 Before we get into specifics of different types of models of power cycles, I want us to practice analyzing a power cycle without any established model, just an arbitrary power cycle. We have to develop a skill set to analyze power cycles in general before we get lost in the minutia of what makes up an auto cycle or a diesel cycle or a Brayton cycle. So in this problem, we are considering a cold air standard cycle with constant specific heats being executed in a closed system comprised of four processes. The process from state one to state two is isentropic compression from 100 kilopascals in 27 degrees Celsius to 800 kilopascals. The process from two to three is isochore heat transfer, doesn't specify in or out, which brings the temperature to 1800 Kelvin. Three to four is isothermal, that's constant temperature expansion to 100 kilopascals, and four to one is isobaric heat transfer again without any indication of direction, which restores the system to its initial state. Using this information, I want us to determine the net work and net heat transfer occurring in the cycle. So let's parse that out for a second. We have a cold air standard cycle. Air standard is referring to the fact that we are analyzing ideal air and using those same four assumptions that are grouped together under the name air standard. The cold in cold air standard implies to us that we are using constant specific heats, which is also explicitly written here, and furthermore that those constant specific heats are evaluated at 300 Kelvin. That's what the cold refers to. Regardless of the temperature of the process, the temperature of the air at any state point, we are evaluating the properties at 300 Kelvin. The fact that it's a closed system implies to us that the mass remains constant. So all four state points are going to have the same mass, which we can just call the mass m. And then we are looking for net work and net heat transfer. We have enough information in the problem to fully define all four state points. We know that because we have two independent intensive, excuse me, we know that because we have two independent intensive properties at all four state points. If we look at state one, we were told T1 and P1. Those two independent intensive properties fully define state one, from which we can determine any other property we want. At state two, we were told a pressure because one to two is isentropic compression from 100 kilopascals in 27 degrees Celsius to 800 kilopascals. That's process at 800 kilopascals. We also know that the process from one to two is isentropic. Isentropic means that the entropy is going to be constant throughout the process. So S2 will equal S1. And if we know S1, because we can look it up because we can look up anything using T1 and P1 theoretically, then S2 being equal to S1 gives us our other independent intensive property at state two. If we follow that same logic, the process from two to three is an isochoric process. That means constant volume. So the total volume at three is going to be the same as the total volume at two. That itself is not an intensive property. But because we know it's a closed system, therefore the mass is constant, that means M3 is also going to be equal to M2. And the fact that the volume doesn't change and the mass doesn't change means that our specific volume not going to change. So if we use P2 and S2 to determine V2, we can fully define state three. Last up, we were told we have isothermal expansion to 100 kilopascals. That isothermal refers to the 1800 Kelvin temperature that we had from two to three. So both three and four are going to be the same temperature. T3 is 1800 Kelvin and T4 is equal to T3. The problem also tells us that the isothermal expansion process in three to four ends at 100 kilopascals, which fully defines state four. So these eight independent intensive properties fully define the four state points. And from that, the world is our oyster. We can look up whatever we want. We can theoretically calculate whatever aspects of the processes we want. To determine the net work and the net heat transfer, we are going to have to calculate all four works, that is the work from one to two, the work from two to three, the work from three to four, the work from four to one, and all four heat transfers, one to two to three, three to four, four to one. Once we analyze all four works and heat transfers, we can assess the cycle as a whole. I'm going to start by taking our independent intensive properties that are specified and breaking them into a table that isn't necessary, nor is it necessary to complete the table once you have it. It just happens to be a convenient way for me to organize the information in my brain. So I'm going to take the information specified and draw a table. And the three relevant independent intensive properties that I care about are temperature, let's try that again, temperature, pressure, and specific volume. And I know what you're thinking, you're thinking, John, well, temperature and pressure I can get behind. Those are useful for looking stuff up or calculating heat transfers, but why this specific volume? Well, specific volume is particularly useful as a quantity in reciprocating engines. That is an engine that operates by compressing and expanding a gas inside of a piston cylinder arrangement. Now, in this problem, it doesn't explicitly say reciprocating engine, but the fact that it's a closed system implies to us that this power cycle is operating on the same