Coupling the Corrosion-and Pressure-Assisted Stress Buildup Within the Zirconium in PWR Pipes

Abstract

We have developed a new real-time framework for calculating the simultaneous accumulation of oxidation-induced and the internal/external fluid stresses during the corrosion of the zirconium metal, Zr. In order to track such interfacial stress when the zirconium metal turns oxide, we quantify the hypothetical real-time infiltration of the oxygen within the metal matrix in the curved boundary, leading to the augmentation in the volume, and we stoichiometrically compute the resulted equivalent oxide thickness. Subsequently, we calculate the accumulated compressive stress in real-time from both irreversible (plastic) and reversible (elastic) events which could be used for anticipation of the onset of failure. The developed analytical framework could quantify the design parameters for the safe operation of high-pressure pipes in corrosive environments.

Appendix

Here, we present the detailed steps for establishing the relationships shown in the article.

1. Elastic Stress $\sigma _{\mathrm{EL}}$ (Eq. 7)

Imposing the internal and external pressures $P_{\mathrm{I}}$ and $P_{\mathrm{O}}$, the radial $\sigma _{r}$ and hoop $\sigma _{\theta }$ elastic stresses develop, as shown in Fig. 1b which are theonly function of radius r due to polar symmetry. Cutting out the infinitesimal element shown in Fig. 2b, the balance relationships in the horizontal direction would be:

$$\begin{aligned} \sum F_{x}=0 \end{aligned}$$

therefore:

$$\begin{aligned} \left( \sigma _{r}+\frac{\partial \sigma _{r}}{\partial r}\mathrm{d}r\right) \cdot \left( r+\mathrm{d}r\right) \mathrm{d}\theta -\sigma _{r}\cdot r\mathrm{d}\theta -2\sigma _{\theta }\cdot \mathrm{d}r\cdot \frac{\mathrm{d}\theta }{2}=0 \end{aligned}$$

where by simplification and ignoring the higher-order terms, we arrive at Lame’s relationship as:^71,72

$$\begin{aligned} \frac{\partial \sigma _{r}}{\partial r}+\frac{\sigma _{r}-\sigma _{\theta }}{r}=0 \end{aligned}$$

Regarding the strain $\epsilon$, if u is the infinitesimal displacement in the radial direction, from the Hook’s generalized law, one gets:⁷¹

$$\begin{aligned} {\left\{ \begin{array}{ll} {\epsilon _{r}}={\displaystyle \frac{\mathrm{d}u}{\mathrm{d}r}}={\displaystyle \frac{1}{E}\left( \sigma _{r}-\nu \sigma _{\theta }\right) }\\ \epsilon _{\theta }={\displaystyle \frac{u}{r}={\displaystyle \frac{1}{E}\left( \sigma _{\theta }-\nu \sigma _{r}\right) }} \end{array}\right. } \end{aligned}$$

Solving for the radial $\sigma _{r}$ and hoop $\sigma _{\theta }$ stresses and replacing in terms of the infinitesimal displacement, u, one gets:

$$\begin{aligned} {\left\{ \begin{array}{ll} \sigma _{r}={\displaystyle \frac{E}{1-\nu ^{2}}}\left( {\displaystyle \frac{\mathrm{d}u}{\mathrm{d}r}}+\nu {\displaystyle \frac{u}{r}}\right) \\ \sigma _{\theta }={\displaystyle \frac{E}{1-\nu ^{2}}}\left( {\displaystyle \frac{u}{r}}+\nu {\displaystyle \frac{\mathrm{d}u}{\mathrm{d}r}}\right) \end{array}\right. } \end{aligned}$$

(39)

replacing the Eq. 39 into the compatibility Eq. 5 and simplifying we get:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}r}\left( \frac{1}{r}\frac{\mathrm{d}}{\mathrm{d}r}\left( u.r\right) \right) =0 \end{aligned}$$

where integrating leads to:

$$\begin{aligned} u=C_{1}r+\frac{C_{2}}{r} \end{aligned}$$

From Fig. 1b, assuming the $\{R_{\mathrm{I}},R_{\mathrm{O}}\}$ are inner and outer radii of the pipe, respectively, and assigning the boundary conditions for the compressive stresses as $\sigma _{r}\left( R_{\mathrm{I}}\right) =-P_{\mathrm{I}}$ and $\sigma _{r}\left( R_{\mathrm{O}}\right) =-P_{\mathrm{O}}$ , the hoop stress $\sigma _{P}$ is finally obtained as:

$$\begin{aligned} {\left\{ \begin{array}{ll} {\displaystyle \sigma _{r}={\displaystyle A}{\displaystyle +\frac{B}{r^{2}}}}\\ {\displaystyle \sigma _{\theta }=A-\frac{B}{r^{2}}} \end{array}\right. } \end{aligned}$$

