 In the last class we were discussing about hybrid rockets in this class let us look at how the burn rate relationship a g o x to the power of n is obtained. So firstly to do that we need to understand how the burning takes place inside a hybrid rocket motor now if this is the hybrid rocket motor then oxidizer is injected here right so a boundary layer develops and the combustion takes place inside the boundary layer. So this is how the combustion takes place inside a hybrid rocket motor now if we look closely into this combustion zone then we will be able to look at what are all the processes that are taking place in that zone. So let us say this is the solid fuel that we have then let us say we have taken somewhere in this region right where the boundary layer is already established so this is the boundary layer and let us say this is the flame zone right now what is happening because of this boundary layer there is a heat transfer to the propellant which is convective in nature right and that heat transfer is being taken up partly for maintaining the temperature profile in the condensed phase and then partly to evaporate the fuel. So if we were to draw the temperature profile inside the boundary layer it would go something like this this is TS this is TC and this is TE the combustion that is achieved here I mean the temperature that is achieved here is the highest simply because there is combustion and after which it is diluted by oxidizer right now the propellant burns in this direction and there is because of that there is a mass addition that is given by rho PR dot okay mass is getting in and there is something known as blowing effect that we had discussed when we were discussing about erosive burning in solid rockets now so there is heat that is being transferred let me call it as QC okay that is heat that is being converted into the propellant surface now this QC I can write it as H into TC-TS okay this is the convective heat transfer coefficient and this is the adiabatic flame temperature or the combustion temperature and this is the surface temperature now this heat as I said is part of it goes to maintaining a temperature profile and part of it is used for evaporation so QC must be equal to let me call this direction as X and this direction as Y so okay this is the temperature this is the part of the heat that is taken up to maintain the temperature profile and this is the QS we had seen earlier is the heat of evaporation at the surface so this we had seen with regards to solid propellant that we can rewrite this part as rho PR dot CPTS-T infinity let me call the temperature here as T infinity so I can rewrite this as QC must be equal to rho PR dot now for ease of our terminology and other things to follow let me call this part as HV okay HV is nothing but CPTS-T infinity plus QS now we know that this problem is a boundary layer driven problem okay so what we are going to use is something known as Reynolds analogy what it says is there is a similarity between the momentum boundary layer and the thermal boundary layer right and we use that it is easier to calculate the momentum boundary layer there are correlations for it and we use that to calculate the thermal boundary layer this can happen when only what Rundle number is close to one which in gases it is always close to one so we can make this assumption so using approximate Reynolds analogy that is there is similarity between momentum and or momentum and energy so we can define something known as a stanton number you is nothing but the velocity here okay so H is the heat transfer coefficient rho e is the density at the beyond the boundary layer and ue is the velocity beyond the boundary layer CP is the specific heat this is a non-dimensional number it is also known as dimensional dimensionless heat transfer coefficient okay now if we use the Reynolds analogy what it says is that this stanton number indicated by CH is equal to CF by 2 where CF is nothing but skin friction coefficient okay and we know that Rundle number is close to one for gases so this term becomes merely CH is equal to CF by 2 now we have defined what a stanton number is and we have found out that by Reynolds analogy we can equate it to this skin friction coefficient as we multiply the numerator and denominator of the stanton number by a ? t that is given by TS-TC-TS okay we will do that and see what we will get so we had stanton number CF is equal to H by rho e ue CP so we will multiply both numerator and denominator by TC-TS so it does not change anything to the left hand side but what is H into TC-TS that is the convective heat transfer right so QS is equal to H into TC-TS so what we will end up getting this sorry this is not C of CH CH is equal to this which one okay sorry sorry sorry QC fine now what we need to this part is that QC so what is this part we can define it as some change in enthalpy so we will call the ? H is equal to CP TC-TS so what we will get is we also had the other term that is the heat that was taken up by the solid let me call that as QS that was easy equal to rho p R.hv right but we know that these two must be equal right that was the heat balance at the surface for the heat balance at the surface these two must be equal so I can now replace in my stanton number definition this here right so and stanton number this also I know is nothing but CF by 2 right so I will use that so I will get rho p or CF by 2 is equal to rho p R.hv fine or in other words I can rewrite expression for R. as R. is equal to CF by 2 into rho e ue ? H by Hv okay sorry divided by rho p now we have been able to get this expression for R. in terms of rho e into ue is nothing but in some sense mass flux okay and our CF by 2 what we need to further determine is what is this CF for the particular geometry that we have considered so skin friction coefficient without blowing we can write it as CF not by 2 is equal to in terms of Reynolds number this indicates the location along the axis okay this is for a turbulent flow now what happens because of blowing is the boundary layer in a sense tends to get thickened okay so the skin friction with blowing will be different and it can be given as CF by CF not where beta varies from 5 less than 1 okay beta is nothing but the blowing coefficient and we can define beta as follows beta is defined as rho p