 air enters the compressor of a stationary gas-driven engine steadily at 100 kilopascals, 27 degrees Celsius, and 5 cubic meters per second. The engine features a pressure ratio of 10 to 1 and is designed to induce a maximum temperature of 1400 Kelvin. Assuming the cold air standard determine or complete the following. First, the temperature and pressure at all state points, then the net power output, then the required mass flow rate of fuel assuming that the engine runs on pure ethanol, and lastly the thermal efficiency of the engine. First, I will recognize that I'm analyzing a simple Brayton cycle. I know it's simple, first of all because it's in the problem statement, but also because there is no indication of any of the devices that we're going to be adding to the Brayton cycle down the road. General rule of thumb, if you aren't told enough information to deduce otherwise, assume it's the least complicated setup, or the most ideal setup. Then I will draw a system diagram. So I have isentropic compression from 1 to 2 because I wasn't told enough information to deduce that the compressor and turbine were operating at anything other than 100% isentropic efficiency. Then I have a combustion chamber from 2 to 3 that operates isobarically, again isentropic expansion from 3 to 4, and then the process from 4 to 1 just completes the cycle. It is a QL process, so I'm calling it the QL box, and that's isobaric heat rejection. Next, I will populate the state point properties that I know and the ones that I'm looking for into a table. Then I will populate the information that I know so far. I know the temperature at state 1 is 27 degrees Celsius, 27 plus 273.15 would be 300.15. The pressure at state 1 was given as being 100 kilopascals, and then I was told the volumetric flow rate. Next, I know that the pressure ratio across the compressor is 10. So what does that mean about my pressure at 2? You're right, it is 10 times the pressure at 1 because the pressure ratio rp represents p2 over p1 in this case. Therefore, p2 is equal to rp times p1, which is going to be 1,000. Do I know any other pressures? Well, I know the pressure at 2 and 3 is constant because the combustion chamber operates isobarically, and I know the pressure at 4 and 1 is constant because the QL box is an isobaric heat rejection process. Cool, I have all my pressures already. That was easy. Then I was told another temperature, it's 1400 Kelvin. Which state point is likely to have the maximum temperature? You're right, it is T3. Why is it T3? Because the heat addition process is from 2 to 3, and generally speaking the highest temperature is going to occur after the heat addition process. Furthermore, we could think through this process. We have a compression process from 1 to 2 that will increase the temperature. Then I have a heat addition process from 2 to 3 that will also increase the temperature. And then I have an expansion process from 3 to 4 which will drop the pressure, and then I have a heat rejection process from 4 to 1 which will also drop the temperature. Therefore, the highest temperature will have to have occurred at state 3. Is the volumetric flow rate the same at 2, 3, 4, and 1? No. Remember that the mass flow rate is the same, but because the gas may have a different density that does not mean that the volumetric flow rate is the same. There is a such thing as the law of conservation of mass, there is no such law for the conservation of volumetric flow rate. So all I really know about the volumetric flow rate is that it is 5 cubic meters per second at state 1. Well, let's leave the volumetric flow rate aside for a moment and think about T2 and T4. How do I come up with those two temperatures? Well, if I look at the process from 1 to 2, I have an isentropic compression process of an ideal gas, and because I'm using the cold air standard I know that the specific heat capacity will be constant. So because it's an isentropic process of an ideal gas with constant specific heats, I'm going to return to our good friends the isentropic ideal gas relations. Now which form is going to be the most useful to get from T1 to T2? It's this one here. I know P2 over P1, that's RP, it's 10. And K for air I can look up. Therefore, T2 would equal T1 times RP raised to the K minus 1 over K. So that would be 300.15. Multiplying by 10 raised to the power of per K at 300 Kelvin, we turn to table A20. On table A20 I have the CP value, the CV value, and the K value for air as a function of temperature. And what I want is 300 Kelvin because that's the temperature used by the cold air standard. So I'm going to use 1.4 as my value. So if I pop up my good friend the calculator here and take 300.15 multiplied by 10 raised to the 0.4 over 1.4, I get 579.499 Kelvin. Then from 3 to 4 I have an isentropic expansion process so I can still use the same equation except this time it'll be T4 over T3 is equal to P4 over P3 raised to the K minus 1 over K. Therefore T4 is going to equal T3 multiplied by the proportion P4 over P3 raised to the K minus 1 over K. Now is P4 over P3 10 or 1 over 10? That's right it's 1 over 10. This is 1 over RP because the pressure ratio is the big pressure divided by the small pressure. So then T4 would equal T3 multiplied by 1 over 10 raised to the 0.4 over 1.4. T3 was given it's 1400 Kelvin. Therefore I'm taking 1400 multiplied