 A pressure cooker has a volume of 0.011 cubic meters and initially contains a two-phase liquid vapor mixture of water at a temperature of 100 degrees and a quality of 10%. The water is isochorically heated until all the water has evaporated. There is a pressure regulating valve which prevents the pressure inside the cooker from exceeding to bar. Neglecting kinetic and potential energy effects determine the initial mass of the water in the cooker, the quality of the water when the pressure leaf valve opens if it opens, the final mass of the water in the cooker if it's different from the initial mass, the total amount of heat required and how long it would take to accomplish this process if the heat being supplied was from a 1 kilowatt heater. So we definitely know that this is a transient problem. We know that we're going to be describing the properties of the water in the pressure cooker at different points in time. The real question is whether we should treat it as a closed system or not. Now, of course, the problem says open system. That implies something about what you have to do in the problem. But let's ignore that for a moment. You don't see the word open. Actually, you don't see the word system nor the comma just as transient analysis pressure cooker. So basically one of two things will happen. Either the pressure relief valve will open allowing some amount of steam to escape in order to keep the pressure from exceeding to bar or it won't. Those are the two options in scenario one. If the valve doesn't open, let's call it scenario one. In scenario one, if the valve doesn't open, we just have a regular isochoric heating process. This is essentially a chapter three problem. We have some initial condition, which we can call state one. We have a quality of 0.1 and an initial temperature of 100 degrees Celsius. Then at state two at the end of the isochoric heating process, we have the same volume that we started with. No mass was allowed to escape because the valve didn't open. Therefore, the specific volume of the water in the pressure cooker did not change. The other thing that we know about state two is that the quality would have reached one or 100%. That is what we define as our end condition. The problem said it is heated until all of the water has evaporated as soon as the last morsel of water evaporates, the process is done. So our end state is defined as the quality hitting one. From this, we should see a pressure at the end of the process that is less than or equal to 2 bar because the pressure relief valve will open and allow mass to escape as soon as the pressure tries to exceed 2 bar. If scenario one occurs, we have a transient analysis of a closed system. In scenario two, we are considering a situation where the valve opens. In scenario two, as a result of the valve opening, we have two processes across three total state points. We have the same processes earlier. We have an isochoric heating process beginning where the quality is 0.1 and the temperature is 100 degrees Celsius. And we are isochorically heating to a state point two. And state point two is the point at which the valve opens. We have isochoric heating from state one to state two, which means that we still have the same volume. The valve hasn't opened yet, so the mass doesn't change yet. Therefore, the specific volume is the same at two as it is at one. And then we know the pressure at state two is 2 bar because that's the point at which the valve opens. We can conclude that the quality at state two must be somewhere between 0.1 and 1 because if it had reached one before it hit 2 bar, we would have ended the process. So we are continuing to heat this time isochorically. Because as more water evaporates, the pressure relief valve is allowing steam to escape to keep the pressure at 2 bar. So P3 is equal to P2. The amount of mass at state three is going to be somewhere less than the mass of state two because the valve has allowed some steam to escape. The other thing that we know about state three is it is still our end condition, which means that the quality will have been one. In order to figure out which of these two scenarios occurs, we can look at what the pressure at the end of the isochoric heating process would have been. If it's less than 2 bar, the relief valve never opened and the scenario ended. That means scenario one had to have occurred. If the valve had to have opened, that means scenario two occurs. So we look at the pressure at the end of the isochoric heating process. Essentially we're looking at P2 in scenario one. If it's less than or equal to 2 bar, scenario one was the one that occurred. The valve never opened. If it's greater than 2 bar, that means the valve would have opened, therefore scenario two occurs. So to answer that question, we are going to jump into our property tables. We need to look up the specific volume of water at a quality of 0.1 and a temperature of 100 degrees Celsius. That information will come from the saturated liquid, excuse me, the saturated water tables by temperature. Once we know the specific volume at state one, we can use it to fix state two and we can determine the pressure. I'll switch colors here to differentiate the setup of the problem from our analysis to answer the question of which scenario occurred. So I am saying that my specific volume is going to be between 0.0010435 and 1.673. Specifically, pun intended, it's going to be 10% of the way between 0.0010435 and 1.673. To determine that number, I can use my calculator, which surely would never let me down. Oh, look at that. It'll let me down. 0.1 times 1.673 minus 0.0010435 and I add to that 0.0010435. So what we're doing is using the definition of quality. We're saying x1 is equal to the specific volume at state one minus vf at t1 divided by vg at t1 minus vf at t1. Then we're using that equation to solve for specific volume one and we get 0.168. 