# How large would the steam explosion at Chernobyl have been?

So the second episode of the HBO series began to cover the risk of a steam explosion that led to them sending three divers into the water below the reactor to drain the tanks.

This occurred after the initial explosion that destroyed the reactor, and after the fire in the core had been put out. But at this point the decay heat and remaining fission reaction kept the core at more than 1200°C, causing it to melt through the concrete floors below the reactor.

And below the reactor were water tanks which contained 7,000 cubic meters of water (according to the TV show. If anyone has a real figure, I'd love to hear). When the lava of the melted core hit it, it would cause an enormous steam explosion.

Finally, my question: About how large would this explosion have been? The character in the show says "2-4 megatons" (of TNT equivalent, I assume). I'm pretty sure this is absurd and impossible. But real estimates are hard to come by. Other sources vary wildly, some repeating the "megatons" idea, and others saying it would've "level[ed] 200 square kilometers". This still seems crazy.

## tl;dr:

I know a lot of it hinges on unknowns and the dynamics of the structures and materials involved, so I can simplify it to a constrained physics question:

Assuming 7,000 cubic meters of water instantly flashes to steam, how much potential energy is momentarily stored in that volume of steam occupying the same volume as the water did?

I don't know what to assume the temperature of the steam is. There were hundreds of tons of core material at temperatures near 1200°C, so worst case scenario you could assume all the steam becomes that temperature as the materials mix. Best case scenario, I guess we could assume normal atmospheric boiling point (100°C)?

• nationalboard.org/Index.aspx?pageID=164&ID=412 works out to "only" 0.4 kilotons of TNT. – JEB May 15 at 0:54
• @JEB This is great, thanks! Would you like to put your calculations in an answer? – Nick S May 15 at 1:10
• Great question - I actually posted the exact same one, but you were first :) I had the same reaction to the show "Wait, megatons? What? – Flambino May 16 at 12:25

In my view the water isn't really the thing to focus on here. The real energy reservoir was the partially-melted core; the water wasn't dangerous because it held energy, but rather because it had the potential to act as a heat engine and convert the thermal energy in the core into work. We can therefore calculate the maximum work which could conceivably be extracted from the hot core (using exergy) and use this as an upper bound on the amount of energy that could be released in a steam explosion. The exergy calculation will tell us how much energy an ideal (reversible) process could extract from the core, and we know from the Second Law of Thermodynamics that any real process (such as the steam explosion) must extract less.

### Calculation

Using exergy, the upper bound on the amount of work which could be extracted from the hot core is

\begin{align} W_\text{max,out} &= X_1 - X_2 \\ &= m(u_1 - u_2 -T_0(s_1-s_2)+P_0(v_1-v_2)) \end{align} If we assume that the core material is an incompressible solid with essentially constant density, then \begin{align} W_\text{max,out} &= m(c (T_1 - T_2) -T_0 c \ln(T_1/T_2)) \end{align} where $$T_0$$ is the temperature of the surroundings, $$T_2$$ is the temperature after energy extraction is complete, and $$T_1$$ is the initial temperature. At this point you just need to choose reasonable values for the key parameters, which is not necessarily easy. I used:

• $$T_1 = 2800\,^\circ\text{C}$$ based on properties of corium
• $$T_2 = T_0$$ as an upper bound (the most energy is extracted when the system comes to the temperature of the surroundings)
• $$T_0 = 25\,^\circ\text{C}$$ based on SATP
• $$c = 300\,\text{J/(kg.K)}$$ based on properties of UO$$_2$$
• $$m = 1000\,\text{tonnes}$$ based on the text in your question.

This gives me $$W_\text{max,out} = 6.23 \times 10^{11}\,\text{J}$$ or 149 tonnes of TNT equivalent. This is several orders of magnitude lower than the "megatons" estimate provided in your question, but does agree with your gut response that "megatons" seems unreasonably high. A sanity check is useful to confirm that my result is reasonable...

### Sanity Check

With the numbers I used, the system weights 1 kiloton and its energy is purely thermal. If we considered instead 1 kiloton of TNT at SATP, the energy stored in the system would be purely chemical. Chemical energy reservoirs are generally more energy-dense than thermal energy reservoirs, so we'd expect the kiloton of TNT to hold far more energy than the kiloton of hot core material. This suggests that the kiloton of hot core material should hold far less than 1 kiloton of TNT equivalent, which agrees with your intuition and my calculation.

