电力拖动基础课后习题答案
- 格式:doc
- 大小:1.02 MB
- 文档页数:17
第二章
2-1解: ① 122053.40.4
0.132/(min )1500
N N a e N N U I R C V r n ---⨯Φ=
== 9.55 1.26/T N e N C C N m A Φ=Φ=
② 67.28N t N N T C I N m =Φ=
29.5563.67260
N N N N N P P
T N m n n π=
== ③ 02 3.61N N T T T N m =-= ④ 01666.67/min N
e N
U n r C =
=Φ 002
1657.99/min a
e t N
R n n T r C C =-
=Φ实际 ⑤ 0000.5()1583.34/min a
N N N
I n n n n n n r I =-
∆=--= ⑥ 2200.1321400
880.4
N e N a a U C n I A R -Φ-⨯=
==
2-2 解:
131333
1222333112212285.2170.4110
0.646170.40.646
1.710.13
1.710.130.2221.710.130.381.710.130.6460.2220.130.0920.380.N N st st a st a st a st a st st a st st st I I A A U R I R R R R R R R R r R R r R R λλλλ==⨯==
===
====⨯=Ω==⨯=Ω==⨯=Ω=-=-=Ω=-=-3322220.1580.6460.380.266st st st r R R =Ω=-=-=Ω
2-3 解:
取12118.32236.6N I I A ==⨯=
式中1I 为最大允许电流,稳态时,负载电流L N I I =,令切换电流
2 1.15 1.15L N I I I ==
则1
2
1.74I I λ=
= 1lg 3.7lg N
a
U I R
m λ
⎛⎫ ⎪
⎝⎭==
取m=4.则4
1 1.679N a
U I R λ=
=,122 1.191.679N N L I I
I I I λ===>
4342312(1)0.0790.1330.2230.374st a st st st st st st R R R R R R R R λλλλ=-=Ω==Ω==Ω==Ω
、
式中,1st R ,2st R ,3st R ,4st R 分别是第一级,第二级,第三级,第四级切除阻值。 2-4解:
①电枢电路不串电阻达稳态时,
0.80.80.8126.8
em L T N a N T N N a N T T T C I T C I I I φφ=====⇒==
0.20415N N a
N N N a e N N N a e N N
U I R U E I R C n I R C n φφ-=+=+⇒=
= 则电枢回路不串电阻时的转速:
12200.1126.8
1015.53/min 0.20415
N a a e N U I R n r C φ--⨯=
==
② 22
()N a s N a s N a s a
a e N e T N e N e N e N
U R R U R R U R R I n T I C C C C C C φφφφφ++-+=
-=-=
220(0.10.3)126.8
829.2/min 0.20415
r -+⨯=
=
③ 1a
a e N a a U E I R C n I R φ''=+=+ (由于机械惯性,转速n 不变)将电源电压降至188V 后的瞬时电枢电流: 1880.204151015.53
193.20.1
a
a U E I A R --⨯'===- 稳态转速:31880.1126.8
858.78/min 0.204150.20415
a a U I R n r --⨯=
==
④ 180%N a
a e N a a U E I R C n I R φ'''''=+=+ (转速n 不突变) 则瞬时电枢电流: 180%22080%0.204151015.53
541.450.1
N e N a
a U C n I A R φ--⨯⨯''===
稳态转速:42200.1126.8
1269.4/min 80%80%0.20415
N a a e N U I R n r C φ--⨯=
==⨯
⑤ 串电阻调速,设所串电阻为5R
555()
600/min
2200.20415600
0.10.669126.8
N a a e N
N e N a a U I R R n r C U C R R I φφ-+=
=--⨯⇒=
-=-=Ω
降压调速
55600/min 135.17a a
e N a a e N
U I R n r U C n I R V C φφ-=
=⇒=+=Ω
2-5解:
① 端电压不变,励磁回路总电阻不变,所以励磁不变。 电枢回路串入电阻、转速不能突变,所以串入瞬间反电势不变。
105.8110105.8
6.460.150.5
11
()9.559.559.550.674
0.674 6.46 4.35a N N a N a a a e T T T em T a E U I R V U E I A
R R E C n C n C n E
C n
T C I N m =-=--=
==++=Φ=
Φ=Φ∴Φ===Φ=⨯=