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1、施工现场临时用电组织设计一、编制依据1、施工现场临时用电安全技术规范JGJ46-20052、低压配电设计规范GB50054-20113、建筑工程施工现场供电安全规范GB50194-20144、通用用电设备配电设计规范GB50055-20115、工业与民用配电设计手册第三版6、建筑施工安全检查标准JGJ59-2011二、参数设置1、总箱参数用电接驳方式由变压器接入总配电箱编号总箱 A总配电箱距接驳点布线距离Lo(m)5总配电箱的同期使用系数Kx0.8分配电箱数量11导线线芯材料铝线总配电箱进线敷设方式空气明敷/架空线路分配电箱进线敷设方式空气明敷/架空线路设备电线敷设方式空气明敷/架空线路干线允
2、许电压降%52、分箱参数3、用电设备参数4、临电设备计算及配置三、初步设计(1)施工现场临时用电安全技术规范（JGJ46-2005）总则中以强制性条文规定，施工现场必须采用三级配电系统、必须采用TN-S接零保护系统。(2)现场采用380V低压供电，设一配电总箱，采用TN-S系统供电。(3)布置位置及线路走向参见临时配电系统图，采用三级配电，三级防护。(4)按照JGJ46-2005规定制定施工组织设计，接地电阻R4。(5)按照JGJ46-2005规定若为架空线路则“根据机械强度要求，绝缘铜线截面不小于10mm2，绝缘铝线截面不小于16mm2”。四、用电负荷1、1号分箱设备用电负荷(1)塔式起重机
3、Kx = 0.7，Cos = 0.65，tg = (1 - Cos2)0.5/ Cos = (1 - 0.652)0.5 / 0.65 = 1.17Pe = n P (e/)0.5 = 2.00 35.00 (0.40 / 0.25)0.5 = 88.54kWIj1 = Kx Pe / (1.732 Ue cos) = 0.70 88.54 1000 / (1.732 380 0.65) = 144.88APjs = Kx Pe = 0.70 88.54 = 61.98kWQjs = Pjs tg = 61.98 1.17 = 72.46kvA(2)高压汞灯Kx = 0.6，Cos = 0.7
4、，tg = (1 - Cos2)0.5/ Cos = (1 - 0.72)0.5 / 0.7 = 1.02Pe = n P (e/)0.5 = 4.00 3.00 (0.40 / 0.40)0.5 = 12.00kWIj1 = Kx Pe / (1.732 Ue cos) = 0.60 12.00 1000 / (1.732 380 0.70) = 15.63APjs = Kx Pe = 0.60 12.00 = 7.20kWQjs = Pjs tg = 7.20 1.02 = 7.35kvA(3)1号分箱计算负荷用电设备同期使用系数Kx=0.60；Ijs = Kx Pe / (1.732 U
5、e Cos) = 0.60 100.54 1000 / (1.732 380 0.75) = 122.21A；同时要求分箱用电负荷不得小于分箱下用电设备最大计算负荷的1.1倍，故：Ijs = max122.21,1.1max(144.88,15.63,) = 159.37A2、2号分箱设备用电负荷(1)地下车库随楼箱Kx = 0.3，Cos = 0.6，tg = (1 - Cos2)0.5/ Cos = (1 - 0.62)0.5 / 0.6 = 1.33Pe = n P (e/)0.5 = 1.00 15.00 (0.40 / 0.40)0.5 = 15.00kWIj1 = Kx Pe /
6、(1.732 Ue cos) = 0.30 15.00 1000 / (1.732 380 0.60) = 11.40APjs = Kx Pe = 0.30 15.00 = 4.50kWQjs = Pjs tg = 4.50 1.33 = 6.00kvA(2)2号分箱计算负荷用电设备同期使用系数Kx=0.60；Ijs = Kx Pe / (1.732 Ue Cos) = 0.60 15.00 1000 / (1.732 380 0.75) = 18.23A；同时要求分箱用电负荷不得小于分箱下用电设备最大计算负荷的1.1倍，故：Ijs = max18.23,1.1max(11.40,) = 18
7、.23A3、3号分箱设备用电负荷(1)生活区Kx = 0.75，Cos = 0.7，tg = (1 - Cos2)0.5/ Cos = (1 - 0.72)0.5 / 0.7 = 1.02Pe = n P (e/)0.5 = 1.00 30.00 (0.40 / 0.40)0.5 = 30.00kWIj1 = Kx Pe / (1.732 Ue cos) = 0.75 30.00 1000 / (1.732 380 0.70) = 48.84APjs = Kx Pe = 0.75 30.00 = 22.50kWQjs = Pjs tg = 22.50 1.02 = 22.95kvA(2)3号分
8、箱计算负荷用电设备同期使用系数Kx=0.60；Ijs = Kx Pe / (1.732 Ue Cos) = 0.60 30.00 1000 / (1.732 380 0.75) = 36.47A；同时要求分箱用电负荷不得小于分箱下用电设备最大计算负荷的1.1倍，故：Ijs = max36.47,1.1max(48.84,) = 53.72A4、4号分箱设备用电负荷(1)塔式起重机Kx = 0.2，Cos = 0.65，tg = (1 - Cos2)0.5/ Cos = (1 - 0.652)0.5 / 0.65 = 1.17Pe = n P (e/)0.5 = 1.00 35.00 (0.40
9、 / 0.25)0.5 = 44.27kWIj1 = Kx Pe / (1.732 Ue cos) = 0.20 44.27 1000 / (1.732 380 0.65) = 20.70APjs = Kx Pe = 0.20 44.27 = 8.85kWQjs = Pjs tg = 8.85 1.17 = 10.35kvA(2)高压汞灯Kx = 0.6，Cos = 0.7，tg = (1 - Cos2)0.5/ Cos = (1 - 0.72)0.5 / 0.7 = 1.02Pe = n P (e/)0.5 = 2.00 3.00 (0.40 / 0.40)0.5 = 6.00kWIj1 =
10、 Kx Pe / (1.732 Ue cos) = 0.60 6.00 1000 / (1.732 380 0.70) = 7.81APjs = Kx Pe = 0.60 6.00 = 3.60kWQjs = Pjs tg = 3.60 1.02 = 3.67kvA(3)4号分箱计算负荷用电设备同期使用系数Kx=0.60；Ijs = Kx Pe / (1.732 Ue Cos) = 0.60 50.27 1000 / (1.732 380 0.75) = 61.11A；同时要求分箱用电负荷不得小于分箱下用电设备最大计算负荷的1.1倍，故：Ijs = max61.11,1.1max(20.70,
11、7.81,) = 61.11A5、5号分箱设备用电负荷(1)施工升降机Kx = 0.3，Cos = 0.6，tg = (1 - Cos2)0.5/ Cos = (1 - 0.62)0.5 / 0.6 = 1.33Pe = n P (e/)0.5 = 2.00 60.00 (0.40 / 0.40)0.5 = 120.00kWIj1 = Kx Pe / (1.732 Ue cos) = 0.30 120.00 1000 / (1.732 380 0.60) = 91.16APjs = Kx Pe = 0.30 120.00 = 36.00kWQjs = Pjs tg = 36.00 1.33 =
12、 48.00kvA(2)5号分箱计算负荷用电设备同期使用系数Kx=0.60；Ijs = Kx Pe / (1.732 Ue Cos) = 0.60 120.00 1000 / (1.732 380 0.75) = 145.86A；同时要求分箱用电负荷不得小于分箱下用电设备最大计算负荷的1.1倍，故：Ijs = max145.86,1.1max(91.16,) = 145.86A6、6号分箱设备用电负荷(1)施工升降机Kx = 0.2，Cos = 0.6，tg = (1 - Cos2)0.5/ Cos = (1 - 0.62)0.5 / 0.6 = 1.33Pe = n P (e/)0.5 =
13、1.00 60.00 (0.40 / 0.40)0.5 = 60.00kWIj1 = Kx Pe / (1.732 Ue cos) = 0.20 60.00 1000 / (1.732 380 0.60) = 30.39APjs = Kx Pe = 0.20 60.00 = 12.00kWQjs = Pjs tg = 12.00 1.33 = 16.00kvA(2)6号分箱计算负荷用电设备同期使用系数Kx=0.60；Ijs = Kx Pe / (1.732 Ue Cos) = 0.60 60.00 1000 / (1.732 380 0.75) = 72.93A；同时要求分箱用电负荷不得小于分
14、箱下用电设备最大计算负荷的1.1倍，故：Ijs = max72.93,1.1max(30.39,) = 72.93A7、7号分箱设备用电负荷(1)钢筋调直机Kx = 0.65，Cos = 0.7，tg = (1 - Cos2)0.5/ Cos = (1 - 0.72)0.5 / 0.7 = 1.02Pe = n P (e/)0.5 = 1.00 7.50 (0.40 / 0.40)0.5 = 7.50kWIj1 = Kx Pe / (1.732 Ue cos) = 0.65 7.50 1000 / (1.732 380 0.70) = 10.58APjs = Kx Pe = 0.65 7.50 = 4.88kW
