Hello everybody! Welcome back to another video! Today I'm coming from my room with a microphone and a computer and I'm going to be talking about what is Asperger's syndrome? What is... What are the symptoms of... symptoms? Not symptoms. Signs. Signs are better one. Sorry, everybody. Let's try that again. Hello everybody! Welcome back to another video! Today we're going to be talking about what is Asperger's syndrome. I realize that I have done a lot of videos about the signs and symptoms, if you want to call them them, about autism, of autism rather. And I haven't really done a video on what it is. Now there are a lot of sort of theories out there and different classifications that psychologists and psychiatrists use to work with autistic people. Things that researchers use as well to characterize whether someone is autistic. But there is no sort of definitive answer to what it is and why autism makes people different. So in this video I'm going to cover the triad of impairments which I believe is one of the most applicable things to Asperger's syndrome. So the triad of impairments is a triangle and each section of this triangle you have three definitive different things that sort of give an overall characterization of what Asperger's syndrome is. It was invented by this lady called Lorna Wing who basically dedicated her life to understanding Asperger's syndrome and autism and she's made a lot of good progress for the research of autism. So the triad of impairments, one of the first signs of it is difficulties with social or emotional interaction. Now this means anything from expressing how you feel, expressing your emotions, but also being able to read social cues, read the room, understand how you should be acting in a certain scenario and these sort of these two factors in this top of this triad basically describe why it's difficult for autistics or people of Asperger's syndrome to communicate with people and to express their emotions and build connections with people which I can honestly say that is one of the biggest things that I believe is in the way of autistics having good mental health and having a good support network and being you know the reason why this is probably the reason why they get bullied so often. It's a very important point and it's the first point but let's move on to the next one. The next one is difficulties with social communication and language. In general, autistics find it quite hard to communicate their thoughts and feelings into speech. I know this seems very similar to the last point but it's a little bit different. It's about how you use your speech to communicate. It's not about being able to perform in a certain scenario or certain social group or being able to emotionally connect with people. It's just about being able to find the words and find the ways of communicating your mental state to other people. Although this does highlight language, it's not just language. It can be anything from non-verbal communication to actual verbal communication. Non-verbal communication being facial expressions, tone of voice, come on Tom. There's a communication fault clicking in. Body language. That was it. Body language. So anything from body language to facial expressions and all that stuff and this can inherently make it very difficult for people with autism to communicate to their parents or teachers making it very difficult for people to help them and for people to understand them. Now this sort of plays into the first point but it really is just a separate thing because it can be anything. It can be general communication, maybe like an interview trying to put your thoughts and your values into words and in general it can be quite disabling for a lot of people. It definitely was for me until I worked on it quite a lot. That is the second point. So going on to the last point. The last point is difficulty in flexible thinking or inflexible thinking, rigid thinking. In general autistic are very sort of logical and regimented in their thinking. If you imagine like a brain connection, one sort of connection would have to branch off into varying different things and then they would branch off into different neurons and stuff. That sounds very scientific. Terrible example but this sort of regimented branching off of ideas is very very very strong in autistic people. With non-autistic people it's a little bit more flexible so they can pick up things that don't necessarily make sense. They don't have to be rigorously explained to them in a way that they can form new connections and understand things but for autistic people it needs to be very regimented and very logical. There needs to be a strong line of connections and understandings for them to fully grasp a certain idea especially when it's emotions because those are tricky things. Abstract. Abstract things are very difficult. The way that this manifests in kids on the spectrum, children on the spectrum rather, they will usually take part in very sort of rigid play so usually kids develop their their abilities, their communication abilities and their imaginations in a very sort of abstract way. This may not be bound by rules so for example when I was younger I like to play games, imaginary games that had rules. For example I'd play this game where there was you know this one person which was like a boss monster and you'd have another sort of group which would team up and try and take them out sort of like a squadron of soldiers and stuff and that was pretty much the extent at which my imagination could extend to because there needed to be explanations for things. When you look at other kids, when they interact with each other and when they play imaginary games they sort of fluidly move from one scenario to another not really confining the actual play by rules more or less just going with the flow and doing what feels fun and that's a very big difference because although autistics can be very imaginative, very creative, there's a lot of people in music and arts that are autistic and they're absolutely brilliant at it but when it comes to sort of fluid creativity that is inherent in a person, neurotypical people, non-autistics they're a lot better at it. In terms of the problems that this can present to children or adolescents or adults even is that this inflexibility of thinking can make it very difficult for them to interact with people in terms of building connections with others. If they don't think something makes sense it doesn't make sense no matter what the other person thinks, the other person needs to justify things to them and this can make them very sort of appear very confrontational and rude and direct and but it's just a difference in the way that they interact and they process things. It can also really make it hard for them to empathize with people there is this test for a concept called mind blindness which basically means if you know something as an autistic child you expect everybody to know it because it's true and you don't take into account the fact that other people may not have had the same experiences as you but because you have had those experiences your opinion is stronger and that if they do anything that contradicts that it's ridiculous and unintelligent and stupid and that can make it very difficult for children to get on with each other and make it difficult for autistic people to empathize with neurotypical people. So I've gone over the triad of impairments there are a lot of sort of aspects to this there is also an idea which is sort of turns the triad into a quadrant into a square typically because most people with Asperger's syndrome have some form of difference in their sensory profile whether it be hyposensitivity sort of dulled senses or hypersensitivity which is heightened senses anything from noise visuals etc. So thank you very much for watching this video today my lovely viewers hope you're doing all right come back from a long arduous task at university four years four years I'm finished so I've got a lot of time to be talking to you guys on the internet and on a regular basis thank you very much for watching peeps like if you liked the video and subscribe to me please see you later guys


Welcome back danglers to the great outdoors and today we're getting an early start to the day to go meet up with my Buddy John B to do some camping and of course some fishing a bit of a science experiment going on here if you if you ever want to like Look up what a dumpster fire of a boat is on Google. Just don't even bother swing by my house It pops up take a gander. Yeah, she pops right up. See I'm trying some new batteries. I'm running 236 Front I'm running 30 six for the shoulder. Yeah, I'm running parallel. So I got like you got monster power I got like a hundred amp hour Basically because of 250s just in case we break down on the water Oh Man, I love it. I let every boat has a story John has John's boat has many stories Trying for help So the next few videos you guys are gonna see me and John traveling around to a few different lakes and doing some camping and Fishing but I'm excited to show you all a transition in the bite from my last few bass fishing videos It's going to a major plastic bag You guys are gonna see a sleigh bass on plastics and this is not a lack of effort for fishing hard bass We really fished a lot of hard bass We're just watching this transition take place where the fish are going up shallow and they're not as aggressive But we still manage to catch a bunch of them No, I I mean, how are we not gonna see one on a bed back in here? This is like clear water This is definitely weightless soft plastic fling it type stuff here So lock in the looking and all of a sudden it's too light Yeah, they're just they're just up there doing their thing Getting the mating rituals going first. Wow. That's my first pairing that I've seen so far this