某辦公樓電氣部分設計計算書

某辦公樓電氣局部設計計算書1.電光源及燈具的選擇一層商鋪選用3×16W的嵌入式柵格熒光燈，室外走廊選用32W的嵌入式直置插拔式節(jié)能筒燈；辦公室選用3×18W的嵌入式柵格熒光燈；走廊燈分別選用2×32W的嵌入式柵格熒光燈和3×18W的嵌入式柵格熒光燈；電梯前室、樓梯間選用3×16的吸頂燈，衛(wèi)生間選用28W的防水防塵吸頂燈；地下室水池、走廊選用2×18W的三防熒光燈；陽臺燈、三層平臺分別選用22W的壁燈和250W的圓形投光燈。2.確定布置燈具方案和進行照度計算〔1〕選擇照度。查照明標準值，辦公樓辦公室Eav300lx，衛(wèi)生間、樓梯間、電梯前室、走廊Eav150lx，商鋪Eav200lx，變配電間、設備間Eav200lx?！?〕計算高度。辦公樓層高為3.0m，工作面為0.75m，所以各計算高度為：辦公室、商鋪、衛(wèi)生間、變配電間、設備間h=3.0-0.75=2.25m走廊、電梯前室h=3.0m〔3〕面積計算。地下室變配電間A=4.4×7.4=32.56㎡設備間A=4.4×8.2=36.08㎡走廊A=〔9.3+6.3+8.7〕×1.6=38.88㎡過道A=4.6×2.5=11.5㎡一層商鋪A1=〔6.3+9.3〕×9.6×1/2=74.88㎡A2=6.8×9.6=65.28㎡A3=6.2×9.6=59.52㎡A4=7.4×6.3=46.62㎡雨棚A1=1.4×30.8=43.12㎡A2=1.4×14.1=19.74㎡電梯前室A=4.6×2.5=11.5㎡二層開放辦公室A1=23×8.3=190.9㎡A2=〔5.3+9.3〕×14.7×1/2=107.31㎡A3=6.3×6.1=38.43㎡電梯前室A=3.7×6.6=24.42㎡三層開放辦公室A1=6.3×20.8=131.04㎡A2=6.3×6.3=39.69㎡電梯前室A=4.7×3.7=17.39㎡四層辦公室A1=6.3×6.1=38.43㎡A2=5.9×9.5=56.05㎡A3=5.9×6.2=36.58㎡A4=7.8×6.8=53.04㎡走廊A=1.7×20=34㎡電梯前室A=3.7×6.6=24.42㎡五至十三層辦公室A1=6.3×6.1=38.43㎡A2=5.9×9.5=56.05㎡A3=5.9×6.2=36.58㎡A4=7.8×6.8=53.04㎡走廊A=1.7×20=34㎡電梯前室A=3.7×6.6=24.42㎡十四層辦公室A1=6.3×6.1=38.43㎡A2=5.9×9.5=56.05㎡A3=5.9×6.2=36.58㎡A4=7.8×6.8=53.04㎡走廊A=1.7×20=34㎡電梯前室A=3.7×6.6=24.42㎡十五層辦公室A1=6.3×6.1=38.43㎡A2=4.9×8.4=41.16㎡A3=4.9×6.2=30.38㎡A4=6.8×6.8=46.24㎡電梯前室A=4.7×3.7=17.39㎡〔4〕照度計算〔采用單位容量法〕。查表《建筑電氣設計手冊》，得各房間的照明單位安裝功率Po為：地下室配電間Po=0.0560W/㎡設備間Po=0.0551W/㎡走廊Po=0.0897W/㎡過道Po=0.0897W/㎡一層商鋪Po1=0.0435W/㎡Po2=0.0483W/㎡Po3=0.0492W/㎡Po4=0.0508W/㎡雨棚Po1=0.0897W/㎡Po2=0.0897W/㎡電梯前室Po=0.0897W/㎡二層開放辦公室Po1=0.0432W/㎡Po2=0.0416W/㎡Po3=0.0541W/㎡電梯前室Po=0.0735W/㎡三層開放辦公室Po1=0.0458W/㎡Po2=0.0537W/㎡電梯前室Po=0.0810W/㎡四層辦公室Po1=0.0541W/㎡Po2=0.0503W/㎡Po3=0.0548W/㎡Po4=0.0504W/㎡走廊Po=0.0897W/㎡電梯前室Po=0.0735W/㎡五至十三層辦公室Po1=0.0541W/㎡Po2=0.0503W/㎡Po3=0.0548W/㎡Po4=0.0504W/㎡走廊Po=0.0897W/㎡電梯前室Po=0.0735W/㎡十四層辦公室Po1=0.0541W/㎡Po2=0.0503W/㎡Po3=0.0548W/㎡Po4=0.0504W/㎡走廊Po=0.0897W/㎡電梯前室Po=0.0735W/㎡十五層辦公室Po1=0.0541W/㎡Po2=0.0542W/㎡Po3=0.0578W/㎡Po4=0.0518W/㎡電梯前室Po=0.0810W/㎡〔5〕計算總安裝容量?！睵=Po×A×E〕地下室變配電間P=0.0560×32.56×200=364.98W設備間P=0.0551×36.08×200=397.46W走廊P=0.0897×38.88×150=523.13W過道P=0.0897×11.5×150=154.8W一層商鋪P1=0.0435×74.88×200=651.46WP2=0.0483×65.28×200=630.60WP3=0.0492×59.52×200=585.68WP4=0.0508×46.62×200=473.66W走廊P1=0.0897×43.12×150=580.18WP2=0.0897×19.74×150=265.60W電梯前室P=0.0897×11.5×150=154.80W二層開放辦公室P1=0.0432×190.9×300=2476.80WP2=0.0416×107.31×300=1339.80WP3=0.0541×38.43×300=624.20W電梯前室P=0.0735×24.42×150=269.23W三層開放辦公室P1=0.0458×131.04×300=1799.15WP2=0.0537×39.69×300=639.39W電梯前室P=0.0810×17.39×150=211.29W四層辦公室P1=0.0541×38.43×300=624.20WP2=0.0503×56.05×300=847.40WP3=0.0548×36.58×300=601.60WP4=0.0504×53.04×300=802.60W走廊P=0.0874×34×150=457.50W電梯前室P=0.0735×24.42×150=269.23W五至十三層辦公室P1=0.0541×38.43×300=624.20WP2=0.0503×56.05×300=847.40WP3=0.0548×36.58×300=601.60WP4=0.0504×53.04×300=802.60W走廊P=0.0874×34×150=457.50W電梯前室P=0.0735×24.42×150=269.23W十四層辦公室P1=0.0541×38.43×300=624.20WP2=0.0503×56.05×300=847.40WP3=0.0548×36.58×300=601.60WP4=0.0504×53.04×300=802.60W走廊P=0.0874×34×150=457.50W電梯前室P=0.0735×24.42×150=269.23W十五層辦公室P1=0.0541×38.43×300=624.20WP2=0.0542×41.16×300=668.97WP3=0.0578×30.38×300=526.80WP4=0.0518×46.24×300=718.00W電梯前室P=0.0810×17.39×150=211.29W〔6〕計算燈具數(shù)量。N=P/p地下室變配電間N=8套設備間N=9套走廊N=9套過道間、樓梯間N=8套水池N1=6套N2=9套一層店鋪N1=14套〔3×16w〕1套〔28w〕N2=11套〔〔3×16w〕1套〔28w〕3套〔3×18w〕N3=12套〔〔3×16w〕1套〔28w〕3套〔3×18w〕N4=11套〔3×16w〕配電間N=1套〔3×16w〕倉庫N=2套〔3×16w〕樓梯間、電梯前室N=7套〔3×16w〕二層開放辦公室N1=34套〔3×18w〕N2=26套〔3×18w〕N3=12套〔3×18w〕衛(wèi)生間N=5套〔28w〕樓梯間、電梯前室N=10套〔3×16w〕8套〔3×18w〕配電間N=1套〔3×16w〕三層開放辦公室N1=28套〔3×18w〕N2=12套〔3×18w〕衛(wèi)生間N=5套〔28w〕樓梯間、電梯前室N=8套〔3×16w〕7套〔3×18w〕配電間N=1套〔3×16w〕陽臺、平臺N=13套〔22w〕4套〔250w〕四層辦公室N1=12套〔3×18w〕N2=16套〔3×18w〕N3=12套〔3×18w〕N4=12套〔3×18w〕衛(wèi)生間N=5套〔28w〕樓梯間、電梯前室N=10套〔3×16w〕7套〔2×32w〕配電間N=1套〔3×16w〕五至十三層辦公室N1=12套〔3×18w〕N2=16套〔3×18w〕N3=12套〔3×18w〕N4=12套〔3×18w〕衛(wèi)生間N=5套〔28w〕樓梯間、電梯前室N=10套〔3×16w〕7套〔2×32w〕配電間N=1套〔3×16w〕十四層辦公室N1=12套〔3×18w〕N2=16套〔3×18w〕N3=12套〔3×18w〕N4=12套〔3×18w〕衛(wèi)生間N=5套〔28w〕樓梯間、電梯前室N=10套〔3×16w〕7套〔2×32w〕配電間N=1套〔3×16w〕十五層辦公室N1=12套〔3×18w〕N2=12套〔3×18w〕N3=10套〔3×18w〕N4=9套〔3×18w〕衛(wèi)生間N=5套〔28w〕樓梯間、電梯前室N=8套〔3×16w〕7套〔2×32w〕配電間N=1套〔3×16w〕陽臺N=14套〔22w〕3.確定照明供電方式和照明路線布置方式。采用380/221V三相四線制供電。電源有地下室2軸線地下電纜穿鋼管引至地下室變配電間，再由變配電間引至配電間，再分別穿鋼管引至各個樓層配電間的配電箱。