电机原理及拖动 (彭鸿才 著) 机械工业出版社 课后答案

P42 1－33 解：
P1 = U N I N = 220 × 81.7 = 17974(W )
PN = P1ηN = 17974 × 0.85 = 15278(W )
I fN
= UN Rf
= 220 = 2.48( A) 88.8
IaN = I N − I fN = 81.7 − 2.48 = 79.22( A)
PM = EaN IaN = 424.7 ×187 = 79419(W )
TN
= 9550 × PM nN
= 9550 × 79.419 = 1011.3(N ⋅ m) 750
3
彭鸿才《电机原理与拖动》习题解答
（3） CeφN
=
EaN nN
=
424.7 750
= 0.566
no
= UN CeφN
=
= 776rpm
P42 1－35 解：
IaN = I N − I fN = 91 − 2.5 = 88.5( A)
Ceφ N
= UN
− IaN Ra nN
=
220 − 88.5× 0.074 1500
= 0.142
n = U N − Ia Ra = 220 − 50× 0.074 = 1523rpm
Ceφ N
= UN
− IaN Ra nN
=
440 − 250 × 0.078 500
=
0.841
Q Ce = π CT 30
∴ CT φ N
=
30 π
×
Ceφ
N
= 9.55 × 0.841 = 8.032
（1） T2N
= 9550× PN nN
= 9550× 96 500
= 1833.6(N ⋅ m)
pcu
=
I
2 a
Ra
=
71.132
× 0.128
=
647.6(W
)
PM = Ea Ia = 239.1× 71.13 = 17007(W )
P1
= PN ηN
16 ×103 =
0.855
= 18713.5(W )
P42 1－29 解： Ia = I N − I fN = 40.6 − 0.683 = 39.92( A)
彭鸿才《电机原理与拖动》习题解答
第一章 直流电机原理
P41
1－3
解： IN
= PN UN
14 ×103 =
230
= 60.87( A)
P1
= PN ηN
=
14 0.855
= 16.37(Kw)
P41 1－4 解： P1 = U N IN = 110 ×13 = 1430(W )
ηN
=
PN P1
1.1×103 =
n = U N − Ia Ra = 220 − 0.75× 305× 0.038 = 1014rpm
Ceφ N
0.2084
或者：
n = UN Ceφ N
−
Ra
CeCT
φ
2 N
×T
=
220 0.2084
−
0.038× 0.75× 607 9.55× 0.20842
= 1014rpm
（3） I = U N − CeφN n = 220 − 0.2084 ×1100 = −243.2( A)
440 0.566
= 777.4rpm
（4）To = TN − T2N = 1011.3 − 955 = 56.3(N ⋅ m)
I ao
= To CT φ N
= To 9.55CeφN
=
56.3 9.55× 0.566
= 10.42( A)
no'
= U N − Iao Ra Ceφ N
=
440 −10.42 × 0.082 0.566
U （2） n =
− I aN Ra = 180 − 41× 0.376 = 1210rpm
Ceφ N
0.136
P106 3－32 解：
QCT = 9.55Ce T = CTφ Ia = 9.55CeφIa
n = UN Ceφ
−
Ra CeCTφ 2
× TN
=
UN 0.8CeφN
−
Ra 9.55(0.8CeφN )2
Ra
0.038
4
P105 3－29 解：
彭鸿才《电机原理与拖动》习题解答
RN
= UN IN
=
220 305
= 0.72Ω
（1） nRN
= n0
−
0.5RN 9.55(CeφN
)2
TN
=
1055.7
−
0.5× 0.72 9.55× (0.2084)2
× 607 = 529rpm
过（ n0 ，0），（ 529，607）两点画直线。
=
n0 = 1055.7 = 2111.3rpm 0.5 0.5
nφ N
=
n0′′
−
Ra 9.55(0.5CeφN
)
2
TN
=
2111.3 −
0.038 9.55× (0.5× 0.2084)2
× 607
= 1889rpm
过（ n0′′ ，0），（ 1889，607）两点画直线。
P105 3－30 解：
× 9.55CeφN IaN
=
UN 0.8CeφN
−
Ra 0.82 CeφN
× IaN
=
220 −
0.8× 0.136
0.376 × 41 0.82 × 0.136
= 1845rpm
或者：QT = CTφ Ia = TN = CTφN IaN
∴ Ia
=
φN φ
I aN
=
φN 0.8φN
I aN
=
41 0.8
pcu
=
I
2 aN
Ra
=
79.222
× 0.1 =
627.6(W )
pf
=
I
2 fN
Rf
= 2.482 × 88.8 = 546(W )
∑ p = P1 − PN = 17974 −15278 = 2696(W )
∑ pm + pFe = p − pcu − p f = 2696 − 627.6 − 546 = 1522(W )
CeφN
0.142
P105 3－28 解：
（1）估算： Ra
=
1 ×UNIN 2
− PN
I
2 N
×103
=
1 2
220 × 305 − 60×103 3052
= 0.038(Ω)
Ceφ N
= UN
− IaN Ra nN
=
220 − 305× 0.038 1000
= 0.2084
no
=
UN Ceφ N
=
220 0.2084
= 1055.7rpm
TN = 9.55CeφN IaN = 9.55× 0.2084× 305 = 607(N ⋅ m)
过（no = 1056rpm，T = 0），（ n N = 1000rpm TN = 607 N.m）两点，可画出固有机械特性。
（2）T = 0.75 TN 时， Ia = 0.75IaN ：
EaN
115
P42 1－31 解：
Q EaN = CeφN nN n = nN φ = φN Ia = IaN ∴ Ea = EaN = U N − I N Ra = 110 −13×1 = 97(V ) U = Ea − I N Ra = 97 −13×1 = 84(V )
P42 1－32 解：
IaN = I N − I fN = 255 − 5 = 250( A)
Rst 3
= UN I1
=
110 2 × 85.2
= 0.65(Ω)
λ = 3 Rst3 = 3 0.65 = 1.7 Ra 0.129
I2
=
I1 λ
=
2 × 85.2 1.7
= 100.2( A)
> 1.1IN
= 93.72( A)
符合要求。
RST1 = λRa = 1.7 × 0.129 = 0.219(Ω)
Ceφ N
0.1
或者： EaN = U N + IaN Ra = 110 + 250.5× 0.02 = 115(V )
Ea = U N − IaN Ra = 110 − 250.5× 0.02 = 105(V )
Q EaN = Ea nN n
∴ n = EanN = 105 ×1150 = 1050rpm
P105 3－31 解：
（1） CeφN
=
UN
− IaN Ra nN
=
220 − 41× 0.376 1500
= 0.136
降压瞬间，转速不变，有： n = nN
U Ia =
− CeφN nN Ra
= 180 − 0.136 ×1500 = −63.83( A) 0.376
T = 9.55CeφN Ia = 9.55 × 0.136 × (−63.83) = −82.9(N ⋅ m)
UN CeφN
−
Ra CeCTφN2
× To
=
523 −
0.078 ×174.4 0.841× 8.032
=
521rpm
或者： Iao
= To CT φ N
174.4 =
8.032
= 21.71( A)
no'
= U N − Iao Ra Ceφ N
=
440 − 21.71× 0.078 0.841
= 521rpm
PM = PN + pm + pFe = 15278 +1522 = 16800(W )
P42 1－34 解：
（1） T2N
= 9550× PN nN
= 9550× 75 750
= 955(N ⋅ m)
（2） IaN = I N − I fN = 191− 4 = 187( A)
EaN = U N − IaN Ra = 440 −187 × 0.082 = 424.7(V )
RST 2 = λRST1 = 1.7 × 0.219 = 0.372(Ω)
5
彭鸿才《电机原理与拖动》习题解答 rst1 = (λ −1)Ra = (1.7 −1) × 0.129 = 0.091(Ω)
rst2 = λrst1 = 1.7 × 0.091 = 0.153(Ω)
rst3 = λrst2 = 1.7 × 0.153 = 0.26(Ω)
Ea = U N − Ia Ra = 220 − 39.92 × 0.213 = 211.5(V )
PM = Ea Ia = 211.5 × 39.92 = 8443.1(W )
pcu
=
I
2 a
Ra
=
39.922
× 0.213
=
339.4(W
)
p f = I fNU N = 220 × 0.683 = 150.26(W )
= 51.25( A)
n = U N − Ia Ra = 220 − 51.25× 0.376 = 1845rpm
Ceφ
0.8× 0.136
Q n = 1845rpm > nN = 1500rpm , Ia = 51.25A > I N = 41A
P1 = U N I N = 220 × 40.6 = 8932(W )
η = PN = 7.5×103 = 83.97% P1 8932
po = PM − PN = 8443.1 − 7.5 ×103 = 943.1(W )
1
彭鸿才《电机原理与拖动》习题解答
TN
= 9550 × PM nN
= 9550 × 8443.1/1000 = 26.88(N ⋅ m) 3000
T2 N
= 9550× PN nN
= 9550× 7.5 3000
= 23.88(N ⋅ m)
To = TN − T2N = 26.88 − 23.88 = 3(N ⋅ m)
或者 To
= 9550 × Po nN
= 9550 × 943.1/1000 3000
= 3(N ⋅ m)
P42 1－30 解：
1430
= 76.92%
∑ p = P1 − PN = 1430 −1100 = 330(W )
P42
1－21
解：
If
= UN Rf
= 230 = 1.53( A) 150
Ia = I N + I f = 69.6 +1.53 = 71.13( A)
Ea = U N + IaRa = 230 + 71.13× 0.128 = 239.1(V )
（2）TN = CTφN IaN = 8.032 × 250 = 2008(N ⋅ m)
2
彭鸿才《电机原理与拖动》习题解答 （3）To = TN − T2N = 2008 −1833.6 = 174.4(N ⋅ m)
（4） no
=
UN Ceφ N
=
440 = 523rpm 0.841
(5)
no'
=
（3） n0′
=
0.5U N CeφN
= 0.5n0
Biblioteka Baidu
= 0.5×1055.7 = 528rpm
nVN
=
n0′
−
Ra 9.55(CeφN
)
2
TN
= 528 −
0.038 9.55× (0.2084)2
× 607
=
472rpm
过（ n0′ ，0），（ 472，607）两点画直线。
（4）
n0′′
=
UN 0.5CeφN
IN
= PN UN
=
27000 110
= 245.5( A)
IaN = I N + I fN = 245.5 + 5 = 250.5( A)
Ceφ N
= UN
+ I aN Ra nN
= 110 + 250.5 × 0.02 1150
= 0.1
n = U N − IaN Ra = 110 − 250.5 × 0.02 = 1050rpm
（2） nR′ N
=
n0
−
2RN 9.55(CeφN )2
TN
2× 0.72 = 1055.7 − 9.55× (0.2084)2
× 607
=
−1052rpm
过（ n0 ，0），（ –1052，607）两点画直线。
或： nR′ N
= U N − 2RN IN CeφN
=
220 − 2× 0.72× 305 = −1052rpm 0.2084