So, we are coming to the last leg of the day. So, this is the last session which is the tutorial session. So, these are the problems I am sure you have all these problems with you. No, I do not intend them. So, I do not intend to flash the problems. So, I am sure all the coordinators have given the problem sets which was uploaded in the morning one set with the each of the participants I am sure about that. So, we had a question between 3 to 330 whether we should be collecting the tutorials and correcting and all. So, we had internal discussion see the way we are planning to conduct the tutorial is that we will give first 15, 20 minutes to solve may be let us say today we have total 10 problems. So, we have 1 and a half hours. So, 1 and a half hours means we can take may be 3 problems in every 15 minutes roughly, but first 15 minutes we will give you time. So, that you can sit down and solve each of these problems. So, first 15 minutes you can solve and then we will perhaps solve problem 1, 2 and 3. So, let us see this way how many we solve. We do not plan to solve all 10 problems today. Yeah and another thing today being the first day we are given 10 problems because most of them you will see are very very straight forward direct application of formulae, but later on tomorrow onwards we will give about 5 to 6 depending on the difficulty level and the time involvement for each of the problems. So, as we go further and further in the course especially problems on heat exchangers you will have 2 or 3 in the given 1 and a half hours time. So, but today things being very straight forward we will try to I mean we would expect you to solve all 10 of them in the allotted time. So, for the next 15 minutes I would request all the coordinators to insist all the participants to solve the problems. So, for next 15 minutes now may be 10 minutes we will take questions from various centers if they have any. So, may be we are going to Bharamati. Yeah, if I remember some centers participants started the quiz that around 340 they will be doing it till about 4 10, 4 15 those guys can join the tutorial session after that, but beyond that let us not discuss anything related to the quiz today. We will have a discussion on the quiz and the solution tomorrow sometime no discussion on the quiz no questions on the quiz today ok. So, over to Bharamati VP COE Bharamati. Ok sir my question is regarding the fins part how to decide the pitch of the fins or design point of fins. So, that is a good question actually see in the previous the question is how to decide the pitch of the fins. So, if we take our problem which we had solved earlier. So, here the pitch was given. So, what we are saying is that see in the previous problem what we said what is the pitch given the pitch given is given to be 3 mm. So, if you ask me how does this pitch affect in the pitch on the efficiency of the fin if you see in the problem in the efficiency the pitch is not coming anywhere, but the pitch how will it decide pitch is decided on the base of on the basis of how can I place. See here if we see the pitch is around 3 mm the pitch is around 3 mm, but this is a very difficult question to answer directly. So, why I say very difficult to answer directly because if I decrease the pitch very close to each other then I am impeding the natural convection within this space. So, my space has to be reasonably large so that the natural convection is aided, but how will I know that whether the heat transfer coefficient is going to get affected or not. So, it is quite difficult to answer that question whether I can just by simple calculations what is the optimal pitch, but if you see as far as the efficiency is concerned it is not dependent on the pitch at all as you can see here in any of these parameters the efficiency is not dependent on the pitch, but the pitch will affect only the heat transfer coefficient. So, we can say if we assume for a minute that heat transfer coefficient is not going to be dependent on the spacing we can it will be decided on the basis of the fabricational difficulty. Here we can see it is 3 mm I cannot get closer than 3 mm because I need to have access either for soldering or welding or any other fabrication procedure. So, that would be the answer for your question. One other thing probably is the pressure drop associated if your pitch becomes very small then your calculated delta P would be much larger and remember velocity square is there and actually I think professor Prabhu can correct or add typically what is done is when you put these enhance surfaces or these geometries for improving the heat transfer you take the ratio of the new heat transfer by the old heat transfer divided by new pressure drop by old pressure drop and that ratio should come out to be some number greater than 2 or something like that which will justify the use of fins or extended surfaces even though there is a pressure drop related penalty. So, basically what we are saying is that the heat transfer coefficient has to be taken along with the pumping power when you decrease the pitch as professor said the pumping power is also going to increase. So, we have to take the pressure drop and the heat transfer coefficient together that will actually that is an indirect way from these calculations it is not going to get decided that has to come independently. I hope you have answered your question. Yes, sir. Thank you. Okay. Over to MK Triple Spune. Good afternoon. In case of two dimensional heat conduction which you had explained yesterday if all the sides of the plate are maintained at constant temperatures which are in general different what can we say about the corners because at the corner I suspect that there might be some sort of mismatch between the two temperatures coming from both sides. Okay, that is a mathematical issue, but physically it is not possible to maintain at the interface all of a sudden temperature jump it has to gradually change. If I have an interface one wall is at 40 degree Celsius and the other one at 150 how can I have all of a sudden 40 become 150 that is not possible. So, it has to gradually change it can be mathematically handled, but not physically. So, that is all I would say definitely as you said at the corners there has to be tapering down. You will have a tapering. Tapering down. Definitely the answer is definitely there will be tapering at the interfaces that is at the corners. Yes. Okay. Thank you. Over. Bhopal. May 90 Bhopal. Over to you. Good evening sir. This is Dr. Suresh from NIT Bhopal. So, sir actually I have one application question regarding TGA and TGTA. Once you are going for find out the temperature exothermic or endothermic reaction. Okay. In that normally when TGA, TDA we will use to advise that minimum amount of material minimum amount of material to minimize the heat transfer effect. We have to take minimum quantity of material to analysis that TGA, TDA the process. So, what is the minimum amount of material to be used? Can you suggest? What is TGA and what is TDA? Over to you. Sir, it is in thermogravimetric analysis. DTA is differential thermogravimetric analysis in which temperature we use a material to analyze its characteristics. So, when we use at various heating rate and with different material quantity the curve is differing. So, it is advised to use minimum amount of material and minimum heating rate to avoid the heat transfer problems. So, that the material can transfer better. There will not be any difference in temperature. Okay. So, can we find what is the method we have to put it over there to solve this problem? Okay. Now, the question asked is for all participants for the benefit of all participants there is something called thermogravity, gramimetric analysis. In thermogravimetric analysis one would try to measure the constants that is the Arrhenius constants in a first order reaction in any exothermic or endothermic reaction. So, the question asked is how much amount of the mass has to be put in a thermogravimetric analyzer. So, that my results are independent of the mass. Honestly, I have no answer for this because I do not know I have never used thermogravimetric analysis. However, here we have a professor, Professor Kananayar. We will get back to him. We will ask him this question and get back with the answer. I would request you to please put this question in the forum with proper formulation. So, that we will definitely answer this question. This being very specific only you would be interested in this. So, we will get back to you with the answer in the forum. Okay. Surely, we will put the in moodle, sir. Over. Question over. Thank you. Okay. We will get started with the, okay. So, let me start off with tutorial problem one. The question asked is tutorial one this is that is first problem. So, let us take what is given here is find. Let us write find the problem number one. So, what is given? T1 is given to be 400. I am saying find in each of the case what we have to find is the heat flux. We need to find heat flux and its direction. In the coordinate system what is being shown and also in addition we need to find dt by dx change over of 10. So, that is what we are saying is what is to be found. We need to find the heat flux okay and temperature gradient dt by dx dt by dx we need to find for each case. So, what are the assumptions involved? One is steady state, second is known properties. All properties are constant that is thermal conductivity all are constant. In this case properties means only thermal conductivity and it is 1D conduction it is 1D conduction and there is no volumetric heat generation. It is not an assumption because it is not given in the problem but still nevertheless we will state that. So, schematic here is given in each case. So, let me take the first schematic that is let me draw that T1, T2. So, what is known? Known is given to be thermal conductivity of the material is given to be 100 watt per meter Kelvin and thickness is the thickness of this is given to be thickness is given to be 100 m and I have T1 that is T1 is equal to 400 Kelvin and T2 is equal to 600 Kelvin. So, these are these are the things which I know. So, I am switching over now to another notebook. Let me rewrite this dt by dx. So, what is known? Let me rewrite this quickly known is T1 is 400, T2 is 400 Kelvin and T2 is 600 Kelvin and thickness is 100 mm that is 10 centimeters. It is little smaller than smaller scale and the figure is given what is to be found out is heat flux and dt by dx. So, let me first take there are three situations. Let me take the first situation this is temperature the y axis and the x axis is x and given temperature is T1 and T2. So, the temperature is increasing with the increase of x. So, T1 is 400 T2 is 600 this is x equal to 0 and this is x equal to L. So, first let us find in each case dt by dx. So, dt by dx equal to T1 minus T2 upon x1 minus x2 that is x1 minus x2 that is 0 minus L or you can write this as T2 minus T1 upon L minus 0 either ways. So, T1 is we have already found T1 is we have been given 400 minus 600 divided by L is given to be 100 mm that is 0.1 minus 0 minus 0.1. So, numerator happens to be minus 200 minus 200 upon minus 0.1 that means 2000 Kelvin per meter 2000 Kelvin per meter that is dt by dx is positive number 1 and number 2 it is 2000 Kelvin per meter. So, that answers dt by dx, but now what is heat flux? Heat flux is given by minus k dt by dx straight forward there is no question here. So, k is minus is there as it is k is 100 Watt per it is a good idea actually in few textbooks they put units even while substituting it is a good idea if we can substitute nothing like it Watt per meter Kelvin into dt by dx is 2000 Kelvin per meter. So, Kelvin Kelvin gets cancelled out. So, I end up with minus 2 into 10 to the power of 5 Watt per meter square. What does this mean? Note here it is 200 kilo Watt per meter square number 1 quite a huge heat flux and if you have to give the feel of the heat flux generally the heat flux is what we come across in laptops and all are of the order of 10000 Watt per meter square. So, this is quite high compared to that now negative sign negative sign means what? It is moving in the negative x direction. So, heat transfer is in heat transfer is in negative x direction. If you are not careful anywhere we are most likely to go wrong. So, what is that we need to do is systematically do dt by dx and then take minus k dt by dx according to coordinate system that is all. So, there is nothing difficult here. Just to add if you look at the sketch itself in the positive direction t 2 is higher than t 1. So, the gradient it has to be positive. So, that sign 2000 with a positive sign itself is a correct answer looking at this low. So, you have an acute angle in the positive x direction. So, mathematically also without knowing heat transfer you can say the slope has to be positive here. That is temperature is increasing with the increase of x. Yeah. So, and then k minus k dt by dx has a negative sign associated with it. So, there will be a minus sign in the answer for heat flux and obviously, heat is going to flow from right to left and that is the physics will tell us because higher temperature to lower temperature. So, if we correlate physics mathematics everything I do not think these things are any difficult. Let us take up next part that is next part we are most likely to get confused. So, let us take that. So, I will do only the second one third one I will skip. So, that is this is temperature t of x this is t 1 t 2, but my x direction unlike the last problem it is in this direction. So, again dt by dx equal to 0 and L. So, dt by dx equal to t 2 minus t 1 upon 0 minus L there is no confusion on this or t 1 minus t 2 upon L minus 0. So, what is t 2 we have 600 minus 400 upon minus 0.1. So, that means I get minus 2000 watt per meter. That means temperature is decreasing with the increase of x I think it is right that is what we have got dt by it is indeed physically that is what is happening temperature is decreasing with the increase. So, I have made a mistake here here Sameer and others are correcting me. So, it is Kelvin per meter you are right. So, that is 2000 Kelvin per meter. So, that is temperature is decreasing with the increase of length x. So, that is physically realizable. So, q double dash is equal to minus k dt by dx. So, what is minus k minus 100 into dt by dx is minus 2000. So, the unit of k is watt per meter Kelvin and this is Kelvin per meter. So, Kelvin Kelvin gets cancelled out. So, I get plus 2 into 10 to the power of 5 watts per meter square. So, you notice here positive sign naturally the heat flux is moving in the positive x direction because it is moving from higher temperature to lower temperature. So, no matter however it is twisted and given now I think you can handle this all that we are saying is find out dt by dx plug in q double dash equal to minus k dt by dx. So, now whichever way it is twisted and given to us we can now simply sit down and calculate it. So, that is why I am skipping the third part ok over to problem 2. What does that problem 2 say? An electrical resistance heater is embedded in a long cylinder of diameter 30 mm when water at a temperature of 25 degree Celsius and velocity 1 meter per second flows crosswise over the cylinder. The power per unit length required to maintain the surface at a uniform temperature of 90 degree Celsius is 28 kilowatt per meter. So, what is given? The cylinder is there let me start sketching. So, let me just say schematic. So, I have a small cylinder that is electrical resistance wire heater I am imagining as a cylinder and there is a flow taking place across the cylinder. And the velocity of the flow is 1 meter per second velocity is equal to 1 meter per second that is known is diameter of the cylinder is 30 mm and T infinity is 25 and velocity is given to be 1 meter per second. And the wire is being maintained at a temperature of wire temperature is being maintained at let me call that is T wall at 90 degree Celsius. And the power generated is per unit length that is I would call that as Q dash is 28 kilowatts per meter. So, when air also at 25 degree Celsius, but with a velocity of 10 meters per second the power this is at 1 meter per second. But when air also at 25 degree Celsius, but with a velocity of 10 meters per second is flowing then the power per unit length this is water temperature is at 25 degree Celsius. So, but in case of in the second case I have air in second case I have air whose velocity is 10 meters per second, but the power I can give is quite high as you can see no quite less in case of air all the velocity is 10 you see in the earlier case it was 28 kilowatt that is 28000, but in this case we are able to give only 400 watt per meter. So, why we can give only 400 watt per meter because my wire will burn out. So, because it is air so air convective heat transfer coefficient is going to be less compared to the tough water let us see whether that does that happen or not. So, here essentially what we are checking is we are trying to check in this problem whether we have understood how to apply Newton's law of cooling. So, Q equal to h A into Q equal to h A into T wall minus T infinity. So, h equal to which implies h equal to Q upon area into T wall minus T infinity. What I have here is this is equal to Q dash L upon Q dash L upon pi into D into L note I am not taking pi D squared by 4 I am taking pi D L this is because it is surface area. Professor Bajan calls this as bathing area. So, it is getting bathed my cylinder is getting bathed. So, pi D L into T wall minus T infinity. So, L gets cancelled out I do not need to know L. So, rest is substitution now for the case of water let me calculate that that is h equal to Q dash is 400 sorry 28000 upon pi into diameter is 30 mm 30 into 10 to the power of minus 3 into T wall is 90 minus T infinity is 25. If you just press calculators we are going to get this as 4.57 kilo watts per meter square Kelvin or 4570 watts per meter square Kelvin remember that this is in thousands. So, now let us do the same problem for air that is which is at 10 meters per second and Q dash is at 400 watts per meter. So, that is we have h equal to Q dash upon pi D into T wall minus T infinity. In this case my Q dash is given to be let me just take Q dash is given to be 400. So, 400 upon pi into my diameter of the wire is 30 mm 30 into 10 to the power of minus 3 and T wall is 90 minus 25. So, if I substitute that I get h of air as 65.3 watt per meter square Kelvin. You see h of water we got 4570 watts per meter square Kelvin, but h of air is found to be 65.3. So, as we go along in convection we will see that this is very natural. So, for water it has to heat carrying capacity of water is very high. So, we should be able to take away more heat in case of water as opposed to that of air. So, we will see that it is actually dependent on Prandtl number higher the Prandtl number higher the heat transfer coefficient. So, for now it would suffice to say that h of heat transfer coefficient of water is always greater than heat transfer coefficient of air. In fact, this is how one measures the heat transfer coefficient also. So, the next question was calculate and compare the convective. So, we have answered all questions. So, all that we are saying is h of water is greater than h of air. So, now let us move on to the third problem. So, what does the third problem say? Problem number 3 that is we have a surface of an area what is known we have a surface of an area 0.5 meter square emissivity that is it is a perfect black body. So, we do not have to worry emissivity is 1. Let us go ahead and write it as 1 and the temperature of the body is 150 degree Celsius. So, one important thing whenever we handle radiation is that we need to convert degree Celsius into Kelvin. So, that is the safest way because in radiation we handle in our Stefan Boltzmann law sigma t to the power of 4 t is in Kelvin. So, I have to convert this temperature to Kelvin. So, 150 plus 273 we get 423 Kelvin and evacuated chamber whose walls are maintained at 25 degree Celsius surface area and the temperature is placed in a large evacuated chamber. Chamber wall temperature is found to be 25 degree Celsius this is the body temperature T s. So, I would say chamber I would call this a surrounding temperature that is 25 degree Celsius that is 25 plus 273 equal to 298 Kelvin. What is the rate at which the radiation is emitted by the surface? So, what is the radiation emitted by the surface? We will just take the find we will take first that is not coming out find that is net energy emitted q emitted that is equal to sigma area of the body into T s to the power of 4 emissivity emissivity is there. So, but here emissivity is 1. So, emissivity is 1 sigma is 5.67 into 10 to the power of minus 8 into area is 0.5 into T s is 423 to the power of. So, I get q emitted equal to 907.64 watts. So, now q net what is q net that is the next part of the question is what is the net rate at which the radiation is exchanged between the surface and chamber walls that is q net is equal to q emitted minus q received. So, that will be equal to sigma A into T s to the power of 4 minus T surrounding to the power of 4. So, if I substitute for sigma 5.67 into 10 to the power of minus 8 into 0.5 into T s is 423 point to the power of 4 minus 298 to the power of 4 I am going to get 684.07 watts naturally this number has to be lower than that of emitted. So, I do not think any comment we can add here just for the simple reason all that I want to carry the message by solving this problem is that temperature should be in Kelvin for radiation this is the point we want to emphasize. So, that we do not commit mistake because in Stefan Boltzmann law T s is in Kelvin T s is in Kelvin. So, what we will do is next 10 minutes we will take questions either on these problems what we have solved or any other questions which we have on the topics which we have covered so far that is now it is 10 minutes to 5 till 5 we will take questions before we start taking the solutions from problem number 4. NIT Trichy any questions? Sir, I have one doubt related to nanoscale and microscale heat transfer. Okay, you ask it? Sir, is it applicable this law? Sir, 4 year law is applicable for nanoscale and microscale. I have no answer for this question. The question asked is is Fourier law of conduction applicable for nanoscale and microscale on the face of it? Yes. In fact, if you see incorpora and David fourth edition not fourth edition sorry sixth edition which is come out the recent edition which has come out there is a problem in which the nanoscale problem has been solved. So, actually there also again parallel circuits are used and again resistance analogy is used and Fourier law of conduction is used. In principle yes even for microscale and nanoscale one can use Fourier law of conduction. So, there is no doubt about that but if you ask me whether thermal conductivity is going to be constant or is it going to be varying I have no answer for that but for usually they are all made of silicon, silicon carbide. So, silicon carbide thermal conductivity is very well documented. The answer is in simple terms even for microscale and nanoscale Fourier law of conduction is still valid with may be some exceptions but I am not aware of those exceptions but in general yes it can be applied. What is the advantage of this micro fins over the conventional fins? The question asked is by one of the participant what is the advantage of micro fins over conventional fins? Micro fins are indeed conventional, there is nothing like very unconventional about micro fins. Since ages micro fins have been used in refrigerating circuits for evaporators and condensers. So, micro fins is indeed conventional but micro fins are used on the inner side of the tube not on the outer side usually. Why because micro fins essentially now it is little early now that you have asked the question I have to answer it. Micro fins actually break the laminar sub layer and make the boundary layer completely turbulent which will enhance the heat transfer coefficient and decrease the convective resistance on the inner side of the wall. So, micro fin tubes are very much used in refrigeration circuits. Sir good evening sir another question sir. So, yesterday in refrigeration for suction line we have to provide insulation and the answer given is that for the compressor the entry to be liquid so that it should be insulated but normally it is a vapour sir I do not know whether sir that point please explain sir. See one of the participant is asking the question that yesterday she is referring back to one of the questions yesterday what we had answered that is the compressor inlet is usually he is claiming that it is vapour it is inside yes the inlet is vapour and by the time it gets out it is going to be superheated. So, yes it is vapour did we say that it is going to be liquid. So, no if we have told that it is liquid we stand corrected we are wrong yesterday if you have told that it is liquid we are wrong yes at the inlet it is going to be little lower than the saturated but it will be saturated at least it will be saturated liquid and as it is coming out of the compressor it is going to be vapour. Now, the answer given to that question was why is the insulating medium given only on the suction side why not on the pressure side honestly I know answer for that may be professor Arun can chip in and answer that. Probably because suction line is the difference between the temperature T infinity or the room temperature and the suction line temperature is substantial meaning 30 40 degrees but for the discharge line the temperature of the fluid is almost close to the surrounding temperature usually because it is at the higher temperature compressor inlet typically is at lower than atmospheric temperature. So, the chances of heat coming in from the atmosphere are high when the delta T is large when the delta T is small the amount of heat transfer is negligible is small therefore you do not really care about insulating the discharge side probably that is the better explanation. Sir, another question shall I ask sir. Go ahead. Sir, another question from the refrigerator itself in cold climate in cold climates the condenser is exposed to the atmosphere whether there is any heat load more heat load given to the compressor sir. The question asked is by one of the participants in cold climate condenser is exposed to atmosphere is there any additional heat load from the ambient to the condenser. So, condenser condenser is generally sitting at a lower temperature at a higher temperature but then if it so the condenser usually is sitting at it is operating it is taking the fluid the refrigerant to the condenser is coming at a higher temperature. Higher temperature and pressure. Higher temperature and pressure ok we can perhaps understand like this. If I imagine a tube in tube condenser inside the tube let us say we have the refrigerant going at a higher temperature but that has to be cooled by chilled fluid which is which is to be at a lower temperature. So, which is already sitting at a lower temperature that is why perhaps it is exposed to the atmosphere that is all I can say but I do not think we can generalize this because it depends on refrigerant to refrigerant if it is R 12 or R 22 I do not think we can afford to expose. But if it is R 134A yes my chiller temperatures are going to be sufficiently low they are almost negative side then perhaps I can expose it to atmosphere provided my atmosphere is also sitting at negative but if my atmosphere is again sitting at higher temperature then there will be heat loss. So, I would not like to expose the condenser heat gain. So, I would not like to have the heat gain from the ambient to the condenser. So, I think the condenser is exposed as you said in the cold condition only for those refrigerants where on the chiller temperature, chiller fluid temperature and the ambient temperature are of the same order that is all I can think of as the answer ok. Thank you. Kolhapur KIT Kolhapur any questions please. Yes sir there is no question when we are required to use the concept of corrected length of fin that was not touched in that session. The question asked is let me let me rephrase the question asked is by one of the participants what is the corrected when do we use the corrected length of the fin. So, when do we use the corrected length of the fin ok ok. So, we understood your question see corrected length is taken here already for example, of course that figure is not there let me see. Yeah, yeah here it is there you see this zeta is already the corrected length L plus t by 2 into square root of h by k t is already the corrected length. In fact, I do not think it can be called as corrected length it is. It is opening the fin it is cutting the fin and opening it into two flat surfaces so that you take care of the rim that is what it is. That is all it is ok. Thank you. Sir one more question. Please go ahead. We have solved problem number 3 in that we have taken that areas surface area A is located placed in an evacuated chamber. Does the heat transfer rate or net heat transfer rate heat transfer will depend upon how that area is placed in the chamber. One of the major assumptions in that problem is that the object is very very small compared to the chamber. So, wherever you place it however you place it all the energy from the chamber walls are coming to the surface. So, till we do radiation heat transfer we do not we do not calculate anything called as view factors at present. We just use the actually it is heat given by the object minus heat received by the object. So, if I have to write heat received by the object it has to be sigma epsilon t surrounding to the power 4 times a view factor associated between the room and the object. So, that factor and all the area multiplication etcetera there is something called as reciprocity rule which ultimately gives you this formula. So, inherent assumption is that the object is very very small compared to the overall dimensions. So, no matter how the object is oriented it is not going to affect the final heat transfer rate. Thank you. Thank you. Over to J N T U Hyderabad any questions? J N T U Hyderabad. The problem is all no there the velocity of water and velocity of air is given. The problem can be we can solve by external flows no. Bulk mean temperature T F is equal to wall temperature average of T W plus T infinity. Sorry to interrupt. So, meanwhile so in order to solve the second problem Reynolds number and Prandtl number we have to calculate no. I would just like to rephrase one participant is asking that in problem number 2 that was taken up. We can calculate Reynolds number Prandtl number etcetera and try to calculate the heat transfer coefficient. Well that we will do once we cover convection. Right now we are at a stage where we just know Newton's law of cooling. We do not know convective heat transfer we are just applying this equation. Later on when we are going to do convective heat transfer such kind of problems obviously will involve computation of H naught by in that case Q will not be given to you. In fact, Q will have to be calculated by you for the wire basically what you would be asked is how much is the heat removal rate if air at 25 meter or 10 meter per second and 25 degree centigrade flows and how much is the heat removal rate when water at 25 degree centigrade and 10 meter per second flows. So, you would not be given Q in that case you would be asked to calculate Q for which the first step would be to calculate the heat transfer coefficient. This is a beautiful problem I would say why beautiful because this is a method of measuring convective heat transfer coefficient. If one is trying to measure the convective heat transfer coefficient this is the method I take a wire put it in water go on increasing the or for a given fluid velocity I will go on increasing the input power that is voltage and current what I am dumping into the wire until it burns out. So, this is this is a very elegant problem in the sense that this is a this is showing me the method to measure the heat transfer coefficient cylinder in case of external flows. Yes, we are going to do that in convection when we do the questions what you asked. Thank you. One more question sir. Sir, in heat exchanger by inserting pin pins we can increase the heat transfer coefficient so the pin pin analysis can be done either analysis can be done with the help of effectiveness or efficiency sir. See question one of the questions asked by the participants is pin pins can be used in heat exchangers whether efficiency or effectiveness is to be used for evaluating the pin pin performance. Pin pins generally pin pins generally in the they are not used under convective conditions. So, if I have to answer this question little more detail I will have to use the draw something. So, generally pin pin cooling where it is being used. So, it is not being used in convective condition if I take a gas turbine blade this is my gas turbine blade which is there in all aircraft gas turbines or fighter aircraft so here in this portion you use convective cooling that is the fluid goes and comes back it is almost like my hand this is the size also will be roughly around 6 centimeters. So, we put fins throughout here the fins are put throughout the pin fins are put here why because the space here is very less the space here is very less. So, instead of convective cooling I cannot make a passage and have convective cooling. So, in order to have effective heat transfer I put fins here vertically sorry I drew it wrong. So, vertically