............專業(yè)專注..專業(yè)專注一、主運帶巷機電硐室移動變電站的整定計算N負荷統(tǒng)計：∑P=340KWN1〕、長時工作電流：I=K∑P/〔√3VcosΦ 〕ca de N N wm=0.54*340/(1.732*6.3*0.85)=19.8Ase 確定整定值I=1.2*I=1.2*19.8≈23.8se 2〕、短路保護整定sb.r st.m 由過載保護確定短路保護整定值 I =(1.2~1.4)*(I +∑Isb.r st.m 〔22+19〕=41A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/800=0.000595ΩSY S1 b UGSP 3*50 1460mR和電阻XRRL/K1 1 011 SC=0.1*1.46/〔6/0.69〕2=0.002ΩX=X*L/K21 01 1 SC=0.06*1.46/〔6/.69〕2=0.001Ωc 變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.63)=0.0334ΩT s ar.TR=△p V2

N.T=4000×0.692/8002=0.005ΩT N.T

N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 d UPQ–1.14 3*70 1400mR和電阻XRRL3 2 033X=X*L=0.078*1.4=0.1092Ω2 03 33 e UP–1.14 3*16 10m電纜的電阻R和電阻XRR3 3 03X=X*L=0.090*0.01=0.0009Ω3 03最遠點電機機的三相短路電流d=690/〔1.732*√0.39932+0.13472 〕=945A兩相短路電流I〔2〕=0.87*I〔3〕=0.87*945=822Ad df 靈敏度校驗Kr=I〔2〕 /I=822/41d sb=20＞1.2 符合要求〔二〕、移動變電站低壓饋電開關(guān)的過載保護整定1〕、長時工作電流：I=K∑P/〔√3VcosΦ 〕ca de N N wm=0.54*340/(1.732*0.69*0.85)=180A確定整定值Ise=1.2*Ica.=1.2*106≈216A2〕、短路保護整定由過載保護確定短路保護整定值 Isb.r=(1.2~1.4)*(Ist.m+∑IN)=1.2*〔169+160〕=329A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/800=0.000595ΩSY S1 b UGSP 3*50 1460mR和電阻1 R=RL/K2 =0.1*1.46/〔6/0.69〕2=0.002Ω1 011 SCX=X*L/K2 =0.06*1.46/〔6/.69〕2=0.001Ω1 01 1 SCc 變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.63)=0.0334ΩT s ar.TR=△p V2

N.T=4000×0.692/8002=0.005ΩT N.T

N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω△pN.T：變壓器功率損耗，4000Wvs〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕d UPQ1.14 3*70 1400m電纜的電阻R3X3R2=R03L3=0.267*1.4=0.3738ΩX2=X03*L3=0.078*1.4=0.1092Ωe UP–1.14 3*16 10m電纜的電阻R3和電阻X3R3=R03L3=1.16*0.01=0.0116ΩX3=X03*L3=0.090*0.01=0.0009Ω最遠點電機機的三相短路電流=690/〔1.732*√0.39932+0.13472 〕=945A兩相短路電流I〔2〕=0.87*I〔3〕=0.87*945=822Ad df 靈敏度校驗Kr=I〔2〕 /I=822/329d sb=2.5＞1.2 符合要求1、KJ25-400饋電整定計算短路保護整定I≥I +∑I=5*185+330=515Asb st.m N靈敏度校驗a X =V2ar/S=0.692/800=0.000595ΩSY Sb UGSP 3*95 60m電纜的電阻R1

和電阻X1R=RL/K2 =0.1*0.046/〔6/0.69〕2=0.002Ω1 011 SCX=X*L/K2 =0.06*0.046/〔6/.69〕2=0.001Ω1 01 1 SCc 變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.63)=0.0334ΩT s ar.TR=△p V2

N.T=4000×0.692/8002=0.005ΩT N.T

N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω△pN.T：變壓器功率損耗，4000Wvs〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕3 d UPQ1.14 3*70 1400m電纜的電阻R3 R=RL=0.267*1.42 022=0.3738ΩX=X*L=0.078*1.4=0.1092Ω2 02 23 e UP–1.14 3*16 10m電纜的電阻R和電阻XRRL3 3 033X=X*L=0.090*0.01=0.0009Ω3 03 3最遠點電機機的三相短路電流d=690/〔1.732*√0.39932+0.13472 〕=945A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*945=822 Af 靈敏度校驗Kr=I〔2〕d/Isb=822/254=3.2＞1.2 符合要求2、 掌握中間巷DW80-200饋電整定計算(1)、短路保護整定Isb≥Ist.m+∑IN=5*41.72+75=284A〔2〕、靈敏度校驗a XSY=V2ar/SS=0.692/800=0.000595ΩUGSP 3*50 1460m電纜的電阻R1和電阻X1R1=R01L1/K2SC=0.1*1.46/〔6/0.69〕2=0.002ΩX=X*L/K2 =0.06*1.46/〔6/.69〕2=0.001Ω1 01 1 SC變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.63)=0.0334ΩT s ar.TR=△p V2

