 A turbojet aircraft is flying with a velocity of 320 meters per second at an altitude of 9,150 meters where the ambient conditions are 32 kPa and negative 32 degrees Celsius. The pressure ratio across the compressor is 12 and the temperature at the turbine inlet is 1,400 Kelvin. Air enters the compressor at a rate of 60 kilograms per second and the jet fuel has a heating value of 42,700 kilojoules per kilogram. Assuming ideal operation for all components and constant specific heats of air at room temperature determine a the velocity of the exhaust gases, b the propulsive power developed, and c the rate of fuel consumption. I will begin with a diagram. So our center three components here are the same devices that make up our simple Brayton cycle. We have a compressor which is increasing the pressure of the gas prior to combustion, then a combustion chamber which adds energy to the gas which expands it, the turbine which harvests some power out of that expansion. But remember that our goal is no longer to produce power. Instead we want thrust which requires a lot of velocity. So we need to tack on a device at the end of our engine to convert that enthalpy that's left over after the air leaves the turbine into kinetic energy. That device that converts enthalpy into kinetic energy is a nozzle. Next we can consider the fact that the air is moving very quickly as it comes into the engine and we want to slow it down to actually go through the engine. And since we're slowing the fluid down we might as well convert that into enthalpy so that our compressor doesn't have to do as much work so that our turbine doesn't have to provide as much work and reduce the energy of the fluid before it goes into the nozzle to produce kinetic energy which is the whole goal. So we add a diffuser to convert kinetic energy at the inlet into enthalpy which we can take advantage of by reducing how much work the compressor needs. So these are our six state points. Next I would normally generate a table for all of our state point properties and I will I just don't want to do that yet. Instead I want us to think through an outline of what we actually need in order to solve this problem. So the first thing I ask for is the velocity of the exhaust gases. That's v6. Now in order to come up with v6 I'm going to have to perform an energy balance on the nozzle. The energy balance on the nozzle is going to require that I know enough information to fix state 5 and state 6 and I don't know enough information to fix state 5. I have a temperature at state 4 and I could figure out how much expansion there is from 4 to 5 if I knew the work of the turbine. And in order to know the work of the turbine I could figure out the work of the compressor which I could do by performing an energy balance on the compressor which would require that I know states 2 and 3. I have a pressure ratio across them so really all I need is state 2 to be fixed. For state 2 to be fixed I have to use our state 1 conditions which are 32 kPa and negative 32 degrees Celsius and how much the increase in energy of the diffuser affects the enthalpy of our air from 1 to 2. Therefore I'm going to have to analyze the process from 1 to 2 and then 2 to 3 in order to be able to analyze from 4 to 5 and then from 5 to 6. So I really need states 1, 2, 3, 5 and 6 to be fully defined. And then for the propulsive power I'm going to have to take our force of thrust which was the mass flow rate of air through the engine which conveniently we know already times v6 which we will have from part a minus v1 which we know because our engine is traveling at a steady velocity and the air is stationary or at least we are treating the air as being stationary at an altitude of 9,150 meters. Therefore the velocity at state 1 is going to be 320 meters per second relative to the engine. So I know v1, I know mass flow rate and I will have v6 by the time we get around to calculating part b. Then for the rate of fuel consumption I'm going to be recognizing that we're assuming all of the heat going into the engine is coming from burning fuel and we are assuming that the combustion of the fuel emits heat and all of that heat 100% of it goes into the air. Therefore q dot in to our engine is going to be q dot fuel which means m dot air times specific u n is equal to m dot fuel times heating value of the fuel. So in order to answer that question for m dot fuel I'm going to need the mass flow rate of the air which I know the specific heat input to the engine which is going to come from an energy balance in the combustion chamber which means I have to know a state 4 anyway. So I do need all six state points to be fully defined and if we're assuming air is an ideal gas then our enthalpy of our air is only a function of temperature. Therefore I need all six temperatures in order to be able to answer the question so that's going to be our goal that is our intermediate goal before we get back to determining all of the things that the problem actually wants. I need all six temperatures. It's table time as promised. You'll notice that I put x's through v2, v3, v4, v5 because I don't actually care about those velocities. I'm going to be assuming that they are relatively small. So small in fact that they are essentially zero because when we're talking about kinetic energy terms we're talking about a function of velocity squared and relative to the bombacity of 320 meters per second squared a small velocity squared is going to be insignificant. I mean even if the velocity