Nuclear Energy Questions that often go unanswered

from Cavendish Press Ann Arbor

Questions about Nuclear Energy that often go unanswered

Q Can a nuclear reactor melt down, or can't it?
On page 209 of the Nuclear Energy book, you say that some writers claim that a meltdown through the concrete floor of the containment building of a nuclear reactor can not happen. Is there a way to know whether that claim is correct?

A    Whether a meltdown occurs depends on whether the heat energy needed to melt down is available. The answer depends on the values of several factors, some of which are known quite accurately, and some of which must be approximated.

    The word "meltdown" is used variously. Some use it to mean the melting of fuel rods in the core (which has occurred) leading to the melting of the bottom of the steel reactor vessel, with the molten core content flowing to the concrete floor of the reactor containment. All of this is generally agreed to be possible.
    What happens from there on is in dispute, some arguing that the molten material will just puddle there, and the containment will remain intact. We will use the term, "meltdown" to refer to the other alternative, a breach of the containment floor, in which the molten material from the reactor core has melted not only through the steel container, but also through the concrete floor of the reactor containment, and into the ground. This is the catastrophe described in the movie, "China Syndrome."

     Whether the heat energy needed to melt down is available depends on the balance between these two quantities:

The amount of heat energy required to melt through the concrete floor of the containment building
The amount of heat energy produced in the pool of molten uranium and fission products that pours onto the containment floor when the reactor vessel bottom melts through.

    The only source of heat energy is the radioactive decay of fission products. If that energy is sufficient to melt or otherwise open the concrete floor, then a meltdown can occur.
    The balance between the heat produced and the heat required depends on some factors that may vary from one reactor to another, and that depend somewhat on individual circumstances. We will make estimates and chart possible outcomes.
    Throughout, our calculations will be "transparent." The numbers used will be given, the basis for estimates will be stated, and the logic will be traced, so that any reader can check the whole argument out, and if so inclined, can repeat the logic using numbers that he or she may regard as better estimates, and see what this does to the conclusion.
    The outcome of our calculations does not prove that a particular consequence will occur. It does tell us what is possible and what is not possible.

    First, some facts:

Meltdown is not a uranium bomb. The conditions for an explosion such as occurs in a uranium bomb are not met in a reactor. The enrichment of the fuel in a reactor is insufficient for bomb requirements.
In a meltdown, fission has ceased. Most likely a SCRAM will have occurred early in the accident, positioning the control rods to absorb all fission neutrons. Even in the unlikely case where a SCRAM failed to occur, the fuel rods and their content begin to melt only after there is uncovering of the core, in which case the absence of moderator (water) in the melted fuel combined with the control rod metal contained in the melt, will prevent any further chain reaction.
With fission having ceased, the heat energy that is produced in the molten content of the fuel rods is from the radioactive decay of the fission fragments that were produced in the reactor prior to the accident. This is described on page 208.
In an accident that has gone out of control, there is without any doubt enough heat generated to melt the steel bottom of the reactor vessel, causing an approximate 200 tons of molten uranium and radioactive fission products (above the melting temperature of uranium oxide, 2500^o, or at least that of uranium, 1130^o) to pour out onto the concrete floor of the containment building in a puddle with an average depth of about ¼ inch.
Core has been uncovered and partially melted in known nuclear accidents. It can occur when the normal process by which heat is removed from the core is interrupted. Heat is normally removed by pumping water (or some other "primary coolant") through the core. The heated primary coolant either directly or by way of a secondary coolant loop, drives the turbines. Emergency features, such as the high pressure injection system (HPI), have limited power to replace the circulating systems, and over the long run can not by themselves cool the containment. The loss of coolant (LOCA - page 217), which can come about in a number of ways, is the typical reason for the failure of the ability to remove heat. (At Three Mile Island a relief valve was left open for several hours, and meltdown was avoided when cooling was partially restored, just in time.)
THE CALCULATION

    (1) Calculation of heat energy required to melt through the concrete floor.

