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