 Good morning. I think this will be the concluding lecture in this course and what we do in this class is the following. We said VJ must be high. Let us see are there some limitations to VJ. After looking at this, we will try to see whether there are other possibilities of using nuclear forces for generating thrust like nuclear rockets, something which we must take a look at. Whether we can use the light intensity like photons or rather the deep space energy which is available in deep space for developing thrust and these are some of the newer areas like we say maybe advanced propulsion is what we would try to take a look at. The question of advanced is something relative because in some cases maybe electrical rockets are also a little bit more advanced. But anyway let us see whether anything more than electrical propulsion something like newer types of propulsion are possible in the years to come and I think all of us must concentrate on this because maybe you all will have a chance to develop something new and something beautiful. Having said that, let us take a look at VJ. Should VJ be extremely large? After all, I have been telling this point a couple of times namely maybe when Cielkowski formulated the rocket equation and in today's class we will again go back to the equation because we would like to conclude something Cielkowski's equation. And what was Cielkowski's rocket equation? We said that the delta V which rocket gets not only comes from its exhaust jet velocity like let us say ISP or VJ but also the logarithm of the initial mass to the final mass. When he was working on this he felt if I have to get a high VJ maybe I should have electrons which can be pulled in a magnetic field and I could get high VJ. But it so happens that electrons have a very low mass and therefore it does not give you force. Well after him we will go to another 10 years maybe 20 years 18 something like 1930s and 1940s. He was also interested in finding out maybe using electrons itself. Can I get the electrons to go at the speed of light and therefore get a high value of VJ? Are there some limitations? Not limitations in getting a high VJ but is there from the basic equations what we considered? Is there some limit to VJ which is possible and that is what we will try to do in the next 5 or 10 minutes. Let us now take a realistic picture of the electrical rockets what we discussed in the last class. See what we needed in this was something little more. We needed propellant and propellant was let us say xenon. We also said I need an electrical field in case of iron rocket. I need a voltage of something like 1.5 kVA. In terms of hall thruster I needed 30 kVA. That means I need something like a power unit. Even if I have solar power I convert it to electricity in my spacecraft. I still have to recondition the power to give me high voltage. That means I have something like a power I can now write as power processing unit. It has to process according to the voltage current what I require. I call it as PPU. Now if I have a battery it is going to be heavy. That means in addition to the usual chemical rockets which had let us say initial mass. Let us go back to what we discussed when we were developing the C-Halcolysis E useful payload that is the space capsule. It had the structural mass. It had the propellant mass. Now what is it I am saying? Now I am having a power processing unit which has some weight and therefore I should add something like MPP in addition to propellant mass, in addition to structural mass. But the structural mass will also go up because this has some weight. I must have a structure to support this weight and therefore I say in the initial mass in the case of an electrical rocket will consist of useful plus structure and this one together and the propellant mass over here. And what is going to be my final mass? Final mass is going to be the useful thing. The propellant is burnt out and I have structural mass plus the propellant mass over here. Therefore I find I am unnecessarily carrying some part of the carrying this power plant which has a certain unit mass also structural mass associated with the power plant. And therefore I can say something like a specific mass of the power unit coming like mass of the power plant plus mass of the structure associated with the power plant divided by power which is generated by the power plant P. And this is denoted by chi which units is kilogram per let us say kilowatt or kilogram per watt is the unit of this. Now what is it I am trying to aim at? When we supply power P and use this power to enhance the velocity VJ what we are essentially doing is the rate at which the energy is getting released namely power gets into the kinetic energy and what is kinetic energy? M into VJ square divided by 2. But then we are talking of energy here whereas it is power that is rate of energy therefore you have something like d by dt of the kinetic energy. Therefore power gets converted into let us say VJ is a constant. Therefore I take VJ square divided by 2 into M dot or rather power gets converted into the rate of kinetic energy. Whenever some conversion takes place there is some loss of efficiency because I put an electrical power converted into something like kinetic energy over here and therefore I have an efficiency and if the efficiency of conversion is let us say eta. Therefore I have eta P which gets converted into VJ square divided by 2 into M dot or rather I can write that power is therefore equal to from this expression I get M dot into VJ square divided by 2 eta. This gives me the power and therefore now I say well my efficiency is low or efficiency is poor that means I need a higher power to get the same jet velocity. Having said that let me now work with this and also you will recall we had some expression for the specific mass what we define as equal to we had it in terms of let me put the expression down. The value of chi was equal to M mass of the power plant plus you had the mass of the structure divided by power and we said that the efficiency of a power source is defined in terms of the specific mass of the power source which is the total mass of the power unit and the structure associated with it derived by power. Now if I can use this value of power to be substituted over here I get chi is equal to M p p plus M s divided by I substitute this I get M dot into VJ square divided by 2 eta over here. Now why is it I am