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World's 'safest' nuclear reactor in India
The Press Trust of India ^ | THURSDAY, AUGUST 25, 2005 03:45:33 PM | The Press Trust of India

Posted on 08/25/2005 4:11:34 AM PDT by CarrotAndStick

NEW DELHI: India unveiled before the international community on Thursday, its revolutionary design of a 'Thorium breeder reactor' that can produce 600 MW of electricity for two years 'with no refuelling and practically no control manoeuvres.'

Designed by scientists of the Mumbai-based Bhabha Atomic Research Centre, the ATBR is claimed to be far more economical and safer than any power reactor in the world.

Most significantly for India, ATBR does not require natural or enriched uranium which the country is finding difficult to import. It uses thorium -- which India has in plenty -- and only requires plutonium as 'seed' to ignite the reactor core initially.

Eventually, the ATBR can run entirely with thorium and fissile uranium-233 bred inside the reactor (or obtained externally by converting fertile thorium into fissile Uranium-233 by neutron bombardment).

BARC scientists V Jagannathan and Usha Pal revealed the ATBR design in their paper presented at the week-long 'international conference on emerging nuclear energy systems' in Brussels. The design has been in the making for over seven years.

According to the scientists, the ATBR while annually consuming 880 kg of plutonium for energy production from 'seed' rods, converts 1,100 kg of thorium into fissionable uranium-233. This diffrential gain in fissile formation makes ATBR a kind of thorium breeder.

The uniqueness of the ATBR design is that there is almost a perfect 'balance' between fissile depletion and production that allows in-bred U-233 to take part in energy generation thereby extending the core life to two years.

This does not happen in the present day power reactors because fissile depletion takes place much faster than production of new fissile ones.

BARC scientists say that the ATBR with plutonium feed can be regarded as plutonium incinerator and it produces the intrinsically proliferation resistant U-233 for sustenance of the future reactor programme.

They say that long fuel cycle length of two years with no external absorber management or control manoeuvres "does not exist in any operating reactor."

The ATBR annually requires 2.2 tonnes of plutonium as 'seed'. Although India has facilities to recover plutonium by reprocessing spent fuel, it requires plutonium for its Fast Breeder Reactor programme as well. Nuclear analysts say that it may be possible for India to obtain plutonium from friendly countries wanting to dismantle their weapons or dispose of their stockpiled plutonium.


TOPICS: Business/Economy; Culture/Society; News/Current Events
KEYWORDS: china; india; iraq; israel; nuclear; nuclearplant; nuclearpower; nuke; thorium; uranium; weapones; wmd
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To: CarrotAndStick

Yes they do, but what role do they play in nuclear fission compared to the larger heavier nuclear fragments?


61 posted on 08/25/2005 10:57:30 AM PDT by Moonman62 (Federal creed: If it moves tax it. If it keeps moving regulate it. If it stops moving subsidize it)
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To: kidd
Mass is not consumed in the electrochemical reaction that takes place in a battery.

The 890 kJ of energy released by the combustion of one mole of methane thus originates from the conversion of 9.89 ng of mass into energy. Such a small change, about 10 ng out of 80 g cannot be detected by balances. It amounts to the loss of 1.0 x 10-7% of the mass. So we ignore Einstein's equation when doing regular chemical stoichiometric calculations. However this equation is very useful in nuclear chemistry.

http://www.ucdsb.on.ca/tiss/stretton/CHEM2/nuc02.htm


62 posted on 08/25/2005 11:22:13 AM PDT by Cboldt
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To: CarrotAndStick
Batteries are chemical-reaction energy systems. No destruction of mass in any way, is what I am aware of.

The change in mass is so small that it isn't directly measureable - but change in mass there is.

http://www.ucdsb.on.ca/tiss/stretton/CHEM2/nuc02.htm
http://www2.slac.stanford.edu/vvc/faqs/faq1.html

FAQ 1: Why is so much energy produced when an atom is split or fused? What is meant here by "so much energy" -- much relative to what? The answer is relative to the mass of the fuel used. (If we burn enough fuel we can make as much electricity in a coal-fired power plant as in a nuclear one.) In any process the energy produced is determined by how much mass is converted to energy, following the rule E=mc2. So what we are really interested in here is the fraction of the mass that is released in the process.

The short answer to your question is that the energy release in nuclear fission or fusion is a larger fraction of the mass of the fuel than in chemical processes because the binding forces (and hence the binding energies) between the protons and neutrons in nuclei are much larger than those for atoms in a molecule or solid.

To understand this first let's look at a coal-burning power plant (that is for combustion or for any other energy-releasing chemical reaction). The usual rule taught in chemistry is that mass is conserved. The precise version of this statement is that the sum of the masses of the atoms is the same before and after any chemical process, since atoms are not created or destroyed in chemical processes. But every stable molecule has a mass that is a tiny bit less than the sum of the masses of the atoms it contains. It is less by an amount (binding energy)/c-squared, where c is the speed of light (m=E/c2 is just another way of writing E=mc2).

The energy released in any chemical combustion process is just the difference in binding energy between the molecules present before burning and those present after the burning. The typical binding energy of a molecule is a few parts in a billion of the mass-energy (mass times c-squared) of the molecule. (That's why we never see a measurable mass change in chemical reactions, our chemistry lab experiments never have a balance that's accurate to that level.)


63 posted on 08/25/2005 11:30:18 AM PDT by Cboldt
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To: Moonman62
"The electrons play no significant role in nuclear (hence the term nuclear) fission. It is the splitting of the nucleus that matters, and all the resulting fragments are going to have positive charge."

No, and I never said they did. But as soon as those two positively charged nuclei form, they start grabbing electrons from whereever they can get them, including the ones floating around the individual atom. So, probably by the time they have reached their "first bounce", they may be either positively or negatively charged.

But all that aside--the electrostatic repulstion between even two all-positively charged is TINY compared to the energy imparted due to the mass difference. We're talking a few eletron volts vs MILLIONS of electron volts.

64 posted on 08/25/2005 12:46:31 PM PDT by Wonder Warthog (The Hog of Steel)
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To: Wonder Warthog
I am going to quote a competent nuclear physicist from one of his books. I believe he also had a PhD, but he didn't go around flaunting it.
... The nuclear forces act mainly between each proton (or neutron) and its nearest neighbor, while the electrical forces act over larger distances, giving a repulsion between each proton and all the others in the nucleus. The more protons in a nucleus, the stronger is the electrical repulsion, until, as in the case of uranium, the balance is so delicate that the nucleus is almost ready to fly apart from the repulsive electrical force. If such a nucleus is just "tapped" lightly (as can be done be sending in a slow neutron), it breaks into two pieces, each with positive charge, and these pieces fly apart by electrical repulsion. The energy which is liberated is the energy of the atomic bomb. This energy is usually called "nuclear" energy, but is really "electrical" energy released when electrical forces have overcome the attractive nuclear forces.

65 posted on 08/25/2005 1:02:57 PM PDT by Moonman62 (Federal creed: If it moves tax it. If it keeps moving regulate it. If it stops moving subsidize it)
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To: Moonman62
"If such a nucleus is just "tapped" lightly (as can be done be sending in a slow neutron), it breaks into two pieces, each with positive charge, and these pieces fly apart by electrical repulsion. The energy which is liberated is the energy of the atomic bomb. This energy is usually called "nuclear" energy, but is really "electrical" energy released when electrical forces have overcome the attractive nuclear forces."

What text is this, and by whom?? It sounds like "science for high school sophomores". Sorry, but it is NOT the electrical energy that provides the velocity, but the mass difference. Anyone who says otherwise is simply wrong.

66 posted on 08/25/2005 3:15:44 PM PDT by Wonder Warthog (The Hog of Steel)
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To: Wonder Warthog

dudeski, when you own a business, you're liable for the professional conduct (or lack thereof) of your employees.


67 posted on 08/25/2005 4:43:35 PM PDT by the invisib1e hand (see my FR page for a link to the tribute to Terri Schaivo, a short video presentation.)
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To: RAY
What am I missing here that the U.S. power industry and the US government have not been on top of this "safe" nuclear power development?

Notice that the reactor requires plutonium. In the early years, they had so much uranium that they didn't want to try to figure out how to move that processed plutonium (weapons material) around in the civilian market.

68 posted on 08/25/2005 4:49:31 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: Phocion
As a result, at this point uranium plants are still not cheaper than coal.

Counting the whole cycle, nuclear plants are competitive with coal plants and a whole lot cleaner.

69 posted on 08/25/2005 4:51:32 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: Blood of Tyrants
All development int he US has been strictly theoretical since the feds have made building a nuke reactor prohibitively expensive through impossible to meet standards.

Wrong.

70 posted on 08/25/2005 4:52:26 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: CarrotAndStick
It consumes 2.2 tonnes of plutonium, not produces.

His comment still applies. See my earlier post.

71 posted on 08/25/2005 4:54:43 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: Wonder Warthog
Actually, as I remember the "thorium cycle", it does NOT specifically require plutonium. You could also do it with enriched Uranium.

Even worse. Still bomb material and even less radioactive than the plutonium making it safer to work with.

72 posted on 08/25/2005 4:56:15 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: Moonman62

Of course the batteries would be lighter, all the light
will be out of them !

Haha.


73 posted on 08/25/2005 4:59:33 PM PDT by tet68 ( " We would not die in that man's company, that fears his fellowship to die with us...." Henry V.)
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To: Moonman62
You're thinking of fusion.

Fission is also an example of conversion of mass to energy:

E=MC2

So is burning gasoline in your car ...

74 posted on 08/25/2005 5:00:37 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: Moonman62
The same can be said about flashlight batteries. There will be a small change in mass between a new one and a depleted one due to energy conversion. It still doesn't change the fact that the main source of joy in a fission explosion is electromagnetic repulsion.

I can't belive it. I mean, I can't believe I actually see someone else posting this!

75 posted on 08/25/2005 5:02:50 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: CarrotAndStick
No, no! Batteries probably weighed less after usage, if they did so, because of escape of gases of reactions and the like. Nuclear energy produced is very obedient to the law E = mc^2. E is the energy produced, m is the ever so tiny change in mass, c is a large constant whose magnitude is the same as the value of the velocity of light, all in SI units.

Yes, Yes, YES! It is all so basic. E=MC^2. No exceptions.

76 posted on 08/25/2005 5:05:28 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: CarrotAndStick
I don't know where you got that information from. Can you lead me to a reputable link on the same? I'd be grateful.

He is correct. But it is not an easy search.

http://www.newton.dep.anl.gov/askasci/gen01/gen01159.htm

77 posted on 08/25/2005 5:08:56 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: kidd
Mass is not consumed in the electrochemical reaction that takes place in a battery. The E=mc^2 equation applies to a closed system.

You are oh so wrong.

78 posted on 08/25/2005 5:10:44 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: CarrotAndStick
Batteries are chemical-reaction energy systems. No destruction of mass in any way, is what I am aware of. I want you to give me a link on how batteries use up mass to give energy. I am pretty confident there is no nuclear reaction happening in an electrolytic battery.

I am sorry that you got a substandard education in physics and thermodynamics.

79 posted on 08/25/2005 5:11:45 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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To: Wonder Warthog
There is no guarantee, even, that the two particles have the same sign of charge, depending on which one gets more or less of the original fissioned atom's electrons. Look dude, my PhD minor subject area was Nuclear Science, this is kindergarten type nuclear knowledge.

Actually we can pretty well guess that both have the same sign and both are positive.

80 posted on 08/25/2005 5:13:09 PM PDT by WildTurkey (When will CBS Retract and Apologize?)
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