Okay, well good afternoon everybody. Just a couple of minutes left. Just a reminder that your second midterm exam is coming up on Monday and as I mentioned last time, well it'll be in here, you'll be allowed to use a single 3x5 inch note card written on both sides. and it's for equations only. So you're not allowed to have diagrams or worked out problems or you know methods of solution outlined or anything like that.
It's just so that you don't have to memorize equations. And as it says in the syllabus it could be anything from section 4.3 through 7.1. We'll talk more about this on Friday.
I'll set time aside at the end of class Friday to answer questions and to talk a little bit more about your exam. But at least you know what the material is going to cover. Also a reminder that you have homework due today.
Don't forget to turn that in. Thank you. The solutions actually I already posted. They're right outside my office so you can take a look at those right away.
And the homework that you've turned in will be part of the material you're responsible for in your midterm. Anything I assign this week is really not going to be included on the midterm. So it's primarily the chapter 5 and 6, maybe a little bit from chapter 4, the end of chapter 4 that deals with ideal gases, solids or liquids as are associated with first law closed system problems. Anyway, we'll talk more about that on Friday.
So let's get back to this particular example problem. So I put this on the board last time. It's 154 from Chapter 7. And I actually talked about it. I set it up. And this is all in your notes from last time.
So keep in mind that in this particular problem we're trying to find the rate of entropy generation. Pardon me. We understand hopefully. that these problems are really all first law problems with a little bit of second law thrown in at the end, right?
The way we solve this problem is no different than any other first law problem. We have to start with the first law problem so that we can find the rate of heat transfer. And then once we have the rate of heat transfer, then we can take the opposite of that to get the heat transfer rate to the surroundings. And then we can find the entropy change in the surroundings.
Of course, we find the entropy change in the system simply by knowing what my. my inlet and exit states are. Entropy, after all, is a property.
So I am given in this particular problem both the inlet and the exit state so I could very easily find the entropy that I need there as well. We're given the mass flow rate. As we talked about last time, we'll assume that there's no changes in potential or kinetic energy. So basically heat transfer minus the work equals the change in the enthalpy.
Or more specifically, rate of heat transfer minus the power of the reactor equals the mass flow rate times the change So this is what we're going to have to solve first. Now again, we have the states so we can find H1, we can find H2, we're given the mass flow rate, we're given the power. So the problem becomes almost trivial, it's just a matter of solving the first law for Q dot and then plugging it in above, again we have all of our state information, we know the entropy so we can very easily find the rate of entropy generation.
So let's just finish this problem off. And I'll just start by finding the data. So at P1 and T1 we have pressure and temperature. These are the water tables.
We'll find that this is superheated so we go into table. And it's just a matter of pulling out the data we need and we need enthalpy and we need entropy, right? So the enthalpy is 3399.5 kilojoules per kilogram. And off the same 8 megapascal subtable, at the same 500 degrees Celsius, the entropy is 6.7266.
And that's kilojoules per kilogram per Kelvin. So we have our data at 1. Now let's look at point 2. We're told that point 2 is a saturated vapor. So at P2, saturated vapor.
We go into table A5 and we just read the data. So we have a very good data. So H2 is Hg and that's 2636.1 kilojoules per kilogram and the entropy is Sg so that's 7.6691 kilojoules per kilogram per Kelvin.
So we have the data we need. Let's just plug it into the first law above. So I'm not going to rewrite the first law.
But, you know, heat transfer rate is the unknown. The power is given. So Q dot minus W dot, we're told that it's an 8.2 megawatt turbine.
The turbine is producing power output so it's a positive number. And. And let's also note that if it gives me megawatts, all the rest of my units are in kilowatts or kilojoules.
So I should throw on a 10 to the third, and that's going to be in kilowatts, which is the same as kilojoules per second. Oops, 10 to the third. So let's just make sure the units are proper. Keep in mind again the minus sign is within the equation.
It's still a positive work production but the minus sign is accurate. All right, on the right hand side of the equation we have the mass flow rate. So we have 40,000 kilograms per hour. And we've got to resolve the units here too.
So 3,600 seconds per hour. So there's the mass flow rate. and then the enthalpy change, so 2636.1 minus 3399.5 kilojoules per kilogram. Anyway, we go through the mathematics and we end up with Q dot of minus 282.6 kilojoules per second. Now let's keep in mind that this is equal and opposite to the heat transfer rate to the surroundings, right?
So if we're rejecting 282 kilojoules per second from the system, well we're rejecting it into the surroundings so that would be a positive 282.6. And lastly we go to our entropy generation equation. So the entropy generation, and I should really put total, is going to equal the entropy change of the system plus the entropy change of the surroundings.
So we just plug in the numbers. All right, so we have 40,000 over 3,600 kilograms per second, 6.7266 minus, oh, it's S2 minus S1. So it should be 7.6691 minus 6.7266. kilojoules per kilogram per Kelvin. And then we add to this 282.6 kilojoules per second.
Again, this is a positive term. And then we're given that the surroundings are at 25 degrees Celsius. So we add 273.15 and that gives me 298.15 Kelvin.