供電方式見豎向干線系統(tǒng)圖、配電箱系統(tǒng)圖，由地下室變配電間的配線柜引出十六條單相支路到各樓層配電箱，再由層配電箱引至各個照明和插座等用電。4.計算照明負荷熒光燈的損耗系數(shù)α=0.2；指示燈的損耗系數(shù)α=0；需要系數(shù)Kd=0.8；按將功率因數(shù)提高到cosψ=0.9計算。地下室照明設備容量WL1支路：熒光燈8只×〔3×16〕W。那么：燈具設備容量Pe1=∑Pa〔1+α〕=8×〔3×16〕×〔1+0.2〕=460.8W計算負荷P30.1=KdPe1=0.8×460.8=368.64W計算電流I30.1=P30.1/〔Upcosψ〕=368.64/〔220×0.9〕=2.1AWL2支路：熒光燈9只×〔3×16〕W。那么：Pe2=∑Pa〔1+α〕=9×〔3×16〕×〔1+0.2〕=518.4WP30.2=KdPe2=0.8×518.4=414.72WI30.2=P30.2/〔Upcosψ〕=414.72/〔220×0.9〕=2.1AWL3支路：三防熒光燈6只×〔2×18〕W。那么：Pe3=∑Pa〔1+α〕=6×〔2×18〕×〔1+0.2〕=259.2WP30.3=KdPe3=0.8×259.2=207.36WI30.3=P30.3/〔Upcosψ〕=207.36/〔220×0.9〕=1.1AWL4支路：三防熒光燈9只×〔2×18〕W。那么：Pe4=∑Pa〔1+α〕=9×〔2×18〕×〔1+0.2〕=388.8WP30.4=KdPe4=0.8×388.8=311.04WI30.4=P30.4/〔Upcosψ〕=311.04/〔220×0.9〕=1.6AWL5支路：三防熒光燈9只×〔2×18〕W。那么：Pe5=∑Pa〔1+α〕=9×〔2×18〕×〔1+0.2〕=388.8WP30.5=KdPe5=0.8×388.8=311.04WI30.5=P30.5/〔Upcosψ〕=311.04/〔220×0.9〕=1.6AWL6支路：指示燈6只×〔2×8〕W；應急燈11只×〔2×3〕W。那么：Pe6=∑Pa〔1+α〕=6×〔2×8〕+11×〔2×3〕=162WP30.6=KdPe6=0.8×162=129.6WI30.6=P30.6/〔Upcosψ〕=129.6/〔220×0.9〕=0.7AWL7支路：熒光燈8只×〔3×16〕W。那么：Pe7=∑Pa〔1+α〕=8×〔3×16〕×〔1+0.2〕=460.8WP30.7=KdPe7=0.8×460.8=368.64WI30.7=P30.7/〔Upcosψ〕=368.64/〔220×0.9〕=1.9AWL8支路：插座5只，每只按1000W。那么：Pe8=∑Pa〔1+α〕=5×1000=5000WP30.8=KdPe8=0.8×5000=4000WI30.8=P30.8/〔Upcosψ〕=4000/〔220×0.9〕=20.2AWL9支路：熒光燈1只×〔3×16〕W。那么：Pe9=∑Pa〔1+α〕=1×〔3×16〕×〔1+0.2〕=57.6WP30.9=KdPe9=0.8×57.6=46.08WI30.9=P30.9/〔Upcosψ〕=46.08/〔220×0.9〕=0.3AL1-1：Pe=∑〔Pe1+Pe2+Pe3+……+Pe9〕=460.8+518.4+259.2+388.8+388.8+162+460.8+5000+57.6=7696.4WP30=KdPe=0.8×7696.4=6157.12WI30=P30/〔Upcosψ〕=6157.12/〔220×0.9〕=31.1A一層照明設備容量WL1支路：熒光燈14只×〔3×16〕W；防水防塵吸頂燈1只×28W；插座5只，每只按100W。那么：Pe1=∑Pa〔1+α〕=[14×〔3×16〕+1×28]×〔1+0.2〕+5×100=1340WP30.1=KdPe1=0.8×1340=1072WI30.1=P30.1/〔Upcosψ〕=1072/〔220×0.9〕=5.4AWL2支路：熒光燈11只×〔3×16〕W，3只×〔3×18〕W；防水防塵吸頂燈1只×28W；插座5只，每只按100W。那么：Pe2=∑Pa〔1+α〕=[11×〔3×16〕+3×〔3×18〕+1×28]×〔1+0.2〕+5×100=1361.6WP30.2=KdPe2=0.8×1361.6=1089.28WI30.2=P30.2/〔Upcosψ〕=11089.28/〔220×0.9〕=5.5AWL3支路：熒光燈12只×〔3×16〕W，3只×〔3×18〕W；防水防塵吸頂燈1只×28W；插座5只，每只按100W。那么：Pe3=∑Pa〔1+α〕=[12×〔3×16〕+3×〔3×18〕+1×28]×〔1+0.2〕+5×100=1419.2WP30.3=KdPe3=0.8×1419.2=1135.36WI30.3=P30.3/〔Upcosψ〕=11089.28/〔220×0.9〕=5.8AWL4支路：熒光燈11只×〔3×16〕W；插座4只，每只按100W。