from top to bottom there will be pin fins if I redraw this portion again. So, there will be fins throughout here one fin the next one fin. So, like that pin fins are going to be there. So, pin fins why are we using this here pin fins because number one there is no space for convective cooling passage. Number two here by simple conduction itself I can effectively remove the heat transfer because the thickness is very small. So, pin fins generally are not used in convective cooling passages that would be the answer for your question. We will take up problem four onwards for now ok. Yeah, I think there is one more question. In the morning we had one problem relating to finding out the length of the fin. We took the condition adiabatic fin tip is equal to infinitely long fin heat transfer rate and we have equated it to find length. So, why did we do that? We considered these boundary conditions for this and we found the length of the fin. Yeah, the question asked is in one of the problems in the morning we took here estimate how long the rods must be there. Of course, in the problem itself it is stated for the assumption of infinite length. Even if it is not stated we would go ahead with this infinite length. See why this is taken as infinite length? By common sense one would think that as I go on increasing the fin length the heat transfer rate will increase that is heat transfer rate that is number of wattage is the amount the wattage will increase with the increase the length, but then that in real life it is not so we solve the problem. In the problem what did we see? Again we have shown this table several times nevertheless I would like to show that table again that is we have plotted M L versus tan hyperbolic length that is infinite length is infinite only for mathematician. How infinite is infinite for an engineer is decided by tan hyperbolic l? I can decide only when I take the infinite boundary condition. So, that is the reason why we have taken infinite boundary condition. I think that is the only logical answer I can look at I can give for this question. Another thing was I mentioned in the morning as the fin length keeps on increasing the delta t available for heat transfer keeps on decreasing. So, ultimately an infinitely long fin would essentially have a tip which is equivalent to a location where t b is almost equal to t infinity. So, that there is no heat transfer there and that is why we are saying that in the limiting condition we can say an infinitely long fin at the tip has a boundary condition which is equivalent to the adiabatic boundary condition that was the reason for taking this equality infinite length. Because we need to get a length. So, to get a length we need to know what is the heat transfer associated if in the limiting case of this infinite fin it behaves as an adiabatic tip condition and that is logical. Thank you. We are going to problem number 4. Problem number 4 says it is radioactive waste are packed in a long thin walled cylindrical container and the waste generate some thermal energy non uniformly according to some equation which is given it is a parabolic equation in r and q dot or that is a volumetric heat generation rate. This is one of the cases where definitely you will have volumetric heat generation and that is a non uniform volumetric heat generation. All through when we did various derivations we said volumetric heat generation is uniform. This is one of the situations where it is non uniform and the equation is given as a function of the radius of the cylinder. Steady state conditions are maintained by submerging the container in a liquid that is a T infinity and provides a uniform heat transfer coefficient h. We are asked to find the total rate at which energy is generated per unit length and from this to get an expression for the surface temperature at the container wall. So, assumptions in this actually what is there in this problem if I say given we are given a cylindrical container which has some radioactive waste. So, they are being transported let us say. So, there is q which is a function of r volumetric heat generation that is given to us. So, q dot of r is given to you as q dot 1 minus r by r naught whole square. So, this is given to us and q naught is a known positive constant. This to keep things cool this is being surrounded by a fluid which is providing your heat transfer coefficient h and the fluid is at T infinity. So, you are asked to find expression for rate at which energy is generated per unit length. Please note q is the heat transfer rate q dot we call as volumetric heat generation watt per meter cube. In nuclear reactor parlance this will be called as q triple prime triple prime corresponding to per unit volume, but here heat transfer textbooks most of them we use q dot q double prime logically is heat flux watt per meter square q prime is the linear heat rate watt per meter. So, logically this should be q triple prime instead of q dot, but anyway we stick with this nomenclature. So, we are asked to find q prime linear heat rate and also obtain an expression for p surface. So, bunch of assumptions will be there. So, first assumption is of course, steady state this we have not dealt any unsteady state related problem. First is steady state given uniform heat transfer coefficient third thing T infinity remains constant. So, you are adding heat means nuclear reactor waste is giving away heat to the surrounding fluid, but it is a sink in thermodynamics we have studied sink temperature will not change. So, T infinity remains constant otherwise we have a problem that is one thing then yeah. So, if I look at the solution it is just a mathematical exercise we want to calculate the volumetric heat generation per unit length. So, let us first do it per unit volume we can write q dot r d v this is going to give me watt per meter cube times meter cube which is going to be watts and d v for a cylinder volume of a cylinder is pi r square l. So, d v is going to be 2 pi r d r times l. So, this l is what is going to come down.  So, q by l which is q by l which is given by volume q naught 1 minus r by r naught whole squared 2 pi r d r this l has already been brought down and the limits of this integration I will replace this by r is equal to 0 to r is equal to r naught that is where the problem ends we can pull out q naught 2 pi out of the integral you have 1 minus r squared by r naught squared multiplied by r d r integrated from 0 to r naught this is going to give me 2 pi q naught integral of r d r would be r squared by 2 integral of r cube this is to be multiplied r cubed is r to the power 4 by 4 r naught squared applying limits 0 to r naught and that is going to just give me q dot by l equal to 2 pi q naught if I apply the limits this will be r naught square by 2 minus r naught so this is going to give me r naught squared by 4. So, this will be r naught squared by 2 minus r naught squared by 4 would be r naught squared by 4 this 2 cancels with this 4 will get me 2 the answer is pi q naught r naught squared by 2. Let us check whether it is correct in terms of units q naught would have been in watt per meter cube times radius square which would be meter squared answer has to be watt per meter which is what we require. So, this is the first part of the problem second part is very straight forward you are asked to find the surface temperature. So, I have to do an energy balance. So, if I take the cylinder heat is being generated e dot g this has to go out by convection e dot out and my favorite equation is there e dot in minus e dot out plus e dot generated is equal to e dot stored steady state this goes to 0 nothing is coming to the control volume e dot in is 0. So, e dot out is equal to e dot g e dot g we already had.  So, I will get h times pi r naught 2 times h pi r naught t s minus t infinity is nothing but q by l which I substitute from the first part as pi by 2 q naught r naught square pi pi r naught r naught square. So, therefore, I get t s is equal to t infinity plus q naught r naught by 4 h that is the answer this is h. So, it is a very good exercise in book keeping and of course, mathematics simple integration and more importantly application of this conservation equation e dot in minus e dot out plus e dot generated equal to e dot stored. So, this equation we are applying after we have obtained an expression for e dot generated and just things can go wrong if you do not take care of the length properly. So, this is already q by l that is what we need to keep in mind. So, once that is done there is no issue about the problem. We are skipping problem 5, 6, 8, 9 and 10. 8, 9 and 10 are very very straight forward application of thermal resistance concept for insulation properties have been given one case you find k other other case find the heat transfer rate. So, I think all of us know these things. So, we will take up problem number 7 which is slightly lengthy it is not difficult it is slightly lengthy. So, that we try to appreciate these kind of problems also. So, what is given is in problem 7 I am doing problem number 7 of this tutorial. It is one dimensional steady state conduction with uniform internal heat generation occurs in a plane wall of thickness 50 mm. So, as I am reading I am going to write plane wall is given to us 1 d steady state conduction. So, this is one of the assumptions let me as I am writing the problem I am writing assumptions also in view of time one dimensional steady state it is a plane wall dimension is given thickness is 50 mm constant thermal conductivity k is constant k is equal to 5 watt per meter kelvin for these conditions temperature distribution is given to me in the form t x is equal to a plus b x plus c x square. And we are also given couple of other boundary conditions surface at x is equal to 0 t is equal to t naught is 120 degree centigrade and at x is equal to l we have insulated boundary condition which I want to translate mathematically as d t by d x at x is equal to l is equal to 0. We are also given another condition for x is equal to 0 that is the left phase that is a convective boundary conditions. So, at x is equal to 0 let me let me draw a schematic. So, this is the coordinate direction x 0 l there is k which is given to me h comma t infinity on this side. And this surface is at t naught which is specified to be as 120 degree centigrade and this is insulated wall. So, d t by d x is equal to 0 at x is equal to 0 we have h is equal to 500 watt per meter square kelvin and t infinity is 20 degree centigrade. So, this is the plane wall which is given to us whatever we supposed to find applying overall energy balance calculate the internal heat generation. So, find internal heat generation 1 and determine coefficients a b and c by applying the boundary conditions to the prescribed temperature distribution. So, for this problem you are supposed to find the temperature distribution and it is expected that it is a polynomial distribution as we have studied in the class. We want to find the coefficients a b and c. So, as we started we wrote couple of assumptions let us see we have any more 1 d steady state constant properties uniform h and t infinity so constant and uniform values and volumetric heat generation rate is uniform. So, that