N.T=4000×0.692/8002=0.005ΩT N.T

N.TX=√Z2-R2 =√0.03342-0.0052=0.0331ΩT T T△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 UPQ1.14 3*70 1000m電纜的電阻RXRRL3 2 022X=X*L=0.078*1.0=0.078Ω2 02 23 UP–1.14 3*16 10m電纜的電阻R和電阻XRRL3 3 033X=X*L=0.090*0.01=0.0009Ω3 03 3最遠點電機機的三相短路電流d=690/〔1.732*√0.28562+0.11422 〕=1295A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*1295=1127 A靈敏度校驗Kr=I〔2〕d/Isb=1127/284=3.97＞1.2 符合要求3、QBZ-80開關(guān)整定計算過載保護整定N 由拉緊電機額定電流I=8.7A，IN 短路保護整定選整定值為Isb。r=4I=4×9=36A校驗短路保護的靈敏度系統(tǒng)電抗值X =V2ar/S=0.692/800=0.000595ΩSY S1 UGSP 3*50 1460m電纜的電阻R和電阻XRRL/K1 1 011 SC=0.1*1.46/〔6/0.69〕2=0.0019ΩX=X*L/K21 01 1 SC=0.06*1.46/〔6/.69〕2=0.00116Ω變壓器的電阻和電抗值Z=vV2 /(100S

)=4.43*0.692/(100*0.63)=0.0334T s ar.T N.TR=△p V2 /S2 =4000×0.692/8002=0.005ΩT

ar.T

N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0331ΩUPQ–1.14 3*70 200mR3和電阻X3R2=R02L2=0.267*0.2=0.0534ΩX2=X02*L2.=0.078*0.2=0.0156Ω3 UP–660 3*16 20m電纜的電阻R和電阻3 R=RL=1.16*0.02=0.0232Ω3 033X=X*L=0.09*0.02=0.0018Ω3 03 3最遠點的三相短路電流d=690/〔1.732*√0.08072+0.02052〕=2762A兩相短路電流I〔2〕=0.87*I〔3〕=0.87*2762=2403Ad d靈敏度校驗Kr=I〔2〕 /I=2403/168d sb=14.3＞1.2 合格負荷統(tǒng)計：∑PN=200KW〔一〕、移動變電站高壓饋電開關(guān)的過載保護整定長時工作電流：IcaKde∑PN/〔√3VNcosΦwm〕=0.54*200/(1.732*6.3*0.85)=12A確定整定值Ise=1.2*Ica.=1.2*12≈14.4A2〕、短路保護整定由過載保護確定短路保護整定值 Isb.r=(1.2~1.4)*(Ist.m+∑IN)=1.3*〔49+12〕=79A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/500=0.0009522ΩSY S1b 3*50和UGSP-103*35 3600m電纜的電阻R和電阻1X1R=RL/K2 =0.1*3.6/〔6/0.69〕2=0.0047Ω1 011 SCX=X*L/K2 =0.06*3.6/〔6/.69〕2=0.0028Ω1 01 1 SCc 變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.50)=0.0422ΩT s ar.TR=△p V2

N.T=4000×0.692/5002=0.00762ΩT N.T

N.TXT=√ZT2-R2T=√0.04222-0.007622=0.0415Ω△pN.T：變壓器功率損耗，4000Wvs〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕d UPQ1.14 3*70 1500m電纜的電阻R3X3R2=R02L2=0.267*1.5=0.4005ΩX2=X02*L2=0.078*1.5=0.117Ωe UP–1.14 3*6 10m電纜的電阻R3和電阻X3R3=R03L3=1.16*0.01=0.0116ΩX3=X03*L3=0.090*0.01=0.0009Ω最遠點電機機的三相短路電流=690/〔1.732*√0.42442+0.16342 〕=876A兩相短路電流I〔2〕=0.87*I〔3〕=0.87*876=762Ad df 靈敏度校驗Kr=I〔2〕 /I=762/79d sb=9.6＞1.2 符合要求〔二〕、移動變電站低壓饋電開關(guān)的過載保護整定1)、長時工作電流：I=K∑P/〔√3VcosΦ 〕ca de N N wm=0.54*200/(1.732*0.69*0.85)=106Ase 確定整定值I=1.2*I=1.2*106≈128A2〕、se sb.r st.m 由過載保護確定短路保護整定值 I =(1.2~1.4)*(I +∑Isb.r st.m 〔95+123〕=218A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/500=0.0009522ΩSY Sb 3*50和UGSP-103*35 3600m電纜的電阻R1

