UNICAMP – UNESP Sorocaba
August 2012
Conservative Power Theory
A theoretical background to understand energy issues of
electrical networks under nonnon-sinusoidal conditions
and
to approach measurement, accountability and control
problems in smart grids
Paolo Tenti
Department of Information Engineering
University of Padova, Italy
1
Seminar Outline
1. Motivation of work
2. Mathematical and physical foundations of the theory
•
•
Mathematical operators and their properties
Instantaneous and average power & energy terms in polyphase networks
3. Definition of current and power terms in single-phase
networks under non-sinusoidal conditions
4. Extension to poly-phase domain: 3-wires / 4-wires
5. Sequence components under non-sinusoidal
conditions
6. Measurement & accountability issues
2
1. Motivation of work
Why do we need to define power terms
Describe physical phenomena
energy transfer,
energy storage,
rate of utilization of power sources and distribution
infrastructure,
unwanted voltage and current terms, ….
Allow unambiguous measurement of quantities
load and source characterization,
revenue metering, …
Compensation
identify provisions which make the equipment or the
plant compliant with standards & regulations in terms
of symmetry, purity of waveforms, power factor …
3
1. Motivation of work
Few basic questions
While the definition and meaning of instantaneous
power and its average value (active power) are
universally agreed, the situation is less clear with other
popular power terms
What is/means reactive power ?
What is/means distortion power ?
What is/means apparent power ?
These power terms are unambiguously defined when at
least the voltage supply is sinusoidal, but are matter of
controversial discussions (since nearly one century) in
case of distorted voltages and currents.
4
1. Motivation of work
Milestones of power theory history
In the frequency domain
Budeanu (1927)
Sheperd & Zakikhani (1971)
Czarnecki (1984 …)
In the time domain
Fryze (1931)
Kusters & Moore (1975)
Depenbrock (1993)
Akagi & Nabae (1983)
No one of these theories was able to target all goals (characterization
of physical phenomena, load & line identification, compensation).
The timetime-domain theory presented here tries to target all goals at the
same time.
It represents an outcome of a long
long--standing cooperation between
UNIPD, UNICAMP and UNESP.
5
5
1. Motivation of work
Need for a revision of power terms
In modern scenarios (e.g., micro-grids) where:
the grid is weak,
frequency may change,
voltages may be asymmetrical,
distortion may affect voltages and currents,
are the usual definitions of reactive, unbalance and
distortion power still valid ?
Which is the physical meaning of such terms ?
Are they useful for compensation ?
To which extent are power measurements affected by
source non-ideality ?
It is possible to identify supply and load responsibility
on voltage distortion and asymmetry at a given network
port ?
6
Conservative
Power Theory
2. Mathematical and physical foundations
•
Definition of mathematical operators and their
properties
•
Definition of instantaneous power and energy terms
•
Conservative quantities
•
Selection of voltage reference
•
Definition of average power terms and their physical
meaning in real networks
7
Mathematical operators for periodic
scalar quantities
Let T be the period of variables x and y, we define:
• Average value
• Time derivative
1
x= x =
T
( dx
x=
dt
∫0 x(t ) dt
T
• Time integral
x∫ = ∫ x (τ ) dτ
• Unbiased time integral
)
x = x∫ − x∫
• Internal product
• Norm (rms value)
• Orthogonality
t
0
1 T
x, y = ∫ x ⋅ y dt
T 0
X = x =
x, x
x, y = 0
8
Mathematical operators for periodic
vector quantities
Let x and y be vector quantities of size N, we define:
• Scalar product
xo y =
N
∑ xn y n
n =1
N
∑ xn2
• Magnitude
x = xo x =
• Internal product
x, y = x o y = ∑ x n , y n
• Norm
• Orthogonality
n =1
N
n =1
X= x =
N
∑
n =1
xn , xn =
N
∑ X n2
n =1
x, y = 0
• The vector norm is also called collective rms value
9
Properties of mathematical operators
(valid for scalar and vector quantities)
The above operators have the following properties:
(
(
x, x = 0
x, x = 0
• Orthogonality
⇒
)
)
x, x = 0
x, x = 0
• Equivalences
(
(
(
(
x
,
y
=
−
x
,y
x, y = − x , y
)
)
)
)
x, y = − x , y
⇒ x, y = − x , y
( )
) (
( )
) (
x, y = − x , y = − x , y
x, y = − x , y = − x , y
• For sinusoidal quantities
1 (
)
X = x =ω x =
x
x, y = XY cosϕ
ω
(2
x
)2
x + ω x = x 2 + 2 = 2X 2
ω
1
)
x , y = XY senϕ
2
2
ω
10
Instantaneous power definitions
(for periodic variables)
Given the vectors of the N phase currents in and voltages
un measured at a generic network port we define:
Instantaneous (active) power:
p = u ⋅i =
N
N
n =1
n =1
∑ u n in = ∑ p n
N
N
)
)
Instantaneous reactive energy w = u ⋅ i = u i = w
∑nn ∑ n
(new definition):
n =1
n =1
• Both quantities do not depend on the voltage reference
• Both quantities are conservative in every real network
11
Conservation of instantaneous
power and reactive energy
For every real network π, let u and i be the vectors of
the L branch voltages and currents, we claim that:
Branch voltages, their time derivative and unbiased
integral are consistent with network π, i.e. they
comply with KLV (Kirchhoff’s law for voltages)
Branch currents, their time derivative and unbiased
integral are consistent with network π, i.e. they
comply with KLC (Kirchhoff’s law for currents)
Thus, according to
Tellegen’s Theorem all
quantities shown here are
conservative
) ( ( )
u ⋅i = u ⋅i = u ⋅i = 0
)
)
u ⋅i = u ⋅i = 0
(
(
u ⋅i = u ⋅i = 0
12
Average power definitions
(valid for periodic quantities)
P = p = u ,i
)
)
Reactive energy: W = w = u ,i = − u ,i
Active power:
Apparent power:
A = u i = UI
Power factor:
P
λ=
A
• All quantities are defined in the time domain.
• Reactive energy is a new definition, whose properties will be
analyzed in the following.
• Active power and reactive energy are conservative
quantities which do not depend on the voltage reference.
• Unlike P and W, apparent power A is non-conservative and
depends on the voltage reference.
13
Skip voltage reference
Selection of voltage reference (1)
Cauchy-Schwartz
inequality:
The equal sign is
possible if:
u, i ≤ u i
u ∝ i
⇒
⇒λ =
u, i = u i
P
A
⇒
≤1
λ =1
We select the voltage reference so as to ensure unity
power factor in case of symmetrical resistive load. This
gives a physical meaning to the apparent power, which
is the maximum active power that a supply line rated
for Vrms Volts and Irms Amps can deliver to a (purely
resistive and symmetrical) load.
14
Selection of voltage reference (2)
N-phase systems without neutral wire
The proportionality condition
u = Ri
between phase voltages and
N
currents for symmetrical resistive
in = 0 ⇒
load determines voltage reference n =1
∑
N
∑ un = 0
n =1
Thus, the voltage reference must be selected to comply
with the zero-sum condition:
N
∑ un = 0
n =1
ref ) = 0
∑ (1un442−4u4
3
N
⇒
n =1
mea sure
un
⇒ u ref
1
=
N
N
∑ un
n =1
measure
This choice minimizes the norm of the voltage vector
15
Selection of voltage reference (3)
N-phase systems without neutral wire
Measurement of voltages and currents
SOURCE
LINE
LOAD
3
i3
u32
uSn
R2
L2
R1
L1
1
i1
RLn
L3
2
u21
n∈ {1,2,3}
R3
LLn
Derivation of phase voltages
1
un =
N
1
u =
2N
2
N
N
∑ j ≠ n un j
j =1
N
∑∑
n =1 j =1
2
j ≠ n un j
u n2
1  N
=
∑
N  j =1
2
u
j≠n n j
2
− u , n = 1÷ N


16
Selection of voltage reference (4)
N-phase systems with neutral wire
In case of symmetrical resistive load the proportionality
condition between phase voltages and currents holds only
if the voltage reference is set to the neural wire.
u ref = u o = 0 ⇒ u n = R in , n = 0 ÷ N
Unity power factor may occur only if the neutral current is
disregarded for apparent power computation (only phase
currents are considered). Thus:
A = P = U I,