0.168 is our specific volume at state one. That gives us our specific volume at state two and we know the quality at state two. It's one. We can use that information to determine the pressure at state two. So what we need to do here is go into our saturation tables by pressure this time and quality of one represents a fully saturated vapor. So we are going to scroll down our saturated vapor specific volume column until we find 0.168. Then we read off the pressure corresponding with that specific volume and that gives us our pressure. So I scroll on down and I see that 0.168 will lie between 0.1944 and 0.1318. So look, I could interpolate for a p2 but that interpolation is going to be somewhere between 10 and 15 bar. That interpolation will result in a number that no matter what is greater than 2 bar which means that the pressure relief valve had to have opened. Just for extra practice here let's interpolate for an actual p2. So I am going to take my specific volume which is 0.168. I am going to subtract the specific volume that is physically above my specific volume which is 0.1944. Then I am dividing that by 0.1318 minus 0.1944. That gives me the proportion of the way between 10 and 15 bar p2 is. I am going to set that equal to x minus 10 divided by 15 minus 10. And I am solving that for x. If we jump back, add a solve out front. We get 12.1. So the pressure at state 2 is 12.11 bar. That's greater than 2 bar. Therefore the relief valve had to have opened. That means scenario 1 didn't happen. Scenario 2 happened. Now we can analyze scenario 2. First we can recognize that our specific volume at state 1 is the same as it was in scenario 1 still 0.168. And then we are analyzing two processes. In the first process we have a closed system from 1 to 2. So closed system undergoing a transient process. Isochorically from 2 to 3 we have an open system undergoing a transient process. Isochorically because the pressure relief valve is maintaining the pressure inside the cooker. Okay. Now that we have our problem set up, we can look at what we are actually trying to do. The first thing I want to know is the initial mass of the water. Well, we know the volume initially. I can use the volume and the mass, excuse me, I can use the volume and the specific volume to solve for mass. So I will draw a big differentiating line here. And then I will switch back to black. And in the interest of being consistent, I will draw a system diagram. From 1 to 2, that was not good. From 1 to 2, I have a heating process. Causes the pressure to increase isochorically. From 2 to 3, a valve allowing some steam to escape. I'm going to define that as a state point. I'm going to call it state point E for exit. Maybe that's confusing because there's a part E to the problem. So maybe I call that state point O for out. I guess that looks like a zero state point. Epsilon, just to be all cool and Greek like, works for me. Now, part A. The mass at state 1 can be represented as the volume at state 1 divided by the specific volume at state 1. Our volume is given as being 0.011 cubic meters, so 0.011 cubic meters. And then we are dividing by the specific volume that we looked up at state 1, which is 0.168 cubic meters per kilogram. 0.168 cubic meters per kilogram. cubic meters, cancels cubic meters, giving me kilograms. And come now calculator 0.011 divided by that number from earlier, we get 0.065383 kilograms of water. 0.065, let's say 4 kilograms. That represents the mass of water inside my pressure cooker at the beginning of the process. Okay. Ask for it in grams, didn't it? Well, easy enough. If I move that decimal place three points to the right, you get 65.4 grams. Cool beans. Next up, I want the quality of the water when the pressure relief valve opens. That's asking for the quality at state 2. So if I move my very artistic system diagram to the right a little bit, too much to the right, slightly less to the right. Remember that step from the cupid shuffle, slightly less to the right. And we use our specific volume at state 2 and our pressure at state 2 to determine a quality at state 2. Remember that we would expect that to be somewhere between 0 and 1. So we're looking for a number between 0 and 1. We're going to use our specific volume and our pressure to determine a quality, which means we have to go back to our property tables. I am going to find 2 bar on my property tables, and then I'm going to interpolate between 0.0010605 and 0.8857. That'll give me my quality at state 2. So that would be specific volume 2, which is equal to specific volume 1 because it's an isochoric closed heating process. Minus VF divided by VG minus VF. So I am taking my specific volume, which was 0.16824, and I am subtracting 0.0010605. Then I am dividing that by 0.8857 minus 0.0010605. And we get a quality at state 2 that is 0.188979. X2 is equal to 0.189, let's say. That is between 0.1 and 1, which is a good sign. We could also describe