(40)

where A and B are constants obtained by the following relationships:

$$\begin{aligned} {\left\{ \begin{array}{ll} A={\displaystyle \frac{P_{\mathrm{I}}R_{\mathrm{I}}^{2}-P_{\mathrm{O}}R_{\mathrm{O}}^{2}}{R_{\mathrm{O}}^{2}-R_{\mathrm{I}}^{2}}}\\ B={\displaystyle \frac{R_{\mathrm{I}}^{2}R_{\mathrm{O}}^{2}(P_{\mathrm{O}}-P_{\mathrm{I}})}{R_{\mathrm{O}}^{2}-R_{\mathrm{I}}^{2}}} \end{array}\right. } \end{aligned}$$

(41)

Since the compressive stress causing the failure is the hoop direction, therefore $\sigma _{P}=\sigma _{\theta }$ , and hence Eq. 7 is obtained as:

$$\begin{aligned} \sigma _{P}={\displaystyle {\displaystyle \frac{P_{\mathrm{I}}R_{\mathrm{I}}^{2}-P_{\mathrm{O}}R_{\mathrm{O}}^{2}}{R_{\mathrm{O}}^{2}-R_{\mathrm{I}}^{2}}}}{\displaystyle +\left( {\displaystyle {\displaystyle \frac{R_{\mathrm{I}}^{2}R_{\mathrm{O}}^{2}(P_{\mathrm{O}}-P_{\mathrm{I}})}{R_{\mathrm{O}}^{2}-R_{\mathrm{I}}^{2}}}}\right) \frac{1}{r^{2}}} \end{aligned}$$

(42)

which is the compressive stress generated by the boundedness in the azimuthal (i.e., hoop) direction.

2. Corrosion Stress (Eq. 17)

$$\begin{aligned} \sigma _{\mathrm{EL}}&=K\frac{\nu }{1+\nu }\frac{\mathrm{d}A}{A}\\&=K\frac{\nu }{1+\nu }\int _{0}^{s}\frac{2\pi \left( R_{\mathrm{I}}-{\displaystyle \frac{R-1}{R}}s\right) {\displaystyle \frac{R-1}{R}}\mathrm{d}s}{\pi \left( R_{\mathrm{O}}^{2}-\left( R_{\mathrm{I}}-{\displaystyle \frac{R_{\mathrm{PB}}-1}{R_{\mathrm{PB}}}}s\right) ^{2}\right) }\\&=K\frac{\nu }{1+\nu }\int _{0}^{s}\left( \frac{-{\displaystyle \frac{R-1}{R}}}{R_{\mathrm{O}}+R_{\mathrm{I}}-{\displaystyle \frac{R_{\mathrm{PB}}-1}{R_{\mathrm{PB}}}s}}+\frac{{\displaystyle \frac{R-1}{R}}}{R_{\mathrm{O}}-R_{\mathrm{I}}+{\displaystyle \frac{R_{\mathrm{PB}}-1}{R_{\mathrm{PB}}}}s}\right) \mathrm{d}s\\&=K\frac{\nu }{1+\nu }\left( \ln \left( R_{\mathrm{O}}+R_{\mathrm{I}}-\frac{R-1}{R}s\right) \Bigg |_{0}^{s} +\ln \left( R_{\mathrm{O}}-R_{\mathrm{I}}+\frac{R-1}{R}s\right) \Bigg |_{0}^{s}\right) \\&=K\frac{\nu }{1+\nu }\left( \ln \left( 1-\frac{R-1}{R}\frac{s}{R_{\mathrm{O}}+R_{\mathrm{I}}}\right) +\ln \left( 1+\frac{R-1}{R}\frac{s}{R_{\mathrm{O}}-R_{\mathrm{I}}}\right) \right) \\&=K\frac{\nu }{1+\nu }\left( \ln \left( 1+\frac{R-1}{R}\frac{s}{R_{\mathrm{O}}-R_{\mathrm{I}}} -\frac{R-1}{R}\frac{s}{R_{\mathrm{O}}+R_{\mathrm{I}}}+\left( \frac{R-1}{R}\right) ^{2}\frac{s^{2}}{R_{\mathrm{O}}^{2}-R_{\mathrm{I}}^{2}}\right) \right) \\&=K\frac{\nu }{1+\nu }\ln \left( 1+\frac{R-1}{R}\left( \frac{2R_{\mathrm{I}}}{R_{\mathrm{O}}^{2} -R_{\mathrm{I}}^{2}}\right) s+\left( \frac{R-1}{R}\right) ^{2}\frac{s^{2}}{R_{\mathrm{O}}^{2}-R_{\mathrm{I}}^{2}}\right) \end{aligned}$$

