110KV35KV变电站继电保护整定计算书

10〔一、XX变继电保护整定计算书XX变35KV母线短路阻抗最大运行方式下为0.24750.3708，20230KVA20230KVA变压器运行分列运行。荷，故计算定值时按一路进线及一台变压器运行计算。一、绘制等值电路图、绘制系统接线图〔见附图〕选取基准容量Sb=100MVA选取短路点所在母线的平均电压为基准电压即：计算K1点时，选取Ub1=37KVK2点时，选取Ub2=6.3KV计算K6302 K6412-点时，选取Ub3=6.3KV二、计算各元件的电抗标么值SU、I、X——一以基准量表示的容量MA、电压(KV(KAΩ)j j j jSe----额定容量(MA)； Ud%----阻抗电压百分数； Xd% 超瞬变电抗百分数；XO-----每千米电抗(Ω/km)； Up-----平均额定电压(KV)； X% 电抗百分数；L 线路长度(km)；1、1#主变：X1=Ud%/100*Sj/Se=7.85/100*100/20=0.39252#主变：X2=Ud%/100*Sj/Se=8.06/100*100/20=0.4032、6KV线路：(集配站6302线路) X3线=XO*L*Sj/Up2=0.4*3.2*100/6.32=3.225X3缆=XO*L*Sj/Up2=0.08*0.19*100/6.32=0.038X3=X3线+X3缆=3.225+0.038=3.263(旭飞公司6303线路) X4线=XO*L*Sj/Up2=0.353*2.2*100/6.32=1.9567X4缆=XO*L*Sj/Up2=0.08*0.1*100/6.32=0.0202X4=X4线+X4缆=1.9567+0.0202=1.9769(恒兴公司6304线路) X5线=XO*L*Sj/Up2=0.347*4*100/6.32=3.4971X5缆=XO*L*Sj/Up2=0.08*0.15*100/6.32=0.0302X5=X5线+X5缆=3.4971+0.0302=3.5273(金能公司6305线路) X6线=XO*L*Sj/Up2=0.347*4.6*100/6.32=4.0217X6缆=XO*L*Sj/Up2=0.08*0.15*100/6.32=0.0302X6=X6线+X6缆=4.0217+0.0302=4.0519〔一采区Ⅰ路线路〕6306线路为相分裂架设，故：线 X7 =XO*L*Sj/U2/2=0.333*2.6*100/6.32线 缆 X7 =XO*L*Sj/U2/2=0.08*0.4*100/6.32/2=0.0403X7=缆 线 缆线 (环兴磁场6307线路) X8 =XO*L*Sj/U2=0.347*4*100/6.32线 缆 X8 =XO*L*Sj/U2=0.08*0.17*100/6.32=0.0171X8=缆 线 缆线 (乳化车间6308线路) X9 =XO*L*Sj/U2=0.347*6.4*100/6.32线 缆 X9 =XO*L*Sj/U2=0.08*0.07*100/6.32=0.0141X9=缆 线 缆p〔6309线路〕X10〔架空〕=XO*L*Sj/U2/2=0.333*1.7*100/6.32=1.4263pX 〔电缆〕=X LS/U2/2=0.080.24100/6.32=0.048410 O* * j p * *10 X10=X 〔架空〕+X 〔电缆〕=1.426310 〔1#动力变6311线路〕

〔电缆〕=X LS/U2/2=0.080.22100/6.32=0.044311 O* * j p * *(三泵房6402线路) 因6402线路为电缆线路，故：XO为0.08Ω/kmX12=XO*L*Sj/Up2=0.08*0.4*100/6.32=0.0806(深井专线6403线路) X13线=XO*L*Sj/Up2=0.353*2.5*100/6.32=2.2235X13缆=XO*L*Sj/Up2=0.08*0.5*100/6.32=0.1008X13=X13线+X13缆=2.2235+0.1008=2.3243(众元洗煤厂6404线路) 因6404线路为电缆线路，故：XO为0.08Ω/kmX14=XO*L*Sj/Up2=0.08*0.5*100/6.32=0.1008〔一采区Ⅱ路线路〕6406线路为相分裂架设，故：X15线=XO*L*Sj/Up2/2=0.325*3.6*100/6.32/2=1.4739X15缆=XO*L*Sj/Up2/2=0.08*0.32*100/6.32/2=0.0302X15=X15线+X15缆=1.4739+0.03225=1.5062(石康铸造6407线路) X16线=XO*L*Sj/Up2=0.365*0.3*100/6.32=0.2759X16缆=XO*L*Sj/Up2=0.08*0.15*100/6.32=0.2653X16=X16线+X16缆=0.2759+0.2653=0.5412(化工厂6408线路) X17线=XO*L*Sj/Up2=0.347*8.4*100/6.32=7.3439X17缆=XO*L*Sj/Up2=0.08*0.11*100/6.32=0.02217X17=X17线+X17缆=7.3439+0.02217=7.366线 (四泵房6409线路) X18 =XO*L*Sj/U2=0.347*4*100/6.32线 缆 X18 =XO*L*Sj/U2=0.08*0.15*100/6.32=0.0302X18=缆 线 缆线 (然尔特6410线路) X19 =XO*L*Sj/U2=0.325*2.1*100/6.32线 缆 X19 =XO*L*Sj/U2=0.08*0.3/2*100/6.32=0.030X19=缆 线 缆线 (三矿物业6412线路) X20 =XO*L*Sj/U2=0.353*1.8*100/6.32线 缆 X20 =XO*L*Sj/U2=0.08*0.1*100/6.32=0.0202X20=缆 线 缆三、计算K2点的短路参数〔K2点前的总阻抗及短路电流〕1、最大运行方式下：XK2max=0.24756+0.3925=0.64006最小运行方式下：XK2min=0.37084+0.3925=0.763342、k2点的超瞬变短路电流I”：(用来作继电保护的整定计算和校验断路器的断流量)3依据：I”k2=Ij/XK2max 设基准容量为：Sj=100MVA33且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p33I”k2=Ij/XK2max=9.165/0.64006=14.319〔KA〕33同理：I” =I/X

且基准电流为：I=S/ U=100/

*6.3=9.165〔KA〕k2 j

K2min

j j pI”k2=Ij/XK2min=9.165/0.76334=12.006〔KA〕四、计算末端K6302-----K6412点的短路参数〔K6302 K6412各点的阻抗及三相短路电流〕31、∵ K6302max=XK2max+X3=0.64006+3.263=3.90306K6302min=XK2min+X3=0.7634+3.263=4.026433且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6302max=Ij/K6302max=9.165/3.90306=2.349〔KA〕I”6302min=Ij/K6302min=9.165/4.0264=2.277〔KA〕32、K6303max=XK2max+X4=0.64006+1.9769=2.617K6303min=XK2min+X4=0.76334+1.9769=2.74033且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6303max=Ij/K6303max=9.165/2.617=3.502〔KA〕I”6303min=Ij/K6303min=9.165/2.74=3.345〔KA〕3、K6304max=XK2max+X5=0.64006+3.5273=4.1673K6304min=XK2min+X5=0.7634+3.5273=4.29133且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6304max=Ij/K6304max=9.165/4.167=2.199〔KA〕I”6304min=Ij/K6304min=9.165/4.291=2.136〔KA〕34、K6305max=XK2max+X6=0.64006+4.0519=4.5755K6305min=XK2min+X6=0.76334+4.0519=4.699633且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6305max=Ij/K6305max=9.165/4.5755=2.003〔KA〕I”6305min=Ij/K6305min=9.165/4.6996=1.950〔KA〕35、K6306max=XK2max+X7=0.64006+1.1313=1.771K6306min=XK2min+X7=0.76334+1.1313=1.89533且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6306max=Ij/K6306max=9.165/1.771=5.175〔KA〕I”6306min=Ij/K6306min=9.165/1.895=4.836〔KA〕36、K6307max=XK2max+X8=0.64006+3.5142=4.154K6307min=XK2min+X8=0.76334+3.5142=4.27833且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6307max=Ij/K6307max=9.165/4.154=2.206〔KA〕I”6307min=Ij/K6307min=9.165/4.278=2.142〔KA〕37、K6308max=XK2max+X9=0.63006+5.6095=6.2396K6308min=XK2min+X9=0.76334+5.6095=6.372833且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6308max=Ij/K6308max=9.165/6.2396=1.469〔KA〕I”6308min=Ij/K6308min=9.165/6.3728=1.438〔KA〕38、K6309max=XK2max+X10=0.64006+1.4747=2.11476K6309min=XK2min+X10=0.76334+1.4747=2.2380433且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6309max=Ij/K6306max=9.165/2.11476=4.334〔KA〕I”6309min=Ij/K6306min=9.165/2.23804=4.095〔KA〕9、K6311max=XK2max+X11=0.64006+0.0443=0.6844K6311min=XK2min+X11=0.76334+0.0443=0.807633且基准电流为：I=S/ U=100/33