the quality here as being 18.9%. Next up, the problem wants to know the final mass of water in the cooker. So at the end of the process, how much mass is there? That's M3. To answer that question, we are going to recognize that we still have the same volume inside of our cooker at state 3. It's still 0.011 cubic meters. If we determine a specific volume at state 3, we can use the volume and the specific volume again to calculate a mass. For that, I need to... Oh man, you guys were unable to see everything that I was doing. Oh no. Well, I wrote 0.189, I put a square around it. I scrolled up, I talked about part C. I went over here to state 3, we talked about it. I indicated some stuff. It was very impressive, so shame you didn't see it. Anyway, at state 3, we want to know the specific volume at state 3 so that we can use that with our volume to determine how much mass is left. For our specific volume, we are going to jump over to our property tables again, this time using 2 bar and a quality of 1. Therefore, our specific volume at state 3 is just Vg at 2 bar, which is 0.8857. 0.8857 cubic meters per kilogram. Then mass at state 3 is volume at state 3 divided by a specific volume at state 3, which means 0.011 cubic meters divided by 0.8857 cubic meters per kilogram. And I get a result in kilograms. So I am going to take 0.011 and divide that by 0.8857. So that is 0.01242 kilograms. 0.01242 kilograms, which is 12.42 grams. And then in an effort to save space, I will rearrange these, be a little bit more horizontal and a little less vertical. Actually, I can leave that one there. That's probably fine. Okay, A through C done. Next up, Part D. Part D wants to know the amount of heat required to accomplish the process. So we have to figure out how much heat had to have been added between 1 and 2, and then how much heat had to have been added between 2 and 3, and then we're going to add those together. So we're doing an energy balance on the process from 1 to 2, and then adding it to the result of the energy balance on the process from 2 to 3. For that, I will draw another vertical line, energy balance on control volume from 1 to 2, and then we'll be doing an energy balance on the control volume. I guess we can do an energy balance and a mass balance. So let's go mass balance on control volume from 2 to 3, and then an energy balance on the control volume from 2 to 3. Yep, very logical abbreviations and layout. I think that's going to look wonderful. I'm sure it won't get cramped at all. You know what? Maybe I'll scoot C over here. Or better yet, I'll move B over here, and then C over here. You look at that. It's like I'm playing Tetris with my own work. Maybe it's more like 10 grams. And I'll draw a list of assumptions over here. Always good practice to establish a list of assumptions anytime you're about to do a mass balance and an energy balance. So we recognize that we are assuming it's a closed system from 1 to 2, so closed from 1 to 2. I could write out that it's a transient process as an assumption. That's not really necessary because that's the default. If we were simplifying it to steady state, that would be an assumption, but transient is the default state of the world. So it's not really an assumption, but good practice is to err on the side of writing it down. Then, assumption number three, no work. And similarly, 4 Q is only in and maybe a 5. Changes in kinetic energy and potential energy are all zero. Let's see how far that gets us. So on my energy balance and the control volume from 1 to 2, I begin with delta E of the control volume is equal to E in minus E out. I have a transient process. The left-hand side of the equation could be delta U, could be delta KE, could be delta PE. The right-hand side of the equation could be Q in or work in or Q out or work out because it's a closed system. Next, I can start neglecting things. I will neglect my changes in kinetic and potential energy because that was assumption number five. I can neglect Q out because I'm assuming Q is only in the inward direction. I can neglect work in and work out because of assumption number three. That leaves me with Q in from 1 to 2 is equal to delta U from 1 to 2, which is big U2 minus big U1. I don't know U1, I don't know U2. I instead could rewrite it in terms of specific U1 and specific U2, which I have the ability to look up. At which point this would be mass, 2 times little U2 minus mass 1 times little U1 because it's a closed system, the mass is the same, so I can factor that out and write that as mass 1. I'm writing 1 because we already have it, times little U2 minus little U1. Since we are evaluating a delta U, we could use option number one where we look up U1 or U2, excuse me, U1 and U2 and subtract them or we could consider specific heat capacity. Is assuming specific heat capacity is constant an appropriate assumption to make? I'll give you a couple seconds. No, it is not. We have a phase change and assuming constant specific heat capacity is never appropriate across a phase change. With that, we can look up U1 and U2. So I will write a description for what 0.168 is over here. Great, I'm switching colors, that's not going to be confusing. U1 then can be determined from the quality and the temperature in the same way that we determined the specific volume earlier. We have a temperature of 100 degrees Celsius, which means we want to go into our property tables, the