 In the last class we had looked at film cooling right and we said it was a very in efficient way of cooling but it is as I said it is still preferred because it gives you that redundancy in the design. Now let us look at perhaps the most efficient way of cooling that is the regenerative cooling. Now in this method as I earlier said if you have a liquid rocket motor there is a coolant jacket okay and the one of the two liquids is made to come in through this end okay and it will come out through the injector right, come in through that end and it will come and get into the rocket motor like this. Now as it moves along it picks up the heat from the hot gases that is here and puts it back in and therefore in a sense this mimics the adiabatic wall that we had made an assumption of okay. As I said this is very efficient but the trouble with this cooling is it is a very very challenging problem simply because I do not have a choice on the coolant type for example in most other applications like IC engines and other engines you are free to choose the type of coolant that you want right and you will choose the best coolant, coolant oils like cast oil and other things but here I do not have that choice I have to use one of the two liquids. So the liquid coolant is fixed and its flow rate is also fixed that is if the engine has a particular thrust then I can back calculate the mass flow rate and depending on the O by F ratio I know what is the flow rate of fuel and oxidizer so it cannot exceed that flow rate right so that is also fixed and due to this the design of such a coolant system is very very difficult. Now there are a few things that are under the control of the designer they are one is the pressure at which coolant is delivered okay the other thing is this is strictly not true simply because you design the engine for to certain pressure right you can decide on how much of over that pressure you can give in the coolant chambers okay it is not a strict this thing but still there is some flexibility in this and the other thing that is under the control is diameter or geometry of coolant pipes if you remember while deriving our equations for heat transfer in turbulent flow the diameter of the pipe played a very crucial role okay so that is under the control of the designer and the thickness of the diameter and thickness is under the control of the designer so with this we have to go ahead and design this coolant pipes now there are many options that are available as such for designing this these are shown in this figure here if you look at the first one it is something known as a drilled construction if you see the view here there are a lot of holes that are drilled in the wall okay and these holes are going right up from here to the head end okay now there is one problem with this as you move from the divergent portion to the throat the diameter is decreasing right so if you have X number of holes here the same X number should be there at the throat which means that the diameters need to shrink at the throat which is not very easy to sometimes fabricate okay or you need to have such a design wherein the size of the holes remains the same but they are more well spaced as it comes out in this thing okay so that is fabrication is a very difficult part in this then you have something known as a tubular construction this is a lot better than the earlier one here what you are doing is you have two layers okay one is the chamber that is the inner layer on top of it you have pipes that are brazed okay so in a sense you can decide on the type size pipe size as you go from the divergent portion to the throat and to the head end okay this is under in some sense a lot more controllable than the other one and then you will have an outer skin on top of this the pipes are brazed or welded together so as to keep them in contact otherwise you will have development of hot spots in certain regions where there is no liquid coolant then the other this is probably the most best of all the designs that is instead of having pipes if you have a corrugated sheet okay it is something like this you have sheets that will make the coolant channels okay the sheets are such that you can make the coolant channels based on this itself and then brazed such that they are in position and then again you have a top covering layer so this shape changes also from the divergent portion to the throat the size shrinks at the throat and then again becomes larger this is probably one of the better designs because it offers you ease of fabrication in some sense then there is something known as wire wrap which is scarcely used that is you wrap wire around the nozzle itself so that the coolant moves along these lines okay but the trouble with this is you cannot even if you wrap the wires there is going to be some gap between them okay that is going to lead to some kind of hotspot generation there so it is not such a good design because you cannot cover the entire region with this kind of an arrangement okay now I like to draw your attention to something that you have already learnt you have learnt that heat flux from the gas phase side is maximum at the throat right so you need to ensure that this region is cooled appropriate otherwise there is a lot of heat coming in and if you do not cool it you might tend to damage this and in addition to this you have learned that the heat transfer coefficient changes from here to here and again here to here it is a maxima here and then it goes down on both sides so Hg is varying right the heat transfer coefficient coming or the heat flux coming