电力系统暂态分析课后答案最新

x7-1X3+x4+x6

X8?110.5(X1?X2)111?X1X2?X8?1X7?0.5(X3?X4?X6)?0.2565;If??0.17826;

X9?1?3.89856X9IG1?If?X8?X21000??57.758

X1?X2?X23?21X7?0.5(X3?X4?X6)X11000???10.340.5(X1?X2)?0.5(X3?X4?X6)X1?X2?X23?21IG2?IG3?0.5If?

3-1-3

(1)

SB?1000MVA,平均额定电压为基准值；X1?250?0.302?X2?0.14?1000?0.274;52521000?0.1947201000X3?0.217??0.32556000.91000X4?0.12??11201X5??0.179411?X1X2?X3X8?X5?X2?0.3734X6?111?X3X2?X5?0.174

i''f?11000?5.75;i'f'?5.75??158.084；X63?21X81000''?3.072;?i2?3.072??84.46 fX3?X83?21X3X11000''??0.9247;?i3?0.9247??25.423; fX3?X8X1?X2?X33?21X3X2?X31000''??1.7533;?i系?1.7533??48.203;

X3?X8X1?X2?X33?21?i'2'?i'f'''?i3?i'f'''?i系?i'f'(2)f2点短路

X7??0.133311?X10.（5X2?X3）1

i''f?11000?1.579;i'f'?1.579??24.639；X7?0.5X43?21X11000''?0.4053;?iG?0.4053??6.324 2X1?0.5(X2?X3)3?21''i'2'?i3?0.5i'f''i'B?0.5i'f'X11000''?0.7684;?iG?0.7684??11.99 3X1?0.5(X2?X3)3?21

3-2-1 应用例3-4已求得Y矩阵因子计算3-1-1,并与已有的计算结果比较。

j10j10???j26.6666??

j10?j33.3333j10解：Y????j10j10?j20.0???因子表中内容为?j0.0375?0.375?0.375j0.033803?0.464791j0.101420?

?0???节点2处注入单位电流，则电流向量I?1;利用已求得的R和D?1计算电压向量，得到2????0??节点自阻抗和互阻抗。

0?1??W1??0??W1????-0.375??W??I??1???W????

1122???????????????0.375?0.4647911????W3???0???W3????0.464791???X1??j0.0375?X????2????X3????j0.0338030??W1?????W???j0.033803? ??2???j0.101420????W3????j0.047139??-0.375??U1??0?1-0.375??U1??Z12??j0.038569????U???j0.033803???U???Z???j0.055713?

1-0.464791???2????2??22??????1????U3????j0.047139???U3????Z32????j0.047139???j0.0729?或者用MATLAB对Y矩阵求逆得到阻抗矩阵Z?j0.0386???j0.0557?Z12??j0.038569??Z???j0.055713?； ?22?????Z32????j0.047139??节点2的短路电流：If?j0.0386j0.0557j0.0471j0.0557?j0.0471??从而得到j0.1014??1??j17.95； Z22??U1??Z12???0.692???????1?；

各节点电压：?U2??Z22If????????????U3???Z32????0.845???U1???U1??0.307??????0?；

U2?1??U2??????????U3????U3????0.155??发电机电流：IG1??????U1??j0.462； j0.15IG2?IG3?

1??U2*??j6.67。 2j0.075第四章

4-1-1 若有三相不对称电流流入一用电设备，试问：

（1）改用电设备在什么情况下，三相电流中零序电流为零？

（2）当零序电流为零时，用电设备端口三相电压中有无零序电压？

IaIbIc???I(0)??用电设备U(0)??Z(0)

答：（1）①负载中性点不接地； ②三相电压对称；

③负载中性点接地，且三相负载不对称时，端口三相电压对称。

（2）由于零序电流为零，故无零序电压。

4-1-2

由不对称分量法变换知：

?1?2?a?a?得：

1aa2&??Z1???Ua(1)a??&??1???Ua(2)????&??1????Ua(0)??Zb??1??2??aZc????a1aa2&?1???Ia(1)??&?1???Ia(2)? ?&?1????Ia(0)?&?Za?aZb?a2Zc???Ia(1)??&? Za?a2Zb?Zc???Ia(2)??&?Za?Zb?Zc????Ia(0)?&???U?Za?Zb?Zca(1)?&?1???Ua(2)?=3?Za?aZb?aZc&???U?Za?a2Zb?aZca(0)???Za?a2Zb?aZcZa?Zb?a2ZcZa?aZb?a2Zc即无法得到三序独立的电压降方程。

4-2-1 解：

?g??E(1)??1ag??1?2E?(2)??3?aa?g??11?E?(0)????g?Ea2??a???g?a??Eb?

?g?1???Ec???a2??1??? a???1??1??j?????1a1?2 ??aa3?11??0.7887?j0.4553??

??0.2113?j0.1220????j0.3333??

gI(0)?0 I(1)?ggE(1)j2E(2)j2gg?0.7887?j0.4553??0.2276?j0.3943

j20.2113?j0.1220??0.0610?j0.1056

j2I(2)???g??Ia??1?g??2?Ib???a?g??a?Ic????gg1aa2g?g?I1??(1)???0.1666?j0.4999??g??? 1??I(2)????0.1667?j0.4999???g??1?0.333??I????(0)??Ung??Ea?Ia(j2)??j0.3332

4-5-1解：

（1） 三绕组开路

① 直接接地

xIxII