ABB n.v Power Quality in LV installations

Transcription

1 ABB n.v Power Quality in LV installations

2 PQ problems in LV installations Volts Amps :25: :25: :25: :25: :25: :25:43.77 CHA Volts CHB Volts CHC Volts CHA Amps CHB Amps CHC Amps Waveform event at 22/11/01 10:25: ABB n.v Volts msec

3 Key elements of poor LV Power Quality Harmonics Reactive power Load imbalance ABB n.v

4 Reasons for investing in Power Quality Traditional reasons Technical problems leading to system downtime Production loss Compliance with regulations (local/iec/company standards) Penalties if no compliance No connection if no compliance But also Energy savings potential Poor Power Quality results in higher system losses A topic which is becoming more important due to increasing energy prices ABB n.v

5 Introduction: Types of linear load Three categories of linear loads RESISTIVE R Ohms Ω INDUCTIVE L Henry H CAPACITIVE C Farad F ABB n.v

6 Introduction: Categories of linear loads Resistive load VAC IR R Ohms (Ω) Example: Traditional resistive heater Phasor representation: ω VAC IR IR VAC ϕ = 0 ϕ = ABB n.v If ϕ = 0, active power is transferred at minimal current If ϕ = 0, cos ϕ = Displacement Power Factor (DPF) = 1 Active Power P = U 2 /R [W]

7 Introduction: Categories of linear loads Inductive load VAC IL L Henry (Η) Example: Ideal reactor Phasor representation: VAC IL ω ϕ = 90 lagging VAC ϕ = 90 IL ABB n.v If ϕ = 90, no active power is transferred although current flows If ϕ = 90, cos ϕ = Displacement Power Factor (DPF) = 0 Reactive Power Q = U 2 /(ωl) [var]

8 Introduction: Categories of linear loads Capacitive load VAC IC C Farad (F) Example: Ideal capacitor Phasor representation: VAC IC IC ω ϕ = 90 leading ABB n.v ϕ = 90 VAC If ϕ = 90, no active power is transferred although current flows If ϕ = 90, cos ϕ = Displacement Power Factor (DPF) = 0 Reactive Power Q = (ωc)*u 2 [var]

9 Real-life linear loads All real-life inductive loads require two kinds of power to function properly: -Active power (kw) F performs the work (useful power) -Reactive power (kvar) F sustains electromagnetic field (non useful power) kw kvar kva ABB n.v Apparent power (kva) F total power consumed

10 Power triangle for real-life inductive loads ABB n.v Reactive power kvar kvar Example: Active or useful power ϕ Apparent power kw kva Displacement Power factor cos ϕ = kw / kva Active power at full load : 410kW Initial power factor : cos ϕ = 0.7 Then Ł Apparent power kva= 410/0.7= 586 kva Ł Reactive power kvar = kva* sin ϕ=586*0.714=419 kvar Ł 846 A current is flowing for 592 A of useful current Ł Inefficient use of the feeding network kva = kw² + kvar²

11 Effects of a poor power factor ABB n.v The feeding transformer losses increase Ł increased temperature, Ł reduced lifetime The transformer/generator is fully loaded Ł not possible to add new loads The feeding cable losses increase Ł increased temperature Ł reduced lifetime Protecting devices may trip Active power P Reactive power Q The voltage output of the transformer may become too low Ł loads may refuse to start Impermissible voltage variations on the feeding network may result Utility requirements for cos ϕ and voltage variations may not be met Ł Penalties may be levied M

12 Examples of loads and their typical cos j ABB n.v Traditional motors Direct On Line cos ϕ when running steady state: Pay attention to start-up: Istart-up is high (e.g. 5 times In), cos ϕ = AC motors controlled by VFD s: cos ϕ = (Power Factor may be lower!) TL/CFL lamps (non-compensated) cos ϕ = 0.5 (compensated lamps will have higher cos ϕ) PCs/laptops/Servers cos ϕ = (compensated units may have capacitive cos ϕ) Conventional welding equipment cos ϕ = 0.4 Thyristor controlled equipment with Iwelding significantly higher than In cos ϕ = (according to operating point of the equipment) PF DPF Starting current of VFD driven AC motor Cos ϕ Power factor TimeSec

13 Improving the power factor: principle A capacitor connected in parallel with a load draws kvar in the same way as the load but in phase opposition (leading) IC kw kvar1 kvar_c kva ϕ = 90 leading IR ω VAC ϕ = 90 lagging IL ABB n.v

14 Improving the power factor: principle Load representation: Load + capacitor representation: kw ϕ1 ϕ1 kw kvar1 kva1 kvar_c kvar1 kva1 Result: kvar2 ϕ1 ϕ2 kw kva2 kvar_c ABB n.v kvar_c = kvar1 kw (tan ϕ1 tan ϕ2) kva1

15 Improving the power factor: example Active power at full load: 410kW Initial power factor: cos ϕ1 = 0.7 Initial Apparent power kva1= 410/0.7 = 586 kva Ł 846 A current is flowing for 592 A of useful current Desired power factor: cos ϕ 2 = 0.95 kw=410 j1 j2 kva1=586 kva2=432 kvar1=419 kvar2= Needed kvar : = 410 ( tan ϕ 1-tan ϕ 2) = 410 ( ) = 284 kvar Final Apparent power kva2 = 410/0.95 = 432 kva ABB n.v Ł 624 A current is drawn from the supply for 592 A of useful current Ł Very efficient use of the power supply network

16 Advantages of a good power factor Current drawn from the network is reduced Transformers and distribution cables unloaded ( I) Reduced Joule losses (RI²) in cables, transformers, protecting devices Current reduction with increased cos ϕ Cable loss reduction with increased cos ϕ % reduction in losses 46 cos f final values 27% ABB n.v cos f initial values

17 Advantages of a good power factor Voltage drop reduced in cables, transformers and feeding network Increased power available at transformer terminals No penalty from electricity supplier Possibly sponsorship of government for good cos ϕ Voltage drop in transformers as a function of load cos ϕ Voltage drop 5.1 Normal losses Low losses ABB n.v cos f

18 Energy saving calculation Consumer's electricity bill: 1 cable 70 mm² 20 m A 400 kva 400 V 2 cables 70 mm² 100 m B Monthly kwh consumed = Monthly kvarh consumed = Assuming an activity of : 340 days/year 15 hours/day Average kw = 70125/425 = 165 (66% of the full load) Average kvar = 63081/425 = 148 Average kva = 165² + 148² = 222 Average cos j = kw/ kw² + kvar² = 0.74 P=80 kw cos f = 0.75 P=170 kw cos f = 0.75 ABB n.v

19 Energy saving calculation Min cos j value to avoid penalty : 0.93 capacitor required = kw (tgϕ1 - tgϕ2) =165 ( ) = 83kvar (in practice cap. size is increased by 5 to 10% to ensure a P.F.> 0.93) New situation kw = 165 (unchanged) kvar = = 65 ( ) kva = 165² + 65² = 177 ( ) (222 previously) Monthly demand (kva) is reduced by 20%! Transformer and cable losses are also reduced! ABB n.v Typical pay-back period for a contactor switched capacitor bank is from a few months to two or so years

20 Effects of load unbalance Neutral current increase Voltage build-up between neutral and earth leading to technical problems Load imbalance leads to voltage imbalance Voltage imbalance leads to increased stress and reduced efficiency in motors etc. ABB n.v

21 What are harmonics? Integer multiples of the fundamental frequency of any periodical waveform are called Harmonics e.g. Acoustic waves Electrical waves ABB n.v For power networks, 50 Hz (60 Hz) is the fundamental frequency and 150 Hz (180 Hz), 250 Hz (300 Hz) etc. are higher order harmonics viz. 3 rd & 5 th Odd Harmonics (5 th, 7 th..) Even Harmonics (2 nd, 4 th.) Triplen Harmonics (3 rd, 9 th, 15 th..) Non-integer multiples of the fundamental frequency of any periodical waveform are called Inter-harmonics e.g. 2.5 th => 125 Hz at 50 Hz base

22 What is the line current from this device?? ABB n.v

23 ABB n.v Current waveform

24 ABB n.v Fundamental only

25 ABB n.v Fundamental + fifth harmonic

26 ABB n.v H1 + H5 + H7

27 Harmonics representation Time domain 25% 20% 15% 10% 5% 0% Frequency domain ABB n.v

28 Total Harmonic Distortion (THD) Relative importance of harmonics regarding to fundamental C 2 k k= 2 ( expressed in %) THD = C 1 THD(U): meaningful THD(I):??? what is the reference??? ABB n.v

29 Example of variable speed drives in oil field LINE VOLTAGES & LINE CURRENTS AT PUMPING CLUSTER Volts Amps :25: :25: :25: :25: :25: :25:43.77 CHA Volts CHB Volts CHC Volts CHA Amps CHB Amps CHC Amps ABB n.v Waveform event at 22/11/01 10:25: Voltage: THDV = 12% Current: THDI = 27%

30 Harmonics dimensionless numbers THDF (Transformer Harmonic Derating Factor) kva derated = THDF*kVA KF (K-Factor) Extra heat brought by harmonics TIF-Factor (Telephone Interference Factor) Describes interference of a power transmission line on a telephone line ABB n.v

31 Where do the harmonics come from? Power electronics, converters, drives... Rectifiers Inverters... ABB n.v

32 Where do the harmonics come from? Uninterruptible power supplies (UPS) Network Accu ABB n.v Load

33 Where do the harmonics come from? Fluorescent lighting systems ABB n.v

34 Where do the harmonics come from? Computers Printers Faxing machines... Small but... ABB n.v If lots of such devices on same transformer

35 Where to find harmonic loads: summarized Harmonic (non-linear) loads are everywhere and in ever increasing number Industrial loads (mainly 3-wire systems) AC and DC drives (UPS systems) Harmonics between phases, imbalance, sometimes reactive power Commercial loads (mainly 4-wire systems) All office equipment such as computers, saving lamps, photocopiers, fax machines, Harmonics in neutral and between phases, imbalance, sometimes reactive power ABB n.v

36 Effects of ever increasing harmonics Total harmonic distortion 0% 33% 39% 44% Peak 100% 133% 168% 204% RMS 100% 105% 108% 110% ABB n.v Modification of the peak value of the waveform Increase of the RMS value of the waveform

37 Problems created by harmonics Nuisance tripping of circuit breaker Increase of RMS Thermally Increase of peak Magnetically Blown fuses ABB n.v

38 Problems created by harmonics Excessive heating of devices Distortion Increase of RMS Losses # R. I 2 RMS = R. I R. Σ I h 2 Extra heat brought by harmonics ABB n.v

39 Problems created by harmonics Excessive harmonic current may lead to overheating (or even burning) of network components ABB n.v

40 Problems created by harmonics Motor problems Additional losses in windings & iron (RMS increase & skin effect) Perturbing torques on shaft (negative sequence harmonics) ABB n.v

41 Harmonics classification Order Group Effects n = 1 Fundamental active power n = 3k+1 + sequence heating n = 3k-1 - sequence heating & motor problems n = 3n 0 sequence heating & neutral problems ABB n.v

42 Problems created by harmonics Damage to electronic sensitive equipments Electronic communications interferences ABB n.v

43 Problems created by harmonics Excessive neutral current (mainly zero-sequence harmonics) ABB n.v

44 Problems created by harmonics Capacitor problems Decrease of impedance with frequency Resonance problems Z C # 1/f & = Frequency Capacitor overload ABB n.v

45 Problems created by harmonics Capacitor problems Due to its lower impedance, capacitors are even more susceptible to higher order harmonics. If not protected from harmonic stress, a capacitor may fail pretty soon ABB n.v

46 Capacitor banks and harmonics: resonance When harmonics are present in the network, special care has to be given to the capacitor bank design. Ł This is due to possible resonance excitation Series resonance path: U Parallel resonance path: M U ABB n.v

47 Capacitor banks and harmonics: resonance Series resonance leads to high capacitor stress due to background harmonic distortion Parallel resonance leads to high voltage and current stress due to the injection of harmonic currents into a high impedance Series impedance Parallel impedance Capacitive behaviour Inductive behaviour R Inductive behaviour w 0 Capacitive behaviour w 0 n 0 = n 0 = S Q sct C Frequency S Q sct C Frequency ABB n.v Ł When harmonics trigger a resonance frequency, a lot of damage may result!

48 Capacitor banks and harmonics: resonance Example: Feeding network: 11 kv, Ssc = 100 MVA Transformer: 11 kv/400v, 600 kva, 5.5 % Case 1: Motor load: 200 kw, cos ϕ = 0.7, desired cos ϕ = 0.92 Ł Install contactor switched capacitor bank 5 steps, 25 kvar/step Ł Resulting cos ϕ = 0.93 and bank stress acceptable OK ABB n.v

49 Capacitor banks and harmonics: resonance Example: Case 2: DC drive load: 200 kw, cos ϕ = 0.7, desired cos ϕ = 0.92 Ł If samecapacitor bank is usedas usedbefore!!! ABB n.v

50 Capacitor banks and harmonics: resonance Solution: Ł When harmonics are present in the network, install capacitor banks fitted with detuning reactors. U ABB n.v Ł Important: Tuning must always be chosen below the first significant harmonic that is present! Applications where single phase loads are present (e.g. office buildings, hotels, ), first harmonic is 3rd, hence tuning should be below 3rd (e.g. 14% or 12.5 %) Applications where only three phase loads are present (e.g. industrial plants), first harmonic is 5th, hence tuning should be below 5th (e.g. 7% or 5.67%)

51 Example Case 2 - Solution 2 Capacitor bank, 5 steps, 25 kvar/step incorporating protecting reactors rated at p = 7% ABB n.v This solution is acceptable: - Power factor is reached - Bank stress is within specifications - Harmonic filtering effect is present Resonance frequency below 5th harmonic

52 Standards and regulations for harmonics Purpose: Ensure that the network distortion does not exceed permissible levels that guarantee proper operation of connected equipment Typical levels and tendencies: THDV 5% and limit for each harmonic component (acceptable even for very sensitive loads) Derive current limits to obtain voltage limits Take into account high order harmonics Standard references: IEC-standards, IEEE (USA), G5/4 (UK) ABB n.v

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