发布时间 : 星期四 文章电力系统稳态分析习题集答案解析2015更新完毕开始阅读69aef3f9876fb84ae45c3b3567ec102de3bddf98
nnPis=ei??Gijej?Bijfj?+fi??Gijfj?Bijej?j?1j?12Uis=ei2+fi2
采用平直启动方法,设
00U2=U4=1
?10=?20=?40=0
j11?13.91?j54.07?9.29?j36.07??23.2?j102.2??j11?j1000?Y????13.91?j54.07032.5?j126.2?18.59?j72.13????9.29?j36.070?18.59?j72.1327.88?j108.16??
不平衡量
?P1=P1-U1?Uj(G1jcos?1j?B1jsin?1j)??0.0945j?144
?P2=P2-U2?Uj(G2jcos?2j?B2jsin?2j)j?14=1
?P4=P4-U4?Uj(G4jcos?4j?B4jsin?4j)j?14=0.93225
?Q2=Q2-U2?Uj(G2jsin?2j?B2jcos?2j)j?14=1.875
?Q4=Q4-U4?Uj(G4jsin?4j?B4jcos?4j)j?1=1.442
雅可比矩阵
J[0]09.52225???103.66811.27536.9718?11.275?11.275?000????36.97180?109.102027.6478???000?8.7250???028.11220?107.218???9.52225?
第一次迭代修正量
?[1][0]???1??????2???P1???0.0153218??????4??J??P2??0?1????0.104014????UU???P4???0.0100091??2/2????Q?2????0.2149???U4/U4????Q4??????0.0147105??
??[1][0][1]?1????1?????2????2????1???2??0.0153218?0.104014???4?????U???4?????????2???U?????4??0.0100091?2?U4??????U2/U?2?0.2149??U4??????U4/U4?????0.0147105??
?[1]??P1????P2??0.208642????P?4????0.213308?????Q???0.014834?20.515659???????Q4??????0.025611??
???106.84413.644237.56640.99868913.6442?13.64420?0.998689J[1]????37.46380?110.840?1.21331?1.213310?13.0673??9.8615027.99150?[1]???[2]1??????2???P1???0.000118851??????4??J0?1?P2?????0.0129513????U???P4???Q??0.00010803??2/U22????U4/U4?????0.0382703???Q4?????0.00024925??
??[2]?1???[1]???2????[2]?1?1?????2??0.0152029????4????U????4?????2??2???U?????4???0.0910623??2?U4??????U??0.0099003??2/U2?U4?????1.1834???U4/U4?????1.01446??
3.3. 习题3.3
对一般系统,基于极坐标潮流方程式,推导
9.3257?0??28.9946??0???110.269??
1)系统各节点电压对PQ节点 j无功负荷的灵敏度; 2)PV节点 i 无功输出对PQ节点 j 无功负荷的灵敏度; 3)系统网损对PV节点 i 电压幅值的灵敏度。 解:
(1) 节点电压i对PQ节点j无功功率灵敏度: 为雅可比矩阵的逆J-1的对应Ui行,对应Qj列 (2) PV节点i无功输出对PQ节点j无功功率灵敏度;
?Qi?Qi=?Qj?Uj?Qj?Uj?J中Qi行,对应Uj列J中Qj行,对应Uj列
(3) 系统网损对PV节点i电压幅值的灵敏度
PLoss??Pi???UiUjGijcos?ij,则:
i?1i?1j?1N?Ploss?2?UjGijcos?ij ?Uij?1nnn3.4. 习题3.4
如图示意,一条单位长度阻抗为 r + jx的配电线路,给单位长度负荷功率为
P + jQ 的电力负荷供电。假设线路长度为一个单位,线路阻抗和负荷功率均沿线连续、均匀的分布,线路始端电压为 U0?0,列写描述线路潮流的稳态方程及其边界条件,并思考1.1.6节图1.17所示电压分布是否准确。(提示:参考2.1.5节电力线路稳态方程的推导思路)
U0?0?r + jxl = 1P + jQ
解:
显然沿线电压为长度的函数,即U?l?。令始端对应于l=0,由于单位长度,故末端对应于l=1。
U?l?dl???
U?l??U?l?dl??(r + jx)dl(P + jQ)dl如图,考虑l处的dl长度,有
(r + jx)dl(P + jQ)dl
(P + jQ)dl??????????????????Ul?dl?UlUl?Ul?dl??U?l?????P?jQ?dl U?l????r?jx?dl???r?jx?dl?????**整理
????????????????Ul?dl?Ul?Ul?Ul?dl??U?l????????P?jQ
?r?jx?dl?dl?泰勒级数展开
??????????22???????????Ul1?Ul?Ul1?Ul22?U?l???dl??...?U?l????U?l??U?l???dl??...?dl?dl???????l2?l2?l2?l2U?l??????P?jQ?r?jx?dl?dl略去3阶及以上,并整理
U?l???U?l???r?jx??P?jQ? 2?l2?边界条件: