Southern Power System Technology >
2020 , Vol. 14 >Issue 2: 10 - 24
DOI: https://doi.org/10.13648/j.cnki.issn1674-0629.2020.02.002
Review on Large-Disturbance Stability Analysis Methods in Power System
Received date: 2020-01-03
Online published: 2026-03-02
Supported by
National Key Research and Development Program of China(2016YFB0900600)
Science and Technology Project of State Grid Corporation of China(52094017000W)
Nowadays,the Chinese power grid exhibits a new architecture of multi-infeed HVDC transmission with large-scale wind and photovoltaic new power generation interconnection. The complex structure and new component characteristics of hybrid power grid make the stability of AC/DC systems more vulnerable after large disturbance. This paper mainly reviews the research progress of large disturbance stability analysis and divides the analysis methods into five classes: the step-by-step integration,the asymptotic expansion,functional approximation,direct method and other methods. Several common large-disturbance stability analysis methods are introduced. The advantages and disadvantages of the methods are discussed in this paper,which are thought to be useful for further study on large-disturbance stability in the future.
Minquan CHEN , Zhuoran KANG , Qian ZHANG , Shengwen LIAO , Junlei LIU , Deqiang GAN , Hao WU . Review on Large-Disturbance Stability Analysis Methods in Power System[J]. Southern Power System Technology, 2020 , 14(2) : 10 -24 . DOI: 10.13648/j.cnki.issn1674-0629.2020.02.002
${\dot{a}}_{0}$(t)+ε[${\dot{a}}_{1}$(t)+a0(t)-Vs]+
ε2[${\dot{a}}_{2}$(t)+a1(t)]+ε3[${\dot{a}}_{3}$(t)+a2(t)]+…=0V1(t)=$\frac{\overline{h}}{1+t}{\int }_{0}^{t} \frac{1}{1+\tau }$[$\frac{1}{(1+\tau {)}^{2}}$+(1-$\frac{1}{1+\tau }$)-1]dτ
=-$\frac{\overline{h}}{2(1+t)}$+$\frac{\overline{h}}{(1+t{)}^{2}}$-$\frac{\overline{h}}{2(1+t{)}^{3}}$=-$\frac{\overline{h}{t}^{2}}{2(1+t{)}^{3}}$V2(t)=V1(t)+$\frac{\overline{h}}{1+t}{\int }_{0}^{t} \frac{1}{1+\tau }$[${\dot{V}}_{1}$(τ)+V1(τ)]dτ
=-$\overline{h}$($\frac{1}{2}$+$\frac{5}{24}\overline{h}$)$\frac{1}{1+t}$+$\overline{h}$(1+$\frac{\overline{h}}{2}$)$\frac{1}{(1+t{)}^{2}}$-
$\overline{h}$($\frac{1}{2}$+$\frac{3}{4}\overline{h}$)$\frac{1}{(1+t{)}^{3}}$+$\frac{5{\overline{h}}^{2}}{6(1+t{)}^{4}}$-$\frac{3{\overline{h}}^{2}}{8(1+t{)}^{5}}$
=-$\frac{\overline{h}{t}^{2}}{2(t+1{)}^{3}}$-$\frac{{\overline{h}}^{2}{t}^{2}(5{t}^{2}+8t+12)}{24(t+1{)}^{5}}$V3(t)=-$\frac{\overline{h}{t}^{2}}{2(t+1{)}^{3}}$-$\frac{{\overline{h}}^{2}{t}^{2}(5{t}^{2}+8t+12)}{24(t+1{)}^{5}}$-
$\frac{{\overline{h}}^{2}{t}^{2}(120\overline{h}+320t+160\overline{h}t+180\overline{h}{t}^{2}+88\overline{h}{t}^{3})}{240(t+1{)}^{7}}$-
$\frac{{\overline{h}}^{2}{t}^{2}(23\overline{h}{t}^{4}+330{t}^{2}+180{t}^{3}+50{t}^{4}+120)}{240(t+1{)}^{7}}$〈ϕj(x),ϕk(x)〉=$\left\{\begin{array}{l}0,j\ne k\\ {A}_{k}>0,j=k\end{array}\right.$,(j,k=0,1,…)
∫Αρ(x)g(x)dx=$\sum _{i=0}^{n-1}$ai〈ϕn(x),ϕi(x)〉+
$\sum _{i=1}^{n-1}$bi〈ϕi(x),1〉+b0〈ϕ0(x),1〉
=b0〈ϕ0(x),1〉=b0∫Α ρ(x)ϕ0(x)dx
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