供配电技术 王玉华 课后答案

Qe?K??Qc0?Qc1?Qc2?Qc3?Qc4??0.9??1080?680?750?700?720?kvar?3537 kvarSe?Pe2?Qe2?44522?35372kVA?5688.36kVA

A?548.01A

1.73?63?Un高压侧P?PSe?4540.3 3kWe?0.015 e1Qe1?Qe?0.06Se?3878.3kvar高压侧

Ice?Se?5688.36Se1?P?QIce1?Se12e12e1?5971.25kVA?98.62A

3?Un答：该总降压变电所高压侧总的计算负荷Pe1=4540.33kW，Qe1=3878.3kvar ，Se1=5971.25 kVA，Ice1=98.62A。

24解：通风机Kd1?0.8,cos?1?0.8,tan?1?0.75

Qc1?Pc1tan?1?48?1.73kvar?83.04kvar点焊机Kd2?0.35,cos?2?0.6,tan?2?1.33

Pe2??2PN2?0.65?10.5kW?8.47kWPc2?Kd2Pe2?0.35?8.47kW?2.96kW

Pc1?Kd1Pe1?0.8?60kW?48kW

Qc2?Pc2tan?2?2.96?1.33kvar?3.94kvar有连锁的连续运输机械Kd3?0.7,cos?3?0.75,tan?3?0.88

Pc3?Kd3Pe3?0.7?40kW?28kWQc3?Pc3tan?3?28?0.88kvar?24.64kvar行车Kd4?0.15,cos?4?0.5,tan?4?1.73

Pe4??4PN4?0.15?10.2kW?3.95kWPc4?Kd4Pe4?0.15?3.95kW?0.59kWQc4?Pc4tan?4?0.59?1.73kvar?1.03kvar同时系数取K?p,K?q?0.9 车间计算负荷为：

Pc?K?p?Pci?0.9?(48?2.96?28?0.59)kW?71.60kWQc?K?q?Pci?0.9?(83.04?3.94?24.64?1.03)kvar?101.39kvarSc?Pc2?Qc2?124.12kVAIc?Sc3UN?124.12A?179.16A3?0.4

25解：需要系数法：查表A-1可得：

冷加工机床： Kd1?0.2,cos?1?0.5,tan?1?1.73

Pc1?0.2?152kW?30.4kWQ?30.4?1.73kvar?52.6kvarPc2?0.8?6kW?4.8kWQ?4.8?0.75kvar?3.6kvar c2

c1 通风机组： Kd2?0.8,cos?2?0.8,tan?2?0.75

车间的计算负荷:

Pc?0.9??30.4?4.8?kW?31.68kWQc?0.9??52.6?4.8?kvar?50.58kvarSc?Pe? Qe?59.68kVAIc?Se3UN?59.68/(1.732?0.38)?90.7A22

二项式法: 查表A-1可得：

冷加工机床:b1?0.14,c1?0.4,x1?5,cos?1?0.5,tan?1?1.73

(bPe?)1?0.14?152kW?21.28kW?cPx?1?0.4??10?7?2?2?4.5?kW?13.2kW因为n=6＜2x2 ,取x2=3.则

通风机组:b2?0.65,c2?0.25,x2?5,cos?2?0.8,tan?2?0.75

(bPe?)2?0.65?6kW?3.9kW?cPx?2?0.25?3kW?0.75kW

显然在二组设备中 第一组附加负荷(cPx)1最大 故总计算负荷为：

Pc?(21.28?3.9?13.2)kW?38.38kWQc?（21.28?1.73?3.9?0.75?13.2?1.73)kvar?62.56kvarSc?Pe2?Qe2?73.4kVAIc?Se/(3UN)?111.55A26解：（1）电热箱的各相计算负荷

查表A-1得，Kd?0.7,cos??1,tan??0，因此各相的有功计算负荷为： A相 PcA1?KdPeA?0.7?3?2kW?4.2kW B相 PcB1?KdPeB?0.7?6?1kW?4.2kW C相 PcC1?KdPeC?0.7?4.5?1kW?3.15kW

（2）单头手动弧焊机的各相计算负荷

查表A-1得，Kd?0.35,cos??0.35,tan??2.68；查表2-4得，cos??0.35时，则

pAB?A?pBC?B?pCA?C?1.27pAB?B?pBC?C?pCA?A??0.27qAB?A?qBC?B?qCA?C?1.05

qAB?B?qBC?C?qCA?A?1.63将各相换算成??100%的设备容量，即：

PAB??N1PN1?0.65?20kW?16.12kWPBC??N2PN2?0.1?6?3kW?5.69kWPCA??N3PN3?0.5?10.5?2kW?14.85kW各相的设备容量为 A相

PeA?pAB?APAB?pCA?APCA?[1.27?16.12?(?0.27)?14.85]kW?16.46kWQeA?qAB?APAB?qCA?APCA?(1.05?16.12?1.63?14.85)kvar?41.13kvarPeB?pBC?BPBC?pAB?BPAB?[1.05?5.69?(?0.27)?16.12]kW?1.62kW

