工厂供电习题答案2

1、 习题 2-9解：2 台 14kw（）电焊机，接 ab 相e 100%=ep = 2 pnab= 2 p = 214 = 28kwe100neen1 台 20kw（）电焊机，接 bc 相e 100%=e100p = pnbc= 20= 20kwe100100ne1 台 30kw（e = 60%）电焊机，接 ca 相e60p = pnca= 30= 23.24kwne e100100等效相电压设备容量为：（查表 2-6）kwp = pnap + pnabp =1.028+ 023.24 = 28ncaab- aab- aca- akvarq = qnap + qnabp = 0.5828+1.1

2、623.24 = 43.2ncaca- akwp = pnbp + pnbcp =1.020+ 028 = 20nabbc -bbc -bab-bq = qp + qnbcp = 0.5820 +1.1628 = 44.08kvarnabnbab-bp = pncp + pncap =1.023.24+ 020 = 23.24kwnbcca-cbc -ckvarq = qp + qncap = 0.5823.24+1.1620 = 36.68nbcncca-cbc -c电热干燥箱 查表得电焊机取 0.5k = 0.4 0.6x1k = 0.35x2a 相负荷：= 0.520+ 0.3528 =

3、19.8kwp = k p + k pa30x1nax2na=5.684+9.346=15.03kvarqa30 = 0.530+ 0.3520 = 22kwp = k p + k p1b30x1nbx2nb=4.06+11.368=15.428 kvarqb30kwp = 0.520 + 0.3523.24 =18.134c30=4.718+8.05=12.768 kvarqc30b 相负荷最大p = 3 p = 3 2 2 = 6k6 w3 0b3 0q = 3 q = 315.428 = 46.284kvar30b3046.284qtg = 0.7cos = 0.819 jj30p306

4、6s = p + q = 66 + 46.284 = 80.61kva222230303080.611.7320.38si =30=122.48a303un习题 2-10解：一.高压线路1. 功率损耗kw-3dp = 3 i r 102wl30wlkvar-3dq = 3 i x 102wl30wl已知 =900kw,=0.86 得cos jp309001.732100.86p60.42a=i =30=303u cosjn(查附表 8 可得 , )r = r l = 0.46 2 = 0.92wwlr x000x = x l = 0.344 2 = 0.688wwl0kwdp = 360.42

5、 0.9210 =10.072-3wlkvardq = 360.42 0.68810 = 7.5342-3wl2. 年电能损耗, 由t查图 2-5t 与 关系曲线= 4500htmax amax a 得t= 3000h则有dw = dp=10.073000 = 30000.21kw htwlwl二.电力变压器1.功率损耗dp = dp + dpdq = dq + dqbb22t0kt0k2 台电力变压器 sl7-800/10p / cosj 900 / 0.86b= 0.654=302s2800n查附表 1,得 =1540kw9.9kwpdp =0k=1.7=4.5u %i %0k则kwdp

6、=1.54 + 9.90.654 = 5.7752ti % u %1 . 7 4 . 5) = 8 0 0 ( + 0 . 6 5 4= k) 2 9 v a rdq = s (+b0k22100 1001 0 0 1 0 0tn2.年电能损耗dw = dp 8760 + dpi =1.548760 + 9.90.65423000b2t0k=26196.06kwh习题 2-11解:查表 2-4。电气开关制造厂=0.35=0.75（ =0.88）cosjtg jkx2044kwp = kx p = 0.355840 =30n 1802.64 kvarq = p tgj = 20440.88 =3

7、03020440.75ps =30= 2725.3kva30jcos 习题 2-6解：e0.4p = p= 39.6= 50.1kwee0.25n2550.10.8pkvarp = e = 62.6kwq = p tgj = 62.61.732 =108.44303030h125.2kvas = p+q=22303030125.21.7320.38s303ui =30=190.25an习题 2-7解：no1 设备组p = 7 . 5+ 4 3+ 2 .2 =7 k3w4 . 9n 1查表 2-2 得 =0.2=0.6=1.33kcosjtgjx111则有p = k p = 0.234.9 =

8、6.98kw301x 1n 1kvarq = p tgj = 6.981.33 = 9.283013011s = p+ q= 6.98+ 9.28=11.61kva222230130130111.611.7320.38s3013ui =301=17.64anno2设备组 查表2-2=0.8=0.8=0.75kcosjtgjx222=3kwpn 2 p = k p = 0.83 = 2.4kw302x 2n 2kvarq = 2.40.75 =1.8302s = p3 0 2+ q= 2 . 43 0 2+1 . 8 = k3v a22223 0 23si =3 0 2= 4 . 5a61 .