mass the whole time. So there's expansion and contraction of that same mass. And it's useful to have a parameter that lets us kind of keep track of the expansion and contraction process. Plus, you know, it's good character building to add more independent intensive properties for us to keep track of. And over here, I'm going to define these properties relative to these four state points. Let's start by plugging in what we know already. T one is 27 degrees Celsius, which is going to be 300.15 Kelvin, right calculator. I mean, that's fine. We don't need the calculator, right calculator. See, okay, 27 plus 273.15. Hooray, it's 300.15. We did it. P one is 100 kilopascals. V one, we don't know yet. P two was 800 kilopascals. And we know that V two is going to be the same as V one. So drawn arrow or something that helps me keep track of that. State three, we know has a temperature of 1800 because the process from two to three ends at 1800. That's not temperature, John. There we go. 1800. And then the process from three to four is also isothermal, which means that three to four is constant temperature. I'll draw my arrow, you know, just to provide some consistency in this table that I'm drawing. So we don't know V one, which means that we can't set up our two quantitative properties at state two to determine whatever else we want. So let's think about that for a second. How could we use T one and P one to determine a specific volume? Well, you're probably going to be thinking property tables, property tables for air, I look up T one and P one, and I determine a specific volume. Well, that's a good instinct. But in our particular appendices, we don't have properties for air that include an actual specific volume. And why don't we? Because those tables are for ideal air and for ideal air, we have a better way of coming up with a specific volume that doesn't waste a whole bunch of ink. We know that it is going to be R times T over P because it's an ideal gas. And therefore, we're going to be using the ideal gas law to keep track of it. So specific volume can be written as the specific gas constant for air multiplied by T one divided by P one. And that would give us specific volume. Furthermore, remember that the specific gas constant for air is going to be the universal gas constant divided by the molar mass of air. Those quantities can be determined in the textbook. And in fact, why don't we just look up the specific heat capacities while we're at it? So I am going to clear some space here. So I'm going to say ideal air, we want, presumably, some combination of CP, CV and K because it's specific heat capacity. And let's look up a molar mass of air. And let's use the universal gas constant, the molar mass of air to determine a specific gas constant. So into our tables, let us adventure. We know from the inside of the front cover of our textbook that the universal gas constant is 8.314 kilojoules per kilo mole Kelvin. So that's going to be 8.314 per kilo mole Kelvin. And then next, let's grab the molar mass of air just so we can calculate the specific gas constant that's going to come from table A1, which is going to be 28.97 kilo moles cancel, which results in a quantity in kilojoules per kilogram Kelvin. Let's wake the calculator up and have it spit out 0.287 for us. Hey, look, it's 0.287 ish. So 0.286987, which I'm going to round to seven. John, you can't talk and write at the same time. Kilograms per kilogram. CP, CV and K will all come from our specific key capacity. Tables, ideal air. So let's cruise on down to, I believe it's table A19. And we can see that it is not table A19. It is in fact table A20. The table A20 gives us CP, CV and K values for air that is ideal air at different temperatures. Now, which temperature do we use? That's right. We use 300. And it's not because T1 is 300. It is because we were told to use the cold air standard, which means we're evaluating these properties at 300 Kelvin. It is what the cold and cold air standard means. There's also a hot air standard where you look up properties at I think it's 600 degrees Celsius, but that's don't quote me. That's off the top of my head. So 1.005, 0.718, and 1.4. That's 1.005 kilojoules per kilogram Kelvin, 0.718 kilojoules per kilogram Kelvin, and 1.4. And note that every time that we use the cold air standard, we are going to be using these same three properties. So you're going to get very proficient at looking them up. And at a certain point, you can just start grabbing them from memory. That's fine, as long as you know where they come from. I should be able to give you a problem on the exam that's of nitrogen or oxygen, and you should be able to determine the CP and CV and K values and not just default to the ones that you've been using out of muscle memory. So now that we have a specific gas constant, we can calculate a specific volume at state 1. So 0.287 being horizontal line, kilojoules per kilogram Kelvin, we are multiplying by 300 Kelvin, 300.15 because it was 27 degrees Celsius. We are dividing by 100 kilopascals, which was the pressure at state 1. And so far Kelvin cancels Kelvin to get the rest of the units to cancel appropriately. I need to break the pressure and energy units into their primary components. I could do that by breaking a kilopascal into a thousand kilo... Oh, here, I'll just show you. So I could break a kilopascal into a thousand Pascals and break a kilojoule into a thousand joules. But when I do that, I just cancel the thousands anyway. So in these