Session 1

Review of Basics

Types of LoadsTypes of Loads

∗∗ Resistive Resistive ∗∗ Inductive Inductive ∗∗ Capacitive Capacitive

Resistive Resistive CircuitCircuit

VI

Current in Phase with Voltage

Inductive LoadsInductive Loads

V

I

Current Lagging Voltage by 900

Capacitive Loads

V

I

Current Leading Voltage by 900

Power factor correctionSome fundamental thoughts

• What is Power factor?• Why power factor is important?• Why improve power factor?• What is the power factor of various loads?• What is the origin of power factor?• How to improve power factor?

Definition of Power Factor

• Power Factor = Active Power (kW)/Apparent Power (kVA)

• Power Factor can never be greater than 1.00• Power Factor at best can be equal to 1.00• Usually P.F is always “Lag” ( Inductive)• Some times P.F can be “Lead” ( Capacitive)

Origin of Low Power Factor

• Electrical Equipment need Reactive Power• Inductive loads draw Reactive Power• Phase difference between current & Voltage

reduces “Displacement PF”.• Reactive Power to maintain magnetic fields

in Motors.• Non-Linear loads reduces “Distortion PF”.• True PF, being product of displacement and

distortion PF is lower than both.

Power Factor Improvement Concept

• Reactive Power flow analogy

• Power Triangle analogy

• Resonance analogy

Reactive Power Flow Analogy

Voltage

Current

Indu

ctiv

e L

oad

V

I

Pi

+

-

Indu

ctiv

e L

oad

Active power

Reactive power

Reactive Power Flow Analogy

Indu

ctiv

e L

oad

Reactive power

Indu

ctiv

e L

oad

Capacitor

Active power

Power Triangle Analogy

φ1P(kW)

S(kVA)

φ2

φ1

Q(kVAr)

S(kVA)

QC

-QCP(kW)

QC = P (Tan φ 1 - Tan φ 2)kVA=√(kW)2 + (kVAr)2

PF = kW/kVA = Cos φ 1

φ 1Q = P .Tan

Cos φ1

φ 2Cos= Initial Power Factor = Final Power Factor

Partially compensated LoadUncompensated Load

Power Factor Correction

Ø2Ø1

V= Line Voltage

I=Active Current

I1

I2

IR(C)

Reactive Current (capacitive)

IR(L)

Reactive Current (inductive)

Resonance Analogy-1Definition:-

Resonance is defined as a condition where Capacitive Reactance becomes equal to Inductive Reactance in magnitude. The frequency at which this occurs is called the Resonance Frequency.

Parallel Resonance

| XL| = |XC |

Inductor

|Z| =Zeq = Z1Z2 / (Z1+ Z2)

|Z| = 0

| XL| = |XC |

Inductor

Series Resonance

Zeq = Z1 + Z2

Resonance Analogy-2Uncompensated load Compensated Load

Indu

ctiv

e L

oad

Indu

ctiv

e L

oad If you make |XL| of Load = |XC|

of Capacitor at Fundamental Frequency, then the PF will be Unity due to Parallel resonance b/w capacitor & load inductor.

Inductor

Resistor

Resistor

Inductor

Resistor

Practical Example

40 W Fluorescent Tube Light

Choke

P N

230 Volts 50 Hz.P = 40W+10W = 50W

Power Factor = 0.6

Calculation of PF correction based on Power Triangle concept

Active Power = 50 W. ; Power Factor = 0.6

Apparent Power = Active Power/ PF = 83.33 VA.

Reactive Pr.= √(VA)2-(W)2 = √(83.33)2-(50)2

= 66.67 VAr.

Capacitive VAr. req. for UPF = 66.67=V2(2πf)C

Hence Capacitor req. for UPF=106x66.67/2302/100π

= 4.01 µF.

Calculation of PF correction based on Resonance concept

Equivalent Circuit of Tube Light

Inductor

Resistor

L R230V

R = V2/W = 2302/50 = 1058 Ω

XL = V2/VAr = 2302/66.67 = 793.5 Ω

L = XL/(2πf) = 793.5/100π = 2.526 H.The value of capacitive reactance required to

Resonate with the inductive reactance at the fundamental frequency is given by,

|XC| = |XL| = 793.5 Ω = 1/100πCC = 106/(793.5x100π) = 4.01µF.

230V

Inductor

ResistorL

RC

Types of Power Factors• “Displacement PF” is defined as the cosine of the angle between fundamental voltage and fundamental current of the load.

• Presence of “harmonics” increases the RMS current and voltage relative to their respective fundamental values. This increases the kVA of the load.

• The PF taking into account the effect of harmonics, called “True PF”, is lower than or at best equal to displacement PF.

• The factor by which the displacement PF is related to true PF is called the “Distortion PF”

True PF = Displacement PF x Distortion PF

• Capacitors can only improve displacement PF.

Mathematical expression of PFDisplacement Power factor = P/(V1I1)Where P = watts and V1 and I1 are fundamental voltage and current

( )√ THDV

1001 +

2Vrms= V1( )√ THDI

1001 +

2Irms= I1

True Power factor = P/ (VrmsIrms)

( )√ THDI

1001 +

2 ( )√ THDV

1001 +

2

P

V1I1

=

= Displacement PF x Distortion PF

( )√ THDV

1001 +

2( )√ THDI

1001 +

2

1Where Distortion PF =

True Power factor

Effect of harmonics on PF

%THD(V) %THD(I) Distortion PF

0 0 1.00

1 20 0.98

2 40 0.93

3 60 0.86

4 80 0.78

5 100 0.71

Three dimensional power triangle

kVA= kW2+kVAr2+kdVA2√

Displacement PF = kW

√kW2+kVAr2

kW2+kVAr2+kdVA2True PF = kW

√

kW

kVAr

kdVA

kVA

Electric Power

Activ

e Po

wer

Reac

tive

Powe

r

Apparent PowerkVA

Power Triangle

Active PowerRea

ctiv

e P o

wer

Apparent Power

kVA = √kW2 + kVAr2

kWP.F. =

kVA

PF of various Industries

Industry Power Factor

Textiles 0.65/0.75Chemical 0.75/0.85Machine shop 0.4 / 0.65Arc Welding 0.35/ 0.4Arc Furnaces 0.7 / 0.9Coreless induction furnaces and heaters 0.15/0.4Cement plants 0.78/0.8Garment factories 0.35/0.6Breweries 0.75/0.8Steel Plants 0.6 / 0.85Collieries 0.65/0.85Brick Works 0.6 / 0.75Cold Storage 0.7 / 0.8Foundries 0.5 / 0.7Plastic moulding plants 0.6 / 0.75Printing 0.55/0.7Quarries 0.5 / 0.7Rolling Mills (i.e. ,Paper, Steel , etc.) 0.3 / 0.75

Inductive LoadsInductive Loads

Induction Motor

0.8 P.F

FloursentLamp

0.5 P.F.

WeldingTransformer

0.5 P.F.

Arc Furnace

0.8 P.F

Induction Furnace

0.8 P.F

Session 2

Benefits of Power factor improvement

Reduction inTransformer Rating

Reduction in KVARDemand

Advantages of P.FCorrection

Reduction in KVADemand

Reduction in LineCurrent

Reduction in Lineloss

Reduction in Cable / Bus-bar

size

Reduction in Switchgear

Rating

REDUCTION IN KVA DEMAND

LOAD - 900 KW

EXISTING P.F. (COS - 0.6

DESIRED P.F. (COS ) - 0.92KW

KVA

Ø.

KVA 1 = 900 / 0.6 = 1500

KVA2 = 900 / 0.92 = 978

Ø 1)

Ø2

kW kVA

COS =

KVA =

Ø.

kW cosØ.

Reduction in KVA

1500 - 978 = 522

REDUCTION IN KVAR DEMAND

KW - 900

KVA1 - 1500

KVA2 - 978

kVA =

KVAR1 =

=

KVAR 2 =

√KW2 + KVAR2

√KVA12 - KW2

√1500 2 - 900 2 = 1200

√978 2 - 900 2 = 382

KW

KVA

Ø.

KVA

R

Reduction in KVAR

1200 - 382 = 818

REDUCTION IN LINE CURRENT

KVA =

I =

I1 =

=

I2 =

=

√3 V I1000

KVA x 1000√3 x 4151500 x 1000√3 x 415

2087 Amp

978 x 1000√3 x 4151361 Amp

KVA1 - 1500

KVA2 - 978

Reduction in Current

2087 - 1361 = 726

% Rise in Current w.r.t. decrease in Power Factor% Rise in I n

0102030405060708090

100

1 0.9 0.8 0.7 0.6 0.5P.F

.

Cable Losses

1- CosCos

Ø1

Ø2

2% of saving in losses = X 100

1- 0.60.92

2X 100

= 57.46

Saving in Cable Losses

0

10

20

30

40

50

60

70

80

0.5 0.6 0.7 0.8 0.9 1

P.F.1.0

0.950.90.85

0.8

Initial P.F.

Transformer Losses

Saving in losses = Wr x K1

Wr = Full load copper loss of the transformer

connected load in Kwk1

KVA rating of the transformer

1CosØ1

1CosØ2

-

Graph Transformer Losses

Transformer Losses

0

5000

10000

15000

20000

25000

30000

35000

40000

0 500 1000 1500 2000 2500 3000 3500

Transformer KVA

cu L

oss

Copper losses

Transformer Losses

Saving in losses = Wr x K11

CosØ1

1CosØ2

-

900 1 10.6 0.92

= x18000 x1500

= 6260 Watts

Annual Saving = 6260 x 300 x 121000

= 22536 KWH

Power Savings

KVA X'MER CURRENT ACBRATNG

1500

978

1500

1000

2086

1360

2500

0

500

1000

1500

2000

2500

3000

KVA X'MER CURRENT ACBRATNGAT 0.6 PF AT 0.92 PF

1600

Workshop - I1. Calculate the pf and kVA demand at the secondary of

a 1000kVA, 11/0.44 kVA transformer supplying the following loads:

– 100kW - UPF– 150kW - 0.9 lag– 250kW - 0.8 lag– 100kW - 0.9 lead

2. A 50HP 440V, 3ph, 50Hz, 1500rpm Induction motor has the following operating conditions:PF = 0.9lag, & efficiency = 90% under full load.PF = 0.6lag, & efficiency = 70% under half load.

If a 3ph. 440V capacitor of rating 12.5kVAr is connected atthe motor terminals, find the pf of the motor and capacitor combination, under: a) Full load b) Half load.

(Use 1HP = 0.746 kW for HP to kW conversion).

Session 3

Evaluating PF from Electricity Bill

ELECTRICITY TARIFFS - I

In India, there are broadly 4 types of Electricity Tariffs.– Single part tariff

• Measurement of kWh only - Energy charges -Generally applicable for LT installations only.

