5.1 The full-load torque angle of a synchronous motor at rated voltage and frequency
is 35 electrical degrees. Neglect the effects of armature resistance and leakage
reactance. If the field current is held constant, how would the full-load torque
angle be affected by the following changes in operating condition?
a. Frequency reduced 10 percent, load torque and applied voltage constant.
b. Frequency reduced 10 percent, load power and applied voltage constant.
c. Both frequency and applied voltage reduced 10 percent, load torque constant.
d. Both frequency and applied voltage reduced 10 percent, load power constant.
中文翻譯:
一個滿載的同步發電機它的定電壓角頻率為 35 度角，忽略所有電樞電阻及漏
電抗，假如電流保持固定，則跟著改變操作情況的話會受到什麼影響。
a. 頻率減少 10%的話，負載轉距和應用實電壓的持續狀況為何。
b. 頻率減少 10%的話，遺失功率和應用實電壓的持續狀況為何。
c. 兩者頻率和實電壓減少 10%的話，負載轉拒的持續狀況為何。
d. 兩者頻率和實電壓減少 10%的話，負載功率的持續狀況為何。
Ans:
由公式(5.1)知
T
 poles
2
(
2
) 2  R F f sin  RF
因此 T   R F f sin  RF
並且塲電流與 F f 同為常數
所以  R 
Vt
V sin  RF
, 則T  t
f
f
故 P   f T  Vt sin  RF
(a)
減少 31.1
(b)
不用改變
(c)
不用改變
(d)
增加 39.6
5.2 The armature phase windings of a two-phase synchronous machine are displaced
by 90 electrical degrees in space.
a. What is the mutual inductance between these two windings?
b. Repeat the derivation leading to Eq.5.17 and show that the synchronous
inductance is simply equal to the armature phase inductance; that is,
Ls  Laa0  La1 , where Laa0 is the component of the armature phase inductance
due to space-fundamental air-gap flux and La1 is the armature leakage
inductance.
中文翻譯:
一個兩相同步機械用 90 度電度角取代一個電樞相位繞阻
a. 這兩個繞阻中的互感為什麼
b. 根據 Eq.5.17 簡單的證明同步繞阻電感相位相等 Ls  Laa0  La1 ， Laa0 是由
基本相位電感所組成的和 La1 是電樞漏電感
Ans:
(a)
由於繞組互為垂直(正交),所以互感為 0.
(b)
因為兩組線圈正交，相位解耦合時磁通在平衡二相的操作,因為在其間電流是
不變的。因此，相等的電感等於相位自感。
5.3 Design calculations show the following parameters for a three-phase,
cylindrical-rotor synchronous generator:
Phase-α self-inductance Laa  4.83 mH
Armature leakage inductance Lal  0.33 mH
Calculate the phase-phase mutual inductance and the machine synchronous
inductance.
中文翻譯:
計算證明根據同步發電機圓柱轉子的三相參數
α 相自感 Laa  4.83 mH
電樞漏電感 Lal  0.33 mH
計算互感及同步電感機
Ans:
1
Lab  Lba  L ac  Lca  Lbc  Lcb   Laa0
2
Laa  Laa 0  Lal  Laa0  Laa  Lal
1
1
Lab   ( Laa  Lal )   (4.83  0.33)  2.25 mH
2
2
Ls 
3
3
1
3
1
( Laa  Lal )  Lal  Laa  Lal  (4.83)  (0.33)
2
2
2
2
2
 7.245  0.165  7.08mH
5.4 The open-circuit terminal voltage of a three-phase, 60-Hz synchronous generator
is found to be 15.4 KV rms line-to-line when the field current is 420 A.
a. Calculate the stator-to-rotor mutual inductance Laf .
b. Calculate the open-circuit terminal voltage if the field current is held constant
while the generator speed is reduced so that the frequency of the generated
voltage is 50 Hz.
中文翻譯:
一個開迴路三相電壓，60HZ 的同步發電機找 15KV 的線性電壓，當電流是
420A 時。
a. 計算一個定子轉子的互感 Laf 。
b. 計算一個開迴路電壓，假如定電流當發電機的速度降低則產生電壓頻率為
50HZ。
Ans:
(a)
eaf 
af
d af
dt
 Laf I f cos( t   0 )
eaf 
d ( Laf I f cos( t   0 ))
dt
sin    cos(  90 0 )
 Laf I f sin(  t   0 )
 eaf  Laf I f cos( t   0  90 0 )
Laf I f
E af ,l l ,rms 
 Laf 
2
 3
2 E af ,l l ,rms
3 I f