is also an assumption we make we do not have any direction or location specific volumetric heat generation. So, I have tried to fill as much as possible here anyway. So, now our we have to find the heat generation rate. So, overall energy balance is what we have to do overall energy balance for the entire wall. So, I take the entire wall as the region of interest and I do e dot in minus e dot out plus e dot generated equal to e dot stored steady state this goes to 0 e dot in is e dot in is 0 how do I know that I know that because t infinity is 20 t surface at x is equal to 0 at x is equal to 0 t surface is 120 degree centigrade. So, heat is not coming from the outside to the inside heat is going from the wall to the outside. So, in fact physically if I look at the problem the left this phase is insulated. So, nothing can go out from the right hand side phase heat is being generated. So, obviously heat has to go out if steady state conditions have to be maintained from right hand side to left hand side from the object to the fluid. So, this surface temperature obviously has to be higher than the free stream temperature. Therefore, if I take this control volume which is the entire available unit of the plane wall there is no e in it is only e dot generated equal to e dot out that is the balance. So, e dot generated is what I have to find that is equal to e dot out and that comes out to be q dot times volume which is a times l is equal to e dot out which is h times a t naught minus t infinity we use t naught because that is the notation we have used. So, areas get cancelled what is this area that I am talking about this area is the area into the plane of the paper. So, this area is what we are cancelling of this thickness is l which is given to us that we have. So, q dot times l is equal to h times t naught minus t infinity therefore, q dot volumetric heat generation rate comes out to be 500 times 120 minus 20 divided by 0.05 meters which is 10 to the power 6 watt per meter cube. So, this is the energy balance and this given as a volumetric heat generation. Now, you are asked to find what is the is a b and c associated we can I will just outline the steps for this you can solve this at home. So, we have to solve the differential equation with volumetric heat generation. So, 1 d steady state equation with volumetric heat generation is d square t by d x square plus q dot by k is equal to 0 and therefore, d t by d x comes out to be minus q dot x by k plus c 1 and t of x comes out to be minus q dot x square by k plus 2 k plus c 1 x plus c 2 this is known to us from class. Now, I will apply the boundary conditions which are there for the problem. So, t of x is a plus b x plus c x square.  So, d t by d x at the right hand wall at x is equal to 0 is 0 which comes out to be b plus 2 c l equal to 0 therefore, I get b is equal to minus 2 c l 1 relationship. Second, convective boundary condition at x is equal to 0 we get h times t naught minus t infinity is equal to minus k d t by d x at x is equal to 0. So, now here we have to be careful about the sign. So, d t by d x at x is equal to 0 let me evaluate at x is equal to 0 comes out to be b from here I substitute x is equal to 0 I get b h t naught minus t infinity everything is known k is known I get b is there a problem with this equation sign related. So, please take a look at this equation in the form of the sign that we are doing because x is equal to 0 is a left hand boundary it is going this way. So, we have to take care of the sign appropriately. So, left hand side is right. So, this one has to be taken care correctly. So, that d t by d x on the right hand side would be a plus value and I will get b appropriately. Next, I can substitute the overall into the temperature distribution I can get a from substituting for the temperature at the surface at t naught is equal to 120 degrees you get from here b you have got c from here t naught at x is equal to 0 substitute in that equation you will get a and that should explain the solution. So, a comes out to be 120 and b is correct b is equal to 10000 and c is equal to minus 1 lakh. So, this is what you would get. So, I know we went fast in the last problem, but straight forward energy balance related concepts only that nothing new or nothing difficult here, but see here we have to take care as I said about the sign because the direction is important. So, if you to the bookkeeping correctly I do not think there is going to be any problem. We urge all of you to do the remaining problems each of them should not take more than 5 or 10 minutes to do. Tomorrow morning before you come at 9 o clock before we start the session I request all coordinators to collect the tutorials and during the day before lunch just check if all the problems have been attempted the solutions of this tutorial will be available from tomorrow morning. So, we will we will upload the solutions the day after the tutorial is over. So, we urge all participants to spend the evening trying to finish this and we will meet tomorrow morning at 9 o clock where we will start transient conduction. Thank you. Any questions if any we can take yeah. Vijay T. I am Mumbai are you with us. Sir I have we have few questions here my first question is that while solving one problem in the morning it was found out that the product M into L was 2.65. Okay. Whereas when you showed one table in that table for for the infinite infinitely long fin you showed that M into L has to be 5. How is this discrepancy? There is. Okay. The question asked is in the problem solved in the morning ML in the problem it has been stated as 2.65 but in the table it is shown that ML of 5 is required. So, let us get back to the table I I think we have flashed the table N number of times nevertheless that shows the importance of the table. So, ML if we see ML variation with tan hyperbolic ML see what we are saying is that see for 2.65 for an ML of 2.65 what is the efficiency? Efficiency is around 0.99 already it is almost 100 percent for an engineer I think 0.99 is equal to 1. So, but then ML of 2.65 and ML of 5 the length of the fin for a given M that is HP by KAC for a given M length of the fin is almost double but I have not gained much in terms of heat transfer rate. So, that is the reason why we are saying in the problem that ML of 2.65 is enough for us I think that makes sense because ML of 2.65 is already reaching 0.995 times over to you. Okay. Have we clarified? Yeah. Sir, one more thing is that while you were solving a problem in the morning you said that K is equal to K 0 plus 80 that problem I am referring to what I understand from this discussion is that K is never then a constant in a real life problem is that correct? The question asked is K is never constant in a real life problem I do not think this statement is right see our operating temperature ranges are not going to be always from 0 to 1000 degree Celsius. So, they are limited for example if I am taking air let us say if I am taking the temperature between 30 to 50 I do not think temperature the thermal conductivity is going to vary significantly between 30 to 50 but let us say I take water and it is entering at 30 degree Celsius and getting out at 100 degree Celsius yes thermal conductivity is going to significantly vary. So, case by case basis the thermal conductivity variation is important but we cannot make a statement that thermal conductivity is always varying or thermal conductivity is always constant it depends on situation to situation and temperature range ok thank you we are ending this session for the day


Villanueva basketball fans were confident the university's men's team would defeat the Kansas J Hawks this weekend. But after a tough loss in New Orleans, the Wildcats came home empty-handed, missing out on bringing back another NCAA championship. Students and alumni back on campus were disappointed about the result, but proud of how far the Wildcats made it in this year's March Madness tournament. They started to give up a little bit when the gap began expanding, but at the same time I feel like they really pushed and they made it really far in the tournament. Looking forward to next year's season, fans are concerned about the team's performance without two outstanding seniors, Colin Gillespie and Jermaine Samuels, who play their last game on Saturday. I feel like I'm very optimistic about it, you know, but the reality is that two of our good seniors are leaving. Colin and Jermaine are graduating now officially, and I don't know if the other two will come back, the other two seniors, so I don't know. I don't know if it'll be the same. Winning two of the last five national championships in 2016 and then again in 2018, Villanova's men's basketball program proved itself to be a powerhouse on the highest levels of college basketball. So fans remain hopeful that the team will bounce back next year. Although the Villanova men's basketball team missed out on another NCAA championship this year, the Nova Nation is still optimistic in looking forward to cheering on the Wildcats during the upcoming season. I know what I got reporting for Butv.


I'm Travis Nielsen. I'm one of the original Rook Maintainers creators. We started this project back in 2016 and it's been a great journey and so glad to be here with you. Yeah, so let each of the panel members introduce themselves as well. I'm Blaine Gardner. I'm also with IBM. I've been a Rook Maintainer for four or five years now. A lot of focus on object store and some multi-stuff and NFS. Yeah, so... Okay, well, I'm Alex Nautros, Founding Engineer of Co-Technology Sync. I've been, I don't know, six, seven years, something around that, I guess nowadays. Maintainer of the project as well. Yeah, thanks. I'm Deepika. Hey, I'm Deepika Upadhyay. I work as a cloud storage engineer in Co-Technology Sync. And I used to work with Redos core engineering team for two or three years. And then I moved to Redos block device, worked there for one year. And now I'm transition to work on Rook itself. Yeah, happy to be here. Okay, let's get rolling here for our agenda today. So as a Rook panel, we're going to do things in a little different format. If you've been to a Rook talk before. Who has been to a Rook talk before? Just out of curiosity. A few people, okay. And maybe you've seen some of our recordings online too. We've done a few of these. But our goal today is really to familiarize you with storage for Kubernetes. What does Rook provide? What does Seth provide? And ultimately, the question I hope each of you asked yourselves is, is Rook a potential storage solution for you? Does it meet your needs? And we'd love to hear any feedback for how it's working for you and what you might think we could add to it. So at the time, at the end of the meeting, we hope to have time for your questions. But the format will be, we'll present, we came with a list of questions. We'll kind of jump around to a number of different topics. But hope all these questions and answers will be interesting. So if you don't get time for your questions, we do have a Rook booth in the Project Pavilion over in that Pavilion area on the far, far side of the conference. And hope to see you there. Okay, first question. So how should someone new