U = ∑ U n2  =

n =1

N

2 
U
∑ n I=
n =0

N

2 
I
∑ n ≠
n =1

N

2 
I
∑ n
n =0

N
17
Selection of voltage reference (5)
N-phase systems with neutral wire
Measurement of voltages and currents
SOURCE
LINE
LOAD
3
i3
u30 u20 u10
RLn
R2
L2
R1
L1
1
i1
n∈ {0,1,2,3}
L3
2
i2
vSn
R3
0
LLn
Collective rms voltage
and current
Homopolar voltage
and current
N
∑
U=
I=
U n2
n =1
1
u =
N
z
N
∑ un
n=1
N
∑ I n2
n =1
1
i =
N
z
N
∑ in = −
n =1
io
N
18
Power terms in passive networks
Resistor
u = Ri
i = Gu
PR = u , i = G u
2
=R i
2
)
)
WR = u , i = R i , i = 0
19
Power terms in passive networks
Inductor
(
di
u=L
= Li
dt
)
u
i=
L
)
u
PL = u , i = u ,
=0
L
Inductor
energy
)
WL = u , i = L i, i = L i
1 2
1
ε L = L i ⇒ ε L = ΕL = L i
2
2
2
2
WL
=
2
20
Power terms in passive networks
Capacitor
du
(
i=C
= Cu
dt
)
i
u=
C
)
i
PC = u , i =
,i = 0
C
Capacitor
energy
)
WC = − u , i = − u , C u = −C u
1 2
1
ε C = Cu ⇒ ε C = Ε C = C u
2
2
2
2
WC
=−
2
21
Active and reactive power absorption of a
linear passive network π
L=N+M+K
Remark: Whichever is
the origin of reactive
energy, including
active and nonlinear
loads, it can be
compensated by
reactive elements
with proper energy
storage capability
π
N resistors
M inductors
K capacitors
Total active power and reactive energy
L
N
l =1
n =1
P = ∑ ul , il =∑ PRn = PRto t
L
W =∑
l =1
M
K
M
K
)
ul , il = ∑ WLm + ∑ WCk = 2 ( ∑ Ε Lm − ∑ Ε Ck ) =2 (Ε Lto t − Ε Ctot )
m =1
k =1
m =1
k =1
22
Seminar Outline
1. Motivation of work
2. Mathematical and physical foundations of the theory
•
•
Mathematical operators and their properties
Instantaneous and average power & energy terms in polyphase networks
3. Definition of current and power terms in single-phase
networks under non-sinusoidal conditions
4. Extension to poly-phase domain: 3-wires / 4-wires
5. Sequence components under non-sinusoidal
conditions
6. Measurement & accountability issues
23
Conservative
Power Theory
3. Definition of current and power terms in
single-phase networks under nonsinusoidal conditions
•
•
•
•
•
Orthogonal current decomposition into active, reactive
and void terms
Physical meaning of current terms
Apparent power decomposition into active, reactive and
void terms
Physical meaning of power terms
Application examples
24
Orthogonal current decomposition in
single--phase networks
single
(voltage and current measured at a generic network port)
Current terms
i = ia + ir + iv = ia + ir + isa + isr + i g
14243
iv
• ia
• ir
• iv
• isa scattered active current
• isr scattered reactive current
• ig generated current
active current
reactive current
void current
Orthogonality
Orthogonality:: all terms in the above equations are orthogonal
i = ia + ir + iv
2
2
2
2
= ia + ir + isa + isr + ig
2
2
2
2
2
25
Orthogonal current decomposition in
single--phase networks
single
(voltage and current measured at a generic network port)
Active current: the minimum current (i.e., with minimum
rms value) needed to convey the active power P flowing
through the port
ia =
uu,, i
u
2
P
u = 2 u = Geu
U
u = port voltage
U = rms value of port voltage
Ge = equivalent conductance
Pa = u, ia = Ge u, u = Ge U 2 = P Active current conveys full
active power and zero
)
)
Wa = u , ia = Ge u , u = 0
reactive energy
26
Orthogonal current decomposition in
single--phase networks
single
(voltage and current measured at a generic network port)
Reactive current: the minimum current needed to
convey the reactive energy W flowing through the port
)
u, i ) W )
)
ir = ) 2 u = ) 2 u = Be u
U
u
Be = equivalent reactivity
)
Reactive current conveys full
Pr = u, ir = Be u, u = 0
reactive energy and no active
)2
)
) )
Wr = u , ir = Be u , u = BeU = W power
)
ia , ir = Ge Be u, u = 0
Active and reactive current
are orthogonal
27
Orthogonal current decomposition in
single--phase networks
single
(voltage and current measured at a generic network port)
Void current: is the remaining current component
iv = i − ia − ir
Void current is not conveying active power or reactive energy
Pv = u, iv = u, i − u, ia − u, ir = P − Pa − Pr = 0
)
)
)
)
Wv = u , iv = u , i − u , ia − u , ir = W − Wa − Wr = 0
Void current is orthogonal to active and reactive terms
iv , ia = Ge iv , u = Ge Pv = 0
)
iv , ir = Be iv , u = BeWv = 0
28
Orthogonal current decomposition in
single--phase networks
single
(voltage and current measured at a generic network port)
The void current reflects the presence of scattered active,
scattered reactive and load-generated harmonic terms
iv = isa + isr + ig
Scattered current terms:
Account for different values of
equivalent admittance at
different harmonics
isa , isr = isa , i g = isr , i g = 0
Load-generated current
harmonics:
Harmonic terms that exist in
currents only, not in voltages
Scattered and load-generated
harmonic currents are
orthogonal
Skip void current components
29
Orthogonal current decomposition in
single--phase networks
single
Scattered active current
For each co
co--existing harmonic components of voltage
and current we define:
define:
Harmonic active current terms
iak =
uk , ik
uk
2
Pk
I k cos ϕ k
uk = 2 uk =
uk = Gk u k
Uk
Uk
Total harmonic active current
iha = ∑ iak
Pha =
k∈K
∑ Pk = Pa = P,
Wha = 0
k∈K
Scattered active current
isa = iha − ia = ∑ (Gk − Ge )uk
k∈K
Psa = Pha − Pa = 0, Wsa = 0
30
Orthogonal current decomposition in
single--phase networks
single
Scattered reactive current
For each co
co--existing harmonic components of voltage
and current we define:
define:
Harmonic reactive current terms
)
u k , ik )
Wk )
ω k I k sin ϕ k )
)
irk =
)
uk
2
uk = ) 2 uk =
Uk
Uk
u k = Bk u k
Total harmonic reactive current
Whr = ∑ Wk = Wr = W , Phr = 0
ihr = ∑ irk
k∈K
k∈K
Scattered reactive current
)
isr = ihr − ir = ∑ (Bk − Be )u k
k∈K
Wsr = Whr − W = 0, Psr = 0
31
Apparent power decomposition in
single--phase networks
single
A = u i = U I = P2 + Q2 +V 2
Active power:
P = u ia = U I a
Reactive power: Q = u ir = U I r
Void power:
V = u iv = U I v = S a2 + S r2 + V g2
Scattered active power:
S a = u isa = U I sa
Scattered reactive power:
S r = u isr = U I sr
Load-generated harmonic V = u i = U I
g
g
g
power:
32
Reactive Power
U and Û can be decomposed in U = U 2 + U 2 = U 1 + [THD(u )] 2
f
h
f
fundamental and harmonic
)
)2 )2 )
components
) 2
(THD means total harmonic distortion)
U = U f + U h = U f 1 + [THD(u )]
)
Uf =ω
Recalling that:
Uf
We have:
1 + [THD (u )]2
U
Q = U Ir = ) W = ω W
)
U
1 + [THD (u )]2
Note that, unlike reactive energy W, REACTIVE POWER Q
IS NOT CONSERVATIVE. In fact, it depends on line
frequency and (local) voltage distortion.
Under sinusoidal conditions, the definition of Q
coincides with the conventional one
33
Void Power Terms
Void Power:
V = U I v = S a2 + S r2 + Vg2
Scattered active power:
2
S a = U I sa
 Pk
P  2
2

= U
−
U
U2 U2  k
k∈{K } k

∑
Scattered reactive power:
S r = U I sr = ω
1 + THD u2
1 + THD u)2
)2
U
2
 Wk W  ) 2
 ) − )  Uk
 2 U2 
k∈{K } U k

