虎石台220 60KV降压变电所电气部分初步设计计算书.doc_第1页
虎石台220 60KV降压变电所电气部分初步设计计算书.doc_第2页
虎石台220 60KV降压变电所电气部分初步设计计算书.doc_第3页
虎石台220 60KV降压变电所电气部分初步设计计算书.doc_第4页
虎石台220 60KV降压变电所电气部分初步设计计算书.doc_第5页
已阅读5页,还剩12页未读 继续免费阅读

下载本文档

版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领

文档简介

第一章 短路电流的计算1.1系统计算电路图和等值电路图取sb = 100mva,vb = up,求得等值电路图中各阻抗标幺值如下:发电机1:x2 = x4 = ud%/100sb/s = 12.4/100100/200/0.85 = 0.0527变压器1:x3 = x5 = xd%/100sb/s = 14/100100/240 = 0.0583 发电机2:x6 = x8 =x10= ud%/100sb/s = 18/100100/125/0.85 = 0.1224变压器:x7 = x9 = x11 = xd%/100sb/s = 10.5/100100/160 = 0.0656线路:2100km: x= 1/21000.4100/2202 = 0.041360km: x”= 600.4100/2202 = 0.0496x1 = x + x = 0.0413 + 0.0496 = 0.090948km线路: x12 = 480.4100/2202 = 0.0397260km线路: x13 = 1/2600.4100/2202 = 0.02482100km线路: x14 = 1600.4100/2202 = 0.041380km线路: x15 = 800.4100/2202 = 0.06611.2 220kv母线上k1点发生短路时的短路计算首先进行网络化简,化简后的网络图见图:x1 = x* + x1 = 0.05 + 0.0909 = 0.1409x2 =(x2 + x3)/2 =(0.0527 + 0.0583)= 0.0555x3 = (x6 + x7)/3 =(0.1224 + 0.0656)/3 = 0.0627 x4 = x12 = 0.0397 x5 = x13 = 0.0248 x6 = x15 = 0.0661 x7 = x14 = 0.0413进一步化简如下:对x5 x6 x7 进行三角变换网络见图1-4所示x”1 = x5x7/(x5 + x6 + x7)= 0.02480.0413/(0.0248 + 0.0661 + 0.0413) = 0.00748x”2 = x6x7/(x5 + x6 + x7)= 0.06610.0413/(0.0248 + 0.0661 + 0.0413) = 0.02066x”3 = x5x6/(x5 + x6 + x7)= 0.02480.0661/(0.0248 + 0.0661 + 0.0413) = 0.01240网络见图1-5所示x = x1x2/x1 +x2 = 0.14090.0555/(0.1409 + 0.0555)= 0.03983x“ = (x + x3)x4/(x + x3 + x4)=(0.03983 + 0.04742)0.07507/(0.03983 + 0.04742 + 0.07507)= 0.04035x# = x + x” = 0.02066 + 0.04035 = 0.06101c3 = x”/(x + x3) = 0.04035/(0.03983 + 0.04742)= 0.46250c4 = 1 - c3 = 1 - 0.4625 = 0.5375c1 = c3 x/x1 = 0.46250.02066/0.14091 = 0.13072c2 = c3 - c1 = 0.4625 - 0.13072 = 0.33178x = 0.06101x1 = x/c1 = 0.06101/0.13072 = 0.46674x2 = x/c2 = 0.06101/0.33178 = 0.18389x4 = x/c4 = 0.06101/0.5375 = 0.11351xjs1 = x1sg1/sb = 0.4667411000/100 = 51.3409xjs2 = x2sg2/sb = 0.18389200/0.85/100 = 0.43268xjs4 = x4sg3/sb = 0.11351125/0.85/100 = 0.16692x = x + 1/2(ud%/100)(100/s变)= 0.06101 +(1/2)(14.3/100)(100/120000)= 0.06107x1 = x/c1 = 0.06107/0.13072 = 0.46719x2 = x/c2 = 0.06107/0.33178 = 0.18407x4 = x/c4 = 0.06107/0.5375 = 0.11362xjs1 = x1sg1/sb = 0.4671911000/100 = 51.39103xjs2 = x2sg2/sb = 0.18407200/0.85/100 = 0.43310xjs4 = x4sg3/sb = 0.11362125/0.85/100 = 0.16709d1点短路有名值:i” (3) = 1/xjs1sg1/uav = 1/51.3408911000/230 = 0.5378250s:i”d1(3) = i”+ 2.411sg2/uav + 6.763sg3/uav= 0.537825 + (2.4112002 + 6.7631253)/0.85/(230)= 10.775900.6s:i”d1(3) = i”+ 1.872sg2/uav + 3.359sg3/uav= 0.537825 + (1.8722002 + 3.3591253)/0.85/(230)= 6.969101.2s:i”d1(3) = i”+ 1.925sg2/uav + 2.7945sg3/ = 0.537825 + (1.9252002 + 2.79451253)/0.85/(230)= 5.906554s:i”d1(3) = i”+ 2.119sg2/uav + 2.49sg3/uav= 0.537825 + (2.1192002 + 2.491253)/0.85/(230)= 5.79850d2点短路有名值:i” (3) = 1/xjs1sg1/uav = 1/51.3910311000/63 = 1.961570s:i”d2(3) = i”+ 2.411sg2/uav + 6.02sg3/uav= 1.96157 + (2.4112002 + 6.021253)/0.85/(63)= 36.698540.6s:i”d2(3) = i”+ 1.872sg2/uav + 3.186sg3/uav= 1.96157 + (1.8722002 + 3.1861253)/0.85/(63)= 22.915991.2s:i”d2(3) = i”+ 1.925sg2/uav + 2.827sg3/uav= 1.96157 + (1.9252002+2.8271253)/0.85/(63)= 21.693104s:i”d2(3) = i”+ 2.119sg2/uav + 2.476sg3/uav= 1.96157 + (2.1192002 + 2.4761253)/0.85/(63)= 21.11063冲击电流:d1点:ich1 = 2.55i”0s = 2.5510.77599 = 27.4785d2点:ich2 = 2.55i”0s = 2.5536.69854 = 93.5813d1(3)点处短路容量:s”1 = 23010.77599 = 4292.8345d2(3)点处短路容量:s”1 = 23010.77599 = 4292.8345名称 短路点基准容量(mva)基准电压(kv)等值电抗短路电流有名值冲击电流(ka)短路容量(kw)d1(3)1002200.0610110.775927.47854292.8134d2(3)100630.0617036.698593.58134004.5110第二章 电气设备的选择2.1母线选择2.1.1 220kv侧母线选择 (1)根据最大长期工作电流选择igmax = 1.05 = 1.0594.5/(230) = 316.3(a)故选择lgj400型钢芯铝纹线,ie = 825(a) (2)热稳定校验查当实际环境温度为38时的温度修正系数为k = 0.85,故有f = 0 + (e - 0)( igmax/ie)2 = 38 + (70 - 38)260/(0.85825) 2 = 44.5查表得c值为c = 97.5根据公式:s smin = i/c由以上短路计算得i = 5.7985(ka),iz”= 10.78(ka).”= iz”/ i = 10.78/5.7985 = 1.86td = tb + tgf = 3 + 0.05 = 3.05s查曲线得tz = 3.2stfz = 0.05”2 = 0.051.862 = 0.17stdz = tz + tfz = 3.2 + 0.17 = 3.37s取kf = 1,s smin = i/c = 5798.553.371/2/97.5 = 109.176 mm2300mm2故所选母线合格。