The day the current disappeared
Johan.Rens@nwu.ac.za 3 May 2019
Contents
• Finding current: What did Maxwell miss?
• Why the Polish knight wanted to climb another
mountain…
• That is why the current went missing!
• Until sanity was restored by the IEEE 1459-2010
• Application in impedance-constrained networks
A brief history of nearly every electron
• Electricity as a matter of physics since BC
• Found application during 1800’s: “electrical engineering” – using this
resource to do “something”, mostly to better life for mankind
• Moving electrically charged electrons and protons = current is flowing
• Mostly done in an alternating manner (ac electricity)
• How well it is done governed by the constant law of misery in electrical
circuits: Ohm
On average, the voltage remains shockingly zero
i(t)
v(t)
sourcev(t)
loadv(t)
lossResistive component of loadReactive component: L, C or both
-1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 V ol ta ge time (ms) p(t) = v(t) * i(t)
= Vmax*Imax * cos(wt +d)cos(wt +b) = 1
2 2Vrms 2Irmscos(d -b)+ VrmsIrmscos(d -b)cos 2(
[
wt +d)]
+ VrmsIrmssin(d -b)sin 2éë(
wt +d)
ùû = VRMSIRMScos(
d -b)
{
1+ cos 2éë(
wt +d)
ùû}
+ VRMSIRMSsin(
d - b)
sin 2éë(
wt +d)
ùûThat is why that single-phase generator remained so small….
i(t)
v(t)
sourcev(t)
loadv(t)
lossv(t)
bv(t)
cv(t)
ai(t)
ai(t)
bi(t)
cp(t)
ap(t)
cp(t)
bp(t)
3-phaseThe War of Currents winner : George Westinghouse; 1886
v
a(
t) = 2V
LNcos(
w
t +
d
)
v
b(
t) = 2V
LNcos(
w
t +
d
- 120
0)
v
a(
t) = 2V
LNcos(
w
t +
d
- 240
0)
i
a(
t) = 2I
Lcos(
w
t +
b
)
i
b(
t) = 2I
Lcos(
w
t +
b
- 120
0)
i
c(
t) = 2I
Lcos(
w
t +
b
- 240
0)
The War of Currents winner : George Westinghouse; 1886
p3j(t) = pa(t) + pb(t) + pc(t)= 3VLNIL cos(
d
-b
) + VLNIL cos(d
-b
) + VLNIL[cos(d
-b
) + VLNIL cos(2w
t +d
+b
)+ ... ....+ cos(2w
t +d
+b
- 2400)+ cos(d
-b
) + VLNIL cos(2
w
t +d
+b
+ 2400)] = 3VLNIL cos(
d
-b
) = P3jAnd if those voltages are no longer symmetrical and
those currents unbalanced?
The time-independent feature of three-phase energy transfer isIt get worse when the load is non-linear (and supply
voltage non-sinusoidal)….
• What does non-linear loading mean?
• More or less the modern way of consuming current
• Such as in a LED lamp, a laptop charger and and and
• Nowadays, MW’s and kV’s
• The energy transfer is impacted • We are used to:
Not anymore…. LED lamp Laptop charger
S= P + jQ
\ S
2=P
2+Q
2 2 2 2S > P + Q
D
2=S
2- P
2- Q
2Now this is where the current got
lost!
• Distortion power?
• What is that?
• Something missing in power theory?
•Budeanu defined the concept of distortion power – as
“distortion” is the cause
• Many contributions to power theory followed
• Modern power phenomena required new definition
• Engineers have to design, specify and operate power system in
which energy phenomena has physical intrepetation
• Why is the term “distortion power” then still in use?
• Discussion became intense………
Sixth International Workshop on Power Definitions and Measurements under Non-Sinusoidal
And then the Polish knight arrived…
Shouldn’t we rather abandon the
knighthood?
But what is the problem?
• Steinmetz (1892): Ratio of active to apparent power decrease when waveform becomes more distorted such as electric arc (lighting application).