∑
Load
Load--generated harmonic power:
Vg = U I g
Skip examples
34
Application Examples
Example # 1
Voltage and Current : Resistive Load
1.5
1.5
1
1
u(t)
u P C C i P C C [p u ]
u P C C i P C C [p u ]
u(t)
0.5
i(t)
0
-0.5
0.5
-0.5
-1
-1
-1.5
0.1
-1.5
0.1
0.11
0.12
0.13
0.14
0.15
0.16
i(t)
0
0.11
0.13
0.14
0.15
0.16
T im e [s]
T im e [s]
Sinusoidal voltage
0.12
Non – sinusoidal voltage
Current = ipu(t)/2
35
Application Examples
Example # 1
Conservative Power Terms:
Terms: Resistive Load
1
P = p = UIcosϕ
p(t) = u(t)i(t)
1
P = p
p(t) = u(t)i(t)
0.5
[p u ]
[p u ]
0.5
0
0
-0.5
-0.5
0.1
q(t) = ω û(t)i(t)
0.105
0.11
0.115
Q = ω w(t) = UIsinϕ
0.12
0.125
T im e [s]
Sinusoidal voltage
0.13
0.135
0.1
q(t) = ω û(t)i(t)
0.105
0.11
0.115
Q
0.12
0.125
0.13
0.135
T im e [s]
Non – sinusoidal voltage
This example shows the correspondences between the
CPT theory and conventional theory
36
Application Examples
Example # 1 – SingleSingle-phase
Current Terms:
Terms: Resistive Load
1
0
-0.5
-1
0.1
0.11
0.12
0.13
0.14
0.15
0.16
i a [p u]
1
0.5
-0.5
-1
0.1
0.11
0.12
0.13
0.14
0.15
Reactive
current
i r (t)= 0
i r [pu ]
0.25
0
-0.25
-0.5
0.1
0.11
0.12
0.13
0.14
0.15
-1
0.1
0.11
0.12
0.13
0.14
0.15
0.16
0.11
0.12
0.13
0.14
0.15
0.16
0.11
0.12
0.13
0.14
0.15
0.16
0.11
0.12
0.13
0.14
0.15
0.16
1
0.5
0
-0.5
0.25
0
-0.25
-0.5
0.1
0.16
0.5
0
-0.25
0.11
0.12
0.13
0.14
T im e [s]
Sinusoidal voltage
0.15
0.16
Void
current
i v (t)= 0
0.25
i v [pu ]
0.25
i v [p u]
-0.5
0.5
0.5
-0.5
0.1
0
-1
0.1
0.16
0.5
0.5
i PCC (t) = ia(t)
Active
current
0
i PCC [p u]
PCC
current
i a [pu ]
0.5
i r [pu ]
i PCC [p u]
1
0
-0.25
-0.5
0.1
T im e [s]
Non – sinusoidal voltage
37
Application Examples
1.5
1.5
1
1
u(t)
0.5
i(t)
u P C C i P C C [p u ]
u P C C iP C C [ p u ]
Example # 2
Voltage and Current : OhmicOhmic-inductive Load
0
-0.5
i(t)
0.5
0
-0.5
-1
-1
-1.5
0.3
u(t)
0.31
0.32
0.33
0.34
T im e [s]
Sinusoidal voltage
0.35
0.36
-1.5
0.3
0.31
0.32
0.33
0.34
0.35
0.36
T im e [s]
Non – sinusoidal voltage
38
Application Examples
Example # 2
Conservative Power Terms
Terms:: OhmicOhmic-inductive Load
P = p = UIcosϕ
p(t) = u(t)i(t)
1
0.5
[p u ]
[p u ]
1
P = p
q(t) = ω û(t)i(t)
Q
0.5
0
0
q(t) = ω û(t)i(t)
0.3
p(t) = u(t)i(t)
0.305
0.31
Q = ω w = UIsinϕ
0.315
0.32
0.325
T im e [s]
Sinusoidal voltage
0.33
0.335
0.3
0.305
0.31
0.315
0.32
0.325
0.33
0.335
T im e [s]
Non – sinusoidal voltage
This example shows the correspondence between CPT
and conventional theory under sinusoidal conditions
39
Application Examples
1
1
0.5
0.5
PCC
current
0
-0.5
-1
0.3
0.31
0.32
0.33
0.34
0.35
i PC C [p u ]
i P C C [p u ]
Example # 2
Current Terms
Terms:: OhmicOhmic-inductive Load
-0.5
0.31
0.32
0.33
0.34
0.35
i a [p u ]
i a [p u ]
Active
current
0
Reactive
current
0
-0.5
0.31
0.32
0.33
0.34
0.35
i r [p u ]
i r [p u ]
0.33
0.34
0.35
0.36
0.31
0.32
0.33
0.34
0.35
0.36
0.31
0.32
0.33
0.34
0.35
0.36
0.31
0.32
0.33
0.34
0.35
0.36
0
-0.5
1
0.5
0.5
0
-0.5
-1
0.3
0.36
0.5
0.5
0
Void
current
-0.25
0.31
0.32
0.33
0.34
T im e [s]
Sinusoidal voltage
0.35
0.36
0.25
i v [p u ]
i v (t)= 0
0.25
i v [p u ]
0.32
0.5
-1
0.3
0.36
1
-0.5
0.3
0.31
1
0.5
-1
0.3
-0.5
-1
0.3
0.36
1
-1
0.3
0
0
-0.25
-0.5
0.3
T im e [s]
Non – sinusoidal voltage
40
Physical meaning of void current
Example # 2
Void Current Terms:
Terms: OhmicOhmic-inductive Load
0.2
i v [pu]
0.1
0
Void current
-0.1
-0.2
0.3
0.31
0.32
0.33
0.34
0.35
0.36
i sa [pu]
0.2
0.1
Scattered active current
0
-0.1
-0.2
0.3
0.31
0.32
0.33
0.34
0.35
0.36
i sr [pu]
0.2
0.1
0
Scattered reactive current
-0.1
-0.2
0.3
0.31
0.32
0.33
0.34
0.35
0.36
0.2
i g [pu]
0.1
Load-generated
harmonic current
0
-0.1
-0.2
0.3
0.31
0.32
0.33
0.34
0.35
0.36
T im e [s]
Non – sinusoidal voltage
41
Application Examples
Example # 3
Voltage and Current
Distorting Load
1.5
Q=ωw
1
u(t)
P = p
0.5
i(t)
0
[p u ]
u P C C i P C C [p u ]
1
Conservative Power Terms
Distorting Load
0.5
-0.5
-1
-1.5
0.3
0
0.305
0.31
0.315
0.32
0.325
0.33
0.335
0.3
q(t) = ω û(t)i(t)
0.305
0.31
0.315
p(t) = u(t)i(t)
0.32
0.325
0.33
0.335
T im e [s]
T ime [s]
Non – sinusoidal voltage
42
Physical meaning of void current
Void Current Terms:
Terms: OhmicOhmic-inductive Load
i PCC [pu ]
1
0.5
PCC current
0
-0.5
-1
0.3
0.31
0.32
0.33
0.34
0.35
0.36
i a [pu]
1
0.5
Active current
0
-0.5
-1
0.3
0.31
0.32
0.33
0.34
0.35
0.36
1
i r [pu ]
0.5
Reactive current
0
-0.5
-1
0.3
0.31
0.32
0.33
0.34
0.35
0.36
i v [p u]
1
0.5
Void current
0
-0.5
-1
0.3
0.31
0.32
0.33
0.34
0.35
0.36
T im e [s]
Non – sinusoidal voltage
43
Physical meaning of void current
Void Current Terms:
Terms: OhmicOhmic-inductive Load
0.8
i v [pu]
0.4
Void current
0
-0.4
-0.8
0.3
0.31
0.32
0.33
0.34
0.35
0.36
i sa [pu]
0.8
0.4
Scattered active current
0
-0.4
-0.8
0.3
0.31
0.32
0.33
0.34
0.35
0.36
i sr [pu]
0.8
0.4
Scattered reactive current
0
-0.4
-0.8
0.3
0.31
0.32
0.33
0.34
0.35
0.36
0.8
i g [pu]
0.4
Load-generated
harmonic current
0
-0.4
-0.8
0.3
0.31
0.32
0.33
0.34
0.35
0.36
T im e [s]
Non – sinusoidal voltage
44
Seminar Outline
1. Motivation of work
2. Mathematical and physical foundations of the theory
•
•
Mathematical operators and their properties
Instantaneous and average power & energy terms in polyphase networks
3. Definition of current and power terms in single-phase
networks under non-sinusoidal conditions
4. Extension to poly-phase domain: 3-wires / 4-wires
5. Sequence components under non-sinusoidal
conditions
6. Measurement & accountability issues
45
Conservative
Power Theory
4. Extension to poly-phase domain:
3-wires / 4-wires
•
•
•
•
Orthogonal current decomposition into active, reactive,
unbalance and void terms
Physical meaning of current terms
Active power decomposition into active, reactive,
unbalance and void terms
Physical meaning of power terms
46
Orthogonal current decomposition
Extension to polypoly-phase: 33-wires / 4
4--wires
In poly-phase systems, the current components (active,
reactive and void) can be defined for each phase:
Active current
ian =
u n , in
un
2
un =
Pn
U n2
u n = Gn u n , n = 1 ÷ N
Pa = u , i a = P
)
Wa = u , i a = 0
Gn= equivalent phase conductance U I = u i ≠ P
a
a
Reactive current
)
u n , in ) Wn )
)
irn = ) 2 u n = ) 2 u n = Bn u n , n = 1 ÷ N
Un
un
Void current
Bn = equivalent phase reactivity
Pr = u, i r = 0
)
Wr = u , i r = W
)
)
U Ir = u ir ≠ W
Pv = u, i v
)
= 0, Wr = u , i v = 0
ivn = in − ian − irn , n = 1 ÷ N
U Iv > 0
47
Orthogonal current decomposition
Extension to polypoly-phase: 33-wires / 4
4--wires
Active and reactive current terms can also be defined
collectively, i.e., by making reference to an equivalent
balanced load absorbing the same active power and
reactive energy of actual load:
Balanced Active currents: minimum collective
currents needed to convey active power P
i ba
=
u, i
u
2
u=
P
U
2
u = Gb u
Gb= equivalent balanced conductance
U I ab = u i ba = P, Qab = 0
Balanced Reactive currents: minimum collective
currents needed to convey reactive energy W
)
u, i ) W )
b
b)
ir = ) 2 u = ) 2 u = B u
U
u
Bb = equivalent balanced reactivity
) b ) b
U I r = u i r = W , Prb = 0
48
Orthogonal current decomposition
Extension to polypoly-phase: 33-wires / 4
4--wires
Unbalanced currents account for the asymmetrical
behavior of the various phases
Unbalanced Active currents
i ua = i a − i ba
(
)
u
⇒ ian
= Gn − G b u n , n = 1 ÷ N
Pau = Pa − Pab = 0
Wau = 0
Unbalanced Reactive currents
(
)
)
u
i ur = i r − i br ⇒ irn
= Bn − B b u n , n = 1 ÷ N
Pru = 0
Wru = Wr − Wrb = 0
49
Orthogonal current decomposition
Extension to polypoly-phase: 33-wires / 4
4--wires
Void currents: as for single-phase systems, they
reflect the presence of scattered active, scattered
reactive and generated terms.
iv = i − ia − ir = i + i + i g
s
a
Scattered current terms:
Account for different values of
equivalent admittance at
different harmonics
s
r
Load-generated harmonic
current:
Harmonic terms that exist in
currents only, not in voltages
Pv = P − Pa − Pr = 0
Wv = W − Wa − Wr = 0
50
Orthogonal current decomposition
Extension to polypoly-phase: 33-wires / 4
4--wires
Summary of current decomposition
i = ia + ir + iv = i + i + i + i + i + i + i g
b
a
u
a
b
r
u
r
s
a
s
r
i a active currents
• i ab balanced active currents
• i au unbalanced active currents
i r reactive currents
• i rb balanced reactive currents
• i ru unbalanced reactive currents
i v void currents
• i as scattered active currents
• i rs scattered reactive currents
• i g load-generated harmonic currents
51
Orthogonal current decomposition
Extension to polypoly-phase: 33-wires / 4
4--wires
Summary of current decomposition
i = ia + ir + iv = i + i + i + i + i + i + i g
123 123 14243
b
a
u
a
b
r
u
r
s
a
ir
ia
s
r
iv
Each current component has a precise
PHISICAL MEANING and is computed in the time domain
Moreover, all current terms defined in the above equation
are ORTHOGONAL
ORTHOGONAL,, thus:
i = ia + ir + iv
2
2
2
2
= i
b 2
a
+ i
u 2
a
+ i
b 2
r
+ i
u 2
r
+ i
s 2
a
+ i
s 2
r
+ ig
2
52
Apparent power decomposition in
poly--phase: 3poly
3-wires / 44-wires
A = U I = u i = P2 + Q2 + N 2 +V 2
Active power:
P = U I ab = u i ba
Reactive power:
Q = U I rb = u i br
Unbalance power
power:: N = U I u = u i u = N a2 + N r2
Void power:
V = U I v = u i v = S a2 + S r2 + Vg2
53
Unbalance Power Terms
Unbalance power
power::
N=
2
Na
+
2
Nr
Unbalance Active Power
N a = U I au = u i ua = U
N
Pn
P2
∑U 2 − U2
n =1
n
Unbalance Reactive Power
2
1
+
[
THD
(
u
)
]
N r = U I ru = ω U
) 2
1 + [THD (u )]
N
Wn W 2
∑ U) 2 − U) 2
n =1 n
Unbalance active and reactive power vanish
if the load is balanced
Skip examples
54
Application Examples
Example # 1 : 3
3--phase 33-wire – Balanced load
(Resistive)
Voltage and Current
Conservative Power Terms
1.5
ia ib
p (t) & P ; w (t) & W [p u ]
0.5
ua ub uc
ia
0
-0.5
u
PCC
& i
PCC
[p u ]
1
-1
-1.5
0.3
0.31
0.32
0.33
0.34
0.35
0.36
1
0.5
0
-0.5
0.3
0.31
T im e [s]
Sinusoidal voltage
0.32
0.33
0.34
0.35
0.36
T im e [s]
Current = ipu(t)/2
55
Application Examples
Example # 1 : 3
3--phase 33-wire – Balanced load
1
0.5
0
-0.5
-1
0.3
1
0.5
0
-0.5
Balanced active currents
0.31
0.32
0.33
0.34
0.35
0.36
i r (t)= 0
Balanced reactive currents
-1
0.3
1
0.5
0.31
0.32
0.34
0.35
0.36
i ua (t)= 0
0
-0.5
-1
0.3
1
0.5
0
-0.5
0.33
0.31
0.32
0.33
0.34
Unbalanced active currents
0.35
0.36
i ur (t)= 0
Unbalanced reactive currents
i
vn
[p u ]
i
u
r n
[p u ]
i
u
an
[p u ]
i
b
r n
[p u ]
i
b
an
[p u ]
(Resistive)
-1
0.3
1
0.5
0
-0.5
-1
0.3
0.31
0.32
0.33
0.34
0.35
0.36
i v (t)= 0
0.31
0.32
0.33
0.34
Void currents
0.35
0.36
T im e [s]
56
Application Examples
Example # 1 : 3
3--phase 33-wire – Unbalanced load
(resistor connected between two phases)
Voltage and Current
Conservative Power Terms
1.5
& i
ia
ia
ib
-0.5
u
PCC
0
ua ub uc
p (t) & P ; w (t) & W [p u ]
0.5
PCC
[p u ]
1
-1
-1.5
0.3
0.31
0.32
0.33
T im e [s]
0.34
0.35
0.36
1
0.5
0
-0.5
0.3
0.31
0.32
0.33
0.34
0.35
0.36
T im e [s]
Sinusoidal voltage
Current = ipu(t)/2
57
Application Examples
Example # 1 : 3
3--phase 33-wire – Unbalanced load
i
vn
[p u ]
i
u
r n
[p u ]
i
u
an
[p u ]
i
b
r n
[p u ]
i
b
an
[p u ]
(resistor connected between two phases)
1
0.5
0
-0.5
-1
0.3
1
0.5
0
-0.5
-1
0.3
1
0.5
0
-0.5
-1
0.3
1
0.5
0
-0.5
-1
0.3
1
0.5
0
-0.5
-1
0.3
Balanced active currents
0.31
0.32
0.33
0.34
0.35
0.36
i r (t)= 0
0.31
0.32
0.33
0.34
0.35
Balanced reactive currents
0.36
Unbalanced active currents
0.31
0.32
0.33
0.34
0.35
0.36
Unbalanced reactive currents
0.31
0.32
0.33
0.34
0.35
0.36
i v (t)= 0
0.31
0.32
0.33
0.34
0.35
Void currents
0.36
T im e [s]
58
Application Examples
Example # 3 : 3
3--phase 33-wire
Three--phase RL + SingleThree
Single-phase R load
Voltage and Current
Conservative Power Terms
1.5
& i
ia
ib
ia
0
-0.5
u
PCC
ua ua uc
p ( t) & P ; w (t ) & W [p u ]
0.5
PCC
[p u ]
1
-1
-1.5
0.3
0.31
0.32
0.33
0.34
0.35
0.36
1
0.5
0
-0.5
0.3
0.31
0.32
T im e [s]
0.33
0.34
0.35
0.36
T ime [s]
Symmetrical nonnon-sinusoidal voltage
59
1
0.5
0
-0.5
-1
0.3
0.5
0.25
0
-0.25
-0.5
0.3
0.5
0.25
0
-0.25
-0.5
0.3
0.5
0.25
0
-0.25
-0.5
0.3
0.5
0.25
0
-0.25
-0.5
0.3
Example # 3 : 3
3--phase 33-wire
Three--phase RL + SingleThree
Single-phase R load
Balanced active currents
i
vn
[p u ]
i
u
r n
[p u ]
i
u
an
[p u ]
i
b
r n
[p u ]
i
b
an
[p u ]
Application Examples
0.31
0.32
0.33
0.34
0.35
0.36
Balanced reactive currents
0.31
0.32
0.33
0.34
0.35
0.36
Unbalanced active currents
0.31
0.32
0.33
0.34
0.35
0.36
Unbalanced reactive currents
0.31
0.32
0.33
0.34
0.35
0.36
Void currents
0.31
0.32
0.33
0.34
0.35
0.36
T im e [s]
Symmetrical nonnon-sinusoidal voltage
60
Sharing of compensation duties
Orthogonal current terms
terms::
i = ia + ir + iv =
b
u
ia + ia
123
ia
b
u
+ i r + i r + i sa
123
ir
+ i sr + i g
14243
iv
Each current component has
a precise PHYSICAL MEANING
Balanced Active currents convey active power P
Balanced Reactive currents convey reactive power Q
Unbalanced Active and Reactive currents account for
asymmetrical behavior of the various phases
Void currents reflect the presence of different behavior at
different frequencies and/or generated current harmonics
61
Sharing of compensation duties
i = ia + ir + iv = i + i + i + i + iv
b
a
Reactive
compensation
b
r
u
a
Unbalance
compensation
requires controllable
reactances (extended
Steinmetz approach)
Stationary Compensators
Quasi(reactive impedances)
Stactionary
&
Compensators
Quasi-Stationary
(SVC)
Compensators (SVC,
Static VAR Compensators)
u
r
Harmonic
compensation
requires highfrequency response
Passive filters &
Switching Power
Compensators
(SPC=APF+SPI)
62
Effect of compensation on
power terms
b
P
=
U
I
Active power (balanced):
(balanced):
a
) b
Reactive power (balanced)
(balanced):: Q = U I r
Unbalance power:
power:
N = UI = U
u
u2
Ia
Compensation
SVC

→ 0
u2
+ Ir
V = U Iv
Void power:
power:
SVC

→ 0
SPC

→ 0
APPARENT POWER
X2 + N
X2 +V
X2
A = U I = P2 + Q
,SPC
SVC

→
A = U I ba = P
63
Seminar Outline
1. Motivation of work
2. Mathematical and physical foundations of the theory
•
•
Mathematical operators and their properties
Instantaneous and average power & energy terms in polyphase networks
3. Definition of current and power terms in single-phase
networks under non-sinusoidal conditions
4. Extension to poly-phase domain: 3-wires / 4-wires
5. Sequence components under non-sinusoidal
conditions
6. Measurement & accountability issues
64
Conservative
Power Theory
5. Sequence components under nonsinusoidal conditions
1. Problem statement
2. Goal of decomposition
3. Derivation of generalized symmetrical components in
the time domain (extension of Fortescue’s approach)
4. Analysis of generalized symmetrical components in
the frequency domain
5. Orthogonality of sequence components
6. Application examples
Skip
65
1. Problem statement (1)
Symmetrical components are very useful to
simplify the analysis of three-phase networks
under sinusoidal conditions
It is important to extend the definition and
application of symmetrical components to
non-sinusoidal periodic operation
66
1. Problem statement (2)
Given periodic three-phase variables fa(t), fb(t), fc(t) of
period T we define the following symmetry
properties:
Homopolarity (zero symmetry)
f a (t ) = f b (t ) = f c (t )
Positive (direct) symmetry
 T
 2T 
f a (t ) = f b  t +  = f c  t +

3 
 3

Negative (inverse) symmetry
 T
 2T 
f a (t ) = f b  t −  = f c  t −

3 
 3

67
2. Goal of decomposition
Given a set of generic three-phase variables:
f a (t )
f = f b (t )
f c (t )
we decompose them in the orthogonal form:
f = fz+fh= fz+f
p
+fn+ fr
where:
f z are zero-sequence (homopolar) components
f h are non-zero sequence (heteropolar) components
f p are positive-sequence components
f n are negative-sequence components
f r are residual components
68
3. Derivation of generalized
symmetrical components
Zero sequence (homopolar) component
1
f = f
z
z
(t ) 1
1
f a (t ) + f b (t ) + f c (t )
f (t ) =
3
z
Heteropolar components
fh
f ah (t )
f a (t ) − f z (t )
= f − f z = f bh (t ) = f b (t ) − f z (t )
f ch (t )
f c (t ) − f z (t )
69
3. Derivation of generalized
symmetrical components
Positive sequence component
1 h
 2T
 T
f (t ) =  f a (t ) + f bh  t +  + f ch  t +
3
3
 3

p
f ap
(t ) =
f
p
(t ),
f bp
(t ) =
p
T
f  t − ,
 3



fc
p
(t ) =
p
2T 
f t −

3 

Negative sequence component
1 h
 T
 2T
f (t ) =  f a (t ) + f bh  t −  + f ch  t −
3
3
 3

n
f an
(t ) =
f
n
(t ),
f bn
n
T
(t ) = f  t + ,
 3



f cn
n
2T 
(t ) = f  t + 
3 

70
3. Derivation of generalized
symmetrical components
Residual components
 T
 2T 
f ah (t ) + f ah  t −  + f ah  t −

3 
 3

3
 T
 2T 
f ar (t )
f bh (t ) + f bh  t −  + f bh  t −

3 
 3

f r = f br (t ) =
3
f cr (t )
 T
 2T 
f ch (t ) + f ch  t −  + f ch  t −

3
3




3
Note: these components are computed independently
for each phase and vanish in sinusoidal operation
71
3. Derivation of generalized
symmetrical components
Resulting decomposition
f a (t )
f =
f b (t )
f c (t )
f
z
(t ) + f p (t ) + f n (t ) + f ar (t )
= f
z
(t ) + f p  t − T  + f n  t + T  + f br (t )
 3
 3
2T 
 2T 
n
r
f z (t ) + f p  t −
 + f t +
 + f c (t )
3 
3 


