This video is brought to you in part by TrueTechTools. Quality tools, essential support. All right guys, so we're on a grocery store here. This is kind of an old one. So what we've got is a freezer. It's not working. As you can see, this thing's a little bit tired as they call it. So this is the one we're actually working on. I can turn it off the breakers and hitting the reset here on the oil switch. I don't know if it's tripped yet or not. So far all I've done is I've came in and have checked to make sure I have voltage contactors. I do have voltage coming out of my defrost block. We have 160-minute section here. I want to see if there's an issue with this oil safety sight glass on here. This thing's just plump, filthy, nasty, dirty. So my thought is I've probably got a screen down here on the bottom that's dirty. Also maybe a screen on the sensor that's dirty. And this is for the freezers. Let's go take a look at that right now. There's our heating systems. Looks like this has got some air handlers with some wires right across the ceiling. Whatever works, I guess. So what we've got here is a total of four little bunkers. These are all on that system. And there's a thermostat controlling all this, I guess, with this one little independent thing right here. Yeah, so all that stuff, all that stuff down, all that stuff down there, all works. This here is completely dead. So I think our main issue, it sounds like when it's running, it's running. It works. There's supposedly a thermostat in here somewhere, which, you know, thermostat or thermometer there definitely is not working. Old husband cases. It's either 404 or 507. I'm not sure which one it is. And it'd be nice to know if that thermostat's in here. We've got wires down in here, but I think it has something to do with the lights. Door seals are not looking the greatest. They are what they are, I guess, right? That ain't our issue here, though. It's a thermostat. I don't even know if it's still hooked up. That's an old writing possibly in there, but that's what we've got. There's no, this is a dual temp case here on this one here. How they're controlling temperature would be nice to know. But at this point, I think we just need to find out why it's shutting off. I think if a thermostat stuff was working before, it should be working still now. All right, let's go back in here and see what's going on. So first thing I'm going to do is I'm going to take this apart. Let's see if we got power to it. As you can see, we have no green light, none, no power to module it. So we have no power to it. So right now it might be a low pressure. The compressor does not seem to be hot. Second and third one up from the mount. So we need a second and third one up. Skip ground. The switch is open. The second one, third one down, tied in with the fourth one down between these two right here. Their port's not opening up or whatever. So even if we hit power here, it's tripping. That's a problem. So we need to make sure that power stays consistent. We've got issues here. We need to check the wiring on this, make sure that's wired right. So we're hearing issues. We're having issues with the pressure. It's saying there's pressure problems. We need to find out what the oil pressure differential is. We're going to check our oil on the suction side. That's going to give us our differential. Usually it's about 45 psi, something like that. So we're on our suction side. We're going to go ahead and open that up. And our suction pressure is, honestly, but it's job.  So either oil levels flow because you can't see it or our screen's plugged up and it's not pulling the oil up into the sump. So we've got an issue here. And it finally tripped out. There we go. So we're going to need to isolate the compressor from the rack or from the refrigerant. So what we're going to do is we're going to pack up on it.  And then we're going to check that screen. We use that rack to help catch some of that. It's not a good sign. The compressor inside there. We'll now just turn the off, go ahead and hook that back on there that it's actually down to zero. Some of this area around the cover, half piece, part of the filter system in here. I'm hoping maybe I can shove it out in the back. A couple things here. I need to find out the oil safety. I mean, yeah, the oil pressure switch over to the compressor contactor there. I want to make sure this fan whether it runs or not. It took the terminals off the compressor except for one, which won't hurt nothing. Got it wired on there, contactor in one, see if that fan motor turns. So it must not have been wired in right or something. Maybe it was on a high pressure switch. I don't know. It'll be wired on so that whenever the compressor runs, it runs. So we'll probably reuse the oil safety with high and low switch. Bring up a new dual switch for that. Mine there is actually a fake. We'll probably need a little bit of help getting it over here. Use the pie chain to kind of go from there with what we've got right now. Let's see what you need to do. Discover back on. We don't want to lose...


Hello and welcome to another session on problem solving on sequence and series and we are discussing general term of a sequence Okay, so the question says if five times the fifth term of an AP is equal to eight times its eighth term So that it's 13th term is zero. So don't get intimidated by the language Language at times will be a little difficult, but then you know, you can again Read the question so that all the data what you are Gathering should be correct. Okay, so if five times the fifth term Of an AP is equal to eight times the eighth term. So you can you know very well say that five times The fifth term that is five times T five is equal to eight times T It okay, and you have to find out T13 and you have to show that T13 is zero to show Okay, so you'll say let the first term let the first term of the given AP first of first term of an AP of this AP that is given AP be a right and and Common difference common difference. There are two parameters in AP. You know that which is vital Common difference common difference is Is D? Correct common difference is D. Okay, you have to say this then P five, you know fifth term is nothing but a plus five minus one D Correct, so hence it is a plus four D Right and then T what eight is equal to a plus eight minus one D that is a plus seven D Correct. Now if that is so then by this equation here, we can write five times T five So what is T five guys a plus four D? right is equal to eight times T eight so a plus seven D. Isn't it? So what do we get? We get five a Plus 20 D is equal to eight a Plus fifty six D. Is it that means if you simplify you'll get eight a minus five a that is three a plus forty thirty six 36 D Is equal to zero Correct, so if you cancel three from you know the equation so you'll get a plus twelve D is equal to zero So a plus twelve D can be written as a plus thirteen minus one D is zero and what is this guys?


