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工厂供电习题答案

工厂供电习题答案 Ting Bao was revised on January 6, 20021习题2-9解:2台14KW (100%ε=)电焊机,接AB 相1002221428NAB N N P P P KW εεεε===⨯= 1台20KW (100%ε=)电焊机,接BC 相1001002020100NBC N P P KW εεε=== 1台30KW (60%ε=)电焊机,接CA 相100603023.24100NCA N P P KW εεε=== 等效相电压设备容量为:(查表2-6)NA AB A NAB CA A NCA P p P p P --=+ 1.028023.2428=⨯+⨯=KW q q NA AB A NAB CA A NCA Q P P --=+0.5828 1.1623.2443.2=⨯+⨯=Kvar NB BC B NBC AB B NAB P p P p P --=+ 1.02002820=⨯+⨯=KW NB BC B NBC AB B NAB Q q P q P --=+0.5820 1.162844.08=⨯+⨯=Kvar p p NC CA C NCA BC C NBC P P P --=+ 1.023.2402023.24=⨯+⨯=KW NC CA C NCA BC C NBC Q q P q P --=+0.5823.24 1.162036.68=⨯+⨯= Kvar电热干燥箱 查表得x1k 0.40.6=取 电焊机 x2k 0.35=A 相负荷:30x1x2k k A NA NA P P P =+’0.5200.352819.8=⨯+⨯=KW 30A Q =+=130x1x2k k B NB NB P P P =+0.5300.352022=⨯+⨯=KW 30B Q =+= Kvar30C P 0.5200.3523.2418.134=⨯+⨯=KW30C Q =+= KvarB 相负荷最大 3030332266B P P KW ==⨯= 30303315.42846.284var B Q Q K ==⨯= 303046.284tg 0.766Q P ϕ=== ∴cos 0.819ϕ=3080.61A S KV ===3080.61122.481.7320.38I A ===⨯ 习题2-10 解:一.高压线路1. 功率损耗 2330310WL WL P I R -∆=⨯KW 2330310WL WL Q I X -∆=⨯ Kvar 已知 30P =900KW, cos ϕ=得309001.732100.86I ===⨯⨯00.4620.92WL R R L ==⨯=Ω(查附表8可得0R ,0X ) 00.34420.688WL X X L ==⨯=Ω23360.420.921010.07WL P -∆=⨯⨯⨯=KW 23360.420.688107.534WL Q -∆=⨯⨯⨯= Kvar2. 年电能损耗, 由max 4500a T h =查图2-5max a T 与τ关系曲线得3000h τ=则有WL WL W P τ∆=∆=10.0730*******.21KW h ⨯= 二.电力变压器1.功率损耗 20T K P P P β∆=∆+∆ 20T K Q Q Q β∆=∆+∆2台电力变压器SL7-800/1030/cos 900/0.860.65422800N P S ϕβ===⨯ 查附表1,得 0P =1540KW K P ∆= 0%I = %K U = 则 21.549.90.654 5.775T P ∆=+⨯=KW 20%%()100100K T N I U Q S β∆=+21.7 4.5800(0.654)29var 100100K =+⨯= 2.年电能损耗208760T K W P P I β∆=∆⨯+∆21.5487609.90.6543000=⨯+⨯⨯= 习题2-11解:查表2-4。

