So just a reminder that this kid here this kid right now basically grinding up against Yamato's but he's he's one of the most important characters in the entire series. A lot rides on him whilst he's riding on him. Hello and welcome to the groundline review your source for everything one piece and today we have a review of chapter 1006 which all in all was a pretty fun catch-up of all of the lower level stuff happening right now on Onigashima. Mostly centered around the performance floor of the Skull Dome there was a very interesting pattern in that most of the characters featured in this chapter are either already there or converging very closely to that very location. So even though it may seem like a random scattering of catch-up scenes here and there there is a general thread here I think which is good because to be honest chapters like this they can be a bit difficult the ones that focus on pretty much as raw action as well as the secondary and even arguably tertiary characters to which I'm very specifically referring to Yogoro and we will get to him but there was plenty to grab on to and speaking of grabbing on we'll start with that Yamato segment eh. This is probably my least favorite group to be following on Onigashima right now I mean I do quite like Yamato and Shinobu and even Momonosuke he has he's grown on me but this feels like the group with no real mission I get that the idea is to protect Momonosuke because he's the chosen one and everything gonna bring about that new dawn and stuff hmm so just a reminder that this kid here this kid right now basically grinding up against Yamato's but he's he's one of the most important characters in the entire series a lot rides on him whilst he's riding on him but basically they're just running and that's all there really is to them right now there's no goal destination or like a person to reach it's just a giant game of a keep away really so I feel that very little if any progress gets made when we see their actions however I do think that this was definitely more of a setup for Sanji and to jump quickly to Sanji I am pretty much 100% convinced that Sanji has chosen to go downstairs in order to find Momonosuke it's just very hard for me to see a situation where Oda carefully crafts this whole interaction of Sanji discovering this information just for him to immediately abandon that idea and then continue upwards to Kinamon which is quite unfortunate for the whole Sanji potentially taking on Jack idea I did very much like that but it does bring up a whole lot of other potential though like I said before most of this chapter seems to be focusing in on that performance floor which is very vaguely near the area of the forbidden store room and happens to be where King Queen and now even Paris Barrow and while it's nice that we have Drake and Hyogoro they're holding the fort alongside Mr. Marko it could definitely probably use some extra Sanji infused power so this perpetual game of who Sanji is going to face will continue but once again I will say they think there's a lot more potential for cool Sanji moments heading in Yamato's direction than I think there is continuing in Jack's direction upstairs you know it would be really cool if Sanji made a decision he's kind of like me trying to write videos but getting distracted by very shiny objects speaking of getting distracted this video is actually sponsored by the Kingsguard project funky youtube channel designed to chronicle one man's quest to become a real life superhero and as part of that he recently made a video where he lit his feet on fire I repeat lit his feet on fire in order to mimic Sanji's diaboljamba technique and it looks like he did it in his living room as well which is a some bold choice sir but I do wish you the best of luck on your journey and you guys can turn fantasy into reality by checking out the Kings guard project through a link in the description below but it's getting to that point where I really do wish that Sanji just had you know one solid direction it's at the point where his focus seems to change every couple of chapters you know at first he was fighting on the performance floor then he was escorting Luffy to the roof then he went to save a woman and got trapped and started heading back up to the roof and now it looks like he's heading right back down to the performance floor which is exactly where he started and just the idea of mapping out Sanji's path during a raid is quite an exhausting thought something very interesting to think about however is that Sanji does not know Yamato that was one of those fun oh yeah moments because we've all known Yamato for what like a year now but within the story Yamato actually hasn't met many of our characters aside from Luffy, Shinobu, Momo and Frankie I think that's it I might be wrong I'm sometimes wrong wrong is me but Sanji is probably one of the more interesting figures to stumble upon due to a bit of a clash of ideals Sanji with his whole little chivalry business and Yamato currently identifying as a man a very specific man but still so it makes me wonder how that interaction would go down though but at the same time I think we'd almost be guaranteed to get that Momonosuke gag I talked about in the last review I just have this incredibly strong feeling I think I stated that either Sanji or Brooke would catch him within Yamato's clothing and get all jelly whilst Momo makes his momosleazy face I would definitely bet on that exact scenario occurring at this stage so if you're a Sanji fan who doesn't like him being in a clowned around or someone like me who was hoping that Momo gags would perhaps come to an end at least during the raid I wouldn't bet on it oh it's gonna happen isn't it Sanji's gonna see Momo he's gonna do the jealous eyes and he's just gonna lament that he is not the one within Yamato's clothes elsewhere on Onigashima we also got to catch up with Karat and Wanda versus Paris Barrow and well that's already over quite a close call though allegedly it would seem that Paris Barrow is quite beloved by fate having survived both Pedro's suicide explosion and a full on assault from two Sulong minks which is very notable because two is more than one but it makes me really interested to see what Otis plans for Paris Barrow are because as the Karat fight started I thought that this was going to be his whole jam like he was brought here very specifically in order for Karat to exact revenge and bring this whole Pedro thing full circle and to be fair I suppose that may still happen I'm sure that that moon will pop out again say hello eventually but Paris Barrow is just such a cool wild card in this raid he has no allegiance to the beast pirates nor the allied forces so he holds a lot of that lovely chaotic potential at any given moment Paris Barrow could be used against anyone or anything and I really like having that more random factor involved but I'm also quite curious to know how people feel about the Paris Barrow Karat fight being conducted off-screen I personally hold the stance that I usually have I don't really care in fact I think it might have definitely been for the better it's the smaller scale battles like this that would really severely throttle the series if we had to explore each and every one of them so I'm pretty satisfied with just seeing that cool opening exchange and then cutting to the outcome I guess it gives the anime something to play with in the future if they need some film material and I think we all know that they most certainly will just like you will most certainly need to subscribe to the ground line review which will result in consistent injections of one piece culture being uploaded straight into YouTube feed please make sure not to off-screen your pressing of this button also in this chapter absolutely brilliant Paris Barrow moment which was when he was lifting up Karat by the neck very very small thing here but I like the deliberate choice to have Paris Barrow lift Karat up with his prosthetic arm oh Paris Barrow is holding Karat with his his candy hand candy handy if you will although actually just having thought about that a bit further and maybe we probably shouldn't but I love it because it's just such a vengeful thing to do there's this whole feel of look what you did to me this is the arm that your precious Pedro stole from me and now it's being used to utterly humiliate you it was a great character moment and Paris Barrow is just such an awful yet surprisingly captivating dude bro with every subsequent chapter he appears in I feel that both my like and disgust levels for him go up completely in tandem especially because he's not entirely wrong in his thinking either Pedro and Karat were the ones who invaded their territory like did you really really expect to stroll on into the base of an emperor of the sea and just not face any consequences I mean maybe Karat actually did she's all young and innocent after all but what do you do if your Paris Barrow just let them do and steal and fight and destroy whatever they want but this brings us properly to the performance floor where we have two main points of interest the first of which is a pineapple named Marco and I think that he is something of a saviour of this chapter having gone and said that I'm all right with skipping Karat's action I did very much enjoy seeing Marco single-handedly holding off both King and Queen but having those single hands being simultaneously quite full mostly because I love seeing Marco's Phoenix powers on page they're very eye-catching the way in which Oda draws these specific flames and there was also some great choreography where King sliced through one of Marco's wings and just Marco conjuring that cool flame attack lots of generally enjoyable stuff that happens with upper fights like this but probably wouldn't have happened with something like Karat versus Paris Barrow also Queen what are you doing mate did you just shoot it has he always been able to do that just like shoot a beam out of his mouth like a Queen beam Queen is a stupidly interesting figure I mean there was that one panel in chapter 999 where he just had a gun pop out of his mouth which was weird and fun but this is some full on pacifista stuff now which is kind of mind boggling because it's one thing to modify your body but when you really think about it Queen needs to modify two bodies because how does he keep everything in check when Queen has to consistently switch back between his human and beast forms like what happens to these huge guns when Queen reverts back to his small but still kind of huge human form is that why he's so massive because he's just stuffed himself with weapons whatever the case I like it a dinosaur shooting a laser beam is pretty much everything that my child self was looking for in life so sure give me more of the Queen beam but as for our main focus of the chapter we do come to Hylgora so this was something that I noticed but quite frankly couldn't be bothered to talk about in the last chapter review but Hylgora did indeed resume his more chonk form in chapter 1005 and I assumed it was some sort of invocation of Sameik Ikan which is life return or the same sort of thing that Algi did on Dress Rosa I think that was called Geo Ken or Geophist in English but it was a martial art style that amongst other things allowed him to somehow literally conserve and store muscle power from his youth in order to call upon it in his old age quite a cool idea actually and I do wish we got and see a bit more of it the Hylgora situation is quite different though it's nothing that he seems to have really done on purpose more the results are being infected by Queen's ice demon shenaniganry but Hylgora serves as the key thread of this chapter which is potentially problematic to me because it requires me to care about him in any way for that to work and look harsh personal truth here I just don't feel strongly in any way about Hylgora he's a kind of cool old man guy but I know very little about him and I have even less of a connection to the Ark as a dude bros that he is admittedly beautifully asking to end his life for him what I will say is it would be interesting if he died though having earlier stated that you really shouldn't expect a waltz into the territory of an emperor and come out unscathed well Hylgora would serve as a good Pedro towards that idea and when you think about it Wano is actually going very very well at the moment despite the fact that we are dealing with double emperors surely someone's got to permanently go down at some stage and an actual casualty would probably help that feeling of desperation to emerge but I just don't know if that's what's going to happen here because it's one piece and we could easily open up on chapter 1007 with Chopper finishing the antidote saving his life and making this more or less you know false drama I wish