– Two part tariff• kWh - Energy charges • kVA or kW - Maximum demand charges• PF Penalty/Incentives - vary from region to region• Applicable for HT installations

ELECTRICITY TARIFFS – II

– Three part tariff• kWh - Energy charges • kVArh - Reactive Energy charges • kVA or kW - Maximum demand charges• PF Penalty- vary from region to region• Applicable for HT installations

– Time of day tariff • Different charges for the various quantities

mentioned above depending on the time of the day -Analogous to STD (Telecom) tariff structure

PF PENALTIES– Different structures are followed - Some common

features • Minimum monthly PF limit - Varies from 0.85 upwards • If PF falls below minimum limit then penalty is levied• The penalty is normally calculated as a %age of the

Energy charges or the full value of the Electricity Bill. • The %age of penalty is normally linked to the

difference between actual monthly PF as calculated by Electricity supply authorities and the minimum PF limit specified.

– For Ex: Minimum PF limit - 0.90– Actual monthly PF as per calculated by Electricity Supply

authorities - 0.82– Penalty is 1% for every 0.01 difference between above PF

values– Hence penalty will be 8% in this case.

INCENTIVES FOR HIGH PF

– Different structures are followed - Some common features

• Minimum Upper monthly PF limit - from 0.92 upwards • If actual monthly PF exceeds this upper limit, then

incentive is offered• The incentive is normally calculated as a %age of the

Energy charges or the full value of the Electricity Bill. • The %age of incentive is linked to the difference between

actual monthly PF as calculated by Electricity supply authorities and the minimum Upper PF limit specified.

– For Ex: Minimum Upper monthly PF limit - 0.95– Actual monthly PF as per Electricity bill - 0.98– Incentive is 1% for every 0.01 difference between above PF

values– Hence Incentive will be 3% in this case.

EVALUATING PF FROM ELECTRICITY BILL

– The value of monthly PF is normally indicated on the Electricity bill

– If this value is not shown on the Electricity bill, then the normal procedure is as follows

• Note down kWh consumed as given in the bill • Note down kVAh consumed as given in the bill • Divide kWh by kVAh • This value should always be less than 1• This gives the monthly PF as considered by the Electricity

Supply authorities.

ESTIMATE kVAr REQUIRED - I

– From Electricity bill data– Note down the value of maximum demand in kVA as

given in the Electricity bill– Convert this value to kW by multiplying the maximum

demand kVA with the monthly PF • For ex: If maximum demand kVA is 375 and monthly PF is

0.80 then, kW = 375 x 0.80 = 300 kW – Monthly PF should be assumed as “Initial PF” - 0.80 – Fix target PF as “Final PF” - Let us assume - 0.96– Note down multiplying factor from table 4.2 on Page 6

of RPM catalogue– This multiplying factor is 0.458

ESTIMATE kVAr REQUIRED - II

– Multiply the kW calculated earlier by this multiplying factor

• kVAr = 300 x 0.458 = 137.4 • kVAr - rounded off to 150 kVAr, since this is easy to offer

– Always recommend fixed compensation for the transformer in the installation - the kVAr required can be estimated from table 4.3 on Page 7 of RPM Catalogue

• For ex: If in the above installation the transformer is 500 kVA then fixed compensation required is 6% of 500 kVA which works out to 30 kVAr

– Out of 150 kVAr we can now subtract this 30 kVAr i.e., leaving a balance of 120 kVAr

ESTIMATE kVAr REQUIRED - III

– Of this 120 kVAr we can recommend additional Fixed compensation only for base load.

• If base load is given as 40% of the installation, 40% of the above kVAr can be provided as fixed compensation.

• Therefore 40% of 120 kVAr = 48 kVAr - about 50 kVAr• Consequently, the balance kVAr can be as an APFC • This works out to 120 - 50 = 70 kVAr - about 75 kVAr

– The final compensation scheme customer can be • Total Compensation - 150 kVAr - of which

– Fixed compensation - 75 kVAr – APFC - 75 kVAr

– This procedure is common to Industrial and Commercial Installations.

Workshop-2– Calculate the kVAr required to improve the pf of an

LT installation to 0.95 lag. You have the following details from the Bill.Billing date = 01.07.2001 to 31.07.2001Units consumed = 13500 kWh.Avg. PF = 0.8 lag.

– Calculate the kVAr required to improve the pf of an HT installation to 0.97 lag. You have the following details from the Bill.Contract Demand = 300 kVARecorded Demand = 270 kVAAvg. PF = 0.8 lag.Units consumed = 75600 kWh

Session 4

Evaluating kVAr for new installations

ESTIMATE kVAr REQUIRED for New Electrical Installations - I

M M M

75 HP, 415V, 3ph,

compressor

75 HP, 415V, 3ph,

compressor

20 HP, 415V, 3ph,

Pump,PF =0.80

Lag

Other loads, total of 25

kW

Refer the SLD below500kVA, 11kV/415V, %Impedance = 4.25%

50 kVA, 440V,

3ph, UPS

Lighting Load 10kW

M

30 HP, 415V, 3ph,

motor

Resistive Load 30kW

ESTIMATE kVAr REQUIRED for New Electrical Installations - I

– We must draw up a load list of the Maximum operating load (including the supply transformer) & fix the target Power Factor as desired by the Customer.

– Let us assume the load list as follows• Supply transformer - 3 Phase, 500 kVA, 11 kV/415V,• 3 Phase, 415 V, Induction motors - Totaling to 200 HP• 3 Phase, 415 V, UPS system - 50 kVA• Resistive heating load - 30 kW • Lighting load (Fluorescent) - 10 kW• Miscellaneous loads - 14 kW

– Let us assume that the target Power Factor as desired by the Customer is 0.95.

ESTIMATE kVAr REQUIRED - for New Electrical Installations - II

– The kVAr can be estimated as follows:• The kVAr required for the supply transformer can be

estimated from Table 4.3 on pg 7 of RPM Cat. – For 500 kVA transformer kVAr = 30 kVAr

• Convert induction motor rating from HP to kW - 200 HP x 0.746 = 150 kW

– Assume that initial PF of motors is likely to be around 0.7, because of the fact that motors are generally oversized due to other considerations.

– Calculate kVAr by using multiplying factor as given in Table 4.2 on Page 6 of RPM Catalogue - Multiplying factor for initial PF of 0.7 and final PF of 0.95 = 0.692.

– Hence, kVAr = 150 x 0.692 = 104 kVAr

ESTIMATE kVAr REQUIRED - for New Electrical Installations - III

– The kVAr can be estimated as follows:• Convert the UPS system kVA to kW by assuming a PF of

0.7 to be on the safer side. Hence, kVAr required for the UPS is 25 kVAr.

• The resistive heating loads are unity Power Factor loads and hence do not consume Reactive Power. Hence, kVAr compensation is not required.

• For other loads i.e., fluorescent lighting - 10 kW and miscellaneous loads of 14 kW assume an average PF of 0.7. Hence kVAr works out to about 17 kVAr.

ESTIMATE kVAr REQUIRED - for New Electrical Installations - IV

– The total kVAr can be estimated as follows:• Transformer - 30 kVAr , Induction motors - 104 kVAr, UPS - 25

kVAr, Other loads - 17 kVAr: Total kVAr = 176 • Round off on the higher side by about 15% since, significant

assumptions have been made in the calculation. • Hence, total kVAr recommended can be 200 kVAr.• Capacitors can be connected at motor terminals.The total kVAr of

such Capacitors may be subtracted from the figure of 200 kVAr. • For calculating the balance fixed compensation and APFC

combination, the procedure given earlier applies.

HOW TO CALCULATE SAVINGS

• If Power Factor is improved cash savingsarise due to the following :-– Reduction in kVA demand charges– Elimination of Penalties for low Power Factor– Incentives for maintaining high Power Factor

• Hence, we must calculate the above savings as given in the workshop 3 problems.

Workshop-3

1. In problem 1 of workshop-2, if the PF penalty is 5% of kWh charges, calculate the savings when the PF is improved to 0.95 lag.Unit charges = Rs. 3.5/ kWh.Unit consumed = 13500 kWh.

2. In problem 2 of workshop-2, the penalty clause is as follows:kVA charges = Rs. 180/kVA (if Demand < CD)

= Rs. 180 x 3 ( if Demand > CD)Unit charges = Rs. 3.4 / kWh.Units consumed = 56700 kWh PF penalty = 5% on energy consumed

CLAUSES, TERMS & CONDITIONS

• All calculations are done on assumptions of certain Electrical loading in the installation.

• If any changes occur due to modifications/ revisions of load data and characteristics the desired PF may not be achieved

• This is particularly important in Indian conditions, since data given by the Customer is not always accurate.

• This issue must be kept in mind when dealing with customers.

Session 5

Methods of improving power factor

Methods of Improving Power Factor

Fixed Compensation– For Steady Loads– No load compensation of Motors– No load compensation of Transformers

Variable CompensationFor Varying Loads

Selection of Capacitor

1. Individual Compensation

2. Group Compensation

3. Central Compensation

Where to install Power Factor correction Equipment

Individual Compensation

• Directly at the Load terminals• Ex. Motors, Steady loads• Gives maximum benefit to user• Not recommended for Drives• Costly solution

Where to install Power Factor correction Equipment-2

Group Compensation

• Single compensation for Group of Loads• Ex. Group of Motors• Gives moderate benefit to user• Few Capacitor Banks• Relatively easy to maintain

Where to install Power Factor correction Equipment-3

Central Compensation

• Directly connected at the incomer• Improves PF at the metering point• Line losses continue to prevail down stream• Least beneficial to user• Extremely easy to maintain

Providing compensation at the main incomer of the installation is called central compensation (pos. No. 1).

This is suitable for installations where the loads are few and situated close to the main supply. (Refer Fig. 3.1)

M MM M

No 1

Transformer

Circuit Breaker

Fig. 3.1

Supply Bus

Central Compensation

Providing compensation at •main incomer bus – central compensation. (pos. No 1)•At power distribution boards – group compensation (pos. No. 2).•At individual load terminals – individual compensation. (pos. No. 3)This is suitable for installations consisting of main receiving station,substations, several load feeders and a wide variety of loads. (refer fig 3.3)

M

Transformer

Circuit Breaker

Fig. 3.3

Supply Bus

No 3M M MNo 3 No 3No 3

No 2 No 2No 1

Central,Group and Individual Compensation

Session 6

Selection of capacitors

TYPES OF CAPACITOR TECHNOLOGIES

• MPP - METALLISED POLYPROPYLENE

• MD - MIXED DIELECTRIC

• FF/ALL PP - FILM - FOIL OR ALL POLY

PROPELENE

• MD -XL - MIXED DIELECTRIC LOW LOSS

METALISED POLYPROPELENE CAPACITOR

• MPP - METALLISED POLYPROPELENE

• METALISATION HAS BEEN DONE ON ONE SIDE OF POLY PROPELENE FILM AND USED FOR CAPACITOR WINDING

• ECNOMICAL AND COMPETITIVE DESIGN

• MPP-S - NORMAL DUTY• MPP-H - MEDIUM DUTY

PP FILM

METALLISED LAYER

SELF HEALING

DURING INTERNAL FAULTS

SELF HEALING IS PROTECTIVE FEATURE

AFTER SELF HEALING

PRESSURE SENSITIVE DEVICE

SELF HEALING PRODUCES GASES, WHICH WILL INCREASE THE PRESSURE INSIDE THE CAN.

THIS WILL CAUSE THE BELLOWS TO EXPAND. BEYOND A POINT POWER SUPPLY WILL BE CUT-OFF.

THUS BURSTING OF CAPACITOR IS PREVENTED.

MIXED DIELECTRIC

• MD - MIXED DIELECTRIC

• PP FILM, FOIL AND

PAPER ARE USED TO

FORM CAPACITOR

WINDING

PP FILMFOIL

PAPER

FILM FOIL OR APP

• FILM FOIL OR APP -

ALL POLY PROPELENE

• METAL LAYER IS

PLACED IN - BETWEEN

PP FILM TO FORM

CAPACITOR WINDING

PP FILMFOIL

PP FILM

MIXED DIELECTRIC -LOW LOSS

• MD-XL - MIXED

DIELECTRIC LOW LOSS

• PP FILM AND DOUBLE

SIDED METALISED FILM

ARE USED TO FORM

CAPACITOR WINDING

PP FILM

DOUBLE SIDE METALLISED PAPER

Film foil/APP verses Mixed dielectric comparison

Film foil/APP Mixed dielectric

• low dielectric watt loss

• Film not impregnable

• More prone to ‘Self healing’

• Inferior long term stability

• Moderate harmonic overload

capability

• High dielectric watt loss

• Paper impregnable

• less prone to ‘Self healing’

• Superior long term stability

• Good harmonic overload

capability

Mixed dielectric verses MDXL Comparison

MDXLMixed dielectric

• High dielectric watt loss

• Paper impregnable

• less prone to ‘Self healing’

• Superior long term stability

• Good harmonic overload

capability

• Lowest dielectric watt loss

• Combines plus points of MD

and APP types

• Excellent long term stability

• Superior harmonic overload

capability

Comparison of TechnologiesMPP-S Rating MPP-H Rating MD Rating MD-XL Rating FF Rating

Life Optimum 3 Long life 5 Long

life 10 Long life 10 Long

life 10

Non-linear loads

Capability

Up to 10 %

1Up to 15 %

6 Up to 25 % 10

Up to25 %

10 Up to 25 % 10

Initial cost Lowest 10 Medium 4 Highest 1 High 1 Highest 1

Operating cost Lowest 10 Lowest 10 Highest 1 Lowest 10 high 4

Total 24 25 22 31 25

Cylindrical verses stand-alone typeStand-alone TypeCylindrical Type

• Compact Size• Better heat dissipation• Discharge resistor in ∆• Minimal internal wires• Suitable only for panels• MPP-S,MPP-H,MDXL• Not repairable

• Bulky• Inferior heat dissipation• Two resistor configuration• More Internal wires• Robust construction• Available in all types• Elements can be replaced

Gas filled capacitors• Only made by EPCOS & Electro-Nicon (Germany)• Considered to be better heat dissipation than oil• Debated by ABB, hence controversial • Equivalent to MPP but EPCOS claiming as APP• Inferior to MDXL (MKV of EPCOS)• SF6 banned for LV application, hence nitrogen• Leakage not noticeable and failure is sudden• Lighter in weight• Generally available in 10 to 25 kVAr. units

Gas filled capacitors from L&T Meher

• To fill the Technological gap • Design improvement over EPCOS• Protective coating on element ensures

longer life even after gas leakage.• Available in the financial year 2004-05

Launch of Resin filled Capacitors

• Jelly Resin has much better di-electric properties compared to Gas.

• Meher is switching to resin filled capacitors.• Oil filled capacitors will also be available

on request.• However MDXL will continue with oil

Competitive edge of MEHER• Comprehensive test facility in Meher works.• Raw materials imported from premium source.• Automatic element winding machine.• Robot spraying machine.• Only Indian capacitor company to transfer technology to

Germany.• Joint Venture in Capacitor manufacturing in Germany

through “MKS Technologies”• ISO revalidated by BVQI from 2004 to 2007.• On the verge of getting “UL” certification for marketing

internationally.

Peak current measurement capability at Meher Works

Session 7

Some basic formulae –capacitance,capacitor currents

Capacitor Connection

R

Y

B

KVAR =

IL =

√3. VL IL1000

KVAR .1000√3 .VL

Capacitor Rated Current

Change in Current Vs. Change in Voltage

VOLTAGE KVAR CURRENT

440 28.10 36.88

415 25.00 34.78

400 23.23 33.52

380 20.96 31.85

360 18.81 30.17

Capacitance

Calculate Capacitance C∆ and CMfor 25 KVAR, 415 V, 50 Hz. capacitor

CM6π f C∆ VL

2 *10 9

KVAR X 10 9

6 π f VL2

1.5. C∆

KVAR =

C∆ =

CM =

µF C∆

* C∆ in µF and VL in Volts

Ip = Peak inrush Current in Amps

Ir = Capacitor Rated Current in Amps.

MVAsc = Short circuit MVA of the System

kVArc = Capacitor Rating in kVAr.

Ip = Ir MVA SC X 103

kVArc√.√2

Peak inrush current of capacitor

Fault Level Calculation

Transformer = %Z x 10 x kV2

impedance kVA= 5 x 10 x 0.4152

1000

= 0.00861 Ohm

1.1 VL

√3 ZT

Maximum Fault Current =

1.1 x 415

√ 3 x 0.00861=

= 30607 Amp

Transformer % Z = 51000 KVA , 22.0kV/415V

ACB

Short Circuit MVA of the System

Short circuit MVA of the System = 10 6

= √3x 415 x 30607

10 6

= 22.0

√3 VL Isc

Peak Current CalculationCapacitor Rating = 25 KVAR, 415V, 50 Hz.

Ip= 34.78. 22.0 X 103

25√.√2

= 1459.1 Amp

Ip = Ir . MVA SC X 103

kVARc√.√2

Parallel Switching of Capacitor

Ip = Peak inrush current in Amps.VL = Line to Line Voltage in VoltsXC = Capacitive Reactance in OhmsXL = Inductive Reactance Between the Capacitors in Ohms.

√ 2

√ 3Ip = VL

1

√ X C X L

Voltage Rise Due To CapacitorVoltage Rise Due To Capacitor

V = Voltage Rise

V = System Voltage Without Capacitors

Q = Capacitors Rating in MVAR

S = System Fault Level In MVA

VV

=QS

Voltage Rise Due To CapacitorVoltage Rise Due To Capacitor

= 0.47 volts

VV

=QS

415 x 0.025=

22V

For a 25 kVAr, 415V capacitor & System fault level of 22 MVA.

Discharge Time

C

Discharge Time < = 60 sec for LT capacitors< = 10 min for HT capacitors

Voltage at the end of Discharge timeshould be < = 50 volts taking into account the plus side tolerances of the Capacitance value and supply voltage.

Discharge Time

R = Discharge Resistance in M Ohm

t = Discharge Time in Sec.

K = 1/3 or 1 or 3 depending upon discharge resistor Configuration.

C = Capacitance in µF

Vn = Capacitor Rated Voltage

VR = Permissible Residual Voltage

R <K C log e )Vn.√2

VR(t

Configuration of Discharge Resistors

K = 1

C

R

K = 1

C R K = 1

RC

C

R

R

C

K = 1/3

R

C

K = 3

CR

K = 3

USEFUL FORMULAE AND TABLES

1. Capacitance in parallel

C = C1 + C2 + C3

Where C = equivalent capacitance of parallel circuit.

2. Capacitance in Series

1 1 1 1C C1 C2 C3

Where C = equivalent capacitance of series circuit.

= + +

3. Calculation of Capacitor kVAr Required for Power-Factor Improvement

Capacitor kVAr = kW (tanϕ1 - tanϕ2)

Where ϕ1 = Cos-1(PF1) and

ϕ2 = Cos-1(PF2)

PF1 and PF2 are initial and final power factor respectively.

Multiplying Factor Table to Calculate kVAr

Final PFPresent PF 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1.00

0.4 1.807 1.836 1.865 1.896 1.928 1.963 2.000 2.041 2.088 2.149 2.2910.45 1.500 1.529 1.559 1.589 1.622 1.656 1.693 1.734 1.781 1.842 1.9850.5 1.248 1.276 1.306 1.337 1.369 1.403 1.440 1.481 1.529 1.590 1.732

0.55 1.034 1.063 1.092 1.123 1.156 1.190 1.227 1.268 1.315 1.376 1.5180.6 0.849 0.878 0.907 0.938 0.970 1.005 1.042 1.083 1.130 1.191 1.333

0.65 0.685 0.714 0.743 0.774 0.806 0.840 0.877 0.919 0.966 1.027 1.1690.7 0.536 0.565 0.594 0.625 0.657 0.692 0.729 0.770 0.817 0.878 1.020

0.75 0.398 0.426 0.456 0.487 0.519 0.553 0.590 0.631 0.679 0.739 0.8820.8 0.266 0.294 0.324 0.355 0.387 0.421 0.458 0.499 0.547 0.608 0.750

0.85 0.135 0.164 0.194 0.225 0.257 0.291 0.328 0.369 0.417 0.477 0.6200.9 0.029 0.058 0.089 0.121 0.156 0.193 0.234 0.281 0.342 0.484

0.91 0.030 0.060 0.093 0.127 0.164 0.205 0.253 0.313 0.4560.92 0.031 0.063 0.097 0.134 0.175 0.223 0.284 0.4260.93 0.032 0.067 0.104 0.145 0.192 0.253 0.3950.94 0.034 0.071 0.112 0.160 0.220 0.3630.95 0.037 0.078 0.126 0.186 0.3290.96 0.041 0.089 0.149 0.2920.97 0.048 0.108 0.2510.98 0.061 0.2030.99 0.142

4. For Single Phase Capacitor

C = Measured capacitance across terminals 1 &2 in µF

XC = Capacitive Reactance in ohms

V = Voltage in Volts

kVAr = Rated output of capacitor

IC = Capacitor Current in Amps.

lC1 2

V

XC =106

2 πfCkVAr =

2 πfCV2

109IC =

kVAr . 103

V

5. For a Balanced Three Phase Delta Connected Capacitor

CM is the measured capacitance across any two terminals with the other terminal left open circuited.

Where XC = Capacitive Reactance per phase in ohms

V = Voltage (line to line) in volts

kVAr = Rated output of capacitor

IL = Line current in Amps

1

23

V VphC C

C

V = VphCM

C = CM

1.5µF

106

2πfcXC / ph = 2πfcV2

109X 33 phase kVAr =

3ph kVAr X 103

√3 VIL =

6. For a Balanced Three Phase Star Connected Capacitor

1

23

>

V

Vph

CC

C

CM is the measured capacitance across any two terminals with the other terminal left open circuited.

CM

Where XC = Capacitive Reactance per phase in ohms

V = Voltage (line to line) in volts

kVAr = Rated output of capacitor

IL = Line current in Amps

Vph = V/ √3

C= 2CM

106

2πfcXC/Ph =

3 Ph kVAr = x 3

IL = 3 ph kVAr X 103

√3 V

2πfcV2

109

IL

7 Inrush transient current

7.1 Switching in single capacitor

Where IP = The peak value of Inrush Capacitor current in Amps

Ir = The rated capacitor current (rms) in Amps

MVASC = the short circuit power in MVA at the point where the capacitor is connected

kVArC = kVAr of the capacitor

Ip = Ir . √2 . √ MVAsc . 103

kVArc

7.2 Switching of capacitors in parallel with energized capacitor

IP =

Where IP = the crest value of Capacitors inrush current in Amps

V = Rated voltage in volts (line to line)

XC = Capacitive reactance per phase in (ohms)

XL = the inductive reactance per phase between the capacitors in (ohms)

√2 . V

√3 . √XcXL

7.3 Frequency of Inrush Current

√fS = fN .

