Boiling and Drying Accident of High-Level Liquid Waste in a Reprocessing Plant: Examination of the Time-Dependent Temperature Increase of the Waste and of the Generation Rates of the Individual Components Released into the Gas Phase

Abstract

Using the amount, composition, and decay power density of high-level liquid waste in a storage tank, the temperature change of the waste up to 600°C and the corresponding vapor and gas release rates of H₂O, HNO₃, NO₂, NO, and O₂ as a function of time after the loss of cooling function were obtained by the following method. The heat balance equations in and around the tank were derived, and the solution of the waste temperature change was numerically obtained using the vaporization rates of H₂O and HNO₃ and the generation rate of NOx, which were both obtained from the experiments using the simulated liquid waste. Utilizing the temperature versus time curve obtained from the equation, the release rates of the components described above were obtained as a function of time. This information on the progress of the accident can be used to study the Leak Path Factor of radioactive materials, especially of volatilized Ru, and further, it becomes basic information when considering accident management and suppressing the impact of a disaster.

Keywords:

• Reprocessing
• high-level liquid waste
• severe accident
• boiling, drying

I. INTRODUCTION

The boiling and drying accident of high-level liquid waste (HLLW) has drawn considerable attention among stakeholders because of high potential radiation exposure to the general public. Accordingly, many experiments and analyses have been conducted.^[1–10] In order to assess the radiation exposure, the amount of radioactive aerosols and volatilized ¹⁰⁶Ru released into the environment is necessary to be known. To do this, it is necessary to understand the behavior of H₂O and HNO₃ vaporization and of NO₂, NO, and O₂ generation, which not only carry the radioactive materials but also affect the Leak Path Factor (LPF),^[11] especially of volatilized ¹⁰⁶Ru.^[9]

Previously, we reported^[10] the NO₂, NO, and O₂ generation rates from the simulated high-level liquid waste (SHLLW) up to 600°C under constant temperature increasing rates of 0.2 and 1°C∙min⁻¹. In this study, using the amount, composition, and decay power density of HLLW in a storage tank, we predict how the waste^a temperature up to 600°C^b and also the individual generation rates of H₂O and HNO₃ vapors and NO₂, NO, and O₂ gases will change with time, based on the heat balance equation related to the inside and outside of the storage tank and experimental results on the H₂O and HNO₃ vaporization and NOx generation using SHLLW.

Similar reports on the study of the waste temperature and the gas release rates as a function of time exist so far.^[2–4,7] However, we consider that the contents of these reports are not sufficient as shown below.

Ishikawa et al. reported that the temperature increase of the waste up to 140°C could be predicted by extrapolating the molar boiling point elevation model.^[4] However, HLLW is a nitric acid solution in which various types of nitrates are dissolved at high concentrations and precipitate with increasing temperature. Therefore, we consider that it is difficult to construct and solve the theoretical model up to 140°C. Indeed, the agreement between the predicted values and the experimental ones is not always good as the temperature approaches 140°C. For the analysis from 100°C to 800°C, Yoshida and Ishikawa^[3] used the MELCOR code.^[12] But, the code cannot handle nitric acid, so they introduced control functions and multiple state input volumes into the code to model the key phenomena of boiling events such as boiling above 100°C, nitric acid vaporization, and NOx gas generation.^[3] However, in a subsequent paper^[6] from the same group, the results are said to be qualitative.

To obtain the temperature versus time curve up to 600°C, it is necessary to solve the heat balance equation around the storage tank. The H₂O and HNO₃ vaporization, nitrate decomposition, and heat transfer to the cell air and cell wall proceed simultaneously. This paper describes an integrated model for simulating time-dependent changes in the waste temperature and the released gas composition, as well as the solution method and the results obtained. There is no other report of such contents.

Note that the values used in the following analysis may not reflect the actual plant values.

II. OUTLINE OF THE MODEL TO BE ANALYZED

One tank is installed in one cell. The tank is filled with HLLW. Cell ventilation air is flowing into the cell but stops due to the accident. The corresponding numerical values are as follows:

1. The amount of HLLW is 120 m³.

2. The properties of HLLW are as follows. The amount of 0.4 m³ of HLLW is produced by reprocessing 1 ton of spent fuel (pressurized water reactor fuel, specific power of 38 MW∙ton⁻¹, burnup of 45 000 MWd∙ton⁻¹, and cooling period of 6 years), and its composition is obtained using the ORIGEN-2 code. The decay power of 120 m³ of HLLW becomes 578 kW.

3. Cell ventilation air is introduced with a flow rate of 57 mol∙s⁻¹ and at 25°C.

4. All six outer cell walls are surrounded by concrete building walls, and the air in the space and the building walls both have a temperature of 25°C.

The temperature of HLLW rises with time after the cooling function loss and boiling begins. As time passes, the liquid waste dries and solidifies. Therefore, in the following analysis, the model was divided into two stages: the initial stage and the late stage. In the initial stage, the waste remains in the liquid phase, and so, the temperature of the tank wall and inside the tank is considered to be uniform due to a large amount of steam generated by the decay heat. On the other hand, in the late stage, the temperature distribution occurs inside the tank. Based on the experimental results (see Sec. IV.A.2), we assumed that the waste was in a liquid phase up to 130°C and above that was in a solid phase.^c

III. DERIVATION OF THE HEAT BALANCE EQUATION AND ITS SOLUTION METHOD IN THE INITIAL STAGE

III.A. Heat Balance Equation

The various heat transfer processes and related parameters affecting the waste temperature θ(t) (in degrees Celsius) are shown in Fig. 1. Although most of the decay energy is dissipated in the tank, part of the gamma rays is released from the tank contributing to the temperature rise of the cell wall (gamma heating). The decay energy dissipated in the tank is used not only for the temperature rise of the waste and the tank but also for vaporization of H₂O and HNO₃, decomposition of nitrates, and heat loss out of the tank (convection and heat radiation), and as a result, the waste temperature is determined.

Fig. 1. Heat transfer processes and parameters affecting the waste temperature at the initial stage.

The amount of heat transfer associated with the individual heat transfer processes shown in Fig. 1 depends on the following parameters. Variable t (in seconds) means the time after the cooling loss. The value θ (in degrees Celsius) is a function of t, that is, θ(t), and sometimes is described as simply θ. Also, note the following parameters 1 through 8:

1. Initial storage amount of the waste M₀ (m³).

2. Energy release rate due to the radioactive decay R_h (kW) and the leak rate of gamma-ray energy out of the tank (gamma heating) R_γ(t) (kW); the power of [R_h – R_γ(t)] is used to heat the waste and the tank.

3. Amount of H₂O in the waste M_W(t) (mol), its vaporization rate Y_W(t) (mol∙s⁻¹), latent heat λ_W(θ) (kJ∙mol⁻¹), and molar specific heat c_W (kJ∙mol⁻¹∙°C⁻¹); c_W was assumed to be independent of temperature.^[13]

4. Amount of HNO₃ in the waste M_N(t) (mol), its vaporization rate Y_N(t) (mol∙s⁻¹), latent heat^d λ_N(θ) (kJ∙mol⁻¹), and molar specific heat c_N (kJ∙mol⁻¹∙°C⁻¹); c_N was assumed to be independent of temperature.^[13]

5. Heat absorption rate due to the thermal decomposition of nitrates (salts) R_S(t) (kW) and the total specific heat c_S(θ) (kJ∙kg⁻¹∙°C⁻¹); the specific heat of each nitrate was assumed to be equal to that of the oxide, and the oxides were assumed to have a constant total mass M_S (kg) irrespective of temperature (adequacy of the assumptions is explained in Sec. VI.D).

6. Mass of the tank including the internal structural material M_T (kg) and its specific heat c_T(θ) (kJ∙kg⁻¹∙°C⁻¹).

7. The temperature of the tank is assumed to be the same as that of the waste.

8. Heat transfer rate from the tank to the cell air by natural convection R_loss1(θ) (kW) and that from the tank to the cell wall by heat radiation R_loss2(θ) (kW), and their sum R_loss(θ) = R_loss1(θ) + R_loss2(θ). The cell ventilation function is assumed to be lost due to the accident.

Using the foregoing parameters, the waste temperature θ(t) is described by the following heat balance equation^e:

(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1)

where Γ(t) = sum of the heat capacities of the waste (H₂O, HNO₃, and the nitrates = the oxides^f ) (kJ∙°C⁻¹) and the tank and is given by the following equation:

(2) $\mathrm{\Gamma }\left(\mathit{t}\right)=C_W{M}_W\left(\mathit{t}\right)+c_N{M}_N\left(\mathit{t}\right)+c_S\left(\mathit{\theta }\right)M_S+c_T\left(\mathit{\theta }\right)M_T.$(2)

The right side of Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) shows the net power to increase the temperature of the waste and the tank.

III.B. Solution Method

In order to obtain the solution of Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) up to 130°C, it is necessary to know the time change of the individual parameters 1 through 8 shown in Sec. III.A. In the present study, the relationship between the individual vaporization rates of H₂O and HNO₃ and the waste temperature was obtained from the experimental results as shown in Sec. IV. The relationship between the heat absorption rate accompanying the decomposition of nitrates and the waste temperature is examined in Sec. V. In this examination, the experimental results of the previous report concerning the NOx generation^[10] were used. The heat capacity Γ(t) in Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) is treated in Sec. VI, and the heat transfer from the tank including gamma heating is dealt with in Sec. VII. These results are integrated to solve Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) by the difference method in Sec. VIII. Then, using the result of Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) , the generation rates of gaseous components are obtained as a function of time.

IV. HEAT ABSORPTION RATE OF H₂O AND HNO₃ VAPORIZATION

The content in this section can be used for both stages. The heat absorption rate due to liquid vaporization is the sum of the corresponding individual values of H₂O and HNO₃. Each heat absorption rate is obtained by multiplying the vaporization rate and the latent heat. The vaporization rates were obtained from the results of experiments using SHLLW as described below. Since H₂O and HNO₃ are vaporized from nitric acid solution, the interaction between H₂O and HNO₃ needs to be taken into account to obtain the individual latent heats. This effect is described in Secs. IV.B.3. and IV.C.3.

IV.A. Measurement of the Vaporization Rates from the Waste

IV.A.1. Experimental

Two types of the simulated wastes, SHLLW-J and SHLLW-K, were prepared. The individual compositions are shown in Table I. SHLLW-J simulates HLLW, and SHLLW-K contains only the main 12 compounds. The details are given in the previous report.^[9] SHLLW-J was used to examine the vaporization rates of H₂O and HNO₃ as a function of time, and SHLLW-K was used to examine the effect of the temperature increasing rate on their vaporization ratios versus temperature curves (see Sec. IV.A.2.b). Two runs were carried out using SHLLW-J and five runs using SHLLW-K. Individual experimental conditions are given in Table II.

TABLE I Compositions of the Simulated Solutions

		Concentration^a
		SHLLW-J	SHLLW-K
Element	Reagent	mol∙L^−1
P	H3PO4	0.0035	—
Cr	Cr(NO3)3	0.0068	—
Fe	Fe(NO3)3·9H2O	0.025	0.025
Ni	Ni(NO3)2·6H2O	0.0033	0.003
Rb	RbNO3	0.015	—
Sr	Sr(NO3)2	0.033	0.034
Y	Y(NO3)3·6H2O	0.018	—
Zr	ZrO(NO3)2·2H2O	0.18	0.180
Mo	Mo^b	0.117	0.117
Mn	Mn(NO3)2·6H2O	0.051	—
Ru	RuNO(NO3)3·1.8H2O	0.091	0.091
Rh	Rh(NO3)3	0.014	—
Pd	Pd(NO3)2·2H2O	0.045	0.043
Ag	AgNO3	0.0022	—
Cd	Cd(NO3)2·4H2O	0.0035	—
Sn	SnO2	0.0024	—
Sb	Sb2O3	0.0006	—
Te	TeO2	0.016	—
Cs	CsNO3	0.068	0.069
Ba	Ba(NO3)2	0.040	—
La	La(NO3)3·6H2O	0.030	—
Ce	Ce(NO3)3·6H2O	0.10	0.099
Pr	Pr(NO3)3·6H2O	0.027	—
Nd	Nd(NO3)3·6H2O	0.096	0.095
Sm	Sm(NO3)3·6H2O	0.017	—
Eu	Eu(NO3)3·6H2O	0.0034	—
Gd	Gd(NO3)3·6H2O	0.090	0.090
H^+	HNO3	2.0	2.0

^aConcentrations were analyzed by inductively coupled plasma–atomic emission spectroscopy or inductively coupled plasma–mass spectrometry except oxides and HNO₃. Oxide concentrations were calculated from the amounts added.

^bMetal Mo powder was dissolved in 4 mol∙L⁻¹ nitric acid.

TABLE II List of the Experimental Conditions for Each Run

		Temperature Increasing Rate
	Up to	104°C to 120°C	120°C to 155°C	155°C to …
Run	°C	°C∙min^−1
J1	273	0.31	0.91	1.7
J2	280	0.38	1.3	1.9
K1	300	0.40	2.5	5.2
K2	300	0.12	0.28	1.5
K3	300	0.23	0.13	0.75
K4	300	0.17	0.065	0.36
K5	300	0.15	0.042	0.47

The apparatus to measure the vaporization rate is shown schematically in Fig. 2. SHLLW-J of 120 mL was charged into a separable flask with an inner diameter of 85 mm and a height of 110 mm (nominal volume of 500 mL). For SHLLW-K, 120 to 200 mL was charged. Two thermocouples were inserted to measure the waste temperature near the bottom and the vapor temperature at the top of the flask. The flask was heated on a hot plate. A 2-mm-thick aluminum plate was placed between the flask and the hot plate so that the heating from the bottom was uniform. In addition, flexible rubber heaters were wound around the side of the separable flask and the lid of the flask. The pipe connecting the steam outlet to the inlet of a Liebig condenser was heated by a ribbon heater. The waste temperature was controlled by the hot plate and the temperatures of the side wall; the lid and the pipe were controlled to be the same as the waste temperature.

Fig. 2. Schematic diagram of the apparatus used for the vaporization experiments.

Steam generated in the flask flowed into the condenser, and the condensate was recovered periodically. The temperature of the cooling water was 15°C. The recovery was performed at appropriate intervals in a batch manner from the boiling point to about 280°C (the sampling interval can be seen from the data points in Fig. 3 shown in the next section). At temperatures above 120°C where the amount of condensate became small, 10 mL of H₂O was injected into the condenser before recovering the condensate through the pipe shown in Fig. 2 to wash and collect the condensate attached on the inner wall.

Fig. 3. Time changes of vaporized amounts of H₂O and HNO₃.

The mass and density of the recovered liquid including the condensate and the washing water were measured for each batch, and then, the liquid was diluted to 1000 to 10 000 times to measure the pH. These results were used to obtain the net amounts of H₂O and HNO₃ recovered in the condenser. It should be noted that the measured amount of HNO₃ is different from the amount vaporized in the flask as described in the next section.

In order to observe the state of the waste, a vertical slit was made in the rubber heater that was wound around the side of the flask. As boiling began and concentration proceeded, it became unclear whether the waste was liquid or solid. At 130°C, no bubbling was observed, and the waste became like a paste. From the previous experiment, the following is known.^[10] When the temperature reached 180°C, the waste was more like a solid than a paste, but the waste taken out from the flask was somewhat moist and gave off a strong HNO₃ odor. The waste heated to 280°C to 300°C was dry. It was solid with many voids.

IV.A.2. Results and Discussion

IV.A.2.a. Time Change of Vaporized Amounts of H₂O and HNO₃

We examined the amount of vaporized H₂O and HNO₃ using SHLLW-J. The initial amount of H₂O in the liquid waste obtained was as follows:

1. Liquid waste density: 1270 kg∙m⁻³ (measured value of SHLLW-J at room temperature).

2. Amount of HNO₃: 126 kg∙m⁻³ = 2000 mol∙m⁻³ (calculated from the concentration of 2 mol∙L⁻¹).

3. Amount of nitrates (including H₂MoO₄, SnO₂, Sb₂O₃, and TeO₂): 276 kg∙m⁻³ (calculated from Table I assuming anhydrous nitrates).

4. Amount of H₂O: 868 kg∙m⁻³ = 48 200 mol∙m⁻³ (calculated by subtracting the contribution of HNO₃ and nitrates from the liquid waste density).

The initial amount of H₂O obtained in this way, 48 200 mol∙m⁻³, became the same as the final vaporized amount shown in Fig. 3, which shows the time-dependent changes of the measured cumulative amounts of H₂O (and HNO₃) obtained from Runs J1 and J2. In the figure, the cumulative amount is shown as molar quantity per 1 m³ of initial liquid waste, and the boiling start point corresponds to t = 0.

