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```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.
ANS:

T
 poles
2
(
2
) 2  R F f sin  RF

V sin  RF
Vt
, 則T  t
f
f

(a)

(b)

(c)

(d)

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.
ANS:
(a)
The windings are orthogonal and hence the mutual inductance is zero.
(b)
Since the two windings are orthogonal, the phases are uncoupled and hence the flux
linkage under balanced two-phase operation is unchanged by currents in the other phase.
Thus, the equivalent inductance is simply equal to the phase self-inductance.
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.
Sol :
1
Lab  Lba  L ac  Lca  Lbc  Lcb   Laa 0
2
Laa  Laa 0  Lal
 Laa0  Laa  Lal
1
1
Lab   ( Laa  Lal )   (4.83  0.33)  2.25 mH
2
2
3
( Laa  Lal )  Lal
2
3
1
 Laa  Lal
2
2
3
1
 (4.83)  (0.33)
2
2
 7.245  0.165
 7.08 mH
Ls 
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.
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.
a.
Sol :
part (a):
eaf 
 af
d  af
dt
 Laf I f cos( t   0 )
eaf 
d ( Laf I f cos( t   0 ))
dt
 Laf I f sin(  t   0 )
sin    cos(  90 0 )
 eaf  Laf I f cos( t   0  90 0 )
E af ,l l ,rms 
 Laf 
Laf I f
2
 3
2 E af ,l l ,rms
3 I f

2  15.4
3  2  60  420
part (b):
 50 
Voltage    15.4 kV  12.8 kV
 60 
 79.4 mH
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 currect I f if the motor is operating at （a）0.85 power
factor lagging （b）unity power factor，and （c）0.85 power factor leaging
Sol：
40 103
（a）a 相電流 E af  Va  jX s I a
Ia 
 59.1A
0.85  3  460



I a  I a e  j  59.1e  j 31.8


E af  Va  jX s I a 
If 
2 E af
Laf
460
3


 j 4.15  59.1e  j 31.8  136  56.8 V
 11.3 A
（b）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 currect I f if the motor is operating at （a）0.85 power factor lagging （b）
unity power factor，and （c）0.85 power factor leaging
Sol：
Z f  jX s  0.084  j 0.82  j 4.15  0.084  j 4.97


I f  1 2.2 A

I f  16.3 A

I f  22.0 A
（a） E af  106  66.6  V
（b） E af  261  43.7  V
（c） E af  416  31.2  V
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 Xal=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
sol :
(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 )

Ls 
3
2
L aa0  L a1
 L aa0 
L aa0
2
( L s  L a1 )    (1)
3
 代表氣隙磁通的基波分量所產生的自感成分
L a1  代表電樞漏磁通所產生的電感分量

L aa  L aa0  l a1

2
3
(4.48  0.43)  0.43  3.40mH
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
sol :
(a)

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)

