Pity no-one did the maths......
One thing that was notable in the Press coverage, (and, to be fair, the TEPCO intial press release) is that no-one bothered doing the numbers on how hot the pools were likely to get (not least since they'd have to be boiled to call cause a problem).
all of the following, btw, is the sort of stuff you'd be taught in the first term of the heat transfer course of an engineering degree (or cold work out from A level physics), so we certainly aren't talking rocket science....
First off, how much heat is being produced. The Unit 4 SFP is the "hottest" so we'll wok around that.
At the time of the accident there were 784 older assemblies in there, making about 400KW of heat in total, and 584 "new" assemblies which had been out of the core for about 2 months, making about 1.87MW.
For the newer fuel, taking an initial average production of (1870/584) = 3.2 KW/assembly. but that was at the time of the accident. The standard decay curve for spent fuel is shown in the slide headed "Decay heat in Light Water Reactor Fuel) from
http://www-ns.iaea.org/downloads/ni/embarking/argonne_workshop_2010/Braun/L.6.2%20Braun%20Operational%20Safety%20of%20Spent%20Nuclear%20Fuel.pdf
60 days is about 6*10^6 seconds. We're now at about 800 days from the fuel being removed from the reactor, so about 7*10^7 seconds. Using the ratio between 10^6 seconds after shutdown and 10^7 seconds (which will be near enough for a rough calculation), that says the newer fuel assemblies will be making about 1/3rd the amount of heat they were at the time of the accident. So about (3.2/3=~ 1000 watts each).
The older ones will also have decayed, but not by much - so we'll ignore that.
total heat production will therefore be about 1MW. (584KW from the newer fuel, 400KW form the older)
OK, what's the heat balance...
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The pond contains about 1240 cubic metres/tonnes of water - there's also racking etc. in there, but we'll ignore that for the sake of simplicity.
If it's heating that mass of water through 1C takes 1.24*10^6*4.2*10^3 joules = 5.2*10^9 joules. 5.5C in 15 hours is 1*10^-4 C/second.
So, 5.2*10^5 joules/second (or 520KW) was going into heating the water on average over the time that the active cooling was off. The rest is being lost to ambient (in the absence of forced cooling) - so about 480KW, or about 50% of the total for this rough calculation
Heat loss from a fluid surface (in the absence of boiling), and by convection/conduction through the walls of the tank is proportional to the temperature differential to ambient.
If ambient is about 12C, then the average delta so far has been about (30-12) = 18C. 2 times that is 36C, which suggests the whole system would be in equilibrium with the water at around 48 - 50C so, no mass boiling.
You'd nee to periodically dribble some water in to make up for evaporation - but that's about it. Any boiling seems highly unlikely.