電路學(二)
Chapter 12
Three-Phase Circuits
12.1 Introduction (1)
Single-Phase Systems
two-wire type
three-wire type
Three-Phase Four Wire
Systems
12.1 Introduction (2)
• Nearly all electric power is generated and distributed in
3-phase.
• Instantaneous power in a 3- system can be constant.
• The 3- system is more economical than single-phase
system.
12.2 Balanced Three-Phase
Voltages (1)
Van
A three-phase generate
Vbn
Vcn
12.2 Balanced Three-Phase
Voltages (2)
Y-connected source
-connected source
12.2 Balanced Three-Phase
Voltages (2)
positive sequence
正相序
negative sequence
負向序
Van  Vp 0
Vbn  V p   120
Van  Vbn  Vcn  0
Vcn  Vp   240  Vp   120
Balanced phase voltages are equal in magnitude and
are out of phase with each other by 120
12.2 Balanced Three-Phase
Voltages (3)
a Y-connected load.
a -connected load.
A Balanced Load is one in which the phase impedance
are equal in magnitude and in phase.
For a balanced Y-connected load Z1 = Z2 = Z3 = ZY
For a balanced -connected load Za = Zb = Zc = Z
12.2 Balanced Three-Phase
Voltages (4)
Y- transformation
1
Z   3ZY or ZY  Z 
3
There are four possible connections in three-phase systems:
• Y-Y connection
• Y- connection
• - connection
• -Y connection
12.2 Balanced Three-Phase
Voltages (5)
Example 1
Determine the phase sequence of the set of
voltages
van  200cos(t  10)
vbn  200cos(t  230)
vcn  200cos(t  110)
12.3 Balanced Y-Y Connection (1)
ZY = Zs + Zl + ZL
12.3 Balanced Y-Y Connection (2)
Assuming the positive sequence,
Van  Vp 0, Vbn  Vp   120
Vcn  Vp   120
The line-to-line voltages (line voltage)
Vab  Van  Vbn
 Vp 0  Vp   120
 3V p 30
Similarly,
Vbc  Vbn  Vcn  3Vp   90
Vca  Vcn  Van  3Vp   210
VL  3Vp
12.3 Balanced Y-Y Connection (3)
Van
Vbn Van   120
Ia 
, Ib 

 I a   120
ZY
ZY
ZY
Vcn Van   120
Ic 

 I a   120
ZY
ZY
I a  Ib  I c  0
I n  (I a  I b  I c )  0
VnN  Zn I n  0
12.3 Balanced Y-Y Connection (4)
Example 2
Calculate the line currents in the three-wire Y-Y
system.
12.4 Balanced Y- Connection (1)
Assuming the positive sequence,
Van  Vp 0, Vbn  Vp   120
Vcn  Vp   120
Vab  3V p 30  VAB
Vbc  3V p   90  VBC
Vca  3V p 150  VCA
I AB
VBC
VCA
VAB

, I BC 
, ICA 
Z
Z
Z
I a  I AB  ICA  I AB 3  30
Ib  I BC  I AB  I BC 3  30
I c  ICA  I BC  I CA 3  30
I L  3I p
I p  I AB  I BC  ICA
I L  I a  Ib  I c
12.4 Balanced Y- Connection (2)
Example 3
A balanced abc-sequence Y-connected source with
Van = 10010 V is connected to a -connected
balanced load (8 + j4)  per phase. Calculate the
phase and line currents.
12.5 Balanced - Connection (1)
Assuming the positive sequence,
Vab  Vp 0, Vbc  Vp   120
Vca  Vp   120
Vab  VAB , Vbc  VBC
Vca  VCA
I AB
VBC Vbc
VCA Vca
VAB Vab


, I BC 

, ICA 

Z
Z
Z
Z
Z Z
I a  I AB  ICA  I AB 3  30
Ib  I BC  I AB  I BC 3  30
I c  ICA  I BC  I CA 3  30
I L  3I p
12.5 Balanced - Connection (2)
Example 4
A balanced -connected load having an impedance
20 – j15  is connected to a -connected, positivesequence generator having Vab = 3300 V.
Calculate the phase currents of the load and the
line currents.
12.6 Balanced -Y Connection (1)
• Using KVL.
• Replacing the connected source
with its equivalent
Y-connected source.
• Transforming the Yconnected load to
an equivalent Yconnected load.
12.6 Balanced -Y Connection (2)
12.6 Balanced -Y Connection (3)
Example 5
A balanced Y-connected load with a phase
impedance 40 + j25  is supplied by a balanced,
positive-sequence -connected source with a line
voltage of 210 V. Calculate the phase currents. Use
Vab as reference.
12.7 Power in a Balanced System
(1)
The advantage of 3-phase systems for power distribution
• The total instantaneous power in a balanced 3-phas
system is constant.
• The 3-phase system uses a lesser amount of wire than
the single-phase system for the same line voltage VL and
the same absorbed power PL.
For a Y-connected load, the phase voltages are
vAN  2Vp cos t , vBN  2Vp cos(t  120)
vCN  2Vp cos(t  120)
12.7 Power in a Balanced System
(2)
If ZY = Z, the phase currents
ia  2 I p cos(t   ), ib  2 I p cos(t    120)
ic  2 I p cos(t    120)
p  pa  pb  pc  vAN ia  vBN ib  vCN ic
 2Vp I p [cos t cos(t   )  cos(t  120) cos(t    120)
 cos(t  120) cos(t    120)]
1
Appling cos A cos B  [cos( A  B )  cos( A  B )]
2
p  Vp I p [3cos   cos   cos  cos 240  sin  sin 240
 cos  cos 240  sin  sin 240]
 3VP I P cos 
12.7 Power in a Balanced System
(3)
The complex per phase
S p  Pp  jQp  VpI*p
The total complex power
S  3S p  3Vp I*p  3I p2 Z p 
3Vp2
*
P
Z
 3VL I L
where Vp, Ip, VL, and IL are all in rms values and  is the angle
of the load impedance.
for Y-connected loads,
for -connected loads,
VL
Vp 
, I p  IL
3
I
Vp  VL , I p  L
3
12.7 Power in a Balanced System
(4)
Ploss
'
Ploss
2
P
 2I L2 R  2R L2
VL
2
2
P
P
 3( I L' )2 R'  3R' L 2  R' L2
3VL
VL
Ploss 2 R 2 /  r 2 2r '2
 2
 ' 
'2
'
 /r
r
Ploss R
Material for 1- 2( r 2 ) 2r 2

 '2
'2
Material for 3- 3( r ) 3r
2
 (2)  1.333
3
12.7 Power in a Balanced System
(5)
Example 6
Determine the total average power, reactive power,
and complex power at the source and at the load.
12.7 Power in a Balanced System
(6)
Example 7
A three-phase motor can be regarded as a
balanced Y-load. A three-phase motor draws 5.6
kW when the line voltage is 220 V and the line
current is 18.2 A. Determine the power factor of
the motor.
12.7 Power in a Balanced System
(7)
Example 8
Two balanced loads are connected to a
240-kV rms 60-Hz line, as shown in the
figure (a). Load 1 draws 30 kW at a
power factor of 0.6 lagging, while load 2
draws 45 kVAR at a power factor 0.8
lagging. Assuming the abc sequence,
determine:
(a) the complex, real and reactive
powers absorbed by the combined load,
(b) the line currents, and
(c) the kVAR rating of the three
capacitors -connected in parallel with
the load that will raise the power factor
to 0.9 lagging and the capacitance of
each capacitor.
12.10 Applications (1)
• Three-Phase Power Measurement.
12.10 Applications (2)
Consider the balanced Y-connected load
P1  Re[Vab I*a ]  Vab I a cos(  30)
 VL I L cos(  30)
P2  Re[Vcb I*c ]  Vcb I c cos(  30)
 VL I L cos(  30)
P1  P2  Vcb I c cos(  30)  Vcb I c cos(  30)
 3VL I L cos 
P1  P2  Vcb I c cos(  30)  Vcb I c cos(  30)
 VL I L sin 
PT  P1  P2
QT  3( P2  P1 )
12.10 Applications (3)
ST  PT2  QT2
QT
P2  P1
tan  
 3
PT
P2  P1
• If P2 = P1, the load is resistive.
• If P2 > P1, the load is inductive.
• If P2 < P1, the load is capacitive.
12.10 Applications (4)
Example 9
The two-wattmeter method produces wattmeter
readings P1 = 1560 W and P2 = 2100 W when
connected to a -connected load. If the line
voltage is 220 V, calculate: (a) the per-phase
average power, (b) the per-phase reactive power,
(c) the power factor, and (d) the phase impedance.
12.10 Applications (5)
Example 10
The three-phase balanced load in the figure has
impedance per phase of ZY = 8 + j6 . If the load
is connected to 208-V lines, predict the readings of
W1 and W2. Find PT and QT.