circumstances, it's actually more convenient to just leave it in the thousands quantity. So a kilopascal could be written as kilonewton per square meter, as opposed to a thousand Pascals, and then a Pascals and Newton per square meter, just fewer steps to write. And then a kilojoule can be written as a thousand joules. Now, kilojoule cancels kilojoules. Kilonewton cancels kilonewtons. John, we talked about this. The whole point was to leave it in the thousands quantity. What are you doing? Kilonewtons cancels kilonewtons kilojoules. Still cancels kilojoules. Kilopascals cancels kilopascals. That leaves me with square meters and meters, which is going to be cubic meters and kilograms in the denominator. So calculator, if you would wake up. We are going to be taking 0.287 multiplied by 300.15 and dividing that by 100. And we get about 0.861 cubic meters per kilogram. So I'll write that here, 0.861. And then because of the arrow, because we have enough information to give us our quantity at state two. Interesting. I didn't write the arrow in the right spot. Guys, you're supposed to yell at your TVs. Okay, let's try that again. Because we have enough information to fully define state two around state one, we have the ability to determine our other properties. Now, how is it that we get T2 and V2? You might be tempted to say the ideal gas law again, but we could only do that if we had another property at state two. We don't. So the ideal gas law itself isn't going to be useful. Now we could use our ability to look up some representations of entropy and use our knowledge of what isentropic means to determine another property at state two using our ideal gas tables. But remember, we're analyzing a problem that has constant specific heats. Furthermore, we are analyzing an isentropic process with constant specific heats and the working fluid is an ideal gas. Do those three assumptions sound familiar? They should. Those three assumptions are what it takes to use the isentropic ideal gas relations. I know it is very exciting. Remember that all four rows of these equations are actually the same. It's just that I don't want to rely on my ability to do algebra in my head. And as a result, I have solved each row relative to a different parameter. So row one is solved for temperature, row two is solved for pressure, row three is solved for volume, row four is solved for density, even though the last two rows are the same, I feel as though I have to complete all four properties. So here we are. We can use any equation on this sheet to refer to properties from state one to state two because it's isentropic. And it's an ideal gas with constant specific heats. But this only works for process one to two doesn't work for the process two to three, three to four or four to one. So if we were to grab this one, for example, we could say t two is equal to t one times p two over p one raised to the k minus one over k. So I could say t two is equal to t one multiplied by p two over p one raised to the k minus one over k t one, we know it's 300 point one five Kelvin p two over p one is 800 over 100 k is 1.4. We have everything we need to calculate t two. So 300 point one five multiplied by 800 divided by 100, raised to the power of 1.4 minus one divided by 1.4. Our t two is 543.706. For completion's sake, I'm going to round it to two decimal places. We can follow that same logic for the specific volume of state two, jumping back to this set of equations and recognizing that if I divide the numerator and denominator by mass in say this equation, I can write that as specific volume two is equal to specific volume one multiplied by p one over p two raised to the one over k. So same logic. I'm taking 0.86139 multiplied by one excuse me, yeah 100 over 800 because it's p one over p two raised to the power of one over 1.4. We get 0.195. So let's pause on that for a moment. Does it make sense that our specific volume went down? Yes, it does because we have a compression process. It's a compression process, which means that it should be ticking up less volume. It's the same mass therefore the specific volume should be decreasing as a result of a compression process. So it totally makes sense that our specific volume went down. Does it make sense that our temperature went up? Yes, because we are compressing the gas isentropically, we are increasing the temperature as a result of pushing all the molecules together. Now specific volume two is equal to specific volume three because of the arrow. It's totally right this time. Double check that it says isocore from two to three it does. Now we have temperature and specific volume. So how do we get pressure? This time around we're going to use our ideal gas law equation again. Now we can do this in one of two ways. We can say pressure is equal to specific gas constant times temperature divided by specific volume. We can plug in t three and v three and output p three. We could also recognize that because specific volume is constant from two to three and the constant gas constant is also constant. I could write these two constants on one side of the equation and then v over r that quantity is the same at state two as it is at state three so I could write oh that'd be t over p. See algebra on my head doesn't work out. t two over p two is equal to t three over p three. Therefore p three would be equal to p two multiplied by the quantity t three over t two. Again either method works. I prefer this one because it is fewer steps so I'm