 A 2 evaporator compression refrigeration system as shown in the figure to the right uses R134A as the working fluid. The system operates evaporator 1 at 0 degrees Celsius, evaporator 2 at negative 26.4 degrees Celsius, and the condenser at 800 kPa. The refrigerant is circulated through the compressor at a rate of 0.1 kg per second, and the low-temperature evaporator serves a cooling load of 8 kW. So let's just take a minute to think about what that means. We have a vapor compression refrigeration system that is pulling heat in from two different places maintained at two different temperatures. In one of them, we are pulling heat in and maintaining an evaporator temperature at 0 degrees Celsius. In the other one, we are pulling heat in and maintaining an evaporator temperature of negative 26.4 degrees Celsius. You can think of that like maybe a refrigerator section and a freezer section of one big cooling unit, maintaining two different temperatures for two different purposes. As a result of those two different evaporator temperatures, we are going to have two different evaporator pressures. We know evaporator 2 will be the low-pressure evaporator. We know that, first of all, because it's going to operate at the saturation pressure corresponding to negative 26.4 degrees Celsius, which is lower than the saturation pressure corresponding to 0 degrees Celsius. We also know that because of this expansion valve. We are mixing together the streams at the outlets of both evaporators and in order to mix them together, they have to be at the same pressure. Because we need an expansion valve before mixing them together, that implies that evaporator 1 is at a higher pressure. The result of that intermediate pressure is that we have three different operating pressures. We have a high pressure, which is true of states 1, 2, and 3. We have a medium pressure, which is true of 4 and 5. And a low pressure, which is true of 6, 7, and 1. I have two different sources of Q in. I'm going to call the rate of heat absorbed by evaporator 1 Q dot in E1. I'm going to call the rate of heat absorbed by evaporator 2 Q dot in E2. I'm going to call the heat rejected by the condenser Q dot out. There's only a single source of work in that is to our compressor. And as a result of the flow being split across the two evaporators, we are going to have multiple mass flow rates. One mass flow rate will be true of 1, 2, and 3. Another mass flow rate will be at 4 and 5. The third mass flow rate will be at 6 and 7. 1, 2, and 3 are all going to have the highest mass flow rate, what I'm calling m dot cycle in these problems. 4 and 5 and 6 and 7 must then sum together to equal m dot cycle. The problem told me how much heat was absorbed by the low temperature evaporator, which means that I can populate Q dot in E2. It told me the mass flow rate through the compressor, which is going to be m dot cycle. And then again, m dot 4 and m dot 6 must sum together to equal m dot cycle. Since I was told the mass flow rate through the compressor, that gives me m dot cycle. I was told the condenser is maintained at 800 kilopascals. And that's all my given information. The next thing I'm going to do before we get into the cycle analysis is add an additional state point. It's not really necessary for reasons that will become a little bit more clear in a second. But it will be convenient for us to talk about it. I'm going to establish another device here, a mixing chamber. And then I'm going to establish a state point between this expansion valve and that mixing chamber. I'm going to call that state point 8. My high pressure will be at 800 kilopascals. My medium pressure will be the saturation pressure corresponding to zero degrees Celsius. My low pressure will be the saturation pressure corresponding to negative 26.4 degrees Celsius. Now let's start parsing out state point properties. Two and three are at the high pressure. That's after the compressor has compressed the fluid. But before it has been expanded at all, remember that the condenser is going to be isobaric. Then we go through one expansion valve to the intermediate pressure at states four and five. And then are expanded down to the low pressure at state eight. Three is also expanded to the low pressure directly for states six and seven. Eight and seven have to mix together to reach state point one, which means that state point one has to have the same pressure as eight and seven. So two and three are the high pressure. Four and five are the medium pressure. Six, seven, eight and one are all the low pressure. That's half of our independent intensive properties. The next thing we can do is populate isentropic and isenthalpic processes. I recognize that one