On the other hand, the measured cumulative amount of HNO₃ at 280°C is about 2600 mol∙m⁻³ for both runs as shown in Fig. 3, which is different from the initial amount of 2000 mol∙m⁻³. This difference is explained as follows. Previously, we reported^[10] that most of the Zr, Ru, and Pd nitrates in SHLLW-J do not produce NOx but produce HNO₃ by 180°C, while HNO₃ left in the drying waste produces Nox mainly above 300°C, and that the difference in the amount between them is 100 mol∙m⁻³. Therefore, the final vaporized amount of HNO₃ should be equal to 2100 mol∙m⁻³. The difference between 2100 mol∙m⁻³ and the measured 2600 mol∙m⁻³ should be due to the amount produced from NOx, mainly NO₂, in the condenser by the following reactions^[14]:

(3) $3\mathrm{N}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{HN}{\mathrm{O}}_3+\mathrm{NO},$(3)

(4) $2\mathrm{N}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}\to \mathrm{HN}{\mathrm{O}}_2+\mathrm{HN}{\mathrm{O}}_3,$(4)

$3\mathrm{HN}{\mathrm{O}}_2\to \mathrm{HN}{\mathrm{O}}_3+2\mathrm{NO}+{\mathrm{H}}_2\mathrm{O},$

where H₂O comes from the condensate and the washing water injected into the condenser. Formulas (3) and (4) show that H₂O is consumed in absorption of NO₂ and that the amount of H₂O in mol∙m⁻³ in the initial waste is about 24 times larger than that of HNO₃, and so, the consumption due to the NO₂ absorption is negligible. The cumulative amount curve of vaporized HNO₃ shown in Fig. 3 was obtained by using the final value of 2100 mol∙m⁻³ and correcting the amount of HNO₃ produced by NO₂ according to its rate of generation.

IV.A.2.b. Effect of the Temperature Increasing Rate on the Vaporization Ratios

In five runs using SHLLW-K, the heating time from 104°C (boiling start temperature) to 300°C was changed variously as shown in Table II, and the effect of the temperature increasing rate on the vaporization ratios of H₂O and HNO₃ was examined. Individual vaporization ratios were obtained using the corresponding final vaporization amounts as the denominator and the cumulative vaporized amounts at that temperature as the numerator. The final amounts were obtained in the same way as for SHLLW-J described above. The vaporization ratio versus temperature curves for H₂O and HNO₃ thus obtained are shown in Figs. 4 and 5, respectively. These curves can be approximated by the individual single curves for H₂O and HNO₃ irrespective of the temperature increasing rate. As shown in Secs. IV.B.1 and IV.C.1, these individual curves are consistent with those obtained from SHLLW-J. By 155°C, almost all H₂O and 80% HNO₃ are vaporized. The temperature increasing rates up to 155°C used in Table II cover the realistic rates caused by the HLLW’s decay power density of 4.82 kW∙m⁻³ as seen in Fig. 13 of Sec. X.A.

Fig. 4. H₂O vaporization ratio versus temperature curves obtained from Runs K1 through K5.

Fig. 5. HNO₃ vaporization ratio versus temperature curves obtained from Runs K1 through K5.

IV.B. Heat Absorption Rate of H₂O Vaporization

The heat absorption rate due to the vaporization is given by the product of the vaporization rate and the latent heat of vaporization. The vaporization rate can be obtained from the above experimental results, and the latent heat can be obtained from the literature.^[15]

IV.B.1. Relationship Between the H₂O Vaporization Ratio and Temperature

The H₂O vaporization ratio ξ_W(θ), that is, the ratio of the cumulative amount at temperature θ to its final amount, is shown in Fig. 6, which was obtained from the experimental results of Runs J1 and J2. When θ < 104°C (before boiling), ξ_W(θ) = 0. Based on the results of the figure, the curve for dξ_W(θ)/dθ was obtained as shown in Fig. 7. Using the figure, the following approximate expressions were obtained. The solid lines in the figure correspond to these expressions:

(5) $\left.\begin{array}{cc}104<\mathit{\theta }\le {150}^{\circ }\mathrm{C},& \mathrm{lo}{\mathrm{g}}_{10}\{\mathit{d}{\xi }_W\left(\mathit{\theta }\right)/\mathit{d\theta }\}=-0.0474\mathit{\theta }+3.952\\ \mathit{\theta }>{150}^{\circ }\mathrm{C},& \mathrm{lo}{\mathrm{g}}_{10}\{\mathit{d}{\xi }_W\left(\mathit{\theta }\right)/\mathit{d\theta }\}=-0.00709\mathit{\theta }-2.094\end{array}\right\}.$(5)

Fig. 6. H₂O and HNO₃ vaporization ratios versus temperature curves obtained from Runs J1 and J2.

Fig. 7. The dξ_W/dθ versus temperature curve obtained from Runs J1 and J2.

Although the data points scatter above 160°C, it is attributed to a small amount of vaporization and is not important since most of the H₂O has already been vaporized.

IV.B.2. Vaporization Rate

The H₂O vaporization rate Y_W(t), which is equal to the time derivative of the cumulative amount of H₂O vaporization M_W(0)ξ_W(θ), is given as follows:

(6) $Y_W\left(\mathit{t}\right)=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{dt}}=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}}.$(6)

The final cumulative amount of vaporized H₂O is equal to the initial value of M_W(0) as explained in Sec. IV.A.2.a. The value dξ_W(θ)/dθ in Eq. (6)(6) $Y_W\left(\mathit{t}\right)=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{dt}}=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}}.$(6) is given by Eq. (5)(5) $\left.\begin{array}{cc}104<\mathit{\theta }\le {150}^{\circ }\mathrm{C},& \mathrm{lo}{\mathrm{g}}_{10}\{\mathit{d}{\xi }_W\left(\mathit{\theta }\right)/\mathit{d\theta }\}=-0.0474\mathit{\theta }+3.952\\ \mathit{\theta }>{150}^{\circ }\mathrm{C},& \mathrm{lo}{\mathrm{g}}_{10}\{\mathit{d}{\xi }_W\left(\mathit{\theta }\right)/\mathit{d\theta }\}=-0.00709\mathit{\theta }-2.094\end{array}\right\}.$(5) .

IV.B.3. Latent Heat of Vaporization

For the latent heat of H₂O vaporization λ_W(θ) (in kJ∙mol⁻¹), the following approximate equations created from data from 50°C to 373.95°C^[16] were used:

$\mathit{\theta }\le {370}^{\circ }\mathrm{C},$

$\begin{array}{rl}& {\mathit{\lambda }}_W\left(\mathit{\theta }\right)=-450{\left(\mathit{\theta }/100\right)}^6+5070{\left(\mathit{\theta }/100\right)}^5\\ & -22,600{\left(\mathit{\theta }/100\right)}^4+50,000{\left(\mathit{\theta }/100\right)}^3\\ & -58,100(\mathit{\theta }/100{)}^2+28,700(\mathit{\theta }/100)\\ & +38,000;\end{array}$

$370<\mathit{\theta }\le {374}^{\circ }\mathrm{C},$

${\mathit{\lambda }}_W\left(\mathit{\theta }\right)=-2.09\left(\mathit{\theta }-370\right)+8.25.$

Since H₂O is vaporized from nitric acid solution in the present case, the interaction between H₂O and HNO₃ needs to be taken into account. This effect was approximated by the heat of dissolution of HNO₃ in water and included in the latent heat of HNO₃ vaporization (see Sec. IV.C.3).

IV.B.4. Heat Absorption Rate

The product of the vaporization rate Y_W(t) and its latent heat λ_W(θ) obtained above is the heat absorption rate.

IV.C. Heat Absorption Rate of HNO₃ Vaporization

IV.C.1. Relationship Between the HNO₃ Vaporization Ratio and Temperature

The HNO₃ vaporization ratio ξ_N(θ), that is, the ratio of the cumulative amount at temperature θ to its final amount obtained from Runs J1 and J2, is also plotted in Fig. 6. When θ < 104°C (before boiling), ξ_N(θ) = 0. Based on the results of the figure, the curve for dξ_N(θ)/dθ was obtained as shown in Fig. 8. Using the figure, the following approximate expressions were obtained. The solid lines in the figure correspond to the following approximate expressions:

(7) $\left.\begin{array}{c}104\mathit{\hspace{0.17em}}<\mathit{\theta }\le {126}^{\circ }\mathrm{C},\mathit{\hspace{0.17em}}\mathrm{lo}{\mathrm{g}}_{10}\left(\mathit{d}{\xi }_N\left(\mathit{\theta }\right)/\mathit{d\theta }\right)=0.0647\mathit{\theta }-\mathit{\hspace{0.17em}}9.59\\ 126\mathit{\hspace{0.17em}}<\mathit{\theta }\le {170}^{\circ }\mathrm{C},\mathit{\hspace{0.17em}}\mathrm{lo}{\mathrm{g}}_{10}\left(\mathit{d}{\xi }_N\left(\mathit{\theta }\right)/\mathit{d\theta }\right)\mathit{\hspace{0.17em}}=-0.0277\mathit{\theta }+\mathit{\hspace{0.17em}}2.055\\ \mathit{\theta }>\mathit{\hspace{0.17em}}{170}^{\circ }\mathrm{C},\mathit{\hspace{0.17em}}\mathrm{lo}{\mathrm{g}}_{10}\left(\mathit{d}{\xi }_N\left(\mathit{\theta }\right)/\mathit{d\theta }\right)\mathit{\hspace{0.17em}}=-0.00202\mathit{\theta }-\mathit{\hspace{0.17em}}2.313\end{array}\right\}.$(7)

Fig. 8. The dξ_N/dθ versus temperature curve obtained from Runs J1 and J2.

IV.C.2. Vaporization Rate

The HNO₃ vaporization rate Y_N(t), which is equal to the time derivative of the cumulative amount of HNO₃ vaporization M_N(0)ξ_N(θ), is given as follows:

(8) $Y_N\left(\mathit{t}\right)=M_N\left(0\right)\frac{d{\xi }_N\left(\mathit{\theta }\right)}{\mathit{dt}}=M_N\left(0\right)\frac{d{\xi }_N\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}},$(8)

where $\mathit{d}{\xi }_N\left(\mathit{\theta }\right)/\mathit{d\theta }$ is given by Eq. (7)(7) $\left.\begin{array}{c}104\mathit{\hspace{0.17em}}<\mathit{\theta }\le {126}^{\circ }\mathrm{C},\mathit{\hspace{0.17em}}\mathrm{lo}{\mathrm{g}}_{10}\left(\mathit{d}{\xi }_N\left(\mathit{\theta }\right)/\mathit{d\theta }\right)=0.0647\mathit{\theta }-\mathit{\hspace{0.17em}}9.59\\ 126\mathit{\hspace{0.17em}}<\mathit{\theta }\le {170}^{\circ }\mathrm{C},\mathit{\hspace{0.17em}}\mathrm{lo}{\mathrm{g}}_{10}\left(\mathit{d}{\xi }_N\left(\mathit{\theta }\right)/\mathit{d\theta }\right)\mathit{\hspace{0.17em}}=-0.0277\mathit{\theta }+\mathit{\hspace{0.17em}}2.055\\ \mathit{\theta }>\mathit{\hspace{0.17em}}{170}^{\circ }\mathrm{C},\mathit{\hspace{0.17em}}\mathrm{lo}{\mathrm{g}}_{10}\left(\mathit{d}{\xi }_N\left(\mathit{\theta }\right)/\mathit{d\theta }\right)\mathit{\hspace{0.17em}}=-0.00202\mathit{\theta }-\mathit{\hspace{0.17em}}2.313\end{array}\right\}.$(7) .

IV.C.3. Latent Heat of Vaporization

The latent heat of HNO₃ vaporization from aqueous nitric acid λ_N (in kJ∙mol⁻¹) was approximated as the sum of the latent heat of HNO₃ vaporization from pure HNO₃ and the dissolution heat of HNO₃ in water:

${\lambda }_N\left(\mathit{\theta }\right)=39.43\times {\left[\frac{1-\left(\mathit{\theta }+273.15\right)/520}{1-293.15/520}\right]}^{0.375}+H_{\mathit{sol}}.$

The first term on the right side is the latent heat of HNO₃ vaporization,^[4] and the second term H_sol (in kJ∙mol⁻¹) is the dissolution heat of HNO₃ in water. The latter was calculated by the following approximate Eq. (9)(9) $\left.\begin{array}{c}0.03<\mathit{\omega }\le 0.265,H_{\mathit{sol}}=-485{\omega }^3+308{\omega }^2-14.3\mathit{\omega }+0.67\\ \mathit{\omega }>0.265,H_{\mathit{sol}}=43.5\mathit{\omega }-2.02\end{array}\right\},$(9) created from the data at 25°C^[15]:

(9) $\left.\begin{array}{c}0.03<\mathit{\omega }\le 0.265,H_{\mathit{sol}}=-485{\omega }^3+308{\omega }^2-14.3\mathit{\omega }+0.67\\ \mathit{\omega }>0.265,H_{\mathit{sol}}=43.5\mathit{\omega }-2.02\end{array}\right\},$(9)

where ω = molar fraction of HNO₃. H_sol does not depend on temperature as is explained below.

The derivative of the dissolution heat with respect to temperature is equal to the difference between the heat capacity of the solution and the sum of the heat capacities of H₂O and HNO₃. Since the heat capacity of nitric acid solution could be assumed to be equal to the sum of the heat capacities of H₂O and HNO₃ as is explained in Sec. VI.A, the dissolution heat does not depend on temperature. Note that the values of ω and H_sol change with time as a result of vaporization.

IV.C.4. Heat Absorption Rate

The product of the HNO₃ vaporization rate Y_N(t) and the latent heat of vaporization λ_N(θ) obtained above is the heat absorption rate accompanying HNO₃ vaporization.

V. HEAT ABSORPTION RATE OF THERMAL DECOMPOSITION OF NITRATES

The content in this section can be used for both stages. The total heat absorption rate R_S(t) associated with the thermal decomposition of nitrates is equal to the sum of products obtained by multiplying the decomposition heats required for the generation of unit amounts of NO₂ and NO by the respective generation rates (see Sec. V.C). In the following, the decomposition heat and the generation rate are examined.

V.A. Heats of Decomposition Generating NO₂ and NO

In the waste, various nitrates are included. The decomposition heat is estimated as follows. Increasing the temperature of the waste, the nitrates generate not only NO₂ and NO but also HNO₃.^[10] However, the decomposition heat generating HNO₃ is small as is explained in Appendix A, and so, it is not taken into consideration.

First, the decomposition heat generating NO₂ is examined taking Nd(NO₃)₃ as an example:

$4\mathrm{Nd}{\left(\mathrm{N}{\mathrm{O}}_3\right)}_3\to 2\mathrm{N}{\mathrm{d}}_2{\mathrm{O}}_3+12\mathrm{N}{\mathrm{O}}_2+3{\mathrm{O}}_2.$

The enthalpy of this reaction Δ_d1H (in kJ∙mol⁻¹) is obtained as

${\mathrm{\Delta }}_{\mathit{d}1}\mathit{H}={\mathrm{\Delta }}_f\mathit{H}\left(\mathrm{N}{\mathrm{d}}_2{\mathrm{O}}_3\right)/2+3{\mathrm{\Delta }}_f\mathit{H}\left(\mathrm{N}{\mathrm{O}}_2\right)-{\mathrm{\Delta }}_f\mathit{H}\left(\mathrm{Nd}{\left(\mathrm{N}{\mathrm{O}}_3\right)}_3\right),$

where Δ_fH(Nd₂O₃), Δ_fH(NO₂), and Δ_fH(Nd(NO₃)₃) are their formation enthalpies.

The decomposition reaction of Nd(NO₃)₃ to generate NO and its enthalpy Δ_d2H are given as follows:

$4\mathrm{Nd}{\left(\mathrm{N}{\mathrm{O}}_3\right)}_3\to 2\mathrm{N}{\mathrm{d}}_2{\mathrm{O}}_3+12\mathrm{NO}+9{\mathrm{O}}_2$

$\mathit{\hspace{0.17em}}{\mathrm{\Delta }}_{\mathit{d}2}\mathit{H}={\mathrm{\Delta }}_f\mathit{H}\left(\mathrm{N}{\mathrm{d}}_2{\mathrm{O}}_3\right)/2+3{\mathrm{\Delta }}_f\mathit{H}\left(\mathrm{NO}\right)-{\mathrm{\Delta }}_f\mathit{H}\left(\mathrm{Nd}{\left(\mathrm{N}{\mathrm{O}}_3\right)}_3\right).$

Therefore, if the formation enthalpies of Nd(NO₃)₃, Nd₂O₃, NO, and NO₂ are given, the enthalpy of the nitrate decomposition is obtained. The temperature dependence of the enthalpy is several percent within the scope of this study,^[13] so it was not considered, and the standard enthalpy values were used. In Table III, the standard enthalpies related to the various nitrates and oxides in the waste are shown. In the table, the heat absorption accompanying the individual nitrate decomposition is also shown. The heat absorption shown in units of kJ∙L⁻¹ is obtained by multiplying Δ_d1H (or Δ_d2H) by the number of moles of nitrate per 1 L of waste (see Table I). Standard formation enthalpies of all oxides and part of the nitrates were obtained from the literature.^[13,17] The enthalpies of the nitrates that are not given in the literature were estimated as follows.

TABLE III Estimated Heat Absorption Associated with Nitrate Decomposition that Produces NO₂ or NO

	ΔfH°		Oxide	ΔfH° kJ∙mol^−1	End Product: NO2				End Product: NO
Compound	kJ∙mol^−1	Reference Number/Value of ΔfH°	Oxide	ΔfH° kJ∙mol^−1	Δd1H kJ∙mol^−1	Δd1H/NO2 kJ∙mol^−1	NO2 mol∙L^−1	Heat kJ∙L^−1	Δd2H kJ∙mol^−1	Δd2H/NO kJ∙mol^−1	NO mol∙L^−1	Heat kJ∙L^−1
HNO3	−174	16	NO2	33	64	64	0.51	33
			NO	90					121	121	0.15	18
Cr(NO3)3	−652	^a	Cr2O3	−1140	182	61	0.016	1.0	353	118	0.004	0.5
Ni(NO3)2	−415	16	NiO	−240	241	121	0.005	0.6	356	178	0.001	0.3
RbNO3	−495	15	Rb2O	−339	359	359	0.012	4.2	246	246	0.003	0.8
Sr(NO3)2	−978	16	SrO	−593	452	226	0.051	11.6	566	283	0.015	4.1
Y(NO3)3	−1172	^a	Y2O3	−1905	319	106	0.042	4.5	491	164	0.012	1.9
Mn(NO3)2	−576	16	MnO	−385	257	129	0.080	10.2	372	186	0.022	4.2
Rh(NO3)3	−351	^a	Rh2O3	−343	279	93	0.033	3.0	450	150	0.009	1.4
Pd(NO3)2	−233	^a	PdO	−85	214	107	0.012	1.3	328	164	0.003	0.6
AgNO3	−124	16	Ag2O	−31	142	142	0.002	0.2	184	184	0.000	0.1
Cd(NO3)2	−456	16	CdO	−258	264	132	0.005	0.7	379	189	0.002	0.3
CsNO3	−506	16	Cs2O	−346	366	366	0.053	19.4	250	250	0.015	3.7
Ba(NO3)2	−992	16	BaO	−548	510	255	0.062	15.9	625	312	0.018	5.5
La(NO3)3	−1256	^a	La2O3	−1794	459	153	0.070	10.7	630	210	0.020	4.2
Ce(NO3)3	−1226	16	Ce2O3	−1796	427	142	0.234	33.3	599	200	0.066	13.2
Pr(NO3)3	−1239	^a	Pr2O3	−1810	434	145	0.063	9.1	605	202	0.018	3.6
Nd(NO3)3	−1097	^a	Nd2O3	−1808	293	98	0.225	22.0	464	155	0.063	9.8
Sm(NO3)3	−1203	^a	Sm2O3	−1823	391	130	0.040	5.2	562	187	0.011	2.1
Eu(NO3)3	−1097	^a	Eu2O3	−1650	372	124	0.008	1.0	543	181	0.002	0.4
Gd(NO3)3	−1182	^a	Gd2O3	−1820	372	124	0.211	26.1	543	181	0.059	10.7
ZrO(NO3)2	—		ZrO2	—	—	120	0.072	8.6	–	176	0.020	3.6
RuNO(NO3)3	—		RuO2	—	—	120	0.048	5.8	–	176	0.014	2.4
						Sum =	1.86	228		Sum =	0.52	91

^aThe value of Δ_fH° was estimated from Δ_fH° of chloride using Eq. (10)(10) ${\mathrm{\Delta }}_f{H}^{\circ }\left(\mathrm{nitrate}\right)=1.17{\mathrm{\Delta }}_f{H}^{\circ }\left(\mathrm{chloride}\right).$(10) .

Many standard formation enthalpies of chlorides are given in the literature.^[13] For Mn, Ni, Sr, Ag, Cd, Cs, Ba, and Ce elements, standard formation enthalpies of chlorides and nitrates are given. There is a good correlation between the two, which can be approximated by the following equation:

(10) ${\mathrm{\Delta }}_f{H}^{\circ }\left(\mathrm{nitrate}\right)=1.17{\mathrm{\Delta }}_f{H}^{\circ }\left(\mathrm{chloride}\right).$(10)

If the standard formation enthalpy for nitrate was not found in the literature, it was estimated from the standard formation enthalpy of chloride using Eq. (10)(10) ${\mathrm{\Delta }}_f{H}^{\circ }\left(\mathrm{nitrate}\right)=1.17{\mathrm{\Delta }}_f{H}^{\circ }\left(\mathrm{chloride}\right).$(10) . The values obtained in this way are listed as superscript a (^a) in Table III.

The decomposition enthalpies of the individual nitrates generating NO₂ or NO were estimated using the standard enthalpies obtained above and are shown in Table III as Δ_d1H and Δ_d2H. The values of Δ_d1H/NO₂ and Δ_d2H/NO show the decomposition heat generating 1 mol of NO₂ and NO, respectively.

However, since the standard formation enthalpies of not only nitrate but also chloride of Zr and Ru are not found in the literature, the following estimation was used. Values of Δ_d1H/NO₂ for nitrates excluding alkali and alkaline earth metal nitrates go into the range of 61 to 153 kJ∙mol⁻¹, and so, the average of these values of 120 kJ∙mol⁻¹ was adopted for ZrO(NO₃)₂ and RuNO(NO₃)₃. Similarly, 176 kJ∙mol⁻¹ was adopted for Δ_d2H/NO of Zr and Ru nitrates.

In the following calculation, the generation ratio of NO₂, i.e., NO = 80:20, reported in the previous paper^[10] was used. The decomposition heat of each nitrate in 1 L of initial waste (in kJ∙L⁻¹) generating NO₂ was calculated as the product of the concentration of the nitrate in the waste (in mol∙L⁻¹), the amount of NO₂ generated from 1 mol of nitrate (in mol-NO₂∙mol⁻¹), and the NO₂ generation heat (in kJ mol-NO₂⁻¹). However, we reported previously^[10] that Zr, Ru, and Pd nitrates mostly decomposed by 180°C but that only 17% of N in these nitrates was released as NOx, and so, 17% of the concentrations of these nitrates given in Table I were used to calculate the above decomposition heat. In Table III, the decomposition heat of each nitrate obtained above is shown as “Heat,” and the amount of NO₂ generated from each nitrate is also shown. In addition, we reported that with rising temperature, 0.66 mol of NOx was generated per 1 L of initial waste by the decomposition of HNO₃.^[10] In Table III, NOx generation from the decomposition of HNO₃ is also included.

The amount of NO₂ generated from 1 L of initial waste and the corresponding decomposition heat are 1.86 mol∙L⁻¹ and 228 kJ∙L⁻¹, respectively, as shown as “Sum” in the last row in Table III. Therefore, the heat absorption per 1 mol of NO₂ generation is obtained as

${\lambda }_{\mathrm{NO}2}=123\mathrm{kJ}\cdot \mathrm{mo}{\mathrm{l}}^{-1}.$

Similarly,

${\lambda }_{\mathrm{NO}}=174\mathrm{kJ}\cdot \mathrm{mo}{\mathrm{l}}^{-1}.$

V.B. NOx Generation Rate

The generation rates of NO₂ and NO, expressed as R_NO2(t) and R_NO(t) (in mol∙s⁻¹), are given by the following Arrhenius-type equations, respectively, which were obtained for a temperature increasing rate of 0.2°C∙min⁻¹ previously.^[10] The amount of NOx in the equations is the amount generated from the initial waste of 120 m³:

(11) $\begin{array}{rl}& R_{\mathrm{NO}2}\left(\mathit{t}\right)=-\sum _{\mathit{i}=1}^5\frac{d{M}_i\left(\mathit{t}\right)}{\mathit{dt}}\\ & =-\sum _{\mathit{i}=1}^5{A}_i\exp \left[-\frac{E_i}{\mathit{R}\left[\mathit{\theta }\left(\mathit{t}\right)+273.15\right]}\right]M_i\left(\mathit{t}\right)& (11)\end{array}$(11)

(12) $R_{\mathrm{NO}}\left(\mathit{t}\right)=-\sum _{\mathit{i}=6}^7\frac{d{M}_i\left(\mathit{t}\right)}{\mathit{dt}}=-\sum _{\mathit{i}=6}^7{A}_i\exp \left[-\frac{E_i}{\mathit{R}\left[\mathit{\theta }\left(\mathit{t}\right)+273.15\right]}\right]M_i\left(\mathit{t}\right)$(12)

$M_i\left(\mathit{t}\right)=M_i\left(0\right)\exp \left\{-A_i\underset{0}{\overset{t}{\int }}\exp \left[-\frac{E_i}{\mathit{R}\left[\mathit{\theta }\left(\mathit{t}{ }^{\mathrm{\prime }}\right)+273.15\right]}\right]\mathit{d}{t}^{\prime }\right\}.$

The value M_i(t) (in moles) shows an amount of NO₃ in group i that generates NOx. It is assumed that the decomposition of 1 mol M_i(t) generates 1 mol of NO₂ (i = 1 to 5) or 1 mol of NO (i = 6 and 7). The terms A_i (in inverse seconds), E_i (in kJ∙mol⁻¹), and R (in kJ∙mol⁻¹∙°C⁻¹) are the frequency factor, activation energy, and gas constant, respectively. Equation (11)(11) $\begin{array}{rl}& R_{\mathrm{NO}2}\left(\mathit{t}\right)=-\sum _{\mathit{i}=1}^5\frac{d{M}_i\left(\mathit{t}\right)}{\mathit{dt}}\\ & =-\sum _{\mathit{i}=1}^5{A}_i\exp \left[-\frac{E_i}{\mathit{R}\left[\mathit{\theta }\left(\mathit{t}\right)+273.15\right]}\right]M_i\left(\mathit{t}\right)& (11)\end{array}$(11) approximates NO₂ generation by five groups and Eq. (12)(12) $R_{\mathrm{NO}}\left(\mathit{t}\right)=-\sum _{\mathit{i}=6}^7\frac{d{M}_i\left(\mathit{t}\right)}{\mathit{dt}}=-\sum _{\mathit{i}=6}^7{A}_i\exp \left[-\frac{E_i}{\mathit{R}\left[\mathit{\theta }\left(\mathit{t}\right)+273.15\right]}\right]M_i\left(\mathit{t}\right)$(12) NO generation by two groups. The values for these parameters previously obtained^[10] are reprinted in Table IV after converting the waste volume 1 to 120 m³ and time minutes to seconds.

TABLE IV List of Constants Used in the Arrhenius-Type Equations

Group i		Initial Amount, Mi (mol)	Frequency Factor, Ai (s^−1)	Activation Energy, Ei (kJ∙mol^−1)
NO2	1	9960	8.3E+15	150
	2	13 900	3.3E+11	140
	3	41 900	3.3E+08	120
	4	149 500	3.3E+10	150
	5	28 000	5.0E+08	140
NO	6	31 900	3.3E+04	100
	7	18 000	1.7E+10	200

Since 0.25 mol of O₂ is generated per 1 mol of NO₂ generation and 0.75 mol of O₂ per 1 mol of NO, the O₂ generation rate R_O2(t) is obtained from R_NO2(t) and R_NO(t) as follows^[10]:

$R_{\mathrm{O}2}\left(\mathit{t}\right)=0.25R_{\mathrm{NO}2}\left(\mathit{t}\right)+0.75R_{\mathrm{NO}}\left(\mathit{t}\right).$

V.C. Heat Absorption Rate

The heat absorption rate R_S(t) accompanying the decomposition of nitrate is obtained as the product of the NOx generation rate described in Sec. V.B and the amount of the heat absorption per 1 mol of NOx generation described in Sec. V.A. Therefore, R_S(t) is given as follows:

$R_S\left(\mathit{t}\right)={\lambda }_{\mathrm{NO}2}{R}_{\mathrm{NO}2}\left(\mathit{t}\right)+{\lambda }_{\mathrm{NO}}{R}_{\mathrm{NO}}\left(\mathit{t}\right).$

VI. HEAT CAPACITY

The content in this section can be used for both stages.

In addition to the vaporization of liquid and the decomposition of nitrates, the heat capacities of the waste and the tank are required as the heat absorption terms. These heat capacities are included in Γ in Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) . The total heat capacity Γ is examined below.

VI.A. H₂O and HNO₃

As shown in Eq. (2)(2) $\mathrm{\Gamma }\left(\mathit{t}\right)=C_W{M}_W\left(\mathit{t}\right)+c_N{M}_N\left(\mathit{t}\right)+c_S\left(\mathit{\theta }\right)M_S+c_T\left(\mathit{\theta }\right)M_T.$(2) , the heat capacity of the waste is the sum of the products of the individual amounts of H₂O, HNO₃, and nitrates in the waste and their corresponding specific heats. The following values were used for the molar specific heats of H₂O and HNO₃:

$c_W=0.076\mathrm{kJ}\cdot \mathrm{mo}{\mathrm{l}}^{-1}{\cdot }^{\circ }{\mathrm{C}}^{-1}$

$c_N=0.110\mathrm{kJ}\cdot \mathrm{mo}{\mathrm{l}}^{-1}{\cdot }^{\circ }{\mathrm{C}}^{-1}.$

The value of H₂O at 100°C and that of HNO₃ at 25°C are shown in the literature.^[16] Since the temperature dependence on these heat capacities is insignificant,^[18] it was not considered.

VI.B. Nitrates

Specific heats of the individual nitrates in the waste were not found in the literature. However, since the dissolved nitrates precipitate and further decompose into oxides with increasing temperature, we assumed that the heat capacity of the nitrate is approximated by that of the oxide even before boiling. Under this assumption, the total amount of the oxides in the initial waste becomes 153 kg∙m⁻³. The specific heat of 18 oxides lies between 0.27 to 0.79 kJ∙°C⁻¹∙kg⁻¹ (at 25°C),^[13,15] and the average value weighted by their concentrations in the initial waste is 0.37 kJ∙°C⁻¹∙kg⁻¹. Using the specific heats of Ba, Cr, Fe, Sr, and Zr oxides at 400 and 800 K,^[13] the increase rates per 1°C of these oxides were found to be 0.0028°C⁻¹ to 0.0033°C⁻¹, averaging 0.0029°C⁻¹. Therefore, the average specific heat of the nitrates c_S(θ) (in kJ∙kg⁻¹∙°C⁻¹) multiplied by the total mass of the nitrates (= oxides) is approximated as follows:

(13) $c_S\left(\mathit{\theta }\right)\times 153\times 120=6800[1+0.0029(\mathit{\theta }-25)].$(13)

VI.C. Tank

The material of the tank (including the internal structural material) was assumed to be Type 316 stainless steel (SUS-316). Its heat capacity C_T(θ) (in kJ∙°C⁻¹) was obtained by the following formula:

(14) $C_T\left(\mathit{\theta }\right)=c_T\left(\mathit{\theta }\right)\times 70,000=15.61\mathit{\theta }+34,200,$(14)

where c_T(θ) = specific heat of SUS-316 at temperatures between 27°C and 727°C^[18]; 70 000 kg = mass of the tank.