• Impact of distortion and unbalance new phenomena • Power factor reduction a concern
-• Unbalance in loading, asymmetry in supply voltages, AND distortion in voltage and/or current contributes to the degradation of power factor (the effiency in the
transfer of real energy)
• Classical power theory can only deal with perfectly sinusoidal voltages, perfectly
symmetrical between phases and perfectly linear loads that withdraw perfectly balanced currents
• Budeanu (1927) described S2 > P2+Q2 when waveforms are non-sinusoidal
IEEE-1459-2010 attempted to further practical formulations universally acceptable for engineers to deal with modern power systems
Effective values: nonsinusoidal waveform conditions (1)
• A non-sinusoidal, single-phase, time-dependent voltage v(t)1ϕwith fixed and repetitive period
T is applied to a load - represented as a finite series of harmonic components: h=1, 5, 7
• Single phase distortion component of v(t)1ϕ can be isolated as vH(t)1ϕ:
• The IEEE 1459-2000 document further practical guidelines on power definitions •The time-domain or the frequency domain can be used for power definitions • Focus will be on frequency domain power definitions in this presentation
( )
1( )
1(
1)
(
5)
(
7)
1
1 1
1sin sin 5 sin 7
5 7 N h h v t v t
w
tw
tw
t = =v
H
( )
t
= - + - + -1=
2
V
hsin(
hwt+ a
h)
h¹1 N
=
1
5
sin 5
(
wt-
5)
+
1
7
sin 7wt-
(
7)
Effective values: nonsinusoidal waveform conditions (2)
• i(t)1ϕ when v(t)1ϕ is applied to frequency dependent impedance load: • Distortion components i
( )
H(t)1ϕ( )
:(
)
[
]
,1 1 1 1 1 1 2 sin 1,5,7 N N h h h h h i t i t I h t h w
b
= = = = + =
( )
1 1 12 sin(
)
N H h h hi t
I
h t
w
b
¹=
+
Active Power: nonsinusoidal waveform conditions
• The effective or RMS values: • The single phase power:
• Classical power theory formulates Time-dependent Active Power (per harmonic order h):
• The Time-dependent Total Active Power of a circuit under distorted waveform conditions:
• Total (or Joint) Average Active Power requires integration over a period T:
2 2 1 ,1 1 ,1 1 1 ; N N h h h h V I = = =
V =
I( ) ( )
( )
( )
1 1 1 1 1 1 1( )
N h N h h hp t
v t
i t
v t
i t
= =
=
=
( )
1Re 2
( )
1Re 2
( )
1 h h hp t
=
ë
é
v
t
ù
û
ë
é
i
t
ù
û
( )
1( )
1 1 N h h p t p t = =
( )
(
)
1 1 ,1 ,1 1 1 cos N h h h h h T P p t dt V I T a b = =
=
-Reactive Power: nonsinusoidal waveform conditions
• Classical power theory formulates Time-dependent Reactive Power (per harmonic order h):
• The Time-dependent Total Reactive Power of a circuit under distorted waveform conditions:
• Total (or Joint) Average Reactive Power requires integration over a period T: Does it make sense?
☞
It does not make sense!q t
( )
h= Re 2v t
é
ë
( )
hù
û Im 2i t
é
ë
( )
hù
û
( )
( )
1 N h h q t q t = =
QB = 1 T T
q t( )
= 1 T q t( )
h h=1 N
= T
Vh,1Ih,1 sin(
a
h -b
h)
h=1 ¥
Summary of Budeanu’s definitions
Total Active Power Total Reactive Power (Budeanu’s Reactive Power) Budeanu’s Distortion Power
( ) ( )
( )
( )
1 1 1 1 1 1 1( )
N N h h h hp t
v t
i t
v t
i t
= ==
=
( )
1 ,1(
1)
,1(
1)
1 1 2 2 2 ,1 ,1 ,1 1 1 , 1( )2
sin
*
2
sin
N N h h h h h h N N N h B h k h h h k h k H B Bp t
V
h t
I
h t
P
Q
P
Q
D
w
a
w
b
= = = = = ¹=
+
+
=
+
+
=
+
+
V
I
What is wrong with Q
B– the Budeanu Reactive Power?
• Physical nature of reactive power follows from the application of field theory (Maxwell’s equations)
• Reactive Power not to contribute to real energy transfer
• Physical nature of reactive power - energy accumulation in electric and magnetic fields of reactive components in the load and source
• Results in oscillatory exchange of energy between these reactive components
• Similar explanation assigned to harmonic reactive power Qh at each harmonic order h
• Is QB (Joint/Total Reactive Power) a useful concept? • Let’s investigate……
( )
( )
,1 ,1(
)
1 1 1 1 sin N B h h h h h h h T T Q q t q t V I T T a b ¥ = = =
=
=
-Application of Q
Bin power factor correction (1)
v(t)1ϕ 10 ohm 100 mH i(t) • PF = 0.3 based on QBv t
( )
1=
100
h
sin 2
(
×π × 50 × h×t
)
;
hÎ 1,5,7,11
[
]
Application of Q
Bin power factor correction (2)
v(t)1ϕ 10 ohm
100 mH
64 μF
i(t)
• Calculate the capacitance in parallel to compensate QB
2 2 2 2 1 1 5 5 7 7 11 11 B
Q
C
V
w
V
w
V
w
V
w
=
+
+
+
Application of Q
Bin power factor correction (3)
• QB completely compensated by capacitor, but- power factor of compensated circuit did not change significantly.