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. The armature resistance is 11 mOhm/phase. The armature
leakage reactance is estimated to be 0.12 p.u. 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 p.u., (c) the saturated synchronous reactance in per unit and in
ohms per phase.
Sol:
part(a)
SCR 
AFNL
 1.14
AFSC
part(b)
Z base  4160 2
Xs 
(5000  10 3 )
 3.46
1
 1.11 pu  3.86
SCR
part(c)
X s ,u 
AFSC
 0.88 pu  3.05
AFNL, ag
5.12 Consider the motor of Problem 5.10.
a. Compute the filed current required when the motor is operating at rated
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.
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.
Sol:
part( a)
The total power is S 
I a  67029.5 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
part(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.
5-14 Loss data for the motor of Problem 5.10 are as
follows：
Open-circuit core loss at 4160V = 37kW
friction and windage loss = 46kW
field-winding resistance at 75℃=0.279Ω
Compute the output power and efficiency when the motor is operating at rated
input power, unity power factor, and rated voltage. Assume the field-winding to be
operating At a temperature of125℃
Ans:
At rated power, unity power factor, the armature current will be Ia =5000 kW/(√3
4160 V) = 694 A. The power dissipated in the armature winding will then equal
Parm = 3× 6942 × 0.011 = 15.9 kW.The field current can be found from
|Eaf | = |Va − ZsIa| = |4160√3 − ZsIa| = 2394 V, line-to-neutral
If = AFNL_2394/4160/√3) = 238 A
At 125◦C, the field-winding resistance will be
Rf = 0.279(234.5 + 125/234.5+ 75)= 0.324 Ω
and hence the field-winding power dissipation will be Pfield = I2f Rf = 18.35
kW.The total loss will then be
Ptot = Pcore + Parm + Pfriction/windage + Pfield = 120 kW
Hence the output power will equal 4882.75 kW and the efficiency will equal
4882.75/5000
= 0.976 = 97.6%.
5-16 What is the maximum per-unit reactive power that can be supplied by a
synchronous machine operating at its rated terminal voltage whose synchronous
reactance is 1.6 per unit and whose maximum field current is limited to 2.4 times
that required to achieve rated terminal voltage under open circuit conditions?
Ans:
For Va = 1.0 per unit, Eaf,max = 2.4 per unit
and Xs = 1.6 per unit
Qmax = (Eaf,max − Va)/ Xs= 0.875 per unit
5.19 A synchronous machine with a reactance of 1.28 per unit is operating as a
generator at a real power loading of 0.6 per unit connected to a system with a series
reactance of 0.07 per unit. An increase in its filed current is observed to cause a
decrease in armature current.
a. Before the increase, was the generator supplying or absorbing reactive power
from the power system?
b. As a result of this increase in excitation, did the generator terminal voltage
increase or decrease?
c. Repeat parts (a) and (b) if the synchronous machine is operating as a motor.
SOL:
part (a) : It was underexcited, absorbing reactive power.
part (b) : It increased.
part (c) : The answers are the same.
5.22 A four-pole, 60-Hz, 24-kV, 650-MVA synchronous generator with a synchronous
reactance of 1.82 per unit is operating on a power system which can be represented
by a 24-kV infinite bus in series with a reactive impedance of j0.21Ω. The
generator is equipped with a voltage regulator that adjusts the field excitation such
that the generator terminal voltage remains at 24kV independent of the generator
a. The generator output power is adjusted to 375MW.
(i) Draw a phasor diagram for this operating condition.
(ii) Find the magnitude (in kA) and phase angle (with respect to the generator
terminal voltage) of the terminal current.
(iii) Determine the generator terminal power factor.
(iv) Find the magnitude (in per unit and kV) of the generator excitation voltage
Eaf.
b.
Repeat part (a) if the generator output power is increased to 600 MW.
part (a):
(i)
(ii) Vt = V∞ =1.0 per unit. P = 375/650 = 0..577 per unit.
Thus
 t  sin 1 (
and
Ve
Iˆ a t
PX 
)  12.6
VtV
j t
 V
jX 
 0.5783.93 per  unit
I base  Pbase /( 3Vbase )  15.64kA
and Thus I a  9.04 kA
(iii) The generator terminal current lags the terminal voltage by  t / 2 and
thus the power factor is
pf  cos 1  t / 2  0.998 lagging
(iv)
Eˆ af  V  j ( X   X s ) Iˆa  1.50 per  unit  36.0 kV , line  to  line
part (b):
(i)
(ii)
Same phasor diagram
Iˆa =0.928∠6.32° per-unit. Ia=14.5 kA.
(iii) pf=0.994 lagging
(iv) Eaf=2.06 per unit =49.4 kV, line-to-line
5.25 Repeat Example 5.9 assuming the generator is operating at one-half of its rated
KVA at a lagging power factor of 0.8 and rated terminal voltage.
( X d  1.0 p.u, X q  0.6 p.u )
Solution:
0
LetVˆa  Va e j 0.0  Va
 power factor = 0.8 lagging
  cos 1 0.8  36.9
Iˆ  I e j  0.8  j 0.6  1.0e  j 36.9
a
a
Eˆ  Vˆa  jX q I q  1  j 0.6(1.0e  j 36.9 )  1.44e j19.4
'
   19.4
     19.4  (36.9)  56.3
I d  Iˆa sin(    )  1.0 sin 56.3  0.832
I q  Iˆa cos(   )  1.0 cos 56.3  0.555
 Iˆd  0.832e j ( 9019.4)  0.832e j 70.6
Iˆq  0.555e j19.4
 Eˆ af  Vˆa  jX d I d  jX q I q
 1  j 0.5(0.832e70.4 )  j 0.6(0.555e j19.4 )
 1.281  j 0.452
 1.358e j19.4
5.30 What maximum percentage of its rated output power will a salient-pole motor
deliver without loss of synchronism when operating at its rated terminal voltage
with zero field excitation ( Eaf  0 ) if X d  0.90 per unit and X q  0.65 per
unit ? Compute the per unit armature current and reactive power for this operating
condition.
Solution:
For E af  0
LetVt  1.0 p.u
 Pmax 
Vt 2
2
 1
1 