in from the gas phase is varying you have to keep that in mind and you have to also know certain things about how coolants behave we will go through that in a minute that is I am sure most of you have done this boiled water sometime or the other right what do you see when you boil water and then have you ever left the vessel on this tau itself for sometime you know and forgotten to take switch off the gas or something what happens then yeah what happens to the vessel it could get burnt right that is the same thing that is going to happen here to if we do not take appropriate care will see how that happens firstly if you look at the boiling of the liquid there are a few regimes of boiling firstly you have something known as convective boiling that is if you have heating from this side let us say this is the hot gas side that is you are looking at some region here and there is liquid in the pipe when you heat it at the beginning you have seen this that this portion becomes hotter and tends to rise up right and therefore it tends to carry the heat from this surface to somewhere inside okay so heat is being transported from the wall to the fluid inside fine this is by convection so this is known as convective boiling then there is something known as nucleate boiling that is again you have the hot gas side and the heat flux coming in okay as you rightly said after sometime bubbles begin to appear at the bottom of the surface so locally the liquid boils and the bubbles rise up okay and as they rise up they transport the heat from the wall to the fluid inside and then they collapse okay because they have become in cooler so therefore they collapse again become a liquid in this way the heat is being transferred from the wall to the fluid inside this is probably a very efficient way of transferring heat from the surface to the fluid inside now there is a third regime of boiling this is known as nucleate boiling okay there is a third regime of boiling namely film boiling again you have heat flux from the bottom so okay what happens is after sometime if you supply the heat beyond this stage right these bubbles tend to get formed more frequently right there are large number of bubbles and these bubbles tend to coalesce right come together and form something known as a film so there is a thin film of vapor that is formed close to the surface now we all know this that the heat transfer or the thermal conductivity of vapor is an order of magnitude lesser than the liquid okay so the heat transferred from the wall to the fluid gets reduced because there is a layer of gas that is there and that in some sense resist the transfer of heat so in a sense if you keep transferring this heat the wall temperature tends to rise up and this is very very detrimental now if you plot this in a graph with Q dot double prime as the y-axis and T wall liquid as the x-axis okay you can see all these three regimes like this firstly you will have connective boiling that is the heat flux rises linearly as the wall temperature that is if you keep on increasing the wall temperature this also arises up linearly beyond some point the rise is very very steep and this is convective this regime is nucleate okay the slope is drastically different here in this regime and if you look at the wall temperatures for this this will be something like T boiling plus 30 to 50 degree centigrade okay very small temperature rise above the boiling point and you will be able to transfer a lot of heat to the fluid okay so what happens beyond this is what is known as a film boiling and there actually the heat flux drops and it only comes back to this level at something like TB plus 400 to 600 centigrade that is you need to increase the temperature a lot more to get the same kind of heat flux okay if we do not transfer the heat from this region to the fluid what happens is the temperature of the wall tends to increase and it could reach the temperature where in remember this is also under a lot of stress because of the high pressure inside the chamber so it could reach a point where in it could break so you would not want that right so you have to ensure that at the throat there is nucleate boil okay or somewhere in this region just before nucleate boiling if it goes into this region it could be very very detrimental right because there is a lot of heat flux that is coming in and that heat is not being taken out so the temperature of the wall keeps on rising so and that could break open the motor so that is why this is such a challenging problem in heat transfer see if you keep on heating right it becomes a film if you look at viscosity of the gases viscosity of the gases increases with temperature okay and if you look at Prandtl number Prandtl number of gases is close to 1 Prandtl number indicates the ratio of momentum transfer to heat transfer right so the thermal conductivity also goes the same way okay this is a molecular diffusion phenomena so the thermal conductivity also keeps on increasing so it only it has to reach very high temperatures for the thermal conductivity to come to the same level right now in addition to this we need to remember that the pressures inside encountered inside the combustion chamber of a rocket motor are very very high right the temperatures are also very high so there is something known as critical pressure and critical temperature right you remember from your thermodynamics studied earlier that if we have a PV diagram for a here substance right it will go something like this and this point is called a critical pressure now at the critical pressure and temperature the there is no distinction that you can make between liquid and gas phase both of them coexist okay so because the pressures encountered in the rocket motor are so high you we need to be aware of this also right so if you look at as I said earlier if you look at space shuttle main engine the combustion chamber pressure is somewhere around 200 bar and somewhere inside the coolant channels it could go beyond 300 bar okay so very very high pressures so if you have temperatures also in that region then you will probably encounter this region and based on this we can make a map of critical temperature versus critical pressure as shown here if you plot critical pressure versus critical temperature this is T critical and let us say this is P critical okay now you divide the region into four parts right and let us call this in this if the pressures are below critical pressure and the temperature is also below critical temperature there is a distinct liquid and a vapor phase right so in this region there is a liquid and vapor this region is supercritical liquid this is supercritical gas this region is gas okay now why do we need to in a sense worry about this is simply because if you are wanting to design the system coolant regenerative cooling system what we need to know is the properties of the liquid in a particular temperature and pressure and we need to know whether it is a gas a supercritical liquid or something else in that region right otherwise we might determine properties somewhere else and use it something else then we could be under designing over designing and we could probably get into trouble so we need to know which region it is there and in the coolant channel what happens to pressure and temperature as you move along the coolant channel the temperature will increase right and there is a pressure drop because these are very very small channels through which the fluid has to flow at very high flow rates so the pressure drop will be there temperature will increase so in a sense you will depending on which region you are your curve will have a negative slope right so let us look at which are the coolants and what is their you know critical points and if you use them at some pressures how do they move okay the last column is the heat transfer coefficient of the liquid to the heat transfer coefficient of a reference liquid what is used as a reference liquid is kerosene okay kerosene is used here as a reference liquid one of the biggest challenges of using a kerosene locks engine right kerosene locks engine is you know to put it differently if you look at the so use craft it has probably more than a thousand machines right thousand machines and this is being used to take people and equipment to the space station now why is that being chosen as simply because it uses for most of its this thing a kerosene lock system which is very very cheap among all the propellants this is probably the cheapest now kerosene locks although it is very cheap one of the trouble with kerosene lock system is that if you use kerosene as a coolant right what happens is there is something known as coaking kerosene remember is a hydrocarbon right it has hydrogen and carbon at some temperature the carbon tends to get deposited and this might block the coolant channels because these coolant channels are themselves very very fine and that is a very serious problem to overcome in development of these engines okay so kerosene is chosen as the reference fluid and let us look at some of the propellants firstly the fuels hydrogen or NH4 then the other one is unsymmetric dimethyl hydrazine UDMH okay then kerosene then liquid hydrogen okay among oxidizers we will pick N2O4 now the critical temperature for hydrazine is around 380 centigrade 380 degree and the critical pressure is somewhere around 145 so if you are using typically with hydrazine and UDMH you do not go to very very high pressures it somewhere in the region of 70 to 80 bar in the chamber so upstream of that could be lower than this critical pressure so the heat transfer regime would be in a that is here okay so there is a distinct liquid and a vapor phase and as the coolant moves from the nozzle end towards the head end it traces a path like this okay the coolant velocities are something like 30 meters per second and the HL by HL reference is 4 that is it is 4 times effective than kerosene then UDMH this is 249 and this comes down to 60 okay so this could be in either a or b and the velocities would be something like 10 to 30 and this is three times as effective as kerosene kerosene itself will be one the HL by HL reference this is 414 kerosene the critical pressure is very low right it is somewhere around 21 so you will most likely end up above this critical pressure so it is in most cases in this regime and moves in this direction then LH to remember liquid hydrogen hydrogen is a gas at ambient conditions and only if you super cool it it will go to liquid state so the critical temperature is very low – 240 centigrade and critical pressure is something like 13 atmosphere this will be sorry in C now it could start somewhere here and go into C okay because it is if you look at the critical temperature this is also very low this is also very low right so you could end up with C and therefore if you notice this is very high because it has changed its state from liquid to a gas so the flow velocities are very very high right and therefore its coolant properties are also very high it is 10 times more effective than kerosene right and into for the numbers are something like this 158 this is 100 it will be a gas