和電阻X1R=RL/K2 =0.1*3.6/〔6/0.69〕2=0.0047Ω1 011 SCX=X*L/K2 =0.06*3.6/〔6/.69〕2=0.0028Ω1 01 1 SCc 變壓器的電阻和電抗值Z=vV2 /(100S )=4.43*0.692/(100*0.5)=0.0422ΩT s ar.T N.TR=△p V2

=4000×0.692/5002=0.00762ΩT N.T

N.TX=√Z2-R2 =√0.03342-0.0052=0.0415ΩT T T△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 d UPQ1.14 3*70 1500m電纜的電阻RXRRL3 2 022X=X*L=0.078*1.5=0.117Ω2 02 23 e UP–1.14 3*6 10m電纜的電阻R和電阻XRRL3 3 033X=X*L=0.090*0.01=0.0009Ω3 03 3最遠點電機機的三相短路電流d=690/〔1.732*√0.42442+0.16342 〕=876A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*876=762Af 靈敏度校驗Kr=I〔2〕d/Isb=762/218=3.5＞1.2 符合要求、開關(guān)整定計算1、45－21DW80-350饋電的整定計算：(1)、短路保護整定Isb≥Ist.m+∑IN=5*95+32=507A〔2〕、靈敏度校驗a X =V2ar/S=0.692/500=0.0009522ΩSY S1 b UGSP 3*50 1460m電纜的電阻R和電阻1 R=RL/K2 =0.1*1.46/〔6/0.69〕2=0.002Ω1 011 SCX=X*L/K2 =0.06*1.46/〔6/.69〕2=0.001Ω1 01 1 SC1 C UGSP-10 3*35 2200mR和電阻1 R=RL/K2 =0.1*2.2/〔6/0.69〕2=0.003174Ω2 022 SCX=X*L/K2 =0.06*2.2/〔6/.69〕2=0.0019Ω2 02 2 SCd 變壓器的電阻和電抗值Z=vV2 /(100S )=4.43*0.692/(100*0.63)=0.0334ΩT s ar.T N.TR=△p V2

=4000×0.692/5002=0.005ΩT N.T

N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω△pN.T：變壓器功率損耗，4000Wvs〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕e UPQ–1.14 3*70 1500m電纜的電阻R3和電阻X3X3=X03*L3=0.078*1.5=0.117Ωf UPQ–1.14 3*6 10mR3和電阻X3R4=R04L4=1.16*0.01=0.0116ΩX4=X04*L4=0.090*0.01=0.0009Ω最遠點的三相短路電流d=690/〔1.732*√0.42222+0.15512〕=886 A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*886=770Ag 靈敏度校驗Kr=I〔2〕d/Isb=770/507=1.50＞1.2 符合要求2、3130646－70BKD4-400Z/1140.660饋電整定計算：負荷統(tǒng)計：∑PN=40KW、過載保護整定ISe＝Kco∑IN/Kre=1.15*49/0.85=66A過載保護牢靠系數(shù)，1.2;，0.85∑IN 饋電所帶負荷額定電流之和。2〕、短路保護整定Isb≥Ist.m+∑IN=5*5+45=70A3〕、靈敏度校驗a XSY=V2ar/SS=0.692/500=0.0009522Ωb UGSP 3*50 1460mR1和電阻X1R1=R01L1/K2SC=0.1*1.46/〔6/0.69〕2=0.002ΩX1=X01*L1/K2SC=0.06*1.46/〔6/.69〕2=0.001Ω1 C UGSP-10 3*35 2200mR和電阻1 R=RL/K2 =0.1*2.2/〔6/0.69〕2=0.003174Ω2 022 SCX=X*L/K2 =0.06*2.2/〔6/.69〕2=0.0019Ω2 02 2 SC變壓器的電阻和電抗值Z=vV2 /(100S )=4.43*0.692/(100*0.63)=0.0334ΩT s ar.T N.TR=△p V2