72
4. Analysis in the frequency domain
Expressing variables fa(t), fb(t), fc(t) in Fourier series:
∞
∞
k =1
∞
k =1
∞
k =1
∞
k =1
∞
k =1
k =1
f a (t ) = ∑ f ak (t ) = ∑ 2 Fak sin (kω t + α ak )
f b (t ) = ∑ f bk (t ) = ∑ 2 Fbk sin (kω t + α bk )
f c (t ) = ∑ f ck (t ) = ∑ 2 Fck sin (kω t + α ck )
we can determine, for each harmonic, the zero, positive,
and negative components f zk(t), f pk(t), f nk(t). Instead,
residual harmonic components are zero because
harmonic quantities are sinusoidal.
73
4. Analysis in the frequency domain
Contribution of harmonic sequence components to
generalized sequence components
Harmonic order : k = 3m + 1
f kp ⇒ f p ,
f kn ⇒ f n ,
Harmonic order : k = 3m + 2
f kp ⇒ f n ,
f kn ⇒ f p ,
Harmonic order : k = 3m
f kp ⇒ f r ,
∀m ∈ [0, ∞ ]
f kz ⇒ f
z
∀m ∈ [0, ∞ ]
f kz ⇒ f z
∀m ∈ [0, ∞]
f kn ⇒ f r ,
f kz ⇒ f
z
74
5. Orthogonality of components
Given two sets of three-phase quantities f and g,
their sequence components obey the following
general rules:
Scalar product
f z o gh = f z o g p = f z o gn = f z o gr = 0
Internal product
f p, gn = f p, gr = f n, gr = 0
f , g = f z, g z + f h, gh = f z, gz + f p, g p + f n, gn + f r, gr
Norm
f
2
= f
z
2
+ f
h
2
= f
z
2
+ f
p
2
+ f
n
2
+ f
r
2
75
6. Application Example
1
1
1
1
+ sin( 3ωt )
+ sin( 5ωt )
+ sin( ωt ) + ;
3
5
10
10
1
2π
1
2π
1
1
) + sin( 3ωt +
) + sin( 5ωt −
) + sin( ωt ) + ;
10
3
3
5
3
10
1
2π
1
2π
1
1
) + sin( 3ωt −
) + sin( 5ωt +
) + sin( ωt ) + ;
3
3
5
3
10
10
va ( t ) = sin( ωt )
2π
3
2π
vc ( t ) = sin( ωt +
3
vb ( t ) = sin( ωt −
Line to Neutral Voltages
1.5
1
V
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
76
6. Application Example
va ( t ) = sin( ωt )
2π
3
2π
vc ( t ) = sin( ωt +
3
vb ( t ) = sin( ωt −
1
1
1
1
+ sin( 3ωt )
+ sin( 5ωt )
+ sin( ωt ) + ;
3
5
10
10
1
2π
1
2π
1
1
) + sin( 3ωt +
) + sin( 5ωt −
) + sin( ωt ) + ;
10
3
3
5
3
10
1
2π
1
2π
1
1
) + sin( 3ωt −
) + sin( 5ωt +
) + sin( ωt ) + ;
3
3
5
3
10
10
Generalized Zero Sequence Voltages
1.5
1
V
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
77
6. Application Example
va ( t ) = sin( ωt )
2π
3
2π
vc ( t ) = sin( ωt +
3
vb ( t ) = sin( ωt −
1
1
1
1
+ sin( 3ωt )
+ sin( 5ωt )
+ sin( ωt ) + ;
3
5
10
10
1
2π
1
2π
1
1
) + sin( 3ωt +
) + sin( 5ωt −
) + sin( ωt ) + ;
10
3
3
5
3
10
1
2π
1
2π
1
1
) + sin( 3ωt −
) + sin( 5ωt +
) + sin( ωt ) + ;
3
3
5
3
10
10
Generalized Direct Sequence Voltages
1.5
1
V
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
78
6. Application Example
va ( t ) = sin( ωt )
2π
3
2π
vc ( t ) = sin( ωt +
3
vb ( t ) = sin( ωt −
1
1
1
1
+ sin( 3ωt )
+ sin( 5ωt )
+ sin( ωt ) + ;
3
5
10
10
1
2π
1
2π
1
1
) + sin( 3ωt +
) + sin( 5ωt −
) + sin( ωt ) + ;
10
3
3
5
3
10
1
2π
1
2π
1
1
) + sin( 3ωt −
) + sin( 5ωt +
) + sin( ωt ) + ;
3
3
5
3
10
10
Generalized Inverse Sequence Voltages
1.5
1
V
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
79
6. Application Example
va ( t ) = sin( ωt )
2π
3
2π
vc ( t ) = sin( ωt +
3
vb ( t ) = sin( ωt −
1
1
1
1
+ sin( 3ωt )
+ sin( 5ωt )
+ sin( ωt ) + ;
3
5
10
10
1
2π
1
2π
1
1
) + sin( 3ωt +
) + sin( 5ωt −
) + sin( ωt ) + ;
10
3
3
5
3
10
1
2π
1
2π
1
1
) + sin( 3ωt −
) + sin( 5ωt +
) + sin( ωt ) + ;
3
3
5
3
10
10
Residual Voltages
1.5
1
V
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
80
9. Summary - 1
An extension of the sequence components in case of
non-sinusoidal periodic operation has been proposed.
It has been shown that three-phase currents (or
voltages) cannot always be derived from generalized
positive sequence, generalized negative sequence and
generalized zero-sequence components. A residual
component may be required.
To compute the generalized positive and negative
sequence components, the zero-sequence components
should first be subtracted from the phase quantities, in
contrast with the sinusoidal case where this is not
necessary.
In the sinusoidal case the residual component is absent
and the other components reduce to the classical
symmetrical components.
81
9. Summary - 2
The generalized positive sequence, negative sequence,
and zero-sequence components have complete phase
symmetry. This implies that the three-phase analysis can
be reduced to a single-phase analysis.
The residual components do not have the same
symmetry, and the corresponding three-phase analysis
cannot be reduced to single-phase analysis. It
corresponds to a periodic time function in each of the
three phases with a period which is 1/3 of the line period;
this simplifies the analysis because only 1/3 of the period
must be studied.
82
1.5
Voltage
phase a
1
0.5
V
0
-0.5
-1
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
1.5
0.5
0
V
Homopolar
component
-0.5
-1
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
1.5
Voltage
phase c
1
0.5
0
V
Derivation of
homopolar
component
Voltage
phase b
1
-0.5
-1
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
83
1.5
1.5
Voltage
phase a
1
Opposite of the
homopolar
component
1
0.5
0
0
V
V
0.5
-0.5
-0.5
-1
-1
.
-1.5
0
1.5
4
6
8
10
ms
12
14
16
18
20
-1.5
0
1
0.5
0.5
0
0
6
8
ms
10
12
14
16
18
20
Opposite of the
homopolar
component
V
-0.5
-0.5
.
Voltage
phase b
-1
2
4
6
8
10
ms
12
14
16
-1
-1.5
18
0
20
2
4
6
8
ms
10
12
14
16
18
20
1
0.5
0.5
0
0
V
1.5
1
Opposite of the
homopolar
component
-0.5
-0.5
Voltage
phase c
-1
-1.5
0
4
V
1
-1.5
0
1.5
2
1.5
V
Derivation of
heteropolar
component
2
.
-1
-1.5
2
4
6
8
10
ms
12
14
16
18
20
0
2
4
6
8
ms
10
12
14
16
18
20
84
1.5
1.5
1
1
0.5
0.5
0
V
V
0
-0.5
-0.5
Voltage
phase a
-1
-1
-1.5
-1.5
20
30
ms
40
50
1.5
1
1
0.5
0.5
0
0
V
1.5
-0.5
-0.5
-1
-1
-1.5
0
10
20
30
ms
40
50
60
4
6
8
10
ms
12
14
16
18
20
18
20
Positive
sequence component
-1.5
0
1.5
1.5
1
1
0.5
0.5
0
0
V
Voltage
phase c
2
60
2
4
6
8
10
12
14
16
ms
V
Voltage
phase b
10
V
Derivation of positive
sequence component
(phase a)
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
0
10
20
30
ms
40
50
60
0
2
4
6
8
10
ms
12
14
16
18
85
20
1.5
1.5
1
1
0.5
0.5
0
V
V
0
-0.5
-0.5
Voltage
phase a
-1.5
-1.5
0
10
20
30
ms
40
50
60
1
1
0.5
0.5
0
0
V
1.5
-0.5
-0.5
-1
-1
0
10
20
30
ms
40
50
60
2
4
6
8
12
14
16
18
20
Negative
sequence component
-1.5
0
2
4
6
8
2
4
6
8
1.5
1.5
10
ms
1.5
-1.5
10
12
14
16
18
20
10
12
14
16
18
20
ms
1
1
0.5
0.5
0
V
V
0
-0.5
-0.5
Voltage
phase c
0
V
Derivation of negative
sequence component
(phase a)
Voltage
phase b
-1
-1
-1
-1
-1.5
-1.5
0
10
20
30
ms
40
50
60
0
ms
86
1
1
0.5
0.5
0
0
V
1.5
V
1.5
-0.5
-0.5
-1
-1
-1.5
2
4
6
8
10
ms
12
14
14
16
18
18
20
20
-1.5
0
1
1
0.5
0.5
0
0
V
1.5
V
1.5
-0.5
-0.5
-1
-1
-1.5
2
4
6
8
10
ms
12
14
16
18
20
Opposite of the positive
sequence comp.
-1.5
0
2
4
6
8
10
ms
12
14
16
18
0
20
2
4
6
8
10
ms
12
14
16
18
20
1.5
Opposite of the negative
sequence comp.
1
0.5
0
V
V
Derivation of residual
component (phase a)
0
Opposite of the
homopolar comp.
-0.5
Residual component
-1
ms
-1.5
0
2
4
6
8
10
ms
12
14
16
18
20
87
Seminar Outline
1. Motivation of work
2. Mathematical and physical foundations of the theory
3. Instantaneous and average power & energy terms in
poly-phase networks
4. Definition of current and power terms in single-phase
networks under non-sinusoidal conditions
5. Extension to poly-phase domain: 3-wires / 4-wires
6. Sequence components under non-sinusoidal
conditions
7. Measurement & accountability issues (basic
approach)
88
Measurement & accountability issues
Active and reactive current (and power) terms are
affected by the presence of negative-sequence, zerosequence and harmonic voltages
A proper accountability approach must be adopted to
depurate the power and current terms from the effects
of voltage non-idealities, which are not under load
responsibility
If we assume that the supply voltages are sinusoidal
and symmetrical with positive sequence, we have:
U n = U fp , n = 1 ÷ 3 ⇒ U = 3 U fp
U fp
3 U fp
)
)
Un =
, n =1÷ 3 ⇒ U =
ω
ω
89
Phase current and power terms
Assuming that the equivalent phase resistance
remains the same irrespective of supply conditions, we
can express the active current and power accountable
to the load in each phase as:
ialn = Gn u ⇒ Pln = u , ialn = Pn
p
fn
p
fn
U
p2
f
2
n
U
1
I al = p
Uf
3
2
P
∑ ln
n =1
=
Similarly, if the equivalent phase reactivity remains the
same irrespective of supply conditions, we can
express the reactive current and power accountable to
the load in each phase are:
irln
)p
)p
= Bnu fn ⇒ Wln = u fn , irln
) p2
Uf
= Wn ) 2
Un
1
I rl = p
Uf
3
2
Q
∑ ln
n =1
90
Balanced current and power terms
The total power terms accountable to the load are:
3
Pl = ∑ Pln ⇒ G =
b
l
n =1
Pl
3U
p2
f
3
Wl
Ql
Wl = ∑ Wln ⇒ B = ) 2 = ω
p
p2
n =1
3U f
3U f
b
l
The balanced current terms accountable to the load
are:
1 Pl
1 Wl
1 Ql
b
b)p
b
i =G u ⇒ I =
)p =
i rl = Bl u fn ⇒ I rl =
p
p
3Uf
3Uf
3Uf
b
al
b
l
p
f
b
al
91
Unbalanced current and power terms
The unbalanced active current and power
accountable to the load are :
(
)
iauln = ialn − iabln = Gn − Glb u fp n
I aul =
∑ (G
3
n =1
n
)
b 2
l
p2
−G U f
2
P
2
l
P
−
∑
ln
3
n =1
3
1
= p
Uf
The unbalanced reactive current and power
accountable to the load are :
u
rln
i
I rul =
= irln − i
b
rln
∑ (B
3
n =1
n
)
b 2
l
−B
(
)
)p
= Bn − B u fn
b
l
) p2
1
Uf = p
Uf
2
Q
2
l
Q
−
∑
ln
3
n =1
3
92
Void current and power
The void currents satisfy the condition:
u , i v = 0 ⇒ u pf , i v + u nf + u zf + u h , i v = 0
)
)p
)n ) z )
u,iv = 0 ⇒ u f ,iv + u f + u f + u h ,iv = 0
The void current terms which can be accounted
to the load are therefore given by:
i vl = i v −
p
f
u , iv
3U
p2
f
)p
u f , iv ) p
p
uf − ) 2 uf
3U fp
In fact, the fundamental component of the void
current has been already accounted for in the
active and reactive current terms.
93
Currents terms accountable to the load
Summary of current decomposition
i l = i a l + i r l + i v l = i ba l + i brl + i ua l + i url + i v l
All current terms are orthogonal, thus:
I l = I ba 2l + I br l2 + I ua l2 + I ur l2 + I v2l
Apparent power accountable to the load
Al = U fp I l = Pl2 + Ql2 + N a2l + N r2l + Vl2
14243
N l2
94
Application Examples: 33-phase 3
3--wire
Case I: Symmetrical sinusoidal voltages
Case II: Asymmetrical sinusoidal voltages
Case III: Symmetrical nonnon-sinusoidal voltages
Case IV: Asymmetrical nonnon-sinusoidal voltages
Case I
Case II
U1 = 127∠
∠0º Vrms
U1 = 127∠
∠0º Vrms
U2 = 127∠
∠-120º Vrms U2 = 113∠
∠-104,4º Vrms
U3 = 127∠
∠120º Vrms U3 = 147,49∠
∠144º Vrms
cases III and IV are the same of
cases I and II with the addition
of 10% of 5th and 7th harmonics
Balanced Load
SOURCE
LINE
PCC
3
i3
u32
uSn
u21
i1
n {1,2,3}
RLn
LLn
LOAD
L3
2
R
C
L2
1
L1
Distorting Load
The line parameters are : RL1= RL2= RL3= 1mΩ
Ω and LL0 = LL1= LL2= LL3= 10 µH.
95
Application Examples: 33-phase 3
3--wire
Example # 1: Balanced Load
CASE I
PCC
A
P
Q
N
V
λ
LOAD
CASE II
PCC
LOAD
CASE III
CASE IV
PCC
PCC
LOAD
LOAD
1,0000 1,0000 1,0000 0,9634 1,0000 0,9840 1,0000 0,9538
0,7985 0,7985 0,7985 0,7693 0,7913 0,7758 0,7945 0,7556
0,6020 0,6020 0,6020 0,5800 0,6023 0,5962 0,6022 0,5770
0,0000 0,0000 0,0000 0,0000 0,0002 0,0002 0,0056 0,0061
0,0000 0,0000 0,0000 0,0000 0,1054 0,1038 0,0782 0,0760
0,7985 0,7985 0,7985 0,7985 0,7913 0,7885 0,7945 0,7922
The load is accounted for less active, reactive and void
power than the PCC
96
Application Examples: 33-phase 3
3--wire
Example # 2: Distorting load
load
CASE I
PCC
A
P
Q
N
V
λ
LOAD
CASE II
PCC
LOAD
CASE III
CASE IV
PCC
PCC
LOAD
LOAD
1,0000 1,0009 1,0000 0,9050 1,0000 0,9832 1,0000 0,8973
0,8275 0,8280 0,8841 0,7668 0,8254 0,8098 0,8808 0,7570
0,2432 0,2451 0,2135 0,2245 0,2422 0,2417 0,2135 0,2232
0,5060 0,5060 0,4156 0,4250 0,5055 0,4980 0,4185 0,4231
0,0000 0,0000 0,0000 0,0000 0,0674 0,0666 0,0581 0,0566
0,8275 0,8273 0,8841 0,8473 0,8254 0,8237 0,8808 0,8437
Again the apparent and active power accounted to the
load are always lower than those computed at PCC due
to the depuration of the effects of voltage asymmetry
and distortion
97
Defects of proposed accountability
approach
The equivalent phase conductance and reactivity are
computed by considering the effect of fundamental and
harmonic supply voltages as a whole.
whole. In practice, the
load can have different response to fundamental and
harmonic voltages.
The void currents are not orthogonal to supply voltages
if these latter become sinusoidal.
Only the load impedance is modeled, while supply
impedance is not estimated and enters the computation
process indirectly, in a way that does not allow to fully
analyze its effect.
98
Seminar Outline
1. Motivation of work
2. Mathematical and physical foundations of the theory
3. Instantaneous and average power & energy terms in
poly-phase networks
4. Definition of current and power terms in single-phase
networks under non-sinusoidal conditions
5. Extension to poly-phase domain: 3-wires / 4-wires
6. Sequence components under non-sinusoidal
conditions
7. Measurement & accountability issues (extended
approach)
99
Extended approach to accountability
Both load and supply are modeled based on
measurements made at PCC
Load modeling is done under sinusoidal conditions
o This makes the load model more reliable, since harmonic effects
are depurated
o Moreover, the harmonic currents generated by the load are
represented separately and their effect can directly be accounted
for accountability purposes
Supply modeling is made for three-phase symmetrical
systems and allows estimation of no-load supply
voltages and line impedances
The extended accountability approach is more reliable
than the basic one, and possibly avoids under- and overpenalization of the loads.
Of course, better results can be achieved if a more
100
accurate modeling of the load is available.
Load modeling (3(3-phase 44-wire) - 1
Single-phase equivalent
circuit of 3-phase load seen
from PCC
The passive parameters of the equivalent circuit are computed to suit
the circuit performance at fundamental frequency, i.e.:
imf =
u mf
Rm
+
û mf
Lm
 f
U mf 2
f
f
 Pm = u m ,im =
Rm