### Limitations

One factor which could increase the maximum available work would be the fact that the core was partially melted. My calculation neglected any change in internal energy or entropy associated with the core solidifying as it was brought down to ambient conditions; in reality the phase change would increase the maximum available work. The other source of uncertainty in my answer is the mass of the core; this could probably be deduced much more precisely from technical documents. A final factor that I did not consider is chemical reactions: if the interaction of corium, water, and fresh air (brought in by an initial physical steam explosion) could trigger spontaneous chemical reactions, then the energy available could be significantly higher.

### Conclusion

Although addressing the limitations above would likely change the final upper bound, I doubt that doing so could change the bound by the factor of ten thousand required to give a maximum available work in the megaton range. It is also important to remember that, even if accounting for these factors increased the upper bound by a few orders of magnitude, this calculation still gives only an upper bound on the explosive work; the real energy extracted in a steam explosion would likely be much lower. I am therefore fairly confident that the megaton energy estimate is absurd, as your intuition suggested.

• Probably tough to say, but as the show suggested: Would 149 tonnes of TNT detonated in reactor 4 be enough to cause a meltdown in the neighboring reactor 3? – y3sh May 16 at 17:57
• Not my area of expertise - this might be a good question for the engineering stack exchange site. The largest conventional weapons have a yield of around 44 tonnes of TNT (en.wikipedia.org/wiki/Father_of_All_Bombs), so this would be like setting off three enormous bombs. Remember though that 149 tonnes TNT is an estimated upper bound, and the real energy release would likely be lower. – user1476176 May 16 at 22:42
• Note that the mass of the corium could be much bigger than the original mass of the fuel, as it incorporates a good deal of material from elsewhere in the core, the concrete around it, and the air-dropped fire suppressants (of which Wikipedia reports 5,000 tons). And adding material need not make the corium cooler - it has its own heat source in its bulk, and the temperature is limited by heat exchange to the surroundings. If there's a factor of 10 there, then the upper bound gets closer to that megaton estimate. – Emilio Pisanty May 22 at 10:59
• The largest mass of corium was The Elephant's Foot which was about 2 ton weight and contained about 7-8% of Uranium. – AlexD May 31 at 11:35
• @DavidWhite The difference with the corium is that it is strongly radioactive and thus ionizing; water could turn to mix of hydrogen + oxygen and this mix then can explode. – Ján Lalinský Jun 7 at 8:29

The top accepted post (user1476176) has already done a thorough job of calculating the thermodynamics for a steam explosion (spoiler: nowhere near the megaton scale -- they were only off by 10,000X to 100,000X).

To compliment that, here's some intuition for what it takes to achieve a megaton-scale explosion, and why it is so laughably unrealistic to think that could happen by accident in even the worst-possible reactor disaster (i.e., Chernobyl):

1) It took our top scientists and engineers multiple years to achieve kiloton-scale bombs, and years more to achieve megaton-scale bombs using fusion. It was not easy, and they were working with billions to trillions of dollars of government resources at their disposal. If you could just drop some melted corium into water ... they would have done that at least once -- the first H-bomb test involved vaporizing a building-scale cryogenic chilling plant that was keeping the liquid deuterium from boiling away.

Making even kiloton-scale explosions requires precision.

Making megaton-scale explosions requires extreme precision, beyond the capabilities of many nation states. And fusion.

2) The largest pure-fission bomb ever tested was on the order of 0.5 megatons. They used huge quantities of weapon grade U$$^{235}$$ (>95% enrichment), surrounded by a neutron reflecting tamper, and almost instantaneously compressed to super-criticality by two different high-explosives precision engineered to produce a perfectly spherical shock wave. Chernobyl used fuel that was less than 2% enriched, meaning that 98% of it was non-fissile U$$^{238}$$, and that is before you account for contamination by fission byproducts, melted concrete, and melted steel.

3) Fusion is the only way that weapons engineers have been able to create megaton-scale explosions. And fusion is completely out of the picture here for at least two reasons:

• Bombs depend on fusing rare hydrogen isotopes like deuterium (H$$^2$$) and tritium (H$$^3$$) that weren't present at Chernobyl; for deliverable bombs, they use lithium-deuteride which contains deuterium and forms tritium under neutron bombardment (by cracking the lithium). The random fire-hose water that was seeping through Chernobyl was almost entirely (99.98%) composed of normal hydrogen (H$$^1$$), which is so hard to fuse that we don't / can't use it in bombs.
• Even to fuse H$$^2$$ and H$$^3$$, they have to use kiloton-scale fission bombs, combined with precision engineering that uses the fission bomb's x-rays to generate compression that's way beyond what's achievable with conventional explosives. This drives the H$$^2$$ and H$$^3$$ atoms together at extreme pressures and temperatures. It's extremely hard to do and, unlike with fission criticality accidents, fusion will never happen by accident. For example, if you moved the tritium from the center of the plutonium pit and just set it next to the bomb, it wouldn't fuse. Fusion is exceedingly difficult to achieve, on a level that's hard to put into words.