*6.3=9.165〔KA〕j j p∴I”6311max=Ij/K6311max=9.165/0.6844=13.391〔KA〕I”6311min=Ij/K6311min=9.165/0.8076=11.348〔KA〕310、K6402max=XK2max+X12=0.64006+0.0806=0.7207K6402min=XK2min+X12=0.76334+0.0806=0.843933且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6402max=Ij/K6402max=9.165/0.7207=12.717〔KA〕I”6402min=Ij/K64028min=9.165/0.8439=10.860〔KA〕311、K6403max=XK2max+X13=0.64006+2.3243=2.964K6403min=XK2min+X13=0.76334+2.3243=3.08833且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6403max=Ij/K6403max=9.165/2.964=3.092〔KA〕I”6403min=Ij/K6403min=9.165/3.088=2.968〔KA〕312、K6404max=XK2max+X14=0.64006+0.1008=0.7409K6404min=XK2min+X14=0.76334+0.1008=0.864133且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6404max=Ij/K6404max=9.165/0.7409=12.370〔KA〕I”6404min=Ij/K6404min=9.165/0.8641=10.606〔KA〕313、K6406max=XK2max+X15=0.64006+1.5062=2.146K6406min=XK2min+X15=0.76334+1.5062=2.269533且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6406max=Ij/K6406max=9.165/2.146=4.271〔KA〕I”6406min=Ij/K6406min=9.165/2.2696=4.038〔KA〕314、K6407max=XK2max+X16=0.64006+0.5412=1.181K6407min=XK2min+X16=0.76334+0.5412=1.30533且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6407max=Ij/K6407max=9.165/1.181=7.760〔KA〕I”6407min=Ij/K6407min=9.165/1.305=7.023〔KA〕15、K6408max=XK2max+X17=0.64006+7.366=8.0063K6408min=XK2min+X17=0.76334+7.366=8.12933且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6408max=Ij/K6408max=9.165/8.006=1.145〔KA〕I”6408min=Ij/K6408min=9.165/8.129=1.127〔KA〕16、K6409max=XK2max+X18=0.64006+3.5273=4.1673K6409min=XK2min+X18=0.76334+3.5273=4.29133且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6409max=Ij/K6409max=9.165/4.167=2.199〔KA〕I”6409min=Ij/K6409min=9.165/4.291=2.136〔KA〕17、K6410max=XK2max+X19=0.64006+3.5273=4.1673K6410min=XK2min+X19=0.76334+3.5273=4.29133且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6410max=Ij/K6410max=9.165/4.167=2.199〔KA〕I”6410min=Ij/K6410min=9.165/4.291=2.136〔KA〕318、K6412max=XK2max+X20=0.64006+1.6212=2.261K6412min=XK2min+X20=0.76334+1.6212=2.38533且基准电流为：I=S/ U=100/3

*6.3=9.165〔KA〕j j p∴I”6412max=Ij/K6412max=9.165/2.261=4.054〔KA〕I”6412min=Ij/K6412min=9.165/2.385=3.843〔KA〕五、整定计算速断动作电流整定公式：Idzj=KkKjxI”d2max/n1式中： dzj----瞬时电流速断保护装置的动作电流整定值A；”d2max 最大运行方式下线路末端三相短路超瞬变电流A；Kk-----1.2；Kjx 1；n1 电流互感器变比；过流淌作电流整定公式：Idzj=KkKjxIgh/Khn1式中： Idzj=KkKjxIgh/Khn1式中： dzj----保护装置的动作电流整定值〔；gh线路的最大负荷电流A；Kk 1.2；h 继电器返回系数，取1.〔微机型；Kjx 1；n1 电流互感器变比；继电保护整定值计算如下：1、集配站6302〔CT变比：400/5〕31、速断动作电流dzj=kjx”d2max/1=1.2*1*2349/80=3A〕 〔取25〕32、带时限过电流淌作电流：∵Igh=407(A) 依据I=P/

U*COSø=〔800+630〕3/35*800/

*6.3*0.9=146 因6302 线路所带最大变压器为800KVA，启动电流为3*6.3*0.9=4073∴Idzj=KkKjxIgh/Khn1=1.2*1*407/1*80=6.1〔A〕 3、灵敏系数校验、电流速断保护灵敏度校验：Km=KmxdI”d1min/Idz≥2 依据《继电保护及二次回路速查速算手册》Km 灵敏系数；Kmxd 0.866min 最小运行方式下线路始端三相短路稳态电流〔A〕Idz-----保护装置一次动作电流〔A〕Idz=Idzjn1/Kjx依据：Km=KmxdId1min/Idz=0.866*12023/25*80/1=5.2>22=Kmxd”d2min/dz≥1.5Kmxd 0.866min 最小运行方式下线路末端三相短路超瞬变电流〔A〕Idz-----保护装置一次动作点流〔A〕Idz=Idzjn1/Kjx依据：Km=KmxdId2min/Idz=0.866*2277/6*80/1=4.1>1.54)、时限整定：63020.8S2、旭飞公司6303〔CT变比：200/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*3502/40=105.0A〕 〔取45〕2、带时限过电流淌作电流：g=162(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*162/1*40=4.8 〔A〕 〔6A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/45*40/1=5.7>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*3345/6*40/1=12.1>1.54)、时限整定：63030.8S3、恒兴公司6304〔CT变比：800/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*2199/160=16.4A〕 〔取11〕2、带时限过电流淌作电流：g=178(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*178/1*160=1.335 〔A〕 〔2.3A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/11*160/1=5.91>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*2136/2.3*160/1=5>1.54)、时限整定：63040.8S4、金能公司6305〔CT变比：300/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*2023/60=4A〕 〔取26〕2、带时限过电流淌作电流：g=120(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*120/1*60=2.4 〔A〕 〔5A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/26*60/1=6.6>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*195/5*60/1=5.6>1.54)、时限整定：63050.8S5、一采区Ⅰ路6306〔CT变比：1000/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*5175/200=3A〕 〔取28〕2、带时限过电流淌作电流：g=1724(A) 依据一采区600水平要上综采工作面，金公司机械动力部报来需增加负荷为：1724A.∴Idzj=KkKjxIgh/Khn1=1.2*1*1724/1*200=10.3〔A〕 〔取因要与一采变协作故取11.8A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/28*200/1=1.8<22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*4836/11.8*200/1=1.8>1.54)、时限整定：63060.8S6、环兴磁厂6307〔CT变比：300/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*2206/60=44.〕 〔取26〕2、带时限过电流淌作电流：g=250(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*250/1*60=5〔A〕 〔7A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/26*60/1=6.7>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*2142/7*60/1=4.4>1.54)、时限整定：63070.8S7、乳化车间6308〔CT变比：300/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*1469/60=29.3〕 〔取20〕2、带时限过电流淌作电流：g=133(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*133/1*60=2.7〔A〕 〔5A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/20*60/1=8.6>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*1438/5*60/1=4.1>1.54)、时限整定：63080.8S8、众元三分公司6309〔CT变比：400/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*4334/80=6A〕 〔取29〕2带时限过电流淌作电流∵g=2830/1.732/6.3=259(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*259/0.85*80=4.5〔A〕 3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/29*80/1=4.48>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*4095/5*80/1=8.8>1.54)、时限整定：63090.8S91#动力变6311〔CT变比：200/5〕31、速断动作电流dzj=kjx”d2max/1=1.2*1*13391/40=401.7A〕 〔取30〕32g=P/变实际运行负荷得出的值

U*COSø=100/1.732*6.3*0.9=10.2〔A) 依据区∴Idzj=KkKjxIgh/Khn1=1.2*1*10.2/1*40=0.306〔A〕 0.7A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/30*40/1=8.6>22=Kmxd”d2n依据：Km=KmxdI”d2min/Idz=0.866*11348/0.7*40/1=350.98>1.54)、时限整定：63090.8S10、三泵房6402〔CT变比：100/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*12717/20=76〕 〔取90〕2、带时限过电流淌作电流：g=70(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*70/1*20=4.2 〔A〕 〔5A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/90*20/1=5.7>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*10860/5*20/1=94>1.54)、时限整定：64020.8S11、深井专线6403〔CT变比：500/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*3092/100=37.〕 〔取16〕2、带时限过电流淌作电流：g=238(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*238/1*100=2.86 〔A〕 〔5A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/16*100/1=6.5>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*2968/5*100/1=5.14>1.54)、时限整定：64030.8S12、众元洗煤厂6404〔CT变比：150/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*12370/30=494.A〕 〔取50〕2、带时限过电流淌作电流：g=160(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*160/1*30=6.4 〔A〕 〔6.3A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/50*30/1=6.9>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*10606/5*30/1=61>1.54)、时限整定：64040.8S13、一采区Ⅱ路6406〔CT变比：1000/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*4271/200=25.6A〕 〔取28〕2、带时限过电流淌作电流：g=1724(A) 依据一采区600水平要上综采工作面，金公司机械动力部报来需增加负荷为：1724A.∴Idzj=KkKjxIgh/Khn1=1.2*1*1724/1*200=10.3〔A〕 〔取因要与一采变协作故取11.8A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/28*200/1=1.86<22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*4038/7*200/1=2.5>1.54)、时限整定：64060.8S14、石康铸造6407〔CT变比：150/5〕1、速断动作电流dzj=kjx”d2max/1=1.2*1*7760/30=31A〕 〔取40〕2、带时限过电流淌作电流：g=120(A) 依据区变实际运行负荷得出的值∴Idzj=KkKjxIgh/Khn1=1.2*1*120/1*30=4.8〔A〕 〔15A〕3、灵敏系数校验(1、电流速断保护灵敏度校验=Kmxd”d1min/Idz≥2 依据《继电保护及二次回速查速算手册》依据：Km=KmxdId1min/Idz=0.866*12023/40*30/1=8.6>22=Kmxd”d2n依据：Km=KmxdId2min/Idz=0.866*7023/15*30/1=13.5>1.54)、时限整定：64070.8