Download Lecture 3: Complex Power, Three

ECE 476
Renewable Energy Systems
Lecture 3: Complex Power, Three-Phase
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
[email protected]
Announcements
• Please read Chapters 2 and 4
• HW 1 is 2.7, 16, 24, 33; due Thursday 9/1
•
•
•
Problem 2.47 moved to HW 2
HW 1 will be turned in (for other homework we may
have an in-class quiz)
For Problem 2.33 you need to use the PowerWorld
Software. You can download the software and cases at
the below link; get Version 19 (August 11, 2016)
www.powerworld.com/gloveroverbyesarma
Power Consumption in Devices
Resistors only consume real power
2
PResistor  I Resistor R
Inductors only consume reactive power
2
Q Inductor  I Inductor X L
Capacitors only generate reactive power
2
QCapacitor   I Capacitor X C
QCapacitor  
VCapacitor
XC
1
XC 
C
2
(Note-some define X C negative)
3
Example
First solve
basic circuit
400000 V
I 
 4000 Amps
1000 
V  400000  (5  j 40) 4000
 42000  j16000  44.920.8 kV
S  V I *  44.9k20.8 4000
 17.9820.8 MVA  16.8  j 6.4 MVA
4
Example, cont’d
Now add additional
reactive power load
and resolve
Z Load  70.7
pf  0.7 lagging
I  564  45 Amps
V  59.713.6 kV
S  33.758.6 MVA  17.6  j 28.8 MVA
5
Power System Notation
Power system components are usually shown as
“one-line diagrams.” Previous circuit redrawn
17.6 MW
28.8 Mvar
-16.0 MW
-16.0 Mvar
59.7 kV
17.6 MW
28.8 Mvar
slack
40.0 kV
16.0 MW
16.0 Mvar
Transmission lines
Generators are
are
shown
as
a
shown as circles
single line
Arrows are
used to
show loads
6
Reactive Compensation
Key idea of reactive compensation is to supply reactive
power locally. In the previous example this can
be done by adding a 16 Mvar capacitor at the load
16.8 MW
16.0 MW
6.4 MVR
0.0 MVR
44.94 kV
16.8 MW
6.4 MVR
40.0 kV
16.0 MW
16.0 MVR
16.0 MVR
Compensated circuit is identical to first example with
just real power load
7
Reactive Compensation, cont’d
• Reactive compensation decreased the line flow
from 564 Amps to 400 Amps. This has advantages
–
–
–
Lines losses, which are equal to I2 R decrease
Lower current allows utility to use small wires, or
alternatively, supply more load over the same wires
Voltage drop on the line is less
• Reactive compensation is used extensively by
utilities
• Capacitors can be used to “correct” a load’s power
factor to an arbitrary value.
8
Power Factor Correction Example
Assume we have 100 kVA load with pf=0.8 lagging,
and would like to correct the pf to 0.95 lagging
S  80  j 60 kVA
  cos 1 0.8  36.9
PF of 0.95 requires desired
 cos 1 0.95  18.2
Snew  80  j (60  Qcap )
60 - Qcap
80
 tan18.2  60  Qcap  26.3 kvar
Qcap  33.7 kvar
9
Distribution System Capacitors
10
Definition of Reactive Power
p (t )  Vmax cos(t  V ) I max cos(t   I )
1
p (t )  Vmax I max [cos(V   I )  cos(2t  V   I )]
2
1
Vmax I max cos(2t  V   I ) 
2
1
Vmax I max  cos(2t  2 I )cos(V   I )  sin(2t  2 I )sin(V   I ) 
2
 P cos(2t  2 I )  Q sin(2t  2 I )
p (t )  P (1  cos(2t  2 I ))  Q sin(2t  2 I )
11
PowerWorld Simulator Overview
• Used for power system analysis and visualization
–
Runs in Windows
• Download free 42 bus educational version at
–
http://www.powerworld.com/gloveroverbyesarma
• Image on right
shows the
problem 2.33
power
system (case)
Balanced Three-Phase () Systems
• A balanced three-phase () system has
–
–
–
three voltage sources with equal magnitude, but with an
angle shift of 120
equal loads on each phase
equal impedance on the lines connecting the generators to
the loads
• Bulk power systems are almost exclusively 3
• Single-phase is used primarily only in low voltage,
low power settings, such as residential and some
commercial
13
Balanced 3 -- No Neutral Current
I n  I a  Ib  I c
V
In 
(10  1   1  
Z
*
*
*
*
S  Van I an
 Vbn I bn
 Vcn I cn
 3 Van I an
14
Advantages of 3 Power
• Can transmit more power for same amount of wire
(twice as much as single phase)
• Torque produced by 3 machines is constant
• Three-phase machines use less material for same
power rating
• Three-phase machines start more easily than
single-phase machines
15
Three-Phase - Wye Connection
• There are two ways to connect 3 systems
–
–
Wye (Y)
Delta ()
Wye Connection Voltages
Van
 V  
Vbn
 V   
Vcn
 V   
16
Wye Connection Line Voltages
Vca
Vcn
Vab
-Vbn
Van
Vbn
Vbc
Vab
(α = 0 in this case)
 Van  Vbn  V (1  1  120

3 V   30
Vbc

3 V   90
Vca

3 V   150
Line-to-line
voltages are
also balanced
17
Wye Connection, cont’d
• Define voltage/current across/through device to be
phase voltage/current
• Define voltage/current across/through lines to be
line voltage/current
VLine  3 VPhase 130  3 VPhase e
j
6
I Line  I Phase
S3
*
 3 VPhase I Phase
18
Delta Connection
For the Delta
phase voltages equal
line voltages
Ica
For currents
Ia  I ab  I ca
Ic

Ib
Ibc
Iab
Ia
3 I ab   
I b  I bc  I ab
Ic  I ca  I bc
*
S3  3 VPhase I Phase
19
Three-Phase Example
Assume a -connected load is supplied from a 3
13.8 kV (L-L) source with Z = 10020
Vab  13.80 kV
Vbc  13.8 0 kV
Vca  13.80 kV
13.80 kV
I ab 
 138  20 amps
 
I bc  138  140 amps
I ca  1380 amps
20
Three-Phase Example, cont’d
I a  I ab  I ca  138  20  1380
 239  50 amps
I b  239  170 amps I c  2390 amps
*
S  3  Vab I ab
 3  13.80kV  138 amps
 5.7 MVA
 5.37  j1.95 MVA
pf  cos 20   lagging
21
Delta-Wye Transformation
To simplify analysis of balanced 3 systems:
1) Δ-connected loads can be replaced by
1
Y-connected loads with ZY  Z 
3
2) Δ-connected sources can be replaced by
VLine
Y-connected sources with Vphase 
330
22
Delta-Wye Transformation Proof
From the  side we get
Vab Vca
Vab  Vca
Ia 


Z Z
Z
Hence
Vab  Vca
Z 
Ia
23
Delta-Wye Transformation, cont’d
From the Y side we get
Vab  ZY ( I a  I b )
Vca  ZY ( I c  I a )
Vab  Vca  ZY (2 I a  I b  I c )
Since
Ia  I b  I c  0  I a   I b  I c
Hence
Vab  Vca  3 ZY I a
3 ZY
Vab  Vca

 Z
Ia
Therefore
ZY
1
 Z
3
24
Three Phase Transmission Line
25
Per Phase Analysis
• Per phase analysis allows analysis of balanced 3
systems with the same effort as for a single phase
system
• Balanced 3 Theorem: For a balanced 3 system
with
–
–
All loads and sources Y connected
No mutual Inductance between phases
26
Per Phase Analysis, cont’d
• Then
–
–
–
All neutrals are at the same potential
All phases are COMPLETELY decoupled
All system values are the same sequence as sources. The
sequence order we’ve been using (phase b lags phase a
and phase c lags phase a) is known as “positive”
sequence; later in the course we’ll discuss negative and
zero sequence systems.
27