going to take 800 and I'm going to multiply by 1800 divided by 543.706 and I get 2648.5. That gives me enough information to move on to state four. The process from three to four is isothermal expansion to 100 kilopascals so I didn't write 100 when I was populating the information I already knew. So knowing p three doesn't particularly healthy in p four I already had p four. I guess there's a lesson to be learned there. Sometimes what you need is inside the problem statement all along. t four and p four can be used to determine specific volume in the same way as we did for state one. We could say v four is equal to r times t four over p four. We could also use the isobaric process from four to one to write out the idogas law as over p is equal to v over t. Therefore v four could be written as t four. v over t is equal to t four times okay. I can't do algebra. v four over t four. v one over t one. Yeah it's fine. So t four times v four over t one. So I could calculate that as 1800 over 300.15 multiplied by where are you there? You should be one. Multiply by 0.861. So just for funsies here let's do it both ways. So I am taking go down slightly. t four that's 1800 divided by t one which is 300.15 and we are multiplying by 0.86139 and we get 5.16576. 5.16577. I can probably stop. Go one six six everything else was three decimal places in the specific volume column. So let it be. And then you could also write it as v four is equal to r times t four over p four. So for good measure we will check our work by writing 0.287 ish multiplied by our t four which was 1800 divided by our p four which was 100 and we get 5.166 as well. Excellent. We now have all four temperatures, pressures and specific volumes and even though that wasn't explicitly asked for in the problem even though that's partially a waste of time if I'm calculating additional properties that I don't necessarily need on an exam for example it is useful as a way of kind of practicing through the process of taking two properties at one state point and a process to a single property at a second state point and then using the information you know about the process to get to the other property at the second state point you should be able to do this for any of the problems that we're working in chapter nine even if you don't necessarily have to that skill set doesn't go away if you need a property you can determine it anyway I'll get rid of my arrows just because they haunt me of that with blunder I made earlier and I will clear up some space here so next we can use these four temperature pressure and specific volume properties to help us determine a work and heat transfer for all four processes so what I want now is the heat transfer from one to two the heat transfer from two to three the heat transfer from three to four and the heat transfer from four to one furthermore I want the work from one to two I want the work from two to three I want the work from three to four and I want the work from four to one I want a magnitude and direction for each of these eight quantities. Let's start with the easy ones. An isentropic process is going to be adiabatic. We know that because we are saying delta S of our system is added to the delta S of the surroundings to equal zero because it's an isentropic process. The entropy doesn't change and there is no generated entropy which means that this being constant and that being constant must give us zero here. So the heat transfer from one to two is zero. Done. The other easy one is going to be the isochoric work because since we're expanding and contracting a closed system the relevant work here is going to be boundary work which means that our work from two to three is zero. Then our process from one to two is going to have some amount of work as a result. Process from two to three is going to have some amount of heat transfer as a result. Process from three to four is going to have both work and heat transfer. Process from four to one is also going to have both work and heat transfer. That makes sense. So start with one to two. How do we get the work you ask? Well let's consider an energy balance. Our energy balance from one to two is of an isentropic compression process. So I'm talking about the worst drawing ever. Okay I gotta redo that. I'm talking about compressing the gas without allowing it to exchange any heat. So I'm going to start with delta e is equal to en minus e out. I recognize that it is a closed system which means that the heat transfer and work would be my only two opportunities for energies across the boundary. It's a transient process which means the left hand side of the equation matters. So delta u is delta ke plus delta pe. Then I can begin to simplify for the actual problem at hand. I have no heat transfer occurring. It's a compression process which means that work is only going to be in the inward direction. And I'm assuming no changes in kinetic or potential energy of the system itself. Which means the work in is going to be simplified down to just delta u. Which I could write as big u two minus big u one or I could write as mass two times little u two minus mass one times little u one. And because it's a closed system the mass doesn't change therefore the mass comes out and I can write that as mass times little u two minus little u one. Note that up here I wrote specific work. That's not total work. That's specific work. The reason I did that was because I have no indication as to a size of this engine. I was just told enough properties to be able to analyze it on an intensive basis. So I don't actually have any ability to involve mass. I don't know the mass. I don't know enough information to involve the mass. Therefore anything that I do is going to be relative to the amount of mass. It's going to be expressed per unit mass. It's going to be a specific quantity. So I want specific work in from one to two. That's equal to total work in here divided by mass which is just little u two minus little u one. Cool. At this point I could take t one and look up u one using my ideal gas tables. I think that's table 822. I could use t two to look up u two and I could subtract them but because it's a colder standard problem I'm assuming constant specific heats which means I'm going to substitute for delta u. Quick question. Do I use cv or cp? Correct. It is indeed cv. Remember that I define I define my cv as my du dt term for ideal gases which because we are assuming it's constant is going to be deltas. I can say du is equal to cv dt. Integrate both sides. cv comes out of the integral because it's constant. I have delta u is equal to cv times delta t. If it had been a problem involving enthalpy I would have been using the hdt and therefore it would be more useful for me to write a specific heat capacity term that involves enthalpy. Again it would come out because it's constant and I would have delta t is equal to cp times delta t. Oh yeah that's what is happening. So instead of writing little u two minus little v one I'm going to going to write cv times t two minus t one. cv we looked up is 0.718 kilojoules per kilogram kelvin and t two and t one are known it's 543.71 and 300 so calculator if you would please 0.718 multiplied by 543.71 minus 300.15 and we get 174.873 and that's the work in to the process from one to two. So I'll probably round that when I write it above but I'll leave it there for now. So the work in to is 174.84 88 excuse me and that is in the inward direction. Process one to two done. The process from two to three is going to look much the same at the beginning. Note this is one of two. I'll do two to three. We have a closed system which means that energy can cross the boundaries, heat or work. It's a transient process which means that I have to include the left hand side of the equation. My diagram is going to be a little bit different this time because the piston isn't moving. It's an isochoric process we were just told that we have heat transfer. Now it doesn't actually matter which direction we analyze the heat transfer for the purposes of the energy balance because if we get a negative number at the end that means we chose in our direction incorrectly. So in some cases it might be useful to just always call it an in or an out and then just use the positive and negative to inform what you're doing. For my purposes here though I don't need to just arbitrarily pick a direction to analyze because I can deduce the direction of the heat transfer. I have an isochoric process the volume remains the same where heat transfer occurs and the temperature goes from 543 to 1800. The temperature increases which means the energy increases which means I'm probably adding heat to the system. So I'm going to leave that as a Q in term and again if I get a negative number at the end of my calculation that just means I was wrong about the direction specifically and nothing else probably. So Q in is equal to delta u this time. It's a little Q in is going to be big Q in divided by mass which is going to be a little u3 minus a little u2 because again this process is from 2 to 3 so n minus beginning would be u3 minus u2. And I make the same substitution as earlier which was cv times t3 minus t2 so I will say 0.718 multiplied by t3 which is 1800 minus 543.7 and then a bunch of decimals. So we get negative all right I wrote 180 I was like what really calculator okay let's try that again. We got a positive number which we totally knew was going to happen the whole time. 902 is going to be our heat transfer in from 2 to 3 we're going to copy that up here so heat transfer from 2 to 3 of 902 and I'll round to two decimal places kilojoules per kilogram in the inward direction. Two processes done for 3 to 4 I have isothermal expansion copy the table no one knows let's copy this this time put that down here we're going from 3 to 4 this time to 4 and we have an isothermal expansion process from 3 to 4 that is going to affect what we cancel let's go back to this step transient process closed system expansion temperature remains the same for the gas so which work do we leave well it's an expansion process which is going to be work in the outward direction those cancel work in now what about heat transfer do we cancel it or do we leave it I'll give you a sec well you're tempted to say we cancel the heat transfers but your temptation to do that is ill advised if you have an isothermal process that means the temperature remains the same that does not mean there's no heat transfer in fact for the process to occur with a constant temperature there's probably going to be more heat transfer than there would have been otherwise I mean think about it if you're expanding something its natural tendency is going to be to try to cool down so in order to maintain a constant temperature there must be heat coming in from the outside to hold that temperature constant does that make sense so as a result of this expansion process I probably have a heat transfer in so I'm going to cancel q out and again if