2  15.4
3  2  60  420
 79.4 mH
(b)
 50 
Voltage    15.4 kV  12.8 kV
 60 
5.5 A 460V，50-KW，60HZ，three-phase synchronous motor has a synchronous
reactance of X s  4.15 and an armature-to-field mutual inductance ，
Laf  83mH .The motor is operating at rated terminal voltage and an input power
of 40KW.Calculate the magnitude and phase angle of the line-to-neutral generated

voltage E af and field current I f if the motor is operating at (a)0.85 power
factor lagging (b)unity power factor，and (c)0.85 power factor leading
中文翻譯:
一個 460V，50-KW，60HZ 的三相同步電動機有一個同步電抗 X s  4.15 和電樞
互感 Laf  83mH ，這個電動機的操作定電壓和輸入功率為 40KW。計算這個

相角所產生電壓 E af 和電流 I f ，假如這個電動機操作在(a) 落後 0.85 功率因
數(b) 單一功率因數 (c) 超前 0.85 功率因數。
Ans:
(a)
40 103
 59.1A
0.85  3  460
功因 0.85 落後，所以相角    cos 1 0.85  31.8 
a 相電流

E af  Va  jX s I a

I a  I a e  j  59.1e  j 31.8


E af  Va  jX s I a 
If 
(b)
2 E af
Laf
 11.3 A
Ia 

460
3

 j 4.15  59.1e  j 31.8  136  56.8 V
同 (a) pf=1 相角   cos 1 1  0 

I a  I a e  j  59.1e  j 0  59.1



E af  Va  jX s I a  266  38.1 V
If 
2 E af
Laf
 15.3 A
(c)
同 (a) pf=0.85 超前 相角   cos 1 0.85  31.8 

I a  I a e  j  59.1e j 31.8



E af  Va  jX s I a  395  27.8  V
If 
2 E af
Laf
 20.2 A
5.6 The motor of Problem 5.5 is supplied form a 460-V，three-phase source through
a feeder whose impedance is Z f  0.084  j 0.82 . Assuming the system(as
measured at the source) to be operating at an input power of 40KW，Calculate the

magnitude and phase angle of the line-to-neutral generated voltage E af and field
current I f if the motor is operating at (a) 0.85 power factor lagging (b) unity
power factor，and (c)0.85 power factor leading
中文翻譯:
這個電動機在問題 5.5 有提供一個 460V 的電壓，三相電源通過一個線路它的阻
抗是 Z f  0.084  j 0.82 ，假設這個系統的輸入功率操作在 40KW，計算這

個相角所產生電壓 E af 和電流 I f ，假如這個電動機操作在(a) 落後 0.85 功率
因數 (b)單一功率因數 (c)超前 0.85 功率因數。
Ans:
Z f  jX s  0.084  j 0.82  j 4.15  0.084  j 4.97
解法同 5.5
(a)

I f  12.2 A

I f  16.3 A

I f  22.0 A
E af  106  66.6  V
(b)
E af  261  43.7  V
(c)
E af  416  31.2  V
5.7 A 50-HZ, two-pole, 750kVA, 2300V, three-phase synchronous machine has a
synchronous reactance of 7.75Ω and achieves rated open-circuit terminal voltage
at a field current of 120 A.
a. Calculate the armature-to-field mutual inductance.
b. The machine is to be operated as a motor supplying a 600KW load at its rated
terminal voltage. Calculate the internal voltage E af and the corresponding
field current if the motor is operating at unity power factor
c. For a constant load power of 600kW, write a MATLAB script to plot the
terminal current vary between a minimum value corresponding to a machine
loading of 750kVA at leading power factor and maximum value corresponding
to a machine loading of 750kVA at lagging power factor. What value of field
current produces the minimum terminal current? What?
中文翻譯：
一個 50HZ，750KW，2300V，，兩極三相的同步機器，和一個末端開迴路電
壓而它的電流為 120A。
a. 計算這個電樞場的互感。
b. 這個機器在它的端電壓操作一個電動機供應 600KW 的負載，計算內部的
電壓和相對應的電流假設這個電動機操作在單一功率因數。
一個 600KW 的固定負載功率，寫一個 MATLAB 程式它的末端電流對應一個
最小值，給這個機器負載 750kVA 超前功率因數和對應一個落後功率因數
的最大值給這個機器負載 750kVA，則產生這個最小電流為多少？為什麼？
Ans:
(a)
Laf 
2V11 , rms
3I f
 49.8 mH
(b)

