Alright, thank you for coming. This is a talk about real-time networking for real-time systems, and as just a note, this is about where the network itself is a part of the real-time system, not just the real-time system using the network, so just keep that in mind as we go along. My name is Henrik, I work with video conferencing in Cisco in Norway. We make big system systems, and I'm part of the audio group, so we actually have quite strict real-time requirements for the processing that we do, which is why I do most of my work in real-time tweaking of the systems, so tweaking and tuning to make sure that we meet all the deadlines. This has vectored me into looking a lot at traces, that generates a lot of data, so I have scripts for everything, so many scripts that I can't find half of them anymore. And for the last couple of years, I've been working with AVB and TSN on and off, and this is what this talk is all about. So first, real-time systems. A real-time system is a system where the end result of what you said, the end result of the system is not only the logical result, but also the time for which the result was produced. Meaning that it doesn't matter if you plan the perfect trajectory for your robotic arms, if you're two seconds late, you run into something. And this also maps into the audio domain that I work in. Another aspect is the consequences of a failure, so when a normal system fails, the user of the spreadsheet gets annoyed. In a real-time system, the user gets dead, and that's kind of bad. Now, we don't typically kill our users in our telepresence systems, but that's sort of the extreme point of real-time systems, which is why you also have strict requirements for the network that we'll get into. And leading into this, it means that you need to certify your system to make sure that you can meet all the real-time requirements. And for simple single core systems, this is actually quite difficult, because you need to take all the small details into consideration. Do you have virtual memory, for instance, or buffers or all the small details matter? And of course you add more cores than it becomes increasingly more difficult. A multi-core real-time system is actually a conference on itself, so I'm not going to go into detail there. And then some bright chap decided to put a bunch of different cores on the chip, which made it worse. But we're going to do it even better than that, because we're going to add network to all of this and see why this is an interesting problem. Because when you do that, you open yourself up to a new set of challenges. One of them being time. It doesn't matter if you produce your result at the correct time if you don't know what time it is. And no two clocks will ever run at the same rate. The software stack will never run at the same rate on the same machine. The temperature will do something to your clocks, the voltage. And you do have the network adding to this complexity. You have network time protocol that you can tune to get sub millisecond accuracy if you do everything right. And then I mean absolutely everything. And then you have precision time protocol which pretty much solves this problem for you. And I bring this up now because this is really important for the next steps we're going to talk about. And you do have to have hardware support for this to have this work properly and have decent accuracy. But we're not going to go into detail of precision time protocol. The next challenge is a reliable packet switch network.  I mean real-time reliable. The frames will actually go through the network deterministically. You won't lose frames. They will be there on time.  If you have collisions or if the buffers go full, the switches will just drop frames. You will also have the problem with you being queued up in an outgoing buffer. And even though you have a high priority and need to get through, the switch will probably just process the entire buffer before it gets to you. You can improve this by using priority in the VLAN tags. But then you'll have jumbo frames or other frames that are in flight. And one of the problems that you're faced with is you cannot express in a decent way to the network that I want my frames to arrive no later than say 2 milliseconds after I send them.  And for a real-time system, that's not good enough. You can also not express to the network card that I want this frame to be sent at that time. And you have no guarantee that it will be sent at that time. If you're lucky, it will not be sent too much later. Perhaps it will be sent sooner. There is no way for you to say that. And there's also really difficult to get an exact bandwidth through the network. I know there are ways to sort of reserve bandwidth through some sort of means, but to get really accurate bandwidth and never lose frames once you have gotten that bandwidth is actually quite difficult. And this is what AVB and TSN and ultimately that net tries to solve for you. And of course you have all the other challenges that you'll be faced with when you make a distributed real-time system over the network.  I'll try to fix the network for you and you can deal with the rest. So, AVB and TSN. Out of curiosity, how many have used AVB or know what AVB and TSN is? Ah, wow, cool. So, a fair amount of you actually. Yeah, good. So, the initiality was that AD and EA's are getting cheap. Max are getting cheap. Why not move digitalization of all of this as close to the source as properly? As possible. And you solve a lot of problems doing that. You don't have to worry about SNR and ground loops in big systems, especially for large AV setups. But if you want to use a network, you need to have a reliable network. Best effort won't help you. But if you can do that, let's just assume that you can get the network going, then you have a very flexible setup. It's easy to route flows from a single microphone to several sources, for instance. You can send it to a mixing desk, to a storage, to a monitor for local feedback, without having to drag extra cables. It also gives you very high capacity on a single cable. If you don't notice, I'm audio, so I think audio. Audio is sort of never an issue for me. I can put a lot of audio through a cable, which is good. The interesting problem we're faced with at Cisco where I work is we're actually limited by the physical dimensions of the backplane. From the first image of the backplane of the codec, you saw a lot of connectors. We're getting to the point now where it's not the internal processing that limits what we can do, but the number of microphones physically connect to the system. You can sort of pull tricks by attaching external mixers and hook into USB and everything. That's another issue. We would like to have control over all the sources that we get audio from. And using AVB, we can expand this greatly, and we're back to the CPU being the bottleneck again. And another benefit, of course, if you have reliable reserve bandwidth, you can coexist with best-effort traffic, so you don't have to run two networks. Initially, they started making simple microphones and speakers, really small, simple units. And there's no point in implementing the entire IP stack on these small units. They just did either frames. They specified a standard profile for PDP, so they could have accurate time stamping. They make some assumptions to the network so they can converge the clocks a bit faster and pick a new grandmaster faster. You use the stream reservation protocol to communicate to the network I want to reserve bandwidth. And they also made this sort of plug-and-play protocol, the discovery and enumeration of AVB units, which means that it's easy when you're on the subnet for AVB to connect a microphone and a speaker, which makes sense in an AV setup. And the downside of that, of course, is that its security is not really, you don't really care who connects a microphone to a speaker, because you run a subnet and it's a professional media setup, so no one outside is going to connect into that subnet, you hope. The focus is reliable low-jitter streams and a way for the admins of that to connect the dots. And to get the reliable stream, they use the credit-based shaper and the only thing you need to know about the credit-based shaper is the idle slope, which is how fast you recharge your credits. And that basically gives you an idea of how fast you can send out frames. So you send a frame and you lose credits, once you're below zero you can't send a new one, and the moment you hit zero or above zero you can send a frame if another frame is not intrinsic. So that's a sort of fairly elegant way of pacing out the frames on the network. Now once they got this working, people started using this for all sorts of crazy things. So suddenly people were using AVB to send sensor data across the network and they were controlling robotic arms. And it became pretty obvious that AVB was not living up to its highest potential, so they renamed it to time-sensitive networking in 2012 and expanded the scope of the project. So not only did they target professional audio and video, they did consumer because things are getting cheaper all the time, and also infotainment systems in cars and trains, messaging boards, all sorts of places where you just want to send a stream and use multicast to send it out simultaneously. Automotive is another interesting use case where they wanted to use this for the control systems for ABS, for engine control, for engine monitoring and you really don't want to lose those frames. So TSN is what you need to guarantee that if you want to use this over a packet switch network. And of course you have the industrial applications where you want to converge the operational network with the information of the IT network to get the whole industry 4.0 thing. There TSN is not the only solution, it's a crucible building block and then once you have that underlying infrastructure you can do all the other cool things. And this really shifted the focus away from giving you a reliable jitter-free or low jitter stream for media to a way to send frames predictably through the network. Not necessarily at a constant rate. It's more important as to be able to send a frame exactly when you want it to and make sure that it's actually sent and propagated through the network in that time slot. Because if you want to replace industrial control protocols, this is really important. So they had a lot of work with time on transmission, frame preemption guard windows and all these things. And also growing to larger networks. Industrial networks will be quite bigger than the average professional media network. And since you grow we also need to look into path redundancy. And by path redundancy it doesn't matter if you quickly discover that you lose a node in the network and you can reroute the flow, then it's too late. So you need to split the flow or duplicate the flows through several paths. So if one path goes down it doesn't matter because you have copies of the frames going through another path. So you will have immediate recovery from that and don't lose any frames. So this was all good, but why stop there? So they formed a working group in ITF to grow this into wide area networks. And one of the things they do, you can do the pseudowire thing or you have IPv6 tags you can use in the network. But that net is more about a set of recommendations for what you should do. This is what a deterministic network should look like. It's not the hard requirements that IEEE sets forth. So it expects that the end stations which can be ITS and subnet if you want to adhere to that and then you will specify over the net network the parameters you want. And of course multi-path routing is even more important in such networks. And they did away with the everyone can configure everything approach and that is actually a requirement I think that you have to have a central controller or a set of specified controllers that those are the only one who can set up and tear down paths. Because if you want to connect to your windmills you don't want just about anyone to connect to those windmills and do stuff. So security is much more important in that net than it was in AVB. The motivation of course broadcasting would like to use this TV if you want to stream to all your home users. People are using this or are interested to use this for blockchains and then it's not the public bitcoin thing but if you want to have large private blockchains for logging or whatever it's interesting to use something like that net because you can have the consensus process go faster. Building automated systems and one of the really active participants in this is the utilities because they want to make sure that you have the smart meters and the smart grid and connect to the power plants to make sure that you have the right grid frequency and you can balance the load. All these bits and pieces is really interested in this and then security is more important of course. But it still has basically the same requirements. You need to have low packet loss ratio you cannot ever say you have zero but they claim that you should be able to get really low. Low end to end latency of course and perhaps even more important for that net is to be able to coexist with best effort traffic because digging new cables is really expensive so if you can reuse the same set of cables for this it will save you a lot of money and that's important. So the important part of the talk which is what's happening in the Linux kernel and last year I gave a talk about a very hackish module I wrote to expose an ALSA interface so you can use that to connect to the network and play audio and it was initially meant to prove a concept. This is really cool. I saw a lot of potential for this at work when video conferencing and we can just pop up a new ALSA device and connect to the network and everything was pitchy. Getting there was a bit of a hack in progress so I sort of did away with all the timing. I hacked directly into the hardware queue on the network card and it broke several times in interesting ways. But the main problem with that approach was basically it called itself TSN but it was basically classic AVB. It did reliable or attempted to do a reliable media stream. It did not have support for sporadic frames for instance. This is really important if you want to support TSN in the kernel. So recently things have been happening quite fast. Before we move on I tend to note or use the word I210 which is a network card from Intel and the reason for that is it's the only network card I know that has a PCI form factor. There's several other cards available from other producers or several other Knicks but they're typically bundled on evaluation boards or made for embedded systems. So if you want something in your computer to test this that's pretty much your only choice. So about a month ago a set of patches from Richard Cochrane which is the PDP maintainer appeared and he had done the time trigger approach and tried to use the I210 Intel card to send frames at a specific time. And the Intel card is it has two modes. It has the CBS mode and it has the launch time mode. Sorry. So he created a new socket option called the SOTX time and that gives you the ability to send a frame at a specific time which is really useful if you want to do industrial applications for instance. And the setup is pretty straightforward. You create a socket and you set the socket option TX time. You have to do a bit more work when you assemble the frame. Basically you have to add a C message and say this is the TX time and then you send it off to the network card and the network card will then add it to its queue and send it off at the appropriate time. This is a working project so there's a few limitations. For instance if you specify the time more than half a second into the future it will crash with how the time is implemented on the I210. So it may be sent in the past and then underflows into the future and everything gets interesting. It also means that if you send frames out of order it won't sort them because you should be really careful sorting descriptors sent to the network card. So you have to do a bit more work and be a bit more careful. This will probably improve greatly in the next set of patches. Richard is aware of all these limitations. He also specified a test which was quite impressive where basically he took two machines and connected them through a crossover cable to avoid interference from the networking system from the network itself. Running preempt RT which he has tweaked to an impressive degree using the I210 card of course and then he had a look at if I send this at predictable rates what's the offset from when I expect it to arrive. And he did this in pure software implementation and he did it in hardware as well to show the difference. And looking at the first column you have the plane approach without using the hardware offload and you have a difference of at most 75 microseconds which in itself is pretty good. For real time requirements it's nowhere near what you need but even so it's pretty good. The problem is the peak to peak variance which means how much the frames vary in distance to each other which is quite large. But if you look at the next column where he actually offloaded this to the network you see that the peak to peak has dropped to 100 nanoseconds which is pretty good. So I'm a bit excited about this. And it really shows that the networking card is able to deliver what we need and it's possible to write a driver that really works. And he also did for a period of 250 microseconds which is the AVB Class B frequency of frames. So it means that you can do proper Class B with this card using this as well. That's good. Sorry. But at the same time another set of patches dropped from Intel they have spun through several iterations of the patches and it's looking very good. They are implementing the credit-based shaper approach. So basically giving you a steady cadence of frames going out. And they also provided a proper Q disk so it's easy to use and you can configure it through traffic control. And it hooks into the multiple Q priority Q disk. I'm not going to cover Q disk in great detail so if people are unfamiliar with Q disk you have a lot of reading to do. Enjoy. But basically you just give your socket a priority and then you start sending frames to it. And if you have configured the CBS scheduler and the MQPRIO scheduler correctly it just works. So to set up this you have a rather nasty looking traffic control command.  You create four Qs and you provide a map. And basically priority 0 and 1 is here map to Q3. Priority 2 to Q number 2. Now priority 1 priority 2 to Q number 1. Priority 3 to Q number 0. And these are the default priorities set in the AVV standard. You want to do class A. That's by default priority 3 unless the network has specified something else. And priority 2 is class B. And all the rest are given to Q2. And then you list the Qs.  And then you can attach the CBS scheduler to one of these Qs. So here we attach that to Q number 0. And we give it an idle slope which is basically your bandwidth. So here you ask for a 20 megabit bandwidth. So that means that that scheduler will not send out more than 20 megabit to the network. They did not specify a test. Oh, sorry. And this is how you use it. Sorry. You create the socket standard way and you just set the socket option SO priority which is already present in the kernel. And give it a priority, so 3. And then all the frames you send to this socket will be fed to the CBS scheduler and paced out onto the network. So I decided to give this a test. And my test is not up to Richard's standard. So there's a lot to improve here. But just to sort of get a feel for does it work and how well does it work. So I had an old graphy computer at home with an I210 card. And it's going through a couple of switches up to my workstation. I tried to improve a thing, a few things by setting priorities and locking things down to improve things. But I'm not using preempt RT. I'm traversing the network. I'm not using a crossover cable. There's other stuff running on these computers. So the test results are not really clean, but hopefully they'll give you an indication to how they work. And I did this with and without load on the system. Basically just a parallel milk make just to induce some load to see how that affected the talker. And the first two ones are when I using hardware offload. And you see there's a few outliers in both ends and they are typically connected. So when you go to 1,000, there's probably one zero below because there are two frames back to back. But all in all, the spread was fairly good. It was centered around 500 microseconds, 514 for the hardware load. A bit lower for software for some reason. Still not quite sure why there's a difference there. And the standard deviation was fairly low. And if you look at the