by 1 over 10 raised to the power of not P 1.4 minus 1 divided by 1.4. We get 725 by 126 neat. And while we're looking at these temperatures I will point out that having a maximum temperature is often a design criteria of a gas turbine engine. You typically want as much power or efficiency as you can get from a given size and that will come from having higher and higher maximum temperatures. So you typically want the highest temperature that you can have reasonably in your engine and the failing point for a lot of these gas turbine engines is the design and material used in the turbine blades. Depending on the design and cost of your turbine you might be able to withstand a higher maximum temperature but regardless the maximum temperature is something that you design around. Anyway now that we have temperature and pressure we could move on to part B but I don't want to. Let's calculate some volumetric flow rates. We don't need them. They're not going to be useful but they will be fun. So how do we do that? Well we know the mass flow rate is going to be the same at all four state points. How do I know that? Because each of these devices in the cycle is a steady flow device. They're all flowing steadily and if each device is operating steadily then there can be no change in the mass of the system with respect to time. Therefore the entering mass must be the same as the exiting mass and because there's only one mass flow rate in and one mass flow rate out for each of these devices that means m dot one is equal to m dot two which is equal to m dot three which is equal to m dot four. Now if we could calculate the volumetric flow rate corresponding to a mass flow rate we could determine the volumetric flow rate for two three and four but in order to do that we have to have a mass flow rate first. So which state point can we calculate a mass flow rate for? That's right state one. Here the useful form is going to be m dot one is equal to density times volumetric flow rate one which is equal to volumetric flow rate divided by a specific volume. Since the air is an ideal gas I can describe the specific volume in terms of temperature and pressure as well as the specific gas constant for air. Therefore m dot one is going to be volumetric flow rate one divided by r of air times t one divided by p one. See I told you this is going to be fun. So volumetric flow rate one was five cubic meters per second p one is 100 kilopascals r of air is 0.287 kilojoules per kilogram kelvin and t one is 300.15 kelvin. Again the specific gas constant for air is going to be the universal gas constant divided by molar mass for air. Universal gas constant is 8.304 kilojoules per kilomole kelvin. The molar mass for air is 28.97 kilograms per kilomole that gives you approximately 0.287 kilojoules per kilogram kelvin. Now kelvin cancels kelvin but nothing else really cancels immediately so I will break apart the kilojoule and the kilopascal into their component parts. A kilojoule could be described as a kilonewton times a meter and the kilopascal could be described as a kilonewton per square meter and now a kilopascals cancels kilopascals kilodjoules cancels kilodjoules kilonewtons cancels kilonewtons square meters and meters cancels cubic meters giving me kilograms per second. Therefore if I pop the calculator back and I take five times 100 divided by 0.287 times 300.15 I will have a mass flow rate so the mass flow rate of air at state one and therefore everywhere is 5.8 kilograms of air per second. Henceforth I will just be calling that m dot air. Now that I have m dot air I know m dot one m dot two m dot three and m dot four I can determine the rest of the volumetric flow rates. Yeah let's do it. I'm going to use this relationship here except solved for volumetric flow rate so volumetric flow rate is going to equal mass flow rate times r of air times temperature divided by pressure. So if I plug in m dot one t one and p one I will get volumetric flow rate one and if I plug in m two t two and p two I will get volumetric flow rate two. All the mass flow rates are the same so I will just call that m dot air and we can start rolling. 5.8 kilograms of air per second multiplied by 0.287 kilojoules per kilogram kelvin times our temperature at state two was determined to be 579.499 and our pressure at state two was a thousand kilobascals and again in order to get kilojoules and kilobascals to cancel their relative components I'm going to break them apart. A kilobascal not to be confused with a squiggle can be written out as a kilonewton per square meter and a kilojoule can be written out as a kilonewton times a meter. Now a kilonewton cancels kilonewton kilopascal cancels kilopascals kilodjoules cancels kilodjoules square meters and meters cancel nothing because they're what I want and kilograms cancels kilograms kelvin cancels kelvin seconds in the denominator. So v dot two is going to be 5.8 0.43 multiplied by 0.287 multiplied by 579.499 divided by a thousand and I get 0.965 and then for state three I would just change state two to state three so v bar dot three is equal to m dot air three times r of air times t three over p three and since p two is equal to p three the only thing I have to change is my temperature instead of 579 I'm going to be using 1400 so the volumetric flow rate at the inlet to the turbine is 2.332 and then one more time for all the beans I change p three to p four I change t three to t four which was 725 wait one two six I suppose I can just scroll up and grab that and I have 12.079 so that was fun nothing builds character like calculating stuff that you don't need I mean we could come up with more categories of things that I could calculate here but we should probably move on I mean we could come up with more columns of stuff to calculate but I think that's enough character building for now now next I don't want to calculate the net power yet I want to calculate the specific work in the specific q in the specific workout and the specific q out again not specifically necessary but a good opportunity to explore the cycle a little bit more and since this is a stand-in for all simple Brayton cycles we might as well be exhaustive in our analysis so if you've been following along with the auto and diesel cycles you probably know what's coming here in order to solve for the specific work in I have to do an energy balance on all of the places where work in appears luckily for me that's only one device the compressor so in my compressor I have an open system operating steadily there's work in and because it's isentropic that implies adiabatic so I'm drawing fuzzy lines to indicate perfectly insulated so it begins like all energy balances with delta e is equal to en minus e out and then I'm going to divide by dt in order to write d e dt is equal to e dot in minus e dot out and then d e dt is zero because of steady state therefore e dot in equals e dot out it's an open system so energy could enter or exit as heat transfer work or the energy associated with a mass crossing the boundary I'll write that as q dot in work dot in and the sum in of m dot theta and energy could leave as q dot out work dot out this sum out of m dot theta then I can begin to cancel irrelevant terms I have an adiabatic process because isentropic implies adiabatic therefore q in and q out disappear I have a compressor so the work is only in the inward direction and then sum in of m dot theta is really just m dot one theta one and the only outgoing mass is state two so that summation becomes m dot two times theta two then remember that theta for the entering or exiting mass is its enthalpy plus its specific kinetic energy plus its specific potential energy therefore I'm writing power input plus m dot one times h one plus v one squared over two plus g z one is equal to m dot two times h two plus v two squared over two plus g z two next I recognize that I don't have enough information to relate the inlet and outlet velocities because I don't know the cross-sectional areas and I can't determine the average velocity of a volumetric flow rate if I don't know the cross-sectional area plus it's reasonable to assume that all of the work in is going into enthalpy and not much is going into kinetic energy because again if we don't know enough information to deduce otherwise we assume ideal operation therefore I'm assuming v one is approximately equal to v two therefore the specific kinetic energy terms disappear again that's not because there is no kinetic energy it's just that whatever it is at state one it's about the same at state two similarly I was given no information as to a change in elevation therefore it's reasonable to assume that the height difference between state one and state two is minimal therefore I'm neglecting the change in specific potential energy as well that leaves me with m dot one h one and m dot two h two then I recognize that I'm looking for work in so I will solve for power input that would be m dot two h two minus m dot one h one and then I recognize that my mass flow rate is the same at one and two so I will factor that out and I can write power input is equal to m dot times h two minus h one but I didn't write power input up here now did I I wrote the specific work in therefore I'm going to write this as a specific work in which would be a power input divided by mass flow rate which is mass flow rate times h two minus h one divided by mass flow rate therefore my specific work in would be h two minus h one one and because I've assumed constant specific heats by using the cold air standard that would be cp times t two minus t one therefore for work in I can write h two minus h one which because of the cold air standard is equal to cp times t two minus t one why don't you try the other three energy balances on your own all three of them will get to right here and the only difference will be which terms you cancel in this step for the compression process we had neglected heat transfer because it was an isotropic process and work out because our work was only in the inward direction for the combustion chamber I'm going to neglect works because it's a control volume there's no opportunity for boundary work and there's no indication of any other types of work occurring and I'm only going to be using q in because it's a combustion chamber therefore I can cancel q dot out therefore my specific q in is going to be uppercase q in divided by mass flow rate which is going to be m dot times h three minus h two because at that point I would have canceled the specific kinetic energy terms again and the specific potential energy terms again and I would have m dot times h three minus h two and then I'm dividing that by m dot