Carlo Rubbia, "New Approach to Nuclear Space Propulsion” - MIT Media Lab Colloquium Series

PRESENTER: Mr. Rubbia has a distinguished high energy physics career, so he needs a little introduction for any physicists who are around here. But, since we're having this talk at the Media Lab, we probably have a more general crowd, so I'll give a quick introduction.

Even if you're not a high energy physicist, you've probably heard about his famous work at CERN in the early 1980s. He's been a major innovator behind the world's first proton and deproton collider there, the CERN P-bar-P collider. There, he led the UA1 experiment, a very large at that time physics collaboration that discovered there are quantum field particles that mediate the weak interaction, the Ws and the Z.

He won the Nobel Prize in physics for this work, shared with Simon van der Meer, the accelerator expert, in 1984. He has been the Higgins Professor of Physics at Harvard during that time. In the early '90s, he was the director general of CERN-- played a prime role in shaping the Large Hadron Collider, the main accelerator of this decade-- turning on in the middle of the decade, we hope.

RUBBIA: Don't fall asleep.

PRESENTER: And he's active in many projects since, including the ICARUS--

[INTERPOSING VOICES]

PRESENTER: --a neutron [INAUDIBLE] neutrino detector under the Gran Sasso, outside of Rome, and other things, including the energy amplifier, a novel and safe way of producing nuclear energy exploiting present-day accelerator technologies, and what he's going to talk to us about today, an extrapolation of these concepts to exciting new prospects for nuclear rocketry. Professor Rubbia.

RUBBIA: Thank you. How do you turn this thing on? It must be--

MODERATOR: [INAUDIBLE] switch.

RUBBIA: Just a simple switch! That's it. Amazing. Things tend to be complicated, around here.

Well, first of all, I don't know what I'm doing here. I don't know what you're doing here, with me. But I had a very nice afternoon, looking at all many, all different, crazy ideas-- and unthinkable ideas. And so I feel rather prepared to tell you about my ideas, which are not that different [INAUDIBLE] one you are tossing around here. Maybe something probably will never happen, it will never be realized, but is worth trying.

I learned that 10% of your ideas come true to reality. Well, it's a good coefficient. If we can do the same with this, I think we'll be in good shape.

So anyway, the problem is [INAUDIBLE] about getting to Mars. And connected with it is the question of whether nuclear power could be used for that. There is a group of people in Europe working quite hard on this. It's about-- it's been growing. At the beginning, we had a few people. Now it's [INAUDIBLE] to, oh, I would say, 50, 60 people, working on these kind of concepts-- very general concepts. They have no practical connection, for the moment, but they show the need for some new ideas.

In fact, the need for new ideas is pretty evident [INAUDIBLE]. You know that the initial description of going to Mars has been done by von Braun, in the famous early book which he wrote in German when he was a war prisoner. And in that he described the standard mechanism to get there.

There are a lot of difficulty, in doing this, at the present moment. Many people, including especially NASA, is trying to get there. It's not very simple. But eventually it would seem to me that this is one of the things mankind would like to do. Surely I will not be there to see that, but I think it's still fun to think about it.

Now, if you wish to do this kind of jump, from going to the moon, 300,000 kilometers, to going on to Mars, where the round trip is 1 billion kilometers, you have to somehow introduce some substantial change. You just cannot use the same technology over again. And the real question is, can you use, or would you like to use--

This projector's not up to MIT standards. Hmm? It's totally out of focus. And there's a lot of operations. Okay. I would say, up to Harvard standards. [LAUGH]

Anyway, so the question is, do you have to use-- can you use-- nuclear power? And I think you can certainly convince yourself nuclear power is [INAUDIBLE] is interesting, because it offers you 3 million times more specific energy than the chemical reactions, when you burn it all. But energy's not the problem, when, you travel in space. What's important, in space propulsion, is impulse, is momentum, is speed.

And, in fact, there is a famous rocket equation which determines how much stuff you can get there when you start with this initial mass. And this is the ratio between-- it's a very simple equation which contains, here, the sum of the speed that you have to add to the various operations and divides up v, speed of exhaust of your gas. And that is essentially fixed by chemical fuels which can reach temperatures of that order and provide the specific impulse of the order of 450 seconds.

I don't know whether you know what specific impulse is. The specific impulse is, essentially, the amount of time as measured in seconds, the amount of time over which a fuel will apply a thrust equal to its weight. So, if I have a kilogram of a certain material, and I burn it, that gives me a kilogram for so much time. And that time is the impulse, specific impulse. So, essentially, if you have a kilogram of chemical fuel, it provide you a kilogram of thrust over 450 seconds. And that is essentially specific impulse, which is proportional to the speed of exhaust.