2.1.2 63kv侧母线选择(1)根据最大长期工作电流选择igmax1 = 0.71.05 = 0.71.05120/(63) = 59.4(a)igmax2 = 1.05 = 1.05120/(63) = 1154.7(a)故选择1008mm2的矩形铝母线,当母线为立放,正常最高发热温度为700c时的,额定电流ie = 1609(a). (2)热稳定校验查当实际环境温度为38时的温度修正系数为k = 0.85,故有f =0 + (e -0)( igmax/ie)2 = 38 + (70 - 38)1214.134/(0.851609) 2 = 63.2查表得c值为c = 89.5根据公式:s smin = i/c由以上短路计算得i = 21.1106(ka),iz”= 36.6985(ka). ”= iz”/ i = 36.6985/21.1106 = 1.74td = tb + tgf = 3 + 0.05 = 3.05s查曲线得tz = 3.15stfz = 0.05”2 = 0.051.742 = 0.15stdz = tz + tfz = 3.15 + 0.15 = 3.3s取kf = 1,s smin = i/c = 211103.31/2/89.5 = 428.56 mm2800mm2故所选母线合格。(3)动稳定校验:跨距取1.2m两回路相间距离0.75mf = 1.73ish210-7/a = 1.73211102111010-7/0.75 = 220.27(n/m)m = f l2/10 = 220.271.22/10 = 31.72(n/m)w = b2h/6 = 0.0080.0080.1/6 = 1.06710-5(m3) = m/w = 31.72/(1.06710-5) = 29.73106(pa)查手册得硬铝导线母线的最大允许应力a = 70106(pa)a 所以动稳定满足要求。2.2 高压断路器的选择:2.2.1 220kv侧断路器的选择(1)根据最大长期工作电流选择igmax = 1.05s/ue 1.05120000/230= 1.05120000/(230) = 316.2875(a)本设计选用sw7-220型断路器,有关技术数据见表2-1表2-1 lw11-220(p)型断路器有关技术数据表安装地点型号额定电压(kv)最高工作电压(kv)额定电流(a)额定开断电流(ka)额定关合电流(ka)热稳定电流(ka)220kv侧sw7-2202202521600205521(4s)(2)热稳定校验热稳定校验公式:ir2tr i2tdz 在验算电器的短路热效应时,应采用后备保护的动作时间,通常取3s。由以上短路计算得i = 5.7985(ka),iz”= 10.7759(ka)。”= iz”/ i = 10.7759/5.7985 = 1.8584td = tb + tgf = 3 + 0.05 = 3.05s查曲线得tzj = 3.3stfz = 0.05”2 = 0.051.85842 = 0.17268stdz = tz + tfz = 3.3 + 0.17268 = 3.47268s短路发热量为:qk = i2tdz = 5.798523.47268 = 116.76054(ka2s)(3)动稳定校验idw icj;55 2.5510.7759(4)按额定开断电流选择iekd izt; 55 10.7759(5)按额定关合电流选择ieg icj;55 10.7759其保证值与计算值的比较见表22。表22 220kv断路器各项技术数据与各项计算数据比较表计 算 数 据lw11-220p断路器保证值电网电压uew = 220kv额定电压ue = 220kv长期最大工作电流igmax = 316.288a额定电流ie = 1600a次暂态短路电流i(3) = 10.7759ka额定开断电流iekd = 20ka短路冲击电流ich = 27.4785ka额定关合电流iegd = 55ka短路冲击电流ich = 27.4785ka动稳定电流idw = 55ka热效应qd = 116.76ka2.s热稳定:qz = 1764ka2.s2.2.2 63kv侧断路器的选择(1)根据最大长期工作电流选择流过主变低压侧的电流:igmax = 1.05se/uav = 1.05120000/63 = 1154.7a流过最大负荷出线的电流igmax = 1.05se/uav= 1.0535000/(0.8563) = 336.7877a本设计选用lw(ofpi)-63型断路器,有关技术数据见表2-3。表2-3 lw(ofpi)-63型断路器有关技术数据表安装地点型 号额定电压(kv)最高工作电压(kv)额定电流(a)额定开断电流(ka)额定关合电流(ka)额定短时耐受电流(ka)63kv侧lw(ofpi)-636372.520004010031.5(4s)其他校验方法同220kv侧断路器的校验。其保证值与计算值的比较见表24。表24 35kv断路器各项技术数据与各项计算数据比较表计 算 数 据lw(ofpi)-63断路器保证值电网电压uew = 63kv额定电压ue = 63kv长期最大工作电流igmax = 1154.7a额定电流ie = 2000a次暂态短路电流i= 36.6985ka额定开断电流iekd = 40ka短路冲击电流ich = 93.5813ka额定关合电流iegd = 100ka短路冲击电流ich = 93.5813ka动稳定电流idw = 100ka热效应:qk = 1471.1605ka2s热效应:qk = 3669ka2s2.3 隔离开关的选择及校验2.3.1 220kv侧隔离开关选择与校验(1)根据最大长期工作电流选择igmax = 1.05se/uav = 1.051200/230 = 316.2875(a)本设计母线隔离开关选用gw6220gd单柱剪刀型,出线隔离开关选用 gw12220d型 ,有关技术数据见表2-7表2-7 220kv隔离开关有关技术数据表安装地点型 号额定电压(kv)最高工作电压(kv)额定电流(a)动稳定电流(ka)热稳定电流 (ka)220kv侧出线220kv侧母线 gw6220220252160012550(3s)隔离开关的动热、稳定校验方法同220kv侧断路器的动热、稳定校验。其保证值与计算值的比较见表28。表28 220kv隔离开关的各项技术数据与各项计算数据比较表计 算 数 据gw6-220技术数据电网电压uew = 220kv额定电压ue = 220kv长期最大工作电流igmax = 316.2875a额定电流ie = 1600a短路冲击电流ich = 27.4785ka动稳定电流idw = 125ka热效应qd =1471.1605ka2.s热稳定:qz = 3969ka2.s2.3.2 60kv侧隔离开关选择与校验(1)根据最大长期工作电流选择流过主变低压侧的电流:igmax = 1.05se/(ue) = 1.05120000/(63) = 1154.7a流过最大负荷出线的电流:igmax = 1.05se/(ue) = 1.0535000/(0.9563) = 354.5133a本设计选用gw463 型隔离开关,有关技术数据见表2-9。隔离开关的动热、稳定校验方法同220kv侧断路器的动热、稳定校验。其保证值与计算值的比较见表210。表2-9 35kv隔离开关有关技术数据表安装地点型 号额定电压(kv)最高工作电压(kv)额定电流(a)动稳定电流(ka)额定短时耐受电流(ka)63kv侧进线gw4636372.5200010040(4s)63kv侧出线gw4636372.5200010040(4s)表210 63kv隔离开关的各项技术数据与各项计算数据比较表计 算 数 据gw463技术数据电网电压uew = 63kv额定电压ue = 35kv长期最大工作电流igmax = 1154.7a(354.5133a)额定电流ie = 2000a(630a)短路冲击电流ich = 93.5813ka动稳定电流idw = 100ka (50ka)热效应qd = 1471.1605ka2.s热稳定:qz = 6600ka2.s(3969 ka2.s)经以上校验,所选隔离开关符合要求。2.4 电流互感器的选择及校验2.4.1 220kv侧电流互感器的选择及校验220kv侧的进出线上电流互感器均用lcwb220w (3935w)电网电压ue = 220kv最大长期工作电流:igmax = 1.05se/(ue) = 1.05120000/(230) = 316.29a 220kv侧的进出线上电流互感器均用lcwb220w (3935w)l电流互感器; c瓷绝缘; w户外型;b保护级; 7设计序号; 220额定电压(kv);w1适用于中污秽地区有关技术数据见表2-11。表211 220kv侧电流互感器有关技术数据表安装地点型 号额定电压(kv)最高工作电压(kv)额定变流(a)二次组合额定动稳定电流(ka)额定短时热电流(ka)220kv侧 lb220w(3935w)220252400/55p/5p/5p/0.5/0.25521(5s)热稳定检验:(ktie1)21 i2tdz;(211.2)25 116.76动稳定校验:icj ielkdw; 2.5510.7759 1.255其保证值与计算值的比较见表212。表212 220kv电流互感器的技术数据与计算数据比较表计 算 数 据lcwb7-220w1的技术数据电网电压uew = 220kv额定电压ue = 220kv长期最大工作电流igmax = 316.2875a一次额定电流i1e = 2*300a准确度级:0.5 准确度:5p,0.5,0.2短路冲击电流ich = 27.4785ka额定动稳定电流55 ka热效应qd = 116.76ka2.s热稳定:qz = 3175.2ka2.s2.4.2 63kv侧电流互感器的选择及校验电网电压ue=63kv流过主变低压侧的电流:igmax = 1.05se/(ue) = 1.05120000/(63) = 1154.70a流过最大负荷出线的电流:igmax = 1.05se/(ue) = 35000/(63) = 320.75a63kv侧的进出线上电流互感器均用lcwb563, 有关技术数据见表2-13。