Thank you. Mr. Director Brahim Masanu, honorable guest, dear colleagues, ladies and gentlemen, good morning. And thank you for inviting me to speak during what has already become a traditional event for regulators, the global dialogue of all those interested in the evolution of the regulatory landscape. I knew that this event is attended by many representatives of the industry, but I have to confess that I am impressed by such a large audience, so thank you for coming in such a large number. The theme for this year, year's global dialogue, is a very generous one, artificial intelligence, IoT and cybersecurity. It goes without saying that our industry is evolving in an incredible pace, and we are limited in this evolution only by our imagination. When I was a child, even a teenager, I used to watch sci-fi movies, even if this was not easy in Romania at that time. And I remember how amazed I was by all those gadgets, and how I thought that I will never enjoy such toys in my lifetime. And yet, here we are surrounded by what seemed impossible only 10 or 20 years ago, a reality that doesn't cease to amaze me every day. And yet, we keep on walking farther and farther towards achieving the impossible. We are, I think, at a point in history when everything we dream of can be put into practice. I said before that we have only our own imagination as a limit for development. Thousands of pages have been written about the advantages and benefits that the technological development will bring in our lives. And there is no question about it. Every citizen, every individual will feel the effects of it. And this is a great opportunity for you as industry representatives, but also for representatives of regularity and policy makers, as we need to ensure that we unlock the full potential of the technological development. But with great opportunities comes great responsibilities. And at this point, when we begin to understand how artificial intelligence and IoT can change our lives for better, we also need to keep an eye on the reverse of the medal, and we need to explore the challenges that change will pose as regards security. And I would like to thank the ITU for opening up this issue and bringing together all stakeholders to discuss it. But it is my strong belief that only through an open and transparent dialogue between all parties involved, we can secure a sustainable and secure development. Thank you very much for your attention and participation. And I wish you to wisely use the opportunities that lie ahead. Thank you very much.


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Good morning and welcome to the next lecture in our course on chemical engineering principles of CVD process. In previous lectures in this course we have covered materials corresponding to the properties of CVD films, different types of CVD films and different structures, ways of analyzing film properties, various types of CVD reactors and operating conditions and so on. From this lecture onwards we are going to focus more on the thermodynamic and transport aspects of CVD processes. And we will begin by looking at a couple of examples of CVD phenomena that we encounter in everyday life. The first example is one that I referred to back in my first lecture which is the tungsten lamps, the tungsten filament incandescent lamps that we use widely in our houses and other places. Although they are being increasingly replaced by fluorescent lamps and LED lamps, the incandescent lighting technology is still around and bulbs are not going to go away very soon. So how do these bulbs work? How do they provide illumination? It is a very interesting application of CVD technology in the sense that we are actually using the CVD aspects of the bulb as a chemical reactor in order to extend its lifetime. So we will see what I mean by that. If you look at the history of lighting, basically lighting has been provided by incandescent sources from the beginning. So if you look at the candle for example or if you look at kerosene lamps or gas lamps, they all work on the same principle. You take some reacting gases and bring them to very high temperature and as the temperature increases the solid particulates actually start glowing and that is really what provides the illumination. So all these early incandescent lamps worked on the principle of bringing these so-called soot particles to incandescent thereby providing lighting. The problem with these sources was that they were short-lived and they were highly polluting and they were creating an atmosphere where it was difficult to function. So filament lamps were introduced. The first filament lamp was actually invented by Edison back in 1876 and that was a carbon filament lamp. Now carbon burns well and it provides good illumination but the problem with carbon is its vapor pressure is very high. So these carbon filament lamps even though they worked quite well their lifetime was of the order of days. Within days the bulb enclosure would completely get coated with carbon and become unusable. Now when you look at this from the viewpoint of a deposition process you have a carbon source and the film that is depositing is also carbon. So it is a classic example of a PVD process right, physical vapor deposition process. Now this problem was addressed in the early 1900s by adopting tungsten filaments instead of carbon. The advantage of tungsten is that it has about a 5000 times lower vapor pressure compared to carbon but it can also be brought to incandescence at a temperature that is fairly close to carbon filament. The melting point of carbon is 5000, the melting point of tungsten is about 4600 Kelvin. So the melting points are reasonably close so you can heat them to very high almost as high a temperature as a carbon filament lamp but because of the lower vapor pressure now a tungsten filament bulb could be used for weeks instead of days. So you are extending the lifetime by a factor of about 10 to 100 by adopting a tungsten filament rather than a carbon filament. However it was still limited by the rate of evaporation of the tungsten. So sooner or later you will get this coating of tungsten around the bulb and therefore again you could not use it anymore. So this is also a PVD process but a lower rate deposition process compared to carbon. The next advancement in incandescent lighting technology came about when people originally the carbon filament was being built in was being burnt in vacuum. The next big advancement was to actually fill the bulb with an inert atmosphere with either inert gases or nitrogen thereby slowing down the rate of evaporation. Now the way that