 0.21  21%
X

X
q
d


Maximum power for   45
V cos 
so I d  t
 0.786 p.u
Xd
Iq 
Vt sin 
 1.088 p.u
Xq
I a  I d2  I q2  1.34 p.u
S  Vt I a  1.34 p.u
Q  S 2  P 2  1.32 p.u
5.31 if the synchronous motor of Problem 5.30 is now operated as a synchronous
Generator connected to an infinite bus of rated voltage, find the minimum Per-unit
field excitation (where 1.0 per unit is the field current Required to achieve rated
open-circuit voltage) for which the generator will remain synchronized at (a) half
Sol:
p
V Eaf
Xd
sin  
V2 1
1
(

) sin 2
2 Xq Xd

Part (a) For P = 0.5, must have Eaf ≥ 0.327 per unit.
Part (b) For P = 1.0, must have Eaf ≥ 0.827 per unit.
5.32 A salient-pole synchronous generator with saturated Synchronous reactances
Xd=1.57 per unit and Xq=1.34 per unit is connected to an infinite bus of rated
voltage
Through an external impedance Xbus =0.11 per unit. The generation is
Supplying its rated MAV at 0.95 power factor Lagging, as measured at the
generator terminals.
a. Draw a phasor diagram indicating the infinite-bus voltage the armature current the
generator terminal voltage the excitation voltage and the rotor angle
b. Calculate the per-unit terminal and excitation voltage and the rotor angle in degrees
Sol:
Part(a) 向量圖如下

p
VVt
sin t
X bus
Ia 
Vt  V
jX t

Vt = 1.02 per-unit; Eaf = 2.05 per-unit; δ = 46.6
5-34. A two-phase permanent-magnet ac motor has a rated speed of 3000 r/min and a
six-pole rotor.Calculate the frequency (in Hz) of the armature voltage required to
operate at this speed.
ANS:
f 
n  poles 3000  6

 150 Hz
120
120
5-35. A 5_kW，three-phase，permanent-magnet synchronous generator produces an
open-circuit voltage of 208 V line-to-line，60-Hz，when driven at a speed of 1800
r/min。when operating at rated speed and supplying a resistive load，its terminal
voltage is observed to be 192 V line-to-line for a power output of 4.5 kW。
A.
B.
Calculate the generator phase current under this operating condition
Assuming the generator armature resistance to be negligible，calculate the
generator 60-Hz synchronous reactance。
C.
Calculate the generator terminal voltage which will result if the motor
generator load is increased to 5kW(again purely resistive) while the speed is
maintained at 1800 r/min
ANS:
(a) I a 
P
4500

 13.5 A
3 *Va
3 *192
(b) Eaf 
208
 120 V
3
Xs 
(c )
Ia 
Xs 
Eaf2  Va2
Ia
P
3 * Va

Eaf  Va  ( X s I a ) 2
2
 3.41
5000
 15
3 * 192
1202  110.82
 3.07 
15
The easiest way to solve this is to use MATLAB to iterate to find the required
load resistance.If this is done,the solution is Va =108 V(line-to-neutral)=187 V
(line-to-line).
```
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