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Three-Phase Circuits
Why are the transmission lines high-voltage, three-phase and AC?
High-Voltage
Average power loss in a
transmission line with
impedance R  jX :
WL  (1 2) RIm2
the transmitted power:
W  (1 2)Vm I m cos(v   I )
Three-Phase
Under a balanced
load, the torque on
the generator does
not fluctuate much.
Therefore the
vibration on the
generator is less.
The loss of average power
Cheap induction
can be written as:
motors can be made
2 RW 2
which use threeWL  2 2
phase directly .
Vm cos ( v   I )
The loss can be reduced
by using a very high
voltage amplitude Vm .
Alternative Current
It is easy to change the
voltage value by
transformers.
Transformers are very
efficient in 60 Hz, and
require almost no
maintenance.
Easier to produce
Under the balanced
than DC.
load assumption, one
can reduce the number
of lines, therefore
three-phase is more
economical.
Three-phase Generator
Rotor- A magnet
Stator- with three wires : aa’, bb’, cc’
The wire aa’ is the same as bb’
except the angle shift of
120  (2 / 3)rad
Magnetic Flux for different
phases :
aa’:  R   m sin 
bb’:  S   m sin(  2 / 3)
cc’: T   m sin(  4 / 3)
 m , constant
L.O. Chua, C.A. Desoer, S.E. Kuh. “Linear and Nonlinear Circuits” Mc.Graw Hill, 1987, New York
Rotor is rotating with constant
angular frequency:
  2 50 rad/sn
E p ˆ  m
 R (t )   m sin t
vR (t )  E p cost
 S   m sin(t  2 / 3)
vS (t )  E p cos(t  2 / 3)
T   m sin(t  4 / 3)
vT (t )  E p cos(t  4 / 3)
Magnetic
Fluxes
Voltages
Phasors of voltages:
VR  E p
VS  E p e  j 2 / 3
VT  E p e  j 4 / 3
E p  E p e j 2 / 3  E p e j 4 / 3  VR  VS  VT  0
L.O. Chua, C.A. Desoer, S.E. Kuh. “Linear and Nonlinear Circuits” Mc.Graw Hill, 1987, New York
Power System Under Balanced Load
VS
VR
VT
I R  , I S  , IT 
ZL
ZL
ZL
I R  I S  IT  0
The connection n-n’
can be removed.
Why?
L.O. Chua, C.A. Desoer, S.E. Kuh. “Linear and Nonlinear Circuits” Mc.Graw Hill, 1987, New York
Reducing the analysis to one-phase!
VR  E p
VS  E p e  j 2 / 3
Let n be the reference node!
IR
VT  E p e  j 4 / 3
VR
VT
VS
IS
IT
1
Y ˆ
Zl  Z L
KCL equation for n’ is
Y (Vn'  VR )  Y (Vn'  VS )  Y (Vn'  VT )  0
Y (VR  VS  VT )  3YVn'  0
=0
Vn '  0
Node voltage of n’ is the
same as the node voltage of
n. We can assume that n=n’
Instantaneous Power:
1 Ep
pR (t )  vR (t )iR (t ) 
[cos Z L  cos(2t   Z L )]
2 ZL
1 Ep
4
pS (t )  vS (t )iS (t ) 
[cos Z L  cos(2t   Z L  )]
2 ZL
3
1 Ep
8
pT (t )  vT (t )iT (t ) 
[cos Z L  cos(2t   Z L  )]
2 ZL
3
2
3 Ep
pR (t )  pS (t )  pT (t ) 
cos Z L
2 ZL
The total instantaneous power delivered to a balanced load is
constant. Hence, the torque required for the rotor is also constant!
Analysis of Balanced Three-Phase Circuits
Star
Triangle
3
Triangle-To-Star
Star-To-Triangle
Z1 
Zb Zc
Z a  Zb  Zc
Z1 Z 2  Z 2 Z 3  Z1 Z 3
Za 
Z1
Z2 
Za Zc
Z a  Zb  Zc
Zb 
z1
Z a Z b z2
Z3 
Z a zZ b  Z c
3
Z1 Z 2  Z 2 Z 3  Z1 Z 3
Z2
z
Z1Z 2  cZ 2 Z 3  Z1Z 3
Zc 
zb
Z 3 za
How can one obtain the conversion relations?
3
Z c (Z a  Zb )
Z a  Zb  Z c
Z ab  Z1  Z 2
Z ab  Z c //(Z a  Zb )  Z ab 
Zbc  Z 2  Z 3
Z a (Zb  Z c )
Zbc  Z a //(Zb  Zc )  Zbc 
Z a  Zb  Z c
Z ac  Z1  Z3
Zb (Z a  Z c )
Z ac  Zb //(Z a  Zc )  Z ac 
Z a  Zb  Zc
Z ab  Z1  Z 2
Zbc  Z 2  Z 3
Z ac  Z1  Z3
2Z1  Z ac  Zbc  Z ab
Z ac  Zbc  Z1  Z 2
2Z1 
Zb (Z a  Z c )  Z a (Zb  Z c )  Z c (Z a  Zb )
Z a  Zb  Zc
2Z b Z c
2Z1 
Z a  Zb  Z c
Zb Zc
Z1 
Z a  Zb  Zc
IR
VR
VT
VS
IS
IT
KVL for the node sequence n-a-a’-n’-n
VR  ( Z L  Z l ) I R
KVL for the node sequence n-b-b’-n’-n
VS  ( Z L  Z l ) I S
KVL for the node sequence n-c-c’-n’-n
VT  ( Z L  Z l ) IT
There is a ................... between three subcircuits for different
phases. If you solve one you can get the others by..........................!
IR
IS
IT
A balanced three-phase
circuit of 220 Volt is shown
where Z=10e-j45. Find the
currents IR, IS and IT!