6.1
在460-V，50－hp，60赫茲，4 相的電感應電動機的指示額定負荷的速度是1755 r /分鐘。
機在額定負荷操作。
假如電動
a。 這條電動機是什麼？
b。 轉子電流的頻率是多少？
c。 stator 產生的與stator有關的氣隙流波的角速度是多少？ 關於轉子嗎？
d。 轉子產生的與stator有關的氣隙流波的角速度是多少？ 關於轉子嗎？
(a)
根據 4.41 式,
 120 
 f e
ns  
 poles 
得
 120 
ns  
  60  1800 r min
 4 
n n
根據 6.1 式, s  s
ns
得
s
1800  1755
 0.025
1800
(b)
根據 6.4 式,
f r  sf e
得
f r  0.025  60  1.5 Hz
(c)
 120 
120
 f e 
n stator  ns  
 60  1800 r
min
4
 poles 
(d)
nrotor  sn s  0.025  1800  45 r
min
6.2
分散漏電場將引起轉子頻率電壓在沿著一就職電動機的表情騎在馬上的一種加速線圈內。 這些感應
電壓的頻率的測量能用來確定轉子速度。
a。 如果感應電壓的頻率是0.89赫茲，在一台50赫茲，6 根杆的電感應電動機的r /分鐘的轉子速度是
多少？
b。 計算符合以1740 r /分鐘速度操作的一台4 根杆，60赫茲的電感應電動機的感應電壓的頻率。 相
應滑動是什麼？
(a)
 120 
 f e
根據 4.41 式, ns  
 poles 
得
 120 
ns  
  50  1000 r min
 6 
根據 6.4 式, f r  sf e
得
0.89
 0.0178
50
n  1  s ns
0.89  s  50  s 
根據 6.2 式,
得
n  1  0.0178  1000  982.20 r
min
(b)
 120 
 f e
根據 4.41 式, ns  
 poles 
得
 120 
ns  
  60  1800 r min
 4 
n n
根據 6.1 式, s  s
ns
得
s
1800  1740
 0.033
1800
6.3
三相感應電動機在1198 r /MIN在無載狀況和1112 r /MIN滿載當三項電源供應60赫茲
a。 這台馬達有多少磁極？
b。 滿載時轉差率是多少？
c。 轉子電流的相應頻率是多少？
d。 轉子的轉子磁場的相對速度是多少？定子方面是多少？
(a)
ns 
120  f
120  60