• Apparent power is less in compensated circuit but unnecessary loading remains (difference between apparent power and real power).
Remark
• Budeanu’s reactive power (QB) not useful for power factor compensation.
• Power factor correction results in “distortion power” (DB) to increase significantly due to increased interaction between uneven harmonic voltage and current components.
Does distortion power D
B
make sense?
v(t)1ϕ 1 ohm i(t) Time V ol ta ge 1.676 1.676 -Vac j 600 0 j Time C ur re nt 1.676 1.676 -Iac j 600 0 j• DB has zero value: waveforms perfectly sinusoidal? • Both voltage and current are distorted!
• DB does not relate to degree of waveform distortion.
v t
( )
1=
1
Summary: Power theory in a modern
power system
• It must, as far as possible consist of a generalisation of the classic single-frequency power theory that has by now been universally accepted.
• It must be as amenable to conventional measurement techniques as possible and require the minimum of sophistication in instrumentation.
• It’s different defined components must be relatable to physically observable or ascribable phenomena and not to hypothetical or abstract mathematical
definition.
• It must present a suitable basis for quantifiable measurement, control, tariff systems and design.
• It must cater for every conceivable practical situation and never violate circuit laws, regardless of which domain it is transformed into.
• It must be useful to the engineer who has to apply these definitions in design, specification and operation of the power system.
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2. The IEEE 1459 made easy
What is a modern Power System? (1)
-400 -300 -200 -100 0 100 200 300400 Line-neutral voltages in a 400 V power system
P h a s e V o lt a g e s ( V )
What is a modern Power System? (2)
-80 -60 -40 -20 0 20 40 60 80Three phase line-currents in a 400 V power system
P h a s e c u rr e n ts ( A )
What is a modern Power System? (3)
Condition 3: Unbalance in loading…
-1000 -800 -600 -400 -200 0 200 400 600 800
1000 Load Currents withdrawn by Arc Furnace
C u rr e n t (A )
What is a modern Power System? (3)
-60000 -40000 -20000 0 20000 40000 60000Asymmetrical voltages at PCC feeding Arc Furnace
L in e -L in e V o lt a g e s (V )
The IEEE 1459-2000: Voltage and Current (1)
• A Three-phase power system under non-sinusoidal waveform conditions, unbalanced loading and asymmetrical supply voltages is considered.
• Sinusoidal, balanced and single-phase power system operation is easily introduced as a simplification
Define nonsinusoidal line-neutral voltages and line currents:
, , 0 0 , , 0 0 , , 0 ( ) 2 sin( ) ( ) 2 sin( 120 ) ( ) 2 sin( 120 ) a a h a h h b b h a h h c c h c h h v t V h t v t V h t h v t V h t h w a w a w a ¥ ¹ ¥ ¹ ¥ ¹ = + = + -= + +
0 , , 0 0 0 , , 0 0 0 , , 0 ( ) 2 sin( ) ( ) 2 sin( 120 ) ( ) 2 sin( 120 ) a a a h a h h b b b h a h h c c c h c h h i t I I h t i t I I h t h i t I I h t hw
b
w
b
w
b
¥ ¹ ¥ ¹ ¥ ¹ = + + = + + -= + + +
The IEEE 1459-2000: Voltage and Current (2)
• DC components in voltage, (Va0, Vb0 and Vc0) should always be zero.
• DC values in line currents (Ia0, Ib0 and Ic0) could be nonzero depending on nature of load.
