© 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

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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

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• 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

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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

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-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

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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

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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