电拖试习题解答

胡幸鸣主编《电机及拖动基础》部分习题解答

1-6 解： A U P I N N N 43.630230

101453

=⨯==

1-7 解： A U P I N

N N N 08.8089

.022*******

=⨯⨯==

η 1-11解：1）单叠绕组 106.4V

1200

105.32

60152

260 22=⨯⨯⨯⨯⨯=Φ=

Φ===-n a pN n Ce E p a a 2）单波绕组 mi n /21.118410

5.315221

60210pN Ea60a n 2,12

r p a =⨯⨯⨯⨯⨯=Φ=

==- 1-14 解：A I I I A R U I f N a f N f 78280 2110

220

=-=-====

1831.38W

149.6-440-219.02-2460 2460149601760060.1491496001.001.01496085.0176001760080220440110202.219036.0781112222

==---=+==-=-==⨯===⨯=⨯==⨯===⨯===⨯==∑∑s

f cua Fe m o

N N S

N N

N N f f f

a a cua P P P p P P ：P

W

P P p ：W P ：p W

P ：P W

I U ：P W

R I ：P

W R I ：P 空载损耗总损耗附加损耗额定输出功率额定输入功率励磁损耗电枢铜耗η

1-15 解：N aN N

N N N N N N N I I A U P I I U P ==⨯⨯==

∴= 51.509

.02201010 3

他励ηη m

N T T T m

N n P T m N n P T W

I E P V R I U E R I E U N N N N N em a a em a N N a a a a ⋅=-=-=⋅=⨯⨯==⋅=⨯=⨯==⨯===⨯-=-=∴+=96.758.7954.87 58.7955.91200101055.9 54.8755.91200

07

.1100055.9

07.1100051.5078.217 78.217044.051.50220 03

2-7 解：1）A R U I a N st 22001.0220

===

2）Ω=-=-=

11.0200

220

a st N st R I U R 设))2至51( ( 5.1L T .T T T T T st N st N st =>=一般保证

17.25V

0.11151.551 5.1725.1 ,0=⨯⨯====a N st N st R I .U A

I I T 则略

2-9 解： 201

.01000

202

.094220 =⨯-=-=

ΦN a a N N e n R I U C 1）Ω=-⨯-=-Φ-=

428.0202.094

800201.0220a a N e pa R I C U R

2）V R I C U a N N e 79.179202.094800201.0=⨯+⨯=+Φ=

2-11 解： 098.01000

065

.0185110 =⨯-=-=ΦN a a N N e n R I U C

0.8时的转速N T mi n /29.1024098

.0065

.01858.01108.0r C R I U n N e a N =⨯⨯-=Φ-=

1）能耗制动瞬时N k -1.8I I 0 U m in;/29.1024===b r n Ω=-⨯-⨯-=-Φ-=

236.0065.0185

8.129.1024098.00a bk N e bk R I n C U R

2）反接制动瞬时N bk N -1.8I I -U U m in;/29.1024===r n

Ω=-⨯-⨯--=-Φ-=

567.0065.0185

8.129

.1024098.0110a bk N e bk R I n C U R

2-12 解：194.01000

847

.03.30220 =⨯-=-=

ΦN a a N N e n R I U C

1) 能耗制动：Ω=-⨯-⨯-=-Φ-=

35.2847.03.308.0)

400(194.00a bk N e bk R I n C U R

倒拉反接：Ω=-⨯-⨯-=-Φ-=

43.11847.08

.03.30)

400(194.0220a bk N e bk R I n C U R

2）min /85.1239194

.0847

.08.03.30220r C R I U n N e a bk -=⨯⨯--=Φ-=

3) 与上面的各小题一样，稳定制动电流：A I bk 3.308.0⨯=

min /308.221194

.0)

10847.0(8.03.30220)(r C R R I U n N e bk a bk -=+⨯⨯-=Φ+-=

解题技巧：应用电动势平衡方程式，无论电动、起动、调速、制动状态，公式是同一个，但n ）、Ia 、（I 、bk U 等对应的值各有特点

3-4

解：A U S I I N N N N 33.32310

10310560033

3111=⨯⨯⨯==

=ϕ A U S I I N

N

N

N 60.592310

15.331056003333

3222=⨯⨯⨯=

=

=ϕ

3-7

解：电压比：3.510

6.610353

3

21=⨯⨯==N N U U k Wb B S m Fe m 22357.0415.11015804=⨯⨯=⨯=Φ-

70522357

.05044.4103544.43

11=⨯⨯⨯=Φ=m f U N

1333

.570512===

k N N 3-8 1）高串、低串：A I N 273.2211005000

1=⨯=

A I N

73.222

11050002=⨯=