電氣工程基礎(chǔ)系統(tǒng)篇-習題統(tǒng)計-20150227

1、1《電力系統(tǒng)工程基礎(chǔ)》習題1.一臺一臺SFL13150035型雙繞組三相變壓器，額定變比為型雙繞組三相變壓器，額定變比為3511kV，，，kWP300??2.1%0?I，。求變壓器歸算到。求變壓器歸算到低壓側(cè)低壓側(cè)參數(shù)的有名值，并作等值電路。參數(shù)的有名值，并作等值電路。kWPk2.177??8%?kuR=0.022，X=0.307，G=2.510^4s，B=3.1210^32.某三相三繞組自耦變壓器容量為某三相三繞組自耦變壓器容量為90

2、9045MVA，電壓為，電壓為22012111kV，短路損耗，短路損耗kWPk325)21(???、、，短路電壓，短路電壓、、kWPk345)31(????kWPk270)32(????10%)21(??ku6.18%)31(???ku，空載損耗，空載損耗，。求變壓器的參數(shù)并作等值電路。。求變壓器的參數(shù)并作等值電路。1.12%)32(???kukWP1040??65.0%0?IRT1=1.87，RT2=7.4710^2，RT3=6.38

3、XT1=61.8，XT2=8.07，XT3=138.2GT=2.14910^6BT=1.20910^53.電路如下圖，圖中參數(shù)都為標么值，求電路如下圖，圖中參數(shù)都為標么值，求1點的電壓和功率。點的電壓和功率。（電壓計算中忽略橫向分量）（電壓計算中忽略橫向分量）1S3=1j0.75U3=0.9350.03j0.040.06j0.08j0.09j0.090.01j0.0423123U23S??2S?1S??11SU?33RjX?3Y2Y11

4、RjX?1Y33SU?解：33333310.060.750.080.1280.935PRQXUU????????22223333322310.75()(0.060.08)0.1070.1430.935zPQSRjXjjU???????????2333330.9350.1281.0631.1070.893zUUUSSSj?????????????????????3j20j10j10j10j20j10j10j10j20???????????

5、???Y(2)列出待求的狀態(tài)變量。潮流計算直接求出系統(tǒng)的狀態(tài)變量，即平衡節(jié)點之外各節(jié)點的電壓相角與模值，然后可以由狀態(tài)變量的值進一步求出各線路的功率等。該系統(tǒng)中節(jié)點1是平衡節(jié)點，節(jié)點2的電壓模值給定，待求的狀態(tài)變量為節(jié)點2的電壓相角以及節(jié)點3的電壓相角與模值：233U?????????????x(3)列寫潮流方程22222222121212222222323232323(cossin)sin()sin()sin()10.5sin10.5

6、sin()jjjjjjPUUGBUUBUUBUUBU????????????????????????333333331313132233233333333332(cossin)sin()sin()sin()10sin10.5sin()jjjjjjPUUGBUUBUUBUUBUU????????????????????????33333332313131322332333332333323(sincos)cos()cos()cos()10

7、cos10.5cos()20jjjjjjQUUGBUUBUUBUBUUU???????????????????????????222sPPP???333sPPP???333sQQQ???(4)形成雅可比矩陣。222323333332333333233PPPUUPPPUUQQQUU??????????????????????????????????????????????????????J其中：221212123232322323cos(