works is when a filament evaporates the molecules leave the filament, get into the gas phase and then they diffuse and they deposit on the bulb around the filament. When you fill the bulb with some kind of gas the gas molecules now actually can retard the motion of the tungsten vapor molecules and direct them back towards the filament. So the net movement or the net flux of tungsten molecules towards the walls of the bulb is significantly reduced by the presence of this inert gas which provides a layer that the tungsten molecules can bounce off of and return to the filament. So the introduction of this inert atmosphere slowed down the onset of what is called as what is known as bulb blackening which is the deposition of tungsten by another factor of I would say around 10. So now instead of weeks a bulb could be used for months before it fails but still within 5 to 6 months these tungsten lamps with inert gas filling would fail because the bulbs were coated with tungsten. Now is this a PVD process or a CVD process? We have a tungsten filament that is burning, we have tungsten that is depositing. What form is the tungsten in the vapor form inside the bulb? It is an inert atmosphere right so there is no chemical reactions going on so the tungsten in the gas phase is also present as tungsten. So this is still a PVD process but slowed down even further. So we went from having a carbon filament which had a very high vapor deposition rate to a tungsten filament which had a lower evaporation rate and therefore deposition rate to a tungsten filament in an inert gas filled bulb which had an even further reduced rate of deposition of the tungsten on the walls. But still it had a finite lifetime and it is still not economical for the industry. People were not happy about having to replace that bulbs every 4 months. Now if you look at today you know where we are still using bulbs what is the typical replacement frequency and why do we replace it? It can come for years right. I mean usually these bulbs last for you know couple of years and even after that they do not fail because tungsten has deposited around the bulb. They fail because the filament breaks. So essentially this whole bulb blackening has now been eliminated as a mode for obsolescence of the bulb by some very, very clever design. So the big advancement what made bulbs a viable industry was to introduce halogens into the bulb and achieve something called the halogen cycle. So essentially we transposed the bulb from having a PVD process to one having a CVD process. So the way this works is when the tungsten evaporates from the bulb and gets into the gas phase is there a way by which you can provide a flux of tungsten back towards the filament by having an appropriately reactive environment inside the bulb. So if you think about it by the way an electric bulb if you just look at it looks very much like a CVD reactor right. It has a source and it has a sink. Of course the big difference here is that the substrate on which deposition is happening is the walls of the reactor itself I mean you can look at the bulb as a reactor and the deposition of the evaporated material is happening around the walls of this reactor which is something we want to avoid in a typical CVD reactor and that is the same case here. What we are really trying to do is minimize or eliminate the rate at which the tungsten that is evaporated from here gets deposited onto the bulbs as a film. In fact we will later on we will talk about filament CVD reactors they are designed exactly like a bulb except that they provide flow of gases inside the bulb this reactor to take these vapors to a specified location so that essentially you convert an electric bulb into a CVD reactor where the deposition is occurring over a well defined substrate area. But anyway coming back to the bulb situation then so the problem is when tungsten evaporates it has a tendency to diffuse and deposit on the bulb walls and as I said by providing these gas molecules inside inert gas molecules you can because of the collision between the tungsten vapor molecules and the inert gas molecules you can direct some of the tungsten vapor molecules back towards the filament and thereby slow down the rate of deposition. But the major revolutionary change that happened was introducing halogens X where X could be Cl, Br could even be oxy bromides and so on into the bulb before it is sealed up. How does that help you? Well when you look at this area suppose a tungsten film starts to deposit but you have a reactive material like Br present the tungsten is going to react with the Br and get converted to tungsten bromide and stay in the gas phase. So the reaction that occurs here is W solid plus X going to WX so this is in the gas phase and this is in the gas phase. So what will this WX do once it is formed? It has a tendency to diffuse back to the filament because it follows fictiffusion and fictiffusion says diffusion occurs down a concentration gradient since there is an excess of WX concentration near the walls of the bulb and there is virtually no WX at the filament the WX will diffuse back towards the filament and what will happen when it gets to the filament WX gas at the temperature of the filament let us call that some Tf and this is happening at the temperature of the bulb let us call that is Tb. This WX gas breaks down into W solid plus X gas so it releases this X gas back into the gaseous environment the tungsten itself is redeposited on to the film in the form of solid tungsten. So this is essentially a cycle right and it is known as the halogen cycle. So if you look at this halogen cycle in principle it can be set up so that the rate at which the tungsten is evaporating from the filament is exactly balanced by the rate at which it is getting replenished by suitably balancing the rates of transport of the corresponding molecules. So that is a clever concept behind the halogen cycle. Now in principle this bulb can have an infinite lifetime it will keep going as long as this cycle is present and in fact that is exactly what is happening today in the bulbs that are manufactured they all have a halogen environment and that is the reason why they are not failing anymore because of a simple film deposition process. So it is a classic example of a CVD reaction that is very elegantly set up in order to achieve a certain purpose but it is very different of course from the commercial CVD applications that we have talked about so far but it actually illustrates the principles quite well in terms of what is happening inside this whether you call it a bulb or a reactor you know the CVD process