工厂供电习题答案

1、工厂供电习题答案习题2-9 解：2台14KW （ = 100%）电焊机，接AB相"2%底=2%=2xI4 = 28KW1台20KW（£ = 100%）电焊机，接BC相= 2QKWPnbc1台30KW（£ = 60%）电焊机，接CA相= 2324KWPg等效相电压设备容量为：（查表2-6）厂厶一讥汕 + *4 叭c =l0x28+0x2324 = 28KWQxa =qm 尽B +qc"叭C =0.58x28 + LI6x23.24 = 43.2KvarPnb = P bjbPnbc + P ab-b'Pnab = 1 *0x20+0x28 = 20

2、 KWQm = %7 叭肚 + /叭AB =o.58x 20 + 1.16x28 = 44.08KvarPnc =p ca-cnca +P bcy凡bc = 1°x2324+0x20 = 2324KWQm =+ fac-cvBc =0.58 x 23.24+1.16x 20 = 36-68 Kvar电热干燥箱查表得匕严0406取电焊机匕2 =035A相负荷J巳30 =匕|叱¥八+匕2叱以=05x20+035x28 = 198KWPb3o =k乂 + kx2£vB =0*5x30+0-35x20 = 22KWKvar30 =0.5x20+035x23.24 = 18

3、J34KWB相负荷最大P=3禺，=3x22 = 66KWfl«> =，Q/押=3x15.428 = 46.284Kvar即=&=竺型“7/. cos 0 = 0.819Sjo = JPj+Q/ = V66' +46.284- = 80.61KUA" ®N 1-732x038=122.48/1习题2-10解：一高压线路1.功率损耗傀乞=3厶o'.Rm xlCr'KWQm = “厶亠心丄 X10“ Kvar己知二900KW, cos 0二得900s/3U costp 1.732x10x0.86心=凡厶=0,46X 2 = 0.92

4、0 （查附表8可得R。，X。）X 帕=XqL = 0.344x2 = 0.6880=3x 60.42' X 0.92 x 1 (T = 10.07 KW =3x6042-x0.688x10"' =7.534 Kvar2.年电能损耗，由和5=4500/?查图2-57；”"与旷关系曲线得丁 = 3000"则有 %，c=AExr = 1007x3000 = 3000021KW力二电力变压器1.功率损耗 号=+耳+2台电力变压器SL7-800/10即心0_900/0.86_06 钏2Sn 2x800查附表1,得P,=1540KW 叭=久 =则 &

5、= 1.54 +99x 0.6542 =5.775 KW07筈+筈仍= 800（益+辭0.6对= 2922年电能损耗叫=化 X 8760 +氓"J/ = 1.54x8760 + 9.9x0.654-x3000习题2-11 解：查表2-4。电气开关制造厂 Kx= cos0=( tg(p=y/5o = Kx/5v2： =O35x584O = 2044KW 030 = P'Q 彳 g® = 2044x0.88 = Kvar习题2-6徹g底®6X思时OZ0)= &)“gp = 626x 1.732 = 108.44 KvarSjO =(/ +030? =/

6、.=几空?一 = 190.254l732xO38习题2-7解：N01设备组张=75 + 4x3 + 22x7 = 349KW查表2-2得匕产 cos%= tg(P =则有 Eoi=02x349 = 698KW0301 = RoWg® =6.98x1,33 = 9.28 Kvar$301= J&f+Q刘 2 = 76,98'+9.28-=11,61 KVA$301G' =1.732x0.38N02设备组査表2-2 k坨cos 化二 tg(p产z2=3KW心2 =心2叭口 =08x3 = 24KW0302 =2.4x0.75 = 1.8 KvarS刘=屁 + 漏=

7、a + 1.82= 3KUA葩Un 1-732x0.38=4.56AN03设备组单台设备k对应为1,查表2-2cos 03=必3 =则有 />o3=2KW2S” 孟=両=2.。4加1.732X0.38*303干线负荷：&严&|+&位+ &+2=(> = 0K>i + O302 + O303 二+二 KvarS30 = J/V+Q3J = Vli38'+1L48' =16.16KVA30S30®N 1-732x038Z6 =25习题2-8解：查表2-5N01：0.4A+0.14%cos (p =NOl：N02：N03：O

8、257+O65仏0.54 +0.5%&j=04x(7 5+3x4+22)COS0 二cos =+0. 14x34.9=+=0301=13566xl732 = 23496 KvarSjui = JeJ + QoJ = Ji 3.5662+23.4962 = 27 J 3KUA2 ®N 1.732x0.3827 13= 41.24/1N02：&J2 =025x3 + 065x3 = 27KW0奶=心Mg輕=2.7x0.5 = 2.02 KvarSjo? = 5/27'+2.02- = 3.37 KVANOS：3 37九 =513A忑Un 1-732x0.38鬆=0

9、5x2 + 05x2 = 2KW03O3 = /03“g® = 2x02 = 04 Kvar0303 = Ao3"g03 =2x0.33 = 0.66 Kvar$303 = J加+= Q+0.662 = 2.1KVA2 疊-=勺= 3.2A®N 1 732 X 0.38干线负荷：召 0=864 + 4886 +1.95 + 1 = 16516KW030= &68xl.732 + 4.886x 1.732 + L95x0.75 +1x0.33 = 25.29 KvarSjo = J 町+qJ = 716,516'+ 25.29" = 30.