saturation tables by temperature. We're going to find 100 degrees Celsius and then we are going to use Uf and Ug to calculate our U1. Oh, that will be 0.1 times 2506.5 minus 418.94 and then I add to that 418.94 and we get 627.696. With U1 in place, we can start to calculate U2. You can populate X2 while I'm over here, we calculated that already. It was 0.189, 0.189. And then U2 is going to come from the fact that the quality at state 2 is also equal to U2 minus Uf at 2 bar minus Ug at 2 bar minus Uf at 2 bar. So I could write out 0.189 or I could scroll up a couple of lines and grab the quality with more decimal places or I could actually just grab this proportion which represents what the quality was. So we're saying V1 minus Vf over Vg minus Vf is equal to U2 minus Uf divided by Ug minus Uf. Okay, that's equal to X minus 418.94 divided by one calculator, that's not divided by, that's divided by 2506.5 minus 418.94. That's not anywhere near what I wanted. No, it's not 100 degrees Celsius. That was close. We almost looked up an incorrect number. That would have been a remarkable change of pace. Okay, I'm looking for the saturation tables by pressure this time and I want to find 2 bar and at the 2 bar position I need Uf and Ug. So I'm interpolating between 504.49 and 2529.5 using my quality of 0.189 which I am representing symbolically because it's just cooler looking obviously. No, it's actually because I'm jumping back to our definition and not relying on our previous calculation even though I actually grabbed it from a previous calculation. It's a very minor point we're solving for X. So our internal energy is 887.175 887.175 and that number going to be helpful for solving for Q in from 1 to 2 and also 2 to 3. And you know, while I'm here I'm going to convert this back to the color black just to try to reduce confusion for the people who grabbed the solutions PDF without watching the video and I'll scratch that over a little bit more. You know, maybe I'll have to make a version of this PDF that has like paragraphs explaining what I'm doing. We'll see if that happens. Okay, now Q in from 1 to 2 is the mass at state 1. We know that mass. We calculated that mass. It was 0.0654 kilograms and then we are multiplying by the difference from U1 to U2. So that was 887.175 887.175 887.175 I guess I could just look at my calculator but you know, go away calculator. 627.696 627.696 That's kilojoules per kilogram. Okay. When I multiply these two things together I will end up with kilojoules. 0.0654 multiplied by 887.175 minus 627.696 Our amount of heat is just under 17 kilojoules. 16.9699 kilojoules and it's fourth referred to as 16.97. That is the amount of heat it takes to get from 1 to 2. Now we can calculate the amount of heat it takes to get from 2 to 3. But first, a mass balance. Change the mass of the control volume is going to be whatever mass enters minus whatever mass exits. There is no mass entering. Add that to our list of assumptions. Mass only leaves valve K. Then the change in mass of the control volume is going to just be whatever mass exits. Or mass 3 minus mass 2 is equal to mass out. Then our energy balance is going to start similarly to how it looked from 1 to 2. The primary difference being we have to account for energy associated with a mass crossing the boundary as well. So E n could be Q n work in or the energy associated with a moving mass E out could be Q out work out or the energy associated with a mass crossing the boundary. So I will say plus some in m theta. Note that that's not m dot theta because it's a magnitude not a rate. I will see if I can recover any more horizontal space. Guess I can't really. So I'm just going to write this down here. Plus some out of m theta. So just like 1 to 2 our energy balance can be simplified by recognizing that we're neglecting changes in kinetic and potential energy on the control volume itself. The pressure cooker isn't increasing in elevation or changing its velocity and even if it were it's unlikely to be as a result of this process. We have Q in in the form of the heating element and then we have entering or exiting mass. We don't have any entering mass. So we only have to account for the exiting mass. That would be m epsilon. I think I called that state point times H epsilon plus kinetic energy salon plus potential energy epsilon and we don't have enough information to figure out the velocity of this team at the outlet and as you noticed on the hairdryer problem the kinetic energy has a very small effect relative to enthalpy. So even if the velocity of the steam is a couple of meters per second the energy associated with this accelerating that steam is negligibly small. So we're neglecting changes in kinetic and potential energy of the steam exiting the valve as well as the control volume itself. So I just have to account for m epsilon times H epsilon. Furthermore this mass that's the amount of mass that exits is this mass m out. So I can actually write that as m3 minus m2. So my simplification for the energy balance is delta u from 2 to 3 is equal to Q in minus the quantity m3 minus m2 times the enthalpy at state point epsilon. Since I'm looking for Q in I can rearrange and solve I'll write that as Q in is equal to first up I have u3 minus u2 again I'm going to substitute in terms of specific internal energy. So that's m3 times little u3 minus m2 times little u2 and then I add to that the quantity m3 minus m2 times H epsilon. So we know m2 is equal to m1 so we know m1 already. We know u2 because we looked it up. We know the mass at state 3 because I made us calculate it. We zoomed out for some reason. We know the mass of state 3. We know the mass of state 1. The only thing we don't know here is the enthalpy at state epsilon and the internal energy at state 3. Well for that I have to go back to my property tables and I recognize that u3 can be looked up using 2 bar and a quality of 1. So u3 is really just ug at 2 bar and let's see if I can recover any more horizontal space. Kind of the theme of the problem so far ironically the space on the problem is experiencing pressure. I need a relief valve there that's better. Okay so ug at 2 bar jump into our property tables ug at 2 bar is 2529.5. So 2529 2529.5 kilojoules per kilogram. Now let's consider state point epsilon. State point epsilon is at 2 bar-ish because we are talking about the steam that's just crossing the boundary so we're not allowing it to have escaped to atmospheric pressure yet and then we have to decide something else about the properties of state epsilon and the assumption that I'm going to make here is that the steam that escapes is entirely saturated vapor. So the only thing going out the top of the pressure cooker here is vapor. It is not a mixed stream. It is only vapor. Therefore the quality of epsilon is 1 and the pressure of epsilon is about 2 bar. So for that h epsilon will be hg at 2 bar which is 2706.7 2706.7 kilojoules per kilogram and I can plug that into my energy balance down here. I recognize that I'm going to be taking a mass in kilograms multiplied by an internal energy in kilojoules per kilogram and then I'm going to be subtracting a mass in kilograms times a specific internal energy in kilojoules per kilogram. Those results will both be kilojoules then I'm going to add a mass difference times an enthalpy in kilojoules per kilogram so as a result I'm going to get kilojoules. I'm not going to write out the numbers on paper because I'm starting to run out of space and those of you who watch the video can see what I'm typing in my calculator anyway. So first up, mass 3. We have 0.0242 0.01242 multiplied by U3 which is 2529.5 and then I'm subtracting M2 U2 M2 is equal to M1 which is 0.0654 multiplied by U2 which was 887.175 that gives me the change in internal energy of the control volume and then I'm also adding how much energy is associated with the mass that crosses the steam. By the way just a little meta note here I'm adding these together because some energy is lost as the steam escapes. So that's heat that we had to add to make the process happen but not heat that manifested in the control volume itself. So that's adding to how much heat we have to supply. Does that make sense? Cool. I heard you all agree all the way through space and time. So I'm multiplying by the difference in mass which is M3 minus M2. So M3 minus M2. Excuse me. That's not right. Okay. I just did a spiel about how I'm adding energy and yet I would have resulted in a negative number. No. I forgot this here negative sign. So delta mass control volume is equal to negative mass out which means that mass out is M2 minus M3 not M3 minus M2. So I have to switch this from A3 to A2 and this from a 2 to a 3 and then this from A2 to A3. Let's try that again. Mass 2 which is the same as mass 1 minus mass 3. So 0.0654 minus 0.01242 multiplied by the enthalpy at state point epsilon which is 2706.7. We get an amount of heat required to get from 2 to 3 of 116.796 kilojoules. So the amount of heat it takes to get from 1 to 2 is 16.9699. The amount of energy it takes to get from 2 to 3 is 116.796. Therefore the total Q in which is Q in from 1 to 2 plus Q in from 2 to 3 is 116.796 plus 16.699 I get 133.766 so it takes 133.766 kilojoules to evaporate all the water in the pressure cooker. Lastly I wanted to know how long this process would take if the heat was supplied from a 1 kilowatt heater. For that I'm going to really test the confines of the margin down here. I recognize that the rate of heat added is equal to the magnitude of heat added over a duration for the entire process. This is going to be delta T is equal to Q in over Q dot in. So I'm going to take 113.766 kilojoules and I'm going to divide by 1 kilowatt. I recognize that a kilowatt is a kilojoule per second which means that if I take 113.766 and divide by 1 I will yield an answer in seconds. Is that what part E wants? It doesn't say. Well 113.766 divided by 1 is 113.766. So it takes 100 and huh I said 113. I wrote 113 but the number is 133. Well let's pretend that didn't happen. So 133.766 divided by 1 is 133.766 which means that it takes 133.766 seconds or two minutes 13 seconds. And by the way, while we're here let's consider the fact that the pressure cooker has a boiling point that is higher than 100 degrees Celsius. If we jump into our beloved property tables we can see that at two bar the saturation temperature is 120.2 The whole reason the pressure cooker is able to cook things faster is because it boils water at a higher temperature than water at atmospheric pressure. So if you're driving heat into your ramen noodles or your whatever a higher temperature of the water means there'll be a greater temperature difference between your ramen noodles and the water therefore more heat actually goes into the ramen noodles. Neat, right? That's pretty neat.