doing all this? The reason being we know that if I have a higher jet velocity and we were looking at the rocket's alkohol scheme equation when I get a higher jet velocity I get a better payload. However, it is not only the higher velocity but now I find if I give a higher value of jet velocity my power goes up, if my power goes up well my mass also goes up and it is quite possible therefore now if I start plotting these values I say well my payload mass which now I call as useful mass as a function of VJ. If I look at VJ alone may be as VJ increases my useful payload will go up because I am having a higher jet velocity over here. But a higher jet velocity results in a higher power and therefore my power that means my value of M p p plus M s t also increases as VJ increases and if this increases it will adversely affect my value of the useful payload and this is what I want to work out. Therefore, to be able to do that let me go again and write the Sielkovsky's equation which I can write as let us say that the useful value I write the Sielkovsky's equation as the incremental velocity delta V as equal to VJ logarithm of the initial mass divided by the final mass. Rather this particular equation I can again write as initial mass to the final mass is equal to e to the power delta V by VJ and now if I invert it and say I am interested in the final mass divided by the initial mass I have the value as e to the power minus delta V by VJ. Now what is it I find as VJ increases my final mass for a given initial mass will go up but then I am also telling another thing over here namely that my mass that is my unproductive mass also goes up. Therefore, let us try to build an equation which results which will give us something like the useful payload mass. Let me write it over here the value of useful payload as a function of VJ which is also a function of the power and this is what I am searching for. Therefore, if I come back to this I have Mf over Mi is equal to e to the power minus delta VJ or rather now I can write the whole thing as M of the power plant plus mass of the structure over here from this particular expression. I write it as equal to chi of the power. Well, power I again write it from an expression as equal to M dot into VJ square which is equal to chi of M dot VJ square divided by 2 eta and therefore, now I get the value of power plant plus the weight of the structure is given by this value. Somehow I want to bring in the value of useful mass over here and therefore, I look at this expression again and want to express the value of useful mass as a function of initial mass as a function of VJ and that is what is my particular aim which I am trying to solve this equation. Therefore, let me again put this over here. I can now write the final mass what we get in a rocket is equal to the mass of the structure and the power plant which is left behind Mpp plus Mst and what is there you also get the value of the useful part of it which is available. Therefore, if I can write this equation now I can write it as Mpp plus Mst is equal to Mf minus Mu and now I substitute the value of the power plant weight and the structural weight through this particular expression namely this becomes equal to chi into VJ square divided by 2 eta into M dot over here is equal to I get the value of Mf minus Mu. But now I would like the initial mass also to come and therefore, I can write this expression as Mf minus Mu in terms of Mi also and for this I simplify the expression over here, write the expression over here. I now write Mf is equal to Mu plus chi into VJ square into M dot divided by 2 eta and immediately I find that as the efficiency decreases that is the efficiency of conversion of the power into velocity decreases well this factor goes up and for a final mass therefore, the value of Mu will decrease which is what we should really expect also because as efficiency of conversion of the power into the velocity decreases I will have a less useful payload which comes over here. Having said that, let me try to again arrange it in some particular form. I will write it again as Mf minus Mu as equal to chi into VJ square divided by 2 eta. What is the rate at which mass is getting depleted? It is the total propellant weight that means Mp is the total propellant mass divided by the duration of burning let us say Tb. Therefore, I get the value of Mf minus Mu is equal to this and therefore, I again find that the duration of burning also affects my value of the useful payload. Now, I want to arrange this in some form and therefore, now I write this as Mf over Mi, I divide throughout by Mi and I write it as Mf by Mi minus Mu by Mi as equal to the value of chi VJ square divided by 2 eta into Mp by Mi and this I can further simplify and express it as Mf over Mi I already know it is equal to e to the power minus delta V by VJ and Mu by Mi is what I am interested in the value of Mu by Mi and this particular expression can again be put. I know that the mass of the propellant and what is the mass of the propellant? Again it is equal to the final mass minus the initial mass and therefore, I can again put it in terms of the value of e to the power of minus delta V by VJ and therefore, I will write the final expression over here as Mu by Mi is equal to e to the power minus delta V by VJ minus I get Mp is equal to Mf by Mi, Mf by Mi divided by Mi is equal to Mf by Mi minus 1, Mf by Mi is equal to again e to the power minus delta V by VJ is equal to we get 1 minus e to the power minus delta V by VJ into chi of the value that is VJ square divided by 2 eta is the value and of course, the value since we said Mp that is Tb comes over here that means, we had the value Mp by Mi I forgot to put this it should have been 1 over Tb and Tb also comes here and therefore, this becomes my final expression for the useful mass. Let me simplify this a little bit and try to point out the dependence what we get. We can write this particular expression as saying that yes, whenever I have the useful mass to the initial mass this expression tells me that the value of Mu by Mi as a function of VJ as VJ increases the negative exponent decreases and therefore, it goes like this whereas, I am subtracting some quantity over here. Therefore, this is my 0 over here as VJ increases this quantity decreases, but then VJ square decreases this goes up like this or rather I can say the Mu by Mi contributing from the first term on the right hand side comes like this. The second term comes like this and the net of the two things put together perhaps this is again VJ may be looking something like if I put both of them together I have VJ Mu by Mi and this may really go up initially when the initial value is small over here it goes up when this value catches up well it will decrease over here and then you have something like this. In other words there is a value of VJ beyond which it will lead to my useful mass coming down rather than going up and that is why I cannot really operate a rocket under high power conditions. Now, one last point when I am at this particular expression is if I look at the value of this particular value what I have here, I have the value of chi VJ square divided by 2 eta Tb and if I look at the dimensions of this you know we said it is specific mass so much kilogram per watt and what is watt, what is joules per second kilogram second VJ square is meter square by second square eta is efficiency does not have unit Tb has units of second and now if I were to put the units again if I put the units of joules what is it I get kilogram second meter square by second cube second cube here and joules joules is again Newton meter Newton is kilogram meter per second square and you have meter here and therefore this net expression becomes a dimensionless number and therefore we will now take a look at the plot of this particular expression as a function of this this is a non-dimensional number and we will try to express the useful mass as a function of VJ as a function of this non-dimensional number and this is this is what I show here the value of chi by eta I keep changing the value of VJ is on this scale the value of Mu for a particular mission which takes 10 days and it gives a delta V of something like 2 into 2 kilometers per second I find that the optimum depends on the specific mass of the system and therefore there is no way in which may be for a higher specific mass there is no point in me having a velocity greater than some limit and therefore there is a limit on the velocity this is all what I wanted to illustrate using this one. In other words the effect of increasing the power is to increase the specific mass or the increase the mass of the system and it does not really help in any way and therefore we say electrical rockets cannot be operated for very high powers that means there is a power limitation that means beyond a power it is not useful anymore. Why is this? This is because as power increases VJ increases and a large value of VJ is therefore detrimental. But when we talk of chemical rockets were there any limitation can you immediately recall any limitation we had? We had a limitation in temperature because chamber could not take any temperature that means the energetics of the rocket or we say energy of the propellants have a limitation that means chemical rockets are energy limited whereas the electrical rockets are power limited. I think this is something which we need to keep in mind and with this we sort of conclude our discussions on electrical rockets. But let us see whether we can really have rockets which are different why not have something like a light which can give you some pressure or maybe something like nuclear or we talked in terms of tri propellant rockets or other mechanisms to generate the thrust and that is what we will be doing in the next few minutes. Let us consider nuclear reactions after all nuclear reactions also generate heat and power. Since the nuclear reactions take place within the nucleus of the atom and the nucleus of the atom consists of protons and neutrons. The total number of protons and neutrons constitute the mass of the atom or its atomic mass. The number of protons in the nucleus decides what the substance is and is known as the atomic number. In some cases what happens is the same atomic number or the same substance is associated with different number of neutrons in it. That means for a given atomic number the substance has different mass and these substances are known as isotopes. The isotopes are not very stable and therefore they sometimes decay or the nucleus of the particular isotope reacts and such reaction in the nucleus is what constitutes the nuclear reaction. The nuclear reaction is not like a chemical reaction wherein the atoms are conserved but in a nuclear reaction the number of protons and neutrons in the nuclei are conserved. That means what happens is the protons and neutrons get conserved and therefore in the products new elements are formed in a nuclear reaction. The change in the nucleus is associated with heat which is generated and this heat could be used to heat a low molecular mass gas such as hydrogen and generate thrust in the case of nuclear rockets. This is the principle of nuclear rockets using nuclear reactions. When we look at the type of nuclear reactions we find there are three types of nuclear reactions which are possible. These are radioactive decay, nuclear fission and nuclear fusion. In the case of radioactive decay what happens is the unstable isotope as such tends towards forming a stable element. While forming the stable element long wave length radiation like alpha radiation is emitted and during this process there is a release of energy which we think could be used for propulsion. However the rate of energy release is very very slow it could take place or it could stretch over a period of several hundred years and therefore it seems difficult a priority to use radioactive decay for propulsion purposes. The second type of nuclear reaction is the nuclear fission in which the nucleus is broken up and when the nucleus gets broken up protons and neutrons and protons and neutrons are emitted and it is these neutrons which give the energy or the fission energy and this is used for propulsion. The third type of reaction nuclear reaction is the nuclear fusion in which nucleus of isotopes are fused together one over the other to form a new element and during that process well energy gets liberated. Let us take a look at some of these features and see whether we can have propulsion or how the nuclear reactions can be used in practice. Let us take a look at the radioactive decay. We just said it is a slow process and let us illustrate this radioactive decay through one particular substance polonium. You will recall polonium was discovered by Madame Curie noble laureate Madame Curie and this polonium has an atomic mass of 84 has a number of protons of 84 here you have polonium number of protons are 84 number of protons and neutrons in it are 210 that means it has 84 protons and something like 126 neutrons making a total of 210 protons and neutrons and this is a radioactive element. Now this decays and when it decays it forms the more stable lead and the lead has 82 protons and compared to 210 protons and neutrons it has 206 protons and neutrons. We are left with 4 protons and neutrons we are left with 2 protons and helium is a substance which has 2 protons and 2 neutrons and therefore it has 2 4 and therefore helium is formed in this particular reaction. Therefore in the case of radioactive decay the substance decays to form more stable substances and in the process new substances are formed and in the process helium which is formed in this particular case is associated with alpha radiation and this is what constitutes it is something like a long wavelength radiation which has something like radio waves and it is known as radioactive decay. The half lifetime of the reaction polonium to form lead and helium is of the order of 128 days very slow rate of heat release and therefore this heat release is not viable to generate thrust in a rocket. Therefore what is done with radioactive decay is the thermal energy is converted to electrical power in the case of orbiting spacecrafts and this electrical power is generated using the thermoelectric principle. You will recall when we measure temperatures the thermoelectric principle is used to generate millivolts of energy using heat. So also the thermoelectric principle is used to generate power and such type of power generation is known as radioactive isotope thermoelectric power generators or radioactive thermoelectric generators RTGs. The RTGs are used in several orbiting spacecrafts to generate power in them but it is not used for propulsion purposes per se. In addition to polonium substances like thorium uranium can also be used in the radioactive decay process. Let us now go back and take a look at nuclear fission. In the case of nuclear fission you have massive isotopes like uranium, thorium, plutonium which have large number of protons and neutrons in the nucleus. The protons and neutrons are held together by very strong nuclear forces but when the nucleus of such elements are hit strongly with a neutron the nucleus absorbs the neutron and thereafter when it after absorbing the neutrons it breaks up and releases the neutrons which releases the neutrons in the particular nucleus and these neutrons further impact on the nucleus of the maybe uranium, thorium or plutonium to generate more neutrons and in this way you have a cascading process of generating neutrons or an avalanche of neutrons which are generated. Let us take a look at it through this particular example. You have uranium which has 92 protons, 238 protons and neutrons, it is the nucleus is hit by a neutron and when it is hit by a neutron it absorbs the neutron to form U239 that is you have added one neutron to it U239 the element is same U92 and this nucleus now breaks up or it goes into a fission or breaks up to form krypton which has 36 protons and 92 protons and neutrons and barium which has 56 protons and 141 neutrons and protons. In the process neutrons are generated these neutrons go back hit the uranium again keep on generating neutrons and you have an avalanche of neutrons generated. It is these neutrons which carry the energy and the avalanche of neutrons in the cascading process generate lot of energy but if you have a cascading process well it keeps on generating it is something like a bomb it is going to burst you should be able to control it and to be able to control it low molecular mass substances like carbon water, beryllium these absorb the neutrons and are used to control it and these are known as moderators they absorb the neutrons and make sure that if you insert these moderators into the reactor they absorb some of the neutrons and keep the neutrons generated at a steady state. When neutrons are generated at a steady state the fission process is said to be critical you know if by chance the moderators absorb too much of neutrons then you could have some reflectors or mirrors which will which can focus the neutrons back into the reactor and you can still achieve a fission process which is critical. The kinetic energy of neutrons are made use in nuclear rockets what is done is you heat a low molecular mass gas such as hydrogen to high temperatures using the neutrons and when you heat the hydrogen you get high jet velocity which is the principle of rocket propulsion. Having said that let us also take a look at fusion in the case of fusion you have nuclear of isotopes which have small mass or could have large mass these are fused together if they are brought together sufficiently close using very high pressures and if the energy is also sufficient well it could be fused together and one such example I will give you you have hydrogen. Hydrogen has normally one proton only and isotope of hydrogen is deuterium and in deuterium you have one proton and you have one proton and one neutron making it two over here this is deuterium and another isotope of hydrogen is tritium wherein you have one proton and two neutrons making proton and neutron as three. Therefore you have deuterium and tritium at very high pressures the nucleus nuclei are brought very close to each other and at very high temperatures they fused together and when they fused together since in a nuclear reaction the protons and neutrons are conserved you have two protons which are formed you have three plus two five protons and neutrons out of which four two four is helium you are left with a neutron and therefore the fusing of the deuterium and the tritium leads to formation of helium and a neutron and it is such change in the nucleus or the binding force in the nucleus is released and it is possible to use the energy of fusion to heat gaseous hydrogen and generate thrust. It is theoretically possible to use the fusion process just like the fusion process to generate heat and use it for heating hydrogen to generate thrust but then we find that you need high pressure and very high temperature to fuse the nuclei of the isotopes and with the result we find well it is difficult to use the fusion process in practice because you need extremely high temperatures and what is done is in general we use only the fusion process and not the fusion process for nuclear propulsion and the energy of neutrons in the fusion process is used to generate thrust. Well the neutrons can be used directly from the fusion process for heating the hydrogen in which case we say let me go back in which case we say we have a nuclear thermal rocket if the nuclear energy is converted into electrical power and this electrical power is used either as electrostatic propulsion or electromagnetic propulsion well you have a nuclear electric propulsion. Therefore the nuclear propulsion or nuclear rocket propulsion could be subdivided into nuclear thermal rockets and nuclear electrical propulsion. We find that the nuclear electric propulsion is somewhat more involved it is it has not been applied so far I will come back into the details in a moment but getting back to nuclear thermal propulsion we find well in the case of fusion process it is difficult to apply because you need extremely high pressure temperature therefore nuclear fusion process has still not been applied only the nuclear fusion process is applied what is it used you have a reactor in the reactor you have moderators you have the nuclear element or the isotope which is hit by a neutron it generates neutrons you have moderators which make the process critical or a steady state of generation of neutrons you introduce hydrogen the hydrogen is heated to very high temperatures in the reactor and out comes the hot hydrogen and this hot hydrogen is expanded in a rocket nozzle to generate high jet velocity which is what is required in rocket. Therefore the thermal energy of high velocity neutrons generated in the fission reactor is transferred to gaseous hydrogen and the resulting high enthalpy hydrogen is generated in the nozzle to generate thrust and what is vj vj is the difference between the under root of two times the specific enthalpy of the hydrogen at the exit at the exit of the reactor minus the specific enthalpy of hydrogen at the exit of the nozzle and this is what gives you the jet velocity for propulsion. Having said this we also take a look at what is this nuclear electric propulsion which we said is little more difficult what is done is you convert the electrical or the nuclear energy from the fission process into electrical power we know yes you have nuclear reactors which generate electrical power so also you could have a small reactor which generates electrical power use this electrical power for powering electrical rockets like electrostatic rockets or electromagnetic thrusters which we learnt just two or three classes earlier and then use the electrical but the type electrical power requirements for electromagnetic thrusters are huge like for instance in the case of electromagnetic thrusters we are talking of magnetic field of the order of almost 0.1 Tesla which requires several megawatts of electrical energy and it is appropriate that nuclear electricity which is generated could be used for such rockets. In fact there is one particular rocket known as Vassimar variable specific impulse electromagnetic rocket which has been proposed which you which is which would use a nuclear reactor for generating an electromagnetic thruster but it has still not been developed so far the only nuclear rocket which has been developed so far is the Narva in USA Narva stands for nuclear energy for rocket vehicle applications which is developed by US which is nuclear thermal rocket. Having said this let us take a look at the nuclear thermal rocket well you have a reactor a solid core fission reactor you have a solid core fission reactor in this you have a series of tubes in the tubes you introduce pellets of uranium and you bundle a series of tubes together to form clusters and several such clusters is what constitutes the reactor you also have a moderator you have reflectors such that when you introduce a neutron from an auxiliary chamber into the into into the tubes here well it generates more and more neutrons it is moderated or it is controlled by the moderator and the reflector you have a steady state of neutron generation and when hydrogen is introduced the hydrogen gas gets heated you have high enthalpy gas at the exit of the reactor and this high enthalpy gas is expanded to generate thrust in the case of a nuclear thermal rocket well this is how a solid core fission reactor works you know in this case you know you find this is the length of the reactor the hydrogen gas gets heated during passage through the through the core or through the through the clusters what are available through the individual tubes and the state time of hydrogen is small with the result the efficiency of this something like a heat exchanger is somewhat poor and therefore the type of efficiencies of nuclear thermal rocket using solid core fission reactor is of the order of 40 to 50 percent which is extremely poor therefore there has been a proposal to use something like a particle bed fission reactor and what is done in this case is you introduce gaseous hydrogen through the sides you have something like a porous tube here through which hydrogen permeates through the pebbles or through the particle bed here and in this particle bed you have the uranium which undergoes the fission process and the hydrogen as it moves through the bed becomes hot and this hot hydrogen is expanded in the nozzle this is very effective because the state time of hydrogen in the bed is much larger but then there is one problem you know the hydrogen tends to take the path of least resistance and places where the hydrogen is not available tends to get overheated and you could end up in problem of thermal oscillations or thermal instabilities this is an inherent problem with particle bed fission reactor and therefore what is used generally is the solid core fission reactor though work goes on in the area of the particle bed fission reactor well this is about what we use in the case of the nuclear reactor but having said that what does the nuclear rocket therefore consists of it consists of a propellant tank containing liquid hydrogen it consists of a pump which pumps the liquid hydrogen at high pressures and mind you whenever we are talking of nuclear thermal rockets we are talking of huge thrust and therefore you need a large amount of liquid hydrogen to be pumped in and when you when you pump in the hydrogen at high pressure you need a turbine to be able to drive the pump and therefore we are talking of cycles feed system cycles just like what we considered in the liquid propellant rockets and therefore we use something like an expander cycle in which case the hydrogen after being pumped it cools the walls of the reactor and in the process it gasifies it also cools the moderator and the mirrors or reflectors it gets gasified and this gas of hydrogen is used to run the turbine and the work done by the turbine drives the pump and powers the pump such that it can pump it and the exhaust from the turbine is now fed into the reactor and this hydrogen is heated to high temperature and you get the exhaust this is the schematic of a net nuclear thermal rocket and therefore what is it you find you find that the liquid hydrogen is pumped from A to B that means to a high pressure at B from B to C it is heated to a gas from C to D it is expanded in the nozzle from D it enters the reactor at state T it is heated to a state E enthalpy high enthalpy at E and it is expanded in the nozzle to F. I show the same process here on the T s diagram wherein you have a series of constant pressure lines you have at A the liquid hydrogen it is pumped in the high pressure pump to B it is heated during the process of heating by the by the walls of the reactor by the moderator by the reflector to the state C it is expanded in the turbine from C to D and thereafter from D to E it is heated in the reactor and from E to F it is expanded in the nozzle to generate the thrust well this is the expanded cycle for configuration of a nuclear thermal rocket you could also have different types of cycles like we had the gas generator cycle maybe you could have a cold bleed cycle in which case the liquid hydrogen is pumped into pumped over the walls of the reactor it gasifies well part of it is supplied into the chamber for heating in the reactor to generate thrust part of it is generated in a turbine and this is exhausted out rather what happens is you have from A to B the pump work and then you have the constant pressure heating in the over the walls of the reactor maybe in the moderator and then part of it is expanded is further heated and it is expanded in the nozzle but part of it which goes into the turbine is expanded from C to D namely in the turbine to provide work for running the pump well this is the cold bleed cycle but as you see you are using only the heat from cooling of the reactor to run the turbine and therefore this cannot be used for very very large thrust and therefore you have a hot bleed cycle which I show here namely what is the difference now you have the same thing liquid hydrogen pumped cooling the walls of the rocket that is the reactor over here and then you allow some of the hydrogen to come here which generates the thrust you bleed some of the hot gases from the reactor mix it with the warm hydrogen coming from cooling of the reactor and therefore you have high enthalpy gas therefore you do more work in the turbine and you can generate further or can pump in lot of liquid hydrogen to generate high thrust in the case of a nuclear thermal rocket well this is the hot bleed cycle and all these cycles are used in practice but as I said there are some problems which have not made the nuclear rockets to see the application which it deserves the major drawback is the is the radiation nuclear radiation nuclear radiation is always a source of concern you will recall even in India in Kerala you have the monosite sand containing theorem available along the coast especially of Kailan and North Kerala and these things generate sort of radioactive decay they generate nuclear radiation causing cancer and therefore you know it is always that use of nuclear rockets is associated with nuclear radiation and therefore unless it is can be used for deep space applications the use of nuclear thermal rockets has really not been possible and even when we talk of radioactive thermal generators what did we say this we said that the radioactive decay is used for generating power in spacecrafts and even powering of spacecrafts using RTGs are concerned and people have been protesting against this for one reason you know we had one reconnaissance satellite cosmos 954 this was in 1978 it crashed over Canada it had an RTG and people were worried that the radiation from this would affect them and therefore well the application of the nuclear rockets has been limited or has not seen the light of the day like the other forms of propulsion because of the radiation problem but it is a very strong contender for space for deep space applications for interplanetary missions and so on and so forth well when we are talking of nuclear propulsion or nuclear rockets it is interesting to see that maybe if you could have a series of directed nuclear explosions like for instance you have a spacecraft at the bottom of the spacecraft you have a base plate and on the base plate you have a series of directed nuclear explosions like you have nuclear bombs which are exploded well you transmit impulse and that impulse could lead to movement of the spacecraft and such type of concepts are known as pulsed nuclear propulsion it is also known as bomb propulsion but in this case the major deterrent is the shock loads from the explosion has to be shielded from the spacecraft you have mechanisms which can you have springs and dashboards which can shield it but it is going to be difficult to apply bomb propulsion for case of the pulsed nuclear propulsion well this is all about nuclear propulsion or nuclear rockets and since we are talking of pulsed behavior there is another form of propulsion which we call as the pulse detonation rocket this is a extremely simple case and I think we should spend some few minutes on this the thing is that you have detonation in pulses and what is done is what is this detonation first you know you have combustion taking place wherein you said well in a liquid propellant rocket you have the fuel vapor which is formed from the liquid fuel you have the oxidizer vapor which is formed from the liquid oxidizer both are mixed and burnt or combustion takes place in comparison to a combustion what happens in a detonation is you introduce combustion not directly but you create a shock wave and behind a shock wave you have strong compression and therefore high temperature and it is this high temperature which causes the reaction and therefore when a chemical reaction is forced to take place behind a shock wave well you have high pressure high temperature which now begins to drive the shock wave and a shock wave driven by a detonation is what we call as a detonation and therefore you can use these detonations which have high pressure behind it to for a rocket what we do is you have fuel and oxidizer vapor introduced in a tube you start detonation in a rocket they dot the detonation like a shock travels at supersonic speeds it has high pressure and the high pressure could be used for propulsion purposes that means you have a series of pulses of detonation formed to provide the required impulse and this is what is done in a pulse detonation rocket what should it consist of well you have to initiate a detonation but to initiate a detonation I just said you have to initiate a strong shock in a tube it requires large energies therefore it is going to be difficult and therefore in practice what is done is in a tube you form a flame and this flame is made to accelerate and form a shock wave ahead of it and how it is accelerated is along the walls of the tube you put blockages you put spirals in the tube you make it form shocks and once the shock is formed ahead of the flame well it transits to a detonation and this is the way we form a detonation in practice and you must make sure that a detonation is formed as early as possible that is the run-up distance or the distance for forming a detonation is small such that you are able to reap the benefits of a detonation and therefore you form a detonation in a tube and use a detonation to be able to have a rocket or a detonation rocket which in a pulse mode becomes a pulse detonation rocket let us see what it should consist of therefore you have a hydrocarbon fuel you have an oxidizer I put maybe air could be used as an oxidizer you have an electrical spark you form a flame here this is the tube a simple tube which we call as a detonation tube you form a flame you put blockages or you put some resistances or spirals on the wall which converts the flame into a into it forms a shock wave and a detonation and the high pressure behind the detonation is what gives you the thrust this is the schematic of what pulse detonation rocket could consist of and what is done is well the processes are you have a detonation tube you introduce fuel and air maybe you make it as a vapor maybe air is an oxidizer in this case you could use liquid oxygen you could use any any any other oxidizer what you want you mix the fuel vapor and oxidizer you gradually fill the tube with the fuel vapor and the oxidizer vapor you form a spark here create a flame and the flame accelerates forms a detonation and when it forms a detonation well you have high pressure behind it these are the products unburned gas the detonation travels the detonation reaches the exit of the tube you have all products here and the sequence is over once the sequence of detonation is over you purge the tube using air or the oxidizer you purge it you flush out the products and then the tube is again available you fill with the fuel and air vapor or fuel and oxidizer vapor in it these are the cycles which are involved in a single pulse of a detonation and now you see well you have to purge it you have to fill it you have the detonation which travels all these things take some time and therefore you have the cycles number of cycles which are possible in a tube is around 5 to 10 cycles per second which is somewhat very low and therefore what is done is you have a series of tubes which are bundled together each is a detonation tube you have detonation traveling in it you have fuel and oxidizer which are introduced into this manifold from the tank you have a valve you have a mixing chamber or it is mixed in the chamber you have vapors coming over here you have the vapor mixture being formed in the tubes you ignited you allow the detonation to travel and when the detonation travel it travels at supersonic speed you have a divergent nozzle and you get the thrust this is the principle of multiple detonation tubes clustered together in the case of a pulsed detonation rocket well the elements of a pulse detonation rocket therefore consist of you need tanks for fuel and oxidizer you need a feed system could be pumps or could be pressurized system you allow the fuel and the oxidizer to come into a mixing chamber or it is mixed in the tubes you have a series of tubes you have blockages in the tubes you have the manifolds for admitting well you have blockages as I just now said and well the detonation travels and it is expanded out in a divergent and this constitutes the different elements of a pulse detonation rockets one major advantage is you know the tubes are at low pressures therefore you do not need high pressure pumps a moderate pump or gas for stored gas for pumping the fuel and oxidizer is sufficient and therefore the as you see a pulse detonation rocket consists of simple tubes and it is a detonation in these tubes in pulses which generate impulse for a particular rocket well this is about the pulse detonation rocket and but there are some issues and we find it is still not totally developed there are claims being made that a pulse detonation rocket has been flown but it is not very clear the difficulties are formation of a steady state detonation in small diameter tubes is difficult normally we use diameters of the order of 4 to 5 inches maybe a few tens of centimeters and to form stable detonation in these small diameter tubes is not that easy second is you need large values of run-up distance if it is going to be there leads to lot of thrust you need immediate transition to detonation from a flame and this is also not not that easy unless you are able to design your blockages and your spirals to immediately transit to a detonation well if you use blockages and transition for getting a detonation they also create pressure losses and lead to loss of thrust and as I said earlier time for filling time for purging time for detonation to travel put some limitations on the number of cycles per second but putting a number of tubes together you can get between 50 to 70 cycles per second or even around 100 cycles per second and this is what is a pulse detonation rocket well further developments are required before it can be applied in practice well this is about the pulse detonation engine well you will recall I also told you about the electromagnetic radiation which can be used to generate pressure and thrust how does it generate thrust well you have energy of photon energy you know radiation consists of photons energy of a photon E is given by H into nu where nu is the frequency H is known as a Planck's constant in joule second and the Planck's constant is 6.624 into 10 to the power minus 34 joule second and what is the frequency frequency for electromagnetic radiation is the speed of light in vacuum determined divided by the wavelength of radiation therefore the energy joules in joules is equal to H into nu which is C by lambda and if you have N photons which are available well you have NHC by lambda joules which is available from the electromagnetic radiation well the electromagnetic radiation could also be quantized in terms of intensity of radiation you could say well the intensity of radiation is so much i joules per meter square second and therefore the same energy could also be written as intensity of radiation into the area over which radiation is incident into the time of radiation because intensity of radiation is joules per meter square second you have meter square into second here which is the energy therefore the energy of the number of photons in the radiation NHC by lambda is equal to IAT or rather I can write this expression as I over C I take the C over here I over C is equal to NH over here divided by lambda AT in the denominator but if I look at I over C you know I is joules per meter square second C has units of meter per second I divide Newton meter by meter square second divided by meter second the unit of I by C works out to be Newton per meter square or rather I by C is something like a pressure which is coming and therefore this type of pressure what you get from photons comes from the energy of the photons which are incident on surfaces and but this pressure could be small because you see pressure is I by C you find the Planck's constant is a small number energy of photons is small energy intensity of radiation is small and C is a large number and therefore if I get a small value of pressure and if the radiation passes through a particular material the pressure I get is the intensity of radiation divided by the speed of light whereas if the radiation is totally reflected well it is totally reflected I get twice this value namely pressure is equal to 2 I over C if the radiation is totally reflected and now this pressure can be made use for generating thrust but you find the pressure is small therefore if I want a thrust then I need a large surface area and if I have a large surface area like let us say I deploy in space a huge surface something of the order of a let us say a football ground or a football play field of that area and then I sort of impinge the photons onto it or the electromagnetic waves onto it well I can generate thrust and this thrust which is generated is something like you have in the case of boat and ship in which case you have a sail and the wind move pushes the sail or pushes the boat because of the pressure on the sail so also the pressure of photons on a particular surface can be used for pushing in space and this is known as sail propulsion therefore you find well sail propulsion can also be used using the electromagnetic radiation which are available you remember we also talked in terms of dark matter in space or space having is at zero Kelvin we say zero point energy and this zero point energy of the dark matter in space could also be used in a similar manner for propulsion purposes therefore you find well we could make use of electromagnetic radiation but to be able to complete the subject let us see you know we talked in terms of mono propellant rockets bi propellant rockets at that time we also said three propellants could be used together like for instance we could use liquid kerosene kerosene as a fuel liquid hydrogen as a fuel we could use liquid oxygen as a fuel and all the three could be used together like for instance in the booster stages or the initial stages I have kerosene with liquid hydrogen both as fuels burning with liquid oxygen and in this case you know kerosene is going to burn with liquid liquid oxygen but if I introduce liquid hydrogen into it liquid hydrogen hydrogen rapidly burns and therefore it stabilizes the combustion I can generate high thrust because kerosene is a dense fuel and in the initial phase I use kerosene and liquid hydrogen as fuels and burn it with liquid oxygen when the rocket moves up well I cut off kerosene and therefore I use only the liquid hydrogen and liquid oxygen which gives me a very high performance and therefore in the case of liquid propellant rocket I do the job of two stages and therefore it is quite possible I can cut down the number of stages or I can even use a single stage to orbit vehicle and this is a priced position because in a single stage if I can go to orbit nothing like it but it is attendant with problems and issues this question of liquid propellant rocket was talked of very vehemently around 10 to 15 years ago but the interest in it has waned because the effort is towards getting better materials of construction you bring down the mass of the innards and still try to aim at a single stage to orbit vehicle therefore you find well you can keep on using different mechanisms you have endless mechanisms of generating thrust and making different types of rockets you have myriads of possibilities of pushing in space therefore you find well imagination is the limit and therefore there are various various mechanisms to get back where we got started we said propulsion comes from the word propellary which is essentially to push forward and where do we push forward in space and therefore we can as long as we get something to push using different fields using different different mechanisms using different substances using different principles well I can push in space and I get what we say is rocket propulsion because we are moving about in space I think with this we finish our subject I must tell you it was nice getting going through the different aspects in this class and I hope you all will go through whatever I have been taught and it will be nice if we can advance ourselves by looking into the newer principles maybe looking more at sail propulsion maybe at photon propulsion and I think there must be a good future in this subject right well thank you then I think that is about it