were a relatively quick moving five meters per second through the engine five meters per second squared is 25 compared to 320 meters per second squared which is something I can't calculate in my head but much much much larger than 25 so large in fact that the delta v is essentially just 320 meters per second. So all we really care about is v1 and v6 and then I'm trying to populate the temperatures so that I can determine all of the results of the energy balances on our various components like the compressor and therefore the turbine and the combustion chamber in order to answer part c and the nozzle and diffuser to determine how much the kinetic energy of the air is affecting our temperature in the case of the diffuser and how much the temperature change is going to say about our increase in kinetic energy in the case of the nozzle. So here we go. First up I have a temperature of negative 32 degrees Celsius at day one. I could try to add a 273.152 that in my head but that would be a laughable experience and since this is being recorded onto the internet I'm going to have my calculator do it so that we don't have to try to do that in our head. It's 127 John. 273 that's much more better. Okay okay since we have a safety net here I'm going to try to do it. Here we go 32 from 273 is like 4241.15. Survey says 241.15. Ah yeah to have more confidence in your mental math John. Okay now how do we get from 1 to 2? Now I'm imagining you saying to your phone or your computer the isentropic ideal gas equations we were told to assume constant specific heats around temperature which means that we are using the cold air standard to analyze this problem so I can say T2 over T1 is equal to P2 over P1 raised to the K minus 1 over K. For air at 300 Kelvin I know the K is 1.4 and because the isentropic efficiency of all of our devices is assumed to be 100% because the problem told us all the devices were ideal we can say this T2 that we're calculating is our state two temperature. So T2 is equal to T1 multiplied by P2 over P1 raised to the K minus 1 over K. Now let me pose a question to you what is P2 over P1? Did you just say 12? I understand why you said 12 it's because the pressure ratio across the compressor is 12 but note we're not analyzing the compressor we are analyzing the diffuser. The diffuser is not the compressor. This is a little bit of a change of pace because every other Brayton-Seckl problem that we've worked through has had the compression process from 1 to 2 but here the compressor is from 2 to 3 which means P3 over P2 is 12. P2 over P1 is not something that we have. So is there another way that I can come up with state 2 without using our isentropic ideal gas equation? And there is and we're going to have to go back to our roots. We need to perform an energy balance. Reason being we know how much the kinetic energy of the air is going to decrease and because the diffuser is operating ideally all of that kinetic energy decrease is going into the enthalpy of our fluid and for an ideal gas that enthalpy change directly corresponds to a temperature change because the enthalpy is only a function of temperature and because we know the specific heat capacity of the air is assumed to be constant we can substitute in Cp delta T for that delta H and we can then say the temperature change as a function of the change in velocity. Energy balance time. We have a diffuser which is assumed to be isentropic therefore we are treating it as being adiabatic because isentropic implies adiabatic I'm going to put a D on here for diffuser. We have an inlet condition at state 1 we have an outlet condition at state 2. I'm identifying a control volume then I can perform an energy balance on that control volume. Our energy balance if we want to skip a few is going to simplify down to E dot in is equal to E dot out. The rate of energy entering our diffuser could be described as Q dot in plus work dot in plus the sum in of m dot theta and the energy out could be described as Q dot out plus work dot out plus the sum out of m dot theta but we recognize that it's adiabatic so there's no heat transfer and we recognize that there's no opportunity for work. Furthermore there's only one entering state point it is state one there's only one exiting state point it is state two which means we're left with just m dot one theta one is equal to m dot two theta two our m dots are the same because m dot one is equal to m dot two which means that if I divide by that mass flow rate I'm left with theta one is equal to theta two and what is theta that's right it's enthalpy plus specific kinetic energy plus specific potential energy which means h one plus specific kinetic energy one plus specific potential energy one is equal to h two plus specific kinetic energy two plus specific potential energy two. Next I'm going to assume the change in potential energy is negligibly small that leaves me with h one plus the specific kinetic energy at state one which I represent as one half times velocity squared because remember total kinetic energy would be one half times mass times velocity squared specific kinetic energy would be total kinetic energy divided by mass so I'm going to write that as v one squared over two and I can do the same thing for state two and then I want to try to write this in terms of a delta h so that I can substitute in cp delta t so I'm going to group together my velocities on one side and my enthalpy's on the other and then because we are told to assume constant specific heats of room