Three Mile Island Nuclear Power Station Accident

Abstract

Let us start with the scientific and engineering basis of the severe accident analysis carried out before the Fukushima nuclear accident. In particular, the accident of the Three Mile Island Nuclear Power Station (TMI) had given a comprehensive view when the severe accident took place due to a small break loss of coolant accident. In addition, it provided us important knowledge on the core melt from the destructive examinations undertaken by the international collaboration. How the core melt took place and how such a huge amount of energy came from could be understood with the help of research work on the behavior of the fuel carried out by the international collaboration in the 1970s. I led these studies including fuel destructive tests by the NSSR that clarified how the Zircaloy cladded fuel of the LWR chemically interact with water and generate intense energy to cause core melt and associated hydrogen. These experimental evidences provided the basis of a scenario of the TMI and Fukushima accidents.

Appendices

Appendices

1.1 Appendix 1.1

The phase of the contact surface between the inner surface of the zirconium cladding tube and the surface of uranium dioxide pellets is extremely complicated from the professional standpoint (in the order of micrometer). That is where a lot of complicated things occur, e.g., uranium entering the zircaloy layer, zirconium entering uranium, and oxygen entering the zircaloy layer to change into a substance called α-zircaloy, or reduced uranium metal solving, etc. However, prior to the core melt, the α-zircaloy layer existed behind this complex boundary surface (i.e., on the zircaloy side). The melting point of α-zircaloy is higher than that of the zircaloy alloy which constitutes the cladding tube main body, so that it also serves as the film for forming a sandwich, just like the zirconium oxide coating on the outer surface of the cladding tube. Using this result, I refer to them in this book collectively as an “oxide coatings” in order to avoid a complex explanation.

1.2 Appendix 1.2

1.2.1 Ball-Park Calculation on the 5.5 MPa Pressure Rise

Let us assume that the total volume of the TMI’s primary coolant system is approximately 300 m³, and 200 m³ of saturated water at 8.5 MPa and 100 m³ of saturated steam turned into L kilograms of saturated water of 14 MPa and S kilograms of saturated vapor by heat generated by a zirconium-water reaction.

First, the zirconium-water reaction formula is:

$$ \mathrm{Z}\mathrm{r}+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{ZrO}}_2+2{\mathrm{H}}_2+586\mathrm{kJ}/\mathrm{mol}. $$

The atomic weight of zirconium is approximately 91.2 and 1 t of zirconium is approximately 1.1 × 10⁴ mol.

When 1 t of zirconium reacts with water, it generates a heat quantity of approximately 6.6 × 10⁶ kJ and approximately 44 kg of hydrogen.

Next, as saturated water of 8.5 megapascal occupies a volume of 0.0014 m³ per kilogram, and saturated steam occupies 0.022 m³, the water volume W that exists in the primary coolant system is:

$$ \mathrm{W}=200/0.0014+100/0.022=1.5\times {10}^5\mathrm{kg} $$

The total internal energy U (8.5 MPa) in the primary coolant system is:

$$ \begin{array}{c}\mathrm{U}\left(8.5\mathrm{M}\mathrm{P}\mathrm{a}\right)=1,328.3\;\left[\mathrm{kJ}/\mathrm{kg}\right]\times 200/0.0014+2,564\;\left[\mathrm{kJ}/\mathrm{kg}\right]\times 100/0.022\\ {}=2.0\times {10}^8\mathrm{kJ}\end{array} $$

As saturated water of 14 MPa occupies a volume of 0.0016 m³ per kilogram, and the specific volume of saturated steam is 0.012 m³ per kilogram,

$$ \begin{array}{l}0.0016\times \mathrm{L}+0.012\times \mathrm{S}=300\hfill \\ {}\mathrm{L}+\mathrm{S}=1.5\times {10}^5\hfill \end{array} $$

Hence, $\mathrm{L}=1.4\times {10}^5\mathrm{kg}, \mathrm{S}=1\times {10}^4\mathrm{kg}$

The total internal energy U (14 MPa) in the primary coolant system is:

$$ \begin{array}{c}\mathrm{U}\left(14\mathrm{M}\mathrm{P}\mathrm{a}\right)=1,548.4\;\left[\mathrm{kJ}/\mathrm{kg}\right]\times 1.4\times {10}^5\left[\mathrm{kg}\right]+2,476.1\left[\mathrm{kJ}/\mathrm{kg}\right]\times 1\times {10}^4\left[\mathrm{kg}\right]\\ {}=2.4\times {10}^8\mathrm{kJ}\end{array} $$

$$ \mathrm{U}\left(14\mathrm{M}\mathrm{P}\mathrm{a}\right)-\mathrm{U}\left(8.5\mathrm{M}\mathrm{P}\mathrm{a}\right)=4\times {10}^7\mathrm{kJ} $$

In order to provide this much energy by a zirconium-water reaction, approximately 6 t of zirconium needs to react with water.

Also, when 6 t of zirconium reacts with water, it generates approximately 260 kg of hydrogen.

In this evaluation, the pressure rise due to the generation of hydrogen gas is not considered so the zirconium-water reaction is overestimated. (The contribution of the hydrogen gas generation to the pressure rising is estimated to be about a half of that of the heat generation.)

While there are various estimations as to the amount of hydrogen gas generated during the TMI accident, it is generally agreed that it was about 400 kg, so that 260 kg as the amount of hydrogen generated at this point is more or less appropriate.