• The RMS line-neutral voltage Va and RMS line current Ia (similarly for phase b and c) are related to harmonic components by:
• Fundamental frequency and the nonfundamental frequency (harmonic frequencies grouped) 2 2 2 2 2 2 2 ,1 , ,1 , 2 2 2 2 2 2 2 2 2 ,1 , ,1 , 2 2
(
)
(
)
a a a h a aH aH a h h h a a a h a aH aH a h h hV
V
V
V
V
V
V
I
I
I
I
I
I
I
¥ ¥ ¹ ¹ ¥ ¥ ¹ ¹=
+
=
+
=
=
+
=
+
=
The IEEE 1459-2000: Voltage and Current (4)
☝
What is the “effect” of three-phase voltages and three-phase currents in terms of the three-phase power system?“Effective” voltage and “effective” current to represent the state of the
three-phase power system:
“Effective” voltage and “effective” current in a 3-wire three-phase power system:
2 2 2 1 2 2 2 1 e e eH e e eH
V
V
V
I
I
I
=
+
=
+
2 2 2 2 2 23
3
a b c e a b c eV
V
V
V
I
I
I
I
+
+
=
+
+
=
Calculating the effective voltage and current (1)
☝
If an artificial neutral point in a 3-wire three-phase system is not used to find the line-neutral voltage values, the effective three-phase voltage can becalculated from the RMS phase-phase voltage values as:
• The fundamental frequency
componens of the effective voltage
and current in a 3-wire three-phase
power system: 2 2 2
9
ab bc ca eV
V
V
V
=
+
+
2 2 2 ,1 ,1 ,1 ,1 2 2 2 ,1 ,1 ,1 ,19
3
ab bc ca e a b c eV
V
V
V
I
I
I
I
+
+
=
+
+
=
Calculating the effective voltage and current (2)
• Non-fundamental frequency components of the effective voltage and current in a 3-wire three-phase power system:
• Non-fundamental frequency components relates to the harmonic components (line-neutral voltages assumed):
2 2 2 2 2 2
9
3
abH bcH caH eH aH bH cH eHV
V
V
V
I
I
I
I
+
+
=
+
+
=
(
)
(
)
2 2 2 , , , 1 2 2 2 , , , 13
3
a h b h c h h eH a h b h c h h eHV
V
V
V
I
I
I
I
¥ ¹ ¥ ¹+
+
=
+
+
=
Calculating the effective voltage and current (3)
• Unbalanced condition in a
4-wire three-phase power system
requries reformulation of the
effective voltage and current :
• The fundamental
frequency components of
the effective voltage and
current in a 4-wire
three-phase power system:
(
2 2 2)
2 2 2 2 2 2 21
3
18
3
e a b c ab bc ca a b c n eV
V
V
V
V
V
V
I
I
I
I
I
é
ù
=
ë
+
+
+
+
+
û
+
+
+
=
(
2 2 2)
2 2 2 ,1 ,1 ,1 ,1 ,1 ,1 ,1 2 2 2 2 ,1 ,1 ,1 ,1 ,11
3
18
3
e a b c ab bc ca a b c n eV
V
V
V
V
V
V
I
I
I
I
I
é
ù
=
ë
+
+
+
+
+
û
+
+
+
=
Calculating the effective voltage and current (4)
• The non-fundamental frequency components of the effective voltage and
current in a 4-wire
three-phase power system:
☞
Implementation straightforward with modern digital instrumentation(
2 2 2)
2 2 2 2 2 2 21
3
18
3
eH aH bH cH abH bcH caH aH bH cH nH eHV
V
V
V
V
V
V
I
I
I
I
I
é
ù
=
ë
+
+
+
+
+
û
+
+
+
=
And power? Not all S is the same S…..
• Arithmetic apparent power SA:
• Active, Budeanu’s reactive and
distortion power is summated over all three phases:
• Vector apparent power Sv:
• The Budeanu power definition for single-phase power systems are applied per phase 2 2 2 2 2 2 2 2 2 a a Ba Ba b b Bb Bb c c Bc Bc A a b c
S
P
Q
D
S
P
Q
D
S
P
Q
D
S
S
S
S
=
+
+
=
+
+
=
+
+
=
+
+
,3 a b c B Ba Bb Bc B Ba Bb BcP
P
P
P
Q
Q
Q
Q
D
D
D
D
=
+
+
=
+
+
=
+
+
2 2 2 ,3 V B B S = P + Q + DAnd the effective apparent power
is…..
