Transcript Line Losses & Unified Power

Line Losses & Unified Power
By
Mark B Bakker
Tech Sales Manager @ FLUKE EUROPE
© FLUKE Europe BV
1
Presenter
• Mark B Bakker - Netherlands
• BSc Electronic design
– Rens & Rens Institute for Advanced electronics
• Brüel & Kjær SA, 14 years
• FLUKE Europe since 2010
– Power Quality Products
– Scopemeter Products
– Mechanical Products
© FLUKE Europe BV
2
Summary
• Power calculations – Recap
– Classical power
– IEEE 1459
– Unified Power
• Determination of line losses in three phase systems
– Focus on Joule effect
• Measurement equipment
– Energy Loss Calculator
– Measurement Equipment + example
© FLUKE Europe BV
3
Classical Power (Steinmetz 1897)
Single phase:
Active:
P = 1/T ∫(u(t) i(t))
= U I cos(φ)
Apparent: S = U I
Reactive:
(W)
(VA)
Q = √(S2 – P2)
= U I sin(φ)
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(var)
Slide 4
Reactive Power (analogy)
S
P
work = force x length
Q
AC network: S2 = P2 + Q2
P flows from the source to the load.
Q bounces between source and load.
S is what the network has to deal with.
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Slide 5
Classical Power
Three phase:
Active:
PT = PA + PB + PC
Apparent: ST = SA + SB + SC
ST = √(PT2 + QT2)
Reactive
Company Confidential
(arithmetic)
(vector)
QT = QA + QB + QC
Fluke 430 Series II
Slide 6
Classical Power
Classical Power works fine if:
• The system is sinusoidal
– Harmonic content is negligible
• Unbalance is negligible
– Amplitude Unbalance
– Phase Unbalance
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Slide 7
IEEE 1459-2010 Power
Originally published in 2000 :
Draft Standard Definitions for
the Measurement of Electric Power
Quantities Under Sinusoidal, NonSinusoidal, Balanced, or Unbalanced
Conditions
Chair: A.E. Emanuel
Updated: 2010
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Fluke 430 Series II
Slide 8
IEEE 1459-2010 Power
• Pro’s:
– Complete
– Mathematically correct
• Con’s:
– Many parameters
– Physical significance not always clear
– Use of a virtual replacement system for unbalance
• Question: Too academic for practical use?
Company Confidential
Fluke 430 Series II
Slide 9
Unified Power
• Developed by V. Leon and J. Montanana
• Unites various power theories
(outcomes are compatible with other theories i.e. IEEE-1459)
• Breaks down the total Power in physical
significant components
(the components can be measured with physical instruments)
• Gives direct insight in Unbalance problems
• Gives direct insight in Power Loss problems
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Fluke 430 Series II
Slide 10
From scientific breakthrough to reality
• In collaboration with the
University of Valencia
– Unified Power
© FLUKE Europe BV
Slide 11
• Unified Power supports the concept of
breaking down the currents into so we know
what countermeasures to take to reduce the
losses.
© FLUKE Europe BV
Slide 12
Three-phase line loss
IA
RLA
IC
RLC
C
U'A
≈
U'C
≈
C
A
In
≈
A
UC
RLn
U'B
Un
UB
UA
B
B
IB
source
RLB
line
load
13
Three-phase line loss – line losses
Line loss:
Ploss = 3 . IL2 . RL + In2 . RLn
IL2 = (I1a+)2 + (I1r+)2 + (I1U)2 + (IN)2
Ploss  PJa  PJr  PJN  PJU  PJn
loss due to:
active current
W 
loss due to:
unbalance
loss due to:
reactive current
loss due to:
neutral current
loss due to:
harmonics &
interharmonics
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So if we know
• If we know
– IA, IB , Ic and the components
– In
– RA, RB, RC
– RLn
We can calculate the Power loss in the lines
P = I2 * R
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Slide 15
Current Decomposition
• Current Decomposition
– Harmonics (Non fundamental)
– Unbalance
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Slide 16
Linear loads
• A linear load will draw a sinusoidal current
when connected to a sinusoidal voltage.
• There are…… only 3 linear elements:
AC Current
AC Voltage
Power Quality Workshop
© Fluke Europe B.V.
17
Linear loads
• Inductive motor:
– The electrical equivalent is a combination of a resistor and
a coil (inductance).
