电力系统自动化习题&答案

ΔP* 2500MW 800MW Δf* 48Hz 50Hz

5、解：

⑴ 两系统都参加一次调频时

△f=0.09434 Hz △Pab=-19.8 MW

⑵ A系统参加一次调频，B系统不参加一次调频时

△f=0.16854 Hz △Pab=43.26 MW

⑶ 两系统都不参加一次调频时

△f=1.6667 Hz △Pab=-16.6667 MW

6、解：

⑴ 两系统都参加一次、二次调频，两系统都增发50MW时

△f=0.03145 Hz △Pab=-23.27 MW

⑵ 两系统都参加一次调频，A系统参加二次调频并增发60MW时

△f=0.0566 Hz △Pab=8.1132 MW

⑶ 两系统都参加一次调频，B系统参加二次调频并增发60MW时

△f=0.0566 Hz △Pab=-51.887 MW

7、解：

?n1由??f?PL?得：?????a?k?1k?????1n?bk?2P??L?ak?1k???? ?由??f?Pk?得：??2ak??Pk???bk2ak??? ?8、解：

?Utmax??Utmin?U1tmax?U1tmin?U1t?PmaxRt?QmaxXT30?2.5?15?40??6.136kV(3分)U1max110PminRt?QminXT12?2.5?7?40??2.719kVU1min114U1max??Utmax110?6.136U2N??6.3?109.06kV(3分)U2max6U1min??Utmin114?2.719U2N??6.3?106.22kVU2min6.6

11(U1tmax?U1tmin)?(109.06?106.22)?107.64(2分)22取U1t?107.25的分接头6.3?(110?6.036)?6.10?6.0(2分)107.256.3??(114?2.719)?6.53?6.6107.25U2max?U2min9、解：

'U2max?U1?P1maxR?Q1maxX15?30?12?150?120??101.25kV(2分)U1120P1minR?Q1minX10?30?8?150?120??107.5kV(2分)U1120

'U2min?U1?'U2107.5Ut?minU2N??11?112.62kV(2分)U2min10.5取U1t?107.25的分接头，那么k?UQC?2cmaxX10、解：

由系统结构可以得到：??P??P??P123????P??P2?P45?P??P??P?34?6(3分)'?U2max?U?2cmax?k?107.25?10.25(2分)11?210.5?101.25?2?k?10.5?10.25?4.57(2分)???15010.25????????????????P??90???P??66???P??26???P??22???P??P??42???P??P??50???P??902?P??662?P??262?P??222?P??422?P??502J?P12345622222232342434?2(2分)?、P?、P?求偏导，得：上式分别对P234?J??P??90?2P??66?2P??P??22?0?2P23224??P2?J??P??90?2P??26?2P??P??50?0?2P23334??P3?J??22?2P??P??22?2P??P??50?0?2P42434??P4??65??26??23解以上三式得：P，P，P234???????

(1分)(1分)(1分)?????????????91??42??49代入数学模型得：P，P，P156(2分)