那么：Pe4=∑Pa〔1+α〕=11×〔3×16〕×〔1+0.2〕+4×100=1033.6WP30.4=KdPe4=0.8×1033.6=826.88WI30.4=P30.4/〔Upcosψ〕=826.88/〔220×0.9〕=4.2AWL5支路：嵌入式直插拔式節(jié)能筒燈13只×32W。那么：Pe5=∑Pa〔1+α〕=13×32×〔1+0.2〕=499.2WP30.5=KdPe5=0.8×499.2=399.36WI30.5=P30.5/〔Upcosψ〕=399.36/〔220×0.9〕=2.1AWL6支路：熒光燈7只×〔3×16〕W。那么：Pe6=∑Pa〔1+α〕=7×32×〔1+0.2〕=403.2WP30.6=KdPe6=0.8×403.2=322.56WI30.6=P30.6/〔Upcosψ〕=322.56/〔220×0.9〕=1.7AWL7支路：指示燈2只×〔2×8〕W；應急燈4只×〔2×3〕W。那么：Pe7=∑Pa〔1+α〕=2×(2×8)+4×〔2×3〕=56WP30.7=KdPe7=0.8×56=44.8WI30.7=P30.7/〔Upcosψ〕=44.8/〔220×0.9〕=0.3AWL8支路：熒光燈2只×〔3×16〕插座2只，每只按100W。那么：Pe8=∑Pa〔1+α〕=2×〔3×16〕×〔1+0.2〕+2×100=315.2WP30.8=KdPe8=0.8×315.2=252.16WI30.8=P30.8/〔Upcosψ〕=252.16/〔220×0.9〕=1.3AWL9支路：熒光燈1只×〔3×16〕W。那么：Pe9=∑Pa〔1+α〕=1×〔3×16〕×〔1+0.2〕=57.6WP30.9=KdPe9=0.8×57.6=46.08WI30.9=P30.9/〔Upcosψ〕=46.08/〔220×0.9〕=0.3AL1-2：Pe=∑〔Pe1+Pe2+Pe3+……+Pe9〕=1340+1361.6+1419.2+1033.6+499.2+403.2+56+315.2+57.6=6485.6WP30=KdPe=0.8×6485.6=5188.48WI30=P30/〔Upcosψ〕=6485.6/〔220×0.9〕=26.2A二層照明設備容量WL1支路：熒光燈34只×〔3×18〕W。那么：Pe1=∑Pa〔1+α〕=34×〔3×18〕×〔1+0.2〕=2203.2WP30.1=KdPe1=0.8×2203.2=1762.56WI30.1=P30.1/〔Upcosψ〕=1762.56/〔220×0.9〕=8.9AWL2支路：熒光燈26只×〔3×18〕W。那么：Pe2=∑Pa〔1+α〕=26×〔3×18〕×〔1+0.2〕=1684.8WP30.2=KdPe2=0.8×1684.8=1347.84WI30.2=P30.2/〔Upcosψ〕=1347.84/〔220×0.9〕=6.8AWL3支路：熒光燈12只×〔3×18〕W。那么：Pe3=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.3=KdPe3=0.8×777.6=622.08WI30.3=P30.3/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL4支路：熒光燈8只×〔3×18〕W，10只×〔3×16〕W。那么：Pe4=∑Pa〔1+α〕=[8×〔3×18〕+10×〔3×16〕W]×〔1+0.2〕=1094.4WP30.4=KdPe4=0.8×1094.4=875.52WI30.4=P30.4/〔Upcosψ〕=875.52/〔220×0.9〕=4.5AWL5支路：防水防塵吸頂燈5只×28W。那么：Pe5=∑Pa〔1+α〕=5×28×〔1+0.2〕=168WP30.5=KdPe5=0.8×168=134.4WI30.5=P30.5/〔Upcosψ〕=134.4/〔220×0.9〕=0.7AWL6支路：指示燈5只×〔2×8〕W；應急燈6只×〔2×3〕W，4只×36W。那么：Pe6=∑Pa〔1+α〕=5×〔2×8〕+6×〔2×3〕+4×36=260WP30.6=KdPe6=0.8×260=208WI30.6=P30.6/〔Upcosψ〕=208/〔220×0.9〕=1.1AWL7支路：插座13只，每只按500W。那么：Pe7=∑Pa〔1+α〕=13×500=6500WP30.7=KdPe7=0.8×6500=5200WI30.7=P30.7/〔Upcosψ〕=5200/〔220×0.9〕=26.3AWL8支路：插座5只，每只按500W。