資源簡介
基于matlab IEEE14節點系統潮流計算(牛拉法和PQ分解法)。
兩種方法求解IEEE14節點的潮流計算程序,并輸出計算結果與迭代次數。經核對,結果計算正確。
另外我也上傳了此程序的原理與14節點數據word版,可以進我空間查看下載,謝謝。

代碼片段和文件信息
clear;clc;
t1=clock;%用于計算程序運行時間
%%%?~~~~~~~~~~~~~~~~輸入原始數據~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
isb=1;???%input(‘請輸入平衡母線節點號:isb=‘)
n=14;????%input(‘請輸入節點數:n=‘)
b=20;????%input(‘請輸入支路數:b=‘)
pr=1e-5;?%input(‘請輸入誤差精度:pr=‘)
n1=n-1;?
%???首端節點編號??????????????末端節點編號??????支路阻抗????????????????支路對地導納??????????變比?????
B1=[????????1???????????????????2?????????????0.01938+0.05917i???????????0.0264*2i??????????1???????????????????????
????????????2???????????????????3?????????????0.04699+0.19797i???????????0.0219*2i??????????1????????????????
????????????2???????????????????4?????????????0.05811+0.17632i???????????0.01870*2i?????????1???????????????
????????????1???????????????????5?????????????0.05403+0.22304i???????????0.02460*2i?????????1
????????????2???????????????????5?????????????0.05695+0.17388i???????????0.01700*2i?????????1????????????????
????????????3???????????????????4?????????????0.06701+0.17103i???????????0.01730*2i?????????1?????
????????????4???????????????????5?????????????0.01335i+0.04211i??????????0.00640*2i?????????1?????????
????????????7???????????????????8?????????????0.17615i??????????????????????0???????????????1??????
????????????7???????????????????9?????????????0.11001i??????????????????????0???????????????1???????
????????????9??????????????????10?????????????0.03181+0.08450i??????????????0???????????????1?????????
????????????6??????????????????11?????????????0.09498+0.19890i??????????????0???????????????1??????
????????????6??????????????????12?????????????0.12291+0.15581i??????????????0???????????????1?
????????????6??????????????????13?????????????0.06615+0.13027i??????????????0???????????????1?
????????????9??????????????????14?????????????0.12711+0.27038i??????????????0???????????????1?
????????????10?????????????????11?????????????0.08205+0.19207i??????????????0???????????????1?
????????????12?????????????????13?????????????0.22092+0.19988i??????????????0???????????????1?
????????????13?????????????????14?????????????0.17093+0.34802i??????????????0???????????????1?
????????????4??????????????????7??????????????0.20912i??????????????????????0??????????????0.978?
????????????4??????????????????9??????????????0.55618i??????????????????????0??????????????0.969
????????????5??????????????????6??????????????0.25202i??????????????????????0??????????????0.932?];??????
????????????
%?????????節點編號 發電機功率 ??????? 節點負荷功率 電壓值 補償導納 節點類型(平衡節點1,PQ節點2,PV節點3)
B2=[????????1????????0?????????????????????0????????????????????1.0600??????????0??????????????1
????????????2????????0.4+0.4904i???????????0.2170+0.1270i???????1.0450??????????0??????????????3
????????????3????????0.2744i???????????????0.9420+0.1900i???????1.0100??????????0??????????????3
????????????4????????0?????????????????????0.4780-0.0390i?????????1?????????????0??????????????2
????????????5????????0?????????????????????0.0760+0.0160i?????????1?????????????0??????????????2
????????????6????????
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????9773??2018-11-28?22:48??14jiedian_zhijiao_xiugai3.m
?????文件????????6127??2018-11-30?21:58??pq_14.m
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