6.6 Describe the effect on the torque-speed characteristic of an induction motor
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 Figure 6.23 shows a system consisting of a three-phase wound-rotor induction
machine whose shaft is rigidly coupled to the shaft of a three-phase synchronous
motor . The terminals of the three-phase rotor winding of the induction machine are
brought out to slip rings as shown. With the system supplied from a three-phase ,
60Hz source , the induction machine is driven by the synchronous motor at the
proper speed and in the proper direction 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
解:
a. 四極
b. 逆時針方向
感應機與同步機連接
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 synchronous 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 to the rotation of the
synchronous motor). The machine has a wound rotor whose terminals are brought
out through slip rings.
a. At what speed does the motor run?
b. What is the frequency of the voltages produced at the slip rings of the induced
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
 6   1000；所以轉子磁通旋轉速度為 2500
之轉速為 3000×(2/poles)= 3000  2
r/min。注意磁通轉速在 1000 r/min 會在 50Hz 產生滑差電壓。轉子頻率
 2500 
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。.
6.10 A three-phase, Y-connect, 460-V(line-line), 25-KW, 60-Hz,four-pole induction
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 may be assumed to be equal to 220 W. With the rotor connected directly
to a 460-V source, compute the speed, output shaft torque and power, input and
power factor and efficiency for slip of 1, 2 and 3 percent. You may choose either to
either to represent the core loss by a resistance Rc connected in parallel with the
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
s 
4f e
poles
Vl  V / 3
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[%]
1.0
2.0
3.0
Speed[r/min]
1782 1764
1746
Tout[N-m]
8.5
16.5
23.4
Pout[kw]
8.5
16.5
23.4
Pin[kw]
45.8
89.6
128
Power factor
0.81
0.87
0.85
Efficiency[%]
93.3
94.4
93.8
6.13 A 15-kW 230-V three-phase Y-connected 60-Hz four-pole squirrel-cage induction
motor develops full-load internal torque at a slip of 3.5 percent when operated at
rated voltage and frequency. For the purposes of this problem, rotational and core
losses can be neglected and cord losses can be neglected . The following motor
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
maximum torque , and the internal starting torque at rated voltage and frequency.
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 voltage and frequency, has a
staring torque of 135 percent and a maximum torque of 220 percent, both with
respect to its rated-load torque. Neglecting the effects of stator resistance 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
所以 s rated
 0.240s max T  0.0824

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
替代值與解：



2



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 

  
R
  2 
2
所以可寫成： Tstart  1.26TA
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 
∴
Iˆ2, fl
Iˆ2,max T

將(2)代入(1)
∴
Iˆ2, fl
Iˆ2,max T

R2 / s  j ( X eq  X 2 )
j ( X eq  X 2 )  R2 / s fl
∴
R2
s max T
 X 1,eq  X 2 ………(2)
s max T
Iˆ2,max T

j 1
j  s max T / s fl
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.
………(1)
 R12,eq  ( X 1,eq  X 2 ) 2
R2
Iˆ2, fl
I
Vˆeq
j ( X eq  X 2 )  R2 / s max T
由(6.34)式可得：
∵ R1,eq  0
2
將(6.33)和(6.36)式整理後可得
2,max T
 3 I 2, fl




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

Iˆ2,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
SA
每單位因此Tmax = 3.12。
)2 ]
 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 apply to a 125-kW﹐2300-V﹐three-phase﹐four pole﹐60-Hz
Squirrel-cage induction motor: Stator-resistance between terminals= 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:
Volts
Three-phase
Motor
(line-to-line)
Amperes
1
2
74.7
99.4
72.9
72.9
power, kW
4.40
11.6
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
又因為馬達跟 I b21 R2 成比例
 
 
所以
I2
R
Tmotor2
(R )
(I 2 )
 22 motor2 2 motor2  ( 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
所以
( 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
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)
從習題 6.15 可得
2
Tmax 0.5[1  ( s max T / s fl ) ]

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
7.1 Consider a separately-excited dc motor. Describe the speed variation of the motor
operating unloaded under the following conditions:
a. The armature terminal voltage is varied while the field current is held constant.
b. The filed current is varied while the armature terminal voltage is held constant.
c. The filed winding is connected in shunt directly to the armature terminals, and the
armature terminal voltage is then varied.
Ans:
(a)
Vt  Va  I a Ra
假 設 電 動 機 感 應 轉 矩 (  ind ) 不 變 下 , 感 應 電 壓 Va  km ， 又 感 應 電 流
Ia 
 ind
, 因為電流 I a 固定，所以  為固定,又 Va  km ,故 V a (轉子電壓)與
k
 m (電動機速度)成正比。
(b)
m 
1
If
(c)
因為他激式直流電動機的場電路是由外部定電壓所供應，所以電動機的轉子末
端電壓是固定的，又 Va  km ，故  m (電動機速度)是固定不變的。
7.2 A dc shunt motor operating at an armature terminal voltage of 125V is observed to
be operating at a speed of 1180 r/min. When the motor is operated unloaded at the
same armature terminal voltage but with an additional resistance of 5Ω in series
with the shunt field, the motor speed is observed to be 1250 r/min.
a. Calculate the resistance of the series field.
b. Calculate the motor speed which will result if the series resistance is increased
from 5Ω to 15Ω.
Ans:
(a)
末端電壓固定,
故If 
(b)
1
,
Rf
84.3

1180
Rf
1180


nf ( n 為馬達轉速)不變
Rf  5
1250
 R f  84.3
8 4 . 3 1 5
 n2  1390r
n2
/ mi n
7.3 For each of the following changes in operating condition for a dc shunt motor,
describe how the armature current and speed will vary:
a. Halving the armature terminal voltage while the field flux and load torque remain
constant.
b. Halving the armature terminal voltage while the field current and load torque
remain constant.
c. Doubling the field flux while the armature terminal voltage and load torque
remain constant.
d. Halving both the field flux and armature terminal voltage while the load power
remains constant.
e. Halving the armature terminal voltage while the field flux remains constant and
the load torque varies as the square of the speed.
Only brief quantitative statements describing the general nature of the effect are
required, for example. ”speed approximately doubled.”
解:
驗證這個
(a)  m :減半， I a :常數。
(b)  m :減半， I a :加倍。
(c)  m :減半， I a :減半。
(d)  m :常數， I a :加倍。
(e)  m :減半， I a :減少四分之一。
7.4 The constant-speed magnetization curve for a 25-kW, 250-V dc machine at a speed
of 1200 r/min is shown in Fig7.27. This machine is separately excited and has an
armature resistance of 0.14  . This machine is to be operated as a dc generator
while driven from a synchronous motor at constant speed.
a. What is the rated armature current of this machine?
b. What the generator speed held at 1200 r/min and if the armature current is limited
to its rated value, calculate the maximum power output of the generator and the
corresponding armature voltage for constant field currents of (i) 1.0A,(ii) 2.0A
and (iii)2.5A.
C. Repeat part (b) if the speed of the synchronous generator is reduced to 900 r/min.
解:
(a) 轉子電流= 25kW
250V
 100 A .
(b) 1200 r/min， Ea 從磁化直接確定圖7.27的曲線。電樞電流計算為:
Va  Ea  I a Ra
輸出功率為: Pout  Va I a ﹒ 這裡 I a  100 A 。
(c) 解決辦法除了用作為被產生的電壓作為部份(b)產生900/1200 = 0.75
做為(b)的所用時間
7.9 When operated from a 230-V dc supply, a dc series motor operates at 975 r/min with
a line current of 90A. Its armature-circuit resistance is 0.11  series-field
resistance is 0.08  . Due to saturation effects, the flux produced by an armature
current of 30A is 48 percent of that an armature current of 90A. Find the motor
speed when the armature voltage is 230V and the armature current is 30A..
解：
從規定的數據來說，在 I a  90 A ， n(90)  975r / min 時產生電壓
Ea (90)  Va  I a ( Ra  Rs )  230  90(0.11  0.08)  212.9V
同樣的，在 I a  30 A 時，所產生的電壓為:
Ea (30)  230  30(0.11  0.08)  224.3V
從 E a  n
E a (30)
n(30)  (30)
(
)(
)
E a (90)
n(90)  (90)
因此當
 (30)
 (90)
n(30)  n(90)(
 0.48 ，我們可以解出 n(30)
E a (30)  (90)
)(
)  2140r / m i n
E a (90)  (30)
7.12 Two adjustable-speed dc shunt motors have maximum speeds of 1800 r/min and
minimum speeds of 500r/min. Speed adjustment is obtained by field-rheostat
control. Motor A drive a load requiring constant power over the speed range; motor
B drive one requiring constant torque. All losses and armature reaction may be
neglected.
a. If the power outputs of the two motors are equal at 1800r/min and the armature
currents are each 125A, what will the armature currents be at 500r/min?
b. If the power outputs of the two motors are equal at 500r/min and the armature
currents are each 125A, what will the armature currents be at 1800r/min?
c. Answer parts (a) and (b) with speed adjustment by armature-voltage control with
conditions otherwise the same.
解：
(a):
電極電壓的常數和速度變動得到場電流控制， 場電流將與速度成反比。
恆定的功率運算(電動機A)需要相同電樞電流，恆定轉拒(電動機B)電樞電
流變化與電動機速度成反比。因此，
motor A: I a  125A
motor B: I a  125(500
)  34.7 A
1800
(b):
motor A: I a  125A
motor B: I a  125(1800
)  450 A
125
(c):
在電樞電壓控制下和相同的場電流，速度將與電樞電流成正比，產生的電
壓將與速度成正比。恆定功率(電動機A)將要求隨著速度反增加電樞電
流，當恆定轉矩時(電動機B)將需要恆定的電樞電流。
對於(a)的條件來說:
motor A: I a  125(1800
)  450 A
125
motor B: I a  125A
對於(b)的條件來說:
motor A: I a  125(500
)  34.7 A
1800
motor B: I a  125A
7.13 Consider a dc shunt motor connected to a constant-voltage source and driving a
load requiring constant electromagnetic torque. Show that if Ea > 0.5 Vt (the
normal situation), increasing the resultant air-gap flux decreases the speed, whereas
if Ea< 0.5Vt (as might be brought about by inserting a relatively high resistance in
series with the armature), increasing the resultant air-gap flux increases the speed.
Ans：
E
V I R
m  a  a a a
K ad
K ad
T
K ad
Ia 
Thus
m 
1 
TRa
Va 
K ad 
K ad




