主运带 移变开关整定计算.docx
《主运带 移变开关整定计算.docx》由会员分享,可在线阅读,更多相关《主运带 移变开关整定计算.docx(39页珍藏版)》请在冰豆网上搜索。
主运带移变开关整定计算
一、主运带巷机电硐室移动变电站的整定计算
(一)、移动变电站高压馈电开关的过载保护整定
负荷统计:
∑PN=340KW
1)、长时工作电流:
Ica=Kde∑PN/(√3VNcosΦwm)
=0.54*340/(1.732*6.3*0.85)=19.8A
确定整定值Ise=1.2*Ica.=1.2*19.8≈23.8A
2)、短路保护整定
由过载保护确定短路保护整定值Isb.r=(1.2~1.4)*(Ist.m+∑IN)=1.4*(22+19)=41A
3)、灵敏度的校验
a系统电抗值
XSY=V2ar/SS=0.692/800=0.000595Ω
bUGSP3*501460m电缆的电阻R1和电阻X1
R1=R01L1/K2SC
=0.1*1.46/(6/0.69)2=0.002Ω
X1=X01*L1/K2SC
=0.06*1.46/(6/.69)2=0.001Ω
c变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/8002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
dUPQ–1.143*701400m电缆的电阻R3和电阻X3
R2=R03L3=0.267*1.4=0.3738Ω
X2=X03*L3=0.078*1.4=0.1092Ω
eUP–1.143*1610m电缆的电阻R3和电阻X3
R3=R03L=1.16*0.01=0.0116Ω
X3=X03*L=0.090*0.01=0.0009Ω
最远点电机机的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.39932+0.13472)=945A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*945=822A
f灵敏度校验
Kr=I
(2)d/Isb=822/41
=20>1.2符合要求
(二)、移动变电站低压馈电开关的过载保护整定
1)、长时工作电流:
Ica=Kde∑PN/(√3VNcosΦwm)
=0.54*340/(1.732*0.69*0.85)=180A
确定整定值Ise=1.2*Ica.=1.2*106≈216A
2)、短路保护整定
由过载保护确定短路保护整定值Isb.r=(1.2~1.4)*(Ist.m+∑IN)=1.2*(169+160)=329A
3)、灵敏度的校验
a系统电抗值
XSY=V2ar/SS=0.692/800=0.000595Ω
bUGSP3*501460m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*1.46/(6/0.69)2=0.002Ω
X1=X01*L1/K2SC=0.06*1.46/(6/.69)2=0.001Ω
c变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/8002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
dUPQ–1.143*701400m电缆的电阻R3和电阻X3
R2=R03L3=0.267*1.4=0.3738Ω
X2=X03*L3=0.078*1.4=0.1092Ω
eUP–1.143*1610m电缆的电阻R3和电阻X3
R3=R03L3=1.16*0.01=0.0116Ω
X3=X03*L3=0.090*0.01=0.0009Ω
最远点电机机的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.39932+0.13472)=945A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*945=822A
f灵敏度校验
Kr=I
(2)d/Isb=822/329
=2.5>1.2符合要求
二、主运带移变馈电、开关整定计算
1、控制主运皮带机的KJ25-400馈电整定计算
(1)短路保护整定
Isb≥Ist.m+∑IN=5*185+330=515A
(2)灵敏度校验
aXSY=V2ar/SS=0.692/800=0.000595Ω
bUGSP3*9560m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*0.046/(6/0.69)2=0.002Ω
X1=X01*L1/K2SC=0.06*0.046/(6/.69)2=0.001Ω
c变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/8002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
dUPQ–1.143*701400m电缆的电阻R3和电阻X3
R2=R02L2=0.267*1.4
=0.3738Ω
X2=X02*L2=0.078*1.4=0.1092Ω
eUP–1.143*1610m电缆的电阻R3和电阻X3
R3=R03L3=1.16*0.01=0.0116Ω
X3=X03*L3=0.090*0.01=0.0009Ω
最远点电机机的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.39932+0.13472)=945A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*945=822A
f灵敏度校验
Kr=I
(2)d/Isb=822/254
=3.2>1.2符合要求
2、控制中间巷DW80-200馈电整定计算
(1)、短路保护整定
Isb≥Ist.m+∑IN=5*41.72+75=284A
(2)、灵敏度校验
aXSY=V2ar/SS=0.692/800=0.000595Ω
bUGSP3*501460m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*1.46/(6/0.69)2=0.002Ω
X1=X01*L1/K2SC=0.06*1.46/(6/.69)2=0.001Ω
c变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/8002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
dUPQ–1.143*701000m电缆的电阻R3和电阻X3
R2=R02L2=0.267*1.0=0.267Ω
X2=X02*L2=0.078*1.0=0.078Ω
eUP–1.143*1610m电缆的电阻R3和电阻X3
R3=R03L3=1.16*0.01=0.0116Ω
X3=X03*L3=0.090*0.01=0.0009Ω
最远点电机机的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.28562+0.11422)
=1295A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*1295=1127A
f灵敏度校验
Kr=I
(2)d/Isb=1127/284
=3.97>1.2符合要求
3、主运带巷机电硐室QBZ-80开关整定计算
1)过载保护整定
由拉紧电机额定电流IN=8.7A,确定整定值为ISe=9A
短路保护整定
选整定值为Isb。
r=4Ise=4×9=36A
校验短路保护的灵敏度
a系统电抗值
XSY=V2ar/SS=0.692/800=0.000595Ω
bUGSP3*501460m电缆的电阻R1和电阻X1
R1=R01L1/K2SC
=0.1*1.46/(6/0.69)2=0.0019Ω
X1=X01*L1/K2SC
=0.06*1.46/(6/.69)2=0.00116Ω
c变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334
RT=△pN.TV2ar.T/S2N.T=4000×0.692/8002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
dUPQ–1.143*70200m电缆的电阻R3和电阻X3
R2=R02L2=0.267*0.2=0.0534Ω
X2=X02*L2.=0.078*0.2=0.0156Ω
eUP–6603*1620m电缆的电阻R3和电阻X3
R3=R03L3=1.16*0.02=0.0232Ω
X3=X03*L3=0.09*0.02=0.0018Ω
最远点的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.08072+0.02052)=2762A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*2762=2403A
f灵敏度校验
Kr=I
(2)d/Isb=2403/168
=14.3>1.2合格
二、06回顺46联巷移动变电站的整定计算
负荷统计:
∑PN=200KW
(一)、移动变电站高压馈电开关的过载保护整定
长时工作电流:
Ica=Kde∑PN/(√3VNcosΦwm)
=0.54*200/(1.732*6.3*0.85)=12A
确定整定值Ise=1.2*Ica.=1.2*12≈14.4A
2)、短路保护整定
由过载保护确定短路保护整定值Isb.r=(1.2~1.4)*(Ist.m+∑IN)=1.3*(49+12)=79A
3)、灵敏度的校验
a系统电抗值
XSY=V2ar/SS=0.692/500=0.0009522Ω
bUGSP-63*50和UGSP-103*353600m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*3.6/(6/0.69)2=0.0047Ω
X1=X01*L1/K2SC=0.06*3.6/(6/.69)2=0.0028Ω
c变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.50)=0.0422Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/5002=0.00762Ω
XT=√ZT2-R2T=√0.04222-0.007622=0.0415Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
dUPQ–1.143*701500m电缆的电阻R3和电阻X3
R2=R02L2=0.267*1.5=0.4005Ω
X2=X02*L2=0.078*1.5=0.117Ω
eUP–1.143*610m电缆的电阻R3和电阻X3
R3=R03L3=1.16*0.01=0.0116Ω
X3=X03*L3=0.090*0.01=0.0009Ω
最远点电机机的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.42442+0.16342)=876A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*876=762A
f灵敏度校验
Kr=I
(2)d/Isb=762/79
=9.6>1.2符合要求
(二)、移动变电站低压馈电开关的过载保护整定
1)、长时工作电流:
Ica=Kde∑PN/(√3VNcosΦwm)
=0.54*200/(1.732*0.69*0.85)=106A
确定整定值Ise=1.2*Ica.=1.2*106≈128A
2)、短路保护整定
由过载保护确定短路保护整定值Isb.r=(1.2~1.4)*(Ist.m+∑IN)=1.2*(95+123)=218A
3)、灵敏度的校验
a系统电抗值
XSY=V2ar/SS=0.692/500=0.0009522Ω
bUGSP3*50和UGSP-103*353600m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*3.6/(6/0.69)2=0.0047Ω
X1=X01*L1/K2SC=0.06*3.6/(6/.69)2=0.0028Ω
c变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.5)=0.0422Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/5002=0.00762Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0415Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
dUPQ–1.143*701500m电缆的电阻R3和电阻X3
R2=R02L2=0.267*1.5=0.4005Ω
X2=X02*L2=0.078*1.5=0.117Ω
eUP–1.143*610m电缆的电阻R3和电阻X3
R3=R03L3=1.16*0.01=0.0116Ω
X3=X03*L3=0.090*0.01=0.0009Ω
最远点电机机的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.42442+0.16342)=876A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*876=762A
f灵敏度校验
Kr=I
(2)d/Isb=762/218
=3.5>1.2符合要求
三、31306回顺46联巷移变所带馈电、开关整定计算
1、控制45-21联巷DW80-350馈电的整定计算:
(1)、短路保护整定
Isb≥Ist.m+∑IN=5*95+32=507A
(2)、灵敏度校验
aXSY=V2ar/SS=0.692/500=0.0009522Ω
bUGSP3*501460m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*1.46/(6/0.69)2=0.002Ω