is actually happening inside here. So how do you provide this balance? I mean there is something called a ZEF which stands for 0 element flux. This is the condition you are trying to achieve. You want 0 net flux of the element in this case is tungsten. You want the evaporative flux of tungsten to be exactly balanced by the re-depositing flux of tungsten so that the net is 0. So how do you achieve this? In order for this to happen the flux that is leaving the filament must be balanced by the flux that is re-depositing or getting re-transported towards the filament an easy way to think about it is in terms of diffusivities. So you have DWX which is the diffusivity of the gaseous species WX that multiplied by P of W at the bulb wall must be equal to D of W times PWF. It is basically a diffusional mass balance. It basically says that if you take the partial pressure of tungsten near the bulb and you multiply it by the diffusivity of the dominant species containing W that should be exactly equal to the diffusivity of tungsten vapour species near the filament multiplied by the partial pressure of tungsten near the filament. These are essentially mass fluxes written in terms of partial pressure instead of concentration but it is the same I mean basically this are the Fick diffusion fluxes. Normally you would write the Fick diffusion flux as some minus d dc by dr in this case assuming it is a radial symmetry that is basically what we have written here except that the dc in this case is basically equal to the partial pressure at one side because the partial pressure on the other side is virtually 0 right that is where this equation came from. So if your partial pressures PWB over PWF is exactly equal to the ratio of the diffusivities which is dw over dx this is called the 0 element flux condition. If the partial pressures are exactly in this ratio in the ratio of their diffusivities then there will be no net flux of tungsten either way it will be in dynamic balance not static it is an equilibrium dynamic balance. So if you can design your bulb in such a way that you achieve this balance then it will operate forever in principle. Now if you look at that ratio dw by dwx is it greater than 1 or smaller than 1? Why is that? Greater than 1. Why is that? At partial pressure bulb will be higher than 1. Just look at the right hand side I am just looking at this ratio dw over dwx you have two molecules one is a tungsten molecule the other is a tungsten halide molecule if you look at the ratio of the diffusivities will it be greater than 1 or smaller than 1? Gas diffusivity. They are all they are all in gas phase w is also in gas phase wx is also in gas phase these are all gas phase molecules right the diffusion process we are talking about is all happening in the gas phase solids do not diffuse. The molecular weight of wx is higher so diffusivity will be affected by the molecular weight. It is not the molecular weight but yeah so dwx will be greater than dw. It is actually size which one will be larger is w larger or wx and as size increases what happens to diffusivity reduces. So which one is lower in this case dwx. So this ratio is always greater than 1 diffusivity of a tungsten halide molecule is always lower than the diffusivity of a tungsten molecule. So this ratio is always greater than 1 if the tungsten pressure partial pressure at the bulb exceeds this value this particular ratio then what happens you get more deposition back towards the filament right in order to achieve balance again and similarly if this becomes larger then what happens you get more transport towards the bulb. Now so what happens today when bulbs fail why do they fail because there is a sudden increase in this value and why does it happen when a filament breaks the area that is exposed is suddenly much larger so the rate of evaporation is much greater. So there is an instantaneous jump in the partial pressure of tungsten in the gas phase that is adjacent to the hot filament and therefore you have a sudden shift in the equilibrium and all of a sudden the bulb becomes very black you know you will see that as soon as the filament breaks the entire bulb will look black right and that is the reason. So very, very basically this is how a bulb works and as long as you can maintain this zero element flux condition inside the bulb it will be in good shape but as soon as you break the condition there is a tendency for it to fail for one reason or the other. Now if you do a transport analysis this what I have shown here is a very simplistic way of looking at it but you can do a more rigorous transport analysis of the system as well as a thermodynamic analysis of the system. In terms of thermodynamics because the filament temperatures are very high you know 4000 Kelvin somewhere in that range the assumption of local thermochemical equilibrium is valid the higher the temperature the more the closer you get to equilibrium conditions. So at the filament you can assume that what is known as LTCE prevails LTCE stands for local thermochemical equilibrium. So you can calculate your gas phase composition by using for example the free energy minimization procedure that we discussed earlier in class. The assumption of chemical equilibrium is valid at the filament and close to the filament but if you look at the bulb wall the outer wall is at room temperature right the inner wall gets heated up because of the radiation from the filament but still this TB is going to be much smaller than TF. So given that is it still okay to assume equilibrium at the bulb wall or are you going to be kinetically constrained? Well strictly speaking yes you would have kinetic constraints that come into play because the temperatures are not high enough to guarantee equilibrium conditions but there is a gradient the temperature changes gradually from TF to TB inside the bulb. So at each prevailing temperature you have to reassess the validity of the equilibrium assumption. Is the equilibrium assumption only valid near the filament? Is it valid till about half way across the bulb or is it valid all the way up to the bulb wall? That is a very important consideration in terms of the thermodynamic analysis because if you can assume chemical equilibrium throughout the bulb then it becomes fairly straightforward to calculate the composition of the gas phase inside the bulb simply by assuming free energy minimization strategies. However if you have to assume kinetic constraints then you have to really understand the dominant chemical reactions that are taking place inside the bulb. So it requires a lot more analysis it requires a lot more monitoring and so if I am a bulb designer I will try to operate the bulb in such a way that I maintain equilibrium conditions as much as possible inside the bulb. Now in terms of transport what happens inside the bulb? You have a heated filament and you have a sealed enclosure. What kind of flow do you expect inside this enclosure? Will you have turbulent flow? Will you have laminar flow? Will you have unidirectional flow? Will you have recirculating flow? Will you have no flow? What do you expect? What are the types of fluid flow that can happen? That is let us say in terms of let us talk about heat you know you are essentially transferring heat from the filament to the bulb. What are the mechanisms of heat transfer that can happen? There is conduction, convection and radiation right. Conduction will certainly happen as long as you have a hot source and there is a temperature gradient. Radiation will happen because it is a you know radiating source that is at a very high temperature. How about convection? Can convection happen inside a bulb? So there will be natural convection right. So in principle all 3 modes of transport can happen. However in terms of the convection part, again if I am a bulb designer I want to design it so that the entire fluid region inside the bulb is stagnant and there is no flow. The reason is as soon as you set up a flow it increases the rate of mass transfer. So if the filament starts to evaporate the rate at which these molecules will be transported away from the filament and deposited on the bulb will be greatly enhanced if there is natural convection going on. So you want to design the bulb in such a way that there is a stagnant layer inside the bulb with no natural convection and the stagnant layer extends all the way to the walls of the bulb. This type of layer is called the Langmuir layer. Within a Langmuir layer surrounding an object the only modes of transport or conduction for energy and diffusion for mass and momentum. Under these conditions the rate of loss of tungsten from the filament can be greatly minimized. So as a design principle you would want this Langmuir layer to extend all the way from the filament to the walls of the bulb. So let us say that you have achieved Langmuir conditions and let us say that you are operating this tungsten lamp under conditions where you have only an inert gas inside the bulb. So we are looking at the case of an inert gas filled tungsten filament bulb. So essentially the deposition process that is going on is a PVD process. The tungsten that evaporates from the filament stays as tungsten. So it just goes from being solid phase tungsten to being gas phase tungsten and then at the walls of the bulb it re-deposits as solid phase tungsten, right. How would you do the transport analysis of this type of a problem? The first you have to write the energy balance. How do you write conservation equations? How many terms are there in a typical conservation equation for mass, momentum, energy, entropy? What are the four, well how many terms and what are they? Inflow is by what mechanism? We just said there are three modes of transport, right. There is convection, diffusion. So when you write a mass balance or a heat balance or you have to say it in terms of accumulation is 1, generation is another but then there is flow but conventionally it is split into convective outflow and diffusive inflow. So essentially any energy balance or any kind of balance equation is written in terms of an accumulation term plus a convective term which balances a diffusive term plus a source or sink term, right. That is a general formulation of a conservation equation. So in the specific case of an inert gas filled tungsten filament bulb let us say that you are doing a steady state analysis, okay. So the accumulation term drops. So we are assuming that it is under steady state conditions. Convection we just said we will design it so that there is no convection inside the bulb. So this term will drop. The source term for heat, there is radiation but again once you have achieved a steady state the radiation term is not going to really affect your heat flux in terms of the time dependent change or the position dependent change in your temperature distribution for example. So essentially you just have to solve a diffusion equation to know the temperature distribution inside the bulb and that is done in terms of a divergence of heat flux equal to 0. You know divergence, right. You have gone through this in your heat transfer course. So the notation that I am using here is a dot on top stands for per unit time and this double prime stands for per unit area. So what this equation means is it is a divergence of the heat flux which is expressed as heat transferred per unit time per unit area. That must be equal to 0, right. That is a diffusion equation in its simplest form where this q dot double prime can be written in terms of, if this is a conduction term how would you write q dot double prime? What is k dT by dx minus k dT minus k times gradient in temperature, right. What is the corresponding mass balance? How would you write the mass balance for in this case let us say tungsten? Again the terms remain the same and let us say that again these 3 terms are dropped and only diffusion is there. So what is the corresponding diffusional equation for mass? What would you substitute instead of q dot double prime you know which is heat flux what will you substitute? Mass flux, right. So the mass balance is simply given by divergence of let us call that Jw dot double prime where J stands for mass transfer and again the dot is per unit time and the double prime is per unit area. So basically what we are, this Jw dot double prime is the mass flux of tungsten expressed in per unit time per unit area and that must be equal to 0 if diffusion is the only process that is inducing transport of the tungsten species and similarly you will write Jw dot double prime as equal to what? What is the mass diffusion flux term? Minus dW actually there is a rho, right density of the gas minus rho dW gradient in omega W where omega is the mass fraction of the tungsten. This is assuming that there is only thick diffusion happening. Fick diffusion is the diffusion that happens because of a concentration gradient. However there is also something known as thermal diffusion. A diffusion process can also happen because there is thermal gradient. When there is a temperature difference the molecules are energized to different levels depending on the temperature that they are exposed to and this difference in energy states can actually drive a diffusional process until