Chap. 1 Basic
1/17
1-1 Single-Phase Power Consumption
v(t )  V
m cos(t  v )
Vmcoslwtt0
i(t )  Im cos(t i )
Imcoscwtti
A. Instantaneous Power（瞬時功率）
X
Y
p  t   v  t  i  t   Vm I Vm
m cos t  v  cos t  i
1
1
cos X cos Y  cos  X  Y   cos  X  Y 
2
2
vltj.it
Imcoswtǜvicoswǜ


Energy flow into the circuit
1
 p  t   Vm I m cos v  i   cos  2t   v  i  
2
A
B
1

 Vm I m cos v  i   cos  2 t2Cwttov
Imlcosloroijtcos  v   v  i or
2
B
cos( A  B )  cos A cos B  sin A sin B
A
Ìh 

州
pR (t )  V I cos [1  cos 2(t  v )]
------(1)
Energy borrowed and returned by the circuit
pX (t )  V I sin  sin 2(t   )
----------(2)
1
 p  t   Vm I m cos v  i   cos 2 t  v  cos v  i   sin 2 t  v  sin v   i 
2
B
A
B
A
Let Vm  2 V ,
 =( v  i )
Im  2 I
p(t )  V I cos [1  cos2(
tICuttov
 v )]  V I 1V11Ilsìnosìniwttov
sin  sin 2(t   )
cos
MIIlcoso It
PR
Px
2/17
資料來源（教科書）：John J. Grainger, William D. Stevenson, Gary W.
(Author), “Power System Analysis” (東華代理) ISBN: 9781259008351
資料來源（教科書）：John J. Grainger, William D. Stevenson,
Gary W. (Author), “Power System Analysis” (東華代理) ISBN:
9781259008351
Energy flow into the circuit
pR (t )  V I cos [1  cos 2(t  v )] ------(1)
Energy borrowed and returned by the circuit
pX (t )  V I sin  sin 2(t   )
----------(2)
B. Average power（平均功率）
功率
1 2
p(t )dt
2 0
2
1

V I  [1  cos 2(t  v )]cos  sin 2(t   v )sin  dt
0
2
P

2
0
cos(t )dt  0,

2
0
sin(t )dt  0
P  V I cos
P  瞬間功率的平均值
功率 平均值
瞬時
的
C. Reactive Power（虛功率）
振幅 頻率
pX (t )  V I sin  sin 2(t   )
Q  V I sin 
Q  脈動功率的振福
功率的振幅
脈動
 V I sin
 (kVAR)
KVAR
lvllIlSho
tj
S  P(kW)
NAR
Q(kVAR)
kW jQ
P
 V I cos ( kWkW
)
lvllIlcoso
3/17
Kilo-watt-hours
4/17
Kilowatthours
Kilo-watt-hours
5/17
Kilo-watt-hours
Kilo-VAR-hours
6/17
7/17
8/17
D. Power Factor（功率因數）
PF  cos 
資料來源（教科書）：John J. Grainger, William D. Stevenson, Gary W.
(Author), “Power System Analysis” (東華代理) ISBN: 9781259008351
P
V I
For a inductive(電感性) circuit
電
感
性
：
電電
壓流
角角
度度
落
後
霽
籩
i   v
Q>0
S
Ptj Q
Lagging Power Factor
（落後功因）
功因
落後
v(t )  2 cos(t  0)
i(t ) 
For a capacitive(電容性) circuit
2 cos(t  30)
 =(v  i )=30
3
t.la38
P  V I cos  1 1  cos30
Q  V I sin  11 sin30
Hschió
電
感
器
吸
收
+Q
霽
電
容
性
：
電電
壓流
角角
度度
超
前
i   v
電
容
器
雷
吸
收
－Q
Q<0
5 RJQ
鸞
竇竇
竇
Leading Power Factor
（超前功因）
超前
功因
v(t )  2 cos(t  0)
i(t ) 
2 cos(t  30)
 =(v  i )= so
30
l.l.us 30
Q  V I sin   1
1  sin30 so
t.l.sn
P  V I cos  1 1  cos30
（負值）
9/17
E. Power Factor Improvement
Q1  P tan1  P tan cos PF1
-1
希望 0⾓度 ⼩⼼ 0較⼤
Q2  P tan 2  P tan cos-1 PF2
 Q  Q1  Q2  Pltanaipkitmcoi
P(tan cos-1 PF1  tanPE
cos-1 PF2 )
Q
S
S
1
Q
Q
1
2
Q
1

1
P
Q
Q
2

2
P
2
P
10/17
1-2 Complex Power
Note: V  Vm / 2,
Instantaneous Power
Complex Power
v(t )  Vm cos(t  v )
V  V⼼

V21V1
i(t )  I m cos(t  i )
I  I i
p(t )  v(t )i(t )
1
= Vm I m cos v  i   cos 2 t   v  cos  v  i 
2
 sin 2 t   v  sin  v   i  
S  V I cos  j V I sin  P(kW)  jQ(kVAR)
1-3 Apparent Power
S  P 2  Q 2 (kVA)
I  Im / 2
v
EIIK0is
NIIIKla
S  V I (v  i )

0il.MU
xlIloi
⼆
 V ( v )  I   i
UI VI *
(OR S *  V * I ) (OR S  V 2 / Z * )
S  VI *
sw
IzU YzT
 S *  (VI * )*  V * I
視在功率
S  VI
1v12⼼
*
 Z
S  V  V
ys
*
*
 V V
Z*

V
UYE
2
Z*
liicsttit.ch
11/17
Example 2.4 (Hadi)
Three loads are connected in parallel a 1400-V rms, 60-Hz single-phase supply as shown in Figure 2.1.
Load 1: Inductive load, 125kVA at 0.28 power factor.
Load 2: Capacitive load, 10kW and 40kVAR.
Load 3: Resistive load of 15kW.
(a) Find the total kW, kVAR, kVA, and the supply power factor.
(b) A capacitor of negligible resistance is connected in parallel with the above loads to improve the power
factor to 0.8 lagging. Determine the kVAR rating of this capacitor and the capacitance in μF.
(a)
S1  125cos 1 (0.28) kVA  (35  j120) kVA  35kW  j120 kVAR
Q
S2 10kW
10 kW - j40 kVAR
S3  15kW
j40WAR
6okwjfǜǜǚo
53.130KVA
S  S1  S2  S3  60 kW  j80 kVAR  100
53.13 kVA
Q
PF  cosθ  cos
cos(53.13
)  0.6
53.130
0.6(lagging)
1
lagging
OR
P 60

 0.6
(lagging)
0.6
lagging
S 100
fl

PF 
PE
Q
2
Ans : d1 60kW
60 kW  2   80kVAR  33 look
100kVA
PF  0.6
 4  PF
VA ⼭
6 (lagging)
⼭ 8WAR
lagy
P
12/17
<sol>
Q
Example 2.4 (Hadi)
Q
1
Q
2
(b)
-1
ΔQ  P(tancos -1 (PF1 ) - tancos -1 (PF2 ))  60(tancos -1 (0.6)-tancostancoico
(0.8))  35 kVAR
22
V
1400
⼼
S  *   j35 kVAR 
Z
Z*
2
1400
*
Z  14⼼  j56()
 j35k
1
1
Z   j56   j
 j
ωC
2 f  C
C  47.37
47.37 F
2
i35k
TEJ56n.io
JÈ
Ans :
NF
zt
60ltmoico6
8
35WAR
P
jlc
(1) ΔQ 35
35KVAR
kVAR
(2) C  47.37 F
47.3UF
13/17
資料來源：電能管理技術與案例分析 台灣綜合研究院 楊正光
14/17
12段式功率因數調整器
15/17
自動功因調整器
資料來源：電能管理技術與案例分析 台灣綜合研究院 楊正光
16/17
17/17
Kilo-watt-hours
280
18/17
Text books
Textbook：
1.
2.
3.
本講義僅節錄部分內容，細節請購買正版書籍
John J. Grainger, William D. Stevenson, Gary W. (Author), “Power
System Analysis” (東華代理) ISBN: 9781259008351
A.R. Bergen, prentice-Hall, “Power System Analysis” (東華代理) ISBN10: 0131784609
Power System Analysis, by Hadi Saadat, McGraw-Hill, 1999.
19/17
Chap. 2 Three-Phase Circuits
& Per Unit System
Example 2.2. Two ideal voltage sources designated as machines 1 and 2 are
connected, as shown in Fig. 1.10. If E1  1000V , E2  10030V ,
and Z  (0  j5) , determine (a) whether each machine is generating or
consuming real power and amount, (b) whether each machine is receiving or
supplying reactive power and the amount, (c) the P and Q absorbed by the
資料來源（教科書）：John J. Grainger, William D. Stevenson, Gary W.
impedance.
(Author), “Power System Analysis” (東華代理) ISBN: 9781259008351
Inducǚe
100 比
10020
Figure 1.10
<Solution>
1650 165)
S1  P1  jQ1  E1 (  I)* 100Lǒx
100010.35
 (  10.35
1000
j268NA