which yields h three minus h two the turbine energy balance looks very similar to the compressor the primary difference being that instead of only looking for work in this time I'm only looking for work out therefore I'm going to be left with total power output is equal to m dot three h three minus m dot four h four I factor out the mass again and specific work would be m dot times h three minus h four divided by m dot that time it's inlet minus outlet because I have work out appearing on the right hand side of my energy balance for q out I'm looking at the q out box I have no works because it's boring there's no change in volume so I have no boundary work there's no indications of any other type of work and I'm neglecting q in because he transfers only in the outward direction it's a cooling process I neglect specific kinetic energy and specific potential energy again and it simplifies down to h four minus h one again it's beginning minus end and all three of these become cp times t three minus t two and cp times t three minus t four and cp times t four minus t one before we move on I will point out that I would discourage you from just writing down these as like the Brayden cycle equations in a notebook somewhere because your state point properties might be different you might have different processes occurring between different state points I mean if we had called state one over here and then two over here and then three over here and four over here all four of those equations would be different furthermore as I start adding more stuff into my Brayden cycle my analysis is going to get more complex and it's much more useful to be able to build these on the fly for any cycle in front of you as opposed to relying on predefined notes all the time so again I would discourage you from just writing down a section in your notes as the Brayden cycle equations similarly don't write down the auto cycle equations or the diesel cycle equations just understand how to build them for the thing in front of you since I know all four temperatures once I look up cp I will be made in the shade so for cp we go back to table a 20 and for air at 300 kelvin cp is 1.005 kilojoules per kilogram kelvin therefore I'm taking 1.005 kilojoules per kilogram kelvin multiplied by a temperature difference in kelvin which will give me a quantity in kilojoules per kilogram if I pop up the calculator again we can get started on this 1.005 multiplied by 579.499 minus 300.15 yields 280.746 and then 1.005 times 1400 minus 579.499 yields a hue in of 824.604 and then 1.005 times 1400 minus 579.499 yields 824.604 and then 1.005 times 1400 minus 725.126 yields a workout of 678.248 and 1.005 times 725.126 minus 300.15 gives me q out 427.101 okay so we had 280.746 and then 824.604 and then 678.248 and then we had 427.101 so from this we can determine a specific network out which would be workout minus work in and a specific heat transfer in which would be q in minus q out that specific network out was 678 minus 280 and I get 397.502 and for q in minus q out I have 678 minus no that's not right hold up that was 824 minus 427 which is 397.502 because they match that's a good indication that I built these four equations correctly yay and you probably know what's coming next I mean after I move these equations out of the way of course just a little bit of a scooch that's right I want to calculate the thermal efficiency so the network out is 397.502 I guess I should write that down is the network out divided by q in okay let's try that again 397.502 divided by q in which was a number that started with an 8 and that was 0.48 okay so again we didn't need all of those steps actually calculated we could have gotten to the actual answers by determining only the quantities that we actually needed but we took a couple of additional steps to just you know get us used to analyzing this particular type of power cycle and the different ways in which we can evaluate parameters but now that I have this information I will actually start working on what I need so part a is done that's up here for part b I don't have the net power output yet but I'm pretty dang close the net power output is just going to be the mass flow rate of air times the specific network out see guys aren't you happy that I made us calculate the mass flow rate of air even though we didn't need it because if we didn't have it now we would have to calculate it now and that would just be chaos so we have 5.8043 5.8043 how far on my calculator do I have to scroll up in order to not just type it in and then we are multiplying by our specific network out which was 397.502 and I get 2307 now what are the units on that that is an excellent question I'm taking a quantity in kilograms per second and I'm multiplying by a quantity in kilojoules per kilogram the kilograms cancel kilograms leaving me with kilojoules per second and that is the definition of a kilowatt therefore I have 2307.2 kilowatts let's try that again much more better okay part c I want the required mass flow rate of fuel assuming the engine runs on pure ethanol so for that we are going to be relating the mass flow rate of fuel to the amount of heat generated by burning the fuel for that I need to use the heating value of our fuel because the amount of heat emitted by burning the fuel is going to be the mass flow rate of fuel