year. Oh Yeah, sandy spots Not full lock on but they're back here. I Gotta switch up. We got it. We got to have that log game Do you have logs? I figured you'd have a couple. I mean, how are you gonna come out here in March not have long So the newest worm that you guys are gonna see us fish today It's called the pinner worm it launches a couple weeks It'll be up on the site, but any other plastics you guys want to get and really anything You can save 10% off using my promo code Lfg at Guggen Squad dot-com oh Gosh, do you watch immediately giant how big And I'm so excited fishing today with John because we are literally watching the transition A fish going from winter into spring shallow mode We're watching male bass come up and make beds and just hunt around and it's such a good time to be on the water You got a seven inch. Oh It's happening though first Okay fishing freaks so last few videos for bass we have been We have been going after this pre-spawn thing. This is this is sort of transitioning right now. So this is good It's good for you guys to see the the full program transition So this is when I like to switch to these plastics. It's calm right now And there's fish up shallow and they're being a little sneaky doodles Being a little sneaky so I'm gonna take a Just a worm hook here I'm gonna take a worm hook and I'm gonna put a plastic on it and then I'm gonna fish that around in this case This is gonna be a six inch dart This is gonna represent any kind of bait fish just subtly moving through whereas Casting a vibe jig or a lipless crank bait or a lot of the hard baits here You're Disturbing that fish there. They're in a love-making mood They're starting to touch tails They're not aggressively feeding so What John is doing is he's throwing a frog which is a nice subtle bait on top and it's a good good bed-finding lure as well and I've got this dart which is You know fishing a fluke basically Like a watermelon red fluke that's been one of my favorite Ways to catch fish when they get in the stage when it's right at low 60s You got any sink goes.  He just ate that. He just came up. You watch when he I Watched him eat it. I Don't know how he didn't have it. I mean he Three-pounder slurp rolled on it three pounder. Yeah flash full flash had him like felt the thump Almost think he was a deep garter well John I think is captured. Are you at three? Are you at I got three you have three. All right You wanted to you wanted to return one of them. I've had two that have escaped me So we got some some fish. We're Approaching a point right now and there's some fish that are sort of staging out here in this grass so we're throwing weightless plastics and Well, we've had about five bites unless 10 15 minutes, I would say There's a lot I know we're just kind of soaking it in I know I do like I do really like what you have on I almost want to try a more natural color Got a couple of big armadillos down there.  Yeah, go This whole combination is not good I'm just a straight worm on this and it's not it's not a good gap. It's just not good. I Love a hook set on a spinning rod. It's just like Bins over in your line like cracks over you know because it's broad spinning so much Just feels good. Oh, I get them on spinning rods. All right. I'm switching What are you going? Daddy didn't raise no chump Where's that where's him pitterworm son right here? I got pink, but I got different color. I want to try another color so I've just lost my third fish and John is Just straight licking them here. So I'm gonna switch it up. I'm gonna switch it up To what is known as a one of our new baits It has not launched yet, but it will We plan on it so popular rigging style the Do you say Niko or Neco? You're your Niko guy I'm gonna follow your lead there.  That's what we just put on Seems to be the deal doesn't it? They're a little bit Come on now. Oh, she's coming around. Oh, jeez. Oh, jeez. You got your hands full I got her on the little rod here Is that the Crappie stick? No when you only catch one every hour you gotta play It is tugging pretty hard though. It's a nice fish. That's nice. Oh, yeah Biggest one of the day for sure. Fairly hooked Is that meat Fish brother up It has been doing some action in the shallows Hmm love to smell you smell you later.   We're trying to hit that 30 pound mark How are we gonna do that 30 inch mark? Maybe We found the grass So how are we gonna get out of here? I don't know. I'm waiting dude.  Oh, yeah See the bed, I mean look at that. Look at the calories. This guy is burned. He's chased off a few worms in his day Wow, this water looks amazing back here Clean I gotta tell you looking at fish. I'm kind of about that right now I'm about that looking at these fish right now This is really fun dude, uh know a Come on, baby. Nice. Let's go. Come on. Don't do that to me now. Why you gotta downgrade my fish mid mid swim That's what you say when it's smaller like it's beautiful. That's a chunky male though That's a chunked up male. He's got a black spot on his tail kind of cool You know, I've been waiting like 12 months to do this since the last time it happened Just dream about it just You dream about when that water hits 64 just going up there to look at Adam We're looking at him boys See a little chunk Oh God oh Gosh, there we go. That's a good one, man. You can help with that guy or not.  You just look at that beautiful, oh My gosh, I'm just seeing black tails swimming everywhere dude. We're in the spawn zone This is awesome This is the bone zone I'm about to catch this one out about he's about to crunch it. He's coming way out of the bed to eat it Oh my gosh, you see that? Yeah. Oh, you know, he's okay. I'm like I'm in his bed now So there's there's animosity and oh Yeah, oh, no, there it is It's going now Like I think we're I think we're watching some males come in and make these beds So fun. I was just looking at like Five fish all together right there Oh Wow spotted lip Do you think they get that during the spawn? No, I think it's a it's a yearly I got it all the time. I Know, you know crappie get those black spots, but I think yeah, I get darkened up. I guess So cool. This is it dude. I mean, I feel like we need a push pole and You've literally read my mind right there Me and John I've been waiting for this this moment for a long time, you know, I think it was actually last year where you and I We went up in a creek. It was either March or whatever and we both experienced our first You know bedding experience for the first time It was like late March and it was still cold Yeah, and it was actually kind of weird that we saw him on that because the water is still pretty chilly Fun times Guys, I got to tell you this worm It's it's actually Amazing in the water this action. This water's so clear out here. You can probably see this black one pretty decent but That twitch Neck of style my goodness, it's perfect size too So if you wanted to skip this or just do whatever I mean, it is nasty I Absolutely more than enough so for anyone that has never seen a bass bed before this is it It's a really good example so all this would be an Algae bottom But because of these fish are coming in here and they're fanning out the bottom Because they want a harder surface to lay their eggs and that's what it ends up looking like So that that's that's what you look for This time of year and if you're lucky you can find a really big bass in one of these shallow ones But usually the bigger bass are gonna bed somewhere out there where you can't really see them They're pretty smart about that But every once in a while you'll see a big one up shallow and I have caught some over the 10 pound mark that that are up there That's pretty exciting. You see one. There's just something about sight fishing Something about sight fishing is just exciting. I don't care if you're catching an eight-inch or an eight-pounder. It's just It's one of the best things and fishing That's a good Hey, right up as the lead in lead in the lead in Good snow and we're in Thank you Beautiful hey, man, we got all freaking day. Let's make the most of it Well, man, dude, there's just beds right on the outside of this grass line deep ones This is obviously not a big one, but we got healthy fish. He's a little less beat up very nice Looking for Tina. I think I've actually gone through a whole pack of black. I need to take a peek Running off like crazy, bro So I need to take a peek and a couple of these little every once in a while just to see if Nadine has moved up Now recording here. We are There's Caleb in the back taking a whiz Getting the full scenery here folks. You don't get to see the behind the scenes very much. I Don't know if you can see these beds, but we are just in the zone.  This is just absolute fun. We're in the fun zone We're at fun mountain at the peak of fun mountain We're summoning the peak right now Folks we are I just switched to a lunker log and as you would expect Caught one It's that time it is that time of year Plastic time what what time did you say it was I said it's plastic plastic time It's plastic 30. Look at your watch folks when that water is hitting 60 I'm gonna say 60 really it really starts kicking in but man, we're hitting 64 65 Now when we get before that I I honestly don't even really fish plastics Really until March usually every year I fish hard baits, you know But jerk baits lipless cranks cranks All the hard things don't ask me why it's such a big switch We're fishing a grass lake and it's more effective because we can get it through there That's that's not it. It's just they just eat them. They want the wigglies and jigglies right now So go to SWAT comm you must get lunker logs get some of the new pen or worms when they come out Just Dude literally you belly landed me Right about belly. I was aggressive use my code LFG good spot dot com And it's a big difference from if we throw a hard bait right now to a plastic. It's big Plastics go on leech watch plastic time I just got to get a shot of these hams because I think they're they're ready to come out of the oven Back Yep, you need to probe those are 185 right It's it's that weather where it's like in between shorts and bounce and Feels like shorts weather, but man early on Early on in the air when your legs aren't ready for it. It can really be Hurtful on those those legs. Okay watermelon red pender worm.  