3. Corrosion Stress (Eq. 21)

$$\begin{aligned} \sigma _{\mathrm{EL}}&=K\frac{\nu }{1+\nu }\int _{0}^{s}\frac{2\pi \left( R_{\mathrm{O}}+{\displaystyle \frac{R-1}{R}}s\right) {\displaystyle \frac{R-1}{R}}\mathrm{d}s}{\pi \left( \left( R_{\mathrm{O}}+{\displaystyle \frac{R-1}{R}}s\right) ^{2}-R_{\mathrm{I}}^{2}\right) }\\&=K\frac{\nu }{1+\nu }\int _{0}^{s}\left( \frac{{\displaystyle \frac{R-1}{R}}}{R_{\mathrm{O}}+{\displaystyle \frac{R-1}{R}}s-R_{\mathrm{I}}}+\frac{{\displaystyle \frac{R-1}{R}}}{R_{\mathrm{O}}+{\displaystyle \frac{R-1}{R}}s+R_{\mathrm{I}}}\right) \mathrm{d}s\\&=K\frac{\nu }{1+\nu }\left( \ln \left( 1+\frac{R-1}{R}\frac{s}{R_{\mathrm{O}}-R_{\mathrm{I}}}\right) +\ln \left( 1+\frac{R-1}{R}\frac{s}{R_{\mathrm{O}}+R_{\mathrm{I}}}\right) \right) \\&=K\frac{\nu }{1+\nu }\ln \left( 1+\frac{R-1}{R}\left( \frac{2R_{\mathrm{O}}}{R_{\mathrm{O}}^{2}-R_{\mathrm{I}}^{2}}\right) s+\left( \frac{R-1}{R}\right) ^{2}\frac{s^{2}}{R_{\mathrm{O}}^{2}-R_{\mathrm{I}}^{2}}\right) \end{aligned}$$

4. Equation 23

Translating Eq. 22 into a 1D radial direction, we get:

$$\begin{aligned} \left( \frac{\mathrm{d}C}{\mathrm{d}t}\right) _{\text {Diff}}&=D\nabla .\left( \nabla C+\frac{f\Omega }{R_{u}T}C\nabla \sigma _{r}\right) \\&=D\frac{\partial }{\partial r}\left( \frac{\partial C}{\partial r}+\frac{f\Omega }{R_{u}T}C\frac{\partial \sigma _{r}}{\partial r}\right) \end{aligned}$$

The $\sigma _{r}$ is obtained from Eqs. 40 and 41, therefore: ${\displaystyle \frac{\partial \sigma _{r}}{\partial r}=-\frac{2B}{r^{3}}}$, replacing gives:

$$\begin{aligned} \left( \frac{\mathrm{d}C}{\mathrm{d}t}\right) _{\text {Diff}}&=D\frac{\partial }{\partial r}\left( \frac{\partial C}{\partial r}-\frac{2f\Omega CB}{R_{u}Tr^{3}}\right) \\&=D\left( \frac{\partial ^{2}C}{\partial r^{2}}-\frac{2f\Omega B}{R_{u}Tr^{3}}\frac{\partial C}{\partial r}+6\frac{f\Omega B}{R_{u}Tr^{4}}C\right) \\&=D\frac{\partial ^{2}C}{\partial r^{2}}+\alpha \left( r\right) \frac{\partial C}{\partial r}+\beta \left( r\right) C \end{aligned}$$

and the coefficients $\alpha$ and $\beta$ are obtained, respectively, as:

$$\begin{aligned} \alpha \left( r\right) =-\frac{2Df\Omega B}{R_{u}Tr^{3}} ,{\beta =6\frac{fD\Omega B}{R_{u}Tr^{4}}} \end{aligned}$$

5. Equation 29

In order to have a stable solution, we need to have $Q_{1}>0$, hence:

$$\begin{aligned} 1-\frac{2D\delta t}{\delta r^{2}}+\frac{\alpha \left( r\right) \delta t}{\delta r}+\beta \left( r\right) \delta t>0 \end{aligned}$$

replacing the $\alpha$ and $\beta$ values gives:

$$\begin{aligned} 1-\frac{2D\delta t}{\delta r^{2}}-\frac{2Df\Omega B}{R_{u}Tr^{3}}\frac{\delta t}{\delta r}+\frac{6fD\Omega B}{R_{u}Tr^{4}}\delta t>0 \end{aligned}$$

which means that:

$$ \delta t < \left( {\frac{{2D}}{{\delta r^{2} }} + \frac{{2fD\Omega B}}{{R_{u} Tr^{3} \delta r}} - \frac{{6fD\Omega B}}{{R_{u} Tr^{4} }}} \right)^{{ - 1}} \checkmark$$