表213 63kv侧电流互感器有关技术数据表安装地点型 号额定电压(kv)最高工作电压(kv)额定变流比(a)准确级额定动稳定电流1s短时热电流63kv侧进线lcwb56363691500/50.51255063kv侧出线lcwb5636369400/5b10040热稳定检验:(ktie1)21 i2tdz;(501.5)21 1471.1605(401.5)21 1471.1605动稳定校验:icj ielkdw; 2.5593.5813 1.51252.5593.5813 1.5125其保证值与计算值的比较见表214。表214 63kv电流互感器的技术数据与计算数据比较表计 算 数 据lcwb5-63的技术数据(变压器侧)lcwb5-63的技术数据(负荷侧)电网电压uew = 63kv额定电压ue = 63kv额定电压ue = 63kv长期最大工作电流 igmax = 1154.7a额定电流ie = 1500a额定电流ie = 400a准确度级:0.5 ,b准确度:0.5,b准确度:0.5,b短路冲击电流ich =93.5813ka动稳定电流倍数125动稳定电流倍数100热效应qd = 1471.1605ka2.s热稳定值5625 ka2.s热稳定值3600ka2.s2.5 电压互感器选择及校验1) 按装置种类及型式选择电压互感器的种类及型式应根据安装地点和使用条件进行选择, 60千伏配电装置采用jdcf63型电压互感器;220kv采用jdcf220(gyw2)型串级磁绝缘式电压互感器。为检查和监视一次回路单相接地,所有电压互感器选用具有第三绕组的单相电压互感器组。2) 按一次回路电压选择 1.1uel u1 0.9uel式中:u1电网电压;uel电压互感器一次绕组额定电压3) 按二次回路电压选择所选电压互感器二次侧额定电压均选择100/ 3伏, jdcf63型电压互感器第三绕组电压选用100/ 3伏,jccf220(gyw2)型串级磁绝缘式电压互感器第三绕组电压选用100伏。 本设计选用电压互感器有关技术数据见表2-17。表217 电压互感器有关技术数据表型式额定变比准确级额定容量(va)0.20.513jdc-220(gyw2)220/100/1001503005005002000型式额定变比准确级额定容量(va)0.20.53p3pjdcf-6366/0.1/0.1/0.1/35010040010020002.6 避雷器的选择在选用避雷器时,应保证避雷器安装点的工频电压升高在任何情况下都不会超过灭弧电压,否则避雷器可因不能灭弧而爆炸,对单纯防雷器来说,只需考虑系统单相接地非故障相对地电压升高,这一升高显然与系统中性点接地方式有关,选定避雷器型号及技术数据见表218表218 避雷器型号及技术数据型号额定电压(kv)灭弧电压有效值(kv)工频电压(kv)冲击电流残压(kv)不大于不小于不大于5kv10kvfcz-606069117133178122fcz-220j220200448536664728fcz-110j110100170195260260注:fcz110j(为磁吹阀式避雷器,j系统为中性点接地时数字为标称电流最大残压下最大残压kv),用于变压器中性点。第三章 避雷针的保护范围计算3.1 避雷针的定位及针距按照规程规定,首先确定避雷针安放的地点。在220kv侧进出线构架上安装三支避雷针,在主变压器附近两侧安装三支独立避雷针。在60kv出线侧安装三支独立避雷针。合计共有九支避雷针。由于220kv侧共有9个间隔,且软母线的跨距一般在3040米,所以在220kv配电装置中共设六个母线构架,中间跨距为915 = 135米,在两边和中间的其中一个构架上装设避雷针。60kv侧共有20个间隔,每个间隔宽度为5米,共需205 = 100米。 本变电所220kv和60kv配电装置设计防雷保护的范围为135126,利用9根避雷针进行保护。9根避雷针的定位及针距见图3-1。图3-1 9根避雷针的定位及针距根据电气工程电气设计手册第一册第十五章第一节中关于避雷针保护范围计算的有关内容,要求针高h d/7p + hx,220kv侧被保护物最大高度为15米,确定避雷针高度时针间距离d取最大,经计算h = 99.1/7 + 15 = 27.57,选220kv侧针高h = 30米;60kv侧被保护物最大高度为7.5米,确定避雷针高度时针间距离d取最大,经计算h = 86.1/7 + 7.5 = 18.41,选60kv侧针高h = 20米。