I get a negative number that means I've chosen correctly now what about delta u I mean right now I have delta u is equal to q in minus work out and yes I could do the same thing that I've done the last couple of times I could write that as mass times little u three excuse me little u four minus little u three and then I could divide everything by mass at which point I would have delta little u that was an interesting scroll could have delta little u u four minus u three is equal to a little q in minus a little work out again that's what happens when you divide everything by mass but u three and u four are going to be the same because for an audio gas the internal energy is only a function of temperature I mean think about if I plug in cv times t four minus t three here t three and t four are both 1800 Kelvin 1800 minus 1800 is zero zero time the number is still zero so this entire quantity is zero therefore work out is going to be equal q in we have one equation and two unknowns which means the energy balance alone isn't going to be enough I have to involve something else to calculate the heat transfer or the work so again I'll give you a minute can you think of any way that we could calculate the work or the heat transfer without using the energy balance the relevant type of work here is boundary work and we have a definition for boundary work boundary work is defined as being the integral of pressure with respect to volume we can use an expression for pressure like for example pressure times specific volume is equal to gas constant times temperature therefore pressure as a function of volume is equal to gas constant times temperature divided by specific volume therefore I could write this as the integral of r times t over specific volume times d volume could bring out the gas constant and the temperature at which point I would be left with a natural log I would actually probably want to write this in terms of total volume instead of specific volume but dividing everything by mass gets rid of the total volume in instead gets specific volume anyway so I would have rt times the natural log of v evaluated from v3 to v4 so I would say r times t times the quantity natural log of v4 minus the natural log of v3 which because of natural log rules I can write that as r times t times the natural log of v4 over v3 the temperature is constant so it doesn't matter if I plug in three or four it's 1800 kelvin the specific gas constant I evaluated already that's a 0.287 kilojoules per kilogram kelvin and v4 and v3 I happen to have already I know the numbers because I calculated them even though at the time it seemed like extra work for nothing so calculator come back we have 0.287 kilojoules per kilogram kelvin multiplied by 1800 kelvin that's going to cancel the kelvins which is going to leave me with kilojoules per kilogram and I'm multiplying by a natural log which I don't have a convenient button for in this calculator emulator so I'm going to write that as natural log and then 5.166 divided by 0.195 and I get 1692.82 and the natural log is going to produce a unitless quantity so kilojoules per kilogram kelvin multiplied by kelvin is going to yield kilojoules per kilogram multiplied by a unitless proportion which is just going to be kilojoules per kilogram in my end result so the boundary work the specific boundary work occurring in this process is 1692.82 kilojoules per kilogram that's going to be a work output because it's positive remember that we defend boundary work as being in the outward direction because our dv term here is positive for the case of expansion therefore the workout in this process and the heat transfer in for this process are both 1692.82 I will go write that down heat transfer in 1692.82 kilojoules per kilogram and that would be in the inward direction and our workout is also 1692.82 this time in the outward direction by the way that boundary work equation would have allowed us to calculate one and two without involving the delta u relationship I could have used the specific boundary work from one to two that's going to be an isentropic process which means we're going to use the poly trick form that'd be p2 v2 minus p1 v1 divided by one minus n and we are plugging in k in place of n so I could say p2 times v2 minus p1 times v1 divided by one minus 1.4 but I'm going to be even lazier than that instead of plugging in pressures in specific volumes I'm going to recognize that pressure times specific volume is gas constant times temperature because of the ideal gas law which means I can write that as r times t2 over t1 so I can say 0.287 multiplied by t2 which was 543.70591565 minus t1 which is 300.15 and divide that by one minus 1.4 and I get negative 174.751 so the boundary work the work output from one to two is negative 174.75 which means that my work in would actually have been 175.75 so I got 174.88 I think 174.75 is within enough rounding errors in the calculation itself to be the same as 174.88 yeah probably and again we could do that for the process from two to three as well I guess two to three would be boring not particularly helpful my work is zero so we can't involve that to be useful in the process from two to three okay armed with all of this skill set let's go back to analyzing one process at a time in four to one my energy balance is going to start off the same I'm going to have a simplification for the closed system energy across the boundaries heat and work and draw a little diagram I have heat transferred but