Ia
600  10 3
3 2300

 151 A

E af  Va  jX s I a  1.77  41.3 V
If 
2 E af
Laf
 160 A
(c)
由圖可知,當場電流為 160A 時,將會得到最小的功率因數及電流。
5.8 The manufacturer’s data sheet for a 26-kv, 750MVA, 60Hz, three-phase
synchronous generator indicates that it has a synchronous reactance Xs=2.04 and
a leakage reactance X al  0.18 , both in per unit on the generator base. Calculate (a)
the synchronous inductance in mH, (b) the armature leaking inductance in mH,
and (c) the armature phase inductance Laa in mH and per unit
Ans:
中文翻譯:
這個製造公司給一個 26kw，750MVA，60Hz，指示三相同步發電機有個同步電抗
Xs=2.04，和一個漏電抗 X al  0.18 ，兩者經由發電機上。計算(a)這個同步電抗在
mH，(b)這個電樞漏電感在 mH，(c)這個電樞相電感 Laa 在 mH 和標么值。
(a)
Z base 
Ls 
2
Vbase
(26  10 3 ) 2

 0.901
Pbase
750  10 6
X s , pu Z base


2.04  0.901
 4.88 mH
2  60
(b)
La1 
X a1, pu Z base


0.18  0.901
 0.43 mH
2  60
(c)
先求三相同步電感
Ls 
Laa0
L s 定義為
3
2
Laa0  La1  Laa0  ( Ls  La1 )    (1)
2
3
 代表氣隙磁通的基波分量所產生的自感成分
La1  代表電樞漏磁通所產生的電感分量
將(1)式代入電樞相電感公式得
2
Laa  Laa0  l a1  (4.48  0.43)  0.43  3.40mH
3
5.9 The following reading are taken from the results of an open-and a short-circuit test
on an 800-MVA,three-phase,Y-connected,26-kV,two-pole,60-Hz turbine generator
driven at synchronous speed:
Field current, A
Armature current, short-circuit test, kA
Line voltage, open-circuit characteristic, kV
Line voltage, air-gap line, kV
1540
9.26
26.0
29.6
2960
17.8
(31.8)
(56.9)
The number in parentheses are extrapolations based upon the measured data. Find
(a) the short-current ratio, (b) the unsaturated value of the synchronous reactance
in ohms per phase and per unit ,and (c) the saturated synchronous reactance in per
unit and in ohms per phase.
Ans:
中文翻譯:
根據這個開和&閉迴路的結果測試一個三相 800-MVA，Y 接，26-Kv，兩極，
60-Hz，同步速度的渦輪發電機。
Field current, A
Armature current, short-circuit test, kA
Line voltage, open-circuit characteristic, kV
Line voltage, air-gap line, kV
1540
9.26
26.0
29.6
2960
17.8
(31.8)
(56.9)
根據這些量過的數據，找(a)短路比。(b) 這個不飽和的同步電抗的總電抗和阻
抗，和(c) 飽和同步電抗和阻抗。
(a)
短路比SCR 
AFNL 1540