VI.D. Examination of the Nitrate Heat Capacity Estimation Error

Since some assumptions are included in the derivation of the heat capacity of the nitrates, the effect of the error was examined. Using the initial amount of the waste of M₀ = 120 m³, the heat capacities of H₂O, HNO₃, nitrates, and tank are given as follows:

1. $H_2\mathit{O}:c_W{M}_W\left(t_b\right)=0.076\times 48,200\times 120=$ $4.40\times {10}^5\mathrm{kJ}$ ${\cdot }^{\circ }{\mathrm{C}}^{-1}$

2. $\mathit{HN}{O}_3:c_N{M}_N\left(t_b\right)=0.110\times 2300\times 120=3.0$ $\times {10}^4\mathrm{kJ}$ ${\cdot }^{\circ }{\mathrm{C}}^{-1}$

3. Nitrates: $c_S({104}^{\circ }\mathrm{C})M_S$ $=8.4\times {10}^3\mathrm{kJ}{\cdot }^{\circ }{\mathrm{C}}^{-1}$ $(M_S=153\times 120\mathrm{kg})$ [see Eq. (13)(13) $c_S\left(\mathit{\theta }\right)\times 153\times 120=6800[1+0.0029(\mathit{\theta }-25)].$(13) ]

4. Tank: $c_T({104}^{\circ }\mathrm{C})M_T=$ $3.6\times {10}^4\mathrm{kJ}{\cdot }^{\circ }{\mathrm{C}}^{-1}$ [see Eq. (14)(14) $C_T\left(\mathit{\theta }\right)=c_T\left(\mathit{\theta }\right)\times 70,000=15.61\mathit{\theta }+34,200,$(14) ]

5. Total heat capacity: $\mathrm{\Gamma }\left(t_b\right)=5.14\times {10}^5\mathrm{kJ}{\cdot }^{\circ }$ ${\mathrm{C}}^{-1},$

where t_b and 104°C are the time and the temperature at the beginning of boiling, respectively.

Since the contribution of the nitrates is only 1.6% before boiling, the effect of the error assuming that the heat capacity of the nitrate is equal to that of the oxide is negligible. At 300°C, at which the waste is almost dry and most of the nitrates convert to oxides, the contribution of the oxides to the total heat capacity becomes 19%, and so, the effect of the heat capacity estimation error of the nitrates (= oxides) is small.

VII. HEAT TRANSFER RATE FROM THE TANK IN THE INITIAL STAGE

The way of handling this part in the late stage is described in Sec. IX.

VII.A. Used Parameters and Values

There are three energy loss paths from the tank: convective heat transfer to the cell air, heat radiation to the cell wall, and gamma-ray release. The following parameters and values were used to determine the heat transfer rate of each path:

Parameter 1. The temperature of the tank is equal to that of the waste, and the initial value is θ(0) = 50°C.

Parameter 2. In the calculation of the convective heat transfer, the following assumptions were used:

a. The tank size is 7-m outside diameter × 4.0-m height, and its surface area S_T = 165 m². The thickness of the tank wall is 0.03 m. The heat transfer coefficients at the top and bottom of the tank are given by the same equation [Eq. (B.1)(B.1) ${\alpha }_{\mathrm{T}}=0.13{\left[\frac{\mathit{g\beta }\left(\mathit{\theta }-{\mathit{\theta }}_{\mathrm{a}}\right)}{{\nu }_{\mathrm{a}}{a}_{\mathrm{a}}}\right]}^{\frac{1}{3}}{\kappa }_{\mathrm{a}},$(B.1) ] as that of its vertical side^[17] (see Sec. B.II).

b. The inside cell size is an 8.1-m cube, and its surface area S_C = 394 m². The thickness of the concrete cell wall is L = 2.0 m.

c. The cell air is heated by natural convective heat transfer on the tank outer wall and cooled on the inner cell wall. Since the heat capacity of air is small and the air is well mixed by convection, this heating and cooling can be regarded as balanced. The air temperature inside the cell, θ_ai(t) (in degrees Celsius), is assumed to be uniform before and after the accident.

Parameter 3. Physical property values used for the calculation of the convective heat transfer are given in Appendix B.

VII.B. Equation for the Heat Transfer Rate

VII.B.1. Convective Heat Transfer

The heat transfer rate from the tank to air R_loss1(θ) is obtained by the following equation:

(15) $R_{\mathit{loss}1}\left(\mathit{\theta }\right)={\alpha }_T{S}_T\left[\mathit{\theta }\left(\mathit{t}\right)-{\theta }_{\mathit{ai}}\left(\mathit{t}\right)\right],$(15)

where α_T = convective heat transfer coefficient of the tank wall (kW∙m⁻²∙°C⁻¹), the details of which are described in Appendix B.

VII.B.2. Heat Radiation

The radiant heat transfer rate from the tank to the cell wall R_loss2(θ) is given by the following equation,^[18] which is expressed in absolute temperature T = θ + 273.15 (in kelvins) to avoid complications in display:

(16) $R_{\mathit{loss}2}\left(\mathit{\theta }\right)={\mathit{\epsilon }}_{\mathit{To}}\mathit{\sigma }{S}_T\left[T^4-{\left({\mathit{T}}_{\mathit{IW}}\right)}^4\right],$(16)

where σ = Stefan-Boltzmann constant (5.67 × 10⁻¹¹ kW∙m⁻²∙K⁻⁴); T = tank (= waste) temperature; T_IW = temperature of the inner cell surface; ε_To = emissivity of the tank outer surface^[19] of 0.4. The emissivity of concrete^[19] is 0.94. In this paper the emissivity larger than 0.9 was approximated as 1, that is, no reflection, in the calculation (see Sec. B.II).

VII.C. Solution Method of the Equations and Calculated Results

To solve Eqs. (15)(15) $R_{\mathit{loss}1}\left(\mathit{\theta }\right)={\alpha }_T{S}_T\left[\mathit{\theta }\left(\mathit{t}\right)-{\theta }_{\mathit{ai}}\left(\mathit{t}\right)\right],$(15) and (16)(16) $R_{\mathit{loss}2}\left(\mathit{\theta }\right)={\mathit{\epsilon }}_{\mathit{To}}\mathit{\sigma }{S}_T\left[T^4-{\left({\mathit{T}}_{\mathit{IW}}\right)}^4\right],$(16) , it is necessary to know the air temperature in the cell and the temperature of the inner cell surface. Since the heat capacity of air is small, the heat transferred from the tank to the air is soon transferred to the cell wall. Therefore, the following equation is obtained:

(17) ${\alpha }_T{S}_T\left[\mathit{\theta }\left(\mathit{t}\right)-{\theta }_{\mathit{ai}}\left(\mathit{t}\right)\right]={\alpha }_{\mathit{Ci}}{S}_C\left[{\theta }_{\mathit{ai}}\left(\mathit{t}\right)-{\theta }_C\left(\mathit{t},0\right)\right],$(17)

where α_ci = convective heat transfer coefficient of the inner cell wall surface (kW∙m⁻²∙°C⁻¹) given by the same equation as Eq. (B.1)(B.1) ${\alpha }_{\mathrm{T}}=0.13{\left[\frac{\mathit{g\beta }\left(\mathit{\theta }-{\mathit{\theta }}_{\mathrm{a}}\right)}{{\nu }_{\mathrm{a}}{a}_{\mathrm{a}}}\right]}^{\frac{1}{3}}{\kappa }_{\mathrm{a}},$(B.1) in Appendix B; θ_C(t,0) = inner surface temperature of the cell wall. Using Eq. (17)(17) ${\alpha }_T{S}_T\left[\mathit{\theta }\left(\mathit{t}\right)-{\theta }_{\mathit{ai}}\left(\mathit{t}\right)\right]={\alpha }_{\mathit{Ci}}{S}_C\left[{\theta }_{\mathit{ai}}\left(\mathit{t}\right)-{\theta }_C\left(\mathit{t},0\right)\right],$(17) , we get

(18) ${\theta }_{\mathit{ai}}\left(\mathit{t}\right)=\frac{{\alpha }_T{S}_T\mathit{\theta }\left(\mathit{t}\right)+{\alpha }_{\mathit{ci}}{S}_C{\theta }_C\left(\mathit{t},0\right)}{{\alpha }_T{S}_T+{\alpha }_{\mathit{ci}}{S_C}_{\mathit{\hspace{0.17em}}}}.$(18)

The inner surface temperature of the cell θ_C(t,0) is determined by the balance of the heat transfer from the gas, the heat radiation from the tank, the heat generation inside the cell by gamma rays, and the heat conduction to the inside of the cell wall. Therefore, it is necessary to find the temperature distribution inside the cell by solving the heat equation. The temperature distribution inside the concrete cell wall denoted by θ_C(t,x) is expressed by the following equation:

(19) $\frac{\mathrm{\partial }{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial t}}={\alpha }_C\frac{{\mathrm{\partial }}^2{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial }{x}^2}+\frac{q_{\gamma }\left(\mathit{t},\mathit{x}\right)}{{\rho }_C{C}_C},$(19)

where a_C = thermal diffusivity (m²∙s⁻¹); ρ_C = density (kg∙m⁻³); C_C = specific heat of the cell concrete (kJ∙kg⁻¹∙°C⁻¹) (see Appendix B); q_γ(t,x) = heat generation density due to the gamma heating (kW∙m⁻³) given by Eq. (C.1)(C.1) $\left.\begin{array}{c}0\le {\xi }_{\mathit{\gamma }}\le 0.66,R_{\mathit{\gamma }}\left(\mathit{t}\right)=1.64{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)+2.18\\ 0.66\le {\xi }_{\mathit{\gamma }}\le 1,R_{\mathit{\gamma }}\left(\mathit{t}\right)=20.8{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)-10.47\end{array}\right\},$(C.1) in Appendix C.

The boundary conditions are

(20) $\begin{array}{rl}& \mathit{x}=0,\mathit{t}>0:{\alpha }_{\mathit{ci}}\left[{\theta }_{\mathit{ai}}\left(\mathit{t}\right)-{\theta }_C\left(\mathit{t},0\right)\right]\\ & +\frac{{\mathit{R}}_{\mathit{loss}2}\left(\mathit{t},\mathit{x}\right)}{S_C}=-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=0}& (20)\end{array}$(20)

(21) $\mathit{x}=\mathit{L},\mathit{t}>0:-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=\mathit{L}}={\alpha }_{\mathit{co}}\left[{\theta }_C\left(\mathit{t},\mathit{L}\right)-{\theta }_{\mathit{ao}}\right]+F_{\mathit{OW}}\mathit{\sigma }\left[T_C{\left(\mathit{t},\mathit{L}\right)}^4-{{\mathit{T}}_{\mathit{OW}}}^4\right],$(21)

where κ_C = thermal conductivity of the cell concrete (kW∙m⁻¹∙°C⁻¹); θ_ao = air temperature outside the cell; α_co = convective heat transfer coefficient of the outer cell wall surface (kW∙m⁻²∙°C⁻¹) given by the same equation as Eq. (B.1)(B.1) ${\alpha }_{\mathrm{T}}=0.13{\left[\frac{\mathit{g\beta }\left(\mathit{\theta }-{\mathit{\theta }}_{\mathrm{a}}\right)}{{\nu }_{\mathrm{a}}{a}_{\mathrm{a}}}\right]}^{\frac{1}{3}}{\kappa }_{\mathrm{a}},$(B.1) ; F_OW = configuration factor between the outer cell surfaces and the building wall surfaces facing the outer cell surfaces. The second term on the right side of Eq. (21)(21) $\mathit{x}=\mathit{L},\mathit{t}>0:-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=\mathit{L}}={\alpha }_{\mathit{co}}\left[{\theta }_C\left(\mathit{t},\mathit{L}\right)-{\theta }_{\mathit{ao}}\right]+F_{\mathit{OW}}\mathit{\sigma }\left[T_C{\left(\mathit{t},\mathit{L}\right)}^4-{{\mathit{T}}_{\mathit{OW}}}^4\right],$(21) is the net radiant heat transfer rate from the outer cell wall surfaces to their surrounding concrete building surfaces at T_OW = 25 + 273.15 K. The outside air temperature θ_ao was assumed to be constant at 25°C. F_OW was set to 0.2 assuming a distance of 8 m between the two surfaces.^[19]

These boundary conditions were rearranged as follows in order to apply the numerical calculation algorithm shown in the literature.^[19] Substituting Eq. (18)(18) ${\theta }_{\mathit{ai}}\left(\mathit{t}\right)=\frac{{\alpha }_T{S}_T\mathit{\theta }\left(\mathit{t}\right)+{\alpha }_{\mathit{ci}}{S}_C{\theta }_C\left(\mathit{t},0\right)}{{\alpha }_T{S}_T+{\alpha }_{\mathit{ci}}{S_C}_{\mathit{\hspace{0.17em}}}}.$(18) into Eq. (20)(20) $\begin{array}{rl}& \mathit{x}=0,\mathit{t}>0:{\alpha }_{\mathit{ci}}\left[{\theta }_{\mathit{ai}}\left(\mathit{t}\right)-{\theta }_C\left(\mathit{t},0\right)\right]\\ & +\frac{{\mathit{R}}_{\mathit{loss}2}\left(\mathit{t},\mathit{x}\right)}{S_C}=-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=0}& (20)\end{array}$(20) to eliminate θ_ai(t) and substituting Eq. (16)(16) $R_{\mathit{loss}2}\left(\mathit{\theta }\right)={\mathit{\epsilon }}_{\mathit{To}}\mathit{\sigma }{S}_T\left[T^4-{\left({\mathit{T}}_{\mathit{IW}}\right)}^4\right],$(16) for R_loss2, the boundary condition at x = 0 can be rewritten as

(22) $\mathit{x}=0,\mathit{t}>0:{\mathrm{\Lambda }}_1\left[\mathit{\theta }-{\theta }_C\left(\mathit{t},0\right)\right]=-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=0}$(22)

$\begin{array}{rl}& {\mathrm{\Lambda }}_{\mathrm{l}}=\frac{{\alpha }_T{\alpha }_{\mathit{ci}}{S}_T}{{\alpha }_T{S}_T+{\alpha }_{\mathit{ci}}{S}_C}\\ & +\frac{S_T}{S_C}{\epsilon }_T\mathit{\sigma }\left[T^2+T_C{\left(\mathit{t},0\right)}^2\right]\left[\mathit{T}+T_C\left(\mathit{t},0\right)\right],\end{array}$

where Λ₁ = overall heat transfer coefficient including the effect of radiant heat transfer.

The boundary condition at x = L, Eq. (21)(21) $\mathit{x}=\mathit{L},\mathit{t}>0:-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=\mathit{L}}={\alpha }_{\mathit{co}}\left[{\theta }_C\left(\mathit{t},\mathit{L}\right)-{\theta }_{\mathit{ao}}\right]+F_{\mathit{OW}}\mathit{\sigma }\left[T_C{\left(\mathit{t},\mathit{L}\right)}^4-{{\mathit{T}}_{\mathit{OW}}}^4\right],$(21) , can also be rewritten as

(23) $\mathit{\hspace{0.17em}}\mathit{x}=\mathit{L},\mathit{t}>0:-{\kappa }_{\mathit{c}}{\left.\frac{\mathrm{\partial }{\mathit{\theta }}_{\mathrm{C}}}{\mathrm{\partial }\mathit{x}}\right|}_{\mathit{x}=\mathit{L}}={\mathrm{\Lambda }}_{\mathrm{L}}\left[T_{\mathrm{C}}\left(\mathit{t},\mathit{L}\right)-{\mathrm{\Theta }}_{\mathrm{L}}\right]$(23)

$\begin{array}{rl}& {\mathrm{\Lambda }}_{\mathit{L}}={\alpha }_{\mathit{co}}+\mathit{\sigma }\left[T_{\mathit{C}}\left(\mathit{t},\mathit{L}\right)+T_{\mathit{ow}}\right]\left[T_{\mathit{C}}{\left(\mathit{t},\mathit{L}\right)}^2+{{\mathit{T}}_{\mathit{OW}}}^2\right]\\ & {\mathrm{\Lambda }}_{\mathit{L}}{\mathrm{\Theta }}_{\mathit{L}}={\alpha }_{\mathit{co}}{T}_{\mathit{ao}}+\mathit{\sigma }\left[T_{\mathit{C}}\left(\mathit{t},\mathit{L}\right)+T_{\mathit{ow}}\right]\left[T_{\mathit{C}}{\left(\mathit{t},\mathit{L}\right)}^2+{{\mathit{T}}_{\mathit{OW}}}^2\right].\end{array}$

Equation (19)(19) $\frac{\mathrm{\partial }{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial t}}={\alpha }_C\frac{{\mathrm{\partial }}^2{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial }{x}^2}+\frac{q_{\gamma }\left(\mathit{t},\mathit{x}\right)}{{\rho }_C{C}_C},$(19) was numerically solved under the boundary conditions (22) and (23) and the initial condition, which will be explained later. The 2-m-thick wall was divided into 150 increments (Δx = 0.0133 m), and the time step Δt was set to be 109.7 s, which was derived from the condition that α_CΔt/Δx² = ½, which enables simplicity and stability of numerical integration.^[17]

The initial temperature distribution inside the cell wall is the distribution before the accident happens. It was determined as follows. Assuming that the cell ventilation air is continuously fed at 25°C (= ${\theta }_a^{in}$ in the following equation) and the temperature of the tank is maintained at 50°C, the temperature of the cell air θ_a is given by

$\mathit{\hspace{0.17em}}{\mathit{\theta }}_{\mathrm{a}}=\frac{{\mathit{c}}_{\mathrm{a}}{F}_{\mathrm{Ca}}{\theta }_{\mathrm{a}}^{\mathrm{i}\mathrm{n}}+S_{\mathrm{T}}{\alpha }_{\mathrm{T}}{\theta }_{\mathrm{T}}+S_{\mathrm{C}}{\alpha }_{\mathrm{C}}{\theta }_{\mathrm{C}}}{c_{\mathrm{a}}{F}_{\mathrm{Ca}}+S_{\mathrm{T}}{\alpha }_{\mathrm{T}}+S_{\mathrm{C}}{\alpha }_{\mathrm{C}}},$

where c_a = molar specific heat of air (kJ∙mol⁻¹∙°C⁻¹); F_Ca = flow rate of cell ventilation air before the accident (mol∙s⁻¹).