I wasn't told which direction so I will have to think about it I'm going from a large specific volume to a small specific volume which means that I'm likely going to be compressing as a result of the heat transfer from four to one as it's cooling and maintaining a constant pressure I'm likely going to have heat transfer in the outward direction I know that because the temperature is dropping from 1800 to 300.15 so I'm going to call my work in from four to one and I'm going to call my heat transfer out and can I get rid of delta u no I can't that means I have to use delta u is equal to work in minus u out so in the process from one to two I had delta u and work in I got rid of q in the process from two to three I got rid of work in the process from three to four I got rid of delta u this time I have to consider all three at the same time oh that means I'm going to be using three equations to solve for these three unknowns the equation for delta u is going to be cv times t1 minus t4 the equation for work in is going to be negative boundary work that's going to be easy because it's the integral of pressure with respect to specific volume and the pressure comes out because I have a constant pressure process from four to one which means this is just pressure times our relationship between specific volume which is going to be v1 minus v4 so the pressure at four is 100 kilopascals I know both specific volumes I could do the same thing I did earlier by the way I could substitute in r times t1 minus t4 I wanted to that would be perfectly valid that might be more convenient in this this case because theoretically I'm solving for q out by rearranging the equation writing it is work in minus delta u so I could say r times t1 minus t4 except negated so it'd be r times t4 minus t1 because of this negative sign r times t4 minus t1 minus cv times t1 minus t4 did you follow all that okay good so I will say 0.287 and I will write times t4 minus t1 so 1800 minus 300.15 and that gives me 430 it's positive which means that I guessed correctly in my directionality 430.46 our heat transfer is going to be 430.46 minus cv times t1 minus t4 minus 0.718 times 300.15 minus 1800 and I get 1507 so 1507.35 kilojoules per kilogram in the outward direction and with that I have all four heat transfers and works then determining the net work is just going to be a combination of the works determining the net heat transfer is going to be the combination of the heat transfers I have 902.02 plus 1692.82 heat transfer in the inward direction which means I have more heat transfer in than out which means I'm going to have a net heat transfer in the inward direction q nut in is equal to q in minus q out which is going to be the heat transfer from 2 to 3 plus the heat transfer from 3 to 4 minus the heat transfer from 4 to 1 so 902.02 plus 1692.82 minus 1507.35 and I get 1087.49 is presumably going to be in the outward direction because it is a power cycle after all but just to double check 174.88 is in 430.46 is in combined that's I don't know about 600 1692.82 in the outward direction is still bigger than that therefore I'm going to have a network in the outward direction so network out is going to be the outs that's only work from 3 to 4 minus the ins which is the work from 1 to 2 plus the work from 4 to 1 so 1692.82 minus 174.88 plus 430.46 so I am going to get 1087.48 now remember this cycle is itself a closed system and it's operating steadily when you consider an entire cycle at a time so it has to get back to where it started every time so my network out should equal my net heat transfer in because the energy doesn't have the opportunity to do anything else so I should have gotten the exact same number for both I'm going to chalk up the differences in the hundreds place to the fact that I rounded all of these numbers to the 100th place presumably if you kept track of your decimals a little bit more carefully you would have gotten exactly the same number for both cool that's all we actually had to do for the problem statement as written but since we're here let's do some more let's say I had asked us for the thermal efficiency of this power cycle well it is a common parameter that we use to evaluate relative performance and remember that thermal efficiency is our desired effect of our power cycle which is the net workout divided by what you put in to make that happen in the case of a power cycle what we are putting in this heat transfer that's why we call it thermal efficiency it's efficiency relative to the heat added so that's q in remember don't use that q in otherwise you should get a 100% efficient turbine every excuse me power cycle every single time we only want the heat transfer in so I'm going to take 1087.48 kilojoules per kilogram and divide it by the heat transfer in the inward direction there are two sources of heat transfer in so I'm going to divide by 1692.82 plus 902.02 and I get a thermal efficiency of 41.9% that's a shame had written it to hit 42 curse you rounding errors so about 42% efficient anything else we can do while we're here well how about for fun let's draw the pv and ts diagrams so I want graphs or pressure with respect to specific volume and temperature with respect to entropy these graphs are useful for keeping track of what's happening in a cycle the pv diagram is going to give us an idea of the work that's happening because movement to the left or right is going to represent work the ts diagram is going to give us an idea of what heat transfers happening because movement to the left or right on the ts diagram represents heat transfer out as a result looking at the pv and ts diagrams can help us keep track of what's actually happening and consider the operation of the cycle as a whole so move into the right on the pv diagram represents work out move into the left represents work in the ts diagram is the opposite move into the left represents heat transfer out because of Beyonce move into the right represents heat transfer in now we could actually use our pressures and specific volumes to plot this to scale but I don't necessarily want that what I want for these sorts of drawings is just a relative idea of what's happening I don't need it to be drawn to scale I just want the positioning of the state points relative to one another to make sense so let's start by considering what's happening on the x-axis for the specific volume on the pv diagram well if movement to the left and right represents work then I should be able to look at the works and heat transfers and deduce what's happening based on the input or output of work from work from one to two I have in three to four I have out four to one I have in so from one to two I'm going to move to the left and then I'm going to stay there from two to three and then I'm going to move to the right and three to four and I'm going to move back to the left to get back to where I started so one is going to be in the middle two and three are going to be to the left and four is going to be to the right again because I have work in from one to two. I have no work from two to three, so there's no horizontal displacement. I have work out from three to four, and then I have work in again to get back to where we started. You should be able to use this reasoning to build a PV and TS diagram even if you don't have these specific properties. So yes, we totally could have looked at this and gone. 0.861 is state one, 0.195 is less than that, that's state two and three, and 5.166 is state four. But I want you to be able to think through these. That's why they're useful after all. The pressures themselves though, how about as well grab 100, 800, 2650. I'm going to grab 100 and 800 and about 2650. Draw a tiny tilde there. That doesn't help. Okay, about 2650. And these pressures are in kilopascals. So state one is going to be here. State two is going to be at 800 kilopascals. So it'll be here. State three will be up here. State four will be down here. And eventually we might get to the point where I ask you to draw these curves for the processes a little bit more accurately. But for now you can just connect them with straight lines if you want. I know that should have a curve like this. So work in from four all the way to two. You can think of the work in as the area under this curve. Work out from three to four. Which would be the area under this entire curve. And then the region enclosed is going to be the network. Because the workout is higher, there's more area going to be a network in the outward direction. Okay, one graph down. Let's do the same thing for heat transfer. So we have heat transfer in from two to three and three to four. And heat transfer out from four to one. And horizontal displacement on the TS diagram represents heat transfer. And outward heat transfer is movement to the left, inward just to the right. So from one to two I stay where I am horizontally. Then two to three I move to the right. And then three to four I move to the right. And then four to one I move back to the left to get back to where I started. So I'm going to have three locations. One and two, three, four and one. Does that make sense? You don't move because there's no heat transfer, one to two. Two to three is movement to the right because it's heat transfer in. Three to four is movement to the right because it's heat transfer in. And then four to one is movement to the left. And it gets us all the way back to where we started. I could also use the relative magnitudes of these heat transfers to place three relative to four a little bit more intentionally. But for now, all I want is general shape. Then again, because we have the temperatures, we might as well just use them 300, 543 and 1800. Those are in Kelvin. So one and two, three and four are up here. Straight line. And there are our PV and TS diagrams. Just like the PV diagram, I can look at the TS diagram and deduce what's happening with the heat transfer. I can see that there's heat transfer in from two to three, that's going to be the area under this curve. Heat transfer in from three to four, that's the area under this line. And then heat transfer out from four to one, that's the area under this curve. And the heat transfer in on a net basis is the region enclosed. Okay. Now let's really be done with the problem. Now that we have that skill set, we can actually think through models of more realistic power cycles. And as we develop these models like the auto cycle and the diesel cycle and the Brayton cycle and the Atkinson cycle and the Stirling cycle and the Ericsson cycle, remember that those sets of assumptions that go along with it are on top of our skill set for analyzing a power cycle in general, you should be able to solve through the processes and the state points without getting into what it is that makes up the auto cycle or the diesel cycle or the Brayton cycle. Does that logic make sense? Don't get too lost in the minutia of those specific assumptions that are layered on top of this analysis.