I cared more about Hylgora but I just don't and look it is a sweet moment there's definitely a somber feeling attached to seeing this old man look back on his life as well as appreciate the future whether he's in it or not I just don't think that we or I guess I should more accurately say me know enough about Hylgora for this to really hit hard let me know if it does hit with you guys though I'm very curious to hear what you all think of it but some pretty cool stuff happened with chapter 1006 outside of the chapter as well firstly one piece was on the cover weekly shonen jump and it's a pretty damn cool cover a truly superb shot of the monster tree of the worst generation very much echoing the whole when you're at sea you fight against pirate scene and this is a cover that I really want to pick up actually I've had to force myself to stop collecting jumps years and years ago actually because they're pretty massive and they just take up far too much space but I may very well need to make an exception because that's not all we also have one of the most interesting color spreads I think this series may have ever seen this is a full-on image of the top players of the beast pirates which is pretty wild because I think this may be the first time we've had a color spread in the series that features no straw hats whatsoever I might be wrong in that but I can't think of any other off the top of my head even the promotional one film said which featured every marine in the series to that point it still had the straw hats in it I do really like it though especially getting see the toby ropa because it's still quite a novelty to have them in color I always forget that who's who has this crimson red scheme going on and he definitely has my favorite design of the six ah so who's who is literally satan and just on a minor artistic note I've pointed this out using a similar spread on this channel before but this is a very basic but effective technique just observe the gaze of all of the characters kaito and king are acting as the central pillar of the spread looking right at you but the characters to the left and to the right are either looking or facing in their respective directions which is a major aspect of why this image is so well balanced and a pure pleasure to look at it's also a legitimately fantastic color spread and it makes me wish that we got more like these for the other antagonistic crews but if you are interested in some more antagonistic emperor action and please do check out this video which ranks each one of our glorious emperors plenty of kaito so I look forward to seeing you there


Hey everyone! Welcome back to Six Sister Stuff. Today we're making three desserts that are no-bake and dump-and-go. So my name is Kristen, my name's Elise, and we're so excited to show you these dump-and-go desserts. We grew up having a lot of family parties in Pollux, and we still do, and everyone's supposed to bring a dish. So the easier the better, and I love that these desserts are no-bakes. If you need to throw something together last minute, these are perfect for that no-cook time. Really simple, really easy. Yeah, and that's usually my problem. I'm like a last-minute person, and so dump-and-go is exactly what I need. So if you guys are ready, let's get cooking. So the first recipe we're making today is no-bake mini Oreo cheesecakes. Now this only takes a few ingredients. You literally throw it all together and then put it in the freezer for a bit. So should we get started? Yeah. Should we start with the filling? Yeah, let's start with the first. So we've got two eight-ounce blocks of cream cheese that have been softened. You'll want these to be softened because you're going to beat them, and it's going to be a lot easier if they're not cold and right out of the fridge. So just blend it until it starts getting fluffy. You'll keep mixing it in a minute, and then you're going to add two and a half cups of powdered sugar, and this whole recipe calls for one package of Oreos, but you're going to set 12 aside to go in the bottoms of the cheesecake. Okay. And then you're going to turn the rest into just Oreo cookie crumbs. So we took the cream out and just did it into the crumbs. Perfect. And then we reserve some of the crumbs to put on top. Yep. So it really is just a few ingredients, but you kind of separate a little bit as you go. Perfect. So that's it for the cream cheese filling. It's all mixed together. So now I think we're ready to assemble the cheesecakes. Okay. So it's just a whole Oreo, right? Yep. Okay. Now the Oreo doesn't fit exactly in the bottom, but that's okay because it's just going to freeze over. Yeah. It really won't make that big of a difference. Okay. So now you can use a spoon and do this, but we're going to just use cookie scoop because then they'll all be even, and it will make our lives a whole lot easier instead of like using your finger to get it out. And then it doesn't get on the edge, too. You don't have to clean it up. Yes. Okay. So after she puts the filling in, I'm just going to spread it around to try and fill up that cup. It's okay if it's not perfect, but it will go a little bit over the top. So just kind of press it down in the middle so it covers the whole Oreo in the base. So my husband loves this recipe and he thinks that it's like a lot of hard work to make this recipe when you release a whole cheesecake, right? It's like three ingredients. Yeah. That's all. That's my kind of cheesecake. Exactly. Because real cheesecake, I don't have the patience to be dealing with it. It's a lot. I know. It's a lot. And this is good for summertime, too, where you just stick it in the freezer instead of having to wait for it to cook for an hour, however long. Exactly. Cheesecake takes quick and easy, cold. Perfect. All right. While she's finishing up there, I'm going to take these lovely Oreo crumbs and just kind of sprinkle them on top of each one just to make them a little fancier.  Yeah. Plus, it covers up if you weren't very good at spreading like I was. You're great. A little garnish. A little garnish. There we go. It's for the garnish. So after you've got all the extra Oreo cookie crumbs on top, we're just going to put them in the freezer probably for about an hour until they set up. Okay. And they're solid. Good to go. Perfect. Oh, I'm excited. All right. So we're done with this recipe on to the next one. So the next recipe we're making is our no-bake-a-clair cake. This is one of our favorites. It also, I do have something to admit. So we've made this recipe before on our YouTube channel, but that was way back in the day when... It was a struggle. We'll just say that. So I'm going to put a link in the description. We were learning. We were learning. We were learning how to edit. We were learning how to even make food. So if you want a good laugh, I'll put it down in the description for you. But we're going to show you how to make it because it really is easy. And it's so good. And it's so good. Like people will think that you slave for hours. So, okay, let's start. So we have three and a half cups of milk that we already poured in just to make our lives a little easier. Next, we're going to add like, it's a 3.4 ounce of the French vanilla jello. So we're going to put it in. It's like the instant pudding. You want to pour that one in for me? Okay. Okay. So we're going to mix these in first. There we go. Then we're just going to mix it up real quick with beaters.  All right. So once this is mixed, we're going to put it in the fridge for a few minutes. Let it set up because we're really making like the pudding out of it. So, you want to get in the fridge for me? Awesome. All right. So our pudding is all nice and stiff. And so we're ready to put in the cool up. Do you want to put that in and I will mix it with the beaters? Okay. So it's just a, what size is this? 8 ounce container of cool whip. Nice. And you can get the light. You can get any kind. It will still work and taste good. So, can't go wrong. Putting cool whip. I think cool whip might be one of my most favorite foods. All right. So once it's all done, it will be a little bit runny. I mean, it's still thick, but it will thicken up as you put it in the fridge or freezer depending on how thick you need it. So, all right. This, I know this is going to sound crazy, but this is part of the eclair cake is graham crackers and this and that's about it. So, I was surprised when I found out that it was a graham cracker. Right. Baked because something happens like the consistency or something changes. It changes. Yeah. And it really makes it, it just has a different flavor to it. So, am I okay to be overlapping these a little bit? Yep. Okay. So it, the recipe calls for a 16 ounce package of graham crackers. So, anywhere like, I would say like four. Yeah, probably four of those. So, we'll do about three layers of graham crackers. Okay. So, as she's finishing that up, we're going to put on about half of this pudding. And it's going to make the graham cracker soft and that's what you want it to do, because it's going to be so good when it's done. So, is this okay as far as a little overlap? Yeah, it's okay, but overlapped. Because I mean, you think of what's going on it, they're going to get softer and then they'll form into the pan. So, as long, I might have to have you hold it a little bit as I dump it in. So, it might. It's a two man job. Actually, we're going to do it so you can see. Okay, here we go. Thank you. Sure, that's half right, that's enough. Mom would make this in the summertime all the time when we were little. Yeah, smells like childhood. It does smell like childhood. These are most recipes we make on here, smells like childhood. Summer and childhood. Our mom was a good cook. She's, I mean, she still is a good cook. Right mom. All right, go ahead. Okay, take two. Take two. I'm going to start my full ones on this side this time. Good idea. And you really won't be able to tell where the breaks are on the recipe because it all. Yeah, because you use a spatula to serve it and it's easy to cut through. So, it all works out. It's nice that it has the pudding on it. I know, it's easier to get the graham crackers that second layer. Yes. And so then we'll just do one more layer of the graham crackers. That will be the top. Okay, so after all the graham crackers are on, you put on your favorite frosting. Now this is my mom's homemade chocolate frosting and it's amazing. I'm not going to show you how I make it, but I'll put a link in the description for you so you can find it there. Or if you don't like to make frosting, if you want to make this recipe really easy, you can just grab a can of chocolate frosting and just carefully spread it on and it will taste delicious. But we love mom's frosting. We do. Mm-hmm. This frosting also is on her chocolate brownie recipe. I'll put a link below there. Yeah, those are good. It's her marshmallow brownies. She's famous for that. She is. She is. Probably my favorite food. Okay. Are we ready? You're just going to spread it around. Now there are some parts where there's no graham cracker. You're just going to put the frosting right over it, like there is graham cracker and it will be just fine. Just going to do a spin a room. So if you are making this, how far in advance do you need to make this? Because it does need to set up a little bit. It does. I love to do it like the night before or you could do it. It needs about an hour, maybe two hours in the refrigerator or freezer depending on how cold you want it. Like some people like it frozen. But I like mine nice and soft so I stick it in the fridge. Okay. All right. I'm not fancy. But that's okay. We're moms. So we don't have to be. Kids are just going to tear through this. Exactly. They're just going to eat it. All right. So let's stick this in the fridge for about an hour or so. Our last recipe is our no-bake peanut butter bars. Probably one of the most popular recipes on our blog. Yes. Now when I was in medical school, well, when my husband was in medical school, we were so poor. You were in medical school. Pretty much. I was, I was supporting. But we were so poor. But this is one recipe that I made all the time because it was quick. It was easy and it wasn't very expensive. And you probably had all these ingredients on. Exactly. Exactly. Perfect for that. Okay. Should we get started? Yeah. What should we do first? So let's, should we make the base first? Yes. Okay. So for the no-bake peanut butter base, we need a cup of melted butter. So it's like, what, two sticks? Yes. Two sticks of butter, just melted. And then a cup of creamy peanut butter. I'm just kind of eyeballing it. It's a forgiving recipe, so it's okay. I think they've learned now that we eyeball everything. Nobody's got time for that. Who measures perfectly. So, and then you've got two cups of graham cracker crumbs. Yep. And two cups of powdered sugar. Perfect. This really is like a dump and mix. Yeah. Dump and go, no-bake. So you're just going to mix it together till it's all combined. And it takes a minute, I remember, it takes a minute to mix because it's, it just feels hard. Yeah. It's kind of hard to mix. Yeah, it does get thick. But where it's no-bake, that's probably for the best. So it's not runny and falling apart. Exactly. Okay, once it's all mixed, we're just going to put this into the bottom of either a 9 by 13 inch pan, or if you want a thicker bar, you could do it in a 9 by 9. But we've just sprayed it with nonstick cooking spray. And it is pretty sticky. So if you tried to spread this or press it into the pan with your hands, it's going to be kind of a nightmare.  