Where fS = the frequency of inrush current in Hz

fN = the Rated frequency in Hz

XC = Capacitive Reactance per phase in (ohms)

XL = the inductive reactance per phase between the capacitors in (ohms)

Xc

XL

Measured verses cell capacitance

where

( )1 1 11Cc = C2 C3 C1+ -( )Cb =

1 1 1C1 C2 C3

+ -1)Ca =1 1 1

C1 C3 C2( + -1

=C1+C2+C3

C1C2C3 C12 C2

2 C32

1 1 1++( )12

-

C1,C2,C3 are the capacitance measured as indicated with the third line open-circuited.

Ca,Cb,Cc are the cell capacitance, internal to the three phase capacitor.

Using following formulae we can calculate cell capacitance, without opening and de-soldering/cutting the capacitor units.

Cc+C1 =CaCbCa+ Cb

Ca+C2 =CbCcCb+ Cc

Cc+ CaCb+C3 =

CcCa

These are derived from the following basic relationships.

C2

Ca

Cb

Cc

C1

C3

Measured verses cell capacitance

Ca

Cb

Cc

1

2

3

C2

Ca

Cb

Cc

2

1

3

C3

Ca

Cb

Cc

2

1

3

C1

C1,C2,C3 are measured capacitances across any line and other two lines short circuited. Full line to line voltages are applied across the cells.

C1 = Ca+Cb C3 = Cb+CcC2 = Ca+Cc

The individual cell capacitances can be computed as follows.

Ca = (C1+C2-C3) ;12 Cb = (C1+C3-C2) ;

12

Cc = (C2+C3-C1)12

Cost Based Selection

The total cost of using a capacitor is a function of

•Purchase Cost • Operating Cost

•While purchase cost is easy to estimate it is necessary to also evaluate operating cost

•The operating cost of a capacitor is a function of the total losses & the operating time of the capacitor.Ex: An installation requires 1000 kVAr which will be operated for about 6000 hrs per year. Calculate the operating cost of MD-XL Capacitors verses MD type Capacitors assuming a life expectancy of 15 years for the Capacitors. The total energy consumed by the Capacitors for its on operation is calculated as follows

An installation requires 1000 kVAr which will be operated for about 6000 hrs per year. Calculate the operating cost of MD-XL Capacitors verses MD type Capacitors assuming a life expectancy of 15 years for the Capacitors. The total energy consumed by the Capacitors due to its internal watt loss is calculated as follows.

Calculation of operating cost of capacitors

Cost Based Selection• MD-XL CapacitorsEnergy Consumed = (Loss per kVAr x Total kVAr x Operating time)/1000

= (0.5 x 1000 x 6000 x 15) / 1000

= 45,000 kWh

Energy Consumed = (Loss per kVAr x Total kVAr x Operating time)/1000

= (1.5 x 1000 x 6000 x 15) / 1000

= 1,35,000 kWhConsequently, the excess energy consumption due to the MD Capacitor shall be

= 1,35,000 – 45,000

= 90,000 kWh

This energy consumed can be converted into cost using a weighted average cost of Rs.5 per kWh. Consequently, the extra cost shall be Rs.5 x 90,000 = Rs.4,50,000.

On a per kVAr base this can work out to Rs.450/-per kVAr. It is obvious that operating cost must be evaluated carefully before taking the final decision on the type of capacitor to be used. It is also self explanatory that lower the losses, lower will be the operating cost.

• MD Capacitors

Session 8

Automatic Power Factor Correction(APFC)

•Modern Power networks cater to a wide variety of electrical and power electronic loads, which create a varying power demand on the supply system.

•In case of such varying loads, the power factor also varies as a function of the load requirements.

•It therefore becomes practically difficult to maintain a consistent power factor by use of Fixed Compensation i.e. fixed capacitors which shall need to be manually switched to suit the variations of the load.

• This will lead to situations where the installation can have a low power factor leading to higher demand charges and levy of power factor penalties.

NEED FOR AUTOMATIC POWER FACTOR CORRECTION

•The use of fixed compensation can also result in leading power factor under certain load conditions.

•This is also unhealthy for the installation, as it can result in over voltages, saturation of transformers, mal-operation of diesel generating sets, penalties by electric supply authorities etc.

•It is therefore necessary to automatically vary, without manual intervention, the compensation to suit the load requirements.

•This is achieved by using an Automatic Power Factor Correction(APFC) system which can ensure consistently high power factor.

•In addition, the occurrence of leading power factor will be prevented.

NEED FOR AUTOMATIC POWER FACTOR CORRECTION

Session 9

Intelligent APFC Relay

POWER FACTOR CONTROL SCHEME

PFCR

MEASURINGUNIT

OUTPUTRELAYS

CAPACITORBANKS

R Y B

TO LOAD 8 STAGE14 STAGE

1 : 1 : 1 : 1 : 1 : …….1 : 2 : 2 : 2 : 2 : …….1 : 2 : 4 : 4 : 4 : …….1 : 2 : 4 : 8 : 8 : …….

Features

• Controls power factor• Protects capacitor banks• Measures & displays various parameters• Records

Controls Power Factor

• Maintains system power factor at a set value• Under varying load conditions• Uses microprocessor techniques for

measurement of reactive current & system power factor

• Can control upto 8 capacitor banks

Optimization of Capacitor Banks

• Constantly selects right combination of capacitor banks to ensure pf is kept very close to the desired value.

Top-up Facility

• Constantly monitors the actual pf & compares with target value.

• Any spare capacitor bank is utilized to push up the pf to unity even after the target value is met.

• Ensures reactive power consumption is kept to the minimum.

• Feature can be enabled or disabled.

Short Time Delay Facility

• This is the time delay between the immediate OFF to ON of a capacitor bank.

• Provides additional delay of 30msec between switching, if enabled

• Ensures longer life of capacitors

Hunt Free Operation

• Capacitor hunting is avoided by providing threshold values.

Automatic Disconnection of Faulty Capacitor Bank

• If any capacitor bank is reduced to 60% of its original rating, it treats the bank as defective, after three successive switching

Protection Against Fault Conditions

• Switches OFF all capacitor banks & provides a safety lockout period of 60 sec when power interruption occurs

• Protects against – over voltage,– under voltage– Under current– Harmonic overloadBy switching of capacitor banks one after another

• Provides alarm for fault conditions & for 5th & 7th

harmonics

Multi-parameter Display

• Voltage• Current• Reactive power• Accurate display of power factor even in

presence of harmonics.

Records

• Keeps updated (every two hours) records for each bank.

• Number of times each bank is switched on for pf compensation.

• Configuration with respect to lowest bank size.

Session 10

Issues in Power Quality

Momentary Voltage Sag

Momentary voltage sag , which is a momentary decrease in voltage outside normal tolerance.

Momentary Voltage Swell

Momentary Voltage swell , which is a corresponding voltage increase often caused by the sudden de – energizing of heavy equipment.

Voltage Loss

Voltage Transient / Impulse

Voltage transient or impulse , which is a very short duration voltage, whose amplitude will be in the range of several tens to thousand volts.

Voltage Spike

Voltage Spike ,which can destroy electronic equipment and damage transformer and motor insulation. They also cause failures in capacitors and indicators.

Notch in the Voltage Waveform

Liner Load Characteristics

Voltage Waveform

Current Waveform

Non-linear Load Characteristics

Voltage Waveform

Current Waveform

Scope

Improvement of power quality in LV networks• To enhance network reliability• To reduce failure of electrical &

electronic equipment• To increase profitability by saving on

energy costs• To achieve energy conservation

Power Quality in LV Networks - I

External power quality• Conditions arising from incoming

power supply source– Voltage fluctuations

• Steady state• Transients

– Frequency variations– Interruptions in power supply– Import of harmonics

Power Quality in LV Networks - II

• Internal power quality– Function of conditions arising due to use of

equipment to overcome external power quality problems

• On-line UPS systems• Voltage regulating devices

– AUTO TRANSFORMER, STABILISERS etc.

• Back-up power supply– DG Sets– Inverters etc.

Power Quality in LV Networks - III

• Internal power quality– Is also a function of types of loads connected

in the network• Non-linear loads

– Rectifiers, converters, drives– Battery chargers, UPS systems– Modern lighting systems

• Rapidly fluctuating loads– Welding machines– Plastic extruders– High speed presses

• Single phase SMPS loads– PC’s, Servers, LAN Networks.

Power Quality Issues

• Reactive power flow • Harmonic currents & voltages• Voltage dips• Voltage flicker• Unbalanced load • High neutral current• Excessive neutral to earth voltage

Problems ! - I

Reactive power flow– Lower PF & Increased kVA Demand.– Overloading of transformers, cables &

switchgear.– Increased energy consumption due to higher

losses.– Financial penalties for low PF.– Loss of financial incentives for high PF.

PROBLEMS ! - IIHarmonic currents & voltages

– Overheating & failure of • Electrical equipment

– Motors, transformers, switchgear– Capacitors

• Power electronic equipment

– Malfunction/failure of• Protective relays• Control & Automation equipment

– Increased energy consumption.– Risks of resonance - current amplification -

extremely dangerous

PROBLEMS ! - III

Voltage dips & flicker– Failure of power electronic equipment– Malfunction/failure of

• Protective relays• Control & Automation equipment

– Increased strain on eyes.• Due to fluctuations in intensity of lighting systems.

PROBLEMS ! - IV

Unbalanced loads– Over & under voltage in the network.– Increased energy consumption by motors

• Due to reduced efficiency.

– Failure of power electronic equipment– Malfunction/failure of

• Protective relays• Control & Automation equipment

PROBLEMS ! - V

High neutral current– Overheating of neutral conductors– Increased energy consumption– If neutral becomes open high voltages will

occur, resulting in• Malfunction/failure of

– Single phase loads– Protective relays– Control & Automation equipment

– Risk of fire & destruction

Three phase system

Time.

R - phase.

Time.

Y - phase.

Time.

B - phase.

Time.

Addition of third harmonics in Neutral conductor

Time.

Wave forms of balanced three phase fundamental currents.

R-Phase current with its third harmonic component.

Y-Phase current with its third harmonic component.

B-Phase current with its third harmonic component.

Third harmonic currents of R,Y&B phases are in phase with each other and hence adds up, without cancellation in the neutral conductor.

Accumulation of 3rd harmonic current in neutral

SOLUTIONS - I

Networks with <20% non linear load• Improve PF, reduce voltage dips/flicker

by the use of– Power capacitors– APFC systems

• Contactor switched• Thyristor switched

– Open loop systems– Closed loop systems

SOLUTIONS - II

Networks with >20% non-linear load• Improve PF, reduce harmonics & voltage

dips/flicker by the use of– Fixed detuned filters– Detuned filters + APFC systems

• Contactor switched• Thyristor switched

– Open loop systems– Closed loop systems

– AHF - Active Harmonic Filters

SOLUTIONS - III

Networks with unbalanced loads.• Improve PF, reduce harmonics & voltage

dips/flicker by the use of– Phase balancing circuits– Electronic VAr Compensation Systems.– AHF - Active Harmonic Filters

SOLUTIONS - IV

• Networks with high neutral current / excessive neutral to earth voltages– Ensure proper EARTHING quality– Oversize all neutral conductors to reduce neutral

heating– If OVERSIZING is not possible, reduce

harmonics by the use of• AHF - active harmonic filters in 4 line configuration

Active filter schematic diagram

ActiveF ilter

Load C urrent withHarm onics

SinusoidalSupplyC urrent Supply

System

C ompensatingC urrent

+ =

(Tim e Domain)

(Frequency D omain)

+ =

Benefits of POWER QUALITY MANAGEMENT

• Improved reliability of installation• Reduced energy consumption• Reduced fuel consumption• Better productivity• Improved profitability• Enhanced equipment life

Session 11

Dynamic Compensation

Scope

• Need for Dynamic Compensation

• Applications

Need for DYNAMIC COMPENSATION

• When load conditions demand rapidly fluctuating reactive power.– Due to nature of load.– Due to process requirements.

• When switching transients are to be eliminated.

• For Optimizing performance & fuel consumption of DG sets.

THYRISTOR SWITCHED APFC

There are certain loads which demand, under certain operating conditions, large amount of reactive power for very short duration of time.

Typical examples are:

•Welding equipment

•Injection moulding equipment

•Starting of large induction motors

•Traction loads such as, lifting cranes, elevators, lifts, etc.

Thyristor Switched APFCThe large demand for reactive power by such loads during operation can cause:

•Rapid voltage fluctuation

•System instability

•Over sizing of electrical installation since kVA capacity will have to be provided for the maximum power demand.

•Malfunctioning of sensitive electrical and electronic equipmentsuch as relays, PLC’s etc.

These ill-effects can be overcome by injecting into the network defined amount of reactive power at a very fast rate which can meet the demand of such loads.

RAPIDLY FLUCTUATING LOADS - I

• Variations in current are sudden & high

• Lower PF & voltage dip• Examples

– Motor starting– EOT cranes, lifts– Rolling mills– Conveyors - Mining etc.– Wind electric generators

RAPIDLY FLUCTUATING LOADS - II

• Current drawn in repetitive pulses

• Lower PF & voltage dips / flicker

• Examples– High speed presses

• Metal working• Plastic processing

– Balanced welding loads

SWITCHING TRANSIENTS

• Capacitor switching by contactors results in transients

• These transients may interfere with operation of modern relays, control & automation equipment– For ex: digital relays, PLC’s etc. can malfunction

• Thyristor switching is a must for eliminating switching transients

• Hence, dynamic compensation

DG SET PERFORMANCE &

FUEL CONSUMPTION• Use of dynamic compensation systems

can– Stabilize DG set output voltage. – Reduce DG set rating for a given load.– Enable better % loading of the DG set. – Reduce fuel consumption.– Enhance life of DG set.

APPLICATIONS - I• Industrial networks

– Automobile & automobile component Mfrg. Plants

– Metal working • Fabrication & press shops• Rolling mills• Forging

– Plastic extrusion & Moulding.– Mining

• Extraction• Polishing, Crushing etc.

APPLICATIONS - II• Industrial networks

– Paper, wood & particle board Mfrg. plants– Plants with CNC machines

• Other networks with sensitive loads– Hospitals– IT parks– Intelligent buildings

DYNAMIC COMPENSATION SYSTEMS

Open loop systems– Suitable for dedicated loads.– Fastest Response. (< 15 msec)– Unique “EPS” logic.– External signal from load can be used for

switching on.– “On” time - externally settable.– Integrated protection for

– Wrong phase sequence.– Phase fall out.– Over temperature.

DYNAMIC COMPENSATION SYSTEMS

Closed loop systems– For groups of diverse loads.– Use advanced programmable controller. – Fast Response. (< 60 msec)– “EPS” logic– Desired PF & specified harmonic distortion

values are settable.– Integrated protection for

– Wrong phase sequence.– Phase fall out.– Over temperature.– Harmonic over load.

DYNAMIC COMPENSATION SYSTEMS

Open & closed loop systems– The system consists of

• Ergonomic metal enclosure.• Incoming switchgear & protection.• Modern Copper Busbar System. (upto 80 kA short

circuit withstand capacity)• Power capacitors.• Harmonic Reactors. (if required)• Thyristor modules

– Firing circuits – Electronic control modules– Protection fuses

• By-pass contactors.

THYRISTORISED SWITCHED APFC

•Conventional power factor correction systems using contactors as switching devices cannot be used in sufficient speed of response to meet the reactive power demand imposed by such loads.

•It is necessary to use a dynamic power factor correction system in which the switching and controlling devices used have a response time in milliseconds.

DYNAMIC COMPENSATION SYSTEMS : Advantage - ILOSSES & TEMPRATURE RISE

– Thyristor Loss = 2 x In wattsEx: 3 Phase, 440 V, 50 Hz. 50 kVAr capacitor has In = 65 A.– watt loss/thyristor module= 130 watts.– for three phase switching two thyristor modules are

required.– total watt loss = 260 watts.– unit watt loss = 5.2 watts/kVAr.– Generally therefore cooling fans are required for each

thyristor step.– Energy consumption is very high.

– INTELLVAr - D : No cooling fans for each Thyristor step.

DYNAMIC COMPENSATION SYSTEMS : Advantage - II

THYRISTOR RELIABILITY – Use of bypass contactors reduces

• Operating losses.• Utilization time of Thyristors.• PIV of Thyristors used = 1800 Volts.• In a 415 V, 50 Hz. system

– peak to peak voltage = 1174 V– Therefore Safety factor > 150% in Voltage

– Lower utilization time + high PIV results in enhanced Thyristor reliability.

DYNAMIC COMPENSATION SYSTEMS : Advantage - III

SWITCHING LOGIC – ZERO CROSSOVER SWITCHING DESIGNS

– Due to various factors exact zero crossover is not consistently achievable in practice.

– consequently use of [di/dt] limiting inductors/coils is common in zero crossover switching designs.

– EQUI-POTENTIAL SWITCHING - EPS LOGIC – Enables continuous sensing of capacitor & line potential.– Switching is done at equi-potential instant.– This reduces the [di/dt] to very safe values.– Hence no current limiting devices are needed.

– EPS LOGIC THUS INCREASES RELIABILITY

Session 12

Power factor improvement of DG sets

Capacitors with Generators

G

G

Prime mover Alternator

100 KVA0.8 P.F.80 kW

Diesel generator set

Connected load P.F. is 0.6

100 kVAP.F. 0.880 kW

P.F. 0.680 kWkVA 133.33

Alternator overloadedby 33.33 %

100 kVAP.F. 0.660 kW

Shortfall of20 kW

Case 1

Case 2

Load P.F. improved up to 0.98

100 kVAP.F. 0.880 kW

100 kVAP.F.0.9898 kW

Prime mover overloaded

• As load kW (Output) increases, input power from prime mover has to be increased.

• Diesel engines can be overloaded by 10 %, for half an hour, within a span of 12 hours.

• Prime movers are matched with alternator to operate at specific P.F.

• Lagging P.F. weakens the flux which links with alternator stator and leading P.F. strengthens it.

DG Set fundamentals

• At low lagging P.F, it is not possible to reach the nominaloperating voltage of the alternator, even at low load.

• With leading P.F, even with low excitation, there could berise in voltage, causing damages to the connected equipment.

• At leading P.F. generator becomes unstable.

• If generator is operated with purely capacitive load voltage increases by 33.33 %

DG Set fundamentals

Selection of DG set rating

• Connected load and demand factor.

• Short duration peak loads like starting of induction motors.

• Allowance for extra kVA for harmonic generating loads.

• Allowance for accommodating future additional loads.

Hence DG sets are always oversized for a given application

and operate at relatively lower percentage loading.

Loading verses Yield curve

Operating at lower % of loading , result in poor yield from DG set.

How to improve % loading in DG

• Do not exceed the current rating of Alternator.

• Do not exceed the BHP/kW rating of the prime mover (Engine).

Golden Rules for safe DG set operation :-

• Load the Alternator by ‘Amperes’.

• Load the diesel engine by BHP/kW.

Improve % loading by operating at higher power factor

• Higher PF reduces current output from DG at a given load.

• Loads can now be added without violating the ‘Golden Rules’.

Operating at highest feasible PF, enables higher loading, resulting in better yield from DG set.

PF CORRECTION IN INSTALLATION WITH CAPTIVE

GENERATION BY DG SETS

The DG set consists of a diesel engine, which is mechanically coupled to an alternator. The engine supplies the mechanical energy to the alternator and the alternator supplies the electrical energy to the load.

The alternator is subjected to certain copper losses, which is proportional to the square of the current delivered by it. The diesel engine has to supply these losses in addition to supplying the load requirements.

Thus by reducing the losses in the alternator the diesel consumption of the diesel engine can be brought down.

ALTERNATOR LOSS REDUCTION BY P.F IMPROVEMENT

Reducing the current output from alternator without altering the loading conditions can reduce the alternator losses.

Improving the power factor at the output of the alternator can conveniently do this.

Hence, improvement in the Power Factor in alternator leads to reduction of fuel consumption in DG sets.

DG sets operate at a relatively low power factor of 0.6 to 0.8. Conventionally capacitors are not used along with DG sets.

Effect of PF improvementOUTPUT CURRENT FROM DG WITH /WITHOUT COMPENSATION

0

100

200

300

400

500

600

1 2 3 4 5 6 7 8 9 10

TIME IN SECONDS

CU

RR

EN

T IN

AM

PS

WITHOUTCOMPENSATION

WITHCOMPENSATION

The following example gives an approximate calculation to show the impact of power factor improvement on reduction of alternator losses and accrued savings in Diesel consumption.

Consider a 3 phase, 415V, 50Hz, 500 kVA DG set used in an industry for 6000 hours/year with an average load of approximately 250kW at 0.65 PF. What is the fuel saving if PF is improved to 0.93? The full load copper loss of the alternator is 12kW and average yield of the DG set is 3kWh/litre of fuel (HSD).

Alternator loss reduction due to P.F improvement

Alternator loss reduction by P.F. improvement

Rated Current of Alternator = 500000/(√3 x 415)= 695.60 A

Current at 0.65 PF = 250000/ √3 x 415 x 0.65

= 535.08 A

Copper loss at this current = (535.08)/695.6)2x12kW

= 7.1 kW

Current at 0.93pf = 250000/ √3 x 415 x 0.93

= 373.98A

Copper loss at this current = (373.98/695.6)2x12kW= 3.47kW

Saving in Copper loss = 7.1 – 3.47 kW= 3.63 kW

Alternator loss reduction by P.F. improvement

Energy saved for 6000 hour Generation = 3.63 x 6000 kWh

= 21780 kWh

DG set Yield = 3 kWh / liter of HSD

Potential savings in HSD fuel = 21780/3

= 7260 liters per year

Potential savings in

Rs @ Rs. 35/liter = 2,54,100 per year

Method of P.F. improvement

Conventional fixed capacitors, should not be used with DG sets. This is because, by using fixed capacitors, there is a danger that the PF can become leading under lightly loaded condition, which is highly undesirable in DG set operation. Hence only Automatic Power factor Correction system (APFC) should be used with DG sets. The target power factor can be set to 0.93 to 0.95 for optimum performance.

APFC Selection

The appropriate APFC system can be selected based upon the harmonic content of the load. If the harmonic generating load is less than 20%, the APFC should be used as mentioned in section-8. If the Harmonic load is greater than 20% then reactor protected APFC should be used as mentioned in section 9.5

The rating of the APFC has to be selected depending upon the kW of the load connected and the minimum power factor in the installation.

= M a in S u p p ly C u rre n t T ra n s fo rm e rC T

F .S .U

C ... . . .C

F . .. . . .F

F a c to r

= D ie s e l G e n e ra to r

= C a p a c i to r S te p P ro te c tio n F u s e s

= F u s e S w itc h U n it (O p tio n a l)

= N o . o f C a p a c ito r S te p C o n ta c to rs = N o . o f C a p a c ito r S te p s

= C o n tro l F u s e s

C o n tro lle r

C ... . . .C

F .S .U

F . . . . . .F K .. . . . .K

2f

G

1

1 n

1

n

n

K . . . . .K1 n 1

P o w e r

f 2 1

n

n

V a r ia b le L o a d sC T ... ./5 A

~G

F IG .3 - B L O C K D IA G R A M O F C O N T A C T O R S W IT C H E D A P F C S Y S T E M

f

Thy 1......Thy n = Thyristor SwitchesS ......S

F ......F

Thy 1

= Main Supply Current TransformerCT

= Fuse Switch Unit (Optional)

= Capacitor Step Protection Fuses

= Diesel Generator

= Control Fuses

= No. of Steps

1 n

F.S.U G

2

1 n

1S 2S nS

F.S.U

FF 1

Thy 2

2 F

Thy n

n

2fReactive

ControllerPower

~G

CTVariable

Load

FIG 5. BLOCK DIAGRAM OF DYNAMIC COMPENSATION SYSTEM IN CLOSED LOOP

Conclusions• The efficiency of the DG set is maximum at UPF.

• The yield of the DG set is maximum at its peak loading.

• By proper use of reactive power management the efficiency of DG set can be improved.

• The loads can be transferred so as to optimize the loading of DG for better yield output.

• The output voltage of DG set can be stabilized under rapidly fluctuating loads by use of “Dynamic compensation systems”

• Saving in fuel is possible resulting in economic benefit to user.

Session 13

Harmonics & Effect of Adding Capacitors in the System

POWER FACTOR CORRECTION IN HARMONIC RICH ENVIRONMENT

••A harmonic rich environment is said to exist when the A harmonic rich environment is said to exist when the percentage of non linear loads in an installation becomes percentage of non linear loads in an installation becomes greater than 20% of transformer rating.greater than 20% of transformer rating.

••Power factor correction by the use of capacitors, in such an Power factor correction by the use of capacitors, in such an environment, must therefore be carried out with certain environment, must therefore be carried out with certain precaution.precaution.

••This is due to the fact that parallel resonance conditions can This is due to the fact that parallel resonance conditions can occur, I.e. the magnitude of the Capacitive reactance of occur, I.e. the magnitude of the Capacitive reactance of capacitors installed and the inductive reactance of the capacitors installed and the inductive reactance of the network can tend to be come equal.network can tend to be come equal.

••If such resonance occurs near to a frequency which is present If such resonance occurs near to a frequency which is present in the network, current amplification takes place. in the network, current amplification takes place.

POWER FACTOR CORRECTION IN HARMONIC RICH ENVIRONMENT

XC XLlh

POWER FACTOR CORRECTION IN HARMONIC RICH ENVIRONMENT

•This current amplification can lead to overloading of capacitors and an increase of the voltage distortion in the network.

•Capacitors drawing higher current i.e. more than the rated current at normal operating voltages is a typical indication of presence of harmonics.

•While it is possible to design the capacitors to withstand the overload conditions, the increase in distortion will cause otherill effects such as :

• Capacitors installed being subjected to severe harmonic overloading, leading to premature failure

POWER FACTOR CORRECTION IN HARMONIC RICH ENVIRONMENT

• Total harmonic distortion in the network increasing beyond the permissible levels, which is harmful to various equipments within the installation.

•The use of capacitors in the conventional manner is therefore not recommended in such situations.

Technical problems experienced in industry

Case – 1

• Type of industry - paperboard Manufacturing industry.

Brief description of installation.Primary power source = grid supply at 33kv.Distribution voltage = 440V.

• Load details:The total induction motor load was 800HP.

Case – 1• There were no non-linear loads installed in

this plant.• 300kvar, 3ph, 440V MPP-H capacitor banks

were installed for power factor correction. Some capacitors were connected across the motor terminals and remaining used as central compensation.

• Problem experienced:• Frequent failure of capacitors.

Case – 1- Analysis

• The system voltage was around 435 – 440V. • The capacitors installed were subjected to severe

and intermittent overload. • For ex:- A 25 kVAr, 440V capacitor was drawing

a current of 90 - 120A for certain periods of time. Compared to its rated current of 33 amps, the capacitors were subjected to an overload greater than 250%.

• This abnormal overloading resulted in frequent capacitor failure.

Case – 1- Analysis

• Following observations were made on the 33kV grid :-

• Only two industries were found to be connected to 33kV line, one of them being the paperboard manufacturing industry and the other was a steel rolling mill.

• The length of the 33 kV transmission line between the two plants was approximately10kms.

• The steel rolling mill had installed a high frequency induction furnace.

Case – 1- Analysis

• It was observed that, whenever the high frequency induction furnace installed in steel rolling mill was operated, the harmonic distortion on the grid abnormally increased and during this period the capacitors were getting severely over loaded.

Case – 1- Conclusion

• Thus the use of conventional capacitors in the network, where the harmonic voltage distortion at the grid was abnormally high resulted in the following:

• Over loading of capacitors due to series resonance.

• Frequent failure of capacitor banks• Increased harmonic distortion at the LV bus• Inability of the customer to maintain the

desired power factor.• Financial losses incurred by customer.

Case – 2• Type of industry- Cement industry• Brief description of installation.

Primary power source = grid at 220kV.Distribution voltage = 6.6kv and 440V.

• The LV power factor correction is done as follows:At 440V bus, around 2000 kVAr APFC panels with conventional capacitors were installed.The APFC panels were distributed on several 6.6/0.440 kV distribution transformers installed in the plant.

Case – 2

Problem experienced at LV bus:

• Malfunction of 350kW, 440V DC drive used for kiln motor installed at cement plant substation, when capacitors are connected in the network.

Case – 2: Analysis• The cement plant sub station was fed by a 1600kVA,

6.6/0.440 kV transformer. • The 350kW, 440V DC drive, was connected to this

transformer, consequently the % non-linear load exceeded 22%.

• A 475kVAr, 440V APFC panel with conventional capacitors, was also connected to this transformer for power factor correction. Other linear loads such as compressors, pumps etc were also connected to this transformer.

• Malfunctioning of the DC drive was co-related to a situation when specific combination of capacitor steps in the APFC were ON.

Case – 2: Conclusion

• The use of conventional capacitor in harmonic rich environment led to high total harmonic distortion on the LV bus.

• As the same distorted sine wave is applied to the 350kW DC drive, sensitive electronic devices used in this drive were mal-functioning.

Case – 3• Type of industry- Steel rolling mill.• Brief description on installation.• Primary power source = grid at 33kV.• Distribution voltage = 440V.• The plant was installed with 2 x 2000 kVA + 1 x 1000

kVA, 33/0.440 kV distribution transformers.• At the secondary of one of the 2MVA transformer

following loads were connected:• 1000HP AC induction motor.• 500HP DC drive.

Case – 3

• Around 900 kVAr of conventional capacitors were connected to this transformer for power factor improvement as shown in the fig.

Case – 3

M

2000kVA, 33/0.44 kV Trafo.

Feeder for 1000kVA Trafo.

Feeder for other 2000kVA Trafo.

900kVAr Capacitor1000HP

Induction Motor

33kV supply from Grid.

440V bus

500HP DC Drive

Case – 3• Problem experienced:• Frequent failure of capacitors installed for

power factor correction.

Case – 3- Analysis

• A team of engineers from MEHER made an analysis at the site. The result of the analysis is as follows:

• The capacitors installed were drawing more than it’s rated current. A 50 kVAr, 440V capacitor was drawing a current of 200A against the rated value of 65amps, thus constituting an overload greater than 300%.

• This abnormal over load resulted in frequent capacitor failure.

• The total harmonic voltage distortion at the secondary of this 2000 kVA transformer with all the capacitors switched on was more than 25%. This value is higher than acceptable levels.

Case – 3- Conclusion

• Thus the use of conventional capacitor in harmonic rich environment resulted in:

• Over loading of capacitors due to parallel resonance.

• Increased harmonic distortion at the LV bus.

• Frequent failure of capacitor banks.• Inability of the customer to maintain the

desired power factor.• Financial losses incurred by customer.

What Are Harmonics ?

• Distorted sine wave cause harmonics.• Distorted current wave cause current harmonics.• Distorted voltage wave cause voltage harmonics.• Fourier expansion result in integral multiples of

fundamental frequency components.• Nth order harmonics is of n.Fs frequency.• Generally odd harmonics are prevalent because of

half wave symmetry.

How Harmonics Are Generated ?

• Non-linear loads generate current harmonics.• Harmonic currents flow largely through capacitors.• Harmonic currents also flows through network.• The flow of harmonic currents cause voltage harmonics.• Harmonics are thus injected to other linear loads

connected in the same bus.• Harmonics injected into the network flow towards other

users connected to the network.

What Loads Generate Harmonics• Equipment using switched mode power supply

- Television- Computers, other IT loads

• Equipment using power electronic devices- AC & DC drives- Frequency converters- Rectifiers- Arc & induction furnaces- UPS- Compact fluorescent & other discharge lamps

Sources of Harmonics

Following are some of the non-linear loads which generates harmonics:

•Static Power Converters and Rectifiers, which are used in UPS, Battery chargers, etc.

•Arc furnaces.

•Power Electronics for motor controls (AC/DC Drives)

•Computers.

•Television receivers

•Saturated Transformers

•Fluorescent Lighting with electronic ballast.

•Telecommunication equipment.

Type of HarmonicsCharacteristic harmonics

- Related to circuit configuration.- Fairly predictable frequency spectrum.- Frequency spectrum given by k*p+1 ; k = 1,2,3….- For ex. 5&7 for 6 pulse, 11 & 13 for 12 pulse.- Magnitude inversely proportional to order.

Non-characteristic harmonics- Caused by frequency converters.- System imbalance (voltage & impedance)

Triplen harmonics- 3.(2n+1) order n = 0,1,2… i.E 3,9,15,21.. Etc.- Zero sequence in nature.- Accumulates as neutral current.

Harmonic Order & Phase Sequence

• Each harmonic order has a particular phase sequence relationship with respect to fundamental.

• By convention the fundamental is assumed to have positive phase sequence.

• All higher order harmonics have either positive,negative or zero phase sequence with respect to fundamental.

How to Determine the Phase Sequence of Each Harmonics ?

Phase Sequence of RYB (+ Seq.)For Fundamental Component

R Y BFundamental +120o 0o -120o

Second +240o 0o -240o

Harmonic -120o 0o +120o

Thus Second Harmonic Behaves Asa Negative Sequence Component.

Y

R

B

Y

R

B

Positive Sequence

Negative Sequence

How to determine the phase sequence of each harmonics ?

R Y BFundamental +1200 00 -1200

Third Harmonic +3600 00 -3600

00 00 00

Fundamental Third Harmonics

R

Y

B

Y

R

B

Positive Sequence Zero Sequence

Thus third harmonic behaves asa zero sequence component.

Three phase system

Time.

R - phase.

Time.

Y - phase.

Time.

B - phase.

Time.

Addition of third harmonics in Neutral conductor

Time.

Wave forms of balanced three phase fundamental currents.

R-Phase current with its third harmonic component.

Y-Phase current with its third harmonic component.

B-Phase current with its third harmonic component.

Third harmonic currents of R,Y&B phases are in phase with each other and hence adds up, without cancellation in the neutral conductor.

Accumulation of 3rd harmonic current in the neutral

How to Determine the Phase Sequence of Harmonics ?

Harmonic order 1 2 3 4 5 6 7 8 9

Phase Sequence + - 0 + - 0 + - 0

3n+33n+23n+112th Harmonic11th Harmonic10th Harmonic9th Harmonic8th Harmonic7th Harmonic6th Harmonic5th Harmonic4th Harmonic

Fundamental

Positive Sequence

3rd Harmonic2nd Harmonic

Zero SequenceNegative Sequence

Divisible by 3Div. by 3 Rem. 2 Div. by 3 Rem. 1

Characteristics of Harmonics

Positive Sequence Negative Sequence Zero SequenceCauses over heating due to ‘Skin effect’

Aids the fundamental Opposes the fundamental Accumulates in the neutral

Moderate heating Excessive heating Creates ‘hot neutral’

Relatively less harmful Most harmfulResponsible for neutral to earth voltage and open neutral condition.

Causes over heating due to ‘Skin effect’

Causes over heating due to ‘Skin effect’

Skin effectCross-section of current carrying conductor

DC current flow Low frequency AC current flow

High frequency AC current flow

The effective area of the conductor, available for current flow,reduces as the frequency of the AC current increases. Hence, the resistance of the conductor increases, at higher frequencies, as it

is inversely proportional to its area of cross-section.

R =ρ LA

Skin effect explanationMagnified view of current carrying conductor

Enhanced impedance due to high mutual inductance. Hence least current flow.

Moderate impedance due to medium mutual inductance.

Hence moderate current flow.

Least impedance due to low mutual inductance.

Hence max. current flow.

Effect of HarmonicsType of equipment• Rotating machines

• Transformer, switch-gear, power cables

• Protective relays

• Power electronics• Control & automation• Power capacitors

Effect of harmonics• Increased losses, over

heating due to skin effect.• Pulsating torque• Over heating, increased

power consumption• Mal-operation, nuisance

tripping• Mal-operation, failure• Erratic operation• High currents & failure

due to overload

Effect of Harmonics on Protective Relays

Mal-operation Nuisance tripping

Trip level set higher than the fundamental value. The relay should not trip as the fundamental

value is lower than the trip level. But the presence of harmonics has increased the peak

value. Hence the protective relay will trip.

Trip level set lower than the fundamental value. The relay should trip as the fundamental

value is higher than the trip level. But the presence of harmonics has reduced the peak

value. Hence the protective relay will not trip.

Circuit configuration of six pulse drive

Current spectrum of six pulse drive for star-star & star-delta configuration

Twelve pulse drive configuration

Current spectrum of twelve pulse drive

How Capacitors & Harmonics Are Related -1

Network behaviour without capacitors• Network do not reveal harmonics.• Most of the harmonic currents internal to

network go to the grid.• No resonance at harmonic frequencies.• Network power factor is unacceptably low.

Network Without Capacitors

Harmonic currents flow towards Grid.

Min. Import of Harmonics from Grid.

No Resonance at harmonic frequencies.

Hence least Harmonic Problem.

Power Factor Very Low.

M

GRID

BUS

Non

Lin

ear

Loa

d

How Capacitors & Harmonics Are Related -2

Network behaviour with capacitors• Network start revealing harmonics• Internally generated harmonic currents may

amplify due to parallel resonance• Externally generated harmonics enter capacitors

due to series resonance• May increase harmonic distortions.• Capacitors draw excessive currents & fail• Network power factor improves

Network With Capacitors

Harmonic currents flow towards Capacitors , due to parallel resonance with load “ZL”Import of Harmonics from Grid towards Capacitors , due to series resonance with network & transformer impedances “ZN”&“ZT”Increase of THD(V) in the Bus Harmonic overloading of Capacitors, leading to its failureImprovement in Power Factor With Harmonic overload

BUS

M

ZNGRID

ZT

Non

Lin

ear

Loa

d

Equivalent Load Impedance “ZL”

Session 14

Harmonic Filters

Harmonic Mitigation Concept

Harmonic TriangleFire Triangle

How to Improve Power Factor Without Causing Harmonic

Problem ?• Conventional capacitors should not be used.• Capacitors should be replaced by harmonic suppression

filters (series combination of suitable series reactor & capacitors) so that,

• It offers capacitive reactance at fundamental frequency for necessary power factor correction.

• It offers inductive reactance at all higher order dominant harmonic frequencies to avoid resonance.

• Its self series resonance frequency “fR” do not coincide with predominant harmonics.

Network With Harmonic Filters

No resonance at harmonic frequencies as filter is inductive at such frequenciesHarmonic currents flow towards Grid , as it offers least impedance compared to filterPredominantly fundamental current flows through Capacitors Moderate THD(V) in the Bus No harmonic overloading of CapacitorsImprovement in Power Factor without Harmonic overload

BUS

M

GRID

ZT

ZN

L

C

Non

Lin

ear

Loa

d

Equivalent Load Impedance “ZL”

Harmonic Filter

• Harmonic filter comprises of a reactor (L) in series with a capacitor (C)

• Such a filter has a unique self series resonance frequency fR at which inductive reactance of reactor equals capacitive reactance of capacitor. Fr = 1/(2π√LC)

• Below fR the filter is capacitive• Above fR the filter is inductive

Characteristics of Harmonic Filter

fR

InductiveCapacitive

Frequency

Impe

danc

e

fR= ResonantFrequency

f < fR - Capacitive

f > fR - Inductive

Harmonic filters are classified based upon how close fR is to a Harmonic frequency

Classification of Harmonic Filters

• Detuned or harmonic suppression filters• Resonance frequency fR< 90% of

lowest dominant harmonic frequency.

• Tuned or harmonic absorption filters• Resonance frequency fR within 10% of

the frequency of the harmonic to be absorbed.

Classification of Harmonic Filters

Harmonic Filters

Passive Harmonic Filters Hybrid Harmonic FiltersActive Harmonic Filters

Detuned Filters

Tuned Filters

3Ф3wire

3Ф4wire

SinglePhase

7 %

14 %

Selection Criteria for Harmonic Filters

• Detuned filters• Power factor correction is of paramount importance.• If ordinary capacitors draw > 130% of its rated

current.• Reduction of THD(V) not relevant.• To prevent capacitors from harmonic overload• Harmonic study not required for installing standard

detuned filters.

Selection Criteria for Harmonic Filters

• Tuned filters• Power factor correction & reduction of THD(V) are

of paramount importance.• Ordinary capacitors draw > 130% of its rated

current.• Harmonic study required for installing tuned filters.• Specifically designed for each location.• More bulky, since it carries large amount of

harmonic currents. Hence expensive.

Standard Detuned Filters-1

• Standard detuned filters have a fixed percentage tuning factor “p”

• Percentage tuning factor is defined asReactor reactance at system frequencyCapacitor reactance at system frequency

• Standard detuned filters are available for 7% tuning factor

• The resonant frequency of the filter fR is related to tuning factor “p” by

Fr = Fs/ √(p/100) = 189 Hz for 7% filter

X 100 %p =

Standard Detuned Filters-2

• Standard 7% detuned filters are suitable for use in majority of installations where the dominant harmonics are higher than 189 Hz like 5th and higher.

• 7% detuned filters should not be used in installations where predominant 3rd harmonics are present like “IT based” industries.

• For “IT based” industries 14% detuned filters (fR=134 Hz) should be used.

Design Features of Detuned Filter

• Detuned filter consists of matched pair of specially designed reactor and capacitor.

• Detuned filter is designed to provide the rated kVAr at the rated voltage at the bus.

• The reactor capacitor combination is designed for the rated tuning factor.

• Standard detuned filters are available for 7% tuning factor rated for 12.5, 25, 50, 75 & 100 kVAr at 440 volts.

Design Features of Detuned Filter

Reactor features.• Reactors are specially designed to carry wide

spectrum of harmonic and fundamental currents without saturating.

• They are rated for operation up to 160°C through use of class “F” insulation.

• Over load thermal cut off provided to protect the reactor.

Design Features of Detuned Filter

Capacitor features• Capacitor is specially designed to carry wide

spectrum of harmonic and fundamental currents without overloading.

• It is designed for higher voltage to allow for increased voltage due to introduction of series reactor.

• The kVAr of the capacitor is suitably designed to deliver the rated kVAr of the filter at the bus.

Calculation to Estimate the Rated Voltage of the Filter Capacitor

I = V/ Xeq = V/ (XC (1-p/100))Voltage across Capacitor VC is given by

VC = I XC = V/ (1-p/100)Allowing 10% for over voltage, the rated voltage of the capacitor is given by

1.1 VC = 1.1 V/ (1-p/100)

XL= (p/100) XC

C

L

XC

I

V

Bus Percentage Voltage Rating Voltage RatingVoltage Tuning Factor of Capacitor Rounded off

415 7% 490.86 500 V415 14% 530.81 550 V

Note on Capacitor for Detuned Filter Application

• It is seen that the voltage rating of the capacitor has to be higher than the system voltage.

• Hence normal capacitor of 415/440 volts rating should never be used in series with reactor.

• Any such attempt would be hazardous to the capacitor and the installation.

Analysis of Detuned Filters

C∆

L L L L L L

CY

L

CY

Actual connectionof Detuned Filter

SLD Representationof Detuned Filter

Star equivalentof Detuned Filter

Analysis of Detuned FiltersAnalysis of Detuned Filter can be done by analysing its single line diagram representation as shownLet the net available kVAr. at Bus = NCLet the System Line Voltage in Volts = VLet the Tuning Factor in % = p Line current of the Filter IL= V/√3/(XC - XL)

= V/√3/XC/(1-p/100)The 3 phase kVAr. At Bus = √3V IL/1000

i.e NC = V2/ XC/(1-p/100)/1000

XL

V/√ 3

C

L

XCY

IL

The XCY of the star eq. Capacitor = V2/ (NC/1000/(1-p/100))The XL of the Reactor = XCY p/100 = V2/(NC/1000/(100/p-1))The kVAr of the Capacitor at its rated voltage VC and theinductance of the reactor can be computed from the above.

