Download Document

Chapter 9 Three-Phase Circuits
三相电路
9.1 Balanced Three-Phase Voltages
9.2 Balanced Three-Phase systems
9.3 Power in a Balanced System
Nearly all electric power is generated and
distributed in three-phase,at frequency of
50Hz(or =314rad/s) in China or 60Hz(or
=377rad/s) in some other parts of the
world.
9.1 Balanced Three-Phase Voltages 对称三相电压
1. Three-phase source
a
b
va + vb +
–
–
–
–
x
y
c
vc +
–
–
z
va (t )  2V p cos t
vb (t )  2V p cos(t  120)
vc (t )  2V p cos(t  240)  2V p cos(t  120)
va(t)
vb(t)
vc(t)
t
Balanced three-phase voltages are equal in magnitude and are
out of phase with each other by 120o.
  V 0
V
a
p
  V   120
V
b
p
V cb

V
a
  V   240  V   120
V
c
p
p

V
bc
 V p 0  V p   120  V p   120
0
• phase sequence 相序
 V
 V

V
a
b
c
 V

V
a
b
---positive sequence （abc sequence) 正序
---negative sequence （acb sequence) 负序
  V 0
V
a
p
  V  120
V
c
p
  V   120
V
b
p
phase voltage 相
电压
phase current
相电流
c
line current 线电流
I a
a
a

+
–

Va
V ab
V ca
z
Vc y
x
n
I b

Vb
line 端线


b

Ic
line voltage
线电压
b
V bc
c
neutral line 中线
Phase voltage相
电压
line current 线
电流
I a
a
a
I ab
phase current 相 V c
电流
line 端线

V a V ab
I ca
b I b
c

Vb
I bc


V ca
b
V bc
c
Ic
line voltage线
电压
2. Three-phase load
A balanced load is one in which the phase impedances are
equal in magnitude and in phase.
a
a
n
ZY
b
Z
ZY
b
c
c
ZY
Z
Z
ZY
1

Z
3
3. Y-connected source or load
I a
a
a
+
–

Vc
Va
x
V ab
V ca
I b

b
 V
 V

V
ab
a
b
 V
 V

V
b
V ca


Vc
V ab
300
n
Vb
bc



z
y
c

Assuming the positive sequence
c

Ic
b

Va

Vb
V bc
c
  3V
 30
V
L
p

V bc
IL  Ip
 V
 V

V
ca
c
a
Thus, the magnitude of the line voltages VL is 3 times the
magnitude of the phase voltages VP
The line voltages lead their corresponding phase voltages by 30°
4.  -connected source or load
Assuming the positive sequence
IL  3Ip   30
I a
a
a

I b
Ic

I ca
b
b
 V

V
L
p
Z
I ab
Z

I bc
Z
c
c
Thus, the magnitude of the line current IL is 3 times the
magnitude of the phase current IP
The line currents lags their corresponding phase currents by 30°
Example 9.1 given that V b  11030 V , find V a and V c ,
assuming a positive sequence.
Solution:
positive sequence gives that V a leads V b by 120°,
 by 120°
V c lags V
b
So
  110150 V
V
a
  110  90 V
V
b
9.2 Balanced Three-Phase Systems
1. Balanced Y-Y Connection
a

Vc
A
+ 
– Va
n
ZY
N

c
Vb
ZY
b
B
three-phase
three-phasethree-wire
four-wire system
system
三相三线制
三相四线制
ZY
C
2. Balanced Y-  Connection
a

Vc
A
+ 
– Va
n
Z
Z

c
Vb
b
B
three-phase three-wire system
三相三线制
C
Z
3. Balanced  - Y Connection
a
A

Vc

ZY
Va
N
b
c
ZY
B

Vb
three-phase three-wire system
三相三线制
ZY
C
4. Balanced  -  Connection
A
a

Vc

Va
Z
Z
b
c
B

Vb
three-phase three-wire system
三相三线制
C
Z
• Balanced Y-Y Connection
I a
a

Vc
A
+ 
– Va
n
I n
N
I b

c
Vb
ZY
b
Ic
ZY
B
ZY
C
Assuming the positive sequence
  V   240  V   120
  V   120 V
  V 0 V
V
c
p
p
b
p
a
p


   120
V
V
I  Va
b

Ib 
 a
 I a   120
a
Z
Z
Z

   240
V
V
c
I c 
 a
 I a   240  I a   120
Z
Z
I c
I n  (I a  I b  I c )  0
I a
Single-phase equivalent circuit
I a
I b
a

Vc
A
+ 
– Va
n

I n
ZY
N
I b
ZY
ZY
C
b
• c A balanced three-phase
circuit
can
be
transformed
to
Y-Y
B
Ic
system and then analyze the single-phase equivalent circuit.
Vb
Example 9.2 Calculate the line currents in the three-wire Y-Y
system.
I a
a
A
+ 1100 V
–
I n  0
n
110  240 V
I b
c
110  120 V
b
Ic
5  j 2
10  j8
N
5  j 2
10  j8
5  j 2
B
10  j8
C
I a
Solution:
A
a
The three-phase circuit is
balanced, we may use the
single-phase analysis
+

110

0
V
–
n
5  j 2
10  j8
N

V
I a  a
ZY
ZY  (5  j 2)  10  j8)
I a
+

– 1100 V
n
 15  j 6  16.1521.8 A

110

0

Ia 

6
.
81


21
.
8
A

16.1521.8
Positive sequence, so
Ib  6.81  141.8 A
I  6.8198.2 A
c
A
a
5  j 2
10  j8
N
9.3 Power in a Balanced System
The single-phase :
Pp  V p I p cos 
S p  Vp I p
Q p  V p I p sin 
   pv   pi   z
In balanced three- phase system:
The average power:
P  3V p I p cos  
3VL I L cos 
The reactive power:
Q  3V p I p sin  
3VL I L sin 
The apparent power:
S  3V p I p 
3VL I L
The complex power:
S  P  jQ

3VL I L   3V p I p 
Example 9.3 A balanced Y-load draws 5.6kW when the line
voltage is 220V and the line current is 18.2A. Determine the
power factor of the load.
Solution:
Since the real power is
P  3VL I L cos 
The power factor is
P
cos  
3VL I L
5.6 103
 0.807

3  220 18.2
部分电路图和内容参考了：
电路基础（第3版），清华大学出版社
电路（第5版），高等教育出版社
特此感谢！