 0.52
AFSC 2960
(b)
Z base
2
Vbase
(26  10 3 ) 2


 0.845
Pbase
800  10 6
X s ,u 
2960 29.6

 2.19 pu  1.85
1540 26
(c)
由課本5 - 29式可發現, SCR是飽合同步電抗標么值得倒數
1
1
Xs 

 1.92 pu  1.62
SCR 0.52
5.10 The following readings are taken from the results of an open and a short circuit
test on a 5000-kW, 4160-V, three-phase, four-pole, 1800-rpm synchronous motor
driven at rated speed.
Field current, A
Armature current, short-circuit test, A
Line voltage, open-circuit characteristic, V
Line voltage, air-gap line, V
169
694
3920
4640
192
790
4160
5270
The armature resistance is 11 mΩ/phase. The armature leakage reactance is
estimated to be 0.12 per unit on the motor rating as base. Find (a) the short-circuit
ratio, (b) the unsaturated value of the synchronous reactance in ohms per phase
and per unit, (c) the saturated synchronous reactance in per unit and in ohms per
phase.
中文翻譯:
根據這個開迴路和閉迴路的結果測試一個三相 5000-kW，4160-V，四極，轉
速每分鐘 1800 轉的定速同步馬達。
Field current, A
Armature current, short-circuit test, A
Line voltage, open-circuit characteristic, V
Line voltage, air-gap line, V
169
694
3920
4640
192
790
4160
5270
這個電樞電抗是 11 mΩ/phase，這個電樞漏電抗是估算 0.12 per 在馬達額定功率
上。找(a))這個短路比。 (b) 這個不飽和的同步電抗的總電抗和阻抗，和(c) 飽
和同步電抗和阻抗。
Ans:
(a)
SCR 
AFNL
 1.14
AFSC
(b)
Z base  4160 2 (5000  10 3 )  3.46
Xs 
1
 1.11 pu  3.86
SCR
(c)
X s ,u 
AFSC
 0.88 pu  3.05
AFNL, ag
5.11 Write a MATLAB script which automates the calculation of Problem 5.9 and
5.10.The following minimum set of data is required:
■ AFNL:The filed current required to achieve rated open-circuit terminal
voltage。
■ The corresponding terminal voltage on the air gap line。
■ AFSC:The field current required to achieve rated short-circuit current on
the short-circuit characterstic。
Your scrip should calculate(a)the short-circuit ratio，(b)the unsaturated value of the
synchronous reactance in ohms per phase and per unit，and(c)the saturated synchronous
reactance in per unit and in ohms per phase。
中文翻譯:
寫一個 MATLAB 自動化計算問題 5.9，5.10 ，必須根據這個資料的最小值。
■ 輔助設施:必須有開迴路定電壓的電流。
■ 這個有相同的線性電壓。
■ 上述:這個電流必須為短路定電流，在短路特性上。
你將要計算(a)這個短路比，(b) 這個不飽和的同步電抗的總電抗和阻抗，和(c) 飽
和同步電抗和阻抗。
5.12 Consider the motor of Problem 5.10.(考慮問題 5.10 的馬達)
a. Compute the filed current required when the motor is operating at rated
voltage, 4200kW input power factor leading. Account for saturation inder load
by the method described in the paragraph relating to Eq. 5.29.
b. In addition to the data given in Problem 5.10, additional points on the
open-circuit characteristic are given below:
Field current, A
Line voltage, V
200
4250
250
4580
300
4820
350
5000
If the circuit breaker supplying the motor of part (a) is tripped, leaving the motor
suddenly open-circuited, estimate the value of the motor terminal voltage
following the trip.
Ans:
(a)
The total power is S 
I a  6 7 02 9.5 0
and
P 4200kW

 4828kVA
pf
0.87
Z s  0.038  j 4.81
4349
I f  A F N (L
)  3 0 6A
4 1 6 /0 3
(b)
If the machine speed remains constant and the field current is not reduced, the
terminal voltage will increase to the value corresponding to 306 A of field current
on the open-circuit saturation characteristic. Interpolating the given data shows
that this corresponds to a value of around 4850 V line-to-line.
中文翻譯:
考慮問題 5.10 的馬達
a. 計算這個電流當馬達操作在定電壓，4200kw 超前輸入功率因數，把飽和負載
描述在 Eq. 5.29 有關這段的方法。
b. 除此之外這個數據給問題 5.10，根據下面的資料額外的指出在開迴路的特性。
Field current, A
200
250
300
350
Line voltage, V
4250
4580
4820
5000
假如電路的開關提供一個馬達在一部分(a)突然關閉這個馬達的開路，計算這個
馬達的端電壓的值。
(b)
假設這個機器的速度不變且不減少它的電流，這個端電壓的數值將增加到
306A，在開迴路的飽和數，改它的數據讓它相當於在線性 4850 V 的值。
5-13 Using MATLAB, plot the field current required to achieve unity-power-factor
operation for the motor of Problem 5.10 as the motor load varies from zero to
full load. Assume the motor to be operating at rated terminal voltage.
中文翻譯：
使用 MATLAB，劃出這個電流必須為單一功率因數操作馬達在問題５．１０上，
修改這個馬達從零到滿載，假設這個馬達操作在定端電壓上。
Ans: