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Power Quality inLV installations

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PQ problems in LV installations

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Key elements of poor LV Power Quality

Harmonics

Reactive power

Load imbalance

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Reasons for investing in Power Quality

Traditional reasonsn Technical problems leading to system downtime

n Production loss

n Compliance with regulations (local/IEC/company standards)n Penalties if no compliancen No connection if no compliance

But alson Energy savings potential

n Poor Power Quality results in higher system lossesn A topic which is becoming more important due to increasing energy prices

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Introduction: Types of linear load

n Three categories of linear loads

n RESISTIVE

n INDUCTIVE

n CAPACITIVE

Ohms Ω

Henry H

Farad F

R

L

C

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Introduction: Categories of linear loads

n Resistive load

Example: Traditional resistive heater

If ϕ = 0°, active power is transferred at minimal current

If ϕ = 0°, cos ϕ = Displacement Power Factor (DPF) = 1Active Power P = U2/R [W]

Ohms (Ω)

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VACIR

VACIR

ω

IRVAC R

Phasor representation:

ϕ = 0°ϕ = 0°

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Introduction: Categories of linear loads

n Inductive load

Example: Ideal reactor

If ϕ = 90°, no active power is transferred although current flows

If ϕ = 90°, cos ϕ = Displacement Power Factor (DPF) = 0

Reactive Power Q = U2/(ωL) [var]

Henry (Η)

Phasor representation:

ϕ = 90° lagging

ILVAC L

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VACIL

VAC

IL

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n Capacitive load

Example: Ideal capacitor

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)*U2 [var]

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VACIC

Introduction: Categories of linear loads

Farad (F)

Phasor representation:

ICVAC C

ϕ = 90° VAC

IC ω

ϕ = 90° leading

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Real-life linear loads

kW

- Active power (kW) F performs the work (useful power)

kvar

- Reactive power (kvar) F sustains electromagnetic field (non useful power)

kVA

- Apparent power (kVA) F total power consumed

n All real-life inductive loads require two kinds of power to function properly:

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Power triangle for real-life inductive loads

kVA

Apparent power

Active or useful powerkW

Reactive power kvar

kvar

Displacement Power factorcos ϕ = kW / kVA

ϕ

Example: Active power at full load : 410kW Initial power factor : cos ϕ = 0.7Thenè Apparent power kVA= 410/0.7= 586 kVAè Reactive power kvar = kVA* sin ϕ=586*0.714=419 kvarè 846 A current is flowing for 592 A of ‘useful current’è Inefficient use of the feeding network

kvar²kW²kVA +=

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Effects of a poor power factor

n The feeding transformer losses increaseè increased temperature, è reduced lifetime

n The transformer/generator is fully loadedè not possible to add new loads

n The feeding cable losses increaseè increased temperatureè reduced lifetime

n Protecting devices may tripn The voltage output of the transformer may become too lowè loads may refuse to start

n Impermissible voltage variations on the feeding network may resultn Utility requirements for cos ϕ and voltage variations may not be metè Penalties may be levied

M

Active power PReactive power Q

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Examples of loads and their typical cos ϕn Traditional motors – Direct On Line

cos ϕ when running steady state: 0.7 – 0.85Pay attention to start-up:

Istart-up is high (e.g. 5 times In),cos ϕ = 0.2-0.3

n AC motors controlled by VFD’s: cos ϕ = 0.97 - 1 (Power Factor may be lower!)

n TL/CFL lamps (non-compensated)cos ϕ = 0.5 (compensated lamps will have higher cos ϕ)

n PCs/laptops/Serverscos ϕ = 0.97 - 1 (compensated units may have capacitive cos ϕ)

n Conventional welding equipmentcos ϕ = 0.4 with Iwelding significantly higher than In

n Thyristor controlled equipmentcos ϕ = 0.3 - 0.9 (according to operating point of the equipment)

0 5 10 15 20 25 30 35 40 45 500

100200300400500600

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Cos ϕ Power factor

Starting current of VFD driven AC motor

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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)

kW

kvar_c

kVAkvar1

ϕ = 90° lagging

VAC

IL

ω

IC

IR

ϕ = 90° leading

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Improving the power factor: principle

kW

kvar1kVA1

ϕ1 ϕ1 kW

kvar1kVA1

kvar_c

ϕ1kW

kvar1kVA1

ϕ2

kVA2

kvar_c

kvar2

)2 tan1 (tankWkvar_c ϕ−ϕ•=

Load representation: Load + capacitor representation:

Result:

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Improving the power factor: example

Active power at full load: 410kWInitial power factor: cos ϕ1 = 0.7Initial Apparent power

kVA1= 410/0.7 = 586 kVA

Desired power factor: cos ϕ 2 = 0.95Needed kvar :

= 410 ( tan ϕ 1-tan ϕ 2)= 410 (1.020-0.329) = 284 kvar

Final Apparent power kVA2= 410/0.95 = 432 kVA

kW=410

kVA1=586

kvar1=419

ϕ1 ϕ2

kVA2=432

kvar2=135284

è 846 A current is flowing for 592 A of ‘useful current’

è 624 A current is drawn from the supply for 592 A of ‘useful current’è Very efficient use of the power supply network

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Advantages of a good power factorCurrent drawn from the network is reduced

n Transformers and distribution cables unloaded ( I)n Reduced Joule losses (RI²) in cables, transformers, protecting

devices

0.70.95

27%

46cos φ final values

cos φ initial values

% reduction in lossesCable loss reduction with increased cos ϕCurrent reduction with increased cos ϕ

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n Voltage drop reduced in cables, transformers and feeding network

n Increased power available at transformer terminals

n No penalty from electricity supplier

n Possibly sponsorship of government for good cos ϕ

Advantages of a good power factor

5.1

Voltage drop

cos φ0.6

Normal lossesLow losses

Voltage drop in transformers as a function of load cos ϕ

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Energy saving calculation

400 kVA400 V

P=80 kWcos φ = 0.75

P=170 kWcos φ = 0.75

1 cable70 mm²20 m

2 cables70 mm²100 m

A B

Consumer's electricity bill: • Monthly kWh consumed = 70.125 • Monthly kvarh consumed = 63.081 • 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 ϕ = kW/ √kW² + kvar² = 0.74

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Energy saving calculationMin cos ϕ value to avoid penalty : 0.93 Õ capacitor required = kW (tgϕ1 - tgϕ2) =165 (0.898 - 0.395) = 83kvar (in practice cap. size is increased by 5 to 10% to ensure a P.F.> 0.93) New situation kW = 165 (unchanged) kvar = 148 - 83 = 65 (â) kVA = √165² + 65² = 177 (â) (222 previously)

Monthly demand (kVA) is reduced by 20%!Transformer and cable losses are also reduced!

⇒ Typical pay-back period for a contactor switched capacitor bankis from a few months to two or so years

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Effects of load unbalance

n Neutral current increase

n Voltage build-up between neutral and earth leading to technical problems

n Load imbalance leads to voltage imbalance

n Voltage imbalance leads to increased stress and reduced efficiency in motors etc.

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What are harmonics?n Integer multiples of the fundamental frequency of any periodical

waveform are called Harmonics e.g.n Acoustic waves n Electrical ‘waves’

n 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. 3rd & 5th

n Odd Harmonics (5th, 7th…..)n Even Harmonics (2nd , 4th ….)n Triplen Harmonics (3rd, 9th , 15th ..)

n Non-integer multiples of the fundamental frequency of any periodicalwaveform are called Inter-harmonics e.g. 2.5th => 125 Hz at 50 Hz base

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What is the line current from this device?

?

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Current waveform

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Fundamental only

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Fundamental + fifth harmonic

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H1 + H5 + H7

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Harmonics representation

0%

5%

10%

15%

20%

25%

5 7 11 13 17 19 23 25

Time domainFrequency domain

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Total Harmonic Distortion (THD)

n Relative importance of harmonics regarding to fundamental

n ( expressed in %)

n THD(U): meaningful

THD(I): ??? what is the reference ???

1

2k

2k

C

CTHD

∑==

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Example of variable speed drives in oil fieldLINE VOLTAGES & LINE CURRENTS AT PUMPING CLUSTER

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Voltage: THDV = 12% Current: THDI = 27%

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Harmonics dimensionless numbers

n THDF (Transformer Harmonic Derating Factor)n kVAderated = THDF*kVA

n KF (K-Factor)n Extra heat brought by harmonics

n TIF-Factor (Telephone Interference Factor)n Describes interference of a power transmission line on a

telephone line

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Where do the harmonics come from?

n Power electronics, converters, drives...

n Rectifiers

n Inverters

n ...

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Where do the harmonics come from?

n Uninterruptible power supplies (UPS)

Accu

Net

wor

k Load

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Where do the harmonics come from?

n Fluorescent lighting systems

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Where do the harmonics come from?

n Computers

n Printers

n Faxing machines

n ...

Small but ...

If lots of such deviceson same transformer

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Where to find harmonic loads: summarized

Harmonic (non-linear) loads areeverywhere

and in ever increasing number

n Industrial loads (mainly 3-wire systems)n AC and DC drives (UPS systems)

n Harmonics between phases, imbalance, sometimes reactive power

n Commercial loads (mainly 4-wire systems)n All office equipment such as computers,

saving lamps, photocopiers, fax machines, …n Harmonics in neutral and between phases, imbalance,

sometimes reactive power

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Effects of ever increasing harmonics

Peak 100% 133% 168% 204%

RMS 100% 105% 108% 110%

à Modification of the peak value of the waveform à Increase of the RMS value of the waveform

0% 33% 39% 44%

Total harmonic distortion

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Problems created by harmonics

n Nuisance tripping of circuit breaker

n Increase of RMS → Thermally

n Increase of peak → Magnetically

n Blown fuses

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Problems created by harmonics

n Excessive heating of devicesDistortion → Increase of RMS

Losses # R . I2RMS = R . I12 + R . Σ Ih2

Extra heat broughtby harmonics

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Problems created by harmonics

n Excessive harmonic currentmay lead to overheating (or even burning) of network components

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Problems created by harmonics

n Motor problems

n Additional losses in windings & iron (RMS increase & skin effect)

n Perturbing torques on shaft (negative sequence harmonics)

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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

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Problems created by harmonics

n Damage to electronic sensitive equipments

n Electronic communications interferences

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Problems created by harmonics

n Excessive neutral current(mainly zero-sequence harmonics)

n …

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Problems created by harmonics

n Capacitor problemsn Decrease of impedance with frequency

n Resonance problems

Frequency

ZC # 1/f & =

Capacitor overload

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Problems created by harmonics

n 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

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Capacitor banks and harmonics: resonance

n When harmonics are present in the network, special care has to be given to the capacitor bank design.

è This is due to possible resonance excitation

M

U

U

Series resonance path:

Parallel resonance path:

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Capacitor banks and harmonics: resonance

n Series resonance leads tohigh capacitor stress due to background harmonic distortion

n Parallel resonance leads to high voltage and current stress due to the injection of harmonic currents into a high impedance

Frequencyω0

Serie

s im

peda

nce

R

Capacitive behaviour

Inductivebehaviour

Frequencyω0

Para

llel i

mpe

danc

e ∞ Capacitive behaviour

Inductivebehaviour

è When harmonics trigger a resonance frequency, a lot of damage may result!

n SQ

scT

C0 =

n SQ

scT

C0 =

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n 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

è Install contactor switched capacitor bank 5 steps, 25 kvar/step

è Resulting cos ϕ = 0.93 and bank stress acceptable OK

Capacitor banks and harmonics: resonance

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n Example:Case 2: DC drive load: 200 kW, cos ϕ = 0.7, desired cos ϕ = 0.92è If same capacitor bank is used as used before…

!!!

Capacitor banks and harmonics: resonance

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n Solution:è When harmonics are present in the network, install capacitor

banks fitted with detuning reactors.

è 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%)

Capacitor banks and harmonics: resonance

U

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Example

n Case 2 - Solution 2

Capacitor bank, 5 steps, 25 kvar/step incorporatingprotecting reactors rated at p = 7%

This solution is acceptable:- Power factor is reached- Bank stress is within specifications- Harmonic filtering effect is present

Resonance frequency below 5th harmonic

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Standards and regulations for harmonics

n Purpose:Ensure that the network distortion does not exceedpermissible levels that guarantee proper operation of connected equipment

n Typical levels and tendencies:n THDV ≤ 5% and limit for each harmonic component

(acceptable even for very sensitive loads)n Derive current limits to obtain voltage limitsn Take into account high order harmonics

n Standard references: IEC-standards, IEEE 519-1992 (USA), G5/4 (UK)