 A rigid tank is divided into two equal parts by a partition. One part of the tank contains two and a half kilograms of compressed water at 400 kilopascals and 60 degrees Celsius while the other part is evacuated. The partition is now removed and the water expands to fill the entire tank. Determine first the entropy change of the water during the process if the final pressure in the tank is 40 kilopascals, then the entropy change of the surrounding air assuming the temperature of that air is constant at 25 degrees Celsius, third the entropy generated by the process, fourth what does that number imply about the operation of the process, and fifth what would the best case scenario pressure be. So in this process I have rapid expansion. You'll remember from our chapter 5 discussion that ideal expansion and compression happens very very slowly. Therefore the fact that this is expanding rapidly with no restraint or restriction whatsoever, it is likely to be relatively inefficient. Exactly how inefficient we will explore by quantifying the losses with entropy. I'm going to define my system as being the mass of water. At the beginning of the process I will call that state 1. I know the initial temperature and pressure are 400 kilopascals and 60 degrees Celsius. And then at state 2 the mass isn't changed. The total volume is doubled because the problem said two equal partitions. The fact that the mass doesn't change but the volume doubles implies to us that the specific volume doubles. And then I know that the pressure at the end of the process is 40 kilopascals. So the first thing I asked for was the entropy change of the system. Well, the entropy change of the system is just going to be the entropy change of the water. The entropy change of the water is just going to be the entropy of the water at state 2 minus the entropy of the water at state 1. Where do we get the entropy of water you ask? Well, where do we get any properties of water from our steam tables? If we go into our steam tables we can see that for a given temperature and pressure we can look up the entropy as a specific quantity, lowercase s. As a result I will factor out the mass and write this as mass 2 times little s2 minus mass 1 times little s1. If I were to forget about the mass I'm assuming my little s is a big s which would leave me in a situation where I had a big s problem. I know the mass is constant so it can be factored out. Little s2 minus little s1 and I know the mass already. s1 and s2 come from property tables. So using t1 and p1 I will need to look up s1. Using v2 and p2 I will need to look up s2. Since I need v2 to look up s2 I will add specific volume 1 to my required lookups at state 1. So you guys ready for some property table lookups? Here we go. First thing I have to do is fix the phase. I'm going to look up the saturation temperature corresponding to my pressure of 400 kPa which is 4 bar. At 4 bar I have a saturation temperature of 143.6 degrees. My temperature is 60 which means that I have not yet boiled which means that I have a compressed liquid. If I go into my compressed liquid tables my lowest pressure subtable is 25 bar so I'm going to be making the approximation that my s1 and v1 are sf and vf at our temperature. I am assuming that the pressure has no effect in the compressed liquid region. So at 60 degrees Celsius s1 is going to be approximately 0.8312. So s1 is approximately sf at T1 because I have a compressed liquid at a low pressure. And that sf at 60 degrees Celsius is 0.8312. 0.8312 kilojoules per kilogram kelvin. And v1 is going to be approximately vf at T1 which is equal to 0.0010172. cubic meters per kilogram. Then I can multiply that number by 2. So calculator if you would please 0.0010172 times 2 is 0.002034. So 0.002034 cubic meters per kilogram. Now I have a pressure and a specific volume I can use that to look up the specific entropy. So first question, what is the phase at state 2? Well I go into my property tables at a pressure of 40 kilopascals. And 40 kilopascals is going to be 0.4 bar. Which means that my specific volume of a saturated liquid is 0.0010265. And my specific volume of a saturated vapor is 3.993. My specific volume at state 2 is between vf and vg. Therefore I have a saturated liquid vapor mixture. Which means that I'm going to be describing the quality at state 2. Which is not the same as the quality of the energy at state 2. As v2 minus vf divided by vg minus vf. Which is equal to s2 minus sf divided by sg minus sf. I could calculate the quality at state 2, but I don't need it right now. So I'm just going to skip directly from the specific volume to entropy. And I'm going to let my calculator do the multiplication. So I'm not going to handle that algebra so that hopefully you can see where my numbers came from a little bit more clearly. So 0.002034 minus vf which at 0.4 bar is 0.0010265. Calculator that is not a zero. What are you doing? Divided by 3.993 minus 0.0010265. And I'm setting that equal to the thing that I'm looking for minus sf which is 1.0259. Divided by sg which is 7.6700 minus 1.0259. Solving that for x and I get 1.02758. Let me double check that that wants it in kilojoules per kelvin. That implies a total entropy not a specific entropy so I'm on the right track. Now I have two and a half kilograms multiplied by the value of little s2 which is 1.02758. Minus the value of s1 which was 0.8312 and that's in kilojoules per kilogram kelvin. Kilograms cancels kilograms leaving me with a quantity in kilojoules per kelvin. So calculator you need it again. That's 2.5 multiplied by 1.02758 minus 