The boundary condition at x = 0 for the time before the accident is

(24) $\mathit{x}=0,\mathit{t}>0:{\mathrm{\Lambda }}_0\left[\mathit{\theta }-{\theta }_{\mathrm{C}}\left(\mathit{t},0\right)\right]=-{\kappa }_{\mathrm{c}}{\left.\frac{\mathrm{\partial }{\mathit{\theta }}_{\mathrm{C}}}{\mathrm{\partial }\mathit{x}}\right|}_{\mathit{x}=0},$(24)

${\mathrm{\Lambda }}_0=\frac{{\mathit{\alpha }}_{\mathrm{ci}}{c}_{\mathrm{a}}{F}_{\mathrm{a}}\left[{\theta }_{\mathrm{a}}^{\mathrm{i}\mathrm{n}}-{\theta }_{\mathrm{C}}\left(\mathit{t},0\right)\right]/\left[\mathit{\theta }-{\theta }_{\mathrm{C}}\left(\mathit{t},0\right)\right]+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{T}}{\alpha }_{\mathrm{T}}}{c_{\mathrm{a}}{F}_{\mathrm{ca}}+S_{\mathrm{T}}{\alpha }_{\mathrm{T}}+S_{\mathrm{C}}{\alpha }_{\mathrm{C}}}+\frac{{\mathit{\epsilon }}_{\mathrm{T}}\mathit{\sigma }{S}_{\mathrm{T}}\left[T^2+T_{\mathrm{C}}{\left(\mathit{t},0\right)}^2\right]\left[\mathit{T}+T_{\mathrm{C}}\left(\mathit{t},0\right)\right]}{S_{\mathrm{C}}},$

where Λ₀ = overall heat transfer coefficient for the time before the accident.

Using Eqs. (21)(21) $\mathit{x}=\mathit{L},\mathit{t}>0:-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=\mathit{L}}={\alpha }_{\mathit{co}}\left[{\theta }_C\left(\mathit{t},\mathit{L}\right)-{\theta }_{\mathit{ao}}\right]+F_{\mathit{OW}}\mathit{\sigma }\left[T_C{\left(\mathit{t},\mathit{L}\right)}^4-{{\mathit{T}}_{\mathit{OW}}}^4\right],$(21) , (22(22) $\mathit{x}=0,\mathit{t}>0:{\mathrm{\Lambda }}_1\left[\mathit{\theta }-{\theta }_C\left(\mathit{t},0\right)\right]=-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=0}$(22) ), and (24(24) $\mathit{x}=0,\mathit{t}>0:{\mathrm{\Lambda }}_0\left[\mathit{\theta }-{\theta }_{\mathrm{C}}\left(\mathit{t},0\right)\right]=-{\kappa }_{\mathrm{c}}{\left.\frac{\mathrm{\partial }{\mathit{\theta }}_{\mathrm{C}}}{\mathrm{\partial }\mathit{x}}\right|}_{\mathit{x}=0},$(24) ), Eq. (19)(19) $\frac{\mathrm{\partial }{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial t}}={\alpha }_C\frac{{\mathrm{\partial }}^2{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial }{x}^2}+\frac{q_{\gamma }\left(\mathit{t},\mathit{x}\right)}{{\rho }_C{C}_C},$(19) was solved numerically. Starting with an initial uniform distribution at 35°C, a steady-state distribution before the accident was obtained after a sufficient number of time steps (about 400 h). The temperatures of the cell air, the inside surface of the cell wall, and the outside surface of the cell wall are 34°C, 39°C, and 32°C, respectively. This temperature distribution is used as an initial condition for the waste temperature calculation after the accident.

VIII. CALCULATION OF THE HEAT BALANCE EQUATION IN THE INITIAL STAGE

The time change of the waste temperature is obtained by integrating Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) over time. The amounts of H₂O and HNO₃, the heat absorption rates due to the liquid vaporization and the nitrate decomposition, and the heat transfer rates are already described in the previous sections as a function of time.

Substituting Eqs. (6)(6) $Y_W\left(\mathit{t}\right)=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{dt}}=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}}.$(6) and (8)(8) $Y_N\left(\mathit{t}\right)=M_N\left(0\right)\frac{d{\xi }_N\left(\mathit{\theta }\right)}{\mathit{dt}}=M_N\left(0\right)\frac{d{\xi }_N\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}},$(8) into Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) , the following equation is obtained:

(25) $\left[\mathrm{\Gamma }\left(\mathit{t}\right)+{\lambda }_{\mathrm{W}}{M}_{\mathrm{W}}\left(0\right)\mathit{d}{\xi }_{\mathrm{W}}/\mathit{d\theta }+{\lambda }_{\mathrm{N}}{M}_{\mathrm{N}}\left(0\right)\mathit{d}{\xi }_{\mathrm{N}}/\mathit{d\theta }\right]\frac{\mathit{d\theta }}{\mathit{dt}}=R_{\mathrm{h}}-R_{\mathrm{s}}\left(\mathit{t}\right)-R_{\mathrm{loss}}-R_{\mathit{\gamma }},$(25)

where R_γ = leak rate of gamma-ray energy out of the tank given by Eq. (C.1)(C.1) $\left.\begin{array}{c}0\le {\xi }_{\mathit{\gamma }}\le 0.66,R_{\mathit{\gamma }}\left(\mathit{t}\right)=1.64{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)+2.18\\ 0.66\le {\xi }_{\mathit{\gamma }}\le 1,R_{\mathit{\gamma }}\left(\mathit{t}\right)=20.8{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)-10.47\end{array}\right\},$(C.1) shown in Sec. C.II.

Rearranging Eq. (25)(25) $\left[\mathrm{\Gamma }\left(\mathit{t}\right)+{\lambda }_{\mathrm{W}}{M}_{\mathrm{W}}\left(0\right)\mathit{d}{\xi }_{\mathrm{W}}/\mathit{d\theta }+{\lambda }_{\mathrm{N}}{M}_{\mathrm{N}}\left(0\right)\mathit{d}{\xi }_{\mathrm{N}}/\mathit{d\theta }\right]\frac{\mathit{d\theta }}{\mathit{dt}}=R_{\mathrm{h}}-R_{\mathrm{s}}\left(\mathit{t}\right)-R_{\mathrm{loss}}-R_{\mathit{\gamma }},$(25) , we get

(26) $\frac{\mathit{d\theta }}{\mathit{dt}}=\frac{{\mathit{R}}_{\mathrm{h}}-R_{\mathrm{s}}\left(\mathit{t}\right)-R_{\mathrm{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right)}{\mathrm{\Gamma }\left(\mathit{t}\right)+{\lambda }_{\mathrm{W}}{M}_{\mathrm{W}}\left(0\right)\mathit{d}{\xi }_{\mathrm{W}}\left(\mathit{\theta }\right)/\mathit{d\theta }+{\lambda }_{\mathrm{N}}{M}_{\mathrm{N}}\left(0\right)\mathit{d}{\xi }_{\mathrm{N}}\left(\mathit{\theta }\right)/\mathit{d\theta }}.$(26)

Here, variables whose values change with time are expressed as a function of time.

Denoting a time step by $\mathrm{\Delta t}$, $\mathit{n}\Delta \mathit{t}$ by $t_n$, and θ(t_n) by θ_n and approximating dθ/dt = (θ_n+1 − θ_n)/Δt and the value of the right side of Eq. (26)(26) $\frac{\mathit{d\theta }}{\mathit{dt}}=\frac{{\mathit{R}}_{\mathrm{h}}-R_{\mathrm{s}}\left(\mathit{t}\right)-R_{\mathrm{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right)}{\mathrm{\Gamma }\left(\mathit{t}\right)+{\lambda }_{\mathrm{W}}{M}_{\mathrm{W}}\left(0\right)\mathit{d}{\xi }_{\mathrm{W}}\left(\mathit{\theta }\right)/\mathit{d\theta }+{\lambda }_{\mathrm{N}}{M}_{\mathrm{N}}\left(0\right)\mathit{d}{\xi }_{\mathrm{N}}\left(\mathit{\theta }\right)/\mathit{d\theta }}.$(26) to be equal to the value at t = t_n, we get

${\theta }_{\mathit{n}+1}={\theta }_n+\frac{{\mathit{R}}_{\mathrm{h}}-R_{\mathrm{s}}\left(t_n\right)-R_{\mathrm{loss}}\left({\theta }_n\right)-R_{\mathit{\gamma }}\left(t_n\right)}{\mathrm{\Gamma }\left(t_n\right)+{\lambda }_{\mathrm{W}}{M}_{\mathrm{W}}\left(0\right){\left[\mathit{d}{\mathit{\xi }}_{\mathrm{W}}\left(\mathit{\theta }\right)/\mathit{d\theta }\right]}_{\mathit{t}=t_n}+{\lambda }_{\mathrm{N}}{M}_{\mathrm{N}}\left(0\right){\left[\mathit{d}{\mathit{\xi }}_{\mathrm{N}}\left(\mathit{\theta }\right)/\mathit{d\theta }\right]}_{\mathit{t}=t_n}}\mathrm{\Delta }\mathit{t}.$

By repeating this operation, θ can be determined as a function of time.

IX. DERIVATION OF THE HEAT BALANCE EQUATION AND ITS SOLUTION METHOD IN THE LATE STAGE

IX.A. Heat Balance Equation

IX.A.1. Description of the Model

Figure 9 shows the modeled temperature distribution inside and outside the tank. The tank is modeled as a cylinder with a diameter of 7 m, a height of 4 m, and a wall thickness of 0.03 m. The thickness of the debris is 0.32 m (see Sec. C.II). The tank wall is separated into the following two parts. One contacts the debris (bottom and a small portion of the side wall of the tank), and the other does not contact the debris (top and most of the side wall), which are named as the tank bottom and the tank top/side, respectively. The debris is evenly divided into three parts: the upper, the middle, and the bottom. The debris volume was assumed to be a constant 12 m³ above 130°C.

Fig. 9. Parameters on temperature θ and on surface area S around the tank in the late stage.

The temperature of each part is displayed as follows. The temperatures of the tank bottom and the tank top/side are θ_Tb(t) and θ_Tt(t), respectively. The temperatures of each piece of the debris are denoted as θ_du(t), θ_dm(t), and θ_db(t) assuming that the temperature of each part is uniform. We considered θ_db(t) = θ_Tb(t) because the heat transfer from the bottom debris to the tank is much larger than that from the tank to the cell. Θ_g(t) denotes the gas temperature in the tank, and θ_ai(t) denotes the inner cell air temperature. For simplicity, the cell wall temperature θc(t,x) is assumed to be the same regardless of the cell’s top, bottom, left, and right positions.

The surface area of each part is displayed as follows and shown in Fig. 9. S₁ is the surface area of each piece of the debris, which is equal to S₂ of the tank top. S₃ is the surface area of the tank side without contacting the debris. S₄ is the area of the side surface of the debris (S₄/S₃ is the area of the side surface of each piece of the debris). Thus, S₁ = S₂ = 38.5 m², S₃ = 80.9 m², and S₄ = 7.0 m². In deriving S₃ and S₄, the debris thickness of 0.32 m was used. S_C is the area of the inner cell surface of 394 m².

The mass of the tank is 39 200 kg, and that of the structural material inside the tank is 30 800 (= 70 000 – 39 200) kg. Assuming that the structural material was distributed uniformly according to the longitudinal length of the tank, the mass of the tank bottom is 13 300 (= 10 800 + 2500) kg, that of the side part is 47 500 (19 200 + 28 300) kg, and that of the top is 9140 (9140 + 0) kg (the latter figure in parentheses is the contribution of the structural materials).

IX.A.2. Heat Balance Equation for Each Part Inside and Outside the Tank

IX.A.2.a. Equation for Each Piece of the Debris

Bottom Debris

It was assumed that the temperature of the tank bottom wall is equal to that of the bottom debris θ_db (in degrees Celsius). Therefore, the heat balance equation that combines the bottom debris and the tank bottom is given as follows:

(27) $\left(C_{\mathrm{db}}+C_{\mathrm{Tb}}\right)\frac{\mathit{d}{\mathit{\theta }}_{\mathrm{db}}}{\mathit{dt}}=R_{\mathrm{hb}}-R_{\mathrm{Sb}}-\left({\lambda }_{\mathrm{Wb}}{Y}_{\mathrm{Wb}}+{\lambda }_{\mathrm{Nb}}{Y}_{\mathrm{Nb}}\right)+Q_{\mathrm{b}}-R_{\mathrm{lossb}},$(27)

where C_db and C_Tb = heat capacities of the bottom debris and the tank bottom (kJ∙K⁻¹); R_hb = heat generation rate due to the decay heat in the bottom debris (kW); R_Sb = heat absorption rate due to the salt (nitrate) decomposition (kW); λ_Wb and λ_Nb = latent heats of vaporization of H₂O and HNO₃ (kJ∙kg¹); Y_Wb and Y_Nb = their vaporization rates (kg∙s⁻¹); Q_b = heat input to the bottom debris from its surroundings (kW); R_lossb = sum of the convective heat transfer from the tank bottom to the cell air and the heat radiation to the cell wall (kW). The equation for Q_b is given as follows:

$Q_{\mathrm{b}}\left(\mathit{t}\right)=S_1{\mathrm{K}}_{\mathrm{d}}\left({\theta }_{\mathrm{dm}}-{\theta }_{\mathrm{db}}\right)+\left(S_4/3\right){\mathrm{K}}_{\mathrm{d}}\left({\theta }_{\mathrm{dm}}-{\theta }_{\mathrm{db}}\right)+\left(S_4/3\right){\mathrm{K}}_{\mathrm{d}}\left({\theta }_{\mathrm{du}}-{\theta }_{\mathrm{db}}\right)-S_{\mathrm{ss}}{\mathrm{K}}_{\mathrm{ss}}\left({\theta }_{\mathrm{db}}-{\theta }_{\mathrm{Tt}}\right),$

where K_d = heat transfer coefficient between two pieces of the debris in contact (W∙m⁻²∙K⁻¹); S_ss = sum of the cross-sectional area of the tank side wall and that of the internal structure (m²); K_ss = transfer coefficient between the tank bottom part contacting with the debris to the tank upper part not contacting with the debris (kW∙m⁻²∙°C⁻¹). The equation for R_lossb is given as follows:

$\begin{array}{rl}& R_{\mathrm{lossb}}\left(\mathit{t}\right)=\left(S_1+S_4\right)[{\alpha }_{\mathrm{l}}\left({\theta }_{\mathrm{Tt}}-{\theta }_{\mathrm{ai}}\right)\\ & \left.+{\epsilon }_{\mathrm{To}}\mathit{\sigma }\left({{\mathit{T}}_{\mathrm{Tb}}}^4-T_{\mathrm{C}}{\left(\mathit{t},0\right)}^4\right)\right],\end{array}$

where α₁ = convective heat transfer coefficient between the tank bottom and the inner cell air; σ = Stefan-Boltzmann constant; ε_To = emissivity of the tank outer surface. It is assumed that the proportion of the heat radiation from the cell wall reaching the outer surface of the tank is equal to the surface area ratio. Although S₄ is not the bottom surface, it has the same temperature as the bottom surface, so it is included in the equation.

