ECE 333
Renewable Energy Systems
Lecture 4: Three-Phase
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
Announcements
•
•
•
Be reading Chapter 3 from the book
Quiz today on Homework 1
Homework 2 is 2.16, 3.5, 3.8, 3.12, 3.13 It will be
covered by the first in-class quiz on Thursday Feb 5
1
Reactive Power Compensation
•
•
•
•
Reactive compensation is used extensively by
utilities
Capacitors are used to correct the power factor
This allows reactive power to be supplied locally
Supplying reactive power locally leads to decreased
line current, which results in
–
–
–
Decrease line losses
Ability to use smaller wires
Less voltage drop across the line
2
Power Factor Correction Example
• Assume we have a 100 kVA load with pf = 0.8 lagging,
and would like to correct the pf to 0.95 lagging
We have:
1


cos
(0.8)  36.9
S  80  j 60 kVA
We want:
desired  cos 1 (0.95)  18.2
Qdes.
Qdes.=?
tan(18.2) 
P
S
18.2
Qdes.  tan(18.2) * 40  26.3 kVAr
P=80
This requires a capacitance of:
P
Q=60
Qdes=26.3
P
Q=-33.7
Qcap  60  26.3  33.7 kVAr
3
Distribution System Capacitors for
Power Factor Correction
4
Balanced 3 Phase () Systems
• A balanced 3 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
5
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
6
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 constrant
Three phase machines use less material for same
power rating
Three phase machines start more easily than single
phase machines
7
Three Phase Transmission Line
8
Three Phase - Wye Connection
•
There are two ways to connect 3 systems
•
•
Wye (Y)
Delta ()
Wye Connection Voltages
Van
 V  
Vbn
 V   
Vcn
 V   
9
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
10
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
11
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
12
Three Phase Example
Assume a -connected load is supplied from a 3
13.8 kV (L-L) source with Z = 10020W
a
a c
Vab  13.80 kV
b
V  13.8 0 kV
bc
c
b
Vca  13.80 kV
13.80 kV
I ab 
 138  20 amps
 W
I bc  138  140 amps
I ca  1380 amps
13
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
14
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
15
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
16
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.
17
Per Phase Analysis Procedure
To do per phase analysis
1. Convert all  load/sources to equivalent Y’s
2. Solve phase “a” independent of the other phases
3. Total system power S = 3 Va Ia*
4. If desired, phase “b” and “c” values can be
determined by inspection (i.e., ±120° degree phase
shifts)
5. If necessary, go back to original circuit to determine
line-line values or internal  values.
18
Per Phase Example
Assume a 3, Y-connected generator with Van = 10
volts supplies a -connected load with Z = -jW
through a transmission line with impedance of j0.1W
per phase. The load is also connected to a
-connected generator with Va”b” = 10 through a
second transmission line which also has an impedance
of j0.1W per phase.
Find
1. The load voltage Va’b’
2. The total power supplied by each generator, SY and
S
19
Per Phase Example, cont’d
First convert the delta load and source to equivalent
Y values and draw just the "a" phase circuit
20
Per Phase Example, cont’d
To solve the circuit, write the KCL equation at a'
1
'
'
'
(Va  10)(10 j )  Va (3 j )  (Va 
   j  
3
21
Per Phase Example, cont’d
To solve the circuit, write the KCL equation at a'
1
'
'
'
(Va  10)(10 j )  Va (3 j )  (Va 
   j  
3
10
(10 j 
60)  Va' (10 j  3 j  10 j )
3
Va'  0.9  volts
Vb'  0.9  volts
Vc'  0.9 volts
'
Vab
 1.56 volts
22
Per Phase Example, cont’d
*
Sygen  3Va I a
*
'
 Va  Va 
 Va 
  5.1  j 3.5 VA
 j 0.1 
"

V
"
Sgen  3Va  a
' *
 Va
  5.1  j 4.7 VA
 j 0.1 
23
Brief Coverage of Magnetics
Ampere’s circuital law:
F
  H dl  I e
F = mmf = magnetomtive force (amp-turns)
H = magnetic field intensity (amp-turns/meter)
dl = Vector differential path length (meters)

Ie
= Line integral about closed path 
(dl is tangent to path)
= Algebraic sum of current linked by 
24
Line Integrals
•Line integrals are a generalization of traditional
integration
Integration along the
x-axis
Integration along a
general path, which
may be closed
Ampere’s law is most useful in cases of symmetry,
such as with an infinitely long line
25
Magnetic Flux Density
•Magnetic fields are usually measured in terms of flux
density
B = flux density (Tesla [T] or Gauss [G])
(1T = 10,000G)
For a linear a linear magnetic material
B =  H where  is the called the permeability
 = 0  r
0 = permeability of freespace = 4  10-7 H m
 r = relative permeability  1 for air
26
Magnetic Flux
Total flux passing through a surface A is
 =
 A B da
da = vector with direction normal to the surface
If flux density B is uniform and perpendicular to an
area A then
 = BA
27
Magnetic Fields from Single Wire
•Assume we have an infinitely long wire with current
of 1000A. How much magnetic flux passes through a
1 meter square, located between 4 and 5 meters from
the wire?
Direction of H is given
by the “Right-hand” Rule
Easiest way to solve the problem is to take advantage
of symmetry. For an integration path we’ll choose a
circle with a radius of x.
28
Single Line Example, cont’d
2 xH  I  H 
B  0 H
I
2 x
0 I
  A 0 H dA  4
dx
2 x
5
I
5
  0 ln
2 4
For reference, the earth’s
magnetic field is about
0.6 Gauss (Central US)
5
 2  10 I ln
4
7
  4.46  105 Wb
2  104
2
B 
T 
Gauss
x
x
29
Flux linkages and Faraday’s law
Flux linkages are defined from Faraday's law
d
V =
where V = voltage,  = flux linkages
dt
The flux linkages tell how much flux is linking an
N turn coil:
 =
N
i
i=1
If all flux links every coil then   N
30
Inductance
For a linear magnetic system, that is one where
B =H
we can define the inductance, L, to be
the constant relating the current and the flux
linkage
 =Li
where L has units of Henrys (H)
31
Inductance Example
Calculate the inductance of an N turn coil wound
tightly on a torodial iron core that has a radius of R
and a cross-sectional area of A. Assume
1) all flux is within the coil
2) all flux links each turn
32
Inductance Example, cont’d
Ie 
 H dl
NI  H 2 R (path length is 2 R)
NI
H 
2 R
  AB
B   H   r 0 H
  N  LI
NI
  NAB  NAr 0
2 R
N 2 Ar 0
L 
H
2 R
33
Transformers Overview
•
•
•
•
Power systems are characterized by many different
voltage levels, ranging from 765 kV down to
240/120 volts.
Transformers are used to transfer power between
different voltage levels.
The ability to inexpensively change voltage levels
is a key advantage of ac systems over dc systems.
In 333 we just introduce the ideal transformer, with
more details covered in 330 and 476.
34