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© FLUKE Europe BV 1
Line Losses & Unified Power
By Mark B Bakker
Tech Sales Manager @ FLUKE EUROPE
Slide 2
• 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
Presenter
© FLUKE Europe BV
Slide 3
• 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
Summary
© FLUKE Europe BV
Single phase:
Active: P = 1/T ∫(u(t) i(t)) (W) = U I cos(φ)
Apparent: S = U I (VA)
Reactive: Q = √(S2 – P2) (var) = U I sin(φ)
Classical Power (Steinmetz 1897)
© FLUKE Europe BV 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.
© FLUKE Europe BV Slide 5
Classical Power
Three phase:
Active: PT = PA + PB + PC
Apparent: ST = SA + SB + SC (arithmetic)
ST = √(PT2 + QT
2) (vector)
Reactive QT = QA + QB + QC
Company Confidential 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
© FLUKE Europe BV Slide 7
IEEE 1459-2010 Power
Originally published in 2000 :
Draft Standard Definitions forthe Measurement of Electric PowerQuantities Under Sinusoidal, Non-Sinusoidal, Balanced, or UnbalancedConditions
Chair: A.E. EmanuelUpdated: 2010
Company Confidential 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
Company Confidential Fluke 430 Series II Slide 10
• In collaboration with the University of Valencia – Unified Power
From scientific breakthrough to reality
© FLUKE Europe BV Slide 11
© FLUKE Europe BV
• Unified Power supports the concept of breaking down the currents into so we know what countermeasures to take to reduce the losses.
Slide 12
Slide 13
AC
B
IA
IC
In
IB
UA
UC
UBA
C
B
U'AU'C
U'B
≈≈
≈
RLA
RLC
RLB
RLn
source loadline
Un
Three-phase line loss
Line loss: Ploss = 3 . IL2 . RL + In
2 . RLn
IL2 = (I1a
+)2 + (I1r+)2 + (I1U)2 + (IN)2
14
Three-phase line loss – line losses
W PPPPPP JnJUJNJrJaloss
loss due to:active
current
loss due to:reactive current
loss due to:harmonics &
interharmonics
loss due to:
unbalance
loss due to:neutral current
© FLUKE Europe BV
• 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
Slide 15
So if we know
© FLUKE Europe BV
• Current Decomposition– Harmonics (Non fundamental)– Unbalance
Slide 16
Current Decomposition
• A linear load will draw a sinusoidal current when connected to a sinusoidal voltage.
• There are…… only 3 linear elements:
© Fluke Europe B.V. 17
Linear loads
Power Quality Workshop
AC Current
AC Voltage
L
IR
IL
IT IT
• 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)
© Fluke Europe B.V. 18
Linear loads
Power Quality Workshop
-400
-300
-200
-100
0
100
200
300
400
1 11 21 31 41 51 61 71 81 91 101 111 121 131 141 151 161 171 181 191 201Volt
50Hz
50Hz I
• 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.
© Fluke Europe B.V. 19
Non linear loads
Power Quality Workshop-400.0
-300.0
-200.0
-100.0
0.0
100.0
200.0
300.0
400.0
1 23 45 67 89 111 133 155 177 199 221 243 265 287 309 331 353
Spanning
Stroom
• The distorted waveform contains Harmonics….
© Fluke Europe B.V. 20
Distortion
Power Quality Workshop
What do we mean by Harmonics?
WAVEFORMS
-400
-300
-200
-100
0
100
200
300
400
0 5 10 15 20
TIME
V (
RM
S)
50 Hz
150 Hz
350 Hz
Sum
© Fluke Europe B.V. 21
Signal decomposition
Power Quality Workshop
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
© Fluke Europe B.V. 22
Fourier analysis
Power Quality Workshop
© Fluke Europe B.V. 23
Fourier transform
Power Quality Workshop
-400-300-200-100
0100200300400
0 5 10 15 20
TIME
V
-400-300-200-100
0100200300400
0 5 10 15 20
TIME
V
-400-300-200-100
0100200300400
0 5 10 15 20
TIME
V
-400-300-200-100
0100200300400
0 5 10 15 20
TIME
V
-400-300-200-100
0100200300400
Frequency
V
50 150100
-400-300-200-100
0100200300400
Frequency
V
50 150100
-400-300-200-100
0100200300400
Frequency
V
50 150100
-400-300-200-100
0100200300400
Frequency
V
50 150100
© FLUKE Europe BV
• Current Decomposition– Harmonics (Non fundamental)– Unbalance
Slide 24
Current Decomposition
-200
-150
-100
-50
0
50
100
150
200
0 30 60 90 120 150 180 210 240 270 300 330 360
-8
-6
-4
-2
0
2
4
6
8
0 30 60 90 120 150 180 210 240 270 300 330 360
25
three-phase motor
InstantaneousvaluesuA(t) : 120 V 0º uB(t) : 120 V -120º uC(t) : 120 V -240º
iA(t) : 5 A -30ºiB(t) : 5 A -150ºiC(t) : 5 A -270º
pX(t) = uX(t) . iX(t)
Slide 26
I1A
60 Hz
three-phase motor
Parametervalues
Reference: U1A
U1A at 0ºU1B at -120ºU1C at -240º
length: relative to U1A or largest U1X
I1A at -30ºI1B at -150ºI1C at -270º
length: relative to I1A or largest I1X
rotation: anti-CW
All values are fundamental!
C
B
A
U1C
U1A
I1C
U1B
-30º
-30º
-30º
I1B
Slide 27
Positive & negative sequence
Positive sequence
A – B – C
(motor runs forward)
I1A
60 Hz
C
B
A
U1C
U1A
I1C
U1B
-30º
-30º
-30º
I1B
Slide 28
Positive & negative sequence
Positive sequence
A – B – C
(motor runs forward)
swapping B – C phase:
Negative sequence
A – C – B(motor runs in reverse)
I1A
60 Hz
C
B
A
U1B
U1A
I1B
U1C
-30º
-30º
-30º
I1C
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
29
Unbalanced system decomposition
( C.L. Fortescue 1918 )
A-
B-
C-
A0
B0
C0
A+
B+
C+
A-
B-
C-
A0
B0
C0
positive sequence
negative sequence
zero sequence
A+
C+
B+
B1
A1
C1
30
Symmetrical components
Slide 31
-200
-150
-100
-50
0
50
100
150
200
0 30 60 90 120 150 180 210 240 270 300 330 360
-10
-8
-6
-4
-2
0
2
4
6
8
10
0 30 60 90 120 150 180 210 240 270 300 330 360
In
-10
-8
-6
-4
-2
0
2
4
6
8
10
0 30 60 90 120 150 180 210 240 270 300 330 360
u1A(t) u1B(t) u1C(t)
i1A(t) i1B(t) i1C(t)
i1n(t)
Unbalanced systemBalanced voltagesU1A=U1B=U1C = 120 V0º, -120º, -240º
Symmetrical components – example
Unbalanced currentsI1A = 5 A -30ºI1B = 6 A -160ºI1C = 4 A -260º
Neutral currenti1n(t) = i1A(t)+i1B(t)+i1C(t)I1n = 2.09 A -163º
• To each component Classical power applies
• Positive sequence– Active current– Reactive current
• Negative component , Zero component
Slide 32
Symmetrical components
Slide 33
AC
B
IA
IC
In
IB
UA
UC
UBA
C
B
U'AU'C
U'B
≈≈
≈
RLA
RLC
RLB
RLn
source loadline
Un
Three-phase line loss
Slide 34
Combined phase currents:
Three-phase line loss
Symmetrical currents:
Currentcomponents:
Fundamental phase currents:
I1a+
active
I1r+
reactive
I1U
unbalance
IN
non-fund.
In
neutral
IA, IB, IC, In
I1A, I1B, I1C
I1+
phase level
system levelI1
- I10
wire
linewireline A
lρR
System line resistance:
• 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 resistance
Line loss: Ploss = 3 . IL2 . RL + In
2 . RLn
IL2 = (I1a
+)2 + (I1r+)2 + (I1U)2 + (IN)2
36
Three-phase line loss – line losses
W PPPPPP JnJUJNJrJaloss
loss due to:active
current
loss due to:reactive current
loss due to:harmonics &
interharmonics
loss due to:
unbalance
loss due to:neutral current
Slide 37
example
© FLUKE Europe BV
• 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
Slide 38
Conclusion
© FLUKE Europe BV
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 Fluke 430 Series II Slide 39Slide 39
© FLUKE Europe BV
Useful kilowatts (power) available
Reactive (unusable) power
Power made unusable by unbalance
Unusable distortion volt amperes
Total cost of wasted kilowatt hours per year
Neutral current
Company Confidential Fluke 430 Series II Slide 40
What you see on Energy Loss
Slide 40
© FLUKE Europe BVCompany Confidential Fluke 430 Series II Slide 41
Where do the numbers come from
These five values are directly calculated according to IEEE 1459.
Slide 41
© FLUKE Europe BVCompany Confidential Fluke 430 Series II Slide 42
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.
Slide 42
© FLUKE Europe BV 43
Thanks for your attention