电气开关制造厂 x K = cos ϕ=(tg ϕ=) 30x 0.355840N P K P ∑==⨯=2044KW 303020440.88Q P tg ϕ==⨯= Kvar 303020442725.3cos 0.75P S KVA ϕ===习题2-6解:2539.650.1NP P KW εεε=== 3050.162.60.8P P KW εη=== 303062.6 1.732108.44Q P tg ϕ==⨯= Kvar30S ==30125.2190.251.7320.38I A ===⨯ 习题2-7解:NO1设备组 17.543 2.2734.9N P KW ∑=+⨯+⨯= 查表2-2得 x1k = 1cos ϕ= 1tg ϕ=则有301110.234.9 6.98X N P K P KW ∑==⨯=3013011 6.98 1.339.28Q P tgϕ==⨯= Kvar30111.61A S KV ===30111.6117.641.7320.38I A ===⨯ NO2设备组 查表2-2 x2k = 2cos ϕ= 2tg ϕ= 2N P ∑=3KW302220.83 2.4X N P K P KW ∑==⨯=302 2.40.751.8Q =⨯= Kvar 3023AS KV === 3023 4.561.7320.38I A ===⨯ NO3设备组 单台设备 x3k 应为1,查表2-2 3cos ϕ= 3tg ϕ=则有 303P =2KW 303303320.20.4Q P tg ϕ==⨯= Kvar 303303322.04cos 0.98P S KVA ϕ===303 2.04 3.11.7320.38I A ===⨯ 干线负荷:30301302303P P P P =++=++2= 30301302303Q Q Q Q =++=++= Kvar3016.16A S KV ===3016.1624.571.7320.38I A ===⨯ 习题2-8解:查表2-5 NO1:50.40.14N P P ∑+ 1cos ϕ= NO2:50.250.65N P P ∑+ 2cos ϕ= NO3: 50.50.5N P P ∑+ 3cos ϕ=NO1: 3010.4P =⨯⨯⨯(7.5+34+2.2)+0.1434.9 =+=301301113.566 1.73223.496Q P tg ϕ==⨯= Kvar30127.13A S KV ===3027.1341.241.7320.38I A ===⨯ NO2: 3020.2530.653 2.7P =⨯+⨯=KW 3023022 2.70.5 2.02Q P tg ϕ==⨯= Kvar302 3.37A S KV ===302 3.375.131.7320.38I A ===⨯ NO3: 3030.520.522P =⨯+⨯=KW303303320.330.66Q P tg ϕ==⨯= Kvar303 2.1A S KV ===303 2.13.21.7320.38I A ===⨯ 干线负荷:308.64 4.886 1.95116.516P =+++=KW308.68 1.732 4.886 1.732 1.950.7510.3325.29Q =⨯+⨯+⨯+⨯= Kvar3030.2A S KV ===30I ==45.91A 习题4-7解:㈠NO1 甲电设备组各支线截面和熔体电流的选择3078.1870.855NNP P KW η=== 3030/cos 8.187/0.8110.107S P KVA ϕ===3010.10715.361.7320.38NM I I A ====⨯ ①选熔断器 NF NM I I ≥=15.36A5.515.3633.792.5stst NMNF I K I I A αα⨯≥=== 选较大者 取NF I =40A选RT0-100/40型号熔断器,熔体NF I =40A②选支线截面 al 15.36NM I I A ≥= al2.5NFI I 〈 查手册 BLV-500-3⨯⨯(14)+(1 2.5)导线穿钢管敷设 在环境温度 00al 3020I A θ=+=时,则al 2.5NFII=40/20=2< A 满足要求故选支线为BLV-500-3⨯⨯(14)+(1 2.5)导线㈡NO2用电设备组支线截面和熔断器选择301011.630.86NNPP KW η===11.6321.51.7320.380.82NMI A===⨯⨯521.5107.73st st NMI K I A==⨯=①选熔断器NF NMI I≥=21.5A107.7343.12.5stNFII Aα≥==取较大者NF I=50A选RT0-100/50型号熔断器,熔体NFI=50A②选支线截面al21.5NMI I A≥=BLV-500-3⨯⨯(16)+(14)导线穿钢管敷设在环境温度00al3026I Aθ=+=时,则al2.5NFII=50/26=1.92< A 满足要求故选支线为BLV-500-3⨯⨯(16)+(14)导线㈢NO3用电设备组支线截面和熔断器选择302022.730.88NNPP KWη===22.7342.11.7320.380.82NMI A===⨯⨯4.542.1189.5st st NMI K I A==⨯=①选熔断器 NF NM I I ≥=42.1A 189.554.143.5stNF I I A α≥== 取NF I =60A 