to two is going to be isentropic. And then three to four and three to six are isenthalpic. Five to eight is isenthalpic. The next thing I can assume is that the condenser condenses the substance and nothing more. And that the evaporators evaporate the substance and nothing more. They aren't super heating nor super cooling the substance. That means that the quality at the outlet of the condenser, state three would be zero. The quality at the outlet of evaporator one would be one. That would be state five. The quality at the outlet of evaporator two is also one, which means state seven. And that leaves me with state point one. So let's think about state point one for a moment. If we knew everything about state points eight and seven, we could use an energy balance on the mixing chamber to figure out our enthalpy at state one. Let's leave state point one aside for a moment and come back to it. I will start with my property of lookups at these state points for which I have quality. At state point three, for example, I can look up H three. And if all I'm looking for for the moment is enthalpy, that will also give me H four. I can look up H five. I will know H six as a result of knowing H three. I can look up H seven. And then as a result of knowing H five, I will also know H eight. So I can get 123456 of the enthalpies by only doing three property lookups. For those, I'm going to have to go into my R134A tables. The relevant tables are going to be tables A11 and A12. And I can also recognize that if I'm at saturation conditions at a pressure for which I have a saturation temperature, I can just jump into the saturation tables by temperature instead of having to interpolate for that pressure. But first things first, I have a high pressure, which is 800 kilopascals. So I'm going to go into my saturation tables by pressure. 800 kilopascals would be eight bar. I'm grabbing the saturated liquid specific enthalpy, which is 93.42. That also gives me H three, excuse me, H four. Then H five is at the medium pressure, which is the saturation pressure corresponding to zero degrees Celsius, which means I can actually just go into my tables by temperature and find zero degrees Celsius and read off the saturated vapor property. Zero degrees Celsius has a saturated vapor specific enthalpy of 247.23247.23. Then because I know H three, I also know H six, 93.42 93.42. At state seven, I'm at the low pressure, which is the saturation pressure corresponding to negative 26.4 degrees Celsius, which means that I can use negative 26.4 degrees Celsius. I also, because I don't want to interpolate any more than I have to, I will just check the pressure tables to see if negative 26.4 happens to be one of my saturation temperatures. Ah, it's pretty close, but not quite. So the low temperature evaporator is operating at a pressure that is around one bar. We could probably get away with just using the one bar row here, but in the interest of some more character building, I think it would be better for us to actually perform the interpolation. So I'm going to use the low pressure, which is the saturation pressure corresponding to negative 26.4 degrees Celsius. So going into my tables, I'm going to be interpolating between negative 26 and negative 28. And I specifically want to determine the specific enthalpy of a saturated vapor. So with my handy dandy calculator, I can take negative 26.4 minus a calculator really wants that to be a negative sign, not a minus sign. So let's try that again. Negative 26.4. And then I'm going to subtract the value that is physically above it, not numerically above it, but physically above it, which is negative 28 divided by negative 26 minus negative 28. And then I'm going to set that equal to the thing that I'm looking for minus the value at negative 28, which is 230.38 divided by the value at negative 26, which is 231.62 minus the value at negative 28, 230.38. I'm looking for x and I get 231.372. Then at state point eight, I already know state point five, 247.23. That gives me my first six enthalpies. Now if I can figure out state one, I can use the enthalpy at state one to look up the entropy at state one. And I can use the entropy at state one to determine the enthalpy at state two. So for the energy balance on the mixing chamber, we would need h eight, which we have h seven, which we have, and the mass flow rates at one, eight, and seven, or the proportion of mass flow rates at eight and seven relative to one. I recognize that I know and bad cycle. I was told the mass flow rate through the compressor, which is the same as the mass flow rate through the condenser. That gives me one of my three required mass flow rates. The other thing that I can figure out