Now, the idea is, of course, to go nuclear. But the nuclear system is limited in the temperature. No, you cannot even get to the temperature of the gas. And because essentially your material-- so your reactor is not standing the high temperature itself.

And so you do with a thing called [INAUDIBLE] developed in Russia, in the United States, 900 seconds. But you do that essentially not because you have a higher temperature but simply because you're using hydrogen. And hydrogen has this lower atomic number, so molecules move faster, and therefore you get higher speed of exhaust.

Now, in other words, this is a little bit the situation. If you look at the standard-- here is the stagnation temperature, temperature which your system is operating. Here is liquid hydrogen and oxygen-- gets you an exhaust speed of about 4 kilometers per second-- So, a specific impulse about 450. And we are talking about machines, or, if you like, systems, which should allow you to get to temperatures which are typical in the order between 10,000 and 15,000 degrees.

This is hydrogen [INAUDIBLE] and the specific impulses, which are in the region between 2,500 and maybe 4,000. So we are talking about an order-of-magnitude improvement in the specific impulse, respect to this system, based with normal chemical system. Factor 10 in speed means a factor of 100 in specific energy. Let's be clear on that.

And that, of course, makes the situation much easier when you try to go to Mars. In fact, the most-used trajectory is so-called minimum-energy trajectory. It's called the Hohmann trajectory, the one which was proposed initially by the original thing-- document-- book I was telling you before.

The travel time is long. It's 280 days it's essentially freefall trajectory, as you'll see in a minute. However, if you get there and then you want to come back, you have to make sure that the planets are in the right place. And that means that you have to add another 460 days residence.

So there's a fixed trajectory, which you go for a year, you stay there a year and a half, and then you come back for a year. And if you miss the opportunity, of course, things are not the right place, and then you don't get anywhere. Now, if you want to get a little faster with this, you have to catch up on the motion of the Earth. And that in fact requires much higher transfer speed. And I will explain you what I mean.

Now, let's look at this curve here. You see here you have essentially represented here the [INAUDIBLE] of residence time. And you have the 460, 640 days there, which correspond-- 640 days, which corresponds to the number I gave you before, 540 days. And then you see, here, the various-- this is the Hohmann trajectory.

And then you see, all these lines are possible. They're characterized, of course, by faster transfer velocity, but they give you the same residence time, which is of the order of a couple of years. If you want to go to a much shorter time mission, you have to make conjunction trajectory. In other words, you have to run after the Earth, which is escaping away from you. And that implies that you have to go to much faster speeds. This is the area which is permitted. There is no other way in between.

So, if you have very slow motions, 5 kilometers per second, then you will be forced to do that. If you want to get to stay there for [INAUDIBLE] less time, you are obliged to go to speeds, transfer speeds, which are in fact quite large-- 25, 30 kilometers per second. Which implies, of course, speeds of your ejected gas [INAUDIBLE] order of magnitude and therefore require new methods.

We are talking about, now, missions of this type for the manned CLIPPER missions. And [INAUDIBLE] stay there for the case of transporting materials. Now, let me show you briefly how this mission can be realized.

The fast mission, as I said, consists of leaving the Earth here and then running on this trajectory and getting here, after 150 days, on Mars. Then you stay on Mars as long as you wish, from 0 to 100 days or whatever is required. And then you decide to come back. If you want to come back, this time the Earth has gone further, so you have to catch up the Earth position. So you have to run much closer to the Sun and then make a turn inwards. And that allows you to get back even something like, as I say, another 179 days, but-- so, therefore, you can close the whole thing in a year. But it requires speeds of, as I said, much higher.

This is for the people which want to get there. Now, if you want to go and send equipment, as you probably need anyway, then you can follow the Hohmann trajectory, which is the one represented here. And you can see what Hohmann trajectory is all about, is half a circle. You start here. You go in 280 days to Mars, 260 days. Then you have to stay all this time, until such a time as the Earth is in the right position. And you leave here, and then you go another Hohmann trajectory, back 180 degrees [INAUDIBLE], and you hit the Earth where it belongs. That is, in fact, the long-time trip.

Now, we are trying to study all kinds of different missions with the potential challenge of this new engine, which I'll describe in a minute. Essentially, nuclear power is going [INAUDIBLE] to be used only for transplanetary trip. Classical method for landing and takes-off, so you use standard fuel to get to orbit and then to come down.

Air braking is being removed, at this level. Air braking is an extremely complicated procedure. Imagine you arrive on Mars and then you have to get into sort of an orbit which is slowing down into the gas of Mars. And the window is very narrow. The difficulties of doing that are considerable.

In fact, many of the problems in getting on Mars material come from that point. We think that, with people, you should avoid it.