Analysis of Detuned Filters

XC

V/√ 3XL

kVAr. of the Capacitor at its rated voltage VC= (VC/V)2 NC (1-p/100)

Inductance per phase of the 3 phase Reactor in mH.L = V2/NC/(100/p-1)/100/πFrom these formulae we can calculate the DetunedFilter elements for standard outputs as under.For 7%, 440 Volts Detuned Filters

Available Bus kVAr.

Inductance per Phase in mH.

Rated Voltage of the Capacitor VC

kVAr.of the Capacitor at VC

12.5255075100

3.711.8550.9280.6180.464

500500500500500

15.0130.0260.0590.07120.09

C

LIL

Section 15Exercise

Parallel Resonance

Description of the ProblemA 500kVA, 415V, 3 phase, 50Hz., 4% impedance drop transformer is feeding a 100kW, 6pulse DC drive. The PF of the DC drive under full load condition is 0.7 lag. Calculate the following :-

• A) Calculate the harmonic current spectrum of 100kW DC drive.

• B) Calculate the kVAr required to improve the PF of the drive above 0.99 lag.

• C) Calculate the harmonic voltage distortion and harmonic overload with the following :-

• 1) Without any capacitors.• 2) With conventional capacitor connected to the

network.• 3) With de-tuned filter connected to the network.

Harmonic Spectrum

• Harmonic spectrum of 100kW, 6 pulse, 0.7 PF DC drive is given by the relation: Ih =i1/h, where h is the harmonic order.Since the drive is 6 pulse, h = 5,7,11,13,17,19.

I1 = 100x1000 / (√3 x 415 x 0.7)= 200 A (Approx.).

Harmonic Spectrum

h Ih Amps57

11131719

4028.618.215.411.810.5

Estimation of kVAr

• kVAr required to improve the PF above 0.99 lag.

• kVAr =kW x (tan (cos-1 (PF1)) - tan (cos-1 (PF2))=100 x (tan (cos-1 (0.7)) - tan (cos-1 (0.99))=87.8 kVAr.

• kVAr =100 kVAr. (standard available)

Network Without Capacitors:

100 kW, 3ph,415V, DC

drive Linear resistive Loads

500 kVA

Network Without Capacitors Equivalent Circuit

XTIh

Calculation of TransformerImpedance Xt

• We use the following formula.

% Z = Transformer Impedance / Base Imp.Base Imp.= Phase voltage / Full load current.∴ Xt = % Z* Phase voltage / Full load current.

= 0.04* 415/√3/(500*1000/(415*√3 )) ohm.= 0.013778 ohm.

Harmonic Simulation Without Capacitors

h Ih Xth = Xt*h

Vh = √3*Ih*Xtotal

5 40 0.0689 4.77287 28.6 0.0964 4.7728

11 18.2 0.1516 4.772813 15.4 0.1791 4.772817 11.8 0.2342 4.772819 10.5 0.2618 4.7728

Total Harmonic Voltage Distortion % Without Capacitors

Total harmonic voltage distortion

THDV% = (√(V52+V7

2+V112+V13

2+V172+V19

2) / V1) x 100

= (√(4.772+4.772+4.772+4.772+4.772+4.772)/415)*100

THDV% = (11.69/415)*100

THDV% = 2.82%

Network With Capacitors:• Calculation of harmonic voltage distortion and

harmonic overload:• With conventional capacitor of rating 100 kVAr.

Capacitive reactance Xc= V2/ (kVAr x 1000) ohms.= 4152 / (100 x 1000).

Xc = 1.72225 ohms.Transformer reactance.Xt = V2 x (%Z / 100 ) / (kVA x 1000) ohms

= 4152 x (4 / 100) / (500 x 1000).Xt = 0.013778 ohms.

Network With Capacitors

100 kW, 3ph,415V, DC

drive Linear resistive Loads

100 kVAr, 3ph,415V

500 kVA

Network With Capacitors: Equivalent Circuit

Ih Xt XC

Harmonic Simulation WithCapacitors

16.112.52810.138660.09060.261810.519

20.733.63790.178530.10130.234211.817

59.0813.55930.508850.13250.179115.413

549.89149.1514.736190.15660.151618.211

18.427.85010.158630.24600.096428.67

10.005.96590.086110.34450.068940.05

Ic = Vh/(√3* Xch)

Vh = √3*Ih*Xtotal

|Xtotal| = Xth*Xch|(Xth-Xch)|

Xch=Xc/hXth=Xt*hIhh

Overloading of Capacitors

Ich = 554 Amps.

Ic1 =139 Amps.

Iceff=571 Amps.

Overload = 411 %

THDV% With Capacitors

THDV% = (√(V52+V7

2+V112+V13

2+V172+V19

2) / V1) x 100

=(√(5.972+7.852+149.152+13.562+3.642+2.532)/415)*100

= (150.16)/415*100

= 36.18%

Calculation With Detuned Filters

100 kW, 3ph,415V, DC

drive Linear resistive Loads

112.5 kVAr, 3ph,440V,DF

500 kVA

XLF

Why 112.5 kVAr• The detuned filters, supplied by L&T/Meher are rated

for a bus voltage of 440V. This is mainly done to provide a standardized solution.

• The kVAr output of a 440V capacitor is reduced if connected to a 415 volts system, given by the relation.

= (Vsystem/Vrated)2 * kVAr.

= (415/440)2 * 100 kVAr.= 88.9 kVAr.

Hence while suggesting a detuned filter for a 415 V system, additional kVAr has to be provided so as to provide the required 100 kVAr at system voltage. .

Why 112.5 kVAr• Hence if the 100 kVAr capacitors have to be

supplemented by detuned filter, the detuned filter should be rated for 112.5 kVAr, calculated by the relation

= (Vrated/Vsystem)2 * kVAr

= (440/415)2 * kVAr= 1.124 * 100= 112.5 kVAr. (approx.)

Network With Detuned Filters: Equivalent Circuit

Ih Xt XLF

XCF

Detuned Filter of 112.5 kVAr, 440V

Capacitive kVAr = 112.5 * (1 – p/100)= 112.5* (1-7/100)= 112.5 *0.93= 104.625 kVAr

XC = 4402 / (104.625 x 1000)XCF = 1.8504 ohms.

Reactance of reactorXLF = p/100* Xc

= 0.07*1.8504= 0.129528 ohm.

Harmonic Simulation

Transformer reactance.

Xt = V2 x (%Z / 100 ) / (kVA x 1000) ohms.

= 4152 x (4 / 100) / (500 x 1000).

Xt = 0.013778 ohms.

Harmonic Simulation WithDetuned Filter

1.054.29690.23572.36360.09742.46100.261810.519

1.184.29250.21072.09310.10882.20200.234211.817

1.604.27600.16051.54150.14231.68390.179115.413

1.964.25910.13531.25660.16821.42480.151618.211

3.734.14980.08390.64240.26430.90670.096428.67

7.953.82380.05520.27760.37010.64760.0689405

Ic= Vh/(√3*XFilter)

Vh =√3*Ih*Xtotal

Xtotal =

Xfilter*Xth( XFilter+Xth)

XFilter = XLFH –XCFH

XCFH = XCF/h

XLFH = XLF*h

Xth = Xt*hIhh

Overloading of Detuned Filter

Ifh = 9.27 Amps.

If1 =139.12 Amps.

Ifeff=139.43 Amps.

Overload = Negligible.

THDV% With Detuned Filters

THDV% = (√(V52+V7

2+V112+V13

2+V172+V19

2) / V1) x 100

= (√(3.822 +4.152+4.262+4.282+4.292+4.302)/415)*100

THDV% = (10.25/415)*100

THDV% = 2.47%

SummaryIt is seen that the introduction of capacitors into a network with non-linear loads not only leads to very high overloading of the capacitors but also increases the harmonic voltage distortion in the network due to parallel resonance.

If the same capacitors are supplemented with detuned filters, the overloading is prevented and also the voltage distortion in the network is reduced to acceptable limits.

Conclusion

Detuned filter is a safe and proven solution to improve power factor in harmonic rich environment.

Limitations of Passive filters

• Sensitive to system frequency change.

• Sensitive to change in network parameters.

• Has location limitation when ‘Drives’ exist in system.

• Mixing of filters having different tuning factor generally not possible due to risk of resonance.

• Not immune from harmonic overloading.

• Generally cannot handle wide harmonic spectrum.

• kVAr. output of filter variable only in steps.

Session 15

Active Compensation

SCOPE

• ACTIVE COMPENSATION

• NEED

• PRINCIPLE

• BENEFITS

• SOLUTIONS

• AHF - ACTIVE HARMONIC FILTER

• INTELLVAr - E

• Electronic VAr Compensation

Need for Active Compensation - I

In networks where HARMONIC FILTERING is to be done

• Independent of PF Improvement

• for complex harmonic frequencies

• for fine control of THD(V): <3%

Need for Active Compensation - IIfor Unsymmetrical Reactive Power Compensation (PF improvement)

• in REAL TIME MODE

• in STEPLESS MODE

• where INFINITE CONTROL is needed

• for real time Voltage Support

Principle of Active Compensation

Involves real time CURRENT INJECTION into a network

• in Variable AMPLITUDE & PHASE ANGLE

• in COMPLEX WAVE SHAPES

• with INFINITE CONTROL

• at any LOCATION (in shunt)

Active Compensation Benefits

• Filtering upto the 50th HARMONIC including Inter-Harmonics

• Unsymmetrical Compensation of Reactive Power

• Real time response < 2 msec

• Independent of network characteristics, voltage & frequency behavior

• INFINITE CONTROL

• Compatibility with conventional compensation installations

Active Compensation Active Compensation -- BENEFITSBENEFITS

Total PF Control

No risk of Resonance

Extremely Flexible

Voltage Stabilty

THD (V) Control

Plug & Play Solution

SOLUTIONS - I

• ACTIVE HARMONIC FILTERS

- for 3 Phase, 415/440 V, 50 Hz. NETWORKS

- In Current Ratings from 32 to 630 Amps

- Optional Reactive Power Compensation

SOLUTIONS - II

• INTELLVAr - E

- Hybrid ELECTRONIC VAr COMPENSATION

- for 3 Phase, 415/440 V, 50 Hz. NETWORKS

- Output from 50 to 1000 kVAr

ActiveFilter

Load Current withHarmonics

SinusoidalSupplyCurrent Supply

System

CompensatingCurrent

+ =

(Time Domain)

(Frequency Domain)

+ =

Active filter schematic diagram

Waveform of current without Active filter

Waveform of current with Active filter

Current harmonics without active filter

Current harmonics with active filter

Cost-Technology Pyramid of Harmonic Filters

Top end solution for wide spectrum of current harmonics & suitable for installations having sensitive equipment

Activefilters

Most common, Base end product suitable for majority of industries having 5th and above harmonics

Suitable for installations having 3rd harmonics and above (IT parks, corporate banks & establishments)

Suitable for installations having high harmonic distortion (cement, sugar & steel plants, etc.)

Hybrid Filters

Tuned Filters

14% Detuned Filters

7% Detuned Filters