⇒ 
f2
Û
m
W f = û f ,i f =
m m
 m
Lm

⇒
⇒
Rm =
Lm =
U mf 2
Pmf
Û mf 2
Wmf
With this assumption current jm is purely harmonic. In fact:
j m = im −
um
Rmf
−
)
um
Lmf
= imf + imh −
+ u mh
Rmf
u mf
−
) f )h
um + um
Lmf
= imh −
u mh
Rmf
−
)h
um
Lmf
101
Load modeling (3(3-phase 44-wire) - 2
Single-phase equivalent
circuit of 3-phase load seen
from PCC
For the validity of the model we must assume that the
equivalent circuit parameters remain the same within
reasonable variations of the voltage supply, both in terms
of asymmetry and distortion.
This is only approximately true in real networks, but it
makes possible an accountability approach based on
measurement at the load terminals, without requiring a
precise knowledge of the load itself.
102
Supply modeling (3(3-phase 44-wire) - 1
Single-phase equivalent circuit
of 3-phase supply seen from
PCC
The passive parameters of the equivalent circuit are the same for all
phases, due to supply lines symmetry. The circuit equations are:
di
e = u + R S i + LS
dt
 p
p
p
e
=
u
+
R
i
+ LS

S

⇒ 
 en = un + R in + L
S
S


di p
dt
d in
dt
Due to its linearity, this equation can be applied separately to the
fundamental positive-sequence voltage and current terms (index p)
and the remaining terms (index n), which represent the unwanted
current and voltage components.
103
Supply modeling (3(3-phase 44-wire) - 2
Single-phase equivalent circuit
of 3-phase supply seen from
PCC
The passive parameters of the equivalent circuit are selected so as to
minimize the unwanted components of the supply voltages en, i.e., the
function:
ϕ = en
2
=
M
∑
emn , emn =
m =1
= u
n
2
+
RS2
i
n
2
+
n
L2S
di
dt
2
+ 2 RS u , i
n
n
+ 2 LS
d in
u , LS
dt
n
104
Supply modeling (3(3-phase 44-wire) - 3
Single-phase equivalent circuit
of 3-phase supply seen from
PCC
The result is expressed as a function of the quantities measured at
PCC in the form:
∂ϕ
=0 ⇒
∂ RS
∂ϕ
=0 ⇒
∂ LS
RS = − u , i
n
LS = − u n ,
n
i
n
di
dt
n
2
n
di
dt
2
Obviously, only positive solutions are acceptable for RS and LS. In
case of negative solution, the corresponding parameter is set to
zero.
105
Supply modeling (3(3-phase 44-wire) - 4
Single-phase equivalent circuit
of 3-phase supply seen from
PCC
Given RS and LS, we may compute the positive-sequence supply
voltages to be included in the equivalent circuit :
di p
e = u + R S i + LS
dt
p
p
p
Note that ep, up, ip are the fundamental positive sequence
components of the related voltages and currents, while en, un, in are
calculated by difference from the original voltages and currents.
106
Accountability – Procedure (1)
Single-phase equivalent
circuit of 3-phase load seen
from PCC
1. From the voltages and currents measured at PCC we estimate the
phase parameters Rm and Lm and the current source jm of the
equivalent circuit.
Rm =
U mf 2
Pmf
jm = imh −
Lm =
u mh
Rmf
−
Û mf 2
Wmf
)h
um
Lmf
107
Accountability – Procedure (2)
Single-phase equivalent circuit
of 3-phase supply seen from
PCC
2. From the voltages and currents measured at PCC we estimate the
supply line parameters RS and LS and the fundamental positivesequence supply voltages ep
RS = − u n , i n
in
2
LS = − u n ,
n
di
dt
n
di
dt
2
di p
e = u + R S i + LS
dt
p
p
p
108
Accountability – Procedure (3)
Equivalent circuit for the
computation of fundamental
voltages at PCC
4. Applying now the positive-sequence supply voltages at the input
terminals of the equivalent circuit, we may determine the
f
fundamental phase currents i l absorbed by the load under these
supply conditions and the corresponding fundamental phase
f
voltages u l appearing at the PCC terminals.
Note that the currents and voltages at PCC may result
asymmetrical due to load unbalance. This non-ideality must
obviously be ascribed to the load, since the voltage supply and
distribution lines are symmetrical
109
Accountability – Procedure (5)
5. Finally, the load voltages and currents at PCC, which are
accountable to the load are given by:
u l = u lf + u lh
i l = i lf + i lh
We can now compute all power terms accountable to the load and
the corresponding performance factors.
Plm = u lm ,ilm
M
)
Wlm = u lm ,ilm
M
Pl = ∑ Plm
Wl = ∑ Wlm
N la = K
N lr = K
m =1
Vl
m =1
Glm , Blm , I la , I lr
Glb , Blb , I bla , I blr
⇒ u u
I l a , I lr
I lv
110
Performance factors
Distortion factor: λ D = 1 −
Unbalance factor: λ N = 1 −
Reactivity factor: λ Q = 1 −
Power factor:
λ=
P
=
A
V2
A
2
P2 + Q2 + N 2
=
A2
N2
P +Q + N
2
2
2
=
P2 + Q2
P2 + Q2 + N 2
Q2
P2 + Q2
P
P 2 + Q 2 + N a2 + N r2 + D 2
= λQ λ N λ D
111
Application Examples: 33-phase 4
4--wire
Load circuits
A. Unbalanced
linear load
B. Unbalanced
nonlinear load
Line parameters: RL1= RL2= RL3= 10.9 mΩ
Ω - LL1= LL2= LL3= 38.5 µH
112
Application Examples: 33-phase 4
4--wire
Supply conditions
Case 1: Symmetrical sinusoidal voltages
Case 2: Symmetrical nonnon-sinusoidal voltages
Case 1
1 = 127∠0 Vrms
2 = 127∠ − 120 Vrms
3 = 127∠120 Vrms
Case 2
1 = + ∑ 1 Vrms
2 = + ∑ 2 Vrms
3 = + ∑ 3 Vrms
In case 2 the terms called ΣH represent the harmonic contents of the
phase voltages.
Each phase voltage includes 2% of 3rd harmonic, 2% of 5th harmonic.
The phase angle of each harmonic term is the phase angle of the
fundamental voltage (as in Case 1) multiplied by the harmonic order.
113
Application Examples: 33-phase 3
3--wire
Case A:
A: Unbalanced
Unbalanced linear load
A [KVA]
P [KW]
Q [KVA]
N [KVA]
D [KVA]
λ
λQ
λN
λD
Case A.1
PCC
Load
95.522
98.722
65.224
67.862
67.698
69.813
15.158
16.334
0.017
0.074
0.6828
0.6874
0.6938
0.6970
0.9872
0.9862
1.0000
0.9999
Case A.2
PCC
95.263
65.231
67.729
15.163
1.627
0.6847
0.6937
0.9872
0.9999
Load
96.571
65.089
69.597
15.582
1.649
0.6740
0.6831
0.9869
0.9999
The load is penalized for its unbalance, especially in case A.1
(sinusoidal and symmetrical supply voltages)
114
Application Examples: 33-phase 3
3--wire
Case B:
B: Unbalanced
Unbalanced nonlinear load
A [KVA]
P [KW]
Q [KVA]
N [KVA]
D [KVA]
λ
λQ
λN
λD
Case B.1
PCC
Load
93.267
94.007
63.909
63.334
33.274
35.376
20.873
21.143
55.421
55.916
0.6852
0.6737
0.8870
0.8730
0.9605
0.9601
0.8043
0.8039
Case B.2
PCC
89.494
62.738
33.599
20.619
50.190
0.7010
0.8815
0.9605
0.8279
Load
91.207
62.763
36.159
21.296
51.171
0.6881
0.8665
0.9594
0.8278
The apparent, reactive and unbalance power
accounted to the load are higher than those computed
at PCC
115