the Narrowing Compressor Gas Path: Jet Engine Mistakes 3

This is one of the most common misunderstandings about how a jet engine compressor works.
Please post in the comment section any questions that you might have after watching this video. It's not the easiest principle to understand, but you can do it!

[@phasm42]

"The pathway is narrowed in order to keep the velocity up, because as it's being compressed it takes up less space, and being smaller, more of it can fit in the same place." - nicely put.

[@milosivankovic9791]

There is no mistake in the book. The bottom probe measures the total pressure (think about it like this - it faces the flow and 'sees' BOTH static pressure of the environment AND dynamic pressure due to velocity of the gas). The upper probe measures only the static portion of the pressure (as it is shielded from the flow velocity due to being recessed). The way you measure dynamic (what you call ram) pressure is by having a so called differential probe. The differential probe is a system of 2 probes like pictured, where you get dynamic pressure by subtracting static from total pressure signal. This is how you measure speed of aircraft (see Pitot tube).

[@AgentJayZ]

Upon review, I am agreeing with your analysis. I may, or may not, remove that section.

[@sergeigarbar1896]

Prandtlsonde, Prandtl tube.

[@ivorphillips7826]

@@AgentJayZ Hey, don't remove it - after all it reinforces the main reason I think you put it in "don't unquestionably trust a single source, even me", words to live by.

[@user-tw9io9nz2m]

I’m glad I scrolled down and read this because I was totally sold on the initial analysis in the video. Now I’ll have to do more research to see which one of you is right

[@milosivankovic9791]

@@sergeigarbar1896 Ludwig Prandtl is really the father of modern aerodynamics. It's a shame nobody outside of Germanosphere is honoring his name by using such a term, cause for sure you don't hear it elsewhere.

[@jonathanrackham171]

I believe the figure in the textbook (as discussed at ~5:15) is correct, as the scoop (like a pitot tube) measures the total/stagnation pressure (static + dynamic) rather than the dynamic pressure alone. You can see this explained in the paragraph (not the note in italics) immediately above the figure.

[@jonathanrackham171]

Obviously didn't read far enough down the comments to see that this had already been discussed!

[@micstonemic696stone]

he admitts this under the video have a look

[@leewood365]

A thing (your presentation) of real beauty Jay, especially the part about not trusting one and only one source. Thanks Man!

[@javierweatherspoon4313]

In one of your videos I watched recently, you said to verify information from multiple sources in relation to understanding new concepts and even go to older manuals that explain more than just the steps. I ordered more textbooks from Jepp, AIAA and even pulled out some old pilots manuals and found a better understanding of what you are explaining and what I am learning in A&P school. I don’t know why it’s taken me this long to open my eyes to this. Thanks for sharing all of this with us.

[@johno9507]

I did my aero engineering apprenticeship with Qantas back in the mid 90s and you described this subject far better and clearer than any of our instructors did. 🙂

[@AG-pt1sg]

Thanks a lot! 6 years in Moscow aviation institute, 2 years like technologist, 4 years b1.3, 2 years head of quality C category and only now I UNDERSTOOD))

[@ivorphillips7826]

With respect AgentJayZ I think you were mistaken in claiming in your text book diagram was wrong in the first few minutes of this vid. The pitot tubes readings (which all read 20 in the diagram) read the sum of the static and the ram (dynamic) pressure, to get the the ram pressure only you need to subtract the nearby static pressure readings from the pitot tube readings. So the diagram makes sense and reinforces your correct assertion that the sum of the static and dynamic pressure remain constant. Thought experiment: if there was no flow through the duct in the diagram what would the the pitot tube (lower row) of pressure gauges show?

[@AgentJayZ]

Yes, already noted in the first 3 minutes, by Milos Ivankovic.

[@corymitchell3228]

Actually, i think that in a real non-idealized jet engine, the stagnation pressure does not remain constant. The compressor blades are doing work on the fluid which means that the energy equation isnt quite so simple. Even for 1D flow with heat addition the total pressure is not constant so you can probly imagine that things get gross here 😂

[@ASJC27]

I'm sorry, Jay, but the mistake at 5:05 is on your part. The book is correct. The 90 degree bent gauges are total pressure sensors, not ram pressure sensors, which is why their reading is constant and equal to the total pressure. To make a ram pressure sensor you have to substract a static pressure reading from a total pressure reading. You can do it mechanically using a chamber that has a flexible diaphragm as one of its walls. Total pressure is fed from the 90 degree bent tube into this chamber, while the static port is connected to the other side of the diaphragm. The diaphragm is stretched in accordance with the difference between the total and static pressures, which is the ram, or dynamic pressure. The diaphragm is linked mechanically to an indicator. This is how a pitot tube works on an aircraft. The same can be achieved electronically.