multiplied by the heating value of the fuel so if I get like 30,000 kilojoules for every kilogram of fuel burned then if I multiply that quantity in kilojoules per kilogram by a number of kilograms per second I will get a rate of energy emitted and then like with our analysis of the reciprocating engines we are going to be assuming ideal combustion here so all of the heat is going into the air we have perfect heating efficiency therefore q dot in is equal to q dot fuel and q dot in is just going to be m dot air times the specific q in and the q dot fuel is again m dot fuel times heating value of the fuel therefore m dot fuel is m dot air times little q in divided by the heating value of the fuel we have m dot air we have specific heat transfer in all we need is the heating value of the fuel for that we go back into our tables and this time I am looking for table a 25 so if I scroll down on this table and find ethanol we again see that we have two options one is the ethanol in gas form in vapor form or in liquid form while I'm treating this as vaporized fuel just like we did in the reciprocating engine analysis and we are going to be using the lower heating value because we don't have some hyper efficient natural gas furnace here we just have a box with a bunch of fuel being sprayed into it with a big fire constantly going so the lower heating value is the right column therefore I'm using 27,720 kilojoules per kilogram that's from table a 25 heating value of ethanol is that was 27,720 yes calculator scroll up thank you very helpful want to do is write a number didn't realize it was that hard okay and again that's kilojoules of energy produced per kilogram of fuel burn so now we can take 5.8 kilograms of air per second and then I'm multiplying by a number that starts with eight I will pop up the calculator so I don't have to scroll up hopefully avoid at least a little bit of nauseation and that would be kilojoules per kilogram of air and then I'm dividing by 27,720 kilojoules of energy produced for every kilogram of fuel burned and kilograms of air cancels kilograms of air and kilojoules cancels kilojoules leaving me with kilograms of fuel per second so I'm taking that 5.8 number which is here is it worth backspacing as opposed to scrolling up probably not so 5.8043 multiplied by 824 let's scroll on up and then we are dividing by 27,720 and we get 0.1726 that's kilograms of fuel burned every second quite a bit of fuel so we knew the volume of the gas tank we could figure out how much mass of fuel could fit in that if we knew the density of ethanol and we could figure out how quickly it would take to drain the fuel tank lastly I wanted the thermal efficiency of the engine which I already calculated hooray we did it we quantified all of the things and we could be done but in the interests of character building it'd be a shame to let this go without drawing a ts diagram you'll notice that I'm saying ts diagram as opposed to pv and ts diagrams because for open analysis like this where the mass lorries are all the same the pv diagram is not particularly helpful instead it's more useful just to look at the ts diagrams so I will draw a ts diagram we've gone over this a few times already but just to recap move into the left on the ts diagram represents q out move into the right represents q in and we can reason through how many horizontal locations we have on our ts diagram by thinking through the heat transfer processes from one to two we have isentropic compression which is going to be no horizontal displacement and then I'm going to erase the parentheses around my s because that was a mistake and then we have the process from two to three which is iso barricade addition so from two to three we are going to move to the right on our ts diagram that doesn't mean directly to the right that just means we have some horizontal displacement to the right and then from three to four I have isentropic expansion again no horizontal displacement there because it's an isentropic process and then four to one I have isentropic heat rejection which is q out which is movement back to the left therefore I have two different horizontal locations one and two and three and four then for the temperatures I recognize that I have a low temperature and I go up a bit for t2 and I go up more for the fire so it'll be one two three and then four is going to be back down because I have isentropic expansion I don't know the relative position of two and four from just thinking through the processes alone luckily for us we calculated an actual number four was higher than two therefore I'll put four here and since I'm drawing these as state points I will get rid of the unit there because it's not one two three and four Kelvin now it's my favorite game just connect the dots one two three four and then we will draw one to two as a vertical line because again isentropic three to four is a vertical line because again isentropic two to three is going to be an isobaric process in the TS diagram which is a curve and one to four is an isobaric process which is also a curve technically speaking I didn't draw three and four quite high enough to be actually aligned with their descriptions over here but you know I think it's fine and then Q in is going to be the area under the curve from two to three Q out is going to be the area under the curve