电机与拖动基础 第二版 刘启新 第二章部分答案

'
I a = (n0 − n)
CeΦ N 0.068 = (1618 − 1530) = 10.5A 0.571 Ra
n = 1625 − 7.196Tem 稳态时， Tem = TL ，n=1369r/min UN 220 = = 4062.5r / min 0.0677 × 0.8 CeΦ Ra 0.315 = = 11.24 2 0.0677 × 0.6466 × 0.82 Ce CTΦ
nN = Ia =
U N − CeΦ nN 220 − 0.8 × 0.193 × 1000 = = 131.2A Ra 0.5
TemN = 0.8CTΦ N I a = 0.8 × 9.55 × 0.193 × 131.2 = 193.5N ⋅ m
（2）
Ia = n = =
220 0.5 − × 67.5 = 1206r / min 0.193 × 0.8 0.193 × 0.8
第二章 习题作业参考答案 15．
1 ⎛ U I − P ⎞ 1 ⎛ 110 × 20.1 − 1.75 × 103 ⎞ （1） Ra = ⎜ N N 2 N ⎟ = ⎜ ⎟ = 0.571Ω IN 2⎝ 20.12 ⎠ ⎠ 2⎝
CeΦ N =
U N − I N Ra 110 − 20.1 × 0.571 = = 0.068 nN 1450
(3)
η=
P2 P1
n 1206.3 = 10 = 12.063kW nN 1000
P2 = PN
P 1 = U N I a = 220 × 67.5 = 14850W = 14.85kW
课
UN Ra Ia − C eΦ C eΦ
后
答
TN TN I 54 = = N = = 67.5A CTΦ 0.8CTΦ N 0.8 0.8
U − CeΦ N nN 176 − 0.193 × 1000 = = −34A Ra 0.5
TemN = 9.55CeΦ N I a = 9.55 × 0.193 × (−34) = −62.67N ⋅ m
(2)
I a = I N = 54A
n = =
(3)
P2 = T2 ⋅ Ω = 95.54 ×
答
案
网
I a = (n0 − n)
CeΦ N 0.068 = (1618 − 1530) = 10.5A Ra 0.571
ww
Ra
w.
Ra 0.571 I a = 1618 − × 0.5 × 20.1 = 1533r / min CeΦ N 0.068
kh
（2）
da
B 点（ Tem = 13.04N ⋅ m， n = nN = 1450r / min ）
n0 '' =
β '' =
n = 4062.5 − 11.24Tem 稳态时， Tem = TL ，n=3733r/min （5）图，略 17． U − E a UN 220 （1） I st = N = = = 3283.6A = 15.8 I N Ra Ra 0.067 （2） I st = 1.5 I N = 1.5 × 207.5 = 311.25A Rst = UN 220 − Ra = − 0.067 = 0.64Ω 311.25 I st
w.
A 点（ Tem = 0 ， n0 = 1618r / min ）
co
固有特性两点坐标为：
m
n = n0 − β 'Tem = 3250 − 52.9Tem
稳态时， Tem = TL ，n=1407r/min （3） UN Ua 2 = 110 = 1625r / min = n0 = CeΦ N CeΦ N 0.0677
CTΦ N = CeΦ N × 9.55 = 1.843 TemN = 99.522N ⋅ m
3
课
（3）加一个问题：如果由 1.2nN 的下降转速开始提升负荷，提升转速要求达到 0.5nN，用什么方法实现？电枢中应串接 多大电阻？画出相应的机械特性曲线。 可用串电阻方法
0.296 + Rad × 0.3 × 179.395 0.2935 × 2.803
（2）
β' =
Ra + Rad 0.315 + 2 = = 52.9 2 Ce CTΦ N 0.0677 × 0.6466
1
课
CeΦ N =
U N − I N Ra 220 − 53.7 × 0.315 = = 0.0677 nN 3000
后
1 ⎛ U I − P ⎞ 1 ⎛ 220 × 53.7 − 10 × 103 ⎞ （1） Ra = ⎜ N N 2 N ⎟ = ⎜ ⎟ = 0.315Ω 2⎝ IN 53.7 2 ⎠ ⎠ 2⎝
案
网
ww
w.
4
UN R − a Ia C eΦ C eΦ
kh
23． （1）
da
w.
原η =
P2 10000 = = 84.17% P1 220 × 54
co
m
η=
η=
P2 12.063 = = 81.2% P1 14.85
24. (1) nN = Ia = Ra U − Ia C eΦ N C eΦ N
m
η=
P2 7719.9 × 100% = = 81.2% P1 9504
课
后
答
案
网
ww
w.
5
2 × 3.14 × 772 = 7719.9W 60
kh
P 1 = UI N = 176 × 54 = 9504W
da
w.
176 − 0.5 × 54 = 772r / min 0.193
co
UN R − a Ia C eΦ C eΦ
da
UN Ra Ia − CeΦ N CeΦ N
w.
co
Ra
m
U N − I N Ra 220 − 64 × 0.296 = = 0.2935 685 nN
（1）n 不能突变, n = nN = 1000r / min
Ia =
U N −CeΦ N nN 220 − 0.193 × 1000 = = 13.5A Ra + Rad 0.5 + 1.5
19．
E aN = U N − I N Ra = 220 − 110 × 0.12 = 206.8V I max = −2.5I N = −2.5 × 110 = −275A Rad = − EaN 206.8 − Ra = − − 0.12 = 0.632Ω I max −275
能耗制动机械特性方程为： n = −2.67Tem
后
答
机械特性方程为： n = −749.6 − 1.34Tem
案
−0.3 × 64
− 0.296 = 0.81Ω
网
( −1.2 × 685 + 749.6 ) × 0.29.35
ww
w.
= −768.9r / min
kh
220 0.296 − × 0.3 × 64 0.2935 0.2935 = −749.6 − 19.36
CeΦ N
Ia
16．
CTΦ N = 9.55CeΦ N = 0.6466 n0 = UN 220 = = 3250r / min CeΦ N 0.0677 Ra 0.315 = = 7.196 2 Ce CTΦ N 0.0677 × 0.6466
β=
n = n0 − β Tem = 3250 − 7.196Tem
n = − n0 − (2) n = −n0 − Rad = =
0.12 × 110 = −1354r / min 0.1723
Ra + Rad Ia CeΦ N −IN − Ra − 0.12 = 0.23Ω
( n + n0 ) × CeΦ N
−110
( −1500 + 1277 ) × 0.1723
Tem = CTφN I a = 24.88N ⋅ m
（2） TL 保持不变且 Tem = TL 稳定状态时， I a = I N = 54A
n=
（3）
UN 2 2 T = 1139.90 − − × 99.522 = 580.315r / min 2 emN CeΦ N Ce CTΦ N 0.193 × 1.843 P − I 2R P2 10000 − 542 × 1.5 × 100% = N a ad = = 47.36% P U N IN 220 × 54 1
20.
（1） n = − n0 − CeΦ N = Ra C eΦ N Ia
EaN 206.8 = = 0.1723 1200 nN
2
课
后
答
案
网
ww
w.
kh
da
w.
co
m
（4）
n0 =
UN 220 = = 1277r / min CeΦ N 0.1723 Ra CeΦ N I a = −1277 −
CTΦ N = 9.55CeΦ N = 0.649 n0 = UN 110 = = 1618r / min CeΦ N 0.068
TemN = CTΦ N I N = 0.649 × 20.1 = 13.04N ⋅ m
机械特性方程为： n = 1618 − 12.93Tem
n = n0 −
（3）
n = n0 −
21．
（1） CeΦ N = n = − n0 − =− =−
CeΦ N
Ia
（2） Rad = =
( n + n0 ) × CeΦ N
−IN