temperature I'm going to substitute in cp times t two minus t one and then on our left hand side I'm going to rewind the conversation back to up here remember we are assuming the velocity at states two three four and five is very small relative to 320 so small in fact that I'm treating it as basically zero so 320 squared minus a small number squared is basically the same as 320 squared therefore I am getting rid of v two so v one squared over two is equal to cp times t two minus t one now we are cooking with whatever it is that you cook with I have v one I have t one I can look up the cp of air at 300 kelvin therefore I can write t two is equal to v one squared over two times cp plus t one cool beans velocity is day one was 320 meters per second cp is 1.005 and t one was an astounding 241.15 kelvin right calculator it was right haha it was so why don't you try calculating t two on your own just pause the video give it a shot did you do it I'm gonna guess that quite a few of you got an incorrect answer and the reason for that is a very common mistake at this point would be for people to take 320 squared divided by two times 1.005 and then add to that 241.15 that doesn't look right does it no it does not the reason that this doesn't work is because meter squared per second squared is not the same as a kilojoule per kilogram they differ by a factor of a thousand so in order to actually add these together I need to perform a unit conversion in our first part of our calculation to get the meter squared per second squared and the kilojoules per kilogram to cancel so we have to actually keep track of our units 320 squared meters squared per second squared divided by 1.005 kilojoules per kilogram kelvin kelvin's what I want so I'm going to leave that my goal is going to be to get the kilograms and the meter squared and the second square to cancel out of the kilojoule and for that I'm going to break the kilojoule into its components a kilojoule is a thousand joules and a joule is a newton times a meter and a newton is a kilogram meter per second square the newton cancels newton kilogram cancels kilogram kilojoule cancels kilojoule joules cancels joules second squared cancels second squared and meters and meters cancel square meters and then the ipad scrolls out so this one thousand right here is where quite a few people mess up this calculation and it's fortunate that right now the answer that you got was probably high enough so as to raise concern but if you make the same mistake when we are analyzing across the nozzle it's not going to be quite as obvious we'll come back to that in probably about 20 minutes whatever the case though with this new unit conversion in place I get a much more reasonable answer of 292.095 kelvin and you know me I'm going to keep track of all those decimal places for now and I will point out that if we needed the pressure ratio across the diffuser we could hypothetically use the isentropic ideal gas equations to determine what that pressure is at state two but let's leave it for now we only are looking for temperatures right right so let's get to our compressor analyzing across the compressor is just going to be the isentropic ideal gas equation that we tried to use earlier t3 over t2 is equal to p3 over p2 raised to the quantity k-1 over k then t3 can be written as t2 times p3 over p2 raised to the k-1 over k we know t2 I got it on my calculator right here I know k I can look that up that's 1.005 and p3 over p2 is that 12 number that you totally convinced us we should try to use earlier but we didn't so I am going to take 292.095 multiplied by 12 and then I'm going to raise that to the power of 1.4 minus 1 divided by 1.4 and we get 594.103 which I don't actually need to write down here I'm gonna write that at the table 594.103 we'll probably read that correctly later fingers crossed now how do we get from three to four we don't we already have t4 it's 1400 kelvin I really should have written that earlier but I didn't that's one much more better 1400 kelvin cool how do I get from four to five well it's going to require an assumption that we had talked about in the previous video that assumption is that the work produced by the turbine is only going into a compressor it is not running an alternator it's not producing some shaft work to power something else it is only powering the compressor any additional extraction of energy is going to hurt our end goal which is producing thrust from the nozzle therefore to get from four to five we are saying the work out is equal to the work in I could write that on a power basis but when I divide by the same mass flow rate I'm left with specific work and for the specific work out we are looking at the turbine an energy balance on the turbine is going to simplify down to the change in enthalpy across the turbine and the energy balance of the compressor is going to yield specific work in is equal to h3 minus h2 because it is isentropic which implies adiabatic and then because of the assumption of constant specific heats I can write cp times t4 minus t5 is equal to cp times t3 minus t2 and then same cp so it cancels therefore I can write t5 is equal to t4 minus t3 plus t2 four minus three plus two let's see if I can remember that for like four seconds four minus three plus two four minus three plus two okay four minus three plus two I'm right that there so that I have reference I don't make the mistake of plugging in the wrong numbers we have 1400 minus 594.103 which I will grab as the result on my calculator then we are adding state point two I don't really know why I had those parentheses but we have them now 1097.99 is the temperature as the gas leaves the turbine and now that we have that we are trying to produce thrust so we are converting all that leftover enthalpy into kinetic energy for that we are going to have to set up an energy balance on our nozzle this time and being as that is so close to our diffuser I will write that to the right of our diffuser so we have a nozzle and I'm going to put a five here and a six here I'm going to draw the requisite fuzzy lines to indicate that it is adiabatic which is a result of the fact that we are treating it as isentropic I'm identifying the control volume for the purposes of using our energy balance and then I'm going to put an n on it or nozzle so just like the diffuser our energy balance is going to simplify all the way down to m dot five theta five is equal to m dot six theta six we have the same mass flow rate which means that our m dots cancel and then I'm neglecting changes in potential energy again therefore I can write h five plus v five squared over two is equal to h six plus v six squared over two and remember from earlier v five is very small relative to v six based on our assumption that v five is very small relative to everything else so small in fact that when I take v six squared v five is basically nothing because even v five squared is very small so we are going to say each five is equal to h six plus v six squared over two then again I want to write this as a delta h so that I can plug in cp delta t and because it is a delta h I'm writing that as cp delta t and then I have t five I'm looking for v six I have cp we have everything we need don't we oh no we don't know the temperature of the exhaust gases so what do we do ah what do we do well I mean we know the ambient conditions will that help us somehow I mean do you think it's reasonable to assume that the exhaust gases are exiting at negative 32 degrees Celsius probably not no but what we can assume is that the nozzle is expanding as much as it possibly can from the pressure on the high pressure side of the nozzle all the way down to the pressure on the outside of the nozzle and that pressure is 32 kPa so the nozzle is expanding from whatever the pressure is as the air exits the turbine down to 32 kPa well for that I could describe the isentropic process across the nozzle as t six over t five is equal to p six over p five raised to the k minus one over k so t six could be calculated by taking t five multiplied by the pressure ratio across the nozzle raised to the k minus one over k now do I know the pressure ratio across the nozzle well I know the pressure at state six and I could figure out p five I mean I know enough to get there first of all I'll ask is it reasonable to assume that the pressure ratio across the turbine is one over 12 because the compressor increased by 12 should we use one over 12 to go across the turbine no we should not in the previous problems we had been able to do that because we had to end at the same pressure because we had a ql box that was operating isobarically so we had to end at the same pressure that the inlet air to the compressor was at here we don't have that same luxury because we don't actually care about p five what we care about is extracting only as much work as is required to operate the compressor so whatever pressure ratio is required to do that is going to be what happens so in order to come up with p five I have to come up with p four and then I could use the isentropic ideal gas equations to get across the turbine because I know t four and t five and in order to get to p four I have to know p three because the combustion chamber is assumed to be isobaric and in order to get to p three I would just take p two multiplied by 12 which brings us back to the diffuser if I can figure out what the pressure is at state two I can multiply that by 12 to get p three and therefore p four and then I could use the isentropic ideal gas equation across the turbine to get to p five which I can use with the isentropic ideal gas equation on the nozzle to get to get to t six and then I could use the energy balance on the nozzle to calculate v six did you follow that so in order to answer t six I have about four intermediate steps here I need to come up with p two which is going to require I add a column to our table that's annoying I'm not even going to bother drawing it all pretty because this is the disgruntled column what a shame as a general rule I would encourage you to come up with only the properties that need and in most cases in the previous analyses that we had performed we only cared about q in work in q out and work out and that had only required the temperature change in the case of the cold air standard with the enthalpy change in the case of the non cold air standard but here when we are talking about energy balances under diffusers and nozzles and therefore kinetic energy terms we are going to need pressures as a general rule of thumb so get into the habit of calculating pressures again when we are talking about used for turbojet applications so I know 32 is the pressure at state one and state six I get from one to two with the isentropic ideal gas equation I get from two to three but multiplying by 12 I get to four because p four is equal to p three I get across the turbine with the isentropic ideal gas equations and then I use p five and p six to get t six which I use to calculate v six which I use to calculate the thrust and the power the propulsive efficiency just for fun and then I can determine q in by knowing t four and t three and determine the mass rate of fuel let's get to it at state two I have a temperature and I know p one and t one therefore I can write p two over p one is equal to t two over t one raised to the k over k minus one now I have a terrible propensity for screwing up that exponent so I always like to have my isentropic ideal gas relations handy so that I can just say this this right here is what I want and not have to try to move the exponent around in my head because as you all know my head cannot be trusted when it comes to doing mental algebra so by double checking here we know t two over t one raised to the k over k minus one gives us p two over p one then I can jump back to our calculation and say p two then is going to equal 32 multiplied by t two which was 292 which I will scroll up for divided by 241.15 raised to the power of 1.4 divided by 0.4 p two then is equal to 62.5865 and I really don't have to write these down because they're only intermediate steps but I'm going to do that anyway just to make sure that if you were trying to follow along with this you have some hope of keeping track of what I'm doing then I'm going to multiply that number by 12 and I get to this which is also the same as this and then for our isentropic ideal gas equations we are going to jump down to wherever we analyze the turbine and write p five over p four is equal to t five over t four raised to the k over k minus one so that number multiplied by t five over t four t five was 1097.99 divided by 1400 raised to the power of 1.4 divided by this for final subtracted and I get 320.856 so the pressure ratio across the turbine is really only about two actually that that is an interesting side tangent here I'm going to pull you over to a chrome tab and I'm going to see if I can find a cross section of a turbojet. Okay this one will work okay ignore all the numbers for now can you actually see the whole thing yeah you can this is just the numbers that are used for reference on the Wikipedia article for jet engines this is a turbojet this is a cross section of an actual turbojet I mean this is still an artist's rendering of a turbojet but it's still a better cross section than the crude system diagram that we're using to keep track of our state points here so the compression process is the process from right about here to right here and in this case the compressor is way bigger than the turbine right look at how few blades there are in the turbine it's rotors and staters there's so few blades relative to the compressor and why is that well as we just demonstrated even for this example problem the pressure ratio across our expansion process was only about two but the pressure ratio across the compressor was 12 I mean 12 is a much bigger number than two and theoretically it would take many more stages of rotor than stator than rotor than stator to come up with a ratio of 12 especially considering that it is a much more adverse direction to have to go in I will point out while we're here this doesn't really have a diffuser as shown it's using presumably a little fan out front or prop but this part right here is our nozzle and while we're going down this rabbit hole let me pose this question remember when we were analyzing nozzles in thermal one we showed them with a converging cross sectional area so even in our drawing here for the energy balance on the nozzle I can pop you over to the iPad again we're drawing this as a converging cross sectional area and in this diagram here it is a diverging cross sectional area this cross sectional area is smaller than this one which means that they're getting further apart how can that still be a nozzle well that dear friends is due to the fact that this is presumably operating with flow that is fast enough so as to be compressible and for compressible flow a nozzle is actually a diverging cross sectional area for incompressible flow a nozzle would be a converging cross sectional area is backwards for incompressible and compressible flow like everything that's why compressible fluid mechanics is basically a class in and of itself because it's everything that you know from incompressible fluid mechanics and except backwards for most of it that's indicating here that this is a high velocity for our purposes we typically use a mock number of about 0.3 is the threshold between incompressible and compressible so anytime the air is moving at a velocity higher than a 30% of the speed of sound through air with those conditions we treat it as being compressible and for our purposes here that's going to be probably only about 200 meters per second so if we have a higher velocity than that then that's going to be treated as compressible flow that's all outside the scope of what we're doing right here right now but i thought you might be interested it is a bit of a fun fact after all so if we jump back to the iPad and we get on with our isentropic ideal gas equation from 5 to 6 i can use that pressure ratio across the nozzle to determine a temperature which i can then use to calculate v6 and under most circumstances i would advocate leaving everything symbolically as long as he can but here it's kind of moot especially considering that we have so many running errors already so i'm going to take t5 t5 was 1097 multiplied by p6 over p5 which is 32 divided by 320 and change raised to the power of k minus over k minus 1 over k this time 0.4 minus 1 divided by close parentheses divided by 1.4 that means t6 is 568.268 that is the temperature of the exhaust gases which as we suspected is much higher than t1 then from our energy balance we can calculate v6 v6 is going to be the square root of cp times t5 minus t6 and again why don't you try that on your own pause the video here if you want give that a shot did you do it did you remember the thousand very common for people to forget the fact that meters per second squared and kilojoules per kilogram differ by a factor of thousand and here we are taking that thousand inside the radical so it's really going to be whatever the square root of a thousand is so we're out of a thousand so if you had forgotten the unit conversion thank you calculator your answer would be off by a factor of 32 would be unhelpful so i'm going to take 1.005 multiplied by the difference in temperature between five and six which is 1097.99 1097.99 1097.99 minus t6 which is 568.268 and that is in kelvin and i forgot my units on the 1.005 what was i doing ah the decimals i know the dark period in our calculation get it the decimals up periods uh-huh kilojoules per kilogram kelvin kelvin cancels kelvin but kilojoules per kilogram even inside of a radical does not yield meters per second so i have to convert into meters squared per second squared when i convert from kilojoules per kilogram into meters squared per second squared and then take a square root then the result is going to be in meters per second i will cancel the case while we're here so for that i'm going to take the kilojoule and i'm going to break it into its components kilojoules a thousand joules a joule is a newton meter and a newton is a kilogram meter per second squared and then if i have enough room whoo-hoo i'm going to take that entire quantity to the one half power and then kilojoules is going to cancel kilojoules joules cancels joules newtons cancels newtons kilograms cancels kilograms and i'm left with meters times meters divided by second squared which when taken to the one half power yields meters per second let's do that now 1.005 no come on i've been cooperating here multiplied by the quantity 1097 which i will scroll up for for precision purposes 568.268 multiplied by a thousand and then that entire quantity is raised to the one half power and we get 729.639 as our velocity at state 6 that doesn't seem high enough equation yielded i wouldn't expect it there to be a two in there because it's a kinetic energy calculation and there's always a two oh there it is okay i got all distracted writing cp delta t instead of delta h never got to write divided by two which means that when i wrote this equation i forgot to multiply by two because we would have had two times cp times t5 minus t6 before we took the square root of both sides so this should all be multiplied by two which is unfortunate because we're out of space over here so i will have to to scooch this on over scooch this on over and write a two out front so let's try that again we had two times come on calculator two times 1.005 times the change in temperature and then we multiplied by a thousand to get the inner conversion correct then we took that to the one half power so earlier when i posed the question did you calculate the number correctly and you said yeah i totally did and technically you didn't if you followed along with my algebra because that was velocity that was too small it's a little bit more but i was expecting okay one thousand thirty one point eight six and our velocity is day one was 32 we knew that already right no it's 320 32 i was thinking of the pressure okay 320 to 1031.86 is a much a larger change in velocity which will produce a larger thrust which is going to be the thrust required to travel at 320 meters per second which is very quick that's why i was expecting a higher velocity at state six anyway now that we have all six temperatures and both velocity one and velocity six we can move on to actually answering the questions and for that i think i'm going to open a new page here so let's start by calculating v6 well we know it already it's 1031 186 look at that we did it go us then i had asked for propulsive power but just for fun here let's calculate the thrust first the thrust is going to be the mass flow rate of air to the engine multiplied by v6 minus v1 and let's say we wanted that in kilonewtons so i'm going to take the mass flow rate of air to the engine which was 60 kilograms per second and then i'm going to multiply by the difference between 1031.86 and 320 and then i'm going to start with my destination and work backwards a kilonewton is a thousand newtons a newton is a kilogram meter per second squared and let's see where we are seconds and seconds cancel second squared meters cancels meters kilograms cancels kilograms newtons cancel newtons and we're left with kilonewtons so 60 multiplied by this number minus 320 divided by a thousand hey look we divided by a thousand again it's interesting it's almost as though the theme of this problem 42.71 kilonewtons i know i didn't actually ask for that but let's imagine that i did hey we did it okay then i actually wanted propulsive power i'm pretty sure yeah i actually want to propulsive power and let's say i want kilowatts i mean that probably seems reasonable for propulsive power i'm going to take the thrust force multiplied by the velocity of the engine which i could write as mass flow rate times v6 minus v1 multiplied by the velocity of the engine or i could just take 42.71 kilonewtons and multiply it by 320 meters per second in that velocity of the engine calculation we are using the velocity relative to the ground here by the way that's why it's 320 zero because we are knowing no longer describing velocity from the reference frame of from the engine's perspective and then a kilowatt could be described as a kilojoule per second and a kilojoule is a kilonewton times a meter which means that we are going to end up with an answering kilowatts if we just multiply this number by our velocity and had jumped to directly calculating the propulsive power instead of calculating the thrust force first then you would have had just taken 60 multiplied by the difference between 1031.86 and 320 then you would have multiplied by 320 and then you would have multiplied by 1000 because you need to account for the fact the meter square per second squared differs from kilojoules per kilogram by a factor of a thousand anyway we get 13,367 excuse me 667.8 kilowatts that there is a powerful engine so large in fact that we're probably going to have a very large fuel flow rate to be able to power that engine which leads us into part c where i had asked for the flow rate of fuel now i will point out that we don't actually have a volumetric flow rate versus mass flow rate defined here but i will say that generally the fuel flow rate for aircraft is described as a mass flow rate instead of a volumetric flow rate and furthermore we don't know the density of the jet fuel that we're using here we only know its heating value so we couldn't even calculate a volumetric flow rate if we wanted to so mass flow rate is the name of the game for part c and again the way that we get there is by assuming that all of the heat emitted by burning the fuel goes directly into the air and that increase in energy of the air and that occurs on a 100 percent efficient basis all of the energy from the fuel goes into the air none of it is wasted out the sides of the engine or anything like that then i could describe q dot fuel as m dot fuel times the heating value of the fuel and q dot in as m dot air i almost said m dot in that would have been inaccurate multiplied by a specific q in to our engine and the specific q in into our engine is going to occur in the combustion chamber and if i were to set up a an energy balance in the combustion chamber that would simplify down to h4 minus h3 and then because of the assumption of constant specific heats i can say m dot air times cp times t4 minus t3 and that is equal to m dot fuel times heating value of the fuel therefore m dot fuel is going to be this entire quantity divided by the heating value of the fuel doesn't look at all like the word fuel job fuel still still know third time's charm fuel there you got it then i know m dot air is 60 kilograms per second pretty sure yep cp of air is 1.005 kilojoules per kilogram kelvin t4 minus t3 is 1400 minus t3 was 594.103 and we are dividing by the heating value of the fuel which was given as 42,740,700 kilojoules per kilogram i have some unit simplification here kilograms cancels kilograms kilojoules cancels kilojoules kelvin cancels kelvin and i'm left with kilograms per second and it's important to note here that the kilograms listed in this calculation are not all the same in the massel rate of air that's kilograms of air per second and in the heating value of the fuel that's kilojoules per kilogram of fuel so the relevant kilograms that are cancelled are air because this specific heat capacity of the air is a description of the amount of energy per unit mass of the air kelvin so the kilograms of air cancel the kilograms of air i'm left with kilograms of fuel per second so that will be 60 multiplied by 1.005 multiplied by 1400 minus 594.103 divided by 42,700 and i get a syntax error which is to be expected at this point and i get 1.13807 and that's the mass flow rate of the fuel so that's everything that i was explicitly asked for but just for fun let's also calculate a propulsive efficiency remember that propulsive efficiency is the turbojet equivalent of our thermal efficiency the only difference is we are taking propulsive power divided by q.in instead of network out or net power out excuse me and i had already described q.in because q.in is mass flow rate of air multiplied by a specific q in which is h4 minus h3 which becomes cp times t4 minus t3 so i'm going to take m dot of air times cp times delta t across the combustion chamber as the denominator and the propulsive power that we calculated already as my numerator so that's going to be 1300 excuse me 13,667.8 kilowatts and then i'm dividing by m dot air 60 kilograms per second times 1.005 kilojoules per kilogram kelvin times the delta t across our combustion chamber which was this delta t right here let's see what cancels so far that's kelvin kelvin cancels kelvin kilograms cancels kilograms and a kilowatt you'll remember is kilojoules per second i would normally write that out but i'm actually out of room here okay okay i yield i'll just bring this down i was just about to say i'm not gonna write it this time because i'm out of room but but this is forever and as you all know you do not stand for simplifications what i'm talking about forever kilojoules kilojoules seconds seconds and i'm left with a unitless proportion which is what i wanted so here we go 13,667.8 divided by the quantity 60 times 1.005 times 1400 minus 594.103 and i get a propulsive efficiency of 28.13 percent