 1198 ,
p
p
∴
p
7200
 6極
1200
(b)
額定負載的轉差率 S 
1200  1112
 0.0733  7.33%
1200
(c)
轉子頻率 f r  s  f e  0.0733  60  4.4 Hz
(d)
同步轉速跟定子有關，因此轉子速度為1200 r/min
非同步轉速跟轉子有關，因此轉子速度為(1200-1112)=88 r/min
6.4
線感應電動機已經被為多種應用包括高速地面運輸。 一台基於電感應電動機原則的電感應電動機的
線性電機由一條軌道的一輛汽車組成。 蹤跡是 一只發展的松鼠籠蜿蜒和汽車， 哪個是4.5米長和
1.25 寬米有發展的三相的12根杆對電樞彎曲的在75 hz的動力被透過槽從延伸的武器到在地面下面的
鐵軌供給到汽車。
a。 同步加快的公里/本壘打是多少？
b。 汽車將到達這速度解釋你的答案嗎？
c。 滑動是什麼如果汽車正旅行95 公里/小時，什麼在這個條件下是蹤跡電流的頻率？
d。 如果控制系統控制大小和汽車彎曲的電流的頻率， 當貓旅行75 公里/小時時，什麼在這個條件
下是蹤跡電流的頻率？
6 . 6 D e s c r i b e t h e e ff e c t o n t h e t o r q u e - s p e e d c h a r a c t e r i s t i c o f a n i n d u c t i o n m o t o r
produced by a. halving the applied voltage and b. halving both the applied voltage
and the frequency.
解：
a. 力矩與電壓的平方成正比，而力矩-速度特性只被降低四分之一。
b. 忽視定子阻抗和漏磁電抗的影響，電壓和頻率維持固定的氣隙磁通。因此力矩-速度特性會跟最
初的特性相同，但是同步速度會減半。
6 . 7 Fi gu re 6.23 s h ows a s yst em cons i st i ng of a t hr ee -p hase wo und - rot o r i nduct i on
m a c hi n e w ho s e sh af t i s ri gi dl y c o u pl ed t o t h e s h a ft o f a t h r e e - ph a s e s yn c h r o n ou s
motor . The terminals of the three-phase rotor winding of the induction machine are
brought out t o sl ip ri ngs as shown. Wi t h the s yst em suppli ed from a t hree -phas e ,
6 0 H z s o u r c e , t h e i n d u c t i o n m a c h i n e i s d r i v e n b y t h e s yn c h r o n o u s m o t o r a t t h e
proper speed and in the proper directi on of rotation so that three -phase , 120Hz
voltages appear at the slip rings. The induction motor has four -pole stator winding.
Interconnected induction and synchronous machines.
a. How many poles are on the rotor winding of the induction motor?
b. If the stator field in the induction machine rotates in the clockwise direction, what
is the rotation direction of its rotor?
c. What is the rotor speed in r/min ?
d. How many poles are there on the synchronous motor?
e. It is proposed that this system can produce dc voltage by reversing two of the
phase leads to the induction motor stator. Is this proposal valid?
圖 6.23
感應機與同步機連接
解:
a. 四極
b. 逆時針方向
c. 120f/p=120*60/4=1800 r/min
d. 四極
e. 不是。將有直流磁通連結感應馬達轉子的繞組，但是在滑環沒有因而發生的
電壓。
6.8 A system such at that shown in Fig.6.23 is used to convert balanced 50 -Hz voltage
to other frequencies. The s ynchronous motor has four poles and drives the
interconnected shaft in the clockwise direction. The induction machine has six poles
and its stator windings are connected to the source in such a fashion as to produce a
counterclockwise rotating field (in the direction opposite t o the rotation of the
s yn c h r o n o u s m o t o r ). T h e m a c hi n e h a s a w ou n d r ot o r who s e t e r m i n al s a r e b r o u gh t
out through slip rings.
a. At what speed does the motor run?
b. What is the frequenc y of the volt ages produced at t he slip rings of the i nduced
motor?
c.What will be the frequency of the voltages produced at the slip rings of the induction motor if two leads of
the induction-motor stator are interchanged, reversing the direction of rotation of the result rotating field?
Fig 6.23 感應電動機與同步電動機的連接
解：
a. n 
120  f 120  50

 1500 r
min
P
4
b. 感應馬達轉子順時針方向轉動之轉速為 1500 r/min. 定子磁通逆時針轉動之轉速為
 6   1000 ；所以轉子磁通旋轉速度為 2500 r/min。注意磁通轉速在 1000 r/min
3000×(2/poles)= 3000  2
 2500 
會在 50Hz 產生滑差電壓。轉子頻率 f r  sf e  50  
  120 Hz .
 1000 
c. 當定磁通再順時針方向轉動速度為1000 r/min的時候，轉子磁通轉動之速度
為500 r/min.
f 
500  6
 25 Hz
120
所以在頻率為25 Hz 可以得到感應電壓。.
6.1 0 A t h r e e - ph a se , Y- co nn e ct , 4 60 - V (l i ne -l i n e ), 25 - KW, 60 - Hz , fo ur-p ol e i n du ct i on
motor has the following equivalent-circuit parameters in ohm per phase referred to the stator:
X m  59.4
R1  0.103
R2  0.225
X 1  1.10
X 2  1.13
The total friction and wind age losses may be assumed constant at 265 W, and the
core loss m a y be assumed to be equal to 220 W. With the rotor connect ed directl y
t o a 4 6 0 - V s o u r c e , c o m p u t e t h e s p e e d , o u t p u t s h a f t t o r q u e a n d p o w e r, i n p u t a n d
power factor and efficiency for slip of 1, 2 and 3 percent. You may choose either to
ei t her t o represent t he core l oss by a resi st ance Rc connect ed i n paral l el wi th t he
magnetizing reactance X m .
解：
這個問題可以直接用第六章的方程式代替解決，可以在 MATLAB 裡實現。
下述的表格可從 MATLAB 書寫獲得。而等效電路方程式以平行的磁化電抗
的核心損失抵抗 Rc s。
R

Z f  R f  jX f   2  jX 2  與 jX  相比
 S

Z in  R1  jX 1  Z f
線對中性點的相對電壓為為
定子電流為
Ns  (
V
Iˆ1  1
Z in
120
)  fe
Poles
Vl  V / 3
s 
4f e
poles
n  (1  s)ns
 s  (1  s) s
Pgap  n ph I 22 (
R2
)
s
只有阻抗被包括在 Zf is R2/s
 Pgap  n ph I12 R f
Pshaft  Pmech  Prot  (1  s) Pgap  Pr o t
Ts h a 
f t
Ps h a f t
n
Pin  n ph Re[Vˆ1 Iˆ1* ]
效率等於  
Pshaft
Pin
Slip[%]
Speed[r/min]
Tout[N-m]
Pout[kw]
Pin[kw]
Power factor
Efficiency[%]
1.0
1782
8.5
8.5
45.8
0.81
93.3
2.0
1764
16.5
16.5
89.6
0.87
94.4
3.0
1746
23.4
23.4
128
0.85
93.8
6.13 A 15-kW 230-V three-phase Y-connected 60 -Hz four-pole squirrel-cage induction
m ot or devel ops ful l -l oad i nt ern al t orq ue at a sl i p o f 3.5 perc ent wh en ope r at ed at
rat ed vol t age and f requenc y. Fo r t he purposes of t hi s probl em , rot at i onal and core
l o s s e s c a n b e n e g l e c t e d a n d c o r d l o s s e s c a n b e n e gl e c t e d . T h e f o l l o w i n g m o t o r
parameters , in ohms per phase , have been obtained ：
R1 = 0.21 X1 = X2 = 0.26 Xm= 10.1
Determine the maximum internal torque at rated voltage and frequency , the slip at
m ax i m um t orq ue , and t he i nt e rn al st art i n g t o rqu e at r at ed v ol t a ge and f re quen c y.
解:
在題目中所給的 R1、X1、X2、Xm 下代入 p339-340 的 matlab 經由 R2 的變化
來達到最大轉矩與啟動轉矩。再把答案中所提供的 R2=0.0953 代入得到最大轉
矩 177N-m 及轉差為 18.2%、起動轉矩為 71.6N-m。
6.14 The induction motor of problem 6.13is supplied form 230-V source through a feeder of impedance Zf =
0.05 + j 0.14 ohms. Find the motor slip and terminal voltage when it is supplying rated load.
解：
利用之前 6.13 所提供 R2 的數值配合題目所提供的饋線阻抗代替先前的 R1 與
X1，而其他參數 X2、Xm 和 R2 值固定之後經由 matlab 的運算後，參數電壓
從 230V 變化成 221.6V，而轉差由先前的 3.5%變化為 3.67%。
6.15 A three -phase induction motor, operating at rated vol tage and frequenc y, has a
staring torque of 135 percent and a m aximum t orque of 220 percent, both with
res pect t o i t s r at ed -l oad t orqu e. N egl ect i n g t he e ffe ct s of st at or resi st an ce and
rotational losses and assuming constant rotor resistance, determine:
a. the slip at maximum torque.
b. the slip at rated load.
c. the rotor current at starting (as a percentage of rotor current at rated load).
解:
a. 根據 6.34 式
R2
S m a Tx
 ( X 1,q  X 2 )
從 6.34 式
Tm a x
0.5n phV1,2eq
 s ( X 1,eq  X 2 )
從 6.33 式中可得 s=1
Tstart 
得
n phV1,2eq R2
 s [ R22  ( X 1,eq  X 2 ) 2 ]
Tmax 2.20

 1.63
Tstart 1.35
R2
)2  1
R  (X  X 2 )
X 1,eq  X 2

可以從上式的方程式： 1.63 
R2
R2 ( X 1,eq  X 2 )
X 1,eq  X 2
2
2
2
1,eq
2
2
0.5( S matT
 1)
 1.63
S max T
可以得到最大轉差率 S max T  0.343
因此
b. 從 6.33 式 得知等效電阻 Req ,1  0 and with s  s rated
Trated 
n PhV1,2eq ( R2 / s rated )
 s [( R2 / s rated ) 2  ( X 1,eq  X 2 ) 2 ]
(
所以
Tm a x
0.5[1  ( s m a Tx / s r a t )e 2d]
 2.1 
Tr a t e d
s m a Tx / s r a t e d
 0.240s max T  0.0824
所以 s rated

c.

I
2 , rated

V

1,eq
R2 / s rated  j ( X eq ,1  X 2 )

V
1,eq
( X eq ,1  X 2 )( s max T / s rated  j )

所以
I
2 , start

I

s max T / s rated  j
s max T  j

4.16  j
 4.05
0.343  j
2 , rated
6.16 When operated at rated voltage and frequency, a three-phase squirrel-cage
induction motor (of
the design classification known as a high-slip motor) delivers
full load at a slip of 8.7 percent and
develops a maximum torque of 230 percent of
full load at a slip of 55 percent. Neglect core and
rotational losses and assume that
the rotor resistance and inductance remain constant, independent of
slip. Determine the torque at starting, with rated voltage and frequency, in per unit based upon its
full-load value.
解：
給 Tmax  2.3TA , s max T  0.55 and s A  0.087 ，可從 6.36 與 6.33 式得：
Tmax
TA
 2  R1,eq   1  2  1
     

0.5
 sf 1  R2   s n   s max T

R1,eq
1

R2
s max T



2



替代值與解：
Req ,1
R2
 1.315
從 6.33 式可以寫成：
Tstart
TA
2
2
  Req ,1
1   X eq ,1  X 2  
 
 
 

sf 1  
R2
  R2
 
 sA 
2
2 
  Req ,1  1    X eq ,1  X 2  
 
 R
s m a Tx  
R2
 
 2
從 6.34 式
 X 1,eq  X 2

R2

2

 1
  
 s m a Tx

2
  R1,eq 

  
  R2 
所以可寫成： Tstart  1.26TA
2
6.17 A 500-kw, 2400-v, four-pole, 60Hz induction machine has the following equivalent-circuit parameters
in ohms per phase Y referred to the stator:
R1 =0.122
R2 =0.317
X1 =1.364
X 2 =1.32
X m =45.8
It achieves rated shaft output at a slip of 3.35 percent with an efficiency of 94.0 percent. The machine is to
be used as a generator, driven by a wind turbine. It will be connected to a distribution system which can be
represented by a 2400-V infinite bus.
a. From the given data calculate the total rotational and the core losses at rated load.
b. With the wind turbine driving the induction machine at a slip of -3.2 percent, calculate (i) the electric
power output in kW, (ii) the efficiency (electric power output per shaft input power) in percent and (iii) the
power factor measured at the machine terminals.
c. The actual distribution system to which the generator is connected has an effective impedance of 0.18
+ j 0.41 Ω/phase. For a slip of -3.2 percent, calculate the electric power as measured (i) at the infinite bus
and (ii) at the machine terminals.
Ans：
(a)
從 MATLAB 得知,所需要的能量為 503.2 kW,喪失的能量為 503.2 -500= 3.2kW; 同理可知輸入的
能量為 528 kW,功率可以達到 500 / 528= 94%,實際能量 500kW/0.94 = 531.9 kW,所以核心所耗損的
能量為 531.9-528 =3.9 kW
(b)
從(a)可知阻抗為 1.47 kW,又輸入能量同等於負端軸所耗損的能量,
使用 MATLAB
得輸出能量為 512 kW,功率為 91.6 % 功率因素 0.89 J
(c)
從(b)可以得到,提供能量 498 kW,總輸出能量 508 kW
6.19 For a 25-kW,230-V,three-phase,60-Hz squirrel-cage motor operating at rated voltage and frequency, the
rotor I2R loss at maximum torque is 9.0 times that at full-load torque, and the slip at full-load torque is
0.023.Sator resistance and rotational losses may be neglected and the rotor resistance and inductance
assumed to be constant. Expressing torque in per unit of the full-load torque, find
a. the slip at maximum torque.
b. the maximum torque.
c. the starting torque.
解：
a.
由題目可以得知： I 2,max T R2  9 I 2, fl R2 ，∴
2
由(6.32)式可得： Iˆ2 
2
Vˆeq
R2 / s  j ( X eq  X 2 )
I
2,max T
 3 I 2, fl
∴
Iˆ2, fl
j ( X eq  X 2 )  R2 / s max T