The three-phase or system effective apparent power Se can be written in terms of the contribution of all harmonic components:
The components in the system effective apparent power Se can be grouped in the
fundamental and nonfundamental frequency voltage and current components:
The non-fundamental frequency apparent power SeN consist of 3 “distortion” components:
S
e2= (V
eI
e)
2= (V
e,1I
e,1)
2+ (V
e,1I
eH)
2+ (V
eHI
e,1)
2+ (V
eHI
eH)
2 2 2 2 ,1 e e eNS
=
S
+
S
S
eN2= V
e,1I
eH(
)
2+ V
(
eHI
e,1)
2+ V
(
eHI
eH)
2= D
eI2+ D
eV 2+ D
eH 2And the “distortion” powers are…..
Ve,1IeH: The current distortion power, DeI
VeHIe,1: The voltage distortion power, DeV
VeHIeH: The harmonic distortion power, DeH
An effective harmonic apparent power SeH is defined:
2 2 2 ,3 eH H eH
S
=
P
+
D
(
) (
2)
2(
)
2 2 ,1 ,1 2 2 2 eN e eH eH e eH eH eI eV eHS
V I
V I
V I
D
D
D
=
+
+
=
+
+
The Total Three-Phase Active Power
The Joint (Total) Harmonic Active Power of three-phase power system:
The Joint Active Power of three-phase power system:
(
)
3 1 3 1 Re N H , ah ah bh bh ch ch h N h, h P P ¹ ¹ = + + =
V I V I V I( )
( )
(
)
3 3 3 1 3 1 Re N ah ah bh bh ch ch h N h, h P t t P = = é ù = ë û = + + =
v ,i V I V I V IAnd how much is that “distortion”?
Quantification on the level of distortion is done with three-phase effective values:
☞
Voltage total harmonic distortion factor: VTHDe☞
Current total harmonic distortion factor: ITHDe☞
The IEEE 1459-2000 write these symbols as THDeV and THDeI,1 , ,1 eH e e e H e e
V
VTHD
V
I
ITHD
I
=
=
These distortion factors are useful!
A shortcut to SeN:
And to DeI; DeV; DeH (The “distortion” powers):
To do what with?
(
)
2 1 eN e e e e eS
=
S VTHD
+
ITHD
+
VTHD ITHD
1 1 1 eI e e eV e e eH e e eD
S ITHD
D
S VTHD
D
S ITHD VTHD
=
=
=
To quantify the pollution….
Harmonic pollution (SeN/Se1):
Unbalance pollution:
☝
The positive sequence voltage (V1+) and current (I1+) in the
three-phase fundamental frequency components have to be found…..
Then calculate:
The “unbalance pollution”:
☝
“Unbalance pollution” includes both the effect on loading unbalance and voltage asymmetryS
eNS
e1
2= ITHD
(
e)
2+ VTHD
(
e)
2+ ITHD
(
eVTHD
e)
2S
1 += 3V
1 +I
1 +S
u1= S
e12- S
1 +( )
2UnbalancePollution(%) =
S
u1S
e1*100
What powers are “useful”?
☞
The positive sequence components in the three-phase fundamental frequency voltages and currents:
The 50 Hz positive sequence active power:
The 50 Hz positive sequence reactive power:
The 50 Hz positive sequence apparent power: And power factor?Various options exist….
Different answers to the same question!
P
1+= 3V
1+I
1+cos
1+Q
1 += 3V
1 +I
1 +sin
1 + S1+( )
2 = P( )
1+ 2 + Q 1 +( )
2Power factor formulations
Different apparent powers have been formulated: • Arithmetic apparent power, SA
• Vector apparent power, Sv
• Three-phase effective apparent power, Se • Positive sequence apparent power, S1+
Different apparent power factors have been formulated:
• Arithmetic apparent power factor • Vector apparent power factor
• Three-phase effective apparent power factor
• Positive sequence apparent power factor 1 1 P PF S + + + = A A P PF S = V V P PF S = PFe = P,3 Se = P1,3 + PH ,3 Se
Which power factor should be used?
If the waveforms are sinusoidal, the loading in perfect balance and the supply voltages in perfect symmetry, then:
In a practical power system with distorted waveforms, unbalanced loading and asymmetrical supply voltages:
• PFe reflects the impact of harmonics and asymmetrical waveforms the best.