4) To achieve even kiloton-scale yields, great care must be taken to assemble the super-critical mass as quickly as possible, and to avoid stray neutrons that might start the chain reaction prior to the maximum compression (i.e., maximum super-criticality). For example, the first North Korean bomb attempt "fizzled" with a sub-kiloton yield ... generally, this happens for one of two reasons: either the implosion was less than perfect, or stray neutrons started the chain reaction before the point of maximum compression. Either way, what happens is that the fissile material, which is heating at an exponential rate, physically blows itself apart before the chain-reaction can achieve kiloton yields.

• Compression, compression, compression. The art of designing a nuclear bomb involves three things: Getting the the bomb into a maximally super-critical state (implosion), starting the chain reaction precisely at the moment of maximum criticality (the polonium/gold neutron initiator), and then keeping the fissile material super-critical state for as long as possible to maximize the yield (the "tamper" material slows the expansion by tens of nanoseconds). Note that none of these components were present at Chernobyl.

• Weapons Grade. To get good bomb yields, you want to use a fissile material that's as pure as humanly possible. Both being as close to 100% fissile material as possible (compared to Chernobyl's 2% fuel), as well as not being contaminated with neutron sources that will trigger an early detonation "fizzle". The Chernobyl corium contained highly active neutron emitters and would have instantly fizzled at the moment of criticality well in advance of achieving the super-criticality required for a kiloton-scale yield.

5) The neutron chain-reactions in a reactor are very different from those used in bombs:

• Thermal Neutrons - the only way to achieve criticality with Chernobyl's 2% enriched uranium is to use a neutron moderator, like graphite, that slows down the neutrons emitted by fission until they are in the "thermal" spectrum (i.e., bouncing around at similar thermal temperatures to the surrounding atoms). This increases U$$^{235}$$'s neutron absorption cross-section and consequently, raises the probability that any given neutron will trigger another fission event rather than leaking out of the reactor core or being absorbed into some other atom. But because they need to bounce through graphite before slowly finding more U$$^{235}$$, thermal neutrons have much longer "doubling times" than do "fast" neutrons, meaning that the bomb-scale chain reactions just aren't possible: the critical mass will thermally blow itself apart as soon as even a tiny fraction of the material fissions.

• Delayed Neutrons - in addition to using "thermal" rather than "fast" neutrons, reactors are designed to operate "prompt sub-critical", meaning that the neutrons that are emitted from U$$^{235}$$ fission are insufficient for sustaining a chain-reaction unless one also includes neutrons generated from secondary decay chain events that occur seconds to minutes later. This is important because it makes reactors much easier to control. One of the key questions I have about Chernobyl is whether during the sheer incompetence that led to the initial reactor explosion, they managed to take the reactor into the "prompt criticality" regime, although with thermal neutrons that have to bounce around before chain-reacting, it becomes a more subtle distinction. I'm not sure if that's globally unknown, or just unknown to me.

A steam explosion between corium at 3000 degC and water would be pretty dramatic, potentially destroying additional containment elements, ejecting highly radioactive material onto the roof and grounds, and generally complicating the already hellish clean-up challenges. So no kidding, they wanted to avoid that.

But a steam explosion is nowhere near the megaton-scale energy release described in the show.

It's highly dubious that the Chernobyl corium, without it's graphite and contaminated by concrete, steel, and especially Boron (a potent neutron absorber), could have even assembled into a critical mass at all.

But even if, by some crazy set of coincidences that did happen, the chain reaction of thermal neutrons in a barely critical configuration would have thermally blown itself apart well before reaching even the kiloton-scale range of energy releases. Megatons is laughable.

The show (which, overall, was AWESOME), was embarrassingly unfounded on this point. Chernobyl was awful enough in reality without the need to fear-monger with ludicrous hypotheses.