0.8312. So my entropy change of the system is 0.49095 kilojoules per kelvin. Part A done. Part B wants us to determine the entropy change of the surroundings. Well, I could look up the entropy of the surroundings at the beginning and end of the process but I don't know enough information to do that. What I have to do instead is go back to our definition of entropy for an internally reversible process like what the surroundings are experiencing. I have del Q over T. T comes out of the integral therefore I'm left with the integral of del Q which is just Q so this becomes Q over T. So the entropy change of the surroundings in this case because it's an isothermal process and therefore internally reversible is Q over T. The T here is the temperature of the surroundings which I believe is 25 degrees Celsius. The heat transfer is going to be the heat transfer exchanged between the system and its surroundings. For that I have to consider how my heat transfer comes about. I have a rapid expansion process. Expansion generally implies a decrease in temperature. I mean if you rapidly expand something it often gets cooler so I'm generally going to have a heat transfer in as a result of that. So for the moment let's just call that Q in. Again if we get a negative number at the end for Q in that just means that I chose the wrong direction. Now Q in is to the system but is from the surroundings. So that quantity of heat transfer is the same from the perspective of the system it's an input from the perspective of the surroundings it's an output. So in order to solve for the amount of heat transferred between the system and its surroundings I'm going to figure out how much heat transfer is exchanged within the system. So if I figure out the Q in to the system that's going to be equal to the Q out from the surroundings. Does that make sense? So even though right now I'm analyzing the surroundings I'm going to do an energy balance on the system. So first up I have a transient process therefore the left hand side of my equation is important or at least not negligible. And it is a closed system so E in could be Q in or work in. E out could be Q out or work out. Next I recognize that I am assuming heat transfer is in for the moment. So I'm going to get rid of Q out and I have no opportunities for work. The only opportunity that I could have is boundary work. I know I hear you in your mind thinking but John we're expanding the volume we have a change in volume does not imply a boundary work. Well boundary work is the work associated with expansion here the expansion process is into nothing nothing is being displaced. The boundary doesn't have to push against anything as it expands therefore there is no boundary work. The change in kinetic and potential energy of the system itself is assumed to be negligibly small. So I'm going to get rid of those terms as well. At which point I have delta U is equal to Q in. So the Q in to my system which is equal to the Q out from our surroundings is equal to total U2 minus total U1 which I can write as mass 2 times U2 minus mass 1 times only U1. I factor out the mass again and I have the mass of the system times little U2 minus little U1. And again we are analyzing the system to figure out the heat transfer into the system because we know it's the same as the heat transfer out from the surroundings. So yes this is U1 and U2 of the system and the mass of the system even though we're thinking about the surroundings right now. So mass I know U1 and U2 I have to look up. So I'll add to my list of things to look up U1 and U2. U1 is going to be assumed to be UF at T1 just like S1 and V1. So jumping back into the saturation tables by temperature for was that 60 degrees Celsius? Yes indeed 60 degrees Celsius. U1 is going to be approximately 251.11. Then at state 2 I'm going to end up with the same interpolation because the interpolation for quality is also equal to U2 minus U1 divided by UG minus UF. Therefore I can say that same integral, excuse me, that same proportion can apply to internal energies. So now I'm going to write U minus 251.13 because that's the value at, nope, try that again. We are at a pressure of 40 kilopascals and at 40 kilopascals I have 0.4 bar. I'm interpolating between 317.53 and 2477.0. So let's try that again. My proportion of the way I am across my specific volume which is equal to 0.002034 minus 0.0010265 divided by 3.993 minus 0.0010265 is equal to the thing that I'm looking for minus UF at 0.4 which is 317.53. Calculator, come on. Minus 317.53. Then we divide that by 2477.0 minus 317.53. And we get an internal energy at state 1 of 318.075. So back down here I'm going to take 2.5 kilograms multiplied by 318.075 minus my U1 value which was 251.11 kilojoules per kilogram. Cancel the kilograms and I'm left with kilojoules. So 2.5 multiplied by 318.075 minus 251.11. We get 167.413 kilojoules. So the heat transfer into the system is 167.413 kilojoules and because that's the same heat transfer as what leaves the surroundings the heat transfer out from the surroundings is also equal to 167.413 kilojoules. If I paid you $5 that's a positive change to your money, a negative change to my money, my money out is 5, your money in is 5. Next I have to plug that into my equation but we have to be real careful here because the value of this delta is important because we are basing our entire simplification for delta S of the system plus delta S of the surroundings is equal to S gen based around the positive and negative aspect of this number. So we have to think through what to use as Q and for the purposes of