– The current is no longer “in phase” with the voltage
– Current can shift up to 90 degree’s “AFTER” the voltage
(Inductive) or 90 degree’s “BEFORE” voltage (Capacitive)
400
300
IR
200
L
100
Volt
IT
IT
50Hz
0
1
11 21 31 41 51 61 71 81 91 101 111 121 131 141 151 161 171 181 191 201
-100
-200
IL
-300
-400
Power Quality Workshop
© Fluke Europe B.V.
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50Hz
Non linear loads
• So…then what are the non linear loads?
• And how does the current waveform look
when we connect this load to a sinusoidal AC
voltage.
400.0
300.0
200.0
100.0
Spanning
0.0
1
23 45 67 89 111 133 155 177 199 221 243 265 287 309 331 353
Stroom
-100.0
-200.0
-300.0
Power Quality Workshop
© Fluke Europe B.V.
-400.0
19
Distortion
• The distorted waveform contains Harmonics….
What do we mean by Harmonics?
Power Quality Workshop
© Fluke Europe B.V.
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Signal decomposition
WAVEFORMS
400
300
V (RMS)
200
50 Hz
100
0
-100 0
5
10
15
20
-200
150 Hz
350 Hz
-300
-400
TIME
Sum
Power Quality Workshop
© Fluke Europe B.V.
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Fourier analysis
Any periodic function can be decomposed as a sum of sinusoidal waveforms,
whose frequencies are integer multiples of the frequency of the analyzed
signal.
Fundamental component. The sinusoidal waveform whose frequency matches
that of the analyzed signal.
Harmonic components. The resulting sinusoidal waveforms with frequencies
multiples of the fundamental frequency.
Jean-Baptiste-Joseph Fourier (March 21, 1768 Auxerre - May 16, 1830 Paris)
French mathematician and physicist known for his work on the decomposition
of periodic functions in trigonometric series convergent called Fourier series
Power Quality Workshop
© Fluke Europe B.V.
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400
300
200
100
0
-100
-200
-300
-400
0
5
10
15
20
V
V
Fourier transform
V
TIME
0
5
10
15
20
V
V
0
5
10
100
150
400
300
200
100
0
-100
-200
-300
-400
50
100
150
Frequency
TIME
400
300
200
100
0
-100
-200
-300
-400
50
Frequency
V
400
300
200
100
0
-100
-200
-300
-400
400
300
200
100
0
-100
-200
-300
-400
15
20
400
300
200
100
0
-100
-200
-300
-400
50
100
150
Frequency
V
400
300
200
100
0
-100
-200
-300
-400
V
TIME
0
5
Power Quality Workshop
© Fluke Europe B.V.
10
15
20
400
300
200
100
0
-100
-200
-300
-400
50
100
150
Frequency
TIME
23
Current Decomposition
• Current Decomposition
– Harmonics (Non fundamental)
– Unbalance
© FLUKE Europe BV
Slide 24
three-phase motor
200
150
100
50
0
0
30
60
90
120
150
180
210
240
270
300
330
360
-50
Instantaneous
values
uA(t) : 120 V  0º
uB(t) : 120 V  -120º
uC(t) : 120 V  -240º
-100
-150
-200
8
iA(t) : 5 A  -30º
iB(t) : 5 A  -150º
iC(t) : 5 A  -270º
6
4
2
0
0
-2
30
60
90
120
150
180
210
240
270
300
330
360
pX(t) = uX(t) . iX(t)
-4
-6
-8
25
three-phase motor
C
Parameter
values
I1C
60 Hz
U1C
Reference: U1A
-30º
U1A at 0º
U1B at -120º
U1C at -240º
U1A
-30º
-30º
I1B
I1A
A
length: relative to U1A or largest U1X
I1A at -30º
I1B at -150º
I1C at -270º
length: relative to I1A or largest I1X
U1B
B
rotation: anti-CW
All values are fundamental!