All right, so run through the mathematics and we end up with 11.4 kilojoules per second per kilogram. Now note that kilojoules per second per kilogram is the appropriate units. The entropy would be kilojoules per Kelvin. The entry rate of generation would be be that amount of entropy change per unit time, in other words per second. So kilojoules per Kelvin per second or if you will, kilojoules per second per Kelvin.
So this is the end result and the fact is that this is a positive number so this tells me. that this is a possible process. This satisfies the second law or if you will, this is an irreversible process. It's a real process. It's a possible process.
So there's your first example. Okay. So are there any questions on this particular problem?
Let me remind you again that the equations that were developed for entropy apply to all problems. They could be open system problems like this one or closed system problems like the one that we worked on last time. But the equations are the same.
So don't worry if it's open or closed. The same equations for the entropy change. That's not quite the same for the first law, right? For the first law, the open systems used enthalpy.
The closed systems used internal energy. They were different, although very similar equations. Here are the equations. equations are identical.
Is there a question? About the sign convention, was it these equations where we can only consider magnitude? So like we have the Q minus.
Right, good question. When we have a problem that involves a thermodynamic cycle and we're trying to find the thermal efficiency, the coefficient of performance, you know, where we have the QH, the QL, you know, the THTL terms and all that, in those problems Then everything is a magnitude. The sign was basically put into the equation.
So heat rejection, you'd see a minus sign within that efficiency or coefficient performance equation. So in that case, everything is a magnitude. Here, that's not true.
Okay? And these problems, you know, these are again just like the first law problems where we do have to pay special attention to the sign and make sure we've got the sign correct. Okay. So any other questions? Great.
Now let us move on. Hopefully you're finding that the use of entropy and problem solving is not really that complicated. What we've done is simply shown that for any process to be possible there has to be a net increase in the total entropy. We call this the entropy generation. These problems are no different than first law problems except you have that one entropy change equation at the end that you have to consider.
Of course you also have to take entropy data out of your tables. But besides that the problems are just first law problems. law problems so don't be confused by these problems. You have other things you'll be confused on and we'll get to that here shortly but so far these are just add-ons to first law problems.
What makes these problems somewhat unique and different is when we start looking at special cases. So let's look at some of these special cases. And the first special case we're going to look at is where we have a process that's adiabatic between the system and the surroundings.
Okay. Now we know what adiabatic means, right? Adiabatic means no heat transfer.
So basically Q is equal to zero. And of course if Q is equal to zero then clearly Q into the surroundings is going to have to equal zero. And if that's the case, then the delta S associated with the surroundings, which after all is just Q over T for the surroundings, well, clearly that has to be 0, right?
So if we were trying to find the total entropy change for this particular system, the system plus the surroundings, well, this delta S total, is just going to be, well, S2 minus S1. I'll just write it in this general form for now for the system. And then plus delta S for the surroundings. So it's just S2 minus S1 for the system. And keep in mind that, again, this has to be greater than or equal to 0, right?
So we don't want to forget our greater than or equal to 0 on the end of this particular equation. Now, what we're going to do is we're going to go We also know this is called entropy generation. So the entropy generation is the same as the total entropy change, which is the same as the entropy change of the system, but again, only for this case where it's adiabatic, right, no heat transfer. In other words, if it's a well-insulated to be a heat-contacted system, then there's not going to be any heat transfer. Nonetheless, the entropy generation equals the total entropy change equals the entropy change of the system, which just equals S2 minus S1 for the system, which is also the mass times the specific entropy change for the system, right?
Lowercase S2 minus S1, and this has to be greater than or equal to zero. Okay. So for problems that are adiabatic, we don't even worry about the surroundings. We can just use the entropy data from our tables in order to find the total entropy change or, if you will, the entropy generation. Now there's a special case of this special case and these are actually very important types of processes.
So a further special case is this. It's still adiabatic. between the system and surroundings.
But besides being adiabatic, it's also internally reversible. So and internally reversible. Now how does that actually change my equation above?
Well hardly at all, right? In fact it's exactly the same equation. Remember the greater than sign means that it's an irreversible process, a real process, right?
The equal to means. that it's the internally reversible process. So basically I get exactly the same equation as above.
I get that the entropy generation, or if you will, the entropy change total. This is going to equal the mass times S2 minus S1, but instead of greater than equal to 0, it's simply equal to 0. Okay? So that's the only difference. And of course we can then divide out the mass and then we would find therefore that S2 has to equal S1. And these processes where we have an internally reversible adiabatic process, it turns out they're also isentropic.
Remember we put the iso prefix on a term and that means that it is unchanging. So we use isentropic to represent no entropy change. So this is called an isentropic process. So reversible processes where there's no heat transfer are also isentropic processes.
Now that may sound like a, you know, pretty simple statement, but there are all sorts of problems that we're going to solve where we have to assume that it's an isentropic process. Let me just give you an example. I do have a couple of example problems here, but just so you know where we're headed with this, let's say we have something like a turbine or a pump or a compressor, some device like that.