equilibration happens. So when you include thermal diffusion as well this becomes minus rho dW gradient of omega W plus omega W times alpha T W gradient of temperature divided by temperature. So this is the thick diffusion part and this is the thermal diffusion part. It is also known as Soray diffusion. So the complete diffusion equation is minus rho dW which is the thick diffusivity of tungsten times gradient in the mass fraction of tungsten plus omega W which is the mass fraction of tungsten times the thermal diffusion coefficient of tungsten which is written as alpha subscript T, W times gradient in temperature divided by temperature which by the way this term can also be written as gradient in ln of T. So essentially you can write for this particular case the energy balance equation the mass balance equation you can apply the appropriate boundary conditions T at the filament is T f T at the wall is T b you have the boundary conditions on the species you know you can basically calculate what will be the mass fraction of tungsten at the filament what will be the mass fraction of the bulb and essentially solve the system of equations to get the temperature distribution and from that you can get the concentration diffusion distribution of tungsten. Once you have the concentration distribution you can calculate the corresponding flux. So you can essentially divide this into various sections and do you know finite difference analysis and you know solve this set of equations and you can calculate a finite rate at which tungsten will evaporate from the bulb and be transported to the I mean to evaporate from the filament and be transported to the bulb wall by a purely diffusional mechanism and then you can do a time integration of this and you can actually predict when either the filament will fail because it becomes too thin or the bulb will fail because the layer of tungsten on the wall becomes too thick that you can establish a relationship between light transmission and the thickness of the tungsten around the bulb wall and you can determine the failure limit in both cases. So this is kind of the classical traditional bulb with a purely PVD process PVD again implies that the tungsten is present in the gas phase in the same form as it is present in the condensed phase and this is a case where because it is a purely PVD process there is no way to reverse the process it will always happen. You can try and slow it down by various methods for example if you operate the filament at a lower temperature that will slow down the rate at which tungsten is evaporating so that is one way or you know you can use heavier gases you know this DW for example is different for the various fill gases. The diffusivity of tungsten in vacuum will be very high but the diffusivity of tungsten in a filled environment will be slower and by changing the gas that you are using to fill the bulb you can actually control the diffusivity of the species. So if you use heavier gases larger molecular sized gases as the filler material the bulb will last longer because the probability that the tungsten molecule will collide with the inert gas molecule and bounce back is greater when the size of the gas molecules is larger. So you can do some things like that to increase the lifetime of the bulb but eventually it will fail okay so that is why the improvement that was made was to go to a halogen cycle lamp halogen cycle tungsten filament bulb where now the process is happening inside a CVD. The reason is as soon as you fill it with some reactive gas the environment becomes chemically reactive so the tungsten that is evaporated from the filament is not going to stay as tungsten it is going to be converted to WCl, WCl2, WOCl2, WO2Cl2 I mean again there are dozens of species that can form but the material that is going to deposit on the bulb is still tungsten. So that again it is a classic definition of a CVD process where the identity of the condensed phase is different from the identity of the gas phase species in which it is present. So if you want to take this analysis that we just did and extend it to a CVD process what how would it be different for example if you now do an energy balance can you still drop all the 4 terms or do you have to keep one of the terms suppose you are doing energy balance for species I. So I still have this equation accumulation plus convection equals diffusion plus source. So now which of these terms can I drop accumulation steady state okay about convection can still drop that because we have not done anything to change the fluid mechanic design the diffusion term should still stay about the source term can you drop it no it is a chemical reactive environment right. So for every species there is a source and a sink either species are getting generated or they are getting destroyed. So now there are 2 terms in the equation and so you would have to write your energy balance as a balance between the rate at which energy is getting transported through a diffusional or conduction process balanced by the source term which is that as chemical reactions happen they are constantly either generating heat or consuming heat. So the chemical reactions themselves cause changes to the energy conservation term. Now if you look at the mass balance term again you now have to keep the diffusional term and the source term intact and write different equations for each species. So for example for the ith species I would write this as divergence of J i dot double prime equals some R i dot double prime. So this is not 0 anymore there is a finite rate of reaction associated with each species I. So the divergence in the flux of the ith species must be balanced by the rate at which that species is being consumed or produced and this is true for all I species and again the diffusional flux of J i dot double prime will be written in the same format as minus rho D i gradient in omega i plus omega i alpha T i grad omega I mean grad T over T and you have to write this equation for each ith species and you will now have to also estimate this you have to know the reaction rate for every species the rate at which is being consumed or produced. If you can do that then you can again I mean once you know this from this you can also calculate the heat source associated with that based on the enthalpy change and you can again self consistently solve this but as you can imagine going from there to here it is a much more complex process because now you have to write this balance for every species that is present in the system and it has to be all self consistent you know the chemical equilibrium that the temperature equation and the mass balance equation cannot be decoupled now. In the previous case we would have first solve the energy balance equation to set the temperature profiles and then use the diffusion equation to calculate the corresponding diffusional fluxes. In this case you cannot do that because the temperature distribution influences the chemical reactions but the chemical reactions are also influencing the temperature distribution. So this has to be solved iteratively in a as a coupled set of equations. Now the only reason we are not including momentum balance here is because we have made a certain assumption that we have a stagnant layer that extends all the way till the bulbs till the walls of the bulb. So this becomes numerically very complex I mean there are papers written there are algorithms that have been produced to do exactly this for various systems of gases filling a bulb. However the numerical complexity involved in solving the set of equations in a self consistent manner is tremendous. So we need a simplified procedure how can we take this and simplify it to this you know how can you apply the same principles that we used for the inert gas filled tungsten filament lamp and apply that to the analysis the transport and thermodynamic analysis of a halogen cycle tungsten filament bulb. Is that a trick that we can use that will enable us to do that. Any thoughts I mean how would you reduce the complexity of this analysis and make it as simple as that. When you think about it we will talk about it in the next class when we start the next lecture okay. There is a very simple and elegant way to do this analysis which reduces almost 90% of the numerical complexity okay. Any questions. Yes how actually I could not understand that a Langmuir layers how these are. A Langmuir layer by definition is one where transport phenomena are dominated by diffusional processes for example a boundary layer around an object is a Langmuir layer. Different density gases are layer by layer they are depositing on the film not like that. No I mean this has nothing to do with CVD.  It has nothing to do with our CVD environment. It is a general fluid mechanic definition that in any physical situation where the domain of interest is dominated by purely diffusive phenomena that is known as a Langmuir layer. An example of that is a boundary layer. Another example of that is an enclosure like this where you by design you ensure that only diffusion happens. For example flow inside pores you know micro pores, meso pores is considered Langmuir flow because it is dominated by diffusive phenomena. So it is a general term that applies to all such classes of flow okay. I will see you in the next lecture.


Ladies and gentlemen, welcome here to the transition of authority ceremony between Task Force Gator and Task Force Orion. Ceremonies like this, transitions of authority, they're a good time to reflect. First, I'll remember your TOA ceremony. It was cold. I remember thinking, how are these Floridians going to handle a cold Ukrainian winter? However, we're thankful that you integrated well and by early January you were beginning your first training in that cold Ukrainian winter training on javelin, stinger and bunker busters. In conclusion, I think it's important to talk about one last memory. For Colonel Murphy and Task Force Orion, know that this command will remember you. As the first unit to be recidivist in the training at JMTGU, it is absolutely great to have you back here. We remember what you did on your previous rotation. You left an impact on all of us here in 7th ATC and you left an impact on the armed forces of Ukraine that you trained. The last nine months have been incredible, emotional, rewarding, difficult, but most of all, it's been an honor. Task Force Orion, today as a team we assume authority of the JMTGU. A mission we stand ready to prepare to execute. A mission whose success is not measured by matrixes, but our ability to increase the proficiency of our Ukraine partners in their lethality and survivability as they defend their country against Russian hostilities. I know each one of us understands the urgency of this mission and the gravity of what the Ukraine people have at stake. Finally, to our Ukrainian brothers and sisters-in-arms, who I know are training in the field, fighting in their homeland and are no longer among us. We remain relentless in our support, focused in our training and tireless in our effort. We will not let you down.


Look man, it's the new year and you're still sucking. You can't make progress. You can't make gains. You're fat. You're weak. You're ugly. You're broke. And you're single.  Unprobably to start on our checklist is why you're not making strength gains or muscular gains. You know, there could be a couple of reasons. Number one, it's not the answer you want, what you're going to hear, but it's going to be patience. It's going to take you a while. Number two, a more applicable one, it's probably a frequency issue. I still see a lot of people squatting once a week or doing arms once a week and you're just doing so much like garbage sets or junk sets.  So when we're talking about strength, we want to practice the lift and the best way to practice it is to do it. If you're a basketball player, golf, you're not golfing one every seven days expecting your stroke to get better, right? We're going to do it multiple times a day even and obviously those sports are different because we don't have external load, but you're going to do it multiple times a day to practice that rhythm. Same with lifting. If we can squat three or four times a week and still progress, eat food, sleep, recover, manage our fatigue with our volume and intensity, now we're going to actually get better at the movement while we're building muscle and neuro, muscular-like activity, so we're getting better at firing our muscles, we're getting better at the skill of it, the coordination of it and we're building muscles are kind of the three ways to get stronger. Now in terms of being fat, I'm not fat, shaming.  I mean it's just true. Yeah, there's a percentage of folks that can be real sexy at a higher body fat and there's a small percentage of folks that can still be healthy, but the chances start to go up when you get to be a little bit leaner. And whether it's self-confidence, something we talk about here, the reason I design clothes is for confidence, the reason I lift weights is for confidence, the reason I strive every day to build a routine to get better is for my confidence and I can build on small wins to feel better about myself.  It doesn't it doesn't feel very good. I'm the same way, right? I gain more weight in this bulk and this off-season strength block than I wanted to, right? And that was just lack of routine, lack of consistency, having two cookies instead of one cookie. Like we all know why until you look in the mirror, you know, you look in the mirror, you tell yourself, all right, I know I overdid it. You can also look in the mirror and say, all right, now I'm not going to not overdo it no more, right? So for me today that's that. I'm going back to more whole foods.   Again, yeah, you can flexible diet and there's a percentage of people that can eat one cookie and get shredded. There's also a percentage of us that can't eat one cookie. So we're going to eat 12 cookies and then we ain't going to be lean no more. So for me, no more cookies. Those trigger foods for me are just going out the window. We're back on the grind. So weights don't feel great right now, but we're showing up and we're doing it. Same thing with your diet. So just literally before I came here, I did a little Amazon Fresh deal. I've got all my ground beef for the week. I've got rice at home. I'm going to go home and cook in between meetings. And the same things for you. It's just literally a matter of do you want it? How bad do you want it? Are you going to do it or are you not going to do it? I ain't going to. There's nothing I could say here.  You're the only one putting the cookie in your face. So we're in the middle of the deadlift series. I guess we're near the end of it. Wrapping up the corners. Kind of a roller coaster. A lot of travel, a little bit of weight fluctuations. Stretch is high. Work stress is really high. So that's definitely affecting me. Like today, it's supposed to be 460 to 470 for dub, RPE7. And I probably was in that range seven to eight, but it just took a little bit more energy out of me to get there rather than that should be kind of like a daily thing and easy thing. It's a little disappointing.  But like, I used to just bury 475 high bar for five, six, seven reps, RPE6, you know, an hour of 460, 470 and everyone else like, Mike, when are you going to compete? I don't compete. It's not like a fire that's within me. So that may be some of the difference now, but I just have other shit going on, man. I got a lot of stresses, a lot of things to take care of, a lot of people to take care of. Every employee and I'm so lucky to have these guys on my team, but that's just like another stress to my plate. Like I got to make sure they eat. I got to make sure everybody's doing good and we're all progressing. And that used to be where I would just wake up and worry about my deadlift. My whole mindset would be, oh, what do I got a deadlift today?   So it just takes the mental toll on me a little bit more than the energy I want to throw at the lifting. I wasn't like super excited to get into a deadlift strength blog, but I knew I needed a break from dieting. And so it was kind of an excuse. I'm still excited to like hit a big deadlift and I think we will. You know, anything over 670 is going to be a win in my book conventional. But after that, we're going to start bodybuilding cut mode. So we're going to go 30 day series, maybe two to three videos a week, more vlog style showing more of my lifestyle on how much fat I can lose in about 30 days.  I'm not going to be throwing needles in my arms. We're just going to clean everything up, add a little cardio and see visually and physically what I can do in 30 days. Hopefully you guys tie along with that. That should start in about two, three weeks. So we'll start that off with a new year and then we'll see what happens from there. Probably a little bit of travel. We're heading to the Arnold, I believe. So more vlogs coming in that, but the big series is 30 day, get sure as we might. Question for future avi, but today's avi is going to answer it and then I need you guys to answer it in the comments. So let's say whatever year we're in, let's say 2024 is what I'm predicting. That 500 pound deadlift is just in the pocket, right? It's maybe an RPE 8. You ain't got to overthink it. You've hit it before. You've hit a 500 pound, maybe 501 in comp, right? So you're smashing 500 pounds, squat, bench, deadlift. I don't care, but for you deadlift, for me deadlift, 500 or even less. Let's say you pull 550, right? So 500 is just a day, another day. So today I got six by two. It was supposed to be 500 pounds, but in my head, I'm not throwing the two and a half on, right? Why not? I've been there. Well, why? I've done been there before. But if you've prescribed it, you just have to have a voice across. Why not? Okay, let's just say you're supposed to be around 500. Let's not go, because I agree if the coach says you do it, but let's say you're just supposed to hit something around 500 or it says like 490 to 505 because I feel like some people got to throw that chip on there for the ego. My ego done been there, right? So who cares? Who cares? I don't care either. But some guys will be like, because I'll post it and I'll say 500, 6 by 2, like 495. Oh, that's different too, though. You're being up another topic.  I'm going to move this so fast. So I said 319 today.  You're like 319. But how did I say it? You mean 319?  That's why I was saying it with the sarcasm, because I would have said 3, whatever, 2. You know what I mean? I gave it a voice. Yeah, I would have said 320 also because you squatted 320. I posted it as 319. I didn't hit like 320. I didn't round up. But when I was talking to Mike, I'm like, I know he doesn't care. Someone's saying it's 320. And then I came with sass. I don't care. I came with sass. Because if you smoked it, it's not like if we would have added a pound on your squat, you would have missed it.  That was 495. Because if 500 was on there, I would have died. That's how my brain works. A window with RPE based. Yeah. Can you round up? I don't know. I think you can round up if you're sub-RPE 9.   I'll do that. I'll whip this around and say 315, 6x2. Yeah, you should do that. Easy day. All right, I'll do it. I'll do it right now. Just a little advice. Hopefully some of that makes sense or gets you unstuck in this bullshit, man. New year, new me or something? Something like that. Whatever the kids say. I'm going to finish this workout. I appreciate you guys, man. New videos Monday, Thursday. And we'll see what Seabass got in the works. We might have some shorts on the way. We might be adding a little more frequency to the channel, a little bit more love. So I appreciate you guys a lot, man. We over me. Be a part of something bigger than yourself. So I'm Mike. Catch you in the next one.