[@AgentJayZ]

Yes, already noted in the first 3 minutes, by Milos Ivankovic.

[@ajs9732]

Brilliant video! I’ve been teaching basic jet engine theory for pilots for 1.5 years now. Not having an engineering background this channel has always been a gold mine of information, especially in those things which are hard to wrap your head around by simply reading it in a book. Keep it up!

[@zaydaslam4805]

Excellent intuitive explanation, honestly answered a lot of questions i had. Keep up the quality videos!

[@farayidarlingtonchaparadza20]

It takes a great man to admit his errors. You sir are incredible at disintergrating information into smaller bits for us to understand. Another reminder why I subscribed to your channel many moons ago.

[@surferdude4487]

Best explanation of how the compressors work that I've ever heard. Thank-you.

[@ssgrigio]

great video, but I believe the book is correct. 20 is the isoentropic stagnation pressure, when all the kinetic energy of the fluid is converted into pressure. it adds to the static pressure. this is what this type of theoretical instrument measures.

[@SumbluddyIdiut]

I agree. Static pressure doesn't disappear, it is still the same for the same point in the duct, and is present in both sets of gauges. A paragraph above the diagram states that a specially shaped probe measures both static and ram components of the pressure, meaning it measures the total pressure. The static gauges only measure static pressure. The following sentence of the above mentioned paragraph also points out that ram pressure is then obtained by comparing the total pressure (measured by the lower gauges) to the measured static pressure of the upper gauges. The bottom gauges SHOULD all show 20 as they measure total pressure, not ram pressure alone. Pitot tubes do the same thing, measuring both static and ram pressure. They can't measure just the ram pressure independently. This is why the static (ambient) air pressure needs to be known and accounted for.

[@ssgrigio]

@@SumbluddyIdiut perfect. If you want to messure Just the ram pressure, you need a "pitot tube".

[@alta9dena]

@@SumbluddyIdiut Assuming no work (in or out) and neglecting friction, Static Pressure plus Kinetic Energy (Vsquared*Density/2g) = Total Pressure throughout the duct. In practise the Steady Flow Energy Equation, (which is similar to the Bernoulli Equation) is used to derive the relationship between Total and Static Pressure, because it takes into account any changes in air density. Nevertheless Bernoulli is useful for visualising what is going on.

[@kerryb2689]

@silvio I agree. Energy must be conserved. The sum of the potential energy of the compressed air plus the kinetic energy of the flowing air plus the thermal energy (which was completely ignored for simplicity) must remain constant. There is an exchange between potential energy, kinetic energy, and thermal energy.

[@pinkdispatcher]

There's a nice mnemonic for remembering what the ram air sensors measure: They are called "pitot" probe, and they measure total pressure, which you could write as "p_tot", even though that's not what the name is derived from. Anyway, it shows greatness to admit to one's mistake, and not try to hide it or otherwise justify what you said. Only idiots would claim they never make mistakes. I was mistaken about a lot of things before I watched your videos, although I was convinced I knew it all. Thanks!

[@cjhodgson3000]

The first part gave me a headache with the question of "where is he going?". Then you said "there is no free energy" and the headache went away and it became very clear. Thank you for all your hard work. It was obvious you were really thinking hard to sort this out so we could understand.

[@whitemonkey7932]

Yes...but think on this...15000hp is being added to the system at this point...but we aren't creating energy...which is why none of this makes sense unless you account for temperature in the equation.

[@GermanTabor]

The jeppesen are a great resource. I went to A&P school with the new books which were ok but I got the old jeppesen books for when I really had a question about something…

[@avj4691]

Excellent explanation. Appreciate your patience in trying to impart difficult-to-understand information.

[@arronhyde]

Well explained... It's a head scratcher that you need to think real hard about the concept of how it works and you did very well to explain it. Most guys i work with wouldn't understand it.

[@orthopraxis235]

24:45 to 25:35 excellent summary of the pressure/velocity/narrowing space enigma.

[@electrospank]

I see sparks coming from my brain when I watch all of your videos. It's wonderful.

[@williamyamm8803]

Listening to you is good for my health ! :-) Thanks ! Greeting from France

[@AneriGS]

I love this channel

[@ronjon7942]

20:46 did it for me, along w the p/v graph. I never conceptualized the compressor/stator blades adding energy to the system. Nice work.