Yeah. It would make it a little easier. Okay. Just going to get it all, get all the goodness. Okay. Here we go. Like you said, if you want it thicker, sometimes I'd make it in a 9 by 9 pan. So the crust is, I guess, the base. The base is really thick. That's how I like my peanut butter bars. It's almost the consistency of like a Reese's Peanut Butter Cup. It's true. After it sets up, it's kind of a soft, yet firm peanut butter layer. Here you go. How's that? Good. Okay. I like it. I'll take that one. So for the topping, we've got cup and a half of milk chocolate chips, right? And we like milk. If you like dark, you can use dark too, but I'm a milk kind of a girl. And then we're just going to add a fourth cup of the creamy peanut butter. I can get another spoon. Into the chocolate chips and melt that until it's totally smooth. So probably a minute or two in the microwave. Yeah. Okay. So I'm just eyeballing about a fourth cup of the creamy peanut butter in with my milk chocolate chips. Nice. And we'll just put this in the microwave. And if you notice, she just like literally just put it on top and you can stick it in. It all melts in. It will, it will. It's about a minute at a time. So we'll do it for a minute then mix the new for another minute. All right. Our chocolate and peanut butter is smooth and microwaved and melted. So we're just going to spread that on top, right? Want me to hold the bowl for you? Oh, you got it. We'll see how this goes. Hope you guys can see. And it should be just enough to cover the top of either size pan. So if you are making this in the nine by nine, it will be a thicker chocolate topping than in a nine by 13. But yeah, you can always add more to like add two cups of chocolate chips and a little more peanut butter if you want more topping. Yeah. But we feel like this, this is a good ratio. Yeah. Chocolate and peanut butter. So when it's all done, you're just going to stick it in the fridge for about an hour, maybe two hours. You can even put it in the freezer to make it really hard. Now you want to let it set up before you cut into it, or it'll be kind of all over the place. All right. If you guys want more delicious desserts, make sure you check out this video right here. And we will see you next week. Bye.


So, Hannah, perhaps you might give us your opening of Oxma, please. Thank you. Thank you very much indeed for this introduction. I'm really happy and honored to be here again. Just building on something you said, Nink, I fully agree. Small states can contribute. They do contribute. I wonder whether they also sometimes feel they have to contribute. I'm speaking about Finland, and I distinctly remember when Finland joined the EU for 25 years ago. There were doubts about Finland being a good member of the Union when it comes to foreign and security policy. And I think Finland did think that it has to contribute much more, it has to be really active to dispel the doubts, to show it is a good member. So, it compensated for its small size by being very active, and this has to some extent continued. I think Finland, if we can talk in terms of a national perception of a country, we do, in any case, Finland wants to be seen as a net contributor to security and defence in Europe, which perhaps sounds a bit much. But it sort of sees that it is a net contributor in the sense that it takes care of its own defence, manages to some extent at least, and it also takes part actively in all co-operative endeavours that there are for increasing security and defence in Europe. We'll be looking at many, many different forms of such endeavours. So, both taking care of its own defence and taking part in co-operation. I have to say these two sides have not always been seen as equally valuable. A certain discourse has been developed. So, as to say that they actually do not contradict each other, but they are mutually supportive. So, the more you participate, the better your own military capabilities will be that you use for the defence of your own country. So, by this kind of discourse formation, if we can say so, people have been educated to look at these two realms as belonging together and reinforcing each other. So, it's not necessarily for altruistic reasons that you take part because it is for your own benefit as well. Now, we also already heard that we are now in a very different situation when we think about defence, particularly because we are in a different situation when it comes to the security threats we are dealing with, cyber threats and hybrid forms of threats in particular, which means, speaking from the Finnish context again, more volatility in the security system, more unpredictability and need for higher readiness, but also need for more thinking about resilience, resilience of the society as a whole. So, threats change, yes, and the system changes as well. I already mentioned the great powers becoming more and more central again, some kind of a new great power politics time that we see. So, all these must have some kind of consequence for how cooperation is developing and how cooperation is understood. And for Finland, the consequences of these big changes are twofold, I would say. Cooperation is even more needed, but there are also many more options, seemingly at least, for cooperating with others. One might add also, when it comes to the specificity of small states, is that in some of these new threats, perhaps the smaller states are slightly more equal with the bigger ones than in others, because they are not encountered by only military means, and so you might have capacities where you actually, whether the size doesn't matter that much, perhaps we can discuss these later. Anyway, Finnish security and defence thinking changes quite a lot. You see a change in the past three years, I would say, into ways in particular. Cooperation becomes even more central for Finnish security and defence. It is part of national defence capabilities, and it means both providing and receiving military assistance, which is sort of new for a military non-aligned country, to speak very clearly and straightforwardly about providing and receiving military assistance. Finland has been rather slowly warming up to this and expressing this so clearly. It was only in 2017 that the legislation was changed so as to allow for proper providing military assistance. In the first place, so quite a long time after the Lisbon Treaty that actually is the basis for such action. So deeper cooperation, if you want, but also much-roaming cooperative initiatives when it comes to the sector of security and defence. What I just saw quickly before coming here was that the Ministry of Defence had just signed an agreement on defence industrial matters with its counterparts in India. And there are so many of these, mostly bilateral agreements, now that there is a long list of them. So cooperation in the field of defence with the UK, with the US, of course with Sweden to a very specific extent, and this we can talk later on about. But also with other countries. Sometimes trilateral forms, Sweden, Finland, the US, is a trilateral example. And then these mini-lateral formats that we all know led by France, Germany, and the UK respectively. Where in all of them you find Finland taking part. You might ask why these initiatives, Mushroom, whose interests they are actually reflecting and who takes the initiative for the other bilateral agreements in the first place. This is I think quite interesting. Finland, perhaps because of its small size, certainly does not say no to these possibilities of new kind of bilateral cooperation. Is this a way of sort of, are these the means to reach the goals of Finnish security and defence policy? If we look at the government's programme, it says that the goal of Finnish security and defence policy is to safeguard the national room for maneuver and keep all options open, which is rather interesting. And here we see that there are a number of options indeed. But it could also be that this kind of new mushrooming of initiatives might be a bit difficult to maneuver. And it could be a bit too rich in different options. Again, particularly for a small state to maneuver. Our research says that small states used to be where the proponents of multilateralism, exactly for that, that they didn't have to maneuver with large countries on a one-to-one basis. But they had these multilateral frameworks where they were safe, predictable, where they could control big powers. Now what is changing? Multilateralism is somehow going down because of the lessening interest by the great powers towards it, perhaps for other reasons. This is a big change for a country like Finland. Are these real alternatives to multilateralism? It might also be that the traditional multilateral settings are becoming more difficult to handle than what they used to be. And I'm thinking about the EU. We just had the Finnish EU presidency. And I think Finland was trying to make a contribution there. I saw ideas about developing European defence in new kinds of ideas, perhaps in this thought leadership way even. The defence ministers I hear were discussing climate change for the first time. Digitalisation and artificial intelligence were of course discussed hybrid threats, of course. But I don't know to what extent this actually was carried on. And when it came to the very concrete points on multi-annual financial framework, the EU's budget, there the end result was not that promising. If you think about a Finnish contribution and you think about a contribution of a country that always has been a very strong proponent of European security and defence policy. The compromise that Finland had to offer on defence on budget regarding defence was actually only about the half of the amounts that the Commission was proposing, both to the European Defence Fund and to the Peace Fund. And this was a big disappointment for many. Of course it was explained as a necessary compromise because you couldn't just achieve more. And one is also wondering whether that is really the case, whether it is so difficult to manage the system. So in the end, it looks like the traditionally so important multilateral setting and particularly the EU is now much more difficult to lead. It's hard to be on the driver's seat. But also it is a bit difficult for a country like Finland at least to choose between the seats. There are so many seats available and you want to be everywhere at the same time. So this might be a problem. Thank you very much. So it's a pleasure to welcome you back and to ask you to make some remarks to us now. Thank you very much for that very kind introduction and it's interesting to see that we go further back in history than we actually think. And thank you very much to the institute here for inviting me to contribute to the seminar and the timely topic on the role of smaller states in defence and security. The focus here is from a Stockholm perspective and the horizon and comments are made here in a personal capacity. Again, it is the first to describe very briefly the past few decades where we are today and what might lie in the near future. So the tremendous changes in European security landscape in the early 1990s led to fundamental changes in Swedish security policy, both in its stated doctrine and in practice. So from an official Cold War declaratory policy of non-alignment in peacetime with the aim to neutrality in wartime, the policy gravity changed to military non-alignment of today. And the declaration of solidarity given by the Swedish government in 2009 where we ought to give and receive military aid to European Union member states and Nordic states is a case in point here and it's very similar to what Finland has done as well. In parallel, this opened up for substantial contributions designed to stabilise Europe towards a new equilibrium pose in the Cold War. EU membership in 1994 was made possible as was cooperation with NATO through partnership of peace the same year. The 1990s was in many ways a slow farewell to the declared neutrality policy of the Cold War while perceptions of Swedish neutrality lingered on for many years. Two seminal developments during the 1990s influenced trends in defence and security policy. Firstly, the Balkan Wars during the first half of the decade where Sweden took part in a great power coalition under war-like conditions on the European continent. The Balkan Wars was indeed a steep learning curve and on the tactical operational as well as on the strategic level. Starting with the UN framework of unprofor cooperation with NATO followed with participation in I-4 and K-4 peacekeeping developed over peace support operations to crisis management operations in open conflicts. The second substantial focus was support for the sovereignty of the three Baltic states who had regained their independence in the early 1990s. With the collapse of the Soviet Empire the border of Western Europe moved 400 kilometres to the east and helped stabilise the Baltic Sea region. It led to a trajectory where civil rights normal economic development and the rule of law was gradually established. Baltic EU and NATO membership followed on this development. The first Gulf War in 1991 also appeared to show that a new world order with the UN at the centre had emerged. The liberal world order was further supported by increased trade. Peace would follow from globalisation. As a consequence of these optimistic trends the Swedish defence spending could be cut similar to many other European nations. Gradually the spending went from about 2.5% of GDP in the late 1980s down to just about 1% of GDP today. The Cold War tasks of defence of the realm in an existential and apocalyptic World War III type scenario shifted to a near exclusive focus on peace support operations and the stabilisation of conflict probe regions far away from home. The substantial civil defence organisation was entirely disbanded. The size and military capability shrunk substantially between 80 and 90% of all the wartime units in the wartime organisation were disbanded along with conscription in favour of small full-time armed forces. All this seemed to work well during the 1990s and well into the first decade of the new millennium. But gradually the security climbed the worsen and it was realised that Russia would no longer be a part of the updated European security order or fall of the agreements it had entered into in the 1990s. Russia was returning to old patterns. The disdain for American dominance, the aim for a Cordon sanitaire and a version of 19th century spheres of interest were smaller neighbours would exist at the mercy of Moscow. The attack on Georgia in 2008 followed. This was noticed in the west but most pressed snooze on the alarm clock and failed to draw the right conclusions of what had happened and went back to sleep. While Sweden prolonged its process towards a new defence build as a result changes were mostly rhetorical. Georgia was in many ways a quarrel in a far away country between people of whom we know nothing to quote Neville Chamberlain. These trends continued and in 2014 a rude awakening followed with the Russian occupation and subsequently legal annexation of Crimea. The war in eastern Ukraine that followed led and supported by Russia continues to this day. For Sweden the war in Ukraine led to new stones in defence and security. The shift in emphasis towards territorial defence was clear international crisis management operations were no longer the primary focus in the defence build of 2015. Unfortunately the defence build was underfunded and difficult political negotiations led to supplementary budgets to help fill the gaps in defence that frequently appeared. As a result the scaling up of defence capabilities could not be fully met within the five year period. Valuable time was lost. The current defence planning process will lead to a new defence build this year covering 2021 to 2026. All indications are that spending will increase substantially and a number of organisations or reforms will be included. In parallel with the process towards a stronger defence an effort that deeper and more cohesive international defence cooperation began. The realisation that the combination of military technological developments meaning longer range, higher precision, network systems and shorter advance warning times plus the returning threat from Russia meant that the Nordic Baltic region was now one strategic space. Engagement in the region from the Obama administration also contributed substantially to this trend. Today Finnish-Swedish defence and security cooperation complemented with a trilateral US Finnish-Swedish detail has developed substantially and is even talk of a Swedish-Finnish formal alliance. In addition a Norwegian-Swedish Finnish cooperation is also on the way. Sweden, the UK and the UK has a programme of defence cooperation, closer defence cooperation since 2016. There's also a closer defence dialogue with Denmark, Germany and Poland. Cooperation with the Baltic states forms yet another part of the pattern. The Nordics have also breathed a new life into Nordefko, Nordic defence cooperation organisation where defence cooperation and material acquisitions are discussed. On top of all this the northern group of nations discusses hard security issues and as well the joint expeditionary force is there as another forum with a membership that goes beyond the Nordic region. Relations to NATO are further enhanced and Sweden entered into the enhanced opportunities partnership a few years ago. Full membership in NATO is currently not on the agenda. So taken together these cooperative efforts can be described as a grapple hook defence and security policy aimed at anchoring Sweden firmly in preparation for possibly worse times to come. So what are future trends? The current strategic situation seems to worsen. The 1945 order, modified in the early 1990s looks distinctly shaky. Geopolitical tensions due to the large scale shifts and relative power on the grand strategic level leads to a lower degree of predictability in the international system. This has clear impacts on the Nordic both the region and elsewhere. As a result the international long-term order and the multilateral system is on the threat. While multilateral arrangements help promote and strengthen rules-based world order with long-term norm-building as a central component the hopes for further developing these in the current international climate seems to be dimming. Focus will probably shift towards ensuring the survival of these institutional arrangements over the longer term. So what conclusions can you draw from this? Smaller nations need to adapt to this lower degree of predictability in the international system. How do we ensure that we sufficiently quickly and clearly observe and react to such events? Firstly, sudden events will have to be managed in direct cooperation with others and not necessarily exclusively through well-established multilateral organizations. The recent tragic shooting down of a Ukraine international airlines flight PS 752 near Tehran with the death of 176 civilians led to an almost immediate formation of an international coalition to manage the aftermath. Canada, Sweden, Ukraine, the United Kingdom, the Netherlands and Afghanistan all pulled together to pressure the Iranian regime for a speedy, transparent and fully cooperative inquiry and compensation. This is one example of what could become a pattern for a less predictable situation issue-focused and temporary arrangements to manage security challenges. Secondly, in defense acquisition I think we are getting closer to a point where the usual routines for the sometimes very long development periods will have to be shortened and simplified. A stronger defense capability and increased number of units is needed urgently, not in 10 to 15 years' time. Thirdly, the grand strategic system which many assume will remain a constant factor cannot be taken for granted. It may suffer further shocks which could have serious follow-on effects for Europe. For example, what if the transatlantic link central to Euroclantic security would be seriously and suddenly weakened? This is just one of several possible events that would negatively impact on the security climate in our part of the world. It is our task as security analysts to ponder such possibilities and what would then be our plan B or plan C or plan D. Lastly, the main conclusion I draw from the Swedish experience from the past three decades is that in international security a too optimistic view of long-term trends can lead to increased risks. A priority must be to avoid becoming a security deficit. The smaller the nation is, the higher the risks of mortgaging the taunt is. Thank you. Our thanks also go to Nick for his press and to all of our panelists who I think really have framed the discussion in a very helpful way, a very practical way with a number of very rich ideas there. I'll just mention a few of them, very briefly pick up a few of them just before we open the floor to Q&A. Beginning with Nick, a couple of points. You referred to Ireland's tradition of international engagement. I think you were focusing very much on the peacekeeping and crisis management aspect which I think is very important. I'd add to that other niche areas such as experience of conflict, conflict prevention and resolution. We're doing a lot of work at the moment in relation to the peace process in Columbia and in other places as well. Drawing on and sharing to the extent that this may be relevant and welcome by the parties concerned on our own experience. Also the developed cooperation program like the root or the source of many of the threats that we face in the EU lies in instability in our neighborhoods. And where you see instability you generally find authoritarian government under oppression, denial of human rights, poverty also threats from climate, poor governance and corruption as well. The programs that Ireland and other smaller states are very much geared at building capacity, trying to promote good governance, respect for the rule of law and these make an important each contribution. You referred to the high tech expertise which is certainly important. I think we just need to note and I think we did in the earlier session this also brings responsibilities and vulnerabilities as well from our point of view. The resilience that we can develop and show in relation to our cyber infrastructure. That goes to a point that Hannah made about the importance of ensuring the resilience of your society in the face of threats. Also a very important point which I think Hannah made was taking care of your own defense looking after your own patch. This will obviously vary in its implications for member states according to their geography but still need to ensure that you look after your own area from a security point of view. Hannah made the point that small states can sometimes have more room for maneuver be more acceptable and comfortable interlocutors for third countries than perhaps larger member states that may have more baggage. Then a series of points from Nicholas I thought was interesting the way you traced the evolution from a policy of military neutrality towards one of non alignment declaration of solidarity in 2009. I think that's an interesting point of reference in terms of our own debate here. Also outlined a veritable mosaic of security and defense cooperative arrangements that Sweden is engaged in. This I think does raise a wider point which is that I think there's a growing consensus or at least awareness that Europe needs to do more for its own security and defense amid the increasing threats it faces and the uncertainties around the commitment of the US administration. But there are questions around how this might best be organized and what the distribution of responsibility might be between the existing organizations whether to some extent we need to envisage new formations or structures. You mentioned ad hoc coalitions on different areas in defense of the international rules based order. I just note as well you talk about the need just note as well just a final point before we open it up in terms of where the focus of small state efforts might be the candidate's speech by commission president Ursula von der Leyen to the European Parliament emphasized a few things first of all that NATO remains the cornerstone of collective territorial defense in Europe and that the EU work on security and defense is embedded in comprehensive security that stabilization always comes with diplomacy reconciliation and reconstruction. So I think this is an interesting answer whether to what extent the EU can remain in this space to what extent there may be vacuums elsewhere which one may have to look at member states may look at filling in different ways whether it's European intervention initiatives or other ideas like the European Security Council.


Hello and welcome to lecture number 5 of this lecture series on turbo machinery aerodynamics. In the last few introductory lectures, we have been discussing about various aspects of turbo machines to begin with in the first lecture. And starting the second lecture onwards, we were discussing about axial flow compressors. And we have had some discussion, quite detailed discussion on the two dimensional aspects of axial compressors. And how is it that one can analyze an axial compressor and from fundamental thermodynamic principles. We have looked at the thermodynamics of the compression process. We have discussed about the simplified versions of axial compressors that is the cascade and how tests can be carried out on cascades and what we can make use of or how we can make use of cascade data in terms of analysis. So, these were some of the topics that we had discussed in the last 3 or 4 lectures. And so, it is about time that we now try and utilize what we have learned in the last several lectures in trying to solve some problems. So, in that in mind I have configured a tutorial for you today, wherein we are going to discuss about different problems. We will have a few example problems which I will try to solve it for you. Towards the end of the lecture, I will also have a few exercise problems for you to solve on your own, which you can solve based on our discussion during the last few lectures as well as our discussion in today's lecture. So, today we are going to basically have a tutorial session and we will be trying to solve a few problems which are related to axial flow compressors. So, today's tutorial is going to be on axial flow compressors and which are also true for cascades. But we will be discussing only the 2 dimensional analysis of axial compressors. We will not really be taking up the 3D analysis in today's tutorial. This will be taken up in the subsequent tutorial after our discussion on 3D flows in axial compressors. So, subsequent to this tutorial, we will have a few lectures on 3 dimensional flows and their analysis. And then we will also have a tutorial from 3D flow analysis. So, let us take up the first problem which we have and let us see how we can go about solving this problem. So, the first problem statement is the following. So, here we have an axial compressor where in air at 1 bar and 288 Kelvin enters an axial flow compressor with an axial velocity of 150 meters per second. There are no inlet guide means. The rotor stage has a tip diameter of 60 centimeter and a hub diameter of 50 centimeter and rotates at 100 revolutions per second. The air enters the rotor and leaves the stator in the axial direction with no change in velocity or radius. The air is turned through 30.2 degrees as it passes through the rotor. Assume a stage pressure ratio of 1.2 and an overall pressure ratio of 6. Find the mass flow rate of air, part b is to find the power required to drive the compressor, part c is to find the degree of reaction at the mean diameter and part d is to find the number of compressor stages required if the isentropic efficiency is 0.85. So, in this question that you have been given now, if you look at the question or the problem statement carefully, there are several parameters or data which has already been provided to you. For example, the ambient conditions are given, the pressure ratio is given and the efficiency is given and the some information about the angle or at least the delta change in angle is also given to you. So, with this information you are required to find a few other parameters like the mass flow rate, the power required, the number of stages and so on. So, the first step towards solving any problem in turbo machines is to start from the basics and where does the basics begin? It begins with the velocity triangle. If you recall when we had discussion on the velocity triangle, I had emphasized the fact that velocity triangles form a very important part of the analysis of axial compressor that is where the design of axial compressor begins with and therefore, it is extremely important that you understand the velocity triangles and how velocity triangles are constructed because that will be required for any axial flow compressor analysis problem and that is the beginning or the fundamental point of starting point of analysis of axial compressor. So, that is why it is very important for you to realize the significance of the velocity triangles. So, let us begin constructing the velocity triangle. So, for that how do we begin construction of the velocity triangle? Now, we will take up the rotor first and what is given to you is that there are no inlet guide vanes and the inflow is axial. Now, when you have an axial inflow and there are no inlet guide vanes, it means that the inlet velocity or the absolute component of inlet velocity will be equal to the axial component itself because there is no guide vanes and the inlet flow is entering axially which means that the absolute component of velocity that is c 1 will be equal to c a 1 which will be in the direction in the axial direction itself. And then we are given the blade speed, the rotational speed is given it is given as 100 revolutions per second which means that we can actually find out the blade speed u because rotational speed is known and if the diameter is also calculated you can find the mean blade speed. So, once you know c a or c 1 and u you can construct the inlet velocity triangle which means that v 1 can be calculated from this. Similarly, we move on to the exit and the exit also we have some information given in terms of the angle which with which the flow turns and since blade speed is a constant and axial velocity is a constant from there we can actually construct the rotor exit velocity triangle as well. So, if we do this kind of a construction this is the kind of velocity triangle you are going to get. Now, let me explain this velocity triangle once again we have already discussed this in the last few lectures but we will take it up once again. So, what are the constituents of this velocity triangle? So, this velocity triangle here begins with at the inlet of the rotor this is u the station which is basically station 1 is the inlet station or entry of the rotor exit of the rotor is denoted by this station and. So, we have the blade speed which is u and I have denoted a direction for u as this. So, you might wonder how is it that a blade speed has a direction this way and why is it not that way the reason why it is this way is because the blade moves in this direction it is a compressor blade and therefore, a compressor blade has to rotate in this direction it does work on the flow. So, this is the blade speed u C A which is also equal to C 1 is in the axial direction. So, this is the axial velocity which is equal to C 1 and the relative velocity therefore, is a resultant of these two we get relative velocity v 1 as this. Therefore, the angle which the relative velocity makes with the axial direction is given by beta 1 since u is known C A is can be calculated we can calculate beta 1 and therefore, v 1 can be known. Now, what about the exit of the rotor at the exit of the rotor v 2 will still leave the rotor tangentially because that is the property of relative velocity that it has to leave the blade enter the blade as well as leave the blade tangentially and therefore, if you draw a tangent to the camber at the trailing edge you get v 2 u is known you add v 2 plus u we get the absolute velocity C 2. So, this is how we construct the exit velocity triangle angle which v 2 makes with the axial direction is given by beta 2. So, this is the first step towards solving this kind of a problem that we construct the velocity triangle and that makes the problem solving a lot more easier because it is always possible to still solve a problem without having to you know construct a velocity triangle and still attempt to solve a problem it is still possible, but the chances of one making an error in calculation is very high. And the fact that construction of a velocity triangle simplifies the problem substantially is enough reason that one should always attempt to solve such a problem with the fundamentals that is a velocity triangle. So, once you get the velocity triangles right the chances of making error in subsequent calculations are reduced and therefore, I urge that when you attempt to solve such a problem please make sure that you have the velocity triangle constructed and then continue to solve the problem. So, having constructed the velocity triangle here now what we will do next is to take up the problem solving one by one therefore, different aspects which are to be found which are to be calculated in this problem will take them up one by one in the process we will also be calculating several other parameters. So, let us begin solving this part as well. So, what is given to us is that the mean and we are given the tip and hub diameters the tip diameter is given as 60 centimeter and the hub diameter is given as 50 centimeter rotational speed is given as 100 revolutions per second. So, based on this we can calculate the mean blade speed that is u mean which will be pi d n pi d mean by into n and here d mean would be d tip plus d hub divided by 2. So, this gives the mean diameter this multiplied by the speed which is in revolutions per second. Now, in some problems you may be given in revolutions per minute in which case you will have to convert it to revolutions per second by dividing it by 60. So, the mean diameter is 0.6 plus 0.5 by 2 which is 0.55 is multiplied by pi and the speed 100 will give us the blade speed as 172.6 0.76 meters per second. So, since c 1 or c a is also given to us we can easily find out the angle blade angle at inlet b to a 1 which is tan inverse u by c a and that will come out to be 49.2 degrees. Now, it is given in the question that the flow is turned by 30 degrees as it passes 30.2 degrees as it passes through the rotor which means that at the exit beta 2 would be equal to 49.2 minus 30.2 because the flow is turned by 30 degrees 30.2 degrees and therefore, at the exit the blade angle will be 49.2 minus 30.2 which is 19 degrees and then we again solve the velocity triangle at the outlet tan alpha 2. Let us go to the velocity triangle alpha 2 is this angle. So, tan alpha 2 would be u which is this component minus what is left here which is given by c a tan beta 2 and why is it so because the flow is turned by this angle and c a is this component c a tan beta 2 is basically this component this part of it. Therefore, u minus c a tan beta 2 is this and that basically is equal to this rest of it divided by c a will be tan alpha 2. So, beta 2 we have already calculated and therefore, we can calculate alpha 2 as 38.92 degrees. So, we have now solved both the velocity triangles here we have only beta 1 at the outlet we have beta 2 and alpha 2 and so we have all the angles required for the velocity triangle. So, once we calculate all the angles we have calculated the inlet angle or blade angle beta 1 exit blade angle beta 2 as well as the exit absolute angle that is alpha 2. Having solved all these angles it will now make it very simple for us to solve for the other components which are not known and therefore, calculate the parameters which we are required to find. The first parameter which we will try to solve is the mass flow rate. Now, mass flow rate as you know is equal to the product of density multiplied by the annulus area through which the mass flow rate is taking place multiplied by the axial velocity. So, rho into the annulus area into c a is basically the mass flow rate out of this c a is known annulus area is known because the diameters are given d t and d hub are given. So, square of the differences between the 2 divided by 4 multiplied by pi gives us the annulus area and this multiplied by density. Now, density is something which we will need to find because we have been given the ambient conditions we can find the absolute with the density with which the flow is passing through the compressor and therefore, calculate the mass flow rate. So, let us try to do that now. Now, mass flow rate as I mentioned is pi by 4 times d t square minus d h square multiplied by c a into density rho 2 or which even otherwise rho 1 is also sufficient. We could use either rho 2 or rho 1 the mass flow rate does not change with that. So, since we have been given the temperatures we can also find out the corresponding values at the exit of the compressor and therefore, t 1 which we will now calculate as t 0 1 minus c a square by 2 c p t 0 1 is given already as 288. So, from there we get 276.8 Kelvin t 0 2 is equal to t 0 1 multiplied by p 0 2 by p 0 1 raise to gamma minus 1 by gamma this is obviously, from the isentropic relations. Since the other parameters are known the pressure ratio is also known on the per stage pressure ratio is also given. We can calculate t 0 2 as 303.41 Kelvin therefore, t 2 is 303.41 minus c 2 square by 2 c p and. So, here we need c 2 and c 2 is basically the cos of c a divided by cos alpha 2 and c a is given as 150 meters per second alpha 2 we have already calculated as 38.92 and therefore, c 2 is 150 divided by cos 38.92 that is 192.79 meters per second. So, once we have calculated c 2 we can calculate t 2 and therefore, density and obviously, the mass flow rate. So, let us do that therefore, static temperature at the exit is t 2 which is p 0 303 minus 0.41 minus 192.79 square divided by 2 into c p that is 1005 that is 2010. So, this is 284.91 Kelvin p 2 is the pressure ratio multiplied by the inlet pressure we get 1.216 bar. Therefore, rho 2 is the pressure ratio which we have converted that to Pascal's 1.216 into 101325 divided by RT and R here is 287 into temperature 284.91. So, the density comes out to be 1.507 kilograms per meter cube. So, this if we substitute in the mass flow rate equation we can calculate mass flow rate we get that as 19.53. So, we have calculated mass flow of course, this is one of the ways of calculating mass flow you can calculate it in different ways. If you calculate the inlet density you can still calculate the mass flow rate using the inlet density as well. And the second component or second aspect that you need to calculate is the power required to drive the compressor. Now, how do you calculate the power required to calculate drive the compressor? Power required is basically the product of the blade speed multiplied by delta C w. If you remember when we had done the analysis of the velocity triangles I had mentioned that delta C w that is the tangential component of velocity is a significant parameter because that is something which will determine the power that is required to drive such a compressor. Therefore, u multiplied by delta C w is the power required to drive the compressor. Alternatively you can use u times m into C p into delta t or you can use the pressure ratio to calculate the power required. But, the simpler way is to use the velocity triangle which is u times delta C w. So, if you look at the velocity triangle let us go back to the velocity triangle now. So, what is the C w component here? C w at the inlet is 0 here there is no C w whereas, at the exit here we have a C w tangential component which is equal to this. And so from the velocity triangle if you look at the velocity triangle and try to simplify the delta C w is basically equal to C a times tan beta 1 minus tan beta 2. Therefore, mass flow rate multiplied by u into C w delta C w which is C a times tan beta 1 minus tan beta 2 is basically equal to the power required to drive the compressor. So, this is 172.76 which is u C a is 150 mass flow rate is 19.53 multiplied by tan into beta 1 49.2 minus tan beta 2 that is 19. So, this gives us the power required to drive the compressor and that comes out to be 412 kilowatts. The third component to find is the degree of reaction. Now, if you recall our discussion on degree of reaction that is one of the performance parameters which are used extensively during design of axial compressors. Degree of reaction refers to the amount of pressure ratio that is developed by a rotor as compared to the pressure ratio developed by a stage. So, that basically tells us what is the total fraction of the power pressure ratio that is developed by the rotor in relation to the entire stage itself. So, degree of reaction is something which is very significant in terms of the design of compressors because that comes up as a very significant parameter which will determine whether the design is feasible or it has a certain problem or not. So, degree of reaction if you recall our discussion on that we had derived an equation for degree of reaction from the fundamental principles which was basically relating the degree of reaction to two components. One is the axial velocity then the blade speed and the angles the blade angles at the inlet and exit. So, degree of reaction is something which you can determine by solving the velocity triangle. And therefore, since we already have all the angles associated with this particular rotor we can easily find out the velocity triangles and therefore, the degree of reaction. So, degree of reaction in this case is 1 minus C A by 2 u multiplied by tan beta 1 plus tan beta 2. So, if you substitute all these values which are already known to us we get 1 minus 150 divided by 2 into u which is 172.76 multiplied by tan beta 1 which is 49.2 plus tan 19. So, this is 1 minus 0.65 that is 0.35. So, here this degree of reaction of 0.35 means that the rotor is responsible for 35 percent of the pressure ratio of a stage that is the rest of the pressure ratio that is 65 percent of the pressure rise is actually contributed by the stator. So, rotor contributes only 35 percent in this particular case and of course, depending upon the design this fraction can be altered and that is why I said degree of reaction is a significant component or significant parameter which is used in design of axial compressors. So, we have now solved three different aspects of this particular problem mass flow rate power required and the degree of reaction. And the last thing that we have to find out is the number of stages that this compressor will require if it has to develop a pressure ratio and a certain isentropic efficiency. So, the way we are going to solve this problem this particular part is by first calculating the total temperature rise taking place over the entire compressor and then we will also calculate the total temperature rise in one stage. So, total temperature rise of the compressor divided by total temperature rise for one stage will give us the number of stages that are required for driving this for developing this kind of a pressure ratio. So, we will determine first the total temperature rise across the whole compressor then we will also determine temperature rise across one stage ratio of that will give us the number of stages. So, for one particular stage the temperature rise across a stage that is T 0 s here represents the stage. So, delta T 0 for the stage is basically given by u times C a divided by C p into tan beta 1 minus tan beta 2 and where does this come from? Well this basically is equal to enthalpy rise. So, delta T 0 multiplied by C p is basically delta H 0 which is enthalpy rise. So, enthalpy rise in one stage is equal to u times delta C w which is what is given by C a times tan beta 1 minus tan beta 2. So, from this we get since all these values are already known to us once we substitute them we get 172.76 C a is given as 150 C p is 1005 this multiplied by tan beta 1 this is 49.2 and 19 we get delta T 0 in one stage yes as 20.99 Kelvin that is this is the amount of pressure ratio temperature rise taking place in one stage. Now, we can also find out the temperature rise taking place across the whole compressor and for that we will make use of the definition of efficiency. Now, in this case the efficiency is given as 85 percent and from fundamental thermodynamics you might recall isentropic efficiency of a compressor is basically equal to the difference in enthalpy for the ideal compression process divided by difference in enthalpy for the actual compression process. And so that is also expressed in terms of temperature. So, you get T 0 2 prime minus T 0 1 divided by T 0 2 minus T 0 1 the numerator can be expressed in terms of pressure ratio and the denominator is what we need. So, what we get is the delta T naught overall as T 1 by efficiency basically this comes from efficiency is equal to delta T naught isentropic divided by delta T naught overall delta T naught isentropic is also equal to T 1 into pi c which is compressor pressure ratio raise to gamma minus 1 by gamma minus 1. So, overall temperature rise is T 1 by isentropic efficiency multiplied by the pressure ratio pi c raise to gamma minus 1 by gamma minus 1. So, if you substitute all these values we get delta T naught overall as 226.5 Kelvin. So, this is the total temperature rise taking place across the whole compressor if it is developing a pressure ratio of 6. Therefore, the number of stages for this compressor would be n which is 226.5 divided by 20.99 which is 10.79 which is approximately equal to 11. Therefore, this compressor requires 11 stages to develop a pressure ratio of 6 with an efficiency of 85 percent. So, this brings us to the end of the first exercise problem which we have solved today which was basically of an axial compressor mean diameter analysis we found out the mass flow rate the power required and the number of stages as well as the degree of reaction. Now, what we have learnt from this problem and for from any other problem which we will solve is the significance of the velocity triangle. So, I am I keep emphasizing this every now and then because of the fact that in order that one understands and develops the ability to analyze and design axial compressors it is essential that one understands the velocity triangles thoroughly because without understanding velocity triangle it is nearly impossible for us to for anyone to analyze a compressor from the fundamentals and also design compressors starting from the fundamental analysis. Let us move on to the next problem now. Now, next problem is again as I said on axial compressors and then we will take a look at what the problem statement is and how we can go about solving this problem. So, the next problem statement is the following an axial compressor is to be designed to generate a pressure ratio of 4 with an overall isentropic efficiency of 0.85 the inlet and outlet blade angles of the rotor are 10 45 degrees and 10 degrees respectively and the compressor stage has a degree of reaction of 50 percent. If the blade speed is 220 meters per second and the work done factor is 0.86 find the number of stages required is it likely that the compressor will suffer from shock losses the ambient air static temperature is 290 Kelvin and the air enters the compressor through guide mains. So, this is a statement of course this is slightly different from the previous problem we have solved one part of it is identical in the sense that we still need to find the number of stages required. But the second problem is quite different in the sense second part of the problem where we are required to find or determine if this compressor will suffer from any shock losses and you might all probably notice that there is another aspect which has been described here or given as a data that is work done factor. So, we have not really discussed about work done factor in our discussions in the previous lectures. Now, what do we mean by work done factor? So, work done factor is basically a parameter which is taking into account the aspect of blockage. Let me explain this in little more detail because you need to understand what work done factor is now in a multi stage axial compressor as we move from the inlet all the way through these stages to the exit. What happens is that there is a growth of boundary layer you might have seen boundary layer development in a pipe flow right that is from the inlet of the pipe all the way to certain distance the boundary layer keeps becoming thicker and thicker till a point it reaches what is known as a fully developed turbulent boundary layer and boundary layer extends all the way to the mid line or center line of the pipe. A very similar aspect happens here as well there is a growth of boundary layer the difference here is of course, that there is also an annulus. So, there is a boundary layer growth on the casing there is also a boundary layer growth on the annulus. So, from the inlet to the exit there is a successive increase in the boundary layer thickness and because of that the effective area that is available for this work generation of work done diminishes by a certain factor because of this so called blockage which is created by the boundary layer. So, work done factor is a parameter which is less than one in actual compressors which will get multiplied by the actual work that the compressor is developing that is that much fraction of work which the compressor should have done is diminished because of this blockage effect by the boundary layer. So, that is where the work done factor comes we will make use of that in calculating the number of stages. So, in this particular problem again as we have done in the previous cases we will begin with velocity triangle and we will see what this particular problem at hand is and how we can use the data that is given to us to solve the velocity triangle and solve the problem in this process. So, the velocity triangle for this case I have combined the inlet and exit velocity triangles. So, at the inlet we have this velocity triangle which has an absolute velocity of c 1 relative velocity of v 1 angles of alpha 1 and beta 1 at the exit of the rotor we have angles sorry velocities c 2 and v 2 with angles alpha 2 and beta 2 the blade speed is u axial velocity is given by c a. Now, in this question towards the end of the statement it is mentioned that the air enters the compressor through guide vanes. Now, the effect of this guide vanes is that there is an alpha 1 that is introduced that is c 1 will no longer be axial unlike the first problem which we solved c 1 was indeed axial because there was mention it was mention that there was no guide vanes. Now, guide vanes are provided to ensure that the flow enters the rotor in a certain direction and that is why the presence of guide vanes will cause a certain amount of alpha 1 and therefore, c 1 is not axial and there is an angle which c 1 makes with the axial direction that is alpha 1. So, this is basically because of the effect of the presence of guide vanes beta 1 is the angle which the relative velocity makes with the axial direction c 2 makes an angle of alpha 2 to the axial direction and beta 2 is the angle which the relative velocity at the exit makes with the axial direction. So, the first part of the problem let me go back to the statement you need to find the number of stages required for which you have been given the angles the efficiency pressure ratio blade speed and the work done factor. So, all the data that you need for calculating the number of stages is already been provided to you and so, that makes it quite easy for us to solve this problem for the first part of it. So, let us do that now. Now, given the blade speed we can calculate the axial velocity from this velocity triangle that is axial velocity divided by the axial velocity will be equal to this blade speed divided by tan of beta 1 plus tan of beta 2. Let me take a look at that once again what is tan beta 1 tan beta 1 is basically this component c a times tan beta 1 is this and c a times tan beta 2 is the remainder part of it. So, some of these 2 will give us the blade speed u therefore, axial velocity is basically equal to u by tan beta 1 plus tan beta 2 and both these angles have been given to us and therefore, axial velocity can be calculated as u which is 150 divided by tan beta 1 plus tan beta 2 which is 187 meters per second. We can also calculate the absolute velocity at the inlet which is c 1 that is c a divided by cos alpha 1 c 1 is this component alpha 1 is cos alpha 1 is basically the ratio of c a to c 1 c i divided by c 1 is cos alpha 1 and therefore, c 1 is c a divided by cos alpha 1 which is 190 meters per second. Now like in the previous problem we will first solve we will first find the per stage temperature ratio we will then use the isentropic efficiency to find and the pressure ratio to find the total pressure total temperature rise taking place over the entire compressor ratio of these 2 should give us the number of stages. So, solving this part is rather straight forward identical to what we have solved in the previous question only difference here is the presence of a work done factor which was not mentioned in the previous problem and thus therefore, assume to be 0 or to be equal to 1. So, here there is a work done factor which has been specified and therefore, that has to be accounted for in our calculation. So, the way that work done factor here works is that because of this blockage effect there is a reduction in the net work done by the compressor and so, that factor gets multiplied to the delta T naught part of it. So, instead of u times c a into the tan beta 1 minus tan beta 2 divided by c p which is delta T naught for a stage this gets multiplied by this work done factor which is usually denoted by a symbol lambda. So, lambda times u times c a into tan beta 1 minus tan beta 2 divided by c p is the stagnation temperature rise in one stage. Now, let us calculate what the stagnation temperature rise is because all these parameters which are required for this calculation is already known to us. So, lambda times u lambda is the work done factor multiplied by u into c a times tan beta 1 minus tan beta 2 this whole thing divided by c p gives us the stagnation temperature rise per stage and this comes out to be 29 Kelvin. Total temperature rise we need to find the total temperature rise across the compressor and for which we will first find the total temperature rise at the compressor inlet which is the static temperature plus c 1 square by 2 c p and that comes out to be 331.8 Kelvin and the isentropic total temperature at the compressor exit which is T 0 3 s or the isentropic total temperature is equal to T 0 2 multiplied by pi c rise to gamma minus 1 by gamma this is 493.9 Kelvin. Therefore, the actual temperature at the compressor exit will be equal to T 0 2 plus the difference between this and T 0 2 divided by efficiency T 0 3 s minus T 0 2 by eta c. Now, if you carefully look at this equation here which is basically equal to the efficiency definition eta c should be equal to T 0 3 s minus T 0 2 divided by T 0 3 minus T 0 2. So, from that we have simplified that for an expression for T 0 3. So, T 0 3 we can calculate and that comes out to be 522.5 Kelvin. Therefore, the total temperature rise across the compressor is this temperature 522.5 minus 331.8. So, that is 190.74 Kelvin this is the total temperature rise across the whole compressor. So, therefore, the number of stages required in this case will be the overall temperature rise across the entire compressor divided by per stage temperature rise that is 190.74 divided by 29 this is 6.6 and that has been rounded off to the next integer and that is 7. So, in this case this compressor requires 7 stages for developing a pressure ratio of 6 and with an efficiency of 85 percent and other parameters as specified in the problem. So, this solves the first part of the problem which was basically to find the number of stages and this is the simpler part because that is something we have already solved in the first question. Now, second part of this question is to determine if this particular compressor is likely to suffer from shock losses. As you probably are aware that shock losses are or shocks are basically present in supersonic flows. That is in the presence of supersonic flows the likelihood of the presence of shocks are quite high that is if the flow is supersonic then it is very likely that there may be shocks present in the flow. So, that is the basic principle which we are going to use in solving this part of the problem to estimate whether the flow is supersonic or not that is if the Mach number exceeds 1 then there is a chance that there could be shocks present in the flow. Now, here there are two ways of determining the Mach number one is based on the absolute velocity and the other is based on the relative velocity. So, we basically going to make use of the relative Mach number and determine whether that is indeed greater than 1. If it is greater than 1 then there is a chance that the there could be shocks present in the flow. So, for this problem we will need to find out the relative Mach number at the tip of the compressor and then compare this and then find out whether this is greater than 1 or not. If it is exceeding 1 then there is a chance that the flow might undergo or might see shocks in the flow. So, let us calculate the relative Mach number and then see if it is greater than 1. So, to determine whether the compressor will suffer from shock losses we will need to find the relative Mach number. Relative Mach number is the ratio of the relative velocity to the blade speed of sound which is gamma R T and V 1 from the velocity triangle is C A divided by cos beta 1 and that is 264.5 meters per second. So, if you take the ratio of V 1 to square root of gamma R T we get C A divided by cos beta 1 which is V 1 that is 264.5. So, relative Mach number comes out to be 0.77. Now, here we get the relative Mach number which is less than 1. So, what it means is that since the relative Mach number is less than 1 the chances that the flow might see shocks are not it is not likely to suffer from shock losses. For this particular speed that is if you change the blade speed rotational speed yes probably you might have shock losses, but given these conditions and data it is unlikely that this particular compressor is unlikely to suffer from any shock losses because the relative Mach number has been calculated and comes to be less than 1 it is only 0.77. So, in this particular problem we have which we had we have two different aspects which we have solved one is of course, the number of stages very similar to what we did in the first problem. Second part was to find out if there were any shocks present in the flow for which we have used the for which we have basically calculated the relative Mach number and determine whether it is exceeding 1 or not. In this case it so happens that it is less than 1 and therefore, the compressor is unlikely to suffer from shock losses. Now, in some problems it might be specified that you might be given data both at the hub mean diameter and the tip. So, you might wonder it under which case you should be finding out this relative Mach number. Now, if the data is been given for both hub for hub mean diameter as well as the tip then it is obvious that we need to calculate this at the tip because it is at the tip that you have the highest rotational speed or the blade speed is the maximum at the tip and therefore, relative Mach number at the tip is what one needs to calculate and ensure that it and see if it is exceeding 1 or not. In many of the modern day compressors the relative Mach numbers and the tip relative Mach numbers can be as high as 1.6 and that is true for many of the so called transonic compressors where the relative Mach numbers are supersonic which is why they are called transonic. So, relative Mach numbers can exceed 1 and often it reaches as high as 1.6 to 1.8 that is the kind of relative Mach numbers that modern day fans transonic fans and compressors are likely to operate and we will of course, be discussing lot more details about transonic blades and transonic compressors and their properties and how they are different from the traditional low speed compressors how the blades are different and how the flow properties are etcetera are different. We will probably discuss that little later during some of the later lectures. So, we have now so far solved two problems on axial compressors discussing various aspects of the two dimensional flow and how we can calculate different parameters associated with an axial compressor. So, let us now move on to the third problem which we have in hand and let us see how we can proceed towards solving this problem. So, we have the third problem statement here the conditions of air at the entry of an axial compressor stage are P 1 is 1 bar and T 1 is 314 Kelvin. The air angles are beta 1 is equal to 51 degrees, beta 2 is 9 degrees, alpha 1 is equal to alpha 3 is equal to 7 degrees. The mean diameter and the peripheral speed are 50 centimeters and 100 meters per second respectively. Given that the work done factor is 0.95, the stage efficiency is 0.88, the mechanical efficiency is 0.92 and the mass flow rate is 25 kgs per second. Determine part a air angle at the stator entry, part b blade height at the entry and hub to tip diameter ratio, part c the stage loading coefficient and part d power required to drive the stage. So, this is a problem where we have lot of data which is given to us all the data are required for determining the velocity triangle like all the angles are being given to us beta 1, beta 2, alpha 1, alpha 3 all the angles have been given to us. We have some of the velocity, the peripheral velocity or u has been given to us. We have the mean blade diameter and also the temperature and pressure at the inlet. We also have the work done factor and stay efficiency. So, with this we need to find out a host of other parameters like the air angle at the stator entry and power required and so on stage loading coefficient etcetera. Now, in this problem as before we will have to draw the velocity triangle first. So, I am assuming that by now you have understood how to construct a velocity triangle and so I am skipping that part I am leaving it to you to find the velocity triangle for this particular problem. I am sure based on what we have discussed for the previous two cases you will be able to calculate determine the velocity triangle for this case. So, I am not solving the velocity triangle right now I am leaving that as an exercise for you. I will straight away go ahead and solve the problem for you assuming that you can construct the velocity triangle on your own. The first part of the problem is to find the air angle at the inlet which is alpha 2 at the stator entry basically for which once you have constructed the velocity triangle it makes it very simple to solve. From the velocity triangle you will see that the ratio u by C a should be equal to tan alpha 1 plus tan beta 1 this is basically because u is equal to C a times tan alpha 1 plus C a times tan beta 1 alpha 1 is known beta 1 is also known and u is known for. So, from this we can calculate C a which comes out to be 73.65 meters per second. Now, since C a is known from the velocity triangle again we have tan alpha 2 plus tan beta 2 again is equal to u by C a since beta 2 is known and this ratio is also known we can calculate alpha 2 which is the angle at the stator inlet alpha 2 is 50.18 degrees. So, this is so as I mentioned if you have the velocity triangle right this is very straight forward you can very easily find out alpha 2. The second parameter to find is the hub to tip diameter ratio. Now, we know the mass flow rate it is given as 25 kgs per second mass flow rate is given as 25 kgs per second. This is basically equal to density times C a times pi times mean diameter into the blade height h. So, mass flow rate is known density is known because P 1 is known T 1 is known. So, P 1 by R T 1 gives us density which is rho 1 C a is known and diameter is known mean diameter and the blade height from this can be calculated because from mass flow rate density hub diameter is mean diameter minus the blade height and that is 31 centimeter. So, the tip diameter is 69 centimeters and the hub diameter is 31 centimeters and therefore, the hub to tip diameter ratio is 0.449. The other way to find this probably is to use this directly in the mass flow rate formula by using the annular area where this ratio will appear as one of the fractions. So, that is another way of finding the hub to tip diameter ratio which is basically boils down to the same thing mass flow rate is density times axial velocity into pi into annular areas that is d 2 square minus d h square by 4 from which we can take a ratio and find out the hub to tip diameter ratio. So, here it comes out to be 0.449. Now, the third part of this problem is to find the loading coefficient. The blade loading coefficient is the work done by the blade divided by u square. So, work done by the blade as we have seen is lambda times C a into u into tan beta 1 minus beta 2. This is basically the work done which is the enthalpy rise across the blade delta h naught multiplied by the work done factor will give us the net work done. All these parameters are known to us now. So, the work done per unit mass is 0.95 into axial velocity 100 blade speed 73.65 into tan beta 151 minus tan beta 2 which is tan 9. So, this is 7534.8 joules per kilogram. Therefore, the loading coefficient is the work done per unit mass divided by u square which is 100 square that comes out to be 0.7535. This is the loading coefficient of this particular geometry. And the last part of course, is to find the power required which is very simple now because that is mass flow rate multiplied by the work done per unit mass divided by the mechanical efficiency. So, in this question it is also mentioned that there is a certain mechanical efficiency. So, mass flow rate times the work done per unit mass gives us the power divided by of course, the mechanical efficiency because the mechanical efficiency is basically an indicator of the amount of power loss taking place between the turbine and the compressor. So, the turbine is the one which provides work done work to the compressor and so because of various losses taking place like frictional losses and so on. There is certain amount of power loss taking place and that is what is given by the mechanical efficiency. So, the actual power required by the compressor will be higher by this amount of mechanical efficiency. So, mass flow rate multiplied by work done per unit mass this divided by mechanical efficiency gives us the power required. So, we have now solved all the aspects required for solving this problem and I hope you will be able to calculate and determine the velocity triangle and use that in solving this problem. So, with this background of what we have discussed in today's lecture and also what we have been discussing in the last few lectures. I have a few exercise problems for which you can solve based on our discussion. So, let me take you straight to the first exercise problem. First exercise problem is axial compressor of 50 percent reaction design that is the degree of reaction is 0.5 has blades with inlet and outlet angles of 45 and 10 degrees respectively. Compressor is to produce a pressure ratio of 6 is to 1 with an efficiency of 0.85 when the inlet static temperature is 37 degree Celsius. The blade speed and axial velocity are constant throughout the compressor assuming a value of 200 meters per second for the blade speed. Find the number of stages required if the work done factor is part A unity and part B 0.87. Answer for this is part A is 8 stages if the work done factor is unity with the work done factor of 0.87 you need more stages to generate the same amount of pressure ratio that is 9 stages. So, this is the first exercise problem Second exercise problem is air at 1 bar and 288 Kelvin enters an axial flow compressor with an axial velocity of 150 meters per second. There are no inlet guidelines the rotor has a tip diameter of 60 centimeter and hub diameter of 50 centimeter and rotates at 100 r p s. The air enters the rotor and leaves the stator with no change in velocity or radius the air is turned through 30 degrees as it passes through the rotor. Determine the blade angles mass flow rate power required and degree of reaction. So, this is very similar to one of the problems which we have solved today. So, answers for this are 49 degrees 19 degrees mass flow rate of 14.3 kgs per second power required of 300.7 kilo watts degree of reaction of 0.65. Third problem is axial compressor stage has the following data degree of reaction of 50 percent mean blade diameter of 36 centimeter rotational speed 18000 rpm blade height at the entry of 6 centimeter air angles at rotor and stator exit of 25 degrees axial velocity 180 meters per second work done factor 0.88 stage efficiency 0.85 mechanical efficiency of 96.7 percent. Determine the air angles at rotor and stator entry mass flow rate power required the stage loading coefficient pressure required pressure ratio developed by the stage and relative Mach number at rotor entry. Answers are the following 54.82 and 25 degrees part B is 14.37, part C is 51.2 kilo joules per kilogram, part D is 0.44, part E is 1.6 and part D is 0.9. Fourth problem is 50 percent reaction axial flow compressor has inlet and outlet blade angles of 45 degrees and 12 degrees respectively. The blade speed at the tip of the rotor is 320 meters per second. If the inlet total temperature is 300 Kelvin, determine the tip relative Mach number. In this case the Mach number is 1.146. And the last exercise problem is a 10 stage axial flow compressor develops an overall pressure ratio of 8 with an isentropic efficiency of 0.85. The absolute velocity component of air enters the rotor at an angle of 27 degrees to the axial direction. The axial component of velocity is constant throughout the compressor and is equal to 150 meters per second. If the ambient air conditions are 15 degree Celsius and 1 bar, determine the angle at which the angle which the relative component of velocity makes with the axial direction at the exit of the rotor. So, in this case the angle is 14 degrees. So, these are a set of 5 exercise problems which you can solve hopefully based on our discussion during today's lecture as well as the theory lectures which we had during the last 3 or 4 lectures. So, with this I would like to conclude today's tutorial session and I hope you have had now some insight into how you can go about solving problems related to axial flow two dimensional analysis of axial flow compressors and cascades and hopefully you will be able to use this knowledge based on the discussions in the last few lectures and today's lecture to solve problems and take up preliminary design of at least the two dimensional case of axial flow compressors. So, that brings us to the end of today's tutorial session. We will continue discussion on axial compressors and three dimensional flow through axial compressors in future lectures.