3.2 单根避雷针的保护半径计算1) 针高30米时单根避雷针在hx = 15m的水平面上保护半径 rx为:hx h/2 , rx = h - hx = 3015 = 15m单根避雷针在hx = 10.5m的水平面上保护半径 rx为:hx h/2, rx = 1.5h - 2hx = 4521 = 24m2) 针高20米时单根避雷针在hx = 7.5m的水平面上保护半径 rx为:hx h/2 , rx = rx = 1.5h - 2hx = 3015 = 15m单根避雷针在hx = 5.6m的水平面上保护半径 rx为:hx h/2, rx = 1.5h - 2hx = 30 - 11.2 =1 8.8m3.3 多根等高避雷针的保护范围计算1) 针高30米时a. 12,23针之间,d12 = d23 = 67.5m, h0 = h-d/7p = 30 - 67.5/7 = 20.36m,在hx = 15m的水平面上:hx h0/2 , bx12 = bx23 = h0 - hx = 20.36 - 15 = 5.36m在hx =10.5m的水平面上:hx h0/2, bx12 = bx23 = h0 - hx = 20.36 - 10.5 = 9.86mb. 14,36针之间,d14 = d36 = 72.5m,h0 = h - d/7p = 30 - 72.5/7 = 19.6m在hx = 15m的水平面上:hx h0/2 , bx14 = bx36 = h0 - hx = 19.615 = 4.6m在hx = 10.5m的水平面上:hx h0/2, bx14 = bx36 = h0 - hx = 19.6 - 10.5 = 9.1mc. 24针之间,d24 = 88m,h0 = h - d/7p = 30 - 99.1/7 = 15.8m在hx = 15m的水平面上:hx h0/2 , bx24 = h0 - hx = 15.815 = 0.8m在hx =10.5m的水平面上:hx h0/2, bx24 = h0 - hx = 15.8 - 10.5 = 5.3md. 45针之间,d45 = 76m,h0 = h - d/7p = 30 -76/7 = 19.1m在hx =15m的水平面上:hx h0/2 , bx45 = h0 - hx = 19.1 - 15 = 4.1m在hx = 10.5m的水平面上:hx h0/2, bx45 = h0 - hx = 19.1 - 10.5 = 8.6me. 56针之间,d56 = 59m,h0 = h - d/7p = 30 - 59/7 = 21.6m在hx =15m的水平面上:hx h0/2 , bx56 = h0 - hx = 21.6 - 15 = 6.6m在hx = 10.5m的水平面上:hx h0/2, bx56 = 1.5h0 - 2hx = 32.421 = 11.4mf. 35针之间,d35 = 93.5m,h0 = h - d/7p = 30 - 93.5/7 = 16.6m在hx = 15m的水平面上:hx h0/2 , bx53 = h0 hx = 16.615 = 1.6m在hx = 10.5m的水平面上:hx h0/2 , bx53 = h0 - hx = 16.6 - 10.5 = 6.1m2) 针高20米时 78,89针之间,d78 = d89 = 50m, h0 = h - d/7p = 20 - 50/7 = 12.86m,在hx = 7.5m的水平面上:hx h0/2 , bx78 = bx89 = h0hx = 12.86 - 7.5 = 5.36m在hx =5.6m的水平面上:hx h0/2

温馨提示

  • 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
  • 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
  • 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
  • 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
  • 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
  • 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
  • 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。

最新文档

评论

0/150

提交评论