How can I find a theoretical maximum amount of radioactively polluted water which could be created by a single nuclear reactor?

Question

This question has a lot of tolerance.

The current context is that I'm writing a story which involves a group of terrorists who are using the threat of radioactive water pollution to deter interference. Sort of like an ecological hostage situation.

They have access to a Nuclear Reactor and any imaginary body of water. I'm trying to find the maximum amount of water they could render non-potable for ~100 years given any typical Nuclear Reactor. Given the time frame, it may be fair to say the water is not moving too much.

I've looked into the terms/units (https://www.epa.gov/radiation/radiation-terms-and-units). I've also found that the max Bq for potable water is

(WHO† Guideline: Radium 226 = 1 Bq/l Radium 228 = 0.1 Bq/l)

(https://www.wqa.org/Portals/0/Technical/Technical%20Fact%20Sheets/2014_Radium.pdf)

Given that WHO lists two different Radiums, it would appear that Bq limits in water are different for separate materials, so perhaps sievert (Sv) would be better?

I would guess that different materials (https://www.nrc.gov/materials.html) are more likely to irradiate water, or have differing half-lives, or are typically stored in different amounts at nuclear facilities. If any type of material stands out in all three fields, that would be the best choice for the story.

What I don't know is whether or not different states of the nuclear material are better/worse for irradiating water.

To be honest, I really don't know much about radioactivity in general, and apologize for that.

Overall, the question I'm looking to answer is: About how much water could be rendered non-potable by a single bad party with access to a nuclear power plant?

Surely the ocean is too big and a typical pond too small. A lake seems small as well, though the great lakes might be right at the cusp. If there's any kind of super hazy/inaccurate means of quickly estimating how much water a gram of each material could make non-potable, that would be interesting.

Any direction at all would be appreciated.

Answer 1

For long-term exposure over 100 years via the water release pathway, you would need radionuclides with a long half-life, high dose coefficient, and good solubility in water. Of the various radionuclides generated in a nuclear reactor, that practically only leaves Cs-137 as the relevant radionuclide. Cs-137 has a half-life of $30.1671\ \mathrm a$ and an effective dose coefficient (ingestion, adult members of the public) of $1.3\times10^{-8}\ \mathrm{Sv/Bq}$.

You are asking for a nuclear reactor near the Pacific. The largest single unit at the Pacific coast has a thermal power of $4590\ \mathrm{MW}$. The equilibrium Cs-137 activity in the reactor core is about $6.4\times10^{17}\ \mathrm{ Bq}$. (That’s about 2.5 times the inventory of the Chernobyl-4 core at the 1986 accident.) Let’s say you also have spent fuel corresponding to three unloaded cores in the spent fuel pool, which gives a maximum Cs-137 activity in the spent fuel pool of about $1.8\times10^{18}\ \mathrm{Bq}$. However, you would not be able to release this inventory to the environment.

Thus, you would have to look for other systems with a relevant radioactive inventory. The reactor coolant system has a Cs-137 activity concentration of about $3.2\times10^{8}\ \mathrm{Bq/Mg}$ during stable conditions and up to $6.4\times10^{9}\ \mathrm{Bq/Mg}$ during shutdown transients. Assuming a total coolant inventory of $321\ \mathrm{Mg}$, the total activity is up to $2.1\times10^{12}\ \mathrm{Bq}$. Some other potentially relevant Cs-137 inventories are:

• mixed-bed filters of the reactor coolant purification system: $1.9\times10^{13}\ \mathrm{Bq}$
• a spent resin storage tank of the spent resin flushing and storage system: $1.3\times10^{14}\ \mathrm{Bq}$
• degasifier of the reactor coolant degasifying system: $8.0\times10^{8}\ \mathrm{Bq}$
• volume control tank of the chemical and volume control system: $2.4\times10^{9}\ \mathrm{Bq}$
• a coolant storage tank of the coolant supply and storage system: $3.1\times10^{10}\ \mathrm{Bq}$
• boric acid evaporator of the coolant treatment system: $6.5\times10^{10}\ \mathrm{Bq}$
• condensate tank of the coolant treatment system: $8.9\times10^{7}\ \mathrm{Bq}$
• boric acid tank of the reactor boron and water make-up system: $3.4\times10^{11}\ \mathrm{Bq}$
• storage tank of the waste water storage and treatment system: $6.7\times10^{9}\ \mathrm{Bq}$
• evaporator of the waste water storage and treatment system: $1.3\times10^{11}\ \mathrm{Bq}$
• concentrate tank of the waste water storage and treatment system: $8.6\times10^{11}\ \mathrm{Bq}$
• monitoring tank of the waste water storage and treatment system: $6.7\times10^{6}\ \mathrm{Bq}$

These activities are design values; typical realistic values would be lower.

Assuming you have someone who knows how to operate the plant, it would be quite easy to release the inventory of a monitoring tank to the sea. However, this operation would take several hours considering that you have to release $70\ \mathrm{m^3}$ of waste water using only a small pump.

For any larger releases, you would have to make technical modifications to the plant. In addition to the paramilitary terrorists needed to force an entry to the inner security area of the plant and maintain it for several hours, you would preferably have nine qualified plant operators (a shift supervisor, a reactor operator, a turbine operator, two mechanical infield operators, two electrical infield operators, and two I&C technicians). Maybe you would also need pipe welders. You might be able to make the necessary modifications to the waste water storage and treatment system in the limited available time so that you could release the inventory of a storage tank to the sea.

Therefore, your Cs-137 source term could be $6.7\times10^{9}\ \mathrm{Bq}$. After $100\ \mathrm a$, the activity would still be about $6.7\times10^{8}\ \mathrm{Bq}$.

However, sea water isn’t used as drinking water in that region. And even if it would be used, it would have to be purified in a desalination plant before use anyway. That process would also effectively remove any relevant Cs-137 contamination. Therefore, you would have to find a different water body for your releases.

Note that a large fraction of the activity would actually end up in the sediment and thus would no longer be available for the contamination of drinking water. The partitioning of radionuclides between water and suspended matter can be described in terms of a distribution coefficients $K_\mathrm d$, expressed as the concentration ratio of the particulate phase to the dissolved phase under equilibrium conditions (in $\mathrm{Bq\ kg^{-1}}$ of suspended particulate matter per $\mathrm{Bq\ l^{-1}}$, i.e. in $\mathrm{l\ kg^{-1}}$). Values for Cs obtained from field measurements are between $1.6\times10^3\ \mathrm{l\ kg^{-1}}$ and $5.2\times10^5\ \mathrm{l\ kg^{-1}}$.

For the assessment of radiological consequences, we may assume that an adult member of public consumes $350\ \mathrm{l}$ of drinking water per year. The annual dose limit for members of the public is $1\ \mathrm{mSv}$. The effective dose coefficient (ingestion, adult members of the public) of Cs-137 is $1.3\times10^{-8}\ \mathrm{Sv/Bq}$. Thus, you could contaminate a water volume of $$V=\frac{6.7\times10^{8}\ \mathrm{Bq}\times1.3\times10^{-8}\ \mathrm{Sv\ Bq^{-1}}\times350\ \mathrm{l}}{1\ \mathrm{mSv}}=3.0\times10^6\ \mathrm l=3.0\times10^3\ \mathrm{m^3}$$

Clearly, the psychological and social impact of such an incident would be larger than any actual radiological consequences.