☞
The smallest numerical value – regulatory application☝
What power factor formulation does your instrument use?A V e
PF = PF = PF = PF+
e A V
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3. Case Study
Application of IEEE 1459-2000 - practical power system
Rs + ωLs 6-pulse controller Heating load RL + ωLL 11kV/400 V 1 MVA 0.01+j0.043 p.e. V and I measurements PCCVoltage and Current waveforms
-400 -300 -200 -100 0 100 200 300 400 Va Vb Vc Line-neutral voltages V o lt ag e ( V ) -800 -600 -400 -200 0 200 400 600 800 Ia Ib Ic Line currents C u rr e n t (A ) • Line-neutral voltages • Line CurrentsWaveform analysis: Fourier transform
The IEEE 1459 power definitions require translating time-domain waveforms to the frequency domain by means of the Fourier transform:
The level of distortion requires quantification: • ITHDe= 48.3%
Isolating the distortion components
• The non-fundamental frequency components of the effective voltage and
current:
• The fundamental
frequency components of
the effective voltage and
current:
(
2 2 2)
2 2 2 2 2 2 21
3
18
17.2 V
3
145.8 A
eH aH bH cH abH bcH caH aH bH cH nH eHV
V
V
V
V
V
V
I
I
I
I
I
é
ù
=
ë
+
+
+
+
+
û
=
+
+
+
=
=
(
2 2 2)
2 2 2 ,1 ,1 ,1 ,1 ,1 ,1 ,1 2 2 2 2 ,1 ,1 ,1 ,1 ,11
3
18
227.6 V
3
301.5 A
e a b c ab bc ca a b c n eV
V
V
V
V
V
V
I
I
I
I
I
é
ù
=
ë
+
+
+
+
+
û
=
+
+
+
=
=
The three-phase effective voltage and current
The effective voltage and current :
The RMS line-neutral voltages per phase:
The RMS line currents per phase:
The RMS neutral current:
=
21+
2= 228.3 V
2=
1 2+
2= 335 A
2 , 1 229.2 V 227.7 V 227.8 V a a h h b c V V V V ¥ = = = = =
2 , 1 315.4 A 342.9 A 345.8 A 8.1 A a a h h b c n I I I I I ¥ = = = = = =
The powers
Current distortion power: Voltage distortion power: Harmonic distortion power:
Effective apparent power: Arithmetic apparent power: Vector apparent power:
Joint (three-phase) active power: Joint harmonic active power:
1
7.5 kVar
eH e e eD
=
S ITHD VTHD
=
199.7 KVA
VS
=
(
)
3 3 1 1 Re N N 387 W H , ah ah bh bh ch ch h, h h P P ¹ ¹ =
V I +V I +V I =
=-( )
( )
(
)
3 3 3 3 1 1 Re N ah ah bh bh ch ch N h, 92 9 kW h h P t t P . = = é ù = ëv ,i û =
V I +V I +V I =
= 1 229.4 KVA e S = 199.7 KVA A S = 199.9 kVar
eI e eD
=
S ITHD
=
115.5 kVar
eV e eD
=
S VTHD
=
And the power factors….
The arithmetic power factor: 0.465 p.u. The vector power factor: 0.465 p.u. The effective power factor: 0.405 p.u
And the “pollution” factors….
Harmonic pollution: Unbalance pollution: Unbalance factors:
( )
(
) (
2) (
2)
2 1 % *100 3.6% eN e e e e e S ITHD VTHD ITHD VTHD S = + + =( )
2 2 1 1 1 1 1 9.1 KVA (%) *100 4.4% u e u e S S S S UnbalancePollution S + = - = \ = = 2 1 2 1 0.4% 3% UB UB V V V I I I = = = =Summary
☞
The three-phase effective power factor is numerically the smallest if unbalance and/or waveform distortion exist.☞
The contribution to apparent power by both voltage asymmetry and unbalance in loading, was shown to be significant even with “low” values in VUB and IUB.☞
The three-phase effective distortion index for voltage (VTHDe) and current (ITHDe) furthers straightforward calculation of distortion powers.☞
It is possible to isolate the contribution of harmonics to useles power by means of voltage distortion power (DV), current distortion power (DI) and harmonic distortion power (DH).☞
It is a helpful reflection on the impact of nonsinusoidal waveforms,unbalanced loading and asymmetrical supply voltages from the point of view from operating such power system.
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Conclusion on Power Theory
Conclusion
• The classical power theory explains phenomena in a power system with a sound physical explanation but is inadequate when waveforms are non-sinusoidal,
loading is unbalanced and voltage waveforms are asymmetrical.
Conform to Electrical Network Laws• Numerous approaches to power theory exists which are formulated in either the time- or frequency domain.
• New contributions are forthcoming as inadequacies are better understood. • In general, a power theory has to: