unified power nader verklaard...from scientific breakthrough to reality • unified power supports...
TRANSCRIPT
Unified Power nader verklaard
Gerard Grashof
Fluke Industrial bv
Almelo
© FLUKE Europe BV 1
• 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
2
Summary
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)
Slide 3
Classical Power (Steinmetz 1897)
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.
Slide 4
Reactive Power (analogy)
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 Slide 5
Classical Power
Fluke 430 Series II
Classical Power works fine if:
• The system is sinusoidal
– Harmonic content is negligible
• Unbalance is negligible
– Amplitude Unbalance
– Phase Unbalance
Slide 6
Classical Power
© FLUKE Europe BV
Originally published in 2000 :
Draft Standard Definitions for
the Measurement of Electric Power
Quantities Under Sinusoidal, Non-
Sinusoidal, Balanced, or Unbalanced
Conditions
Chair: A.E. Emanuel
Updated: 2010
Company Confidential Slide 7
IEEE 1459-2010 Power
Fluke 430 Series II
• 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 Slide 8
IEEE 1459-2010 Power
• Developed by Vicente León-Martínez and
Joaquín Montañana-Romeu
• 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
Slide 9
Unified Power
Fluke 430 Series II
• In collaboration with
the University of
Valencia
– Unified Power
Slide 10
From scientific breakthrough to reality
• Unified Power supports the concept of
breaking down the system currents into
– Active current
– Reactive current
– Unbalance current
– Non fundamental current..
• so we know what countermeasures to take
to reduce the losses.
© FLUKE Europe BV Slide 11
Decomposing Current
A
C
B
IA
IC
In
IB
UA
UC
UB A
C
B
U'A U'C
U'B
≈
≈
≈
RLA
RLC
RLB
RLn
source load
line
Un
12
Three-phase line losses
Line loss: Ploss = 3 . IL2 . RL + In
2 . RLn
IL2 = (I1a
+)2 + (I1r+)2 + (I1U)2 + (IN)2
13
Three-phase 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
• 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 14
So if we know
• Current Decomposition
– Harmonics (Non fundamental)
– Unbalance
© FLUKE Europe BV Slide 15
Current Decomposition
• A linear load will draw a sinusoidal current
when connected to a sinusoidal voltage.
• There are…… only 3 linear elements:
16
Linear loads
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)
17
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.
18
Non-linear loads
-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….
19
Distortion
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
20
Signal decomposition
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
21
Fourier analysis
22
Fourier transform
-400 -300
-200 -100
0 100 200 300 400
0 5 10 15 20
TIME
V
-400 -300 -200 -100
0 100 200 300 400
0 5 10 15 20
TIME
V
-400 -300 -200 -100
0 100 200 300 400
0 5 10 15 20
TIME
V
-400 -300 -200 -100
0 100 200 300 400
0 5 10 15 20
TIME
V
-400 -300 -200 -100
0 100 200 300 400
Frequency
V
50 150 100
-400 -300 -200 -100
0 100 200 300 400
Frequency
V
50 150 100
-400 -300 -200 -100
0 100 200 300 400
Frequency
V
50 150 100
-400 -300 -200 -100
0 100 200 300 400
Frequency
V
50 150 100
• Current Decomposition
– Harmonics (Non fundamental)
– Unbalance
© FLUKE Europe BV Slide 23
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
24
Three-phase motor
Instantaneous
values
uA(t) : 120 V 0º
uB(t) : 120 V -120º
uC(t) : 120 V -240º
iA(t) : 5 A -30º
iB(t) : 5 A -150º
iC(t) : 5 A -270º
pX(t) = uX(t) . iX(t)
I1A
60 Hz
25
Three-phase motor
Parameter
values
Reference: U1A
U1A at 0º
U1B at -120º
U1C at -240º
length: relative to U1A or largest U1X
I1A at -30º
I1B at -150º
I1C 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
26
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
27
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
28
Unbalanced system decomposition
( C.L. Fortescue 1918 )
A+
B+
C+
A-
B-
C-
A0
B0 C0
positive sequence
negative sequence
zero sequence
B1
A1
C1
29
Symmetrical components
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
-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 system
Balanced voltages
U1A=U1B=U1C = 120 V
0º, -120º, -240º
31
Symmetrical components – example
Unbalanced currents
I1A = 5 A -30º
I1B = 6 A -160º
I1C = 4 A -260º
Neutral current
i1n(t) = i1A(t)+i1B(t)+i1C(t)
I1n = 2.09 A -163º
• To each symmetrical component
Classical power applies
• Positive sequence
– Active current
– Reactive current
• Negative component , Zero component
Slide 32
Symmetrical components
Combined phase currents:
33
Three-phase line loss
Symmetrical currents:
Current
components:
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 level I1
-
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
34
Three-phase line loss – line resistance
Line loss: Ploss = 3 . IL2 . RL + In
2 . RLn
IL2 = (I1a
+)2 + (I1r+)2 + (I1U)2 + (IN)2
35
Three-phase 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
36
Example
• 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 37
Conclusion
Thank you for your attention
Questions ????
© FLUKE Europe BV 38