太原市城南污水处理厂施工临时用电方案 - 策划方案(3)

太原市城南污水处理厂（EPC）工程 临时用电专项方案 6+24）=185.76kW

总的无功功率Qjs = K∑q×ΣQjs = 0.9

（27.10+3.33+3.33+4.26+5.5+5.51+4.04+35.1+3+6.05+61.18+9.69+21.12）=167.05 Kvar

总的视在功率Sjs = (Pjs2+Qjs2)1/2 =249.83kVA

总的计算电流计算Ijs = Sjs/(1.732×Ue) = 379.59A 根据总的视在功率选取功率为315 kVA三相电力变压器。 3、确定2#变压器用电负荷: (1)塔吊组

Kx=0.30 Cosφ=0.70 tgφ=1.02 将Jc=40%统一换算到Jc1=25%的额定容量

Pn=70kW

Pe=(Jc/Jc1)1/2×Pn = (0.40/0.25)1/2×70=88.55 kW Pjs1=Kx×Pe =0.30×88.55=26.56 Kw

Qjs1= Pjs1× tgφ=26.56×1.02=27.10kvar (2)钢筋切断机组

Kx=0.6 Cosφ=0.7 tgφ=1.02 Pjs5=0.6×9=5.4kW

Qjs5=Pjs5×tgφ=5.4×1.02=5.5 kvar (3)钢筋弯曲机组

Kx = 0.6 Cosφ = 0.7 tgφ = 1.02 Pjs6 = 0.6×9 =5.4kW

10

太原市城南污水处理厂（EPC）工程 临时用电专项方案 Qjs6=Pjs6×tgφ=5.4×1.02=5.51 kvar (4)钢筋调直机组

Kx = 0.6 Cosφ = 0.7 tgφ = 1.02 Pjs7 = 0.6×6.6 = 3.96kW

Qjs7=Pjs7×tgφ=3.96×1.02=4.04 kvar (5)交流电弧焊机组

Kx = 0.45 Cosφ = 0.75 tgφ = 0.88 将Jc =60%统一换算到Jc1 =100%的额定容量 Pn=S×Cosφ=152×0.75=114Kw

Pe=(Jc/Jc1)1/2×Pn = (0.60/1)1/2×114= 88.31Kw Pjs8=Kx×Pe =0.45×88.31 =39.8kW

Qjs8=Pjs8×tgφ=39.8×0.88=35.1 kvar (6)平板振动器组

Kx=0.70 Cosφ=0.68 tgφ=1.08 Pjs6=0.70×4.4 = 3.08 kW

Qjs6=Pjs6×tgφ=3.08×1.08=3.33 kvar (7)插入式振动棒组

Kx = 0.70 Cosφ = 0.68 tgφ = 1.08 Pjs7=0.70×4.4 = 3.08kW

Qjs7=Pjs7×tgφ=3.08×1.08=3.33kvar (8)镝灯组

Kx = 0.4 Cosφ = 0.55 tgφ = 1.52

11

太原市城南污水处理厂（EPC）工程 临时用电专项方案 Pjs8 = 0.4×3.5= 1.4kW

Qjs8=Pjs8×tgφ=1.4×1.52=2.13kvar (9)砂浆搅拌机组

Kx=0.7 Cosφ= 0.68 tgφ=1.08 Pjs9=0.7×4 = 2.8kW

Qjs9=Pjs9×tgφ=2.8×1.08=3.03Kvar (10)打夯机

Kx = 0.7 Cosφ = 0.68 tgφ = 1.08 Pjs=0.8×6=4.8ＫＷ

Qjs=Pjs×tgφ=4.8×0.7=3.36kvar (11)木工圆锯机组

Kx=0.6 Cosφ=0.7 tgφ=1.02 Pjs9=0.6×4=2.4kW

Qjs9=Pjs9×tgφ=2.4×1.02=2.45Kvar （安1)交流电弧焊机组

Kx=0.45 Cosφ=0.75 tgφ=0.88 将Jc=60%统一换算到Jc1=100%的额定容量 Pn=S×Cosφ=380×0.75=285Kw

Pe=(Jc/Jc1)1/2×Pn =(0.60/1)1/2×285=220.8Kw Pjs11=Kx×Pe =0.45×220.8=99.4kW

Qjs11=Pjs11×tgφ=99.4×0.88=87.5kvar （安2）电动工具组

12

太原市城南污水处理厂（EPC）工程 临时用电专项方案 Kx = 0.8 Cosφ = 0.75 tgφ = 0.88 Pjs=0.8×30=24ＫＷ

Qjs=Pjs×tgφ=24×0.88=21.12kvar 因此计算负荷为（K∑P=0.9，K∑q=0.9）

总的有功功率Pjs = K∑P×ΣPjs=0.9×

（26.56+5.4+5.4+3.96+39.8+3.08+3.08+1.4+2.8+99.4+24+4.8+2.4）=199.86kW

总的无功功率Qjs = K∑q×ΣQjs=0.9×

(27.1+5.5+5.5+4.04+35.1+3.33+3.33+2.13+3.03+87.5+21.12+3.36+2.45)=183.15Kvar

总的视在功率Sjs = (Pjs2+Qjs2)1/2 =268.00kVA 总的计算电流计算Ijs = Sjs/(1.732×Ue) =407.20A 根据总的视在功率选取功率为315 kVA三相电力变压器. 4、确定3#变压器用电负荷: (1)塔吊组

Kx=0.30 Cosφ=0.70 tgφ=1.02 将Jc=40%统一换算到Jc1=25%的额定容量

Pn=70kW

Pe=(Jc/Jc1)1/2×Pn = (0.40/0.25)1/2×70=88.55 kW Pjs1=Kx×Pe =0.30×88.55=26.56 Kw

Qjs1= Pjs1× tgφ=26.56×1.02=27.10kvar (2)钢筋切断机组

13

太原市城南污水处理厂（EPC）工程 临时用电专项方案 Kx=0.6 Cosφ=0.7 tgφ=1.02 Pjs5=0.6×9=5.4kW

Qjs5=Pjs5×tgφ=5.4×1.02=5.5 kvar (3)钢筋弯曲机组

Kx = 0.6 Cosφ = 0.7 tgφ = 1.02 Pjs6 = 0.6×9 =5.4kW

Qjs6=Pjs6×tgφ=5.4×1.02=5.51 kvar (4)钢筋调直机组

Kx = 0.6 Cosφ = 0.7 tgφ = 1.02 Pjs7 = 0.6×6.6 = 3.96kW

Qjs7=Pjs7×tgφ=3.96×1.02=4.04 kvar (5)交流电弧焊机组

Kx = 0.45 Cosφ = 0.75 tgφ = 0.88 将Jc =60%统一换算到Jc1 =100%的额定容量 Pn=S×Cosφ=224×0.75=168Kw

Pe=(Jc/Jc1)1/2×Pn = (0.60/1)1/2×114=130.14Kw Pjs8=Kx×Pe =0.45×130.14 =58.56kW

Qjs8=Pjs8×tgφ=58.56×0.88=51.53 kvar (6)平板振动器组

Kx=0.70 Cosφ=0.68 tgφ=1.08 Pjs6=0.70×4.4 = 3.08 kW

Qjs6=Pjs6×tgφ=3.08×1.08=3.33 kvar

14

百度搜索“77cn”或“免费范文网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读，免费范文网，提供经典小说教育文库太原市城南污水处理厂施工临时用电方案 - 策划方案(3)在线全文阅读。