That's tasty This technique's been around actually a really long time back with wacky rigging was really popular when I was a guide People would often take like a zoom Trick worm, I got it the nest worm actually see look at that like I'm hooked up Just I'm just hooked up. I switched to the pender worm. It's it's on dude. I'm gonna join you What I was gonna say was the Fishing like a regular, you know Cinco style bait Sometimes it's it's not exactly the the action that they don't want. It's just the I Don't know. It's the profile so this is just a little smaller profile same type of action though same type of deal and You just get them Sometimes with those Wacky rigs with a bigger thicker plastic It just folds up and you you don't get a good hook set the fish will fold it up in their mouth This is almost a guarantee. You're gonna get a hook in there. Yeah And that's what it is.  Oh, dude.  I think it's bigger than that. Oh Let's go, baby Just thump it hard. Oh, yeah, just a donk. Donk. Yeah, just directly put so eloquently get back behind this wall do a little Chunk and can't be wait a minute might be up there on the bonk. Oh, yeah We got baby. It's just long Oh gosh, your boat is deep Oh Pinning them on the pinner This little spot right here is just uh It's a little nook behind this grass wall and it got super clear They just had to be in there That's really do like clear water for their spawning. It's like they like to see each other, you know They like to keep the lights on Nice fish. Let it go See you buddy Bye Right on the bank Is it a different color? Oh, this is not the pink.  That's her. That's her. That's her. No, I'm just kidding It's on the back side. Oh look it's gonna it's gonna jazz that other one up It's so deep. Oh, yeah, look at look at look. Oh, yeah. Yeah. Yeah. I'll look at that dude. It's a good one Look at this. I'm gonna I'm gonna flip in there. There's two. Look at that. I see you know, I see. Oh nice That's cool. They do not like that. Oh boy. Oh, that's a good one That's like my father.   Yeah Come on, baby watch this watch this watch this Oh, yeah, beat it.  Oh Jesus did he smack that look at what just happened there. I don't know Little too much tWitch just a twight little tight little Okay, I'm sorry. I'm so excited. You're good. I'm trying to get a big reaction. I Twitched my rod this thing went for I know Oh Hit it. Yeah, see the line see the line I don't know which one it was but got cranked dude. Get that flat bill out son. We are here gliding It's out here gliding Gliding with the boys spanker one right there up here in the reeds. Yeah, well spanky two of them. Oh my gosh They're roll. They're spanking each other right now. Oh my gosh Oh gosh, just taking me in the reeds Oh I'm looking at three site fish back here Yeah, just this one was on the bed with another one There's another solo guy And then there's you know four-pounder over here and three-pounder They're just oh, I watched this little morning dawn worm just disappear into its mouth Just be gone. The pinner was gone. I think it's a bigger one Sure. I'm gonna follow up. She's strong. She's with her man She's standing by her man No, we're not we don't need to get that Technical. Oh, she's jerking over at the jig man. She didn't like it. She came right to it. Oh She's on it.  Yeah, you can put it back. No way. Mm-hmm. She picked it up again.  Okay, let's switch to this one. Oh trusty. No trusty. We'll try it in true Okay That's her that's it I couldn't boat flip it man.  So we had literally two fish here at the end of the day fired up John caught the male. I thought they were gonna eat that glidey glide and Just one of my all-time favorites for fishing bed fish When they're contankerous The jig I mean she was just thumping it so hard We'll let this fish go nice fish. We're gonna let it go right back to its bed It's always cool to see that behavior interacting.  I don't know you got to watch every fish and they're all different, but I'm telling you I threw that that jig in there and we had blue three times And finally Yeah I just want to end the day on that that just Just completes my day. Oh my gosh. I love the spawn. I love texas Hey guys There we go Oh, man On the morning dawn. Look at him. I mean just Do you think it's gonna be like this getting on the worms, dude? No, I did not I thought it was going to be more of a moving bait kind of grind it out Maybe catch a giant, but Honestly What's more fun, you know, I We're just catching I like catching them. Hmm. It's probably the best, uh Weightless worm bite been on in a minute. That was weightless.  Cheers Cheers I love this. This is fun. I love this so much.  What'd you catch it on? Fishing in the spring is sort of weird when you get like You'll get fish that are kind of going up to spawn And then they'll they'll stay up there, but then they don't want to eat a lot of the other stuff This is when the plastics really come into play and This rig right here is just the ultimate finesse weightless plastic wind You're fishing that shallow water and what's really cool about this is you can also skip it under cover So you could skip this under docks. You could skip it under Trees and brush lines and things like that and get it up under where you might see fish Spawning or maybe they're just sitting. I'm fishing the size two or size That's the size two. This is size two John's fishing the size one. This is the gold series wide gap finesse hook I could fish both the lunker log and this bait on this one Uh, the size one would be better suited for this in in my opinion John, do you like fishing? I don't like fishing. I love fishing I just want more of those pink worms. How come you got a ton? I I got like not they just sent it They like the crew just sent it to me a couple days. Come on. I got to bring these one more fishing I've honestly been waiting on this day Like for a month. I've been going out and like oh, it's almost water's almost 65 and then it's like finally here We witnessed it. It was a great great time out here with John B. By the way, go and check out my man here he's actually got a uh a recent video that I thought was a Excellent excellent video and throws a little perspective on fishing and electronics and everything right now So he'll be linked here. Go go check that video out. We're going to be doing another adventure We're actually going to be heading over to John's camper. I've never Been in that thing we're going to get a little tour of that and we're going to be going to a completely Different lake that I've been wanting to fish for years that could have some monsters for site fishing So subscribe to the channel Smash that like button. We'll see you on the next


Jeff, tell me about what happened on Friday here. I was traveling southbound on Broadway. I was in the farthest left lane. I caught movement out of the proof of vision of my eye. And I saw a bicycle go between myself and another car. He swung out and made a left turn. The crosswalk was full of people. And he hit this young lady. When he hit her, he flew over his handlebars and he landed on his head. He was not wearing a helmet. So you could tell that hurt a little bit. So I pulled over and immediately contacted the young lady that he hit and asked her if she was OK. She said she wasn't hurt. She actually kind of thought it was her phone. But then I told her, well, you're in the crosswalk. The light to cross was on. You're perfectly fine. A lot of people may not know. Is the bicyclist at fault here? The bicyclist has to follow all traffic rules, which would be yielding to a pedestrian in the crosswalk. I did cite him for a bicyclist failing to obey all traffic laws since he did not have the right of way. And he ran into the young lady. So we've talked about the responsibilities of bicyclists out here that they have to obey all traffic laws. But what about pedestrians? What are their responsibilities? Pedestrians have to cross at the crosswalk. And they have to obey the crosswalk signals. If there is no signal there, they still have to cross within the crosswalk. The laws are designed for the safety of the people.


So we're here at the Linaro Connect, so what's the latest that you've been working with? Personally at this Connect I've spent so much time in steering committee meetings, supervising all the segment groups, discussing with all our members, trying to provide as much value as possible because Linaro is all about collaboration with our members. So it has been a pretty intense week, we're all tired, all got cold, their conditioning is pretty strong here, spend very valuable time together. Even we had social events in the evening and those are very good opportunities for networking regardless of being in a nice venue with good food, but you continue discussing, talking about priorities, talking about things to improve or things that have gone really well, so very intense week. I'm very pleased with the keynotes, lots of good stuff. We tried also to have something new, different. We had Qualcomm present artificial intelligence into drones, self-driving, everything and it was really exciting, forward-looking. We had Hadamica discussing about software licenses, open source. How the open source license is going to stay irrelevant forever basically because if some people are excited about it, who knows what will happen with the next generation. That was exactly the topic of Hadam keynote. All the open source contributors, they own a copyright and within the licenses, even within the GPL, they grant use of that code to the developers who receive the code. But now the question is when the first generation of developers, of contributors, will they move to a better phase of life, somewhere up? When they get upstreamed. They get upstreamed there. Yes, that's a nice way. Their hairs will own those copyrights for 70 years, at least according to the US law. You hold a copyright until you get upstreamed and for 70 years. Will their hairs be as good as favorable with the open source community? Or will there be any rogue contentions trying to make money out of it? This is what Hadamica was raising in her keynote in a very clear and positive way. Was she suggesting to cancel all copyrights? No, she was not. No, she was not. She proposed a few solutions. I'm not an expert in the domain. I'm very interested in the software licenses, but I'm not an expert. So I'm not the one who can explain those proposed solutions. But there will be fixes, I'm sure. And today there was a keynote from Yan Rabe. That was also one of my favorites. I hope people enjoyed it. If not, I'm the guilty one. Because five years ago I attended a keynote by Yan Rabe at the University of Lund in Sweden. And I was completely fascinated by his keynote, by his communication style as well, his vision. Five years ago he was envisioning connecting IoT devices, sensors and more, not just as we do today in a one-to-one, a sensors giving you a temperature or counting how many steps and telling it to you, and that's it. No, his vision five years ago and he was already working on it. So this comes to ten years ago. He was envisioning connecting all these sensors and distributing tasks across the sensors and possibly having the sensors coordinate by themselves just like bees in a swarm. And this is the swarm, his vision. And now Linaro is making this real. So today, fast forward, and today this is for real, the swarm OS. This was the demo on the opening keynote. So the swarm OS is a kind of coordination framework, software framework between the IOT nodes and there are students and professors like Professor Zufo from Brazil who are prototyping these on the Linaro 96 boards. And so since at Linaro we started working on Internet of Things, my dream was to have Jan Rabe come and connect and give a talk about the swarm and about the future. And the vision is going forward. This morning Jan was talking about expanding this network of nodes, the swarm, even more and having humans part, humans part of the swarm. We all carry our phones, but we should do more and be connected, which should be smart humans connected to the swarm. That was very fascinating. Hopefully it will be a positive swarm, another aggressive swarm. Yes. It will be our friends. Our friends, as Jan said. And as my boss says, yes, we love the idea, but my boss says he loves the idea but he wants to become a smarter, connected human. Well, maybe the next generation when he's not anymore. So it's fascinating. The keynote was really great. So thank you, Jan, if you're watching this. Professor Jan Rabe, thank you for coming and giving such a fascinating keynote. So what's the other stuff you're working on the next few months? What's going to happen? We had a great day with the ARM data center today. We had a lot of companies coming and talking about their products. There's some pretty cool chips coming out. Thunder X2, the new Qualcomm chipset. I'm sure you can add links under this video with links to those companies talking about their chips. It's been many years in the making, many years. Now, seven years. Correct. Have you worked even longer than that on the server ecosystem stuff? Linaro started seven years ago. And you? I started with Linaro seven years ago. I was with SDRyxon in the Linaro board and then moved to Linaro five years ago. And we started working on servers five years ago. And so now it's maybe the year that things are going to shake up a lot. And all this work that Linaro's been doing is maybe going to pay off one more. It's all coming together. So if you see, we have LEG. We have the developer cloud. The developer cloud is a full blown cloud instance running complete full open source. So at Linaro, we have collected all the open source packages. We have made sure that they all run on ARM and all the patches are upstream. It shall just work. No local quirks. It shall just work. And so the developer cloud is running these open source software with all the patches pulled down, for example, from the OpenStack Color project. It's upstream. There's no hidden patches. And so we have a complete data center that is running on ARM servers full open source. That's the best proof that we are ready. So it's coming. You're ready? It's coming. It's not just the swarm. It's also the servers. Everything. And as soon as the chips are qualified for mass production, they will go in. That's the hardware side. And the home group is pretty busy. The mobile group is still busy. They're all busy. They're all busy. We're all pleased about the way they are progressing. There's a lot of interest in the home group in continuing on unifying the media acceleration set of boxes, smart TVs, OTT, digital home devices. And Linaro with OptiE, the open platform trusted execution environment. We are in a unique position to have a complete open source, secure solution. We have integrated Microsoft Whitevine, Microsoft Play Raid in Google Whitevine. Apologies. We have integrated them with OptiE. We have a complete open source solution for the security and the DRM. Of course, deploying these into products requires the right licenses with Microsoft and Google, of course. So that's not just their legal implications that people should comply with. But we can implement it with open source. And so we can combine the unified work on media acceleration with a secure solution and we can achieve a complete open source, secure data plate solution. These can be applicable to smart TVs, OTT. These can be applicable to both Linux. We have an open SDK, which is a collection of open source components into a complete media reference framework with open embedded Yocto. But also Android TV. We can help our members interpret the secure solution within the media acceleration for Android TV products. So all the industry is super happy with the work that Linaro is doing. So, like, you're probably watching this video and saying thanks a lot for what you're doing. No, thanks. Thanks to you, Charlie. And you're probably telling them you're welcome. Just come over. And maybe you can't just solve everything. Maybe you're too good and you solve so much. But you always have new challenges. Another very exciting announcement was from our CEO last Monday, the associate program. So we want to widen the base of companies collaborating with Linaro. It's beyond the Silicon vendors, the commercial district vendors. We want to help to have the OEMs, ODMs, building products. They can be part of what we are doing. They can help share valuable feedback so that the Silicon vendors and the ODMs, they can build better products, better time to market. They can help influence the... Yes, correct. It's the kind of the voice of the market. So there should be hundreds of associate members by next connect. I don't know. I don't know. Potentially we have this in every segment group and it looks exciting. Cool. And then there's also a lot of cool stuff happening with the networking. Now every base station is going to be on par, running some of the cool new networking solutions. And this has been working for years also. I let those companies who are providing those demos. I'm sure you have filmed the demos here. So the demos are pretty cool. Whether base station will run, I let you check with those companies. I cannot speak for the companies. But it's exciting. It's exciting that the open data plane acceleration is showing the full potential. And this shall not be just an ARP only thing. We really want to break this assumption that ARM is ODP and Intel is DPDK. It's wrong. It's a wrong setting. We do work on DPDK as well, on ARM. And we would be pleased to share best practices from ODP to DPDK. It's about collaboration. So Llinaro, we want to collaborate with any project. And ODP has been proving a lot of increased value. Already at last connect in Budapest, last March, we demonstrated a complete high PSEC acceleration with ODP. Integrated with open fast path. And that is tremendous. So you can expect more to come. Networking is looking at SparkNIC, Edge devices. It's exciting. There's a lot that will be coming. So that risk idea really paid off. And Sophie Wilson did a nice little implementation of the instruction set. And everything is great, right? Now, you know my next stream? Last Tuesday, you saw the announcement of the social next, the developer box. That is super cool desktop machine based on the social next 24 core chip. That is running Linux. You want to put that inside the DBC computer? I want to port RISCOS on that machine. Port RISCOS? RISCOS. RISCOS is the heritage. It was developed by Sophie, Sophie Wilson together with the first ARM chips. So RISCOS in 1988 had anti-aliasing and vector rendering and drag and drop. We discussed this last few months ago. RISCOS now is an open source project with the RISCOS Open Limited, if I'm not wrong. Some friends in UK, they parted it on the, I think, on... Dragon board? No, I'm not sure. I would love that. I would love that. So I would like to see RISCOS ported on the social next PC. That would be the new RISC PC or the new... Can you invite Sophie Wilson to do the keynote at the next Linux? That will be a dream come true as well. Why not? Why not? Hopefully, Hong Kong loves a nice place. Why not? Why not? That's a great idea. Thank you.