I had the same question and found an old Guardian articla from 2005 saying that the explosion would come from:

"There was a moment when there was the danger of a nuclear explosion, and they had to get the water out from under the reactor, so that a mixture of uranium and graphite wouldn't get into it - with the water, they would have formed a critical mass. The explosion would have been between three and five megatons. This would have meant that not only Kiev and Minsk, but a large part of Europe would have been uninhabitable. Can you imagine it? A European catastrophe."

https://www.theguardian.com/environment/2005/apr/25/energy.ukraine

• Well this is a good point. Were they thinking the water could moderate the reaction and cause another power jump, adding energy to the steam explosion? But I thought because of the large amounts of graphite already moderating the reaction, water was an overall inhibitor to reactivity in RBMK cores. – Nick S May 18 at 8:18
• @NickS Water does work as a moderator for the sort of fuel used in this case. However, I'm still skeptical that a lump of molten corium would become a megaton nuclear device upon contact with water. The ingredients might be there, but still... I'd imagine the steam explosion would eject a lot of material (horrible) or maybe a fizzle explosion where the critical mass blows itself apart as much it blows up (also horrible, and maybe what already happened in the first explosion)... well, bad either way – Flambino May 21 at 14:13
• @Flambino Yeah, clearly a megaton explosion is impossible. But the water moderation could help us get to an explosion size that would disperse the material in the other 3 reactors. 150 tons of TNT is only about 3x the original explosion, which didn't fully disperse the fuel in the reactor it happened in. – Nick S May 22 at 19:10
• @NickS Good point. I think what's confusing is that they kept saying steam explosion when the greater risk - whatever its scale - might have been a nuclear reaction. Not just "lava" and water. What strange is that I've seen that "3-5 megatons" estimate repeated verbatim in some documentaries I've watched since watching the episode. But never with an explanation attached, and always talking about a steam explosion. One that would somehow rival the total munitions (incl. nukes) fired in WW2... So it might just be a single bad source that's been quoted enough to become unexamined "fact". – Flambino May 23 at 11:41
• That article is flat out wrong. Thanks for posting it so that we know where the mis-information originated. – Dave Dopson Jun 14 at 16:18

I had the exact same question when I saw the episode. Building on user1476176's answer, I have thoughts as to how someone might have derived a much higher upper bound, though until someone pulls the actual source they used we won't know. They did explicitly say in the episode that other reactors would be engulfed, presumably with their water. That only multiplies the amount of useful fluid a few times, but then there is the exterior reservoir that is itself connected to internal pumps that might also be part of someone else's calculation, since that reservoir would have had nearly 10^6 tons of water judging from Wikipedia's map, which would make the yield upper bound 150 kT TNT. Of course it's not like an open reservoir would magically behave like a steam vessel when lava touches it, but it might be the source of this worst-case calculation.

Another possibility is that our unknown source made the common mistake of assuming that the material in these nuclear reactors can explode like an atomic bomb. If the mass of fuel in a core is about 200 tons (for RBMK reactor) and the yield/weight ratio for the earliest fission bombs is about .05 MT/ton, then we're getting close. I really hope a show as good as this didn't source this datum from somewhere that would make so bad a mistake though.

• Note that Stack Exchange posts are version controlled, so you don't need to add in "Edit" into the post, just seemlessly integrate the new material into the original post. There is a version history for interested parties who want to see what changed. – Kyle Kanos May 17 at 2:37

I had heard this scenario many years ago and its primary source I believe was an interview with Gorbachev were he mentioned it (I cant find the source though, so take with a pint of salt).

I too considered it to be without much foundation (Given the known facts its out right impossible unless they had stored nuclear weapons hidden under the cores foundation) and given that it comes out of a man that isn't a scientist but a politician, my best guess would be that the 3 megaton figure should not be considered as the yield of an explosion event but more likely the fallout equivalent of the radiation that would have been released after the steam explosion and the subsequent destruction of the remaining 3 cores at the vicinity

• Ohh, this is a really good point. I could easily see the amount of fallout being equivalent to a megaton-range bomb. This is the best way I've heard of making sense of the people saying this. – Nick S Jun 18 at 23:30

Guys you need to read more papers on steam explosions. When talking about steam explosions with corium, you need to note the velocity of energy exchange between the heated mass and the water. That needs intermixing of really tiny superhot particles and water in very very short time(hard to achieve). The efficiency of this steam explosions ( ratio of thermal energy to converted mechanival eneegy) is very very low. Even below 1%.

Read steam explosions in light water reactors, swedish comité.

About criticality, i certanly doubt that it was a threat at that moment. Probably impossible to have with corium.

The upper-voted answer has the right aim, but it still doesn't give the right estimate. The true source of energy under consideration is indeed the core, not the water, but the calculation in that post is essentially assuming cooling the core from its standing temperature to normal temperature - and as Emilio Pisanty pointed out in the comments, this won't happen as the core is actually its own energy source, capable of maintaining an elevated temperature.

Hence, what you get is effectively a heater that will apply to whatever is applied against it a thermal transfer power equal to the wattage being created by the ongoing fission process within the core material. As such, it is legitimate, as that poster also mentioned, to surmise in theory that an upper limit of megatonnes of total core potential energy is available. In particular, if you have roughly (using figures floating here) 1000 Mg of nuclear fuel that is maybe 5% fissile uranium ($$^{235}_{92}\mathrm{U}$$), that is 200 Mg of such and this fuel has an energy content of about $$86 \times 10^9\ \mathrm{MJ/Mg}$$, so the total available energy is on the order of $$1.3 \times 10^{11}\ \mathrm{MJ}$$, while a megatonne equivalent TNT is roughly $$4 \times 10^9\ \mathrm{MJ}$$, hence easily tens of megatonnes of available potential fission energy.

But this energy cannot turn into an explosion of that same size under these condition because the core is not releasing that energy fast enough. If it were, it would have already exploded in the manner of a gigantic pure-fission nuclear weapon of that yield. The rate of the fission reaction depends on the composition of the core melt mix, and to get such a reaction would require extreme concentration of the fissile $$^{235}_{92}\ \mathrm{U}$$ (basically, so that the nuclei are close together and there are few to no obstacles to absorb the neutrons that are needed to propagate the chain reaction), but melting and mixing the material can only serve to dilute it at best. Increasing fissile concentration is the definition of "uranium enrichment" and as we all know, that's HARD! Dumping water on it won't change that. Instead, you a better model would be a a thermal terminal that maintains a constant temperature of 2800 C against anything that hits it, or, at least, something suitably well above the boiling point of water.

Thus, in fact, the question asker is right to imagine this instead as asking for the energy required to vaporize all the water, and this is the maximum energy that can be released in a steam explosion. Energy is contact-transferred - hence once converted to steam, it is very difficult to absorb more from the core.

And this is relatively simple to obtain. With $$7000\ \mathrm{m^3}$$ of water volume, that's $$7000\ \mathrm{Mg}$$ of water mass, and the heat of vaporization for water is $$2260\ \mathrm{kJ/kg} = 2260\ \mathrm{MJ/Mg}$$ (hence my use of megajoules as the unit above), but we also need to take into account the energy to heat the water to the boiling point, which means we should use $$4.184\ \mathrm{\frac{kJ}{kg \cdot K}} = 4.184\ \mathrm{\frac{MJ}{Mg \cdot K}}$$ times the temperature rise (75 K) which gives $$314\ \mathrm{\frac{MJ}{Mg}}$$ and hence $$2574\ \mathrm{\frac{MJ}{Mg}}$$ of total energy to vaporize each megagram (tonne) of water starting at the given staid temperature 25 °C. With 7000 Mg of water, thus, the total potential energy is thus about

$$1.8 \times 10^7\ \mathrm{MJ}$$

maximum possible steam explosion energy. In terms of tonnes equivalent of TNT, it is ~4 kilotonnes TNT equivalent, and so still well below the range given (though also well in excess of the present top answer's figure).

• There are reasonable estimates and unreasonable estimates, but there is no "right estimate." My estimate neglects decay heat because we know that a real explosion would be fast ($\lim_{\Delta t \rightarrow 0}\ \dot{Q} \Delta t = 0$) - IMO this is quite reasonable. Your estimate uses a different, reasonable model and reaches a different answer. Both are reasonable; neither is "right." – user1476176 Jul 10 at 22:47

Considering the character making the 3 megaton steam explosion assertion, Ulana Khomyuk, was a composite of multiple Soviet scientists and the fact that the state was making every attempt to cover up, divert blame and threaten those who were trying to bring the problem out in the open, I do not think it would be amiss to consider one other option. The scientists involved in containing the accident may have exaggerated the gravity of the situation in order to make an impression on the political apparatchiks. It wouldn't be unreasonable to expect that many in the government would have understood what the meaning of a megaton was given the general knowledge of nuclear weapon yields. Telling a member of the politburo a megaton equivalent, even if it was a vast exaggeration may have been seen as a way to get the political authority to commit the type of resources required to deal with the problem. Recall that for nearly 30 hours after the accident, the operators on scene were insisting that it had not happened, that it couldn't happen, and that it was "no big deal". Were I a responsible scientist living within the culture that was the late cold war Soviet Union I think I would have done whatever was needed to get the problem under control, including exaggeration.

• Welcome to Physics.SE! Since you are new here I wanted to leave a comment explaining my downvote: While your answer is perhaps possible (I am no expert on the political realities facing Soviet scientists), it does not actually provide a physical answer to the question at hand. – gabe Jun 15 at 18:45
• You might want to post this as a comment (or broken into a couple of comments) under the question. Otherwise I see it being deleted quite soon. – Helen Jun 15 at 19:14

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