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: 由公式(5.1)知

RFfR FpolesT δφπ sin)

2(

22=

因此 RFfR FT δφ sin∝

並且塲電流與 fF 同為常數

所以 f

VtR ∝ϕ , 則

fV

T RFt δsin∝

故 RFtf VTP δω sin∝=

(a) 減少 °1.31 (b) 不用改變 (c) 不用改變 (d) 增加 °6.39

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,

10 aaas LLL += ,where 0aaL is the component of the armature phase inductance due to space-fundamental air-gap flux and 1aL 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 mHLaa 83.4= Armature leakage inductance mHLal 33.0=

Calculate the phase-phase mutual inductance and the machine synchronous inductance.

Sol :

mHLLL

LLLLLL

LLLLLLL

alaaab

alaaaa

alaaaa

aacbbccaacbaab

25.2)33.083.4(21)(

21

21

0

0

0

−=−−=−−=

−=⇒+=

−======

mH

LL

LLLL

alaa

alalaas

08.7165.0245.7

)33.0(21)83.4(

23

21

23

)(23

=−=

−=

−=

+−=

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.

Sol : part (a):

)90cos()90cos(sin

)sin(

))cos(()cos(

00

0

0

0

0

++=⇒

+−=

+−=

+=

+=

=

θωω

αα

θωω

θω

θωλ

λ

tILe

tILdt

tILde

tILdt

de

fafaf

faf

fafaf

fafaf

afaf

mHI

EL

ILE

f

rmsllafaf

fafrmsllaf

4.794206023

4.1523

2

32

,,

,,

=⋅⋅⋅

⋅==⇒

⋅=

−

−

πω

ω

part (b):

kV 12.8 kV 15.4 6050 Voltage =

=

5.5 A 460V，50-KW，60HZ，three-phase synchronous motor has a synchronous reactance of Ω= 15.4sX and an armature-to-field mutual inductance，

mHLaf 83= .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 afE∧

and field currect fI if the motor is operating at （a）0.85 power

factor lagging （b）unity power factor，and （c）0.85 power factor leaging Sol：

（a）a相電流 asaaf IjXVE −=∧

AIa 1.59460385.0

1040 3

=××

×=

功因 0.85落後，所以相角 o8.3185.0cos 1 −=−= −α

o8.311.59 jj

aa eeII −−∧

== α

VejIjXVE jasaaf

oo

8.561361.5915.43

460 8.31 −∠=×−=−= −∧∧

ALE

Iaf

aff 3.11

2==

ω

（b）同（a）pf=1相角 o01cos 1 == −α

1.591.59 0 === −−∧ ojj

aa eeII α

VIjXVE asaafo1.38266 −∠=−=

∧∧

ALE

Iaf

aff 3.15

2==

ω

（c）同（a）pf=0.85超前 相角 o8.3185.0cos 1 == −α

o8.311.59 jj

aa eeII == −∧

α

VIjXVE asaafo8.27395 −∠=−=

∧∧

ALE

Iaf

aff 2.20

2==

ω

5.6 The motor of Problem 5.5 is supplied form a 460-V，three-phase source through

a feeder whose impedance is Ω+= 82.0084.0 jZ f . 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 afE∧

and

field currect fI if the motor is operating at （a）0.85 power factor lagging （b）

unity power factor，and （c）0.85 power factor leaging Sol：

Ω+=++=+ 97.4084.015.482.0084.0 jjjjXZ sf

解法同 5.5

（a） VE afo6.66106 −∠=

∧

AI f 2.12=

（b） VE afo7.43261 −∠=

∧

AI f 3.16=

（c） VE afo2.31416 −∠=

∧

AI f 0.22=

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

mHZX

L

b

mHZX

L

PVZ

asol

basepuaa

basepuss

base

basebase

43.0602901.018.0

)(

88.4602901.004.2

901.010750

)1026(

)(:

,11

,

6

232

=×

×==

=×

×==

Ω=×

×==

πω

πω

mH

lLL

LL

LLL

LLL

c

aaaaa

a

aa

asaa

aaas

40.343.0)43.048.4(32

)1(

)1()(32

23

L)(

10

1

0

10

10

s

=+−=

+=

=

=

−−−−=⇒

+=

得式代入電樞相電感公式將

的電感分量代表電樞漏磁通所產生

量所產生的自感成分代表氣隙磁通的基波分

定義為先求三相同步電感

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 1540 2960 Armature current, short-circuit test, kA 9.26 17.8 Line voltage, open-circuit characteristic, kV 26.0 (31.8) Line voltage,air-gap line,kV 29.6 (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

Ω====

Ω==×=

Ω=×

×==

===

62.192.152.011

SCR,29-5(c)

85.119.226

6.2915402960

845.010800

)1026(

(b)

52.029601540

)(:

,

6

232

puSCR

X

puX

PVZ

AFSCAFNLSCR

asol

s

us

base

basebase

得倒數是飽合同步電抗標么值式可發現由課本

短路比

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)

14.1==

AFSCAFNLSCR

part(b)

Ω=×

= 46.3)105000(4160 32

baseZ

Ω=== 86.311.11 puSCR

X s

part(c)

Ω=== 05.388.0,, puagAFNL

AFSCX us

5.12 Consider the motor of Problem 5.10. 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.

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 kVAkWpfPS 4828

87.04200

===

05.29670∠=aI and Ω+= 81.4038.0 jZ s

AAFNLI f 306)3/4160

4349( ==

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 125C, 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 loading. 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

and

o6.12)(sin 1 ==∞

∞−

VVPX

ttδ

unitperjX

VeVItj

ta −∠=

−=

∞

∞ o93.3578.0ˆδ

kAIThusand

kAVPI

a

basebasebase

04.9

64.15)3/(

=

==

(iii) The generator terminal current lags the terminal voltage by and thus the power factor is

(iv) part (b): (i) Same phasor diagram (ii) =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

2/tδ

laggingpf t 998.02/cos 1 == − δ

linetolinekVunitperIXXjVE asaf −−=−=++= ∞∞ ,0.3650.1ˆ)(ˆ

aI

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.

( upXupX qd .6.0,.0.1 == )

Solution:

aj

aa VeVVLet ==00.0ˆ

Qpower factor = 0.8 lagging

°=°−−°=−⇒°=⇒

=+=+=

=−==

°−==∴

°°−

°−

−

3.56)9.36(4.194.19

44.1)0.1(6.01ˆˆ0.16.08.0ˆ

9.368.0cos

4.199.36'

9.36

1

φδδ

φφ

jjqqa

jjaa

eejIjXVE

ejeII

555.03.56cos0.1)cos(ˆ

832.03.56sin0.1)sin(ˆ

=°=−=

=°=−=

φδ

φδ

aq

ad

II

II

°−°+°− ==⇒ 6.70)4.1990( 832.0832.0ˆ jjd eeI

°= 4.19555.0ˆ jq eI

qqddaaf IjXIjXVE ++=⇒ ˆˆ

°

°°−

=

+=++=

4.19

4.194.70

358.1452.0281.1

)555.0(6.0)832.0(5.01

j

j

ej

ejej

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 ( 0=afE ) if 90.0=dX per unit and 65.0=qX per

unit ? Compute the per unit armature current and reactive power for this operating condition.

Solution:

For 0=afE

upLetVt .0.1=

%2121.0112

2

max ==

−=⇒

dq

t

XXVP

Maximum power for °= 45δ

so upX

VId

td .786.0

cos==

δ

upPSQ

upIVS

upIII

upX

VI

at

qda

q

tq

.32.1

.34.1

.34.1

.088.1sin

22

22

=−=

==

=+=

==δ

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 load and (b) full load

Sol: 發電機要保持同步只要 Pmax＞P就可以，但我們可以用MATLAB就可以輕易的解題 而且可以滿足任何特定的條件 所以

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.

δδ 2sin)11(2

sin2

dqd

af

XXV

XEV

p −+= ∞∞

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

在 P=0.95之下必須先滿足下列兩公式 同時帶入 在指定功因角為提供一定功率，而且要考慮其發電機端電壓。 使用MATLAB來快速求解算出 Vt = 1.02 per-unit; Eaf = 2.05 per-unit; δ = 46.6

tX

VVpbus

t δsin∞=t

ta jX

VVI ∞−=

tVV tt δ∠=

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: 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. Calculate the generator phase current under this operating condition B. 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:

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).

AV

PIaa

a 5.13192*3

4500*3

)( ===

VEb af 1203

208)( ==

Ω=−

= 41.322

a

aafs I

VEX

15192*3

5000*3

)( ===a

aV

PIc

Ω=−

= 07.315

8.110120 22

sX

22 )( asaaf IXVE +=

Hzpolesnf 150120

63000120

=×

=×

=