Using the evaporation rates Y_Wb and Y_Nb given in Eqs. (6)(6) $Y_W\left(\mathit{t}\right)=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{dt}}=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}}.$(6) and (8)(8) $Y_N\left(\mathit{t}\right)=M_N\left(0\right)\frac{d{\xi }_N\left(\mathit{\theta }\right)}{\mathit{dt}}=M_N\left(0\right)\frac{d{\xi }_N\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}},$(8) , Eq. (27)(27) $\left(C_{\mathrm{db}}+C_{\mathrm{Tb}}\right)\frac{\mathit{d}{\mathit{\theta }}_{\mathrm{db}}}{\mathit{dt}}=R_{\mathrm{hb}}-R_{\mathrm{Sb}}-\left({\lambda }_{\mathrm{Wb}}{Y}_{\mathrm{Wb}}+{\lambda }_{\mathrm{Nb}}{Y}_{\mathrm{Nb}}\right)+Q_{\mathrm{b}}-R_{\mathrm{lossb}},$(27) can be transformed as follows:

(28) $\frac{\mathit{d}{\mathit{\theta }}_{\mathrm{db}}}{\mathit{dt}}=\frac{{\mathit{R}}_{\mathrm{hb}}-R_{\mathrm{Sb}}+Q_{\mathrm{b}}-R_{\mathrm{lossb}}}{{\mathrm{C}}_{\mathrm{db}}+C_{\mathrm{Tb}}+{\lambda }_{\mathrm{W}}{M}_{\mathrm{Wb}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Wb}}/\mathit{d}{\theta }_{\mathrm{db}}\right)+{\lambda }_{\mathrm{N}}{M}_{\mathrm{Nb}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Nb}}/\mathit{d}{\theta }_{\mathrm{db}}\right)}.$(28)

In this section, suffixes u, m, and b denote the upper, middle, and bottom debris, respectively.

Middle Debris

Similar to Eq. (28)(28) $\frac{\mathit{d}{\mathit{\theta }}_{\mathrm{db}}}{\mathit{dt}}=\frac{{\mathit{R}}_{\mathrm{hb}}-R_{\mathrm{Sb}}+Q_{\mathrm{b}}-R_{\mathrm{lossb}}}{{\mathrm{C}}_{\mathrm{db}}+C_{\mathrm{Tb}}+{\lambda }_{\mathrm{W}}{M}_{\mathrm{Wb}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Wb}}/\mathit{d}{\theta }_{\mathrm{db}}\right)+{\lambda }_{\mathrm{N}}{M}_{\mathrm{Nb}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Nb}}/\mathit{d}{\theta }_{\mathrm{db}}\right)}.$(28) , the temperature change of the middle debris θ_dm is derived as follows:

(29) $\frac{\mathit{d}{\mathit{\theta }}_{\mathrm{dm}}}{\mathit{dt}}=\frac{{\mathit{R}}_{\mathrm{hm}}-R_{\mathrm{Sm}}+Q_{\mathrm{m}}}{{\mathrm{C}}_{\mathrm{dm}}+{\lambda }_{\mathrm{W}}{M}_{\mathrm{Wm}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Wm}}/\mathit{d}{\theta }_{\mathrm{dm}}\right)+{\lambda }_{\mathrm{N}}{M}_{\mathrm{Nm}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Nm}}/\mathit{d}{\theta }_{\mathrm{dm}}\right)}$(29)

$Q_{\mathrm{m}}\left(\mathit{t}\right)=-S_{\mathrm{l}}{\mathrm{K}}_{\mathrm{d}}\left({\theta }_{\mathrm{dm}}-{\theta }_{\mathrm{db}}\right)-S_{\mathrm{l}}{\mathrm{K}}_{\mathrm{d}}\left({\theta }_{\mathrm{dm}}-{\theta }_{\mathrm{db}}\right)-\left(S_4/3\right){\mathrm{K}}_{\mathrm{d}}\left({\theta }_{\mathrm{dm}}-{\theta }_{\mathrm{db}}\right),$

where Q_m = heat input to the middle debris from its surroundings (kW).

Upper Debris

Similarly, the temperature change of the upper debris θ_du is derived as follows:

(30) $\frac{\mathit{d}{\mathit{\theta }}_{\mathrm{du}}}{\mathit{dt}}=\frac{{\mathit{R}}_{\mathrm{hu}}-R_{\mathrm{Su}}-Q_{\mathrm{u}}-R_{\mathrm{rad}}}{C_{\mathrm{du}}+{\lambda }_{\mathrm{W}}{M}_{\mathrm{Wu}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Wu}}/\mathit{d}{\theta }_{\mathrm{du}}\right)+{\lambda }_{\mathrm{N}}{M}_{\mathrm{Nu}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Nu}}/\mathit{d}{\theta }_{\mathrm{du}}\right)}$(30)

$Q_{\mathrm{u}}\left(\mathit{t}\right)=-S_1{\mathrm{K}}_{\mathrm{d}}\left({\theta }_{\mathrm{dm}}-{\theta }_{\mathrm{du}}\right)-S_1{\alpha }_{\mathrm{dg}}\left({\theta }_{\mathrm{du}}-{\theta }_{\mathrm{g}}\right)-\left(S_4/3\right){\mathrm{K}}_{\mathrm{d}}\left({\theta }_{\mathrm{du}}-{\theta }_{\mathrm{db}}\right),$

where Q_u = heat input to the upper debris from its surroundings (kW); R_rad = net radiation heat transfer rate from the upper debris surface to the tank inner surface and is expressed as follows:

$R_{\mathrm{rad}}=\left(F_{12}+F_{13}\right)S_{\mathrm{l}}\mathit{\sigma }{{\mathit{T}}_{\mathit{du}}}^4-\left(F_{21}{S}_2+F_{31}{S}_3\right)\mathit{\sigma }{{\mathit{T}}_{\mathrm{Tu}}}^4.$

In the equation, the emissivity of the upper surface of the debris and the inner surface of the tank are both treated as 1 (see Sec. B.II). F₁₂, F₁₃, F₂₁, and F₃₁ are the configuration factors (suffixes 1, 2, and 3 correspond to the upper debris surface, the upper tank surface, and the side tank surface), and their values are 0.365, 0.635, 0.365, and 0.302, respectively.

IX.A.2.b. Equation for the Gas in the Tank

The gaseous components are H₂O vapor, HNO₃ vapor, NO₂, NO, and O₂. The gas temperature θ_g is described by the following heat balance equation:

$Q_{\mathrm{g}}{c}_{\mathrm{g}}{\theta }_{\mathrm{du}}+S_1{\alpha }_{\mathrm{dg}}\left({\theta }_{\mathrm{du}}-{\theta }_{\mathrm{g}}\right)=\left(S_2+S_3\right){\alpha }_{\mathrm{g}2}\left({\theta }_{\mathrm{g}}-{\theta }_{\mathrm{Tu}}\right)+Q_{\mathrm{g}}{c}_{\mathrm{g}}{\theta }_{\mathrm{g}},$

where Q_g = sum of the gas generation rates from three pieces of the debris (mol∙s⁻¹), each rate being given by equations similar to Eqs. (6)(6) $Y_W\left(\mathit{t}\right)=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{dt}}=M_W\left(0\right)\frac{d{\xi }_W\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}}.$(6) , (8)(8) $Y_N\left(\mathit{t}\right)=M_N\left(0\right)\frac{d{\xi }_N\left(\mathit{\theta }\right)}{\mathit{dt}}=M_N\left(0\right)\frac{d{\xi }_N\left(\mathit{\theta }\right)}{\mathit{d\theta }}\frac{\mathit{d\theta }}{\mathit{dt}},$(8) , (11)(11) $\begin{array}{rl}& R_{\mathrm{NO}2}\left(\mathit{t}\right)=-\sum _{\mathit{i}=1}^5\frac{d{M}_i\left(\mathit{t}\right)}{\mathit{dt}}\\ & =-\sum _{\mathit{i}=1}^5{A}_i\exp \left[-\frac{E_i}{\mathit{R}\left[\mathit{\theta }\left(\mathit{t}\right)+273.15\right]}\right]M_i\left(\mathit{t}\right)& (11)\end{array}$(11) , and (12)(12) $R_{\mathrm{NO}}\left(\mathit{t}\right)=-\sum _{\mathit{i}=6}^7\frac{d{M}_i\left(\mathit{t}\right)}{\mathit{dt}}=-\sum _{\mathit{i}=6}^7{A}_i\exp \left[-\frac{E_i}{\mathit{R}\left[\mathit{\theta }\left(\mathit{t}\right)+273.15\right]}\right]M_i\left(\mathit{t}\right)$(12) ; c_g = its molar heat capacity; α_dg = convective heat transfer coefficient between the upper debris surface and the tank gas; α_g2 = convective heat transfer coefficient between the tank top/side surface and the tank gas; c_a = specific heat of the air. The first term on the left side of the above equation is the carry-on enthalpy of the gas generated from the debris, and the temperature of the generated gas leaving the debris is assumed to be equal to the temperature of the upper debris. The second term is the heat transfer rate from the debris to the gas in the tank. The first term on the right side is the heat transfer rate from the gas to the top/side wall of the tank, and the second term is the take-out enthalpy of the gas flowing out of the tank.

From the above equation, the gas temperature is described as follows:

(31) ${\theta }_{\mathrm{g}}=\frac{{\mathit{Q}}_{\mathrm{g}}{c}_{\mathrm{g}}{\theta }_{\mathrm{du}}+S_1{\alpha }_{\mathrm{dg}}{\theta }_{\mathrm{du}}+\left(S_2+S_3\right){\alpha }_{\mathrm{g}2}{\theta }_{\mathrm{Tu}}}{Q_{\mathrm{g}}{c}_{\mathrm{g}}+S_1{\alpha }_{\mathrm{dg}}+\left(S_2+S_3\right){\alpha }_{\mathrm{g}2}}.$(31)

IX.A.2.c. Equation for the Tank Top/Side Wall

The heat source of the top/side wall is the convective heat transfer from the gas in the tank (the first term on the right side of the following equation), the heat radiation R_rad between the upper debris and the top/side wall, and the conductive heat transfer from the bottom wall to the side wall (the third term on the right side of the following equation). The heat sink R_losst is the heat release rate from the tank to the cell. Therefore, the following equation is obtained:

$C_{\mathrm{Tt}}\frac{\mathit{d}{\mathit{\theta }}_{\mathrm{Tt}}}{\mathit{dt}}=\left(S_2+S_3\right){\alpha }_{\mathrm{g}2}\left({\theta }_{\mathrm{g}}-{\theta }_{\mathrm{Tt}}\right)+R_{\mathrm{rad}}+S_{\mathrm{SS}}{\alpha }_{\mathrm{SS}}\left({\theta }_{\mathrm{db}}-{\theta }_{\mathrm{Tt}}\right)-R_{\mathrm{losst}}$

$R_{\mathrm{losst}}=\left(S_2+S_3\right){\alpha }_2\left({\theta }_{\mathrm{Tt}}-{\theta }_{\mathrm{ai}}\right)+\left(S_2+S_3\right){\epsilon }_{\mathrm{To}}\mathit{\sigma }\left[{{\mathit{T}}_{\mathrm{Tt}}}^4-T_{\mathrm{C}}{\left(\mathit{t},0\right)}^4\right],$

where C_Tt = heat capacity of the tank top/side wall.

IX.A.2.d. Equation for the Cell Air

The following equation is obtained from the conditions described in parameter 2 in Sec. VII.A:

${\alpha }_1\left(S_1+S_4\right)\left[{\theta }_{\mathrm{Tb}}\left(\mathit{t}\right)-{\theta }_{\mathrm{ai}}\left(\mathit{t}\right)\right]+{\alpha }_2\left(S_2+S_3\right)\left[{\theta }_{\mathrm{Tu}}\left(\mathit{t}\right)-{\theta }_{\mathrm{ai}}\left(\mathit{t}\right)\right]={\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}\left[{\theta }_{\mathrm{ai}}\left(\mathit{t}\right)-{\theta }_{\mathrm{C}}\left(\mathit{t},0\right)\right],$

where α₂ = convective heat transfer coefficient between the tank upper/side and the inner cell air; α_ci = inner cell wall and the inner cell air. From this, we obtain the following equation, which gives the cell air temperature:

(32) ${\theta }_{\mathrm{ai}}\left(\mathit{t}\right)=\frac{{\alpha }_1\left(S_1+S_4\right){\theta }_1\left(\mathit{t}\right)+{\alpha }_2\left(S_2+S_3\right){\theta }_2\left(\mathit{t}\right)+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}{\theta }_{\mathrm{C}}\left(\mathit{t},0\right)}{{\alpha }_1\left(S_1+S_4\right)+{\alpha }_2\left(S_2+S_3\right)+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}}.$(32)

IX.A.3. Examination of Each Parameter Value in the Heat Balance Equation

IX.A.3.a. Heat Generation Rate R_h of Each Piece of the Debris

The equation

$R_{\mathrm{hb}}=R_{\mathrm{hm}}=R_{\mathrm{hu}}=\left(578-10\right)/3=189\mathrm{kW},$

where the second suffixes b, m, and u of R_h correspond to the bottom, middle, and upper debris, respectively. In the equation, 10 kW is the gamma-ray release out of the tank [see Eq. (C.1)(B.1) ${\alpha }_{\mathrm{T}}=0.13{\left[\frac{\mathit{g\beta }\left(\mathit{\theta }-{\mathit{\theta }}_{\mathrm{a}}\right)}{{\nu }_{\mathrm{a}}{a}_{\mathrm{a}}}\right]}^{\frac{1}{3}}{\kappa }_{\mathrm{a}},$(B.1) ].

IX.A.3.b. Heat Transfer Coefficient Between Two Pieces of the Debris in Contact

The coefficient K_d (in kW∙m⁻²∙°C⁻¹) is given as follows:

$K_{\mathrm{d}}={\kappa }_{\mathrm{d}}/\left(0.32/3\right)=0.0038,$

where κ_d = effective thermal conductivity of debris given as 0.00037 at 150°C to 270°C and 0.00040 at 400°C.^[20] We adopted 0.00040 over the entire temperature range.

IX.A.3.c. Heat Transfer Coefficient Between the Tank Bottom to the Tank Top/Side

The coefficient K_ss is obtained as follows. If the thermal diffusivity of the material is a (in m²∙s⁻¹) and if the distance is d, the timescale τ (in seconds) required for heat conduction over the distance d (in meters) is given by $\mathit{\tau }\approx d^2/\mathit{a}$. The duration of the accident targeted in this study is 100 to 200 h, so the timescale of the temperature change would be about 10 h. Using the thermal diffusivity a = 4.1 × 10⁻⁶ m²∙s⁻¹ of stainless steel (SUS), it is considered appropriate to set the distance scale for heat conduction in SUS to $\sqrt{\mathit{a\tau }}$ = 0.4 m. Therefore, using the thermal conductivity of SUS, which is 0.0165 kW∙m⁻¹∙K⁻¹,^[18] the heat transfer coefficient is calculated as K_ss = 0.0165/0.4 = 0.040 kW∙m⁻²∙°C⁻¹. The sum of the cross-sectional area of the tank side wall and that of the internal structure is 0.66 + 0.98 = 1.64 m². When the temperature difference between θ_db and θ_Tt is 200 K, the heat transfer rate from the bottom surface to the side surface is 1.64 × 0.040 × 200 = 13 kW, which is a small value in the overall heat balance of 568 kW (heat generation inside the debris). Therefore, the value of K_ss considered in this paper has little effect.

IX.A.3.d. Convective Heat Transfer Coefficient and Emissivity

As is explained in Sec. B.II, Eq. (B.1)(B.1) ${\alpha }_{\mathrm{T}}=0.13{\left[\frac{\mathit{g\beta }\left(\mathit{\theta }-{\mathit{\theta }}_{\mathrm{a}}\right)}{{\nu }_{\mathrm{a}}{a}_{\mathrm{a}}}\right]}^{\frac{1}{3}}{\kappa }_{\mathrm{a}},$(B.1) can be used for all coefficients, α_ci, α_co, α₁, α₂, α_dg, and α_g2.

Emissivities, ε_Ti, ε_To, ε_d, and ε_c, are also treated in Sec. B.II.

IX.A.3.e. Heat Capacity

The heat capacity of debris C_d is given by Eq. (13)(13) $c_S\left(\mathit{\theta }\right)\times 153\times 120=6800[1+0.0029(\mathit{\theta }-25)].$(13) . The heat capacity of each piece of the debris is C_d/3.

The specific heat c_T of SUS-316, which makes up the tank and the internal structural material, is given in the literature.^[18] Therefore, the heat capacities of the tank bottom C_Tb and its top/side C_Tt (J∙K⁻¹) are expressed as follows:

$C_{\mathrm{Tb}}=c_{\mathrm{T}}\times 13,300$

$C_{\mathrm{Tt}}=c_{\mathrm{T}}\times \left(9100+47,500\right)=c_{\mathrm{T}}\times 56,600.$

The temperature-dependent molar heat capacity of each gas (NO₂, NO, O₂, and H₂O) composing the gas in the tank was obtained from Ref. 15. The molar heat capacity of HNO₃ vapor was assumed to be the same as that of H₂O. The heat capacities of the nitrates (= oxides) and tank (including the internal structural material) are given by Eqs. (13)(13) $c_S\left(\mathit{\theta }\right)\times 153\times 120=6800[1+0.0029(\mathit{\theta }-25)].$(13) and (14(14) $C_T\left(\mathit{\theta }\right)=c_T\left(\mathit{\theta }\right)\times 70,000=15.61\mathit{\theta }+34,200,$(14) ), respectively.

IX.B. Solution Method

IX.B.1. Temperature Inside the Concrete Cell Wall

To obtain a numerical solution from Eq. (19)(19) $\frac{\mathrm{\partial }{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial t}}={\alpha }_C\frac{{\mathrm{\partial }}^2{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial }{x}^2}+\frac{q_{\gamma }\left(\mathit{t},\mathit{x}\right)}{{\rho }_C{C}_C},$(19) in the late stage, it is necessary to know θ_C(t,x) at t = t_d when the waste temperature reaches 130°C. According to the results of the initial stage, t_d = 4.64 × 10⁵ s (129 h). Therefore, the various parameters in Eq. (19)(19) $\frac{\mathrm{\partial }{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial t}}={\alpha }_C\frac{{\mathrm{\partial }}^2{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial }{x}^2}+\frac{q_{\gamma }\left(\mathit{t},\mathit{x}\right)}{{\rho }_C{C}_C},$(19) at t = t_d are obtained from the results at the initial stage (see Sec. X.A).

The boundary condition at x = 0 for t > t_d can be obtained using the fact that the heat flux into the wall due to convection and radiation on the inner wall surface is equal to the conductive heat flux inside the wall:

$\mathit{x}=0,$

(33) $\begin{array}{rl}{\mathit{\alpha }}_{\mathrm{Ci}}\left[{\theta }_{\mathrm{ai}}\left(\mathit{t}\right)-{\theta }_C\left(\mathit{t},0\right)\right]& +\frac{\left(S_1+S_4\right)}{S_{\mathrm{C}}}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_1{\left(t\right)}^4\\ & +\frac{\left(S_2+S_3\right)}{S_C}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_2{\left(t\right)}^4\\ & -\frac{\left(S_1+S_2+S_3+S_4\right)}{S_{\mathrm{C}}}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_{\mathrm{C}}{\left(t\right)}^4\\ & =-{\kappa }_{\mathrm{c}}{\left.\frac{\mathrm{\partial }{\mathit{T}}_{\mathrm{C}}}{\mathrm{\partial }\mathit{x}}\right|}_{\mathit{x}=0}.& (33)\end{array}$(33)

The temperature of the cell air ${\theta }_{\mathit{Ai}}$ is given by Eq. (32)(32) ${\theta }_{\mathrm{ai}}\left(\mathit{t}\right)=\frac{{\alpha }_1\left(S_1+S_4\right){\theta }_1\left(\mathit{t}\right)+{\alpha }_2\left(S_2+S_3\right){\theta }_2\left(\mathit{t}\right)+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}{\theta }_{\mathrm{C}}\left(\mathit{t},0\right)}{{\alpha }_1\left(S_1+S_4\right)+{\alpha }_2\left(S_2+S_3\right)+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}}.$(32) . When obtaining the numerical solution of Eq. (33(33) $\begin{array}{rl}{\mathit{\alpha }}_{\mathrm{Ci}}\left[{\theta }_{\mathrm{ai}}\left(\mathit{t}\right)-{\theta }_C\left(\mathit{t},0\right)\right]& +\frac{\left(S_1+S_4\right)}{S_{\mathrm{C}}}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_1{\left(t\right)}^4\\ & +\frac{\left(S_2+S_3\right)}{S_C}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_2{\left(t\right)}^4\\ & -\frac{\left(S_1+S_2+S_3+S_4\right)}{S_{\mathrm{C}}}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_{\mathrm{C}}{\left(t\right)}^4\\ & =-{\kappa }_{\mathrm{c}}{\left.\frac{\mathrm{\partial }{\mathit{T}}_{\mathrm{C}}}{\mathrm{\partial }\mathit{x}}\right|}_{\mathit{x}=0}.& (33)\end{array}$(33) ), it was transformed into the following equation:

${\mathrm{\Lambda }}_2\left[\mathrm{\Theta }-{\theta }_C\left(\mathit{t},0\right)\right]=-{\kappa }_{\mathrm{c}}{\left.\frac{\mathrm{\partial }{\mathit{\theta }}_{\mathrm{C}}}{\mathrm{\partial }\mathit{x}}\right|}_{\mathit{x}=0}.$

In the above equation, Λ₂ and Θ are given as follows:

${\mathrm{\Lambda }}_2={\alpha }_{\mathrm{ci}}\frac{{\alpha }_1\left(S_1+S_4\right)+{\alpha }_2\left(S_2+S_3\right)}{{\alpha }_1\left(S_1+S_4\right)+{\alpha }_2\left(S_2+S_3\right)+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}}+\frac{\left(S_1+S_2+S_3+S_4\right)}{S_C}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_{\mathrm{C}}{\left(\mathit{t},0\right)}^3$

$\mathrm{\Theta }=\frac{{\mathit{\alpha }}_{\mathrm{ci}}}{{\mathrm{\Lambda }}_2}\frac{{\alpha }_1\left(S_1+S_4\right)T_1+{\alpha }_2\left(S_2+S_3\right)T_2}{{\alpha }_1\left(S_1+S_4\right)+{\alpha }_2\left(S_2+S_3\right)+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}}+\frac{{\mathit{\epsilon }}_{\mathrm{so}}\mathit{\sigma }}{\mathrm{\Lambda }{S}_{\mathrm{C}}}\left[\left(S_1+S_4\right){T_1}^4+\left(S_2+S_3\right){T_2}^4\right].$

The boundary condition at x = L for t > t_d is the same as that in the initial stage given by Eq. (21)(21) $\mathit{x}=\mathit{L},\mathit{t}>0:-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=\mathit{L}}={\alpha }_{\mathit{co}}\left[{\theta }_C\left(\mathit{t},\mathit{L}\right)-{\theta }_{\mathit{ao}}\right]+F_{\mathit{OW}}\mathit{\sigma }\left[T_C{\left(\mathit{t},\mathit{L}\right)}^4-{{\mathit{T}}_{\mathit{OW}}}^4\right],$(21) .

IX.B.2. Calculation of the Heat Balance Equation

The temperatures that need to be obtained in the late stage are the temperatures of the bottom, middle, and upper debris, the temperatures of the tank top/side wall and the bottom wall (the bottom temperature is assumed to be equal to the bottom debris), the gas temperature in the tank, the air temperature in the cell, and the temperature distribution inside the cell wall. Time changes in the debris temperatures and the tank temperatures are obtained by solving numerically the heat balance equations, Eqs. (28)(28) $\frac{\mathit{d}{\mathit{\theta }}_{\mathrm{db}}}{\mathit{dt}}=\frac{{\mathit{R}}_{\mathrm{hb}}-R_{\mathrm{Sb}}+Q_{\mathrm{b}}-R_{\mathrm{lossb}}}{{\mathrm{C}}_{\mathrm{db}}+C_{\mathrm{Tb}}+{\lambda }_{\mathrm{W}}{M}_{\mathrm{Wb}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Wb}}/\mathit{d}{\theta }_{\mathrm{db}}\right)+{\lambda }_{\mathrm{N}}{M}_{\mathrm{Nb}}\left(0\right)\left(\mathit{d}{\xi }_{\mathrm{Nb}}/\mathit{d}{\theta }_{\mathrm{db}}\right)}.$(28) through (31(31) ${\theta }_{\mathrm{g}}=\frac{{\mathit{Q}}_{\mathrm{g}}{c}_{\mathrm{g}}{\theta }_{\mathrm{du}}+S_1{\alpha }_{\mathrm{dg}}{\theta }_{\mathrm{du}}+\left(S_2+S_3\right){\alpha }_{\mathrm{g}2}{\theta }_{\mathrm{Tu}}}{Q_{\mathrm{g}}{c}_{\mathrm{g}}+S_1{\alpha }_{\mathrm{dg}}+\left(S_2+S_3\right){\alpha }_{\mathrm{g}2}}.$(31) ), with the values at time t_d obtained in the initial stage as the initial conditions. The gas temperature and the air temperature can be calculated using the above results and Eqs. (31)(31) ${\theta }_{\mathrm{g}}=\frac{{\mathit{Q}}_{\mathrm{g}}{c}_{\mathrm{g}}{\theta }_{\mathrm{du}}+S_1{\alpha }_{\mathrm{dg}}{\theta }_{\mathrm{du}}+\left(S_2+S_3\right){\alpha }_{\mathrm{g}2}{\theta }_{\mathrm{Tu}}}{Q_{\mathrm{g}}{c}_{\mathrm{g}}+S_1{\alpha }_{\mathrm{dg}}+\left(S_2+S_3\right){\alpha }_{\mathrm{g}2}}.$(31) and (32(32) ${\theta }_{\mathrm{ai}}\left(\mathit{t}\right)=\frac{{\alpha }_1\left(S_1+S_4\right){\theta }_1\left(\mathit{t}\right)+{\alpha }_2\left(S_2+S_3\right){\theta }_2\left(\mathit{t}\right)+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}{\theta }_{\mathrm{C}}\left(\mathit{t},0\right)}{{\alpha }_1\left(S_1+S_4\right)+{\alpha }_2\left(S_2+S_3\right)+{\alpha }_{\mathrm{ci}}{S}_{\mathrm{C}}}.$(32) ). The surface temperature of the cell wall is required for radiant heat calculation. This is obtained by numerically solving Eq. (19)(19) $\frac{\mathrm{\partial }{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial t}}={\alpha }_C\frac{{\mathrm{\partial }}^2{\theta }_C\left(\mathit{t},\mathit{x}\right)}{\mathrm{\partial }{x}^2}+\frac{q_{\gamma }\left(\mathit{t},\mathit{x}\right)}{{\rho }_C{C}_C},$(19) under the boundary conditions Eqs. (33)(33) $\begin{array}{rl}{\mathit{\alpha }}_{\mathrm{Ci}}\left[{\theta }_{\mathrm{ai}}\left(\mathit{t}\right)-{\theta }_C\left(\mathit{t},0\right)\right]& +\frac{\left(S_1+S_4\right)}{S_{\mathrm{C}}}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_1{\left(t\right)}^4\\ & +\frac{\left(S_2+S_3\right)}{S_C}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_2{\left(t\right)}^4\\ & -\frac{\left(S_1+S_2+S_3+S_4\right)}{S_{\mathrm{C}}}{\epsilon }_{\mathrm{so}}\mathit{\sigma }{T}_{\mathrm{C}}{\left(t\right)}^4\\ & =-{\kappa }_{\mathrm{c}}{\left.\frac{\mathrm{\partial }{\mathit{T}}_{\mathrm{C}}}{\mathrm{\partial }\mathit{x}}\right|}_{\mathit{x}=0}.& (33)\end{array}$(33) and (21(21) $\mathit{x}=\mathit{L},\mathit{t}>0:-{\kappa }_C{\left.\frac{\mathrm{\partial }{\theta }_C}{\mathrm{\partial x}}\right|}_{\mathit{x}=\mathit{L}}={\alpha }_{\mathit{co}}\left[{\theta }_C\left(\mathit{t},\mathit{L}\right)-{\theta }_{\mathit{ao}}\right]+F_{\mathit{OW}}\mathit{\sigma }\left[T_C{\left(\mathit{t},\mathit{L}\right)}^4-{{\mathit{T}}_{\mathit{OW}}}^4\right],$(21) ) at each time step and calculating the temperature distribution of the cell wall. The numerical calculation formulas are the same as those of the initial stage model except for the boundary condition at x = 0, and the time step value is also the same (see Sec. VII.C).

X. CALCULATED RESULTS AND DISCUSSION OVER BOTH STAGES

X.A. Transition of the Accident with Time

Figure 10 shows the temperatures of the waste, tank walls (tank top/side and tank bottom, both including the internal structure), tank gas, cell air, and inner and outer cell surfaces as a function of time, obtained by solving the heat balance equations. The temperature of the tank bottom is the same as that of the bottom debris. Elapsed time 0 h corresponds to the time when the cooling function loss occurred. Boiling begins after 14 h. Because Ru volatilization is reported^[1] to begin when the waste temperature reaches 119°C, it starts after 111 h. The waste reaches 130°C after 129 h (= t_d), when the model has changed from the initial stage to the late stage. The temperatures of the upper and bottom debris rise at almost the same rate, but that of the middle debris increases quickly. When those of the upper and bottom debris reach 600°C, the middle debris exceeds 1200°C. This quick temperature increase is explained by the low effective thermal conductivity of the debris (see Sec. IX.A.3.b) and the sandwich effect between the upper and bottom debris. The heat of the upper debris is transferred to the upper part of the tank, and the heat of the lower debris is transferred to the tank bottom. At time t_d, it can be seen that the temperature of the tank side temporarily drops. This is explained by the abrupt model change, i.e., from the homogeneous liquid state to the heterogeneous dry state.

Fig. 10. Temperatures of the waste, tank walls (tank top/side and the tank bottom), tank gas, cell air, and cell inside and outside wall surfaces versus time curves (temperature of the tank bottom is the same as that of the bottom debris). At time t_d = 129 h, HLLW temperature reached 130°C, and the stage was changed from the initial stage to the late stage.

When the bottom debris reaches 180°C after 142 h, the inner cell surface reaches 81°C, and when it reaches 600°C after 456 h, the surface reaches 482°C. The outer cell surface is assumed to be in contact with air at 25°C, and the outer surface temperature increases slowly and reaches 62°C when the bottom debris reaches 600°C.

Figure 11 shows the NO₂ and NO generation rates as a function of time, which was calculated by the Arrhenius-type model explained in Sec. V.B. The bottom debris temperature is shown for reference. A small amount of NO₂ is generated when boiling begins, but most of it is generated at temperatures of the debris between 150°C and 350°C. The NO generation curve shows several peaks. This is mainly because the temperature of the middle debris is much higher than those of the upper and bottom debris.

Fig. 11. NO₂ and NO generation rates and the bottom debris temperature (dotted line for reference) versus time curves.

Figure 12 shows the heat absorption rates due to the liquid vaporization and the nitrate decomposition, the heat transfer rate from the tank to the cell, and the sensible heat accumulation rate inside the tank, all as a function of time. The sum of these heat absorption terms should be equal to the value obtained by subtracting the energy release rate of leaked gamma rays from 578 kW due to the decay of radioactive materials. Abrupt discontinuous changes of vaporization and heat accumulation curves at around 129 h are due to the model change from the initial stage to the late stage. The latent heat of vaporization is the main heat absorption source until 142 h (180°C bottom debris temperature) when liquid evaporation from the waste is almost gone, and then, the heat absorption source of nitrate decomposition becomes main until 167 h (280°C bottom debris temperature). After that, the heat transfer to the cell wall becomes the main heat absorption source. The sensible heat accumulation rate curve shows two peaks at 130 to 150 h and 155 to 190 h, and a trough around 150 h. This phenomenon is explained by the fact that much of the decay heat is spent on nitrate decomposition at around 150 h, as can be seen in the figure.

Fig. 12. Vaporization heat absorption rate, nitrate decomposition heat absorption rate, heat transfer rate from the tank, sensible heat accumulation rate in the tank versus time curves (up to 129 h, the waste remains in the liquid state).

Figure 13 shows the temperature increasing rate of each debris as a function of time. For reference, the temperature of the bottom debris is indicated by a dotted line. Before boiling, the rate of the liquid waste is 0.06°C∙min⁻¹. After boiling, it decreases to 0.0012°C∙min⁻¹ and then increases gradually as the concentration progresses. When the time exceeds 129 h (130°C), the increasing rates of the upper and bottom debris come within the range of 0.02 to 0.2°C∙min⁻¹ and that of the middle debris within 0.02 to 1.0°C∙min⁻¹. Each rate curve shows two peaks. The rise of the first peak is explained by the substantial end of the vaporization of H₂O and HNO₃. Then, since the nitrate decomposition generating NO₂ begins, the peak decreases forming the first peak. The second peak is formed by the completion of NO₂ generation and the increased heat loss mainly due to heat radiation. Humps seen around 200°C are caused by NO generation. After that, the debris temperatures increase under the balance of the internal heat generation and the heat loss through heat radiation, conduction, convection, and sensible heat accumulation.

Fig. 13. Temperature increasing rates of the individual debris and the temperature of the bottom debris (dotted line for reference) versus time curves (up to 129 h, the waste remains in the liquid state).

Figure 14 shows the changes in the released gas composition and the total gas flow rate corresponding to the bottom debris temperature. At the initial stage between 104°C and 130°C, the major component of the gas is H₂O vapor. Though it decreases with the increasing temperature, its percentage is still 70% at 180°C, when the waste seems almost solid with a slight nitric acid smell.^[9] The figure shows that even at 300°C, the HNO₃ vapor is contained in the gas by 17% although the H₂O vapor is zero. It should be noted that in this temperature region, these errors in vapor amount can be large. However, the errors do not affect the waste temperature rise because of their small heat absorption rates (see Fig. 12). Gas flow above 400°C consists of NO and O₂ produced by the decomposition of nitrates.

Fig. 14. Changes in the released gas composition and the total gas flow rate corresponding to the bottom debris temperature.

In order to examine the validity of the present model, we attempted to apply it to the experimental results of other papers. However, no comparable results could be found.

X.B. Application of the Results to a Postulated Accident of a Plant

The above results were obtained using the initial waste volume of 120 m³, the decay power density of 4.82 kW∙m⁻³, and the composition of the initial waste given in Table I. These values will change from time to time in a reprocessing plant. However, information necessary to solve Eq. (1)(1) $\begin{array}{rl}& \mathrm{\Gamma }\left(\mathit{t}\right)\frac{\mathit{d\theta }}{\mathit{dt}}=R_h-\left\{{\lambda }_w\left(\mathit{\theta }\right)Y_w\left(\mathit{t}\right)+{\lambda }_N\left(\mathit{\theta }\right)Y_N\left(\mathit{t}\right)+R_S\left(\mathit{t}\right)\right\}\\ & -R_{\mathit{loss}}\left(\mathit{\theta }\right)-R_{\mathit{\gamma }}\left(\mathit{t}\right),& (1)\end{array}$(1) is the decay power density; the initial waste volume; and the initial amounts of H₂O, HNO₃, and nitrates. Such information will be obtained from the daily HLLW management. Accordingly, the waste temperature and the release rates of the components transferred to the gas phase are predictable as a function of time after the cooling loss using the HLLW conditions in a plant.

The ultimate goal of our study is to predict the release behavior of radioactive aerosol and volatile Ru into the outside atmosphere under accidental conditions. The results obtained above can be used to study the LPF^[11] of radioactive materials, especially volatilized Ru, because the Ru and NO₂ are absorbed into the condensate of vaporized H₂O and HNO₃ formed in the release path. Since aerosol release will cease by 130°C^[8] and volatile Ru by 300°C,^[6] the experimental results up to 300°C shown in this section are important.

In the present study, since Excel calculation was used to obtain the solution, some assumptions and approximations were used for simplicity. For example, although the waste changes gradually from liquid phase to solid phase, it was assumed to change suddenly at 130°C. Although the solid phase was continuous, it was divided into three pieces, and the temperature of each piece was made uniform. The solid-phase thermal conductivity will vary with temperature due to the presence of some residual liquid, nitrate decomposition, and volume change, but it was approximated to be constant. In the future, the use of more a sophisticated modeling and numerical solution will bring these assumptions and approximations closer to reality, enabling more accurate predictions.

XI. CONCLUSIONS

We proposed a method for systematically predicting temperatures of the waste, tank, and cell concrete up to 600°C and individual gas release rates as a function of time under accidental conditions. This method would be applicable to HLLW with different decay power densities, volumes, and compositions.

The results obtained in this study are useful not only for examining the LPF but also for developing measures to reduce it. Further, they will become the basic information to examine the accident management measures and to provide the disaster prevention system since the release of radioactive materials and their carrier gases can be quantitatively predicted as a function of time.

Acknowledgments

The analysis using the MCNP code was performed by Yuri Tajimi, Daisuke Takano, and Tami Mukohara of TEPCO Systems Corporation. The authors are very grateful for their contributions.

Disclosure Statement

No potential conflict of interest was reported by the author(s).

Correction Statement

This article has been corrected with minor changes. These changes do not impact the academic content of the article.

Notes

^a The liquid waste changes from a solution state to a dry state with increasing temperature. Hereinafter, “waste” covers both states.

^b By 600°C, the generation of vapors and gases that transports radioactive materials outside the building will cease. Further, since it takes more than 10 days to reach 600°C as shown in Sec. X, it is possible to take measures during this time such as introducing water into the cell and submerging the storage tank.

^c Sections III, VII, and VIII are applicable to the initial stage, and Sec. IX is applicable to the late stage. Sections IV, V, VI, X, XI, and the Appendixes are applicable to both stages.

^d Since H₂O and HNO₃ are vaporized from nitric acid solution, the interaction between H₂O and HNO₃ needs to be taken into account. The influence of this effect is explained in Sec. IV.C.3.

^e The tank is made of SUS with a thickness of 0.03 m. The thermal conductivity of SUS is 17 W∙m⁻¹∙K⁻¹, and the heat transfer coefficient through the tank wall is 17/0.03 = 570 W∙m⁻²∙K⁻¹, which is about 100 times larger than the natural convection heat transfer coefficient (2 to 5 W∙m⁻²∙K⁻¹), and therefore, the heat transfer resistance from the tank to the cell air is controlled by natural convection.

^f The meaning of this equal sign is explained in Sec. III.A, No. 5.

^g 2440 is obtained from the values in Table IV.

^h The shape of the top and bottom is slightly mortar-like rather than flat (see Fig. 1).

ⁱ A debris thickness becomes 0.32 m.

APPENDIX A

HEAT OF HNO₃ GENERATION FROM NITRATES IN THE WASTE

It is reported that Fe nitrate decomposes and releases HNO₃ by the following reaction during the temperature increase process^[21]:

$\mathrm{Fe}{\left(\mathrm{N}{\mathrm{O}}_3\right)}_3+\mathit{\hspace{0.17em}}\left(3/2\right){\mathrm{H}}_2\mathrm{O}\to \left(1/2\right)\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3+3\mathrm{HN}{\mathrm{O}}_3.$

The standard formation enthalpy of Fe(NO₃)₃ was estimated from that of the chloride compound using Eq. (10(10) ${\mathrm{\Delta }}_f{H}^{\circ }\left(\mathrm{nitrate}\right)=1.17{\mathrm{\Delta }}_f{H}^{\circ }\left(\mathrm{chloride}\right).$(10) ), and those of the other compounds were cited from the literature.^[16] The heat of the reaction is thus obtained as follows:

$\begin{array}{rl}& \mathrm{\Delta }\mathit{H}=\left\{-824/2+3\times \left(-174\right)\right\}\\ & -\left\{\left(-399\times 1.17\right)+\left(-286\times 1.5\right)\right\}\\ & =-38\mathrm{kJ}\cdot \mathrm{mo}{\mathrm{l}}^{-1}.\end{array}$

Therefore, the reaction is exothermic, and the calorific value of HNO₃ generation is 13 kJ∙mol⁻¹.

A previous report^[10] indicates that 30% of the N in nitrates shown in Table I is released as HNO₃ instead of NOx. If the calorific value per mol of HNO₃ generation from the nitrates is assumed to be equal to that from Fe nitrate, the calorific value of HNO₃ generation from the nitrates in 1 m³ of waste becomes 9500 kJ∙m⁻³ (= 2440 × 0.30 × 13).^g Assuming that this exothermic reaction occurred during 100 h beginning from boiling, the exothermic density was as small as 0.026 kW∙m⁻³, so the heat of HNO₃ generation was not considered in the calculation of the heat of NOx generation.

APPENDIX B

PHYSICAL PROPERTY VALUES USED FOR CONVECTIVE AND RADIATIVE HEAT TRANSFER CALCULATIONS

B.I. VALUES RELATED TO THE CELL WALL

The wall material was assumed to be solely concrete, and its physical property values used here are the following:

1. Thermal conductivity^[22] κ_c at 25°C = 0.00175 kW∙m⁻¹∙°C⁻¹.

2. Specific heat^[18] c_c at 25°C = 0.90 kJ∙kg⁻¹∙°C⁻¹.

3. Density^[18] ρ_c at 25°C = 2400 kg∙m⁻³.

The thermal diffusivity is a_c = κ_cc_c⁻¹ρ_c⁻¹ = 8.1 × 10⁻⁷ m²∙s⁻¹. The cell wall is made of reinforced concrete, but it is difficult to estimate the overall thermal properties. Concrete at high temperatures would deteriorate resulting from decomposition of the components or thermal deformation. Analysis of these phenomena is beyond the scope of this paper, and we used the above values as constant for calculation of the wall temperature.

B.II. CONVECTIVE HEAT TRANSFER COEFFICIENT

Convective heat transfer occurs between the tank outer surface and the cell air. Since the length of the tank is as long as several meters, the Rayleigh number becomes 10⁹ or more even if the temperature difference is 1°C, and so, the convection becomes turbulent.^[19] Although the heat transfer coefficient differs slightly depending on the direction of the surface (top, side, and bottom^h ), the difference is small.^[19] The following equation of the turbulent natural convection heat transfer coefficient along the vertical plate and the surrounding air^[19] was used for all points, α₁ and α₂. Assuming that the transport properties of the gas in the tank are the same as those of air, the equation is applicable to the tank gas, α_dg, and α_g2:

(B.1) ${\alpha }_{\mathrm{T}}=0.13{\left[\frac{\mathit{g\beta }\left(\mathit{\theta }-{\mathit{\theta }}_{\mathrm{a}}\right)}{{\nu }_{\mathrm{a}}{a}_{\mathrm{a}}}\right]}^{\frac{1}{3}}{\kappa }_{\mathrm{a}},$(B.1)

where β = coefficient of thermal expansion of air expressed in (θ + 273.15)⁻¹ (°C⁻¹); ν_a = kinematic viscosity (m²∙s⁻¹); a_a = thermal diffusivity (m²∙s⁻¹); κ_a = thermal conductivity of air (kW∙m⁻¹∙°C⁻¹); g = gravitational acceleration (m∙s⁻²). The value of κ_a was calculated from the following approximate equation, which was derived using the literature values^[15]:

${\kappa }_{\mathrm{a}}=3.26\times {10}^{-7}{\left({\mathit{\theta }}_{\mathrm{ai}}+273.15\right)}^{0.773}.$

The values of ν_a, a_a, and κ_a used are those at the average temperature of the tank and the cell air. Since the tank temperature changes from 50°C to 600°C, the parameters are also affected by temperature. Using the temperature dependence of ν_a and a_a obtained from the literature,^[18] the following equation (in s²∙°C⁻¹∙m⁻⁴) was obtained:

$\mathit{\beta }/\left({\nu }_{\mathrm{a}}{\alpha }_{\mathrm{a}}\right)=4.66\times {10}^{17}/{\left({\mathit{\theta }}_{\mathrm{Ta}}+273.15\right)}^{4.33},$

where θ_Ta = average temperature of the tank and the cell air.

Similarly, the heat transfer between the cell air and the cell wall surface is dominated by turbulent natural convection, and the heat transfer coefficient can be handled in the same manner as described above. Therefore, Eq. (B.1)(B.1) ${\alpha }_{\mathrm{T}}=0.13{\left[\frac{\mathit{g\beta }\left(\mathit{\theta }-{\mathit{\theta }}_{\mathrm{a}}\right)}{{\nu }_{\mathrm{a}}{a}_{\mathrm{a}}}\right]}^{\frac{1}{3}}{\kappa }_{\mathrm{a}},$(B.1) is applicable to α_ci and α_co.

B.III. EMISSIVITIES USED FOR THE CALCULATION OF HEAT RADIATION

The following emissivities were assumed from the literature values:

1. 0.92 for the tank inner surface ε_Ti (assumed from SUS oxidized black surface values).^[19]

2. 0.4 for the tank outer surface ε_To (SUS brown surface value due to heating).^[19]

3. 0.98 for the upper debris surface ε_d (assumed to be equal to that of black leather).^[18]

4. 0.94 for the cell concrete surface ε_c.^[19]

Emissivities above 0.9 were approximated as 1 in the calculations. Therefore, except for the reflection from the outer surface of the tank, the reflection from the other three surfaces was not considered.

APPENDIX C

CALCULATION OF GAMMA-RAY LEAKAGE RATE OUT OF THE TANK AND CORRESPONDING GAMMA HEATING OF THE CELL

The content in this section can be used for both stages.

Of the radiation generated by the decay of radionuclides, all alpha rays and beta rays are absorbed in the tank, but some of the gamma rays leak out of the tank. The gamma-ray leakage rate to the total decay power and the corresponding gamma heating of the concrete cell were examined.

C.I. CALCULATION CONDITIONS

The tank and the cell sizes are given in Sec. VII.A. The amount and the composition of the initial waste are given in Sec. III. According to the progress of the accident, the waste volume changes from 120 to 12 m³. 12 m³ corresponds to the debris (dried waste) volumeⁱ assuming its density is 1.5 g∙cm⁻³, which was obtained from the experiment shown in Sec. IV.A. Therefore, the liquid volume in the waste changes from 120 to 0 m³.

In order to calculate the gamma-ray leakage rate and the gamma heating of the cell, the Monte Carlo code MCNP5-1.60 and library MCPLIB84 were used.^[23] The energy and generated number of gamma rays + X-rays per unit time were obtained from ORIGEN-2. Bremsstrahlung X-ray was taken into account assuming the surrounding medium was UO₂ as the maximum condition, but its influence on the results was less than 3%. Therefore, hereinafter, we do not consider X-rays.

In calculating the gamma heating of the cell wall, we did not distinguish a ceiling, a floor, and four side walls. Therefore, the gamma heating in the depth direction of the cell walls was the same regardless of the position of each wall.

C.II. GAMMA-RAY LEAKAGE RATE FROM THE TANK

Before the start of boiling, the gamma-ray leakage rate was 2.18 kW out of 578 kW. When the waste became an oxide, it was 10.3 kW. For the intermediate time from the boiling start and the dryup, the gamma-ray leakage rate (gamma heating) R_γ(t) (in kilowatts) was approximated by the following expressions:

(C.1) $\left.\begin{array}{c}0\le {\xi }_{\mathit{\gamma }}\le 0.66,R_{\mathit{\gamma }}\left(\mathit{t}\right)=1.64{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)+2.18\\ 0.66\le {\xi }_{\mathit{\gamma }}\le 1,R_{\mathit{\gamma }}\left(\mathit{t}\right)=20.8{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)-10.47\end{array}\right\},$(C.1)

where ξ_γ(t) = mass vaporization ratio of liquid from the waste.

C.III. GAMMA HEATING OF THE CELL

Figure C.1 shows the heating rate per unit depth within the cell concrete as a function of depth from the cell surface. It is the average value in all directions. Using this result and Eq. (C.2)(C.2) $\left.\begin{array}{c}0\le {\xi }_{\mathit{\gamma }}\le 0.66,q_{\mathit{\gamma }}\left(\mathit{t},\mathit{x}\right)=\left(0.584{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)+0.0786\right)\exp \left(-14.2\mathit{x}\right)\\ 0.66\le {\xi }_{\mathit{\gamma }}\le 1,q_{\mathit{\gamma }}\left(\mathit{t},\mathit{x}\right)=\left(0.765{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)-0.393\right)\exp \left(-14.2\mathit{x}\right)\end{array}\right\}.$(C.2) , the heating density q_γ(t,x) (in kW∙m⁻³) corresponding to the progress of the boiling accident is obtained by the following approximate equations:

(C.2) $\left.\begin{array}{c}0\le {\xi }_{\mathit{\gamma }}\le 0.66,q_{\mathit{\gamma }}\left(\mathit{t},\mathit{x}\right)=\left(0.584{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)+0.0786\right)\exp \left(-14.2\mathit{x}\right)\\ 0.66\le {\xi }_{\mathit{\gamma }}\le 1,q_{\mathit{\gamma }}\left(\mathit{t},\mathit{x}\right)=\left(0.765{\xi }_{\mathit{\gamma }}\left(\mathit{t}\right)-0.393\right)\exp \left(-14.2\mathit{x}\right)\end{array}\right\}.$(C.2)

Fig. C.1. Total heating rate per unit depth inside the cell concrete as a function of depth from the cell surface.