software approach, you see that without load it's not bad, but it's not perfect either. And when they added load then it got really bad. The interesting thing or the good thing here though is I got a constant 20 megabit outgoing bandwidth and the frames were pretty evenly spaced even though I had crappy hardware to test this on. But the load on the system was zero. This did not cost me more than zero cycles, but the load was fairly low, which means that using this in an embedded system for instance is really desirable. So I'm eager to test this in one of our products. And finally I just plotted all the samples. In the lower right corner you can see the software with load. Can you spot where I read from disk? And you see there's a few spikes. So this is basically not a real-time system where you have all these spikes here. But all in all it was fairly stable. So future, I need to redo the evaluation and do proper pre-empt RT and do away with all the mistakes that I didn't have time to do before the conference. This arrived a month ago. So I've been happily playing with this. I would also like to do a CBS-like Q-disk based on Richard's transmit time. Because then you can really compare apples and pears. I hope both get merged into the kernel. Richard requires a bit more work on his patches. I think the Intel set will be merged. It's looking really good. And also as I said the load on the systems were next to nothing. But I need to make a proper load measurement for that. And of course I want to use my old also shim TSN driver for this to see if I can hook into it. Sorry. So yeah, that was my talk and my journey through the latest patches in TSN. And it's looking really good. So if anyone wants to use TSN in a Linux kernel, you actually have patches that works really well now. And it will hopefully be merged soon. So thank you. Questions? Yes. I can give you or I can just repeat the question. I don't know what's easiest. Just ask and I'll repeat. It's fine. Is there a fallback for the txt time as well? In the patches so far? No, not at the moment. So that is one of the things that will probably hold the merge back. Because they're really strict about having both the software approach and the hardware approach when you do hardware offloading. There's a few other things that needs to be worked out in that as well. But currently this only works for the i210 driver for the txtime. Yes, the i210 has the hardware has dedicated support for both CBS, continuous cadence of outgoing frames, but also for launch time. So you can specify exactly when a frame should be sent. You have a half a second window basically where you can specify. And it's tied into the pdp hardware as well. So once you synchronize through pdp the time on the network card will be the pdp time. Yes. The microphone is approaching. How about support for all our network cars? Is this emerging or is the only one now? Currently this is only for the i210 card and that's basically because that's the easiest car to test with. But there are other vendors that have, especially the launch capability. So NXP has one, Synopsys is working on one, Renesos has one. Yes, Renesos H3 has, yeah, so Renesos will. So more support is coming, but currently there hasn't been a central framework for doing this. So this is sort of a chicken and egg problem. So there are hardware support and currently there's a vendor specific. You can use their proprietary things, but we hope to move this into the kernel. So it will hopefully grow a lot more. Yes, all the way back. Sorry, I didn't catch up. What are the limits? Limits from multiple streams. Yeah, so currently the Intel approach, the CBS, does not care about streams. It only paces out frames. So both of these approaches doesn't really look into the packets. The SLTX time just looks at the attached C message for when to send the frame and it's up to user space or some other kernel driver to multiplex into several streams. So if you have several streams you need to do some juggling in user space. Acudisk is like a, so Acudisk is basically a software queue or a scheduler for outgoing traffic. So you have several different queue disks that can do different things. So the MQ prior that you tie into here is basically just using priorities to map to a hardware queue. So you can stare traffic to a specific queue and then you tie into that with a credit-based shaper because we have available hardware for that. So Acudisk is basically just a scheduler for packets and attaches to a separate queue. I'm not sure if that answered your question. Yeah, I wanted to ask another one. Okay. It's very high. I'm doing embedded software not related to voice or anything for satellite and I'm using multiple queues. And we have constant time queues that now it's in a high priority. But now if I can make sure that the Linux will transmit to the hardware or to the interrupt, doesn't matter what, any constant delay. So I can use it instead of priority which doesn't make sure that it will be constant delay. Yeah. So this will give you the hardware support and support in the kernel for setting a specific time for transmitting a packet. Yeah. But you still need to create your frame properly if you wanted to traverse the network. Right. So both of these approaches doesn't really care about AVB or TSN as long after the frame has left the port. But still I can open a socket with the right attribute to a constant delay. Or I'm mixing. Are you thinking through the network or just transmitting from your application to the network card? No, from the Linux kernel to my driver that is forwarding into the hardware. Yeah. So now you can set a specific time for transmitting and then the network card will send it. What point is if you want... To the binded API, right? When I'm... Okay, I'll talk to you later. Okay. Yeah, that's fine. Thank you. Any other questions? Okay. Everybody happy? Okay. Thank you. Thank you.


Hello and welcome to lecture number 34 of this lecture series on introduction to aerospace propulsion. We have our the last two lectures had a chance to understand and analyze different types of gas turbine engine cycles, which are primarily operating on the Brayton cycle. So, the different cycles that are different types of engines that we have looked at over the past two lectures involved the simple turbojet engine, which is the simplest well not really the simplest form the simplest form of a jet engine is the ramjet engine. But the turbojet engine includes all the different components that is essentially constitute a jet engine like a compressor combustion chamber turbine nozzle and intake and so on. So, turbojet includes all these different components and what we had discussed in the last lectures were the simple forms of these engines in the form in the sense that we had considered the idealized version of these different engines. We had not really looked at what are the different losses that could take place or what are the efficiencies associated with different components and so on. So, what we had discussed were the idealized versions of these jet engine forms and then subsequently we also looked at variance of the turbojet engine. One of these immediate variations of the turbojet engine is the turbojet with after burning, which is burning additional fuel after the turbine and so as to achieve extra thrust. So, that is primarily used in cases where the turbojet engine is required to give additional thrust like when it has to accelerate to supersonic mach numbers and cruise at supersonic mach numbers. Then besides the turbojet with after burning there are different other variants of the cycle. One of the other variations or the most popular variation is the turbofan engine. Turbofan engine constitutes an extra duct over the core engine, where in mass flow rate is driven by a fan which has a larger diameter and so turbofan engines have been used in all the commercial most of the commercial engines. The advantage of using a turbofan engine is that it gives an additional thrust because of the presence of the fan duct. So, the fan duct and the mass flow rate associated with the fan generates an additional thrust and also it has other benefits like it has better fuel efficiency lower noise and so on. Then the other form of the jet engine is what is known as a turboprop engine. Turboprop engine constitutes a propeller which generates a thrust which is driven by a power turbine and also some amount of thrust is generated by the nozzle. So, turboprop engines may have thrust components majority of the thrust being generated by the propeller. It may also have some thrust generated by the nozzle and a slightly different version of that is known as a turbo shaft engine. Turbo shaft engines are used in helicopters where the major thrust is generated by the main rotor blade and turbo shaft engines do not generate any thrust because of the nozzle. So, the entire power that is generated the net power that is generated by the engine goes into driving the main rotor blade. Then we also discussed about a very simple form of a jet engine which is known as a ramjet engine. Ramjet engine does not have any of the turbo machines in the sense that does not have compressor it does not have turbine and so on. But it has an intake which has to be very efficient then there is a combustion chamber and a nozzle. So, there are these just three components and does not have any rotating components in a ramjet which makes it very simple in terms of construction. But ramjets have an inherent disadvantage in the sense that they cannot be used for generating static thrust. So, ramjets can generate thrust only after it has reached a certain mach number which means that it has to be first flown to a certain mach number after which it can start generating thrust. So, ramjets do not generate any static thrust. So, what we shall discuss today is that we will try to solve some problems which are related to some of these types of engines and basically carry out an ideal cycle analysis of some of these engine types of engines. Ideal cycle analysis as we have discussed involves using some of the known parameters the design parameters like the compressor pressure ratio turbine inlet temperature and so on. And also the operating conditions like mach number temperature and pressure to analyze each of the components which constitute the engine like the intake the turbine the compressor combustion chamber turbine nozzle etcetera. And then in the end determine what is the thrust fuel consumption and the efficiencies associated with the engine. So, we will carry out cycle analysis in a systematic way for different types of engine. We will begin with a turbojet then we will move on to a turbo fan engine and then we will take up one example of a turbo shaft engine. And then also I will give you some examples or exercises which you can solve at your leisure time later on based on whatever we have discussed. So, in today's lecture we shall primarily be discussing about cycle analysis as applied to different types of air bathing engines. So, let us take a look at the first problem. So, the problem number one is that it is basically a turbojet problem. So, the problem statement is the following. The following data applied to a turbojet flying at an altitude where the ambient conditions are 0.458 bar and 0.248 Kelvin. Speed of the aircraft is 805 kilometers per hour. The compressor pressure ratio is 4 is to 1, turbine temperature is 1100 Kelvin. Nozzle outlet area is 0.0935 meter square. Heat of reaction of the fuel is or calorific value of the fuel is 43 mega joules per kilogram. And we are required to find the thrust the specific fuel consumption assuming C P as 1.005 kilo joules per kilogram Kelvin and gamma as 1.4. So, this particular problem applies to a pure turbojet without any after burning. And some of the data that is given to us are the compressor pressure ratio, the turbine, the temperature, these are design parameters of the engine. The ambient conditions are given as 0.458 bar and temperature of 248 and the flight speed is also given to us. So, based on this we are required to find the thrust and fuel consumption for this particular turbojet cycle. So, before we start to analyze and try to solve the problem, let us first understand the basic turbojet construction in the form of what are the different components which constitute a turbojet. We already discussed that in the last couple of lectures. Let us take a relook at the different components and what are the station numbers associated with these components. And then we will also take a very quick look at the cycle ideal cycle of a turbojet before we start solving the problem in terms of each of these components. So, turbojet cycle as we have seen or turbojet engine constitutes of certain salient components. One of them the first component is the diffuser and then following the diffuser we have a compressor. It could be an axial compressor or a centrifugal compressor. And then after the compressor we have a combustion chamber where heat is added or the fuel is added there. And following the combustion chamber we have the turbine where the combustion products are expanded. After the turbine we may or may not use an after burner where additional fuel can be added to generate additional thrust. And then we have the nozzle which generates the thrust. So, in a turbojet engine the sole objective of the turbine is to drive the compressor. And so the turbine work will be equal to the compressor work which is not necessarily true in different other types of engines like in a turboprop or a turboshaft where the turbine generates only part of the work to drive the compressor rest of the work is used for either driving the propeller or for driving the main rotor blade. And so whereas in a turbojet the turbo work done by the turbine will be equal to the work done by the compressor. So, if you look at a schematic of the turbojet engine turbojet engine as I mentioned constitutes of a diffuser. Following the diffuser we have a compressor then the combustion chamber. So, compressor delivers air at high pressure and temperature reasonably high temperature to the combustion chamber. Here additional fuel is added combustion takes place and maximum cycle temperature takes place right here at the turbine inlet. Air in the turbine the combustion products in turbine inlet are expanded through the turbine. And then it passes through the nozzle which generates the thrust. So, for cycle analysis we have designated different numbers for each of these components. The subscript A would denote properties in the free stream that is away from the engine influence subscript these are the different other subscript subscript 2 would be used for the compressor inlet or diffuser exit 3 for compressor exit or combustion chamber inlet 4 for combustion chamber outlet or turbine inlet 5 is for the turbine exit 6 is for nozzle entry and 7 is for the nozzle exit. So, air as it passes through these different components where experience changes in its pressure and temperature. And those pressures and temperatures are denoted by some of these subscript suppose for example, we write the stagnation temperature as T 0 3 that denotes the stagnation temperature at the compressor exit or combustion chamber inlet. And T 0 4 in denotes stagnation temperature at the turbine entry or the combustion chamber outlet and so on. Now, if you look at the cycle diagram that is the Brayton cycle diagram for this particular turbojet engine the ideal turbojet cycle without any after burning would look like this on a T s diagram. We have temperature on the y axis and entropy on the x since it is an ideal cycle we have all the processes which are idealized. So, process begins at state A which is the ambient condition then it the first process is the compression process in the intake. So, A to 2 is the compression in the intake. So, 2 denotes the combustion compressor entry 2 to 3 is the process the ideal compression or isentropic compression in the compressor. So, process all the way from A to 3 is isentropic compression then we have a constant pressure heat addition which is in the combustion chamber. So, process from 3 to 4 is the heat addition or combustion process taking place in the combustion chamber which is in an ideal cycle takes place at constant pressure. Process 4 to 5 is the turbine which is an ideal cycle an isentropic expansion 5 to 7 is isentropic expansion in the nozzle. So, this completes the entire Brayton cycle starting from state A all the way from state 2, 3 and so on. So, first process is the isentropic compression in the intake then we have an isentropic compression in the compressor then there is a constant pressure heat addition. At the end of heat addition process that is the turbine entry we first expand the combustion products through the turbine followed by expansion through the nozzle. So, the net thrust that is generated by the turbo jet engine is purely because of the expansion taking place through the nozzle. The compressor is driven by the turbine turbine does not generate an additional power and so it generates just enough power to drive the compressor and so the entire power or entire thrust which is generated by the turbojet is because of the nozzle which is not true for some of the other types of engines like turbo props or turbo shafts and so on. Where there are other components in fact the turbine generates more power than what is required to drive the compressor and that extra power is required is used for driving the propeller in a turbo prop and so on. And so for a cycle analysis what we shall do is to analyze the turbojet engine in a very systematic fashion looking at each of these components and seeing what are the parameters which are known and how do you analyze the next component based on the parameters which are known to us now. And so let us start the cycle analysis process by first looking at the parameters which are known to us. So, what are known to us is firstly the speed of the aircraft is specified it is 805 kilometers per hour. So, let us convert that to Mach number. So, 805 kilometers per hour can be converted to meters per second it is 805 into 1000 divided by 3600 converts kilometer per hour to meters per second. So, this corresponds to 223.6 meters per second. So, how do you convert now this into Mach number? Mach number we know is the speed of the aircraft divided by speed of sound and speed of sound is defined as for an ideal gas square root of gamma R T where T is the static temperature. Now, the temperature that is given to us the ambient temperature is 248 Kelvin and so square root of gamma R T square root of 1.4 into R for air is 287 joules per kilogram Kelvin and T is the temperature ambient temperature. Therefore, the Mach number of the aircraft is 223.6 divided by square root of gamma R T which is 1.4 into 287 into 248. So, this Mach number comes out to be 0.708. So, the Mach number at which this aircraft is operating is 0.708. So, the first component we will analyze is the intake and so for intake exit we need to now calculate the stagnation temperature and stagnation pressure. We know the intake entry temperature and pressure the ambient conditions are given to us and so from isentropic relations we can calculate this stagnation temperature at the exit of the intake and how do we calculate that it is basically using the isentropic relations which relate the ratio of stagnation temperature to the stagnation static temperature that is T naught by T is 1 plus gamma minus 1 by 2 m square. Similarly, we can use the same expression for pressure as well and so using these isentropic relations we can calculate the intake and exit stagnation pressure and stagnation temperature which will form the entry conditions for the compressor. So, for the intake we have T02 which is the intake exit temperature stagnation this is Ta into 1 plus gamma minus 1 by 2 m square Ta is given as 248 Kelvin. So, this multiplied by 1 plus gamma is 1.4 1.4 minus 1 by 2 into m square where m is Mach number is 0.708 we have just calculated that. So, this temperature comes out to be 272.86 Kelvin similarly, stagnation pressure T02 can be calculated from the isentropic relations P02 is Pa into T02 by Ta raised to gamma by gamma minus 1 and Pa is given as 0.458 bar. So, this multiplied by 272.86 divided by 248 raised to 1.4 by 1.4 minus 1. So, this is equal to 0.639 bar. So, these are parameters at the intake exit and so these form the entry conditions for the compressor and in the compressor we have the compressor pressure ratio which is a design parameter we have been given the compressor pressure ratio. So, using that we can calculate the compressor exit stagnation pressure and then again from the isentropic relations we will be calculating the compressor exit stagnation temperature. So, compressor exit stagnation temperature and pressure will form the entry conditions for the combustion chamber and at the exit of the combustion chamber we know the stagnation temperature which is the turbine inlet temperature. So, if from an energy balance we can calculate the fuel to air ratio which is used in the combustion chamber. Then by balancing the compressor work and the turbine work we find the turbine exit conditions in terms of pressure and temperature which form the nozzle entry conditions because there is no after burning and then from the nozzle conditions we can find the nozzle exit velocity and hence the thrust. So, let us take up the next component which is the compressor. So, the compressor exit stagnation pressure is equal to the compressor pressure ratio multiplied by inlet stagnation pressure. So, the compressor pressure ratio is given as 4 this multiplied by the compressor inlet stagnation pressure which we have just now calculated which is 0.639 bar. So, this is equal to 2.556 bar. Compressor exit stagnation temperature is T02 into pi c which is the compressor pressure ratio raise to gamma minus 1 by gamma. So, this is equal to 272.86 into 4 raise to 1.4 minus 1 by 1.4. So, this temperature comes out to be 405.63 Kelvin. So, we have calculated both the compressor exit stagnation pressure as well as the exit stagnation temperature. We also know the combustion chamber exit stagnation temperature which is the turbine inlet temperature. So, all we have to do is to carry out an energy balance. Compressor exit combustion chamber exit enthalpy h 0 4 is equal to h 0 3 which is n t enthalpy plus f times q r where f is the fuel to air ratio q r is the heat of reaction in of the fuel. And therefore, f if we simplify this from assuming air to be ideal gas we get c p into t 0 4 is c p into t 0 3 plus f into q r. If you simplify that we get t 0 4 by t 0 3 minus 1 divided by q r by c p t 0 3 minus t 0 4 by t 0 3. So, all these parameters are known to us. So, we have to just substitute for these different values. So, we have t 0 4 which is 1100 divided by t 0 3 that is 405.63 minus 1 divided by q r which is 43 mega joules per kilogram. So, we have converted that to joules. So, 43 into 10 raise to 6 divided by c p which is 1005 joules per kilogram Kelvin multiplied by 405.63 minus the temperature ratios. So, if we calculate simplify this we get the fuel to air ratio as 0.017. So, fuel to air ratio as we have defined is ratio of mass flow rate of fuel to mass flow rate of air. And so, we see the number to be 0.017 which means that the amount of fuel that is added as compared to the mass flow rate of air is very very small. And in fact, in some of the earlier analysis we had assumed that fuel to air ratio will be very very small it can be neglected. In fact, if you look at many of the text books very often you would see that in many numerical problems f is considered to be very small and it is sometimes in fact, many a times is neglected. The basic reason is that this ratio f is much less than 1. And so, if you neglect this and calculate thrust the error that is introduced because of neglecting fuel to air ratio is usually very very small. But of course, calculating fuel to air ratio always is a good idea because you will get much better more accurate estimate of the thrust that is generated. At the same time if you have to calculate specific fuel consumption or thrust specific fuel consumption you need to know the fuel to air ratio. And so, it is always a good practice to calculate the fuel to air ratio though in many books you might notice that it is not always done. So, now that we have calculated properties at the fuel combustion chamber exit that is we now have the fuel to air ratio. Let us move on to the turbine. Now, in the turbine as I have mentioned for a turbojet engine the turbine generates work only to drive the compressor. Therefore, turbine work is equal to compressor work, turbine work is mass flow rate through the turbine into C p into temperature difference that is T 0 4 minus T 0 5. This is equal to m dot a which is mass flow rate of air into C p into T 0 3 minus T 0 2. So, if you simplify this we get T 0 5 is equal to T 0 4 minus T 0 3 minus T 0 2 divided by 1 plus f. So, all these numbers are known to us T 0 4 is 1100 Kelvin T 0 3 is 405.63 T 0 2 is 272.86 this divided by 1 plus 0.017. So, this temperature once we calculate we get 969.45 Kelvin. So, the turbine exit temperature stagnation temperature is 969.45 Kelvin. Now, how do you calculate the turbine exit stagnation pressure? Since the expansion process is assumed to be isentropic we use the isentropic relations again here. P 0 5 is equal to P 0 4 multiplied by T 0 5 by T 0 4 raise to gamma by gamma minus 1. So, this will be equal to P 0 4 which is 2.556 we have calculated that from the combustion chamber inlet. So, combustion chamber there is no loss taking place for the pressure and. So, P 0 3 is equal to P 0 4 which is 2.556 bar and. So, this multiplied by 969.45 which is T 0 5 divided by 1100 which is T 0 4 raise to 1.4 divided by 1.4 minus 1. So, the turbine exit stagnation pressure is 1.642 bar. So, you have seen that in the case of a turbine we need to just equate if it is a turbojet engine we can equate the compressor work to the turbine work and therefore, determine the turbine exit conditions from the equality of the compressor work to the turbine work. One of the later examples we will see the third one. We cannot really equate the turbine work to the compressor work because the turbine is generating more work than in what is required to drive the compressor and that extra work is actually reused to drive the propeller and so on. So, after we have determined the properties for a turbine the next component is the nozzle and in the case of a nozzle we need to take a little precaution here in one sense that we need to first determine whether the nozzle is operating under choked condition or is it operating under un choked condition because depending upon whether the nozzle is choked or not the exit conditions of the nozzle will be different. That is if the nozzle is operating under choked condition this means that the nozzle exit Mach number is 1 and if Mach number is 1 then we know that the exit velocity will be equal to the speed of sound and therefore, u e will be equal to square root of gamma or temperature static temperature whereas, if it is un choked then the nozzle exit velocity is what we had derived in our cycle analysis discussion earlier which is square root of 2 into C p into T 0 6 1 minus P a by P 0 6 raise to gamma minus 1 by gamma. So, how do you find out whether the nozzle is choked or not? So, to determine whether the nozzle is choked or not we need to find the stagnation pressure ratio across the nozzle as well as the critical pressure ratio across the nozzle which is the function of the ratio of specific heats and then if the actual pressure ratio is greater than the critical pressure ratio then it means that the nozzle is choked. If it is less than the critical pressure ratio it means the nozzle is un choked. So, depending upon whether it is choked or un choked the exit conditions can be different. So, to check for the choking of the nozzle let us find out the nozzle pressure ratio P 0 5 that is turbine exit stagnation pressure is 1.642 this divided by the ambient pressure which is 0.458. So, this ratio comes out to be 3.58 and what is the critical pressure ratio? If you go back to some of the earlier lectures we had discussed on compressible flows we had derived this expression for critical pressure ratio. Critical pressure ratio was derived as P 0 by P star is equal to gamma minus well gamma plus 1 by 2 raise to gamma minus gamma by gamma minus 1. So, it is purely a function of the ratio of specific heats. So, since gamma is 1.4 here we get critical pressure ratio P 0 5 by P star where P star is the critical pressure as 1.4 plus 1 divided by 2 raise to 1.4 divided by 1.4 minus 1. So, this pressure ratio is 1.893. So, what you see here is that the actual pressure ratio is 3.58 whereas, the critical pressure ratio is 1.893. So, the actual nozzle pressure ratio is much greater than the critical pressure ratio and therefore, this means that the nozzle is choking. So, what if the nozzle is choking? This means that if the nozzle is choking then the exit conditions will be fixed accordingly that is now the exit pressure will be equal to the critical pressure exit temperature will be equal to the critical temperature density will be equal to critical density and so on. And so from the critical temperature you can actually find out the exit velocity because exit velocity will now be equal to square root of gamma r t star. And so how do you find out the critical pressure temperature and density? So, we have to now first find these critical parameters because that is now the exit condition and then accordingly we can find the exit velocity and therefore, the thrust and fuel consumption and so on. So, let us now first find the critical pressure and to find the critical pressure we already have the ratios with us and from those ratios we can easily find out the critical pressure. So, the critical pressure temperature and density will need to be determined to calculate the nozzle exit conditions. So, t 7 which is the exit temperature is equal to t star is 2 by gamma plus 1 into t 0 5 and t 0 5 is known it is fixed as 9 69.5 Kelvin. So, 2 by 1.4 plus 1 into 9 69.5. So, this comes out to be 807.92 Kelvin. So, t 7 that is the exit temperature static temperature is equal to 807.92 Kelvin. Similarly, exit pressure will be equal to p 0 5 into 1 by p 0 5 by p star and p 0 5 by p star is already known that is 1.893 we had calculated that already this is 1.893 and p 0 5 is known as 1 is already calculated as turbine exit stagnation pressure 1.642 bar and so this exit pressure comes out to be 0.867 bar. So, the exit pressure of the nozzle is 0.867 and so density now would be equal to p by r t because there is an ideal gas here. Therefore, p is 0.867 bar, bar is converted to Pascal's. So, 0.867 into 10 raise to 5 divided by r that is gas constant for air 287 into the temperature that is 807.92. So, the exit density is 0.374 kilogram per meter cube. So, since Mach number at the exit of the nozzle is 1 the exit velocity is equal to square root of gamma r t 7 which is equal to 1.4 into 287 into 807.92. So, this is equal to 569.75 meters per second. Now, we know the exit velocity, the density and the area, area has already been given in the question we can calculate the mass flow rate. The total mass flow rate is now equal to rho 7 a 7 into u e which is 19.92 that is rho is 0.374 area is given exit velocity is also given. So, it is 19.92. So, the mass flow rate at the exit of the nozzle is also calculated and so now we know all the parameters required for calculating the thrust.  So, all these parameters are now known to us and so we can calculate the thrust generated by the turbojet engine and subsequently also the thrust specific fuel consumption that is T S F C.   So, this comes out to be 10.912 kilo newtons. So, the total thrust developed by the turbojet engine here is 10.912 kilo newtons. Therefore, fuel flow rate is equal to which is m dot f is equal to f into m dot a that is 0.017 into 19.92 which is equal to 0.3387 kilograms per second. Therefore, thrust specific fuel consumption T S F C is m dot f divided by thrust and m dot f is 0.3387 divided by thrust which is 10.912. So, if you divide it by the thrust we get 3.1 into 10 raise to minus 5 kilograms per newton second which is in turn equal to 0.111 kilograms per newton hour. So, in most of the books as well as in industry standards the fuel consumption is usually expressed in terms of a kilograms per newton hour because that is the newtons per second kilograms per newton second comes out to be a very very small quantity. So, they convert that to hour by multiplying it by 3600. So, which is why I have expressed it also in terms of kilograms per newton hour because that is in terms of some form of an industry standard where they would express the fuel flow rate as in the form of kilograms per newton hour. And so once we calculate kilograms per newton second it just have to be multiplied by 3600 to convert it to kilograms per newton hour. So, this completes our cycle analysis for the first problem where we have looked at turbojet engine and we have systematically carried out cycle analysis by considering each of these components in steps that is starting from intake then proceeding towards compressor combustion chamber turbine and then the nozzle. Only thing is that the nozzle we need to now check we have to check for choking that is if the nozzle is choked then the nozzle exit conditions are fixed by the choking conditions and therefore, the exit velocity temperature etcetera needs to be calculated based on the choked properties or the critical properties. And so if it is un choked of course, nozzle exit velocity is determined from the enthalpy and so from their enthalpy drop we can calculate the exit velocity. And the thrust is basically a function of the mass flow rate 1 plus f the exit velocity flight speed and also the pressure thrust term that is if the pressure thrust term is significant that will also contribute towards the thrust in some form or the So, this was the cycle analysis this is how we calculate or carry out a cycle analysis for a turbojet engine. And in the next problem that we shall be looking at we shall discuss about cycle analysis to be carried out for a turbofan engine. So, the analysis is very similar to that of turbojet just that we have an additional set of calculations to be carried out for the bypass duct that is the fan and the secondary nozzle. And the total thrust will be equal to sum of these two separate nozzles which are there. And let us look at the second problem statement. The second problem states that the following data apply to a twin spool turbofan engine with the fan driven by the LP turbine and the compressor driven by the HP turbine. Separate hot and cold nozzles are used that means that in this turbofan the exhaust streams do not mix before exiting the nozzles. So, you have separate hot and cold nozzles. The overall pressure ratio is given as 19 the fan pressure ratio is 1.65 the bypass ratio is 3 and turbine inlet temperature is 1300 Kelvin air mass flow rate is given as 115 kilograms per second. We need to find the sea level static thrust the specific fuel consumption when the ambient pressure and temperature are 1 bar and 288 Kelvin heat of reaction of the fuel is 43 mega joules per kilogram. So, this is a problem which involves a turbofan it is basically a twin spool turbofan engine. And the fan is here driven by the LP turbine, compressor is driven by the HP turbine and we have separate hot and cold nozzles which are used that is the two streams do not mix before exiting the nozzle. And so, we have thrust components because of the core nozzle as well as the bypass nozzle. So, some of the parameters of the engine which are the overall pressure ratio, fan pressure ratio, bypass ratio, the mass flow rate and turbine inlet temperature these have been specified. And so, we can based on this we need to carry out the cycle analysis basically find the sea level static thrust. Now, sea level static thrust means that the engine is stationary. So, it is generating thrust while it is stationary which means Mach number is 0. And so, the ambient conditions are also specified like temperature and pressure which correspond to the sea level conditions. So, this is a schematic of the turbofan engine which we have seen in the earlier lectures. So, turbofan engine as we know consists of these additional components that is the fan and the bypass duct and also the secondary nozzle. Other components are identical to that of a turbojet like the compressor, combustion chamber, turbine and the primary nozzle. And so, for cycle analysis we shall be using additional symbols which denote the bypass duct that is symbols with the prime symbol that is 2 prime, 3 prime and so on which correspond to properties in the bypass duct. 2 prime for example, corresponds to the fan inlet, 3 prime corresponds to fan exit and also the compressor exit inlet. And 7 prime is the secondary nozzle exit. So, other symbols are same as what we have used for the turbojet. So, we will begin the cycle analysis by calculating properties across the intake and since the Mach number here is 0 because we are required to find the static thrust that means Mach number is 0. So, the intake exit properties will be same as the ambient conditions. So, intake exit stagnation temperature T naught 2 prime is equal to ambient stagnation temperature which is 288 Kelvin and P 0 2 prime is equal to 1 bar. Now, the first component that we will take up for analysis is the fan after the intake and in the case of fan the pressure ratio is fixed it is known to us that is 1.65 bar, pressure ratio is 1.65. And therefore, P 0 3 prime which is the fan exit stagnation pressure is equal to the pressure ratio that is pi f multiplied by P 0 2 prime which is 1 bar and therefore, fan exit stagnation pressure is 1.65 bar. To calculate stagnation temperature at the fan exit we use the isentropic relation. Therefore, T 0 3 prime is equal to T 0 2 prime into pi f raise to gamma minus 1 by gamma. So, this is equal to 288 which is the stagnation temperature multiplied by 1.65 raise to 1.4 minus 1 by 1.4. So, fan exit stagnation temperature is 332.35 Kelvin. Now, in the case of the compressor in this question we have been given the overall pressure ratio as well as the fan pressure ratio. So, the compressor pressure ratio will be equal to the overall pressure ratio divided by the fan pressure ratio and so overall pressure ratio is given as 19 and fan pressure ratio is 1.65. And therefore, the compressor pressure ratio will be equal to 19 divided by 1.65 and once we know the compressor pressure ratio we can find the compressor exit conditions that is the exit stagnation pressure as well as stagnation temperature. And then we shall take up the combustion chamber the turbine. Turbine will be in two stages now we have the HP turbine first and then the LP turbine. So, HP turbine properties are determined by equating the HP turbine work to the compressor work. And LP turbine we equate that to the fan work and from there we can find the exit conditions of the turbine and then the nozzle that is the primary as well as the secondary nozzle. And we add up the thrust which is generated by the primary nozzle as well as the secondary nozzle that gives us the total thrust developed by the turbofan. So, for the compressor we have the pressure ratio which is equal to overall pressure ratio divided by 1.65. So, this comes out to be 11.515 p 0 3 is compressor exit stagnation pressure this is equal to pi c which is compressor pressure ratio into p 0 2 which comes out to be 19 bar because we are looking at static thrust. Therefore, p 0 2 is actually equal to 1.65. T 0 3 which is the compressor exit stagnation temperature is equal to T 0 2 which is inlet stagnation temperature multiplied by pressure ratio raise to gamma minus 1 by gamma. So, this is 332.35 which is T 0 2 fan exit stagnation temperature multiplied by 11.515 raise to 1.4 minus 1 by 1.4. So, this is 668.53 Kelvin. In the combustion chamber we carry out the energy balance as we have discussed for the turbine. Turbine inlet temperature is fixed that is 1300 Kelvin and the other parameters are already known that is T 0 3 and the heat of reaction of the fuel. So, we substitute all these different values here. So, we get 1300 divided by 668.53 minus 1 divided by 43 into 10 raise to 6 divided by C p which is 1005 joules per kilogram Kelvin into 668.53 minus 1300 by 668.53. So, the fuel to air ratio comes out to be 0.01522. So, after the combustion chamber we shall first take up the HP turbine and in the case of HP turbine we equate the HP turbine work to the compressor work because HP turbine is driving the compressor. And therefore, the enthalpy across the HP turbine which is mass flow rate times C p into the temperature difference T 0 4 minus T 0 5 dash this is equal to mass flow rate through the compressor into C p into T 0 3 minus T 0 2. So, the HP turbine exit stagnation temperature can now be determined which is T 0 5 prime this is equal to T 0 4 minus T 0 3 minus T 0 2 divided by 1 plus f. So, if you substitute for all these values we get the HP turbine exit temperature as 969.04 Kelvin. Similarly, we can calculate the exit pressure that is HP turbine exit stagnation pressure which is P 0 5 prime this is equal to P 0 4 into T 0 5 prime divided by T 0 4 raise to gamma by gamma minus 1. The combustion chamber inlet pressure is known that is P 0 3 which is equal to P 0 4 that is 19 bar multiplied by 969.04 divided by 1300 raise to 1.4 divided by 1.4 minus 1. So, this is equal to 6.79 bar. So, the turbine exit HP turbine exit stagnation pressure is 6.79 bar. Now similarly, we can also