[@warrenbosch3581]

Thank you for this excellent tutorial. I thought I knew just a little until I watched this. I have helped design numerous flying things in the toy industry and even have one aerospace patent. I now realize I don't know anything! BUT. Now you have made me more interested in learning than ever before. Great job, Professor! You are a damn good teacher!

[@sionekulu7912]

Thanks very much J , its very clear and bright understand what you just explained about the compressed air along this pathway which become narrower and the air become smaller and more dense with this high rates of compressions . I also got the idea of when changing of pressure from the angle of the blade to the angle of the status

[@garym8348]

Thank you for taking the time to explain all of this. I've watched a ton of your videos (including the jet boat ones) but I think, knowing myself how a jet engine works, think this video just confused the shtuff out of some people. I've seen you do it before too but I'm just letting you know. I do enjoy your videos and the knowledge you share. Thanks.

[@kayhankurul4113]

The best explanation about how aircraft compressor works! Greetings from Turkey!🙂

[@albertcoia184]

Great presentation Zed!

[@governmentsnizzinspector2305]

I am in love with the way your mind works

[@ZenWithKen]

The rotor forces the air back, the stator compresses it, the path narrows to maintain the speed, check...I am definitely the brick in your explanation, lol. Thanks for sharing.

[@nickmiller76]

Compression also occurs in the rotor stages. In fact the rotor/stator split is usually about 50:50.

[@paemt6220]

You explained it well, I know because I understand know. The breakthrough was adding in the information I was not thinking about. The 10,000+hp being added into the system. I think people ARE thinking of this as a passive Bernoulli duct.

[@greghelms4458]

When you said towards the end that the air was compressed and got smaller that’s when I truly understood. Thanks.

[@AgentJayZ]

Oh, you've got potential and promise!

[@abdullahmohamed2083]

one of your best explanation videos thank you

[@grahamj9101]

Your J65 GA reminded me that my mentors in my early days in design at R-R IMD were some of the guys who designed the Sapphire 6.

[@jamyers1971]

SO GLAD that I bought that book a couple years ago, it really is great. Also, Ramjet. (Ducking for cover)

[@AgentJayZ]

Subsonic aerodynamics only.

[@jasondansie9970]

WHEN YOU INFORMED ME THAT THE COMBUSTION OF THE FUEL DOESN'T INCREASE PRESSURE BUT RATHER THE SPEED OF THE EXAUST INCREASES BY A HUGE FACTOR, I THOUGHT I FINALY UNDERSTOOD ALL THE IN'S AND OUT'S OF HOW A GAS TURBINE WORKS. "WELL YOU JUST STUFFED UP THAT NOTION" THANKS FOR OPENING MY MIND MR JZ

[@Kiera_Jackson74]

Honestly you are the best explainer... James Burke and Asimov level of understanding

[@AgentJayZ]

Thanks. The original Connections is one of my faves.

[@Kiera_Jackson74]

@@AgentJayZ Yeah... I want to start a language channel like yours but I'm years too late to the game I think I don't have the motivation

[@paramtrx9558]

26:36 That's pretty cool thx! I enjoyed this video. Have a Happy New Year JayZ!

[@user-tw9io9nz2m]

Thanks for the video uncle Jay

[@11Petrichor]

That was incredible information! Thank you!

[@briancox2721]

Another good place to look in the book if something doesn't make sense: the errata section. Not all textbooks have them, and of the ones that do, some list things corrected at that printing, while others list things which will be corrected at the next printing / edition. But if the book has it, and it lists errors that will be fixed later, the errata section is a good first place to look for clarity.

[@martinmendoza920]

You are awesome,, I'm a little slow in comprehension 😂, and I understood your expectations, that being said, I think you need not to worries much on your technique or methods you use, they are great, you always pull off the message you intend to share with your viewers, I ought to say it takes a brave person to try and explain the things you explain , great job, love your channel, Thank you for all your hard work and knowledge. ✌️👍👍😎

[@AgentJayZ]

Thank you! 😃

[@RCHeliMartin]

Like your explanation, is a good refresher for me after 45 years ago being in school 😁

[@jesser9134]

Each set of blades acting as diverging nozzles really clarified everything

[@baconbots]

very cool, your videos are so informational!

[@SpiralDiving]

I think you made the mistake of thinking the depicted gauge pressure was ram pressure. It's not it's Pitot pressure. Ram pressure is only a function of fluid velocity which is Pitot pressure minus static pressure via P(ram)=rho x v^2. The other mistake is talking about isothermal compression but its not -when you compress a gas it gets hot... You need to use P.V/T = a constant not P.V =const. As you say the jet engine compressor also adds energy to the airstream before the combustion section to raise pressure and _temperature_ further.

[@AgentJayZ]

description

[@sorgfaeltig]

Yes, "SpiralDiving" is correct to point out AgentJayZ's mistake. The illustration in the book is correct and does NOT contradict the text descripion or the calculations seen below the illustration. The gages as shown there will measure TOTAL PRESSURE = PITOT PRESSURE = STATIC+DYNAMIC PRESSURE = 20 PSI for each shown position. If you wanted to measure the dynamic pressure at the different station then each gage would need two connections, one like the one shown and another for static pressure to measure the differential pressure between static and pitot pressure for each station.

[@AgentJayZ]

description !

[@georgew.5639]

As soon as you mentioned Bernoulli I understood what you’re saying. The pressure does not increase as the area decreases because of the Venturi principle. As the space decreases in the engine, the flow of air will decrease in pressure. That’s provided that there wasn’t the compressor working there.

[@AgentJayZ]

This is not a Venturi duct. It's a compressor. That means work is being done and energy is being added to the inlet air. The Bernoulli effect is for a fluid stream moving through a passive duct, not a compressor. The pressure is increased by the interaction of the rotor and stators, requiring work. Completely independently of that process, the pathway is reduced in size in order to keep the velocity up. As air gets compressed, it takes up less space. If the pathway was not reduced, the velocity of the air would go down as it got more dense, and took up less space. Edit: I think I misunderstood your comment at first. I think we are both saying the same thing. Cheers!

[@mp6756]

Brilliant description you make it simple to follow despite your artwork disability.

[@shanakaliyanage6875]

Great explanation!

[@kerryh8er04]

Well that just answered a question I had today looking at a diffuser, why aren't they typically really eroded away.

[@faizalraazy91]

Ohhhh.. Great class , I enjoyed.

[@pamersiel]

I'm not sure whether silent omitting the temperature in the expression around 21:21 brings more harm or more benefit.

[@whitemonkey7932]

A great deal of harm.

[@HunterNapier]

This was fantastic, thanks!

[@steffinsokanovic4606]

Love your vids

[@torhacs]

Man what a great pratical learning video, THANK YOU VERY MUCH!!!

[@TeemarkConvair]

dude! you are So good!

[@karlanm925]

Love the channel, been watching for years. I've always dreamt of making a miniature Jet Engine, I've got a resin 3d printer and a Dental Lab with an ability to cast chrome cobalt from prints. Could you point me in the direction of any open plans for a simple kerosene jet engine? Would like to make something silly with it like turning it into a USB charging device, but am not sure where to get started. Also, my biggest concern would be balancing the blade assembly after casting. Thanks!

[@ddegn]

This would be an amazing project to follow. Maybe you could get something working without the need for super high strength and high temperatures. I have a couple 3D printers myself and as I watched this video, I kept wondering about making some sort of turbine engine model. Being able to cast parts in chrome cobalt may be enough to make functional parts. AgentJayZ has a video titled *"Turbine Blade Production Techniques"* which shows a lot of the blades up close. I read your comment before watching the production techniques video and as I watched it was thinking of ways you could make some of the parts. At 21:29 of the above mentioned video, he shows on blade with an internal steel pipe. Maybe you could reinforce your cast blades in a similar manner. Before reading your comment I wondered if there's any way one could make a turbine blade out of Porcelite or other 3D printable ceramic. A Porcelilte blade could probably handle the temperatures of a small turbine but I think the blades would be too brittle to work. I don't know if you're familiar with this material or not but you might want to keep in mind. The channel *Integza* has some videos where he attempts to make a pulse jet. That's where I learned about the 3D printable ceramic resin. If you attempt to make a miniature jet, I hope you document your efforts. I just subscribed to your channel in hopes you give it a try.

[@mrdarho4718]

the figure of the book is correct since it shows the static pressure measurement "tube in above" and the total pressure measurement "tube in below", there is no drawing of "pitot tube, that measure the dynamic pressure" in the book's figure. and I'm wondering where the writing of: P1V1=P2V2 (21:04) come from?

[@AgentJayZ]

Yes, covered in the first couple of minutes in the comments.

[@mrdarho4718]

@@AgentJayZ I didn't find down comment that discussed the written expression P1V1=P2V2 (21:04) where this comes from?

[@stgrundmann]

@@mrdarho4718 This is Avogadro's law (en.wikipedia.org/wiki/Avogadro%27s_law) first stated in 1811. It basically says that for a given amount of "ideal" gas (e.g. 1kg of air) p * V / T is constant. (air is approximately "ideal", water steam isn't.). Under the assumption that the temperature T is constant we get to p*V = const. So we could take an amount of air, compress it to half the volume and (after it cooled down to the initial temperature) it will have twice the pressure. That is called isothermic state change. But it is more complicated. The mechanical energy of the compressor blades goes into both pressure and rising temperature. If we assume that no heat is dissipated (in our case because the compression happens rather quickly for air passing the 1m compressor length at over 100 knots speed) then the state change model is called adiabatic state change. en.wikipedia.org/wiki/Adiabatic_process#Derivation_of_P%E2%80%93V_relation_for_adiabatic_heating_and_cooling might look scary but it basically says that p*V = const becomes (p*V) to the power of gamma = const Most importantly all 3 values (p, V and T) change . But , I assume for the sake of simplicity and to explain things step by step the temperature part has been left out.

[@randomtux1234]

what manufacturing process is used to make the compressor case halves with the blades on the inside face? might hazard a guess-shell cast?

[@ahmedtarek8983]

Thank you a lot sir

[@chilly856]

Excellent video!, I always thought until just now that the narrowing of the inlet was for compression. what is the limit to which air can be compressed for use in a turbine engine?

[@AgentJayZ]

More compression means more efficiency, but it's hard to achieve using a compressor with "seals" that don't actually touch. Modern turbofans have compressor ratios of over 40 to 1, and research engines at approaching 60.

[@tomshelton435]

Another great vid with tons of information. One observation, wish I had a Jetcity/Orenda T shirt or jacket. Lol

[@randomtux1234]

AgentJayZ what 3 books is at the top of your list of books on jet enjines

[@kayhankurul4113]

Thank you for great videos! Hello from Turkey! Intake air is getting pressurized by compresor stage by stage ,right? How high pressure air doesn't go back to low pressure zones? Also, do you have any video explain about co-axial shaft bearing arrangement? Regards!

[@AgentJayZ]

The air is moved by the compressor blades. Why does a propeller not make the air move forward? Same answer.

[@jaysonlee8625]

Excellent video as usual. However I would like to point out that the static pressure vs ram pressure diagram in the book is not a mistake. The reason those gauges read 20 is that they are adding the static pressure to the ram pressure as the gauges are physically exposed to the static pressure. Just like why pitot tubes have both a static and ram pressure port. There is no way to directly measure ram pressure without subtracting static.

[@AgentJayZ]

Got it. My mistake was pointed out 3 min after the vid was uploaded.

[@drawingboard82]

Good talk, well explained. I couldn't see for sure but the sawtooth Pressure-Velocity chart at 27:03 looks to be incorrect. I would expect, based on my own book, "The jet engine", published by Rolls Royce, that the velocity wont rise instantly at the rotor stage, but rather will build up slowly. The velocity would then drop smoothly over the stator stage where it will be converted to pressure. Something like an equilateral triangle wave ^^^^. Also, do you have an email address where I can send you anything? I might have a few large-format air system diagrams kicking about the house that may be of interest. Thanks so much for sharing!

[@rainerkinzinger555]

Thanks for your videos! I am learning allot from you :-)

[@henrikgormsen1020]

Great video

[@alte9751]

Hi. I have a question about the bell shaped intake. You mentioned before that they're used in the test cell to optimize airflow to the engine, but they're not necessary in airplanes due to ram air pressure during flight. How do you get them for the test? Do you have a bunch lying around? Do you order them from the manufacturer? I'm curious about this detail. Sorry if you've mentioned it before

[@AgentJayZ]

They are rare and difficult to obtain. They are also in the hundreds of thousands of dollars. So we don't do that. We design and make our own.

[@alte9751]

@@AgentJayZ Cool thanks! Love the videos

[@Eashlo88]

Hello AgentJayZ I am doing a Senior design project at my university where we are designing and building a small scale turbojet engine. I was wondering if you would offer any advice and answer some questions. Thanks!

[@jamesmason2228]

Just saw some pictures of RR's new test cell. It's impressive - but I wonder what they could do in a canvas tent? Half way to the arctic circle? In winter? :)

[@grahamj9101]

They do their Arctic testing, in the Arctic winter, with two engines mounted on a real airliner's wings, when they have already done a massive amount of test bed running. Try checking out what they did with the Trent 1000 at a place called Iqaluit

[@dickgoesinya4773]

Sounds like there is a extreme amount of engineering in the actual shape of the rotor and stater blades

[@khaledyousfi6819]

Hi AgentJayZ, I hope my question this time won't be a ridiculous . How does the temperature of the air change through the different stages of the compressor or at least in the last stage??

[@AgentJayZ]

Compressing air increases its temperature. Compressor discharge temp of modern airliner engines can be over 600 degrees F, and that's at altitude, where inlet air temp can be -50F. Compressor discharge air is as hot as a pizza oven, and is used as cooling air in the combustors.

[@grahamj9101]

Greetings from the UK on Burns Night - and yes, we do jump on the bandwagon and celebrate the occasion in England. My haggis is in the oven and will be ready in an hour or so. Let me assure your readers that, after being fattened up during our lockdown, it was dispatched humanely and didn't suffer: we don't treat them like lobsters. But I digress. I did make a point of reading your intro and I did get out my copy of 'Aircraft Gas Turbine Powerplants', which I bought (expensively), at your suggestion, from the USA, second hand. You will, therefore, get no criticism from me, unlike some who didn't take the trouble to read it. My only suggestion is that it would be useful if you explained that the energy imparted to the air in a rotor stage is effectively turned into a velocity increase in the circumferential direction. That unit of air in Fig. 3-40, which is grabbed in the space between two blades, is being whirled around and spat out further around the annulus. It is then grabbed by a pair of stator vanes, which turn that increase in velocity into pressure.

[@markdavis2475]

Luckily I got my copy, used (actually ex RAF Cosford Trenchard Library) 2nd edition courtesy of Amazon UK. It has the same mistake! First came across Bernoulli 25 years ago when I started selling D.P. flowmeters. Sold stuff to Filton where BMW were air flow testing plastic turbine blade models.

[@grahamj9101]

@@martinda7446 Delicious with tatties and neeps

[@grahamj9101]

@@martinda7446 I assume you already understand that tatties are potatoes. Neeps are turnips.

[@forloop7713]

the compressor gets narrower so the air velocity stays constant. the air velocity has to stay constant so the compressor blades can meet the air at the correct angle of attack or else they stall.

[@andrejradulovic5666]

Bottom pressure gauges are measuring total pressure (dynamic + static), not dynamic only ...

[@whitemonkey7932]

From which we calculate mass flow.

[@merrittkwarner8475]

I have a question about the burnooey thing. Years ago...30 or so....the tech I was apprenticeing under ( yes, I'm one of those piston guys ) had a gadget that when connected to an air line would produce colder than ambient air if he hooked it up one way and warmer than ambient air if he hooked it up opposite direction. The thing had a sort of venturi in the middle of it with a large funnel type cone on one side and a smaller one on the other. Would the burnooey effect come in to play here? And I understand that rapid decompression creates a drop in temp.

[@Skunkhunt_42]

Yes, a venturi tube is applying Burnoullis principal

[@fascistpedant758]

You're probably referring to cooling devices used in mines. It didn't use a venturi. Compressed air was injected tangentially into the tube to create a vortex. The air on the outside of the vortex was hotter and vented out one port while the cooler air in the center of the vortex was vented to a different port.

[@jonathanrackham171]

Sounds like you're describing a vortex tube, check out This Old Tony's video about it kzbin.info/www/bejne/fp-bmXePaceppqs

[@ajuc005]

Aren't the bottom gauges just showing the total pressure? Static pressure works in all directions so it should add to the dynamic pressure when you try to measure it (I think?). Then everything checks out on that diagram.

[@AgentJayZ]

description

[@andrejsgelins9296]

Good material to listen

[@slikerdet]

Yooohooo. I have seen some videos of LOCKWIRE AND NOT SAFETY lock wire. What’s the point of benefits to lock 6 or 8 or 10 bolts with one long lockwire vs 2 and 2 or 3 and 3 bolts together..?

[@clintnesbitt2704]

With the reduced tapering in the last few stages of the compressor, would expansion due to heating of the air charge not offset some of this slowing effect?

[@AgentJayZ]

The air might normally expand upon heating, but it's in a compressor, which is compressing it, by force, which causes the discharge air to be even hotter.

[@clintnesbitt2704]

@@AgentJayZ I understand that, the question was more about flow rate through the last few stages of the compressor

[@AgentJayZ]

I answered your question. The mass flow rate through the entire compressor does not change. What goes in, comes out. Any of the books I recommend would really help with understanding how compressors work.

[@randomtux1234]

AgentJayZ is there a jet engine noise generator app? I cant sleep without loud jet engine noise

[@raffaelefilardo170]

Thanks AgentJayZ. Great videos, informative and entertaining. Thank you. I have a question: All things equal, increasing the compression ratio increases power. If you double the CR then you double the amount of power needed by the compressor but the engine power output more than doubles. Why? (I did search your channel .... you mention increased CR in Questions 14 but didn't state why.)