R. that is the mass flux that is coming in from the surface to the mass that is going through the port okay this is also known as non-dimensional so we have defined CF in terms of CF not into beta and we have beta so we can now get the expression that we were looking for that is we had said that CH is equal to this is nothing but rho p R. HV by UE that is this expression right so we will we can use this expression if you look at this expression and the expression for beta right there is some similarity between the two and we can write ? H by HV it is nothing but rho p R. that is equal to beta right so now I have this expression for R. here where in I have defined CF I have defined all other terms so I can get plug them in and get my expression for R. which we will do R. is equal to CF by 2 RUE by rho p now we know that CF is nothing but CF not into beta so we will use that here and get our expression for R. so R. is equal to 1.27 by 2 into CF not into beta to the power of 0.77 into RUE by RUE into beta fine because we know that ? H by HV is nothing but beta so CF not also we know in terms of Reynolds number as this expression here so if we substitute that it will get R. is equal to so beta is minus 0.77 and 1 so I can write it as beta to the power of 0.23 Reynolds number is nothing but RUE into X divided by ? so I get R. is equal to so we have RUE here and here so we can club those two and rewrite our expression as R. is equal to and I can take the rho p that is here this rho p I can multiply it back with the R. so I will get rho p R. is equal to what is RUE RUE is nothing but the mass flux in the port right. So mass flux let us define G as m. by AP so that will be nothing but RUE so you have we can replace this with G right and we will get minus yes thank you so we come to this expression there is a 0 here so we get this expression wherein the burn rate is related to the mass flux through the port right and if you look at when we do experiments it is easier to characterize this in terms of the oxidizer mass flux itself because that is something that we have control of and we know what it is so which is why you will find that if you take a look at any hybrid literature you will find that the burn rate is expressed as a function of G O X okay. So we can recast this in terms of G O X and that is what you will find in most literature as I dot is a function of G O X to the power of N and if you look at this expression you will find that it depends very weakly on X the axial location and beta but it is a very strong function of the mass flux right now let us look at the problem that we had said yesterday with regards to what happens to burn rate as we proceed in from the head end to the nozzle if you look at this expression here it looks like what should happen to the burn rate as you increase X it should decrease but I said the opposite yesterday right is that correct or is that wrong if you look at this yes you are saying but you are forgetting that there is a greater dependence on G if you look at what is happening as you move from the head end of the port to the nozzle end there is continuous mass addition and this is increasing as you go from head end to the nozzle end and this is raised to a power of 0.8 so therefore you will find that the burn rate as you go from head end to the nozzle end will always increase now people have done people have conducted experiments and found out that if you take the oxidizer mass flux the dependence of burn rate varies this N varies from 0.4 to 0.7 one of the things that we have kind of made an assumption right at the beginning was we had assumed something like a flat plate if we assume a pipe flow kind of a situation and redo the calculations we will probably get closer to what we what is observed experimentally what we can get is in terms of the oxidizer mass flux and not the overall flux overall flux involves that you take a cross section and find out what is the fuel plus oxidizer that is coming which is a lot more difficult to do if you are doing an experiment experiment you are better of knowing what is your mass flow rate of oxidizer and then calculating the oxidizer mass flux there was another problem that I said hybrid rockets suffer from because of this N that is if you look at holding even even when you hold the oxidizer mass flux constant we find that the fuel mass flow rate was changing now let us look at is there any condition at which the oxidizer mass flux if you hold constant or if you hold the oxidizer mass flow rate constant will the fuel flow rate also become constant we know that the mass flow rate of fuel is given by rho P AB into I dot so if you are taking a port configuration rho P into this is the diameter of the port and this is the length of the port this is the burning surface area into our dot is nothing but a into okay now let us determine what value of N will give us something that when you hold the oxidizer mass flow rate constant you can also get a fuel mass flow rate constant if you have to engineer such a hybrid then that is a very good situation because you can operate it at a set specific impulse even though the diameters are changing so G O X we know is nothing but fine this is G O X so if we plug it into this equation we will get M dot F now this is you have 2 into N right in the denominator and if you look at this equation you have D to the power of 1 and D to the power of 2 N as very obvious that if you have a value of N that is equal to 0.5 for N equal to 0.5 M dot F becomes independent of the port diameter so therefore you can operate it at a set O by F ratio so that you can choose that O by F ratio such that it gives the maximum ISP now what happens if N is greater than 0.5 so it is desirable to have M equal to 0.5 now for N greater than 0.5 what will happen to the mass of fuel as diameter increases it will reduce because you will have to N and you will have DP here so the denominator will be greater than the numerator so therefore as diameters increase mass flow rate of fuel will decrease and similarly or for the other case that is for N less than 0.5 M dot F this finishes our discussions on hybrid rockets so in a sense we have learnt 3 kinds of rocket engines in this course that is one is solid propellant the other one is liquid propellant and then lastly the hybrid propellant okay thank you.