if you look at liquid hydrogen liquid oxygen injectors the hydrogen is coming in as a gas which is why we are going to have swirl injectors for this kind of a system we will discuss that when we discuss injectors it will be a gas now it is also important for us to kind of know that what is the properties that are desired of a coolant you will have two fluids on board right one an oxidizer one a fuel if you are looking at a by propellant system which one among them to use as a coolant is the question typically the choice is invariably a fuel simply because if you look at its properties what is it that you desire of a coolant firstly its CP should be very high right its specific heat capacity should be very high that is if you have a higher CP let us compare air and water okay for the same amount of heat that you pump into both these systems the CP of water and the density of water is also very high so therefore it is something like CP is four times density is 1000 times so if you pump in 4000 kilo joules the water temperature will go up by 1 degree whereas the air temperature will go up much much more right so that is something that you need to keep in mind when you look at what kind of coolant to use CP and thermal conductivity are something that is important and similarly you have a table for so you have to choose a liquid that has higher CP and probably higher thermal conductivity and which obviously in most cases happens to be a fuel now if we have to look at the analysis of the heat transfer let us say this is the wall and this is hot gas side so there is TC this is hot gas this is liquid and this here is the wall there is heat transfer by convection on this side in the wall there is conduction and here again you have convection right so there are there is one that is TC in the hot gas side and let me call this temperature as T wall gas T wall liquid and let me call this temperature as TL okay so what are the temperatures that are known and what are the that are unknown okay and what are the equations to use again this is a series connection right so the heat flux is going to be the same so you have Q dot double prime must be equal to Hg into TC – T wall G this must be equal to Kw by TW is equal to HL TW L – TL okay this we had seen we could write it as HW right now there is a radiative heat transfer also coming in from the gas phase side which could be significant you could absorb this in this but you need to be careful in deciding on the edge it has to also account for if the radiation part is significant it also has to account for the radiation part okay fine now how many unknowns are there we will know this part TC we will know what about T wall G we do not know this and T wall L we do not know and TL also we do not know right so there are three unknowns okay and we have only two equations so how do we solve this before we get into that we have seen how to estimate this part right Hg we know how to get it we said it is turbulent flow and therefore we based it on Reynolds number right and this part is conduction if we know the wall property thermal conductivity of the wall we can evaluate this this part if you look at this this is again flow through a channel but this is a liquid okay prunel number of the liquid is not something close to one that you can strike off okay so you need to have prunel number also so HL would be again the flow is turbulent because this is a confined flow and the transition Reynolds number is somewhere around 2300 right flow through tubes and the tube diameter is a very very small although if you look at this here the velocities may not be too large but the channel diameter is a very very small so therefore the flow would be turbulent okay so we can estimate the HL in this fashion and then solving these two equations we can rewrite expressions for T wall G as and T wall L also we can get and okay so we got expressions for T wall L and T wall G now we do not know how to take care of the other one right if you remember your basic energy transfer fundamentals this is nothing but a pipe flow right you have a pipe flow right and there is a coolant jacket fine so if I consider a very small elemental area DX right then what I know is let us say this is TL at X then I will call this TL X plus ? X okay so here you have hot gas and this is liquid now there is a certain heat that is being transferred from here to here and you can write the energy balance for that as m.L CPL TL X plus DX-TL of X this must be equal to the heat flux coming in that is Q. double prime into the DS is the surface area right that is perimeter into DX fine ? D into DX so if you take a very small elemental area and you need to also remember that at the entry to the coolant pipe you know the temperature right so you will know the temperature there and that is let me assume that to be this point if you know this then from this equation right you know at the entry point you can calculate T wall L corresponding to that okay and if you know T wall L from this you can calculate Q. double prime and you will know the area so you can calculate the heat that is coming in fine you know m. So you know CP so the only unknown would be this so you can evaluate this then if you evaluate this this becomes the next entry temperature so you can go on and proceed in steps from the nozzle end towards that is you need to divide the motor into sections from the nozzle end towards the head end okay and proceed in this direction then you will be able to get all the temperatures okay fine this clear we will stop here we will continue in the next class thank you.