We estimate the floor of the containment to be 140 feet in diameter, and ten feet thick. This is a volume of 3,750 cubic meters of concrete. Using the density of quartz (2.65 g/cc -- concrete is made of sand (quartz) and gravel), this is a mass of 10 million kg, or 10,000 tons.
     In the extreme case, if everything were to hold still until every last chunk of the containment floor is heated to its melting point (1500^oC) and melted, it would require 2x10¹³ (20 trillion) Joules of heat energy. (The specific heat of sand, the main ingredient of concrete, (the book value for quartz is used) goes from .17cal/g^o at low temperatures to .3cal/g^o. Its heat of fusion is 50cal/g)
    2x10¹³ Joules is an absolute upper limit of heat energy needed for meltdown. It is unlikely that the entire floor would need to be melted in order to breach the containment and release the molten mix of uranium and fission products into the ground.
    To begin with, the floor of the containment is not a flat slab. Prominent in any design is a "sump," a depression several feet deep and in area anywhere from 5 to 25 per cent of the floor area. If it is to serve as a sump, the floor must slope towards it.
    One must not imagine, however, that the molten fuel rod contents will flow towards and into the sump, as water would. The molten material will first solidify wherever it lands, melting some of the concrete, possibly producing new depressions in the floor.
    Imagine molten lead dripping down onto a large cake of ice. Wherever it drips it will make its own hole, which may gather other drippings.
    A major flaw in the lead&ice analogy is that the solidified lead would remain solid, because it does not contain within it the source of the heat that melted it in the first place, as the radioactive fission products do. The molten core material that has solidified will continue to be heated from within by the continuing decay of the radioactive fission products, and so will solidify and melt over and over, as it progresses down through the melting concrete, and even after it melts through and hits the ground below. One must resist the intuitive picture that, once through the floor, and into the cold ground, the core material will be chilled solid and just remain in place until it cools off.
    The 10 cubic meters of molten fuel rod contents, however they reach the floor, are likely to puddle and concentrate in smaller areas. Some will collect in the sump, some will create its own depressions. Several things would happen before the entire floor would melt:

Regardless of the geometry of the melting of the concrete, heat would spread downward and sideways through the concrete floor. Eventually, it would begin to heat up the earth below. This would produce first superheated water (under pressure) eventually vaporizing the moisture in the earth into steam above its critical temperature. Steam confined with no place to go, at its critical temperature of 370^o, exerts a pressure of about 218 atmospheres, or 1.6 tons per square inch. Over the entire concrete slab, this is 3 million tons, enough to lift the containment building up in the air many times over.
It is unlikely that this steam would lift the containment building, because there are ways in which the pressure can be relieved.
The pressure can be relieved by pushing sideways on the ground, allowing it to expand out around the side of the building and up into the free atmosphere.
This sideways movement of ground with the entrapped high pressure steam, however, would most likely leave the floor unevenly supported, with the enormous upward force helping to buckle the concrete slab. Even ten feet of reinforced concrete is likely to develop cracks in such an event, allowing the molten uranium (which is 7 times denser than concrete) to flow down into the ground.
The other way for the floor of the containment to be breached is for the heating to be localized, melting a hole in the concrete allowing the molten uranium and fission products to flow toward the ground.

In either of these ways, a sideways explosion of steam, clay, and rock remains the only way to relieve the pressure. The high pressure steam consists of the moisture imbedded in the clay or soil in the ground, and is not likely to be able to collect into pockets of pure steam and be released cleanly. The escaping steam will take with it the clay and soil and rock, as well as the highly radioactive fission products in the molten content of the fuel rods. The release of the high pressure steam carrying all this debris from the ground around the reactor building is likely to eject this material upward with considerable force. This is the meltdown scenario.

The suggestion by authors like Arthur Nero (in the book edited by Kaku and Trainer, listed in the bibiliography) that the molten mass of fuel rod contents would dig itself a nice clean hole downward and come to rest there seems quite unlikely. The displaced earth would have to find someplace to go (sideways is the only available way to open space) and would take with it some of the molten fission products.

    (2) The calculation of the amount of heat energy released by the radioactive fission products in the fuel rods after fission has been stopped.

    Suppose the reactor to be of average size, 1GW (1 Giga watt, or 1000 Million Watts). Since electric power plants are typically about 33% efficient, tossing 67% into the cooling towers and the water passing through the condensers, a 1GW power plant will need 3GW of heat energy to be released in the core.
    The 7% of full energy that is initially produced by the radioactivity of the fission products (page 208) lasts only briefly, but 1% lasts a day, and ½% declines to ¼% only after 11 days. Assume we are down to 1% when the reactor vessel releases the molten mass to the floor.
    1% of 3GW is 3x10⁷ or 30,000,000 Joules per second. In one day this produces 2.5x10¹² Joules. It would take another 14 days at ½% of 3GW to completely melt the concrete slab. The heat required to transmit a temperature sufficient to produce steam at the critical temperature (370^o) in the supporting moist earth would, according to these calculations, occur in 3½ days. To melt a hole in the concrete slab 25 feet across at the top and 12 feet across at the bottom would take 10 hours.

SOME QUESTIONS

    Wouldn't this timetable give opportunity to intervene to stop the meltdown?
    This would require that something can be done to either stop or slow the production of heat, or to remove some or all of it.
    The production of the heat can not be stopped or slowed. Unfortunately, radioactive dissociation is a nuclear process that is governed solely by the time table of the half lives involved, and can not be manipulated by any human activity.
    Various mechanisms in fact are installed to provide cooling in the containment building. Schemes to refrigerate the walls of the containment are among them. The thought is that the fission products would boil away as a gas that could be condensed on the containment walls, and would remain confined in the containment building. Other schemes include blowing with fans across pipes with water circulated from outside, spraying pressurized refrigerant, and using a pre-deposited supply of ice. These are all either short-term remedies or are subject to failure under accident conditions. It is quite a stretch to imagine that electric fans can continue to function in the same environment in which uranium is melting.
    The ultimate fact that has to be reckoned with is that in order to remove heat at the rate required (15,000,000 Joules/sec) on a continuing basis, cold material would have to be introduced into the containment, used to absorb heat, and removed hot. Some kind of circulating system would have to be functioning. This involves working equipment inside the containment in which uranium is melting, and is subject to malfunction.
    In theory, the condenser coolant, the water that is used to condense the steam in the normally-operating reactor, could be re-routed to cooling coils built strategically into the concrete floor slab. (If they were inside the containment, they would be in danger of melting.) If water is pumped in at 20^o and taken out at 80^o, the quantity required would be about 1000 gallons per minute. As with other schemes, pumping is subject to failure, and maintenance of imbedded cooling coils is problematic. From minor cracks in the concrete due to long-term settling of the foundation to uneven heating over the concrete area resulting from un-evenness in the melted layer, there are many ways for such a cooling system to fail just when it is needed.

    If the normal pumping of coolant through the reactor has failed, only minimal amounts of heat can be removed from the molten mass of uranium and fission products, except by the concrete and ultimately the ground. Once the reactor vessel bottom has been breached (melted), or if the accident began with a rupture of the reactor vessel, it is not possible to use the coolant circulation system even if the pumping could be restored. So, although 10 hours, or 3½ days or 15 days may seem like a long time, there is virtually nothing that anyone can do to affect the steady accumulation of heat energy or to cool or remove or scatter the uranium mix.

    Could enough heat energy be conducted through the concrete floor of the containment and be absorbed by the ground?
    At ½% of full power, 1.5x10⁷ Joules per second (15 Million Watts) of heat would have to be removed.
    The rate of transmission of heat through a material is given by the heat conductivity, which for concrete is 0.008Joule/cm²/sec for a temperature gradient of 1 degree per cm of thickness.
    Supposing under ideal conditions that the temperature gradient in the concrete floor is uniform and is therefore equal to a temperature difference between top and bottom of 1800 degrees divided by the thickness of 300 cm, or about 6 deg/cm. For a floor slab area of 1,250 m² (140 foot diameter), the rate of heat transfer from top to bottom of the slab would be 600,000 Joules/sec. This is less than 5% of the rate that would be required to dispose of the heat produced by the radioactive fission products. There is absolutely no chance at all that this mechanism might provide alternative cooling for the melted-down reactor core.

CONCLUSION

    The calculations we have done here do not allow us to say what would happen in the case of melting or rupture of the reactor vessel and the release of molten core to the containment floor. The calculations suggest that the possibility of a meltdown and subsequent release of vast quantities of radioactive fission products by upheaval of the ground under the reactor can not be ruled out as a possible consequence of a LOCA (Loss of Coolant Accident) in a common uranium reactor.

Q A reader's question: "It is said that energy has mass. Quite simple really! But does mass 'have' energy?"

A    Let's give it a shot. The plain answer is, of course, "yes." The problem really is with the words, 'has,' and 'have.' Energy IS mass, and so, of course, mass IS energy. The purist would simply say, Friend, you've got to use better words, then there would be no problem. But that, of course, is why the purists are so notoriously unsuccessful in communicating with the non-expert. It is also why the purists rarely understand the problem themselves.
    Anyway. When we say, energy HAS mass, we mean, operationally, that all those things that we have identified as being part of the "Work-Energy Theorem," that is, those things that obey the energy conservation laws -- all of those things HAVE the PROPERTIES of mass.
    Kinetic energy does, electric potential energy does, gravitational potential energy does, heat energy does, radiant energy does, etc.
    The properties of mass are: (1) inertia, as in Newton's Law, and (2) response to gravitational field, as in "weight." That's operationally understandable, which is why that part came easy to you, I suppose.
    Now, does mass HAVE energy? The mass that you refer to in that question probably refers to those "mass objects" whose mass is not there by virtue of being energy of the kinds referred to in the list above. The "stuff" of a little steel ball has never appeared in our energy conservation laws. You may think, "That stuff is the REAL mass. Eh?" To ask, "Does it HAVE energy?" means, "Does it have the properties of energy?"
    The properties of energy are not as easily defined, operationally, as the properties of mass. Chunks of Energy are not quantified by a measurement such as mass is, by doing something like measuring acceleration when subjected to force, or weighing.
    Energy is quantified by being in the list of things that, when added together in the conservation equation, remains the same after as before. Well, so we ask, "Should we have put the little steel ball in the energy conservation equation?"
    The answer is, emphatically, yes. Theoretically, properly, absolutely, yes. But practically speaking, the answer is usually, no, because usually the steel ball is there before and after, its mass is judged to be the same before and after, and so the steel ball would contribute equally to both sides of an energy equation, and may as well be left out.
    Of course, this is not always so. As you saw from the example of the two aluminum cookie sheets (in chapter 2), if the steel balls are electrically charged (oppositely), and you bring them together, their total mass changes because of the decrease in electic field as they approach each other. Well, you say, that's not the steel balls, but their electric field that is becoming kinetic energy.
    But, the electric field is part of the steel ball. It is part of its inertia, because the mass of the electric field has to be accelerated too if you accelerate the steel ball. If you weigh the steeel ball, you automatically weigh the mass of the electric field along with it.
    And, now, your instinct says to separate the two. Talk about the steel ball without its electric field, and then separately, its electric field. Then the electric field part of it 'has energy.' But the steel ball (minus its electric field) may not have energy, because it will always be there, before and after, and can not become kinetic energy. (Unless it is made of uranium, where some of the mass can become kinetic energy if there is a nuclear reaction. Whereupon you are into the mystery of 'where do they shave off the little bits of matter?')

The electron
    To illustrate this issue, I find the electron a very instructive example. I assume you accept the idea that when you weigh the electron, or measure its inertia, you include the electron's electric field in its mass. Now, how much electric field does a solitary un-bound electron have? Because, you can calculate the energy of that electric field with the energy-density equation [Eq 2.03]. The field goes infinitely far outward, but diminishes sufficiently rapidly, that a space integral of the field energy does not become infinite. The trouble comes from trying to define the inner boundary of the electric field.
    For that, one relies on the 'steel ball' picture of the electron. From the center out to a certain boundary, is the 'actual' electron. From there out it is the electron's electric field. Nobody knows where that boundary is.
    Nobody knows what the radius of the actual electron is. And now we are not talking quantum chemistry, with orbitals and all that. Just plain, how close can you get to the electron before you bump into the 'steel ball' part of it?
    If you pursue this line of thinking, you find that as you get closer to the center, the part of the electron's mass that is in the electric field farther out than you are, gets greater and greater. You can not get all the way to the center with your calculation, because the electric field close in gets so great that the integral becomes infinite as you approach zero radius. (This is one reason often cited for the electron having to have finite size, and not being a 'point' particle.)
    Obviously, before you get to zero radius there comes a point where the energy of the electric field farther out than you are gets to be pretty darned close to the energy equivalent of 9.11x10^-31kg, the mass of the whole electron! That radius turns out to be 1.405x10^-15m, quite a respectable size.
    The point is that if you fill the space outside that radius all the way to infinity with electric field, of intensity determined by the Coulomb equation, you have no mass left for the 'little steel ball' of 'actual electron.'
    Conceivably, then, the electron is nothing at all except its electric field. A sort of three dimensional donut made up of electric field with a hole in the middle. No one has a scientific verdict on that possibility, except that more than a few physicists are describing the matter of the universe as simply 'field' energy of various sorts.
    The mass of particles is nowadays conventionally given in MeV's, which are energy units. If you take an electron, pictured this way, and bring it together with a positron, pictured the same way, except with positive electric field, the fields merge into a madly oscillating electric field that goes off as the 'annihilation' energy of the two particles, in the form of a photon or two.
    The quantum folks complicate this a little bit, with their spin angular momentum, and so on. So I'm not suggesting that this very classical-like picture of an electron is what an electron is like. I'm suggesting that here is a way of understanding how the mass of the electron can be the mass equivalent of a certain easily calculated electric field energy.
    Same with nucleons, and their quark field energies. Nuclear force fields are both energy and mass, and it is from the shrinkage of the nuclear force fields that nuclear energy is obtained. (Of course, the mass shrinks along with the fields.)

Q This space is reserved for YOUR so-far unanswered question.
Send it to "Cavendish Press, PO Box 2588, Ann Arbor, MI 48106"
or submit it by e-mail to Cavendish@worldnet.att.net

Back to the NUCLEAR ENERGY BOOK home page
BOOKS home page
Return to the Web site home page.

CAVENDISHSCIENCE.ORG