to Kubernetes think about storage? Blaine? Yeah, I love starting with this question. I think like a lot of answers, it's two-parted. Kubernetes easily manages distributed applications. And for a user, for the user side, for an application, the storage should just be easy. I want, you know, if my application fails over, if it scales out, then that storage has to follow my pod. This has to be across nodes, across partitions. And that means for administrators, this like distributed aspect is pretty challenging. These are big shoes to fill. If I have external storage, who's going to manage that? If I'm using a cloud provider, am I locking myself in? Myself in? And if I'm going to run my own, can I handle that? So I think this really segues well into the rest of that presentation. Thanks, Blaine, for that introduction. So as we started to explore storage for Kubernetes early in the Kubernetes days, we had some questions that really led us to this project. First of all, as Blaine said, storage, I mean, we all have to worry about it. It's all in our mind. Storage is commonly provided by cloud providers, but if I'm not running in a cloud provider, what about storage in my own data center? And even if I am running in the cloud, there are certain limitations that we'd like to overcome. We'll talk more about that. And then at the end of the day, Kubernetes really doesn't treat storage as a first class citizen. It's external. You have CSI drivers to connect to it. But why not manage storage as any other Kubernetes application? Why shouldn't it be running natively as part of my Kubernetes cluster so I can manage it with the same tools that I'm managing other applications? And then the next question is obviously, well, which storage platform can we trust? Enterprises don't generally go full on board with a new data platform. They want a data platform that's been proven. It's been running in production for some time. Data is sensitive. It's valuable. We have to protect it. So we made the decision, first thing to build on Ceph, and we'll talk more about why we did that and what it provides. So what is Rook, Alexander? Yeah, so Rook, I think most people of you have heard about an operator, maybe I guess. Who knows what an operator is? I still have two slides or so to really quick catch everyone else up, but thanks. Yeah, so what is Rook? The idea in the end is with Ceph being, well, quite the complex system in itself as well. One might see it as even more complex as Kubernetes. Rook is basically there to make it at least easy to run and to some degree maintain. There are simply certain aspects as well to just catch that as well, which you can just have an operator take care of the certain settings in Ceph, the same with Kubernetes, there's certain tweaks you can make to make it more responsive, for example, to node failure. Failure, the same applies to Ceph from Rook as well. But the point is that we have custom objects for which we can easily, for example, upgrade your cluster if you just change the CRD with the new version. Operator sees that, takes care of that. We'll go into that in a second as well. And the integration, for example, the Ceph sees iDriver, Rook sets that up so that he can immediately start using the storage in the cluster. Or even for object storage, you have the currently, what is it called again, object storage claim, right? Object bucket claim. Object bucket claim, exactly, where if you need S3 for an application, it's kind of the same concept as like a persistent volume claim at the moment. These things, that's what Rook takes care of and like sets up in your cluster. Okay. Did you comment about open source? It's open source. I'll patch you 2.0 license. Perfect. Oh, you have your... So an operator, you know, orchestration at the top, I guess the example that I would simply like to use there if you go one further. It's a custom object, you operate it, observes it, detects any changes or any deviation from the desired state. Reconciliation, if anyone wants to kind of Google it, but that isn't too deep into operators yet, attacks it, tries to see what it needs to do, and acts upon that. Just to have the operator kind of, that's what an operator does basically. Yeah. Okay. So Rook makes storage happen in your cluster with its operator. So just a quick note on what Rook's CNCF status is. We started our journey with CNCF early on. CNCF defined their graduation status as, oh, you need to go through this process so that people know that, hey, this is a stable project. It's a project that the community trusts. It's a project where we try to do the right thing for the community. It's not just a product we want to sell you. It's what's right for the community. So we are happy that as of October 2020, we are a graduated project. So thank you for all your support and the community support for making this happen. Lots of people running in production. It's just been a great journey to get through this. So yeah, just wanted to make that quick comment. So let's talk more about the Rook community who's involved in it and why exactly is this our focus. From the start, we wanted to build a project that's open source. It's what the community needs. Apache 2.0 license really gives us that ability to just say, it's yours, deploy it how you need to. And we do have maintainers across four companies currently. So we're us two from IBM and Coore. And a couple we don't have here in this meeting, it's CyboZoo and UpBound. So we really are, we have a steering committee and maintainers and everything set up according to a healthy CNCF project where we want cross company collaboration. We have had over 400 contributors to the GitHub project and according to Docker Hub metrics, 280 million container downloads now. So I guess a few people are using it. Just curious for a quick survey. Are you here to learn about Rook for the first time? A few of you? Okay, a lot of you. All right. How many have experimented with Rook? Okay, a lot. And who's deployed Rook in production? Great. A number of you. Wonderful. Anybody deployed it for longer than three years in production? A few. Okay. Yes. So awesome. Let's get into what is Ceph, why we chose Ceph and where that gets us. Deepika? Yeah. Can you move a slide ahead? So Ceph is, again, an open source project. It's around, I think, 10 plus years now for Ceph to be kind of in open source world, even being used in production clusters. One of the trusted product in open source. It's, it provides a kind of block shared file system, as well as object store, all three at one place. And you can check out for more details. Ceph.io covers the case studies and what is covered. We also have services like dashboard and monitoring everything in place for Ceph. Thanks. And maybe I'll just jump in and add a translation to Kubernetes terms. So block storage, that's where you get your RWO volumes. Read, write once. Shared file system. You get your RWX volumes if you really need to share them. And then object, of course, that S3 endpoint. So thanks. And Deepika, why is that? Yeah. So again, as I said, why somebody should use Ceph. So just moving to that, it provides an all in one unified solution for block file and object storage. And along with that, again, 10 plus years of it being used. Many production clusters and case studies. You can check that out. It was first released in July 2012. And yeah, my favorite one is it's being used in CERN and in large withdrawn collider project as well. And it's working quite efficiently there. We also have talks around it in an event called CephlaCon. It was co-located with KubeCon. So if you are interested in learning more, you can check out Ceph talk from CephlaCon to learn more about it. Thanks, Deepika. Next question. Does Rook support storage providers other than Ceph? So early on in the Rook project, the maintainers, we wanted to do what was right for the community. So we explored this question and put quite a bit of effort into it. We did explore several providers and, you know, Minio and CockroachDB and a number of them and NFS. And we tried to create a common framework where we could bring them all together and share something kind of like you've got the operator SDK for creating operators. Could we create some sort of storage operator SDK? And at the end of the day, what we found was that, well, Ceph today is the only storage provider that we support. The truth of it was that the other ones either had their own community or we just didn't have the community support come to join Rook.  They've got their separate operators and Rook's focus is really on Ceph. And since I grew up with Star Wars, I just have to say the force is strong with Ceph. And Yoda's helping us out here. So, okay. How stable is Rook? Just a few comments on that. We've already said some things. So we're in our third year since the CNCF graduation. It was about five years ago. We declared Rook stable for production. And, you know, we have had several longtime users using it for that long, which is just a great testament to, you know, how Ceph does provide enterprise quality storage. We have many upstream users. I wish we knew how many upstream users. The nature of the upstream is go use it and we just don't know about it. We should do a survey at some point and try and get some more of that feedback. We just haven't for a while. And then also there are products built around Ceph and Rook downstream that we don't have metrics for either. So anyway, it's out there. It's in production. People are using it. I just wanted to give one case study. I was chatting with someone from the Rook community that has been using Rook for at least the past five years. He's working with the National Research Platform based on National Science NSF funding. Anyway, he just said, yep, I've got a cluster with three petabytes, almost 240 OSDs for those familiar with Ceph. It's a pretty big cluster. Bigger clusters do exist. Six petabytes of storage. That's not the limit by any means. It's just an example I wanted to show that, yes, you can deploy large storage platforms with this. And I just asked him to say, hey, what can you say about Rook for future users? And I appreciated his quotes here. Rook significantly simplifies our persistent storage needs in Kubernetes by automating Ceph, essentially. And then with Rook, adding and using a new Ceph cluster requires almost no efforts and becomes a trivial task. And that really comes back to this model of how Kubernetes does things where it's desired state. You tell Kubernetes what you want it to look like, and then Rook makes it happen with our operator. You can codify it. You can put it in your Helm charts, your YAML, and anyway, just make it happen. And if you need to deploy multiple of them, you can replicate it very easily. On to next. How do you install Rook? Yeah, first of all, the Rook operator, the Ceph cluster, they do have Helm charts. We have, especially for more common and complex configurations, we have example manifests. But really going to the Rook website, which is Rook.io, and clicking Get Started is the easiest way to kind of get in on the ground floor with Rook. Last year, Travis and I gave a demonstration of installing Rook on a multi-node AWS cluster, and this took about 12 minutes. The link is not great, but it is in the PDF slides for afterwards. All right. And in what environments can Rook be installed? Yeah, in short, anywhere Kubernetes runs. This can be in the cloud. This can be on-premises, virtual hardware, bare metal hardware. And the underlying storage could be disks attached to my nodes. It could also be PVCs. And recently, we've put in support for loopback devices as well for testing. And with the kind of key point to drive home here is that Rook really helps enable this cross-cloud support. All right. Thank you. So what should you know about the architecture? We just want to provide a brief overview. We don't have time to get into a lot of details here, but it's important to note, so Rook as the operator really owns the management of the storage in the cluster, the management of Ceph. At the next layer, we have the CSI driver, so the Ceph CSI driver, which then will provision and mount the storage to your user application pods. So just as you may be familiar with other CSI drivers, that's the way Kubernetes defines that applications plug in their storage for RWX volumes. The CSI driver can actually be used independently of Rook even, but it's just not as integrated as. So Rook just makes all this integrated. And then Ceph provides the data layer. Again, back to this, Ceph already existed. It just wasn't built for Kubernetes initially, but Rook manages it for you, so you don't have to worry about all the details of getting Ceph going. So Ceph provides that hardened data layer. Just a brief overview of what it looks like when you deploy Rook after you have it running. So this is just a view of three nodes, a three node system with pods color coded depending on those different three layers that I had in the previous slide. So the blue pods would be really what we call the Rook pods, where you've got the operator that's the brains of Rook deploying everything. We have an optional discovery component that helps find devices. And then the green pods would be for the CSI driver. So there's a provisioner, there's the plug-ins. So Ceph RBD is for those RWO volumes, Ceph FS for RWX. All those red pods are the various Ceph components. Ceph has a number of demons to, you know, provide the Ceph mons, provide a quorum, and the brains kind of like SCD, they are the brains of the system. The OSDs manage individual disks and store the data on disk. Anyway, lots of those components working together. And it's a lot to, if you have to deploy it individually in Ceph, it's more work manually, but Rook takes care of that for you. So that's just a brief look at that. We have more slides at the end, or we can talk offline about more architecture if you're interested. So Alexander, how can we monitor Ceph once it's up and running? Well, who's heard about Prometheus? Okay. Well, starting off with Ceph patch to that. So Ceph itself has kind of like a dashboard built-in. If anyone wants to look at it, it's, well, I think the more important point as I asked for Prometheus. Oh, the next slide. Yeah, there we go. Prometheus metrics. Thanks. You have Prometheus metrics, but like what else I guess you want to hear? There's Kavana dashboards pre-prepared. You have the metrics alerts are also available from, well, Ceph project. And technically you can also even toggle the flag and they will be automatically deployed in your cluster. You just need Prometheus operator for that. We try to make it as convenient as possible there. So, well, Prometheus metrics. Yeah. And I'll just add that all these metrics come from the Ceph project. Ceph that's been building in these metrics for, you know, since early in the project, they're always looking to add these metrics, make sure we can keep track of what's happening. So you know it's healthy. All right. How can I troubleshoot Rook clusters? Deepika? So now comes the question. I mean, Ceph is hard at times and Rook tries to make life easier there. So as stated, we have troubleshooting guide available at Rook website. There we discuss some scenarios where you can, I mean, if there is some problems at the CSI layer or if there are some problems with Ceph layer. We have scenarios covered. Most of the scenarios that are common are documented in our website. And yeah, you can also reach out to us on kind of Slack channel or at last we can cover that. But we are available and we are there to help you with the troubleshooting of Rook in general. Apart from that, there is an interesting thing that we worked on. There is a CUBE CTL plugin support called crew plugin. So we kind of added Rook Ceph support in it. So if you install the crew plugin, you can do everything that you used to do or you can do with Ceph layer. So any Ceph commands, you can run that. You can check the Ceph status from it. You can check the Rook status from it. You can check the status of the CRDs, everything. And apart from that, you can also do the maintenance operations. If you think that Ceph is hard, we are trying to make it easier with Rook to provide advanced troubleshooting scenarios for Ceph in one line command by using this plugin. So if you run into any scenarios and you want a simplified solution to be available for we are working on the development of this project but it... Thank you, Deepika. So how safe is my data in Ceph? You put your data there and, I think, Blaine, you have this? Very. Very safe. Yeah, Ceph is designed to be more consistent than available. So this means that if the data is not safe, Ceph will sometimes not work. But that is also one of its greatest strengths.  And the amount of replication that you have is configurable, and over the 12-plus years Ceph's been available, it's proven highly, highly durable, and even in really extreme disasters with the Rook we've seen users recover, like with manual intervention. But for the group, again, speaking, we're trying to add certain of these more complicated recovery steps as well that they are fully automated. In the worst case, you still need to manually run a command, but that most things after that are just automatically done by a script that runs automatically using the crew plugin. Yeah, very rare scenarios, but yes, we want to make sure it just works. All right, what Rook features are we most excited about? So we've each kind of chosen a feature to talk about, so we'll go through these slides now. I think, Blaine, you're first. I'm here to evangelize Kosi, which is the container object storage interface. So this is effectively just the container storage interface, but for object storage. So this allows pods to request and access buckets or blob storage if you're an Azure fan, and aims to make object storage as cloud agnostic, as block and file have been for so long. The alpha release of Kosi was in Kubernetes 1.25. We're continuing to make progress on beta designs. And in Rook, well, in Seth, we have a Kosi driver, and then we're working on adding that to Rook for our v1.12 release. One quick thing to add, as I mentioned, the object bucket claims in the beginning, that's the better version. This is the better version. Yeah. All righty. Deepika. Yeah. Even Alexander covered some scenarios already about the disaster recovery, and in case of failure. But for me, it's the crew plugin. It's simple. It's fundamentally solving the problem of maybe having the expertise, Seth expertise, if you are running a Seth cluster. Rook is simply finding the installation and maintenance and upgrading a Seth cluster aspects of it. But what about we, in rare case, get stuck in some problems with Seth demons. Not everybody is familiar with it at the first glance. So we are automating solutions. For example, there is a unit in Seth called Seth monitors, which does the talking and maintains the network. It's important for them to be in quorum, and if even two of them are out, Seth cluster is drastically affected. So how to restore them by just having one of the good ones available, we have one line command for it. Earlier, we used to kind of have to do manual intervention and kind of bring back the monitors back from there. So those complexities with Seth, we are trying to simplify it with crew plugin, even kind of have a debug pod, where it will have all the debugging tools that we use with Seth available. So you can, it will spin up a debug pod, and it will have the monitor debugging tool. You do the magic with the monitor debug tool, and then you kind of, after the debugging, can again bring back the normal pod, and everything is running smooth. You do not have to manually go and debug things, along with kind of checking the status, any kind of automation around Seth, we can write our own code around it, simplify that for Rook. So for me, it's crew simple, and it's good for Seth. Thanks, Topeka. There's something else about the crew plugin is, you might ask, why do we need a crew plugin if we have an operator? And the pattern we really found is that the operator is really good at desired state operations, like make this happen, but there are just some operations that are, you know, you need to do something one time, you need to do some maintenance operation, and so the crew plugin is really where maintenance operations just make sense. Just tell the operator to stop, even sometimes for some of these operations, and let us do this one time operation, and then start up again. Okay, Alexander. More encryption is good. It's that simple for that, but making the OSDs, which is daughter data in the end, be encrypted as well, and that's happening automatically. More encryption. So keeping your data safe and secure. Yeah, the one that came to mind for me was really the ability to recover from disaster. We've already talked about it some. This picture is from a data center fire that was in France, I believe, a couple of years ago that affected even millions of websites, something like that. Was anybody affected by this fire? Probably some of you, yes. Not a good experience.  You can spread your data across, replicated across multiple data centers, and even if a whole data center survives, you can design it so that you can survive those outages. Even at a simpler layer, CEP is designed to keep your data available while you have a loss of a node or individual disks. To some degree, CEP can even keep your data available and online, and you won't even notice if it's down because of those multiple replicas. And then similarly, kind of flipping around the title here, instead of recovery of disaster scenarios, it's the DR features, which CEP supports. So this is the ability basically to mirror your data across clusters.  