Iˆ2,max T
j ( X eq  X 2 )  R2 / s fl
由(6.34)式可得：
∵ R1,eq  0
∴
Iˆ2, fl

Iˆ2,max T
s max T
 R12,eq  ( X 1,eq  X 2 ) 2
R2
s max T
Iˆ2, fl
將(2)代入(1)
∴
R2
Iˆ2,max T
………(1)

j 1
j  s max T / s fl
 X 1,eq  X 2 ………(2)
j 1
j  s max T / s fl

2
1  ( s max T / s fl ) 2
s max T  3.24 s fl  0.0745  7.45%
b.
將(6.33)和(6.36)式整理後可得




0.5 ( R2 / s fl ) 2  ( X 1,eq  X 2 ) 2
0.5 1  ( s max T / s fl ) 2
Tmax


 1.77
T fl
( X 1,eq  X 2 )( R2 / s fl )
( s max T / s fl )
Tmax = 1.77
c.
1  ( s max T / s fl ) 2
Tstart
 s fl (
)
T fl
1  s 2max T
Tstart = 0.263
6.20 A squirrel-cage induction motor runs at a full-load slip of 3.7 percent．The rotor current at starting is 6.0
times the rotor current at full load．The rotor resistance and
inductance is independent of rotor frequency and rotational losses，stray-load losses，stray-load losses
and stator resistance may be neglected．Expressing torque in per unit of the full-load torque，compute
a.
the starting torque．
b.
the maximum torque and the slip at which the maximum torque occurs．
解：
(a)： T 
因此
I 22 R2
S
．
I
Tstart
 S  ( 2, start ) 2  1.32
T
I 2, A
每單位因此 Tstatr =1.32 。
(b)：作為在這個解決問題6.15的方法方面，忽視 R1 的影響
Iˆ2, s t a r t

Iˆ
2,r a t e d
S m a Tx
S  j
S m a Tx  j
這可能被求出 S max T
1 (
S max T  S A
(
I 2,start
IA
S A I 2,start
)2
 0.224  22.4% .
) 1
2
IA
再次，從這個解決問題6.15的方法，
Tmax

Trated
0.5[1  (
S max T
S max T
SA
)2 ]
 3.12
SA
每單位因此Tmax = 3.12。
6.21 A  -connected﹐ 25-kW﹐230-V﹐three-phase﹐six-pole﹐50Hz squirrel-cage induction motor has the
following equivalent-circuit parameters in ohms per phase Y:
R1  0.045 R2  0.054 X 1  0.29 X 2  0.28 X m  9.6
a. Calculate the starting current and torque for this motor connected directly to a 230-V source﹒
b. To limit the starting current﹐ it is proposed to connect the stator winding in Y for starting and then to
switch to the  connection for normal operation (i) What are the equivalent-circuit parameters in ohms per
phase for the Y connection？ (ii) With the motor Y-connected and running directly off of a 230-V source﹐
calculate the starting current and torque﹒
解：
(a)：
解決方程式如第 6 章，有 s=1 為啟動，用 MATLAB 產生
I start  233 (A)
Tstart  79.1 (N-m)
(b)：
(i) 當電動機連接在 Y，相等電路參數將是連接  的 3 倍。因此
R1  0.135()
R2  0.162()
X 1  0.87()
X 2  0.84()
X m  28.8()
(ii)
I start  77.6 (A)
Tstart  26.3 (N-m)
6.22 The following data appl y to a 125 -kW﹐2300-V﹐three-phase﹐four pole﹐60-Hz
S q u i r r e l - c a g e i n d u c t i o n m o t o r : S t a t o r - r e s i s t a n c e b e t w e e n t e r m i n a l s = 2.23
No-load test at rated frequency and voltage: Line current=7.7A three -phase power=2870W
Blocked-rotor test at 15Hz: Line voltage=268V current=50.3A
Three-phase power=18.2kW
a. Calculate the rotational losses.
b. Calculate the equivalent-circuit parameters in ohms. Assume that X 1  X 2 .
c. Compute the stator current, input power and power factor, output power and efficiency when this
operating at rated voltage and frequency at a slip of 2.95 percent.
解:
(a):
Pr o t Pn1  3I n21 R1  2 6 7W
2 .
(b):
參數的計算正好遵循著例 6.5 結果是:
R1  1.11
X 1  3.90
R2  1.34
X 2  3.90
X m  1.68
(c):
解決這條方程式, 使用第6章等效電路方程式,由(b)
I a  29.1A
P in  106kW
功率因素  0.91落後
pout  100kW
效率  94.5%
6.23 Two 50-kW, 440-V, Three-phase, six-pole, 60-Hz squirrel-cage induction motors have identical stators.
The dc resistance measured between any pair of stator terminals is 0.204Ω. Blocked-rotor tests at 60-Hz
produce the following results:
Motor
Volts
(line-to-line)
Amperes
1
2
74.7
99.4
72.9
72.9
Three-phase
power, kW
Ans:
因為 blocked-rotor , 所以馬達的輸入阻抗為: Zin ≈ R1 + R2 + j(X1 + X2)
Bb1
R2 
 R1
3I b21
馬達 1: R2  0.174 ,
馬達 2:
R2  0.626
4.40
11.6
又因為馬達跟 I b21 R2 成比例
 
 
所以
I 22 motor2 R2 motor2
Tmotor2
(R )
(I 2 )
 2
 ( 2 motor2 )( 22 motor2 )
Tmotor1 ( I 2 ) motor1 ( R2 ) motor1
( R2 ) motor1 ( I 2 ) motor1
得
Tmotor2
 0.278
Tmotor1
從上列可以得知電流比為
( I 2 ) motor2
 99.4 / 74.7  1.39
( I 2 ) motor1
Tm o t 2o r
 0.4 9 2
Tm o t1o r
所以
6.25 A 230-V, three-phase, six-pole, 60Hz squirrel-cage induction motor develops a maximum internal
torque of 288 percent at a slip of 15 percent when operated at rated voltage and frequency. If the effect of
stator resistance is neglected, determine the maximum internal torque that this motor would develop if it
were operated at 190V and 50Hz. Under these conditions, at what speed would the maximum torque be
developed?
解:
忽略 R1 和 Req ,1 ，因此從方程式 6 .35
S m a Tx 
R2
X 1,eq  x2
和從方程式 6 .36
Tm a x
0.5n phV1,2eq S m a Tx
 s ( X 1.eq  X 2 )

0.5n phV1,2eq S m a Tx
 s R2
如果頻率從60Hz降到50Hz， X 1,eq  X 2 將會下降
5
6
， 因此 S max T 將是原本的
6
5
S max T  18% ，相同的速率為 1000(1  0.18)  820r / min .
Tm a x將增加為:
190
)2 (6 )
(Tmax ) 50 (
230
5  0.983

5
(Tmax ) 60
6
或則 (Tmax ) 50  283%
6.26 A 75-kW, 50Hz, four-pole, 460-V, three-phase, wound-rotor induction motor develop Full-load torque
at 1438 r/min with the rotor short-circuited. An external non-inductive resistance of 1.1  is placed in
series with each phase of the rotor, and the motor is observed to develop its rated torque at a speed of 1405
r/min. Calculate the rotor resistance per phase of the motor itself?
解:
S m a Tx  R2 , 因此
R2 
1.1
[( S max T ) Re xt 1.1
( S max T ) Re xt 0
] 1
 2.07
6.27 A 75-kW, 460-V, Three-phase, four-pole, 60-Hz, wound-rotor induction motor develops a maximum
internal torque at 225 percent at a slip of 16 percent when operated at rated voltage and frequency with its
rotor short-circuited directly at the slip rings. Stator resistance and rotational losses may be neglected, and
the rotor resistance and inductance may be assumed to be constant, independent of rotor frequency.
Determine.
a. the slip at full load in percent.
b. the rotor I 2 R loss at full load in watts.
c. the starting torque at rated voltage and frequency in per unit and in N‧m. If the rotor resistance is
doubled (by inserting external series resistance at the slip rings) and the motor load is adjusted for such that
the line current is equal to the value corresponding to rated load with no external resistance, determine.
d. the corresponding slip in percent and
e. the torque in N‧m.
Ans:
(a) 從習題 6.15 可得
2
Tmax 0.5[1  ( s max T / s fl ) ]

