1. Line Losses & Unified Power By Mark B Bakker Tech Sales Manager @ FLUKE EUROPE © FLUKE Europe BV1

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 2 Presenter © FLUKE Europe BV

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 3 Summary © FLUKE Europe BV

4. Single phase: Active:P = 1/T ∫(u(t) i(t))(W) = U I cos(φ) Apparent:S = U I(VA) Reactive:Q = √(S 2 – P 2 ) (var) = U I sin(φ) Classical Power (Steinmetz 1897) © FLUKE Europe BV Slide 4

5. Reactive Power (analogy) S P work = force x length Q AC network: S 2 = P 2 + Q 2 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

6. Classical Power Three phase: Active:P T = P A + P B + P C Apparent:S T = S A + S B + S C (arithmetic) S T = √(P T 2 + Q T 2 ) (vector) ReactiveQ T = Q A + Q B + Q C Company ConfidentialFluke 430 Series II Slide 6

7. 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

8. IEEE 1459-2010 Power 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 ConfidentialFluke 430 Series II Slide 8

9. 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 ConfidentialFluke 430 Series II Slide 9

10. 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 ConfidentialFluke 430 Series II Slide 10

11. In collaboration with the University of Valencia – Unified Power From scientific breakthrough to reality © FLUKE Europe BV Slide 11

12. 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

13. A C B IAIA ICIC InIn IBIB UAUA UCUC UBUB A C B U' A U' C U' B ≈ ≈ ≈ R LA R LC R LB R Ln sourceload line UnUn 13 Three-phase line loss

14. Line loss:P loss = 3. I L 2. R L + I n 2. R Ln I L 2 = (I 1a + ) 2 + (I 1r + ) 2 + (I 1U ) 2 + (I N ) 2 14 Three-phase line loss – line losses loss due to: active current loss due to: reactive current loss due to: harmonics & interharmonics loss due to: unbalance loss due to: neutral current

15. If we know – I A, I B, I c and the components – I n – R A, R B, R C – R Ln We can calculate the Power loss in the lines P = I 2 * R © FLUKE Europe BV Slide 15 So if we know

16. Current Decomposition – Harmonics (Non fundamental) – Unbalance © FLUKE Europe BV Slide 16 Current Decomposition

17. 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

18. L IRIR ILIL ITIT ITIT 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

19. 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

20. The distorted waveform contains Harmonics…. © Fluke Europe B.V. 20 Distortion Power Quality Workshop What do we mean by Harmonics?

21. WAVEFORMS -400 -300 -200 -100 0 100 200 300 400 05101520 TIME V (RMS) 50 Hz 150 Hz 350 Hz Sum © Fluke Europe B.V. 21 Signal decomposition Power Quality Workshop

22. 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

23. © Fluke Europe B.V. 23 Fourier transform Power Quality Workshop -400 - 300 -200 - 100 0 100 200 300 400 05101520 TIME V -400 - 300 -200 - 100 0 100 200 300 400 05101520 TIME V -400 - 300 -200 - 100 0 100 200 300 400 05101520 TIME V -400 - 300 -200 - 100 0 100 200 300 400 05101520 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

24. Current Decomposition – Harmonics (Non fundamental) – Unbalance © FLUKE Europe BV Slide 24 Current Decomposition

25. 25 three-phase motor Instantaneous values u A (t) : 120 V  0 º u B (t) : 120 V  -120 º u C (t) : 120 V  -240 º i A (t) : 5 A  -30 º i B (t) : 5 A  -150 º i C (t) : 5 A  -270 º p X (t) = u X (t). i X (t)

26. I 1A 60 Hz 26 three-phase motor Parameter values Reference: U 1A U 1A at 0º U 1B at -120º U 1C at -240º length: relative to U 1A or largest U 1X I 1A at -30º I 1B at -150º I 1C at -270º length: relative to I 1A or largest I 1X rotation: anti-CW All values are fundamental! C B A U 1C U 1A I 1C U 1B -30º I 1B

27. 27 Positive & negative sequence Positive sequence A – B – C (motor runs forward) I 1A 60 Hz C B A U 1C U 1A I 1C U 1B -30º I 1B

28. 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) I 1A 60 Hz C B A U 1B U 1A I 1B U 1C -30º I 1C

29. 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: U 1A, U 1B, U 1C → U +, U -, U 0 I 1A, I 1B, I 1C → I +, I -, I 0 29 Unbalanced system decomposition ( C.L. Fortescue 1918 )

30. A-A- B-B- C-C- A0B0C0A0B0C0 A+A+ B+B+ C+C+ A-A- B-B- C-C- A0B0C0A0B0C0 positive sequence negative sequence zero sequence A+A+ C+C+ B+B+ B1B1 A1A1 C1C1 30 Symmetrical components

31. u 1A (t)u 1B (t)u 1C (t) i 1A (t) i 1B (t) i 1C (t) i 1n (t) Unbalanced system Balanced voltages U 1A =U 1B =U 1C = 120 V  0º,  -120º,  -240º 31 Symmetrical components – example Unbalanced currents I 1A = 5 A  -30º I 1B = 6 A  -160º I 1C = 4 A  -260º Neutral current i 1n (t) = i 1A (t)+i 1B (t)+i 1C (t) I 1n = 2.09 A  -163º

32. To each component Classical power applies Positive sequence – Active current – Reactive current Negative component, Zero component Slide 32 Symmetrical components

33. A C B IAIA ICIC InIn IBIB UAUA UCUC UBUB A C B U' A U' C U' B ≈ ≈ ≈ R LA R LC R LB R Ln sourceload line UnUn 33 Three-phase line loss

34. Combined phase currents: 34 Three-phase line loss Symmetrical currents: Current components: Fundamental phase currents: I 1a + active I 1r + reactive I 1U unbalance ININ non-fund. InIn neutral I A, I B, I C, I n I 1A, I 1B, I 1C I1+I1+ phase level phase level system level I1-I1- I10I10

35. System line resistance: l line : length of the line A wire : cross area of the wire  wire : specific wire resistance Estimation: R line = 1% of (P nom ) / (I nom 2 ) 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

36. Line loss:P loss = 3. I L 2. R L + I n 2. R Ln I L 2 = (I 1a + ) 2 + (I 1r + ) 2 + (I 1U ) 2 + (I N ) 2 36 Three-phase line loss – line losses loss due to: active current loss due to: reactive current loss due to: harmonics & interharmonics loss due to: unbalance loss due to: neutral current

37. 37 example

38. 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 Conclusion

39. 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 ConfidentialFluke 430 Series II Slide 39 © FLUKE Europe BV Slide 39

40. 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 ConfidentialFluke 430 Series II Slide 40 What you see on Energy Loss © FLUKE Europe BV Slide 40

41. Company ConfidentialFluke 430 Series II Slide 41 Where do the numbers come from These five values are directly calculated according to IEEE 1459. © FLUKE Europe BV Slide 41

42. Company ConfidentialFluke 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. © FLUKE Europe BV Slide 42

43. Thanks for your attention © FLUKE Europe BV43