## 3 Results
### Thermal Effect Verification  
Based on the analysis and the results of the simulation of the short-circuit variation process described in Section 2, the Joule integral can be accurately calculated.  
Assuming that the short-circuit duration is \(\mathrm{T_{k}}\), according to the process of change in the short-circuit current shown in Figure 6, the calculation of the Joule integral can be divided into two stages: 0-\(\mathrm{T_{1}}\) is the capacitor discharge stage (\(\mathrm{i_{c}}\)), and \(\mathrm{T_{1}}\)-\(\mathrm{T_{k}}\) is the steady-state current stage (\(\mathrm{i_{s}}\)). After the two stages are calculated, the total Joule integral is obtained through summation:  
\[\int_{0}^{\mathrm{T_{K}}}\mathrm{i^{2}dt}=\int_{0}^{\mathrm{T_{1}}}\mathrm{i_{ c}^{2}dt}+\int_{\mathrm{T_{1}}}^{\mathrm{T_{K}}}\mathrm{i_{\overline{5}}dt} \tag{10}\]  
It is difficult to directly calculate the Joule integral of the capacitor discharge process, but it can be easily obtained from an energy perspective analysis. Figure 7 shows the capacitor discharge circuit.  
The energy stored in the capacitor is eventually consumed by the resistor \(\mathrm{R}\). Therefore [18],  
\[\frac{1}{2}\mathrm{CV_{0}^{2}}=\int_{0}^{\infty}\mathrm{i_{c}^{2}Rdt}\]  
According to the analysis in Section 2.3.1 and Figure 6, when the short-circuit current enters Stage 3, uncontrolled rectification occurs, and the AC-link starts to inject the short-circuit current into the DC-link. At this time, the discharge current of capacitor \(\mathrm{C}\) and the freewheeling of inductor \(\mathrm{L}\) can be neglected, and the value of \(\mathrm{T_{1}}\) can be obtained as \(2\) ms, as shown in Figure 6; therefore,  
\[\int_{0}^{\mathrm{T_{1}}}\mathrm{i_{c}^{2}dt}= \frac{1}{2\mathrm{R}}\mathrm{CV_{0}^{2}} \tag{11}\]  
The Joule integral of the steady-state stage \(\mathrm{T_{1}}\)-\(\mathrm{T_{k}}\) is  
\[\int_{\mathrm{T_{1}}}^{\mathrm{T_{K}}}\mathrm{i_{\overline{5}}^{2}dt}= \mathrm{i_{s}^{2}}(\mathrm{T_{k}}-\mathrm{T_{1}})\]  
Figure 7: Capacitor discharge circuit.  
As shown in Figure 6a, due to the DC component of the short-circuit current on the AC side, at the beginning of Stage 3, \(\mathrm{i_{s}}\) has a fluctuation of the grid's synchronous frequency. This will make the Joule integral larger, and it can be corrected by referring to Equation (108) in the IEC 60909-0(2016) standard [33]:  
\[\int_{\mathrm{T_{1}}}^{\mathrm{T_{K}}}\mathrm{i_{s}^{2}dt=i_{s}^{2}(m+n)(T_{k} -T_{1})} \tag{12}\]  
where m is taken according to Figure 18 in the IEC 60909-0 (2016) standard [33], and n is taken as 1 according to Figure 19 in the IEC 60909-0 (2016) standard and the discussion in Section 2.1. By bringing Formulas (11) and (12) into Formula (10), the total Joule integral of \(0\mathrm{-T_{k}}\) can be obtained as follows:  
\[\int_{0}^{\mathrm{T_{K}}}\mathrm{i^{2}dt=}\frac{1}{2R}\mathrm{CV_{0}^{2}+i_{s} ^{2}(m+1)(T_{k}-T_{1})} \tag{13}\]  
The rest of the verification steps are as specified in the IEC61660-2 (1997) standard [32]. The thermal effect verification of electrical equipment shall hold the following relation:  
\[\int_{0}^{\mathrm{T_{K}}}\mathrm{i^{2}dt<I_{thr}^{2}T_{k}}\]  
where \(\mathrm{I_{thr}}\) is the rated short-term current of the electrical equipment that can be withstood.  
The thermal effect verification of conductors shall hold the following relation [32]:  
\[\int_{0}^{\mathrm{T_{k}}}\mathrm{i^{2}dt<(S_{thr}A)^{2}T_{k}}\]  
where \(S_{thr}\) and A denote the rated short-term current density that can be withstood and the cross-section of the conductor, respectively.