X1=X01*L1/K2SC=0.06*1.46/(6/.69)2=0.001Ω
CUGSP-103*352200m电缆的电阻R1和电阻X1
R2=R02L2/K2SC=0.1*2.2/(6/0.69)2=0.003174Ω
X2=X02*L2/K2SC=0.06*2.2/(6/.69)2=0.0019Ω
d变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/5002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
eUPQ–1.143*701500m电缆的电阻R3和电阻X3
R3=R03L3=0.267*1.5=0.4005Ω
X3=X03*L3=0.078*1.5=0.117Ω
fUPQ–1.143*610m电缆的电阻R3和电阻X3
R4=R04L4=1.16*0.01=0.0116Ω
X4=X04*L4=0.090*0.01=0.0009Ω
最远点的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.42222+0.15512)=886A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*886=770A
g灵敏度校验
Kr=I
(2)d/Isb=770/507
=1.50>1.2符合要求
2、控制31306回顺46-70联巷BKD4-400Z/1140.660馈电整定计算:
负荷统计:
∑PN=40KW
1)、过载保护整定
ISe=Kco∑IN/Kre=1.15*49/0.85=66A
Kco----过载保护可靠系数,取1.2;
Kre-----过载保护的返回系数,取0.85
∑IN-----馈电所带负荷额定电流之和。
2)、短路保护整定
Isb≥Ist.m+∑IN=5*5+45=70A
3)、灵敏度校验
aXSY=V2ar/SS=0.692/500=0.0009522Ω
bUGSP3*501460m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*1.46/(6/0.69)2=0.002Ω
X1=X01*L1/K2SC=0.06*1.46/(6/.69)2=0.001Ω
CUGSP-103*352200m电缆的电阻R1和电阻X1
R2=R02L2/K2SC=0.1*2.2/(6/0.69)2=0.003174Ω
X2=X02*L2/K2SC=0.06*2.2/(6/.69)2=0.0019Ω
d变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/5002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
eUPQ–1.143*701350m电缆的电阻R3和电阻X3
R3=R03L3=0.267*1.35=0.3604Ω
X3=X03*L3=0.078*1.35=0.1053Ω
fUPQ–1.143*610m电缆的电阻R3和电阻X3
R4=R04L4=1.16*0.01=0.0116Ω
X4=X04*L4=0.090*0.01=0.0009Ω
最远点的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.38222+0.14342)=693A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*693=603A
g灵敏度校验
Kr=I
(2)d/Isb=603/70
=8.6>1.2符合要求
3、06回顺23联巷QJZ-300/1140开关整定计算
1)、过载保护整定
由水泵的电机额定电流IN=114A,确定整定值为ISe=115A
2)、短路保护整定
选整定值为Isb。
r=4Ise=4×114=456A
校验短路保护的灵敏度
a系统电抗值
XSY=V2ar/SS=0.692/500=0.0009522Ω
bUGSP-63*501460m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*1.46/(6/0.69)2=0.0019Ω
X1=X01*L1/K2SC=0.06*1.46/(6/.69)2=0.0012Ω
CUGSP-103*352200m电缆的电阻R1和电阻X1
R2=R02L2/K2SC=0.1*2.2/(6/0.69)2=0.003174Ω
X2=X02*L2/K2SC=0.06*2.2/(6/.69)2=0.0019Ω
d变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334Ω
RT=△pN.TV2ar.T/S2N.T=4000×0.692/5002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
△pN.T:
变压器功率损耗,4000W
vs(%):
变压器的阻抗电压百分数,4.43(%)
eUPQ–1.143*701350m电缆的电阻R3和电阻X3
R3=R03L3=0.267*1.35=0.36Ω
X3=X03*L3=0.078*1.35=0.105Ω
fUPQ–1.143*7010m电缆的电阻R3和电阻X3
R4=R04L4=0.267*0.01=0.00267Ω
X4=X04*L4=0.078*0.01=0.00078Ω
最远点的三相短路电流
I(3)d=Var/(√3*√ΣR2+ΣX2)
=690/(1.732*√0.3732+0.14242)=998A
两相短路电流
I
(2)d=0.87*Id(3)=0.87*998=868A
g灵敏度校验
Kr=I
(2)d/Isb=868/456
=1.90>1.2符合要求
4、06回顺45联巷QC83-80开关整定计算
1)过载保护整定
由水泵电机额定电流IN=41.72A,确定整定值为ISe=42A,根据实际经验整定值取48A。
短路保护整定
选整定值为Isb。
r=5Ise=5×42=210A
校验短路保护的灵敏度
a系统电抗值
XSY=V2ar/SS=0.692/500=0.0009522Ω
bUGSP3*501460m电缆的电阻R1和电阻X1
R1=R01L1/K2SC=0.1*1.46/(6/0.69)2=0.0019Ω
X1=X01*L1/K2SC=0.06*1.46/(6/.69)2=0.00116Ω
CUGSP-103*352200m电缆的电阻R1和电阻X1
R2=R02L1/K2SC=0.1*2.2/(6/0.69)2=0.003174Ω
X2=X02*L1/K2SC=0.06*2.2/(6/.69)2=0.0019Ω
d变压器的电阻和电抗值
ZT=vsV2ar.T/(100SN.T)=4.43*0.692/(100*0.63)=0.0334
RT=△pN.TV2ar.T/S2N.T=4000×0.692/5002=0.005Ω
XT=√ZT2-R2T=√0.03342-0.0052=0.0331Ω
fUP–6603*1660m电缆的电阻R3和电阻X3
R3=R03L3=