The space station ALPHA is used as a departure. And [INAUDIBLE] mission needs a reusable shuttle, since many trips are necessary and-- could be necessary. And this device will in fact operate on a periodicity of Earth-Mars orbits, which is 740 days.

And the question is, the presence on Mars. So the man is necessary for the complexity of the task, but the harsh environment limits a stay to a short time. It's inconceivable you can leave a person for first time on Mars one and a half years eventually to make his own fuel and come back and so forth. So he has to be able to go there, to get the equipment going, but then he has to come back. And you need no sharp departure window.

And then you have two kind of missions-- the CLIPPER, for fast manned transit, and the TUG, for material transportation, which can be slow. And the key, of course, all that is a higher specific impulse.

Now, the present state of nuclear propulsion is not a very good state. Essentially, Nerva is the design which is now the best in town. And it's an essentially gas-core fast reactor.

The maximum temperature is determined by the temperature of this uranium zirconium niobium carbide, which has to remain solid. It runs at high pressure, because the gas has to go from here to there. Here is the low temperature. Here is 3,000 degrees, so it has to warm up, going through the rods of the reactors-- so, high operating pressure.

The mass of the fuel must be large. It's 1 ton of highly enriched uranium, which is put into orbit in order to get criticality. But, however, during the burning of all this system, only 5 grams are burned. The rest is-- it just goes there. You have to deliver-- get rid of it. And there are particular-- many people getting very nervous when you put a ton of highly enriched, bomb-grade uranium on orbit.

Neutron leakage is very large-- 10 to the 19 neutrons per second. So it makes it very hard for people to stay around. It has to be shielded. And even when it gets shielded, it is not trivial. You can avoid that kind of irradiation from [INAUDIBLE] these fluxes.

And the real problem is essentially that the fuel is, in fact, the hottest part in the system. Now, the basic idea is, can we keep the fuel cold and the gas hot? It's what you would like to do.

The answer to that is a very simple idea which we are working on, which is direct heating by the fission fragments. Now, let me remind you that 88% of nuclear energy due to fission is carried by two fission fragments, the rest being additional neutrons and gamma rays. You have essentially uranium 235. You swallow a neutron. Uranium 235 starts to vibrate, to move. It separates, like two drops of water.

You get two little droplets. They're both positively charged. Therefore, they repel each other. No longer nuclear forces are present, because you have severed the connection with the two. And the 200 MeV you recover is just simply the coolant repulsion of these two parts. So two fission fragments come off, and they carry essentially 98% of the total kinetic energy.

So why not use-- in addition, there are a few gammas and neutrons, but they carry only a few MeV. So most of the energy is carried by these two fission fragments, which are back to back, so to speak. Now the question is, how can you make use of that fission fragment directly? They have an enormous enthalpy inside. They represent tremendous high temperature. But you never see that mini reactor, because it stop into the reactor. Then you get warming up of your material. You don't get access to that beauty that nature is providing you, there, with.

And so the question is, can you use them directly to heat up the gas? The penetration range of those fission fragments is extremely short. It's 10 microns, in the fuel. And usually the energy is converted in heat inside the fuel. And now the question-- can you use directly these fission fragments in order to heat the gas? That will, of course, give you an enormous advantage, as you'll see in a minute.

So we assume a very thin fissile deposit. Let's take a foil-- put on a foil a layer of fissionable material. And then make it thin enough so that the fission fragment can come out. And then see what happens, as a function of the thickness of the foil. Here you have a plot which says 1 micron, 1 milligram per square centimeter, 10 milligram per square centimeters.

These foil are made out of uranium or americium or whatever you want. And you see here the energetic fraction of the energy coming out. Obviously you get 44% if you had 0 thickness, because 88% is fission fragment and 44 are going one direction, 44 go to the other.

And then, when the thickness becomes bigger, you get a loss. For instance, if you have 1-micron thickness, you get 34% instead of 44%. If you go to 3 microns milligrams per square centimeter, you get 24%.

Now, what are we talking about, here? We are talking about 1 milligram per square centimeter, which is 0.8 microns in thickness, which is 10 grams of a meter-squared surface. And we are talking about 1,100 atomic layers.

This is certainly not what you'd call a nuclear fuel. A nuclear fuel is a very heavy material, hundreds of tons of stuff inside the system. Here is rather a radioactive source, if you like, deposited on a thin layer.

Now, how do you get criticality? Now, for that you need to do some clever things, which have to do with neutronics. Let me anticipate to you that I do succeed in having criticality with that little bit of americium.

Now, let me be clear about numbers. If it take this americium, which is the largest fission cross-section known around, 5,300 barns, and you get the neutron through that, it is 1% probability of interacting. So, obviously, you would like to have 100% probability of interacting, to keep the reaction going. And therefore you need about a factor of 100 enhancement.

So can you enhance the neutron flux, with some kind of storage [INAUDIBLE] device, so that you can get the neutrons over and over again through your americium foil and enhance, this way, the probability of fission? The answer is yes. Well, we have done lots of experiment at CERN on that. And, in fact, this is not just a theory. It's just an experimental results.

The basic idea is that you take a highly diffusing but neutron-transparent medium, and you let in that thing a neutron travel. Only elastic cross-sections is there. There is no inelasticity.

So this neutron moves around inside this material, in the same way that, for instance, an electron moves in an argon gas. Well, the physical reason's very similar. Normally, you're talking about nuclei which are double-magic. We have here the great experts of these things. And a double-magic nucleus is like a noble gas. The shell is closed.

So, in the same way that an electron cannot capture in argon of helium, and it will go around as a free electron forever, here we are talking about a nucleus. And a neutron will not capture the nucleus, because the shell is closed. And the cross-section is very small, and the result is, this thing will just go around.

Look here-- there's [INAUDIBLE] tortuous path inside the highly efficient medium. This is computer simulated, but the reality is not any different. The neutron is essentially stored, stored not by, like, a storage ring, it's stored by diffusion. It's a random process but, effectively, it keeps the neutrons together.

And that is, in fact, has been showed, that, in fact, 60-meters-long path is wound in 35 centimeters, 1 foot in length. And this is due to the fact that this movement [INAUDIBLE].

Now, in such a storage medium-- so you have a volume of this type of high-purity graphite or high-purity lead. In that volume, you have this phenomenon. Such a storage medium, because of multiple trials, even a very tiny amount of impurity can absorb a significant amount of capture elements.

Suppose you have a ball full of water. And you put there a little bit of ink, just a little drop of ink, the whole box will have color of the ink, even if the amount of ink is infinitesimal. Because the water is transparent and is highly absorptive. And sooner or later light will find the molecule ink and will send back the color which you put in.

And same thing happens, here. You have a very small amount of impurities, inside this medium, because you go around and around and around, over and over again, you will eventually hit an impurity, and then you'll get a factor, because of this multiple trial effect. And therefore, 20 grams of americium inside this box will make it critical.

So now you can make devices which are critical with grams of material, not kilograms. Remember, in the Nerva, you needed one ton of highly fissile bomb-grade uranium. Here, with 20 grams of americium, you're critical. In this continued medium.

Now, of course, you don't want to have continued medium. You want to have, essentially, the way to do this fission. So we introduce a statement which we call Hohlraum, because it reminds of a black body, which is essentially a box which [INAUDIBLE] high-purity [INAUDIBLE] diffuser, on which walls you put a thin layer-- 1 micron, 1,000 molecules layer of americium. And then what happens-- neutrons will ping-pong between the walls of this container, will traverse many times the walls. And [INAUDIBLE] enhance the interaction probability, as we will see in a minute, about 100-fold.

And the ultra-thin fissile layer requires a finite fission probability, because of these many traversals. And the new fission produced neutrons, of course, will get thrown into the box, and it will start again. More precisely, we assume the empty cavity is covered with a layer. The configuration is similar to a black-body radiator. And I'll remind you, the black-body radiation in flux is uniform independent of the shape. And the flux is determined essentially by the properties of the material, not by the geometry. The same thing will also happen here.

And identifying independent shape of the cavity, provided, of course, it's well-behaved. And therefore it doesn't matter what the shape is. You will get this result.

And also neutron temperature, and therefore neutron energy, will be relayed to the center by the fuel but the quality of the container. So temperature will be fully thermalized with the temperature of your container. So you'll know very precisely what the energy neutrons you're dealing with.

And, of course, no neutron will come out, because, being a black body, it will contain the neutrons inside. If they're inside, they're not outside.

Now, the theory of enhancement can be easily done. I will not spend much time with this. It tells you that, if you have any shape, and you take the reflector away, and then you put it back and you leave the same source, and you ask to change enhancement of the flux, because of the reflector on, that enhancement factor is equal to that expression which is square root of elastic scattered [INAUDIBLE] capture cross-section of scattering.

Elastic is typical 10 barns. In a capture cross cross-section is order of millibarns. So we are talking about a ratio 10 to the 4, 10 to the 5, between the two numbers. And therefore, when you take the square root, you get good results.

And there are two parameters which are important, here. One is the diffusion length, diffusion length, which means the size of your sphere of containment-- in other words, the thickness of your walls that you have to use to do the job. And the second one is enhancement factor-- by how much you increase the flux.

Now, if you get graphite, you get 27 centimeters diffusion length. And you get 94 enhancement. Beryllium will do fine, as well. It is much smaller. It's 12 centimeters, but your enhancement factor only 73.

Lithium will be completely transparent. And liquid deuterium will give an enormous advancement factor, but you also have very long length. So you'll have to build a huge device. It's not practical. And liquid hydrogen is also excluded.

So, essentially, you are ending up with beryllium and beryllium oxide and carbon as interesting materials. And then you try to calculate with this system criticality. And you do achieve criticality rather rapidly.

Here is the thickness of the carbon wall, which is increased. Sorry-- here. I increase the thickness of the carbon wall. And then, here, I put the k value-- which means k equal 1 means criticality, means you get back as many neutrons as you made.

And if I have a layer of 1 milligram per square centimeter, 1,100 layers, you can see I get critical with about 250 grams per square centimeter, which means 1 meter 20 of carbon. It's a bit too much. But if I go to 3 milligrams, you see I'm already running at 75 here, grams per square centimeter, which means about 35 centimeters, which is that much.

This is 22 centimeters. [INAUDIBLE] flying over the Atlantic. That's [INAUDIBLE] 22 centimeters.

And so it's a quite reasonably small amount of reflector. And beryllium oxide can do it fine, even better. And you can see that, indeed, criticality is obtained with layers which are only a few milligrams per square centimeter.

So it is possible, with this technology, to make a critical reactor which will operate with a very small amount of material. Remember, 1 milligram a square centimeter is 10 grams per square meter of surface. We are talking about probably 100 square meters, for this mission, so we're talking about between 5 and 10 kilograms [INAUDIBLE] of americium to burn. Which is not very much.

In fact, americium has the same properties as uranium 230-- plutonium 238, which is used now on Cassini mission and was on Voyager. And usually you had more than these kilograms on it. So it's deja vu, as one says in French.

Now, the problem is, which fuel element you like to use very quickly? You have to have a high fission probability, exceptionally long lifetime, and not-too-extravagant production procedure. Now, of course, americium 242m is an ideal device. This is a cross-section.

Notice the cross-section drops, when a temperature goes up. This is the temperature of your Hohlraum, 1,000 degrees 2,000 degrees. So you're running here in this region that correspond to those energies. Capture is very high, fission very high, capture very small. Big ratio between the two.

Now, if the other element which is not so bad is plutonium 239, which is well-known as a resonance here, at about 0.3, 0.24 electrovolts. So that could be used if you want to make a container which had high temperature, for instance of the order of 3,000 degrees. Then, in this case, you could also use plutonium. But you'll have problem of temperature coefficient.

Here you have very nice negative temperature coefficient, because of this cross-section. So plutonium is second choice. americium is the first choice.

So I've done a very complete burn-up through a reactor of this type. I'm not spending time to discuss that. Here is what happens when you take 1.4 kilograms of that americium and you burn it.

You can have a fairly long burnup. "Burnup" means megawatt multiplied by day, per kilogram. Means one kilogram will give you that many megawatt per that many days. So 1 kilogram, here, will give you 400 megawatts for one day, or 1 megawatt for 400 days, which is much bigger than what you get for a normal reactor. A normal reactor, you get a burnup of 20, 30, in those units.

Essentially comes from the fact that the system is much more efficient, when it comes to burning. For instance, here, you can see, during this 500-kilogram, 500-megawatt day per kilogram, the americium coming down from 1.4 kilograms to about 0.7, americium being burnt, americium 243 produced by the captures, since there's a clean situation. And here it stops, because it gets below critical.

And so you can get a lot of energy out of this. You can burn about half. Even more than half of your americium can go. And 3 milligram per square centimeter layer delivers that much power.

And, of course, if you have a much longer operation, you have to remake your layer. It's a very thin layer. It can be deposited in some kind of a way. This is-- the whole technology has to be developed. But obviously, periodic refueling may have to be envisaged, eventually.

Now, how do you reduce the americium very rapidly? Well, essentially, you start from americium 241, which is naturally grown in aged reactor-spent fuel. Plutonium 241 has a lifetime of 14 years and goes into americium. And then you make a neutron capture. And here is to explain how you do the neutron capture.

And then you get an extra neutron, and you end up to 242m. 242-not-m is not stable. And so the only thing left is that one.

So let's come back to the question of heating of the gas with those fission fragments. And so let's assume we have a tube. Typically, we are talking about a tube which is 2 and 1/2 meters long, 30 centimeter in diameter.

This is like a fuel rod in a reactor, only a fuel rod in a reactor is full of material and the cooling is outside. Here, we are talking about a fuel rod, which is empty. And the fuel, instead of being deposited as the volume, is deposited to the surface, inside.

And then we get gas in, and then the gas go through this thing, and it gets heated up. Let's see what happens, here. You have a tube, 30 centimeters in diameter-- in this particular case, 2 meter 50 long. There is a thin wall which is cooled from liquid from outside and produces 2 watts per square centimeter of heat and produces 3.95 megawatts of power. This rod is immersed, as you will see, in a lithium bath.

On this wall, there is americium, 3 milligram per square millimeter. This total gram is [INAUDIBLE] 71 gram, so it's not very much. And there is a little bit of gas flowing through this system, 1.2 grams per second-- so, 58 microgram per square centimeter. So it's very slow-flow gas. The whole damn thing is taking 1 gram of gas per second.

And the gas is flowing very slowly. It leaves the wall. It departs from the wall, and it [INAUDIBLE] warms up. And when it comes down here, you will see, from exact calculation, there is temperatures of the order of 10,000, 15,000 degrees. And, in fact, it's the wall are cold, and the gas is hot.

So it's maintained. This is my statement maintained [INAUDIBLE] foil reasonable temperature cooling from outside the wall, the sheet, the foil. And the gas is moving from the wall. And when it gets in, it gets hot. And, by the time it comes down there, it goes between 10,000 and 15,000 degrees [INAUDIBLE] which limits there are and how can you get there.

Now, temperature limitation due to mechanical strength of nuclear fuel are removed, because the fuel is cold and the gas is hot. Now, there is another-- there are several gifts of nature in here, which are worth reminding you. First of all, the range of the fission fragments exiting from 3 milligram material has been calculated for the actual Z and A distribution. And it turns out that the specific ionization per unit of mass is, in hydrogen, much greater than 1 in americium. There is almost a factor 20 higher specific ionization loss in hydrogen, with respect to americium. What's the cost of that?

The cost of that is the following-- that ionization losses only work when the particle traveling is faster than the electron bound in the atom. And the sufficient fragment is beta equal 0.04, 4%. So many of the electrons bound in the americium foil have a speed which is higher than the speed of the fission fragments. And therefore they are essentially not participating in the ionization loss.

If you take hydrogen, well, the velocities is 1 over alpha, and therefore it's much smaller than the velocity of the fission fragment, so that you have high ionization losses. You can see here that the range of these fission fragments gives about 10 microns, as I said, in americium, but it's less than 0.5 micron in gas. So you get an enhancement of a factor 20 in ionization losses in the gas, which is [INAUDIBLE] extremely important.

And so even 1/2 a milligram per square centimeter gas will absorb the fission fragment completely. So you can operate very low pressure. Actually, in reality, the fission fragments go in all possible direction, so the range is only the upper limit to the travel.

So let me take two infinite planes coated with americium, defining a volume filled with hydrogen gas. So you have two planes, here. And these two planes are now filled with gas. And the fission fragment traversing it.

The thickness is 1/2 a milligram per square centimeter, so it's about 20 centimeter, 3 atmospheres, just to get [INAUDIBLE] understood. Even at high temperature. And then you have, this is the kind of heat produced by the wall to the left. This is the kind of heat produced by the wall to the right. And when you put them together, you get a reasonably flat distribution.

The important number here is this very simple one dimension of this drawing, which in fact will be-- is worse than the case of the tube, which I'll discuss two dimension in a minute. You get, with 400 watts per square centimeter on this foil, you get about half a megawatt per gram in the gas. Now, a megawatt is enormous power. A megawatt in a gram is just putting a megawatt on a mosquito. And obviously, if you put it a megawatt on a mosquito, it will fry very rapidly.

And therefore the idea is that you can keep the foil under very normal condition of a reactor but get real hell inside the gas. As I said, you can get a megawatt per gram with hundreds of watt per square centimeter on the wall. Now, this whole thing is played the following way.

Here you have, in this curve, here, the watts-- unfortunately, it's hard to see. I cannot even see, myself. I wonder whether you can read, yourself, over there.

Here you have one megawatt per gram, 100 kilowatt per gram, 10 megawatts per gram. And here you have [INAUDIBLE] surface power density watts per square centimeter, 100 watts per square centimeter, 400 watts per square centimeter, 800 watts per square centimeter. This is the kind of region which normal reactor is operating.

And you can see that you can go all the way up to about 1 megawatt per gram, in a gas, with reasonable heating on the walls. And that is, of course, transform into temperature, through the enthalpy diagram for hydrogen, where you have-- here, you have enthalpy 10 to the 5, 10 to the 6. This is in joule per gram. And you have temperature [INAUDIBLE] 10 to the 3, 10 to the 4, 4 10 to the 4.

Now, you have to pay a little attention to this curve. Here, up to 3,000 degrees, the enthalpy is rising linearly. This is a perfect gas. Enthalpy specific heat is constant. It's essentially a temperature of a single hydrogen molecules.

At about 3,000 degrees, the molecule disassociates, becomes [INAUDIBLE]. Then you get the enthalpy to do the destruction. That is, like, a chemical reaction. You put energy in it; eventually it will come back, after you've done it.

And then you start with another slope, which is twice this sharp, because you have twice as many atoms there. And therefore you have a higher specific heat contained for a given gas. And then you get to about 10,000 degrees. And when you get to 10,000 degrees, the gas gets ionized. And then you are getting the 13 electrovolts of ionization. Remember, 10,000 degrees kelvin is one electrovolt, but here we are talking about having recovering something like of the order of 10 electrovolts. And therefore the enthalpy grows tremendously, even if the temperature doesn't go very high.

And essentially, you are now talking about enthalpy, joule over gram, is going to power watts per gram times dwell time, time which you are in the exposure, time gas takes to come out of the container. Typically, we're talking about of several seconds, in this machine. So we are talking about seconds, multiply megawatt per gram, equal to gigajoule per gram. Gigajoules per gram is right here. So we're talking about temperatures 10,000, 15,000 degrees.

And, when Nerva, you have fast motion, short dwell, here you have slow motion, long dwell. And this gives you the impulse which you can get with 10,000 degrees. You have about 25 kilometers a second exhaust speed and about 3,000 in specific impulse.

How much time do I have left?

MODERATOR: So 20 minutes.

RUBBIA: 20 minutes. Okay, fine. Now, then, to do that more clearly, we are talking about parameters of this tube, 20-centimeter radius, 2 and 1/2 meters in length. And we are talking about 1,000 degrees for the wall. Final temperature in the gas of 10,000 degrees.

And we are talking a very small amount of gas coming through. We are talking about essentially speed of the gas of 0.75 centimeter per second in the beginning, 3 from 5 grams per second through the system. And this is the heating stage. Then you have an expansion cone, which brings you down to about 25,000, 2,500 in specific impulse, so 25,000 meter per second, and with a thermodynamic efficiency of about 17%.

Now, there is one technical point which I'd quickly like to mention, which is a substantial limitation to this method, to this system, which are the radiative losses of the gas which gets heated. When that gas gets very hot, it starts to radiate-- emits power away. Eventually, if the power radiated is more than the fission power coming in, you are in trouble.

Essentially, it's a lot of work to calculate all this. There are type of losses-- line, band, and continuum. Now, remember, [INAUDIBLE] we only have atomic hydrogen. Therefore, the molecular levels are not there. And essentially you're talking about theoretical hydrogen [INAUDIBLE] 1, 2 electrons [INAUDIBLE]. [INAUDIBLE] then [INAUDIBLE] we know all the numbers, so all the calculation can be done very precisely.

And it turns out that the radiated power there is essentially, the line emission is quickly absorbed. Essentially, the line looks black because of the absorption is, in fact, quite strongly in the line. And therefore, you don't have--

Essentially, it's like the sun. When you look at the sun, you see the sun, the plasma of the sun. You see the lines dark. You don't see the light coming out of it. So there is no radiation from the lines.

There is, however, a substantial band emission, which is there. It has been all calculated. I'm not going to get into details with this.

Essentially it turns out that what happens is that the light is growing with the temperature of the gas. And this is frequency, and this is the power, after a certain finite distance. And it's becoming a lamp, a big lamp.

And here is the surviving radiated power, as a function of the distance of the gas. This is about 1 centimeter of gas at 3 atmosphere. So most of the light emitted by the band is absorbed immediately in the region where it's emitted. And therefore it contributes only to heat conductivity.

But there is a [INAUDIBLE] component, which is this flat line here, which is, in fact, having a very constant, a very long attenuation path. And this is the light you can see on the walls, yourself. This is all taken into account in the calculation. And essentially it tells you that, up to about 10,000 or 12,000 degrees, you can operate, and above that you have a problem of radiation which you have to solve.

The other question which is important is the recombination of the hydrogen. The hydrogen is starting from being-- well, let me show you, here, the curves, which is probably the best way of showing it.

The hydrogen starts as a molecule. This is this line, here, is function of temperature. This is logarithmic scale. Very quickly becomes an atom. And then, at about 10,000 degrees, it starts to ionize. Here, for instance-- 3 atmospheres. Here 10,000 degrees [INAUDIBLE] ionized. And at 15,000 degrees, it's about 50% ionized.

So essentially what happens is that you are dominated with atomic gas. And above a certain value, you're also dominated by plasma which is substantially ionized. The plasma recombination electron proton-- in other words, the combination of the plasma-- it's a very fast reaction, because it's a gamma plus neutral hydrogen equal to electron plus proton. It's a perfectly large cross-section.

The recombination molecule to atom, it's very slow. Remember, two atoms cannot make a molecule, because there is no momentum energy conservation. So a single collision of two atoms will not make a molecule, because they have to emit a gamma, and the gamma is not there, at those temperatures.

And therefore the thing is metastable. You remember the so-called hydrogen torch, plasma torch, in which you take atomic molecular hydrogen and you make it atomic hydrogen, and then you throw it on a wall. And when it hits the wall, it becomes molecular again, because it finds recoil. And therefore, it gives you heat. And, in fact, you can melt a lot of stuff in this way.

Now, that is essentially what's happening in our case, here. However, as I showed at the beginning, first of all, the gas is essentially-- therefore, our gas is frozen. In other words, we expect this gas to come out essentially as an atomic gas. However, atomic gas means the A is 1 instead of being 2. And therefore you pick up a square root of 2 in the number of the speed of your gas.

And therefore, although you will lose part of the energy because you're dealing with an atomic system rather than it being a molecular system, the effect on the specific impulse-- as I showed in the first transparency, which I hope I can find again, here-- is quite small, essentially because of the fact that the gas is reducing its A. And so we have essentially--

This gas comes out very cold. When it hit-- so the problem of the funnel is not a very important point. Now, since we are in computer science, here, let me tell you some computer simulation. So I'll feel a little bit more at home.

So what we've done-- we have developed a lot of programs which do the following. This is doing fluid-dynamics calculation. This is making a fission-fragment transport heat release. This is worrying about radiative effects of the gas. And this guy, here, is combining the energy losses and energy radiation to close the loop with the program, which is the program on fluid dynamics.

So this is a fluid dynamics program. Then it goes into a heating from fission fragment [INAUDIBLE] neutronics providing the supply. Radiation, you lose power, you make power, you take a power balance, and you put it back, and you make [INAUDIBLE] current calculations.

And then we have done several models, here, with a typical 20-centimeter, 2 and 2/2 meters machine-- simulation, rather sophisticated. We have hydrogen inflow at 100 kelvin, which is in fact uniformly distributed. And-- this thing is really horrible.

And then you get essentially various intensity of flow. Let me show you some results. Here is, we are increasing the temperature of the heater. This is half of the cylinder. You can see that there's another half coming on the other side.

So you start from the wall, here. When you move on, you heat up the gas slowly, until the gas gets very hot. And then, eventually, down here, the gas is the hottest. And it comes out from the funnel. You see the same thing here, case number 3, which is the one we consider the most appropriate. And you do get about 10,000 degrees out of that. And you do get essentially all the gas is nicely hot, at the end.

So the fluid dynamics works extremely well in this story. Actually, you can see, here, global power, power, the position and temperatures. And you can see, in fact, that these kind of calculations are perfectly adequate. So we have a good fluid-dynamic model of this.

The heating process of the fission fragment in the two dimensions is quite remarkable. Here, we had 200 watts per square centimeter on the tube. And here, we are collecting, here, 900 watts per milligram. So we are collecting 900 kilowatts-- so, 1 megawatt per gram of the gas.

And you see here is the radius of the container. In the central part, it's very uniform. These two lines correspond to the top and the bottom, where there is no coverage, on one hand, because you're going out into the nozzle, the top, because there is nothing on the top. The americium is only the positive of the sides.

If you were to increase the pressure from 2 atmospheres to 12 atmospheres, then you will observe this. Your fission fragment would just stop before they get to the center, and only the sides is heated. A few more minutes left. I can describe you, here, some of the design consideration.

As I was saying, this is the kind of tubes we are talking about. And then these many tubes are inserted inside a box. So let's look at this. This is a realistic engine.

This will carry emission, according to what [INAUDIBLE] calculating, over to Mars and back again, in the way he was describing it before. It's made essentially by a series of tubes, 2 and 1/2 meters long. Those tubes are equivalent to the fuel elements. They have, on the bottom, the nozzle, and therefore they can be clamped on a baseline, if you like.

They are immersed inside a sea of lithium, liquid lithium bath, which is the coolant And these works as a boiling-lithium reactor, if you like. The lithium is all around here. And it touches the outside of those tubes, so that the outside wall is cooled with this lithium. And the lithium is traveling all over, here.

There are control bars coming into the Hohlraum, so you take away the neutrons to control the process. The system is essentially a lithium-cooled system. You have lithium which boils. Lithium becomes vapor.

The vapor lithium goes over to the panels, which are the cooling panels. There it condenses again. It becomes liquid and then is recollected and put back again. And so it operates at constant temperature. But there is the heating-- the fraction of the heat which is used by the boiled-off heat, which is used for cooling.