那么：Pe8=∑Pa〔1+α〕=5×500=2500WP30.8=KdPe8=0.8×2500=2000WI30.8=P30.8/〔Upcosψ〕=2000/〔220×0.9〕=10.1AWL9支路：插座9只，每只按500W。那么：Pe9=∑Pa〔1+α〕=9×500=4500WP30.9=KdPe9=0.8×4500=3600WI30.9=P30.9/〔Upcosψ〕=3600/〔220×0.9〕=18.2AWL10支路：熒光燈1只×〔3×16〕W。那么：Pe10=∑Pa〔1+α〕=1×〔3×16〕×〔1+0.2〕=57.6WP30.10=KdPe10=0.8×57.6=46.08WI30.10=P30.10/〔Upcosψ〕=46.08/〔220×0.9〕=0.3AL2-1：Pe=∑〔Pe1+Pe2+Pe3+……+Pe10〕=2203.2+1684.8+777.6+1094.4+168+260+6500+2500+4500+57.6=19745.6WP30=KdPe=0.8×19745.6=15796.48WI30=P30/〔Upcosψ〕=15796.48/〔220×0.9〕=79.8A三層照明設備容量WL1支路：熒光燈28只×〔3×18〕W。那么：Pe1=∑Pa〔1+α〕=28×〔3×18〕×〔1+0.2〕=1814.4WP30.1=KdPe1=0.8×1814.4=1451.52WI30.1=P30.1/〔Upcosψ〕=1451.52/〔220×0.9〕=7.4AWL2支路：熒光燈12只×〔3×18〕W。那么：Pe2=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.2=KdPe2=0.8×777.6=622.08WI30.2=P30.2/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL3支路：熒光燈7只×〔3×18〕W,8只×〔3×16〕W。那么：Pe3=∑Pa〔1+α〕=[7×〔3×18〕+8×〔3×16〕]×〔1+0.2〕=914.4WP30.3=KdPe3=0.8×914.4=731.52WI30.3=P30.3/〔Upcosψ〕=731.52/〔220×0.9〕=3.7AWL4支路：防水防塵吸頂燈5只×28W。那么：Pe4=∑Pa〔1+α〕=5×28×〔1+0.2〕=168WP30.4=KdPe4=0.8×168=134.4WI30.4=P30.4/〔Upcosψ〕=134.4/〔220×0.9〕=0.7AWL5支路：指示燈5只×〔2×8〕W；應急燈6只×〔2×3〕W，4只×36W。那么：Pe5=∑Pa〔1+α〕=5×〔2×8〕+6×〔2×3〕+4×36=260WP30.5=KdPe5=0.8×260=208WI30.5=P30.5/〔Upcosψ〕=208/〔220×0.9〕=1.1AWL6支路：壁燈13只×22W；投光燈4只×250W。那么：Pe6=∑Pa〔1+α〕=〔13×22+4×250〕×〔1+0.2〕=1543.2WP30.6=KdPe6=0.8×1543.2=1234.56WI30.6=P30.6/〔Upcosψ〕=1234.56/〔220×0.9〕=6.3AWL7支路：插座11只，每只按500W。那么：Pe7=∑Pa〔1+α〕=11×500=5500WP30.7=KdPe7=0.8×5500=4400WI30.7=P30.7/〔Upcosψ〕=4400/〔220×0.9〕=22.3AWL8支路：插座5只，每只按500W。那么：Pe8=∑Pa〔1+α〕=5×500=2500WP30.8=KdPe8=0.8×2500=2000WI30.8=P30.8/〔Upcosψ〕=2000/〔220×0.9〕=10.1AWL9支路：熒光燈1只×〔3×16〕W。那么：Pe9=∑Pa〔1+α〕=1×〔3×16〕×〔1+0.2〕=57.6WP30.9=KdPe9=0.8×57.6=46.08WI30.9=P30.9/〔Upcosψ〕=46.08/〔220×0.9〕=0.3AL1-2：Pe=∑〔Pe1+Pe2+Pe3+……+Pe9〕=1814.4+777.6+914.4+168+260+1543.2+5500+2500+57.6=13535.2WP30=KdPe=0.8×13535.2=10828.16WI30=P30/〔Upcosψ〕=10828.16/〔220×0.9〕=54.7A四層照明設備容量WL1支路：熒光燈16只×〔3×18〕W。那么：Pe1=∑Pa〔1+α〕=16×〔3×18〕×〔1+0.2〕=1036.8WP30.1=KdPe1=0.8×1036.8=829.44WI30.1=P30.1/〔Upcosψ〕=829.44/〔220×0.9〕=4.2AWL2支路：熒光燈12只×〔3×18〕W。那么：Pe2=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.2=KdPe2=0.8×777.6=622.08WI30.2=P30.2/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL3支路：熒光燈12只×〔3×18〕W。那么：Pe3=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.3=KdPe3=0.8×777.6=622.08WI30.3=P30.3/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL4支路：熒光燈12只×〔3×18〕W。那么：Pe4=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.4=KdPe4=0.8×777.6=622.08WI30.4=P30.4/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL5支路：熒光燈7只×〔2×32〕W,10只×〔3×16〕W。那么：Pe5=∑Pa〔1+α〕=[7×〔2×32〕+10×〔3×16〕]×〔1+0.2〕=1113.6WP30.5=KdPe5=0.8×1113.6=890.88WI30.5=P30.5/〔Upcosψ〕=890.88/〔220×0.9〕=4.5AWL6支路：防水防塵吸頂燈5只×28W。那么：Pe6=∑Pa〔1+α〕=5×28×〔1+0.2〕=168WP30.6=KdPe6=0.8×168=134.4WI30.6=P30.6/〔Upcosψ〕=134.4/〔220×0.9〕=0.7AWL7支路：指示燈7只×〔2×8〕W；應急燈6只×〔2×3〕W，3只×36W。那么：Pe7=∑Pa〔1+α〕=7×〔2×8〕+6×〔2×3〕+3×36=2656WP30.=KdPe7=0.8×256=204.8WI30.7=P30.7/〔Upcosψ〕=204.8/〔220×0.9〕=1.1AWL8支路：插座5只，每只按500W。那么：Pe8=∑Pa〔1+α〕=5×500=2500WP30.8=KdPe8=0.8×2500=2000WI30.8=P30.8/〔Upcosψ〕=2000/〔220×0.9〕=10.1AWL9支路：插座4只，每只按500W。那么：Pe9=∑Pa〔1+α〕=4×1500=2000WP30.9=KdPe9=0.8×2000=1600WI30.9=P30.9/〔Upcosψ〕=1600/〔220×0.9〕=8.1AWL10支路：插座5只，每只按500W。那么：Pe10=∑Pa〔1+α〕=5×500=2500WP30.10=KdPe10=0.8×2500=2000WI30.10=P30.10/〔Upcosψ〕=2000/〔220×0.9〕=10.1AWL11支路：插座4只，每只按500W。那么：Pe11=∑Pa〔1+α〕=4×500=2000WP30.11=KdPe11=0.8×2000=1600WI30.11=P30.11/〔Upcosψ〕=1600/〔220×0.9〕=8.1AWL12支路：熒光燈1只×〔3×16〕W。那么：Pe12=∑Pa〔1+α〕=1×〔3×16〕×〔1+0.2〕=57.6WP30.12=KdPe12=0.8×57.6=46.08WI30.12=P30.12/〔Upcosψ〕=46.08/〔220×0.9〕=0.3AL2-2：Pe=∑〔Pe1+Pe2+Pe3+……+Pe12〕=1036.8+777.6+777.6+777.6+1113.6+168+256+2500+2000+2500+2000+57.6=13964.8WP30=KdPe=0.8×13964.8=11171.84WI30=P30/〔Upcosψ〕=11171.84/〔220×0.9〕=56.5A五至十三層照明設備容量WL1支路：熒光燈16只×〔3×18〕W。那么：Pe1=∑Pa〔1+α〕=16×〔3×18〕×〔1+0.2〕=1036.8WP30.1=KdPe1=0.8×1036.8=829.44WI30.1=P30.1/〔Upcosψ〕=829.44/〔220×0.9〕=4.2AWL2支路：熒光燈12只×〔3×18〕W。那么：Pe2=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.2=KdPe2=0.8×777.6=622.08WI30.2=P30.2/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL3支路：熒光燈12只×〔3×18〕W。那么：Pe3=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.3=KdPe3=0.8×777.6=622.08WI30.3=P30.3/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL4支路：熒光燈12只×〔3×18〕W。那么：Pe4=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.4=KdPe4=0.8×777.6=622.08WI30.4=P30.4/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL5支路：熒光燈7只×〔2×32〕W,10只×〔3×16〕W。那么：Pe5=∑Pa〔1+α〕=[7×〔2×32〕+10×〔3×16〕]×〔1+0.2〕=1113.6WP30.5=KdPe5=0.8×1113.6=890.88WI30.5=P30.5/〔Upcosψ〕=890.88/〔220×0.9〕=4.5AWL6支路：防水防塵吸頂燈5只×28W。那么：Pe6=∑Pa〔1+α〕=5×28×〔1+0.2〕=168WP30.6=KdPe6=0.8×168=134.4WI30.6=P30.6/〔Upcosψ〕=134.4/〔220×0.9〕=0.7AWL7支路：指示燈7只×〔2×8〕W；應急燈6只×〔2×3〕W，3只×36W。那么：Pe7=∑Pa〔1+α〕=7×〔2×8〕+6×〔2×3〕+3×36=2656WP30.=KdPe7=0.8×256=204.8WI30.7=P30.7/〔Upcosψ〕=204.8/〔220×0.9〕=1.1AWL8支路：插座5只，每只按500W。那么：Pe8=∑Pa〔1+α〕=5×500=2500WP30.8=KdPe8=0.8×2500=2000WI30.8=P30.8/〔Upcosψ〕=2000/〔220×0.9〕=10.1AWL9支路：插座4只，每只按500W。那么：Pe9=∑Pa〔1+α〕=4×1500=2000WP30.9=KdPe9=0.8×2000=1600WI30.9=P30.9/〔Upcosψ〕=1600/〔220×0.9〕=8.1AWL10支路：插座5只，每只按500W。那么：Pe10=∑Pa〔1+α〕=5×500=2500WP30.10=KdPe10=0.8×2500=2000WI30.10=P30.10/〔Upcosψ〕=2000/〔220×0.9〕=10.1AWL11支路：插座4只，每只按500W。那么：Pe11=∑Pa〔1+α〕=4×500=2000WP30.11=KdPe11=0.8×2000=1600WI30.11=P30.11/〔Upcosψ〕=1600/〔220×0.9〕=8.1AWL12支路：熒光燈1只×〔3×16〕W。那么：Pe12=∑Pa〔1+α〕=1×〔3×16〕×〔1+0.2〕=57.6WP30.12=KdPe12=0.8×57.6=46.08WI30.12=P30.12/〔Upcosψ〕=46.08/〔220×0.9〕=0.3ALa-b：Pe=∑〔Pe1+Pe2+Pe3+……+Pe12〕=1036.8+777.6+777.6+777.6+1113.6+168+256+2500+2000+2500+2000+57.6=13964.8WP30=KdPe=0.8×13964.8=11171.84WI30=P30/〔Upcosψ〕=11171.84/〔220×0.9〕=56.5A十四層照明設備容量WL1支路：熒光燈16只×〔3×18〕W。那么：Pe1=∑Pa〔1+α〕=16×〔3×18〕×〔1+0.2〕=1036.8WP30.1=KdPe1=0.8×1036.8=829.44WI30.1=P30.1/〔Upcosψ〕=829.44/〔220×0.9〕=4.2AWL2支路：熒光燈12只×〔3×18〕W。那么：Pe2=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.2=KdPe2=0.8×777.6=622.08WI30.2=P30.2/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL3支路：熒光燈12只×〔3×18〕W。那么：Pe3=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.3=KdPe3=0.8×777.6=622.08WI30.3=P30.3/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL4支路：熒光燈12只×〔3×18〕W。那么：Pe4=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.4=KdPe4=0.8×777.6=622.08WI30.4=P30.4/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL5支路：熒光燈7只×〔2×32〕W,10只×〔3×16〕W。那么：Pe5=∑Pa〔1+α〕=[7×〔2×32〕+10×〔3×16〕]×〔1+0.2〕=1113.6WP30.5=KdPe5=0.8×1113.6=890.88WI30.5=P30.5/〔Upcosψ〕=890.88/〔220×0.9〕=4.5AWL6支路：防水防塵吸頂燈5只×28W。那么：Pe6=∑Pa〔1+α〕=5×28×〔1+0.2〕=168WP30.6=KdPe6=0.8×168=134.4WI30.6=P30.6/〔Upcosψ〕=134.4/〔220×0.9〕=0.7AWL7支路：指示燈7只×〔2×8〕W；應急燈6只×〔2×3〕W，3只×36W。那么：Pe7=∑Pa〔1+α〕=7×〔2×8〕+6×〔2×3〕+3×36=2656WP30.=KdPe7=0.8×256=204.8WI30.7=P30.7/〔Upcosψ〕=204.8/〔220×0.9〕=1.1AWL8支路：插座5只，每只按500W。那么：Pe8=∑Pa〔1+α〕=5×500=2500WP30.8=KdPe8=0.8×2500=2000WI30.8=P30.8/〔Upcosψ〕=2000/〔220×0.9〕=10.1AWL9支路：插座4只，每只按500W。那么：Pe9=∑Pa〔1+α〕=4×1500=2000WP30.9=KdPe9=0.8×2000=1600WI30.9=P30.9/〔Upcosψ〕=1600/〔220×0.9〕=8.1AWL10支路：插座5只，每只按500W。那么：Pe10=∑Pa〔1+α〕=5×500=2500WP30.10=KdPe10=0.8×2500=2000WI30.10=P30.10/〔Upcosψ〕=2000/〔220×0.9〕=10.1AWL11支路：插座4只，每只按500W。那么：Pe11=∑Pa〔1+α〕=4×500=2000WP30.11=KdPe11=0.8×2000=1600WI30.11=P30.11/〔Upcosψ〕=1600/〔220×0.9〕=8.1AWL12支路：熒光燈1只×〔3×16〕W。那么：Pe12=∑Pa〔1+α〕=1×〔3×16〕×〔1+0.2〕=57.6WP30.12=KdPe12=0.8×57.6=46.08WI30.12=P30.12/〔Upcosψ〕=46.08/〔220×0.9〕=0.3AL1-2：Pe=∑〔Pe1+Pe2+Pe3+……+Pe12〕=1036.8+777.6+777.6+777.6+1113.6+168+256+2500+2000+2500+2000+57.6=13964.8WP30=KdPe=0.8×13964.8=11171.84WI30=P30/〔Upcosψ〕=11171.84/〔220×0.9〕=56.5A十五層照明設備容量WL1支路：熒光燈12只×〔3×18〕W。那么：Pe1=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.1=KdPe1=0.8×777.6=622.08WI30.1=P30.1/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL2支路：熒光燈10只×〔3×18〕W。那么：Pe2=∑Pa〔1+α〕=10×〔3×18〕×〔1+0.2〕=648WP30.2=KdPe2=0.8×648=518.4WI30.2=P30.2/〔Upcosψ〕=518.4/〔220×0.9〕=2.6AWL3支路：熒光燈9只×〔3×18〕W。那么：Pe3=∑Pa〔1+α〕=9×〔3×18〕×〔1+0.2〕=583.25WP30.3=KdPe3=0.8×583.2=466.56WI30.3=P30.3/〔Upcosψ〕=466.56/〔220×0.9〕=2.4AWL4支路：熒光燈12只×〔3×18〕W。那么：Pe4=∑Pa〔1+α〕=12×〔3×18〕×〔1+0.2〕=777.6WP30.4=KdPe4=0.8×777.6=622.08WI30.4=P30.4/〔Upcosψ〕=622.08/〔220×0.9〕=3.2AWL5支路：熒光燈7只×〔2×32〕W,8只×〔3×16〕W。那么：Pe5=∑Pa〔1+α〕=[7×〔2×32〕+8×〔3×16〕]×〔1+0.2〕=998.4WP30.5=KdPe5=0.8×998.4=798.72WI30.5=P30.5/〔Upcosψ〕=798.72/〔220×0.9〕=4.1AWL6支路：防水防塵吸頂燈5只×28W。那么：Pe6=∑Pa〔1+α〕=5×28×〔1+0.2〕=168WP30.6=KdPe6=0.8×168=134.4WI30.6=P30.6/〔Upcosψ