10、2/CVA9/和=-=4591A习题4-7 解：NOl甲电设备组各支线截面和熔体电流的选择p 7出厂亠亦EE"S旳= &o/cos0 = & 187/0.81 = 10.107 KM1 = 15.36Ad "代®“1.732x0.38选熔断器 L 2 /丽=15. 36ANF a a> ZiL = Kg.NM _ 55x 15.36 = 33 79A选较大者 取/m=40A 选RT0-100/40型号熔断器，熔体G =4OA选支线截ffl打 I粉=15.364 也2.5Lq査手册 BLV-500-3(lx4) +(1x2. 5)导线穿钢管敷设

11、在环境温度&。=+30。时人=20人则L=4o/2O=225 A 满足要求故选支线为 BLV-500-3C1X4) +(1x2. 5)导线(2)N02用电设备组支线截面和熔断器选择11.63P 10I =出 0= =力 C A如"®nCos0_1732xO38xO825 =5x21.5 = 107.73/1选熔断器>1补,=21. 5AL h =畔害= 43.M 取较大者心二50ACt2 A选RT0-100/50型号熔断器，熔体/呼=50A选支线截面打斗=21.54BLV-500-3(lx6) +(1x4)导线穿钢管敷设在环境温度 = +30°时,人

12、严26A则1=50/26=1. 9225 A 满足要求故选支线为BLV-500-3(lx6) +(1x4)导线N03用电设备组支线截面和熔断器选择22.73弓厂铲卷= 22.73KW= Ro = 42 AN" y/3U cos<p L732x0.38x0.82/刃=心丿"=4.5 X 42=189.5 ANM选熔断器I心=42. 1A1厝2 =聲¥ = 54.I4A 取=6OA选RT0-100/60型号熔断器，熔体/朋二60A选支线截面ZInm =42aBLV-500-3(lxl6) +1(1x10)导线穿钢管敷设在环境温度% = +30°时,厶=

13、46A则 ZiiL =60/46=1. 3<2.5A 满足要求故选支线为 BLV-500-3(lxl6) +1(1x10)导线C段干线：= 32KW S30 =厶产75Apk =/刃 +人-/丽| =75+ 84.48-15.36 = 144.2A选熔断器/沖/3o=75AZ 2于= 57.65A 取较大者=80A选RT0-100/S0型号熔断器C段干线截面 几/3o=75A 选BLV-500-3(lx50) +(1x25)导线穿钢管敷设在环境温度6*0= +30°时,厶=93A则如:二8O/93<1.5A'al"嵌m见表4-4生产车间动力干线）满足要求

14、B段干线：厶产123A/pk =123 + l0773-215 = 20923A选熔断器/网厶0=123人NF号=警=沁A取较大者加150A选RTO-200/150型号熔断器B段干线截面/注 > 厶0 = 123A选BLV-500-3(lx95) +(1x50)导线穿钢管敷设在环境温度q)=+30°时厶= 142力则 如L二 150/142二 1. 05<1.5A 满足要求(A 段干线：/,o=194A/pk =194 + I895-421=3414A选熔断器L>/3o=194ANP>- =冷芋=97.54A 取较大者心=2OOA选RT0-400/200型号熔

15、断器A段干线截面/川 > 厶。=194A 选BLV-500-3( 1x185) +(1x95)导线穿钢管敷设在环境温度&。=+30°时,仃=215?1则如1=2OO/2151.5A满足要求 人I保护装置动作电流与导线，电缆允许电流的借数关系:回路名称导线，电缆种类 及敷设方式电流倍数熔断器动开关动力支线裸线，穿管线及电缆05/呃 <1.0长延时脱扣心 <4. 5打瞬时脱扣动力干线515动力支线明设单芯绝缘线1.5照明电路人F <1.0习题4-8 解；初设Xo=O35G/km10台 10x10-(800x1.5 + 600x2 + 500x2.5) =

16、1.28Ur % fU % -山X % = 5 -1.28 = 3.723 72必=必5=而"。詡2W则导线截面s=yPi厶r4J科 i-iLcmsmUE® 沁 2+i(xx)6=47 mm'取 5=50 nun选择铝绞线LJ-50 如下图:L5Kma,5Kn0,5弘V1#1000+j800V 2#800+j6004/ 3#lOOO+jBOO习题4-9解；P| =1000x0.8 = 800KWq, = 800 X tg 0 = 800 x 0.75 = 600 Kvar p2 =400x0.9 = 360KW q? = 360 X 0.484 = 174.3 Kvar查表 LJ-35mm-导线 Q = 1 时 凡=0.920/km X,= 03660/km则山=吋召叭厶+Sq儿厶= _800x0.5x0.92 + 360x1.5x0.92 + 600x0.366x0.5 + I74.3x 1.5x0.366如下图:500in1#lODomlOOOm2#4oom习题4-10解 S _ (3900+ 73000)x3 + (3380+ 72210)x5 爭厶(3380+ )2210)x2 +