 In the last class we had dealt with how to obtain a grain design for a given thrust time curve right and towards the end of the last class we kind of found out what is it that is required of a grain design that is what are the desired features we discuss this that for a solid propellant system to be very good you need to have as high a loading as possible because then you are letting you you are making an optimal use of the given volume right so first is high propellant loading then we realize that it is better to have neutral burning right if you have neutral burning then your structural utilization is much better. So for better structural utilization then lastly it should have low sliver loss in addition to meeting the thrust time curve if there are three or four designs that meet the thrust time curve then we should look at one which has a high propellant loading and then near neutral burning and low sliver loss okay there is a flip side to high propellant loading namely erosive burning which we will discuss in the future class okay we will come to that little later now if you remember in the last class if you look at this figure here we had designed the grain in terms of Y or the extent of the propellant burnt and if you look at this graph then from this graph you can get for any given Y the burning perimeter and from the burning perimeter you can calculate the burning surface area AB right. So in a sense we had got what we had got at the end of the last class was how to get burning surface area AB to meet our requirements now how do we cross check whether what we have designed is going to give us what we wanted in the first place right we have to cross check our design so to do that let us say we were given a pressure versus time curve like this right if you were given a pressure versus time curve like this now having gone through the grain design how do we know that we have been able to get as close to this as possible all the information that we know is we know how the burning surface area evolves right as some extent of web burnt we know Y versus AB right from the graph now we also know that from equilibrium relations equilibrium pressure PC is equal to rho P AB by 80 one my innocent so if we know AB right we can calculate what is PC right the other terms are fixed 80 density of the propellant A and C star and N all these are fixed so we will get to know what is the chamber pressure variation with AB so we know Y versus AB from there we can get Y versus PC so knowing Y versus AB we get Y versus PC but we are still not there because we need Y versus PC versus T right this is what we want so how do we get that very simple we know burn rate right burn rate we know is nothing but dy by dt is nothing but burn rate so this is nothing but a PC to the power of N so using this I can integrate for Y and then get or this equation and get time so you can rewrite this as T is equal to 0 to Y DZ APC to the power of N so we now know time we now know PC so we can get PC versus time so from this we can get PC versus time so after this we can evaluate whether the design that we have proposed is a is going to agree with what was given to us what was desired of us right you might not be able to meet with precisely at all the locations but in an overall sense if you are going to meet it that is good enough okay so there is one doubt still that is sometime back somebody post this aren't we using equilibrium pressure here this relationship right but if you look at this portion right this portion is not constant pressure are we right and using this equation even to get something for this portion right this is changing with time is not a constant pressure so let us see how to find out that okay before we do that we have to derive the unsteady equation so let us go back to a few classes back wherein we had this control volume of the rocket motor if we have a rocket motor like this then we know that there is mass coming in because of the burning of the propellant right now it is a Z is a dummy variable that is all if you have why you will integrate it and it has no meaning here if this is a dummy variable so here it is burning in this direction so this is adding mass and the mass is going out through the throat so if we write the unsteady equation right then the accumulation of mass with time must be equal to mass flow in minus mass flow out now how do we express equation for this accumulation of mass with time this is nothing but m m dm by dt mass flow rate in m dot in minus m dot out m dot in and m dot out we had learnt what they are this is rho p a b r dot minus pc at by c star r dot I can write it as a pc to the power of n so I will get rho p a b a c star now what is this m is nothing but density into volume right so we can write okay rho c is nothing but density of gas and vc is the chamber volume now we can write d rho c by rho c vc by dt as equal to rho c into d vc by dt plus vc into d rho c by dt by chain rule right what is d vc by dt if we come over here and look at this finger how is the VC increasing a b into r dot is what is giving the increase in the chamber volume right so d vc by dt I can write it as a b into r dot so I write this as r dot I can write it as a pc to the power of n now if I substitute this back in this equation I can rewrite that equation as VC d rho c by dt is equal to this goes to the other side rho p minus rho c now before we go further let us get an estimate of what is rho p and what is rho c rho p is density of the propellant right depending on the kind of propellant it varies between 1,600 to 1,800 okay what about rho c how do you calculate gas density PC upon RTC so what typical chamber pressures can go up to 100 okay and chamber temperatures of the order of 3000 so PC around 100 atmospheres then TC around 3000 kelvin molecule of 8 around 26 so you will get typically row variation for this case would be around 12.5 rho c that is and typically depending on this pressure if the pressure is higher you are going to get a higher density and if the temperatures are higher you are going to get a lower density and pressure lower also you are going to get so typical variation of rho c is from somewhere between 8 to 15 kg per meter cube compare this with rho p it is somewhere around one less than 100 right so if you neglect this term what is the error that you are going to make less than 1% right so from an engineering perspective it makes sense to neglect this term okay. So this rho c we can write it as PC by RTC right so here if we make the assumption that assuming the gases to be following the perfect gas law which we have already used and also making the assumption that TC and C star are constant which is valid assumption in a sense even if pressure varies quite a bit beyond some level TC and C star do not change by a large magnitude they do change but by not by a large magnitude so if you make this assumption then we can write rho c is nothing but PC by RTC so if you include in that you will get VC these two one is gas constant the other one we have assumed for the for this case to be a constant okay so then I can take out VC by RTC and I will get DPC by DT is equal to rho P AB APC to the power of n now we also know that C star what is C star C star is nothing but 1 by ? of ? RTC right or if you are going to use the universal gas constant Ru Tc by weight so RTC that I have here I can replace it with C star into ? of ? there is a reason for doing that we will see that in a little time we see by I can take this to the other side right and read this expression as yeah where is it oh VC VC sorry fine between these two right I can what is rho P AB AC star if I divide this by 80 and multiply this term by C star what do I get this entire term ? P AB AC star by 80 what is this this is PC to the power of 1 by N right so now I can do that here there is a C star by 80 that I can take out and rewrite this expression as follows so if I take out 80 by C star right this term will be PC this PC is nothing but PC equilibrium pressure right PC equilibrium to the power of 1-n so I will get there was a reason for doing this which you will notice in a short time what is the unit of this this is pressure right and the left hand side is pressure by time so in some sense this is 1 by time right so let us look at how this is 1 by time we will call VC by 80 right VC by 80 is chamber volume by throat area we will call that a quantity named as L star okay sorry so if we call VC by 80 as L star okay this is the length scale okay if you look at this this has meter cube and meter square so this is a length scale this is the characteristic length scale of the motor now if you divide length by velocity what do you get time so L star by C star ? square is nothing but TC or the characteristic time scale this is also known as the residence time this gives us an idea about what is the time that the gases are going to be there in the chamber okay and L star as I said is a length scale L star assumes importance in all the three rocket motors that we are going to decide even in liquids and in hybrids and this is related to there is an instability called as L star instability also now if you look at this time characteristic time if you have a short motor right we have a small motor this should be smaller and if you have a large motor they should be larger so typical variation of TC is it is around 1.5 to 2 milliseconds for small motors and 20 to 40 milliseconds for large motors this has a implication on what is known as combustion efficiency that is how much of the combustion is being completed inside the rocket motor also depends on the residence time right if the residence time is very small then you would expect if the reactions take a little longer those would not be completed right so the slowest process in solid propellant combustion is aluminum oxidation okay and that takes something like 5 to 20 milliseconds so if you have a very small motor then probably your combustion efficiencies are not going to be very high okay whereas if you have a large motor your combustion efficiencies are going to be very high because there is ample residence time to complete the reactions okay the typical combustion efficiencies for large motors would be of the order of 99% so most of the reactions are complete before they go out of the throat now coming back to the question that we had in our mind right that is when we were trying to look at PC versus T variation for a given grain design we had assumed that it is equilibrium pressure relations we had used is that valid is what we asked ourselves right so let us look at whether that is so or not so to do that we will take this equation here okay when we take this equation and if you how do we decide whether this term is large or small one way to look at it is if these two terms are nearly the same then this term would be small right so one way to get a good estimate is to divide this term by one of these two and find out if the fraction is large or small if it is very small then the error that we are going to make by neglecting it is going to be small to okay so let us take the term VC by RTC DPC by DT is equal to ?p now we will only need to consider the left hand side of the equation okay as I said we will divide the left hand side of the equation by this term okay what happens to the right hand side of the equation is not our concern right now we are trying to estimate this term in reference to one of the terms here right so I will take the let me call this term as R LR okay so now what is DPC by DT I can rewrite this as DPC by DY x DY by DT right and DY by DT we know is nothing but R dot so we will get this is equal to R dot x DPC by DY so if you substitute this back into this equation you will get R dot VC by RTC R LR is equal to into so I can take out this R dot sorry thank you now I need to get a relationship for DPC by DY right let us use the equilibrium relation we know that PC is equal to ?p AB by AT a if you take the logarithm of this you will get log PC is equal to 1-1 by 1-n we know that AB is a function of y right so differentiating both sides with respect to y we get 1 by 1-n into DAB by AB sorry AB DY okay so or in other words I can get the expression for DPC by DY from this right and which is as follows PC by AB okay so if I substitute this back in the equation here right DAB by DY then I will get R LR would be VC by RTC to PC there is an AB here so I will get AB2 1-n we still need to know how the burning surface area varies with why let us take progressive burning rain that is let us take a cylindrical green burning from inside to the outside okay so if you take a cylindrical green burning from inside to the outside so at any given let me call this small distances why so at any given instant why if DP is the port diameter right so the burning surface area this is nothing but for any instantaneous burning surface area you need to add to y the diameter is DP plus to y so the burning surface area would be pi DP plus to y into L the length of the grain into the board and AB at y equal to 0 would be okay and therefore you can calculate DAB by DY for this what will it be 2 pi L right take the derivative of this you will get okay so let us substitute that back into the equation that we have here this equation again we can simplify it further let us see how to do it and substitute it back into that we know here PC by RTC is nothing but density right so I can rewrite this as RLR is equal to ? C VC divided by ? P AB square into 1-n into DAB by DY and DAB by DY we know is nothing but 2 ? L VC ? C remains the same VC is nothing but what is the chamber volume it is the burning surface area into the length so AB at 0 that is ? DL ? DP square by 4 this burning surface this port area into the length is the volume and DAB by this is 2 ? L ? P remains as this AB square sorry this is ? DP okay so we can cancel out DP square DP square here L square and L square here right so after counseling out and you have ? square in ? square again here so you have 2 and 4 so you will have 1 by 2 so you will get 1 by 2 1-n ? C by ? P okay we had calculated what is ? C in term and ? right we had seen that a little earlier that ? C varies from 8 to 15 kg per meter cube whereas ? P is around 1600 to 1800 kg per meter cube now if you take n even if you take a large n let us say of n of 0.7 what you will get is RLR will vary from 0.01 to okay so what it tells us is even if the port is burning in a progressive manner or in any other manner even if you take equilibrium relations there you are not making a very significant error primarily because of the density difference between the gas and the solid being of the order of around more than 100 right and that is the reason why you can afford to take a steady case also and you will get very accurate result okay thank you we will stop here we will continue in the next class.