this equation Q is in. This equation is defined with heat transfer in the inward direction. Just like how boundary work is defined with a delta V that is positive that means boundary work is defined with a work output. So if you get a positive boundary work that means that you have a work output. If you have a negative boundary work that means you have a work input. Same is true here. The Q in this equation is specifically Q in. If I have a Q out I have to plug in negative Q in. So what I plug in here is negative 167.413 kilojoules divided by 25 plus 273.15 Kelvin. For the purposes of this equation Q is always in. Another way to think through that is to think about the PV and TS diagrams. On a plot of pressure with respect to specific volume, movement to the right represents work out. Movement to the left represents work in. The equivalent graph for heat transfer is temperature with respect to entropy. And our directions are reversed. Now movement to the right represents Q in. Movement to the left represents Q out. You can remember that as being opposite just by remembering that it's opposite. You can also remember that by saying well Q out is heat is fire. You know who else is fire. Beyonce is fire. So Beyonce is Q out. Therefore any movement to the left to the left because everything you own is in a box to the left is Beyonce and therefore fire and therefore Q out. Whatever means you use to remember the directionality of heat transfer and how it affects changes in entropy. You do you. So we're saying the entropy change of the surroundings is equal to negative 167.413 divided by 25 plus 273.15. And we get a syntax error because I used a subtraction instead of a negative. So the entropy change of our surroundings is negative 57.1506 kilojoules per kelvin. And that's our answer to part B. The negative is important here because for part C we're going to be adding together these two numbers. Our entropy change of our system was a positive change of 0.49095. Our entropy change of our surroundings is negative 0.561506. So when I add those two numbers together I get the entropy generated by the process which is negative 0.07. Now what does that number imply about this process? If you said the fact that S gen is negative implies that this process cannot happen you are correct. Remember a positive S gen implies that that process is irreversible. An S gen of 0 implies that the process is perfectly reversible. There are no losses. An S gen that is less than 0 implies that the process is impossible. It would be violating the second law of thermodynamics. So for part D we say S gen is less than 0. Therefore this process is impossible. A better way of saying that would be to look at this and say, well this process cannot occur in such a way that the pressure is 40. So we're starting at 400 kPa and we are expanding to 40 kPa and we are determining that that process is impossible. So now if there were no losses, if everything were perfect, what would our pressure be? That's useful because we can figure out what the best case scenario pressure is and then because we know it can't be less than that we know that the pressure at the end of the process must be greater than or equal to whatever that result is. So for part E we want the best case scenario which means that we want a reversible process which means that we want S gen is equal to 0 and for S gen to be equal to 0 that means delta S of the system must equal delta S of the surroundings. So we could guess and check pressures and observe that as we increase the pressure this number goes down and this number goes down. However, we would observe that the only point where they are equal to each other is when they are 0 because they don't intersect anywhere else. They intersect only at 0. Therefore delta S of the surroundings is 0 and delta S of the system is 0. So what does that mean? Well, for this to be operating perfectly that means that there must be no heat transfer because delta S of the surroundings is 0 therefore it's an adiabatic process and delta S of the system is 0 which implies this is an isentropic process. So note that ideal compression and expansion is isentropic. So now in our new end state v2 is still whatever we got earlier it is still 0.002034 that's cubic meters per kilogram and now the other property that we know at the end of the process is S2. So we're saying S2 is equal to S1 that's the ideal case which would yield the best case scenario pressure at the end of the process and that specific entropy was 0.8312 so now we have two independent intensive properties now we can proceed to look up the resulting pressure the problem with this particular lookup is that our property tables are built using temperature and pressure as one of the driving properties to figure out what the pressure corresponding to this entropy and specific volume are using the tables is very difficult so difficult that I don't want us to spend any time on it so what we do instead is let a computer do the work for us if we let a computer do the work for us we end up with this MATLAB code and the pressure at the end of this loop where we are converging on an answer is 0.1987 bar so we can say that pressure is 0.1987 bar or 19.87 kPa so what does that imply? well it means that the problem with this pressure was not that it was too low it was that it was too high so we know 40 is impossible we know the best case scenario is 19.87 kPa therefore our actual pressure must be less than or equal to 19.87 kPa if everything were perfect the best case scenario pressure is 19.87 for everything else it's less than that