In a newspaper Op Ed last year, I wrote the following:
Imagine someone handed you a lump of silvery metal the size of a golf ball. They said you might wish to put on some plastic gloves to hold it, although that would not be necessary if you washed your hands afterwards. You look down at the metal resting on your palm. It feels heavy, because it’s very dense. You are then told that this metal golf ball can provide all the energy you will ever use in your life. That includes running your lights, computer, air conditioner, TV, electric car, synthetic jet fuel. Everything. Using 1 kilogram of uranium (or thorium, take your pick). That is what modern nuclear power offers. An incredibly concentrated source of energy, producing a tiny amount of waste.
I’d like to explain this statement here in a little more detail.
In earlier IFR Fad posts, I’ve explained that 1 tonne of depleted uranium (or you can also use mined uranium or used nuclear fuel) has sufficient energy to run a 1,000 megawatt electrical power station for 1 year — if run through a fast spectrum reactor. (I’ve also explained, in more detail, some key differences between the fast reactor and light water reactor fuel cycles). What does this mean in personal terms? Time to crunch some numbers…
The Australian population of 21 million currently consumes about 250,000 GWh of electricity per year. That works out to be 12 MWh per person, or 33 kWh per day. (This is similar to the figure David Mackay worked out for the British). A 1 GWe IFR (integral fast reactor nuclear power plant), running at 90% capacity factor, would produce 7,884 GWh of electricity per year. This would, therefore, be enough to satisfy the current electricity needs of 657,000 Australians. Or, to put it another way, one Aussie would require 1.5 grams of uranium per year. If they lived to be 85 years old and consumed electricity at that rate throughout their life, they’d require 130 g of uranium.
Australia’s total energy consumption is about 5,500 petajoules per year (1 PJ = 278 GWh). This includes electricity, non-electrical residential and commercial energy, transport fuels, mining, manufacturing and construction. What if this entire energy consumption had to be met by electricity? It would require the production of 1,530,000 GWh per year, or 6 times Australia’s current electricity generation. Referring back to the figure above, this would require 9 g of uranium per person per year, or ~0.8 kg of uranium for an 85 to 90 year lifespan.
The figure of 0.8 kg of uranium is a little less than the 1 kg figure I cited in the quote at the top of this post, but given the margins of uncertainty we’re dealing with here, it’s close enough. Without getting into complications, and despite being fully cognizant of the first law of thermodynamics, not all energy is ‘created’ equal. Electricity is a particularly convenient and flexible way to package energy, and could, in the future, more efficiently substitute for less efficient energy uses (including electric vehicles displacing oil combustion, electrically driven heat pumps replacing gas, etc.). In short, it’s not that difficult to justify a more conservative figure of 1 kg rather than the 20% smaller value calculated above, especially if population growth, the need to adapt to climate change (e.g. desalination), etc., is also considered. But really, whether it’s 500g, 1 kg or even 2 kg of uranium (or thorium) consumed over a lifetime, it’s still a tiny amount of fuel (and waste).
Out of interest, I’d also like to share some discussions I had last year on this topic, with George Stanford, Tom Blees, Steve Kirsch and Yoon Chang (all members of SCGI — bios here). The figures/units may differ a little from the above, but the bottom line remains the same.
The ping-pong ball (or half-ball) is the volume of uranium that was fissioned to release the lifetime’s worth of energy. It weighs a kilogram or less. The resulting fission products necessarily weigh the same (minus the 0.09% of the original mass that was converted to energy in accordance with E=mc^2). The fission-product elements all have a density much less than that of uranium, so their volume (if frozen into a solid ball and not radioactive) would necessarily be bigger than a ping-pong ball, before any vitrification. But that knowledge really has no practical utility. In an underground repository (e.g. Yucca Mtn), the distance between the tunnels (called “drifts”) and between the waste canisters (whether spent fuel or fission products) has to be such that the temperature in the soil between the drifts does not get too high over the years.
The activity of the fission products is dominated for the first few hundred years by just two isotopes — Cs-137 and Sr-90 (each with a half life of about 30 years). Just how the fission products will be disposed of is to be determined. But their heat will have to be managed somehow, which could mean storage for a while in surface facilities with forced-air cooling (although the sensible thing to do with them is to vitrify them and drop them into the silt at the bottom of the ocean). That’s why a calculation of the fission products’ “volume” has no useful meaning, and reveals nothing about the ease or difficulty of disposal.
What’s the volume of the waste without the glass? Shouldn’t it be smaller than a ping pong ball?
Yoon and I ran some figures one time for total energy consumed per person, for everything: heating and cooling, electricity, transport, etc. After I crunched the numbers I came to the conclusion that an entire person’s energy needs could be met by a chunk of depleted uranium the size of half a ping-pong ball. The waste product would be larger because you embed a little bit of it in a lot of glass, so it wouldn’t surprise me if it might take up half a liter, of which most would be glass.
If I remember correctly, the amount of fission products slightly exceeds the size of the DU, because the smaller atoms into which the DU/Pu has split will be greater in number, albeit smaller, and require more space because as we all know, most of the volume of an atom is space. So you can’t pack twice as many light atoms into the same space as a given number of heavy atoms. As George said, it’s heat, not space, that’s the issue. So you have to have enough glass to keep the heat below the level that will melt glass, I would think, requiring a substantial amount of glass.
The density of metallic uranium is 19 g/cc. Thus the volume of 1 kg is 52.6 cc. That’s a sphere with a diameter of 4.6 cm (1.8 inches) — slightly bigger than a ping-pong ball (4.0 cm). The volume is proportional to what you use for the per-person average energy consumption. The “waste volume” will be larger — maybe about the size of a soda can. But remember, the waste volume calculated that way is irrelevant — for disposal on land, it’s the heat generation that has to be managed, so the volume of the disposal facility is orders of magnitude larger than a soda can.
In the U.S., the energy per person per year is given as 8.25 TOE (tonnes of oil equivalent). If a lifetime is 85 years, that’s 700 TOE. One TOE = 11,630 kilowatt hours, for a lifetime total of 8 million kW-hr. As a rule of thumb, fissioning one gram of heavy metal (uranium) releases 1 MWth-day, or roughly 8 MWe-h = 8,000 kWe-h. So 8M/8k = 1,000 grams — one kg of fissions — which, of course, means one kilogram of fission products. If the average density of the waste form were 2 g/cc, then the volume would be 0.5 liter. Ball-park calculation only, assuming all energy comes from nuclear-generated electricity.
Note that that’s total US energy consumption divided by the number of people (not counting calories derived from food). That’ makes it 850 grams per 85-year lifetime instead of 1 kg. Still in the soda-can range for the fission products (a very crude approximation at best, since the waste form is not defined). Note that burning 700 tonnes of oil produces about 2,000 tonnes of CO2, for a waste-weight ratio >2 million to one (for whatever that’s worth).
When Yoon and I did this, we also used figures for total national energy consumption divided by the number of people in the country, so the energy that goes into food and items imported minus the energy that goes into exports gives you a little shortfall (balance of trade and all that). But I think I’ve allowed enough of a cushion in these old calculation I did back then:
A cube of DU sufficient for an American lifetime would be about 0.92 on a side. Figuring a little extra, at .95 on a side, give a volume of .9 cu. in. If in the form of a sphere, the diameter of the sphere would be about 1.2 in, or 30.5 mm, considerably smaller than a ping pong ball. (A ping pong ball is 40 mm, or 1.57″ in diameter.) The volume of a ping pong ball, then, is 2 cu in, more than enough for the lifetime of two people (it would weigh a little over a pound).
Here are Yoon’s numbers that he wrote me about on July 31, ’07, upon which I based the above:
Yoon: Regarding your question, this is my rough estimate. EIA reports that the energy consumption in 2006 was about 100 guads (1quad=10 to 15th power Btu). This divided by 300 million population and multiplied by 70 years of lifetime gives 23 trillion Btu. One gram of uranium fission (direct or after converted into fissionable isotopes) yields one MW-day energy, or 86.4 million Btu. Therefore about 260 gram (9 ounces) of uranium required per person’s lifetime. The uranium density is 19 g/cc, so 260 gram is about 14 cc in volume or somewhat less than one cubic inch. The above calculation assumes all thermal energy equivalent. All electric energy society will require more energy to account for the efficiency loss in converting the electric energy back to thermal energy needs.
The first thing you’ll notice is that Yoon isn’t planning to live as long as George. Then you see that I failed to account for the electric to thermal conversion he mentioned in the last paragraph for space heating. Nevertheless, I should think it’ll still be way less than a ping-pong ball per person, even assuming we all live as long as George plans to, and then some.
If I take your 260 grams, multiply it by 85/70 for lifetime estimate, and by a factor of 3 to account for efficiency of conversion to electricity, I get 947 grams — just shy of the1 kg I came up with. Our calculations therefore agree very well, when using the same assumptions. By the way, I said “not counting calories derived from food.” Energy required to make the food is of course an important part of the picture. I do hope to make it to 85 — that’s only 4 years — but the envelope is indeed closing in.