B相 C相

QeB?qBC?BPBC?qAB?BPAB?(1.27?5.69?1.63?16.12)kvar?33.50kvarPeC?pCA?CPCA?pBC?CPBC?[1.27?14.85?(?0.27)?5.69]kW?17.32kW

QeC?qCA?CPCA?qBC?CPBC?(1.05?14.85?1.63?5.69)kvar?24.87kvar

各相的计算负荷为

A相 B相 C相

PcA2?KdPeA?0.7?16.46kW?11.52kWQcA2?KdQeA?0.7?41.13kvar?28.79kvarPcB2?KdPeB?0.7?1.62kW?1.13kWQcB2?KdQeB?0.7?33.50kvar?23.45kvarPcC2?KdPeC?0.7?17.32kW?12.12kWQcC2?KdQeC?0.7?24.87kvar?17.41kvar

（3）各相总的计算负荷为（设同时系数为0.95） A相 B相 C相

PcA?K?(PcA1?PcA2)?0.95?(4.2?11.52)kW?14.93kWQcA?K?(QcA1?QcA2)?0.95?(0?28.79)kvar?27.35kvarPcB?K?(PcB1?PcB2)?0.95?(4.2?1.62)kW?5.53kWQcB?K?(QcB1?QcB2)?0.95?(0?33.50)kvar?31.83kvarPcC?K?(PcC1?PcC2)?0.95?(3.15?12.12)kW?14.51kWQcC?K?(QcC1?QcC2)?0.95?(0?17.41)kvar?16.54kvar

（4）总的等效三相计算负荷

因为A相的有功计算负最大，所以Pc?m?PcA?14.93kW,Qc?m?QcA?27.35kvar，则

Pc?3Pc?m?3?14.93kW?44.79kWQc?3Qc?m?3?27.35kvar?82.05kvarSc?44.792?82.052kVA?93.48kVA Ic?Sc93.48?A?142.03A3UN3?0.38

27解：（1）查表A-1

Kd1?0.2,cos?1?0.5,tan?1?1.73小批量金属切削机床：Pc1?Kd1Pe1?0.2?920kW?184kW

Qc1?Pc1tan?1?184?1.73kvar?318.32kvarKd2?0.8,cos?2?0.8,tan?2?0.75通风机： Pc2?Kd2Pe2?0.8?56kW?44.8kW

Qc2?Pc2tan?2?44.8?0.75kvar?33.6kvarKd3?0.2,cos?3?0.5,tan?3?1.73起重机： Pc3?2Kd3?3Pe3?0.4?0.15?76kW?11.78kW

照明：

Qc3?Pc3tan?3?11.78?1.73kvar?20.38kvarPc4?42kWQc4?0kvar

总的计算负荷为（设同时系数为0.95）

Pc?K?(Pc1?Pc2?Pc3?Pc4)?0.95?(184?44.8?11.78?42)kW?268.45kWQc?K?(Qc1?Qc2?Qc3?Qc4)?0.95?(318.32?33.6?20.38?0)kvar?353.69kvarSc?Pc2?Qc2?444.03kVAcos?1?为

变压器的功率损耗

Pc?0.60Sc

?PT?0.015Sc?0.015?444.03kW?6.66kW?QT?0.06Sc?0.06?444.03kvar?26.64kvar变电所高压侧总的计算负荷为

Pc??Pc??PT?275.11kWQc??Qc??QT?380.33kvarSc??Pc'2?Qc'2?469.40kVAS'469.40Ic??c?A?27.10A3UN3?10变电所高压侧的功率因数为

cos?'?（2）

Pc'?0.586 Sc'Qcc?Pc?(tan?1?tan?0)?268.45?[tan(arccos0.60)?tan(arccos0.95)]?268.45kvar268.45n??1028实际补偿容量为280kvar

（3）变电所低压侧视在计算负荷为

Sc'1?Pc2?(Qc?Qcc)2?278.38kVA'?PT?0.015?278.38kW?4.18kW'?QT?0.06?278.38kvar?16.70kvar

变电所高压侧总的计算负荷为

Pc'2?Pc'1??PT'?268.45?4.18?272.63kW'Qc'2?Qc'1??QT?353.69?280?16.70?90.29kvar

S?P?Q?287.19kVA?S?469.4?287.19?182.21kVA变电所高压侧的功率因数为

'c2'2c2'2c2Pc'2cos?'?'?0.95

Sc228解：tan?av1?tan(arccos0.67)?1.1080 tan?av2?tan(arccos0.9)?0.4843

Qcc?Pav(tan?av1?tan?av2)?KaLPc(tan?av1?tan?av2)=0.75?320?(1.1080-0.4843)kvar =149.688kvarn=Qcc/qcN?149.688/10=15考虑到三相均衡分配，每相装设5个，此时并联电容器的实际补偿容量为150kvar。选择BW-10.5-10-1型电容器15个。

29解：Pav?Wa/t?2.5?10/8760?2853.88kW

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