Hello and welcome to lecture number 36 of this lecture series on turbo machinery aerodynamics. The last class we had introduced a very new topic which is quite different from what we have discussed in some sense that is on radial flow turbines. Of course, we have already discussed its counterpart the centrifugal compressors, but in the context of turbines radial flow turbines operate in a quite different manner as compared to axial flow turbines. So, last lecture was exclusively devoted towards understanding of radial flow turbines, the basic working of radial flow turbines, the thermodynamics of radial flow turbine that is as the flow passes through a radial flow turbine, what in a thermodynamic sense happens to the flow as it passes through the different components of the turbine. We have discussed the different types of radial flow turbines which are possible, the two broad classes are the outward flow radial turbines and the inward flow radial turbines. We have seen that though the development the original development of radial flow turbines was in the out flow mode, very soon it was realized that the inflow mode or the inflow radial flow turbine is a much more efficient way of managing the flow and achieving work output than the out flow type of turbine. So, majority of the turbines which are being used currently are inward flow radial turbines. The radial turbines as a concept was originally developed towards meeting the hydraulic power requirement that is to convert hydraulic power into work output and then that continued for a very long time. In fact, that even continues till date and the hydroelectric power plants that we are aware of use one or more types of radial flow turbines. Of course, some of them also have axial flow types, but apparently radial flow turbines are very commonly used and the use of radial flow turbines for gas turbines was much later that it was also being considered as one of the options for use in a gas turbine engine. And the modern day gas turbine technology has restricted use for radial flow turbines it is primarily used in the smaller engine of smaller thrust class engines. And it is the main disadvantage when it comes to use of radial flow turbines for gas turbine applications is the fact that in order to achieve very high efficiencies and power output the turbine performance is a strong function of the inlet temperature. And in axial turbines we have already discussed that modern day gas turbines use different cooling technologies which can be used to which can actually permit us to use turbine inlet temperatures which are much higher than the material limits themselves. This is not it is still possible in radial flow turbines, but it is much more complicated to introduce and use some sort of cooling mechanisms in radial flow turbines. There are very complicated methods which are being proposed and probably being used in some types of engines. And this is one major disadvantage by radial turbines are not really used in gas turbine technology in gas turbine engines for larger thrust class engines because the turbine inlet temperature is sort of restricted. We saw that there are two distinct types of inward flow turbines the cantilever type and the 90 degree inward flow turbines. And we discussed in the previous class we devoted most of the time towards discussion on the 90 degree inward flow turbine because that is more commonly used. The cantilever type turbine is very similar to the impulse turbine that we have already discussed in the axial turbine context. And therefore, there are certain disadvantages associated with impulse turbine as we have already seen. And so the inward flow 90 degree inward flow turbine is like a reaction turbine. And in geometrical terms a 90 degree inward flow turbine looks exactly same exactly similar to a centrifugal compressor. But of course, just that the direction of flow and the rotation of the impeller or the rotor is exactly opposite in each cases. We also discussed about the governing equations which are used when we analyze the flow in these different components starting from the outlet flow the volute or the scroll. And then the nozzle blades and then it goes into the rotor or impeller. And then there is the exducer which forms the later part of the impeller it is counterpart and centrifugal compressor was the inducer. And then in typical turbine we may also have a diffuser downstream of the rotor to recover part of the kinetic energy which would otherwise be lost. And I think we also discussed two other aspects of radial flow turbines the efficiency and performance parameters. We define different forms of efficiency the total to static efficiency and how it is related to total efficiency and the work output and so on. We also discussed or spent quite some time discussing about the different loss parameters which are used in radial turbines like the nozzle flow coefficient and the rotor flow coefficient. We also looked at the incidence losses that is the flow entering the rotor need not necessarily be at zero incidence. And off design conditions the incidence angles could be quite high which leads to a substantial increase in the losses that is an additional component of loss that comes into picture when a turbine is operating in an off design condition. So, these are the different loss mechanisms that we discussed in little bit detail we did not of course spent lot of time discussing them because I think it is fairly out of the scope of this course to discuss the loss mechanisms and design optimization techniques in detail that is a vast subject on its own. So, which is why I decided not to spend too much time on that we will discuss some aspects of design and performance in the next class. But today's class I thought it is a good idea to have a tutorial session we shall be discussing a few problems which I will solve for you here. And then I also have a couple of exercise problems which I think you should be able to solve based on what we had discussed today as well as in the previous class. So, today's lecture is going to be a tutorial session and so let us take a look at the first tutorial problem that we have let us take a look at what the problem statement is. So, the problem number one that we have for today is the following the rotor of an inward flow radial turbine which is designed to operate at the normal condition is 23.76 centimeters in diameter and rotates at 38140 rpm. The design at the design point the absolute flow angle at the rotor entry is 72 degrees. The rotor mean exit diameter is one half of the rotor diameter and the relative velocity at the rotor exit is twice the relative velocity at the rotor inlet determine the specific work done. So, in this particular problem it is basically a very simple problem that of course the first problem that I normally solve is a very simple problem. We have some dimensions of the rotor we have the rotor diameter we have the rotational speed and the nozzle inlet angle. We also have been given that the rotor mean diameter is one half of the rotor exit diameter and the relative velocity at rotor exit is twice the relative velocity at the rotor inlet. So, based on this data that we have we need to find a specific work done. So, this is I would say very simple problem but again as I keep emphasizing every time I have a tutorial session is that we start solving a problem with the velocity triangles. So, let us construct the velocity triangles in this particular case and then we shall see and we shall proceed towards solving this problem because velocity triangle will help us understand what are the no parameters which are known to us and what are those parameters which we need to estimate and calculate. So, for a 90 degree IFR turbine we have the velocity triangles as shown here. We had already discussed this in the last class let me again just quickly explain the construction of a 90 degree IFR turbine. We have a volute or a scroll which sort of acts like a plenum chamber through which the flow enters a stagnation chamber here where the flow is stagnant and then that gets accelerated through the nozzle blades. So, these are the nozzle blades nozzle exit station is denoted by station 2 the flow then passes through the rotor blades or the impeller and exits at station 3 before going into the diffuser which is exit of the diffuser denoted by station 4. You can see that as I mentioned this is very similar to a centrifugal turbine a centrifugal compressor. In that case of course, the velocity the flow direction is the other way around it is reverse the flow actually proceeds in this direction. And instead of nozzle blades we have diffuser wings and therefore, these airfoil orientation will also be the reverse. Let us look at the velocity triangles for this case we normally have v 2 which is entering the rotor radially. So, the nozzle flow leaves relative velocity entering the nozzle is in the 90 degrees and the c 2 is at an angle of alpha 2 in this case it is given as 72 degrees. So, alpha 2 in this particular case is 72 degrees u 2 is the blade speed at the station that is station 2. Now, as the flow leaves the rotor the absolute velocity leaving the rotor is axial under design condition relative velocity is that an angle of beta 3 with the axial direction and that is v 3 and u 3 is the rotor speed at station 3. So, this is a typical inward flow radial turbine and the corresponding velocity triangles at the rotor inlet and rotor exit. So, based on the data that we have for this case we basically have the rotational speed we have the exit diameter. So, I think we should be able to find out u 2 and then subsequently we also have been given some ratio of the blade speed at the rotor exit to the mean diameter and the relative velocity at rotor inlet and exit. So, with this data we should be able to find the specific work done. Now, let me recall what we had discussed in the last class when we had derived a very general expression for a 90 degree I of R turbine where the specific work was if you recall a function of three distinct parameters. One is a difference between the blade speed at the inlet and outlet u 2 square minus u 3 square. The second term was a function of the relative velocity and the third term function of the absolute velocities. So, we are going to do exactly the same thing here to calculate the specific work done. Let us calculate these three individual components and then add up all of these and that gives us the specific work done. So, specific work done was 1 by 2 into 3 different terms u 2 square minus u 3 square plus w 2 v 2 square minus v 3 square and the third term was the absolute velocity. So, let us get these individual terms first add them up and then we get the specific work done. So, blade speed at the tip is u 2 which is basically pi d into n divided by 60 and so n has been given as 38, 140 rpm the diameter is given as 23.76 centimeter. So, it is 2.2376 this divided by 60. So, here u 2 comes out to be if you substitute these values we get u 2 as 474.5 meters per second. Now, this is where the velocity triangles come into picture that if we need to calculate v 2 if we have the velocity triangles in front of us it is very straight forward. It is ratio between v 2 and u 2 is related by tan alpha 2 or cot tangent alpha 2 cot alpha 2. So, let us look at the velocity triangle v 2 and u 2. So, tan alpha 2 is u 2 by v 2 and therefore, v 2 is u 2 into cot alpha 2 alpha 2 is given as 72 degrees and therefore, v 2 the relative velocity at rotor inlet is 154.17 meters per second. Similarly, c 2 is u 2 sin alpha 2 c 2 is this and this. So, sin alpha 2 is u 2 by c 2 and therefore, c 2 is u 2 sin alpha 2 and since alpha 2 is 70 degrees we get c 2 as 498.9 meters per second.  So, 2 into v 2 and 0.5 times u 2. So, what we get is that the square of this we basically get c 3 square c 3 square is 38786 meters square per second square. Similarly, let us find out the 3 individual components u 2 square minus u 3 square this is u 2 square into 1 minus 1 by 4 this is 1 by 2 square and so, it is 1 by 4 this should be 168863 meter square per second square.  So, this is 71305 meter square second square per second square. The third term is c 2 square minus c 3 square we already know c 2 and we know c 3 square already. So, that difference is 210115 meter square per second square. So, these are the 3 individual components which contribute towards the specific work done. So, the next specific work done would be 1 by 2 times the sum of these 3 components. So, we add up all the 3 divide that by 2 we get the specific work done. Now, this is one way well probably the direct way of calculating specific work done. We can also approximate specific work done without having to undergo any of this, but of course, this is an approximate estimate of the specific work done that is simply equal to the square of the blade speed at the exit. So, delta w is u 2 square and why is it u 2 square? That is because the flow enters the rotor and leaves the rotor in the axial direction. So, c w 3 is 0 c w 2 will basically be equal to u 2 specific work done is u 2 c w 2 minus u 3 c w 3. The second term would become 0 the first term is equal to u 2 square. So, delta c w should also be equal to simply u 2 square. So, if we do that if we calculate work done in both ways let us see what happens. So, if we add up all the 3 different terms and divide by 2 we get the specific work done as 225 142 meter square per second square. And what we see is that the fractional contribution of each of these 3 terms the first term which is u square is 0.375 v square is 0.158 and c square is 0.467. That is you can see that there is a fair share for all these 3 components in the specific work done. So, we can calculate specific work done also by the second method which is basically because in this particular case beta 2 is 0 and alpha 3 is also 0 in which case delta w simply becomes u 2 square. So, if you simply square 474.5 the whole square you get 225 150 meter square per second square you can see that they are very close. Specific work done calculated in either of these ways should give you the correct answer both these methods give you identical specific work done. So, one way is to calculate the individual components and add up all of them which is a more general method because irrespective of how the velocity triangles are you could that is still valid. Whereas, delta w is equal to u square u 2 square is valid only if in this case like in this case the incidence is 0 the deviation is also 0 in which case you can directly calculate delta w. And one would get the same delta w excepting the round of errors which you see in both these calculations. So, this first problem as you can already as we have seen is a very simple problem which is involves simple application of mind to solving the velocity triangle to calculate the different components or constituents of the specific work done. One is the blade speed the other is relative velocity and the absolute velocity we add up all the 3 divided divide them by 2 we should get the specific work done. So, let us now move on to the next problem that I have for you and it is a slightly more involved problem. But of course, again I normally keep these problems limited to very fundamental aspects of the particular aspect that we design working on in this case it is the radial turbine. So, we are just looking at very fundamental thermodynamics of radial turbine and how we can apply some of these principles to calculate some parameters associated with radial flow turbines. Let us take a look at the second question we have for today. Problem statement number 2 is a radial flow real inflow turbine develops 60 kilowatts power when running at 60,000 rpm. The pressure ratio P 0 1 by P 3 of the turbine is 2 the inlet total temperature is 1200 Kelvin. The rotor has an inlet tip diameter of 12 centimeters and an exit tip diameter of 7.5 centimeters. The hub to tip ratio at the exit is 0.3 mass flow rate is 0.35 kg per second. The nozzle angle is 70 degrees and the rotor exit blade angle is 40 degrees. If the nozzle loss coefficient is 0.07 determine the total to static efficiency of the turbine and the rotor loss coefficient. So, here you can see of course, that the problem involves lot more data than what we had for the first case. We have the power input or power developed by the turbine that is 60 kilowatts the rotational speed 60,000 rpm the pressure ratio is 2 turbine inlet temperature is 1200 the dimensions of the rotor the tip diameter 12 centimeter exit tip diameter 7.5 hub to tip ratio 0.3 mass flow rate and the angles. And additionally the fact that the nozzle has a loss coefficient of 0.07. We need to find total to static efficiency and the rotor loss coefficient. So, this is the problem statement for this second question that we have as always we will first start with the velocity triangle. It is exactly the same as we have seen in the first problem. Nevertheless, let us just quickly look at the velocity triangles and understand what are the data that has been provided for us in this question and what is it that we need to find. So, in this case the velocity triangle again is the same as we have seen in the previous question. This angle is given to us as 70 degrees and the exit blade angle beta 3 is given as 40 degrees. We have the dimensions at station 2 and station 3 and hub to tip diameter ratio is also given to us. We have the power output and the rotational speed. So, quite a bit of data is given to us rotational speed is given means u 2 and u 3 are already known. Since angles are specified the other components can also be calculated. So, let us start calculating some of these parameters. The rotor tip rotational speed we can calculate from pi d n by 60 which is 377 meters per second and here d 2 is given in this question. It is given as 12 centimeters rotational speed is 60000 rpm. So, if we substitute pi and the rotational speed and the diameter divide that by 60 we get 377 meters per second. Now, from the velocity triangle we know that at the rotor inlet beta 2 is 0 and therefore, sin alpha 2 is simply u 2 by c 2 or c 2 is u 2 into cosecant alpha 2. Alpha 2 is given as 70 degrees and u 2 we have already calculated as 377 meters per second. Therefore, c 2 is equal to 377 into cosecant 70 degrees that is 401.185 meters per second. T 0 2 is given to us thermally temperature is 1200 Kelvin. Since c 2 is now calculated we can calculate t 2 static temperature that is t 0 2 minus c 2 square by 2 c p that should come out to be 1130 Kelvin. So, this is the preamble of the question that is we are solved some single parameters which may be required for solving the rest of the problem. So, first part of the question is to find the total to static efficiency and that is where we will make use of the pressure ratio that we have been given to we have been given as 2.0. From the pressure ratio we should be able to use that data to calculate the total to static efficiency and so that the next part of the question we going to solve is to find the stagnation temperature drop across the turbine and from the power output we can actually find the isentropic or the actual power output actual temperature drop across the turbine. If you take the ratio of the 2 we get the total to static efficiency. So, the stagnation temperature drop which is given by t 0 1 minus t 3 s is equal to t 0 1 into 1 minus t 3 s by t 0 1. We can convert this ratio into the pressure ratio because this ratio has been given as 1 by 2 that is 0.5 t 0 1 is known to us and therefore, this temperature drop is required for calculating the total to static efficiency because total to static efficiency is t 0 1 minus t 0 3 divided by t 0 1 minus t 3 s. So, t 0 1 is 1200 this multiplied by 1 minus 0.5 raise to gamma minus 1 by gamma you can calculate this and this should come out to be 190.92 Kelvin. Turbine power output as we know is m dot c p into t 0 1 minus t 0 3 power output is given as 60 kilo watts mass flow rate is 0.35 kgs per second c p we are assuming for gases as 1.148 therefore, t 0 1 minus t 0 3 is 149.33 Kelvin. So, if you simply take the ratio of this total to static efficiency is 149.33 divided by 190.92 and that is 0.782. So, total to static efficiency in this case is 0.782. So, that solves the first part of the question where we were required to calculate the total to static efficiency. Now, the next part of the question involves or requires us to calculate the rotor loss coefficient we have been given the nozzle loss coefficient as 0.07 and. So, we will need to make use of that rather complex formulae that we had derived in the last class if you remember there was a very long formulae which was relating the loss coefficient to the radius ratio and the static temperature ratios. So, we will make use of that formulae to calculate the nozzle well the rotor loss coefficient and that formulae involved both nozzle loss coefficient as well as rotor loss coefficient multiplied by the angles and so on. So, we know that in this case the radius ratio because this will be required in that formulae. Radish ratio R 3 by R 2 is basically the hub diameter at station 3 plus the shroud diameter divided by 2 that is the mean diameter divided by diameter at station 2 and this we know the hub to tip ratio is given as 0.4. So, we substitute zeta here and that is basically 0.4 in this question this multiplied by d 3 s plus d 3 s divided by 2 into d 2. So, if you substitute these diameters which have been given as well as the hub to tip ratio we can get the radius ratio as 0.406. Now, for efficiency this was the long expression I was mentioning total to static efficiency is 1 plus 1 by 2 into zeta n the nozzle loss coefficient t 3 by t 2 into cosecant square alpha 2 plus R 3 by R 2 square into zeta R which is the rotor loss coefficient into cosecant square beta 3 plus cot square beta 3 this whole raise to minus 1. Now, this though it looks rather complex formulae