from four back to one therefore the difference between the two this region enclosed is going to be the net heat transfer in the inward direction now it could be done but there's probably more stuff we could calculate for this oh I've got one how about the back work ratio sometimes when we are talking about axial devices in this case a compressor and a turbine that are driven along the same shaft it is useful to describe the proportion of work out that goes back in to power the input devices so in this case that would be the proportion of work out that goes back to power the compressor you can think of that like how much of your raw revenue is going back into the business to make that revenue happen that back work ratio which is abbreviated BWR is just work in over work out and it is a percentage that represents how much of your workout has to go back and power the compressor that should be easy because we have both of those quantities calculated already so I will take the work in divided by the work out that work in term was 280 the work out term was I know at this point I definitely could have just re-typed the numbers but you know what this is what we're doing we're dividing by 678 not just putting two numbers back to back and we get 0.4139 therefore my back work ratio is about 41% that's how much of the work produced by the turbine goes into powering the compressor so I guess that's pretty much everything that we can calculate reasonably for this so I'll just pose a couple of questions for you what if I were to ask you a question like how would changing the pressure ratio affect the thermal efficiency let's say or how would it affect the power output or how would changing T1 and P1 affect our power output or thermal efficiency or how would changing the maximum temperature affect the power output or the thermal efficiency well most of those you can reason through on the TS diagram for example that pressure question if I change the pressure ratio from 10 to 11 how would that affect the thermal efficiency and I think that the easiest way to think through that is by visualizing the proportions on the TS diagram our lines of constant pressure this is below 100 kilobascals and P high appear as curves in the TS diagram and they increase their vertical position as the pressure increases so 1,000 kilobascals is higher than 100 kilobascals so 2,000 kilobascals would be even higher than that therefore 1,100 would be a little bit higher than 1,000 now remember the region enclosed represents our net Q in which is the same as our network out and the area under this curve from 2 to 3 is our Q in network out divided by Q in which would be Q net in divided by Q in is our thermal efficiency it's the visual proportion that this region takes up relative to this region maybe that'll make more sense if I draw it with a different color here so the very crude representation of the red hatches divided by the very crude representation of the blue hatches is our thermal efficiency so if you move the top line up you are increasing Q net in therefore increasing network out and also increasing Q in and you're increasing them by the same quantity so how does that affect the thermal efficiency it increases it I mean if you had 1 divided by 2 that's one half and if you had the same amount added to both that would be a higher number therefore moving the top line up increasing the maximum pressure is going to improve the thermal efficiency then we could think through something like well what about increasing the maximum temperature well with a higher maximum temperature we are essentially moving the right line further right and because the lines of constant pressure get further apart the further to the right you go on the graph they are diverging from each other you are adding more region here than you are proportionally anywhere to the left therefore you are also improving the thermal efficiency or what if I ask you a question like what if you were to increase the low pressure if the incoming pressure was 125 kPa instead of 100 how would that affect the thermal efficiency well Q net in decreases Q in stays the same therefore a lower thermal efficiency I think thinking through this on a TS diagram is the easiest way to answer those sorts of questions quickly if you want to be more exact about it you probably know where I'm going next that's right you should open this in MATLAB this analysis here is the same analysis we went through by hand except performed by MATLAB and the advantage of this is I can change a quantity and observe the effects immediately so instead of a pressure ratio of 10 you can change that to 11 and instead of a thermal efficiency of 48.2% I get 49.6% and if I were to change rp back to 10 and then increase the pressure but remember that I said moving the bottom line up and not affecting the top line so I have to change my pressure ratio to keep it at the same pressure so change that from 100 to 200 and then change rp to 5 to keep p2 at a thousand still that's 36.85% the thermal efficiency has dropped neat right right we could even make MATLAB plot the thermal efficiency as a function of rp but since we've done that for the last couple of examples I will leave that as an exercise for you guys if you are so interested again this MATLAB code is posted on D2L for you to play with I think it's a good time