-
大小: 2KB文件類型: .zip金幣: 2下載: 1 次發(fā)布日期: 2021-05-15
- 語言: Matlab
- 標(biāo)簽: 可靠度??電力系統(tǒng)??matlab??
資源簡介
蒙特卡洛模擬法是一種模擬法,由電子計算機(jī)來模擬一個過程的實(shí)現(xiàn),重復(fù)
一定次數(shù),然后計算系統(tǒng)的風(fēng)險指標(biāo)。它是用隨機(jī)數(shù)來模擬數(shù)學(xué)與物理問題以求其近似解的一種通用方法。它可以解決帶有隨機(jī)性的問題和確定性問題;處理具有指數(shù)型分布和非指數(shù)型分布的問題;以解決易于建立數(shù)學(xué)模型的問題和不能建立數(shù)學(xué)模型的問題或建立了數(shù)學(xué)模型而難以求得數(shù)值解的問題。

代碼片段和文件信息
clear;clc;
%%?系統(tǒng)數(shù)據(jù)
%??編號??類型??幅值??相角??發(fā)電有功??發(fā)電無功??負(fù)荷有功??負(fù)荷無功
NODE=[?1??3??1.04??0??0????????0??0????0
???????2??2??1.025?0??1.63?????0??0????0
???????3??2??1.025?0??0.85?????0??0????0
???????4??0??1?????0??0????????0??0????0
???????5??0??1?????0??0????????0??1.25?0.50
???????6??0??1?????0??0????????0??0.9??0.30
???????7??0??1?????0??0????????0??0????0
???????8??0??1?????0??0????????0??1????0.35
???????9??0??1?????0??0????????0??0????0
????];
%??編號1?編號2??類型??R??X??B/2??K
BRANCH=[?2??7??1??0??????0.0625??0??????1
?????????3??9??1??0??????0.0586??0??????1
?????????1??4??1??0??????0.0576??0??????1
?????????4??5??0??0.010??0.085???0.088??1?
?????????4??6??0??0.017??0.092???0.079??1
?????????5??7??0??0.032??0.161???0.153??1
?????????6??9??0??0.039??0.170???0.179??1
?????????8??9??0??0.0119?0.1008??0.1045?1
?????????7??8??0??0.0085?0.072???0.0745?1
????];
%發(fā)電機(jī)參數(shù):??發(fā)電機(jī)編號????H???????xd‘???????xd?????????xq???????D?
GEN=[??????????1???????23.64???0.0608????0.146??????0.0969?????0
???????????????2????????6.4????0.1198????0.8958?????0.8645?????0
???????????????3????????3.01???0.1813????1.3125?????1.2578?????0?????????];
%
%%?節(jié)點(diǎn)導(dǎo)納矩陣形成
a=size(NODE);?num_node=a(1);
b=size(BRANCH);?num_branch=b(1);
G=zeros(num_node);?B=zeros(num_node);
for?ii=1:num_branch
????ST??=?BRANCH(ii1);???
????ED??=?BRANCH(ii2);???
????R???=?BRANCH(ii4);
????X???=?BRANCH(ii5);
????BB??=?BRANCH(ii6);
????K???=?BRANCH(ii7);???
????RX2?=?R^2+X^2;
????G(STED)?=?-K^2*R/RX2;
????G(EDST)?=?-K^2*R/RX2;
????B(STED)?=??K^2*X/RX2;
????B(EDST)?=??K^2*X/RX2;
????G(STST)?=?G(STST)?+?K^2*R/RX2;
????B(STST)?=?B(STST)?-?K^2*X/RX2;
????B(STST)?=?B(STST)?+?BB;
????G(EDED)?=?G(EDED)?+?K^2*R/RX2;
????B(EDED)?=?B(EDED)?-?K^2*X/RX2;
????B(EDED)?=?B(EDED)?+?BB;
end
clear?ii?R?X?BB?RX2?ST?ED;
%%?穩(wěn)態(tài)計算
time=0;
while(time==0||flag>1e-5)
u=NODE(:3);?th=NODE(:4);?pg=NODE(:5)-NODE(:7);qg=NODE(:6)-NODE(:8);
kp=1;kq=1;sum_pi=0;sum_qi=0;sum_qi1=0;dp=zeros(81);dq=zeros(61);pi=zeros(91);qi=zeros(91);
%功率平衡方程
for?ii=2:9
????for?jj=1:9
????????sum_pi=sum_pi+u(jj)*(G(iijj)*cos(th(ii)-th(jj))+B(iijj)*sin(th(ii)-th(jj)));
????????if?ii>3
????????????sum_qi=sum_qi+u(jj)*(G(iijj)*sin(th(ii)-th(jj))-B(iijj)*cos(th(ii)-th(jj)));
????????end
????end
????dp(kp1)=pg(ii)-u(ii)*sum_pi;
????kp=kp+1;
????if?ii>3
????????dq(kq1)=qg(ii)-u(ii)*sum_qi;
????????kq=kq+1;
????end
????sum_pi=0;sum_qi=0;
end
clear?ii?jj?kp?kq?sum_pi?sum_qi;
%雅克比矩陣
sum_pi1=0;sum_qi1=0;
for?ii=1:9
????for?jj=1:9
????????sum_pi1=sum_pi1+u(jj)*(G(iijj)*cos(th(ii)-th(jj))+B(iijj)*sin(th(ii)-th(jj)));
????????sum_qi1=sum_qi1+u(jj)*(G(iijj)*sin(th(ii)-th(jj))-B(iijj)*cos(th(ii)-th(jj)));
????end
????pi(ii1)=u(ii)*sum_pi1;
????qi(ii1)=u(ii)*sum_qi1;
????sum_pi1=0;sum_qi1=0;
end
clear?ii?jj?sum_pi1?sum_qi1;
H=zeros(8);N=zeros(86);K=zeros(68);L=zeros(6);
for?ii=2:9
????for?jj=2:9
????????if?ii==jj??H(ii-1ii-1)=u(ii)^2*B(iiii)+qi(i
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????6800??2018-11-06?15:49??WSCC.m
評論
共有 條評論