here is the mass flow rate through evaporator two. I can relate the specific heat transfer in evaporator two to the enthalpy difference between six and seven, then I can use eight kilowatts to figure out what the mass flow rate of refrigerant has to be. So I go over here and set up an energy balance on evaporator two. What we are going to be left with is q dot in to e two is going to be m dot evaporator two, multiplied by the difference in enthalpy. Because the evaporator is pulling heat into it, into it from its surroundings, we should end with a higher energy than we started. Therefore, we should take h seven minus h six, then m dot e two is going to be q dot in e two divided by h seven minus h six, which would be eight kilowatts divided by h seven minus h six. H seven was 231.372, 231.372 minus h six, which is 93.42 kilojoules per kilogram. I recognize that a kilowatt is a kilojoule per second. Therefore, kilojoules will cancel and I'll be left with kilograms per second. I can formalize that a little bit more if you would like. Kilowatt is a kilojoule per second. Kilowatts cancels kilowatts, kilojoules cancels kilojoules, leaving me with kilograms per second. So come back, calculator eight divided by, that's an h calculator, we need an eight, much more better, divided by 231.372 minus 93.42 and we get a mass flow rate of about 0.058 kilograms per second. And that's all we need to figure out the last mass flow rate, because we know the mass flow rate at state point one has to equal the mass flow rate at seven plus the mass flow rate at eight from a mass balance on the mixing chamber, seven and eight have to combine to one. And I know that m dot one is 0.1 kilograms per second. And I now know m dot e two, which is my m dot seven is 0.058. That means m dot eight has to make up the difference between 0.058 and 0.1, which means that m dot eight is going to be 0.042. I can do that math on a calculator and probably get a better result than in my head, 0.058 and we get 0.042. Okay, fair. How about 0.057991? Yeah, see that's more decimal places, 0.042. Therefore, m dot e one, which you could also call m dot eight, or m dot five or m dot four is 0.042. Now that's everything we need to be able to actually do our energy balance on the mixing chamber to solve for H one energy balance on mixing chamber. We have a box of two inlets, state seven and state eight. We have one outlet, it is state one. The mixing chamber is assumed to be adiabatic because in reality it's operating very quickly. This wouldn't be so much a box as it probably would just be a t joint. And then because we have no opportunity for heat transfer, we're assuming no work, we're assuming no changes in kinetic nor potential energy. We have steady state operation of an open system. Our energy balance will simplify all the way down to the sum in of m dot h this time, because we're neglecting changes in kinetic and potential energy within the theta term is equal to the sum out of m dot h. Therefore, m dot eight h eight plus m dot seven h seven will equal m dot one h one. Therefore, h one will equal the mass flow rate through evaporator one times h eight plus the mass flow rate through evaporator two times h seven, because I have all of those enthalpies and all of those mass flow rates, I can compute a result m dot eight over m dot one will be 0.042. And just to be extra precise, I'll grab those numbers. So I have all sorts of decimal places, we're divided by 0.1 and multiplying that by our enthalpy at state eight, which was 247.23. Then we are adding to that the other proportion of mass flow rate divided by 0.1 times h seven, which is 231.372. And we get 238.034 238.034. And just to help indicate where that number came from, I'm going to highlight that in blue. And then we are going to jump up to here write h one is equal to 238.034. And we can use our enthalpy at state one and our low pressure, which is the saturation pressure corresponding to negative 26.4 degrees Celsius, to determine an entropy. Once we have that entropy, we can use the entropy at state two, which is the same as state one, and the high pressure to determine an enthalpy at state two. But before we get too far away from it, let me take a minute to point out the fact that earlier I had said, we don't really need a state point eight. And the reason we don't need an additional state point is because m dot eight is equal to m dot five, and h eight is equal to h five. So in our mass and energy balance down here, we could have just written state point five here instead of state point eight. That would have changed nothing. Because the enthalpy and the mass flow rate doesn't change across the expansion valve. And those two terms are the only