calculate the properties at the exit of the LP turbine because, now we know the LP turbine inlet conditions. So, for the LP turbine which drives the fan enthalpy across LP turbine is equal to enthalpy across the fan. So, this is mass flow rate through the LP turbine into C P into T 0 5 prime minus T 0 5 this is equal to m dot a c which is mass flow rate through the fan multiplied by C P into T 0 3 prime minus T 0 2 prime. So, from this we can determine the LP turbine exit stagnation temperature. So, this can be simplified in terms of the bypass ratio here we also have a bypass ratio because, that will be equal to fan mass flow divided by the core mass flow. So, we have 969.05 minus 3 which is bypass ratio into 332.35 minus 288 divided by 1 plus 0.01522. So, this is 837.98 Kelvin. Similarly, the pressure we can calculate as 4.08 bar. So, the primary nozzle we again check for choking as we did for the turbo jet. We shall check for choking for the primary nozzle here the primary nozzle pressure ratio is 4.08 bar whereas, the critical pressure ratio is 1.893 which means that the primary nozzle is choking and therefore, the primary nozzle exit conditions will be determined by the critical properties. Therefore, T 7 is equal to T star which is 2 by gamma plus 1 into T 0 5 this is 2 by 1.4 plus 1 into T 0 5 comes out to be 698.32 Kelvin. P 7 is equal to P star this is again from the pressure ratios we get 2.155 bar. Therefore, primary nozzle exit velocity is square root of gamma R T 7 this is 1.4 and 280 into 698.32 that is 529.7 meters per second. Now, for the secondary nozzle we again check for choking here the nozzle pressure ratio is less than the critical pressure ratio which means the secondary nozzle is not choking and therefore, the secondary nozzle exit velocity is calculated as this square root of 2 C p T 0 3 prime into 1 minus P a by P 0 3 prime raise to gamma minus 1 by gamma. So, this comes out to be 298.52 meters per second. Therefore, the total thrust generated is equal to sum of the primary nozzle thrust and the secondary nozzle thrust.  Now, from the mass flow ratios that is m dot a c divided by m dot a h is equal to 3 bypass ratio sum of these two is given as 115 kilograms per second. Therefore, the core mass flow is m dot a h should be equal to 28.75. So, if you substitute it for all these values we get the total thrust which is 40.74 kilo newtons. So, I leave it as an exercise for you to calculate the pressure thrust term as well because you now know the exit velocity the density and mass flow. So, you can calculate the exit area and therefore, the pressure thrust. So, the fuel consumption is equal to the fuel flow rate fuel to a ratio divided by mass flow rate. This is 0.4376 kilograms per second from there we can find the thrust specific fuel consumption which is 0.0388 kilograms per Newton hour. So, this was a problem where we were analyzing turbofan engine which is a twin spool turbofan engine with two separate streams. Third problem which we shall quickly look at is that of a turbo shaft engine we will need to calculate the specific power that is generated and the efficiency. So, let us look at the third numerical problem we have which is a helicopter using a turbo shaft engine is flying at 300 kilometers per hour at an altitude where the ambient temperature is 5 degree Celsius. Determine the specific power output and the thermal efficiency the specifications of the engine are compressor pressure ratio is 9 and the turbine inlet temperature is 800 degree Celsius. So, in this case we first calculate the speed flight speed which is 300 kilometers per hour. So, we convert that to meters per second that is 83.33 meters per second ambient temperature is 278 therefore, Mach number is 0.25. So, at the intake exit we have stagnation temperature and pressure which we can calculate as 288, 1.48 and pressure as 0.835 bar. For the compressor similarly, we calculate exit stagnation pressure which is 7.52 exit stagnation temperature which is 527.67 and therefore, specific work required to drive the compressor is C p into T 0 3 minus T 0 2 which is 247.42 kilo joules per kilogram. Combustor we get the fuel to air ratio which we can calculate as 0.013. Now, for the turbine we cannot here equate turbine work to the compressor work because this is no longer a turbo jet engine. So, for calculating turbine exit conditions we have the turbine exit turbine pressure ratio P 0 4 by P 0 5 this should be equal to the cycle pressure ratio which is P 0 3 by P a which is in turn equal to P 0 3 by P 0 2 into P 0 2 by P a. This is equal to 9 into 0.835 divided by 0.8 that is 9.394. So, from this we can calculate the turbine exit stagnation temperature from isentropic relations which comes out to be 565.63 Kelvin. Therefore, work done by the turbine is equal to 1 plus F into C p into T 0 4 minus T 0 5 this is 516.54 kilo joules per kilogram. Therefore, specific work output is W net that is W t minus W c comes out to be 269.12 kilo joules per kilogram. Then, how do you calculate thermal efficiency? Thermal efficiency as by definition is net work output by heat input heat input is C p into T 0 4 minus T 0 3 that is across the combustion chamber. Since, both these temperatures are known we can calculate heat input as 548.05 kilo joules per kilogram. Therefore, the thermal efficiency is the net work output 269.12 divided by heat input 548.05. So, this is 49 percent. So, I have now three exercise problems for you which you can solve based on what we have discussed in today's lecture. So, the first exercise problem for you is on a turbojet engine. A turbojet engine index 51 kilograms of air per second and propels in aircraft with a uniform flight speed of 912 kilometers per hour. The enthalpy change for the nozzle is 200 kilo joules per kilogram. The fuel to air ratio is 0.0119 and the heating value of the fuel is 42 mega joules per kilogram. Determine the thermal efficiency, the fuel consumption and the propulsive power. So, answer for this is thermal efficiency is 0.34. The fuel consumption is 0.1034 kilograms per Newton hour and the propulsive power is 8012 kilowatts. The second exercise problem is of a turbofan. A twin spooled mixed turbofan engine operates with an overall pressure ratio of 18. The fan operates with a pressure ratio of 1.5 and the bypass ratio is 5. The turbine inlet temperature is 1200 Kelvin. If the engine is operating at a Mach number of 0.75 at an altitude where the ambient temperature and pressure are 240 Kelvin and 0.5 bar, determine the thrust and specific fuel consumption. So, in this case the thrust is 74 kilo Newton's and fuel consumption is 0.027 kilograms per Newton hour. The third exercise problem is of a turbo prop engine. An aircraft using a turbo prop engine is flying at 800 kilometers per hour at an altitude where ambient conditions are 0.567 bar and 20 degree Celsius. Compressor pressure ratio is 8, turbine inlet temperature is 1100 Kelvin. Assuming that turbo prop does not generate any nozzle thrust, determine the specific power output and the thermal efficiency. So, here the power output comes out be 311 kilo Joules per kilogram. Thermal efficiency is 0.44. So, these are three exercise problems for you based on cycle analysis on exactly the same way we had discussed in in today's tutorial session on three different cases. A turbo jet engine, turbofan engine and a turbo prop engine. And in the second problem of course, it is a mixed turbofan. So, you need there is only one nozzle which generates thrust. So, you have to calculate nozzle entry properties determined based on an enthalpy balance between the two streams. So, I hope you have been able to at least understand the basic procedure involved in the ideal cycle analysis of jet engines. And so in today's lecture we have had discussions on cycle analysis, ideal cycle analysis of turbojet, twin spool turbofan engine and a turbo shaft engine. So, this brings us to the end of today's lecture which was basically a tutorial session on ideal cycle analysis of jet engines.


zero accounting software export to excel profit and loss or income statement reports get ready to be an office hero with zero here we are in our custom zero homepage we set up in a prior presentation scrolling in a bit holding down control up on the scroll wheel currently at 175 percent zoom in we're gonna open the demo company but do so by resetting the data resetting the data and open in the demo at the same time well then open up a couple tabs to put our reports in as we've done in every presentation right-click in the tab up top to duplicate it right-click in the duplicated tab to duplicate it again back to the tab to the middle accounting drop-down we want that balance sheet report go into the tab to the right accounting drop-down we're looking for the income statement profit and loss back to the tab to the middle we'll hit the drop-down on the date customizing it and take it on up to 2022 the end of it and update that's the setup process we've been doing every time support accounting instruction by clicking the link below giving you a free month membership to all of the content on our website broken out by category further broken out by course each course then organized in a logical reasonable fashion making it much more easy to find what you need then can be done on a YouTube page we also include added resources such as Excel practice problems PDF files and more like QuickBooks backup files when applicable so once again click the link below for a free month membership to our website and all the content on it we've been focusing in on the income statement and thinking about how we can compile our data in order to give it to a client and do so as efficiently and nicely as possible so last time we created some new reports with the income statement that we might bundle together and group them along with what we put together for the balance sheet report so now we have all these reports that we might want to be adding so the next strategy we might have is to continue exporting our reports to an Excel sheet and then use Excel and a PDF printer to make one PDF report with all of our reports on it so we'll continue with our strategy with this now we've got the Excel report that we started last time in a prior presentation of the balance sheet reports if you don't have this that's okay you could start a new one and just practice this technique we're gonna add then a couple other reports income statement reports and then use a PDF printer to print them all on one PDF file so let's go back on over I'm gonna just create these reports quickly again so the first one let's just do a standard income statement this time so I'm just gonna say a standard income statement I'm gonna right click on this this tab duplicate it before I do that so I'm gonna have my standard income statement on the left and then the one we will modify the only modification I'm gonna make to start out with is to say that I want to get rid of the decimals so I'll get rid of the decimals and there we have it so now that looks good let's save this one as just the standard income statement so I'm gonna export it to Excel I'll say open up Excel export it you need Excel in order to do this I'm gonna open it in Excel and then I'm just gonna basically copy this information onto the report that I want it to include it on or the workbook I want it in to copy it I'm gonna select the entire thing with the triangle and then I'm gonna say control C or right click and copy go on over to our Excel file we want to put it in add a new tab at the bottom I'm just gonna double click on the name and call it an IS or income statement put my cursor in either a one or select the entire workbook or worksheet and control V or right click and paste just normally and there it is I'm holding control and scrolling in a bit so there we have it I'm not gonna do a lot of more other fancy formatting but of course once you have it in here you can do other formatting once in here then I could go to the file tab I can print it I can print it using a PDF printer this one's a free printer that you could check out I'm not trying to advertise for them or anything but that's an option if you don't have one and then we can print the entire workbook which will include all the reports including the last one we put in notice it's left aligned right now so you could do some work to try to center the report on the page and so on with some formatting I won't go into that in detail at this time I just want to focus in on this technique because it's not an Excel course but there's a useful tool they can be used all right I'm gonna close the other report that we had here and do it again so let's go back on over and I'll save this one as a as a custom I'll say income statement number one income statement for for I'll just call it income statement and then we'll save it there so that's in the custom reports so if I wanted to line them up if I went to the first tab here and I went to my accounting and reports dropdown then I go to the custom area we have our custom report then I'm gonna create another one so this time I'm gonna go back to my normal profit and loss right click on it duplicate it that will be my starting point and this time I'm gonna create that vertical analysis but I think there's an easier way to do it we were doing it with the with the layout tab and if you hit this dropdown here they actually have a percentage of income in here already so that's a nice deal so I'm gonna select that and get rid of the decimals and update it and so now we've got like our vertical analysis I'm gonna call this vertical analysis something like that I probably spelled it wrong and we're gonna say you've done it a hundred times you should spell it right by this I know but whatever so there we have that let's update it and then I will open up that in Excel just a bit put that one there and I'll also save it let's save it in our custom reports report number two vertical analysis let's say and I'll paste that let's say vertical analysis analysis now this is incomes income I know how to spell that one statement okay save it open this one up and then I'll copy the entire sheet on on that one so I'm gonna say enable I'll select the triangle control C or right click and copy and then I'm gonna open a new tab where I want to put it hitting the plus button down below I'm gonna say an income statement vert or something like that put it in a one control V or right click and pay special you have to be in a one or select the entire work book worksheet I mean and there we have it if you go back to the file tab and we were to print this now we can print the entire workbook using the cute PDF printer and put this on a PDF so that's our end goal let's make one more over here just to to put the last one in place I'm gonna go back on over we're gonna go to another one let's do an income statement so let's go to the original income statement right click duplicate it pull that on over to the right and we'll start to build one from this one this time let's do a one that has multiple periods and so let's do it with the quarters again and then maybe a total on it so I'm gonna say let's go to the fiscal drop down and I want to say this quarter so let's start with this quarter and then also add to it the prior quarter update it so there we have it and then I'm gonna I'm gonna add them two up this time for the two quarters just to show that one so I'm gonna say let's edit the layout and let's say I just add another column and I do it with a form you love clicking on it we're gonna say I want this one plus this one I'm doing this fast because we've seen it before but notice how how nice that is and I could just get the total this is the tow towel and I'll go ahead and update that one and then we'll call us the income statement income statement quarter or something like that by quarter so now we've got first quarter second or the last quarter the prior quarter and then the total right we could do that we could add more quarters if we wanted to do four quarters on it so so but I won't I won't do that now so let's go ahead and export let's save that one I'm gonna say custom and save that one I'll just call it that and say save it I can export it as well to excel export it as well to excel we'll open that one up and copy it over enable select the entire worksheet control c or right click and copy and then back on over to where we want it add another tab double click on it income statement I'm gonna say cute quarter or something for the quarters I know I could name these better I'm gonna select the entire worksheet and right click and paste it so there we have that now this one I've got to expand this last column a bit does it all fit on one page I can check that looks like it does so that looks good so we'll keep it there I'll save it and let's go back on over I'm gonna delete this one I don't need that I'm gonna delete this one don't need that and then let's do one more starting with the original income statement right click duplicate it pull that to the right and this one I'm gonna do that difference kind of thing this time so let's do it with the quarters again let's go to this quarter let's say we want to see the prior quarter two boom then I want to see the difference between the quarters and I would like to see a percentage change horizontal analysis type of format edit the layout to do it and I'm gonna say delete changes add a column with with a form you line and I want it to be this one minus this one and then I'm gonna call that the difference and then add another column with a formula I'm gonna select that and say this is gonna be the difference divided by the prior period boom I want that showing as a percent please and this is the percent difference something like that update this one and so now we've got another fancy report done fairly quickly I could I should be removing the decimals on all of them because I that's what I've been my it's my custom so then I'm gonna say this is going to be a comparative I'm gonna call it a comparative comparative comparative income statement I'll copy that name and then go on down and say that I want to custom that one comparative income statement boom export it to excel export to excel open it up in excel add the final report to our excel worksheet that we're going to be doing here enable the editing triangle up top copy the whole thing minimize add another worksheet to our workbook double click on the name income statement comparative or something like that cop income statement cop comparative comparative and then put our cursor in a one or select the whole sheet control v or right click and paste so there it is there it is so now once we have these set up we could do more formatting on them but just note that now we have a fairly significant amount of reports that all relate to roughly the same period that we might want to put on one pdf file that we can give to someone at one time which has the reports in order that we want to put the order in them and we can give uh easily to someone so I could go to the file tab print it we could go then to the entire workbook that's the key component and you need a pdf printer the cute pdf printer I believe is free so you could check that one out then we have all this information up top it's not as centered as when we print the reports uh to a pdf but we could format them more if you wanted to do that and correct that fairly easily so there's clearly pros and cons to any technique but now we can get them all on one pdf file here which is pretty nice so let's go ahead and print it so there it is and let's uh minimize and and minimize and I'm going to save it over writing the reports here and save it boom and then open up the reports so now we've got our balance sheet that we did in a prior presentation we've got our comparative balance sheet we've got our vertical analysis we've got our our uh income statement and then we've got the vertical analysis income statement the quarter by quarter which I should fix because it's best up down there comparative income statement all in one area and people can then of course sort around in the pdf file pretty pretty nicely these days because of the the easier use of programs these days makes things a little bit nicer okay so let's open back up uh here also note if I go to the first tab that we have now saved these in our reports I didn't do it as meticulously as I have before but you know we've got the custom reports over here only the ones we created this time because we reset the data but if I have all of my custom reports in here then I can I can just open them up whenever I need them which is which is quite nice I might be providing these reports monthly quarterly yearly for example and uh if we have it set up cleanly then we could just set the the process one time and possibly assign someone else to print the reports or something like that fairly easily we can sign out the task possibly