9、7 32 0 . 3 83 0 23unno3 设备组 单台设备 应为 1，查表 2-2kx3=0.98=0.2cosjtgj33则有 =2kwkvarpq = p tgj = 20.2 = 0.4303303303322.041.7320.38ps3033us =303= 2.04kva= 3.1ai303jcos0.983033n干线负荷：=6.98+2.4+2=11.38kwp = p + p + p30301302303=9.28+1.8+0.4=11.48 kvarq = q + q + q30301302303s = p+ q= 11.38+11.48=16.16kva222230

10、303016.16s303ui =30= 24.57a1.7320.38n习题 2-8解：查表 2-5 no1：no2：=0.5=0.8=0.950.4p + 0.14pc o sj5n 10.25p + 0.65pc o sj5n 2no3：0.5p + 0.5pcosj5n 3no1：p = 0.4（7.5+34+2.2）+0.14 34.9301=8.68+4.886=13.566kwkvarq = p tgj =13.566 1.732 = 23.4963013011 s = p+q= 13.566+ 23.496 = 27.13kva222230130130127.131.7320.

11、38s303ui =30= 41.24anno2：kwp = 0.253+ 0.653 = 2.7302kvarq = p tgj = 2.70.5 = 2.023023022s = p+ q= 2.7+ 2.02 = 3.37kva22223023023023.37s3023ui =302= 5.13a1.7320.38nno3：kwp = 0.5 2 + 0.5 2 = 2303kvarj 2 0.33 0.66q = p tg = =3033033s = p+ q= 2+ 0.66= 2.1kva22223033033032.1s3033ui =303= 3.2a1.7320.38n干线

12、负荷：kwp = 8.64 + 4.886 +1.95 +1 =16.51630kvarq = 8.681.732 + 4.8861.732 +1.950.75 +10.33 = 25.2930s = p+ q= 16.516+ 25.29 = 30.2kva2222303030s303u45.91ai =30=n习题 4-7解：no1 甲电设备组各支线截面和熔体电流的选择7pp =30= 8.187kwn0.855hns = p / cosj = 8.187 / 0.81 10.107=kva3030 10.107s303ui = i =15.36a= 33.79a1.7320.38nm30

13、n选熔断器15.36ai i =n fn m5.515.36ik isti =stanma2.5nf选较大者 取 =40ainf选 rt0-100/40 型号熔断器，熔体 =40ainfi选支线截面i i =15.36a2.5nfalnmial查手册 blv-500-（3 14）+(12.5)导线穿钢管敷设在环境温度= +30 时,i = 20aq00ali则=40/20=22.5a 满足要求nfial故选支线为 blv-500-（3 14）+(12.5)导线no2 用电设备组支线截面和熔断器选择10pp =30=11.63kwn0.86hn11.63pi =nm= 21.5a303u cos

14、j 1.7320.380.82ni = k i = 5 21.5 =107.73aststnm选熔断器21.5ai i =n fn m107.732.5i取较大者 =50ai = 43.1aistanfnf选 rt0-100/50 型号熔断器，熔体 =50ainf选支线截面i i = 21.5aalnm blv-500-（3 16）+(14)导线穿钢管敷设在环境温度q30 时,i = 26a= +00ali则a 满足要求=50/26=1.922.5nfial故选支线为 blv-500-（3 16）+(14)导线no3 用电设备组支线截面和熔断器选择20pp =30= 22.73kwn0.88h

15、n22.731.7320.380.82p303u cosni =nm= 42.1aji = k i = 4.5 42.1 =189.5aststnm选熔断器42.1ai i =n fn m189.53.5i取 =60ai = 54.14aistanfnf选 rt0-100/60 型号熔断器，熔体 =60ainf选支线截面blv-500-（3 116）+1(110)导线穿钢管敷设i i = 42.1aalnm在环境温度q30 时,i = 46a= +00ali则a 满足要求=60/46=1.32.5nfial故选支线为 blv-500-（3 116）+1(110)导线c 段干线：32kw49.3

16、kva=75ai30p =30s =30i = i + i - i= 75 + 84.48 -15.36 = 144.2apk30st1nm 1选熔断器75ai i =nf30 144.132.5i取较大者 =80ai pk= 57.65aianfnf选 rt0-100/80 型号熔断器c 段干线截面选 blv-500-（3 150）+(125)导线穿钢管敷设i i = 75aa l3 0在环境温度= +30 时,i = 93aq00ali则a=80/931.5nfial1（a见表 4-4 生产车间动力干线）满足要求=1.50.66b 段干线：=123ai30i =123 +107.73 - 21.5 = 209.23apk选熔断器123ai i =nf30209.233.5i取较大者 =150ai pk= 59.78aianfnf选 rt0-200/150 型号熔断器b 段干线截面选 blv-500-（3 195）+(150)导线穿钢管敷设i i =123aa l3 0在环境温度= +30 时,i =142aq00ali则=150/142=1.051.5a 满足要求nfiala 段干线：=194ai30i =194 +189.5 - 42.1 = 341.4 apk 选熔断器194ai i =nf30341.43.