(1,000
 j268) VA
*
 1000  10030 
*
10.3521650  10.35165
I

I


j5


Machine No. 1 consumes
consumes 1,000 W
where
where
0iio
f
Machine No. 1 supplies
supplies 268 VAR
S2  P2  jQ 2  E 2共
1003521650
30  10.35165
 I 1⼼⼼⼼
SEBtjQEE.CI
*
⼼⼼⼼⼼
NO2consumes now
Ans a Machine ⼼
bMachineNo1supplies268VAR
NO 2supplies268VAR
c Norealpowerabsorbedbytheimpedance
Ans:(a) Machine
No.absorbed
1 consumes
1000 W
536UAR
theimpedance
looo.pt8
 j268  VA
 1000 VA
Machine No. 2 generates
1,000 W
generates
Machine No. 2 supplies
supplies 268 VAR
g
No. 2 generates
1000 W
(b) Machine No. 1 supplies 268 VAR
No. 2 supplies 268VAR
(c) No real power absorbed by the impedance
536 VAR absorbed by the impedance
2.2 VOLTAGE AND CURRENT IN BALANCED
THREE-PHASE CIRCUITS
戴维 等效
heyff
Vcn
Van
Vbn
序
abc phase sequence
正相
Vcn
Van
Figure 1.11
Example:
Van  1100 V, Vbn  110240 V, Vcn  110120 V
CBACBA 負相序
BCABCA 正相序
CAB CAB 正相序
Vbn
2.3 VOLTAGE AND CURRENT IN BALANCED
THREE-PHASE CIRCUITS
Vcn
Van
Vbn
abc phase sequence
V
Vcn
cn
iin
V
Van
an
⾯
V
Vbn
bn
(or
3Van 150)
P
a  120
Vab  Van  Vnb  Van－Vbn
(or
3Van   90)
Vi E Up430
8
b
全国中
Iip
Vab  3Van 30
I a  I ab－I ca ⼆百Iab⼈300
 3I ab－30
Ii-
有⼈30
吉咨
Summary
Ｉa =ＩL
+
Vab
Ｉab
Vca  VP  VL
+ Ｉca =ＩP
－ Ｉbc
+
VL  Vab
－ Ｉb Ｉbn
Ｉc
：相壓  線壓：相同
相同
相流  線流：不同
不同
I a  Ib  I c  I L
Ｉan =ＩP
+
Van  VP
－
+
I ab  I bc  I ca  I p
1
IP 
IL
3
Vbc －
Ｉa =ＩL
說明：Y：相壓  線壓：不同
不同
相流  線流：相同
相同
Vab  Vbc  Vca  Vp  VL
－
－－
Ｉcn
V
V
cn +
+ bn
I an  Ibn  I cn  I p  I L
Van  Vbn  Vcn  Vp
Vab  Vbc  Vca  VL
VP 
1
VL
3
資料來源（教科書）：John
J. Grainger, William D.
Stevenson, Gary W.
(Author), “Power System
Analysis” (東華代理) ISBN:
9781259008351
Example 2.3. In a balanced three-phase circuit the voltage Vab
is 173.20V . Determine all voltages and the currents in Yconnected load having Z R  1020 . Assume that phase
sequence is abc . ZL =0
A’ V
 173.20(V)
Z =0
a 'b '
L
Vb ' c '  173.2
1200
 120
v (V)
173.2 上
Vc ' a '  173.2
 ⼼⼼
240
173.242
VP
VL
b’
ZL =0
Ia 'n
Ib 'n
100  30

上
10
 A 50 (A)
50
10
Ǜtiǒ
1020
 10
  170 Iciwlhó
I c ' n  10
1051700
A 70 (A)
173.2 ⼼ V1
 173.2
120
173.2L1200
⼼(V)
173.2
Va ' n 
⼼⼼
 30(V)
上
學3
173.2
 150
150
v (V)
上
c’ Vb ' n 
可3
173.2
Vc ' n  ÈL900
90
V
3
9/30
=100
90(V)9/25
loouǒcn
Example 2.4 The terminal voltage of a Y-connected load consisting of
three equal impedances of 2030 is 4.4kV line to line. The
impedance of each of the three lines connecting the load to a bus at a
substation is Z L  1.475 . Find the line-to-line voltage at the
substation bus.
Z =1.475
資料來源（教科書）：
John J. Grainger,
William D. Stevenson,
Gary W. (Author),
“Power System
Analysis” (東華代理)
ISBN: 9781259008351
匯 _ZǛ44750
L
2030
露
a
+
?
4.4kV
Ia
a
2030
+
2030
4.4kV
Vab
－
b
ii
ti
－
n
⼝
Example 2.4 The terminal voltage of a Y-connected load consisting of
three equal impedances of 2030 is 4.4kV line to line. The
impedance of each of the three lines connecting the load to a bus at a
substation is Z L  1.475 . Find the line-to-line voltage at the
substation bus.
Z =1.475
L
+
+
Van
Va n
a
+
Ia
a
+
VL  4.4kV
－
－
Vab
－
b
2030
－
n
Example 2.5
Va ' n 4100
4, 400
 2,540⼼
0(V)
何 ⼆32540200
I L  2540
2,540
0lzo
   30
2030 
LO
+
+
Van
Va n
－
－
VL  4, 400(V),
I  127(A)
Sb (3 )  3  VL  I
 3  4, 400 127(VA)
 127
3030A(A)
2
127
Van  I L Z L  Va ' n
 127
  30
1.4
  2,5400
L75752540
x1.4
127430
 2668
2, 668270
2.7(V)
VL  El
3 VVan
⼆4622
an I 4,622(V)
Va ' n 4400
4, 400 3⼆  2,540  ⼼
30(V)
何 2540430
I L 2540430
 2,540  3020L30
 2030
IE
600
A (A)
127
127L
 60
Van  I L Z L  Va ' n
 127
60XL4
1.4
75  2,540
  30
275012540
⼈⼼
127260
 2668427.30
2, 668  27.3
⼼(V)
VL 可3 Van 4622
4,622(V)
⼼
Nank
Ans:
VL  4,622(V)
Ans
Mk 46221
2.6 POWER IN BALANCED THREE-PHASE CIBCUITS
For a Y-connected load
Ib
Ia
n
Vp  Van  Vbn  Vcn
I p  I an  Ibn  I cn
1
Vp 
VL
3
P 33Npll
Vp IIplcoso
p cos 
Q 33Npll
VpIplsino
I p sin 
P ENLHIn.la
3 VL I Lsocos
Q Elhllhlsino
3 VL I L sin 
S  P2  Q2  可3 ⼼
VL 到
IL
Ic
If the load is connected 
Vp  VL
and
P 3
3Npll
Vp Iplcoso
I p cos 
Ip 
IL
3
P  3 VL I L cos
Q 31
3Vpll
Vp Iplsìno
I p sin 
Q  3 VL I L sin 
S  P2  Q2  3 VL I L
PT 河川 1到
2.7PER
PER-UNIT
QUANTITIES
么值
UNIT標
1元港幣＝4元台幣
8元港幣＝?? 元台幣
320元港幣＝1,280 元台幣
1元美金＝32元台幣
2元美金＝?? 元港幣
40元美金＝ 320 元港幣
Base 值
Real price
p.u. 值
美金
港幣
台幣
11
88
32
32
32
32
11
1/1=1
11 1
美金
Base 值
Real price
p.u. 值
11
40
40/1=40
kolklto
88
8/8=1
32132 1
港幣
台幣
88
32
32
818 1
320
320
32/32=1
1280
1280
港幣
1
1
p.u. 值
1
1
Base 值
1
1
8
8
Real price
p.u. 值
4040
320/8=40
1280/32=40
⼆比
32018
40 1280132
美金
Base 值
Real price
HH
1
1x8
88
以
台幣
1
1
32
32
32232
It1x32
美金
港幣
台幣
40
40
40
40
4
11
40x1=40
40
88
40x8=320
Do
32
32
40x32=1280
1280
2.7 PER-UNIT QUANTITIES
#1 Choose Vb
Sb

I

 b V

b
#2 Compute I b 
 I b  Sb

3Vb
Sb

Vb 2
Zb 
1

Sb

#3 Compute Z b 
2
2
(
V
/
3)
V
b( L)
 Z  b( L)

 b
Sb (3 ) / 3
Sb (3 )

#4 Compute I pu Vpu
#5 Back to I ( A)
3
V (V )
OR Zb 
Vb ( p - p )
Ib
1
3
For Y connected
Z pu ( new) 
Z pu ( old ) 
Z()
Zbase ( new)
Z()
Zbase ( old )


Z()
2
Vbase
( new ) Sbase ( new )
Z()
Z pu ( new)
2
Vbase
( old ) Sbase ( old )
Z pu ( old )
Sbase ( new)  Vbase ( old ) 



Sbase ( old )  Vbase ( new) 
2
Example 2.6. Find the solution of Example 1.3 by working in per unit on a base of
4.4kV, 127 A so that both voltage and current magnitudes will be 1.0 per unit. Current
rather than kilovoltamperes is specified here since the latter quantity does not enter the
problem.
資料來源（教科書）：John J.
ZLine =1.475
a
+
Ia
a
+
4.4kV
Vab
－
Grainger, William D. Stevenson,
Gary W. (Author), “Power
System Analysis” (東華代理)
ISBN: 9781259008351
－
b Z =1.475
Line
Z Load  2030
n
ZLine =1.475
a
+
Ia
a
+
4, 400V
Choose Vbase( phase)  4, 400 / 3(V)

Sbase(3 )  3  4, 400
Z Load
+ V
 2030
'
an
－
b
－
b'

 3  4, 400
Computer Zbase 
Z Load ( pu ) 
Z Line ( pu ) 
I a ( pu ) 
Z Load ( pu )

( Phase ) base
Vbase ( Phase )
I base

Van ( pu ) 
1.00
 1.0  30( pu )
1.030

3  4, 400 3 
4, 400 / 3
()  20 ()
127
Z Load 2030

 1.030( pu )
Zbase
20
Z Line 1.475

 0.0775( pu )
Zbase
20
Van ( pu )

3 V
3 127(VA)
 127(A)
n
Vab
－
Computer I base  Sb (3 )

3 127(VA)
Phase
Van 4, 4000 / 3

 1.00( pu )
Vbase
4, 400 / 3
Van( pu )  Van ( pu )  I a ( pu ) Z Line ( pu )  1.00  1.0－30 0.0775  1.05062.7( pu)
Vab ( pu )  3Van j 30  3 1.050632.7( pu)
 4, 400 
 VL  Vab ( pu )  Vbase  3 1.0506  
  4, 622(V)
 3 
ZLine =1.475
a
+
Ia
a
+
4, 400V
Choose Vbase( Line)  4, 400(V)
Sbase(3 )  3  4, 400 127(VA)
Z Load
+ V
 2030
'
an
－
n

3  4, 400 127(VA) 
3  V( Line )base

3  4, 400  127(A)
Computer Zbase 
Vab
Z Load ( pu ) 
Vab '( pu ) 
－
b
－
b'
Van ( pu ) 
Vbase ( Phase )
I base
Z Load
 1.030( pu )
Zbase

4, 400 / 3

()  20 ()
127
Z Line ( pu ) 
Z Line
 0.0775( pu )
Zbase
Vab ' 4, 4000

 1.00( pu)
Vbase
4, 400
Vab '( pu )
3
Van ( pu )
  j 30 
1.00  j 30 1.0


  30( pu )
3
3
1.0  30 3 1.0

  60( pu )
Z Load ( pu )
1.030
3
1.0
1.0

  30 
  60 0.0775  0.6066  27.3( pu)
3
3
I a ( pu ) 
Van ( pu )  Van ( pu )  I a ( pu ) Z Line ( pu )

Computer I base  Sb (3 )

Vab ( pu )  3Van j 30  3  0.60662.7( pu)
 VL  Vab ( pu ) Vbase  3  0.6066  4, 400  4,622(V)
另解
20/25
ZLine =1.475
a
+
Ia
a
+
Choose Vbase ( Line)
4, 400

(V) Sbase (3 )  3  4, 400 127(VA)
2
Computer I base  Sb (3 )
Z Load

+ V
 2030
'
an
－

4, 400V
Vab
n
Z Load ( pu ) 
Vab '( pu ) 
Vbase ( Line )
Van ( pu ) 
－
b
－
b'
I a ( pu ) 
Van ( pu )  Van ( pu )  I a ( pu ) Z Line ( pu ) 
Vab '( pu )
3
Van ( pu )
Z Load ( pu )

Z Line ( pu ) 
Z Line
 0.07  475( pu )
Zbase
4, 4000
 1.0  20( pu )
2, 200
  j 30 



Z Load 2030

 430( pu )
Zbase
2
0
4


Vab '
3  V( Line )base
4, 400 

3


  127  2(A)
2 

Vbase ( Phase )  4, 400 / 2  3 20


 ()
I base
127  2
4
3  4, 400 127
Computer Zbase

1.0  20  j 30 1.0  2


  30( pu )
3
3
1.0  2  30
430
3

1.0
  60( pu )
2 3
1.0  2
1.0
  30 
  60 0.07  475  0.6066  2  27.3( pu)
3
2 3
Vab ( pu )  3Van j 30  3  0.6066  22.7( pu)
 VL  Vab ( pu ) Vbase  3  0.6066  2  2, 200  4,622(V)
另解
21/25
ZLine =1.475
a
+
Ia
a
+
4, 400V
Vab
Choose Vbase ( Line)
Computer I base  Sb (3 )
Z Load
+ V
 2030
'
an
－
n
4, 400
4, 400
S

3

127(VA)

(V) base (3 )
2
2

3  V( Line )base

1
4, 400
4, 400 

 
  3
127   3 

127

2

(A)

2
2 
2

 
Vbase ( Phase )  4, 400 / 2  3 20
Computer Zbase 

  2()
I base
127  2 / 2
4
Z Load
2030
1

 4  30
Zbase  20 4   2
2
Z
1
Z Line ( pu )  Line  0.07  4  75( pu )
Vab '
4, 4000
Vab '( pu ) 

 1.0  20
Zbase
2
V
2, 200
Z Load ( pu ) 
base ( Line )
Van ( pu ) 
－
b
－
b'
I a ( pu ) 
Van ( pu )  Van ( pu )  I a ( pu ) Z Line ( pu ) 
Vab '( pu )
3
Van ( pu )
Z Load ( pu )
  j 30 

1.0  20  j 30 1.0  2


  30( pu )
3
3
1.0  2  30 3 1.0

 2  60( pu )
430  1 2 
2 3
1.0  2
1.0
1
  30 
 2  60 0.07  4  75  0.6066  2  27.3( pu)
2
3
2 3
Vab ( pu )  3Van j 30  3  0.6066  22.7( pu)
 VL  Vab ( pu ) Vbase  3  0.6066  2  2, 200  4,622(V)
另解
22/25
Example 2.9
ARTHUR R. BERGEN
Consider a system with one-line diagram shown in Figure E5.14(a). The three-phase
transformer nameplate ratings are listed. The transformer reactances are given in
percent. The transmission line and load impedances are in actual ohms. Calculate the
generator current, the transmission-line current, the load current, the load voltage, and
the power delivered to the load.
Figure E5.14(a)
<sol>
Vb1  13.2kV
Zb 2
Z b1
Let
Sb  10MVA
ZT1
ZL
ZT2

Vb12 (13.2k ) 2

 l7.4
17.4()
 Z b1 
Sb
10 M


Vb 2 2 (132k ) 2

⼆
1742.4(
⼼
)
1742.4
Zb 2 
S
10
M
b

2
2 2

V
(66k )
435.6
 Z b 3  b 3  66⼼ 435.6(
)
Sb
10 M

T
t
n
66N
VbVb3
3  66kV
Zb3
lpu
Zload
#
Vhi132kV
b 2  132kV
2

V

S
 ZT 1  j 0.1 base ( new)   base ( old )   j 0.1  10   j 0.2( pu )
jo.tl 5  jo2Cpu
Sbase ( old )  Vbase ( new) 


10  j100 10  j100

-3
Z



(5.74

j
57.4)

10
( pu )
xP

L
5.74tj57.4
pu
Zb 2
1742.4

2

10
69
 ZT 2  j 0.08        j 0.0874(
pu )
0874cpu
jo

 10   66 

Zload ( pu ) 
300
⼆0.689Cpu
0.689( pu )
Zb3
ZL
ZT1
ZT2
Zload
I 
#
10
10
10


 1.29－26.4( pu)
-3
j 0.2  (5.74  j57.4) 10  j0.0874  0.689 0.695  j0.345 0.77626.4

Sb
10M
I


 437.4(
A)
 b1 荒⼭⼆
437
4A
3Vb1
3 13.2k


Sb
10 M
I


⼀
 b2
⼀  43.74( A)
3Vb 2
3 132k43.74


Sb
10M

 87.48(AA)
 Ib3 
⼆ 87，48
⼆
点3 k66k
3V的

b3
⼀点
点
荒 前是
荒 器
Ans:
Ans
A
⼆56.43
到
⼭
1
I

564.3(A)
(2)
I

56.43(A)
3
A
⼭ lI(1)
1
2
i564
⼼
A ⼭ Mk58.66
(3) I 3  112.8(A)
(4) VL  58.66(kV)
3 1Ijkll28
(5) PL  11.45(MVA)
45mA
5
kill
 I1  I  I b1  1.29  437.4  564.3(A)

 I在
1.29 43.74
43.7456.43A
 56.43(A)
⼆1.29
 I 2  IIIK
b2 
 I  I  I  1.29  87.48  112.8(A)
b3
 3
iii
𣊭
VL ( pu )  IIx
 Z  1.29
 0.689⼆
0.89cpu
0.89(pu)
Z 1.29x0.689
VL  VL ( pu )  Vb  0.89(pu)  66(kV)
58.66(kV)
58.66CNJ
x66CN⼆
Vwadcpilhzofacpu
S L( pu )  I 2 Z  1.292 0.68
0.689 1.145cpu
1.145(pu)
ÌZ
1.2收
⼼
S  Sb  S L( pu )  10Mx
10M  S L( pu )  10  1.145⼆
 11.45(MVA)
Swadcpuiloxl415 11.45 NA
名 的以⼆ 江
0.89 ㄨ 1.29
1.145 m
SzsbxSwadkpu⼆10以 Sloadcpus
疵品坐蕊筘⼀
Sway 3SLoaald ll.IS CMVM
Text books
Textbook：
1.
2.
3.
本講義僅節錄部分內容，細節請購買正版書籍
John J. Grainger, William D. Stevenson, Gary W. (Author), “Power
System Analysis” (東華代理) ISBN: 9781259008351
A.R. Bergen, prentice-Hall, “Power System Analysis” (東華代理) ISBN10: 0131784609
Power System Analysis, by Hadi Saadat, McGraw-Hill, 1999.
26/10
Power Flow
o
Chap-3
o o
V2  

V10
P12  jQ12
R  jX
o
V200
V1  
P21  jQ21
R  jX

 V 0  V2   
 V10  V2 
S12  V1I   V1   1

V

1

 R  jX

R  jX
*




 V20  V1   
 V20  V1 
*
S

V
I

V


V

21
2
2
2


 R  jX

R  jX
2
R

jX




 2
V

V
V
cos


j
V
V
sin

1 1 2

1
2
R  X2
R  jX
 2
V 2  V1 V2 cos  j V1 V2 sin  
2  2
R X
1
2
P12  2
RV

R
V
V
cos


X
V
V
sin

 1 Rlullnlcosotxlullnkìno

1
2
1
2
R  X2
1

P

RV2 2  R V1 V2 cos  X V1 V2 sin  
21
2
2 


1
R X
2
Q12  2
XV

X
V
V
cos


R
V
V
sin




1
1
2
1
2
1
R  X2
2
Q 
Isin
XV
 X V1 V2 cos  R V
RN.nu

21
2
1 V2 sin  
2
2 CXVEXNillhlcoio


R X
⼝1
TǛCRVF
1
Tǜliixlullnlao_Rlullnlsinoj
_ǖinllnlcoottsina
Fb
1
Chap-3
Pr ove P12   P21
Q12  Q21
V1  V1 0
when R=0
<Proof>

 V  V2 
S12  P12  jQ12  V1I   V1   1

 R  jX 

R  jX
2
 2
V
 V1 V2 
1
2
R X
I
R
V2  V2   
jX
P21  jQ21
P12  jQ12
S21  P21  jQ21  V2   I 





 V V 
 V2   2 1 
 R  jX 

R  jX
2
V
2  V2 V1   
R2  X 2
0 By
Mind
2 Rlullnlcosotxlullnkìn
RVERN.IN
i  R2  jXd
V

V
V
cos


j
V
V
sin

R  jX
2
1
1
2
1
2
 2
V
 V1 V2 cos  j V1 V2 sin 
R  X2
2
2
R X
since
RN.IM
XVRXN.llnlcoso
⼝
1
2
1
2
P12  2
R
V

R
V
V
cos


X V1 V2 sin 
1
1
2
2
P

R
V
21
2  R V1 V2 cos  X V1 V2 sin 
R X
R2  X 2
1
1
2
2
Q12  2
X
V

X
V
V
cos


R
V
V
sin

Q

X
V
1
1
2
1
2
21
2  X V1 V2 cos  R V1 V2 sin 
R  X2
R2  X 2
1
1
P21    V1 V2 sin  
P12  V1 V2 sin 
X 1川 ⼼⼼0
X
lullvzlsino
i.RE
 P12   P21 Q12  Q21
1
1
2
2
Q12 
V1  V1 V2 cos Q21 
V2  V1 V2 cos
cosol
MRMINHcoo
X
X
sis
RUE



必
⼆⼀

蕤
zlootxlu

 Qiizcxw_XNHNLIcoio




RNMnlsinoP.it
Bit
ÌNIRMINH B.Q.si

Qzij

2

Chap-3
For the system shown in figure, all quantities are per phase values.
Ref : ARTHUR R. BERGEN
S G1
jQG 2
V1  10
 S21
S12
S D1  1
(a) Pick QG 2 so that V2  1 .
Z  j 0.5
ko
V2
SD 2  1
(b) In this case, what is V2  ?
(c) If QG 2  0 , can we supply the load S D 2  ?
(d) If yes, what is V2  ?
3
Assume : V1  V1 0 V2  V2    Chap-3
Solution.
*
 V22  V2V1*  V22  V2 V1 cos  j V2 V1 sin  
 V2  V1 
V2*  V1*
S21  P21  jQ21  V2 I  V2 
 V2
 j

 j

jX

jX
X
X



 

0
R
1
1

(b) P21  V2 V1 sin    1 sin   1  030300
X
0.5
Èixlxsino
1 2
1
Q

V

V
V
c
os


os30   0.268
1  cLos30
0.268


(a) 21
⼆
2
2
1
X
0.5
*
Èx
(C)
1
Q21   V2  V2  V1 cos   0
0
x
V
P21  2 sin   ⼀1
X
(1)+(2)
4
2
2
V

V

0.5
0
2
2
V24_Vito.io
P21  1 
(D)
Ans:
1 1
 sin 
2 0.5
 V22  V2 V1 cos  0
 V22 sin 2   X 2
 V22 
1
2
 V24  V22 cos2   0
  V22 sin 2    X 2
 V2 
Ì
(1)
(2)
1
2
È
   45
Yes, we can supply the load
1
V2  V2    
  45
2
yoej
V1  10
(aa)Qzl
Q21⼆
0.268
0.268 (bb
) N
V2  30
30
1 450
(c)Yes
Yes,we
wecan
cansupply
supply the
the load
load (da) V2
V2 
  45
⼼
2
ÈL
V2  V2   
4
叫4
Chap-3
Ref : ARTHUR R. BERGEN
SG1  PG1  jQG1
10  jQG 2
V1  10
Z  j 0.03
SD1  10  j3
Find
( a) 
S12
(b) QG 2 ,
V2  1.0  
I
SD 2  20  j10
S 21
(c) PG1 , QG1
5
Chap-3
SG1  PG1  jQG1
Solution.
S12  P12  jQ12
0
R
1

 P12  x V1 V2 sin θ

Q  1 V 2  V V cos θ
1
2
 12 x 1
S 21  P21  jQ21


1

Rio
P

V
V
sin
θ
21
1
2

x

Q  1 V 2  V V cos θ
1
2
 21 x 2
1

P


10


1 1sin θ 
 21 to
Clxlsìno
0.03

θ0⼆
17.460
17.46


1
Q21 
1.54
1  1 1cos θ  1.54
0.03



È
P
V1  10
10  jQG 2
w
V2  1.0  
I
Z  j 0.03
S D1  10  j3
SD 2  20  j10
S 21
S12
QG 2  10
Q21 ⼆10
101.54
 1.5411.54
 11.54
lotQzl
PG1  10
lot P12 1010w 10
20  20
Rz
1
where P12  1 ⼀l.si
1 no
1  sin    P21  10
0.03
QG1  3  Q12  4.54
Pzilols


1 Ctlxcoso 1.54
1  1  cos  1.54
0.03
Ans (a)θ  17.46
(b)QG 2  11.54
where Q12 
(c)PG1  20,
QG1  4.54
6
Chap-3
Ref : ARTHUR R. BERGEN
7
Chap-3
Solution.
(a)

1
2
P21  2
R
V
 R V1 V2 cos θ  X V1 V2 sin θ
2
2
R X
1
0.5 
0.01cos θ  0o.ls.io
.1sin θ 
 0.01  0.01coso
0.5⼆0.012  0.12
θ  2.9
V2  2.90
2.9
i.LU 2.90
V1  10

Fhzlool
0
0.5  j 0.5
V2  1  
0.5  j 0.5
(b) QG 2  0.063  0.5  0.563
1
2
Q21  2
X V2  X V1 V2 cos θ  R V1 V2 sin θ (c) S12  P12  jQ12
2
R X
1
2
P

R
V
1
12
1  R V1 V2 cos θ  X V1 V2 sin θ
2
2
cz
R

X
们

0
.
1

0
.
1cos
2
.
9


0
.
01sin
2
.
9

in
0




0
tools
 l lcosk.ci

0.012  0.12
1

0.01
 0.01cos(2
.9)  0.1sin(2
.9)   0.502
to.lsincz.ci
o_olcoscz.ci
 0.063
⼆ 2
2 
Cool
0
.
01

0
.
1
up
S21  P21  jQ21  0.5  j  QG 2  0.5 
1
2
Q12  2
X
V
1  X V1 V2 cos θ  R V1 V2 sin θ
R  X2
1
Hint : cos 2 α  sin 2 α  1

(0.1 o 0lcoscz.ci
.1cos(2.9)  0o.01sina
01sin(2.990D
))  0on
.037
Hint
0.012  0.12
0.01cos θ  0.1sin θ  0.015
 SG1  S D1  S12  (0.5  j 0.5)5 (0.502
- j 0.037)
Mo 0.015
o.olaso o.IS
SD 5⼝ lo 5tjo
⼼50270037
cos θ  1.5  10sin θ
(1.5  10sin θ ) 2  sin 2 θ  1
 1.002  j 0.463
2 5
loshocl.IN
co
soil
looztjo463
101sin θ  30sin θ  1.25  0 θ  2.9(θ  14.3)

in


Ǜap

Shoititl
30sinotl.si⼆0

ifeng.l
aixtsnil
1015
⼼0

0 2.90
yat
Ans:
(a)
V2 ⼩Qaz
2.9 0.563
(b)QG 2  0.563
⼆ 
290
⼈
a
Ahs
以
對
(c) SaGSci1.002tj0.463
1  1.002  j 0.463
8
Chap-3
Example 3.13 Two transformers are connected in parallel to supply an impedance to neutral per
phase of 0.8+j0.6 per unit at a voltage of V2  1.00 per unit. Transformer Ta has a voltage ratio
equal to the ratio of the base voltages on the two sides of the transformer. This transformers has an
impedance of j0.1 per unit on the appropriate base. The second transformer Tb also has an
impedance of j0.1 per unit on the same base but has a step-up toward the load of 1.05 time that of
Ta (secondary windings on 1.05 tap).
Figure shows the equivalent circuit with transformer Tb represented by its impedance and the
insertion of a voltage V . Find the complex power transmitted to the load through each
transformer.
I
I
1 ⼼⼼ 8⼀ 加
o 8go6J W 8
8 jo 6
il
資料來源（教科書）：John J.
Grainger, William D. Stevenson,
Gary W. (Author), “Power
System Analysis” (東華代理)
ISBN: 9781259008351
1 105
f
TDIUAO
8tjabio
9
Chap-3
Solution.
in
Eb
I circ 
0.05
o 05
  j⼆
0.25 per
25 unitunit
j 0.2juz 70 per
So that
⼼ per unit
0 4
ITa  0.4  j 0.3    j 0.25
 j7
0.05
25  0.4
oujoj.ci
Fwit
joipenunito
Pēmi
5
4
⼀下
o.DE0
IT  0.44 j 0.3


j
0.25

0.4

j
0.55
per
unit
jo3 tl i 
b
STa  V2 IT*a  0.40  j 0.05 per unit
0.4
STb  V2 IT*b  0.40
 j 0.55pernit
per unit
o 4tjo.TT
10
Example 3.14
Chap-3
Repeat Example 3 except that Tb includes both a transformer having the same turns
j


ratio as Ta and a regulating transformer with a phase shift of 3  t   60  1.03  . The


impedance of the two components of Tb is j0.1 per unit on the base of Ta .
資料來源（教科書）：John J.
Grainger, William D. Stevenson,
Gary W. (Author), “Power
System Analysis” (東華代理)
ISBN: 9781259008351
Solution.
t  1  1.03  1.00   2sin1.5  91.5  0.052491.5
0.0524
91.5
So that
4
0
0
I circ 
 0.262tT
 j 0.0069
_o.us
*
S

V
I
 j 0.307
0.290
1384ju.in
Ta
a哈
Ta  0.138
per unit
27
i
0077203if
ITa  0.44
 j 0.3   0.262

j
0.007

0.138

j
0.307
*
a
STb  Vb ITb  0.662
加
j 0.293
⼼
jot lo.zh.to 
喊
per unit
ITb  0.4  j 0.3   0.262  j 0.007   0.662
 j 0.293
ehjzg
呂
器
00691
0 4
did
jhtlo.tn
11
Chap-3
Text books
Textbook：
1.
2.
3.
本講義僅節錄部分內容，細節請購買正版書籍
John J. Grainger, William D. Stevenson, Gary W. (Author), “Power
System Analysis” (東華代理) ISBN: 9781259008351
A.R. Bergen, prentice-Hall, “Power System Analysis” (東華代理) ISBN10: 0131784609
Power System Analysis, by Hadi Saadat, McGraw-Hill, 1999.
12/10
Chap-4
200kg
報價
甲
100元/時
200kg
乙
200元/時
0kg
勝出 ---> 甲
500kg
報價
300kg+1kg
甲
乙
報價
6元/時/kg 5元/時/kg
200kg
100kg
200kg+1kg 100kg+1kg
200kg
100kg+1kg 勝出 ---> 乙
300kg+1kg
甲
乙
報價
6元/時/kg 6元/時/kg
200kg
100kg
200kg+1kg 100kg+1kg
200kg+1kg
100kg 勝出 ---> 均勝出
200kg
100kg+1kg 勝出- ---> 均勝出
甲
400kg
乙
100kg
500kg+1kg
甲
乙
報價
2元/時/kg 10元/時/kg
400kg
100kg
400kg+1kg 100kg+1kg
400kg+1kg
100kg
勝出 ---> 甲
Chap-4
Economic Dispatch
2
Chap-4
Case 1 :
C1 (馬)  C2 (牛)
馬  牛  pull  700kg
馬  pull  350kg
馬 
pull 
700kg
牛  pull  350kg
Case 2 :
牛 
pull 
0kg
2
2
C1 (馬) 
900

45
馬

0.01
馬
($/hr)
45
⼆900
⾺ tool ⾺
2 牛2
C2 (牛) 2500t43
2500 年
43too03
牛  0.003
($/hr)
年
(a).Calcul ate
馬  pull  ?kg
牛  pull  ?kg
(b).Find the lowest cost3
C1 (馬)  900  45馬  0.01馬2
Chap-4
($/hr)
C2 (牛)  2500  43牛  0.003牛2 ($/hr)
喝
43 0.0064
45

0.02
馬

43

0.006
牛

dC1
700
7
⽜
(a). IC1 ⼆

45

0.02
馬
($/hr/kg)
⾺

⼆
0.0
45
Mhrlkg
嗎
IG
d馬
馬  牛  700
死 之 ⼆ dC2⼆ 43too 64cpyg
IC2 
 43  0.006牛 ($/hr/kg) 馬  pull  84.6kg
84tkg
d牛
⼼
Solution
鴦
槔
(b). IC1  45  0.02  84.6
IC2  43  0.006  615.4
⼆
牛  pull  615.4
615.4kg
可
($/hr/kg)
($/hr/kg)
C1 (馬)  900
45 84.6toolx8462
84.6  0.01 84.62
90 25
($/hr)
0 003xb15.42
C2 (牛) 2500
2,500213x615.4
 43  615.4
 0.003  615.42
($/hr)
C1 (馬)  C2 (牛)  34877
34,877($/h
hrlr)
C2 (牛)  2,500  43  700  0.003  7002
4
($/h
r)
Chap-4
Ref : ARTHUR R. BERGEN
Solution The incremental costs are
IC1 
dC1
 45  0.02 PG1
⼗O02Pa
dPG1
IC2 
dC2
 43  0.006 PG 2
dPG 2 43too06PGL
45
?? MW
G1
G2
?? MW
 0.02
PGF
PG1 
430.006Pa
 0.006  PG 2
0.02
4545
43

+PG 2 =700700
 PG1PGHPGE
PG1  84.6
MW )
84.6(CMU
PG 2  615.4
615.4 (Mw
MW )
Ans： PG1  84.6 (MW )
Di 84.6CMD
PG 2  615.4 (MW )
Pli 615.4CMWJ
700MW
5
Chap-4
18
C1 ( P1 )  P12  P1
16
14
(3,12)
(3,10)
12
12
10
$
$
$
8
(2, 6)
⼀
8
zi
6
4
(1, 4)
4
(1, 2)
2
2
0
0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
0.0
MW
kW
9
7
7
6
5
5
iii
4
$ / kW
6
(2, 5)
$/MW
(1, 3)
3
1.0
1.5
2.0
dC2 ( P2 )
 2 P2  3
dP2
8
(3, 7)
0.5
2.5
3.0
MW
kW
9
dC1 ( P1 )
 2 P1  1
dP1
8
$ / kW
(2, 10)
10
6
$/MW
C2 ( P2 )  P22  3P2
16
14
$
(3, 18)
18
(2, 7)
ni
(3, 9)
(1, 5)
4
3
(0, 3)
2
2
1
1
(0, 1)
0
0
0.0
0.5
1.0
1.5
kW
MW
2.0
2.5
3.0
0.0
0.5
1.0
1.5
kW
MW
2.0
2.5
6
3.0
Chap-4
4-1 classical economic dispatch
(Line losses neglected)
F($/hr)
C($/hr)
F     PG   PG ($/hr)
xtp Pair Pi4 ⼼
2
or C     PG   PG
xtp Patrpi
2
P(MW)
7
Chap-4
IC($/MWh)
Incremental cost, IC
F     PG   PG 2 ($/hr)
B
4
A
IC 
F
   2 PG ($/MW - hr)
PG
hr
ftp.GLNMW
2
20
40
P(MW)
A : Power output is 20MW.The cost of increasing 1MW is 2($)
2
B : Power output is 40MW.The cost of increasing 1MW is 4($)
4
Note : It costs much money if generators have to be increased power
output when the generators are on the higher power output.
8
($/MWh)
($/MWh)
2
2
60
60MW
Chap-4
G1
60
40
MW
F1 F2

P1 P2
G2
MW
40MW
100MW
Note : The incrementa l costs are equal when the load demands are known
which is called classical economic dispatch.
9
F1
Load Demand
G1
PTotal  P1  P2  P3  .. .  PN
Rt Rt Btcntk
坑
... ... ... ...
G2
P1
Total cost
Chap-4
F2
FN
GN
PN
P2
FT  片
F1  F2  F
t.cat
3  .. .  FN
PTotal
Fǐxd
dFN
dF2
dPTotal  0 (The demand is constant)  dF1
dP

dP

......

dPN  0...............(1)

1
2
dP2
dPN
 dP1
 dP1  dP2  .........  dPN  0
dP  dP  .......  dP  0....................................(2)
2
N
 1
Total cost FT  F1  F2  F3  . ..  FN
dF
dF1
dF
dP1  2 dP2  ... ...  N dPN
dP1
dP2
dPN
dFT  0 will minimize the total cost
 (dP1  dP2  ...  dPN )  0
FT
FT
FT

dP1 
dP2  ..... . 
dPN  0  dF
 dFN


 dF2

1
P1
P2
PN


dP



dP

.
..




 dPN  0

 1 
 2
RtdhidlbtutdPio
y
歲Mtilpztmtdpio
dF
dF1
dF
dP1  2 dP2  ......  N dPN  0
dP1
dP2
dPN
Hint : Lagrange Multipliers
y0
x y
x y 0
x0
 dP1

 dP2

 dPN
dFN
dF1
dF2
 ,
  , ...

dP1
dP2
dPN
dF
dF dF
i 1  2  .......  N  
入
dP1 dP2
dPN
器器 器

10
Chap-4
Ref : ARTHUR R. BERGEN
Solution The incremental costs are
IC1 
dC1
 45  0.02 PG1
dPG1
IC2 
dC2
 43  0.006 PG 2
dPG 2
45  0.02  PG1  43  0.006  PG 2

 PG1 +PG 2 =700
?? MW
G1
G2
?? MW
PG1  84.6 (MW )
PG 2  615.4 (MW )
Ans： PG1  84.6 (MW )
PG 2  615.4 (MW )
700MW
11
Chap-4
Ref : ARTHUR R. BERGEN
增量成本
Solution
2
1067
(a) Fuel input rate  175
175⼗
8MW
8.7 100to
 002
0.0022
 100
 1,067
MBtu/hr
MBtulhr
以1002
(b) Fuel cost  1,067MBtu/hr
 5($/MBtu)  $5,335/hr
067M
l
BtulhrXJXMBtu
53331hn
(c) IC  55ㄨ
 (8.7
0.00441
0.0044PG )⼆5 ㄨ
(8.7 0.004
0.0044 100)
 45.7($/MWh)
8.7⼗
45.7
到 5 8.7
收⼼
(d) Approx cost  ($5,335
$45.7)/hr ⼆
5335 45.7
5380.7 lhr
1比 $5,380.7/hr
狐⼭
12
Chap-4
Ref : ARTHUR R. BERGEN
Solution
 
 
C  900  45P  0.01P 2
$
1
1
1
hr


C2  2500  43P2  0.003P22 $
hr

IC1 
dC1
 45  0.02 PG1
dPG1 45 0.02DH
IC2 
dC2
 43  0.006 PG 2
6
dPG 2 4了
Dnftt
50  P1  200
50  P2  600
⼈
PG1 ⼆440.006
43  0.006  PG 2
45  0.02
O02
𠮿
红

2 =700
 PG1 +PGPain
Put 700
PG1  84.6
84.6(MW )
PG 2  615.4
⽇ (MW )
6 4
PD  P1  P2  700MW
PG2  600MW
PG1  700
600o_o
 100MW
no 6
Ans：
Anhiomw
P  600MW
G2
hicoomw
P  100MW
G1
13
Chap-4
 
 
C (P )  900  45P  0.01P 2
$
1
G1
1
1
hr


C2 (PG2 )  2500  43P2  0.003P22 $
hr

2
50  P1  200 PD  P1  P2  700MW

50  P2  600
IC

1
PD  P1  P2
IC

P
(a)
3
14
PD  P1  P2
(b)
Chap-4
 
 
C (P )  900  45P  0.01P 2
$
1
1
hr
 1 G1

C2 (PG2 )  2500  43P2  0.003P22 $
hr

此?


50  P1  200
PD  P1  P2  700MW
50  P2  600


 IC (P )  45  0.02 P
$
1
MW  hr
 1 G1

 IC2 (PG2 )  43  0.006 P2 $
MW  hr

蚍
?
?
器
?
?
?
50
so
?
?
? boo
15
coooo cao⼼
Chap-4
⼼
550，5
批以
⼼⼼
no
 
 
C (P )  900  45P  0.01P 2
$
1
1
hr
 1 G1

C2 (PG2 )  2500  43P2  0.003P22 $
hr

cbfo.to
50  P1  200
PD  P1  P2  700MW
50  P2  600
16
?
?比
 
 
C (P )  900  45P  0.01P 2
$
1
1
hr
 1 G1

C2 (PG2 )  2500  43P2  0.003P22 $
hr

?46.6
50  P1  200
PD  P1  P2 Chap-4
 700MW
50  P2  600
80，491
168，46.6
5501461
?
ecoo 4的
?
?
43.3
?
?
17
Chap-4
Solution
 
 
 $ MW  hr 
 $ MW  hr 
C (P )  900  45P  0.01P 2
$
1
G1
1
1
hr


C2 (PG2 )  2500  43P2  0.003P22 $
hr

 IC (P )  45  0.02 P
1
 1 G1

 IC2 (PG2 )  43  0.006 P2

50  P1  200
50  P2  600
PD  P1  P2  700MW
Ex 11.8 Ex 11.9
 IC1  50   46

 IC1  200   49
 IC2  50   43.3

 IC2  600   46.6
46.6  45  0.02 P1 P1  80  MW 

46  43  0.006 P2 P2  500  MW 
18
 
 
 $ MW  hr 
 $ MW  hr 
C  900  45P  0.01P 2
$
1
1
hr
 1

C2  2500  43P2  0.003P22 $
hr

 IC  45  0.02 P
1
 1

 IC2  43  0.006 P2

50  P1  200
PD  P1  P2  700MW
50  P2  600
 IC1  50   46

 IC1  200   49
 IC2  50   43.3

 IC2  600   46.6
IC1  45  0.02 P1
Chap-4
Ref : ARTHUR R. BERGEN

46.6  45  0.02 P1 P1  80  MW 


46  43  0.006 P2 P2  500  MW 
 $ MW  hr 
IC2  43  0.006 P2
 $ MW  hr 
19
(680, 600)
(800, 600)
Chap-4
(550, 500)
(800, 200)
(100, 50)
(680, 80)
(550, 50)
20
Chap-4
(800, 49)
tho un
60.46.01
(550, 46)
T 46
(680, 46.6)
⼼ 43.5
(100, 43.3)
21
Chap-4
22
Chap-4
Solution
8
8
IC1 (1200
)  8  0.003 0.003
1200  11
6 ⼆11.6
1200
x.1200
IC2 (1200)  8  0.0010
1200  2002
1200
oolxl9.2 9.2
IC3 (1200)  7.5  0.002 1200  9.9
1200 7.5⼗O0以1200
IGC
IGC
IG
9.9
(1200,11.6)
(1200,9.9)
(1200,9.2)
23
Chap-4
Solution
IC1 (PG1 )  8  0.003  PG1
(a) From the graph we estimate (the common) IC  8.1.
1
Calculating we find IC  8.1  PG1  33 , PG2  100, PG3  300
3
low7 (too low).
too66.6
less66.67
500
33.33
300  66.67
less
了  100 66.67
500 33.3
Try IC  8.15  PG1  50, PG2  150, PG3  325.25
⼀⽇
325.25
500
 50
 150
 325.25
 over
25.25
overto25.25
25.25
high (too high)
25.25
too50
In this way after a few iterations, we find
136  136,
318 P  318
PGE
Ici8.136
PuP⼆45.33
IC
8.136 
G1  45.33, PG2Pci
G3
⼀⽇
318 o.nu
50043.33
6
500
45.33
 136
318  0.67 (O.K.)
Calc. of CT . We find
C1 (PG1 )  $665.75/hr 

C2 (PG2 )  $1547.25/hr   CT  $5399.12/hr
C3 (PG3 )  $3186.12/hr 
IC2 (PG2 )  8  0.001 PG2
IC3 (PG3 )  7.5  0.002  PG3
(b) From the graph we estimate IC  8.4. After a few
iterations we find IC  8.409  PG1  136.33,
PG2  409, PG3  454.5.
P 999.83
999.83.Close
Close
enough!
enough
Pa
Then C  1, 418.55  3,805.64
 4,315.32
 $9,539.51
⼆ 9539
⼗4315.32
tl / hr.
Then G 141855 385.64
Gi
T
(c) From the graph IC  9.0. After a few iterations
we have IC  8.955  PG1  318.33, PG2  955, PG3  727.5
hrIPci2000.83G
hn
P
 2, 000.83. Close enough. Then
l9 / hr.
Mt8546.0lt6685iTl
CT  2998
2,998.67
 8,546.01  6, 685.51 lf23o
$18, 230.19
Gi
24
(2)   (1)=0
4.2 Line losses considered
Chap-4
 dFN
 dF1
 dF2
 PL  
 PL  
 PL  


1

dP



1

dP

...





1 
  dPN  0

 1 

 2
dP

P
dP

P
dP

P
1 
2 
N 



 1
 2
 N

losses
line
line losses
t 傳輸缐
PpiPHB.tltmtPN PL

Load Demand
PR  P1  P2  P3  ...  PN  PL
Total cost
FT  F1  F2  ...  FN
 dPR  0

 P 
dF1
  1  L 
dP1
 P1 
dF1
1


dP1  PL 
1 

 P1 
optimal :
 P 
dF2
  1  L  ...
dP2
 P2 
 P 
dFN
  1  L 
dPN
 PN 
dF2
1


dP2  PL 
1 

 P2 
dFN
1


dPN  PL 
1 

 Pn 
dF
dF1
dF
1
1
1

 2
 ...  N 

dP1  PL 
dP2  PL 
dPN  PL 
1 

1 

1 

 P1 
 P2 
 Pn 
黸 澨巚 囇 𦿞
 dP1  dP2  dP3  ...  dPN  dPL  0
where
P
P
P
dPL  L dP1  L dP2  ...  L dPN
P1
P2
PN
 P

P
P
 dP1  dP2  dP3  ...  dPN   L dP1  L dP2  ...  L dPN   0
P2
PN
 P1

 P 
 P 
 PL 
 P 
1 
dP1  1  L dP2  1  L dP3  ...  1  L dPN  0.......(1)
 P1 
 P2 
 P3 
 PN 
x
 dFT  0
 dFT 

FT
F
F
dP1  T dP2  ...  T dPN  0
P1
P2
PN
dF
dF1
dF
dP1  2 dP2  ...  N dPN  0 .......(2)
dP1
dP2
dPN
y
25
Chap-4
26

Chap-4
設  5時
dF
1
 1
dP1 1  PL
P1
dF2
1

dP2 1  PL
P2
 dF1
 0.007 PG1  4.1

O
dP
 1
 dF2
 0.007 PG 2  4.1

dP
 2
 PL
PL
 0,
 0.002( PG 2  50)  0.002 PG 2  0.1

P2O0⼝ Pci九 ⼆0.002肛 ⼀
 P1
1

P

 5  4.1  128.6MW
G1

0.007

1.1 5  4.1

 82.4 MW
PG 2 
0.007

0.002

5

P  0.001  82.4  50 2  1.047
 L

007Pat4.1
PG1  PG2  PL  210( MW )  350( MW )
設  6時
o 0MPcitlhl
0.007 PG1  4.1

1



  4.1
P

G
1


1
0.007


0
.
007
P

4
.
1
G2
 
 P  1.1  4.1
 G 2 0.007  0.002

1.1  0.002 PG 2
PGF
fPtsi
Ans：
PG1  234.3MW
PG2  119.8MW
PL  4.87MW
6  4.1

271.4MW
4MW
PG1  0.007  271.⼆

1.1 6  4.1

131.58MW

 131
.58MW
PG 2Paz
0
.
007

0
.
002

6

2 6.66MW
2
PL  0.001O
⼆MW
131.58131.58
001
 50 06.66


PGF 器
f
不4 ly
Ri
ǜh芘品
得P1  P2  PL  396.32(MW )  350(MW )
Ǜ
設  5.74時
⼆5.74
PG1  234.3MW

PG 2  119.8MW
P  4.87 MW8MW
 L Pmilk
設入
fi
時
i
P1  P2  PL  349.2(MW )  350( MW )
RtRtPli349.2 ⼆350
27
Chap-4
28
with Penalty factor
Ex 11.11
PG1  234.3

PG2  119.8
Chap-4
optimal :
dF
dF1
dF
1
1
1

 2
 ...  N 

dP1  PL 
dP2  PL 
dPN  PL 
1 

1 

1 


P

P
1 
2 


 Pn 
假設忽略線路損失項 (neglect line losses item)... Penalty factor
optimal : ICIG
IC1  IC2
2

C1  α  4.1PG1  0.0035PG1 ($/hr)

2

C 2  α  4.1PG2  0.0035PG2 ($/hr)
 2α
1694($/hr)
2  1694
⼩
慧管
 PG1  PG2 at classic economic dispatch
PD  PPGHPGL
G1  PG2  PL  2PGz
R 2PG2  P2 L
5072
zpGz
2PG2  0.001(P
 50)
 350
350
o.co I G2
Pcn
Phi
Rzxtu.li
2
2
to0035
CT ''  2α  4.1(PG1  PG2 )  0.0035(P
G1  PG2 )
Path
dF1 dF2

dP1 dP2
PG2  183.97MW  PG1
1hrlzxt4
PL  0.001(183.9  50)2 MW
PGF183.9MW
Ans：
PuzPG1183.9MW
 183.9MW
 183.9MW
PLPG217.9MW
 17.9MW
hr
GPL 51
CT  51($/hr)
2
2 CPGD
CT '  2α  4.1
2PG1  0.0035  2(P
lx2Thto.co35
G1 )
Nn
1745($/hr)
 2α
 237 22
2α⼈
 1745
1508.5t237
以 1508.5
CT'  CT''  51($/hr)
29
Chap-4
Text books
Textbook：
1.
2.
3.
本講義僅節錄部分內容，細節請購買正版書籍
John J. Grainger, William D. Stevenson, Gary W. (Author), “Power
System Analysis” (東華代理) ISBN: 9781259008351
A.R. Bergen, prentice-Hall, “Power System Analysis” (東華代理) ISBN10: 0131784609
Power System Analysis, by Hadi Saadat, McGraw-Hill, 1999.
30/10
Chap 5-1
Stability
• The ability of the power system
to remain in synchronism and
maintain the state of equilibrium
following a disturbing force
– Steady-state stability: analysis of
small and slow disturbances gradual
power changes
• gradual power changes
– Transient stability: analysis of large
and sudden disturbances
• faults, outage of a line, sudden
application or removal of load
1
Chap 5-1
複習第三章第一頁
V2  
V10
P12  jQ12
R  jX

 V 0  V2   
 V10  V2
S12  V1I   V1   1

V

1

 R  jX
R  jX



@R  0
 R  jX
V 2  V1 V2 cos  j V1 V2 sin  
 2
2  1
 R X
1
2
P12  2
RV
1  R V1 V2 cos  X V1 V2 sin  
2 
R X
⼼ V1 V2 sii Pmaxshó
sin   Pmax sin 
PowerAngle
RiP12 
X
1
2
Q12  2
XV
1  X V1 V2 cos  R V1 V2 sin 
2
2
R X

舉

Chap 5-1
H d 2
 Pm  Pe
2
f 0 dt
Synchronous Machine Model
(power - angle equation )
PeiPmaxsmó
靠器
Pm Pnnxshó
where Pm  shaft power input to the machine less rotational losses 轉軸功率
Pe  electrical power crossing its air gap 電功率
H  inertia constant 慣性常數
3
Fig 13.4
Example 13.3
V
Vt  1.00LŌ
VEI

Pnil
Vt  1
j0.2 Mj0.1
Tm
帷
+
E
-
資料來源（教科書）：John J. Grainger, William D.
Stevenson, Gary W. (Author), “Power System Analysis” (東
 華代理)
1.00ISBN:
 9781259008351
1 1
sin   1
X
0.3
δ  17.45  Vt  1.017.45
P
violin
I
j0.2
M7
雨
V  10
I
→
Chap 5-1
Vt V
sin  
450
Vt  V 11⼭
17.45
  11200
0
I

 1.012
8.729
⼆
⼼2Lf7290
j0.3
j 0.3
jo 3 0522840
E  j0.2 tI 
V  1.0528.4
Veil
PXI
EV
t
I 1
1.05
sinδ  2.10sinδ  pu 
x
0.5
H d 2δ
H d 2δ 比 0 210sinócp
(b)
 Pm
Pm  P
 1.0  2.10sinδ  pu 
1e
2
2
πf dt
 f dt
(a) Pe 

sinδ 
_shfziosinólp
汫崑瑟
4
Exercise 11.6(Hadi)
•
Chap 5-1
H
一部 60 Hz 同步發電機其暫態電抗為0.2標么，慣量常數
5.66MJ/MVA。 此發電機經由一個變壓器和雙回路傳輸線連接到無
限匯流排，如圖所示。忽略所有電阻值，且將以共同的MVA基準的
電抗標么值標示在圖上。發電機送出0.77標么的實功率到匯流排1
，匯流排1上電壓值為1.1，而無限匯流排電壓V = 1.00標么。試
決定發電機激磁電壓及搖擺方程式。
Vt  1.1
課堂練習題
5
Exercise 11.6(Hadi)
Chap 5-1
N
⼼
N
doin
皆
lot jo758lonn jo.it
H d 2δ
Pm.Pe
Pm  Pe
2
πf dt
器
V1圳 ⼭6.26
o.nn ju4 on826L
a
V1  1.116.26
w.si A
1.25⼈27.8190
器 想
5
sìnó
e
OM
sinó
管管
6
Chap 5-1
7
Example 13.4
j0.2
Example 13.3
j0.4
j0.2
j0.2
iii
器

j0.2

j0.1
j0.05
簽

V

0.2  0.2
 0.05
0.4  0.2  0.2
⼀
j0.1
ÉǛ

V
j0.05

0.4  0.05  0.05  0.1  0.4  0.1
 1.3
0.05
V
j0.1
j0.1
0.4  0.2
 0.1
0.4  0.2  0.2
j0.4
+
E
-
j0.1
iiiji
Chap 5-1
氧還
+
E
-
j1.3
0以 ⼼
0以
o05

1.3
V

E  V
1.05
⼼  1.0
s.info
sinδ 
sinδ 88sìnócpul
0.808sinδ (pu)
X
1.3
H d 2δ
5 d 2δ
8
(b)

P

P

1.0

0.808sinδ(pu)
m
e
πf dt 2
πf dt 2 1.0 a808pu
(a) Pe 
苦
ǙǛ
Chap 5-1
Example 13.5
Example 13.3
資料來源（教科書）：John J. Grainger, William D.
Stevenson, Gary W. (Author), “Power System Analysis” (東
華代理) ISBN: 9781259008351
+
E
-
j0.2
j0.1
j0.4

V

E  V
1.05  1
sinδ 
sin   1.5sinδ (pu)
X
0.7 shflisinōcpu
H d 2δ
5 d 2δ
 Pm  Pe
 1.0 
1.5sinδ
(pu)
1 5smōcpul
1
2
2
πf dt
πf dt
Pe 
舉
急悲
9
Chap 5-1
Before
During
After
5
πf
⼆Pmaxshó
r2
⼝Pmaxshó
πf
per unit
πf
2
If
l0
r
Pmaxshó
Pe''
Pe 210
n
o
I
想
0
21
Pe '
shōvf
贬1.5shó
⼆0.808
Pmzfr
δ Pe
138.19
shó
max
28.440
90
Óma⼆138.190180
10
Chap 5-1
Pmax  2.1
2.1
0.808
r1  0808
 0.385
⼀ ⼆0.385
2.1
21
1.5
r2  l 5  0.714
0 nl4
2.1
2.1sinδ0  1
δ 0  28.44(or 0.496 rad)
1.5sinδ max  1
Pe  Pmax sinδ
Pe ' '  r2 Pmax sinδ
Pe '  r1Pmax sinδ
IN 5⼼有
δ max  138.19
(or
rad)
138.190
2 4⼝ 叭
⼼ 2.412

1  Pm
 cos  cr 
(



)

r
cos


r
cos

max
0
2
max
1
0

r2  r1  Pmax

In
Before : Pe 2.1
2.1sin
  PPmaxshó
max sin
shf
During : Pe '  0
0.808
808sin
  rPmaxsin
1Pmax sin
shfr
After : Pe ' '  1.5sin  r2 Pmax sin
1.5shfbihaxshó
1
1.0

138.190
0.496
Cos
0.714
0.385
⼝
⼆
2.
412

0.496

0.714cos138.19


0.385cos
28.4

cos


2.4
28.47


0.714  0.385  2.1
Thus,  cr  82.726
Tj
無
資料來源（教科書）：John J. Grainger, William D. 11
Stevenson, Gary W. (Author), “Power System Analysis” (東
華代理) ISBN: 9781259008351
Chap 5-1
Pe  Pmax sin 
r1Pmax sin 
r1  0
r2  1
r2 Pmax sin 
Pm
cos  cr 
( max   0 )  cos  max
Pmax
12
Chap 5-1
1
t=0
2
t=0+
3
t=t1
Close
13
Chap 5-1
Example 13.7
f  60Hz
Example 13.3
Pe  Pmax sinδ  2.1sinδ
δ0  28.44(0.496 rad), δmax
151
151.56(or
2.645 rad)
Ómax⼆
or 2645rad
I
cosδcr 
Pm
 δmax  δ0   cosδmax
Pmax
56
1.0
0.496   cos151.56 0.144
 0.144
 2.645
2645 0.496Has151.560
2.1
δcr 81
8170
.7Corh426radl
(or 1.426 rad)

兴
4  5  1.426  0.496 
tcr  45 ⼼426 0496
 0.222(sec)
0.222 Sec
377  1
37以
1
資料來源（教科書）：John J. Grainger, William D. 14
Stevenson, Gary W. (Author), “Power System Analysis” (東
華代理) ISBN: 9781259008351
Chap 5-1
15
Chap 5-1
16
Chap 5-1
Text books
Textbook：
1.
2.
3.
本講義僅節錄部分內容，細節請購買正版書籍
John J. Grainger, William D. Stevenson, Gary W. (Author), “Power
System Analysis” (東華代理) ISBN: 9781259008351
A.R. Bergen, prentice-Hall, “Power System Analysis” (東華代理)
ISBN-10: 0131784609
Power System Analysis, by Hadi Saadat, McGraw-Hill, 1999.
17/10
Chap 5-2
Prove
Proof
H d 2
 2  Pm ( pu )  Pe ( pu )
 f dt
Total moment of inertia  kg  m
rotor acceleration
dm
J
 Tm  Te
dt
J s
2

electrical torque
Chap 5-2
contigency
 m  rotor speed less than syn. speed
 N  m
dm d m2
 2
dt
dt
electrical power output
1
F  mv 2
2
mechanical power output
synchronous speed
H
Assume : m  s
2 HSb dm

 Pm  Pe
s
dt
dm d 2 m

dt
dt 2

 2  4
f 
f
p
p


s  2  
 m  ct
 m  rotor speed over syn. speed
mechanical torque N  m 
dm
 s Tm  Te   Pm  Pe
dt
d
m  s  m
dt
1 2
J s stored kinetic energy in MJ
2

Sb
Mathine rating in MVA
 J s 
2 HSb
s
2 H dm Pm Pe


  Pm ( pu )  Pe ( pu )
s dt
Sb S b
2 H d 2 m


 Pm ( pu )  Pe ( pu )
s dt 2
2 
d 2  e 
p  H d 2 e
2H




 2  Pm ( pu )  Pe ( pu )
2
4
dt

f
dt
f
p
資料來源（教科書）：John J. Grainger, William D.
Stevenson, Gary W. (Author), “Power System
Analysis” (東華代理) ISBN: 9781259008351

1  Pm
cos  cr 
 max   0   r2 cos  max  r1 cos  0 

r2  r1  Pmax

Prove
Chap 5-2
Proof
 cr
  P
0
m
 r1Pmax sin   d  
 max
 cr
 r2 Pmax sin   Pm  d
Pm  cr   0   r1Pmax  cos  cr  cos  0   r2 Pmax  cos  cr  cos  max   Pm  max   cr 
 r1Pmax  r2 Pmax  cos  cr  Pm  0   max   Pmax  r2 cos  max  r1 cos  0 

1  Pm
 cos  cr 
 max   0   r2 cos  max  r1 cos  0 

r2  r1  Pmax

資料來源（教科書）：John J. Grainger, William D.
Stevenson, Gary W. (Author), “Power System Analysis” (東
華代理) ISBN: 9781259008351
Chap 5-2
1
t=0
2
t=0+
3
Close
t=t1
tc 
Prove
2 H ( c   0 )
 f 0 Pm
Chap 5-2
3
t=t1
H d 2
 Pm
 f 0 dt 2
d 2
dt
2

 f0
H
Pm
f
t
d  f 0

Pm  dt  0 Pm t
0
dt
H
H
 
tc 
 f0
2H
Pm t 2   0
2 H ( c   0 )
 f 0 Pm
資料來源（教科書）：John J. Grainger, William D.
Stevenson, Gary W. (Author), “Power System Analysis” (東
華代理) ISBN: 9781259008351
Chap 5-2
Chap 5-2
Text books
Textbook：
1.
2.
3.
本講義僅節錄部分內容，細節請購買正版書籍
John J. Grainger, William D. Stevenson, Gary W. (Author), “Power
System Analysis” (東華代理) ISBN: 9781259008351
A.R. Bergen, prentice-Hall, “Power System Analysis” (東華代理) ISBN10: 0131784609
Power System Analysis, by Hadi Saadat, McGraw-Hill, 1999.
7/10