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单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,例:,某,220/380V,三相四线制线路上,装有,220V,单相电热干燥箱,6,台、单相电加热器,2380V,单相对焊机,6,台。其在线路上的连接情况为:电热干燥箱,2,台,20kW,接于,A1,台,30kW,接于,B3,台,10kW,接于,C,相;电加热器,2,台,20kW,分别接于,BC,相;对焊机,3,台,14kW,(,=100%,)接于,AB2,台,20kW,(,=100%,)接于,BC1,台,46kW,(,=60%,)接于,CA,相。试求该线路的计算负荷。,解,1.,电热干燥箱及电加热器的各相计算负荷,6,1,得,Kd=0.7,,,cos=1,,,tg=0,,因此只要计算有功计算负荷,A,相,PcA1=KdPeA=0.7202=28kW,B,相,PcB1=KdPeB=0.7(301+201)=35kW,C,相,PcC1=KdPeC=0.7(103+201)=35kW,QcA1,QcB1,QcC1,0,1,2.,对焊机的各相计算负荷,查表,6,1,得,Kd=0.35,,,cos=0.7,,,tg=1.02,查表,2-4,得,cos=0.7,时,pAB-A=pBC-B=pCA-C=0.8,pAB-B=pBC-C=pCA-A=0.2,qAB-A=qBC-B=qCA-C=0.22,qAB-B=qBC-C=qCA-A=0.8,2,将接于线电压的单相设备容量换算为相电压的设备容量的换算公式为,3,先将接于,CA,相的,46kW,(,=60%,)换算至,=100%,的设备容量,即,各相的设备容量为,A,相,P,eA,=p,AB-A,P,AB,+p,CA-A,P,CA,=0.8143+0.235.63=40.73kW,Q,eA,=q,AB-A,P,AB,+q,CA-A,P,CA,=0.22143+0.835.63=37.74kvar,B,相,P,eB,=p,BC-B,P,BC,+p,AB-B,P,AB,=0.8202+0.2143=40.4kW,Q,eB,=q,BC-B,P,BC,+q,AB-B,P,AB,=0.22202+0.8143=42.4kvar,C,相,P,eC,=p,CA-C,P,CA,+p,BC-C,P,BC,=0.835.63+0.2202=36.5kW,Q,eC,=q,CA-C,P,CA,+q,BC-C,P,BC,=0.2235.63+0.8202=39.84kvar,4,电焊机各相的计算负荷为,A相,P,cA2,=K,d,P,eA,=0.35,40.73=14.26kW,Q,cA2,=K,d,Q,eA,=0.35,37.74=13.21kvar,B相,P,cB2,=K,d,P,eB,=0.35,40.4=14.14kW,Q,cB2,=K,d,Q,eB,=0.35,42.4=14.84kvar,C相,P,cC2,=K,d,P,eC,=0.35,36.5=12.78kW,Q,cC2,=K,d,Q,eC,=0.35,39.84=13.94kvar,各相总的计算负荷为(设同时系数为,0.95),A相,P,cA,=K,(,P,cA1,+P,cA2,),=0.95,(,28+14.26)=40.15kW,Q,cA,= K,(,Q,cA1,+Q,cA2,),=0.95,(,0+13.21)=12.55kvar,B相,P,cB,= K,(,P,cB1,+P,cB2,),=0.95,(,35+14.14)=46.68kW,Q,cB,= K,(,Q,cB1,+Q,cB2,),=0.95,(,0+14.84)=14.10kvar,C相,P,cC,= K,(,P,cC1,+P,cC2,),=0.95,(,35+12.78)=45.39kW,Q,cC,= K,(,Q,cC1,+Q,cC2,),=0.95,(,0+13.94)=13.24kvar,5,总的等效三相计算负荷为,:,因为,B相的有功计算负荷最大,所以,P,cm,= P,cB,=46.68kW,Q,cm,=Q,cB,=14.10kvar,P,c,=3P,cm,=3,46.68=140.04kW,Q,c,=3Q,cm,=3,14.10=42.3kvar,6,
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