clc
clear
Bus=[100;210060;39040;412080;56030;66020;7200100;8200100;96020;106020;114530;126035;136035;1412080;156010;166020;176020;189040;199040;209040;219040;229040;239040;24420200;25420200;266025;276025;286020;2912070;30200600;3115070;32210100;336040;];
Branch=[1120.09220.0407;2230.49300.2511;3340.36600.1864;4450.38110.1941;5560.81900.7070;6670.18720.6188;7780.71440.2351;8891.03000.7400;99101.04400.7400;1010110.19660.065;1111120.37440.1238;1212131.46801.1550;1313140.54160.7129;1414150.59100.5260;1515160.74630.5450;1616171.28901.7210;1717180.73200.5740;182190.16400.1565;1919201.50421.3554;2020210.40950.4784;2121220.70890.9373;223230.45120.3083;2323240.89800.7091;2424250.89600.7011;256260.20300.1034;2626270.28420.1447;2727281.05900.9337;2828290.80420.7006;2929300.50750.2585;3030310.97440.9630;3131320.31050.3619;3232330.34100.5302;];
[busnumrow]=size（Bus）;
[branchnumrow]=size（Branch）;
soubus=Branch（:2）;
mobus=Branch（:3）;
Vbus=12.66*ones（busnum1）;
Vbus1=Vbus;
Ploss=zeros（busnum1）;
Qloss=zeros（busnum1）;
e=1;
k=0;
Branch1=Branch;
n=1;
%%?精髓
%%%%%%%%%%%%%%%支路重新排序，各個分支線同時進行計算
while?~isempty（Branch1）%%%%T1為排好的支路矩陣。
????m=1;
????[srow]=size（Branch1）;
????????while?s>0
????????????t=find（Branch1（:2）==?Branch1（s3））;%判斷是否是葉子節點
????????????????if?isempty（t）%空，則是葉子節點
????????????????????T1（n:）=?Branch1（s:）;
????????????????????%將葉子節點放入T1中，
????????????????????%且從節點系統末端向首端進行；
????????????????????%第二次由于Branch1排序問題從首端向末端，無差別，因為同事進行
????????????????????n=n+1;
????????????????else
????????????????????T2（m:）=?Branch1（s:）;%非葉子節點
?????????????????????m=m+1;
????????????????end;
???????????????s=s-1;
????????????end;
????????Branch1=T2;
????????T2=[];
end;
%%?
%%%%%%%%%%%%%%%%%%%%%%%
while?e>1.0e-05%%%%收斂條件是精度
????%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
????%前推功率
????%從末端向首端推
????%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
????P=zeros（busnum1）;%%存放后續功率情況
????Q=zeros（busnum1）;
????????for?s=1:branchnum
????????????????i=T1（s2）;
????????????????j=T1（s3）;
????????????????R=T1（s4）;
????????????????X=T1（s5）;
????????????????%按尾節點讀取Bus數據
????????????????Pload=Bus（j2）;
????????????????Qload=Bus（j3）;
????????????????II=（（Pload+P（j））^2+（Qload+Q（j））^2）/（Vbus（j）^2*1000）;
????????????????Ploss（ij）=II*R;
????????????????Qloss（ij）=II*X;
????????????????P（ij）=Pload+Ploss（ij）+P（j）;
????????????????Q（ij）=Qload+Qloss（ij）+Q（j）;
????????????????P（i）=P（i）+P（ij）;
????????????????Q（i）=Q（i）+Q（ij）;
?????????end;
?????%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
????%后推電壓
????%從首端向末端推
????%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
?????????for?s=branchnum:-1:1
????????????????i=T1（s2）;
????????????????j=T1（s3）;%存儲推導電壓的順序
????????????????R=T1（s4）;
????????????????X=T1（s5）;
????????????????Vbus（j）=（Vbus（i）-（P（ij）*R+Q（ij）*X）/（Vbus（i）*1000））^2

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件??????35328??2018-07-03?16:07??IEEE33配電網潮流計算\html\PF33.doc

?????文件??????35328??2018-06-24?15:20??IEEE33配電網潮流計算\html\powerflow33_.doc

?????文件???????3400??2018-07-03?16:06??IEEE33配電網潮流計算\PF33.m

?????文件??????12397??2018-06-20?20:16??IEEE33配電網潮流計算\完成分層的T1.xlsx

?????文件??????13189??2018-06-20?10:26??IEEE33配電網潮流計算\工作簿1.xlsx

?????文件???????6323??2018-06-20?10:20??IEEE33配電網潮流計算\新建?Microsoft?Excel?工作表.xlsx

?????文件??????27689??2018-06-24?14:01??IEEE33配電網潮流計算\新建?Microsoft?Visio?繪圖.vsdx

?????目錄??????????0??2018-07-03?16:07??IEEE33配電網潮流計算\html

?????目錄??????????0??2018-07-08?22:38??IEEE33配電網潮流計算

-----------?---------??----------?-----??----

???????????????133654????????????????????9