1
dm
1  2TRa

2 I a Ra  Va   1 2 Va  2 Ea 

 Va  
2
2 
K ad
dd
K ad  K ad
 K ad
我們可以看見這 Ea > 0.5Va, dωa/dΦd < 0 and for Ea < 0.5Va, dωa/dΦd > 0. Q.E.D.
7.14 A separately-excited dc motor is mechanically coupled to a three-phase, four-pole,
30-kVA, 460-V, cylindrical-pole synchronous generator. The dc motor is connected
to a constant 230-V dc supply, and the ac generator is connected to a 460-V,
fixed-voltage, fixed-frequency, three-phase supply. The synchronous reactance of
the synchronous generator is 5.13Ω/phase. The armature resistance of the dc motor
is 30 mA. The four-pole dc machine is rated 30 kW at 230V. All unspecified losses
are to be neglected.
a. If the two machines act as a motor-generator set receiving power from the dc
source and delivering power to the ac supply, what is the excitation voltage of
the machine in volts per phase (line-to-neutral) when it delivers 30kW at unity
power factor? What is the internal voltage of the dc motor?
b. Leaving the 15eld current of the ac machine at the value corresponding to the
condition of part (a), what adjustment can be made to reduce the power transfer
between the two machines to zero? Under this condition of zero power transfer,
what is the armature current of the dc machine? What is the armature current of
the ac machine?
c. Leaving the field current of the ac machine as in parts (a) and (b), what
adjustment can be made to cause the transfer of 30 kW from the ac source to the
dc source? Under these conditions what are the armature current and internal
voltage of the dc machine? What will be the magnitude and phase of the current
of the ac machine?
Ans：
(a)
I a ,dc 
30 103
 37.7 A
3 460
Eaf  Va ,ac  jX a I a ,ac  460
3
 j5.13  37.7  328.4V , l  n
P  E a I a ,dc  30kW
Ea  Va ,dc  I a ,dc Ra
E 
2
a
Va ,dc  Va2,dc  4 PRa
2
 226V
(b)
I a ,ac 
E af  Va ,ac
Xs
 12.2 A
(c)
P  E a I a ,dc  30kW
E a  Va ,dc  I a ,dc Ra
E 
2
a
Va ,dc  Va2,dc  4 PRa
I a ,dc 
2
E a  Va ,dc
Ra
 128 A
 226V