And we're going to have to figure out how to do that. if that device were built the best it possibly could, if that were truly built as a reversible adiabatic process, then there'd be no entropy change. So what we want to do is we want to compare this ideal pump turbine compressor to the actual pump turbine compressor.
And as you'll see as we get into Chapter 7, we've got this new type of efficiency that's called isentropic efficiency. What we will do is we'll solve a problem assuming that it's isentropic. and then we'll throw on this isentropic efficiency term in order to get the actual values of something like work. So for instance, if we're solving a turbine problem like the problem that we just finished, the net result is for most of those problems that we have to find out how much work is done by that turbine or how much power is done by that turbine. So the way we're going to approach these problems again in the future, this will be next week, is we're going to assume that they're ideal and then we're going to apply this efficiency term in order to compare to what would be the ideal work that could be developed by this turbine.
And we'll do the same thing for pumps. We'll do the same thing for compressors. So I'm just giving you a bit of a heads up.
But for now, let's look at a couple example problems that deal with isentropic processes and then we'll move on and. Take it from there. So this next example problem is from Chapter 7, 7-42. Okay.
Seventh edition. Yeah, this is from the seventh. So here, I'll write this down as I start.
Alright, so this is 742 from the seventh edition. All right, so I'll read the problem and then I'll also illustrate it. All right, so what we have is a heavily insulated piston cylinder device.
Please note when it says heavily insulated it means no heat transfer, right, adiabatic. And it contains. of steam at 300 kilopascal, 200 Celsius. It says steam is now compressed in a reversible manner to a pressure of 1.2 megapascals, determine the work done during this process. So it sounds like a lot of work.
like a pretty similar problem to many of the problems we saw in chapter 4, right? It's a closed system problem. It's a piston cylinder device. We're taking some substance and we're compressing it, but this is different than the other problems we've seen because we're not given enough thermodynamic properties at the final stage.
in order to determine what that final state is. I mean we actually are and that's where entropy comes in. But if we were to look at this problem and if it didn't say that it was reversible, then we would have needed some additional information at the final state point in order to determine, well, our final properties, like our final internal energy.
So let's just illustrate the problem. All right, so we have a piston cylinder device and it says it contains steam and it says that the volume initially is.02 cubic meters. It also gives me the initial. Pressure and temperature, 300 kilopascal, 200 degrees C. It says it's heavily insulated, so I'll put insulation around it and I'll just note the Q is equal to zero.
It also says it's compressed reversibly. So remember, if it's adiabatic and reversible, by the way, technically it only has to be internally reversible, but if something is reversible, then that means it's completely reversible. In other words, both internally and externally reversible. So certainly a reversible process. is also an internally reversible process.
Anyway, this means it is isentropic, right? So whenever you see anything that implies no heat transfer and that it's reversible, we know that it's isentropic and we know that the initial and final entropies are going to be the same. Nonetheless, they also give you the final pressure of 1.2 megapascals and we're asked to find the work done for this particular process.
Okay, so work is my one and only unknown. Okay, so. Where do we begin this process? Well, again, these are really first law problems, right?
The only way we have to find the work is through the first law. We'll note that there's no heat transfer. We're So we will assume like all these closed system problems that there's no change in kinetic potential energy, right? There's no velocities or velocity changes. There's nothing measurable as far as the height change.
I mean.02 cubic meters. is kind of small and there can't be much of a height change associated with that, right? So it certainly makes sense to neglect those terms. So we end up with the first law that says that the heat transfer is going to equal the change in internal energy plus the work. And I'll put the little subscripts 1, 2. Okay, now again if you want to you can throw in the kinetic and potential energy changes, but we've already noted that they're zero.
So in order for us to finish this problem, all we really need to know is what the initial and final state is. The mass is something we could figure out. Remember that mass is just volume over specific volume and we're given the initial and final state. initial volume and we're given the initial state so we can find the initial specific volume.
So there's no problem finding the mass. We know again P1 and T1 so we can find the initial internal energy. What about U2? That might seem problematic because it appears that we only have one property, but we really don't.
Okay? If this were a problem from Chapter 4, you would have probably been told something like assume the pressure remains constant in this piston cylinder device. Or after compression, you end up with a substance at this pressure and this temperature, or at this pressure with this volume.
And then you could use the volume and the mass to find specific volume. I mean, there'd be a lot of ways to solve the problem out of Chapter 4. But here from Chapter 7, we only are given one thermodynamic property. Where's the second one?
You need two, right? You always need two. And that comes from the fact that it's isentropic.
So we really do have a second property. Anytime it's isentropic, you know that the initial and final entropies are the same. That's what gives you the second property. So essentially we've just finished the problem. We know that we have data to find all the properties of both state one and state two and the mass.
It's already been noted that there's no heat transfer. so we can very easily solve this just by pulling a little bit of data out of our tables. So let's do that.
All right, so first of all, at P1 and T1, it's steam. This is actually superheated, so we're in Table A6. So we need three things here, right?