选RT0-100/60型号熔断器,熔体NF I =60A ②选支线截面 al 42.1NM I I A ≥=BLV-500-3⨯⨯(116)+1(110)导线穿钢管敷设 在环境温度 00al 3046I A θ=+=时, 则al2.5NFI I =60/46=1.3< A 满足要求 故选支线为BLV-500-3⨯⨯(116)+1(110)导线 ㈣C 段干线:30P =32KW 30S = 30I =75Apk 30st117584.4815.36144.2NM I I I I A =+-=+-=①选熔断器 30NF I I ≥=75A pk144.1357.652.5NF I I A α≥== 取较大者NF I =80A 选RT0-100/80型号熔断器 ②C 段干线截面 al 3075I I A ≥=选BLV-500-3⨯⨯(150)+(125)导线穿钢管敷设 在环境温度 00al 3093I A θ=+=时, 则al1.5NFI I =80/93< A (11.50.66α== 见表4-4 生产车间动力干线)满足要求 ㈤B 段干线: 30I =123Apk 123107.7321.5209.23I A =+-=①选熔断器 30NF I I ≥=123Apk209.2359.783.5NF I I A α≥== 取较大者NF I =150A 选RT0-200/150型号熔断器 ②B 段干线截面 al 30123I I A ≥=选BLV-500-3⨯⨯(195)+(150)导线穿钢管敷设 在环境温度 00al 30142I A θ=+=时, 则al1.5NFI I =150/142=1.05< A 满足要求 ㈥A 段干线: 30I =194Apk 194189.542.1341.4I A =+-=①选熔断器 30NF I I ≥=194A pk341.497.543.5NF I I A α≥== 取较大者NF I =200A 选RT0-400/200型号熔断器 ②A 段干线截面 al 30194I I A ≥=选BLV-500-3⨯⨯(1185)+(195)导线穿钢管敷设 在环境温度 00al 30215I A θ=+=时, 则al1.5NFI I =200/215< A 满足要求保护装置动作电流与导线,电缆允许电流的倍数关系:回路名称导线,电缆种类及敷设方式电流倍数熔断器自动开关动力支线裸线,穿管线及电缆al0.5NFII<oplal.0II<1长延时脱扣oplalII<4.5瞬时脱扣动力干线al.5NFII<1动力支线明设单芯绝缘线al.5NFII<1照明电路al.0NFII<1习题4-8解:初设0.35/kmX=Ω3x i2i1%q10iNXU LU==∑=20.35(800 1.56002500 2.5) 1.281010⨯+⨯+⨯=⨯%%%5 1.28 3.72R XU U U=-=-=3.72%10000372100R R NU U U V=⨯=⨯=则导线截面3ii11piR NS LU Uγ==∑=1[1000 1.580021000 2.5]0.03237210⨯+⨯+⨯⨯⨯=472mm取S=502mm 选择铝绞线LJ-50 如下图:习题4-9解: 1p 10000.8800KW =⨯=1q 800tg 8000.75600ϕ=⨯=⨯=Kvar 2p 4000.9360KW =⨯= 2q 3600.484174.3=⨯= Kvar查表LJ-352mm 导线 ar 1D =时 00.92/km R =Ω 00.366/km X =Ω则220i 0i 2i 1i 11%[p q ]10i i N U R L X L U ===+∑∑ =21[8000.50.92360 1.50.926000.3660.5174.3 1.50.366]1010⨯⨯+⨯⨯+⨯⨯+⨯⨯⨯ =如下图:习题4-10解:2'ii 1(39003000)3(33802210)57i A S Lj j S L =∑+⨯++⨯==∑=+2ii 1(33802210)2(39003000)47i B S Lj j S L =∑+⨯++⨯==∑=+A B c d S S S S +=+ 正确2A c S S S =-=+-(3380+j2210)=+查附表8 : 00.335/km X =Ω 00.34/km R =Ω22ii22i 1i 1%p q 1010i i N N R X U L L U U ===+∑∑=220.340.335(33802705.74)(22102654.34)10101010⨯+⨯+⨯+⨯⨯⨯ => 不满足要求应将导线的截面放大一级选LJ-120 如下图:。

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