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Positive & negative sequence
C
Positive sequence
I1C
U1C
60 Hz
A–B–C
(motor runs forward)
-30º
U1A
-30º
A
-30º
I1B
I1A
U1B
B
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Positive & negative sequence
C
Positive sequence
I1B
U1B
60 Hz
A–B–C
(motor runs forward)
-30º
swapping B – C phase:
U1A
-30º
-30º
I1C
I1A
A
Negative sequence
A–C–B
(motor runs in reverse)
U1C
B
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Unbalanced system decomposition
A three-phase unbalanced phasor system
can be decomposed into three balanced phasor systems:
• three-phase Positive Sequence System
• three-phase Negative Sequence System
• single-phase Zero Sequence System
Decomposing the fundamental components:
U1A, U1B, U1C
→ U+, U-, U0
I1A, I1B, I1C
→ I+, I- , I0
( C.L. Fortescue
29 1918 )
Symmetrical components
positive sequence
C-
C+
C1
C+
A+
B+
AC-
B1
AA+
A0
B0
C0
negative sequence
A1
B+
Bzero sequence
A0
B0
C0
B-
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Symmetrical components – example
200
u1A(t)
150
100
u1B(t)
u1C(t)
50
0
0
30
60
90
120
150
180
210
240
270
300
330
360
-50
-100
-150
Unbalanced system
Balanced voltages
U1A=U1B=U1C = 120 V
0º, -120º, -240º
-200
10
i1A(t)
8
6
i1C(t)
i1B(t)
4
2
0
-2 0
30
60
90
120
180
150
210
240
270
300
330
360
-4
-6
Unbalanced currents
I1A = 5 A -30º
I1B = 6 A -160º
I1C = 4 A -260º
-8
-10
10
In
8
6
i1n(t)
4
2
0
-2 0
30
60
90
120
150
180
210
240
270
300
330
360
Neutral current
i1n(t) = i1A(t)+i1B(t)+i1C(t)
I1n = 2.09 A -163º
-4
-6
-8
-10
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Symmetrical components
• To each component Classical power applies
• Positive sequence
– Active current
– Reactive current
• Negative component , Zero component
Slide 32
Three-phase line loss
IA
RLA
IC
RLC
C
U'A
≈
U'C
≈
C
A
In
≈
A
UC
RLn
U'B
Un
UB
UA
B
B
IB
source
RLB
line
load
33
Three-phase line loss
phase level
Combined phase currents:
IA, IB, IC, In
Fundamental phase currents:
Symmetrical currents:
Current
components:
I1A, I1B, I1C
I1+
I1-
system level
I10
I1a+
I1r+
I1U
IN
In
active
reactive
unbalance
non-fund.
neutral
34
Three-phase line loss – line resistance
R line 
System line resistance:
•
•
•
ρ wire  l line
A wire
lline : length of the line
Awire : cross area of the wire
wire : specific wire resistance
Estimation: Rline = 1% of (Pnom) / (Inom2)
This assumes each line is laid out in such a way that at maximum
rated load the loss in each phase is ≈ 1% of nominal power
35
Three-phase line loss – line losses
Line loss:
Ploss = 3 . IL2 . RL + In2 . RLn
IL2 = (I1a+)2 + (I1r+)2 + (I1U)2 + (IN)2
Ploss  PJa  PJr  PJN  PJU  PJn
loss due to:
active current
W 
loss due to:
unbalance
loss due to:
reactive current
loss due to:
neutral current
loss due to:
harmonics &
interharmonics
36
example
37
Conclusion
• Unified Power decomposites the currents in
the system
• Calculate the line losses with currents and line
resistance
• Identify the source of biggest losses
– Unbalance
– harmonics
© FLUKE Europe BV
Slide 38
Energy Loss Calculator
• Settings for feeder conductor length and
diameter are entered for resistive losses
calculation.
• Up to four electricity tariffs can be entered
for different times of day.
• Losses arise from a number of sources
including resistive, harmonic, unbalance,
and neutral losses.
Company Confidential
430Europe
SeriesBV
II
©Fluke
FLUKE
Slide 39
39
Slide
What you see on Energy Loss
Useful kilowatts
(power) available
Reactive (unusable)
power
Power made unusable
by unbalance
Unusable distortion
volt amperes
Neutral current
Total cost of wasted
kilowatt hours per year
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430Europe
SeriesBV
II
©Fluke
FLUKE
Slide 40
40
Slide
Where do the numbers come from
These five
values are
directly
calculated
according to
IEEE 1459.
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430Europe
SeriesBV
II
©Fluke
FLUKE
Slide 41
41
Slide
Where do the numbers come from
• These loss values are
dependent on the totals of
each of the measured
values.
• These values are derived
using the Unified Power
method to discover the
waste energy in the system.
• The calculation method
used is Fluke’s patented
method.
Company Confidential
430Europe
SeriesBV
II
©Fluke
FLUKE
Slide 42
42
Slide
Thanks for your attention
© FLUKE Europe BV
43