We need the specific volume so that we can find the mass, and that's.71643 cubic meters per kilogram. We also need the specific internal energy. So that's 2651 kilojoules per kilogram.
And we also need the entropy. And the entropy is 7.3132 kilojoules per kilogram per Kelvin. All right, so we have the data at point 1. Now what about state point 2?
Well, at P2 and at S2, which again equals S1, well, here it may not be entirely clear. whether it's a two-phase mixture, whether it's saturated, whether it's superheated. So we would have to do the usual test.
I mean, again, you can do it, what I've always called the brute force method, just hunt around in the tables until you find the data. But it would be a lot easier just to compare. this to the saturation conditions.
If you go into table A5, you'll actually find that S2 is greater than SG at the pressure of P2. So clearly that means that it's a superheated vapor, right? So therefore, superheated. So we now have to go into table A6 and find the data that we're looking for.
Now, as you might expect, We're not going to find exactly what we want. Let me just go to the property tables real quickly here. All right, so we know that we have to go to the appropriate subtable. So here's my table A6 and here's 1.2. So we're going to interpolate but we have to interpolate on entropy now.
And this is what's going to make this just a little more complicated compared to others. So we go to the 1.2 megapascal subtable. It's the entropy data that we're going to have to interpolate on. I would note that we need internal energy and because of the way the data is presented, we have temperature data so we're actually going to be between the readings at 350 and at 400, right?
So just look at the entropy. We have... I'm sorry I said 350, yeah that's right, between 350 and 400. So my entropy is 7.3132 and you can see that we're somewhere in here, right, greater than 7.21 but less than 7.37 so clearly we have to do an interpolation.
So we end up with S. here of 7.3132 and then the internal energy, let's see, we have 2872.7 and 2955.5. And then my value of U2 is right in between, right?
So I just have to do the interpolation. I think I better switch to another page. Okay, so we do our interpolation.
And we find that U2 is 2921.6 kilojoules per kilogram. Okay? So, by the way, I probably should have written down the entropy numbers here as well. But again, we've done this many times now.
I mean, believe it or not, we're two months into this course, so it may not seem like it's been two months since you started, but it has. So we're all pretty familiar. with the use of these tables. We've all done interpolations before. Anyway, we're done with the interpolation.
Now it's just a matter of calculating first the mass and then we can find the work. So the mass is V1 over V1. So 0.02 cubic meters divided by 0.71643 cubic meters per kilogram, so that gives me 0.02792 kilograms.
And then lastly, the first law. So we have heat transfer minus the work. I'm just going to move the work from the right-hand side over to the left-hand side. And then this is going to equal the mass times the change and internal mass. Energy.
If you prefer, we can just swap the U1 and U2 terms and get rid of the minus sign. And then we could just go through the mathematics. And if you go through the math, again, I'm not going to bother to plug in the numbers, we end up with minus 7.55 kilojoules.
Okay? So there's an example of an isentropic process. I will note that the minus sign makes sense, right? We are compressing.
The system is not doing work to the surroundings, right? The surroundings are pushing on the piston and doing work to the system. So it's definitely a minus sign. And there's your example.
So are there any questions on this example? Great. Let's look at another example problem.
And this is also going to be for an isentropic process, but this time it's going to be for a compressor. In other words, it's a steady flow process. And this will be number 734 in your textbook.
All right. So this is actually from the eighth edition. All right, so let me read this to you and I'm actually going to add on a little bit to it. So it says that we have saturated refrigerant 134A vapor, it enters a compressor at 6 degrees Fahrenheit.
At the exit, the specific entry is the same as that at the inlet and the pressure is 80 psia. Now why they put it this way I have no idea but you know the author likes to kind of mix it up with his wording. If it says that the specific entropy is the same as that at the inlet, and it's an isentropic process, right? They could tell me it's adiabatic reversible.
They could tell me that it's isentropic. They can word it the way they've done, but it all means the same thing. So we have the inlet conditions and we have one property at the exit that says determine the exit temperature and the change in enthalpy of the refrigerant. Let me add a part B to the problem. If the volumetric flow rate at the inlet is 1 cubic foot per second, find the power required by this compressor.
So I'm just adding that little bit to it. All right, so I think that's legible. All right, so let's start by just setting up the problem. And this is a compressor. So we would generally show compressors getting smaller.
After all, we're compressing the fluid. Its volume gets smaller as its density increases. So we always show these as trapezoids with a decreasing size. And it tells me that we've got refrigerant 134A in the compressor. It says that it's a saturated vapor at 6 degrees.
So T1 is 6 degrees F, sat vapor. So that's. That's my inlet. It says that the entropy is the same. So it's isentropic, S1 equals S2.
And we're given the outlet pressure P2 of 80 PSIA. Okay. Now, again, I would note that for these flow problems, these open system problems, I could use I and E for inlet and exit, but it's customary and the author does the same thing. So just use 1s and 2s to represent inlet and exit.
Remember, 1 and 2 does not represent initial and final state within the control volume. represents the inlet and exit state for these single stream steady flow processes. And this is a single stream steady flow process so it doesn't hurt to just write SSSF. And we're trying to find the exit temperature and the enthalpy change.
So T2 is unknown and H2 minus H1 is unknown. Okay. So how do we approach this problem?
Well, pretty much just like the last problem. Ultimately, to find the enthalpy change, what we really need to do is just note that the inlet and the exit, pardon me, the inlet and the exit entropies are the same. You can see that we have enough information at the inlet state so that we can find whatever properties we need. And because S1 equals S2, we know the entropy and pressure at the exit state so we can get those properties as well.
Once we know the that we can find each state, then it's just a matter of going into the property tables to get H1 and H2. And in fact, to solve the problem the way the book has presented it, I don't even need the first law. I'm going to need the first law for my little add-on, part B. But otherwise, I simply have to recognize that I know the initial and final, I'm sorry, the inlet and exit states, so I can look up the inlet and exit properties, including the enthalpy.
So I'll just do that. So at T1, so I'm going to look at the inlet and exit. Saturated vapor. We know this is refrigerant, so we're given temperature, so we're going to table A11. We would need enthalpy and entropy.
So H1 is Hg at 6 degrees Fahrenheit, so that's 103.98. And this is all in British units, right? So BTUs per pound mass. And we also need the entropy. So S1 is equal to SG.
And that's just.22477 BTUs per pound mass per degree R. So that we know. Now let's move on. We're given an exit pressure.
So at P2 and S2, which equals S1, well, here now, it's not clear whether. This is going to be two phase. I guess we should do the usual test, right?
So we'll go into A12 and we actually find, like what we did in the previous problem, that S2 is going to be greater than SG at this particular pressure and therefore it's superheated. So we really have to go into the superheat table. So we go into A13 and once again it's going to require some interpolation. So we're going to go to the 1.2 megapascal, is that right? No.
Now the last problem. We go to the 80 PSIA subtable. And we're going to need temperature. Our interpolation is going to be on entropy, but we're also going to need enthalpy data. So I know S2 is the same as S1,.22477.
And now I just have to go onto the sub table and make sure I know where I am. So I'm going to have to go back. Okay, by the way, it should really say 812E, 813E for English units. All right, so here's A12E. We already looked at this table and concluded that it was superheated.
And by the way, it doesn't exactly, whoops, it was A11. It doesn't exactly list 6 degrees Fahrenheit, but you can approximate it, right? We know that at 6 degrees Fahrenheit.
We look for our value of entropy, which is what,.22 something? And if we look at SG over here in the last column, it turns out my value is going to be greater than that. that right I mean mine is 0.22477 even at 10 degrees is 0.22437 so clearly the entropy of the saturated vapor at 6 degrees is going to be 0.224 something something less than I've value but nonetheless you would have to check it yourself and you know you have to do a little interpolation on here just to verify that it's superheated it is superheated now that we know it's superheated we can go to the superheat tables I'm going to the 80 psi sub table which is that one there okay You know, I just realized I did not explain that properly. Six degrees Fahrenheit was the inlet conditions.
At the exit conditions, we're at 80 PSIA. I wrote it down properly, but I didn't describe it properly. All right, here ADPSIA, here's the value of SG,.22045.
My value is.22477. So again, my value of entropy is greater than SG. I don't have to do any interpolation, so please just ignore what I was saying two minutes ago. Now that we know it's superheated, now we go to A13.
ADPSIA sub table, and you can see that my value of entropy is somewhere. in here, somewhere between the saturation conditions and 80 degrees, right? So here let's note that the temperature T sat is 65.89.
It's actually listed at the top of that table. My value is going to be between that and 80 degrees. So T2 is my unknown. And the same thing with regards to the enthalpies, right? So the enthalpy is going to be 112.22.
That's the saturated vapor conditions. And then 115.51. That's the conditions at 80 PS, I'm sorry, 80 degrees Fahrenheit. And then my value of H2 is unknown, it's in between there somewhere. Now again, I should write down the entropy,.22045, and then.22663.
All right, so we just do like we always do, we interpolate. And as usual, I'm not going to go through all the mathematics, but what we find is that T2 is going to be 75.7 Fahrenheit and H2 is going to be 114.52 BTUs per pound mass. So there's the final temperature. I'm sorry, the discharge temperature. And as far as the enthalpy change, you know, we have H2.
We have H1 above. Just take the difference between the two and it gives you 10.5 approximately BTUs per pound mass. That's actually 10.46, but 10.5 is close enough. So there's the enthalpy change.
Now what about part B of the problem? First of all, I might note that I would not likely give you a problem like this one. without the part B.
I mean, why are you trying to find the enthalpy change? It's not just so that you have practice finding enthalpy changes. I mean, that's good for your first example problem.
In this case, second example problem. But these are all first law problems and. I'm going to have to use the first law. So just think of what I've done so far as just kind of the prelude to the first law problem.
So now let's look at the first law. Remember, I'm trying to find the power required. So heat transfer minus work, really, rate of heat transfer minus rate of work or power is going to equal the mass flow rate times the enthalpy change.
Now, again, I could include kinetic and potential energy change. if I had the ability to determine them, but I don't, right, I have no information at all about the height of this particular turbine, I'm sorry, compressor, I'll call it a turbine, and I know nothing about the velocities. I mean, if I were given the inlet area and an appropriate volumetric flow rate, then I could find the velocity and then I could include the kinetic energy change term here, but I can't, right, I just don't have that information, so we'll just have to assume that both kinetic and potential energy changes are.
of zero. Now this is an isentropic process, right? And the only way to have an isentropic process is if it's an adiabatic reversible process. I mean they mean the same thing. So Q is also zero, right?
It's isentropic. So therefore in order for me to find the work, and again, I'm just going to change the sign and therefore reverse the order of the enthalpy terms. In order for me to get the work, really I should say power I've got to solve this particular equation.
All right, so the question now is do I have enough information? Almost. Yes? The problem is?
Well, the only thing it said was that the specific entropy is the same as that at the inlet. So if the entropy is the same, I mean the. The way for the entropy to be the same is if it's an adiabatic reversible process. If it's not adiabatic reversible, it's not isentropic.
So if it's isentropic, then how does that figure with the different temperatures at the inlet and exit? Well, isentropic has nothing to do with temperature, nor does adiabatic, right? Adiabatic means that there's no heat transfer. Isentropic means that there's no change in entropy, but there's still energy being transferred, right?
You start with a high enthalpy, I'm sorry, you start with a low enthalpy vapor and you're doing work to it. And all that energy that's being added is going to do certain things like raise the pressure and raise the temperature. So one should understand that work has just as much ability to change properties as does heat. I mean it makes sense for us. We know that if you add heat to something, then the temperature rises.
But if you add work to something, the temperature rises too. And for those that don't believe me, just take out a paper clip and do work to it. Just, just.
Just keep bending it back and forth, back and forth. That's mechanical work. And after a while you'll have to drop the paperclip because it'll get so hot. You haven't added any heat to it.
All you've done is done work to it. Yet there's clearly a temperature change. So again, that's just to illustrate for yourself if you're interested that, you know, heat, work, temperature, they're certainly related but you can't draw conclusions on one just because there's a conclusion on the other, right? Just because entropy is constant doesn't mean enthalpy is constant.
doesn't mean just because entropy is not changing that temperature is not changing. They're different properties. They behave differently.
Any other questions? All right, so let's find the mass flow rate and then we can finish the problem. So the mass flow rate is the volumetric flow rate over the specific volume.
Now I should have found the specific volume earlier when I was up at the top of the page. We knew that we needed the initial specific volume in order to use the initial volumetric flow rate to get the mass flow rate. So if you would, if you could just go all the way back to the top and just add another line here and just say V1 equals VG and that value is 2.1126 cubic feet per pound mass.
Okay. So with this in mind, we go back to the bottom. We're told that the volumetric flow rate at the inlet is 1. So 1 cubic foot per second divided by 2.112.
six cubic feet per pound mass. And if we do the mathematics we get.4734. And that would be what? Pounds mass per second.
And now that we have the mass flow rate, now we can finish it up and find the power. So. .4734 pounds mass per second.
And then just multiply by this enthalpy change. So 103.98 minus 114.52 and that's BT. per pound mass. You can see that the pounds of mass cancels.
We get BTUs per second and it's a negative number. So minus 4.989. Btu's per second and the minus sign again makes sense, right?
It's a compressor, work is being done. Just like the previous example, it was also a problem of compression, right? You're compressing in a closed system for the previous example. Here you're compressing in an open system but it's still work input, the negative sign makes sense. I might note further that usually we wouldn't leave these answers with the units of Btu's per second.
If you divided by the appropriate conversion factor, factor then this would be minus 7.06 horsepower. So horsepower is generally considered the appropriate units when you're dealing with the British system. All right.
Yeah. To find the specific volume you went to the A11? Yeah, yeah. I mean the very first thing I did was find my enthalpy and entropy at the inlet point one.
I should have just found the specific volume at the same time. But isn't it six degrees Fahrenheit? Did I use the wrong numbers? It's very possible.
2.1126 Okay, so looking for a specific volume, saturated vapor. So for the test would you press power? You know, on an exam if I asked you for a specific set of units then I'd like you to use it, otherwise it doesn't matter.
Oh, it looks like I have a math error, don't I? Well, that's no good. Yeah, I mean clearly I have a math error.
At zero degrees Fahrenheit it's 2.15 for VG. At five degrees Fahrenheit it's 1.93. It's actually declining.
So at six degrees Fahrenheit it would probably be closer to what, 1.93? 1.96, 1.97, something like that. So it would appear that I have a math error and I'm not going to bother to redo the problem. I'll make a note to myself so I don't make the same mistake next year.
But there's a slight error. It doesn't change anything. things a lot.
If the specific volume is a little bit smaller, the mass flow rate is a little higher, and therefore the power is going to be a little bit higher. And we would have to, since it's given at 6 degrees Fahrenheit, we would have to interpolate all the beginning stuff too, right? Right.
And, you know, I'm just beginning to wonder if I pulled a lot of incorrect data. Well, at the inlet you were told that it's a saturated vapor at 6 degrees Fahrenheit. So, you know, I would have to use all the data at 6 degrees Fahrenheit.
And I'm just wondering if you could tell us It could very well be that I used a bunch of incorrect numbers. I'll have to look at it again to see. Or you could just look at it and see. But, you know, clearly V1 is wrong. It's not the right number.
Maybe H and S are wrong too. We would have to interpret it. Yeah, yeah, you'd have no choice but to interpolate. We'll just pretend it's a 5. Well, you know, often on exams I won't give you numbers that require you to interpolate, you know, just to save time.
But, you know, as long as you know the basic process. Okay, so yeah, sorry about that math error. You know, in many ways it's good making errors because it kind of forces you to understand the processes a little bit better and you know, what my thought process was as I fixed my problems.
So it never really bothers me to make errors. Oh well. So let us move on.
Let me just talk a little bit about entropy from the standpoint of diagrams. Now we've actually looked at some diagrams that are. Well, involving the specific volume, right? PV diagrams, TV diagrams.
But it makes sense now to start looking at diagrams that include entropy on one of the axes. Keep in mind that if the process is isentropic, that is if there's no entropy change, then that represents the ideal process, the best possible process, right? So it would make a certain amount of sense to show a diagram with an entropy axis because then I could just put either a vertical or horizontal line and that line will represent the ideal process.
So one useful diagram is the TS diagram. And the TS diagram, believe it or not, doesn't look a whole lot different than a TV diagram. You're still going to have your characteristic dome, your saturation region if you will. You're still going to have lines of constant pressure.
And of course pressure is constant. as long as temperature is constant. So I mean this would be a constant pressure line that would look something like that. There will be lines of constant specific volume, but they're not going to be quite the same as we've seen previously because quite frankly on a.
or TV diagram constant volume is just a vertical line. So I'm not even going to bother to mess up this diagram with the volume line. But nonetheless a TS diagram is a useful diagram for us. I believe that there's a TS diagram diagram in your book, I think it's figure A9 if I'm not mistaken.
Nonetheless, when we solve problems involving entropy, the second law types of problems that we're looking at now, often it would be a TS diagram that we would utilize. So for instance, the last problem, the one that we just finished, we would have noted that state point one was right on the saturated vapor line. And then when we compressed it, the pressure actually increased. After all, at.
At 6 degrees the saturation pressure is about 25 psia and we compressed all the way up to 80 psia. So in that particular problem we could have just shown a vertical line that goes from state point 1 to state point 2. That would have made us. Certain amount of sense. Anyway, this is just one useful diagram.
We just call this the TS diagram. Another useful diagram is a HS diagram, which is often called a Mollier diagram. Now the Mollier diagram used to be the preferred thermodynamic property data of choice.
As we talked about in the first week, we can get our property data off of tables, we can get it off of figures, we can use equations of state. Back when I started working with the started here at Cal Poly Combustion Engineering, which is one of the big boiler and power plant manufacturers, they used to give these away, just little steam tables. And within the steam tables was your Mollier diagram.
It was the easiest way to find property data. No interpolation necessary. You just knew two properties. There's lines of temperature. There's lines of pressure.
There's lines somewhere of specific volume. Anyway, you just go into the diagram at your known conditions and you read across. to get enthalpy on the left and entropy on the bottom. So the Mollier diagram is pretty useful.
Now it's not useful for compressed liquids, nor is it really useful for low quality two-phase mix. It's primarily useful for superheated vapors and high quality two-phase mixtures. This dot I put over here is the critical point.
In other words, that would be the very top on a TS diagram. or the top on a PV or the top on a TV diagram. That's a critical point.
Here the critical point is way over here on the side. So, you know, there's no space at all for compressed liquids. And, you know, once you get down to lower qualities, you basically go off the bottom of the diagram.
But nonetheless, I still like to show the Mollier diagram. And honestly, it's still a very useful way to get thermodynamic property data, especially if you would otherwise have to do an interpolation. Or sometimes you actually have to do double interpolations.
I mean what if that previous problem you were told that the exit pressure was 75 psia. Well you have an 80 psia and a 60 psia chart. So you're going to have to interpolate to create a brand new sub table at 75 psia.
And then you're going to have to interpolate again on entropy to find the enthalpy data. Now you can do that if you want, but I would much rather go into a diagram and just pick the point. Anyway, these are diagrams that are useful.
useful to us. The nice thing about both of these really is that S is on the horizontal axis. So for either of them, you know, any isentropic process is basically a vertical line. So it makes it real easy to illustrate on this particular diagram. Well, again, that was just something I wanted to mention and let us move on.
All right. Well. The next topic, actually we're moving on to the next section, 7-7. The next topic is something that I need to cover before we start getting into solids and liquids and gases.
You probably expected this was coming, right? Any of the problems we solve, we first solve them for a substance where we have a property table, right? Like water or R134A, but then very quickly we move on and we start using specific heat data in order to find our enthalpy or internal energy changes. for either ideal gases or for solids and liquids.
So that's where we're headed to now. We're going to try to find the way to solve problems for entropy change for a solid or liquid and for an ideal gas. Now unfortunately what this requires is a relationship that we do not have yet. In fact, it's a relationship that will eventually be derived, wow, I guess it's like chapter 15 or 16, it's not that high, 13, 14, 15, it's not that high, 13, 14, 15, 15, which we'll get to in ME 302. But unfortunately, we need those equations right now. We can't wait until the end of the second quarter of this class to get them.
So what I'm going to do is I'm just going to present these equations. And if it makes you feel any better, the author does exactly the same thing in the textbook. This comes from section 7-7 and these are called the TDS relationships. Now, I need these equations and next time we'll get into how we're going to use these equations. I don't think I'm going to have time today.
And maybe a little bit, we'll see. But there's two relationships that are of interest to us. One is this, and these are differential equations. Tds equals du plus pdv.
And please note that these are all lower case, su and v. These are all intensive properties. So this is one Tds relationship. And we'll see what happens. And the other TDS relationship is this one, TDS equals DH minus VDP. And again, this is lowercase S, H, and V. So these two relationships are going to be used in analysis of ideal gases and liquids.
and solids. Okay. So you should just be aware of that. Now how they're going to be used is, well, not entirely obvious, but I'm going to derive some relationships for you and let's just see if I can derive this first one.
So we now move on to section 7-8. So we're actually about, oh, it's beeping later in the day than last time. Usually it's about 1-30 that that happens. Okay. So let us begin by looking at solids and liquids.
Okay. And keep in mind that what I'm trying to do is I'm trying to find an equation for entropy. Remember, I don't have property tables at this point. I need some sort of equation of state to compare and relate. different properties, so that's what I need.
And we start with the first of the two TDS relationships. So from TDS equals DU plus PDV. Let me just make the following. following note that for solids and liquids first of all the specific volume is very small, right?
We know that the density of solids and liquids is high and therefore the specific volume is very small. And also solids and liquids are incompressible. And therefore, any change in specific volume is going to be approximately equal to zero.
So therefore, can't we just write that TDS is equal to DU? And we can further write that ds is du over t. And further, you may recall from when we first discussed all this back in, I guess it was chapter 3, right?
Maybe it was chapter 4. But when we were talking about solids and liquids, it was shown earlier that du. to what we call the heat capacity times DT. So for a solid or liquid, DU equals heat capacity DT. And therefore, DS is going to equal CDT over T. Okay.
Now, you may recall that when we were dealing with solids or liquids, we did have the option of integrating, now granted we were looking at internal energy or enthalpy change then, but we had an equation for those properties as a function of heat capacity and temperature. And. And I don't really want to integrate.
First of all, we would have to have a functional relationship between heat capacity and temperature and we just don't. All we have is heat capacity data for a variety of temperatures. So what we want to do is this.
If we assume that C is a constant, so for C, which remember, heat capacity. For c equal to a constant, well, we still have to integrate, but it makes the integral a lot easier. So first of all, c is a constant, so therefore the integral of ds, that's just going to be, that's going to be the entropy change.
Let me rewrite that. S2 minus S1 is the integral of DS, right, from 1 to 2. And this is going to be, therefore, the integral from 1 to 2 of CDT over T. But if C is a constant, it comes out of the integral.
And the integral of DT is just a natural log. Sorry, integral of dt over t is just a natural log. So this equals c times the natural log of t. And evaluated between t1 and t2, this just becomes c times natural log t2 over t1.
Okay. So this, believe it or not, as simple as it is, is how we're going to find the entropy change for a solid or liquid. No more complicated than that.
I will note that A3 is where you get your data for solids and liquids. I think we've seen that already. And I'll also note that we should use C at the average temperature of the process. Now I might just note that that's if it's available, right? We've already seen that for some substances we only have a single value of the heat capacity at room temperature, right?
But for other substances we have a range of heat capacities over a range of temperatures. So if you have a range of heat capacities, you can have a range of heat capacities over a range of temperature. So if you If we have the ability, if we know C at the average temperature of the process, then we'll use C at the average temperature of the process. Okay.
So this is it. Now all I really have to do is give you an example problem. So is there a question? Well, one thing that was also identified way back again in chapter three or so was that the specific heat at constant volume is equal to the specific heat at constant pressure for a solid or liquid.
And therefore we just give it one name. We call it C or heat capacity. In table A3 it actually says CP, you know, specific heat at constant volume.
But the fact is that's the heat capacity. It's exactly the same. All right.
So any other questions? All right, fantastic. So that's it for the day. Remember, I'll set aside time on Friday, the last 10 minutes or so, to talk about your midterm. You can use a 3 by 5 note card on your midterm with equations only, so you can start working on those now.
See you guys on Friday.