T fl
s max T / s fl
然而 Tmax / T fl  2.25 ， s max T  0.16 ，可得 s fl  0.0375 =3.75%
(b) 軸損失能量為
s fl
Protor  Prated (
)  2.9 kW
1  s fl
(c) 從習題 6.19 可得
1  ( smax T / s fl ) 2
Tstart
 s fl (
)  0.70
2
T fl
1  smax
T
軸速比 75 kW/ωm,fl ωm,fl = 60π(1−sfl) = 181.4 rad/sec.
所以 Trated = 413 N·m , Tstart = 0.70 per unit = 290 N·m.
(d)
s = 2sfl = 7.50%
(e)
T = 413 N·m
6.29 A 100-kW, three-phase, 60-Hz, 460-V, six-pole wound–rotor induction motor develop its rated full-load
output at a speed of 1158 r/min when operated at rated voltage and frequency with its slip rings
short-circuited. The maximum torque it can develop at rated voltage and frequency is 310 percent of
full-load torque. The resistance of the rotor winding is 0.17Ω/phase Y. Neglect any effects of rotational any
stray-load loss and stator resistance.
2
a. Compute the rotor I R loss at full load.
b. Compute the speed at maximum torque in r/min.
c. How much resistance must be inserted in series with the rotor windings to produce maximum starting
torque? with the rotor windings short-circuited, the motor is now run from a 50-Hz supply with the applied
voltage adjusted so that the air-gap flux wave is essentially equal to that at rated 60-Hz operation.
d. Compute the 50.Hz applied voltage.
e. Compute the speed at which the motor will develop a torque equal to its rated 60-Hz value with its
slip-rings shorted.
Ans:
(a)
s fl
Protor  Prated (
)  3.63 kW
1  s fl
(b)
2
Tmax 0.5[1  ( s max T / s fl ) ]

從習題 6.15 可得
T fl
s max T / s fl
S fl = (1200 − 1158)/1200 = 0035
可得 S max T = 0.211 = 21.1%
Tmax / Tfl = 3.10 ,
速度就相當於 1200(1 − 0.211) = 947 r/min.
(c)
阻抗 S max T from 0.211 to 1.0
 R2 ,tot = 0.17/.211 = 0.806 Ω
所以 Rext = 0.806 − 0.211 = 0.635 Ω
(d)
因為是線性系統所以依照比值 5/6 所以電壓應為 383V
(e)
從 Eq 6.35 可得 S max T = R2/(X1,eq + X2),因為是線性系統所以依照比值 5/6 , s max T =0.042, 轉速為
1000(1 − 0.042) = 958 r/min
6.31 The resistance measured between each pair of slip rings of a three-phase, 60-Hz, 250-kW, 16-pole,
wound-rotor induction motor is 49 mΩ. With the slip rings short-circuited, the full-load slip is 0.041. For the
purposes of this problem, it may be assumed that the slip-torque curve is a straight line from no load to full
load. The motor drives a fan which requires 250kW at the full-load speed of the motor. Assuming the torque
to drive the fan varies as the square of the fan speed, what resistance should be connected in series with the
rotor resistance to reduce the fan speed to 400r/min?
Ans:
由 題 目 可 得 轉 速 為 3600/8 = 450 r/min, 在 全 速 中 450(1 − 0.041) = 431.6 r/min,
得
(250 *10 3 )[ 47.12(1  0.041)]  5.53 *10 3 N‧m。
在 400r/min 的 速 度 下 得 到
5.53 *10 3 (400 / 431.6) 2 = 4.75 *10 3 N‧m, 在 沒 有 其 他 的 阻 抗 下
5.53 *10 3 / 431.6  12.81 ;
而所要求的角速度特性則為: 4.75 *10 3 / 400  11.88 ,則
12.81
Rtot  (
) * 24.5  26.4 mΩ , 得 26.4-24.5 = 1.9 mΩ
11.88