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1、肯尼亚母线跨线短路动态张力计算工程描述：本期肯尼亚220kV为单母线接线，母线采用双分裂铝绞线型号为2* AAC-800,母 线最大跨距34.8m。Description of the project: 220kv is Single bus connection, the bus is 2* AAC-800, the maximum distance between the bus is 34.8m.计算依据：IEC_60865-1-1993短路电流效应的计算The basis of the calculation: calculation of short-circuit current
2、in IEC_60865-1-1993计算断面及平面图如下：The section and plan of the calculation:1.1计算输入参数:1.1 Common data220kV三相短路电流：=31.5kAThree-phase initial symmetrical short-circuit current(r.m.s)【=31.5kA短路电流开断时间：T=1sDuration of the first short-circuit current T=1s母线跨度（考虑绝缘子串长度）l=33.8mDistance between supports l=33.8m耐张串
3、和可调串长度分别为：3. 3m及3.6mThe length of one insulator chain and the adjustable length: 3. 3m and 3.6m导线短路工况时计算跨度：二=33.8-3.6-3.3=26.9mLength of the conductor carrying short-circuit current =33.8-3.6-3.3=26.9m母线相间距离a=4.5m。Centre-line distance between phase conductors a=4.5m 单跨弹性模量：S=500N/mmResulting spring
4、constant of both supports of one span S=500N/mm双分裂导线2*AAC-800参数：Twin conductor 2 AAC-800分裂数：n=2Number of sub-conductors n=2分裂间距：a、=200mm单导线截面积：As=0.00080209 m2单导线重量：m=2.218 kg/mSub-conductor mass per unit length m =2.218 kg/m 单导线弹性模量：e =5.50E+10 N/m2Youngs modulus E =5.50E+10 N/m2单导线外径：ds=0.0368 mSu
5、b-conductor diameter=0.0368 m母线跨静态张力（考虑大风时最严重工况）Fst= 10.454 kNStatic conductor tensile force （local maximum wind speed） Fst= 10.454 kN引下线导线及金具参数：Additional concentrated masses representing the connections of pantograph disconnectors 引下线数量：nc=2Number nc=2引下线设备重量：mc= 30 kgMass of one connection mc= 30
6、 kg引下线距离：Ls1=0.4m ;Ls2=17.4m;Ls3=9.1mDistances Ls1=0.4m ;Ls2=17.4m;Ls3=9.1m重力加速度：gn= 9.81 m/s2Conventional value of acceleration of gravity gn= 9.81 m/s21.2短路时三相导线电磁力及特性参数计算1.3 electromagnetic force and characteristic parameters1）三相导线电磁力：= 50.7S*K*=*Q.75*3L5*m|4#MT =26.32tca L 2te4-5338N/mThe electr
7、omagnetic force per unit length is对应 IEC_60865-1 公式 19 ff . nc$cmc（考虑金具附加重）导线重量：】 =吃一 77 =2.218+2*30/2*26.9=3.33 kg/mThe resulting mass per unit length of one sub-conductor is f _ t . nc：mc _=】L - =2.218+2*30/2*26.9=3.33 kg/mF _参数 r= =26.3/9.8*2*3.33=0.402gnnmscF _The Parameter r is r= =26加8*2*3.33=
8、0.4022）电磁力作用下导线偏角5 = arctan r = 21.9。12）The direction of the resulting force is 5 = arctanr = 21.9电磁力作用下中相导线等效静态弧垂：b = n *m * gn*12 = 2*3.33*9.8*33.8 2 = 0 89c8*F8*10.454*103. mstThe equivalent static conductor sag at mid span isb = n *m * gn* 12 = 2*3.33*9.8*33.8 2 = 0 89 c81*?8*10.454*103. mst3）导线
9、震荡时间:1.691 申=1.64 s1.04*0.99T = 2兀：0.8 -b = 1.69s gnfTT =1* （京I 64 903）the periods of the conductor oscillation areT = 2k |0.8 ；- = 1.69sgnres1.691.04*0.99=1.64 s4)应力计算4) Stress calculationF因一s =nA10.454*10002*0.00080209=6.52*106 v5*107BecauseF 10.454*1000s =nA 2*0.00080209=6.52*106 v 5*107乒(n * m *
10、 gn * l)224F 3NstThe stress factor is*sc24 F 3 Nst匕 (n * m * gn * l)2 S =(2*3.33*9.8*33.8) 224*(10.454*103)3 *8.48*10 -8= 2.09则根据IEC_60865-1公式26得出以下参数FE广 E 0.3 + 0.7*sin(泌 *5* *90。) =2.43*1010 N/m2 sSo E = E 0.3 + 0.7*sln(F*90。)=2.43*1010 N/m2st|_nA *5*107应力计算结果如下：N = + 1=1+1= 8.48*10 -8 1/nSl n * E
11、 * A 5*105*33.8 2*2.43*10 10 *0.00080209The stiffness norm isN = + 1=1+1= 8.48*10 -8 1/nSl n * E * A 5*105*33.8 2*2.43*10 10 *0.00080209应力系数:(2*3.33*9.8*33.8) 2=2.0924*(10.454*103)3 *8.48*10 -8因 T = = 0.61 0.5T 1.64T 11.64Because = 0.61 0.5Tres则根据IEC_60865-1公式29得出应力摇摆角：6 k = 261 = 2*21.9 = 43.8。The
12、span swing-out angle is6 k = 261 = 2*21.9 = 43.8。则根据IEC_60865-1公式30, 31得出短路时最大摇摆角:6 = 10。+ arccos(1- rsin6 ) = 53.8。The maximum swing-out angle is6 = 10。+ arccos(1- rsin6 ) = 53.8。1.3短路状态下导线拉伸力计算1.3 Tensile force during the short circuit因 T /4 v TBecause T、/4 v T则根据IEC_60865-1公式32得出负载参数：甲=3(顼1 + r2
13、-1) =0.23The load parameter is甲=3(*1 + r2 -1) =0.23根据IEC_60865-1附图表格7如下：可得出参数W =0.78According to figure 7 of IEC_60865-1, the factor W =0.78Figure 7 一 Factor 甲 for tensile force in flexible conductors短路状态下导线拉伸力：F = 1.1F (1+甲 *W) = 1.1*10.454*(1 + 0.23*0.78) = 3.56kNThe tensile force during the short
14、 circuit isF =1.1F (1+甲 *w) = 1.1*10.454*(1 + 0.23*0.78) =3.56kN1.4短路后导线下落时拉伸力计算1.4 Tensile force after the short circuit因参数r=0.402V0.6,并且短路时最大摇摆角5皿=53.8。 70。，此工况下不需验算导线下 落时的拉伸力。Because r=0.402V0.6, and 5 = 53.8。 70 during the short circuit, so the tensile force after the short circuit is not signif
15、icant.1.5短路工况导线摇摆安全距离校验1.5 Horizontal span displacement and minimum air clearance弹性膨胀系数：七=(F - F )N = (13.56-10.454)*103*8.48*10-8 = 0.26*10-3The elastic expansion is七=(F - F )N = (13.56-10.454)*103*8.48*10-8 = 0.26*10-热膨胀系数：The thermal expansion因 T /4 v TBecause T、/4 v T则根据IEC_60865-1公式37得出热膨胀系数:8 th - Cth nA*Ts = 0.27*10 -18
