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Quick question. An article was posted last night that showed that radioactive decay would continue to produce .5% of the original reactor power for at least ten days after the reactor was shut down. Since that determines the amount of cooling the reactors need it would seem to be important. How many MW of heat did those reactors produce at peak power. And how much heat does it take to break containment. Get those numbers and we don't need speculation, just mathematics to figure out where this is going.

NRC lists the megawatt thermal (MWt) ratings for operating US reactors. Dresden 1 and 2 are GE Mark 1 BWR - apparently similar to the Fukushima Daiichi plant.

They are rated at 2957 MWt. However, I’m not sure this is a useful comparison, since the figure listed is an operating rating - basically an upper limit of the licensed power for the plant.

With a loss of core structural integrity, loss of the geometry, loss of moderator control, I’m not sure we can calculate with any degree of confidence how much heat the plant will produce in its current state, at least not without a lot of data that even the plant operators themselves are struggling to determine.

If it were just a matter of the reactor being scrammed, all rods in place, all systems normal, then yes we could I’m sure reasonably estimate how much decay heat will be produced. Right now, there are just too many wildcards for that sort of computation.

But then, I’m not a nuclear engineer.

I think the issue here is "shut down" is no longer the operative term. Shut down means that the reactor vessel is completely intact and the control rods fully inserted with the core completely covered. I believe also implicit in the definition is that sufficient cooling is taking place. Let's refer to that as complete reactor integrity. And that has nothing to do with the spent fuel storage issue but more on that in a moment.

Once complete reactor integrity is breached, the cool-down-over-time factor is no longer operative. Eventually the heat will compromise the integrity of the rods, the precise geometric relationship between the fuel and control rods is lost, and you are on your way to a melt down. There are remedies including the boron injection, which is an attempt to make the water surrounding the failing rods into a control rod equivalent.

We are told that in the event of a meltdown, the mass will pool at the bottom of the 3rd stage containment building, spread out and naturally cool over time. Radiation will be emitted but particles will not. And that's the good news!

Someone posted a cutout diagram of the BWR when the discussion was about losing the first sacrificial roof of the outer containment structure (3rd level containment). At the time, I and a few others saw the spent fuel pools situated well up in the structure and asked the question "what about Bob?" Well, that is turning out to be the billion dollar question.

The spent fuel pools are uncontained and subject to all the problems occurring outside the pressure vessel that is doing its level best to keep the unspent fuel rods in their shut down status. Fire in the 3rd containment - did I say that meant the 4 walled structure around the reactor, the ones that lost their roof - might not hurt the first two containment structures but the unprotected spent fuel pool? Oh mama, there's the problem and we've heard precious little about that kettle of neutrons.

I am not a nuc nor did I sleep at a Holiday Inn last night. I am remaining clam but am more than a little concerned that not enough people see the elephant in the outer containment zone. I pray that the folks on the ground see it but are too busy to talk about it.