So each of those mirroring technologies, it's a whole topic to dive into that we don't have time for today, but those are there. So what if I have a CEP cluster deployed outside of Kubernetes? Can I connect to it from Kubernetes? Well, yes, you can. If you have an existing self-cluster, Rook can basically connect to it, take care of setting up the CSI driver or even in certain, well, there's different modes to accessing an external cluster from Rook's perspective to, for example, even start running certain components of a self-cluster in your Kubernetes cluster. And it's, for the most part, as simple as that, like you give the operator the credentials for the other cluster, tell it where it is. And for the main part, for the CSI driver, at least configuration, that's taken care of. And if you have RJW in the other cluster as well, that's happening automatically as well in regards to the object bucket claims and future cozy. And you can kind of think about it as well if you have a Rook self-cluster that you want to share with multiple Kubernetes clusters, you can do it as well. Maybe a bit obvious to some people, but because one Kubernetes cluster can't directly talk with the network of another Kubernetes cluster, you need to interconnect them. And there's several ways. Either you have a CNI, which allows you to do that. There's what was called, I think, like... Yeah, well, Cilium, for example, with the mesh, yeah, there's other projects as well that mainly focus on just bridging two clusters that even have overlapping networks. Submariner is another one. And for those cases as well, we're looking more and more into them. Submariner, for example, to make that easy as well. Okay, yeah. Just regarding that one central, for example, Rook self-cluster, that's one way some people do it as well. They have one central Rook self-cluster for the storage because, again, with the aspect of Kubernetes abstraction, making it quite easy to run self, like a mon fails, just start up a new one. It's that easy for something like Kubernetes to do that, but in, like, the real-world self-case you don't necessarily have additional nodes where you can just install a component on it. You can do that, but it would most of the time involve manual work, and that's, again, where Kubernetes has this abstraction layer with an operator on top, shines. All right, so can I provision a bucket with an S3 endpoint? Yeah, I feel like this is something we've mentioned and alluded to before. So the current implementation we have for creating a bucket and getting access to it in Rook is with object bucket claims, and this is a similar pattern to PVCs. The bucket gets created when you request it. You get access via a secret. This is, this really ended up kind of being a prototype, and it's still in an alpha API for what became Cozy. It's sort of morphed into Now Cozy, which is a Kubernetes special interest group. And I'm excited that Cozy is finally, finally almost really here. Awesome. Can Rook be configured for clients to access storage outside the cluster? So in other words, you have Rook running and you have clients not a part of this Kubernetes cluster. Alex? Yes? You need to, again, you need to have connectivity from the outside clients with the cluster. So, host network modes or maltosevenous, like a bit of a more and more preference as it can be pretty specific like what network interface or network ranges you want to be added to the self-component containers. As with most things, you need to be able to reach it to access it. And for example, just to elaborate a bit more on what you would need to make accessible depending on the storage type. For example, for block storage, you would need to expose the monitors, the OSDs as well. So there's not just one component. You would have a gateway to access certain types of storage unless for, well, basically, object storage gateway which is a different case because you would, for example, for that case, just expose it via an Ingress node pod service as it's S3 APIs, basically HTTP or slash HTTPS, whatever you make it there. But for example, for the file system type storage, the client also needs direct access to the OSDs, the months and even another component, the MDS, the metadata service. But with, well, host node kind of the quick way to do it, multi-stay more specific way to do it, you can achieve that. All right, thanks. I think we're down to the last couple of questions here real quick and then hopefully we'll have time for your questions. How does Rook keep data available during Kubernetes upgrades? Just to skim over this, Rook does manage pod disruption budgets so that even during a Kubernetes upgrade where nodes are taken offline, we make sure that the nodes are managed so that stuff stays completely available, your data is all available, not just durable but available during these upgrades. How often does Rook have releases? We try to have minor releases on a similar cadence to what Kubernetes has about every four months. We just had 1.11 in March and potentially 1.12 will be in July. Whenever there's a need for a patch, we try to get those out as soon as we can bi-weekly at least just to have that cadence and when there's a critical need, we always get patches out as soon as we can. Fun question, where did the Rook name originate? CASL was the original project name and CASL is a secure place where we wanted to secure data. That's where the theme came from with the Knights and Shining Armor protecting the CASL and of course Rook is the chess piece representing the CASL. Just a little fun background there. Is Rook your next move? So now we'd like to open it up for your questions. I think we have a little time. Orange? Thanks. As I'm totally new to this Rook or any other solution like Rook and Seth, multiple questions, really, really brief. NFS support first of all, as I've seen that the architecture is pretty distributed so there are a lot of components around. The resource footprint would be nice to know because it looks like many components could be awesome, many resources. It depends if the storage is critical, mission critical for your application, for your project or company or not. Like, why should I taking into account that I don't know anything about storage in Kubernetes, why should I choose Rook and Seth over other solutions like maybe Minio or something like that? Rook, do you want to take NFS first? Yeah, I can take the NFS one definitely. So Rook, Seth does have NFS support at least from a cursory level and I'm actually on a team and we're working with Seth folks to make the NFS support more kind of enterprise standard as well. Okay. For another question, why should people choose Rook over something else? I'd say the biggest reasons I hear from people is that you have block object and file all in the same storage platform instead of getting a different solution from different platforms and it's fully open source. You know, Seth belongs to the Linux Foundation Rook's with the CNCF. Fully open source, open community we're open to contributions all the time. And then I think you had one more question. Which one? Resource footprint. Yeah, I mean, Seth is definitely requires some resources and for example, each for each disk that the Seth needs to manage we have say four gigs of memory and so there it would keep you to go along with that. It would take some time to dive into how many resources but yes, it does require some non-trivial resources to run storage. All right. Here in front. So I have also three questions. One it's about upgrading. What do you recommend always upgrade Rook first and then Kubernetes after or take a snapshot before to be better safe than sorry or is it possible and the other one it's about encryption as I know dmcrypt it's only for block device it's also possible for object storage. Yeah. Who wants it? Do you want to handle the dmcrypt file? Yeah. So for the OSD speaking as that's where the data is stored even like for object storage that's the kind of like the from architecture for you as safe works those object storage demons are the w also just start the data into the OSDs so the OSDs need a block device be it well a loop device or a disk or just a partition it's basically where the data is stored in the end. So that's what the encryption there is basically and that's where dmcrypt comes into play as well. Did you have any questions or regarding upgrading more for like a question for example the last time I think what was it again the pot security policies went away in like 124 25 or something and if you didn't upgrade root before or at least disable the option for pot security policies in the chart you wouldn't be able to upgrade any further but that's more of like a how health does upgrade so to say issue well and as far as keeping your data safe during upgrades and should you be worried about it and take snapshots first you can do that I mean snapshots probably a good idea to have take periodically anyway just to sleep better at night right you always want as many to keep your data as safe as possible it is safe with having backups to the backup is even it's always right but I'd say whether or not to do it before upgrades or after or worry about it I'm not sure I've heard that as a significant reason to need to do snapshots first the upgrades have been reliable from what we've seen and and we always yeah I'd say that's a decision you can make if you need if you feel like you need to do snapshots but yeah good question others sure right here and what about performance oh back here performance compared to other things so seph is a software defined storage network so it's going to be it's distributed it performs well when there are a lot of clients communicating with all of the demons that are spread spread across the the cluster so you can get that nice distributed performance as far as how much performance you get it's going to depend on well are you backing the cluster with SSD or NVMe how could your network it is since it is software defined storage you're not going to get the same local volume access as if you're just writing directly to a disk right because there is a layer in between it but people for who need that software defined storage and the consistency and reliability there everybody would always love more performance right but there is a performance cost there that but people are generally it's acceptable for performance for what they need databases is where I'd say it's more of the challenge where the higher performance needs are and databases sometimes have the replication at their layer too so in those cases I'd usually tell people look at the replication at the database layer and not use stuff just use a local disk but all right just from a performance maybe Deepika you could have some insights from RBD site for block storage like that's one of the most common use cases for people running databases on stuff yeah I think so you have to also see that there is replication and consistency being kind of worked out in SEF in general as well but still I think I would say if you are using NVMEs I do not recall but Alexander do you recall I generally but it was performant enough for HDDs I was able to see 100 MVs per second for 3 replicas but there are numbers you can actually use a tool called Redos Bench you can go to SEF website and kind of if you want to see how your disk is performing you can use there is a FS what tool we use FIO sorry FIO on your regular disk you can compare it by kind of using there is a RBD extension for that it's completely documented in SEF so you can run FIO directly on your disk and then also run FIO with RBD and try to compare how it's performing and those numbers are also published on SEF website so you can check that out well thanks I know there's a lot more there we'd be happy to talk to you after and come to the Rook booth in the pavilion but I think we better stop there officially thanks everyone