terms that we actually care about in that relationship. But now that we have the enthalpy at state one, we can proceed to an entropy lookup. So what I'm going to do is recognize that my low pressure is going to be the saturation pressure at negative 26.4 degrees Celsius, which we know already is going to be about one bar close to one bar. As a result of that, I can go to my saturation tables by pressure and use the one bar information to make an educated guess about the phase of my substance at state one. If I look at one bar, I see that the saturated liquid specific enthalpy is 16.29. And the saturated vapor specific enthalpy is 231.35. The fact that my enthalpy at state one is higher than h g at one bar implies that I likely have a superheated vapor. So in order to actually evaluate my entropy at state one, I'm going to have to determine an actual pressure at state one. For that, I will go back to my saturation tables by temperature and interpolate for a saturation pressure corresponding to a temperature of negative 26.4 degrees Celsius. We would expect it to be real real close to one bar. So I want my saturation tables by temperature. And I'm going to set up essentially the same interpolation as last time. I'm going to take negative 26.4 minus negative 28 divided by negative 26 minus negative 28. I'm going to set that equal to the thing that I'm looking for minus 0.9305 divided by 1.0199 minus 0.9305. I'm looking for x and I get 1.002. So the pressure at state one is not one bar, it is instead 1.002 bar. So let's see if we have a pressure subtable corresponding to 1.002 bar in the superheated vapor region. For R134a, I have a pressure subtable at 0.6 bar and a pressure subtable at 1 bar and a pressure subtable at 1.4 bar. Unfortunately I don't have a pressure subtable corresponding to 1.002. So the best thing to do here would be to generate our own pressure subtable using an interpolation between the 1.0 bar information and the 1.4 bar information and then interpolate on that resulting table to come up with an entropy at state one. But I'm not going to do that. What I'm going to do is assume it's close enough to 1 bar that my property lookups won't be greatly affected by that minor difference. In that case, I can just use the 1 bar table to determine my value. I see 238.034 is going to occur between 236 and 244 which means that my entropy value is going to be between 0.96 and 0.99. So I'm going to be interpolating again. This time I'm taking 238.034 minus 236.54 divided by 244.7 minus 236.54 and we're setting that equal to the thing that we're looking for minus 0.9602 divided by 0.9918 minus 0.9602 and we get an entropy at state one of about 0.966. So again, we are accepting a little bit of error by recognizing that we're not actually at 1 bar. We are slightly off of 1 bar. So again, the bestest thing to have done would have been to set up a pressure sub table interpolate the for those values at 1.002 bar. But our linear interpolation is already introducing some error and the error introduced by that interpolation is probably going to be bigger than the error introduced by using 1 bar instead of 1.002 bar. Also, if you were performing this interpolation on an exam, I would be perfectly fine with you just using the 1 bar sub table. So that's what we're doing here. But now that we have a high pressure and an entropy, we can perform our lookup for H2. So our entropy is 0.966 and our pressure is 8 bar. If we go into our saturation tables by pressure and we find 8 bar, we can compare our entropy to the saturated liquid and saturated vapor specific entropies at 8 bar. I have 0.966, which is greater than SG, which means that I must have a superheated vapor. So now I'm going to go back to my superheated vapor tables. I'm going to find 8 bar. And fortunately for me, I happen to have a pressure sub table corresponding directly to 8 bar. So now I'm just going to be interpolating between 0.9374 and 0.9711 to get my enthalpy value at state two. So that interpolation would go 0.966 or perhaps 0.965986, you know, unnecessarily precise on that approximated value in the first place, minus 0.9374 divided by 0.9711 minus 0.9374 is equal to x minus 273.66 divided by 284.39 minus 273.66. We get an enthalpy at state two of 282.76. Now that we have all eight enthalpies, we can continue on to whatever the problem actually wants us to do. We've defined all of our state points. So I want to know first the cooling rate of the high temperature evaporator. Well, that cooling rate in kilowatts is going to be the mass flow rate through evaporator one times the enthalpy difference between four and five. Because it's gaining energy, I recognize that five is going to end at a higher enthalpy than four. Therefore I should write m dot e1 times h5 minus h4. So q dot e1 is equal to m dot e1 multiplied by h5 minus h4. m dot e1 was 0.042 kilograms per second. I can pop my calculator back up here 0.042. It's a less than or equal to sine calculator 0.042 multiplied by 0.042. Now we are grabbing h5 which was 247.23 and we are subtracting h4 which was 93.42 and we get 6.46 kilowatts. In the interest of being a little bit more illustrative with my calculation here, let me write out what I just multiplied. That way the people that didn't watch the video can see where those numbers were multiplied together. 