=4000×0.692/5002=0.005ΩT N.T

N.TX=√Z2-R2 =√0.03342-0.0052=0.0331ΩT T T△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 UPQ–1.14 3*70 1350m電纜的電阻R和電阻XRRL3 3 033X=X*L=0.078*1.35=0.1053Ω3 03 33 UPQ–1.14 3*6 10mR和電阻XRRL3 4 044X=X*L=0.090*0.01=0.0009Ω4 04 4最遠點的三相短路電流d=690/〔1.732*√0.38222+0.14342〕=693A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*693=603A靈敏度校驗Kr=I〔2〕 /I =603/70d sb=8.6＞1.2 符合要求3、0623QJZ-300/1140開關(guān)整定計算、過載保護整定N 由水泵的電機額定電流I=114A，確定整定值為I=115A2〕、N 選整定值為Isb。r=4I=4×114=456A校驗短路保護的靈敏度a 系統(tǒng)電抗值X =V2ar/S=0.692/500=0.0009522ΩSY S1 b UGSP-6 3*50 1460mR和電阻1 R=RL/K2 =0.1*1.46/〔6/0.69〕2=0.0019Ω1 011 SCX=X*L/K2 =0.06*1.46/〔6/.69〕2=0.0012Ω1 01 1 SC1 C UGSP-10 3*35 2200mR和電阻1 R=RL/K2 =0.1*2.2/〔6/0.69〕2=0.003174Ω2 022 SCX=X*L/K2 =0.06*2.2/〔6/.69〕2=0.0019Ω2 02 2 SC變壓器的電阻和電抗值Z=vV2 /(100S )=4.43*0.692/(100*0.63)=0.0334ΩT s ar.T N.TR=△p V2

=4000×0.692/5002=0.005ΩT N.T

N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω△pN.T：變壓器功率損耗，4000Wv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 UPQ–1.14 3*70 1350m電纜的電阻R和電阻XRRL3 3 033X=X*L=0.078*1.35=0.105Ω3 03 33 UPQ–1.14 3*70 10mR和電阻XRRL3 4 044X=X*L=0.078*0.01=0.00078Ω4 04 4最遠點的三相短路電流d=690/〔1.732*√0.3732+0.14242〕=998A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*998=868A靈敏度校驗Kr=I〔2〕d/Isb=868/456=1.90＞1.2 符合要求4、 06回順45聯(lián)巷QC83-80開關(guān)整定計算過載保護整定IN=41.72A，ISe=42A，依據(jù)實48A。短路保護整定選整定值為Isb。r=5Ise=5×42=210A校驗短路保護的靈敏度系統(tǒng)電抗值X =V2ar/S=0.692/500=0.0009522ΩSY S1 UGSP 3*50 1460m電纜的電阻R和電阻1 R=RL/K2 =0.1*1.46/〔6/0.69〕2=0.0019Ω1 011 SCX=X*L/K2 =0.06*1.46/〔6/.69〕2=0.00116Ω1 01 1 SC1 C UGSP-10 3*35 2200mR和電阻1 R=RL/K2 =0.1*2.2/〔6/0.69〕2=0.003174Ω2 021 SCX=X*L/K2 =0.06*2.2/〔6/.69〕2=0.0019Ω2 02 1 SCd 變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.63)=0.0334T s ar.TR=△p V2

N.T=4000×0.692/5002=0.005ΩT N.T

N.TX=√Z2-R2 =√0.03342-0.0052=0.0331ΩT T T3 f UP–660 3*16 60mR和電阻XRRL3 3 033X=X*L=0.09*0.06=0.0054Ω3 03 3最遠點的三相短路電流d=690/〔1.732*√0.076872+0.01062〕=5139兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*5139=4471Af 靈敏度校驗Kr=I〔2〕 /I=4471/48d sb=93＞1.2 符合要求N負荷統(tǒng)計：∑P=187KWN〔一〕、移動變電站高壓饋電開關(guān)的過載保護整定=K∑P/〔√3VcosΦ 〕ca de N N wm=0.54*187/(1.732*6.3*0.85)=11Ase 確定整定值I=1.2*I=1.2*11≈13.2A2〕、se sb.r st.m 由過載保護確定短路保護整定值 I =(1.2~1.4)*(I +∑Isb.r st.m 〔60+9〕=90A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/630=0.000756ΩSY S1UGSP-63*50和UGSP-103*35 5000m電纜的電阻R和電阻1X1R=RL/K2 =0.1*5.0/〔6/0.69〕2=0.0066Ω1 011 SCX=X*L/K2 =0.06*5.0/〔6/.69〕2=0.0040Ω1 01 1 SC變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.50)=0.0422ΩT s ar.TR=△p V2

N.T=4000×0.692/6302=0.00762ΩT N.T

N.TX=√Z2-R2 =√0.04222-0.007622=0.0415ΩT T T△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 UPQ1.14 3*70 900m電纜的電阻RXRRL3 2 022X=X*L=0.078*0.9=0.0702Ω2 02 23 UP–1.14 3*16 120m電纜的電阻R和電阻XRRL3 3 033X=X*L=0.090*0.12=0.0108Ω3 03 3最遠點電機機的三相短路電流d=690/〔1.732*√0.39372+0.12772兩相短路電流I〔2〕=0.87*I〔3〕=0.87*963=838Ad d靈敏度校驗Kr=I〔2〕 /I =838/90d sb=9.3＞1.2 符合要求〔二〕、移動變電站低壓饋電開關(guān)的過載保護整定

〕＝963A1)、長時工作電流：I=K∑P/〔√3VcosΦ 〕ca de N N wm=0.54*187/(1.732*0.69*0.85)=99A確定整定值Ise=1.2*Ica.=1.2*106≈119A2〕、短路保護整定sb.r st.m 由過載保護確定短路保護整定值 I =(1.2~1.4)*(I +∑Isb.r st.m 〔480+70〕=660A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/630=0.000756ΩSY S1b UGSP-63*50和UGSP-103*35 5000m電纜的電阻R和電阻1X1R=RL/K2 =0.1*5.0/〔6/0.69〕2=0.0066Ω1 011 SCX=X*L/K2 =0.06*5.0/〔6/.69〕2=0.0040Ω1 01 1 SCc 變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.5)=0.0422ΩT s ar.TR=△p V2

N.T=4000×0.692/6302=0.00762ΩT N.T

N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0415Ω△pN.T：變壓器功率損耗，4000Wvs〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕d UPQ1.14 3*70 900m電纜的電阻R3X3R2=R02L2=0.267*0.9=0.2403ΩX2=X02*L2=0.078*0.9=0.0702Ωe UP–1.14 3*16 120mR3和電阻X3R3=R03L3=1.16*0.12=0.1392ΩX=X*L=0.090*0.12=0.0108Ω3 03 3最遠點電機機的三相短路電流d=690/〔1.732*√0.39372+0.12732 〕＝963A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*963=838Af 靈敏度校驗Kr=I〔2〕d/Isb=838/660=1.3＞1.2 符合要求：1、76110KWBKD6—400Z/1140(660)F饋電的整定計算：負荷統(tǒng)計：∑PN=110KW、過載保護整定ISe＝Kco∑IN/Kre=1.2*120/0.85=169A過載保護牢靠系數(shù)，1.2;，0.85∑IN 饋電所帶負荷額定電流之和。2〕、短路保護整定Isb≥Ist.m+∑IN=5*120=600A3〕、靈敏度校驗a XSY=V2ar/SS=0.692/630=0.00075612Ω1b UGSP-63*50和UGSP-103*35 5000m電纜的電阻R和電阻1X1R=RL/K2 =0.1*5.0/〔6/0.69〕2=0.0066Ω1 011 SCX=X*L/K2 =0.06*5.0/〔6/.69〕2=0.0040Ω1 01 1 SCd 變壓器的電阻和電抗值Z=vV2 /(100S )=4.43*0.692/(100*0.63)=0.0334ΩT s ar.T N.TR=△p V2

=4000×0.692/6302=0.005ΩT N.T

N.TX=√Z2-R2 =√0.03342-0.0052=0.0331ΩT T T△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 e UPQ–1.14 3*70 400mR和電阻XRRL3 2 022X=X*L=0.078*0.4=0.0312Ω2 02 2最遠點的三相短路電流d=690/〔1.732*√0.11842+0.06912〕=2906A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.872906=2528Ag 靈敏度校驗Kr=I〔2〕d/Isb=2528/600=4.2＞1.2 符合要求2、70聯(lián)巷――切眼電源KBZ—400/1140(660)Z饋電的整定計算：N負荷統(tǒng)計：∑P=57KW1)、過載保護整定NI＝K∑I/K=1.2*72/0.85=102ASe co N reK----過載保護牢靠系數(shù)，1.2;co-----過載保護的返回系數(shù)，0.85re∑I 饋電所帶負荷額定電流之和。N2〕、短路保護整定I≥I +∑I=5*42+30=240Asb st.m N3〕、靈敏度校驗a X =V2ar/S=0.692/630=0.000756ΩSY S1b UGSP-63*50和UGSP-103*35 5000m電纜的電阻R和電阻1X1R=RL/K2 =0.1*5.0/〔6/0.69〕2=0.0066Ω1 011 SCX=X*L/K2 =0.06*5.0/〔6/.69〕2=0.0040Ω1 01 1 SC變壓器的電阻和電抗值Z=vV2 /(100S )=4.43*0.692/(100*0.63)=0.0334ΩT s ar.T N.TR=△p V2

=4000×0.692/6302=0.005ΩT N.T

N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω△pN.T：變壓器功率損耗，4000Wv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 UPQ1.14 3*70 900m電纜的電阻RXRRL3 2 022X=X*L=0.078*0.9=0.0702Ω2 02 23 UP–1.14 3*16 120mR和電阻XRRL3 3 033X=X*L=0.090*0.12=0.0108Ω3 03 3最遠點的三相短路電流d=690/〔1.732*√0.39112+0.11892〕=966A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*966=840Ag 靈敏度校驗Kr=I〔2〕d/Isb=840/240=3.5＞1.2 符合要求3、0676QJZ-300/1140開關(guān)整定計算、過載保護整定由水泵的電機額定電流IN=116A，ISe=120A,依據(jù)實130A。2〕、短路保護整定選整定值為Isb。r=5Ise=5×116=580A校驗短路保護的靈敏度系統(tǒng)電抗值X =V2ar/S=0.692/630=0.000756ΩSY S1UGSP-63*50和UGSP-103*35 5000m電纜的電阻R和電阻1X1R=RL/K2 =0.1*5.0/〔6/0.69〕2=0.0066Ω1 011 SCX=X*L/K2 =0.06*5.0/〔6/.69〕2=0.0040Ω1 01 1 SC變壓器的電阻和電抗值Z=vV2

/(100S

)=4.43*0.692/(100*0.63)=0.0334ΩT s ar.TR=△p V2

N.T=4000×0.692/6302=0.005ΩT

ar.T

N.TX=√Z2-R2 =√0.03342-0.0052=0.0331ΩT T T△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 UPQ1.14 3*70 400m電纜的電阻RXRRL3 2 022X=X*L=0.078*0.4=0.0312Ω2 02 2最遠點的三相短路電流d=690/〔1.732*√0.11842+0.06912〕=2906A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*2906=2529Ag 靈敏度校驗Kr=I〔2〕 /I =2529/580d sb=4.4＞1.2 符合要求4、 06回順84聯(lián)巷QC810-60開關(guān)整定計算、過載保護整定N 由水泵電機額定電流I=41.72A，I=42A,依據(jù)實際N 2〕、短路保護整定選整定值為Isb。r=5I=5×42=210A校驗短路保護的靈敏度系統(tǒng)電抗值X =V2ar/S=0.692/630=0.000756ΩSY S1UGSP-63*50和UGSP-103*35 5000m電纜的電阻R和電阻1X1R=RL/K2 =0.1*5.0/〔6/0.69〕2=0.0066Ω1 011 SCX=X*L/K2 =0.06*5.0/〔6/.69〕2=0.0040Ω1 01 1 SC變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.63)=0.0334T s ar.T R=△p V2

/S2 =4000×0.692/6302=0.005ΩT

ar.T N.TXT=√ZT2-R2T=√0.03342-0.0052=0.0331ΩUPQ–1.14 900mR3和電阻X3R=RL=0.267*0.9=0.2403Ω2 022X=X*L=0.078*0.9=0.0702Ω2 02 23 UP–660 3*16 10m電纜的電阻R和電阻XRRL3 3 033X=X*L=0.09*0.01=0.0009Ω3 03 3最遠點的三相短路電流d=690/〔1.732*√0.25852+0.077062〕=1477A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*1477=1285A靈敏度校驗Kr=I〔2〕d/Isb=1285/210=6.1＞1.2 符合要求六、07回順1聯(lián)巷移動變電站的整定計算負荷統(tǒng)計：∑PN=157KW〔一〕、移動變電站高壓饋電開關(guān)的過載保護整定長時工作電流：IcaKde∑PN/〔√3VNcosΦwm〕=0.54*157/(1.732*6.3*0.85)=10A確定整定值Ise=1.2*Ica.=1.2*10≈12A2〕、短路保護整定由過載保護確定短路保護整定值 Isb.r=(1.2~1.4)*(Ist.m+∑IN)=1.2*〔168+150〕*0.69/6.3=37A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/500=0.0009522ΩSY S1 b UGSP-6 3*50 1500m電纜的電阻R和電阻1 R=RL/K2 =0.1*1.5/〔6/0.69〕2=0.00198Ω1 011 SCX=X*L/K2 =0.06*1.5/〔6/.69〕2=0.00119Ω1 01 1 SCc 變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.50)=0.0422ΩT s ar.TR=△p V2

N.T=4000×0.692/5002=0.00762ΩT N.T

N.TXT=√ZT2-R2T=√0.04222-0.007622=0.0415Ω△pN.T：變壓器功率損耗，4000Wvs〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕d UPQ1.14 3*70 2400m電纜的電阻R3X3R2=R02L2=0.267*2.4=0.6408ΩX2=X02*L2=0.078*2.4=0.1872Ωe UP–1.14 3*6 5mR3和電阻X3R3=R03L3=1.16*0.005=0.0058ΩX3=X03*L3=0.090*0.005=0.00045Ω最遠點電機機的三相短路電流=690/〔1.732*√0.65622+0.23082 〕=573A兩相短路電流I〔2〕=0.87*I〔3〕=0.87*573=499Ad df 靈敏度校驗Kr=I〔2〕 /I=499/37d sb=13＞1.2 符合要求〔二〕、移動變電站低壓饋電開關(guān)的過載保護整定1)、長時工作電流：I=K∑P/〔√3VcosΦ 〕ca de N N wm=0.54*157/(1.732*0.69*0.85)=84Ase 確定整定值I=1.2*I=1.2*84≈101A2〕、se sb.r st.m 由過載保護確定短路保護整定值 I =(1.2~1.4)*(I +∑Isb.r st.m 〔168+150〕=382A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/500=0.0009522ΩSY S1 b UGSP-6 3*50 1500m電纜的電阻R和電阻1 R=RL/K2 =0.1*1.5/〔6/0.69〕2=0.00198Ω1 011 SCX=X*L/K2 =0.06*1.5/〔6/.69〕2=0.00119Ω1 01 1 SCc 變壓器的電阻和電抗值Z=vV2 /(100S )=4.43*0.692/(100*0.50)=0.0422ΩT s ar.T N.TR=△p V2

=4000×0.692/5002=0.00762ΩT N.T

N.TX=√Z2-R2 =√0.04222-0.007622=0.0415ΩT T T△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 d UPQ1.14 3*70 2400m電纜的電阻RXRRL3 2 022X=X*L=0.078*2.4=0.1872Ω2 02 23 e UP–1.14 3*6 5mR和電阻XRRL3 3 033X=X*L=0.090*0.005=0.00045Ω3 03 3最遠點電機機的三相短路電流d=690/〔1.732*√0.65622+0.23082 〕=573A兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*573=499Af 靈敏度校驗Kr=I〔2〕d/Isb=499/382=1.3＞1.2 符合要求、開關(guān)整定計算1、1－60DW80-350饋電的整定計算：(1)、短路保護整定Isb≥Ist.m+∑IN=5*42+150=360A〔2〕、靈敏度校驗系統(tǒng)電抗值X =V2ar/S=0.692/500=0.0009522ΩSY S1 UGSP-6 3*50 1500m電纜的電阻R和電阻1 R=RL/K2 =0.1*1.5/〔6/0.69〕2=0.00198Ω1 011 SCX=X*L/K2 =0.06*1.5/〔6/.69〕2=0.00119Ω1 01 1 SC變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.50)=0.0422ΩT s ar.TR=△p V2

N.T=4000×0.692/5002=0.00762ΩT N.T

N.TXT=√ZT2-R2T=√0.04222-0.007622=0.0415Ω△pN.T：變壓器功率損耗，4000Wvs〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕UPQ1.14 3*70 2400m電纜的電阻R3X3R2=R02L2=0.267*2.4=0.6408ΩX2=X02*L2=0.078*2.4=0.1872ΩUP–1.14 3*6 5mR3和電阻X3R3=R03L3=1.16*0.005=0.0058ΩX3=X03*L3=0.090*0.005=0.00045Ω最遠點電機機的三相短路電流=690/〔1.732*√0.65622+0.23082 〕=573A兩相短路電流I〔2〕=0.87*I〔3〕=0.87*573=499Ad d靈敏度校驗Kr=I〔2〕 /I=499/360d sb=1.4＞1.2 符合要求2、313072BQD10ZD/660.1140開關(guān)整定計算過載保護整定N I=41.72A，I=42A，依據(jù)實N 短路保護整定選整定值為Isb。r=5I=5×42=210A校驗短路保護的靈敏度系統(tǒng)電抗值X =V2ar/S=0.692/500=0.0009522ΩSY S1 UGSP-6 3*50 1500mR和電阻1 R=RL/K2 =0.1*1.5/〔6/0.69〕2=0.00198Ω1 011 SCX=X*L/K2 =0.06*1.5/〔6/.69〕2=0.00119Ω1 01 1 SC變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.50)=0.0422ΩT s ar.TR=△p V2

N.T=4000×0.692/5002=0.00762ΩT N.T

N.TXT=√ZT2-R2T=√0.04222-0.007622=0.0415Ω△p ：變壓器功率損耗，4000WN.Tv〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕s3 UPQ1.14 3*70 100m電纜的電阻RXRRL3 2 022X=X*L=0.078*0.1=0.0078Ω2 02 23 UP–1.14 3*16 20m電纜的電阻R和電阻XRRL3 3 033X=X*L=0.090*0.02=0.0018Ω3 03 3最遠點的三相短路電流d=690/〔1.732*√0.05952+0.05322〕=4990兩相短路電流I〔2〕d=0.87*Id〔3〕=0.87*4990=4341A靈敏度校驗Kr=I〔2〕d/Isb=4341/210=21＞1.2 符合要求負荷統(tǒng)計：∑PN=157KW〔一〕、移動變電站高壓饋電開關(guān)的過載保護整定長時工作電流：IcaKde∑PN/〔√3VNcosΦwm〕=0.54*157/(1.732*6.3*0.85)=10A確定整定值Ise=1.2*Ica.=1.2*10≈12A2〕、短路保護整定sb.r st.m 由過載保護確定短路保護整定值 I =(1.2~1.4)*(I +∑Isb.r st.m 〔168+150〕*0.69/6.3=37A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/630=0.000756ΩSY S1 UGSP-6 3*50 5000mR和電阻1 R=RL/K2 =0.1*5.0/〔6/0.69〕2=0.0066Ω1 011 SCX=X*L/K2 =0.06*5.0/〔6/.69〕2=0.0040Ω1 01 1 SC變壓器的電阻和電抗值Z=vV2

)=4.43*0.692/(100*0.50)=0.0422ΩT s ar.TR=△p V2

N.T=4000×0.692/6302=0.00762ΩT N.T

N.TXT=√ZT2-R2T=√0.04222-0.007622=0.0415Ω△pN.T：變壓器功率損耗，4000Wvs〔%〕：變壓器的阻抗電壓百分數(shù)，4.43〔%〕UPQ1.14 3*70 1700m電纜的電阻R3X3R2=R02L2=0.267*1.7=0.4539ΩX2=X02*L2=0.078*1.7=0.1326ΩUP–1.14 3*6 120mR3和電阻X3R3=R03L3=1.16*0.12=0.1392ΩX3=X03*L3=0.090*0.12=0.0108Ω最遠點電機機的三相短路電流d=690/〔1.732*√0.60732+0.18972 〕＝626A兩相短路電流I〔2〕=0.87*I〔3〕=0.87*626=545Ad d靈敏度校驗Kr=I〔2〕 /I =545/37d sb=14.7＞1.2 符合要求〔二〕、移動變電站低壓饋電開關(guān)的過載保護整定1)、長時工作電流：I=K∑P/〔√3VcosΦ 〕ca de N N wm=0.54*157/(1.732*0.69*0.85)=83Ase 確定整定值I=1.2*I=1.2*83≈100A2〕、se sb.r st.m 由過載保護確定短路保護整定值 I =(1.2~1.4)*(I +∑Isb.r st.m 〔168+150〕=382A3)、靈敏度的校驗a 系統(tǒng)電抗值X =V2ar/S=0.692/630=0.000756ΩSY S1b UGSP-63*50和UGSP-103*35 5000m電纜的電阻R和電阻1X1R1=R01L1/K2SC=0.1*5.0/〔6/0.69〕2=0.0066ΩX1=X01*L1/K2SC=0.06*5.0/〔6/.69〕2=0.0040Ωc 變壓器的電阻和電抗值Z=vV2