[@ASJC27]

Several things: 1. The term used in jets is pressure ratio, not compression ratio. They are not the same - compression ratio is a piston term - the ratio of volume expansion, so really it is a density ratio. Pressure ratio is self explanatory. 2. If you double the pressure ratio, you don't double the power needed by the compressor. It is a non-linear relationship. If you start with a 10:1 ratio and double it to 20:1, you increase the compressor power by 45%, assuming that the compressor efficiency is the same. If you start at 30:1 and double to 60:1, you increase the compressor power by 31%. 3. Engine power will not necessarily increase if you just increase the PR. It depends on turbine inlet temperature as well. 4. What happens when you increase the PR, is that the thermal efficiency increases, so you will either burn less fuel for the same power, or burn the same fuel for much more power (and if the elevated turbine inlet temperature can be tolerated).

[@raffaelefilardo170]

@@ASJC27 Thanks for the informative reply. 1. How did calculate the power increase for the PR? I assume there must be some additional factor to P1V2 = P1V2? 2. How does the PR increase the thermal efficiency? I assume that increase in efficiency offsets the increase in power required by the higher PR compressor?

[@ASJC27]

@@raffaelefilardo170 1. a. Using Volume (V) is more of a closed system thing, like in a piston engine. It doesn't apply directly to an open system, like a jet, since volume is continuously varying and isn't clearly defined in that case. In open systems you can instead use specific volume (lower case "v"), or equivalently and more prevalently, density. b. P1V1 = P2V2 is incorrect. Temperature varies greatly with pressure and is also a factor. The correct relationship (for a closed system) is P1V1/T1 = P2V2/T2. In an open system, like a jet, the equivalent expression is P1/(rho1*T1) = P2/(rho2*T2). c. The power required by the compressor, normalized by compressor inlet temperature, specific heat capacity and mass flow, is given by 1/eta_c*(r^((gamma-1)/gamma) - 1), where eta_c is the isentropic compressor efficiency, r is the pressure ratio, and gamma is the specific heat capacity ratio (1.4 in air). 2. Jet engines utilize the Brayton cycle. The thermal efficiency of the ideal Brayton cycle is eta_th = 1 - r^-((gamma-1)/gamma). A higher thermal efficiency means more expansion is done following the expansion through the turbine (driving the compressor). A greater pressure ratio means there is more expansion possible (going from a higher pressure back to ambient), and that is proportional to the pressure ratio. The power consumed by the compressor increases at a much slower rate so more residual combustion heat can be recovered to shaft power in a turboshaft or thrust in a jet -> higher efficiency.

[@raffaelefilardo170]

@@ASJC27 Thank you again! You've taken me back to college chemistry. I can see how the power consumed by the compressor at much slower rate than the efficiency of combustion. I'm trying to visualise it at a molecular level. The increased pressure from the compressor is released (ignoring losses) so the net win comes from the combustion process. There must be something about the increased pressure/density of the working gas where more of the heat goes into the gas than is lost to the environment?

[@Craefter]

I sort of imagined that the centrifugal force of the air whipping around would also play a factor in the compression but I now see that is wrong (at least in this compressor design) because it is an interplay between the rotors and stators. If I understood it correctly the air takes a (sort of) straight path through the compressor. This would save weight on added structural design to compensate for centrifugal air rotating at 10.000rpm.

[@tysimpson1035]

I was with ya till gutter. Are we not in Canada?

[@JoeMama-nr9zi]

20:10 So does that mean that the last five-ish rows of compressor blades have a noticeably different angle compared to each other? That would be really interesting to see

[@iaeroespacial]

Hi AgenJayZ! Could you make a video about auto throttle? Expanding the info of broncolirio in his last video.

[@AgentJayZ]

That's an aircraft thing. Not an engine thing.

[@iaeroespacial]

@@AgentJayZ ok. Thank you!

[@thomasfsan]

If I say that in a high speed region, most of the molecules are busy moving in the direction of motion (ram pressure), thus can't put so much pressure on the walls? (static pressure) Does that adequately explain it? Vice versa in an expanded, slower moving area.

[@AgentJayZ]

Some books do explain it that way.

[@zoka7108]

Exactly. Gas pressure is gas molecules hitting the walls. Duct increases their movement in the direction of motion at the cost of movement in other directions.

[@micstonemic696stone]

AgentJayZ I have a question I've heard from 2 sauce's the forward thrust comes from the gas-turbine the moment it light's, I've not been comfortable with this because of thrust vectoring types like the harrier VSTOL, awesome fighter jet, one of my favourite's, can you help please ?

[@AgentJayZ]

Most of the thrust comes from the jet nozzles. If you don't start from there, you are already lost, and will never reach an understanding.

[@micstonemic696stone]

@@AgentJayZ the 4 stroke is often Suck, Squeeze, Bang, Blow, l am vehicle mechanic, but hear jet technicians say the same thing, but Bang implies high pressure explosion, you taught me the pressure is low, so Suck Squeeze, Burn, Blow is this true ?

[@AgentJayZ]

Sure. It's just a colloquial saying, and you can't take it too seriously. If you want to be more accurate, you need to realize that there is no such thing as suction. Not a topic for discussion; a topic for further reading.

[@micstonemic696stone]

@@AgentJayZ indeed, thankyou for your time, forever your student,

[@anthonyxuereb792]

Can you do a video showing the differences between a three shaft and two shaft engine please and the pros and cons? Thanks.

[@AgentJayZ]

I think I just did that recently.

[@anthonyxuereb792]

@@AgentJayZ Ok thanks

[@LordCogordo]

i have a question, sorry fornmy english, when u talk about v1*p1=v2*p2 (i studied chemistry so i saw this) u dont talk of the temperature rise when compressing the air, work is done there too right? thats why we use an intercooler after the turbo in the cars, the pressure shouldnt be higher?

[@LordCogordo]

besides that i think that this video was beautifully explanatory

[@AgentJayZ]

Intercoolers do help with compression, but a jet engine used so much air that the surface area of the cooler is so great that increased engine power is more than offset by the increased drag. Intercoolers used on stationary industrial engines are larger than the engine itself.

[@petermuller3995]

Thank you for your great content! Do you know the channel "Greg's Airplanes and Automobiles"?

[@stormeagle28]

As you're working on the LM2500 and this is derived from the GE CF6 airliner engine - can you please show its HPT 2nd stage disc? How big is it with the blades attached? In October 2016 one of these discs failed in the right hand engine (no.2) of American Airlines flight 383 (B767-300ER) on its takeoff run at Chicago O'Hare airport. It caused an uncontained engine failure and a fire due to rupture of the main fuel line to the engine, which damaged the aircraft beyond repair. One piece of the disc pierced the aircraft's wing, went over the fuselage and came down on an UPS building almost 3.000ft away, smashed right through the roof and ended up laying on the building's floor. Luckily only one passenger was seriously injured and no one died during the accident.

[@AgentJayZ]

I've shown the HPT of the LM2500 in several vids. It's a solid chunk of steel that contains a huge amount of energy whenever the engine is in operation. I believe most of the damage to that aircraft occurred from fragments and debris that bounced up after a large portion of the ruptured turbine disc impacted the concrete under the aircraft.

[@stormeagle28]

@@AgentJayZ Yep that might have been like running over a mine. Maybe you can show this turbine (stage) again the next time you get it in your hands... you have so many videos, I'll try to find it but it might be difficult.

[@AgentJayZ]

On my channel page there is a search bar. I just tried it with "HPT disc".

[@lox_5017]

As a brand new viewer of this channel I'm confused. In the compressor as the air is being squeeze doesn't the temperature of the air increase before fuel is injected into the combustor?

[@AgentJayZ]

The air in the compressor does increase in temp. Compressor Discharge Temp of modern airliner engines can be over 800 degrees F, even though the inlet temp czn be -50 F or even lower.

[@milize1528]

nice education sir

[@joshmellon390]

If I hadn't already subscribed for your cat a few videos back, I would have when you said your channel gets bothered with nonsense about free energy lol. There are some people out there who really don't understand how energy works.

[@AgentJayZ]

Yes. If you ever hear anybody try to explain how something works using the term "over unity", just smile and back away. That's a big red flag, indicating the speaker is an idiot.

[@juliussokolowski4293]

There is nothing wrong in that diagram. The bottim guages show total pressure. You can’t measure ram (dynamic) pressure directly. See pitot-static sytems.

[@AgentJayZ]

Yes, I left that in the video. It's my mistake, not the author's.

[@ChitranjanBaghiofficial]

just wanted to know, what if we replace gas combustion area with pulse jet engine combustion.

[@AgentJayZ]

Pulse jets are junk. That's why nobody uses them.

[@andreavico6198]

What about the centrifugal force applied to the air? How much it contribute to compress the air?

[@Miata822]

There is none. The air does not rotate with the compressor blades. It would if there were no stator blades, but immediately behind each ring of spinning compressor blades is a stationary ring of stator blades that redirects the air straight back. OK, not exactly true. each ring of stator blades _more_ than straightens out the air flow for it to be at the ideal angle for the next row of compressor blades to 'get a good bite.'

[@AgentJayZ]

In axial compressors like these, none. In centrifugal compressors, all.