we have derived this expression right from the first principle. So, if you know if you can if you can actually relate the first principles formulae of total to static efficiency which is t naught 1 minus t naught 3 by t naught 1 minus t 3 s the numerator gets expressed in terms of the nozzle and the stator or the rotor loss coefficients numerator is expressed in terms of static temperature ratios and so on. So, from that we can actually derive this without much difficulty the numerator actually becomes C p times U into C p times C w delta C w and then delta C w we express in terms of the angles and so on. So, it is a very simple 2 3 step derivation from which we can derive this rather longish expression that we see here for total static efficiency. Now, in this expression we still have an unknown that is t 3 by t 2 t 2 we have already calculated, but we do not know the value of t 3 which also we can in fact calculate provided we know the exit stagnation temperature and the exit absolute velocity. So, if that is known we can actually we should be able to calculate because in this case beta 3 is given and from the velocity triangle since beta 3 is given we can and U 3 is known we can calculate C 3 and what about t naught 3 for t naught 3 the power output is given inlet stagnation temperature is given. So, we can calculate t naught 3 from there t naught 3 minus c 3 square by 2 C p will give us t 3 and then we can take ratio t 3 by t 2 that would be deriving the whole thing from the first principle or the whole thing can be expressed in a single definition term which also had I think mentioned in the last class which is basically in terms of some of these parameters which we know and a t 3 by t 2 is actually defined in terms of the velocities and the loss coefficients 1 minus U 2 square by 2 C p t 2 into 1 plus r 3 by r 2 the whole square multiplied by 1 plus zeta r cosecant square beta 3 minus 1 into cot square alpha 2 and. So, if you substitute for these values here the only unknown is zeta r. So, we should get 0.9396 minus 0.02187 zeta r. So, this if you substitute in this expression where zeta n is also known you get a quadratic equation and then we can solve that zeta r which is the rotor loss coefficient can be calculated as 0.62. So, this is one way of calculating zeta r the other way of course, is to calculate t 3 by t 2 using what I had mentioned t 2 we have already calculated we know what the value of t 2 is. For calculating t 3 it involves 2 3 steps 1 is to calculate the stagnation temperature at exit t naught 3 which can be calculated from the power expression power is is equal to mass flow rate into C p into delta t t naught 1 minus t naught 3. So, there all the three all the parameters are known except t 0 3 we can calculate stagnation temperature then to be able to calculate t 3 we also need to know the velocity at the exit that is c 3 and to calculate velocity at the exit c 3 we know beta 3 we also know u 3 and how do we know u 3 because the rotational speed is given to us and the diameter at the hub or at the exit of the rotor is also known. So, from there we can calculate u 3 since u 3 is known beta 3 is known we can calculate c 3 that is basically u 3 cot alpha 3 should be equal to c 3. So, once c 3 is also known the static temperature is at exit t 3 is equal to stagnation temperature t 0 3 minus c 3 square by 2 C p and so that is a rather easier way rather than this like longish formula as I just shown and then substitute for that in the expression and we should be able to calculate zeta zeta r which is the rotor loss coefficient nozzle loss coefficient is already known to us it is 0.07. So, what you can see is that the loss coefficient for the nozzle in this case has been given as 0.07 and for the rotor it is 0.62 and so I think I had also given some range of these values in the last class I had mentioned typically the rotor loss coefficients are on the higher side because of the fact that there are rotational effects coming to picture losses associated with the rotor are much more than the losses associated with the stationary component like a nozzle especially when the flow is accelerating. So, in this case the rotor loss coefficient is in fact close to one order magnitude higher than the nozzle loss coefficient. So, this solves our second problem which required us to calculate the total to static efficiency as well as the rotor loss coefficient. So, you can clearly see that this is a slightly more involved question in this of course I have taken the easier route of directly substituting this in the formulae what I would strongly urge you to do and I probably leave that as an exercise for you is to derive these equations from the first principles and not simply use the direct longish formulae it is very easy to derive the equations from the first principles. In the total to static efficiency definition term it is basically T 0 1 minus T 0 3 divided by T 0 1 minus T 3 s or let us express that in terms of enthalpy h 0 1 minus h 0 3 divided by h 0 1 minus h 3 s the denominator gets expressed in two separate forms one is to do with the nozzle loss coefficient second is the rotor loss coefficient numerator gets expressed in terms of mass flow rate C p and delta T and so on. So, from this you can actually denominator has nozzle and rotor loss coefficient term numerator is already known to us. So, this can be simplified and you can actually calculate the nozzle loss coefficient the rotor loss coefficient given the nozzle loss coefficient in a much more simpler less confusing manner than simply substituting them in the formulae. I have picked up this method because if you take up any text book you would normally see this kind of a method where they would refer to the derivation which was discussed earlier on like what I have done and the problem we just simply substitute plug in the values and calculate the corresponding efficiency and other terms required. Now, that brings us to the third question let us now proceed towards the third problem that we have for us to solve and then we will see how this problem is different from the previous ones. An inward flow turbine with 12 vanes is required to develop 230 kilowatts at an inlet stagnation temperature of 1050 Kelvin and a flow rate of 1 kilograms per second. Using the optimum design efficiency design method and assuming total to static efficiency of 0.81 determine the absolute and relative flow angles at the rotor inlet, part b the overall pressure ratio p 0 1 by p 3 and part c the rotor tip speed and the inlet absolute Mach number. So, this is a three part question where we have required to calculate three different aspects one is the angles absolute and relative flow angles the pressure ratio and of course, the rotor tip speed and the Mach number. And of course, you can see what is mentioned here is that we can assume optimum efficiency design method. I mentioned some preliminary aspects of this in the last class I would urge you to take up slightly more detailed reading of this in any of the text books that we have mentioned. You can take up pick up any book on the turbo machines and you will find a small section on optimum efficiency design methodology wherein a few formulae would basically be derived again from the first principles and you can see what it is basically trying to tell us. So, in this question it is a question which involves optimum efficiency design methodology which can be assumed and then we can we are required to calculate the angles and Mach number and so on. So, the first part of the question is to find the flow angles the relative and absolute flow angles. For optimum design it is known that the absolute angle at the nozzle exit that is alpha 2 is simply related to the number of blades and this comes from what is known as the white field width fields formulae which basically equates or which basically relates the alpha 2 to the number of plates. So, cos square alpha 2 is equal to 1 by n. So, that is where n is the number of blades. So, for optimum design it actually has been shown that by width field that cos square alpha 2 is 1 by n 1 by n where n is the number of blades. So, in this question we have 12 vanes and since we have been told to assume optimum design we can simply substitute the number of vanes here and calculate alpha 2. So, alpha 2 would be 73.22 degrees also as another consequence of the width field equation we have beta 2 is equal to 2 into 90 minus alpha 2. Beta 2 is the blade angle at the inlet of the rotor and that is equal to 2 into 90 minus alpha 2 and so this comes out to be 33.56 degrees. So, the next part of the question is to find the pressure ratio well basically the total to static efficiency and for which of course, we will use the total to static efficiency basically is T 0 1 minus T 0 3 by T 0 1 minus T 3 or T 3 as and from there we can express the denominator in terms of P 3 by P 0 1. And in this question we know that the power developed is given as 230 Kelvin inlet stagnation temperature is given mass flow rate is given and the total to static efficiency is given. So, we need to find the pressure ratio in this case. So, all we have to do is we just substitute the power output here which is m dot into C p into delta T C p is known assumed and the stagnation temperature is also can be calculated or it is given at the inlet. Since the efficiency is known we can calculate the pressure ratio P 3 by P 0 1 as 0.32165 the inverse of this is the turbine pressure ratio which is P 0 1 by P 3 and that is 3.109. So, this basically comes from the efficiency definition eta T s is T 0 1 minus T 0 3 divided by T 0 1 minus T 3 s the denominator gets expressed in terms of the pressure ratio from the isentropic relation. Numerator is simply C p times delta T and therefore, that is the power output of the turbine. Now, the third part of the question is to find the Mach number and then we also need to find the blade speed at the tip of the rotor. So, we will first find the absolute Mach number at the inlet for which we will find first the Mach number corresponding to stagnation conditions. So, m 0 2 this again is coming from the very basic definition for delta w m 0 2 square is equal to delta w by gamma minus 1 into 2 cos beta 2 by 1 plus cos beta 2. Now, where does this equation come from again I will urge you to try to derive this equation from the fundamentals delta w is h 0 1 minus h 0 3 which is m dot C p into T 0 1 minus T 0 3 and there the temperatures can actually be expressed in terms of the corresponding velocities. And therefore, we can express the Mach number in terms of the well part of the temperature gets expressed in terms of Mach number wherein these flow angles alpha 2 and beta 2 will also come into picture. So, from those fundamental equation you can actually derive the stagnation Mach number which is m 0 2 Mach number at the inlet of the rotor m 0 2 square is delta w by gamma minus 1 into 2 cos beta 2 by 1 plus cos beta 2. So, all the parameters on the right hand side are known to us we just substitute these different values and we get m 0 2 is 0.7389. The absolute Mach number which is basically based on the static conditions m 2 is related to m 0 2 as we know.  This again follows from the isentropic relations we have already seen the relation between stagnation temperature to static temperature, stagnation pressure to static pressure and so on. We can also relate the corresponding Mach numbers in this way. So, since we have calculated m 0 2 we substitute that here and then we get the static Mach number based on static conditions that the absolute Mach number as 0.775. So, this is again this can be calculated in multiple ways. The other way which I suggest you can try to calculate would be to take the ratio of the absolute velocity that is C 2 divided by square root of gamma R T 2. Stagnation temperature is known at the inlet and to find static temperature we need T naught minus C square by 2 C p. So, you basically need to calculate C square at the inlet and to calculate C square we will of course need the blade speed because you need to calculate because blade angle is known and so if you know the blade speed at the tip that is U 2 you can solve C 2 and then therefore calculate Mach number. In this case of course we are doing it the other way around we are calculating Mach number first and then now we are going to calculate the blade speed at the tip but it can also be done the other way around. So, let us now calculate the blade speed at the exit of the nozzle or that is the entry rotor entry tip of the rotor delta W by C p T 0 1 as we know is equal to gamma minus 1 by 2 gamma minus 1 into cos beta 2 U 2 square by A 0 1 square. So, here A 0 1 is speed of sound based on stagnation temperature square root of gamma R T 0 1. This can be calculated as 633.8 meters per second assuming T 0 1 is equal to T 0 2. Since in this case we know the power output and stagnation temperature and beta 2 we can simply substitute for these values and then we can calculate the blade speed at the tip U 2 and that comes out be 538.1 meters per second. So, the other approach to calculate Mach number would be to actually calculate the blade speed first and then since blade speed is known and alpha 2 is known you can calculate C 2 and from C 2 you can calculate the static temperature T 2 T 0 2 minus C 2 square by 2 C p and therefore, Mach number would be C 2 by square root of gamma R T 2. So, that is another way of calculating the Mach number of we have calculated that in a slightly different way. So, that completes the third problem that we had set aside for today's tutorial. So, I have one more problem to solve which is a very simple problem where basically not involving any calculation, but this is just to compare performance or operation of two different types of turbines that we have discussed about. One is the axial turbine which we had rather detailed discussion several lectures earlier on and of course, the radial turbine. So, let us compare under certain given operating conditions how the work output of these two different turbines can be calculated and how do they compare with each other. So, the fourth problem statement is the following. Compare the specific power output of axial and radial turbines in the following cases. Axial turbine in this case is given as alpha 2 is equal to beta 3 is 60 degrees and alpha 3 is equal to beta 2 is 0 degrees. For the radial turbine alpha 2 is given as 60 degrees and beta 3 and alpha 3 beta 2 all of them are 0. If the rotational speed is the same in both these cases, we are required to calculate the specific power output. So, what we can see is that immediately for the axial turbine we have we can see that alpha 2 is equal to beta 3 and alpha 3 is equal to beta 2 immediately tells us that this is a 50 percent reaction turbine which means that the velocity triangles would be symmetrical. And for the radial turbine we have already seen the velocity triangle which is for the nominal operation condition where alpha 2 is given beta 2 is 0 alpha 3 and beta 3 are respectively 0 at the exit. Let us take a look at the velocity triangles first and so here we have the velocity triangle for both these cases the axial as well as the radial turbines. Let us look at the axial turbine first velocity triangle at the inlet is given by this alpha 2 is given as 60 degrees and so alpha 2 at 60 degrees and that is equal to beta 3 that is also 60 degrees and beta 2 is 0 as you can see and alpha 2 is 0 that is C 3 makes 0 degrees with the axial direction. And since it is an axial turbine we will assume u 2 is equal to u 3 is equal to u blade speed it is for the same circumferential radial location. So, u 3 and u 2 are the same that is equal to u and these are the corresponding velocities C 2 at the inlet the absolute velocity v 2 is the relative and at the rotor exit we have C 3 and correspondingly v 3 which is the relative velocity. For the radial turbine we have been given that alpha 2 is again 60 degrees and beta 3 is also 60 degrees and this is a typical velocity triangle for an inward flow radial turbine. Since alpha 2 is 60 degrees we have C 2 which is at 60 degrees to the radial location here v 2 is the relative velocity which is entering in the radial direction u 2 and u 3 are not the same they are different in this case. And at the exit of the rotor we have beta 3 which is again 60 degrees and the axial velocity which is C 3 and C A 3 where they are the same v 3 is the relative velocity at the exit of the rotor. So, for the axial turbine we have alpha 2 and beta 3 as 60 degrees alpha 3 and beta 2 as 0 degrees specific work for this case in this 50 percent reaction case would be u times delta C w u times C w 2 and C w 3 here as you can see C w 3 would be equal to 0 for the axial turbine because the flow is leaving in the axial direction. So, C w 3 is 0 and C w 2 is equal to u 2 C w 2 is the tangential component of C 2 which is equal to u 2 and therefore, the specific work done for the axial turbine would be simply u square. For the radial turbine alpha 2 and beta 3 are 60 degrees beta 2 and alpha 3 are equal to 0 and specific work done is u 2 C w 2 minus u 3 C w 3. In this case of course, C w 3 is again 0 like in the case of axial turbine and C w 2 is the inlet is equal to u 2 and of course, at the exit u 2 and u 3 are not the same, but since C w 3 is 0 the radial turbine also develops a work which is equal to u square in this case it is u 2 square. So, specific work done in this particular case one of them is for an axial turbine and the other is for radial turbine for given these conditions for the same rotational speed both of these turbines generate the same work output. They are both functions of square of the blade speed. So, in this specific case that our example that we have looked at because of the conditions that has been specified to us for the same blade angles and rotational speeds both these turbines generate the same work output. So, this just to give you an idea of how the work done can be calculated for different types of turbine configurations of course, we have seen axial turbines in greater detail including a tutorial session earlier on and this is this was just to make a comparison between the work output of these two different types of turbine configurations. So, that completes the fourth problem as well that we have solved today and what I have for you are two exercise problems which I would leave it for you to solve and of course, I had also left a few exercises in between in couple of problems where I had requested that you should try to solve it in a different way. The method I had solved is one of the ways of solving the problem. You can also attempt to solve the problem in a different way for which I had given some hints. So, I suggest that you would also solve those problems using the other alternative approach that I had suggested. So, let us take a look at the first exercise problem that I have a small inward radial flow gas turbine comprising a ring of nozzle blades radial vane rotor and an axial diffuser operates at nominal design point with the total to total efficiency of 0.9. At the turbine entry the stagnation pressure and temperature of the gas is 400 kilo Pascal and 1140 Kelvin respectively. The flow leaving the turbine is diffuse to a pressure of 100 kilo Pascal and has negligible final velocity. Given that the flow is just choked at the nozzle exit determine the impeller peripheral speed and the flow outlet angle from the nozzles. So, this question has an additional component that is the diffuser and it is given that at the exit of the diffuser the flow has negligible velocity. So, you can assume that the flow exits the diffuser with almost 0 velocity and the flow at the nozzle exit is just choked. Which means that Mach number at nozzle exit is 1 velocity there would be equal to square root of gamma R T and stagnation temperature is given to us and. So, that should help you in finding out the parameters at the rotor entry and since rotor exit conditions are sort of fixed you can also calculate the conditions at the rotor exit using the fact that the diffusion pressure is given and the fact that the velocity at the exit is close to 0. So, in this case the tip speed comes out to be 586 meters per second and the angle alpha 2 is 73.75 degrees. Now, the second question exercise problem is if the mass flow rate of the gas through the turbine given in this problem the previous problem is 3.1 kgs per second the ratio of rotor axial width to rotor tip radius is 0.1 and the nozzle isentropic velocity ratio is 0.96. Assuming that the space between the nozzle exit and rotor entry is negligible and ignoring the effects of blade blockage determine the static pressure and static temperature at the nozzle exit the rotor tip diameter and rotational speed the power transmitted assuming a mechanical efficiency of 93.5 percent. So, we need to use part of data which is given in the first question to be able to solve this question as well. So, answer to part 1 is the pressure static pressure is 205.8 kilo Pascal temperature is 977 Kelvin the rotor tip diameter is 125.4 millimeters and rotational speed is 89200 rpm power transmitted with this mechanical efficiency would come out to be 1 megawatts. So, these are two exercise problems that I have for you you can solve this based on what we have discussed in the last couple of lectures including today's. And I would also suggest that you solve couple of those problems which we solved in today's lecture that is problem number 2 and 3 in a different approach from what we have solved in the tutorial. So, I would suggest that you also solve those problems using a slightly different approach. So, we would have just one more lecture on radial turbines where we would discuss some aspects of performance and some preliminary design aspects related to radial flow turbines. So, these we will take up in the next class which would be lecture number 37.