247.23 minus 93.42. The kilograms cancel the kilograms. Leave me with kilojoules per second which is our definition of a kilowatt. Part A done. Part B, the power required by the compressor. The power required by the compressor is going to be the mass flow rate through the compressor which I'm just going to call m dot 1 for now times the enthalpy difference across the compressor which is going to be h2 minus h1. That would be 0.1 kilograms per second this time multiplied by h2 which was 282.762 282.762 minus h1 which was 238.034 238.064 034 kilograms, cancels kilograms, leave me with kilojoules per second which again is a kilowatt. So 0.1 multiplied by the difference between 282.762 minus 238.034 we get 4.4728. Then part C, I want to know the coefficient of performance. So the first question we have to answer before we calculate that number is is this operating as a cooling device or a heating device? Well the fact that I have two different evaporators maintaining two different refrigeration loads is a strong indication that we are operating in a cooling mode. That is cooling is the intention that's why we are going through all the effort of having multiple evaporators to accommodate different cooling loads. So I'm going to consider this a COPR. COPR is the amount of heat absorbed divided by the network that you had to put in to make this happen because this doesn't have any opportunities for work out that's just going to be work in. I could also write that as q.in divided by power input. I recognize that I have two different sources for q.in so I could write that as q.in to evaporator 1 plus q.in to evaporator 2 divided by the work in which is going to be the power input to the compressor. Now I can calculate. I have 8 plus 6.46 which gives me 14.46 kilowatts of total heat absorption. We are taking that number divided by 4.4728 and we get 3.23 so our coefficient of performance for this system is 3.23. The last thing that I wanted was a TS diagram for this device and the easiest way to start that other than drawing the saturation lines themselves is going to be to indicate three lines of constant pressure. I'll start there. My three lines of constant pressure are going to be the pressure corresponding to a saturation pressure at negative 26.4 degrees Celsius which we know is about one bar specifically 1.002 bar. The saturation pressure corresponding to a temperature of zero degrees Celsius which is going to be 4.3 bar and then the high pressure side which was 8 bar. Now that I have those three lines of constant pressure I can indicate my state points that are on the saturated liquid or saturated vapor lines. So first of all I recognize that the outlet of my condenser is a freshly condensed substance i.e. a quality of zero. That was state point three and that's on the high pressure line. Next up evaporator one outlets a freshly evaporated substance at the medium pressure evaporator two outlets a freshly evaporated substance at the low pressure then we can consider our isenthalpic processes three and four and six are all at the same enthalpy so I can indicate that on a line of constant enthalpy like this that gives me state points four and six then four is evaporating to five and six is evaporating to seven and then five is evaporating to eight. I know that eight is going to be to the right of seven because the enthalpy at eight is higher than the enthalpy at seven and then eight and seven mixed together to produce one so perhaps that could be better about indicating the line of constant enthalpy is going over there it's not quite horizontal but I don't have much space so I'm exaggerating that a little bit they mixed together to produce one which is right here in the middle and then we have an isentropic compression process from one all the way up to state point two and then two gets condensed to three this line goes this way this line goes that way that one gets an arrow and none of the other ones did like that a little bit more consistent that way and then for the people that really zoom in here beautiful and that TS diagram rounds out my problem