資源簡(jiǎn)介
對(duì)含儲(chǔ)能和風(fēng)電的電力系統(tǒng)進(jìn)行了可靠性評(píng)估,利用序貫蒙特卡洛法,把風(fēng)儲(chǔ)系統(tǒng)接入IEEE-RBTS系統(tǒng)來仿真,探討了風(fēng)電場(chǎng)、儲(chǔ)能系統(tǒng)、儲(chǔ)能容量和儲(chǔ)能最大充放電功率對(duì)系統(tǒng)可靠性的具體影響。
代碼片段和文件信息
clear;
clc;
tic
year=2
%模擬的年限
for?l=1:year;
????
????%%%%%%%%%%%%%%%%%%%%%%%%%??第一步:數(shù)據(jù)導(dǎo)入與預(yù)處理??%%%%%%%%%%%%%%%%%%%
????
????SW0=load(‘windspeed.txt‘);????%載入原始風(fēng)速數(shù)據(jù)
????SW0=SW0‘;?????????????????????%將風(fēng)速數(shù)據(jù)從橫列轉(zhuǎn)置成豎列
????SW0=SW0/10*36;????????????????%原始數(shù)據(jù)的風(fēng)速單位為m/s,這里轉(zhuǎn)化為km/h
????N=size(SW02);????????????????%把SW0的列數(shù)賦值給N
????mu=mean(SW0);?????????????????%求SW0的平均值
????sigma=var(SW0);
????sigma=sigma^0.5;??????????????%求SW0的標(biāo)準(zhǔn)差
????y=(SW0-mu)./sigma;????????????%數(shù)據(jù)預(yù)處理
????
????N1=8736;??????????????????????%N1為模擬的小時(shí)數(shù)
????WTGnum=25;????????????????????%WTGnum為風(fēng)力發(fā)電機(jī)數(shù)量
????
????%%%%%%%%%%%%%%??第二步:根據(jù)AIC準(zhǔn)則確定ARMA模型的階數(shù)??%%%%%%%%%%%%%%%%%%
????
????for?n=2:7;
????????m=armax(y‘[nn-1]);
????????fai=-m.a;
????????theta=m.c;????????????????%把a(bǔ)rmax函數(shù)得到的參數(shù),取出來
????????
????
?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????文件????????251??2017-04-09?20:59??windspeed.txt
?????文件??????69317??2013-01-19?19:43??load.txt
?????文件??????11176??2017-04-17?17:03??wind.m
-----------?---------??----------?-----??----
????????????????80744????????????????????3
評(píng)論
共有 條評(píng)論