right. Is it reasonable to neglect internal energy? No, it is not. I need my change in internal energy because the properties of this team are changing. And I'm probably going to want to relate work to them. But I can neglect changes in kinetic and potential energy because the problem told me nothing that would indicate a change in velocity or change in elevation. So we are assuming that those changes in kinetic and potential energy are negligibly small. What about heat transfer and work? Well, the problem told me the cylinder was perfectly insulated. Perfect insulation means I have no heat transfer. And we indicate an adiabatic process, which is one where there's no heat transfer, with these fuzzy lines on our system diagram. So we have no heat transfer in either direction. Therefore, I can get rid of q in and q out. And that's not even based on an assumption because the problem told us it was perfectly insulated. This state-of-the-art perfect insulation that doesn't allow any heat transfer in any direction. Okay, what about work in? Do I include work in or do I neglect it? That's right, I keep it. I keep it because I have work in in the form of electrical work. What about work out? Do I have any opportunities for work out? Well, easiest way to determine that would be to go down the list of works that we had talked about. Do we have shaft work? Do we have spring work? Do we have electrical work? Do we have boundary work? Do any of those occur in this problem? Well, I'm going to have boundary work because I have a changing volume. The volume goes from 0.25 to 0.5, which means I have an expansion process which consumes some of the energy that the electric resistance heater is applying. So electricity comes in, it enters the steam, some of the energy stays there, some of it leaves as boundary work. So I will plug in boundary work for work out. And that, remember, is the integral of pressure with respect to volume. And that's going to be pressure times change in volume because we have an isobaric process. The pressure comes out of the integral. So I can say delta U is equal to work in minus pressure times V2 minus V1. And I can say work in then would be U2 minus U1 plus pressure times V2 minus V1. And before we continue, let me just take a moment to have a conversation about this quantity here. We have internal energy 2 minus internal energy 1 plus pressure times volume 2 minus volume 1. We could determine the mass. We can look up specific U1 and specific U2 and multiply each of those by mass to get a total U1 and a total U2. And then I could plug in U2, U1, pressure, volume 2 and volume 1 and get an answer. And that would be fine. Or I could recognize that this is a quantity that we see a lot. We have internal energy pressure and volume. And I could write this as U2 minus U1 plus, if I distributed my pressure, I could write that as either P1 or P2 because they're the same. So I could write P2 V2 minus P1 V1 if I wanted to. And if I were to group together the state one and state two properties, I could write this as U2 plus P2 V2 minus U1 plus P1 V1. And I have a convenient representation for that. That quantity appears so frequently that we have a special name for it. It's enthalpy. Enthalpy is the combination of internal energy plus pressure times volume. So I can write that as H2 minus H1, which if I wanted to write it in terms of specific quantities, which are more conveniently accessible to me, I can write this as mass two times little H2 minus mass one times little H1. And if I were to have calculated mass by now, I could plug in little H1 and little H2 and have an answer with way fewer numbers being written in my calculator. Either way would produce the same answer. So I'm going to look up H1 and I'm going to look up H2 and that will give me everything that I need to be able to complete the problem. Pro tip, your life is going to be easiest if you work through your solution symbolically first before you get caught up in the nitty gritty of looking up properties and calculating numbers. If you can determine what you need to look up before you actually start looking stuff up, then you can pack all your property lookups together, do them all at the same time and knock out what you need before getting back to solving your problem. If you look up properties right away, it can be real easy to just go down a rabbit hole of looking up everything that you think might be useful and wasting a bunch of time that way. Or if you look up properties as you encounter them, you might end up in a situation where you're having to go back and restart an interpolation over or you might waste time trying to get your head wrapped around what you're interpolating between or something like that. So now that we know exactly what we are looking up, we can actually look it up. So at day one, I wanted a temperature and pressure, which I'm going to use to determine a specific volume and specific enthalpy. And the first step is still to fix the phase. I have to know what phase it is so that I can go into the correct property table. So I will look up the saturation temperature corresponding to my pressure, compare my temperature to it and determine a phase. So I'm looking for the saturation temperature corresponding to 150 kPa. So I will jump into our saturation tables by pressure. I will find 150 kPa, which is 1.5 bar. I see the saturation temperature is 111.4, and then I'm going to compare that to my temperature and deduce something about the phase. I see that our temperature at day one is 200 degrees Celsius, which is higher than 111.4. Therefore, we have a superheated vapor. So I'm going to go into my superheated vapor tables to find our properties for state one. I don't remember what our pressure was. It was 1.5 bar. So I'm going to be looking for a pressure sub table of 1.5 bar, and as luck would have it, we happen to have one. That's convenient. I don't have to interpolate or nothing. And I have 200 degrees Celsius, which means I can just grab a specific volume of 1.444 is that times 10 to the negative third is not just 1.444. So our specific volume at state one is 1.444 cubic meters per kilogram. And we can grab a specific enthalpy while we're at it. That is 2872.9, 2872.9. Another tip while we're here, if you can't see internal energy and specific enthalpy anymore, because you've zoomed way in on the PDF, and you don't remember which column is which, remember that enthalpy is always greater than or equal to internal energy, because enthalpy is defined as internal energy plus pressure times volume. Specific enthalpy would then be specific internal energy plus pressure times specific volume and pressure can never be a negative quantity because we're talking about absolute pressure. So it's always going to be greater than or equal to specific internal energy. So if you see these two columns, the bigger of the two is going to be your specific enthalpy. Anyway, so we have V1 and H1. So my next step is to calculate a mass, which I'm going to do by taking 0.25 cubic meters and dividing by 1.444 cubic meters per kilogram cubic meters, cancels cubic meters, giving me a mass. And that mass is if my calculator would cooperate, 0.25 divided by 1.444. And we get 0.17313. We get 0.17313. You like the way that I said that without any sort of confusion? I just said the number right the first time. That was impressive. Okay, I probably have too many decimal places from trying to cover for the fact that I said one before I said zero. The point is, we have a mass of 0.173. And I could use that to determine a specific volume at state two from which I can interpolate for whatever I want. So specific volume two would be volume two divided by mass two. Mass two is equal to mass one. So I'm going to take 0.5 divided by that number we just calculated, and we get a specific volume of state two of 2.888 cubic meters per kilogram. And note, we could have gotten there faster. I mean, we needed the mass anyway. But we could have recognized that our mass is constant and our total volume is doubling. Therefore, our specific volume would also be doubling. So we could have taken 1.444. We could have multiplied it by two to get 2.888 and written that down. And that would have been a much faster way to get to specific volume two. But we are halfway through our property lookups now. So I will proceed back into our tables to try to fix the state point that has a specific volume of 2.888 and a pressure of 150 kilopascals. So what I'm going to do is I'm going to look up VF and VG corresponding to 150 kilopascals. And I'm going to compare V2 to VF and VG. So I'm going to go to my saturation tables in order of pressure. I'm going to find 1.5 bar. I'm going to grab my highlighter tool because I forgot it was there. And I see VF is 0.0010528 and VG is 1.159. My specific volume of 2.888 is higher than both of those, which means that I have a superheated vapor again. So I'm going to go back to my superheated vapors and I'm going to try to highlight everything apparently. And I'm going to stop highlighting. I'm going to stop. I just want to stop highlighting. Go away. I want the hand tool back. Okay, I guess fine. This is how life is. This is why we don't highlight. But well, okay, I'm going to try to find a pressure sub table corresponding to our pressure, which was 1.5 bar. We have the same pressure sub table as earlier, which I probably can't navigate to because I've got rid of my hand tool, stupid highlighter. And I'm going to try to find 2.888 on this sub table. And I see that 2.888 does not appear on this sub table. So, okay, I swear I'll stop moving the screen in a second. What that means is it's at a temperature that is higher than 600 degrees Celsius. This table in the appendix is not exhaustive. It is not all possible temperatures. It's just a selection. And 600 degrees Celsius is what they considered at the time, the point at which to stop printing the little table because they had a limited amount of room, because they have to print a whole bunch of these two dimensional cross sections of a three dimensional surface. So just like any other time that we have a temperature that doesn't appear on this chart, we are going to assume that the relationship between all the data points is a line. And this time, instead of interpolating, we are extrapolating. So anytime you're going between known data points, that's interpolating. Anytime that you're going outside of known data points, that's extrapolating. And the approach is exactly the same. We are extrapolating instead of interpolating, but the equation doesn't change. I'm still going to take the proportion of the way we are between two data points here, that's going to be 500 and 600. And I'm going to extrapolate out for our actual temperature. So for example, if I wanted to determine the temperature at state two, like I do, I'm going to be taking 2.888 minus the higher physically higher on the physical table, not numerically higher, but the higher up row, which is 500, excuse me, which is the specific volume corresponding to 500, which is 2.376. And then we are dividing by the value at the lower row, which is 2.685 minus the value at the higher row. So we are taking our specific volume minus the value at 500 divided by the value at 600 minus the value at 500. It's just that this proportion is now greater than one, which is fine. And we are multiplying that by the quantity 600 minus 500 and adding it to 500. And that tells us our temperature at state two is 665.696. Quick aside here, extrapolation is dangerous. Generally speaking, extrapolation is to be avoided where possible. If you can interpolate, that's always better than extrapolating. Extrapolation is real easy to come up with real unrepresentative results. Like if I said 40 people are in this class today, and one person drops it tomorrow, tomorrow there are 39 people. If we extrapolate from that, in 40 days there will be no one left in this class. That's probably not a representative way of assessing student population over time. Extrapolation can yield very unrepresentable results. And best practice is to interpolate wherever you can. So we can't because we have nothing to interpolate between. So our only option was to extrapolate. I will also point out that generally speaking, it's considered safe to extrapolate as long as you stay within the same quantity of distance that represents the distance between your data points. So like, if we were interpolating between a value of 100 degrees Celsius higher than 600, like we are, that's considered safe. Does that make sense? Because the distance between 500 and 600 is still greater than the distance between our highest data point and our data point. However, if we interpolated for like 1200 degrees Celsius, that would be considered a risky result. Anyway, we have one more interpolation to do and that is for enthalpy. So I will take that same interpolation and replace 500 and 600 with the enthalpy at 500 and 600. Remember, it's the higher of the two columns. So I'm taking 3704.3 minus 3487.6. And I'm adding that to 3487.6. So my enthalpy at state two is 3846.66. So we've completed all the lookups we needed to actually answer everything in the problem. I have enough information to be able to proceed now. All that's left coincidentally is part C, which is going to involve calculating a work. So I'm going to take this relationship that we had simplified. And I'm going to factor out mass at which point we have mass times little h2 minus little h1. Our mass we calculated as 0.17313. And we are multiplying that by h2 minus h1, which is 3846.66 minus h1, which was 2872.9, 2872.9. And then we are getting kilojoules per kilogram for that delta kilograms, cancels kilograms, leaving us with kilojoules. So I'm taking 0.17313, which is this number here, multiplying by this number here, minus 2872.9. And we get a required work input of 168.588. So now we are going to use our power and our work together. Because remember, power represents d work d time, which over the course of the entire process would be the work occurring divided by the duration. Therefore, duration is going to represent the work required divided by the rate at which that work is required or applied, excuse me. So that's 168.588 kilojoules divided by 600 watts. Remember that a watt is defined as a joule per second. And also remember that a kilojoules 1000 joules, the joules cancels joules, watts cancels watts, kilojoules cancels kilojoules. And I'm made in the shade. I got this quantity multiplied by 1000 divided by 600 and I get 200.98 seconds. Therefore, this process would take, what is that, three minutes, 20.98 seconds.