1. 1 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction BASICS OF POWER FACTOR CORRECTION

2. 2 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Harmonics Reactive power Unsymmetrical load Flicker GRID LOAD Loads create disturbances

3. 3 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Caused by: Reactive power Inductive loads, Power Electronics Harmonics Power electronics, non-linear loads Commutation Converter and drives Voltage sags an swells Load variations, high inrush currents Unsymetric grids Unbalanced single phase loads Radio frequencies Ripple control Voltage interruptions Lightning, over load, switching operations Different aspects of electrical power quality

4. 4 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction What are the different types of loads? Ohmic loads Lighting bulbs Iron Resistive heating Capacative loads Capacitors Underground cables Over excited synchronous generators Inductive loads Electrical Motors Transformers Reactors/chokes Overhead lines Under excited Synchronous generators Discharge lamps Power electronic GRID

5. 5 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Three different types of loads: 1. OHMIC-LOADS Ohmic loads U and I in phase Phase shift = 0 No penalty  In resistive circuits the voltage and current waveforms reach their peaks and troughs as well as the electrical zeros at the same instant of time.  The voltage and current are said to be in phase (  = 0°) and the entire input power is converted into active power. Thus, resistive circuits have a unity power factor.  The ohmic resistance does not depend on frequency. U - Voltage I - Current  =0°

6. 6 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Three different types of loads: 2. INDUCTIVE-LOADS Inductive loads U is 90° ahead of I 90° phase shift Penalty!  Most of the industrial loads are inductive in nature e.g. motors, transformers etc. Due to inductive reactance of the load, the current drawn by the load lags behind the voltage waveform electrically by an angle .  The magnitude of  is proportional to the inductive reactance. Since the current lags behind the voltage, inductive loads are said to have a lagging power factor.  Impedance-X L = 2 * 3.14 * f * L U - Voltage I - Current  =90°

7. 7 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Three different types of loads: 2. INDUCTIVE-LOADS  Inductive loads cause a phase shift between current and voltage.  A positive as well as a negative power can be observed.

8. 8 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Three different types of loads: 3. CAPACITIVE-LOADS Capacative loads I is 90° ahead of U 90° phase shift Over compensation is risky!  Due to capacitive reactance of the load, the current drawn by the load is ahead the voltage by an angle .  The magnitude of  is proportional to the capacitive reactance.  Impedance U - Voltage I - Current  =90°

9. 9 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Three different types of electrical power S = Apparent Power P = Active Power Q = Reactive Power P Q1Q1 QCQC S1S1 Q 2 =Q 1 -QCQC  1 S2S2 2 

10. 10 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Three different types of electrical power S = Apparent Power P = Active Power Q = Reactive Power

11. 11 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction What is Active Power? The amount of input power which is converted into output power, is termed as “active power” and is generally indicated by P. The active Power is defined by the following formula. [W] Ideally, entire input power i.e. apparent power should get converted into the useful output, i.e. heating of an oven, movement of an motor, light of an bulb.

12. 12 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction What is Reactive Power? Electrical machines work on the principle of conversion of electromagnetic energy.(e.g. electric motors, transformers). A part of input energy is consumed for creating and maintaining the magnetic field. This part of the input energy cannot be converted into active energy and is returned to the electrical network on removal of the magnetic field. This power is known as “reactive‘‘ power Q and is defined as follows. [VAr]

13. 13 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction What is Apparent Power? Applications of electrical equipment are based on conversion of electrical energy into some other form of energy. The electrical power drawn by an equipment from the source is termed as Apparent Power, and consists of active and reactive power. The current measured with a clamp amp indicates the apparent power. It is defined as follows: [VA]

14. 14 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction What is the power factor? Power factor = cos  cos-phi = P (kW) / S (kVA)

15. 15 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Typical power factors in industries

16. 16 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Why to improve the power factor?  System kVA- release  Reduction of power bill (short pay back time: 6-18 month usually)  Reduction of ohmic losses  Power Quality improvement (harmonics, voltage sags..)  Higher kW loading of transmission and distribution equipment and/or smaller dimensioning of this equipment (cable, transformer, bus bars,...)

17. 17 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction How to improve the power factor?  PFC Capacitors  Over-excited synchronous generators  Active (real time) compensation  Reduce amount of inductive load  Usage of modern converter technology

18. 18 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Principle of PFC

19. 19 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Principle of PFC Mechanical or thermal work Generation of magnetic field Active EnergyReactive Energy Capacitor Supply Corriente 0 Load 95 Current 65

20. 20 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC  Individual compensation  Group compensation  Centralised automatic compensation  Combined compensation  Active (real time, by means of Semiconductors) PFC

21. 21 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 1. Individual (fixed) Compensation Disadvantages  Many small capacitors are more expensive than one single capacitor of total equivalent rating  Low utilization factor of capacitors for equipment not often in operation In fixed compensation, capacitors are directly connected to the terminals of the individual load (e.g. motor, transformer), and switched by means of the load contactor or CB together with the load. Advantages  kvar produced on the spot  Reduction of line losses  Reduction of voltage drops  Saving of switch gear

22. 22 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 1. Individual Compensation - lighting

23. 23 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 1. Individual Compensation - motor

24. 24 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 1. Individual Compensation - motor For compensating of asynchronous motors the capacitor output should be maximum 90 % of no load reactive power of the motor. Higher kvar ratings lead to self excitation of the motor after disconnection from the grid. Risk of over voltage > 1,1 * U nominal ! Recommended kvar size ensures a PF of 0,9 in low load as well as full load operation of the motor. A thumb rule recommends: kvar = 35% of active power (kW) of a motor Active power can be found on the rating plate of a motor.

25. 25 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 1. Individual Compensation - transformer PFC on LV bus bar Compensation of no load reactive power of the transformer Voltage increase on LV side

26. 26 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 1. Individual Compensation - transformers For compensation of no-load reactive power of transformers the kvar output of the capacitor is based on the reactive power consumption of the transformer itself. The recommended values compensate the magnetizing power of a not loaded transformer only. The following approximation formula can be used: Q o = S o = i o x S N / 100 Q o = Transformer no-load reactive power in kvar S o = Transformer no-load apparent power in kVA i o = Transformer no-load current in % of the nominal current S N = Transformer nominal power in kVA

27. 27 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 1. Individual Compensation - transformers

28. 28 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 2. Group Compensation Disadvantages  No load reduction, loss reduction, voltage drop reduction on individual load lines In group compensation, capacitors are connected to a group of loads (e.g. motors), and switched by means of the main load contactor or CB together with the load. Advantages  Reduction of capital investment  Losses reduced in distribution lines  Voltage drops reduced in distribution lines  Higher utilization factor of capacitors

29. 29 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 3. Centralized Compensation Disadvantages  Load not lightened on distribution lines within a factory In factories with many loads of different output and operating times fixed compensation is usually to costly and non-effective. The most economic solution for complex applications is usually a centralized automatic capacitor bank, controlled by a automatic PF controller. Point of connection is usually in the main distribution panel close to the transformer. Advantages  Best utilization of the capacitors  Most cost effective solution  Easier supervision  Automatic control

30. 30 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 3. Centralized Compensation

31. 31 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Frequently asked questions  What is the thumb rule for selection of kvar size for motor fixed compensation?  How to find the active load of a motor for calculating the capacitor size?  In factories with many loads it is problematic to calculate the required capacitor output during planning status. 1) Why? 2) How to select a suitable kvar size?  When to select: A) FixedB) GroupC) Centralised - compensation

32. 32 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 4. Active Compensation Disadvantages  Requires high capital investment  High engineering efforts required  Higher losses of electronic switches Typical applications:  Cranes, Lifts  Spot welding, punching... e.g. car industry  Paper mills, semiconductors,....  All kind of short term loads Advantages  Real time compensation  Reduction of reactive energy costs  Opens new fields of applications  Improvement of power quality

33. 33 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 4. Active Compensation

34. 34 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Caledonian Paper plc Irvine, Scotland Production of 325 tons deposited paper per year year Total load: 47 MVA 10MVA sensitive load 11kV (50 Hz) factory grid supplied from 132 kV HV Scottish Power Lowest voltage sag 34% of nominal voltages 37 voltage sags per year Methods of PFC: 4. Active PFC - Reference project

35. 35 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Methods of PFC: 5. Active harmonic filter

36. 36 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction

37. 37 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Capacitor bank Qc = 10 * 25 kVAr HV Grid Transformer 630 kVA, u k = 5 % Current = 666 A 300 kW cos  = 0.65 M 3 ~ HV Grid Transformer 630 kVA, u k = 5 % Current = ??? 300 kW cos  = 0.65 M 3 ~ Current reduction: ??? Example: Current reduction in main supply cable?

38. 38 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Capacitor bank Qc = ??? kvar HV Grid Transformer 630 kVA, u k = 5 % Current = 666 A 300 kW cos  = 0.65 M 3 ~ HV Grid Transformer 630 kVA, u k = 5 % 300 kW cos  = 0.65 M 3 ~ Example: Required kvars for target PF of 0.98?

39. 39 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Capacitor bank Qc = 4 * 25 kVAr HV Grid Transformer 630 kVA, u k = 5 % Current = 666 A 300 kW cos  = 0.65 M 3 ~ HV Grid Transformer 630 kVA, u k = 5 % Current = ??? 300 kW cos  = 0.65 M 3 ~ System power factor, cos  = ?? Example: PF improvement?

40. 40 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Question:  An electricity bill of a petrochemical factory shows a monthly demand of 720 000 kvarh reactive work  Monthly billing: 720 000 kvarh * x $  Daily operation, 24 hours a day  What remedial measures have do be carried out to reduce the electricity bill? Example for PFC: Kvar calculation based on electricity bill (kvarh)

41. 41 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Example for PFC: Kvar calculation based on electricity bill (kvarh) Answer:  Additional capacitor output has to be installed.  According the following formula the required kvar output can be calculated: Q in kvar = W in kvarh / time in h  Time = 30 days * 24 hours  Q = 720 000 / 30 / 24 = 1000 kvar  By installing a 1000 kvar (2*50+9*100) capacitor bank the customer will eliminate the kvarh consumption down to ZERO.

42. 42 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Capacitor bank Qc = ?? kvar HV Grid Transformer 630 kVA, u k = 5 % 300 kW cos  = 0.65 M 3 ~ Question: A textile factory with a total load of 300 kW shows an actual power factor of 0.65 (phi=49,5°) The local power utility asks for a target PF=0.95 (phi=18,2 °) What capacitor output is required to avoid surcharges for low PF? Example

43. 43 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction Solution:  Qc = P * (tang phi1 - tang phi 2) = 300*(tan (49.5) - tan(18,2)) = 252 kvar  For a proper fine tuning of the target PF we recommend a capacitor bank design: 25 + 25 + 50 + 50 + 50 + 50 kvar  Depending on types of loads, e.g. frequency converters, de-tuned capacitor banks should be used Example

44. 44 March 2001 EPCOS FK PM PFC Basics of Power Factor Correction 35 kV 110 kV 16 MVA Station 10 1.6 MVA 4 % 320 kW (Converter) 1x500 kW 2x250 kW (Converter) 16 MVA Station 15 1.6 MVA 4 % 5x300 kW (Converter) 1.6 MVA 4 % 4x300 kW (Converter) n. o. 3x5 MVA Power uility S k "=310...360 MVA Harmonic Filter Single line diagram essential for system study